BASIC REd4&r
Cornerstones Series Editors Charles L. Epstein, University of Pennsylvania, Philadelphia Steven G. Krantz, University of Washington, St. Louis
Advisoqi Board Anthony W. Knapp, State University of New York at Stony Brook, Emeritus
Anthony W. Knapp
Basic Real Analysis Along with a companion volume Advanced Real Analysis
Birkhauser Boston Base1 Berlin
Anthony W. Knapp 81 Upper Sheep Pasture Road East Setauket, NY 11733-1729 U.S.A. e-mail t o :
[email protected] http://www.math.sunysb.edu/-aknapp/books/basic.html
Cover design by Mary Burgess. Mathematics Subject Classicification (2000): 28-01, 26-01,42-01, 54-01,34-01 Library of Congress Cataloging-in-PublicationData Knapp, Anthony W. Basic real analysis: along with a companion volume Advanced real analysis i Anthony W. Knapp p. cm. (Cornerstones) Includes bibliographical rererences and index. ISBN 0-8176-3250-6 (alk. paper) 1. Mathematical analysis. I. Title. 11. Cornerstones (Birkhauser) -
ISBN- 10 0-81763250-6 ISBN-13 978-0-8176-3250-2
eISBN 0-8 176-4441-5
Advanced Real Analysis Basic Real Analysis and Advanced Real Analysis (Set)
Printed on acid-free paper.
ISBN 0-8 176-4382-6 ISBN 0-8176-4407-5
02005 Anthony W. Knapp All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Birkhauser Boston, c/o Springer Science+Business Media Inc., 233 Spring Street, New York, NY 10013, USA) and the author, except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of infomation storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. Ihe use in this publication of trade names, trademarks, service marks and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. Printed in the United States of America. 987654321
SPIN 10934074
(MP)
To Susan and To M y Real-Analysis Teachers: Salomon Bochner, William Feller, Hillel Fursten berg, Harish-Chandra, Sigurd ur Helgason ,John Kemeny, John Lamperti, Hazleton Mirkil, Edward Nelson, Laurie Snell, Elias Stein, Richard Williamson
CONTENTS
Preface Dependence Among Chapters Guide for the Reader List of Figures Acknowledgments Standard Notation
I.
THEORY OF CALCULUS IN ONE REAL VARIABLE 1. Review of Real Numbers, Sequences, Continuity 2. Interchange of Limits 3. Uniform Convergence 4. Riemann Integral 5. Curnplcx-Valued Furi~.tiuns 6. Taylor's Theorem with Integral Remainder 7. Power Series and Special Functions 8. Summability 9. Weierstrass Approximation Theorem 10. Fourier Series 11. Problems
11.
METRIC SPACES 1. Definition and Examples 2. Open Sets and Closed Sets 3. Continuous Functions 4. Sequences and Convergence 5. Subspaces and Products 6. Properties of Metric Spaces 7. Colnpactness and Col~~pleteness 8. Connectedness 9. Baire Category Theorem 10. Properties of C ( S ) for Compact Metric S 11. Completion 12. Problems
xi xiv xv xviii xix xxi
111. THEORY OF CALCULUS IN SEVERAL REAL VARIABLES 1. Operator Norm 2. Nonlinear Functions and Differentiation 3. Vector-Valued Partial Derivatives and Riemann Integrals 4. Exponential of a Matrix 5. Partitions of Unity 6. Inverse and Implicit Function Theorems 7. Definition and Properties of Rierrianri Integral 8. Riemann Integrable Functions 9. Fubini's Theorem for the Riemann Integral 10. Change of Variables for the Riemann Integral 11. Problems IV. THEORY OF ORDINARY DIFFERENTIAL EQUATIONS AND SYSTEMS 1. Qualitative Features and Examples 2. Existence and Uniqueness 3. Dependence on Initial Conditions and Parameters 4. Integral Curves 5. Linear Equations and Systems, Wronskian 6. Homogeneous Equations with Constant Coefficients 7. Homogeneous Systems with Constant Coefficients 8. Series Solutions in the Second-Order Linear Case 9. Problems V. LEBESGUE MEASURE AND ABSTRACT MEASURE THEORY 1. Measures and Examples 2. Measurable Functions 3. Lebesgue Integral 4. Properties of the Integral 5. Proof of the Extension Theorem 6. Completion of a Measure Space 7. Fubini's Theorem for the Lebesgue Integral 8. Integration of Complex-Valued and Vector-Valued Functions 9. L ~L , ~Lm , , and Normed Linear Spaces 10. Problems VI. MEASURE THEORY FOR EUCLIDEAN SPACE 1. Lebesgue Measure and Other Borel Measures 2. Convolution 3. Borel Measures on Open Sets 4. Comparison of Riemann and Lebesgue Integrals
VI. MEASURE THEORY FOR EUCLIDEAN SPACE (Continued) 5. Change of Variables for the Lebesgue Integral 6. Hardy-Littlewood Maximal Theorem 7. Fourier Series and the Riesz-Fischer Theorem 8. Stieltjes Measures on the Line 9. Fourier Series and the Dirichlet-Jordan Theorem 10. Distribution Functions 11. Problems VII. DIFFERENTIATION OF LEBESGUE INTEGRALS ON THE LINE 1. Differentiation of Monotone Functions 2. Absolute Continuity, Singular Measures, and Lebesgue Decomposition 3. Problems
320 327 334 339 346 350 352 357 357 364 370
VIII. FOURIER TRANSFORM IN EUCLIDEAN SPACE 1. Elementary Properties 2. Fourier Transform on L l , Inversion Formula 3. Fourier Transform on L ~Plancherel , Formula 4. Schwartz Space 5. Poisson Summation Formula 6. Poisson Integral Formula 7. Hilbert Transform 8. Problems
373 373 377 38 1 384 389 392 397 404
IX.
LPSPACES 1 . Inequalities and Completeness 2. Convolution Involving LP 3. Jordan and Hahn Decompositions 4. Radon-Nikodym Theorem 5. Continuous Linear Functionals on LP 6. Marcillkiewicz Interpolation Theol-em 7. Problems
409 409 417 418 420 424 427 436
X.
TOPOLOGICAL SPACES 1. Open Sets and Collstluctiolls of Topologies 2. Properties of Topological Spaces 3. Compactness and Local Compactness 4. Product Spaces and the Tychonoff Product Theorem 5. Sequences and Nets 6. Quotient Spaces
44 1 44 1 447 45 1 458 463 47 1
X.
TOPOLOGICAL SPACES (Continued) 7. Urysohn's Lemma 8. Metrization in the Separable Case 9. Ascoli-Arzela and Stone-Weierstrass Theorems 10. Problems
XI. INTEGRATION ON LOCALLY COMPACT SPACES 1. Setting 2. Riesz Representation Theorem 3. Regular Bore1 Measures 4. Dual to Space of Finite Signed Measures 5. Problems XII. HILBERT AND BANACH SPACES 1 . Definitions and Examples 2. Geometry of Hilbert Space 3. Bounded Linear Operators on Hilbert Spaces 4. Hahn-Banach Theorem 5. Uniform Boundedness Theorem 6. Interior Mapping Principle 7. Problems APPENDIX A1 . Sets and Functions A2. Mean Value Theorem and Some Consequences A3. Inverse Function Theorem in One Variable A4. Complex Numbers A5. Classical Schwarz Inequality A6. Equivalence Relations A7. Linear Transformations, Matrices, and Determinants AS. Factorization and Roots of Polynomials A9. Partial Ordcrings and Zorn's Lcrnma A10. Cardinality Hints for Solutions of Problems Selected References Index of Notation Index
PREFACE
'I'his book and its companion volume Advanced Keal Analysis systematically develop concepts and tools in real analysis that are vital to every mathematician, whether pure or applied, aspiring or established. The two books together contain what the young mathematician needs to know about real analysis in order to communicate well with colleagues in all branches of mathematics. The books are written as textbooks, and their primary audience is students who are learning the material for the first time and who are planning a career in which they will use advanced mathematics professionally. Much of the material in the books corresponds to normal course work. Nevertheless, it is often the case that core mathematics curricula, time-limited as they are, do not include all the topics that one might like. Thus the book includes important topics that may be skipped in rcquircd courscs but that thc profcssional rnathcmatician will ultimatcly want to learn by self-study. The content of the required courses at each university reflects expectations of what students need beforebeginning specialized study and work on a thesis. These expectations vary from country to country and from university to university. Even so, there seems to be a rough consensus about what mathematics a plenary lecturer at a broad international or national meeting may take as known by the audience. The tables of contents of the two books represent my own understanding of what that degree of knowledge is for real analysis today. Key topics and features of Basic Real Analysis are as follows: a
a
a a
Early chapters treat the fundamentals of real variables, sequences and series of functions, the theory of Fourier series for the Riemann integral, metric spaces, and the theoretical underpinnings of multivariable calculus and ordinary differential equations. Subsequent chapters develop the Lebesgue theory in Euclidean and abstract spaces, Fourier series and the Fourier transform for the Lebesgue integral, point-sct topology, mcasurc thcory in locally compact Hausdorff spaccs, and the basics of Hilbert and Banach spaces. The subjects of Fourier series and harmonic functions are used as recurring motivation for a number of theoretical developments. The development proceeds from the particular to the general, often introducing examples well before a theory that incorporates them.
More than 300 problems at the ends of chapters illuminate aspects of the text, develop related topics, and point to additional applications. A separate 55-page section "Hints for Solutions of Problems" at the end of the book gives detailed hints for most of the problems, together with complete solutions for many. Beyond a standard calculus sequence in one and several variables, the most important prerequisite for using Basic Real Analysis is that the reader already know what a proof is, how to read a proof, and how to write a proof. This knowledge typically is obtained from honors calculus courses, or from a course in linear algebra, or from a first junior-senior course in real variables. In addition, it is assumed that the reader is comfortable with a modest amount of linear algebra, including row reduction of matrices, vector spaces and bases, and the associated geometry. A passing acquaintance with the notions of group, subgroup, and quotient is helpful as well. Chapters I-IV are appropriate for a single rigorous real-variables course and may be used in either of two ways. For students who have learned about proofs from honors calculus or linear algebra, these chapters offer a full treatment of real variables, leaving out only the more familiar parts near the beginning-such as clcmcntary manipulations with limits, convcrgcncc tcsts for infinitc scrics with positive scalar terms, and routine facts about continuity and differentiability. For students who have learned about proofs fro111 a first junior-senior course in real variables, these chapters are appropriate for a second such course that begins with Riemann integration and sequences and series of functions; in this case the first section of Chapter I will be a review of some of the more difficult foundational theorems, and the course can conclude with an introduction to the Lebesgue integral from Chapter V if time permits. Chapters V through XI1 treat Lebesgue integration in various settings, as well as introductions to the Euclidean Fourier transform and to functional analysis. Typically this material is taught at the graduate level in the United States, frequently in one of three ways: The first way does Lebesgue integration in Euclidean and abstract settings and goes on tu consider the Euclidean Fourier transform in some detail; this corresponds to Chapters V-VIII. A second way does Lebesgue integration in Euclidean and abstract settings, treats LP spaces and integration on locally compact Hausdorff spaces, and concludes with an introduction to Hilbert and Banach spaces; this corresponds to Chapters V-VII, part of IX, and XI-XII. A third way combines an introduction to the Lebesgue integral and the Euclidean Fourier transform with some of the subject of partial differential equations; this corresponds to some portion of Chapters V-VI and VIII, followed by chapters from the companion volume Advanced Real Analysis. In my own teaching, I have most often built one course around Chapters I-IV and another around Chapters V-VII, part of IX, and XI-XII. I have normally
assigned the easier sections of Chapters 11 and X as outside reading, indicating the date when the lectures would begin to use that material. More detailed information about how the book may be used with courses may be deduced from the chart "Dependence among Chapters" on page xiv and the section "Guide to the Reader" on pages xv-xvii. The problems at the ends of chapters are an important part of the book. Some of them are really theorems, some are examples showing the degree to which hypotheses can be stretched, and a few are just exercises. The reader gets no indication which problems are of which type, nor of which ones are relatively easy. Each problem can be solved with tools developed up to that point in the book, plus any additional prerequisites that are noted. Two omissions from the pair of books are of note. One is any treatment of Stokes's Theorem and differential forms. Although there is some advantage, when studying these topics, in having the Lebesgue integral available and in having developed an attitude that integration can be defined by means of suitable linear functionals, the topic of Stokes's Theorem seems to fit better in a book about geometry and topology, rather than in a book about real analysis. The other omission concerns the use of complex analysis. It is tempting to try to combine real analysis and complex analysis into a single subject, but my own experience is that this combination does not work well at the level of Basic Real Analysis, only at the level of Advanced Real Analysis. Almost all of the mathematics in the two books is at least forty years old, and I make no claim that any result is new. 'I'he books are a distillation of lecture notes from a 35-year period of my own learning and teaching. Sometimes a problem at the end of a chapter or an approach to the exposition may not be a standard one, but no attempt has been made to identify such problems and approaches. In the rcvcrsc dircction it is possiblc that my carly lccturc notcs havc dircctly quotcd some source without proper attribution. As an attempt to rectify any difficulties of this kind, I have included a section of "Acknowledgements" on pages xix-xx of this volume to identify the main sources, as far as I can reconstruct them, for those original lecture notes. I amgrateful to Ann Kostant and Steven Krantz for encouraging this project and for making many suggestions about pursuing it, and to Susan Knapp and David , Kramer for helping with the readability. The typesetting was by A M S - T ~and the figures were drawn with Mathematica. I invite col-1-ectionsand other conments from readers. I plan to nlaintain a list of known corrections on my own Web page. A. W. KNAPP May 2005
DEPENDENCE AMONG CHAPTERS
Below is a chart of the main lines of dependence of chapters on prior chapters. The dashed lines indicate helpful motivation but no logical dependence. Apart from that, particular examples may make use of information from earlier chapters that is not indicated by the chart. r--------------
I I, 11, IIIin order I L
------ -
r------I
GUIDE FOR THE READER
This section is intended to help the reader find out what parts of each chapter are most important and how the chapters are interrelated. Further information of this kind is contained in the abstracts that begin each of the chapters. The book pays attention to certain recurring themes in real analysis, allowing a person to see how these themes arise in increasingly sophisticated ways. Examples are the role of interchanges of limits in theorems, the need for certain explicit formulas in the foundations of subject areas, the role of compactness and complctcncss in cxistcncc thcorcms, and thc approach of handling nicc functions first and then passing to general functions. All of these themes are introduced in Chapter I, and already at that stage they intcract in subtlc ways. For cxamplc, a natural investigation of intcrchangcs of limits in Sections 2-3 leads to the discovery of Ascoli's Theorem, which is a fundamental compactness tool for proving existence results. Ascoli's Theorem is proved by the "Cantor diagonal process," which has other applications to compactness questions and does not get fully explained until Chapter X. The consequence is that, no matter where in the book a reader plans to start, everyone will be helped by at least leafing through Chapter I. The remainder of this section is an overview of individual chapters and groups of chapters. Chapter I. Every section of this chapter plays a role in setting up matters for later chapters. No knowledge of metric spaces is assumed anywhere in the chapter. Section 1will be a review for anyone who has already had a course in realvariable theory; the section shows how compactness and completeness address all the difficult theorems whose proofs are often skipped in calculus. Section 2 begins the development of real-variable theory at the point of sequences and series of functions. It contains interchange results that turn out to be special cases of the main theorems of Chapter V. Sections 8-9 introduce the approach of handling nice fullctiolls before general functions, and Section 10 introduces Fourier series, which provided a great deal of motivation historically for the development of real analysis and are used in this book in that same way. Fourier series are somewhat limited in the setting of Chapter I because one encounters no class of functions, other than infinitely differentiable ones, that corresponds exactly to some class of Fourier coefficients; as a result Fourier series, with Riemann integration in use,
xvi
Guide for the Reader
are not particularly useful for constructing new functions from old ones. This defect will be fixed with the aid of the Lebesgue integral in Chapter VI. Chapter 11. Now that continuity and convergence have been addressed on the line, this chapter establishes a framework for these questions in higherdimensional Euclidean space and other settings. There is no point in ad hoc definitions for each setting, and metric spaces handle many such settings at once. Chapter X later will enlarge the framework from metric spaces to "topological spaces." Sections 1-6 of Chapter I1 are routine. Section 7, on compactness and completeness, is the core. The Baire Category Theorem in Section 9 is not used outside of Chapter I1 until Chapter XII, and it may therefore be skipped temporarily. Section 10 contains the Stone-Weierstrass Theorem, which is a fundamental approximation tool. Section 11 is used in some of the problems but is not otherwise used in the book. Chapter 111. This chapter does for the several-variable theory what Chapter I has done for the one-variable theory. The main results are the Inverse and Implicit Function Theorems in Section 6 and the change-of-variables formula for multiple integrals in Section 10. The change-of-variables formula has to be regarded as only a preliminary version, since what it directly accomplishes for the change to polar coordinates still needs supplementing; this difficulty will be repaired in Chapter VI with the aid of the Lebesgue integral. Section 4, on exponentials of matrices, may be skipped if linear systems of ordinary differential equations are going to be skipped in Chapter IV. Some of the problems at the end of the cliapter introduce 1ia1-1i1oiiic fuiictioiis; harii~oiiicfuiictioiis will be coiiibiiied witli Fourier series in problems in later chapters to motivate and illustrate some of the development. Chapter IV provides theoretical underpinnings for the material in a traditional undergraduate course in ordinary differential equations. Nothing later in the book is logically dependent on Chapter IV; however, Chapter XI1 includes a discussion of orthogonal systems of functions, and the examples of these that arise in Chapter IV are helpful as motivation. Some people shy away from differential equations and might wish to treat Chapter IV only lightly, or perhaps not at all. The subject is nevertheless of great importance, and Chapter IV is the beginning of it. A minimal treatment of Chapter IV might involve Sections 1-2 and Section 8, all of which visibly continue the themes begun in Chapter I. Chapters V-VI treat the core of measure theory-including the basic convergence theorems for integrals, the develop~ilentof Lebesgue measure in one and several variables, Fubini's Theorem, the metric spaces L' and L~ and L m , and the use of maximal theorems for getting at differentiation of integrals and other theorems concerning almost-everywhere convergence. In Chapter V Lebesgue measure in one dimension is introduced right away, so that one immediately has the most important example at hand. The fundamental Extension Theorem for
Guide for the Reader
xvii
getting measures to be defined on o -rings and o -algebras is stated when needed but is proved only after the basic convergence theorems for integrals have been proved; the proof in Sections 5-6 may be skipped on first reading. Section 7, on Fubini's Theorem, is a powerful result about interchange of integrals. At the same time that it justifies interchange, it also constructs a "double integral"; consequently the section prepares the way for the construction in Chapter VI of n-dimensional Lebesgue measure from 1-dimensional Lebesgue measure. Section 10 introduces normed linear spaces along with the examples of L' and L2 and L w ,and it goes on to establish some properties of all normed linear spaces. Chapter VI fleshes out measure theory as it applies to Euclidean space in more than one dimension. Of special note is the Lebesgue-integration version in Section 5 of the changeof-variables formula for multiple integrals and the Riesz-Fischer Theorem in Section 7. The latter characterizes square-integrable periodic functions by their Fourier coefficients and makes the subject of Fourier series useful in constructing functions. Differentiation of integrals in approached in Section 6 of Chapter VI as a problem of estimating finiteness of a quantity, rather than its snlallness; the device is the Hardy-Littlewood Maximal Theorem, and the approach becomes a routine way of approaching almost-everywhere convergence theorems. Sections 8-10 arc of somcwhat lcss importance and may bc omittcd if timc is short; Section 10 is applied only in Section IX.6. Chapters VII-IX are continuations of measure theory that are largely independent of each other. Chapter VII contains the traditional proof of the differentiation of integrals on the line via differentiation of monotone functions No later chapter is logically dependent on Chapter VII; the material is included only because of its historical importance and its usefulness as motivation for the Radon-Nikodym Theorem in Chapter IX.Chapter VIII is an introduction to the Fourier transform in Euclidean space. Its core consists of the first four sections, and the rest may be considered as optional if Section M.6 is to be omitted. Chapter IX concerns LP spaces for 1 5 p 5 oo;only Section 6 makes use of material from Chapter VIII. Chapter X develops, at the latest possible time in the book, the necessary part of point-set topology that goes beyond metric spaces. Emphasis is on product and quotient spaces, and on Urysohn's Lemma concerning the construction of real-valued functions on normal spaces. Chapter XT contains one more continuation of measure theory, namely special features of measures on locally compact Hausdorff spaces. It provides an example beyond LP spaces in which one can usefully identify the dual of a particular normed linear space. These chapters depend on Chapter X and on the first five sections of Chapter IX but do not depend on Chapters VII-VIII. Chapter XI1is a brief introduction to functional analysis, particularly to Hilbert spaces, Banach spaces, and linear operators on them. The main topics are the geometry of Hilbert space and the three main theorems about Banach spaces.
LIST OF FIGURES
Approximate identity Fourier series of sawtooth function Dirichlet kernel An open set centered at the origin in the hedgehog space Open ball contained in an intersection of two open balls Graphs of solutions of some first-order ordinary differential equations Integral curve of a vector field Graph of Bessel function Jo ( t ) Construction of a Cantor function F Rising Sun Lemma
ACKNOWLEDGMENTS
The author acknowledges the sources below as the main ones he used in preparing the lectures from which this book evolved. Any residual unattributed direct quotations in the book are likely to be from these. The descriptions below have been abbreviated. Full descriptions of the books and Stone article rnay be found in the section "Selected References" at the end of the book. The item "Feller's Functional Analysis" refers to lectures by William Feller at Princeton University for Fall 1962 and Spring 1963, and the item "Nelson's Probability" refers to lectures by Edward Nelson at Princeton University for Spring 1963. This list is not to be confused with a list of recommended present-day reading for these topics; newer books deserve attention. CHAPTERI.
Rudin's Principles of Mathematical Analysis, Zygmund's
I rigonometric Series.
CHAPTERII. Feller's Functional Analysis, Kelley's General Topology, Stone's "A generalized Weierstrass approximation theorem." CIIAPTER111. Rudin's Principles of Mathenzatical Analysis, Spivak's Calculus on Manifolds. CHAPTERIV. Coddington-Levinson's Theory of Ordinary Differential Equations, Kaplan's Ordinary Differential Equations. CHAPTERV. Halmos's Measure Theory, Rudin's Principles of Mathematical Analysis. CHAPTERVI. Rudin's Principles of Mathematical Analysis, Rudin's Real and Complex Analysis, Saks's Theory of the Integral, Spivak's Calculus on Manifolds, Stein-Weiss's Introduction to Fourier Analysis on Euclidean Spaces. CHAPTERVII. Riesz-Nagy's Functional Analysis, Zygmund's Trigonometric Series. CHAPTERVIII. Stein's Singular Integrals and Differentiability Properties of Functions, Stein-Weiss's Introduction to Fourier Analysis on Euclidean Spaces. CHAPTERIX. Dunford-Schwartz's Linear Operators, Feller's Functional Analysis, Halmos's Measure Theory, Saks's Theory of the Integral, Stein's Singular Integrals and Differentiability Properties of Functions.
xx
Acknowledgments
CHAPTERX. Kelley's General Topology, Nelson's Probability. CHAPTERX I . Feller's Functional Analysis, Halmos's Measure Theory, Nelson's Probability. CHAPTERXII. Dunford-Schwartz's Linear Operators, Feller's Functional Analysis, Riesz-Nagy's Functional Analysis.
APPENDIX. For Sections 1 , 9, 10: Dunford-Schwartz's Linear Operators, Hayden-Kennison's Zernzelo-Fraenkel Set Theory, Kelley's General Topology.
STANDARD NOTATION
Itcm
Mcaning
#S or IS1
number of elements in S empty set the set of x in E such that P holds complement of the set E union, intersection, difference of sets union, intersection of the sets E, E is contained in F , E contains F products of sets ordered n-tuple, unordered n-tuple function, effect of function composition of f following g ,restriction to E the function x H f ( x , y ) direct and inverse image of a set Kronecker delta: 1 if i = j, 0 if i # j binomial coefficient n >O,n ( 0 integers, rationals, reals, complex numbers maximum of finite subset of a totally ordered set sum or product, possibly with a limit operation finite or in one-one correspondence with Z greatest integer 5 x if x is real real and imaginary parts of complex z complex conjugate of z absolute value of z multiplicative identity idcntity matrix or opcrator dimension of vector space spaces of column vectors determinant of A transpose of A diagonal matrix is isomorphic to, is equivalent to
u Ix
E
E
I
PI
EC E U F , E ~ F E, - F U, E,, ELF, E Z F E x F , X ,,sX, (al,...,a,), {al,...,a,I
n,
t3
n positive, n negative
max (and similarly min)
c om
countable [xI Re z,Imz Z
IzI 1 1 or I dim V Ihpn ,
cn
det A Atr diag(al, . . . , a,) w -
-
CHAPTER I Theory of Calculus in One Real Variable
Abstract. This chapter, beginning with Section 2, develops the topic of sequences and series of functions, especially of functions of one variable. An important part of the treatment is an introduction to the problem of interchange of limits, both theoretically and practically. This problem plays a role repeatedly in real analysis, but its visibility decreases as more and more results are developed for handling it in various situations. Fourier series are introduced in this chapter and are carried along throughout the book as a motivating example for a number of problems in real analysis. Section 1 makes contact with the core of a first undergraduate course in real-variable theory. Some material from such a course is repeated here in order to establish notation and apoint of view. Omitted material is summarized at the end of the section, and some of it is discussed in a little more detail in an appendix at the end of the book. The point of view being established is the use of defining ~ to prove the Bolzan~WeicrstrassThco~em,followed by the properties of the real n u n ~ b esystem use of that theorem to prove some of the difficult theorems that are usually assumed in a one-variable calculus course. The treatment makes use of the extended real-number system, in order to allow sup and inf to be defined for any nonempty set of reds and to allow lim sup and lim inf to be meaningful for any sequence. Sections 2-3 introduce the problem of interchange of limits. They show how certain concrete problems can be viewed in this way, and they give a way of thinking about all such interchanges in a common framework. A positive result affirms such an interchange under suitable hypotheses of monotonicity. This is by way of introduction to the topic in Section 3 of uniform convergence and the role of uniform convergence in continuity and differentiation. Section 4 gives a careful development of the Riemann integral for real-valued functions of one variable, establishing existence of Riemann integrals for bounded functions that are discontinuous at only finitely many points, basic properties of the integral, the Fundamental Theorem of Calculus for continuous integrands, the change-of-variables formula, and other results. Section 5 examines complex-valued functions, pointing out the extent to which the results for real-valued functions in the first four sections extend to complex-valued functions. Section 6 is a short treatment of the version of Taylor's Theorem in which the remainder is given by an integral. Section 7 takes up power series and uses them to define the elementary transcendental iunctlons and establish thelr properties. 'lhe power serles expansion of ( I +x)p for arbitrary complex p is studied carefully. Section 8 introduces Ceskro and Abel summability, which play a role in the subject of Fourier series. A converse theorem to Abel's theorem is used to exhibit the function Ix I as the uniform limit of polynomials on [- 1, I]. The Weierstrass Approximation Theorem of Section 9 generalizes this example and establishes that every continuous complex-valued function on a closed bounded interval is the uniform limit of polynomials. Section 10 introduces Fourier series in one variable in the context of the Riemann integral. The main theorems of the section are a convergence result for continuously differentiable functions, Bessel's inequality, the Riemann-Lebesgue Lemma, Fejkr's Theorem, and Parseval's Theorem.
2
I. Theory o f Calculus in One Real Variable
1. Review of Real Numbers, Sequences, Continuity This section reviews some material that is normally in an undergraduate course in real analysis. The emphasis will be on a rigorous proof of the BolzanoWeierstrass Theorem and its use to prove some of the difficult theorems that are usually assumed in a one-variable calculus course. We shall skip over some easier aspects of an undergraduate course in real analysis that fit logically at the end of this section. A list of such topics appears at the end of the section. The system of real numbers R may be constructed out of the system of rational numbers Q,and we take this construction as known. The formal definition is that a real number is a cut of rational numbers, i.e., a subset of rational numbers that is neither Q nor the empty set, has no largest element, and contains all rational numbers less than any rational that it contains. The idea of the construction is as follows: Each rational number q determines a cut q * , namely the set of all rationals less than q . Under the identification of Q with a subset of R,the cut dcfining a rcal numbcr consists of all rational numbcrs lcss than thc givcn rcal number. The set of cuts gets a natural ordering, given by inclusion. In place of c,we write 5 . For any two cuts r and s, we have r 5 s or s 5 r , and if both occur, then r = s. We can then define <, >, and in the expected way. The positive cuts r are those with 0* < r , and the negative cuts are those with r < O*. Once cuts and their ordering are in place, one can go about defining the usual operations of arithmetic and proving that R with these operations satisfies the familiar associative, commutative, and distributive laws, and that these interact with inequalities in the usual ways. The definitions of addition and subtraction are easy: the sum or difference of two cuts is simply the set of sums or differences of the rationals from the respective cuts. For multiplication and reciprocals one has to take signs into account. For example, the product of two positive cuts consists of all products of positive rationals from the two cuts, as well as 0 and all negative rationals. After these definitions and the proofs of the usual arithmetic operations are complete, it is customary to write 0 and 1 in place of O* and I * . An upper bound for a nonempty subset E of R is a real number M such that x 5 M for all x in E . If the nonempty set E has an upper bound, we can take the cuts that E consists of and form their union. This turns out to be a cut, it is an upper bound for E , and it is 5 all upper bounds for E. We can summarize this result as a theorem.
Theorem 1.1. Any nonempty subset E of R with an upper bound has a least upper bound. The least upper bound is necessarily unique, and the notation for it is sup,,, x or sup {x I x E E}, "sup" being an abbreviation for the Latin word "supremum,"
1. Review of Real Numbers, Sequences, Continuity
3
the largest. Of course, the least upper bound for a set E with an upper bound need not be in E; for example, the supremum of the negative rationals is 0, which is not negative. A lower bound for a nonempty set E of R is a real number m such that x > m for all x E E. If m is a lower bound for E , then -m is an upper bound for the set -E of negatives of members of E. Thus -E has an upper bound, and Theorem 1.1 shows that it has a least upper bound sup,,-, x . Then -x is a greatest lower bound for E . This greatest lower bound is denoted by infyeEy or inf {y I y E E}, "inf" being an abbreviation for "infimum." We can summarize as follows.
Corollary 1.2. Any nonempty subset E of R with a lower bound has a greatest lower bound. A subset of R is said to be bounded if it has an upper bound and a lower bound. Let us introduce notation and terminology for intervals of R, first treating the bounded ones.' Let a and b be real numbers with a 5 b. The open interval from a to b is the set (a, b) = {x E R I a < x < b}, the closed interval is the set [a, b] = {x E R I a 5 x 5 b}, and the half-open intervals are the sets [ a , b ) = {x E R I a 5 x < bland (a,b] = {x E R I a < x 5 b}. Eachofthe above intervals is indeed bounded, having a as a lower bound and b as an upper bound. These intervals are nonempty when a < b or when the interval is [a, b] with a = b, and in these cases the least upper bound is b and the greatest lower bound is a . Open sets in R are defined to be arbitrary unions of open bounded intervals, and a closed set is any set whose complement in R is open. A set E is open if and only if for each x E E , there is an open interval (a, b) such that x E (a, b) E . In this case we of course have a < x < b. If we put c = minix - a , b - x}, then we see that x lies in the subset (x - t , x + t ) of (a, b). The open interval (x-t,x+c)equals{y -xi < t } .T h u s a n o p e n s e t i n R i s a n y s e t E such that for each x E E , there is a number t > 0 such that { y E R 1 y - x I < t } lies in E. A limit point x of a subset F of R is a point of R such that any open interval containing x meets F in a point other than x . For example, the set [a, b) U {b + l} has [a, b] as its set of limit points. A subset of R is closed if and only if it contains all its limit points. Now let us turn to unbounded intervals. To provide notation for these, we shall make use of two symbols +m and -oo that will shortly be defined to be "extended real numbers." If a is in R,then the subsets (a, +m) = {x E R I a < x}, (-00, a) = {x E R I x < a}, (-oo, +m) = R , [a, +oo) = {x E R I a 5 x}, and (-oo,a ] = {x E R I x 5 a } are defined to be intervals, and they are all unbounded. The first three are open sets of R and are considered to be open
EIWI~~
' ~ o u n d e dintervals are called "finite intervals" by some authors
I
4
I. Theory o f Calculus in One Real Variable
intervals, while the last three are closed sets and are considered to be closed intervals. Specifically the middle set R is both open and closed. One important consequence of Theorem 1.1 is the archimedean property of R,as follows.
Corollary 1.3. T f n and h are real nilmhers with n > O, then there exists an integer n with na > b. PROOF.If, on the contrary, na 5 b for all integers n , then b is an upper bound for the set of all na . Let M be the least upper bound of the set {nu I n is an integer}. Using that a is positive, we find that a-l M is a least upper bound for the integers. Thus n 5 a 1M for all integers n , and there is no smaller upper bound. However, the smaller number a-I M - 1 must be an upper bound, since saying n 5 a-'M for all integers is the same as saying n - 1 5 a-l M - 1 for all integers. We arrive at a contradiction, and we conclude that there is some integer n with na > b. The archimedean property enables one to see, for example, that any two distinct real numbers have a rational number lying between them. We prove this consequence as Corollary 1.5 after isolating one step as Corollary 1.4.
Corollary 1.4. If c is a real number, then there exists an integer n such that 12
I c i?2+1.
PROOF.Corollary 1.3 with a = 1 and b = c shows that there is an integer M with M > c , and Corollary 1.3 with a = 1 and b = -c shows that there is an integer rn with rn > -c. Then -m < c < M, and it follows that there exists a greatest integer n with n 5 c. This n must have the property that c < n + 1, and the corollary follows.
Corollary 1.5. If x and y are real numbers with x < y , then there exists a rational number r with x < r < y . ~ O O F By . Corollary 1.3 with a = y x and b = 1, there is an integer N such that N ( y - x) > 1. This integer N has to be positive. Then < y - x. By Corollary 1.4 with c = Nx, there exists an integer n with n 5 Nx < n 1 , hence with $ 5 x i Adding the inequalities $ 5 x and $ iy - x yields < y . Thus x 5 X < < y . Since $ < < %,the rational number r = ?$ lras h e required pruperlies. -
%. %
+
A sequence in a set S is a function from a certain kind of subset of integers into S. It will be assumed that the set of integers is nonempty, consists of consecutive integers, and contains no largest integer. In particular the domain of any sequence is infinite. Usually the set of integers is either all nonnegative integers or all
1. Review o f Real Numbers, Sequences, Continuity
5
positive integers. Sometimes the set of integers is all integers, and the sequence in this case is often called "doubly infinite." The value of a sequence f at the integer n is normally written f, rather than f (n) ,and the sequence itself may be denoted by an expression like { f,},,l, in which the outer subscript indicates the domain. A subsequence of a sequence f with domain {m, m + 1, . . . } is a composition f o n, where f is a sequence and n is a sequence in the domain of f such that nk < nk+l for all k . For example, if (a,],>1 is a sequence, then (a2k]k>l is the subsequence in which the function n is given by nk = 2k. The domain of a subsequence, by our definition, is always infinite. A sequence a, in R is convergent, or convergent in R,if there exists a real number a such that for each t > 0, there is an integer N with la, - a1 < t for all n > N . The number a is necessarily unique and is called the limit of the sequence. Depending on how much information about the sequence is unambiguous, we may write limn,, a, = a or limn a, = a or lima, = a or a, + a . We also say a, tends to a as n tends to infinity or oo. A sequence in R is called monotone increasing if a, ( a,+l for all n in the domain, monotone decreasing if a, 2 a,+l for all n in the domain, monotone if it is monotone increasing or monotone decreasing.
Corollary 1.6. Any bounded monotone sequence in R converges. If the sequence is monotone increasing, then the limit is the least upper bound of the image in R of the sequence. If the sequence is monotone decreasing, the limit is the greatest lower bound of the image. REMARK. Often it is Corollary 1.6, rather than the existence of least upper bounds, that is taken for granted in an elementary calculus course. The reason is that the statement of Corollary 1.6 tends for calculus students to be easier to understand than the statement of the least upper bound property. Problem 1 at the end of the chapter asks for a derivation of the least-upper-bound property from Corollary 1.6.
PROOF. Suppose that {a,} is monotone increasing and bounded. Let a = sup, a,, the existence of the supremum being ensured by Theorem 1.l, and let t r 0 be given. If there were no integer N with a~ r a - t ,then a - t would be a smaller upper bound, contradiction. Thus such an N exists. For that N , n > N impliesa c < a~ 5 a , 5 a < a I c . Thusn > Nimpliesla, a1 < c . Since t is arbitrary, lim,,, a, = a . If the given sequence {a,} is monotone decreasing, we argue similarly with a = inf, a,. In working with sup and inf, it will be quite convenient to use the notation supxEEx even when E is nonempty but not bounded above, and to use the notation
I. Theory o f Calculus in One Real Variable
6
infXELx even when E is nonempty but not bounded below. We introduce symbols +cc and -cc,plus and minus infinity, for this purpose and extend the definitions of sup,,, x and infXEEx to all nonempty subsets E of R by taking supx = +cc
if E has no upper bound,
XCK
inf x = -cx
xtE
if E has no lower bound.
To work effectively with these new pieces of notation, we shall enlarge R to a set R* called the extended real numbers by defining
An ordering on R* is defined by taking -cc < r < +cc for every member r of R and by retaining the usual ordering within R . It is immediate from this definition that inf x 5 sup x XEE
xEE
if E is ally noiieii~ptysubset of R . In fact, we call enlarge tlie defiiiitioiis of iiif,,, x and sup,,, x in obvious fashion to include the case that E is any nonempty subset of R*,and we still have inf 5 sup. With the ordering in place, we can unambiguously speak of open intervals (a,b) ,closed intervals [a,b] ,and halfopen intervals [a, b) and ( a , b] in R* even if a or b is infinite. Under our definitions the intervals of R are the intervals of R* that are subsets of R , even if a or b is infinite. If no special mention is made whether an interval lies in R or R*, it is usually assumed to lie in R. The next step is to extend the operations of arithmetic to R*. It is important not to try to make such operations be everywhere defined, lest the distributive laws fail. Letting r denote any member of R and a and b be any members of R*, we make the following new definitions:
+m if r s- 0, Multiplication:
r (+co) = (+oo)r = -m i f r ( 0 ,
1. Review of Real Numbers, Sequences, Continuity
Addition:
r+(+a)=(+m)+r=+m, r + (-00) = ( - m ) + r = -00, (+00) (-00)
Subtraction: Division:
a
-
+(+a) =+m,
+ (-00) b = a + (-b)
= -00.
whenever the right side is defined.
a/b=O
ifa ~Randbis*oo,
a / b = b-'a
if b
E
R with b # 0 and a is *oo
The only surprise in the list is that 0 times anything is 0. This definition will be important to us when we get to measure theory, starting in Chapter V. It is now a simple matter to define convergence of a sequence in R*. The cases that need addressing are that the sequence is in R and that the limit is +m or -oo. We say that a sequence {a,} in R tends to +oo if for any positive number M , there exists an integer N such that a, > M for all n > N . The sequence tends to -m if for any negative number -M , there exists an integer N such that a, 5 -M for all n > N . It is important to indicate whether convergence/divergence of a sequence is being discussed in R or in R* . The default setting is R , in keeping with standard terminology in calculus. Thus, for example, we say that the sequence {n},21 diverges, but it converges in R* (to + m ) . With our new definitions every monotone sequence converges in R*. For a sequence {a,} in R or even in R*,we now introduce members lim sup, a, and lim inf, a, of R*. These will always be defined, and thus we can apply the operations limsup and liminf to any sequence in R*. For the case of lim sup we define b, = sup,,,- ak as a sequence in R*. The sequence {b,} is monotone decreasing. Thus it converges to inf, b, in R*. We define2 lirn sup a, = inf sup ak n
n k2n
as a member of R* ,and we define lirn inf a, n
=
sup inf ak n
k>n
as a member of R* . Let us underscore that lirn sup a,, and lirn inf a,, always exist. However, one or both may be *oo even if a, is in R for every n.
Proposition 1.7. The operations lim sup and lim inf on sequences {a,} and {b,} in R* have the following properties: (a) if a, 5 b, for all n , then lirn sup a, 5 lirn sup b, and lirn inf a, 5 lirn inf b, , he notation
was at one time used for lim sup, and ~ !J I was used for liminf.
8
I. Theory o f Calculus in One Real Variable
(b) lim inf a, 5 lim sup a,, (c) {a,} has a subsequence converging in R* to lirn sup a, and another subsequence converging in R* to lim inf a,, (d) lim sup a, is the supremum of all subsequential limits of {a,} in R*, and lirn inf, is the infimum of all subsequential limits of {a,} in R*, (e) if lirn sup a,, (: +on, then lirn sup a,, is the infimum of all extended real numbers a such that a, 2 a for only finitely many n , and if lirn inf a, > - oo , then lirn inf a, is the supremum of all extended real numbers a such that a, 5 a for only finitely many n , ( f ) the sequence {a,} in R* converges in R* if and only if liminfa, lirn sup a,, and in this case the limit is the common value of lirn inf a, and lirn sup a,.
-
REMARK.It is enough to prove the results about lim sup, since lim inf a, lim sup (-a,).
=
PROOFSFOR lirn sup. (a) From a1 5 bl for all I, we have a1 5 sup,,, bk if I 2 n . Hence supl,,- a1 5 supk2, bk. Then (a) follows by taking the limit o n n . (b) This follows by taking the limit on n of the inequality infh,, a h 5 supk>, a h . (c) We divide matters into cases. The main case is that a = lim sup a, is in R. Inductivcly, for cach 1 > 1, choosc N > 121-1 such that I SUPkrN a k - a 1 iI-'. Then choose nl > nl-1 such that la, - sup,,, akl < I-'. Together these a,, = a . 'I'he inequalities imply la, - a1 < 21-' for all I , and thus liml,, second case is that a = limsup a, equals +a.Since supkrn a, is monotone decreasing in n , we must have sup,,,- a, = +on for all n . 1nductively for 1 > 1, we can choose nl > nl-1 such that a,, ? I. Then liml,, a,, = +m. The third case is that a = lirn sup a, equals o o . The sequence b, = supk2, a k is monotone decreasing to -oo. Inductively for I 1, choose nl > nl-1 such that = -oo. b,, 5 -1. Thena,, 5 b,, 5 -1, andliml,,a,, (d) By (c), lirn sup a, is one subsequential limit. Let a = lim,,, a,, be another subsequential limit. Put b, = supl,, al. Then {b,} converges to lim sup a, in R*, and the same thing is true of every subsequence. Since a,,, 5 suplznka1 = b,, for all k , we can let k tend to infinity and obtain a = limhioo a,, 5 limk,, b,, = lirn sup a,. (e) Since limsup a, < +oo, we have sup,,,- a, < +on for n greater than or equal to solne N . For this N and any a > supkrN a,, we then have a, 3 a only finitely often. Thus there exists a E R such thaca, 2 a for only finitely many n . On the other hand, if a' is a real number < lirn sup a,, then (c) shows that a, 3 a' for infinitely many n . Hence lirn sup a, 5 inf {a I a,
> a for only finitely many a}.
1. Review o f Real Numbers, Sequences, Continuity
9
Arguing by contradiction, suppose that iholds in this inequality, and let a" be a real number strictly in between the two sides of the inequality. Then sup,,, ak < a" for n large enough, and so a, > a" only finitely often. But then a" is inthe set {a I a,
> a for only finitely many a},
and the statement that a" is less than the infimum of this set gives a contradiction. (f) If {a,) converges in R*, then (c) forces lirn inf a, = lirn sup a,. Conversely suppose lirn inf a, = lirn sup a,, and let a be the common value of lirn inf a, and lim sup a,. The main case is that a is in R.Let t > 0 be given. By (e), a, > a t only finitely often, and a, 5 a - t only finitely often. Thus la, - a1 < t for all n sufficiently large. In other words, lima, = a as asserted. The other cases are that a = +m or a = -m, and they are completely analogous to each other. Suppose for definiteness that a = +m. Since liminf a, = +GO, the monotone increasing sequence b, = infk,, ak converges in R*to + X I .Given M, choose N such that h, 1M for n 1N. Then also a, 1M for n 1N, and a, converges in R*to foe. This completes the proof.
+
With Proposition 1.7 as a tool, we can now prove the Bolzano-Weierstrass Theorem. The remainder of the section will consist of applications of this theorem, showing that Cauchy sequences in R converge in R,that continuous functions on closed bounded intervals of lK are uniformly continuous, that continuous functions on closed bounded intervals are bounded and assume their maximum and minimum values, and that continuous functions on closed intervals take on all intermediate values.
Theorem 1.8 (Bolzano-Weierstrass). Every bounded sequence in R has a convergent subsequence with limit in R. PROOF.If the given bounded sequence is {a,}, form the subsequence noted in Proposition 1 . 7 that ~ converges in R*to lirn sup a,. All quantities arising in the formation of lirn sup a, are in R,since {a,) is bounded, and thus the limit is in R. A sequence {a,} in R is called a Cauchy sequence if for any exists an N such that la, - a, I < t for all n and m that are > N .
t
> 0, there
EXAMPLE.Every convergent sequence in R with limit in R is Cauchy. In fact, let a = lirn a,, and let c > 0 be given. Choose N such that n > N implies la, - a1 < t . Then n , m N implies la,
-
a,I 5 la,
Hence the sequence is Cauchy.
-
a1 + la
-
a,I < t
+ t = 2t
10
I. Theory o f Calculus in One Real Variable
In the above example and elsewhere in this book, we allow ourselves the luxury of having our final bound come out as a fixed multiple M t of t , rather than t itself. Strictly speaking, we should have introduced t' = t / M and aimed for t' rather than t . Then our final bound would have been Me' = t . Since the technique for adjusting a proof in this way is always the same, we shall not add these extra steps in the future unless there would otherwise be a possibility of confusion. This convention suggests a handy piece of terminology-that a proof as in the above example, in which M = 2 , is a " 2 t proof." That name conveys a great deal of information about the proof, saying that one should expect two contributions to the final estimate and that the final bound will be 2 t . Theorem 1.9 (Cauchy criterion). Every Cauchy sequence in R converges to a limit in R. PROOF.Let {a,} be Cauchy in R. First let us see that {a,} is bounded. In fact, for t = 1, choose N such that n , m 2 N implies la, - a, < 1 . Then N , and M = max{lall,. . . , a N - l l , laNl 1 ) is a la,l 5 laNl 1 for m common bound for all la, 1. Since {a,,} is bounded, it has a convergent subsequence {a,,,}, say with limit a , by the Bolzano-Weierstrass Theorem. The subsequential limit has to satisfy 1 a 1 5 M within R*,and thus a is in R. Finally let us see that lima, = a . In fact, if r > 0 is given, choose N such that ?zk N implies la,, - a1 it . Also, choose N' N such that ?z, nz N' implies la, - a, 1 < t. If n 2 N ' , then any nk 2 N' has la, - a,, I < t, and hence la, - a1 5 la, - a,,I + la,, - a1 < t + t = 2t.
+
+
,
This completes the proof. Let f be a function with domain an interval and with range in R. The interval is allowed to be unbounded, but it is required to be a subset of R. We say that f is continuous at a point xo of the domain of f within R if for each t 1 0, there is some S 1 0 such that all x in the domain of f that satisfy Ix - xo I < S have If ( x ) - f ( x o )I < t . This notion is sometimes abbreviated as lim,,,, f ( x ) = f ( x o ) . Alternatively, one may say that f ( x ) tends to f (xo) as x tends to xo, and one may write f ( x ) + f ( x o ) as x + xo. A mathematically equivalent definition is that f is continuous at xo if whenever a sequence has x, + xo within the domain interval, then f (x,) + f ( x o ) . This latter version of continuity will be shown in Section 11.4 to be equivalent to the former version, given in terms of continuous limits, in greater generality than just for R,and thus we shall not stop to prove the equivalence now. We say that f is continuous if it is continuous at all points of its domain.
1. Review o f Real Numbers, Sequences, Continuity
11
We say that the a function f as above is uniformly continuous on its domain if for any t > 0, there is some 6 > 0 such that I f (x) - f (xo) 1 < t whenever x and xa are in the domain interval and Ix - xo I < 8. (In other words, the condition for the continuity to be uniform is that 6 can always be chosen independently of xo .>
EXAMPLE.The function f (x) = x2 is continuous on (-m, foe), but it is not uniformly continuous. In fact, it is not uniformly continuous on [ I , f m ) . Assuming the contrary, choose S f o r t = 1. Then we musthave ( ~ + i ) ~ - . x<~ 1 forallx > 1. ~ u I(x+ t $ j 2 - x 2 = 8 x + $ > Sx,andthisis > 1 forx > s-'.
I
1
Theorem 1.10. A continuous function f from a closed bounded interval [ a ,b] into R is uniformly continuous. PROOF.Fix t > 0. For xo in the domain of f ,the continuity of f at xo means that it makes sense to define
I
6,,(t) = min 1, sup 6' > 0
Ix - xo I < 8' and x in the domain o f f imply If (x) - f (xo)I < t
If IX - xol < Jn,(t)> then I f (x) - f (xo) l < t . h t J ( t ) = infx,g[a,blJx,(t). Let us see that it is enough to prove that S(t) > 0. If x and y are in [ a ,b] with Ix - yl < S(t), then Ix - yl < S(t) I S,(t). Hence I f (x) - f(y)I < t as required. Thus we are to prove that S ( t ) > 0. If S(t) = 0, then, for each integer n > 0, we can choose x, such that SXn(t)< By the Bolzano-Weierstrass Theorem, there is a convergent subsequence, say with x,, + x'. Along this 1 subsequence, SXnk (t) + 0. Fix k large enough so that Ix,, - x'l < zSxl (5). Then I ,f (x,,) - ,f (XI) < Also, Ix - x,, I < iSx,(;) implies
i.
5.
so that
If
(x) - f (xf)I <
5 and
Consequently our arbitrary large fixed k has SXnk> ;SXI(5), and the sequence {Jxnk (t )} cannot be tending to 0.
Theorem 1.11. A continuous function f from a closed bounded interval [ a ,b] into R is bounded and takes on maximum and minimum values. PROOF. Let c = SUP^^,,,^, f (x) in R*. Choose a sequence x, in [ a ,b] with f (x,) increasing to c. By the Bolzano-Weierstrass Theorem, {xn}has a convergent subsequence, say x,, + x'. By continuity, f (x,,) + f (x'). Then f (x') = c, and c is a finite maximum. The proof for a finite minimum is similar.
12
I. Theory of Calculus in One Real Variable
Theorem 1.12 (Intermediate Value Theorem). Let a < b be real numbers, and let f : [ a ,b ] + R be continuous. Then f ,in the interval [ a ,b ], takes on all values between f ( a ) and f ( b ) . REMARK. The proof below, which uses the Bolzano-Weierstrass Theorem, does not make absolutely clear what aspects of the structure of R are essential to the argument. A conceptually clearer proof will be given in Section 11.8 and will bring out that the essential property of the interval [ a ,b] is its "connectedness" in a sense to be defined in that section.
PROOF.Let f ( a ) = a and f ( b ) = p , and let y be between a and p . We may assume that y is in fact strictly between a and p . Possibly by replacing f by - f , we may assume that also a i/I.Let and A = [ x E [ a ,bl I f ( x ) i y } B = Ix E [ a ,61 I f ( x ) > Y 1. These sets are nonempty, since a is in A and b is in B, and f is bounded as a result of Theorem 1.11. Thus the numbers yl = sup [ f ( x ) I x E A } and y2 = inf { f ( x ) I x E B } are well defined and have yl 5 y 5 y2. If yl = y ,then we can find a sequence [x,} in A such that f (x,) converges to y . Using the Bolzano-Weierstrass Theorem, we can find a convergent subsequence [x,,} of [x,}, say wit11 liiiiit xo . By coiitiiiuity of f , [ f (x,,)} coilverges to f ( x o ). Then f ( x o ) = yl = y , and we are done. Arguing by contradiction, we may therefore assume that yl < y . Similarly we may assume that y < y2, but we do not need to do so. Let t = y2 - yl ,and choose, by Theorem 1.10 and uniform continuity, b' > O such that Ixl - x21 < 8 implies If ( x l ) - f (xz)I < t whenever xl and x2 both lie in [ a ,b ] . Then choose an integer n such that 2-n ( b - a ) i8 , and consider the value of f at the points pk = a k2-"(b - a ) for 0 5 k 5 2-". Since pk+l pk = Z p n ( b a ) < 8 , we have I f ( p k + l ) f(pk)I < t = y2 YI. Consequently if f ( p k ) ( y1, then
+
-
-
-
-
~ )y1. NOWf ( p O )= f ( a ) = a 5 y1. Thus induction shows and hence f ( P ~ + 5 that ,f ( p k ) i y1 for all k 5 2-". However, for k = 2", we have p2-n = b, and f (b)= l j ? y > y1, and we have arrived at a contradiction. Further topics. Here a number of other topics of an undergraduate course in real-variable theory fit well logically. Among these are countable vs. uncountable sets, infinite series and tests for their c u ~ ~ v c i g c ~illc ~ c cIaci , il~aicvciy i c r n i a ~ g c ~ rur ~ call ~ ~illIi~liic i bciicb ul pubiiivc ic1111bllab illc balIlc
sum, special sequences, derivatives, the Mean Value Theorem as in Section A2 of the appendix, and continuity and differentiability of inverse functions as in Section A3 of the appendix. We shall not stop here to review these topics, which are treated in many books. One such book is Rudin's Principles of Mathematical Analysis, the relevant chapters being 1 to 5. In Chapter 2 of that book, only the first few pages are needed; they are the ones where countable and uncountable sets are discussed.
2. Interchange of Limits
13
2. Interchange of Limits Let {bij}be a doubly indexed sequence of real numbers. It is natural to ask for the extent to which lirn lirn bi = lirn lirn bij, i
j
j
i
more specifically to ask how to tell, in an expression involving iterated limits, whcthcr wc can intcrchangc thc ordcr of thc two limit opcrations. Wc can vicw matters conveniently in terms of an infinite matrix
The left-hand iterated limit, namely limi limj bij ,is obtained by forming the limit of each row, assembling the results, and then taking the limit of the row limits down through the rows. The right-hand iterated limit, namely limj limi bij, is obtained by forming the limit of each column, assembling the results, and then taking the limit of the column limits through the columns. If we use the particular infinite matrix
then we see that the first iterated limit depends only on the part of the matrix above the main diagonal, while the second iterated limit depends only on the part of the matrix below the main diagonal. Thus the two iterated limits in general have no reason at all to be related. In the specific matrix that we have just considered, they are 1 and 0, respectively. Let us consider some examples along the same lines but with an analytic flavor.
(1) Let bij
Then limi limj bij = 1, while limj lim, b, = 0. i+j (2) Let F, be a continuous real-valued function on R, and suppose that F ( x ) = lim F, ( x ) exists for every x . Is F continuous? This is the same kind of question. It asks whether lirn,,, F ( t ) 2 F ( x ) , hence whether =j .
lirn lirn F, (t)
t+x n+oo
? nlim +m
lim F, ( t ) .
t+x
14
I. Theory o f Calculus in One Real Variable
x'
for k 10 and define Fn ( x ) = fk ( x ) ,then (1 x2Ik each F, is continuous. The sequence of functions {F,} has a pointwise limit x2 The series is a geometric series, and we can easily F(x) = k=O (1 + x2)k calculate explicitly the partial sums and the limit function. The latter is If we take f k ( x ) =
+
Ern
'
F(x) =
ifx = O ifx#O.
l+x2
It is apparcnt that thc limit function is discontinuous. (3) Let { f,} be a sequence of differentiable functions, and suppose that f ( x ) = lim , f , ( x ) exists for every x and is differentiable. Is lim ,fA ( x ) = f f ( x ) ? This question comes down to whether lirn lirn
n-oo
t-x
fn
(t)
-
fn(x)
t -X
lim lim fn(t>- fn(x> n+rn t -x
t-x
An example where the answer is negative uses the sine and cosine functions, which are undefined in the rigorous development until Section 7 on power series. sin nx The exan~plehas fn ( x ) = -for n 3 1. Then limn f, ( x ) = 0 , so that
fi
f ( x ) = 0 and f f ( x ) = 0 . ~ l s hfA(x) , = not tend to 0 = f ' ( 0 ) .
f i cosnx,so that f,I(O) = &does
Yet we know many examples from calculus where an interchange of limits is valid. For example, in calculus of two variables, the first partial derivatives of nice functions -polynomials, for example-can be computed in either order with the same result, and double integrals of continuous functions over a rectangle can be calculated as iterated integrals in either order with the same result. Positive theorems about interchanging limits are usually based on some kind of uniform behavior, in a sense that we take up in the next section. A number of positive results of this kind ultimately come down to the following general theorem about doubly indexed sequences that are monotone increasing in each variable. In Section 3 we shall examine the mechanism of this theorem closely: the proof shows that the equality in question is sup, supj bij = supj sup, bij and that it holds because both sides equal sup,, bij .
Theorem 1.13. Let bij be members of R* that are > 0 for all i and j . Suppose that bzj is monotone increasing in i , for each j , and is monotone increasing in j , for each i . Then J lim - ~lim bij, lim lim b.Z. with all the indicated limits existing in R*.
15
3. Uniform Convergence
PROOF.Put Li = limj bij and L$ = limi bij. These limits exist in R* , since the sequences in question are monotone. Then Li 5 Li+1 and Li 5 L$+, , and thus L = lim Li and L' = lim Lj j
i
both exist in R*. Arguing by contradiction, suppose that L < L'. Then we can choose jo such that Li0 > L . Since Li0 = limi bijo, we can choose io such that bioj, > L . 'I'hen we have L < biojo i Li, i L , contradiction. Similarly the assumption L' iL leads to a contradiction. We conclude that L = L'.
Corollary 1.14. If alj are members of R* that are 10 and are monotone increasing in j for each I , then
in R*, the limits existing. REMARK. This result will be generalized by the Monotone Convergence Theorem when we study abstract measure theory in Chapter V. PROOF.Put bij =
alj
in Theorem 1.13.
Corollary 1.15. If cij are members of R" that are 10 for all i and j , then
in R*, the limits existing. REMARK.This result will be generalized by Fubini's Theorem when we study abstract measure theory in Chapter V. PROOF.This follows from Corollary 1.14.
3. Uniform Convcrgcncc Let us examine more closely what is happening in the proof of Theorem 1.13,in which it is proved that iterated limits can be interchanged under certain hypotheses of monotonicity. One of the iterated limits is L = limi limj bij, and the claim is that L is approached as i and j tend to infinity jointly. In terms of a matrix whose
16
I. Theory o f Calculus in One Real Variable
entries are the various bij's, the pictorial assertion is that all the terms far down and to the right are close to L:
... All terms here are close to L To see this claim, let us choose a row limit Li, that is close to L and then take an entry bioj, that is close to Lie. Then bioj0is close to L , and all terms down and to the right from there are even closer because of the hypothesis of monotonicity. To relate this behavior to something uniform, suppose that L < +m, and let some t > 0 be given. We have just seen that we can arrange to have I L - bij I < t whenever i 1io and j 1jo. Then I Li - bij 1 < r whenever i 1io, provided j > jo.Also, we have 1irn.i bii = Li for i = 1 , 2 , . . . , io - 1. Thus I Li - bii I < t for all i , provided j > ji,, where ji, is some larger index than jo. This is the notion of uniform convergence that we shall define precisely in a moment: an expression with a parameter ( j in our case) has a limit (on the variable i in our case) with ail estimate iiidepeiideiit of tlie parameter. We call visualize matters as in the following matrix:
.i
J;
All terms here on all rows. The vertical dividing line occurs when the column index j is equal to ji,,and all terms to the right of this line are close to their respective row limits Li . Let us see the effect of this situation on the problem of interchange of limits.
...
The above diagram forces all the terms in the shaded part of
i::p K )
be close to one number if lim Li exists, i.e., if the row limits are tending to a limit. If the other iterated limit exists, then it must be this same number. Thus the interchange of limits is valid under these circumstances. Actually, wc can gct by with lcss. If, in thc displaycd diagram abovc, wc assume that all the column limits L,; exist, then it appears that all the column limits with j ji have to be close to the Li's. From this we can deduce that the column limits have a limit L' and that the row limits Li must tend to the limit of the column limits. In other words, the convergence of the rows in a suitable uniform fashion and the convergence of the columns together imply that both
3. Uniform Convergence
17
iterated limits exist and they are equal. We shall state this result rigorously as Proposition 1.16, which will become a prototype for applications later in this section. Let S be a nonempty set, and let f and f,, for integers n 2 1, be functions from S to R. We say that f,(x) converges to f (x) uniformly for x in S if for t for any t > 0, there is an integer N such that n > N implies I fiz(x)- f (x)I all x in S . It is equivalent to say that sup,,, I fn(x) - f (x)I tends to 0 as n tends to infinity. Proposition 1.16. Let bij be real numbers for i > 1 and j > 1. Suppose that (i) Li = limj bij exists in R uniformly in i , and (ii) L,; = limi bij exists in R for each j . Then (a) L = limi Li exists in R, (b) L' = limj L j exists in R, (c) L = L', (d) the double limit on i and j of bij exists and equals the common value of the iterated limits L and L',i.e., for each t > 0, there exist io and jo such that lbij - LI ir whenever i 1io and j jo, (e) Li = limi bij exists in R uniformly in j. REMARK. In applications we shall sometimes have additional information, typically the validity of (a) or (b). According to the statement of the proposition, however, the conclusions are valid without taking this extra information as an additional hypothesis. PROOF.Let r > 0 be given. By (i), choose jo such that lbij - Li I < r for all > jo. With j > jo fixed, (ii) says that lbij - LS I < c whenever i is > some io = io(j). For j > jo and i > io(j),we thenhave
i whenever j
ILi - LiI 5 ILi - bijl
z
+ lbij
-
L)I < t
If j' 3 jo and i io (j'),we similarly have I Li j' > jo,and i > max{iu(j),io (j')},thcn
-
+ t = 2t.
Li, I < 2 t . Hence if j
z jo,
In other words, {L)}is a Cauchy sequence. By Theorem 1.9,L' = limj Li exists in R.This proves (b) . Passing to the limit in our inequality, we have ILi - L'I 5 46 when j jo and in particular when j = jo. If i 2 io(jo),then we saw that Ibijo - Li I < t and I bijo - L)o I < t . Hence i > io (jo)implies
z
18
I. Theory o f Calculus in One Real Variable
Since t is arbitrary, L = limi Li exists and equals L'. This proves (a) and (c). Since limi Li = L, choose i 1 such that I Li - L I < t whenever i > i 1. If i > il and j > jo, we then have
This proves (d). Let il and jo be as in the previous paragraph. We have seen that I L(I - Li, I i4t for j > jo. By (b), I Li - L'I 5 4t whenever j > jo. Hence (c) and the inequality of the previous paragraph give
whenever i > i l and j > jo. By (b), choose jl > jo such that lbij - Lj I < 6t whenever i E ( 1 , . . . , i l - l } and j 1jl.Then j 1jl implies lbij - LiI < 6t for all i whenever j 1jl. This proves (e). In checking for uniform convergence, we often do not have access to explicit expressions for limiting values. One device for dealing with the problem is a uniform version of the Cauchy criterion. Let S be a nonempty set, and let { f,},>l be a sequence of functions from S to R.We say that { f , ( x ) ]is l~niformlyCallchy for x E S if for any t 0, there is an integer N such that n > N and m > N together imply I f, (x) - f, (x) I < t for all x in S.
Proposition 1.17 (uniform Cauchy critcrion). A scqucncc { f,] of functions from a nonempty set S to R is uniformly Cauchy if and only if it is uniformly convergent.
PROOF.If { f,} is uniformly convergent to f ,we use a 2t argument, just as in the example before Theorem 1.9: Given r > 0, choose N such that n 1N implies I fn (x) - f (x)1 < t . Then n > N and m > N together imply
Thus (f, J is uniformly Cauchy. Conversely suppose that { fn] is uniformly Cauchy. Then { f, (x)} is Cauchy for each x . Theorem 1.9 tl~erefol-eshows that there exists a function f : S -t R such that limn f, (x) = f (x) for each x . We prove that the convergence is uniform. Given t > 0, choose N , as is possible since { f,} is uniformly Cauchy, such that n > N and m > N together imply f n(x) - fm (x)1 < t . Letting m tend to oo shows that I fn (x) - f (x) I 5 t for n > N. Hence limn fn (x) = f (x) uniformly for x in S.
19
3. Uniform Convergence
In practice, uniform convergence often arises with infinite series of functions, and then the definition and results about uniform convergence are to be applied to the sequence of partial sums. If the series is CEl ak ( x ),one wants C ; = , a k ( x ) to be small for all m and n sufficiently large. Some of the standard tests for convergence of series of numbers yield tests for uniform convergence of series of functions just by introducing a parameter and ensuring that the estimates do not depend on the parameter. We give two clear-cut examples. One is the uniform alternating series test or Leibniz test, given in Corollary 1.18. A generalization is the handy test given in Corollary 1.19.
I
I
Corollary 1.18. If for each x in a nonempty set S, { a , ( x ) } , > l is a monotone decreasing sequence of nonnegative real numbers such that limn a , ( x ) = 0 uniformly in x , then CEl (- l ) " a , ( x ) converges uniformly.
I
1
PROOF. The hypotheses are such that C ; = , (- l ) k a k ( x ) 5 sup, lam(x)1 whenever n 1 m , and the uniform convergence is immediate from the uniform Cauchy criterion. Corollary 1.19. If for each x in a nonempty set S , { a , ( x ) } , > l is a monotone decreasing sequence of nonnegative real numbers such that limn a , ( x ) - 0 uniformly in x and if { b , ( ~ ) -} , ,is~ a sequence of real-valued functions on S whose partial sums B , ( x ) = bk ( x ) have I B, ( x )I 5 M for some M and all n and x , then C Z ln, ( x ) b , ( x ) converges uniformly. PROOF. If n ? m , summation by parts gives
Let t > 0 be given, and choose N such that ak ( x ) 5 I f n ? m ? N,then
t
for all x whenever k
> N.
and the uniform convergence is immediate from the uniform Cauchy criterion.
I. Theory o f Calculus in One Real Variable
20
A third consequence can be considered as a uniform version of the result that absolute convergence implies convergence. In practice it tends to be fairly easy to apply, but it applies only in the simplest situations.
Proposition 1.20 (Weierstrass M test). Let S be a nonempty set, and let { f,} be a sequence of real-valued functions on S such that If, (x) I i M , for all x in S . Suppose that EnMn < +m. Then f,(x) converges uniformly forx in S.
Xgl
I
PROOF.If n > m > N , then Xi=,+l fk(x)I i X i = , Ifk(x)l i X i = , Mk, and the right side tends to 0 uniformly in x as N tends to infinity. Therefore the result follows from the uniform Cauchy criterion.
EXAMPLES. (1) The series
converges uniformly for - 1 (2) The series
x
1 by the Weierstrass M test with M , = l / n 2 .
converges uniformly for -1 5 x 5 1, but the M test does not apply. To see that the M test does not apply, we use the smallest possible M,, which is M , = n x2+n I = The series X diverges, and hence the M test sup, (- 1) cannot apply for any choice of the numbers M,. To see the uniform convergence of the given series, we observe that the terms strictly alternate in sign. Also,
9.
I
x2
+n
n2
+ ( n + 1) because (n + I,2
x2
x x - >n2 - (n
+
and
1 -
n
1 >n f l '
Finally
uniformly for - 1 ( x 1. Hence the series converges uniformly by the uniform Leibniz test (Corollary 1.18). Having developed some tools for proving uniform convergence, let us apply the notion of uniform convergence to interchanges of limits involving functions of a real variable. For a point of reference, recall the diagrams of interchanges of limits at the beginning of the section. We take the column index to be n and think
3. Uniform Convergence
21
of the row index as a variable t , which is tending to x . We make assumptions that correspond to (i) and (ii) in Proposition 1.16, namely that { f , ( t ) } converges uniformly in t as n tends to infinity, say to f ( t ) ,and that f n ( t ) converges to some limit f, ( x ) as t tends to x . With f, ( x ) defined as this limit, f, is continuous at x . In other words, the assumptions are that the sequence { f,} is uniformly convergent to f and each f,, is continuous.
Theorem 1.21. If { , f n } is a sequence of real-valued functions on [ a ,b] that are continuous at x and if { f,} converges to f uniformly, then f is continuous at x . REMARKS.This is rcally a conscqucncc of Proposition 1.16 cxccpt that onc of the indices, namely t ,is regarded as continuous and not discrete. Actually, there is a subtle simplification here, by comparison with Proposition 1.16, in that { f, ( x ) } at the limiting parameter x is being assumed to tend to f ( x ) . This corresponds to assuming (b) in the proposition, as well as (i) and (ii). Consequently the proof of the theorem will be considerably simpler than the proof of Proposition 1.16. In fact, the proof will be our first example of a 3 t proof. In many applications of Theorem 1.21, the given sequence { f,} is continuous at every x , and then the conclusion is that f is continuous at every x . PROOF.
We write
+
+
I f ( t > - f ( x ) l 5 I f ( t ) - fn(t>l Ifn(t) - fn(x)I Ifn(x> - f ( x ) I . Given t > 0, choose N large enough so that If, ( t ) - f ( t )I < t for all t whenever n ? N . With such an n fixed, choose some 6 of continuity for the function f n , the point x , and the number t-. Each term above is then < t, and hence I f ( t ) - f (x)I < 36. Since c is arbitrary, f is continuous at x . Theorem 1.21 in effect uses only conclusion (c) of Proposition 1.16, which concerns the equality of the two iterated limits. Conclusion (d) gives a stronger result, namely that the double limit exists and equals each iterated limit. The strengthened version of Theorem 1.21 is as follows.
Theorem 121'. If { f,} is a sequence of real-valued functions on [ a ,b] that are continuous at x and if { f,} converges to f uniformly, then for each t > 0, there exist an integer N and a number 6 > 0 such that Ifil(t> -
whenever n
> N and It X I -
f (-u)l
t
< S.
PROOF.If t > 0 is given, choose N such that I f,(t) - f ( t ) 1 < t / 2 for all t whenever n > N , and choose 6 in the conclusion of Theorem 1.21 such that It - xl < S implies I f ( t ) - f ( x ) l < t / 2 . Then Ifn(t>- f ( x ) I
whenever n
> N and It
5
Ifn(t>- f ( t > l
-
xI <
+ If@>f ( x ) I < ;+; -
8.Theorem 1.21' follows.
=t
22
I. Theory o f Calculus in One Real Variable
In interpreting our diagrams of interchanges of limits to get at the statement of Theorem 1.21,we took the column index to be n and thought of the row index as a variable t, which was tending to x . It is instructive to see what happens when the roles of n and t are reversed, i.e., when the row index is n and the column index is the variable t, which is tending to x . Again we have f,(t) converging to f (t) and lim,,, f,, (t) = f,, (x) , but the uniformity is different. This time we want the uniformity to be in n as t tends to x . This means that the S of continuity that corresponds to t can be taken independent of n . This is the notion of "equicontinuity," and there is a classical theorem about it. The theorem is actually stronger than Proposition 1.16 suggests, since the theorem assumes less than that f, (t) converges to f (t) for all t . Let F = {fa I a E A} be a set of real-valued functions on a bounded interval [ a ,b]. We say that F is equicontinuous at x E [ a ,b] if for each t > 0, there is some6 >OsuchthatIt-xi <6impliesIf(t)- f(x)l < tforall f ~ F . T h e s e t F o f functions is pointwise bounded if for each t E [a,b ] ,there exists a number M, such that I f (t) I 5 M, for all f E F.The set is uniforn~lyequicontinuous on [ a ,b] if it is equicontinuous at each point x and if the 6 can be taken independent of x . The set is uniformly bounded on [ a ,b] if it is pointwise bounded at each t E [a, b] and thc bound Mt can bc takcn indcpcndcnt of t.
Theorem 122 (Ascoli's Theorem). If {.f,} is a sequence of real-valued functions on a closed bounded interval [ a ,b] that is equicontinuous at each point of [ a ,b] and pointwise bounded on [ a ,b ] ,then (a) { f,} is uniformly equicontinuous and uniformly bounded on [ a ,b ] , (b) { f,} has a uiiiforiilly coilvergelit subsequence. PROOF.Since each f, is continuous at each point, we know from Theorems 1.10 and 1.11 that each f, is uniformly continuous and bounded. The proof of (a) amounts to an argument that the estimates in those theorems can be arranged to apply simultaneously for all n . First consider the question of uniform boundedness. Choose, by Theorem 1.11, some x, in [ a ,b] with I f,(x,) I equal to K, = sup,E,,, bl I fn(x) . Then choose a subsequence on which the numbers K, tend to sup, K, in R*. There will be no loss of generality in assuming that this subsequence is our whole sequence. Apply the Bolzano-Weierstrass 'I'heorem to find a convergent subsequence {x,, } of {x,}, say with limit xo . By pointwise boundedness, find Mxo with I f , (xo)I 5 Mx, for all n . Then choose some S of equicontinuity at xo for t = 1. As soon as k is large enough so that Ix,, - xo I < 6, we have
Thus 1 + M,, is a uniform bound for the functions f,.
3. Uniform Convergence
23
The proof of uniform equicontinuity proceeds in the same spirit but takes longer to write out. Fix t > 0. The uniform continuity of f, for each n means that it makes sense to define
I f ( x ) - f ( y ) I < twhcncvcr Ix-yl < S f and x and y are in the domain of f If Ix - y 1 < 6, ( t ) ,then If, ( x ) - f, ( y )1 < t . Put 6 ( t )= inf, 6, ( t ). Let us see that it is enough to prove that S ( t )> O: If x and y are in la, 61 with Ix - y I < S ( t ), then Ix - y 1 < S ( t ) 5 S,(t). Hence I f,(x) - f,(y)I < t as required. Thus we are to prove that S ( t ) > 0. If S ( t ) = 0, then we first choose an increasing sequence Ink} of positive integers such that 6,, ( t )< and we next choose xk and yk in [ a ,b] with Ixk - yk 1 < S,, ( t )and I f k ( x k ) - fk ( y k )I > t . Applying the Bolzano-Weierstrass Theorem, we obtain a subsequence { x k l }of { x k }such that { x k l }converges, say to xo. Then
i,
so that { y k l }converges to xo. Now choose, by equicontinuity at xo, a number S f > 0 sucll tliat I f,(x) - fn(xo)1 i for all V L wl.leiiever Ix - xo 1 iS f . The convergence of { x k l }and { y k l }to xo implies that for large enough I , we have lxkl - xo I < 8/12 and 1 yk, - xo 1 < 6/12, Therefore I fk, (xk,)- fk ( x o )1 < and I f k , (yk,) - fk, ( X O ) I < from which we conclude that I f k , (xk,) - f k , (yk,)I < t . But we saw that I j k ( x k ) - jk(yk)l > t for all k , and thus we have arrived at a
5
5,
5
contradiction. This proves the uniform equicontinuity and completes the proof of (a). To prove (b), we first construct a subsequence of { f,} that is convergent at every rational point in [ a ,b ] .We enumerate the rationals, say as x l , xz, . . . . By the Bolzano-Weierstrass Theorem and the pointwise boundedness, we can find a subsequence of { f,} that is convergent at x l , a subsequence of the result that is convergent at x2, a subsequence of the result that is convergent at x s , and so on. The trouble with this process is that each term of our original sequence may disappear at some stage, and then we are left with no terms that address all the rationals. The trick is to form the subsequence { f,,} of the given { f,} whose kth term is the kth term of the kth subsequence we constructed. Then the kth, (k ( k 2)nd,. . . terms of { f,,} all lie in our kth constructed subsequence, and hence { f,,} converges at the first k points x l , . . . , xk. Since k is arbitrary, { f,,} converges at every rational point. Let us prove that { f,,} is uniformly Cauchy. Let t > 0 be given, let 6 be some corresponding number exhibiting equicontinuity, and choose finitely many rationals r l , . . . , rl in [ a , b] such that any member of [ a ,b] is within 6 of at least one of these rationals. Then choose N such that If, ( r j )- f,(rj) 1 < t for
+
+
24
I. Theory o f Calculus in One Real Variable
1 5 j 5 I whenever n and in are > N. If x is in [ a , b], let r ( x ) be an r j with Ix - r ( x )I < 6. Whenever n and m are > N , we then have
Hence { f,,} is uniformly Cauchy, and (b) follows from Proposition 1.17. REMARK. The construction of the subsequence for which countably many convergence conditions were all satisfied is an important one and is often referred to as a diagonal process or as the Cantor diagonal process. EXAMPLE. Let K and M be positive constants, and let F be the set of continuous real-valued functions f on [ a ,b] such that If (t)l K for a t b and such that the derivative f ( t ) exists for a < t < b and satisfies I f ( t ) I 5 M there. This set of functions is certainly uniformly bounded by K , and we show that it is also uniformly equicontinuous. To see the latter, we use the Mean Value Theorem. If x is in the closed interval [ a , b ] and t is in the open interval ( a , b ) , then there exists 6 depending on t and x such that
'
''
From this inequality it follows that the number S of uniform equicontinuity for t and F can be taken to be t / M . The hypotheses of Ascoli's Theorem are satisfied, and it follows that any sequence of functions in F has a uniformly convergent subsequence. The estimate of S is independent of the uniform bound K , yet Ascoli's Theorem breaks down if there is no bound at all; for example, the sequence of constant functions with f , ( x ) = n is uniformly equicontinuous but has no convergent subsequence. We turn now to the problem of interchange of derivative and limit. The two indices again will be ail integer n tliat is tending to iiifiiiity and a parameter t tliat is tending to x . Proposition 1.16 takes away all the surprise in the statement of the theorem, and it tells us the steps to follow in a proof. What the proposition suggests is that the general entry in our interchange diagram should be whatever quantity we want to take an iterated limit of in either order. Thus we expect not a theorem about a general entry f n ( t ) , but instead a theorem about a general entry -
f n ( t ) - f n ( x ) . The limit on n gives us for a limiting function f , t-x t-x and then the limit as t + x gives us f f( x ). In the other order the limit as t + x gives us f A ( x ) , and then we are to consider the limit on n. If Proposition 1.16 is
3. Uniform Convergence
25
to be a guide, we are to assume that the convergence in one variable is uniform in the other. The proposition also suggests that if we have existence of each row limit and each column limit, then uniform convergence when one variable occurs first is equivalent to uniform convergence when the other variable occurs first. Thus we should assume whichever is easier to verify.
Theorem 1.23. Suppose that { f,} is a sequence of real-valued functions continuous for a 2 t 2 b and differentiable for a < t < b such that { f ; } converges uniformly for a < t < b and { f , ( x o ) }converges in R for some xo with a 5 xo 5 b. Then { f,} converges uniforiilly for a 5 t 5 b to a functioi~f , and f ' ( x ) = limn f A ( x ) for a < x < b, with the derivative and the limit existing. REMARKS.Thc convcrgcncc of { f ( x o ) } cannot bc droppcd complctcly as a hypothesis because f,(t) = n would otherwise provide a counterexample. In practice, { f,} will be known in advance to be uniformly convergent. However, uniform convergence of { f,} is not enough by itself, as was shown by the example sin n x f n ( x ) = -in Section 2.
&
PROOF.The first step is to apply the Mean Value Theorem to f, - f m , estimate f; - f k , and use the convergence of { fn ( x o ) }to obtain the existence of the limit function f . The Mean Value Theorem produces some 6 strictly between t and xo such that
Our hypotheses allow us to conclude that { f,(t)} is uniformly Cauchy, and thus { f,} converges uniformly to a limit function f by Proposition 1.17. The second step is to apply the Mean Value Theorem again to f , - f m , this time to see that
converges uniformly in t (for t # x ) as n tends to infinity, the limit being q ( t ) = (I) . In fact, the Mean Value Theorem produces some strictly between t-x t and x such that 'Pn( t ) - 'Pm ( t ) =
[ f n ( t ) - fm(t)l - [ f n ( ~-) fm(x)I = f; t-x
0 )- f:,
0 )
9
and the right side tends to 0 uniformly as n and m tend to infinity. Therefore {rp,(t)} is uniformly Cauchy for t # x , and Proposition 1.17 shows that it is uniformly convergent.
26
I. Theory o f Calculus in One Real Variable
The third step is to extend the definition of rp to x by rp,(x) = f,'(x) and then to see that rp, is continuous at x and Theorem 1.21 applies. In fact, the definition of rp, ( t ) is as the difference quotient for the derivative of f, at x , and thus rp,(t) + f,' ( x ) = rp, ( x ) . Hence rp, is continuous at x . We saw in the second step that rp,( t ) is uniformly convergent for t # x , and we are given that (nlz ( x ) = f; (x) is convergent Therefore (n, ( t ) is uniformly convergent for all t with f ( t ) - f(r-1 f o r t + x, lim rp, ( t ) = t x for t = x. lim f,'(x)
{
Theorem 1.21 says that the limiting function lim rp, ( t ) is continuous at x . Thus
In other words, f is differentiable at x and f '(x) = lirn,
f; ( x ).
4. Riemann Integral
This section contains a careful but limited development of the Riemann integral in one variable. The reader is assumed to have a familiarity with Riernann sums at the level of a calculus course. The objective in this section is to prove that bounded functions with only finitely many discontinuities are Riemann integrable, to address the interchange-of-limits problem that arises with a sequence of functions and an integration, to prove the Fundamental Theorem of Calculus in the case of continuous integrand, to prove a change-of-variables formula, and to relate Riemann integrals to general Riemann sums. The Riemann integral in several variables will be treated in Chapter 111, and some of the theorems to be proved in the several-variable case at that time will be results that have not been proved here in the one-variable case. In Chapters VI and VII, in the context of the Lebesgue integral, we shall prove a much more sweeping version of the Fundamental 'I'heorem of Calculus. First we give the relevant definitions. We work with a function f : [ a ,b] + R with a 5 b in R,and we always assume that f is bounded. A partition P of [ a ,b] is a subdivision of the interval [ a ,b] into subintervals, and we write such a partition as a = xo 5 xl 5 . . . 5 x, = b. The points xj will be called the subdivision points of the partition, and we may abbreviate the partition as P = {xi};="=,In order to permit integration over an interval of zero length, we allow partitions in which two consecutive xi's are
4. Riemann Integral
27
equal; the multiplicity of xj is the number of times that xj occurs in the partition. For the above partition, let Axi
= xi
Mi=
-
xipi,
sup
p(P)
f(x),
= mesh
mi=
of P
inf
= max
Axi,
i
f(x).
X,-l5X5X,
X,-l5X5X,
Pul n
U(P, f )
=
upper Riemann sum for P , C Mx~,. A i=l -
n
mi Axi = lower Riemann sum for P ,
L(P, f ) = i=l -b
f dx = inf U ( P , f ) = upper Riemann integral of f , P
f dx = sup L ( P , f ) = lower Riemann integral of f. P
We say that f is Riemann integrable on [a, b] if
Z
f dx =
/.h f d x , and in
-a
this case we write j;lb f dx for the common value of these two numbers. We write R [ a , b] for the set of Riemann integrable functions on [a, b]. If f 2 0 , an upper Riemann sum for f may be visualized in the traditional way as the sum of the areas of rectangles with bases [xipl, xi] and with heights just sufficient to rise above the graph of f on the interval [xi-1, xi], and a lower sum may be visualized similarly, using rectangles as large as possible so that they lie below the graph. EXAMPLES. (1) Suppose f (x) = c for a 5 x 5 b. No matter what partition P is used, we have Mi = c and mi = c. Therefore U ( P , f ) = L ( P , f ) = c(b - a ) ,
-b
1,
f dx
=
1b f dx
-a
=
c(b - a ) , and f is Riemann integrable on [a, b] with
hbf dx = c ( b - a ) . (2) Let [a, b] be arbitrary with a < b, and let f be 1 on the rationals and 0 on the irrationals. This f is discontinuous at every point of [a, b]. No matter what partition is uscd, wc havc Mi = 1 and mi = 0 whcncvcr Axi 0. Thcrcforc -b
b
U ( P , f ) = b - a a n d L ( P , f)=O.Hence!" f d x = b - a a n d 1 f d x = O , -a and f is not Riemann integrable.
28
I. Theory o f Calculus in One Real Variable
Let us work toward a proof that continuous functions are Riernann integrable. We shall use some elementary properties of upper and lower Riemann sums along with Theorem 1.lo, which says that a continuous function on [ a ,b] is uniformly continuous.
Lemma 1.24, Suppose that f : [ a ,b ] + R has m i f ( x ) i M for all x in [ a ,b ] . Then m ( b - a ) i L ( P , f ) i U ( P ,f ) i M ( b - a ) ,
PROOF.The first conclusion follows from the computation
5
MA C M i a i = U ( P ,f ) 5 i=l i=l
X. -
M(b - a).
If we concentrate on the first, third, and last members of the above inequalities and take the supremum on P , then we obtain the second conclusion. Similarly if we concentrate on the first, sixth, and last members of the above inequalities and take the infimum on P , then we obtain the third conclusion. A refinement of the partition P is a partition P* containing all the subdivision points of P , with at least their same multiplicities. If P1 and P2 are two partitions, then P1 and P2 have at least one common refinement: one such common refinement is obtained by taking the union of the subdivision points from each and repeating each such point with the maximum of the multiplicities with which it occurs in PI and P2. We use this notion in order to prove a second lemma.
Lemma 125. Let f : [ a ,b ] + R satisfy rn f (x) M for all x in [ a ,b ] . Then (a) L ( P , f ) i L ( P * , f ) and U ( P * ,f ) i U ( P , f ) whenever P is apartition of [ a ,b] and P* is a refinement, (b) L ( P I ,f ) 5 U ( P 2 ,f ) whenever P1 and P2 are partitions of [ a ,b],
4. Riemann Integral
29
(d) ~ b f d x - l ~ f d x 5 ( ~ - n l ) ( b - a ) , (e) the function- is Riemann integrable on [a, b] if and only if for each
7
PROOF.In (a), it is enough to handle the case in which P* is obtained from P by including one additional point, say x* between xi-1 and xi. The only possible difference between L ( P , f ) and L ( P * , f ) comes from [xi-l, xi], and there we havc inf
f(x)(xZ-xZ-l)=
inf
f(x)(xZ-x*)+
5
inf
inf
f(x)(~*-x,-~)
x~[x,-l,xz1
x~[xz-l,xz1
xE[xz-l,xzl
f(x)(xi-x*)+
inf
f(x)(~*-x~-~).
xE[x*,xzl
xE[x,-I,x*I
Hence L ( P , f ) i L(P*, f ) , and similarly U ( P * , f ) i U ( P , f ) . This proves (a). Let P* be a common refinement of P1 and P2. Combining (a) with Lemma 1.24 gives L(P1, f ) 5 U P * , f ) 5 U(P*, f ) 5 U(P2, f ) . This proves (b). Conclusion (c) follows by taking the supremum on P1 and then the infimum on P2, and conclusion (d) follows by subtracting the second conclusion of Lemma 1.24 from the third. For (e), we have
for any partitions P1and f i of [a, b]. Riemann integrability means that the center two members of this inequality are equal. If they are not equal, then there certainly -b
canexistnoPwithU(P, f ) - L ( P , f ) < t i f c = I a f d x - ~ ' ~ f d x . O n t h e -a other hand, equality of the center two members, together with the definitions of the lower and upper Riemann integrals, means that for each t- > 0, we can choose PI and P2 with U (P2, f ) - L ( P I , f ) < t . Letting P be a common refinement of P1 and P2 and applying (a), we see that U ( P , f ) - L ( P , f ) < t . This proves (e).
Theorem 1.26. If f : [a, b] + R is continuous on [a, b ] ,then f is Riemann integrable on [ a ,b]. PROOF.From Theorem 1.10we know that f is uniformly continuous on [a,b ] . Given t > 0, we can therefore choose some number 6 > 0 corresponding to f and t on [a, b ] . Let P = be a partition on [ a ,b] of mesh p ( P ) < 8. On any
I. Theory o f Calculus in One Real Variable
30
subinterval [xi-1, xi] corresponding to P , we have mi = f (ti) and Mi = f (ri) forsometi andqi in[xi-l,xi],byTheoreml.ll. Since Iqi 5 Ixi-xi-lI = Axi 5 p ( P ) < S, we obtain Mi - mi = f (qi)- f (ti) < t. Therefore
and thc thcorcm follows from Lcmma 1 . 2 5 ~ . We shall improve upoii Theore111 1.26 by allowing finitely illany poiiits of discontinuity, but we need to do some additional work beforehand.
-
Lemma 1.27. If f is bounded on [a, b] and a 5 c 5 b, then
1
dx =
1'.
7;
dx, and similarly for Consequently f is in R [ a , b] if and f dx + -a only if f is in both R [ a ,c] and R[c, b], and in this case,
1,
REMARKS.After one is done developing the Riemann integral and its properties, it is customary to adopt the convention that J; f dx = - Jab f dx when b ia . One of the places that this convention is particularly helpful is in applying the displayed formula of Lemma 1.27: the formula is then valid for all real a , b, c without the assumption that a , b, c are ordered in a particular way. PROOF.If P1 and P2 are partitions of [ a ,c] and [c, b], respectively, let P be their "union," which is obtained by using all the subdivision points # c of each partition, together with c itself. The multiplicity of c in P is to be the larger of the numbers of times c occurs in P1 and P2. This P is a partition of [a, b]. Then
Taking the infimum over PI and then the infimum over P2, we obtain
For the reverse inequality, let
1
t
> 0 be given, and choose a partition P of
[a.b] with U ( P , f ) f dx iC . Let P* be the refinement of P obtained by adjoining c to P if c is not a subdivision point of P or by using P itself if c is a
4. Riemann Integral
31
1
subdivision point of P Lemma 1.25a gives U (P*,f ) f dx c t . Because c is a subdivision point of P * , the subdivision points c give us a partition P1 of
[ a ,c ] and the subdivision points > c give us a partition P2 of [ c , b ] . Moreover, P* is the union of P1 and P2. Then we have
Since t is arbitrary, the lemma follows.
Lemma 1.28. Suppose that f : [ a ,b] + R is bounded on [ a ,b] and that a 5 c 5 b. If for each 6 > 0 , f is Riemann integrable on each closed subinterval of [ a ,b] - { x Ix - cl > G} ,then f is Riemann integrable on [ a ,b]. PROOF.We give the argument when a < c < b , the cases c = a and c = b being handled similarly. Since f is by assumption bounded, find m and M with m 5 f (x)5 M for all x E [ a ,b ] . Choose S > 0 small enough so that a < c - S < c < c + S < b. To simplify the notation, let us drop "f dx" from all integrals. Since f is by assumption Riemann integrable on [ a ,c - S ] and [ c + 6 , b],Lemma 1.27 gives
-b
Since G is arbitrary, la = l bThe . lemma follows. -
a
Proposition 1.29. If f : [ a ,b] ; . R is bounded on [ a ,b] md is continuous at all but finitely many points of [ a ,b ] ,then f is Riemann integrable on [ a ,b]. REMARK.There is no assumption that f has only jump discontinuities. For example, the proposition applies if [ a ,b] = [O, 11 and ,f is the function with f (x)= sin for x # 0 and f (0)= 0. PROOF.By Lemma 1.27 and induction, it is enough to handle the case that f is discontinuous at exactly one point, say c . Since f is bounded and is continuous at all points but c , Theorem 1.26 shows that the hypotheses of Lemma 1.28 are satisfied. Therefore Lemma 1.28 shows that f is Riemann integrable on [ a ,b].
32
I. Theory o f Calculus in One Real Variable
We shall now work toward a theorem about interchanging limits and integrals. The preliminary step is to obtain some simple properties of Riemann integrals.
Proposition 130. If f , f l , and f 2 are Riemann integrable on [a, b], then
+
+
+
(a) f l f2 is in W a , bl and J~~ if1 f2) dx = iab f~dx lab fidx. (b) cf is in R [ a , b] and /ab cf dx = c J;I~f dx for any real number c, (c) f l i f2 on [a, bl implies J~~ f l dx i lab f~d x , (d) m 5 f 5 M on [a, b] and rp : [ m , MI + R continuous imply that rp o f is in %?[a, b], (el l f l i s i n R [ a , b l , a n d I ~ ~ ~ fi d lxa b l f l d x , (0 f 2 a n d f 1 f 2 a r e i n R [ a , b ] , (g) f l is in R [ a , b] i f f ? 0 on [a, b], (h) the function g with g(x) = f (-x) is in R[-b, -a] and satisfies 11;R d x = f d x .
lab
REMARK.The proof of (c) will show, even without the assumption of Riemann integrability, that f l dx 5 f2 dx and lb f l dx 5 lb f2 dx . We shall make -a -a use of this stronger conclusion later in this section. ~ O O F For .
inf
XE[X~-I,X~I
(a), write f
fl(x)+-
= fl
+ fi, and let P be a partition. From
inf xE[xz-l,"zl
and a similar inequality with the supremum, we obtain L ( P , fl)
+ L ( P , f2) 5 L ( P , f ) 5 U ( P , f ) 5 U ( P , fl) + U ( P , f2).
Let t > 0 be given. By Lemma 1.25e, choose P1 and P2 with
If P is a common refinement of P1 and P2, then Lemma 1.25a gives U ( P , fl)
-
L ( P , fl) < 1 and
U ( P , f2)
-
L ( P , f2) < 1.
Hence U ( P j fi) i U ( P j fd i
l l
b
+ 21, f2dx + i L i p , f2) + 21, fl
b
dx
+
1i U
1
P , fl)
(*)
4. Riemann Integral
33
and (*) yields U ( P , f ) - L ( P , f ) 5 4t. Since t is arbitrary, Lemma 1.25e shows that f is in R [ a ,b]. From the inequalities for U ( P , f l ) and U ( P , f2), combined with the last inequality in (*), we see that
while the first inequality in (*) shows that
+
+
(f l fi) dx = Since t is arbitrary, we obtain J~~ f l dx f2 d x . This proves (a). For (b), consider any subinterval [xipl, xi] of a partition, and let mi and Mi be the infimum and supremum of f on this subinterval. Also, let mi and Mi be the infimum and supremum of cf an this subinterval. If c > 0, then Mi/ = cMi and mi = cmi ,so that ( M : - m:)Axi = c(Mi - mi)Axi. If c 5 0 , then Mi = -cmi andmi = -cMi, so thatwe still have (M,'-mi)Axi = c(Mi -mi)Axi. Summing oni,weobtain U ( P , cf) L ( P , cf) = c ( U ( P , f ) L ( P , f)),and(b)follows from Lcmma 1.2%. For (c), we have f l d x 5 U ( P , f l ) 5 U ( P , f2) for all P. Taking the -
-
zb
infimum on P in the inequality of the first and third members gives -b
laf 2 d x .
-b fl
dx 5
(Similarly l b f l dx 5 l b f 2 d x . but this is not needed under the hypothesis that fi and ?;are ~ieman: integrable .) For (d), let K = sup,,,,,,, I p(t) 1. Let t > 0 be given, and choose by Theorem 1.10 some S of uniform continuity for cp and t . Without loss of generality, we may assume that S 5 t . By Lemma 1.25e, choose a partition P = of [ a ,b] such that U ( P , f ) - L ( P , f ) < ~ 7 On ~ . any subinterval [xipl,xi] of P , let mi and Mi be the infimum and supremum of f , and let mi and Mi be the infimum and suprernum of y, o f . Divide the set of integers { I , . . . , n } into two subsets- the subset A of integers i with Mi - mi < 6 and the subset B of integers i with Mi - mi > 8 . If i is in A , then the definition of S makes Mi - mi 5 t-. If i is in B, then the best we can say is that Mi' - mi 5 2 K . However, on B we do have Mi - mi 3 8 , and thus
34
I. Theory o f Calculus in One Real Variable
Thus Ci,, Axi < S and
Since t is arbitrary, the Riemann integrability of rp o f follows from Lemma 1.25e. Fur (e), the first cunclusiun fullows fro111(d) with ~ ( 1 = ) 11 I . Fur the asserted f dx 5 inequality we have f 5 If 1 and - f 5 I f 1, so that (c) and (b) give lab If 1 dx and - Jab f dx 5 J~~ 1 f 1 dx. Combining these inequalities, we obtain
lab
I ~ ~ ~ ~5J)2Dlfldx. f d x l For (f), the first conclusion follows from (d) with rp(t) = t2. For the Riemann integrability of f l f2, we use the formula f l f 2 = +(( fl f212 - f; - f22) and the earlier parts of the proposition. Conclusion (g) follows from (d) with rp(t) = ,,h. For (h), each partition P of [ a ,b ] yields a natural partition P' of [-b, -a] by using the negatives of the partition points. When k' and P' are matched in this way, U(P, f ) = U ( P f , g ) a n d L ( P ,f ) = L ( P f , g ) .Itisimmediatethat f E R [ a , b ] implies g E R[-b, -a] and that g dx = I ; ~ f dx. This completes the proof.
+
11;
The next topic is the problem of interchange of integral and limit. EXAMPLE. On the interval [0, 11, define f,(x) to be n for 0 < x < l / n and to be 0 otherwise. Proposition 1.29shows that f, is Riemann integrable, and Lemma 1 1.27 allows us to see that Jb f, dx = 1for all n . On the other hand, limn f, (x) = 0 1 1 forallx E [0, 11. Since& Odx =O,wehaveJb f,dx = 1 # O = l i m , l f,dx. Thus the interchange is not justified without some additional hypothesis.
Theorem 1.31. If { f,} is a sequence of Riemann integrable functions on [ a ,b ] and if { f,} converges uniformly to f on [ a ,b], then f is Riemann integrable on b [ a ,b ] ,andlim, lof,dx = lob f dx. REMARKS.Proposition 1.16 suggests considering a "matrix" whose entries are the quantities for which we are computing itelated limits, and these quantities are U ( P , f,) here. (Alternatively, we could use L(P, f,) .) The hypothesis of uniformity in the statement of Theorem 1.3 1,however, concerns f, ,not U ( P , f,) . In fact, the tidy hypothesis on f, in the statement of the theorem implies a less intuitive hypothesis on U (P, f,) that has not been considered. The proof conceals these details.
4. Riemann Integral
35
PROOF.Using the uniform Cauchy criterion with t = 1 , we see that there exists N such that I fn ( x )I 5 MN 1 for all x whenever n > N . It follows from the boundedness of f1, . . . , fN-1 that the I f,l are uniformly bounded, say by M. Then also If (x)l i M for all x . Put en = sup, I f,(x) - f ( x ) I ,so that fn - E , 5 f 5 f , E,. Proposition 1 . 3 0 and ~ the remark with the proposition, combined with Lemma 1 . 2 5 , then yield
+
+
1
Hence f d x - / f d x 5 k , ( b - a ) for all n . The uniform convergence of -a { f,} to f forccs E , to tcnd to 0, and thus f is in R [ a ,b]. Thc displaycd cquation, in light of the Riemann integrability of f , shows that b
The right side tends to 0, and therefore limn
L~f, d x = hbf d x .
EXAMPLE. Let f : [O, 11 + R be defined by
f (XI
=
l/q
if x is the rational p / q in lowest terms if x is irrational.
This function is discontinuous at every rational and is continuous at every irrational. Its Riemann integrability is not settled by Proposition 1.29. Define 1/ q
if x is the rational p / q in lowest terms, q 5 n if x is the rational p / q in lowest terms, q > n if x is irrational.
Propositivn 1.29shows that f, is Riemann integrable, and Lemma 1.27shows that 1 f, d x = 0. Since If, ( x ) - f ( x )1 (: l / n for all x , { f,} converges uniformly 1 to f . By Theorem 1.3 1, f is Riemann integrable and So f d x = 0 .
1,
Theorem 1.32 (Fundamental Theorem of Calculus). If f : [ a ,b] + R is continuous, then (a) the function G ( x ) = f dt is differentiable for a < x < b with derivative f ( x ) ,and it is continuous at a and b with G ( a ) = 0 , (b) any continuous function F on [ a ,b] that is differentiable for a < x < b with derivative f ( x ) has f dt = F ( b ) - F ( a ) .
lax
s,~
I. Theory o f Calculus in One Real Variable
36
REMARK.The derivative of G ( x ) on ( a ,b ) , namely f ( x ) ,has the finite limits f ( a ) and f ( b ) at the endpoints of the interval, since f has been assumed to be continuous on [ a ,b ] . Thus, in the sense of the last paragraph of Section A2 of the appendix, G ( x ) has the continuous derivative f ( x ) on the closed interval [ a ,b ]. PROOFOF (a). Riemann integrability of f is known from Theorem 1.26. For h > 0 small enough to make x h < b , Lemma 1.27 and Proposition 1.30 give
+
G( x
+ 1%)
-
h
G(x) -
f(x) =
g fdt
J~~~~ fdt
-
h
f
(x)
x+h
and hence If t > 0 is given, choose the S of continuity for f and t at x . Then 0 < h 5 S implies that the right side is 5 t . For negative h , we instead take h > O and consider G(x - h) - G(x) 1 1 fdt- f(x)= -h Then
as required. PROOFOF (b). The functions F and G are two continuous functions on [ a ,b ] with equal derivative on ( a ,b ) . A corollary of the Mean Value Theorem stated in Section A2 of the appendix implies G = F + c for some constant c . Then
lb
f dt
=
G ( b ) - 0 = G ( b )- G ( o ) = F ( b )
+c
-
F(a) - c = F(b) - F(a).
Corollary 1.33 (integration by parts). Let f and g be real-valued functions defined and having a continuous derivative on [ a ,b ]. Then
REMARK.The notion of a continuous derivative at the endpoints of an interval is discussed in the last paragraph of Section A2 of the appendix.
4. Riemann Integral
37
PROOF.We start from the product rule for differentiation, namely
and we apply lab to both sides. Taking Theorem 1.32 into account, we obtain the desired formula.
Theorem 1.34 (change-of-variables formula). Let f be Riemann integrable on [ a ,b] ,let rp be a continuous strictly increasing function from an interval [A, B] onto [a, b] ,suppose that the inverse function q-' : [a, b] + [A, B] is continuous, and suppose finally that rp is differentiable on (A, B) with uniformly continuous derivative p'. Then the product (f o p)p' is Riemann integrable on [A, B], and
REMARKS.The uniform continuity of rp' forces rp' to be bounded. If rp' were also assumed positive on (A, B), then the continuity of p-' on (a, b) would be automatic as a consequence of the proposition in Section A3 of the appendix. The result in the appendix is not quite good enough for current purposes, and thus we have assumed the continuity of q-' on [a, b]. It will be seen in Section 11.7 that the continuity of q-' on [a, b] is automatic in the statement of Theorem 1.34 and need not be assumed. PROOFIF f 2 0. Given t > 0, choose some q of uniform continuity for cp' and r ,and then choose, by Theorem 1.lo, some S of uniform continuity for q-' and 7 . Next choose a partition P = {xi on [ a ,b] such that U ( P , f ) - L (P, f ) < t . Possibly by passing to a refinement of P , we may assume that p ( P ) < 6. Let Q be the partition {yi};=2=o of [ A , B] with yi = q-' (xi). Then p ( Q ) < q . The Mean Value Theorem gives Axi = (Ayi)pf(Ci)for some C between yi-1 and yi . On [A, B], q' is bounded; let in; and Mi* be the infimum and supremum of rp' on [yi-1, yi], so that mT i rpl(() i M: and mTAyi i Axi i M:Ayi. Siiice p ( Q ) < 11, we liave Mjc - rrl; 5 r . Tlieii we have
Wllc11cvt:r F arid G art: 3 0 un a ~.urrirnurldvrrrairl and x is in llial clurrrain, (inf G) F(x) 5 G (x) F (x); taking the supremum of both sides gives the inequality (inf G) (sup F ) 5 sup(FG). Also, sup(FG) 5 sup(F) sup(G) . Applying these inequalities with G = rp' and F = f o rp yields
I. Theory o f Calculus in One Real Variable
38
Subtraction of the right-hand inequality of the first display and the left-hand inequality of the second display shows that
while subtractivrl of the right-hand inequality of the secorid display arld the lefthand inequality of the first display gives
Therefore
Similarly
IL(P, f
)
-
L(Q,(
f 0 v>vf>l it M ( B - A),
and hence
Since t is arbitrary, Lemma 1.25e shows that ( f o q ) p 1 is in R [ A , B ] . Our inequalities imply that
Ijabf d x IU(P3 and
lu(Q,( f
I
0u
f)
-
? ) ~ ' )-
-
U(Q3 (
U ( P , f)I 5 t ,
f 0 ~ > v 5~ c) M l (B - A),
1: ( f P ) P ' ~ Y I
5 2 t M ( B - A).
0
I
Addition shows that lab f d x - 1: ( f o p ) p f d y 5 c b is arbitrary, laf d x = ( f o p)pl d y .
1;
+3cM(B
-
A ) . Since c
PROOF FOR GENERAL f . The special case just proved shows that the result holds for f c for a suitable positive constant c, as well as for the constant function c. Subtracting the results for f c and c gives the result for f , and the proof is complete.
+
+
4. Riemann Integral
39
Iff is Riemann integrable on [ a ,b],then U ( P , f ) and L ( P , f ) tend to /ab f d x as P gets finer by insertion of points. This conclusion tells us nothing about finelooking partitions like those that are equally spaced with many subdivisions. The next theorem says that the approximating sums tend to J~~f d x just under the assumption that p ( P ) tends to 0. Relative to our standard partition P = ( x iJy=o, let t, for 1 5 i 5 n satisfy xi-1 5 ti 5 x i , and define
This is called a Riemann sum of f
Theorem 135. If f is Riemann integrable on [ a ,b],then lim S ( P , { t i } f, )
=
f dx.
fi(P)+'J
Conversely if f is bounded on [ a ,b] and if there exists a real number r such that for any t > 0, there exists some 6 > 0 for which I S ( P , {ti}, f ) - r 1 < t whenever p ( P ) < 8, then f is Riemann integrable on [ a ,b]. PROOF. For the direct part the function f is assumed bounded; suppose If (x)l 5 M on [a,b]. Let t > 0 be given. Choose a partition P" of [ a ,b] with b U ( P *, f ) 5 /a f d x t . Say P* is a partition into k intervals. Put 81 = & , and suppose that P is any partition of [ a ,b] with p ( P ) 5 S1. In the sum giving U ( P , f ) , we divide the terms into two types -those from a subinterval of P that does not lie within one subinterval of P* and those from a subinterval of P that does lie within one subinterval of P*. Each subinterval of P of the first kind has at least one point of P* strictly inside it, and the number of such subintervals is therefore 5 k - 1 . Hence the sum of the corresponding terms of U ( P , f ) is
+
5 (kThe sum of the terms of the second kind is 5 U ( P *, f ) . Thus u ( p ,f ) 5
t
+ U ( P * ,f ) 5
Lb
f'dx
+ 2t.
Similarly we can produce 62 such that p ( P ) 5 62 implies
I. Theory o f Calculus in One Real Variable
40
If S = min{S1,S2}and p ( P ) 5 8, then
lab
andhence S ( P , f ) f d x 5 2c. For the converse let t > 0 be given, and choose some S as in the statement of the
I
-b
I
theorem. Next choose a partition P = {xi}:=, with U ( P , f ) - Jh f' dx < r and f dx - L I P , f ) < r : possibly by passing to a refinement of P . we may assume without loss of generality that p ( P ) < 6 . Choosing suitably for the partition P , we can make I U ( P , f ) - S ( P , {ti}, f ) 1 < t . For a possibly different choice of the set of intermediate points, say {ti}, we can make IS(P, It:}, f ) - U P , f)l -=z t. Then
1
Since c is arbitrary, the Riemann integrability o f f follows from Lemma 1.25e. With integration in hand, one could at this point give rigorous definitions of the logarithm and exponential functions log x and exp x , as well as rigorous but inconvenient definitions of the trigonometric functions sin x ,cos x ,and tan x . For each of these functions we would obtain a formula for the derivative and other information. We shall not pursue this approach, but we pause to mention the idea. We put logx = ft-' d t for 0 < x < +m and see that log carries (0, + m ) one-one onto (-m, + m ) . The function log x has derivative 1/x and satisfies the functional equation log(xy) = log x +log y . The proposition in Section A3 of the appendix shows that the inverse function exp exists, carries (-m, + m ) one-one onto (0, + m ) ,is differentiable, and has derivative exp x . The functional equation of log translates into the functional equation exp(a + b) - exp a exp b for exp, and we readily derive as a consequence that exp x = ex,where e = exp 1. For the trigonornctric functions, thc starting points with this approach arc thc dcfinitions arctanx = (1 t2)-I d t , arcsinx = (1 - t2)-1/2dt, and TC = 4arctan 1. Instead of using this approach, we shall use power series to define these functions and to obtain their expected properties. We do so in Section 7.
Jbx +
JbX
5. Complex-ValuedFunctions
41
5. Complex-Valued Functions
Complex numbers are taken as known, and their notation and basic properties are reviewed in Section A4 of the appendix. The point of the present section is to extend some of the results for real-valued functions in earlier sections so that they apply also to complex-valued functions. The distance between two members z and w of C is defined by d ( z , w ) = 1 z - ui 1 . This has the properties (i) d ( z l , z2) > 0 with equality if and only if zl = 22, (ii) d ( z l , z 2 ) = diz2, z l ) , (iii) d ( z l , z 2 ) 5 d ( z l , z 3 ) + d ( z g , z ~ ) . The first two are immediate from the definition, and the third follows from the triangle inequality of Section A4 of the appendix with z = z1 -23 and w = z3 -22. For this reason, (iii) is called the triangle inequality also. Convergence of a sequence { z n }in @ to z has two possible interpretations: either {Rez,} converges to Re z and {Imz,} converges to Imz, or d (z,, z ) converges to 0 in R. These interpretations come to the same thing because
Then it follows that uniform convergence of a sequence of complex-valued functions has two equivalent meanings, so does continuity of a complex-valued function at a point or everywhere, and so does differentiation of a complexvalued function. We readily check that all the results of Section 3, starting with Proposition 1.16 and ending with Theorem 1.23, extend to be valid for complexvalued functions as well as real-valued functions. The one point that requires special note in connection with Section 3 is the Mcan Valuc Thcorcm. This thcorcm is valid for rcal-valucd functions but not for complex-valued functions. It is possible to give an example now if we again allow ourselves to use the exponential and trigonometric functions before we get to Section 7, where the tools will be available for rigorous definitions. The example is f (x) = eix for x E [O, 2x1. This function has f (0) = f ( 2 n ) = 1 , but the derivative f '(.\-) = ieixis never 0. The Mean Value Theorem was used in the proof of Theorem 1.23, but the failure of the Mean Value Theorem for complex-valued functions causes us no problem when we seek to extend Theorem 1.23 to complex-valued functions. The reason is that once Theorem 1.23 has been proved for real-valued fullctiolls, one simply puts together conclusions about the real and imaginary parts. Next we examine how the results of Section 4 may be extended to complexvalued functions. Upper and lower Riemann sums, of course, make no sense for a complex-valued function. It is possible to make sense out of general Riemann sums as in Theorem 1.35, but we shall not base a definition on this approach.
I. Theory o f Calculus in One Real Variable
42
Instead, we simply define definite integrals of a function f : R + C in terms of real and imaginary parts. Define the real and imaginary parts u = Re f and b u = I m f by f(x) = u ( x ) + i u ( ~ ) , a n d l e t [ ~f d x = ~ ~ ~ u d x + i l ~ We ~udx. can then redefine the set R[a, b] of Riemann integrable functions on [a, b] to consist of bounded complex-valued functions on [a,b] whose real and imaginary parts are each Riemann integrable. Most properties of definite integrals go over to the case of complex-valued functions by inspection; there are two properties that deserve some discussion: (i) If f is in R [ a , b] and c is complex, then cf is in R [ a , b] and cf dx = c f dx. (ii) I f f i s i n R [ a , b ] , t h e n I f l i s i n R [ u , b ] a n d I ~f ~d~x 5 1 a b l f l d x . To see (i),write f = u iv and c = p iq. Then cf = (r is)(u iv) = (ru-sv)+i(rv+su). Thefunctionsru-sv andrvfsu areRiemannintegrable on [a, b] , and hence so is cf . Then
lab
+
=r
b la (u + i u ) d x
+
+isJab (u
+
+ iu)dx = c[ab
+
f dx.
To see (ii), let f be in R [ a ,b]. Proposition 1.30 shows successively that (Re f ) 2 and (Im f ) 2 a r e i n R [ a , b],that (Re f ) 2 + ( I m f ) 2 = I f12isinR[a, b], and that = 1 f 1 is in R [ a , b]. For the inequality with Jab f dx 1, choose c c C with Icl = 1 such that c Jab f dx is real and nonnegative, i.e., equals f dx . Using (i), we obtain (ii) from
I
I sub I
Finally we observe that Theorem 1.35 extends to complex-valued functions f . The definition of Riemann sum is unchanged, namely S ( P , {ti},f ) = C:=lf (ti)Axi,and the statement of Theorem 1.35 is unchanged except that the number r is now allowed to be complex. The direct part of the extended theorem follows by applying Theorem 1.35to the real and imaginary parts off separately. For the converse we use that the inequality I S(P, {ti},f ) - cI ir with c complex implies IS(P, {ti},Ref ) - Recl < t and IS(P, {ti},Im f ) - Imcl < t. Theorem 1.35 for real-valued functions then shows that Re f and Im f are Riemann integrable, and hence so is f .
6. Taylor's Theorem with Integral Remainder
43
6. Taylor's Theorem with Integral Remainder There are several forms to the remainder term in the one-variable Taylor's Theorem for real-valued functions, and the differences already show up in their lowest-order formulations. Let f be given, and, for definiteness, suppose a < x . Tf o(1) denotes a term that tends to O as x tends to n , three such lowest-order formulas are
f
( x ) = f ( a )f 4 1 )
f ( x ) = f (a)
+(x
f (x) = f ( a )
+
-
if f is merely assumed to be continuous, with a < 6 < x if f is continuous on [a, x] and f ' exists on (a, x ) ,
a )f ' ( 6 )
Sx
if f and f' are continuous on [a, XI.
f ' ( t )dt
a
The first formula follows directly from the definition of continuity, while the second formula restates the Mean Value Theorem and the third formula restates part of the Fundamental Theorem of Calculus. The hypotheses of the three formulas increase in strength, and so do the conclusions. In practice, Taylor's Theorem is most often used with functions having derivatives of all orders, and then the strongest hypothesis is satisfied. Thus we state a general theorem corresponding only to the third formula above. It applies to complex-valued functions as well as real-valued functions.
Theorem 1.36 (Taylor's Theorem). Let n be an integer 1 0, let a and x be points of R, and let f be a complex-valued function with n 1 continuous derivatives on the closed interval from a to x . Then
+
where ifa 5 x ,
x) =
-la n!
x
-
t
tt
i f x 5 a.
REMARKS.The notion of a continuous derivative at the endpoints of an interval is discussed for real-valued functions in the last paragraph of Section A 2 of the appendix and extends immediately to complex-valued functions; iteration of this definition attaches a meaning to continuous higher-order derivatives on a closed interval. Once the convention in the remarks with Lemma 1.27 is adopted, namely
I. Theory o f Calculus in One Real Variable
44
that JXa f d t = becomes tidier:
f d t when x < a , the formula for the remainder term
- JaX
with no assumption that a 5 x . PROOF.We give the argument when a 5 x , the case x 5 a being handled analogously. The proof is by induction on n. For n = 0 , the formula is immediate from the Fundamental Theorem of Calculus (Theorem 1.32b). Assume that the formula holds for n - 1. We apply integration by parts (Corollary 1.33) to the remainder term at stage n - 1, obtaining
Substitution gives
I"
Kn-1 ( a ,x ) = (n - I)! - -
n!
[ (
1
= - (X
n!
-
-
(X -t
) l Z f1 ( n ) ( t d) t
t ) n ( t ) ]+
LX
(x - i). f '"+"(t) d t
+
a)' f @ ) ( a ) & ( a , x ) ,
and the induction is complete.
7. Power Series and Special Functions A power series is an infinite series of the form cnzn. Normally in mathematics, if nothing is said to the contrary, the coefficients c, are assumed to be complcx and thc variablc z is allowcd to bc complcx. Howcvcr, in thc contcxt of real-variable theory, as when forming derivatives of functions defined on intervals, vne is inkresLed vrlly in real values vf 2 . In Illis bvvk 1l1e~.unlcxlwill gerlerally make clear whether the variable is to be regarded as complex or as real. f ("1 ( 0 1 ~ " One source of power series is the "infinite Taylor series" of n=O n! a function f having derivatives of all orders, with the remainder terms discarded. In this case the variable is to be real. If the series is convergent at x , the series has sum f (x) if and only if limn R, (0, x) = 0. Later in this section, we shall see examples where the limit is identically 0 and where it is nowhere 0 for x 0 .
EN
+
7. Power Series and Special Functions
45
Theorem 1.37. If a power series C:=ocnzn is convergent in C for some complex zo with Izo I = R and if R' < R , then C,"==, Icnzn1 is uniformly convergent for complex z with IzI i R', and so is CEO( n 1) l ~ , + ~1 . z ~
+
REMARKS.The number
R
= sup
{R'
I Cz,cnznconverges for some zo with lzol = R')
is called the radius of convergence of C z ocnz? The theorem says that if R' < R , then C z olcnznl converges uniformly for lzl 5 R', and it follows from the uniform Cauchy criterion that cnzn converges uniformly for lzl 5 R'. The definition of R carries with it the implication that if zo has lzo 1 > R , then C,"=ocnz{ diverges.
PROOF.The theorem is vacuous unless R > 0. Since cnzz is convergent, the terms cnzt tend to 0 . Thus there is some integer N for which Icnl Rn 5 1 whenn > N . Fix R' < R. For 121 5 R'andn > N,wehave
(g)"
Since 2 < +m, the Weierstrass M test shows that C,", c,z'konverges uniformly for lz 1 5 R'. For the series (n 1 ) lc,+lzn 1 , the inequalities lzl 5 R' and n > N together imply
xr=o +
To see that the Weierstrass M test applies here as well, choose r' with R'/R < r' < 1 and increase the size of N so that 5 r' whenever n > N . For such n , the ratio test and the inequality
#
+
(g)n
show that C ( n 1 ) ~ ~ ' converges. Thus the Weierstrass M test indeed applies, and the proof is complete.
Corollary 1.38. If Cr=o cnxn converges for Ix I < R and the sum of the series for x real is denoted by f (x), then the function f has derivatives of all orders for Ix 1 < R . These derivatives are given by term-by-term differentiation of the series for f , and each differentiated series converges for 1x1 < R . Moreover, f 'k'= (0). c k k!
I. Theory o f Calculus in One Real Variable
46
REMARK. When a function has derivatives of all orders, we say that it is infinitely differentiable. PROOF.The corollary is vacuous unless R > 0 . Let R' < R . The given series certainly converges at x = 0, and Theorem 1.37 shows that the term-byterm differentiated series converges uniformly for Ix I 5 R'. Thus Theorem 1.23 gives f l ( x ) = ( n + l ) ~ , + ~ xfor " 1x1 < R'. Since R' < R is arbitrary, f l ( x ) = C,"=, ( n l ) c , + l x n for 1x1 < R . We can iterate this result to obtain the corresponding conclusion for the higherorder derivatives. Evaluating the derivatives at 0, we obtain f ( k j ( ~ ) = ckk !, as asserted.
+
Corollary 1.39. If Czocnxn and Cr==, dnxnboth converge for Ix 1 < R and if their sums are equal for x real with x I < R , then c , = d , for all n . PROOF. This result is immediate from the formula for the coefficients in Corollary 13 8 . If f : R + C is infinitely differentiable near x
xx n!
=
a , we call the infinite
DO
series
j
a
-
tlre (infinite) Taylor series of f . We call a series
n=O
Cr==O c n ( x - a)" a power series about x = a ; its behavior at x = a + t is the same as the behavior of the series C,"==, c,xn at x = t . In applications, one usually ad,justs the function f so that Taylor series expansions are about x = 0 . Thus we shall concentrate largely on power series expansions about x = 0 . Had we chosen at the end of Section 4 to define log x as J;" t r l d t and exp x as k the inverse function of log x ,we would have found right away that (&) exp x = exp x for all k . Therefore the infinite Taylor series expansion of exp x about x = 0 is CZo This fact does not, however, tell us whether exp x is the sum of this series. For this purpose we need to examine the remainder. Theorem 1.36 shows that the remainder after the term x n/ n ! is
5.
Between 0 and .r ,et is bounded by sorne constant M(.r) depending on .r ,and thus 1 R, (0, x ) 1 5 j; ( x - t)" d t = e x " + ' . With x fixed, this tends to 0 as n tends to infinity, aiid tllus limn Rn ( 0 , x ) = 0 for each x . The coiiclusion is tliat exp x = In a similar fashion one can obtain power series expansions of sin x and cos x if one starts from definitions of the corresponding inverse functions in terms of Riemann integrals. Instead of using this approach, we shall define exp x , sin x , and cos x directly as sums of standard power series. An advantage of using series in the definitions
I
5.
1
7. Power Series and Special Functions
47
is that this approach allows us to define these functions at an arbitrary complex z , not just at a real x . Thus we define
The ratio test shows immediately that these series all converge for all complex z . Inspcction of all thcsc scrics givcs us thc idcntity cxp iz
= cos z
+ i sinz.
Corollary 1.38 shows that the functions exp z , sin z , and cos z , when considered as functions of a real variable z = x , are infinitely differentiable with derivatives given by the expected formulas
d dx
- expx = expx,
d dx
- sin x = cos x ,
&
d dx
- cosx = -sinx.
+
From these formulas it is immediate that (sin2x cos2x) = 0 for a11 x . Therefore sin2x + cos2x is constant. Putting x = 0 shows that the constant is 1. Thus sin2 x cos2 x = I.
+
In order to prove that exp x = ex, where e = exp 1, and to prove other familiar trigonometric identities, we shall do some calculations with power series that are justified by the following theorem.
Theorem 1.40. If f ( 2 ) = with lzl < R , then f ( z ) g ( z )=
x;==,
a,zn and g(z) = bnzn for complex z cnzn for IzI < R , where
CzO
REMARK.In other words, the rule is to multiply the series formally, assuming a kind of infinite distributive law, and reassemble the series by grouping terms with like powers of z. The coefficient c, of zn in the product comes from all products akzkblzzfor which the total degree is n , i.e., for which k + I = n. Thus c, is as indicated. PROOF.The theorem is vacuous unless R > 0. Fix R' < R. For 121 5 R', I bnzn1. These series are uniformly put F ( z ) = lanzn1 and G ( z ) = convergent for 1 z 1 i R' by Theorem 1.37, and also If ( z )I i F ( z ) and Ig ( z )I 5 G ( z ) . By the uniform convergence of the series for F and G when lzl 5 R', there exists M i+oo such that F ( 2 ) 5 M and G ( z ) 5 M for 121 5 R'. Given
xr=o
I. Theory o f Calculus in One Real Variable
48 t
> 0, choose an integer N' such that lzl 5 R' implies
En>N, lbnznl < t. If IzI 5 R' and N > 2 N f ,then
En>N, lanznl < t and -
Call the two terms on the right side TI and T2. Then we have
and also, with [ N / 2 ]denoting the greatest integer in N / 2 ,
+
Since G(z) 5 M and F(z) 5 M for lzl 5 R', the total estimate is that TI T2 5 4 t M . Since t is arbitrary, we conclude that limN cnzn = f (z)g(z) for 121 5 R'. Since R' is an arbitrary number < R , we conclude that C,"=o cnzn = f (z)g(z) for lzl < R.
c:=~
Corollary 1.41. For any z and w in C, exp(z+w) exp z =
m.
= exp z exp w .Furthermore,
PROOF.Theorem 1.40 and the infinite radius of convergence allow us to write 00
00
exp z exp w = s=o oo
N
">
=
2
N =O
(2 + wIN N!
r!s!
x x,, o " l N
ZkwN-k
= N=O k=O k!(N- k)!
s
=
= exp(z
+ w).
(;)zkw~-k
7. Power Series and Special Functions
For the second formula, write z expi
= exp(x
-
iy)
=x
+ iy . Then
= expx exp(-iy)
= (expx)(cos y
=
(expx)(cos y
-
i sin y)
+ i siny) = expx exp(iy) = exp(x + iy) = expz.
Corollary 1.42. The exponential function exp x ,as afunction of areal variable, has the following properties: (a) exp is strictly increasing on (-oo, foo) and is one-one onto (0, +oo), (b) expx = ex, where e = exp 1 , (c) exp x has an inverse function, denoted by log x , that is strictly increasing, carries (0, + m ) one-one onto (-oo, foo), has derivative l / x , and satisfies log(xy) = logx + logy. REMARKS. The three facts that exp x = ex for x real, exp z satisfies the functional equation of Corollary 1.41for z complex, and ez is previously undefined for z nonreal allow one to define ez to mean exp z for all complex z . We follow this convention. In particular, elx = exp(ix) = cos x + i sin x . PROOF.For x > 0, wc certainly havc cxpx > 1. Also, cach tcrm of thc series for exp x is strictly increasing for x > 0, and hence the same thing is true of the sum of the series. From Corollary 1.41, exp(-x) exp x = exp 0 = 1, and thus exp x is strictly increasing for x 5 0 with 0 < expx 5 1. Putting these statements together, we see that expx is strictly increasing and positive on (-oo, +oo). Hence it is one-one. This proves part of (a). Since exp x > 0, it makes sense to consider rational powers of exp x . Iteration of the identity exp(z w ) = exp z exp w shows that (exp p)q= exp px = 9 (expx)P, and application of the qth root function gives exp = (expx)Plq. Taking x = 1 yields exp(p/q) = eP/q for all rational p / q . The two functions exp x and ex are continuous functions of a real variable that are equal when x is rational, and hence they are equal for all x . This proves (b) . From the first two terms of the series for exp 1, we see that e > 2. Therefore en > 2n > n for all positive integers n , and exp x has arbitrarily large numbers in its image. The Intermediate Value Theorem (Theorem 1.12) then shows that [O, + m ) is contained in its image. Since exp(-x) exp x = 1, the interval (0, 11 is contained in the image as well. Thus exp x carries (-oo,+ m ) onto (0, foe). Tliis proves the remainder of (a). Consequently exp x has an inverse function, which is denoted by log x . Since exp x has the continuous everywhere-positive derivative exp x , the proposition in Section A3 of the appendix applies and shows that logx is differentiable with derivative 1/ exp(1og x). Since exp and log are inverse functions, exp(1og x) = x . Thus the derivative of log x is 1/x .
+
5
I. Theory o f Calculus in One Real Variable
50
+
Finally exp(1ogx logy) = expilog x) expilog y) = xy, since exp and log are inverse functions. Applying log to both sides gives log x log y = log(xy). This proves (c) .
+
Corollary 1.43. The trigonometric functions sin x and cos x , as functions of a real variable, satisfy
+ y ) = sinx cos y + cos x sin y, + y) = cos x cos y sinx sin y. . . P ~ o o r By . Corollary 1.41, cos(x + y) + i sin(x + y) = ei("+y) = erne" = (cos x + i sin x) (cos y + i sin y) . Multiplying out the right side and equating real (a) sin(x
(b) cos(x
-
and imaginary parts yields the corollary. The final step in the foundational work with the trigonometric functions is to define n and to establish the role that it plays with trigonometric functions.
Proposition 1.44. The function cos x , with x real, has a smallest positive xo for which cos xo = 0. If n is defined by writing xo = n/2, then (a) sinx is strictly increasing, hence one-one, from [O, $1 onto [0, 11, and cosx is strictly decreasing, hence one-one, from [O, 51 onto [0, 11, (b) sin(-x) = - sin x and cos(-x) = cos x , (c) sin(x + = cosx and cos(x + 5 ) = - sinx, (d) sin(x + n ) = - sin x and cos(x + n ) = - cos x , (e) sin(x + 2n) = sin x and cos(x + 2 n ) = cos x .
5)
PROOF.The function cos x has cos 0 = 1. Arguing by contradiction, suppose that cos x is nowhere 0 for x > 0. By the Intermediate Value Theorem (Theorem 1.12), cos x > 0 for x > 0. Since sin x is 0 at 0 and has derivative cos x ,sin x is strictly increasing for x > 0 and is therefore positive for x > 0 . Since cos x has derivative - sin x ,cos x is strictly decreasing for x > 0. Let us form the function f (n) = (cos x - cos 1) + (sin 1) (x - 1). If there is sorne xl > 1 with f (xi) > 0, then the Mean Value Theorem produces some 6 with 1 4 6 < x l such that 0 < f (xi) = f (xl) - f (1) = (xi - 1)f'(6) = (xl - I)(- sin$
+ sin 1) < 0,
and we have a coiltradictioll. Thus f (x) 5 0 for all x 3 1. I11 other words, cos x 5 cos 1 - (sin 1)(x - 1) for all x > 1. For x sufficiently large, cos 1 - (sin 1)(x - 1) is negative, and we see that cos x has to be negative for x sufficiently large. The result is a contradiction, and we conclude that cosx is 0 for some x > 0. Let xo be the infimum of the nonempty set of positive x's for which cos x = 0. We can find a sequence {x,} with x, + xo and cos x, = 0 for all n . By continuity
7. Power Series and Special Functions
51
cos xo = 0. We know that xo > 0, and we must have xo > 0, since cos 0 = 1. This proves the existence of xo. Since sin x has derivative cos x , which is positive for 0 5 x < n/2, sinx is strictly increasing for 0 5 x 5 n/2. From sin2x cos2x = 1, we deduce that sin(n/2) = 1. By the Intermediate Value Theorem, sinx is one-one from [O, onto [0, 11. In similar fashion, cos x is strictly decreasing and one-one from [0, 21 onto [0, 11. This proves (a). Conclusion (b) is immediate from the series expansions of sinx and cosx. Conclusion (c) follows from Corollary 1.43 and the facts that sin 5 = 1 and cos - 0. Conclusion (d) follows by applying (c) twice, and conclusion (e) follows by applying (d) twice.
+
51
5
x
z
Corollary 1.45. The function eix, with x real, has e i x = 1 for all x , and elx is one-one from [O,2n) onto the unit circle of C , i.e., the subset of ~Cwithlz= l 1. H
+
+
i sin x l 2 = cos2 x sin2x = 1 and therefore PROOF. We have I cos x leiX1 = 1. If eiX1= eix2with xl and x2 in [0, 2n), then ei("1-"2) = 1 with t = xl - x2 in (-2n,2n). So cost = 1 and sin t = 0. From Proposition 1.44 we see that the only possibility for t E (-2n, 2n) is t - 0. Thus xl - x2 - 0, and x H eix is one-one. Now let x I iy have x2 I y2 = 1. First suppose that x > 0 and y > 0. Since 0 5 y 5 1, it follows that there exists t E [O, 51 with sin t = y . For this t , the numbers x and cos t are both 2 0 and have square equal to 1 - y2. Thus x = cos t and eit = x iy. For a general x + iy with x2 + y2 = 1, exactly one of the complex numbers ir'(x iy) with 0 5 n 5 3 has real and imaginary parts > 0. Then in(x iy) = eit for some t . Since i = cos + i sin 2 = ezn/2, we see that + iy = eite-inr/2 - eit-inn/2 F~~~ei(t*2n) - ei t , we can adjust it - inn/2 additively by a multiple of 2ni so that the result it' lies in i [O,2n), and then eit' = x + iy, as required.
+
+
+
5
"
Corollary 1.46. (a) The function sin x carries (5 ) onto (- 1, 1) ,has everywhere-positive derivative, and has a differentiable inverse function arcsin x carrying (- 1, 1) one-one oilto (5 ) . The derivative of arcsin x is 1/1/-. (b) The function tan x = (sin x)/ cos x) carries (onto (- m, +m) , has everywhere-positive derivative, and has a differentiable inverse function arctan x carrying (-oc, +oc) one-one onto (5 ) . The derivative of arctan x is 1/(1 x2), and (1 x2)-' dx = n/2.
5,
5,
+
I!, +
5, 5)
5,
-&
PROOF.From Proposition 1.44 we see that (sin x) = cos x and -&(tanx) = ( c o ~ x ) - ~ The . first of these is everywhere positive because of (a) and (b)
52
I. Theory o f Calculus in One Real Variable
in the proposition, and the second is everywhere positive by inspection. The image of sinx is (- 1, 1) by (a) and (b), and also the image of tanx is (-oo, foo) by (a) and (b). Application of the proposition in Section A3 of the appendix yields all the conclusions of the corollary except the formula for (1 + x2)-' dx. This integral is given by arctan 1 - arctan(-1) by Theorem 1.32. Since tan(n/4) = sin(n/4)/ cos(n/4), (c) in the proposition gives tan(n/4) = 1, and hence arctan 1 = n / 4 In addition, tan(-n/4) = =
I!,
cos(n/4) = -1, and hence arctan(-1) = -n/4. Therefore (n/4) - (-n/4) = n/2.
-S1n(n'4)
1
(1 +xL)-' dx =
A power series, even a Taylor series, may have any radius of convergence in [O, foal. Even if the radius of convergence is > 0, the series may not converge to the given function. For example, Problems 20-22 at the end of the chapter ask one to verify that the function
is infinitely differentiable, even at x = 0, and has f ("1 (0) = 0 for all n. Thus its infinite Taylor series is identically 0, and the series evidently converges to f (x) only for x = 0. Because of Corollary 1.38,one is not restricted to a rote use of Taylor's formula ill order to coiilpute Taylor series. If we are interested ill the Taylor expaiisioii of f about x = 0, any power series with a positive radius of convergence that converges to f on some open interval about x has to be the Taylor expansion of f . A simple example is ex', whose derivatives at x = 0 are a chore to compute. However, eU = C:=l"=o for all u . If we put 1.1 = x 2 , we obtain ex' = Cz0$ for all x. Therefore this series must be the infinite Taylor series of ex'. Here is a more complicated example.
5
EXAMPLE. Binomial series. Let p be any complex number, and put F(x) = (1 + x ) P for -1 < x < 1. We can compute the nthderivative o f f by inspection, and we obtain F ( ~ ) ( x )= p ( p - 1) . . . (p - n + 1)(1 + x)pPn. Therefore the infinite Taylor series of F about x = O is
This series reduces to a polynomial if p is a nonnegative integer, and the series is genuinely infinite otherwise. The ratio test shows that the series converges for Ix I < 1; let f (x) be its sum for x real. The remainder term Rn (0, x) is difficult to
53
8. Summability
estimate, and thus the relationship between the sum f ( x ) and the original function ( 1 x)P is not immediately apparent. However, we can use Corollary 1.38 to obtain
+
for Ix 1 < 1. We compute ( 1 + x ) f l ( x ) by multiplying the first series by x , the second series by 1,and adding. If we write the constant term separately, the result is cc
(l+x)ff(x)=
P + CP ( P n=l
-
+
1 ) . .. ( P - n l ) [ n n!
+( p
-
n)l
xn = p f ( x ) .
Therefore
+
and (1 x ) - p f ( x ) has to he constant for Ix 1 < 1 . From the series whose sum is f ( x ),we see that f ( 0 ) = 1, and hence the constant is 1 . Thus f ( x ) = (1 + x ) p , and we have established the binomial series expansion
8. Summability Summability refers to an operation on a sequence of complex numbers to make it more likely that the sequence will converge. The subject is of interest particularly with Fourier series, where the ordinary partial sums may not converge even at points where the given function is continuous. ~ a sequence in C, and define its sequence of Cesaro sums, or Let { s , } , ~be nrithmctic mcans, to bc givcn by
for n 2 0 . If limn on = o exists in C, we say that Isn}is Cesaro summable to the limit o . For example the sequence with sn = (- l ) n for n 2 0 is not convergent, but it is CesBro summable to the limit 0 because onis 0 for all odd n and is for all even n .
5
54
I. Theory o f Calculus in One Real Variable
Theorem 1.47. If a complex sequence {s,},?~ is convergent in C to the limit s ,then Isn} is CesBro summable to the limit s .
REMARK.The argument is a 2 t proof, and two things are affecting a,. For k small and fixed, the contribution of sk to a, is s k / ( n+ 1 ) and is tending to 0. For k large, any sk is close to s , and the average of such terms is close to s . PROOF.Let t > 0 be given, and choose N1 such that k 2 Nl,then
> N 1 implies Isk
-
sI <
t. Ifn
so that la,
-
sl 5
l s o l + . . . + l ~ ~ , I +( N I + ~ ) I S I n - N 1 n+l n+l
+-
t
lsOl+"'+I~Nllf (Nl+l)lsl t. n+l The numerator of the first term is fixed, and thus we can choose N > N1 large enough so that the first term is < t whenever n > N . If n > N , then we see that 1 a, - s 1 < 2t. Since t is arbitrary, the theorem follows.
+
5
Next let {a,},lo be a complex sequence, and let {s,},~o be the sequence of partial sums with s, = ELoak. Form the power series o; = anrn. We say that the sequence {s,} of partial sums is Abel summable to the limit s in C . . if l i m t l or = s , i.e., if for each t > 0, there is some ro such that ro 5 r < 1 implies that 10, s 1 < t . For example, take ak = ( l ) k ,so that s, equals 1 if n is even and equals 0 if n is odd. The sequence {s,} of partial sums is divergent. The rth Abel sum or is given by the geometric series Ego(- l ) k r kwith sum 1 / ( 1 f r ) . Letting r increase to 1, we see that Is,) is Abel summable with limit -
i.
Theorem 1.48 (Abel's Theorem). Let {a,},lo be a complex sequence, and let {s,},?~ be the sequence of partial sums with s, = ELoak. If {s,},?o is convergent in @ to the limit s, then {s,,} is Abel summable to the limit s. REMARK.The proof will proceed along the same lines as in the previous case. It is first necessary to express the Abel sums or in terms of the sk's. PROOF.Since {s,) converges, {s,} and {a,} are bounded, and thus C z os,rn and CEO akrk are absolutely convergent for 0 5 r < 1. With s-1 = 0, write 00
0,. =
00
C anrn = C (s, n= O
n=O
00
-
snPl)rn=
00
C s,rn C snrn+' -
n= O
n=O
55
8. Summability
Let t > 0 be given, and choose N such that k
> N implies s k
-
sI <
t. Then
With N fixed, the coefficient of (1 - r ) in the first term is fixed, and thus we can choose ro close enough to 1 so that the first term is < t whenever ro 5 r < 1. If ro 5 r < 1, we see that 10, - s 1 < 2 t . Since t is arbitrary, the theorem follows. EXAMPLE.For Ix 1 < 1, the geometric series x z o ( - l ) n x n converges and has sum (1 + x)-'. The Fundamental Theorem of Calculus gives log(1 + t ) = dt = (- l)ntnd t for Ix 1 < 1, and Theorem 1.31 allows us to interchange sum and integral as long as Ix 1 < 1. Consequently
&
xz0
for Ix 1 i1. The sequence of partial sums on the right converges for x = 1 by the Leibniz test, and Theorem 1.48 says that the Abel sums must converge to the same limit. But the Abel sums have limit limxtl log(1 + x) = log 2, since log(1 + x) is continuous for x > 0. Thus Abel's Theorem has given us a rigorous proof of the familiar identity = log 2.
Theorems 1.47 and 1.48, which say that one kind of convergence always implies another, are called Abelian theorems. Converse results, saying that the second kind of convergence implies the first under an additioilal hypothesis, are called Tauberian theorems. These tend to be harder to prove. We give two examples of Tauberian theorems; the first one will be applied immediately to yield an important special case of the main theorem of Section 9; the second one will be used in Chapter VI to prove a deep theorem about pointwise convergence of Fourier series.
56
I. Theory o f Calculus in One Real Variable
be a complex sequence with all terms > 0, and Proposition 1.49. Let let { s , } ~ ?be~ the sequence of partial sums with s, = ak. If {s,},?~ is Abel summable in C to the limit s , then Isn}is convergent to the limit s .
xi=,
PROOF.Let {rj}j>obe a sequence increasing to the limit 1. Since anry 1 0 is nonnegative and since it is monotone increasing in j for each n , Corollary 1.14 any; = Er=o limj a,rJ", the limits existing in R*. applies and gives limj The left side is the (finite) limit s of the Abel sums, and the right side is lims,, which Corollary 1.14 is asserting exists. EXAMPLE. The birlurriial series expansiun in Se~Liun7 sliuws, fur any ~urriplcx p , that (1 - r)P is given for - 1 < r < 1 by the absolutely convergent series
For p real with 0 < p < 1 , inspection shows that all the coefficients in the sum on the right are 5 0. Therefore
has all coefficients 0 if 0 < p < 1. For 0 5 r < 1, the sum of the series is 1 - ( 1 - r)P and is 2 0. The fact that lim,. [ 1 - ( 1 - r ) p ] = 1 means that the sequence of partial sums sk = (- l)k+l P ( P - 1 ) I<..(p-k+l) is ~ b summable ~ l ! to 1. Proposition 1.49 shows that the series (*) is convergent at r. = 1, and the Weierstrass M test shows that (*) converges uniformly for -1 5 r 5 1 to 1 - ( 1 - Y I P .If we now take p = we have
xEl
i,
the series on the right being uniformly convergent for r = 1 - x2 therefore gives
-
1 5 r 5 1. Putting
57
8. Summability
the series on the right being uniformly convergent for - 1 5 x 5 1. Consequently Ix 1 is the uniform limit of a sequence of polynomials on [- 1, 11, and all these polynomials are in fact 0 at x = 0.
Proposition 1.50. Let be a complex sequence, and let {s,},>~ be the sequence of partial sums with .x, = C:=,a,,. If {.x,} is CesBro summable to the limit s in C and if the sequence {na,} is bounded, then {s,} is convergent and the limit is s. The rate of convergence depends only on the bound for {na,} and the rate of convergence of the Cesaro sums. REMARK.In our application in Chapter VI to pointwise convergence of Fourier series, the sequence of partial sums will be of the form {s,(x)}, depending on a parameter x , and the statement about the rate of convergence will enable us to see that the convergence of {s, (x)} is uniform in x under suitable hypotheses. PROOF.Let Isn} be the sequence of partial sums of {an},and choose M such that Inanl 5 M for all n. The first step is to establish a useful formula for s, - on.Let m be any integer with 0 5 m < n . We start from the trivial identity -(n - m)o, = (m 110, - (n l)on,add (n - m)s, to both sides, and regroup
+
+
as
Dividing by (n - m) yields S, - 0,
in + 1 n-m
= -(on-om)
+-n -1m .
2
(sn -sj), ,l=m+l
which is the identity from which the estimates begin. Form+l 5 j in,wehave
Substituting into our identity yields lsn
-
on1
+
m 1 i -I o n n -m
-
om1
+ ( n - mm -t 21)M
58
I. Theory o f Calculus in One Real Variable
Let t > 0 be given, and choose N such that lok- SI 5 t 2whenever k > N . We may assume that t < and N > 4 . With t fixed and with n fixed to be > 2N, define m to be the unique integer with
m5-
n-t l+t
Then 0 5 m < n ,and our inequality for Is, - on1 applies. From the left inequality 712 5 dcfining 711, wc obtain nz + i i z t 5 72 t and hcncc (711 + 1)t 5 72 711 and 5 t-' . From the right inequality < m + 1 defining m , we obtain n - t < m + 1 + t m + t andhencen - m - 1 < t ( m + 2 ) and < t. Thus our main inequality becomes -
Is,
-
onl 5 t-'lo,
-
l , o
-
+ Mt.
To handle om,we need to bound m below. We have seen that n - m - 1 < t ( m + 2 ) , and wehave assumedthat t < Thenn - m - 1 < i ( m +2),and
i.
this simplifies to m > - $,which is > if n > 8, thus certainly if N > 4. In other words, N 1 4 and n 12N makes m 1 1N . Therefore lo, - sI < t 2 , and lo, - om1 < 2 t 2 . Substituting into our main inequality, we obtain
Since t is arbitrary, the proof is complete.
9. Weierstrass Approximation Theorem We saw as an application of Proposition 1.49 that the function Ix I on [- 1 , 11 is the uniform limit of an explicit sequence {P,} of polynomials with P,(O) = 0. This is a special case of a theorem of Weierstrass that any continuous complex-valued function on a bounded interval is the uniform limit of polynomials on the interval. The device for proving the Weierstrass theorem for a general continuous complex-valued function is to construct the approximating polynomials as the result of a smoothing process, known as the use of an "approximate identity." The idea of an approximate identity is an important one in analysis and will occur several times in this book. If f is the given function, the smoothing is achieved by "convolution"
/f
(x - t ) v ( t )d t
of f with some function q ,the integrals being taken over some particular intervals. The resulting function of x from the convolution turns out to be as "smooth as the smoother of f and rp. In the case of the Weierstrass theorem, the function
9. Weierstrass Approximation Theorem
59
p will be a polynomial, and we shall arrange parameters so that the convolution will automatically be a polynomial. To see how a polynomial f ( x - t ) p ( t )dt might approximate f , one can think of rp as some kind of mass distribution; the mass is all nonnegative if p 0. The integration produces a function of x that is the "average" of translates x H f (X - t ) off ,the average being computed according to the mass distribution p. If p has total mass 1, i.e., total integral 1, and most of the mass is concentrated near t = 0, then f is being replaced essentially by an average of its translates, most of the translates being rather close to f , and we can expect the result to be close to f . For the Weierstrass theorem, we use a single starting pl at stage 1, namely c l ( l - x 2 ) on [- 1, 11 with cl chosen so that the total integral is 1. The graph of rpl is a familiar inverted parabola, with the appearance of a bump centered at the origin. The function at stage n is c,(l - x 2 ) " ,with c, chosen so that the total integral is 1. Graphs for n = 3 and n = 30 appear in Figure 1.1. The bump near the origin appears to be more pronounced at n increases, and what we need to do is to translate the above motivation into a proof.
1
z
T,emma 1.51. Tf r , is chosen so that r, for n sufficiently large.
1
(1
-
.x2)"d.x = 1, then r, 5
p f i
PROOF.We have
k),
3 e-' , we have (1 - ;)n 2 ie-' for n large enough (actually Since (1 e-'/,,h for n large enough, and hence c, 5 e,,h for n 2). Therefore c;' for n large enough. This proves the lemma.
z
z
/",
FIGURE 1. l . Approximate identity. Graphs of c, (1 - x21ndx for for n = 3 and n = 30 with different scales used on the vertical axes.
60
I. Theory o f Calculus in One Real Variable
Let y,(x) = c, (1 - x2), on [- 1, 11,with c, as in the lemma. The polynomials q, have the following properties: (1) rpnix) > 0 , (ii) J!~v,(x)dx = 1, (iii) for any S > 0 , sup {pn(x) I S 5 x 5 1) tends to 0 as n tends to infinity. Lemma 1.51 is used to verify (iii): the quantity sup {q,(x) 1 6 5 x 5 l } = c,(l
-
~7~)"
tends to 0 because lim, &(1 - a2)" = 0. A function with the above three properties will be called an approximate identity on [- 1, 11.
Theorem 1.52 (Weierstrass Approximation Theorem). Any complex-valued continuous function on a bounded interval [ a ,b] is the uniform limit of a sequence of polynomials.
PROOF.In order to arrange for the convolution to be a polynomial, we need to make some preliminary normalizations. Approximating f (x) on [ a ,b] by P(x) uniformly within t is the same as approximating f (x + a ) on [0, b - a ] by P (x + a ) uniformly within t ,and approximating g(x) on [0, c] uniformly by Q(x) is the same as approximating g(cx) uniformly by Q(cx). Thus we may assume without loss of generality that [a, b] = [O, 11. If h : [O, 11 3 C is continuous and if r is the function defined by r(x) = h(.u) - h(0) - [h(l) - h(0)l.u , then r is continuous with r(0) = r(1) = 0 . Approximating h(x) on [0, 11 uniformly by R(x) is the same as approximating r (x) on [0, 11 uniformly by R(x) h(0) [h(l) h(0)Ix. Thus we may assume without loss of generality that the function to be approximated has value 0 at 0 and 1. Let f : [O, 11 3 C be a given continuous function with f (0) = f (1) = 0; the function f is uniformly continuous by Theorem 1.lo. We extend f to the whole line by making it be 0 outside [0, 11, and the uniform continuity is maintained. 1 Now let q, be the polynomial above, and put P,(x) = f (x - t)q,(t) d t . Let us see that P, is a polynomial. By our definition of the extended f , the integrand is 0 for a particular x E [O, 11 unless t is in [x - 1, x] as well as [- 1, 11. We change variables, replacing t by s x and making use of Theorem 1 34, and the integral becomes P,(x) = f (-s)q,(s + x) d s , the integral being taken for s in[-1,0]n[-1-x, 1-XI. Sincexisin[O, l],theconditiononsisthatsisin [- l,O]. Thus P,(x) = f (-s)p, (s + x ) d s . In this integral, y, (x) is a linear -
-
-
I-,
1
+
/'01
0
+
combination of monomials x k , and xk itself contributes jPlf (-s)(x slkd s , which expands out to be a polynomial in x . Thus P,(x) is a polynomial in x .
61
10. Fourier Series
By property (ii) of rp, ,we have
Then property (i) gives
and two further uses of property (ii) show that this is
Given t > 0, we choose some 6 of uniform continuity for f and t , and then the first term is 5 t. With S fixed, we use property (iii) of cpn and the boundedness of f , given by Theorem 1.1 1 , to produce an integer N such that the second term is < c for n 1N. Then n 1N implies that the displayed expression is < 2 t . Since t is arbitrary, P,, converges uniformly to f .
10. Fourier Series A trigonometric series is a series of the form C,"=-,cneinxwith complex coefficients. The individual terms of the series thus form a doubly infinite sequence, but the sequence of partial sums is always understood to be the sequence {sN}& with s ~ ( x = ) ~ f = cneinx. - ~ Such a series may also be written as
5 + 7
00
(a, cos n x
+ bn sinnr)
by putting einx
e-inx
cg = ~1 a 0 ,
C,
1
= cos nx = cos nx
= T(an - ib,),
+ i sin n x -
i sin n x
for n > 0,
1 and c-, = T(an
+ ib,)
forn > 0.
Historically the notation with the an's and b,'s was introduced first, but the use of complex exponentials has become quite common. Nowadays the notation with
I. Theory o f Calculus in One Real Variable
62
an's and b,'s tends to be used only when a function f under investigation is real-valued or when all the cosine terms are absent (i.e., f is even) or all the sine terms are absent (i.e., f is odd). Power series enable us to enlarge our repertory of explicit functions, and the same thing is true of trigonometric series. Just as the coefficients of a power series whose sum is a ft~nctionf have to be those arising from Taylor's formtlla for f , the coefficients of a trigonometric series formed from a function have to arise from specific formulas. Let us run through the relevant formal computation: First we observe that the partial sums have to be periodic with period 2n.The question then is the extent to which a complex-valued periodic function f on the real line can be given by a trigonometric series. Suppose that 00
f (x)=
c,einx.
Multiply by ePikxand integrate to get
I J" f(x)e-ikxdx 2n -,
1 2~
=-
00
einx e -ikx
J-nn ,=-,
dx.
If we can interchange the order of the integration and the infinite sum, e.g., if the trigonometric series is uniformly convergent to f ( x ) ,the right side is
because
dx =
{
1
ifm = 0 ,
0
ifm # O .
Let f be Riemann integrable on [-n, n],and regard f as periodic on R.The trigonometric series C,"=-,c,einx with
is called the Fourier series of f . We write f (x)
-E
cneinx
and
.vN(f;x)= Ecneinx.
The numbers cn are the Fourier coefficients o f f ,and the functions s~ ( f ;x ) are the partial sums of the Fourier series. The symbol is to be read as "has Fourier
-
10. Fourier Series
63
series," nothing more, at least initially. The formulas for the coefficients when the Fourier series is written with sines and cosines are
In applications one encounters periodic functions of periods other than 2 n . If f is periodic of period 21, then the Fourier series o f f is f (x) - ~ . cneinnx/' with c, = ( 2 ~ )f ~1 ' f (x)e-in"xl' dx. The formula for the series written with sines and cosines is f (x) ao/2+ (a, cos(nnx/ I) b, sin(nnx/ I)) with a, = I-' f (x) cos(nnx/ 1 ) dx and b,, = 1-' f (x) sin(nnx/ 1) dx. In the present section of the text, we shall assume that our periodic functions have period 2n. The result implicit in the formal computation above is that if f (x) is the sum of a uniformly convergent trigonometric series, then the trigonometric series is the Fourier series of f ,by Theorem 1.31. Wc ask two qucstions: Whcn docs a gcncral Fouricr scrics convcrgc? If thc Fourier series converges, to what extent does the sum represent f ? We begin will1 an illurnirlalirlg cxarriple brings Lugelllcr a rlurribcr uf Lc~.llniqucsfrurn this chapter.
xz-,
-
I!,
xZl
I!,
+
EXAMPLE.As in the example following Theorem 1.48,we have
We would like to extend this identity to complex z with IzI < 1 but do not want to attack the problem of making sense out of log as a function of a complex variable. What we do is apply exp to both sides and obtain an identity for which both sides make sense when the real x is replaced by a complex z: exp (z
1 + i z 2 + iz3+ . . .) = 1-2
for
121 <
1.
In fact, this identity is valid for z complex with lzl < 1, and Problems 30-35 at the elid of tlie chapter lead to a proof of it. Corollary 1.45 allows us to write z = reis and z + Z1Z 2 + 1 3 + . . . = ,oeiP. Equating real and imaginary parts of the latter equation gives us DO
pcosq
=
Er n=l
cos nQ n
DO
and
psinB=E-. n=l
r PI sin nQ n
64
I. Theory o f Calculus in One Real Variable
We shall compute the left sides of these displayed equations in another way. We have e p cos rpeip sin rp -
exp ( p cos rp .
+ ip sin q ) = exp @eirp)= (1
-
z)-'
.
and therefore also eP C O S ~ e - z P S 1(1 n ~- ?)-I. Thus
Taking log of both sides gives 2p cos cp = log ((1 - 2r cos 8 we have 1 log 1 - 2r cos 8 r 2
t (
+ r2)-I), and thus (*I
+
,
Handling p sin rp is a little harder. From e ~ ~ ~ " e ~ ~(1 "-~z)-l, v we have 1-rcos0 irsinO e i p sinrp , and hence - (1 - z)-l/ll - 2 1 - I = (1 - i ) / l l - 21 = 11-z1 cos(psinrp)=(l-rcos8)/11-zl
and
sin(psinrp)=(rsin8)/11-21.
Thus tan(p sinrp) = r sin8/(1 - r cos 8). Since 1 - r cos 8 is > 0, cos(p sin rp) is > 0, and p sin rp = arctan ((r sin8)/(1- r cos 8)) + 2 n N(r, 8) for some integer N (r, 8) depending on r and 8 . Hence
For fixed r , the first term on the left is continuous in 8 , and the series on the right is uniformly convergent by the Weierstrass M test. By Theorem 1.21 the right side is continuous in 8 . Thus N (r, 8) is continuous in 8 for fixed r ; since N (r, 0) = 0, N (r, 8) = 0 for all r and 8 . We conclude that arctan
( 1 r sin8 r.cos8 -
r n sin n8 12
(**I
Problem 15 at the end of the chapter observes that the partial sums c= :~ cos nQ arid ~ f sillno = ~are uriiforrrlly bourlded on ariy set e 5 8 < n - e if e > 0. Corollary 1.19 therefore shows that the series
are uniformly convergent for t 5 8 < n - t if t > 0. Abel's Theorem (Theorem 1.48) shows that each of these series is therefore Abel surnmable with the same
65
10. Fourier Series
limit. We can tell what the latter limits are from (*) and (**), and thus we conclude that
and
arctan(
sin 8 1- C
sin no ) = c-, n 00
O S ~
The sum of the series with the cosine terms is unbounded near 8 = 0, and Riemann integration is not meaningful with it. We shall not be able to analyze this series further until we can treat the left side in Chapter VI by means of Lebesgue integration. The sum of the series with the sine terms is written in a way that stresses its periodicity. On the interval [-n, n ] , we can rewrite its left sideas ;(-n - 8 ) for-n 5 8 < 0 , O f o r 8 =O,and;(n - 8 ) for0 < 8 5 n . The expression for the left side is nicer on the interval (0, 2n), and there we have 00
i(n
-
Q) =
sin no
m
forO< 8 < 2 n
The function i ( n - 8) is bounded on (O,2n), and we can readily compute its Fourier coefficients from the formula bn = n-' (n - 8) sin no dB, using integration by parts (Corollary 1.33). The result is that b, = l l n . Hence the displayed series is the Fourier series. Graphs of some of the partial sums appear in Figure 1.2.
102"
FIGURE1.2. Fourier series of sawtooth function. Graphs of (sin nx)/n for n = 3,5, 10, 30.
c:=~
I. Theory o f Calculus in One Real Variable
66
The sawtooth function in the above example has a discontinuity, and yet its Fourier series converges to it pointwise. The recognition of the remarkable potential that Fourier series have for representing discontinuous functions dates to Joseph Fourier himself and caused many of Fourier's contemporaries to doubt the validity of his work. Although the above Fomier series converges to the function, it cannot do so uniformly, as a consequence of Theorem 1.21. In any such situation the Fourier coefficients cannot decrease rapidly, and a decrease of order l / n is the best that one gets for a nice function with a jump discontinuity. This example points to a general heuristic principle contrasting how power series and trigonometric series behave: whereas Taylor series converge very rapidly and may not converge to the function, Fourier series are inclined to converge rather slowly and they are more likely to converge to the function. We come to convergence results in a moment. First we establish some elementary properties of them. Taking the absolute value of c , in the definition of Fourier coefficient, we obtain the trivial bound I cn I i n:J I f ( x )I d x .
&
Theorem 1.53. Let f be in R[-n,n].Among all choices of d P N , .. . , d N , the expression
is minimized uniquely by choosing d n , for all n with In 1 5 N , to be the Fourier coefficient cn = f (x)e-lnxd x . The minimum value is
&
PROOF.PUt dn = c,
The result follows.
+ e n . Then
10. Fourier Series
67
Corollary 1.54 (Bessel's inequality). Let f be in R[-n, n ] ,and let f ( x ) E n = - , c,einx. Then w
-
In particular, C,"=-,lc, l 2 is finite REMARK.In terms of the coefficients a, and b,, the corresponding result is
PROOF.The theorem shows that the minimum value of a certain nonnegative quantity depending on N is I f ( x )l 2 d x - ~ f =I c , 12. - Thus, ~ for any N , 2 5 If (x)12d x . Letting N tend to infinity, we obtain the ~ f =I c , -I ~ corollary.
& ifn
& I:,
Corollary 1.55 (Riemann-Lebesgue Lemma). If f is in R[-n, n ] and has Fourier coefficients {cn}r=-, ,then limlnl,, cn = 0.
&
REMARK.This improves on the inequality lcnl i :1 If ( x )I d x observed is a bounded above, which shows, by means of an explicit estimate, that {crL} sequence. PROOF.This is immediate from Corollary 1.54. We now turn to convergence results. First it is necessary to clarify terms like "continuous" and "differentiable" in the context of Fourier series of functions on [-n, n ] . Each term of a Fourier series is defined on all of R and is periodic with period 2 n and is really given as the restriction to [-n, n ] of this periodic function. Thus it makes sense to regard a general function in the same way if one wants to form its Fourier series: a function f is extended to all of R so as to be periodic with period 2 n , and if we consider f on [-n, n ] ,it is really the restriction to [ n, n ]that we are considering. In particular, it makes sense to insist that f (-n) = f ( n ) ;if f does not have this property initially, one or both of these endpoint values will have to be adjusted, but that adjustment will not affect any Fourier coefficients. Similarly continuity of f will refer to continuity of the extended function on all of R,and similarly for differentiability. That being said, let us take up the matter of integration by parts for the functions we are considering. The scope of integration by parts in Corollary 1.33was limited
I. Theory o f Calculus in One Real Variable
68
to apair of functions f and g that have a continuous first derivative. In the context of Fourier series, it is the periodic extensions that are to have these properties, and then the integration-by-parts formula simplifies. Namely,
i.e., the integrated term drops out because of the assumed periodicity. The simplest convergence result for Fourier series is that a periodic function (of period 2n) with two continuous derivatives has a uniformly convergent Fourier 0 and use the above integration-by-parts series. To prove this, we take n formula twice to obtain
+
f " ( ~ ) e - ~d~x '.
&ITn
Then ~ c = Icnl ~ 2 cC/n2,where ~ ~ C =~ ~ I f l ' ( x ) I d x , and theFourier series converges uniformly by the Weierstrass M test. The argument does not say that tlie convergence is to f ,but that fact will be proved ill Theorem 1.57 below. Adjusting the proof just given, we can prove a sharper convergence result.
Proposition 1.56 . If f is periodic (of period 2n)and has one continuous derivative, then the Fourier series of f converges uniformly.
&
Ifn
PROOF.As in the above argument, c, = - (&) f ' ( x ) e P i n xd x , and this equals d, ,where d, is the nthFourier coefficient of the continuous function f ' . In the computation that follows, we use the classical Schwarz inequality (as in Section A5 of the appendix) for finite sums and pass to the limit in order to get the first inequality, and then we use Bessel's inequality (Corollary 1.54) to get the second inequality:
&
The right side is finite, and the proposition follows from the Weierstrass M test.
69
10. Fourier Series
The fact that the convergence in Proposition 1.56 is actually to f will follow from Dini's test, which is Theorem 1.57 below. We first derive some simple formulas. The Dirichlet kernel is the periodic function of period 2n defined by N
D N ( ~=)
7 einx=
sin ( ( N + i ) x ) .
Y
1
9
sin $ x
n=-N
the second equality following from the formula for the sum of a geometric series. For a periodic function f of period 2n,the partial sums of the Fourier series of f are given by
5
f (t) n=-N
gin("-'1
dt
+
the last two steps following from the changes of variables t H x s (Theorem 1.34) and s H -s (Proposition 1.30h) and from the periodicity of f and D N .
FIGURE 1.3. Dirichlet kernel. Graph of DN for N
=
30.
This is the kind of convolution integral that occurred in the previous section. Term-by-term integration shows that DN ( x ) d x = 1 . However, DN is not an approximate identity, not being everywhere > 0. Figure 1.3 shows the graph
& l:n
I. Theory o f Calculus in One Real Variable
70
of DN for N = 30. Although DN (x) looks small in the graph away from x = 0 , it is small only as a percentage of DN(0); DN (x) does not have limN DN(x) equal to 0 for x # 0. Thus DN (x) fails in a second way to be an approximate identity. The failure of DN to be an approximate identity is what makes the subject of convergence of Fourier series so subtle.
Theorem 1.57 (Dini's test). Let f : R + C be periodic of period 2 n and Rienlarin integrable on [-n, n ] . Fix n in [-n, n 1. If tliere are constants 13 > 0 and M < +oo such that
If
(x
+ t>
-
f (x>l 5 Mltl
for It1 < 6 ,
REMARK. This condition is satisfied if f is differentiable at x . Thus the convergence of the Fourier series in Proposition 1.56 is to the original function f . By contrast, the Dini condition is not satisfied at x = 0 for the continuous periodic extension of the function f (x) = Ix 1 defined on (-n, n ] . PROOF.With x fixed, let
lu
for t = 0.
Proposition 1.30d shows that (sin t/2)-' is Riemann integrable on t ( It I 5 n for any t > 0, and hence so is g (t) . Since g(t) is bounded near t = 0, Lemma 1.28 shows that g(t) is Ricmann intcgrablc on [-n, n]. Sincc fz DN (x) dx = 1, wc have
f( x t ) =
sin ((N
+ i)t)
sin((~+i)t)~~
1
1
sin 2.t
sin T t
I Jzg(t) sin ((N + i ) t ) dl 2n
-,
[g(t)cos
$1 sinNt d t + 2n
S"-,
[g(t) sin;] cosNt dt,
and both terms on the right side tend to 0 with N by the Riemann-Lebesgue Lemma (Corollary 1.55).
71
10. Fourier Series
Dini's test (Theorem 1.57) has implications for "localization" of the convergence of Fourier series. Suppose that f = g on an open interval I , and suppose that the Fourier series of f converges to f on I . Then Dini's test shows that the Fourier series of f - g converges to 0 on I , and hence the Fourier series of g converges to g on I . For example, f could be a function with a continuous derivative everywhere, and g could have discontinuities outside the open interval I . For f , the proof of Proposition 1.56 shows that lcnl < fm. But for g , the Fourier series cannot converge so rapidly because the sum of a uniformly convergent series of continuous functions has to be continuous. Thus the two series locally have the same sum, but their qualitative behavior is quite different. Next let us address the question of the extent to which the Fourier series of f uniquely determines f . Our first result in this direction will be that if f and g are Riemann integrable and have the same respective Fourier coefficients, then f (x)= g(x) at every point of continuity of both f and g. It may look as if some sharpening of Dini's test might apply just under the assumption of continuity of the function, and then this uniqueness result would be trivial. However, as we shall see in Chapter XII, the Fourier series of a continuous function need not converge to the function at particular points, and there can be no such sharpening of Dini's tcst. Instcad, wc shall handlc thc uniqucncss qucstion in a morc indircct fashion. The technique is to use an approximate identity, as in the proof of the Weierstrass Approximation Theorem in Section 9. Although the partial sums of the Fourier series of a continuous function need not converge at every point, the CesBro sums do converge. To get at this fact, we shall examine the FejCr kernel
The Nth Ceshro sum of s, ( f ;x) is given by
KN(x
1:
-
K N (x - t)f (t)dt because
t)f (t)dt.
The remarkable fact is that the FejCr kernel is an approximate identity even though Llle Diri~hlelkenlcl is nul, arid h e resull will be lllal llle Ccsku surns uf a Fvuricr series converge in every way that they have any hope of converging.
Lemma 1.58. The FejCr kernel is given by
I. Theory o f Calculus in One Real Variable
72
PROOF.We show by induction on N that the values of KN(x) in the definition 1-cos l x and in the lemma are equal. For N = 0, we have KO(x) = Do (x) = 1 = as required. Assume the equality for N - 1. Then
-
-
-
-
1 - cos x
+ i ) x ) .-sin i x + sin ((N sin ~x sin ?x
1 - cos Nx
+ 2 sin ((N + i)x) sin i x
1 - cos NX
1
1
by induction
1c o s x
1 - cos Nx
-
[ cos ((N +i ) x + i x ) - cos ((N + i ) x - i x ) ] 1 - cosx
+
1 - cos(N 1)x 1 - cosx
as required. In line with the definition of approximate identity in Section 9, we are to show that KN(x) has the following properties: (i) K N ( ~l) o , KN(x) dx = 1, (ii) KN(x)tends to 0 as n tends to infinity. (iii) for any 6 > 0,
&
Property (i) follows from the definition of K N (x) , since cos x 5 1 everywhere; (ii) follows from the definition of KN(x) and the linearity of the integral, since Dn(x) dx = 1 for all n ; and (iii) follows from Lemma 1.58, since 1 - c o s ( N + l ) x 12everywhereandl-cosx? 1 - c o s 6 i f 6 I 1x1 I n .
&
Theorem 1.59 (Fejtr's Theorem). Let f : R -r C he periodic of period 2n and Riemann integrable on [-n, n ] . If f is continuous at a point xo in [-n, n ] , then
If f is uniformly continuous on a subset E of [-n, n ] , then the convergence is uniform for xo in E .
10. Fourier Series
73
PROOF.Choose M such that If ( x )1 5 M for all x . By (ii) and then (i) ,
Given t > 0, choose some S for t and continuity of f at xo or for t and uniform continuity of f on E . In the first term on the right side, we then have I f ( x ) - f ( x O I) 5 t on the set where Ix - xo I 5 8. Thus a second use of (i) shows that the above expression is
With S fixed, property (iii) shows that the right side is large, and the theorem follows.
2t-if N is sufficiently
i
Corollary 1.60 (uniqueness theorem). Let f : IR 3 C and g : IR 3 C be periodic of period 2n and Riemann integrable on [-n, n].If f and g have the same respective Fourier coefficients, then f ( x ) = g ( x ) at every point of continuity of both f and g .
REMARK.The fact that f and g have the same Fourier coefficients means that s, ( f ; x) = s, ( g ; x ) for all n ,hence that
for all n . Then the same formula applies with D, replaced by its Ceshro sums KN . PROOF.Apply Theorem 1.59 to f and g .
-
g at a point xo of continuity of both f
74
I. Theory of Calculus in One Real Variable
Our second result about uniqueness will improve on Corollary 1.60, saying that any Riemann integrable function with all Fourier coefficients 0 is basically the 0 function- at least in the sense that any definite integral in which it is a factor of the integrand is 0. We shall prove this improved result as a consequence of Parseval's Theorem, which says that equality holds in Bessel's inequality. The proof of Parseval's Theorem will be preceded by an example and some lemmas. Theorem 1.61 (Parseval's Theorem) Let f . R i (C be periodic of period 2n and Riemann integrable on [-n, n ] . If f (x) -- zyrn c,emX ",hen
and
REMARK.In terms of the coefficients a, and b,, the corresponding result is
EXAMPLE. We saw near the beginning of this section that the periodic function 00 sin nx f givenby f(x) = i ( n - x ) o n ( 0 , 2 n ) h a s f(x) C -. The formulation
-
n=l
of Parseval's Theorem as in the remark, but with the interval (O,2n) replacing 2 = the interval (-n, n ) , says that C:=, ( n - x) dx . The right side i s = I/" 4n -n x'd*.= - =4n $.T~US
Jo"" I
I
This formula was discovered by Euler by other means before the work of Fourier. For the purposes of the lemmas and the proof of Parseval's Theorem, let us introduce a "Hermitian inner productn3 on R[-n, n ] by the definition
3The term "Hermitian inner product" will be defined precisely in Section 11.1. The form (f, g)2 comes close to being one, but it fails to meet all the conditions because (f,f)2 = 0 is possible without f = 0.
10. Fourier Series
as well as a "norm" defined by
and a "distance function" defined by
The role of the function d2 will become clearer in Chapter 11, where "distance functions" of this kind will be studied extensively.
Lemma 1.62.
If f
is in R[-n,n]and
1:
l f ( x ) 1 2 d x = 0 , then
JLIf (x)I d x = 0 and also J_"n f ( x ) g ( x ) d x = 0 for all g PROOF.Write M = sup,,[-,,,] If ( x ) I , and let partition P = xi}^=, with U ( P , 1 f 12) < t 3 ,i.e.,
E
R[-n,n].
> 0 be given. Choose a
t
Divide the indices from 1 to n into two subsets, A and B , with
A = [ ~ I sup
~ f ( x ) l ~ t ] and
B = { i
sup
if(x)lcc].
x€[xz-,,X,]
x€[xz-,,Xz]
The sum of the contributions from indices i E A to U ( P , If 1 2 ) is > e2 CiEA Axi, and thus CiEA A x i 5 t . Hence CiEA ,, If ( x ) l ) A x i 5 M E . Also,
+
C i E B ( s u ~ n ~ [ n , - l , n , ] i f ( x ) l ) h x i 5 2776. Therefore U ( P , If 1) 5 (277 M)t. Since t is arbitrary, If ( x )1 d x = 0. This proves the first conclusion. For the second conclusion it follows from the boundedness of Ig 1, say by M ' , that f (x)g(x)dxI 5 I f (x)llg(x)I d x 5 I f @ ) I d x = 0.
l:n
I ":1
g Sf"
& Sf"
Lemma 1.63 (Schwarz inequality). If f and g are in R[-n,n], then ( f , 8121 5
I l f llzllgll2.
REMARK.Comparc this rcsult with thc vcrsion of thc Schwarz incquality in Section A5 of the appendix. This kind of inequality is put into a broader setting in Section 11.1. PROOF.If Ilg 11, = 0 , then Lemma 1.62 shows that ( , f ,g ) , = 0 for all f . Thus the lemma is valid in this case. If 11 g 11 # 0 , then we have
and the lemma follows in this case as well.
I. Theory o f Calculus in One Real Variable
76
Lemma 1.64 (triangle inequality). If f , g , and Iz are in R[-n, n], then d2(f, h ) 5 d2(f, g ) d2(g, h ) . PROOF.For any two such functions F and G , Lemma 1.63 gives
+
Taking the square root of both sides and substituting F = f we obtain the lemma.
-
g and G
=g
-
h,
Lemma 1.65. Let f : R + C be periodic of period 2 n and Riemann integrable on [-n, n ] , and let t > 0 be given. Then there exists a continuous periodic g : R + @ o f period 2n such that 11 f - gII, < t . PROOF.Because of Lemma 1.64,we may assume that f is real-valued and is not identically 0. Define M = s ~ p ~ , [ - , , , ~I f (t) 1 > 0, let c > 0 be given, and let P = {xi}:=o be a partition to be specified. Using P , we form the function g defined by
The graph of g interpolates the points (xi, f (xi)),0 5 i 5 n , by line segments. f (t) and S = Fix attention on a particular [xipl, xi], and let I = inft,[xz-l,xzl s ~ p , , , , ~ ~f (t). For t E [xi-1 , xi], we have I i g(t) i S. At a single point t in this interval, f (t) > g(t) iiilplies I 5 g(t) 5 f ( t ) 5 S, while g(t) > f (t) implies I g(t) 5 f (t) 5 S. Thus in either case we have I f ( t ) - g (t) I i S - I. Taking the supremum over t in the interval and summing on i , we obtain U ( P , I f - g ) 5 U ( P , f ) - L ( P , f ) . sinceif -g12= If -glIf +gl,wehave
for1 5 i 5 n . S u m m i n g o n i g i v e s ~ (I f~ ,-g12) 5 2M(U(P, f ) - L ( P , f ) ) . Now we can specify P ; it is to be any partition for which U ( P , f ) - L ( P , f ) t2/(2hrl) and no Axi is 0. Then
as required.
10. Fourier Series
77
PROOFOF THEOREM1.61. Given t > 0, choose by Lemma 1.65 a con00 f inx tinuous periodic g with 11 f - gll, < t. Write g ( x ) -- E n = - , cne , and put g N ( x ) = K N ( x - t ) g ( t )d t , where K N is the FejCr kernel. FejCr's Theorem (Theorem 1.59) gives supx,,-,,,, Ig(x) -gN ( x )I < c for N sufficiently Ih(x) 1, we obtain large. Since any Riemann integrable h has I hll, 5 sup,,,-,,,, llg - g ~ 112 < t fur N sufficienlly large. Fixing sucll an N and subsliluling frurn the definition of K N , we have
& ST,
for suitable constants dn . Theorem 1.53 and Lemma 1.64 then give
and the result follows
Corollary 1.66 (uniqueness theorem). Let f : R + C be periodic of period 2n arid Ricrrrarlr~integrable on [-n, n].If f lras all Fuurier cuefficienls 0, tlleri If ( x )1 d x = 0 and ST, f ( x ) g ( x ) d x = 0 for every member g of R[-n,n].
1:
PROOF. If f has all Fourier coefficients 0, then l:n i f (x)12d x = 0 by Theorem 1.61. Application of Lemma 1.62 completes the proof of the corollary. It is natural to ask which sequences { c n }with C lcnI2 finite are the sequences of Fourier coefficients of some f E R[-n,n]. To see that this is a difficult question, one has only to compare the two series C z l n-' sinx and C z l n-' cos x studied at the beginning of this section. The first series comes from a function in R[-n,n],but a littlc argumcnt shows that thc sccond docs not. It was an carly triumph of Lebesgue integration that this question has a elegant answer when Llle Rierrrarlrl imlepal is replaced by h e Lebesgue integral: tlle ariswer wller~llle Lebesgue integral is used is given by the Riesz-Fischer Theorem in Chapter VI, namely, any sequence with C lc, l 2 finite is the sequence of Fourier coefficients of a square-integrable function.
I. Theory o f Calculus in One Real Variable
78
11. Problems 1. Derive the least-upper-bound property (Theorem 1.l) from the convergence of bounded monotone increasing sequences (Corollary 1.6). 2. According to Newton's method, to find numerical approximations to & when a > 0, one can set xo = 1 and define x,+l = i(x: a)/x, for n > 0 . Prove that {x,} converges and that the limit is &.
+
3. Find limsupa, and liminfa, when a, is defined by a1 = 0, a2, = 1 azn+l - 2. az,. Prove that your answers are correct.
+
4. For any two sequences {a,} and {b,} in R,prove that limsup(a, + b,) < lim sup a, lim sup b,, provided the two terms on the right side are not +cc and -oo in some order.
+
5 . Which of the following limits exist uniformly for 0 < x < 1: (i) lim,,, xn, (ii) lim,,, x n / n , (iii) lim,,, xi=lx k / k ? Supply proofs for those that do converge uniformly. For the other ones, prove anyway that there is uniform convergence on any interval 0 5 x 5 1 - t ,where t > 0. 6. Let a,(x) = (-l)"xn(l
formly and that 7.
-
x) on [0, 11. Show that
a,(x) converges uni-
zZola, (x) I converges pointwise but not uniformly.
(Dini's Theorem) Suppose that f, : [a, b] -. R is continuous and that f l 5 f2 5 f3 5 . . . . Suppose also that f (x) = lim f,(x) is continuous and is nowhere +GO. Use the Bolzano-Weierstrass Theorem (Theorem 1.8) to prove that f, converges to f uniformly for a 5 x < b .
8. Prove that
for all x > 0. 9. Let f : (-GO, +oo) -. R be infinitely differentiable with I f(")(x)l n and x . Use Taylor's Theorem (Theorem 1.36) to prove that
< 1 for all
for all x . 10. (Helly's Selection Principle) Suppose that { F , } is a sequence of nondecreasing functions on [-I, 11 with 0 5 F,(x) 5 1 for all n and x . Using a diagonal process twice, prove that there is a subsequence {F,, } that converges pointwise on [- 1, 11. 11. Prove that the radius of convergence of
zZoanxnis 1/ lim
SUP
m.
11. Problems
79
12. Find a power series expansion for each of the following functions, and find the radius of convergence: (a) 1/(1 - x12 = (1 - x)-l, (b) log(1 - x) = - ./ X1 l-t '
&
+
(c) l l ( l x2), (d) arctan x = ;J f $ . 13. Prove, along the lines of the proof of Corollary 1.46a, that cos x has an inverse function arccos x defined for 0 i x i n and that the inverse function is differentiable. Find an explicit formula for the derivative of arccos x . Relate arccosx to arcsinx when 0 i x i n/2. 14. State and prove uniform versions of Abel's Theorem (Theorem 1.48) and of the corresponding theorem about Cesaro sums (Theorem 1.47),the uniformity being with respect to a parameter x .
zK1
15. Prove that the partial sums cos nQ and onanysetEIQ<2n-cifc>O.
zL1sin nQare uniformly bounded
16. Verify the following calculations of Fourier series: +1 f o r O < x i n has f (x) -1 for - n < x < O (b) f (x) = ePiax on (O,2n) has f (x)
-
- 2 4
-
sin rm
n
sinChIll)x
n=l
einx
-, provided
n--oon+a
a is not an integer. 17. Combining Parseval's Theorem (Theorem 1.61) with the results of Problem 16, prove the following identities:
Problems 18-19 identify the continuous functions f : IR + C with f (x) f (y) = f (x y) for all x and y as the 0 function and the functions f (x) = eCX, using two different kinds of techniques from the chapter.
+
1;
18. Put F (x) = f ( t )dt . Find an equation satisfied by F , and use it to show that f is differentiable everywhere. Then show that f '(y) = f '(0) f (y), and deduce the form of f . 19. Proceed without using integration. Using continuity, find xo > 0 such that the expression I f (x) - 1I is suitably small when Ix 1 5 Ixol. Show that f ( 2 ~ ~ xis0 ) then uniquely determined in terms o f f (xo) for all k > 0. I f f is not identically 0, use xo to define c. Then verify that f (x) = cCxfor all c. Problems 20-22 construct a nonzero infinitely differentiable function f : having all derivatives equal to 0 at one point.
IR + R
I. Theory o f Calculus in One Real Variable
80
20. Let P ( x ) and Q ( x ) be two polynomials with Q not the zero polynomial. Prove that
21. With P and Q as in the previous problem, use the Mean Value Theorem to prove that the fuilctioi~g : R -t R with for x # 0, for x
= 0,
has gf ( x ) = 0 and that gf is continuous.
22. Prove that the function f : JR + JR with for x # 0, for x
= 0,
is infinitely differentiable with derivatives of all orders equal to 0 at x
= 0.
Problems 23-26 concern a generalization of Cesbo and Abel summability. A Silverman-Toeplitz summability method refers to the following construction: One starts with a system {MiJ}i,j,nof nonnegative real numbers with the two properties that (i) C jMij = 1 for all i and (ii) limi,, Mij = 0 for all j. The method associates to a complex sequence {.r,},,O the complex sequence {tn},,O with ti = CjZ0 MiJ.yj as if the process were multiplication by the infinite square matrix { M i j }on infinite column vectors.
23. Prove that if {s,} is a convergent sequence with limit s , then the corresponding sequence {t,} produced by a Silverman-Toeplitz summability method converges and has limit s .
24. Exhibit specific matrices { M i j } that produce the effects of CesBro and Abel summability, the latter along a sequence ri increasing to 1 . 25. Let ri be a sequence increasing to 1, and define Mij = ( j + l ) ( r i ) j ( l - T , ) ~ . Show that { M i j }defines a Silverman-Toeplitz summability method.
26. Using the system { M i j }in the previous problem,prove the following: if abounded sequence {s,} is not necessarily convergent but is CesBro summable to a limit a , then {s,} is Abel summable to the same limit a. Problems 27-29 concern the Poisson kernel, which plays the same role for Abel sums of Fourier series that the FejCr kernel plays for Ceskro sums. For 0 5 r < 1, define the Poisson kernel P, ( Q )to be the r" Abel sum of the Dirichlet kernel D,(Q) = = 1 and 1 (eike e p i k e ) .In the terminology of Section 8 this means that ak = eike for k 1 0, so that the sequence of partial sums C t = o ak is exactly
+
+
+
. p i k e
11. Problems
81
the sequence whose nth term is D,(O). The rth Abel sum expression
a,rn is therefore the
27. For f in R[-n, n ] , verify that the rth Abel sum of s,(f; x ) is given by the expression z:J P,. (Q - p) f (p) d p .
&
28. Verify that Pr (8) = properties:
1-r2 Deduce that Pr (8) has the following 1 -2rcosQ+r2'
(i) pr (Q) 0, (ii) P, (0) do = 1, (iii) for any S > 0, sups,lBl5zPr (0) tends to 0 as r increases to 1.
&1 :
-
29. Let f : R + C be periodic of period 2 n and Riemann integrable on [-n, n ] . (a) Prove that if f is continuous at a point 00 in [-n, n ] , then limr f l 2n
S"
P,(80-8)f(8)d8=f(80).
-n
(b) Prove that if f is uniformly continuous on a subset E of [-n, n ] , then the convergence in (a) is uniform for 00 in E . Problems 30-35 lead to a proof without complex-variable theory (and in particular without the complex logarithm) that exp (z i z 2 5z3 . . .) = 1/(1 - z) for all complex z with lz 1 < 1.
+
+
+
30. Suppose that R > 0, that fk(x) = CEO c,,kxn is convergent for Ix 1 < R , that cn,k > 0 for all n and k, and that limk,, fk(x) = f (x) uniformly for 1x1 5 r whenever r < R. Prove for each r < R that some subsequence { fk,}of { fk}has liml,, fLl (x) existing uniformly for Ix 1 5 r .
3 1. In the setting of the previous problem, prove that f is infinitely differentiable for 1x1 I R . 32. In the setting of the previous two problems, use Taylor's Theorem to show that f (x) is the sum of its infinite Taylor series for 1x1 < R . 33. I f 0 < r < 1,proveforlzl ' r t h a t I $ z N + & z N 1 + . . . I
+
+
34. Why is it true that if a power series CZocnznwith complex coefficients sums to 0 for all real z with lzl < R , then it sums to 0 for all complex z with lz 1 < R ? 35. Prove thatexp (z+ i z 2 + 5z3
+. ..) = 1/(1
-
z) forallcomplexz with lzl < 1.
CHAPTER I1 Metric Spaces
Abstract. This chapter is about metric spaces, an abstract generalization of the real line that allows discussion of open and closed sets, limits, convergence, continuity, and similar properties. The usual The other ~ ~ o t i oare ~ l sdefi~ledin distance function for the real line beco~nesa11 exrurlple of a 111eti.i~. terms of the metric. The advantage of the generalization is that proofs of certain properties of the real line immediately go over to all other examples. Section 1 gives the definition of metric space and open set, and it lists a number of important examples, including Euclidean spaces and cerhin spaces o l Iunctions. Sections 2 through 4 develop properties of open and closed sets, continuity, and convergence of sequences that are simple generalizations of known facts about R. Section 5 shows how a subset of a metric space can be made into a metric space so that the restriction of a continuous function froill the whole space to the subset renlains continuous. It also shows that three natural metrics for the product of two metric spaces lead to the same open sets, continuous f~mctions,and convergent sequences. Section 6 shows that any metric space is "Hausdorff," "regular," and "normal," and it goes on to exhibit three different countability hypotheses about a metric space as equivalent. A metric space with these properties is called "separable." Section 7 concerns compactness and completeness. A metric space is defined to be "compact" if every open cover has a finite subcover. This property is equivalent to the condition that every sequence has a convergent subsequence. The HeineBorel Theorem says that the compact sets of Rn are exactly the closed bounded sets. A number of the results early in Chapter I that were proved by the Bolzano-Weierstrass Theorem in the context of the real line are seen to extend to any compact metric space. A metric space is "complete" if every Cauchy sequence is convergent. A metric space is compact if and only if it is complete and "totally bounded." Section 8 concerns connectedness, which is an abstraction of the property of an interval of the line that accounts for the Intermediate Value Theorem. Section 9 proves a fundamental result known as the Baire Category Theorem. A sample consequence of the theorem is that the pointwise limit of a sequence of continuous complex-valued functions on a complete metric space must have points where it is continuous. Section 10 studies the spaces of real-valued and complex-valued continuous functions on a compact metric space. A generalization of Ascoli's Theorem from the setting of Chapter I provides a characteiization of coillpact sets in either of these spaces of continuous functions. A generalization of the Weierstrass Approximation Theorem, known as the Stone-Weierstrass Theorem, gives sufficient conditions for a subalgebra of either of these spaces of continuous functions to be dense. One consequence is that these spaces of continuous functions are separable. Section 1I constructs the "completion" of a metric space out of Cauchy sequences in the given space. The result is a complete metric space and a distance-preserving map of the given metric space into the completion such that the image is dense.
1. Definition and Examples
83
1. Definition and Examples Let X be a nonempty set. A function d from X x X , the set of ordered pairs of members of X, to the real numbers is a metric, or distance function, if (i) d ( x , y ) > 0 always, with equality if and only if x = y , (ii) d ( x , y ) = d ( y, x ) for all x and y in X , (iii) d ( x , y ) 5 d ( x , z ) d ( z , y ) for all x , y , and z , the triangle inequality. In this case the pair ( X , d ) is called a metric space. The real line IR1 with metric d ( x , y ) = Ix - y 1 is the motivating example. Properties (i) and (ii) are apparent, and property (iii) is readily verified one case at a time according as z is less than both x and y , z is between x and y , or z is greater than both x and y. We come to further examples in a moment. Particularly in the case that X is a space of functions, a space may turn out to be almost a metric space but not to satisfy the condition that d ( x , y ) = 0 implies x = y . Accordingly we introduce a weakened version of (i) as
+
(i') d ( x , y ) > 0 and d ( x , x ) = 0 always, and we say that a function d from X x X to the real numbers is a pseudometric if (i') , (ii) , and (iii) hold. In this case, ( X ,d ) is called a pseudometric space. Let ( X , d ) be a pseudometric space. If r > 0, the open ball of radius r and center x , denoted by B ( r ; x ) , is the set of points at distance less than r from x , namely B ( r ;x ) = { y E X I d ( x , y ) < r } . The name "ball" will be appropriate in Euclidean space in dimension three, which is part of the Example 1below, and "ball" is adopted for the corresponding notion in a general pseudometric space. A subset U of X is open if for each x in U and some sufficiently small r > 0, the open ball B ( r ; x ) is contained in U . For the line the open balls in the above sense are just the bounded open intervals, and the open sets in the above sense are the usual open sets in the sense of Chapter I.
Lemma 2.1. In any pseudometric space ( X , d ) ,every open ball is an open set. The open sets are exactly all possible unions of open balls. PROOF.Let an open ball B ( r ; x ) be given. If y is in B ( r ; x ) ,then the open ball B(r - d ( x , y ) , y ) has center y and positive radius; we show that it is contained in B ( r ; x ) . In fact, if z is in B(r - d ( x , y ) , y ) , then the triangle inequality gives d ( x , ~5) d ( x , y) + d ( y , 2 ) < d ( x , y )
and the containment follows.
+ (r - d ( x ,
y ) ) = r,
II. Metric Spaces
84
For the second assertion it follows from the definition of open set that every open set is the union of open balls. In the reverse direction, let U be a union of open balls. If y is in U , then y lies in one of these balls, say in B(r ; x) . We have just shown that some open ball B(s; y) is contained in B(r; x), and B ( r ; x) is contained in U . Thus B(s; y) is contained in U , and U is open. EXAMPLES.
(1) Euclidean space Rn.Fix an integer n > 0. Let Rn be the space of all n-tuples of real numbers x = (xl,. . . , x,). We define addition of n-tuples componentwise, and we define scalar multiplication by cx = (cxl,. . . , cx,) for real c. Following the normal convention in linear algebra, we identify this space with the real vector space, also denoted by Rn,of all n-component column vectors of real numbers x
. Generalizing the notion of absolute value when
= xn
n = 1, we let x 1 = ('j& x f ) li2 for x = ( x l ,. . . , x,) in Wn. The quantity ix 1
is the Euclidean norm of x. The Euclidean norm satisfies the properties
>
0 always, with equality if and only if x equals the zero tuple (0, . . . , O), (b) 1~x1= IcIIxI forallx andforallrealc, (c) Ix +yl i 1x1 +Iyl forallx and y. Properties (a) and (b) are apparent, but (c) requires proof. The proof makes use of the familiar dot product, given by x . y = C;=,xjyj if x = (xl, . . . , x,) and y = ( y l , . . . , y,). I11 terms of dot product, the Euclidean 1101111 is nothing more than 1x1 = (x . x)lI2. The dot product satisfies the important inequality Ix . y 1 5 Ix 1 1 y 1, known as the Schwarz inequality and proved for this context in Section A5 of the appendix. A more general version of the Schwarz inequality will be stated and proved in Lemma 2.2 below. The Schwarz inequality implies (c) above because we then have
(a) 1x1
o=
We make X = Rninto a metric space (X, d) by defining
Properties (i) and (ii) of a metric are immediate from (a) and (b), respectively; property (iii) follows from (c) in the form la bl 5 la1 Ibl if we substitute a = x - z and b = y - z . For n = 1, this example reduces to the line as
+
+
1. Definition and Examples
85
discussed above. For n = 2, open balls are geometric open disks, while for n = 3, open balls are geometric open balls. For any n , the open sets in the metric space coincide with the open sets as defined in calculus of several variables. (2) Complex Euclidean space C n . The space C of complex numbers, with distance function d ( z , w ) = lz - w I as in Section 1.5,can be seen in two ways to be a metric space. One way was carried out in Section 1.5 and directly uses the properties of the absolute value function 121 in Section A4 of the appendix. The other way is to identify z = x + i y with the member ( x , y ) of R 2 , and then the absolute value 1 z 1 equals the Euclidean norm I ( x , y ) I in the sense of Example 1 ; hence the construction of Example 1 makes the set of complex numbers into a metric space. More generally the complex vector space C n of n-tuples
becomes a metric space in two equivalent ways. One way is to define the norm 1 ; = ( E : = , lzj 1') ' I 2 as a generalization of the Euclidean norm for R n ; then we put d (2, w ) = 12 - w 1. The argument that d satisfies the triangle inequality is a variant of the one for R n : The object for C n that generalizes the dot product for R n is the Hermitian inner product
The Euclidean norm is given in terms of this expression by lzl = ( 2 , z ) ' I 2 ,and the version of the Schwarz inequality in Section A5 of the appendix is general enough to show that I(z, w ) I 5 Iz I I w 1 . The same argument as for Example 1 shows that the norm satisfies the triangle inequality, and then it follows that d satisfies the triangle inequality. The other way to view C n as a metric space is to identify C n with R2, by(z1,. . . , z,) H ( x l , . . . ,x,, y l , . . . , y,) andthentousethemetriconR2"from Example 1. This is the same metric, since ELl Izj l 2 = E y = l X; E y = l yj2 . We still get the same metric if we instead use the identification ( z l ,. . . , z,) H ( x l , yl , . . . , x,, y,) . With either identification the Hermitian inner product (2, w ) for C n corresponds to thc ordinary dot product for EX2,. (3) System R* of extended real numbers. The function f ( x ) = x / ( l x ) carries LO, +m) into [O, 1) and has g ( y ) = y / ( l - y ) as a two-sided inverse. Therefore f is one-one and onto. We can extend f so that it carries ( - X I , + X I ) one-one onto (- 1, +1) by putting f ( x ) = x / ( l + x 1 ) . We can extend f further by putting f (-oo) = - 1 and f (+m)= +l , and then f carries [-m, + X I ] , i.e., all of R*,one-one onto [- 1 , + l ] . The function f is nondecreasing on [-oo, f o o l . For x and x' in R * , let
+
+
+
II. Metric Spaces
86
We shall show that d is a metric. By inspection, d satisfies properties (i) and (ii) of a metric, and we are to prove the triangle inequality (iii), namely that d ( x , x') 5 d ( x , x")
+ d ( x f ' ,x').
The critical fact is that f is nondecreasing. Since d satisfies (ii), we may assume that x 5 x', and then d ( x , x') = f ( x ' ) - f (x). We divide the proof into three cases, depending on the location of x" relative to x and x'. The first case is that x" 5 x , and then d ( x , x")
+ d(xl',x ' ) = f
( x ) - f ( x " ) + f ( x ' ) - f (x")
Thus the question is whether
hence wlietller
2f ( x u )
2f ( x ) .
This inequality holds, since f is nondecreasing. The second case is that x 5 X" 5 x', and then d ( x , x")
+ d ( x f ' ,x') = f ( x " )
-
f (x)
+ f (x')
-
f (x") = f ( x ' ) - f ( x )
Hence equality holds in the triangle inequality. The third case is that x' 5 x", and then d ( x , xl')
+ d(xl',x ' ) = f
(x") - f ( x )
+ f (x")
-
f (x')
The triangle inequality comes down to the question whether ?
2 f ( x ' ) 5 2 f (x").
This inequality holds, since f is nondecl-easing. We collclude that ( R * ,d ) is a metric space. It is not hard to see that the open balls in R* are all intervals ( a ,b), [-oo, b ) , ( a , +m], and [-oo, +oo] with -oo 5 a < b 5 +oo. Each of these open balls in R*intersects R in an ordinary open interval, bounded or unbounded. The open sets in R therefore coincide with the intersections of R with the open sets of R* .
1. Definition and Examples
87
(4) Bounded functions in the uniform metric. Let S be a nonempty set, and let X = B ( S ) be the set of all "scalarn-valued functions f on S that are bounded in the sense that If ( s )1 5 M for all s E S and for a constant M depending on f . The scalars are allowed to be the members of R or the members of C , and it will ordinarily make no difference which one is understood. If it does make a difference, we shall write B ( S , R) or B ( S , (C) to be explicit about the range. For f and g in B ( S ), let d ( f 9 f , ) = sup I f ( s ) g(s)l. sES
It is easy to verify that ( X , d ) is a metric space. Let us not lose sight of the fact that the members of X are functions. When we discuss convergence of sequences in a metric space, we shall see that a sequence of functions in this X converges if and only if the sequence of functions converges uniformly on S . (5) Generalization of Example 4. We can replace the range R or C of the functions in Example 4 by any metric space (R, p ) . Fix a point ro in the range R . A function f : S + R is bounded if p ( f ( s ) ,ro) 5 M for all s and for some M depending on f . This definition is independent of the choice of ro because p is assumed to satisfy the triangle inequality. If we let X be the space of all such boundcd functions from S to R , wc can makc X into a mctric spacc by dcfining d ( f , g ) = sup,,s p ( f g(s)). (6) Sequence space f 2 . This is the space of all sequences {cn}r=-, of scalars with C lc, l 2 im. A metric is given by ($13
In the case of complex scalars, this example arises as a natural space containing n], all systems of Fourier coefficients of Riemann integrable functions on [-n, in the sense of Chapter I. Proving the triangle inequality involves arguing as in Examples 1 and 2 above and then letting the number of terms tend - to infinity. The role of the dot product is played by ({cn},idn}) = C r = - , cndn.
(7) Indiscrete space. If X is any nonempty set and if d ( x , y) = 0 for all x and y , then d is a pseudometric and the only open sets are X and the empty set 0. If X contains more than one element, then d is not a metric. (8) Discrete metric. If X is any nonempty set and if d(x,y)
=
1
ifx # y,
0
ifx = y,
then d is a metric, and every subset of X is open.
11. Metric Spaces
88
(9) Let S be a nonempty set, fix an integer n > 0, and let X be the set of n-tuples of members of S. For n-tuples x = ( x l ,. . . , x,) and y = ( y l , . . . , y,), define d ( x ,Y ) = # { j I X j f yjI, the number of components in which x and y differ. Then ( X , d ) is a metric space. The proof of the triangle inequality requires a little argument, but we leave that for Problem 1 at the end of the chapter. Every subset of X is open, just as with the discrete metric in Example 8. (10) Hedgehog space. Let X be IR2, and single out the origin for special attention. Let d be the metric of Euclidean space, and define p(x3 Y ) =
{
if x and y are on the same ray from 0,
+(
0 y)
otherwise.
Then p is a metric. Every open set in (X, d ) is open in ( X , p ) , but a set like the one in Figure 2.1 is open in ( X , p ) but not in (X, d ) .
FIGURE 2.1. An opcn sct ccntcrcd at thc origin in thc hcdgchog spacc (1 1) Hilbcrt cubc. Let X be the set of all sequences ( x , satisfying 0 5 xm 5 1 for all m , and put
of real numbers
Then (X, d ) is a metric space. To verify the triangle inequality, we can argue as follows: Let { x m } ,{y,}, and { z m }be in X . For each m ,we have
Thus
for each N. Letting N tend to infinity yields the desired inequality.
1. Definition and Examples
89
(12) L 1 metric on Riemann integrable functions. Fix a nontrivial bounded interval [a,b] of the line, let X be the set of all Riemann integrable complexvalued functions on [a,b] in the sense of Chapter I, and define
for f and g in X. Then (X, dl) is a pseudometric space. It can happen that b 1 f (x) - g(x)l d x = 0 without f = g ;for example, f could differ from g at a single point. Therefore d l is not a metric. (13) L2 metric on complex-valued R[-n,n]. This example arose in the discussion of Fourier series in Section 1.10, and it was convenient to include a in front of integrals. Let X = R[-n,n],and define factor
&
Then ( X ,d2) is a pseudometric metric space. The triangle inequality was proved in Lemma 1.64 using the version of the Schwarz inequality in Lemma 1.63; that version of the Schwar7 ineclilality needed a special argilment given in 1,emma 1.62 in order to handle functions f whose norm satisfies 1 1 f 1 , = 0. The constructions of metric spaces in Examples 1,2,6, and 13 are sufficiently similar to warrant abstracting what was involved. We start with a real or complex vector space V , possibly infinite-dimensional, and with a generalization ( . , . ) of dot product. This generalization is a function from V x V to EX in the case that V is real, and it is a function from V x V to C in the case that V is complex. We shall write the scalars as if they are complex, but only real scalars are to be used if the vector space is real. The function is written ( . , . ) and is assumed to satisfy the following properties: (i) it is linear in the first variable, i.e., (xl x2, y) = ( x l ,y) (x2,y) and (cx,Y) = 4 x 9 Y), y2) = (ii) it is conjugate linear in the second variable, i.e., (x,yl (x,yl) (x,y2) and (x,cy) = c(x, y), (iii) it is symmetric in thc rcal casc and Hermitian symmetric in thc complcx case, i.e., (y,x) = (x,y ) , (iv) it is definite,i.e., (x, x) s- 0 if x # 0. The form ( . , . ) is called an inner product if V is real or complex and is often called also a Hermitian inner product if V is complex; in either case, V with the form is called an inner-product space. Two vectors x and y with (x, y) = 0 are said to be orthogonal; the notion of orthogonality generalizes perpendicularity in the case of the dot product.
+
+
+
+
90
II. Metric Spaces
For either kind of scalars, we define Ilx 11 = (x, x) and the function 11 . 11 is called the associated norm. We shall see shortly that a version of the Schwarz inequality is valid in this generality, the proof being no more complicated than the one in Section A5 of the appendix. In many cases in practice, item (iv) is replaced by the weaker condition that (iv') ( . , . ) is semidefinite, i .e., (x , x) 0 if x f 0. This was what happened in Example 13 above. In order to have a name for this kind of space, let us call V with the semidefinite form ( . , .) a pseudo inner-product space. It is still meaningful to speak of orthogonality. It is still meaningful also to define Ilx I = (x, x) ' I 2 ,and this is called the pseudonorm for the space. The Schwarz inequality is still valid, but its proof is more complicated than for an inner-product space. The extra complication was handled by Lemma 1.62 in the case of Example 13 in order to obtain a little extra information; the general argument proceeds along different lines.
,
Lemma 2.2 (Schwarz inequality). Let V be a pseudo inner-product space with f o r m ( . , .). I f x and y arein V,then I(x, y)I 5 IlxIIIIyII. PROOF.First suppose that llyll f 0 . Then
and the inequality follows in this case. Next suppose that 11 y 11 = 0. It is enough to prove that (x, y ) = 0 for all x . If c is a real scalar, we have Ilx+cyll
2
= ( x f c y , x f c y ) = 11 x112+2~e(x,cy)+lc1211y112= 11x112+2c~e(x,y).
The left side is 0 as c varies, but the right side can be < 0 unless Re(x, y) = 0. Thus we must have Re(x, y) = 0 for all x . Replacing x by ix gives us Im(x, y) = - Re i (x, y) = - Re(ix, y), and this we have just shown is 0 for all x . ThusRe(x,y) = I m ( x , y ) = O , a n d ( x , y ) = O .
Proposition 2.3 (triangle inequality). If V is a pseudo inner-product space with form ( . , . ) and pseudonorm 11 . 11, then the pseudonorm satisfies (a) IlxII ?Oforallx E V, (b) llcxII = Icl Ilxll for all scalars c and all x E V, (c) Ilx yll i llxll Iyll for all x and y in V. Moreover, the definition d(x, y) = Ilx - y 11 makes V into a pseudometric space. The space V is a metric space if the pseudo inner-product space is an inner-product space.
+
+
2. Open Sets and Closed Sets PROOF.
91
Properties (a) and (b) of the pseudonorm are immediate, and (c) follows
because
+
Putting x = a - c and y = c - b gives d ( a , b) 5 d ( a , c ) d ( c , b ) , and thus d satisfies the triangle inequality for a pseudometric. The other properties of a pseudometric are immediate from (a) and (b). If the form is definite and d ( f , g ) = 0, then ( f - g , f - g ) = 0 and hence the definiteness yields f - g = 0 . EXAMPLES, continued. 14) Let us take double integrals of continuous functions of nice subsets of IR2 as known. (The detailed study of general Riemann integrals in several variables occurs in Chapter 111.) Let V be the complex vector space of all power series cnzn with infinite radius of convergence. Since any such F ( z ) F(z) = is bounded on the open unit disk D = { z E C lzl < 11, the form (F,G ) = ,[.- - F ( z )G ( z )d x d y is meaningful and makes V into an inner-product space. The proposition shows that V becomes a metric space with metric given by d ( F , G ) = (S,, I F ( Z )- G ( Z ) I ~ ~ ~ ~ Y ) " ~ .
I
2. Open Sets and Closed Sets In this section we generalize the Euclidean notions of open set, closed set, neighborhood, interior, limit point, and closure so that they make sense for all pseudometric spaces, and we prove elementary properties relating these metricspace notions. In working with metric spaces and pseudometric spaces, it is often helpful to draw pictures as if the space in question were IR2, even computing distances that are right for IR2. We shall do that in the case of the first lemma but not afterward in this section. Let (X, d ) be a pseudometric space.
Lemma 2.4. If z is in the intersection of open balls B ( r ; .u) and B ( s ; y ) , then there exists some t > 0 such that the open ball B ( t ; 2 ) is contained in that intersection. Consequently the intersection of two open balls is open. REMARK.Figure 2.2 shows what B ( t ; 2 ) looks like in the metric space IR2. PROOF.Take t = min{r triangle inequality gives
-
d ( x ,z ) , s
-
d ( y , z ) } . I f w is in B ( t ; z ) , then the
and hence w is in B ( r ; x). Similarly w is in B ( s ; y )
II. Metric Spaces
FIGURE2.2. Open ball contained in an intersection of two open balls.
Proposition 25. The open sets of X have the properties that (a) X and the empty set M are open, (b) an arbitrary union of open sets is open, (c) any finite intersection of open sets is open. PROOF.We know from Lemma 2.1 that a set is open if and only if it is the union of open balls. Then (b) is immediate, and (a) follows, since X is the union of all open balls and M is an empty union. For (c), it is enough to prove that U f' V is open if U and V are open. Write U = U , B, and V = U gBg as unions of open balls. Then U f' V = (B, f' B g ) ,and Lemma 2.4 shows that U f' V is exhibited as the union of open balls. Thus U f' V is open. A neighborhood of a point in X is any set that contains an open set containing
the point. An open neighborhood is a neighborhood that is an open set.' A neighborhood of a subset E of X is a set that is a neighborhood of each point of E. If A is a subset of X , then the set A" of all points x in A for which A is a neighborhood of x is called the interior of A. For example, the interior of the half-open interval [ a ,b) of the real line is the open interval ( a ,b).
Proposition 2.6. The interior of a subset A of X is the union of all open sets contained in A; that is, it is the largest open set contained in A. PROOF. Suppose that U 5 A is open. If x is in U , then U is an open neighborhood of x , and hence A is a neighborhood of x . Thus x is in AO,and A0 contains the union of all open sets contained in A. For the reverse inclusion, let x be in A". Then A is a neighborhood of x ,and there exists an open subset U of A containing x . So x is contained in the union of all open sets contained in A.
Corollary 2.7. A subset A of X is open if and only if A = A0 A subset F of X is closed if its complement is open. Every closed interval of the real line is closed. A half-open interval [ a , b) on the real line is neither open nor closed if a and b are both finite. 'some authors use the term "neighborhood" to mean what is here called "open neighborhood."
2. Open Sets and Closed Sets
93
Proposition 2.8. The closed sets of X have the properties that (a) X and the empty set M are closed, (b) an arbitrary intersection of closed sets is closed, (c) any finite union of closed sets is closed. PROOF.This result follows from Proposition 2.5 by taking complements. In (a),the complements of X and M are M and X ,respectively. For (b) and (c), we use Fa)' = U, F,' and (U, Fa)' = F,' for the complements the formulas of intersections and unions.
(nu
n,
If A is a subset of X , then x in X is a limit point of X if each neighborhood of x contains a point of A distinct from x . The closure2 A" of A is the union of A with the set of all limit points of A . For example, the limit points of the set [a,b) U {b + 1) on the real line are the points of the closed interval [a,b], and the closure of the set is [ a , b] U {b+ 11.
Proposition 2.9. A subset A of X is closed if and only if it contains all its limit points. PROOF. Suppose A is closed, so that AC is open. If x is in AC,then AC is an open neighborhood of x disjoint from A, so that x cannot be a limit point of A. Thus all limit points of A lie in A. In the reverse direction suppose that A contains all its limit points. If x is in A", then x is not a limit point of A, and hence there exists an open neighborhood of x lying conlpletely in A'. Since x is arbitrary, ACis open, and thus A is closed.
Proposition 2.10. The closure AC' of a subset A of X is closed. The closure of A is the intersection of all closed sets containing A; that is, it is the smallest closed set containing A. PROOF. We shall apply Proposition 2.9. If x is given as a limit point of A"', we are to see that x is in A". Assume the contrary. Then x is not in A, and x is not a limit point of A. Because of the latter condition, there exists an open neighborhood U of x that does not meet A except possibly in x . Because of the former condition, U does not meet A at all. Since x is a limit point of A"', U contains a point y of A". Since U does not meet A, y has to be a limit point of A. Since U is an open neighborhood of y , U has to contain a point of A, and we have a contradiction. We conclude that x is in A"', and Proposition 2.9 shows that A ~ is ' closed. Any closed set F containing A contains all its limit points, by Proposition 2.9, and hence contains all the limit points of A. Thus F 2 A". Since A" itself is a closed set containing A, it follows that A"' is the smallest closed set containing A. 'some authors write 2 instead of A'' for the closure of A.
94
II. Metric Spaces
Corollary 2.11. A subset A of X is closedif and only if A = A''. Consequently (AC1)"= A" for any subset A of X . Two remarks are in order. The first remark is that the proofs of all the results from Proposition 2.6 through Corollary 2.1 1 use only that the family of open subsets of X satisfies properties (a), (b), and (c) in Proposition 2.5 and do not actually depend on the precise definition of "open set." This observation will be of importance to us in Chapter X, when properties (a), (b), and (c) will be taken as an axiomatic definition of a "topology" of open sets for X , and then all the results from Proposition 2.6 through Corollary 2.11 will still be valid. The second remark is that the mathematics of pseudometric spaces can always be reduced to the mathematics of metric spaces, and we shall normally therefore work only with metric spaces. The device for this reduction is given in the next proposition, which uses the notion of an equivalence relation. Equivalence relations are taken as known but are reviewed in Section A6 of the appendix.
Proposition 2.12. Let ( X , d ) be a pseudometric space. If members x and y of X are called equivalent whenever d ( x , y ) = 0, then the result is an equivalence relation. Denote by [x] the equivalence class of x and by Xo the set of all equivalence classes. The definition d o ( [ x ] ,[ y ] )= d ( x , y ) consistently defines a function do : Xo x Xo + R,and ( X o , 4)is a metric space. A subset A is open in X if and only if two conditions are satisfied: A is a union of equivalence classes, and the set A. of such classes is an open subset of X o . PROOF. The reflexive, symmetric, and transitive properties of the relation "equivalent" are immediate from the defining properties of a metric. Let x and x' be equivalent, and let y and y' be equivalent. Then d ( x , y ) 5 d ( x , x')
+ d ( x f ,y') + d ( y l ,y ) = 0 + d ( x f ,y') + 0 = d ( x f ,y'),
and similarly d ( x l ,y l ) i d ( x , y ) .
Thus d ( x , y ) = d ( x l ,y ' ) , and 4 is well defined. The properties showing that 6 is a metric are immediate from the corresponding properties for d. Next let x be in an open set A, and let x' be equivalent to x . Since A is open, some open ball B ( r ; x ) is contained in A . Since x' has d ( x , x ' ) = 0, x' lies in B ( r ; x) . Thus x' lics in A , and A is thc union of cquivalcncc classcs. Finally let A be any union of equivalence classes, and let A. be the set of those classes. If x is in A , then the set of points in some equivalence class lying in B ( r ; [ x ] )is just B ( r ; x ) , and it follows that A is open in X if and only if A. is open in Xo .
3. Continuous Functions
95
3. Continuous Functions Before we discuss continuous functions between metric spaces, let us take note of some properties of inverse images for abstract functions as listed in Section A1 of the appendix. If f : X + Y is a function between two sets X and Y and E is a subset of Y, we denote by f - l ( E ) the inverse image of E under f , i.e., { x E X I f (x) E E } . The properties are that inverse images of functions respect unions, intersections, and complements. Let ( X , d) and ( Y , p) be metric spaces. A function f : X + Y is continuous at a point x c X if for each c > 0, there is a S > 0 such that p ( f ( x ) ,f ( y ) ) < c whenever d ( x , y) < 6 . This definition is consistent with the definition when ( X ,d ) and ( Y , p ) are both equal to R with the usual metric.
Proposition 2.13. If ( X ,d) and ( Y , p) are metric spaces, then a function f : X + Y is continuous at the point x E X if and only if for any open neighborhood V off ( x )in Y ,there is aneighborhood U of x such that f (U) V .
PROOF.Let f be continuous at x and let V be given. Choose t > 0 such that B ( t ; f ( x ) ) is contained in V , and choose S > 0 such that p( f ( x ) ,f ( y ) ) < t whenever d ( x , y ) iS. Then y E B ( S ; x ) implies f ( y ) E B(t-; f ( x ) ) 5 V . Thus U = B(S; x ) has f ( U ) 5 V . Conversely suppose that f satisfies the condition in the statement of the proposition. Let t > 0 be given, and choose a neighborhood U of x such that f ( U ) c B ( t ; f (.\-)). Since U is a neighborhood of x , we can find an open ball B(6; x ) lying in U . Then f ( B ( 6 ;x ) ) g B ( t ; f ( x ) ), and hence p ( f ( x ) , f ( y ) ) < c whenever d ( x , y) < 8. Corollary 2.14. Let f : X + Y and g : Y + Z be functions between metric spaces. If f is continuous at x and g is continuous at f ( x ) ,then the composition g o f , given by ( g o f ) ( y ) = g( f ( y ) ),is continuous at x . PROOF.Let W be an open neighborhood of g( f ( x ) ) . By continuity of g at f ( x ) ,we can choose a neighborhood V of f ( x ) such that g ( V ) g W . Possibly by passing to a subset of V , we may assume that V is an open neighborhood of f ( x ) . By continuity of f at x , we can choose a neighborhood U of x such that f ( U ) c V . Then g ( f ( U ) ) c W . Taking Proposition 2.13 into account, we see that g o f is continuous at x .
Proposition 2.15. If ( X ,d ) and ( Y , p) are metric spaces and f is a function from X into Y , then the following are equivalent: (a) the function f is continuous at every point of X , (b) the inverse image under f of every open set in Y is open in X , (c) the inverse image under f of every closed set in Y is closed in X .
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PROOF.Suppose (a) holds. If V is open in Y and x is in f ( V ) ,then f ( x ) is in V . Since f is continuous at x by (a), Proposition 2.13 gives us a neighborhood U of x , which we may take to be open, such that f ( U ) 5 V . Then we have x E U g f - l ( V ) . Since x is arbitrary in f ( V ), f ( V ) is open. Thus (b) holds. In the reverse direction, suppose (b) holds. Let x in X be given, and let V be an open neighborhood of f (x). By (b), U = f ( V ) is open, and U is then an open neighborhood of x mapping into V . This proves (a), and thus (a) and (b) are equivalent. Conditions (b) and (c) are equivalent, since f ( V ) " = f ( V c ).
-'
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A functioi~f : X -t Y that is coiltiiluous at eveiy point of X , as ill Proposition 2.15, will simply be said to be continuous. A function f : X -t Y is a homeomorphism if f is continuous, if f is one-one and onto, and if f :Y +X is continuous. The relation "is homeomorphic to" is an equivalence relation. Namely, the identity function shows that the relation is reflexive, the symmetry of the relation is built into the definition, and the transitivity follows from Corollary 2.14. If (X, d ) is a metric space and if A is a nonempty subset of X , then the distance from x to A , denoted by D ( x , A ) , is defined by
-'
D ( x , A ) = inf d ( x , y ) . YEA
Proposition 2.16. Let A be a fixed nonempty subset of a metric space (X, d) Then the real-valued function f defined on X by f ( x ) = U ( x , A ) is continuous. PROOF.If x and y are in X and z is in A , then the triangle inequality gives
Taking the infimum over z gives D ( x , A ) 5 d ( x , y ) + D ( y , A ) . Reversing the roles of x and y , we obtain D ( y , A ) 5 d ( x , y ) + D ( x , A ) , since d ( y , x ) = d ( x , y ) . Therefore
Fix x , let t > 0 be given, and take S = t . If d ( x , y ) < S = t ,then our inequality gives us If ( x ) - f ( y )I < t. Hence f is continuous at x . Since x is arbitrary, f is continuous.
Corollary 2.17. If ( X , d ) is a metric space, then the real-valued function d ( x , y) for fixed x is continuous in y. PROOF.This is the special case of the proposition in which A is the set { y } .
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Corollary 2.18. Let ( X , d ) be a metric space, and let x be in X . Then the closed ball { y E X I d ( x , y ) r } is a closed set. REMARK.Nevertheless, the closed ball is not necessarily the closure of the opcn ball B ( r ; x ) = { y E X I d ( x , y ) < r } . A countcrcxamplc is providcd by any open ball of radius 1 in a space with the discrete metric. PROOF.If f ( y ) = d ( x , y ) , the set in question is f ([o,r ] ) . Corollary 2.17 says that f is continuous, and the equivalence of (a) and (c) in Proposition 2.15 shows that the set in question is closed.
A''
Proposition 2.19. If A is a nonempty subset of a metric space ( X , d ) , then = { x I D ( x , A) = 01.
PROOF.The set { x I D ( x , A) = 0 } is closed by Propositions 2.16 and 2.15, and it contains A. By Proposition 2.10 it contains A". For the reverse inclusion, suppose x is not in A"', hence that x is not in A and x is not a limit point of A . These conditions imply that there is some t > 0 such that B ( t ; x ) is disjoint from A , hence that d ( x , y ) 1t. for all y in A. Taking the infimum over y gives D ( x , A) r t > 0. Hence D ( x , A) f 0 .
4. Sequences and Convergence For a set S , we have already defined in Section 1.1 the notion of a sequence in S as a function from a certain kind of subset of integers into S. In this section we work with sequences in metric spaces. A sequence { x n }in a metric space ( X , d ) is eventually in a subset A of X if there is an integer N such that x, is in A whenever n > N . The sequence { x , } converges to a point x in X if the sequence is eventually in each neighborhood of x . It is apparent that if { x , } converges to x , then so does every subsequence {xnk}.
Proposition 2.20. If ( X , d ) is a metric space, then no sequence in X can converge to more than one point. PROOF.Suppose 011 the contl-arythat { x , } converges to distillct points x and y . The number m = d ( x , y) is then > 0 . By the assumed convergence, x, lies in both open balls B ( y ; x ) and B ( y ; y ) if n is large enough. Thus x, lies in the intersection of these balls. But this intersection is empty, since the presence of a point z in both balls would mean that d ( x , y ) 5 d ( x , z ) d ( z , y ) < = m, contradiction.
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If a sequence {x,} in a metric space (X, d) converges to x , we shall call x the limit of the sequence and write limn,, x, = x or limn x, = x or limx, = x or x, + x . A sequence has at most one limit, by Proposition 2.20. If the definition of convergence is extended to pseudometric spaces, then sequences need not have unique limits. Let us identify convergent sequences in some of the examples of metric spaces in Section 1. EXAMPLES OF CONVERGENCE IN METRIC SPACES. (0) The real line. On R with the usual metric, the convergent sequences are the sequences convergent in the usual sense of Section 1.1. (1) Euclidean space R n . Here the metric is given by
if x = (xl, . . . , x,) and y = (yl, . . . , y,). Another metric df(x,y) is given by df(x,y) = max Ixk likin
-
ykl,
and we readily check that
From this inequality it follows that the convergent sequences in (Wn,d) are the same as the convergent sequences in (Rn,d'). On the other hand, the definition of d' as a maximum means that we have convergence in (Wn,dl) if and only if we have ordinary convergence in each entry. Thus convergence of a sequence of vectors in (Rn,d) means convergence in the kth entry for all k with 1 5 k 5 n. (2) Complex Euclidean space Cn. As a metric space, Cn gets identified with W2". Thus a sequence of vectors in Cn converges if and only if it converges entry by entry. (3) Extended real line R*. Here the metric is given by d(x, y) = If (x) f (y) I with f (x) = x / ( l 1x1) if x is in R , f (-oo) = -1, and f (foo) = + l . We saw in Section 1 that the i~ltersectio~ls with R of the open balls of R* are the open intervals in R. Thus convergence of a sequence in W* to a point x in R means that the sequence is eventually in (-oo, foo) and thereafter is an ordinary convergent sequence in W. Convergence to +oo of a sequence {x,} means that for each real number M, there is an integer N such that x, > M whenever n > N . Convergence to -oo is analogous.
+
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(4) Bounded scalar-valued functions on S in the uniform metric. A sequence { f,} in B(S) converges in the uniform metric on B ( S ) if and only if { f,} converges uniformly, in the sense below, to some member f of B(S). The definition of uniform convergence here is the natural generalization of the one in Section 1.3: { f,} converges to f uniformly if for each r > 0, there is an integer N such that n > N implies I f,, (s) - f (s) 1 < t for all s simultaneously. An important fact in this case is that the sequence { f,} is uniformly bounded, i.e., that there exists a real number M such that I fn (s) 5 M for all n and s. In fact, choose some integer N for t = 1. Then the triangle inequality gives
+
for all s if n 2 N , so that M can be taken to be m a x ~ , , ~ ,{ ~sup,,s I f , (s) I } 1. (5) Bounded functions from S into a metric space (R, p ) . Convergence here is the expected generalization of uniform convergence: { f,} converges to f uniformly if for each r > 0, there is an integer N such that n > N implies p (f, (s), f (s)) < r for all s simultaneously. As in Example 4, a uniformly convergent sequence of bounded functions is uniformly bounded in the sense that p (.f, (s) , ro) 5 M for all n and s , M being some real number. Here ro is any fixed member of R . (7) Indiscrete space X . The function d (x , y) in this case is a pseudometric, not a metric, unless X has only one point. Every sequence in X converges to every point in X . (8) Discrete metric. Convergence of a sequence {x,} in a space X with the discrctc mctric mcans that {x,] is cvcntually constant. (1 1) Hilbert cube. For each n , let ({xm}~=l)n be a member of the Hilbert cube, and write x,,,, for the mth term of the nth sequence. As n varies, the sequence of sequences converges if and only if limn x,, exists for each in. (12) L metric on Riemann integrable functions. The function d (f, g) defined in this case is a pseudometric, not a metric. Convergence in the corresponding metric space as in Proposition 2.12 therefore really means a certain kind of convergence of equivalence classes: If { f,} and f are given, the sequence of classes b {[ fn]} converges to the class [ f ] if and only if limn I fn(x) - f (x)I dx = 0. The use of lasses in llle rivkalivrl is raher ~.ur~ibcrsvrne arid nu1 very Iiclpful, and consequently it is common practice to treat the L' space as a metric space and to work with its members as if they were functions rather than equivalence classes. We return to this point in Chapter V. Let us elaborate a little on Examples 4 and 5 , concerning the space B(S) of bounded scalar-valued functions on a set S or, more generally, the space of bounded functions from S into a metric space (R, p ) . Suppose that S has
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the additional structure of a metric space (S, d). We let C(S) be the subset of B ( S ) consisting of bounded continuous functions on S, and we write C(S, R)or C(S, C) if we want to be explicit about the range. More generally we consider the space of bounded continuous functions from S into the metric space R . All of these are metric spaces in their own right.
Proposition 2.21. Let (S, d) and (R, p) be metric spaces, let xo be in S, and lct fn : S + R bc a scqucncc of boundcd functions from S into R that convcrgc uniformly to f : S + R and are continuous at xo. Then f is continuous at xo. In parlicular, llie unifunn lirrlil uf cunlinuuus fu~~cliuns is cunlinuuus. PROOF.For x in S. we write
Given F > 0, we choose an integer N by the uniform convergence such that the first and third terms on the right side are < t for n 1N . With N fixed, we choose 6 > 0 by the continuity of f N at xo such that p (f N (x) , fN (xO)) < t whenever d (x, xo) < 8 . Then the displayed inequality shows that d(x , xo) < S implies ,o (f (x) , f (x0)) < 3t,and the proposition follows. We conclude this section with some elementary results involving convergence of sequences in metric spaces.
Proposition 222. If ( X , d) is a metric space, then (a) for any subset A of X and limit point x of A, there exists a sequence in A - {x}converging to x , (b) any convergent sequence in X with limit x E X either has infinite image, with x as a limit point of the image, or else is eventually constantly equal to x . REMARK.This result and the first corollary below are used frequently-and often without speciiic reference. ~ O O OF F (a). For each n 2 1,the open ball B(l/n; x) is an open neighborhood of x and must contain a point x, of A distinct from the limit point x . Then d(x,, x ) < l / n , and thus limx, = x . Hence {x,} is the required sequence. PROOFOF (b). Suppose that {x,} converges to x and has infinite image. By discarding the terms equal to x , we obtain a subsequence {x,,} with limit x . If U is an open neighborhood of x , then {x,,} is eventually in U , by the assumed convergence. Since no term of the subsequence equals x , U contains a member of the image of {x,} different from x . Thus x is a limit point of the image of {x,}.
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Now suppose that {x,} converges to x and has finite image { p l, . . . , p, }. If x, is equal to some particular pj, for infinitely many n , then { x n }has an infinite subsequence converging to pj, . Since {x,} converges to x , every convergent subsequence converges to x . Therefore pj, = x . For j # jo ,only finitely many x, can then equal pj, and it follows that {x,} is eventually constantly equal to p JU' = x . Corollary 2.23. If ( X ,d) is a metric space, then a subset F of X is closed if and only if every convergent sequence in F has its limit in F. PROOF.Suppose that F is closed and {x,} is a convergent sequence in F with limit x . By Proposition 2.22b, either x is in the image of the sequence or x is a limit point of the sequence. In either case, x is in F; thus the limit of any convergent sequence in F is in F. Conversely suppose every convergent sequence in F has its limit in F. If x is a limit point of F, then Proposition 2.22a produces a sequence in F - { x } converging to x . By assumption, the limit x is in F. Therefore F contains all its limit points and is closed. Corollary 224. If ( S , d ) is a metric space, then the set C ( S ) of bounded continuous scalar-valued functions on S is a closed subset of the metric space B ( S ) of all bounded scalar-valued functions on S . PROOF.Proposition 2.21 shows for any sequence in C ( S ) convergent in B ( S ) that the limit is actually in C (S) . By Corollary 2.23, C ( S ) is closed in B ( S ). Proposition 2.25. Let f : X + Y be a function between metric spaces. Then f is continuous at a point x in X if and only if whenever {x,} is a convergent sequence in X with limit x , then { f (x,)} is convergent in Y with limit f ( x ) . REMARK.In the special case of domain and range R , this result was mentioned in Section I. 1 after the definition of continuity. We deferred the proof of the special case until now to avoid repetition. PROOF.Suppose that f is continuous at x and that {x,,}is aconvergent sequence in X with limit x . Let V be any open neighborhood of f ( x ). By continuity, there exists an open neighborhood U of x such that f ( U ) 5 V . Since x, > x , there exists N such that x, is in U whenever n > N . Then f (x,) is in f ( U ) V whenever n N . Hence { f (x,)} converges to f ( A ) . Conversely suppose that x, + x always implies f (x,) + f ( x ) . We are to show that f is continuous. Let V be an open neighborhood of f ( x ) . We are to show that some open neighborhood of x maps into V under f . Assuming the contrary, we can find, for each n > 1, some x, in B ( l / n ; x ) such that f (x,) is not in V . Then x, + x , but the distance of f (x,) from f ( x ) is bounded away
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from 0. Thus f (x,) cannot converge to f ( x ) . This is a contradiction, and we conclude that some B ( l / n ; x ) maps into V under f ; since V is arbitrary, f is continuous.
5. Subspaces and Products
When working with functio~lson the real line, one frequently has to address situations in which the domain of the function is just an open interval or a closed interval, rather than the whole line. When one uses the t - 6 definition of continuity, the subject does not become much more cumbersome, but it can become more cumbersome if one uses some other definition, such as one involving limits. The theory of metric spaces has a device for addressing smaller domains than the whole space-the notion of a subspace-and then the theory of functions on a subspace stands on an equal footing with the theory of functions on the whole space. Let ( X , d) be a metric space, and let A be a nonempty subset of X . There is a natural way of making A into a metric space, namely by taking the restriction d l A x Aas a metric for A . When we do so, we speak of A as a s~~hspace of X . When there is a need to be more specific, we may say that A is a metric subspace of X . If A is an open subset of X , we may say that A is an open subspace; if A is a closed subset of X , we may say that A is a closed subspace.
Proposition 2.26. If A is a subspace of a metric space ( X , d), then the open sets of A are exactly all sets U 1-1 A, where U is open in X , and the closed sets of A are all sets F n A, where F is closed in X . ~ O O F The . open balls in A are the intersections with A of the open balls of X , and the statement about open sets follows by taking unions. The closed sets of A are the complements within A of all the open sets of A, thus all sets of the form A - ((In A ) with U open in X . Since A - ( U n A ) = A n U c ,the statement about closed sets follows.
Corollary 2.27. If A is a subspace of ( X , d) and i f f : X + Y is continuous at a point a of A, then the restriction f A , mapping A into Y, is continuous at a . Also, f is continuous at a if and only if the function fo : X + f ( X ) obtained by redefining the range to be the image is continuous at a .
I
PROOF.Let V be an open neighborhood o f f ( a ) in Y. By continuity of f at a as a function on X , choose an open neighborhood U of a in X with f ( U ) V . Then U f' A is an open neighborhood of a in A , and f ( U f' A ) c V . Hence f is continuous at a .
1,
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The most general open neighborhood of f (a) in f (X) is of the form V f' f (X) with V an open neighborhood of f ( a ) in Y. Since f ( v ) = f i l ( V n f (X)) , the condition for continuity of fo at a is the same as the condition for continuity off ata. We now turn our attention to product spaces. Product spaces are a convenient device for considering functions of several variables. If (X, d ) and (Y, d f ) are metric spaces, there are several natural ways of making the product set X x Y , the set of ordered pairs with the first member from X and the second from Y , into a metric space, but all such ways lead to the same class of open sets and therefore also the same class of convergent sequences. We discussed an instance of this phenomenon in Example 1 of Section 4. For general X and Y, three such metrics on X x Y are
Each satisfies the defining properties of a rnetric. Sirnple algebra gives
whcncvcr a and b arc nonncgativc rcals, and thcrcforc
Let us check that this chain of inequalities implies that the neighborhoods of a point (xo, yo) are the same in all three metrics, hence that the open sets are the same in all three metrics. For any r > 0, the open balls about (xo, yo) in the three metrics satisfy
The first and second inclusions show that open balls about (xo, yo) in the metrics ,ol and pw are neighborhoods of (xo, yo) in the metric pl . Similarly the second and third inclusions show that open balls in the metrics p, and pl are neighborhoods in the nletric p2, and the third and first inclusions show that open balls in the metrics pl and p2 are neighborhoods in the metric p,. We shall refer to the metric pw as the product metric for X x Y . If X x Y is being regarded as a metric space and no metric has been mentioned, p, is to be understood. But it is worth keeping in mind that pl and p2 yield the same open sets. In the case of Euclidean space, it is the metric p2 on Rm x Rn that gives the
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Euclidean metric on Rmfn;thus the product metric and the Euclidean metric are distinct but yield the same open sets. A sequence {(x,, y,)} in the product metric converges to (xo, yo) in X x Y if and only if {x,} converges to xo and {y,} converges to yo. Since the three metrics on X x Y yield the same convergent sequences, this statement is valid in the metrics pl and p2 as well. It is an elementary property of the arithmetic operations in R that if {x,} converges to xo and {y,} converges to yo, then {x, y,} converges to xo yo. Similar statements apply to subtraction, multiplication, maximum, and minimum, and then to absolute value and to division except where division by 0 is involved. Further similar statements apply to those operations on vectors that make sense. Applying Proposition 2.25, we obtain (a) through (e) in the following proposition. Conclusions (a') through (el) are proved similarly.
+
+
Proposition 228. The following operations are continuous: (a) addition and subtraction frorn Rn x Rn into Rn, (b) scalar multiplication from R x Rn into Rn , (c) the map x H x-I from R - {0}to R - {O}, (d) themapx H 1x1 fromRn to R , (e) the operations from R2 to R of taking the maximum of two real numbers and taking the minimum of two real numbers, (a') addition and subtraction from Cn x Cn into Cn, (b') scalar multiplication from @ x cCn into C n, (c') the map x H x-I from C - {0}to @ - {O}, (dl) the map x I > I x 1 from Cn to R , (e') the map x H X from C to C . Corollary 2.29. Let (X, d) be a metric space, and let f and g be continuous functions from X into Rn or C n . If c is a scalar, then f + g , cf , f - g ,and I f are continuous. If n = 1, then the product f g is continuous, and the function l / f is continuous on the set where f is not zero. Tf n = 1 and the functions take values in R, then maxi f, g} and mini f ,g} are continuous. If n = 1 and the functions take values in @, then the complex conjugate f is continuous. REMARKS.If ( S , d) is a metric space, then it follows that the metric space C(S) of bounded continuous scalar-valued functions on S is a vector space. As such, it is a vector subspace of the metric space B(S) of bounded scalar-valued functions on S , and it is a metric subspace as well.3 'The word "subspace" can now be used in two senses, that of a 111etricsubspace of a metric space and that of a vector subspace of a vector space. The latter kind of subspace we shall always refer to as a "vector subspace," retaining the word "vector" for clarity. A "closed vector subspace" of B ( S ) then has to mean a closed metric subspace that is also a vector subspace.
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PROOF.The argument for f + g and for functions with values in Rn will illustrate matters sufficiently. We set up x H f ( x ) g ( x ) as a suitable composition, expressing the composition in a diagram:
+
- - (x,y)-(f
x-(x,x)
(u,v)-u+v
(x),g(y))
X XxX RnxRn Rn. Each function in the diagram is continuous, the last of them by Proposition 2.28a, and then the composition is continuous by Corollary 2.14.
We conclude this section with one further remark. When ( X , d ) is a metric space, we saw in Corollary 2.17that x H d ( x , y ) and y H d ( x , y ) are continuous functions from X to R. Actually, ( x , y ) H d ( x , y ) is a continuous function from X x X into R if we use the product metric. In fact, if p, denotes the product metric with p, ( ( x ,y ) , ( X OY, O ) ) = max{d( x , X O )d, ( y , Y O ) ) ,then we have d ( x, y ) 5 d ( x , xo) d (xo, yo) d ( y o ,y ) and therefore
+
+
+
d ( x , y ) - d@o, yo) i d ( x , xo) d ( y , yo). Reversing the roles of (x, y ) and (xo, yo), we see that
+
Id(x, Y ) - d(x0, yo)l 5 d ( x , xo) d(?, yo) i 2maxId(x, xo), d ( y , yo)} = 2pw ( ( x , Y ) , ( X O , Y O ) ) . From this chain of inequalities, it follows that d is continuous with 6 = t/2.
6. Properties of Metric Spaces This section contains two results about metric spaces. One lists a number of "separation properties" of sets within any metric space. The other concerns the completely different property of "separability," which is satisfied by some metric spaces and not by others, and it says that separability may be defined in any of three equivalent ways. Proposition 230 (separation properties). Let (X, d) be a metric space. Then (a) every one-point subset of X is a closed set. i.e., X is T I , (b) for any two distinct points x and y of X , there are disjoint open sets U and V with x E U and y E V ,i.e ., X is Hausdorff, (c) for any point x E X and any closed set F 5 X with x $ F, there are disjoint open sets U and V with x E U and F g V , i.e., X is regular, (d) for any two disjoint closed subsets E and F of X , there are disjoint open sets U and V such that E 5 U and F 5 V , i.e., X is normal, (e) for any two disjoint closed subsets E and F of X , there is a continuous function f : X + [O, 11 such that f is 0 exactly on E and f is 1 exactly on F.
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PROOF. For (a), the set { x }is the intersection of all closed balls B ( r ; x ) for r > 0 and hence is closed by Corollary 2.18 and Proposition 2.8b. For (e), the function f ( x ) = D ( x ; E ) / ( D ( x ;E ) + D ( x ; F ) ) is continuous by Proposition 2.16 and Corollary 2.29 and takes on the values 0 and 1 exactly on E and F, respectively, by Proposition 2.19. For (d) ,we need only apply (e) and Proposition 2.15b with U = f -l ((-on, and V = f , + X I ) ) Conclusions . (a) and (d) imply (c) , and conclusions (a) and (c) imply (b). This completes the proof.
-'((i
4))
A base B for a metric space ( X ,d) is a family of open sets such that every open set is a union of members of t?. The family of all open balls is an example of a base.
Proposition 2.31. If ( X ,d) is a metric space, then a family B of subsets of X is a base for ( X , d ) if and only if (a) every member of B is open and (b) for each x E X and open neighborhood U of x ,there is some member B oft? such that x is in B and B is contained in U . PROOF.If B is a base, then (a) holds by definition of base. If U is open in X , then U = U , B, for some members B, of B, and any such B, containing x can be Lake11as Lllc scl B in (b). Conversely suppose that B satisfies (a) and (b). By (a), each member of B is open in X . If U is open in X , we are to show that U is a union of members of 8. For each x E U , choose some set B = Bx as in (b). Then U = U X E B,, u and hence each open set in X is a union of members oft?. Thus t? is a base. This book uses the word countable to mean finite or countably infinite. It is then meaningful to ask whether a particular metric space (X, d ) has a countable base. On the real line R , the open intervals with rational endpoints form a countable base. A subset D of X is dense in a subset A of X if DC' 2 A; D is dense, or everywhere dense, if D is dense in X . A set D is dense if and only if there is some point of D in each nonempty open set of X. A family U of open sets is an open cover of X if the union of the sets in U is X . An open subcover of U is a subfamily of U that is itself an open cover.
Proposition 2.32. The following three conditions are equivalent for a metric space ( X , d ) : (a) X has a countable base, (b) every open cover of X has a countable open subcover, (c) X has a countable dense subset.
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107
PROOF.If (a) holds, let t? = {Bn},>l be a countable base, and let U be an over cover of X . Any U E U is the union of the B, E t? with B, g U . If B0 = { B , E B I B, 5 U for some U E U } ,then it follows that UBnEBO = UUEU = X . For each B, in Bo, select some U, in U with B, g U,. Then U , U, 2 U,a,Bo = X , and {U,} is a countable open subcover of U . Thus (b) holds. If (b) holds, form, for each fixed n 2 1, the open cover of X consisting of all open balls B ( l / n ; x ) . For that n , let { B ( l / n ;x,,)},>~ be a countable open subcover. We shall prove that the set D of all x,, ,wit11 rn and n arbitrary, is deiise in X . It is enough to prove that each nonempty open set in X contains a member of D, hence to prove, for each n , that each open ball of radius l / n contains a member of D. Thus consider B ( l / n ; x ) . Since the open balls B ( l / n ; x,,) with m > 1 cover X , x is in some B ( l / n ; x,,). Then that x,, has d(x,,, x ) < l / n , and hence x,, is in B ( l / n ; x ) . Thus D is dense, and (c) holds. If (c) holds, let {x,},?~be a countable dense set. Form the collection of all open balls centered at some x, and having rational radius. Let us use Proposition 2.3 1 to see that this collection of open sets, which is certainly countable, is a base. Let U be an open neighborhood of x . We are to see that there is some member B of our collection such that x is in B and B is contained in U . Since U is a neighborhood nf x , we can find an open hall R(r; x ) such that R(r; x ) 5 11; we may assume that r is rational. The given set {x,},>~being dense, some x, lies in B ( r / 2 ;x ) . If y is in B ( r / 2 ; x,), then d ( x , y ) 5 d ( x , x,) + d(x,, y ) < + = r . Hence x lies in B ( r / 2 ; x,) and B ( r / 2 ; x,) 5 B ( r ; x ) 5 U . Since r / 2 is rational, the opcn ball B ( r / 2 ; x,) is in our countablc collcction, and our countablc collcction is a base. This proves (a).
5
A metric space satisfying the equivalent conditions of Proposition 2.32 is said to be separable. Among the examples of metric spaces in Section 1 , the ones in Examples 1, 2 , 3, 6, 8 if X is countable, 9, 11, 12, 13, and 14 are separable. A countable dense set in Examples 1 , 2 , and 3 is given by all points with all coordinates rational. In Example 6, one countable dense set consists of all sequences with only finitely many nonzero entries, those being rational, and in Exampled and 9, X itself is a cnnntahle dense set. Tn Example 1 1 , the serlilences that are 0 in all but finitely many entries, those being rational, form a countable dense set. In Example 13, the set of finite linear combinations of exponentials em""using scalars in @ i@ is dense as a consequence of Parseval's equality. In Examplc 12, whcn [ a ,b] = [-n, n], thc samc countablc sct as for Examplc 13 is dense by Proposition 2.25 because the sets of functions in Examples 12 and 13 coincide and the inclusion of L~ into L' is continuous. In Example 14, the set of is countable and can be shown to be dense. polynomials with coefficients in @+i@ Example 10 is not separable, and Example 8 is not separable if X is uncountable.
+
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7. Compactness and Completeness
In Section 6 we introduced the notions of open cover and subcover for a metric space. We call a metric space compact if every open cover of the space has a jinite subcover. A subset E of a metric space ( X , d) is compact if it is compact as a subspace of the whole space, i.e.,if every collection of open sets in X whose union contains E has a finite subcollection whose union contains E. Historically this notion was embodied in the Heine-Bore1 Theorem, which says that any closed bounded subset of Euclidean space has the property that has just been defined to be compactness. As we shall see in Theorem 2.36 and Corollary 2.37 below, the Heine-Bore1 Theorem can be proved from the BolzanoWeierstrass Theorem (Theorem 1.8)and leads to faster, more transparent proofs of some of the consequences of the Bolzano-Weierstrass Theorem. Even more important is that it generalizes beyond metric spaces and produces useful conclusions about certain spaces of functions when statements about pointwise convergence of a sequence of functions are inadequate. Easily established examples of compact sets are hard to come by. For one example, consider in a metric space ( X ,d) a convergent sequence {x,} along with its limit x . The subset E = {x} U U,{x,}of X is compact. In fact, if U is an open cover of E , some member U of U has x as an element, and then all but finitely many elements of the sequence must be in U as well. Say that U contains x and all x, with n > N . For 1 5 n < N , let U , be a member of U containing x, . Then {U, U1, . . . , U N - l } is a finite subcover of U . It is easier to exhibit noncompact sets. The open interval (0,1)is not compact, 1)).Nor is an infinite discrete space, since as is seen from the open cover one-point sets form an open cover. A subtle dramatic example is the closed unit ball C of the hedgehog space X ,Example 10 in Section 1; this set is not compact. In fact, the open ball of radius 1/2 about the origin is an open set in X , and so is each open ray from the origin out to infinity. Let U be this collection of open sets. Then U is an open cover of C . However, no member of U is superfluous, since for each U in U ,there is some point x in C such that x is in C but x is in no other member of U. Thus U does not contain even a countable subcover. Let us now work directly toward a proof of the equivalence of compactness and the Bolzano-Weierstrass property in a metric space.
{(i,
Proposition 233. A compact metric space is separable. PROOF.This is immediate from equivalent condition (b) for the definition of separability in Proposition 2.32.
Proposition 234. In any metric space (X, d), (a) every compact subset is closed and bounded and (b) any closed subset of a compact set is compact.
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109
PROOF.For (a), let E be a compact subset of X , fix xo in X , and let U, for n 2 1 be the open ball { x E X d ( x o , x ) < n } . Then {U,} is an open cover of E . Since the U,'s are nested, the compactness of E implies that E is contained in a single U N for some N . Then every member of E is at distance at most N from xo, and E is bounded. To see that E is closed, we argue by contradiction. Let .uh be a limit point of E that is not in E . By the Hausdorff property (Proposition 2.30b), we can find, for each x c E ,open sets Uxand V, with x c U,,xh c V, , and Uxf' Vx = O.The sets Uxform an open cover of E. By compactness let {U,,, . . . , Uxn} be a finite which is disjoint froiii the neighborl~ood subcover. Then E 5 Ux,U . . . U UXn, V,,n . . . n Vx,, of xl,.Thus x; cannot be a limit point of E , and we have arrived at a contradiction. This proves (a). For (b), let E be compact, and let F be a closed subset of E . Because of (a), F is a closed subset of X . Let U be an open cover of F. Then U U { F c }is an open cover of E. Passing to a finite subcover and discarding FC,we obtain a finite subcover of F. Thus F is compact. A collection of subsets of a nonempty set is said to have the finite-intersection property if each intersection of finitely many of the subsets is nonempty.
Proposition 2.35. A metric space ( X , d ) is compact if and only if each collection of closed subsets of X with the finite-intersection property has nonempty intersection. PROOF. Closed sets with the finite-intersection property have complements that are open sets, no finite subcollection of which is an open cover.
Theorem 2.36. A metric space ( X , d ) is compact if and only if every sequence has a convergent subsequence. PROOF. Suppose that X is compact. Arguing by contradiction, suppose that
{x,},? is a sequence in X with no convergent subsequence. Put F = U E l{x,}. The subset F of X is closed by Corollary 2.23, hence compact by Proposition 2.34b. Since no x, is a limit point of F ,there exists an open set U, in X containing x, but no other member of F. Then {U,},,l is an open cover of F with no finite subcover, and we have arrived at a contradiction. Coilversely suppose that evely sequence has a coilvergent subsequence. We first show that X is separable. Fix an integer n. There cannot be infinitely many disjoint open balls of radius l l n , since otherwise we could find a sequence from among their centers with no convergent subsequence. Thus we can choose a finite disjoint collection of these open balls that is not contained in a larger such finite collection. Let their centers be xl, . . . , x ~ The . claim is that every point of X is
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at distance < 2 / n from one of these finitely many centers. In fact, if x E X is given, form B(:; x ) . This must meet some B(:; x i ) at a point y , and then d ( x , xi) i d ( x , y )
+ d ( y ,xi) < ;+ 1
=
i.
Thus x is at distance < 2 / n from one of the finitely many centers, as asserted. Now let n vary, and let D be the set of all these centers for all n . Then every point of X has members of D arbitrarily close to it, and hence D is a countable dense set in X . Thus X is separable. Let U be an open cover of X having no finite subcover. By the separability and condition (b) in Proposition 2.32, we may assume that U is countable, say U = { U 1 ,U2,. . . }. Since U 1U U2 U . . . U Un is not a cover, there exists a point x, not in the union of the first n sets. By hypothesis the sequence { x n }has a convergent subsequence {xn,},say with limit x . Since U is a cover, some member UN of U contains x . Then {x,,} is eventually in U N ,and some nk with nk > N has x,, in U N . But x,, is not in U 1 U . . . U U,, by construction, and this union contains UN, since nk > N . We have arrived at a contradiction, and we conclude that U must have had a finite subcover.
Corollary 2.37 (Heine-Bore1 Theorem) I11 Euclidean space R n ,evely closed bounded set is compact.
REMARK.Conversely we saw in Proposition 2.34a that every compact subset of any metric space is closed and bounded. PROOF.Let C be a closed rectangular solid in R n ,and let x ( ~=) (xl(k),. . . , x,( k ) ) be the members of a sequence in C. By the Bolzano-Weierstrass Theorem (Theorem 1.8) for R1,we can find a subsequence convergent in the first coordinate, a subsequence of that convergent in the second coordinate, and so on. Thus { x ( ~ ) } has a convergent subsequence. By Theorem 2.36, C is compact. Applying Corollary 2.34b, we see that every closed bounded subset of Rn is compact. Thc ncxt fcw rcsults will show how thc usc of cornpactncss both sirnplifics and generalizes some of the theorems proved in Section I. 1.
Proposition 2.38. Let ( X , d) and ( Y , p ) be metric spaces with X compact. If f : X + Y is continuous, then f ( X ) is a compact subset of Y .
-'
PKOOP.If {U,} is an open cover of f ( X ), then { f (U,)} is an open cover of X . Let { f - l ( ~ ~ ) } j nbe_ ~a finite subcover. Then {Uj}jn=lis a finite subcover of
f (XI. Corollary 2.39. Let ( X ,d ) be a compact metric space, and let f : X + R be a continuous function. Then f attains its maximum and minimum values.
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111
REMARK.Theorem 1.11 was the special case of this result with X = [ a ,b ] . This particular space X is compact by the Heine-Bore1 Theorem (Corollary 2.37), and the corollary applies to yield exactly the conclusion of Theorem 1.11. PROOF.By Proposition 2.38, f ( X ) is a compact subset of R. By Proposition 2.34a, f ( X ) is closed and bounded. The supremum and infimum of the members of f ( X ) in R* lie in R, since f ( X ) is bounded, and they are limits of sequences in f ( X ) . Since f ( X ) is closed, Proposition 2.23 shows that they must lie in f ( X ) .
Corollary 2.40. Let ( X , d ) and ( Y , p) be metric spaces with X compact. If + Y is continuous, one-one, and onto, then f is a homeomorphism.
f :X
REMARK.In the hypotheses of the change of variables formula for integrals in R 1 (Theorem 1.34), a function q~ : [ A , B] + [ a ,b] was given as strictly increasing, continuous, and onto. Another hypothesis of the theorem was that p-' was continuous. Corollary2.40 shows that this last hypothesis was redundant.
-')-'
( E ) = f ( E ). PROOF.Let E be a closed subset of X , and consider ( f The set E is compact by Proposition 2.34b, f ( E ) is compact by Proposition 2.38, and f (E) is closed by Proposition 2.34a. Proposition 2.15b thus shows that f is continuous.
-'
If ( X ,d ) and ( Y , p) are metric spaces, a function f : X + Y is uniformly continuous if for each t > 0, there is some 6 > 0 such that d ( x l ,x 2 ) < 6 implics p( f ( x l ) ,f ( x z ) ) < t . This is thc natural gcncralization of thc dcfinition in Section 1.1 for the special case of a real-valued function of a real variable.
Proposition 2.41. Let (X, d ) and ( Y , p ) be metric spaces with X compact. If f : X + Y is continuous, then f is uniformly continuous. REMARK. This result generalizes Theorem 1.10, which is the special case X = [ a ,b] and Y = R . PROOF. Let t > 0 be given. For each x E X , choose 8, > 0 such that d ( x f ,x ) < S , implies p ( f ( x ' ) , f ( x ) ) < t / 2 . The open balls ~ ( $ 3 , x; ) cover X ; let the balls with centers x l , . . . , x, be a finite subcover. Put S = rnin{b,,, . . . , 6,a}. Now suppose that d ( x f ,x ) < 6. The point x is in some ball in the finite subcover; suppose x is in ~ ( $ 3;,x~J ) . Then d ( x , xJ) < ;SxJ , so that d ( x f ,x i ) 5 d ( x f ,X )
+ d ( x , xi) < S + ;Sxj 5 Sxj.
By definition of SXJ, p ( f ( x ' ) , f ( x j ) ) < t / 2 and p ( f ( x j ) ,f ( x ) ) < t/2. Therefore
and the proof is complete.
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One final application of compactness is the Fundamental Theorem of Algebra, which is discussed in Section A8 of the appendix in the context of properties of polynomials. Theorem 2.42 (Fundamental Theorem of Algebra). Every polynomial with complex coefficients and degree > 1 has a complex root. PROOF.Let P : C + C be the function P ( z ) = Cj"=, a j z j , where ao, . . . , a, are in C with a, # 0 and with n > 1. We may assume that a, = 1. Let m = infZEcIP(z)I. Since P ( z ) = z n ( l +an-lz-' +. . .+alz-("-') +aozPn),we have lim,,, P ( z ) / z n= 1 . Thus there exists an R such that I P ( z )I > lzln whenever lzl > R . Choosing R = Ro such that R: > 2 m , we see that I P ( z )I > 2m for lzl r Ro . Consequently m = inflz15Ro I P ( z )1. The set S = { z E C IzI 5 R o } is compact by the Heine-Bore1 Theorem (Corollary 2.37), and Corollary 2.39 shows that I P ( z )1 attains its minimum on S at some point zo in S. Then I P ( z ) 1 attains its minimum on C at 20. We shall show that this minimum value m is 0. Assuming the contrary, define Q ( z ) = P ( z zo)/ P (zo),so that Q ( z ) is a polynomial of degree n > 1 with Q (0) = 1 and 1 Q ( 2 ) > 1 for all z . Write
I
+
Q ( z )= 1
+ bkzk + bk+lzk + . . . + bnzn I
with bk # 0.
Corollary 1.45produces a real number 0 such that eikobk= - Ibk I. For any r > 0 with r k lbk I < 1, we then have
For such r and that B ,this equality implies that
For sufficiently small r > 0, the expression in parentheses on the right side is positive, and then I Q (reis)I < 1 , in contradiction to hypothesis. Thus we must have had m = 0, and we obtain P (zo)= 0 . Another theme discussed in Section 1.1 is that Cauchy sequences in R 1 are convergent. This convergence was proved in Theorem 1.9 as a consequence of the Bolzano-Weierstrass Theorem. Actually, many sequences in metric spaces of importance in analysis are shown to converge without one's knowing the limit in advance and without using any compactness, and we therefore isolate the forced convergence of Cauchy sequences as a definition. In a metric space (X, d), a sequence {xn} is a Cauchy sequence if for any t- > 0, there is some integer N
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113
such that d ( x , , x,) it whenever 172 and n are > N . A familiar 2 t argument shows that convergent sequences are Cauchy. Other familiar arguments show that any Cauchy sequence with a convergent subsequence is convergent and that any Cauchy sequence is bounded. We say that the metric space ( X , d) is complete if every Cauchy sequence in X converges to a point in X . We know that the line R' is complete. It follows that Rn is complete because a Cauchy sequence in Rn is Cauchy in each coordinate. A nonempty subset E of X is complete if E as a subspace is a complete metric space. The next two propositions and corollary give three examples of complete metric spaces.
Proposition 2.43. A subset E of a complete metric space X is complete if and only if it is closed. REMARK.In particular every closed subset of Rn is a complete metric space. PROOF.Suppose E is closed. Let {x,} be a Cauchy sequence in E. Then {x,} is Cauchy in X , and the completeness of X implies that {x,} converges, say to some x E X. By Corollary 2.23, x is in E . Thus {x,} is convergent in E . The converse is immediate from Corollary 2.23.
Proposition 2.44. If S is a nonempty set, then the vector space B ( S ) of bounded scalar-valued functions on S , with the uniform metric, is a complete metric space. PROOF.Let { f,} be a Cauchy sequence in B ( S ) . Then { f n ( x ) } is a Cauchy sequence in C for each x in S . Define f (x) = limn f,(x). For any t > 0, we know that there is an integer N such that I f n ( x ) - f m ( x ) 1 < t whenever n and m are > N. Taking into account the continuity of the distance function on C, i.e., the continuity of absolute value, we let m tend to infinity and obtain I f, ( x ) - f ( x )I 5 e for n 1N. Thus { f,} converges to f in B ( S ).
Corollary 2.45. Let ( S , d) be a metric space. Then the vector space C ( S ) of bounded continuous scalar-valued functions on S , with the uniform metric, is a complete metric space. REMARK.C ( S ) was observed to be a vector subspace in the remarks with Corollary 2.29. PROOF.The space B ( S ) is complete by Proposition 2.44, and C ( S ) is a closed metric subspace by Corollary 2.24. Then C ( S ) is complete by Proposition 2.43. Now we shall relate compactness and completeness. A metric space ( X , d) is said to be totally bounded if for any t > 0, finitely many open balls of radius t cover X .
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Theorem 2.46. A metric space ( X , d ) is compact if and only if it is totally bounded and complete. PROOF.Let ( X , d ) be compact. If t > 0 is given, the open balls B ( t ; x ) cover X . By compactness some finite number of the balls cover X . Therefore X is totally bounded. Next let a Cauchy sequence { x , } be given. By Theorem 2.36, { x , } has a convergent subsequence. A Cauchy sequence with a convergent subsequence is necessarily convergent, and it follows that X is complete. In the reverse direction, let X be totally bounded and complete. Theorem 2.36 shows that it is enough to prove that any sequence { x , } in X has a convergent subsequence. By total boundedness, find finitely many open balls of radius 1 covering X . Then infinitely many of the x,'s have to lie in one of these balls, and hence there is a subsequence {x,,} that lies in a single one of these balls of radius 1. Next finitely many open balls of radius 112 cover X. In the same way there is a subsequence {x,,,} of { x , , } that lies in a single one of these balls of radius 112. Continuing in this way, we can find successive subsequences, the mth of which lies in a single ball of radius 1/m. The Cantor diagonal process, used in the proof of Theorem 1.22, allows us to form a single subsequence {xjz} of { x , } such that for each m, { x j ,} is eventually in a ball of radius 2-,. If t > 0 is given, find m such that 2 "\ t , and let c, be the center of the ball of radius l / m . Choose an integer N such that xjz lies in B ( l / m ; c,) whenever ji > N . If ji > N and j i t 2 N , then d ( c , , x j z ) < t and d ( c , , x j z , ) < t , whence d ( x j z ,x j z , ) < 2 t . Therefore the subsequence {xjz} is Cauchy. By completeness it converges. Hence { x , } has a convergent subsequence, and the theorem is proved. Let ( X , d ) and ( Y , p ) be metric spaces, and let f : X + Y be uniformly continuous. Then f carries Cauchy sequences to Cauchy sequences. In fact, if { x , } is Cauchy in X and if t- > 0 is given, choose some S of uniform continuity for f and t , and find an integer N such that d ( x , , x,,) < 6 whenever n and n' are 2 N . Then p ( f ( x , ) , f ( x , , ) ) < t- for the same n's and n"s, and hence { f ( x , ) } is Cauchy.
Proposition 2.47. Let ( X , d ) and ( Y , p ) be metric spaces with Y complete, let D be a dense subset of X , and let f : D 3 Y be uniformly continuous. Then f cxtcnds uniqucly to a continuous function P' : X + Y , and P' is uniformly continuous. PROOF OF UNIQUENESS. If x is in X , apply Proposition 2.22a to choose a sequence { x n } in D with x , + x . Continuity of F forces F ( x , ) + F ( x ) . But F ( x , ) = f (x,) for all n . Thus F ( x ) = lim, f ( x , ) is forced.
PROOFOF EXISTENCE. If x is in X , choose x , E D with x , x . Since { x , } is convergent, it is Cauchy. Since f is uniformly continuous, { f ( x , ) } is
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Cauchy. The completeness of Y then allows us to define F ( x ) = lim f (x,), but we must see that F is well defined. For this purpose, suppose also that {y,} is a sequence in D that converges to x . Let { z n }be the sequence X I , yl, x2, y2, . . . . This sequence is Cauchy, and {x,} and {y,} are subsequences of it. Therefore lim f (y,) = lim f (2,) = lim f (x,), and F ( x ) is well defined. For the uniform continuity of F, let t :, 0 be given, and choose some S of uniform continuity for f and t/3. Suppose that x and x' are in X with d ( x , x') < 613. Choosex, in D with d(x,, x ) < S/3 and p( f (x,), F ( x ) ) < t/3, and choose xl, in D with d(xA, x') < 6/3 and p ( f (xl,),F ( x ' ) ) < t/3. Then d ( x , , x;) iS by the triangle inequality, and hence p ( f (x,) , f ( x ; ) ) it/3. Thus p (F( x ), F ( x ' ) ) < t by the triangle inequality.
8. Connectedness Although the Intermediate Value Theorem (Theorem 1.12) in Section 1.1 was derived from the Bolzano-Weierstrass Theorem, the Intermediate Value Theorem is not to be regarded as a consequence of compactness. Instead, the relevant property is "connectedness," which we discuss in this section. A metric space ( X , d ) is connected if X cannot be written as X = U U V with U and V open, disjoint, and nonempty. A subset E of X is connected if E is connected as a subspace of X, i.e., if E cannot be written as a disjoint union ( E f' U ) ILJ ( E n V ) with U and V open in X and with E n U and E f' V both nonempty. The disjointness in this definition is of E n U and E f' V ; the open sets U and V may have nonempty intersection.
Proposition 2.48. The connected subsets of R are the intervals-open, closed, and half open. PROOF.Let E be a connected subset of R, and suppose that there are real numbers a , b, c such that a < c < b, a and b are in E, and c is not in E . Forming the open sets U = (- m, c ) and V = ( c , +oo) in R,we see that E is the disjoint union of E n U and E n V and that these two sets are nonempty. Thus E is not connected. Conversely suppose that I is an open, closed, or half-open interval of R from a to b, with a # b but with a or b or both allowed to be infinite. Arguing by contradiction, suppose that I is not connected. Choose open sets U and V in R such that I is the disjoint union of I n U and I n V and these two sets are nonempty. Without loss of generality, there exist members c and c' of I n U and I n V , respectively, with c < c'. Since U is open and c has to be < b, all real numbers c t with t > 0 sufficiently small are in I f' U . Let d = sup { x [c,u ) c I n u } , so that d > c .
I
+
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If d < b , then the fact that U is open implies that d is not in I n U . Thus d is in I n V . Since V is open and d > a , d - t is in I n V if t > 0 is sufficiently small. But then d - t is in both I f' U and I f' V for t sufficiently small. This is a contradiction, and we conclude that d = b. If d = b is in I n V , then the same argument shows that b - t is in both I n U and I n V for t positive and sufficiently small, and we again have a contradiction. Consequently all points from c to the right end of I are in I f' U . This is again a contradiction, sincc c' is known to bc in I f' V .
Proposition 2.49. The continuous image of a connected metric space is connected. PROOF. Let ( X , d ) and ( Y , p) be metric spaces, and let f : X + Y be continuous. We are to prove that f (X) is connected. Corollary 2.27 shows that there is no loss of generality in assuming that f ( X ) = Y ,i.e., f is onto. Arguing by contradiction, suppose that Y is the union Y = U U V of disjoint nonempty open sets. Then X = f (u)U f ( v )exhibits X as the disjoint union of nonempty sets, and these sets are open as a consequence of Proposition 2.15a. Thus X is not connected.
Corollary 2.50 (Intermediate Value Theorem). For real-valued functions of a real variable, the continuous image of any interval is an interval. PROOF. This is immediate from Propositions 2.48 and 2.49. Further connected sets beyond those in R are typically built from other connected sets. One tool is a path in X , which is a continuous function from a closed bounded interval [ a ,b] into X . The image of a path is connected by Propositions 2.48 and 2.49. A metric space ( X , d ) is pathwise connected if for any two points xl and x2 in X , there is some path p from xl to x2,i.e., if there is some continuous p : [ a ,b] + X with p(a) = xl and p(b) = xz. A pathwise-connected metric space ( X ,d) is necessarily connected. In fact, otherwise we could write X as a disjoint union of two nonempty open sets U and V . Let xl be in U and x2 be in V , and let p : [ a ,b] + X be a path from xl to x2. Then p ( [ a ,b ] ) = ( p ( [ a ,b ] )n U ) U ( p( [ a ,b ] ) f' V ) exhibits p ( [ a ,b ] ) as a disjoint union of rclativcly opcn scts, and thcsc scts arc noncmpty, sincc xl is in the first set and x2 is in the second set. Consequently p ( [ a , b ] ) is not connected, in contradiction to the fact that the image of any path is connected. We can view a pathwise-connected metric space as the union of images of paths from a single point to all other points, and such a union is then connected. The following proposition generalizes this construction.
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117
Proposition 2.51. If (X, d) is a metric space and {E,} is a systemof connected subsets of X with a point xo in common, then U , E, is connected. PROOF.Assuming the contrary, find open sets U and V in X such that U , E, is the disjoint union of its intersections with U and V and these two intersections are both nonempty. Say that xo is in U . Since E, is connected and xo is in E, nU , the decomposition E, = ( E , f l U ) U ( E , il V ) forces E, f l V to be empty. Then ( U aE,) n V = U , ( E , n V ) is empty, and we have arrived at a contradiction.
Proposition 2.52. If (X, d) is a metric space and E is a connected subset of X, then the closure EC' is connected. PROOF.Suppose that U and V are open sets in X such that EC' is contained in U U V and E"' n U n V is empty. We are to prove that E"' n U and E"' n V cannot both be nonempty. Arguing by contradiction, let x be in E" f' U and let y be in EC' n V . Since E is connected, E n U and E n V cannot both be nonempty, and thus x and y cannot both be in E . Thus at least one of them, say x,is a limit point of E . Since U is a neighborhood of x , U contains a point e of E different from x . Thus e is in E n U . Since y cannot then be in E n V , y is a limit point of E . Since V is a neighborhood of y , V contains a point f of E different from y . Thus f is in E n V , and we have arrived at a contradiction.
EXAMPLE. The graph in IR2 of sin(l/x) for 0 < x 5 1 is pathwise connected, and wc havc sccn that pathwisc-conncctcd scts arc conncctcd. Thc closurc of this graph consists of the graph together with all points (0, t) for - 1 5 t 5 1, and this closure is connected by Proposition 2.52. One can show, however, that this closure is not pathwise connected. Thus we obtain an example of a connected set in IR2 that is not pathwise connected. 9. Baire Category Theorem A number of deep results in analysis depend critically on the fact that some metric
space is complete. Already we have seen that the metric space C ( S ) of bounded continuous scalar-valued functions on a metric space is complete, and we shall see as not too hard a consequence in Chapter XII that there exists a continuous periodic function whose Fourier series diverges at a point. One of the features of the Lebesgue integral in Chapter V will be that the metric spaces of integrable functions and of square-integrable functions, with their natural metrics, are further examples of complete metric spaces. Thus these spaces too are available for applications that make use of completeness. The main device through which completeness is transformed into a powerful hypothesis is the Baire Category Theorem below. A closed set in a metric space
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is nowhere dense if its interior is empty. Its complement is an open dense set, and conversely the complement of any open dense set is closed nowhere dense. EXAMPLE. A nontrivial example of a closed nowhere dense set is a Cantor set4 in IR.This is a set constructed from a closed bounded interval of IR by removing an open interval in the middle of length a fraction 1-1 of the total length with 0 < rl < 1, removing from each of the 2 remaining closed subintervals an open interval in the middle of length a fraction r 2 of the total length of the subinterval, removing from each of the 4 remaining closed subintervals an open interval in the middle of length a fraction r 3 of the total length of the interval, and so on indefinitely. The Cantor set is obtained as the intersection of the approximating sets. It is closed, being the intersection of closed sets, and it is nowhere dense because it contains no interval of more than one point. For the standard Cantor set, the starting interval is [0, 11, and the fractions are given by rl = I-2 = . . . = -31 at every stage. In general, the "length of the resulting set5 is the product of the length of the starting interval and (1 - r,) .
Theorem 2.53 (Baire Category Theorem). If (X, d) is a complete metric space, ale11 (a) the intersection of countably many open dense sets is nonempty, (b) X is not the union of countably many closed nowhere dense sets. PROOF.Conclusions (a) and (b) are equivalent by taking complements. Let us prove (a). Suppose that U , is open and dense for n > 1. Since U 1 is nonempty and open, let E l be an open ball B(rl ; x l ) whose closure is in U 1and whose radius is rl 5 1. We construct inductively open balls E, = B(r,; x,) with r, 5 such that En 5 U1 n . . . n U , and E:' 5 E n P 1 . Suppose En with n > 1 has been constructed. Since Un+1 is dense and En is nonempty and open, U,+l n En is not empty. Let x,+l be a point in U,+l n En. Since U,+l n En is open, we can and center the find an open ball E,+l = B(r,+l; x,+l) with radius r,+l 5 point x,+l in Un+1 such that E:+~ 5 Un+1 n E n . Then En+1 has the required properties, and the inductive construction is complete. The sequence { x n }is Cauchy because whenever n > m , the points x , and x , are both in E, and thus have d (x,, x,) i Since X is by assumption complete, let x, + x . For any integer N ,the inequality n > N implies that xn is in EN+1.Thus the limit x is in E:+~ 5 E N 5 U1 n . . . n U N . Since N is arbitrary, x is in Un.
--&
A.
40ften a mathematician who refers to "the" Cantor set is referring to what is called the "standard Cantor set" later in the present paragraph. 5 ~ be o precise, the length is the "Lebesgue measure" of the set in the sense to be defined in Chapter V.
9. Baire Category Theorem
119
REMARK.In (a), the intersection in question is dense, not merely nonempty. To see this, we observe in the first part of the proof that since U 1is dense, El can be chosen to be arbitrarily close to any member of X and to have arbitrarily small radius. Following through the construction, we see that x is in El and hence can be arranged to be as close as we want to any member of X . The corresponding conclusion in (b) is that a nonempty open subset of X is never contained in the countable union of closed nowhere dense sets. EXAMPLES. (1) The subset Q of rationals in R is not the countable intersection of open sets. In fact, assume the contrary, and write Q = U , with U, open. Each set Un contains Q and hence is dense in R.Also, for q E Q,the set R - {q} is U, implies that open and dense. Thus the equality Q =
n,"=,
nZ1
is empty, in contradiction to Theorem 2.53. (2) Let us start with a Cantor set as at the beginning of this section. The total interval is to be [O, I], and the set is to be built with middle segments of fractions r l , r2, . . . . Within the closure of each removed open interval, we insert a Cantor set for that interval, possibly with different fractions 1-1,1-2, . . . for each inserted Cantor set. This insertion involves further removed open intervals, and we insert a Carllvr scl irllu each uf tllese. We curllirlue his pruwss i11Clelirlilely. The u~liuri of the constructed sets is dense. Can it be the entire interval [0, I]? The answer is "no" because each of the Cantor sets is closed nowhere dense and because by Theorem 2.53, the interval [0, 11 is not the countable union of closed nowhere dense sets. A subset E of a metric space is said to be of the first category if it is contained in the countable union of closed nowhere dense sets. Theorem 2.53 and the remark after it together imply that no nonempty open set in a complete metric space is of the first category.
Theorem 2.54. Let (X, d) be a complete metric space, and let U be an open subset of X . Suppose for n ? 1 that f, : U + C is a continuous function and that f, converges pointwise to a function f : U + C.Then the set of discontinuities of f' is of the first category. The proof will make use of the notion of the oscillation of a function. For any function g : U + C ,define oscg(XO)
= lim
sup
6 $ O x€B(G;xo)
Ig(x) - g(x0) I,
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so that g is continuous at xo if and only if oscg(xo) = 0. At first glance it might seem that the sets {x osc,(x) > r } are always closed, no matter what discontinuities g has. Actually, these sets need not be closed. Take, for example, the function g : R + R that is 1 at every nonzero rational, 0 at every irrational, and 112 at 0. Then oscg(x) is 1 at every x in R except for x = 0, where it is 112. Thus, in this example, the set {x oscx(x) 2 1) is R - {0}and is not closed.
I
I
Lemma 2.55. Let (X, d) be a complete metric space, and let U be an open subset of X . If g : U + @ is a function and t > 0 is a positive number, then
PROOF.We need to see that the limit points of the set on the left are in the set on the right. Thus suppose that oscg(x,) > 2t for all n and that x, + xo. For each n ,choose x,,, such that lim, x,,, = x, and Ig(x,,,) - g(x,) I > t for all m . Because of the convergence of x,,, to x,, we may choose, for each n, an integer m = rn, such that d(x,,, , x,) < d (xo,x,) , and then limn x,,n = xo by the triangle inequality. From lg(~,,,~)- g(x,)l > t, the triangle inequality forces Ig(xn,m,>- g(xo)l > 5 or lg(xn) - ~ ( x o ) I> (*I
5.
Dcfining y, to bc x,,,,, or x, according as thc first or sccond incquality is thc casc in (*), we have y, + xo and I g(y,) - g(xo)I ? 5 . This proves the lemma. PROOFOF THEOREM2.54. In view of Lemma 2.55 and the fact that U is not uf firs1 calcgvry (Tlicurern 2.53 arid h e remark aflerwarcl), il is crluugli lu pruve for each c 0 that {x oscf (x) > t ) does not contain a nonempty open subset of X . Assuming the contrary, suppose that it contains the nonempty open set V . Define Amn = {X E V Ifm(x) - fn(x)I 5 5).
I
I
n,?,
This is a relatively closed subset of V . Then A, = A, is closed in V . If x is in V , the fact that { fn(x)} is a Cauchy sequence implies that there is some m such that x is in A, for all n > rn. Hence Uz==l A, = V . Again by Theol-em 2.53 and the remark after it, some A, has nonempty interior. Fix that in, and let W be its nonempty interior. Since
5
every point of W has I f,(x) - f (x)l 5 and oscf(x) > t . Let xo be in W and choose x, tending to xo with f (x,) - f (xo)I > $. From I f, (x,) - f (x,) I 5 5 and I f, (XO)- f (xo) I 5 5, we obtain I f, (x,) - f, (xo)1 > 5. Since x, converges to xo, this inequality contradicts the continuity of f, at xo.
10. Properties of C (S) for Compact Metric S
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10. Properties of C ( S ) for Compact Metric S If ( S , d ) is a metric space, then we saw in Proposition 2.44 that the vector space B ( S ) of bounded scalar-valued functions on S , in the uniformmetric, is a complete metric space. We saw also in Corollary 2.45 that the vector subspace C ( S ) of bounded continuous functions is a complete subspace. In this section we shall study the space C ( S ) further under the assumption that S is compact. In this case Propositions 2.38 and 2.34 tell us that every continuous scalar-valued function on S is automatically bounded and hence is in C ( S ). The first result about C ( S ) for S compact is a generalization of Ascoli's Theorem from its setting in Theorem 1.22for real-valued functions on a bounded interval [ a ,b] . The generalized theorem provides an insight that is not so obvious from the special case that S is a closed bounded interval of R. The insight is a characterization of the compact subsets of C ( S ) when S is compact, and it is stated precisely in Corollary 2.57 below. The relevant definitions for Ascoli's Theorem are generalized in the expected way. Let F = { f aI a E A} be a set of scalar-valued functions on the compact metric space S. We say that F is equicontinuous at x E S if for each t- > 0, there is some S > 0 such that d(t, x ) < 6 implies I f (t) , f (x)l < t for all f E F. The set F of functions is pointwise bounded if for each t E [ a ,b] , there exists a number Mt such that If ( t )1 5 Mt for all f E F. The set is uniformly equicontinuous on S if it is equicontinuous at each point x E S and if the S can be taken independent of x . The set is uniformly bounded on S if it is pointwise bounded at each t E S and the bound M, can be taken independent o f t ; this last definition is consistent with the definition of a uniformly bounded sequence of functions given in Section 4. -
Theorem 2.56 (Ascoli's Theorem). Let ( S , d ) be a compact metric space. If { f,} is a sequence of scalar-valued functions on S that is equicontinuous at each point of S and pointwise bounded on S , then (a) { f,} is uniformly equicontinuous and uniformly bounded on S , (b) { f,} has a uniformly convergent subsequence.
REMARKS.The proof involves only notational changes from the special case Theorem 1.22;there are enough such changes, however, so that it is worth writing out the details. Inspection of this proof shows also that the range R or @ may be replaced by any compact metric space. We shall see a further generalization of this theorem in Chapter X, and the proof at that time will look quite different. PROOF.Since each f, is continuous at each point, we know from Propositions 2.38, 2.34a, and 2.41 that each f , is uniformly continuous and bounded. The proof of (a) amounts to an argument that the estimates in those theorems can be arranged to apply simultaneously for all n .
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First consider the question of uniform boundedness. Choose, by Corollary 2.39, some x, in S with I f, (x,) I equal to K, = supxGsI f, ( x )1 . Then choose a subsequence on which the numbers K, tend to sup, K , in R*. There will be no loss of generality in assuming that this subsequence is our whole sequence. By compactness of S , apply the Bolzano-Weierstrass property given in Theorem 2 36 to find a convergent subsequence { x l z , }of {x,}, and let xo be the limit of this subsequence. By pointwise boundedness, find Mxowith I fn(xo)l 5 Mxofor all n. Then choose some 6 of equicontinuity at xo for t = 1. As soon as Ic is large enough so that d (x,, , x o ) < 6 , we have
Thus 1 + Mxois a uniform bound for the functions f, . For the uniform equicontinuity, fix t > 0. The uniform continuity of f, for each n , as given in Proposition 2.41, means that it makes sense to define
I
6,(t) = min 1 , sup 6' > 0
If ( x ) - f ( y )I < t whenever d ( x , y ) < S f and x and y are in S
1f d ( x , y ) < 8, ( t ) ,then I f, ( x ) - fn ( y )1 < t . h t S ( t )= inf, J , ( t ) . Let us see that it is enough to prove that 6 ( t )> 0: If x and y are in S with d ( x , y ) < 6 ( t ), thcn d ( x , y) iS ( t )5 8, ( t ). Hcncc I f,(x) - f,(y) I it as rcquircd. Thus we are to prove that S (r) > 0. If S ( t )= 0, then we first choose a strictly increasing sequence I n k } of positive integers such that S n k ( t ) < and we next choose xk and yk in S with d ( x k ,y k ) < Snk ( t )and I fk ( x k )- f k ( y k )1 > t . Using the Bolzano-Weierstrass property again, we obtain a subsequence { x k 1 }of { x k } such that {xkl} converges, say to a limit xo . Then
i,
lim sup d(yk,, xo) 5 lim sup d (ykl,xkl) I
I
+ limI sup d (xkl,x0) = 0 + 0 = 0,
so that { y k l }converges to xo. Now choose, by equicontinuity at xo, a number for all n whenever d ( x , x o ) < 6'. The convergence of { x k l }and { y k l }to xo implies that for large enough I , we have d ( x k l ,x o ) < S f and d ( y k l ,x o ) < S f . Therefore I fkl ( x k l )- fkl (xO)I < and I fkl Cykl) fkl(x0) I ( 5 , from which we conclude that I fkl (xkl) fkl ( Y ~ JI )< C . But we saw that I fk(xk) - fk(yk)l > t for all k , and thus we have arrived at a contradiction. This proves the unifor~llequicontinuity and con~pletesthe proof of (a). To prove (b), let R be a compact set containing all sets image( f,). Choose a countable dense set D in S by Proposition 2.33. Using the Cantor diagonal process and the Bolzano-Weierstrass property of R , we construct a subsequence { fnk} of { f,} that is convergent at every point in D. Let us prove that { f,,} is 6' > 0 such that I f , ( x ) - f, ( x o )1 <
10. Properties of C (S) for Compact Metric S
123
uniformly Cauchy. Let t > 0 be given, and let S be some corresponding number exhibiting equicontinuity. The balls B(6; r ) centered at the members r of D cover S, and the compactness of S gives us finitely many of their centers rl , . . . , rl such that any member of S is within 6 of at least one of r l , . . , rl . Then choose N w i t h I f n ( r J ) - fm(rJ)I < r f o r l 5 j 52whenevernandmare1 N . I f x i s i n S , let r ( x ) be an rJ with d ( x , r ( x ) ) r S Whenever n and m are > N , we then have
I fn(x>-
fm(x>I
5 Ifn(x>- fn(r(x>>I+ I f n ( ~ ( x )> fm(r(x>>I+ Ifm(r(x>)- fm(x)I
Hence { fn,} is uniformly Cauchy, and (b) follows since the metric space C ( S ) is complete.
Corollary 2.57. If ( S , d) is a compact metric space, then a subset E of C ( S ) in the uniform metric has compact closure if and only if E is uniformly bounded and uniformly equicontinuous.
PROOF.First let us see that if E is uniformly bounded and uniformly equicontinuous, then so is EC'. In fact, if If ( x )1 5 M for f E E , then the same thing is t n ~ of e any uniform limit of such functions. Hence E'' is uniformly bounded. For the uniform equicontinuity of E"' ,let t be given, and find some 6 of equicontinuity for t and the members of E . If f is a limit point of E , we can find a sequence { f n } in E converging uniformly to f . If d ( x , y ) < 8 , then the inequality
I f ( x >- f'(y)I 5 I f ( x >- fn(x>I + Ifn(x>- fn(y>l + Ifn(y>- ~ ( Y > I and the uniform convergence show that we obtain If ( x ) - f ( y )I < 3c by fixing any sufficiently large n . Thus E"' is uniformly equicontinuous. Now suppose that E is a closed subset of C ( S ) that is uniformly bounded and equicontinuous. Then Theorem 2.56 shows that any sequence in E has a subsequence that is convergent in C ( S ) . Since E is closed, the sequence is convergent in E . Theorem 2.36 then shows that E is compact. Conversely suppose that E is compact in C ( S ). Distance from 0 in C ( S ) is a continuous real-valued function by Corollary 2.17, and this continuous function has to be bounded on the compact set E . Thus E is uniformly bounded. For the uniform equicontinuity, let t > 0 be given. Theorem 2.46 shows that E is totally bounded. Hence we can find a finite set f ; , . . . , f t in E such that each member f' of E has supxESI f ( x ) - fJ ( x )1 < t for some j . By uniform continuity of each fi ,choose some number S > 0 such that d ( x , y ) < S implies I fi ( x )- fi ( y )I < t for 1 5 i 5 I . If f , is the member of the finite set associated with f , then d ( x , y ) < S implies I f ( x ) - f (Y>I 5 I f ( x >- fj(x>I + I f j ( x ) - fj(y>I + I f j ( y >- f Hence E is uniformly equicontinuous.
(Y>I
< 3t.
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The second result about C ( S ) when S is compact generalizes the Weierstrass Approximation Theorem (Theorem 1.52) of Section 1.9. We shall make use of a special case of the Weierstrass theorem in the proof- that Ix I is the uniform limit on [- 1, 11 of polynomials P, ( x ) with Pn (0) = 0. This special case was proved also by a direct argument in Section 1.8. Let us distinguish the case of real-valued functions from that of complexvalued functions, writing C ( S ,R) and C ( S ,C ) in the two cases. The theorem in question gives a sufficient condition for a "subalgebra" of C ( S , R) or C ( S , C ) to be dense in the whole space in the uniform metric. Pointwise addition and scalar multiplication make C ( S , R) into a real vector space and C ( S , C ) into a complex vector space, and each space has also the operation of pointwise multiplication; all of these operations on functions preserve continuity as a consequence of Corollary 2.29. By a subalgebra of C ( S ,R) or C ( S , C ),we mean any nonempty subset that is closed under all these operations. The space C ( S ,C) has also the operation of complex conjugation; this again preserves continuity by Corollary 2.29. We shall work with a subalgebra of C ( S , R) or of C ( S , C ) , and we shall assume that the subalgebra is closed under complex conjugation in the case of complex scalars. The closure of such a subalgebra in the uniform metric is again a subalgcbra. To scc that this closurc is a subalgcbra rcquircs chccking cach operation separately, and we confine our attention to pointwise multiplication. If sequences { f, } and {g, } converge uniformly to f and g , then { f,g, } converges uniformly to f g because
with supxEsIg(x) I finite and supxEs I f n ( x )I convergent to supxEsI f ( x )1. We say that a subalgebra of C ( S ,R) or C i s , C ) separates points if for each pair of distinct points x l and x2 in S , there is some f in the subalgebra with f(x1) # f(x2).
Theorem 2.58 (Stone-Weierstrass Theorem). Let ( S , d) be a compact metric space. (a) If A is a subalgebra of C ( S , R) that separates points and contains the constant functions, then A is dense in C ( S , R) in the uniform metric. (b) If A is a subalgebra of C ( S , C ) that separates points, contains the constant functions, and is closed under complex conjugation, then A is dense in C ( S , C ) in the uniform metric.
10. Properties of C (S) for Compact Metric S
125
PROOFOF (a). Let A"' be the closure of A in the uniform metric. We recalled above from Chapter I that It I is the limit of polynomials t H P, ( t ) uniformly on [- 1, 11. It follows that It I is the limit of polynomials t H Q, ( t ) = M p n ( M p 1 t ) uniformly on [ - M , M I . Taking M = sup,,, If ( x ) I ,we see that If 1 is in A"' whenever f is in A. Since A"' is a subalgebra closed under addition and scalar multiplication as well, the formulas
show that k' is closed under pointwise maximum and pointwise minimum for two functions. Iterating, we see that A"' is closed under pointwise maximum and pointwise minimum for n functions for any integer n 2 2. The heart of the proof is an argument that if f E C ( S ,R),x E S , and t > 0 are given, then there exists g, in A"' such that g x ( x ) = f ( x ) and
for all s E S . Thc argumcnt is as follows: For cach y E S othcr than x ,thcrc cxists a function in A taking distinct values at x and y . Some linear combination of this lun~.Liu11arid llle curlslarll funcliun 1 is a fun~.liunhy in A will1 hy( x ) = f ( x ) and h, ( y ) = f ( y ) . To complete the definition of h, for all y E S , we set h, equal to the constant function f ( x )1. The continuity of h, and the equality hy( y ) = f ( y ) imply that there exists an open neighborhood Uy of y such that h , ( s ) > f ( s ) - t for all s E U,. As y varies, these open neighborhoods cover S , and by compactness of S , finitely many suffice, say Uyl,. . . Uyk. Then the function g, = max{hyl,. . . , h,} has g,(s) > f ( s ) - t for all s E S . Also, it has g, ( x ) = f ( x ) ,and it is in k',since k' is closed under pointwise maxima. To cornplete the proof of (a), we continue with f E C ( S , R) and t- > 0 as ahove. We shall prodnce a memher h of A"' such that I h (s) - f ( s )I < c for all s E S . For each x , the continuity of g, and the equality g, ( x ) = f ( x ) imply that there is an open neighborhood V, of x such that g, (s) < f ( s ) + t for all s E Vx.As x varies, these open neighborhoods cover S , and by compactness of S , finitcly many sufficc, say Vxl, . . . V, . Thc function h = min{g,, , . . . , g,, } has h ( s ) < f ( 3 ) t for all s E S , and it is in (Ac')"'= A"',since each gXJis in A"'. Since each gxJ has gxJ( s ) > f ( s ) - t for all s E S , we have h ( s ) > f ( s ) - t as well. Thus Ih(s) - f (s)l < t for all s E S . Since t is arbitrary, we conclude that f is a limit point of k'.But AC' is closed, and hence f is in A"'. Therefore A"' = C ( S , R).
+
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II. Metric Spaces
PROOFOF (b). Let Ae be the subset of members of A that take values in R. Then AR is certainly closed under addition, multiplication by real scalars, and pointwise multiplication, and the real-valued constant functions are in AR.If f = u i v is in A and has real and imaginary parts u and v , then f is in A by assumption, and hence so are u = ( f + f ) and v = $ ( f - f ) . We are given that A separates points of S If xl and ~2 are distinct points of S with f ( x l ) # f ( x 2 ) ,then either u ( x l ) # u ( x 2 ) or v ( x l ) # v ( x 2 ) ,and it follows that Slmf separates points. By (a), ARis dense in C ( S , R ) . Finally let f = 1.1 + i v be in C ( S , C ) , and let {u,} and {v,} be sequences in ARconverging uniformly to u and v , respectively. Then { u , + i v,} is a sequence in A converging uniformly to f . Hence A is dense in AR.
+
EXAMPLES. (1) On a closed bounded interval [ a ,b] of the line, the scalar-valued polynomials form an algebra that separates points, contains the constants, and is closed under conjugation. The Stone-Weierstrass Theorem in this case reduces to the Weierstrass Theorem (Theorem 1.521, saying that the polynomials are dense in C ( [ a ,61). (2) Considcr thc algcbra of continuous complcx-valucd pcriodic functions on [-n, n] and the subalgebra of complex-valued trigonometric polynomials ~ f =rnpznx; - ~here N dependson the trigonometric polynomial. Neither the algebra nor the subalgebra separates points, since all functions in question have f (-n) = f ( n ) . To make the theorem applicable, we consider the domain of these functions to be the unit circle of C , parametrized by em ";his parametrization is pcrmissiblc by Corollary 1.45, and continuity is prcscrvcd. Thc StoncWeierstrass Theorem then applies and gives a new proof that the trigonometric polynomials are dense in the space of complex-valued continuous periodic functions; our earlier proof was constructive, deducing the result as part of FejCr's Theorem (Theorem 1.59). be the unit sphere { x E Rn Ix 1 = l } in R n . The restrictions (3) Let to sn-' of all scalar-valued polynomials P ( x l , . . . , x,) in n variables form a subalgebra of C that separates points, contains the constants, and is closed under conjugation. The Stone-Weierstrass Theorem says that this subalgebra is dense in c ( s n - l ) . (4) Let S be the closed unit disk {z lzI 5 1) in C . The set A of restrictions to S of sums of power series having infinite radius of convergence is a subalgebra of C ( S , C) that separates points and contains the constants. However, the continuous function Z is not in the closure, because it has integral 0 over S with every member of A and also with uniform limits on S of members of A.This example shows the need for some hypothesis like "closed under complex conjugation" in Theorem 2.58b.
I
(s"-')
I
11. Completion
127
Corollary 2.59. If (S, d) is a compact metric space, then C(S) is separable as a metric space. PROOF.It is enough to consider C(S, C), since C(S, R) is a metric subspace of C (S, C). Being compact metric, S is separable by Proposition 2.33. Let B be a countable base of S. The number of pairs (U, V) of members of B such that u"' 5 V is countable. By Proposition 2.30e, there exists a continuous function f u v : S + R such that f u v is 1 on u"' and fuv is 0 on Vc. Let us show that the system of functions f U separates points of S. If xl and x2 are given, the T1 property of S (Proposition 2.30a), when combined with Proposition 2.3 1, gives us a member V of B such that xl is in V and V c { x ~ }Since ~ . the set Vc is closed and does not contain xl, the property that S is ) us disjoint open sets U1 and Vl with xl E U1 regular (Proposition 2 . 3 0 ~ gives and Vc g Vl. The latter condition means that V 2 V,". By Proposition 2.31 let U be a basic open set with xl E U and U 5 U1. Then we have xl E U 5 U1 g u,"' g V," g V and hence also xl E U g UC' g V. The function fuv is therefore 1 on xl and 0 on x2, and the system of functions f u v separates points. The set of all finite products of functions f u v and the constant function 1 is countable, and so is the set D of linear combinations of all these functions with coefficients of the form ql + i q z with ql and q z rational. The claim is that this countable set D is dense in C (S, C) . The closure of D certainly contains the algebra A of all complex linear combinations of the function 1 and arbitrary finite products of functions f u v , and A is closed under complex conjugation. By the Stone-Weierstrass Theorem (Theorem 2.58), k'= C (S, C) . Since D"' contains A, we have C(S, C) = AC' 5 ( D ~ ' ) ~=' DC'. In other words, D is dense. 11. Completion
If (X, d) and (Y, p ) are two metric spaces, an isometry of X into Y is a function rp : X + Y that preserves distances: p (rp(xl), rp(x2)) = d(xl, x2) for all xl and x 2 i n X . Forexample,arotation(x, y) H (xcose-ysinB, x s i n 8 + y c o s Q ) i s an isometry of R2 with itself. An isometry is necessarily continuous (with 6 = t-). However, an isometry need not have the whole range as image. For example, the map x H (x, 0) of R1 into R2 is an isometry that is not onto R2. In the case that there exists an isometry of X onto Y , we say that X and Y are isometric. Theorem 2.60. If (X, d) is a metric space, then there exist a complete metric space (X*, A) and an isometry p : X + X* such that the image of X in X* is dcnsc. REMARK. It is observed in Problems 25-26 at the end of the chapter that (X*, A) and rp : X + X* are essentially unique. The metric space (X*, A) is
128
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called a completion of (X, d), or sometimes "the" completion because of the essential uniqueness. There is more than one construction of X*, and the proof below will use a construction by Cauchy sequences that is immediately suggested if X is the set of rationals and X* is the set of reals.
PROOF.Let Cauchy(X) be the set of all Cauchy sequences in X . Define a relation -- on X as follows: if {p,) and {q,) are in Cauchy(X), then {p,) -- {q,} means lim d (p, , q,) = 0. Let us prove that -- is an equivalence relation. It is reflexive, i.e., has {p,} -{ pn} ,because d (p, , p,) = 0 for all n . It is symmetric, i .e.,has the property that {p,} -- {q,} implies {q,} -- {p,}, because d(p,, q,) = d(q,, p,). It is transitive, i.e.,has the property that {p,} -- {q,} and {q,} -- {r,} together imply {p,} -- {r,}, because 0 i d(pn, rn) i d(pn, qn) d(qn, rn)
+
and each term on the right side is tending to 0. Thus -- is an equivalence relation. Let X* be the set of equivalence classes. If P and Q are two equivalence classes, we set A ( P , Q) = limd(p,, q,), (*I where {p,} is a member of the class P and {q,} is a member of the class Q . We have to prove that the limit in (*) exists in R and then that the limit is independent of the choice of representatives of P and Q . For thc cxistcncc of thc limit (*), it is cnough to provc that thc scqucncc {d(p,, q,)} is Cauchy. The triangle inequality gives
+
and hence d(p,, y,) - d(p,, y,) 5 d ( p n , pm) d(qm,q,). Reversing the roles of m and n. we obtain
The two terms on the right side tend to 0, since {pk}and {qk}are Cauchy, and hence {d(p,, q,)} is Cauchy. Thus the limit (*) exists. We have also to show that the limit (*) is independent of the choice of representatives. Let {p,} and {ph}be in P , and let {q,} and {q;} be in Q. Then
Since the first and third terms on the right side tend to 0 and the other terms in lim, d (pk, q;). Reversthe inequality have limits, we obtain limn d (p, , q,) , 5 ing the roles of the primed and unprimed symbols, we obtain l i m d ( ~ hq;)
11. Completion
129
lim d ( p , , q,) . Therefore lim d ( p , , q,) = lim d (PA, q;) , and A ( P , Q ) is well defined. Let us see that ( X * , A) is a metric space. Certainly A ( P , P ) = 0 and A ( P , Q ) = A ( Q , P ) . To prove the triangle inequality
let {p,} be in P , {q,} be in Q , and {m}be in R. Since
we obtain (**) by passing to the limit. Finally if two unequal classes P and Q are given, and if {p,} and {q,} are representatives, then limd(p,, q,) # 0 by definition of --. Therefore A ( P , Q ) > 0. Thus (X" , A ) is a metric space. Now we can define the isometry rp : X + X * . If x is in X , then p ( x ) is the equivalence class of the constant sequence {p,} in which p, = x for all n . To see that rp is an isometry, let x and y be in X , let p, = x for all n , and let q , = y for all n . Then A(rp(x), rp(y)) = lim d ( p , , q,) = limd(x , y ) = d ( x , y ) , and rp is an isometry. Let us prove that p ( X ) is dense in X * . In fact, if P is in X* and {p,} is a representative, we show that rp (p,) + P . If rp(p,) = P for all sufficiently large n , then P is in rp(X); otherwise this limit relation will exhibit P as a limit point of rp(X), and we can conclude that P is in rp(x)" in any case. In other words, rp(p,) -z P implies that q ( X ) is dense. To prove that we actually do have rp(p,) + P , let t 0 be given. Choose N such that k 2 m 2 N implies d (p,, pk) < t . Then A (rp(p,) , P ) = limk d ( p , , pk) ( t for m 2 N . Hence linl, A(rp(p,), P) = 0 as required. Finally let us prove that X* is complete by showing directly that any Cauchy sequence { P,} converges. Since rp ( X ) is dense in X* ,we can choose x , E X with A(rp(x,), P,) < l / n . First let us prove that {x,} is Cauchy in X . Let t > 0 be given, and choose N large enough so that A(P,, P,)) < t / 3 when n and n' are - N . Possibly by taking N still largcr, wc may assumc that 1 / N < t / 3 . Thcn whenever n and n' are > N , we have
Thus {x,} is Cauchy in X . Let P E X* be the equivalence class to which { x r } belongs. We prove completeness by showing that P, + P . Let t > 0 be given,
II. Metric Spaces
130
and choose N large enough so that r > n > N implies d(x,, x,) it/2. Possibly by taking N still larger, we may assume that < 5. Then r > n > N implies
Thus P, + P . Hence every Cauchy sequence in X" converges, and X" is complete.
An important application of Theorem 2.60 for algebraic number theory is to the construction of the p-adic numbers, p being prime. The metric space that is completed is the set of rationals with a certain nonstandard metric. This application appears in Problems 27-31 at the end of this chapter.
12. Problems 1. As in Example 9 of Section 1, let S be a nonempty set, fix an integer n > 0, and let X be the set of n-tuples of members of S. For n-tuples x = (xl, . . . , x,) and y = (y1, . . . , yn),defined(x,y) = #{jI xj f yj},thenumberof componentsin which x and y differ. Prove that d satisfies the triangle inequality, so that (X, d) is a metric space. 2. Prove that a separable metric space is the disjoint union of an open set that is at most countable and a closed set in which every point is a limit point. 3. Give an example of a function f : [O, 11 + R for which the graph of f , given by {(x, f (x)) 0 5 x 5 11, is a closed subset of Kt2 and yet f is not continuous.
I
4. If A is a dense subset of a metric space (X, d) and U is open in X, prove that u c (A n ulC1. 5. Let (X, d) be a metric space, let U be an open set, and let E l 3 E2 2 . . . be a E, 5 U. decreasing sequence of closed bounded sets with (a) For X equal to Rn, show that E N C U for some N. (b) For X equal to the subspace Q of rationals in R1,give an example to show that E N C U can fail for every N.
nz,
6. Let F : X x Y + Z be a function from the product of two metric spaces into a mekic space. (a) Suppose that (x, y) H F i x , y) is continuous and that Y is compact. Prove that F (x, .) tends to F (xo, .) uniformly on Y as x tends to xo. (b) Conversely suppose H F i x , y) is continuous except possibly at points (x, y) = (xo, y), and suppose that F (x, .) + F (xo, .) uniformly. Prove that F is continuous everywhere.
12. Problems
131
7. Give an example of a continuous function between two metric spaces that fails to carry some Cauchy sequence to a Cauchy sequence. 8. (Contraction mapping principle) Let ( X , d ) be a complete metric space, let r be a number with 0 5 r < I , and let f : X + X be a contraction mapping, i.e.,afunctionsuchthatd(f(x), f ( y ) ) < rd(x, y) forallx and y i n X . Prove that there exists a unique xo in X such that f (xo) = xo.
9. Prove that a countable complete metric space has an isolated point. 10. A metric space (X, d) is called locally connected if each point has arbitrarily small open neighborhoods that are connected. Let C be a Cantor set in [0, 11, as described in Section 9, and let X c IK2 be the union of the three sets C x [0, 11, [O, 11 x {O},and [O, 11 x {l}. Prove that X is compact and connected but is not locally connected. Problems 11-1 3 concern the relationship between connected and pathwise connected. It was observed in Section 8 that pathwise connected implies connected. A metric space is called locally pathwise connected if each point has arbitrarily small open neighborhoods that are pathwise connected. 11. Prove that a metric space (X, d) that is connected and locally pathwise connected is pathwise connected.
12. Deduce from the previous problem that for an open subset of Rn, connected implies pathwise connected. 13. Prove that any open subset of JR1 is uniquely the disjoint union of open intervals. Problems 14-17 concern almost periodic functions. Let f : EX1 + C be a bounded uniformly continuous function. If c > 0, an c almost period for f is a number t such that If (x t ) - f ( x )1 5 t for all real x . A subset E of IK1 is called relatively dense if there is some L > 0 such that any interval of length 2 L contains a member of E. The function f is Bohr almost periodic if for every t > 0, its set o f t almost periods is relatively dense. The function f is Bochner almost periodic if every sequence of translates { f t , } ,where f t ( x ) = f (x + t ) , has a uniformly convergent subsequence. Any function x H elCXwith r real is an example.
+
14. As usual, let B(R', C ) be the metric space of bounded complex-valued functions on IK1 in the uniform metric. Show that the subspace of bounded uniformly continuous functions is closed, hence complete. 15. Show that a bounded uniformly continuous function f : P S I + C is Bohr almost periodic if and only if the set { f , t E K t 1 } is totally bounded in B ( R ' , C ) . 16. Prove that a bounded uniformly continuous function f : JR' + C is Bohr almost periodic if and only if it is Bochner almost periodic. Thus the names Bohr and Bochner can be dropped.
132
II. Metric Spaces
17. Prove that the set of almost periodic functions on R 1is an algebra closed under complex conjugation and containing the constants. Prove also that it is closed under uniform limits. Problems 18-20 concem the special case whose proof precedes that of the StoneWeierstrass Theorem (Theorem 2.58). In the text in Section 10, this preliminary special case was the function 1x1 on [-I, 11, and it was handled in two ways-in Section 1.8 by the binomial expansion and Abel's Theorem and in Section 1.9 as a special case of the Weierstrass Approximation Theorem. The problems in the present group handle an alternative preliminary special case, the function f i on [0, 11. This is just as good because Ix I = 0. 18. (Dini's Theorem) Let X be a compact metric space. Suppose that fn : X + R is continuous, that fl < f2 < f3 < . . . , and that f (x) = lim fn(x)is continuous and is nowhere foe. Use the defining property of compactness to prove that fnconverges to f uniformly on X . 19. Define a sequence of polynomial functions Pn : [O, 11 + R by Po(x)= 0 and Pn+1(x)= Pn(x)+ i ( x - P,(x)~). Prove that 0 = Po < P1 < P2 < . .. < f i 1 and that limn Pn(x)= ,& for all x in [0,11. 20. Combine the previous two problems to prove that polynomial functions on LO, 11.
f i is the uniform limit of
Problems 21-24 concem the effect of removing from the Stone-Weierstrass Theorem (Theorem2.58) the hypothesis that the given algebracontains the constants. Let (S,d) be a compact metric space, and let A b e a subalgebra of C(S,R)that separates points. There can be no pair of points {x,y} such that all members of A vanish at x and y. 21. If for each s E S,there is some member of A that is nonzero at s, prove in the following way that A is still dense in C(S,R): Observe that the only place in the proof of Theorem 2.58a that the presence of constant functions is used is in the construction of the function hy in the third paragraph. Show that a function hy still exists in Ad with hy(x) = f (x)and hy(y)= f (y)under the weaker hypothesis that for each s E S,there is some member of A that is nonzero at s .
+
22. Suppose that the members of A all vanish at some so in S. Let B = A R1, so that Theorem 2.58a applies to B. Use the linear function L : C(S,R) + R given by L( f)= f (so), together with the fact that BC' = C(S,R),to prove that A is uniformly dense in the subalgebra of all members of C(S.R) that vanish at so. 23. Adapt the above arguments to prove corresponding results about the algebra C (S,C) of complex-valued continuous functions. +m), R) be the algebra of continuous functions from [O,+m) into 24. Let Co([O, R that have limit 0 at foe. (a) Prove that the set of all finite linear combinations of functions epnx for +a)R). , positive integers n is dense in Co([O,
12. Problems
133
0 for x > 6 , and that J~~ f ( x ) e P n xd x = 0 for all integers n > 0. Prove that f is the 0 function.
(b) Suppose that f is in Co([0, feu), R),that f ( x )
=
Problems 25-26 concern completions of a metric space. They use the notation of Theorem 2.60. The first problem says that the completion is essentially unique, and the second problem addresses the question of what happens if the original space is already complete; in particular it shows that the completion of the completion is the completion. 25. Suppose that ( X , d ) is a metric space, that ( X T , A l ) and ( X z , A 2 ) are complete metric spaces, and that rpl : X x XT and : X x X ; are isometries such that rpl ( X ) is dense in XI and rp2(X) is dense in X;. Prove that there exists a unique isometry IJI of XT onto Xg such that rp2 = IJI o rpl. 26. Prove that a metric space X is complete if and only if X* = X , i.e., if and only if the standard isometry p of X into its completion X * is onto. Problems 27-31 concern the field Q , of p-adic numbers. The problems assume knowledge of unique factorization for the integers; the last problem in addition assumes knowledge of rings, ideals, and quotient rings. Let Q be the set of rational numbers with their usual arithmetic, and fix a prime number p. Each nonzero rational number r can be written, via unique factorization of integers, as r = m p k / n with p not dividing m or n and with k a well-defined integer (positive, negative, or zero). Define Ir 1, = p P k . For r = 0 , define 101, = 0 . The function I . 1, plays a role in the relationship between Q and Q, similar to the role played by absolute value in the relationship between Q and R. 27. Prove that I . 1, on Q satisfies (i) Ir 1, > 0 with equality if and only if r = 0, (ii) I - r 1, = Irl,, (iii) lrsl, = Irl,lsl,, and (iv) Ir sl, 5 max{lrl,, Isl,}. Property (iv) is called the ultrametric inequality.
+
28. Show that ( 0 , d) is a metric space under the definition d (r, s) = Ir
-
s 1,
29. Let (Q,, d ) be the completion of the metric space ( 0 , d ) . Since Irlp can be recovered from the metric by lr 1, = d(r, O), the function I . 1, extends to a coiltinuous fuilction I . 1, : Q p R. (a) Using Proposition 2.47, show that addition, as a function from Q x Q to Q,, extends to a continuous function from Q, x Q,to Q,. Argue similarly that the operation of passing to the negative, as a function from Q to Q,, extends to a continuous function from Q, to 0 , . Then prove that Q , is an abelian group under addition. (b) Show that multiplication, as a function from Q x Q to Q,,extends to a continuous function from Q, x Q, to Q,. (This part is subtler than (a) because multiplication is not uniformly continuous as a function of two variables .)
-
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II. Metric Spaces
(c) Let Q X = Q - {O} and Q,X = Qp - {O}. Show that the operation of taking the reciprocal, as a function from Q Xto Q,X,extends to a continuous function from Q,X to itself. Then prove that Q; is an abelian group under multiplication. (d) Complete the proof that Qp is a field by establishing the distributive law t(l- s ) - tl- t s within Q,.
+
+
I
30. (a) Prove that the subset { t E Q, Itl, 5 1 ) of Q, is totally bounded. (b) Prove that a subset of Q, is compact if and only if it is closed and bounded. 3 1. Prove that the subset Z,of Q, with 1.~1, 5 1 is a commutative ring with identity, that the subset P with Ix 1, 5 pp' is an ideal in Z,,and that the quotient Z,/P is a field of p elements.
CHAPTER I11 Theory of Calculus in Several Real Variables
Abstract. This chapter gives a rigorous treatment of parts of the calculus of several variables. Sections 1-3 handle the more elementary parts of the differential calculus. Section 1 introduces an to Cf" into a metric operator norm that makes the space of linear functions from RPLto R" or from CPL space. Section 2 goes through the definitions and elementary facts about differentiation in several variables in terms of linear transformations and matrices. The chain rule and Taylor's Theorem with integral remainder are two of the results of the section. Section 3 supplements Section 2 in order to allow vector-valued and complex-valued extensions of all the results. Sections 4-5 are digressions. The material in these sections uses the techniques of the present chapter but is not needed until later. Section 4 develops the exponential function on complex square matrices and establishes its properties; it will be applied in Chapter IV. Section 5 establishes the existence of partitions of unity in Euclidean space; this result will be applied at the end of Section 10. Section 6 returns to the development in Section 2 and proves two important theorems about differential calculus. The Inverse Function Theorem gives sufficient conditions under which a differentiable function from an open set in R n into Rn has a locally defined differentiable inverse, anrl the TmplicitF~~nctinn Theorem givev vl~fficicntconditionv for the local volvahility of m nonlinear equations in n m variables for m of the variables in terms of the other n . The Inverse Function Theorem is proved on its own, and the Implicit Function Theorem is derived from it. Sections 7-10 treat Riemann integration in several variables. Elementary properties analogous to those in the one-variable case are in Section 7, a useful necessary and sufficient condition for Riemann integrability is established in Section 8, Fubini's Theorem for interchanging the order of integration is in Section 9, and a preliminary change-of-variables theorem for multiple integrals is in Section 10.
+
1. Operator Norm This section works with linear functions from n-dimensional column-vector space to m-dimensional column-vector space. It will have applications within this chapter both when the scalars are real and when the scalars are complex. To be neutral let us therefore write IF for R or C. Material on the correspondence between linear functions and matrices may be found in Section A7 of the appendix. Specifically let L(IFn,IFm) be the vector space of all linear functions from IFn into IFm. This space corresponds to the vector space of m-by-n matrices with entries in IF, as follows: In the notation in Section A7 of the appendix, we let
136
111. Theory o f Calculus in Several Real Variables
( e l , .. . , en) be the standard ordered basis of IFn, and ( u l ,. . . , u,) the standard ordered basis of IFm. We define a dot product in IFm by
with no complex conjugations involved. The correspondence of a linear function T in L ( I F n , IFm) lu a rrralrix A will1 cnlries in IF is llicrl give11by Aij = T ( e j ) . ui . Let I . I denote the Euclidean norm on IFn or IFm, given as in Section 11.1 by the square root of the sum of the absolute values squared of the entries. The Euclidean norm makes IFn and IFm into metric spaces, the distance between two points being the Euclidean norm of the difference.
Proposition 3.1. If T is a member of the space L (IFn, IFm) of linear functions from IFn to IFm, then there exists a finite M such that I T ( x )1 5 M lx I for all x in IFn. Consequently T is uniformly continuous on IFn. PROOF.Each x in IFn has x ELl ( x . e j )T (ej). Thus
=
x;,,( x . e j ) e j ,and linearity gives T ( x )
=
The expression x . ej is just the jthentry of x ,and hence Ix .ej I 5 Ix I. Therefore IT ( x )1 5 (E;=, ~ ~ ( e , j ) l1,) land x the first conclusion has been proved with M = ELl IT(ej)1. Replacing x by x - y gives IT(x) - T(y)l = IT(x
-
Y ) i Mlx
-
YI,
and uniform continuity of T follows with S = t / M . Let T be in L(IFn, IFm). Using Proposition 3.1, we define the operator norm of T to be the nonnegative number
11 T 11
Then
IT(x)l5llTllIxl
forallx~F'"
Since I T ( c x )1 = Icl Ix I for any scalar c, the inequality IT (x) 1 5 M lx 1 holds for all x # 0 if and only if it holds for all x with 0 < 1x1 5 1, if and only if it holds for all x with Ix 1 = 1. Also, we have T (0) = 0 . It follows that two other expressions for 11 T 1 1 are
1 . Operator Norm
Proposition 3 2 . The operator norm on L(IFn,Fm) satisfies (a) llT 11 > 0 with equality if and only if T = 0, (b) llcT 11 = Icl IT 11 for c in I F , (c) IIT + SII 5 IITll + IlSII, (d) IITSI ( IITIIIISII if S is in L(IFn, IFm) and T isin L(Fm,IFk), (e) 11 111 = 1 if 12 = n z and 1 denotes the identity function on IFn. PROOF.All the properties hut (d) are immediate. For (cl), we have
Taking thesupremumfor 1x1 5 1 yields IITSll 5 IITIIIISII.
Corollary 3.3. The space L(Fn,IFm) becomes a metric space when a metric d is defined by d ( T , S ) = 11 T - S I . PROOF.Conclusion (a) of Proposition 3.2 shows that d ( T , S ) > 0 with equality if and only if T = S , conclusion (b) shows that d ( T , S) = d ( S , T ),and conclusion (c) yields the triangle inequality because substitution of T = T' - V' and S = V' - U' into (c) yields d ( T f ,U ' ) 5 d ( T f ,V ' ) d ( V f ,U ' ) .
+
Suppose that F = C . If the matrix A that corresponds to some T in L (Cn,C m ) has real entries, we can regard T as a member of L ( R n ,Rn" as well as a member of L ( C n ,C m ) . Two different definitions of IIT 1 1 are in force. Let us check that they yield the same value for 11 T 11 . ),
Proposition 3.4. Let T be in L (C", C"), and suppose that the vector T ( e j ) lies in Rm for 1 5 j 5 n . Then T carries Rn into Rm , and 11 T 11 is consistently defined in the sense that
PROOF. The first conclusion follows since T is IR linear. For the second conclusion, let 11 T I I R and 11 T I I c be the middle and right expressions, respectively, in the displayed equation above. Certainly we have I I T l l 5 I I T I I . If z is in C n, write z = x I i y with x and y in Rn. Since T ( x ) and T ( y ) are in Rn and T is C linear,
Hence IT(z)l i llTll,lzl, conclusion follows.
and it follows that IITII, i ITII,.
The second
111. Theory o f Calculus in Several Real Variables
138
We shall encounter limits of linear functions in the metric d given in Corollary 3.3, and it is worth knowing just what these limits mean. For this purpose, let T be in L(IFn,IFm),and define the Hilbert-Schmidt norm of T to be
This quantity has an interpretation in terms of the m-by-n matrix A that is associated to the linear function T by the above formula Aij = T (ej) . ui . Namely, 1 T 1 equals 1 Aij 12) "', which is just the Euclidean norm of the matrix A if we think of A as lying in IFnm. This correspondence provides the license for using the notation of a Euclidean norm for the Hilbert-Schmidt norm of T . The Hilbert-Schmidt norm has the same three properties as the operator norm that allow us to use it to define a metric: (i) IT I > 0 with equality if and only if T = 0, (ii) IcTI = Icl IT1 forc in F , (iii) IT +Sl i I T ISI. Let us write Q(T, S) = IT - SI for the associated metric. Parenthetically we might mention that the analogs of (d) and (e) for the Hilbert-Schmidt norm are (iv) ITS1 ( IT1 IS1 i f S i s i n L(Fn,Fm)andT i s i n ~ ( I F ~ , ~ ~ ) , (v) I 1I = f i if n = m and 1 denotes the identity function on Fn . We shall have no need for these last two properties, and their proofs are left to be done in Problem 1 at the end of the chapter.
(xi,
+
Proposition 3.5. The operator norm and Hilbert-Schmidt norm on L(Fn,Fm) are related by IITll i lTl i f i I I T I I Consequently the associated metrics are related by
PROOF. If 1x1 5 1, thcn thc trianglc incquality and thc classical Schwarz inequality of Section A5 give
2. Nonlinear Functions and Differentiation
Taking the supremum over x yields 11 T 11 5 I T I . In addition,
and the second asserted inerlilality follows. Proposition 3.5 implies that the identity map between the two metric spaces ( L(IFn, IFm) , d ) and ( L(IFn, IFm), d 2 )is uniformly continuous and has a uniformly continuous inverse. Therefore open sets, convergent sequences, and even Cauchy sequences are the same in the two metrics. Briefly said, convergence in the operator norm means entry-by-entry convergence of the associated matrices, and similarly for Cauchy sequences. 2. Nonlinear Functions and Differentiation
We begin a discussion of more general functions between Euclidean spaces by defining the multivariable derivative for such a function and giving conditions for its existence. Let E be an open set in Rn, and let f : E + Rn"e a , where
function. We can write f ( x ) = ,
( x ) = f (x) .
Then
f ( x ) = C:', f i ( x ) u i . The functions fi : E + R are called the components of f . The associated partial derivatives are given by
We say that f is differentiable at x in E if there is some T in L ( R n ,Rm) with lim
hi0
If(x +A)
f(x) Ihl
-
-
T(h)l
= 0.
The linear function T is unique if it exists. In fact, if TI and T2 both serve as T in this limit rclation, thcn wc writc
and find that
111. Theory o f Calculus in Several Real Variables
140
If TI # T2,choose some v E R"with I v 1 = 1 and T I(v) # T2( v ) . As a nonzero real parameter t tends to 0, we must have
Since t does not appear on the left side but the right side tends to 0, the result is a contradiction. Thus T I = T 2 ,and T is unique in the definition of "differentiable." If T exists, we write f ' ( x ) for it and call f f ( x ) the derivative of f at x . If f is differentiable at every point x in E , then x e f ' ( x ) defines a function f ' : ,5' + L (R'"R'"). We deal with the differentiability of this function presently. A differentiable function is necessarily continuous. In fact, differentiability at x implies that ( f ( x h ) - f ( x ) - T ( h ) (+ 0 as h + 0. Since T is continuous, T ( h ) + 0 also. Thus f ( x + h ) + f ( x ), and f is continuous at x .
+
Proposition 3.6. Let E be an open set of R n , and let f : E + Rm be a
ah
function. If f ' ( x ) cxists, thcn -( x ) cxists for all i and j, and 8x1
afi
-( x ) = f ' ( x )( e j ) . ui
8x1
REMARKS.In otller words, if f ' ( A ) exists at svrrle poirit n, tlierl it llas to be the linear function whose matrix is [ g ( x ) ] . This matrix is called the Jaeobian
matrix of f at x . PROOF.We are given that lim
If
(x
+ h)
-
h+O
f ( x ) - f t ( x ) ( h ) l = 0. Ihl
Dot product with a particular vector is continuous by Proposition 3.1. Take h = tej with t real in the displayed equation, and form the dot product with ui . Then we obtain lim
t+O
The result follows.
I fi ( X
+ tej)
-
fi
( x ) - t f ' ( x )( e j ) . ui I = 0. It l
2. Nonlinear Functions and Differentiation
141
The natural converse to Proposition 3.6 is false: the first partial derivatives of a function may all exist at a point, and it can still happen that f is discontinuous. If f ' ( x )exists at all points of the open set E in R n ,then we obtain a function f' : E + L (Rn,Rm), and we have seen that we can regard L (Rn,Rm) as a Euclidean space by means of the Hilbert-Schmidt norm. Let us examine what continuity of f ' means and then what differentiability of f' means.
Theorem 3.7. Let E be an open set of R n ,and let ,f : E -+ Rm be a function. If f' ( x ) exists for all x in E and x H f' ( x ) is continuous at some xo, then x
H
afi
afi
ax
axi
-( x ) is C . U I ~ ~ ~ I ~ U at U Uxo S for all i M~LI j . Cv~ivcrselyif each - ( A ) J
exists
ah
at every point of E and is continuous at a point xo, then f f ( n o )exists. If all ax, are continuous on E , then x e f ' ( x )is continuous on E. PROOFOF DIRECT
afi
The partial derivative -( x ) is one of the entries of ax; J f ' ( x ),regarded as a matrix, and has to be continuous if f '(x) is continuous. PART.
[z
PROOF OF CONVERSE PART. For the moment, let x be fixed. Regard h as ( h l , . . . , h,), and for 1 5 j 5 n , ~ u h(.i) t = ( h l ,. . . , h j , O , . . . , 0 ) . Define T
to be the member of L ( R n ,B m ) with matrix Theorem gives
=
d C" h j J; ( x + h(jP1)+ t h j e j )I t = , dt j=1
and hence
Consequently
( I ) ] . Use of the Mean Value
with 0 < tij < 1 1
142
111. Theory o f Calculus in Several Real Variables
Let t > 0 be given, and recall that the partial derivatives are assumed to be continuous at xo. If 6 > 0 is chosen such that I h 1 < 6 implies
then we see that I h 1 < 6 implies
Thus f '(xo) exists. Now assume that all the partial derivatives are continuous on E . Since
ah
L(Rn,Rm) is identified with Rnm,the continuity of the entries - ( x ) of the axi matrix of f ' ( x )implies the continuity of f '(x) itself. This completes the proof. J
If x H f ' ( x )is continuous on E , we say that f is of class C' on E or is a C' function on E . Let us iterate the above construction: Suppose that E is open in Rn arid lllal f : E + Rm is uf class C' ,su aial x H f ' ( x ) is c u r ~ t i r i u fru111 ~~s E into L ( R n ,R m ). We introduce second partial derivatives of f and the derivative of f ' . Namely, define
ah
Since the entries of the matrix of f ' ( x )are -( x ) = f ' ( x ) e j . ui ,the expression axi is the partial derivative with respect to xk of an entry of the matrix of axkaxi f ' ( x ). Thus we can say that f is of class C' from E into Rm if f ' ( x )is of class C' , and so on. We say that f is of class C" or is a C m function if it is of class all k . A C" function is also said to be smooth. We write C ~ E and ) C W ( E )for the sets of ckfunctions and C" functions on E .
or
Corollary 3.8. Lct E bc an opcn sct of R'" and Ict f : E + R " 3 c a function. The function f is of class ckon E if and only if all 1"-order partial derivatives uf cacli fi exis1 arld art: curilinuuus on E fur 1 5 k . This is immediate from Theorem 3.7 and the intervening definitions. The definition of a second partial derivative was given in a careful way that stresses the order in which the partial derivatives are to be computed. Reversing the order of two partial derivatives is a problem involving an interchange of limits. In
2. Nonlinear Functions and Differentiation
143
addressing sufficient conditions for this interchange to be valid, it is enough to consider a function of two variables, since n - 2 variables will remain fixed when we consider a mixed second partial derivative. The different components of the function do not interfere with each other for these purposes, and thus we may assume that the range is R 1 .
Proposition 3.9. Let E be an open set in IR2. Suppose that f : E + IR1 is
af af
a2f
a2f
a function such that -, -, and -exist in E and -is continuous at ax a~ a~ax a~ax a2f " a 2f b ). ( x , y ) = ( a ,b ) . Then -( a ,b ) exists and equals -(a, axay ayax PROOF.Put
and let u ( t ) = f ( t ,b + k ) - f ( t, b ) . The function u is a function of one variable t whose derivative is ( t , b + h ) $( t , b ). Use of the Mean Value Theorem produces 6 between a and a h , as well as q between b and b k , such that
+
-
+
Let t > 0 be given. By the assumed continuity of a2f / a y ax at ( a ,b ) , choose 6 > 0 such that I ( h ,k ) I < 6 implies
Then (*) shows that I ( h , k ) 1 < S implies
Letting k tend to 0 shows, for Ihl < S/2, that
a 2f
a2f
Since t is arbitrary, -( a ,b ) exists and equals -(a, axay ayax
b).
111. Theory o f Calculus in Several Real Variables
144
Now that the order of partial derivatives up through order k can be interchanged arbitrarily in the case of a scalar-valued ckfunction, we can introduce the usual notation
to indicate the result of differentiating f a total of k ax,I l . . .ax!? times, namely k l times with respect to x l , etc., through k, times with respect to x,. Sirnplcr notation will bc introduccd latcr to indicatc such itcratcd partial derivatives.
Theorem 3.10 (chain rule). Let E be an open set in R n , and let f : E + Rm be a function differentiable at a point x in E. Suppose that g is a function with range IEk whose domain contains f (E) and is a neighborhood of f ( x ). Suppose further that g is differentiable at f ( x ) . Then the composition g o f : E + IKk is differentiable at x , and ( g o f ) ' ( x ) = g'( f ( x ) )f '(x). PROOF.With x fixed, define y = f ( x ), T = f ( x ), S = g f ( y ), and also u(h) = f ( x
+h)
-
f ( x ) - T ( h ) and v ( k ) = g ( y
+k)
-
g(y) - S(k).
Continuity of f at x and of g at y implies that
) to 0 as h tends to 0 and with ~ ( ktending ) to 0 as k tends to 0 . with ~ ( htending Given h # 0,put k = f (.u + h ) - f (.u). Then
and g ( f (x +h))
-
g(f (x))
-
+
( S T ) @ )= g ( y k ) g ( y ) S ( T ( h ) ) = v(k) S(k)- S(T(h)) = S(k - T ( h ) ) ~ ( k )
+
= Siuih))
-
-
+
+ vik).
Therefore
the last inequality following from the upper bound obtained in (*) for Ikl. As h tends to 0, k tends to 0, by that same bound. Thus e ( h ) and q ( k ) tend to 0. The theorem follows.
145
2. Nonlinear Functions and Differentiation
Let us clarify in the context of a simple example how the notation in Theorem 3.10 corresponds to the traditional notation for the chain rule. Let f and g be given by
( 5 ) (A) (:?,") =f
=
r
and
=g
(;)
=x
2
-
2
Y .
In traditional notation one of the partial derivatives of the composite function is computed by starting from
and then substituting for x and y in terms of r and 8 . In notation closer to that of the theorem, we replace derivatives by Jacobian matrices and obtain
=
(21
-2y)
1
x=r cos 0 ,
cos0
-r sin0
The formula above for az/ar is just the first entry of this matrix equation. The chain rule in several variables is a much more powerful result than its one-variable prototype, permitting one to handle differentiations when a particular variable occurs in several different ways within a function. For example, consider the rule for differentiating a product in one-variable calculus. The function x e f ( x ) g ( x ) can be regarded as a composition if we recognize that oiie of the ingredients is the iiiultiplicatioii fuiictioii from IK2 to R' . Thus let fix) = uu,then u = f ( x ) and v = g ( x ) . Ifwedefine F ( x ) = ( g ( x ) ) and G
(:)
(G
o
F ) ( x ) = f ( x ) g ( x ). Theorem 3.10 therefore gives
111. Theory o f Calculus in Several Real Variables
146
Theorem 3.11 (Taylor's Theorem). Let N be an integer > 0, and let E be an open set in Rn. Suppose that F : E + R' is a function of class cN+l on E and that the line segment from x = ( x l , .. . , xn) to x + h , where h = ( h l ,. . . , h,), lies in E . Then
a N f ' F ( X +sh) ds. ax; . . .xk
C all 1, >O
PROOF.Define a function f of one variable by f ( t ) = F ( x Theorem in one variable (Theorem 1.36) gives f ( t ) = f (0)
+
N
( K ! ) - I f ("'(0) t K K=l
t!l'
+-
N
(t - s )
f
+ t h ) . Taylor's
(N+l) ( s )dss
and we put t = 1 in this formula. If g ( t ) = G ( x + t h ) , the function g is the composition of t H x + t h followed by G , and the chain rule (Theorem 3.10) allows us to compute its derivative as
Taking G cqual to any of various itcratcd partial dcrivativcs of F and doing an easy induction, we obtain
(
f(")cS) =
=K, k l j . . . j k n '>o
k l + ...+k,
dl k,
)h:l
. . . h?
+
a K ~ ( x sh) axfi . . . ax? '
where ( k , , K,kn) is the multinomial coefficient *. Substitution of this expression into the one-variable expansion with t = 1 yields the theorem.
3. Vector-Valued Partial Derivatives and Riemann Integrals It is useful to extend the results of Section 2 so that they become valid for functions f : E 3 C m ,where E is an open set in R n . Up to the chain rule in Theorem
3. Vector-Valued Partial Derivatives and Riemann Integrals
147
3.10, these extensions are consequences of what has been proved in Section 2 if we identify Cmwith R2m.Achieving the extensions by this identification is preferable to trying to modify the original proofs because of the use of the Mean Value Theorem in the proofs of Theorem 3.7 and Proposition 3.9. The chain rule extends in the same fashion, once we specify what kinds of functions are to be involved in the composition We always want the domain to be a subset of some R z ,and thus in a composition g o f , we can allow g to have values in some c',but we insist as in Theorem 3.10 that f have values in R'" . Now let us turn our attention to Taylor's Theorem as in Theorem 3.1 1. The statement of Theorem 3.11 allows R1 as range but not a general Rm. Thus the above extension procedure is not immediately applicable. However, if we allow the given F to take values in Rm,a vector-valued version of Taylor's Theorem will be valid if we adapt our definitions so that the formula remains true component by component. For this purpose we need to enlarge two definitions-that of partial derivatives of any order and that of 1-dimensional Riemann integration- so that both can operate on vector-valued functions. There is no difficulty in doing so, and we may take it that our definitions have been extended in this way. In the case of vector-valued partial derivatives, let f : E + Rmbe given. Then
af
- is now dcfincd without passing to componcnts. Thc cntrics of this vcctor-
ax, valued partial derivative are exactly the entries of the jthcolumn of the Jacohian matrix off . Thus the Jacobian matrix consists of the various vector-valued partial derivatives of f ,lined up as the columns of the matrix. Riemann integration is being extended so that the integrand can have values in Rm or Cm,rather than just R1.Among the expected properties of the extended version of the Riemann integral, one inequality needs proof because it involves interactions among the various components of the function, namely
The Riemann integral on the left side is that of a vector-valued function, while the one on the right side is that of a real-valued function. To prove this inequality, let ( . , . ) be tlie usual inner product for the range space-the dot product if the range is Euclidean space Rm or the usual Hermitian inner product as in Section 11.1 if the range is complex Euclidean space Cm. If u is any vector in the range space with 1 u 1 = 1, then linearity gives
(lb
F(t)dt,
Hence
U )
=
l
b
( F ( f ) ,u ) d f .
111. Theory o f Calculus in Several Real Variables
148
the two inequalities following from the known scalar-valued version of our inF ( t )dt is the 0 vector, then our equality and from the Schwarz inequality. If desired inequality is trivial. Otherwise, we specialize the above computation to -1 b u = lab F ( t ) dtl laF ( t )d t , and we obtain our desired inequality.
2
I
4. Exponential of a Matrix In Chapter IV, we shall make use of the exponential of a matrix in connection with ordinary differential equations. If A is an n-by-n complex matrix, then we define M
.
This definition makes sense, according to the following proposition.
Proposition 3.12. For any n-by-n complex matrix A, e A is given by a convergent series entry by entry. Moreover, the series X H ex and every partial derivative of an entry of it is uniformly convergent on any bounded subset of matrix space (= I W ~ ~and ' ) , therefore X FP ex is a Cm function. REMAKK.The proof will be tidier if we use derivatives of n-by-n matrix-valued functions. If F and G are two such functions, the same argument as for the usual ) )F f ( t )G ( t ) F ( t ) G f( t ). product rule shows that $ ( ~ ( t ) G ( t=
+
PROOF.Let us define 11 All for an n-by-n matrix A to be the operator normof the member of L ( C n ,C n )with matrix A. Fix M > 1. On the set where 11 All 5 M , we have
and the right side tends to 0 uniformly for 11 All < M as N1 and N2tend to infinity. Hence the series for eAis uniformly Cauchy in the metric built from the operator norm and therefore, hy Proposition 7 5,ilniformly Cailchy in the metric hililt from the Hilbert-Schmidt norm. Uniformly Cauchy in the latter metric means that the series is uniformly Cauchy entry by entry, and hence it is uniformly convergent. The matrices that are 1 or i in one entry and 0 in all other entries form a 2n2-memberbasis over R of the n-by-n complex matrices. Call these matrices by the names E,, 1 5 j 5 2n2. To compute the partial derivative in the E, direction of a function f ( A ) ,we form $ f ( A t E,) . We need to estimate the operator norm of a succession of partial derivatives applied to a term (N !)-I AN of the
+
It=,
4. Exponential o f a Matrix
149
exponential series. Thus suppose that we have a product f i ( A ) . . . f N ( A ) with each fi(A) equal to A or to a constant, i.e., a matrix that does not depend on A. The partial derivative in the Ej direction of this product is
Thus we get a sum of N terms, each involving a sum of the kind of product we are considering. If we repeat this process for partial derivatives in the directions of Ejl , . . . , Ejk,we get a sum of Nk terms, each iiivolviiig a suii~ of the kind of product we are considering. For a factor E j ,Proposition 3.5 gives 11 Ej 11 5 E j I = 1 5 M . For a factor of A, we have 11 A 11 5 M . Thus the operator norm of one such product is 5 M ~ The . operator norm of the sum of all such products for a kth-orderpartial derivative is therefore 5 N ~ M Taking ~ . into account the coefficient 1 / ( N !) for the original A ~we, see that the operator norm of terms N1 through N2 of the tenn-by-term k-tirnes differentiated series is
We see as a consequence that the term-by-term k-times differentiated series obtained from C ( N ! ) - I is uniformly convergent entry by entry. By the complex-valued version of Theorem 12 3 , applied recursively to handle kth-order partial derivatives, we conclude that exp A is of class ck and that the partial derivatives can be computed term by term. Since k is arbitrary, the proof is complete.
Proposition 3.13. The exponential function for matrices satisfies (a) eXeY= ex+' if X and Y commute, (b) ex is nonsingular, d t X - xetX (c) z(e 1 > (d) p W - I X W = w - l p X w if W is nclnsingnlar, (e) det ex = eTrX,where the trace Tr X is the sum of the diagonal entries of X. REMARKS.The conclusion of (a) fails for general X and Y, as one sees by and Y = . Relevant properties of the determinant taking X = function det that appears in the statement of (e) are summarized in Section A7 of thc appcndix.
(: A)
(A :)
PROOF.The rate of convergence determined in Proposition 3.12 is good enough to justify the manipulations that follow. For (a), we have
111. Theory o f Calculus in Several Real Variables
Conclusion (b) follows by taking Y = -X in (a) and using e0 = 1. For (c), we have
Conclusion (d) follows from the computation
For conclusion (e), define a complex-valued function f of one variable by f (t) = det e" . By (a), we have
d d (det esx) = f (tIz (det esx) ds Now eSX = 1 s X ;s2x2 ... = 1 + SX + s2F(s) for some smooth matrix-valued function F with entries Fij. If X has entries Xij, then
Is,
= (det etx)-
+
+
I$,
+
for some smooth function G . Thus -$(detesx)ls=o f '(t) = (Tr X) f (t) for all t . Consequently
-ddt (
(TrX)t
.f (t)) = e-(TrX)t .f '(t)
-
=
TrX, and we obtain
( ~~ r ) e - (.f ~(t) ~= ~0 ) ~
for all t , and e-(TrX)tf (t) is a constant. The constant is seen to be 1 by putting t = 0. Therefore f (t) = e(TrX)t.Conclusion (e) follows by taking t = 1.
5. Partitions of Unity
151
5. Partitions of Unity In Section 10 we shall use a "partition of unity" in proving a change-of-variables formula for multiple integrals. As a general matter in analysis, a partition of unity serves as a tool for localizing analysis problems to a neighborhood of each point. The result we shall use in Section 10 is as follows.
Proposition 3.14. Let K be a compact subset of Rn,and let { U 1 ,. . . , U k }be a finite open cover of K . Then there exist continuous functions 91, . . . , q k on Rn with values in [0, 11 such that (a) each pi is 0 outside of some compact set contained in Ui , q~iis identically 1 on K .
(b)
c:=~
REMARKS.The system { p l ,. . . q k } is an instance of a "partition of unity." For a general metric space X , a partition of unity is a family of continuous functions from X into [0, 11 with sum identically 1 such that for each point x in X , there is a neighborhood of x where only finitely many of the functions are not identically 0. The side condition about neighborhoods ensures that the sum C,,, p(x) has only finitely many nonzero terms at each point and that arbitrary partial sums are well-defined continuous functions on X . If U is an open cover of X, the partition of unity is said to be subordinate to the cover U if each member of CJvanishes outside some member of U. Further discussion of partitions of unity beyond the present setting appears in the problems at the end of Chapter X. The use of partitions of unity involving continuous functions tends to be good enough for applications to integration problems, but applications to partial differential equations and smooth manifolds are often aided by partitions of unity involving smooth functions, rather than just continuous functions.' We require a lemma
Lemma 3.15, In RN, (a) if L is a compact set and U is an open set with L 5 U , then there exists an opcrl scl V w i h VC' curnipacL and L c V c VC1c U , (b) if K is a compact set and { U 1 ,. . . , U,} is a finite open cover of K , then there exists an open cover { V 1 ,. . . , V,} of K such that v,"' is a compact subset of U , for each i . Partitions of unity involving smooth functions play no role in the present volume, but they occur in several places in the companion volume Advanced Real Analysis, and their existence is addressed there.
111. Theory o f Calculus in Several Real Variables
152
PROOF.In (a), if L = @, we can take V = @. If L # @, then the continuous function x H D ( x , UC) on RN is everywhere positive on L since L g U . Corollary 2.39 and the compactness of L show that this function attains a positive minimum c on L. If R is chosen large enough so that L g B(R; 0) and if we take V = {x E U I D ( x , UC) > f' B ( R ; 0), then L 5 V, v"' is compact (being closed and bounded), and vC' 5 {x c RN I D ( x , UC) 2 5 U. For(b),since{U1,. . . , U,}isacoverofK,wehaveK-(U2U...UU,) 5 U1. Part (a) produces ail open set Vl with vlC'coiiipact sucli tliat
ic}
ic}
The first inclusion shows that { Vl , U2, . . . , U,} is an open cover of K . Proceeding inductively, let Vi be an open set with
K
-
(Vl U . . . U ViPl U Ui+1 U . . . U U,) 5 Vi 5
v,"' 5 Ui.
At each stage, {V1,. . . , Vi , Ui+1, . . . , U,} is an open cover of K , and Thus {Vl, . . . , V,} is an open cover of K ,and v,"' 5 Ui for all i .
v,"' G Ui .
PROOFOF PROPOSITION 3.14. Apply Lemma 3.15b to produce an open cover ( WI , . . . , W k Jof K such that is compact and 5 Ui for cach i . Thcn apply it a second time to produce an open cover {Vl, . . . , V k } of K such that v,"' is compact and v,"' g Wi for each i . Proposition 2.30e produces a continuous function gi > 0 that is 1 on v,"' and is 0 off Wi . Then g = x r = l gi is continuous and 2 0 on Rn and is > 0 everywhere on K . A second application of Proposition 2.30e produces a continuous function h 0 that is 1 on the set where g is 0 and is 0 on K . Then g h is everywhere positive on Rn, and the functions
w:'
w':
+
pi = gi /(g
+ h) have the required properties.
6. Inverse and Implicit Function Theorems The Inverse Function Theorem and the Implicit Function Theorem are results for working with coordinate systems and for defining functions by means of solving equations. Let us use the latter application as a device for getting at the statements of both thc thcorcms. In the one-variable situation we are given some equation, such as x2 + y2 = a', and we are to think of solving for y in terms of x ,choosing one of the possible y's for each x . For example, one solution is y = - d n , -a < x < a ; unless some requirement like continuity is imposed, there are infinitely many such solutions. In one-variable calculus the terminology is that this solution is
6. Inverse and Implicit Function Theorems
153
"defined implicitly" by the given equation. In terms of functions, the functions F ( x , y ) = x 2 y2 - a 2 and y = f ( x ) = -,/= are such that F ( x , f ( x ) ) is identically 0. It is then possible to compute d y / d x for this solution in two ways. Only one of these methods remains within the subject of one-variable calculus, namely to compute the "total differential" of x 2 + y 2 - a', however that is defined, and to set the result equal to 0. One obtains 2x d x + 2 y d y - 0 with x and y playing symmetric roles. The declaration that x is to be an independent variablc and y is to bc dcpcndcnt mcans that wc solvc for d y / d x , obtaining d y / d x = - x / y . The other way is more transparent conceptually but makes use of rnultivariable calculus: it uses tlie chair1 rule in two-variable calculus to compute d / d x of F ( x , f ( x ) ) as the derivative of a composition, the result being set equal to 0 because ( d / d x )F ( x , f ( x ) ) is the derivative of the 0 function. This aF aF second method gives - -f ' ( x ) = 0, with the partial derivatives evaluated ax ay where ( x , y ) = ( x , f ( x ) ) . Then we can solve for f ' ( x ) provided a ~ / a is y not zero at a point of interest, again obtaining f '(x) = - x / y . It is an essential feature of both methods that the answer involves both x and y ; the reason is that there is more than one choice of y for some x ' s , and thus specifying x alone does not determine all possibilities for f ' ( x ). In the general situation we have m equations in n + m variables. Some n of the variables are regarded as independent, and we think in terms of solving for the other m . An example is
+
+
with x and y regarded as the independent variables. The classical method of implicit differentiation, which is a version of the first method above, is again to form "total differentials"
xyz d w
+ x y w d z + y z w d x + x z w d? = 0,
and then to solve the resulting system of equations for d w and d z in terms of d x and d y . The system is
and the solution is of the form d w = coefficient d x d z = coefficient d x
+ coefficient d y , + coefficient d y .
154
111. Theory o f Calculus in Several Real Variables
Here the coefficients are the various partial derivatives of interest. Specifically
a2 d z = -d x ax
a2 +dy. ay
The analog of the second method above is to set up matters as a computation of the derivative of a composition. Namely, we write
We view the given equations as saying that a composition of
followed by F is the 0 function, i.e.,
We apply the chain rule and compute Jacobian matrices of derivatives, keeping the variables in the same order x, y , w , z . The Jacobian matrix of the 0 function is a 0 matrix of the appropriate size, and the other side of the differentiated equation is the product of two matrices. Thus
Tn other words,
6. Inverse and Implicit Function Theorems
155
and we have the same system of linear equations as before. Comparing the two methods, we see that we have computed the same things in both methods, merely giving them different names; thus the two methods will lead to the same result in general, not merely in this one example. The theoretical question is whether the given system of equations, which was F ( x , y , w , 2) = 0 above, can in principle be solved to give a differentiable function; the latter was (:) = f above. The two computational methods show what the partial derivatives are if the equations can be solved, but these methods by themselves give no information about the theoretical question. The theoretical question is answered by the Implicit Function Theorem, which says that there is no problem if the coefficient matrix of our system of linear equations, namely (2wy3 ' z ' ~ ) in the above example, is invertible at a point of interest.
(;)
xyz x y w
Theorem 3.16 (Implicit Function Theorem). Suppose that F is a C' function from an open set E in Rn+"into Rm and that F ( a , b) = 0 for some ( a , b) in E , with a understood to be in Rnand b understood to be in Rm.If the matrix a Fi is invertible, then there exist open sets U 5 Rnfmand W 5 Rn lx=a, y = b
with ( a ,b) in U and a in W with this property: to each x in W corresponds a unique y in Rm such that ( x , y ) is in U and F ( x , y ) = 0. If this y is defined as f' ( x ),then f' is a C function from W into Rmsuch that f ( a ) = b , the expression F ( x , f ( x ) )is identically 0 for x in W , and
We shall come to the proof shortly. In the example above, f '(x) is the matrix
( ) OW
aw
[I
is
(
XYZ
'XzY'W~ ) ,
and
[z] is
(z3+2y YZW
3w2y2+2x). xzw
' ~ e t i use s the same approach to the question of introducing a new coordinate system in place of an old one. For example, we start with ordinary Euclidean coordinates ( u , v) for R2, and we want to know whether polar coordinates (r, 0) define a legitimate coordinate system in their place. The formula for passing from r case one system to the other is (: ) = ( F sin, ,but this formula does not really define r and 8. Defining r and 8 entails solving for r and 8 in terms of u and v. Thus let us set up the system
)
This is a system of the kind in the Implicit Function Theorem, and the considerations in that theorem apply. The independent vector variable is to be
156
111. Theory o f Calculus in Several Real Variables
x = (:), and the dependent vector variable is to be y itself is F (u, v, r, 0) = 0, where F ( u , v , r, 8 ) =
Fl ( u , V , I, 0 ) ) ( F 2 ( u . V. r. 8)
-
=
(a).
( r cos 0 - U rsin8-v
The system
)
.
The sufficient condition for solving the equations locally is that the matrix be invertible at a point of interest. This is just the matrix cos8 sin8
[z]
-r sin8 r cos 8
The determinant is r ,and hence the matrix is invertible except where r = 0. The Implicit Function Theorem is therefore telling us in this special case that r and 8 give us good local coordinates for R2 except possibly where r = 0. The Implicit Function Theorem gives no information about what happens when r = 0. The general result about introducing a new coordinate system in place of an old one is as follows.
Theorem 3.17 (Inverse Function Theorem). Suppose that rp is a c1function from an open set E of Rn into Rn,and suppose that p l ( a ) is invertible for some a in E. Put b = ~ ( 0 )Then . (a) there exist open sets U c E c Rn and V in V , rp is one-one from U onto V , and (b) the inverse f : V + U is of class C l .
c Rn such that a is in U , b is
Consequently, fl(cp(x)) = cpl(x)-I for x in U . REMARKS.Theorems 3.16 and 3.17 are closely related. We saw in the context of polar coordinates that the Implicit Function Theorem implies the Inverse Function Theorem, and Problem 6 at the end of the chapter points out that this implication is valid in complete generality. Actually, the implication goes both ways, and within this section we shall follow the more standard approach of deriving the Implicit Function Theorem from the Inverse Function Theorem and subsequently proving the Inverse Function Theorem on its own. PROOF OF THEOREM3.16 IF THEOREM3.17 IS KNOWN. Let n , rn , E , F , and ( a , b ) be given as in the statement of Theorem 3.16. We define a function to which we shall apply Theorem 3.17 in dimension n m . The function is
+
6. Inverse and Implicit Function Theorems
This satisfies rp(a, b)
=
(a, F ( a , b))
=
157
(a, O), and its Jacobian matrix at (a, b)
a Fi Since Theorem 3.16 has assumed that is invertible, rp'(a, b) is ,yj . , , y=b invertible. Theorem 3.17 therefore applies to rp and produces an open neighborhood W' of rp(a, b) = (a, 0) such that rp-' exists on W' and carries W' to an open set. Let U = rp-'(W'). Define W to be the open neighborhood W' n (Rn x {0}) o f a i n R n , a n d d e f i n e f(x) f o r x in W by (x, f(x)) = r p - ' ( x , ~ ) . Then f is o f c l a s s ~ l o nW,and f ( a ) = b because (a, f(a)) = rp-'(a,0) = (a,b). The identity
[
II
shows that F(x, f (x)) = 0 for x in W. The latter equation and the chain rule (Theorem 3 .lo) give the formula for f '(x) . Finally we are to see that y = j' (x) is the unique y in RnYor which (x, y) is in U and F (x, y) = 0. Thus suppose that x is in W and that yl and y2 are in Rmwith (x, yl) and(x, y2)in Uand F(x, yl) = F(x, y2) = 0. Thenwehaverp(x, yl) = (x, F(x, yl)) = (x,O) = (x, F(x, y2)) = r p ( ~y2). , Since (x, y ~ and ) (x, y2) are in U, we can apply rp-' to this equation and obtain (x, yl) = (x, y2). Therefore yl = y2. This completes the proof of Theorem 3.16 if Theorem 3.17 is known. Let us turn our attention to a direct proof of the Inverse Function Theorem (Theorem 3.17). When the dimension n is 1, a nonzero derivative at a point yields monotonicity, and the theorem is greatly simplified; this special case is the subject of Section A3 of the appendix. For general dimension n , it may be helpful to begin with an outline of the proof. The first step is to show that p is one-one near the point a in question; this is relatively easy. The hard step is to prove that rp is locally onto some open set; this uses either the compactness of closed balls or else their co~~~pleteness, and we return to a discussion of this step in a moment. The argument for differentiability of the inverse function depends on the continuity of the inverse function; this dependence was already true in the 1-dimensional case in Section A3 of the appendix. Continuity of the inverse function amounts to the fact that small open neighborhoods of a get carried to open sets, and this is part of the proof that rp is locally onto some open set. Finally the chain rule gives
111. Theory o f Calculus in Several Real Variables
158
( c l ) ' ( x )= ( q ~ ' ( ~ ~ ( x )and ) ) the - ~ continuity , of ( r p - ' ) I follows. Thus rp-' is of class C l . In carrying out the hard step, one has a choice of using either the compactness of closed balls or else their completeness. The argument using completeness lends itself to certain infinite-dimensional generalizations that are well beyond thc scopc of this book. Sincc thc argumcnt using compactncss is thc casicr onc, we shall use that. Tlle first step and the llard step rnentiuned above will be carried out in tllree lemmas below. After them we address the continuity and differentiability of the inverse function, and the proof of the Inverse Function Theorem will be complete.
Lemma 3.18. If L : Rn -z Rn is a linear function that is invertible, then there exists a real number m z 0 such that I L ( y )1 > ml y 1 for all y in R" . REMARK.We shall apply this lemma in Lemma 3.19 with L = y l ( a ) . PROOF. The linear inverse function L - l : Rn + Rn is one-one and onto. Thus if y is given, there exists x with y = L-l ( x ), and we have 1 y 1 = I L-' ( x )1 5 I I L - I 11 1x15 I I L - I 11 lL(y)l. The lemma follows with m = IIL-' 1 1 - l .
Lemma 3.19. In the notation of Theorem 3.17 and Lernma 3.18, choose rn > 0 such that Irpl(a)(y)l 1ml y 1 for all y E Rn, and choose, by continuity of rp', any S > 0 for which x E B(S; a ) implies 11 yl(x) - rpl(a)11 i Then
c.
Iv(xl) - r ~ ( x )1 l
Ix'
-
x I whenever x' and x are both in B(S; a ) .
REMARKS.This proves immediately that rp is one-one on B(S; a ) ,and it gives an estimate that will establish that cp-l is continuous, once y-' is known to exist. It proves also that the linear function q l ( x )is invertible for x E B(S; a ) because mlvl 5 Irpl(a)(y)l
5 Ivl(x)(y)l+ I Y ' ( ~ ) ( Y > - cpl(a)(y)l 5 Ivl(x)(y)l+ Ilvl(x) - vl(a)II I Y I
if y l ( x ) were not invertible, then any nonzero y in the kernel of y l ( x ) would contradict this chain of incqualitics. PROOF. The line segment from x to x' lies within B(S; a ) . Put z = x' - x , write this line segment as t H x + t z for 0 5 t 5 1, and apply the Mean Value Theorem to each component rpk of rp to obtain
yk(xl) - ~ k ( x= ) ~ k (f x tz)lt=l - Y ~ ( f X tz)It=O
+ t k z ) ( z ). ek
=
pl(x
=
p l ( a ) ( z ). ek
with 0 < tk < 1
+ (rp1(x+ tkz)
-
p l ( a ) ) ( z ). ek.
6. Inverse and Implicit Function Theorems
Taking the absolute value of both sides allows us to write
Therefore
Lemma 3.20. With notation as in Lemma 3.19, q ( B ( 6 ;a ) ) is open in Rn PROOF.Let c = m l ( 2 f i ) be the constant in the statement of Lemma 3.19. Fix xo in B(S; a ) and let yo = q ( x o ) ,so that yo is the most general element of q ( B ( 6 ;a ) ) . Find 61 > 0 such that B(61; xo)"' c B(6; a ) . It is enough to prove that B(cS112; yo) 5 rp(B(S;a ) ) . Evcn bcttcr, wc provc that B(cS112; yo) 5 rp(B(81;xn)"'). Thus lcl yl have lyl - yol < c S 1 / 2 . Choose, by ~.ornpaclncssof B(S1;x ~ ) ~ ' , a member x = xl of B(S1;xo)" for which Ip(x) - y1 l2 is minimized. Let us show that xl is not on the edge of B(al ; xo)", i.e., that 1x1 - xo 1 < 8 1 .In fact, if 1x1 - x0I = 81, thenLemma 3.19 gives
in colltradictioll to the fact that Ip(x) - yl13 is millimized 011 B(S1;xO)" at x = xl . Thus 1x1 -xo I < S l . In this case the scalar-valued function ( ~ ( x-)y l ) .( q ( x )- y l ) is minimized at an interior point of B(S1; xo)", and all its partial derivatives must be 0. Therefore q' ( x l )( z ) . ( q ( x l )- y l ) = 0 for all z in Rn. Since the linear function q l ( x l ) is onto Rn, we conclude that q ( x l ) - yl = 0 , and the lemma follows.
111. Theory o f Calculus in Several Real Variables
160
COMPLETION OF PROOF OF THEOREM3.17. Lemma 3.19 showed that the restriction of rp to B(6; a ) is one-one, and Lemma 3.20 showed that the image is an open set in Rn. Let f : rp(B(S;a ) ) + B(S; a ) be the inverse function. To complete the proof of Theorem 3.17, we need to see that f is differentiable on rp(B(S;a ) ) . Fix x in B(S; a ) , and suppose that x + h is in B(S; a ) with h # 0. Define y and k by y = q(x) and y + k = (n(x + h ) Since (n is one-one on B(S; a ) , k is not 0. in fact, Lemma 3.19 gives
where c f (y
=m
+k)
l ( 2 f i ) . The definitions give
-
f ( y ) - rpf(x)-'(k) = ( x =h
+h)
-
-
x
-
rpf(x)-'(k)
+
rpf(x)-' ( ~ ( x h ) - rp(x))
= -rp'(x)rl (rp(x
+h )
-
rp(x) - rpf(x)(h)).
Combining this identity with (*) gives
If t > 0 is given, choose
rj
> 0 small enough so that
as long as Ihl < y. If Ikl < c y , then Ihl < q by (*) and hence
In other words, f is differentiable at y , and f f ( y ) = rpf(x)-' . Suppose that the given function rp in the Inverse Function Theorem is better than a C' function. What can be said about the inverse function? The answer is carried by the formula f f ( r p ( x ) ) = rpf(x)-l for the derivative of the inverse function f . This formula implies that the partial derivatives of f are quotients of polyno~l~ials in partial derivatives of rp by a nonvanishing polyllolllial (the determinant) in partial derivatives of rp. Thus the iterated partial derivatives of f can be computed harmlessly in terms of the iterated partial derivatives of rp and this same determinant polynomial. Consequently if rp is of class Ck with k > 1, then so is f . If rp is smooth, so is f . In the case that rp and f are both smooth, we say that rp is a diffeomorphism. Let us summarize these facts in a corollary.
7. Definition and Properties o f Riemann Integral
161
Corollary 3.21. Suppose, for some k > 1, that rp is a ckfunction from an open set E of Rn into R n , and suppose that rpf(a)is invertible for some a in E . Put b = rp(a). Let U and V be open subsets of R n as in the Inverse Function Theorem such that a is in U , b is in V , and rp is one-one from U onto V . Then the inverse function f : V 3 U is of class ck.If rp is smooth, then rp is a diffeomorphism of U onto V .
7. Definition and Properties of Riemann Integral
Section 1.4contained a careful but limited development of the Riemann integral in one variable. The present section extends that development to several variables. A certain amount of the theory parallels what happened in one variable, and proofs for that part of the theory can be obtained by adjusting the notation and words of Section 1.4 in simple ways. Results of that kind are much of the subject matter of this section. In later sections we shall take up results having no close analog in Section 1.4. The main results of this kind are (i) a necessary and sufficient condition for a function to be Riemann integrable, (ii) Fubini's Theorem, concerning the relationship between multiple integrals and iterated integrals in the various possible orders, (iii) a change-of-variables formula for multiple integrals. We begin a discussion of these in the next section. The one-variable theory worked with a bounded function f : [ a ,b] 3 R , with domain a closed bounded interval, and we now work with a bounded function f : A + R with domain A a "closcd rcctanglc" in Rn. For this purposc a closed rectangle (or "closed geometric rectangle") in R n is a bounded set of the form
with rrj 5 bj for all j . Let us abbreviate [ a j ,b j ] as A j . In geometric terms the sides or faces are assumed parallel to the axes or coordinate hyperplanes. We shall use the notion of open rectangle in later sections and chapters, an open rectangle being a similar product of bounded open intervals ( a j ,b j )for 1 5 j 5 n . IIowever, in this section the term "rectangle" will always mean closed rectangle. If Pj is a one-variable partition of Aj ,then we can form an n-variable partition P = ( P I ,. . . , Pn) of the given rectangle A into component rectangles [cl, d l ] x . . x [c,, d,], where cj and dl are consecutive subdivision points of Pj . A typical component rectangle is denoted by R , and its n-dimensional volume (dj- c j ) is denoted by I RI. The mesh p ( P ) of the partition P is the maximum of the
n;=,
111. Theory o f Calculus in Several Real Variables
162
meshes of the one-dimensional partitions Pj ,hence the largest length of a side of all component rectangles of P . Relative to our given function f and a given partition P , define M R ( f ) = supxERf ( x ) and r n ~f () = inf,,~ f ( x ) for each component rectangle R of P . Put
L ( P , f ) = x m R ( f ) RI = lower Riemann sum for P , -
R
U( P ,f )
= upper
Riemann integral of f ,
f d x = sup L ( P , f ) = lower Riemann integral of f. - A
P
We say that f is Riemann integrable on A if f d x = /' A f d x ,and in this case we write f d x for the common value of these two numbers. We write R ( A ) for the set of Riemann integrahle fi~nctionson A. The following lemma is proved in the same way as Lemma 1.24.
Lemma 3.22. Suppose that f : A + R has m 5 f ( x ) 5 M for all x in A. Then for any partition P of A ,
A refinement of a partition P of A is a partition P* such that every component rectangle for P* is a subset of a component rectangle for P . If P = ( P I ,. . . , P,) and P f = ( P ; , . . . , PA) are two partitions of A, then P and P f have at least one common refinement P* = (P;",. . . , P l ) ; specifically, for each j, we can take P;" to be a coiiiiiioii refinenient of Pj and Pj . Arguing as in Leiiiiiia 1.25 and Theorem 1.26, we obtain the following two results. The key to the second one of these is the uniform continuity of any continuous function f : A + IR;for the uniform continuity we appeal to the Heine-Bore1 Theorem (Corollary 2.37) and Proposition 2.41 in several variables, the corresponding one-variable result being Theorem 1.10.
7. Definition and Properties o f Riemann Integral
163
Lemma 3.23. Let f : A + R satisfy in 5 f (x) 5 M for all x in A. Then (a) L ( P , f ) 5 L(P*, f ) and U(P*, f ) 5 U ( P , f ) whenever P is apartition of A and P* is a refinement, (b) L (PI, f ) 5 U (PI, f ) whenever P1 and PI are partitions of A, (c) - *f dx 5 dx,
Lf
I
Ef I A f
(4 dx dx 5 ( M - m ) A , (e) the function f is Riemann integrable on A if and only if for each t > 0, there exists a partition P of A with U ( P , f ) - L ( P , f ) < t .
Theorem 3.24. If f : A + R is continuous on A , then f is Riemann integrable on A. Next we argue as in Proposition 1.30 and Theorem 1.31 to obtain two more generalizations to several variables. The several-variable version of uniform continuity is needed in the proof of Proposition 3.25d.
Proposition 3.25. If f i and fi are Riemann integrable on A , then
+
lA +
+
(a) f l fi is in W A ) and (fl fi) dx = f i dx fi dx , (h) rfl is in %,(A) anrl JA rfi d.x = r JA f l dx for any real nilmher r , (c) f l 5 f2 on A implies f l dx 5 f2 dx, (d) m 5 f l 5 M on A and q : [m, 11/11 ; . R continuous imply that q o f l is in R(A), (e) If11 isinR(A),and IJAfidxI 5 L I f i I d x , (D ff and f l f i are inR(A), (g) f i i s i n R ( A ) if f l > Oon A.
Theorem 3.26. If { f,} is a sequence of Riemann integrable functions on A and if { f,} converges uniformly to f on A, then f is Riemann integrable on A, andlim, 1, f,dx = Ja f dx. There is also a several-variable version of Theorem 1.35, which says that Riemann integrability can be detected by convergence of Riemann sums as the mesh of the partition gets small. Relative to our standard partition P = (PI, . . . , P,) , select a member tR of each component rectangle R relative to P , and define
This is called a Riemann sum of f .
164
111. Theory o f Calculus in Several Real Variables
Theorem 327. If f is Riemann integrable on A , then r
lim S ( P , ItR}, f ) = @(P)+O
Conversely if f is bounded on A and if there exists a real number r such that for any t > 0, there exists some S > 0 for which IS(P, ItR}, f ) - r 1 < t whenever p ( P ) < 6,then f is Riemann integrable on A. REMARK.The proof of the direct part is more subtle in the several-variable case than in the one-variable case, and we therefore include it. The proof of the converse part closely imitates the proof of the converse part of Theorem 1.35,and we omit that.
PROOF. For the direct part the function f is assumed bounded; suppose If (x) 1 ( M on A. Let t > 0 be given. Choose a partition P* = (P;, . . . , P,*) of A with U ( P * ,f ) 5 f d x + t-. Fix an integer k such that the number of component intervals of P; is ( k for 1 ( j ( n . Put
1,
and suppose that P = ( P I ,. . . , P,) is any partition of A = Al x . . . x A, with p ( P ) 5 S1. For each j with 1 5 j 5 n , we separate the component intervals of Pj into two kinds, the ones in F ( j )being the component intervals of Pj that do not lie completely within a single component interval of P>nd the ones in G(.i)being the rest. Similarly we separate the component rectangles of P into two kinds, the ones in 3 being the component rectangles that do not lie completely within a single component rectangle of P* and the ones in G being the rest. If R = R1 x . . . x Rn is a member of 3,then Rj is in 3 ( j ) for some j with 1 5 j 5 n; let j = j (R) be the first such index. Let Fj be the subset of R's in F with j (R) = j ,so that F = Uy=lFj disjointly. Then we have
For the first term on the right side,
7. Definition and Properties o f Riemann Integral
165
Each member Rj of ~ ( jcontains ) some point of the partition Py in its interior, and two distinct Rj's cannot contain the same point. Thus the number of Rj's in F(j)is 5 k . Also, 1 Rj 1 5 p ( P ) . Consequently we have
The contribution to U (P, f ) of the second term on the right side of (*) is
Thus
S,
U(P,f)it+U(P*,f)i
fdx+2t.
Similarly we can define 62 such that w ( P ) 5 62 implies L(P,f)?
If S = min{S1,S2}and p ( P ) 5 8, then
L I
fdx-2€.
for any choice of points t ~, and hence S ( P , { t ~ f} ), completes the proof of the direct part of the theorem.
-
I
f d x 2 2c. This
Finally we include one simple interchange-of-limits result that is handy in working with integrals involving derivatives.
Proposition 3.28. Let f be a complex-valued C1 function defined on an open set U in IRm, and let K be a compact subset of U . Then ( x ), as h tends to 0, is (a) the convergence of f ( x + h e j ) - f ( x ) ]to uniform for x in K , b (b) the function g(x2, . . . , x,) = iuf ( X I ., . . , x,) d x ~is of class C' on the set of all points y = ( x 2 ,. . . , x,) for which [ a ,b] x {y} lies in U , and [ab f ( x ) d x l = z ( x ) d x l for j 1 as long as the set [ n , h] x {(.x2,. . . , .xn)}lies in [I.
i[
&
C
+
PROOF.In (a), we may assume that f is real-valued. The Mean Value Theorem gives
for some t between 0 and h , and then (a) follows from the uniform continuity of af / a x j on K . Conclusion (b) follows by combining (a) and Theorem 1.31.
166
111. Theory o f Calculus in Several Real Variables
As we did in the one-variable case in Sections 3 and 1.5, we can extend our results concerning integration in several variables to functions with values in Rm or Cm E R2m.Integration of a vector-valued function is defined entry by entry, and then all the results from Theorem 3.24 through Proposition 3.28 extend. The one thing that needs separate proof is the inequality fl d x i Ja I fi I d x of Proposition 3 25e, and a proof can be carried out in the same way as at the end of Section 3 in the one-variable case.
I iA I
8. Ricmann Intcgrablc Functions Let E be a subset of Rn.We say that E is of measure 0 if for any E > 0, E can be covered by a finite or countably infinite set of closed rectangles in the sense of Section 7 of total volume less than t . It is equivalent to require that E can be covered by a finite or countably infinite set of open rectangles of total volume lcss than t . In fact, if a systcm of opcn rcctanglcs covcrs E , thcn thc systcm of closures covers E and has the same total volume; conversely if a system of closed rectangles covers E, then the system of open rectangles with the same centers and with sides expanded by a factor 1 + 6 covers E as long as 6 > 0. Several properties of sets of measure 0 are evident: a set consisting of one point is of measure 0, a face of a closed rectangle is a set of measure 0, and any subset of a set of measure 0 is of measure 0. Less evident is the fact that the countable union of sets of measure O is of measure 0. Tn fact, if F > O is given and if E l , E z , . . . are sets of measure 0, find finite or countably infinite systems Rj of closed rectangles for j > 1 such that the total volume of the members of Rj is < t/2". Then R = U j Rj is a system of closed rectangles covering U j E j and having total volume < t . The goal of this section is to prove the following theorem, which gives a useful necessary and sufficient condition for a function of several variables to be Riemann integrable. The theorem immediately extends from the scalar-valued case as stated to the case that f has values in Rm or Cm.
Theorem 3.29. Let A be a finite closed rectangle in R ' k f positive volume, and let f : A + R be a bounded function. Then f is Riemann integrable if and only if the set B = {x f is not continuous at x }
I
has measure 0. Theorem 3.29 supplies the reassurance that a finite closed rectangle of positive volume cannot have measure 0. In fact, the function f on A that is 1 at every point with all coordinates rational and is 0 elsewhere is discontinuous everywhere on
8. Riemann Integrable Functions
167
A. By inspection every U ( P , f ) is I A1 for this f ,and every L ( P , f ) is 0; thus f is not Riemann integrable. The theorem then implies that A is not of measure 0. The proof of the theorem will make use of an auxiliary notion, that of "content 0," in order to simplify the process of checking whether a given compact set has measure 0. A subset E of Rn has content 0 if for any t > 0, E can be covered by a$tinite set of closed rectangles in the sense of Section 7 of total volume less than t . It is equivalent to require that E can be covered by a finite set of open rectangles of total volume less than t. A set consisting of one point is of content 0, a face of a closed rectangle is a set of content 0, any subset of a set of content 0 is of content 0, and the union of$nitely many sets of content 0 is of content 0. Every set of content 0 is certainly of measure 0, but the question of any converse relationship is more subtle. Consider the set E of rationals in [0, 11 as a subset of R1. Since this set is a countable union of one-point sets, it has measure 0. However, it does not have content 0. In fact, if we were to have E c u:=~ [ a j ,b j ] with (bj - a j ) < t ,then we would have EC' ~ f[ a j= ,b j ]by ~ Proposition 2.10, since u:=~ [ a j ,b j ]is closed. Then E'' would have content 0 and necessarily measure 0. This contradicts the fact observed after the statement of the theoremthat a closed rectangle of positive volume, such as EL' = 10, 11in R1,cannot have measure 0. We conclude that a bounded set of measure 0 need not have content 0.
Lemma 3.30. If E is a compact subset of Rn of measure 0, then E is of content 0. PROOF.Let E be of measure 0, and let t > 0 be given. Choose open rectangles E j l < t . By compactness, E 5 ~ y E=j for~ E j with E 5 E j and
Upl
some N . Then ~
y E=j 1 -C~ t . Since t is arbitrary, E has content 0.
Recall from Section 11.9 the notion of the oscillation of a function. For a function f : A + R,the oscillation at xo is given by oscf (XO)= lim
sup
6 4 0 xcR(S;ro)
I f (x) - f (xo)I .
The oscillation is 0 at xo if and only if f is continuous there. Lemma 2.55 tells us that oscg(x) > 2t}C' G {x t U oscg(x) > f} {x t
u1
I
foranyt > 0.
Lemma 331. Let A be a nontrivial closed rectangle in Rn,and let f : A + R be a bounded function with oscf (x) < t for all x in A. Then there is a partition P of A with U ( P , f ) - L ( P , f ) 5 2 t I A .
111. Theory o f Calculus in Several Real Variables
168
PROOF.For each xo in A, there is an open rectangle U,, centered at xo such that I f ( x ) - f ( x o )1 5 t on A n U::. Then Mu;; ( f ) -mu;; ( f ) 5 2c. These open rectangles cover A. By compactness a finite number of them suffice to cover A. Write A c U,, U . . . U Uxmaccordingly. Let P be the partition of A generated by the endpoints in each coordinate of A and the endpoints of the closed rectangles u;: we discard endpoints that lie outside A. Each component rectangle R of P then lies completely within some u::, and we have M R ( f ) - m R ( f ) 5 2 t for each component rectangle X of P . Therefore
I
PROOFOF THEOREM3.29. Define B, = { x oscf ( x ) 2 t } for each t > 0, so that B = UzlB1/,. For the easy direction of the proof, suppose that f is Riemann integrable. We show that B1/, has content 0 for all n. Since content 0 implies measure 0, B, will have measure 0 for all n . So will the countable union, and therefore B will have measure 0. Given t > 0 and n , use Lemma 3.23e to choose a partition P of A with I J ( P , f ) - L ( P , f ) 5 ~ l nLet .
R.= {component rectangles R of P I R0 f' B 1 / , #
M},
where R0 is the interior of R . Then B1/, is covered by the closed rectangles in R and the boundaries of the component rectangles of P. The latter are of content 0 . For R in R,lct us scc that M R (f ) i n R (f ) 2 1/12. In fact, if xo is in R0 f' BIl,, then oscf ( x o ) > l l n , so that lim, suplx-xol,sIf ( x ) - f (xo) l ? l l n and -
Therefore M R ( f ) - m ( f ) > 1In. Summing on R 1 -
E IRI
RER
5
E
R gives
E ( ~ R ( f ) - ~ R ( f I RlI ) 5 E ( M R ( ~ -)m R ( f ) )
R ER
lR1
all R
and thus CREE RI 5 t. Consequently B 1 / , has content 0, as asserted. For the converse direction of the proof, suppose that B has measure 0. We are to prove that f is Riemann integrable. Let t > 0 be given. The inclusion g B , and thus B,"' has measure 0 . The set of Lemma 2.55 gives B,"' c BE' is compact, and Lemma 3.30 shows that it has content 0. Hence the subset
9. Fubini's Theorem for the Riemann Integral
169
B, has content 0. Choose open rectangles U 1 ,. . . , Urnsuch that B, 5 Uy=l Uj and ELl lUj 1 < t . Form the partition P of A generated by the endpoints in each coordinate of A and the endpoints of the closed rectangles u:;;we discard endpoints that lie outside A. Then every component closed rectangle R of P is in one of the two classes
RI= { R R
~,C]forsome j},
In fact, our definition is such that R f' Uj # M implies R C u;'. If R n B, # M, let xo be in R f' B, . Then xo is in some Uj, and R f' Uj D for that j . Hence R is in R1. We shall construct a particular refinement P' of P in a moment. Let R' be a typical component rectangle of P ' . For any refinement P' of P , we have
+
R E R I R'gR
R E R ZR'gR
since ECl I Uj I < t^ . For R in R2, we have oscf (x) < t^ for all x in R . Lemma 3.3 1 shows that there is a partition PR of R such that U ( P R ,f ) - L ( P R, f ) 5 2tlRI. In other words, ( M R ) ( ~) m R 1 ( f ) ) IR'I 5 2tlRI if P' is fine enough to include all the n-tuples of PR . If P' is fine enough so that this happens for all R in R2, then we obtain
x,,,,
and the theorem follows.
9. Fubini's Theorem for the Riemann Integral Fubini's Theorem is a result asserting that a double integral is equal to an iterated integral in either order. A11 unfoltullate feature of the Riemall~lintegral is that when an integrable function f ( x , y) is restricted to one of the two variables, then the resulting function of that variable need not be integrable. Thus a certain amount of checking is often necessary in using the theorem. This feature is corrected in the Lebesgue integral, and that, as we shall see in Chapter V, is one of the strengths of the Lebesgue integral.
111. Theory o f Calculus in Several Real Variables
170
Theorem 3.32 (Fubini's Theorem). Let A 5 Rn and B 5 Rm be closed rectangles, and let f : A x B + R be Riemann integrable. For x in A let f, be the function y e f ( x , y ) for y in B, and define L(x)=
1
fx(y)d. =
- B
-
W x )= / .Bf x ( y ) d y =
1
f(*.y)dy,
- B
-
/ .Bf ( x j y ) d y j
as functions on A. Then L and U are Riemann integrable on A and
PROOF.Let P be a partition of the form ( P A , P B ) ,and let R = R A x R B be a typical component rectangle of P . Then
For x in RA,m ~. R, ~( f ) i mRB(f X ). Hence x in RA implies
Taking the infimum over x in R A and summing over R A gives L(P,f
)
C n z R A ( ~I R) A I
( C n l R A x R (Bf ) I R ~ II R) ~c I
= RA
RE
=
L).
RA
Similarly
U ( P A ,U ) i U ( P , f ) . Thus Since f is Riemann integrable, the ends of the above display can be made close together by choosing P appropriately. The second and third members of the display will then be close, and hence
The result for L follows. The result for U follows in similar fashion immediately from the inequalities
L ( P , f ) i L ( P A , L ) i L ( P A , U ) i U ( P . 4 , U ) i U ( P ,f ) . This proves the theorem.
10. Change o f Variables for the Riemann Integral
171
REMARKS. (1) Equality of the double integral with the iterated integral in the other order is the same theorem. Thus the iterated integrals in the two orders are equal. (2) If f is continuous on A x B, then ,fx is continuous on B as a consequence of Corollary 2.27, so that f ( x , y ) d y = f ( x , y ) d y . Hence
1
- B
when f' is continuous on A x B. This result is isolated as Corollary 3.33 below. Evidently it immediately extends to continuous functions with values in Rk or
ck. (3) In practice one often considers integrals of the form Ju f ( x , y ) d x d y for some open set U , where f is continuous on some closed rectangle A x B containing U . Then the double integral equals , f ( x , y ) Iu ( x , y ) d x d y , where Iu is the indicator function2 of U equal to 1 on U and 0 off U . In many applications the functions ( I " ) , have harmless discontinuities and ( f Iu), is therefore Riemann integrable as a function of x . In this case, the upper and lower integrals can again be dropped in the statement of Theorem 3.32.
Corollary 3.33 (Fubini's Tlleoreii~for coiitiiiuous integralid) . Let A 5 Rn and B g Rm be closed rectangles, and let f : A x B + R be continuous. Then
10. Change of Variables for the Riemann Integral The goal in this section is to prove a several-variables generalization of the onevariable formula rb
given in Theorem 1.34. In the one-variable case we assumed in effect that rp was a strictly increasing filnction of class C'l on [ n , h] and that f was merely Riemann integrable. The several-variables theorem in this section will be only a preliminary result, with a final version stated and proved in Chapter VI in the '~ndicatorfunctions are called "characteristic functions" by many authors, but the term "characteristic function" has another meaning in probability theory and is best avoided as a substitute for "indicator function" in any context where probability might play a role.
111. Theory o f Calculus in Several Real Variables
172
context of the Lebesgue integral. In particular we shall assume in the present section that f is continuous and that it vanishes near the boundary of the domain, and we shall make strong assumptions about rp. To capture succinctly the notion that f vanishes near the boundary of its domain, we introduce the notion of the support of f ,which is the closure of the set where f is nonzero.
Theorem 3.34 (change-of-variables formula). Let rp be a one-one function of class C' from an open subset U of Rnonto an open subset q ( U ) of Rnsuch that det rpf(x)is nowhere 0. Then
for every continuous function f : q ( U ) + R whose support is a compact subset of r p ( ( J ) . Before a discussion of the sense in which this result has to regarded as preliminary, a few remarks are in order. The function rpf is the usual derivative of rp, and rpf(x) is therefore a linear function from Rn to Rn that depends on x . The ~natrixof the linear functio~lrpf(x) is the Jacobian ~natrix[3pi/ a x j ] , and det cpf(x) is the determinant of this matrix. In classical notation, this determinant . a(y1, . . . , yn) is often written as , and then the effect on the integral of changing a(xl,...,xn) ~ ( Y I , . . . , Yd~x ), The variables can be summarized by the formula d y = a(Xl,..,xn)l absolute value signs did not appear in the one-variable formula in Theorem 1.34, but the assumption that rp was strictly increasing made them unnecessary, rpf(x) being > 0. Had we worked with strictly decreasing y ~ we , would have assumed rpf(x) < 0 everywhere, and the limits of integration on one side of the formula would have been reversed from their natural order. The minus sign introduced by putting the limits of integration in their natural order would have compensated for a minus sign introduced in changing 9' ( x ) to Iqf ( x )1 . The hypotheses on cp make the Inverse Function Theorem (Theorem 3.17) applicable at every x in U . Consequently rp(U) is automatically open, and rp has a local1y defined c inverse function about each point p ( x ) of the image. Since rp has been assumed to be one-one, rp : U + rp(U) has a global inverse function r p - I of class c ' . We can use rp-' to verify that f o rp has compact support in U . To the equality rp({x E U f (rp(x))# 01) = { Y E rp(U) f ( Y ) # o ) , we apply r p - I and obtain { x t u f (rp(x)>Z o} = rp-l({Y rp(U) f ( Y ) Z 01). Hence ~,
I
I
I
I
10. Change o f Variables for the Riemann Integral
173
The identity F(EC') 5 (F(E))" holds whenever F is a continuous function between two metric spaces, by Proposition 2.25. When E" is compact, equality actually holds. The reason is that Propositions 2.34 and 2.38 show F(E") to be closed; since F(Ec') is a closed set containing F (E) , it contains (F(E))". Applying this fact to the displayed equation above, we obtain
In other words, support( f
o p) = p-'
(support( f )).
Applying Proposition 2.38 a second time, we see that f o q~ has compact support. As a result, we can rewrite the formula to be proved in Theorem 3.34 as
and the supports will take care of themselves in the proof. The result of Theorem 3.34 has to be regarded as preliminary. To understand the sense in which the result is limited, consider the case of polar coordinates in Kt2. In this case we can take
We readily compute that det q'
(i) = r , and the desired formula is
S
f (r cos 8 , r sin 8) r d r do.
O
At first glance this formula seems fine. But if we refer to the precise hypotheses, we see that f is assu~nedto vanish in a neighborl~oodof the set of points (x, 0) with x 2 0, as well as when (x, y) is sufficiently far from the origin. Without some sort of passage to the limit, the theorem therefore settles few cases of interest. This passage to the limit will be accomplished easily with the Lebesgue integral, and we therefore postpone the final form of the change-of-variables formula to Chapter VI.
174
111. Theory o f Calculus in Several Real Variables
In any event, we shall use Theorem 3.34 in proving the final change-ofvariables formula, and thus a proof is warranted now. Before coming to the formal proof, it is well to understand the mechanism of the theorem. The proof will then flow easily from the analysis that is done for motivation. The motivation for the theorem comes from taking f to be the constant function 1 and from thinking of (o as of the form p(y) - 310 + L ( y - 370) with L linear. In I t 3 , if we take U to be the cube { y = ( y l , y2, y3) 0 5 yi 5 1 for all i ), along with f = 1, thc formula asscrts that p(U) has volumc I dct L I . This is just the well-known fact about 3-by-3 matrices that the volume of the parallelepiped wit11 sides u , u , w is the scalar I(u x v ) . w l . Fur a correspundirig result in R n , where vector product is not available, the relationship between the determinant and a volume has to be argued differently. One way of proceeding in Rn is to use row or column reduction to write the given matrix as the product of elementary matrices (those corresponding to the effect of a single step in the reduction), to check the change of variables for each factor, and to use the multiplication formula det(AB) = det A det B to obtain the result. This argument can be adjusted so as to work with a function f in place; the elementary matrices that interchange two variables are handled by Fubini's Theorem (Theorem 3.32 or Corollary 3.33), and the other elementary matrices are handled by the one-variable change-of-variables formula (Theorem 1.34). That being the case, one can envision a proof of Theorem 3.34 that proceeds by approximation, using Taylor's Theorem (Theorem 3.11), at least if f is of class c 2 . The contribution to the integrand from the integral remainder term in the Taylor expansion of rp is to be estimated as an error term. The approximation generates an additional error term because the image of U under (o does not match the image of U under the approximating first-order expansion of q . Of course, one cannot expect the approximation to be very good far away from the point where the Taylor expansion is centered, and thus the argument needs to be carried out locally. The local contributions can then be pieced together by using a partition of unity. Such an argument can actually be carried out, but the argument is lengthy. A more economical argument comes by finding a nonlinear analog of row or column reduction. The Inverse Function Theorem will allow us to prove that a general q decomposes into suitably defined nonlinear elementary transformations, but the decomposition is valid only locally. A partition of unity is used to piece together the local resillts and ohtain the theorem. We introdilce two kinds of nonlinear elementary transformations:
I
(i) a flip p , which interchanges two coordinates. This is a linear function, and it satisfies I det pf(x)I = 1 for all x . Application of Fubini's Theorem in the form of Corollary 3.33 shows that Theorem 3.34 is valid when rp is a flip.
10. Change o f Variables for the Riemann Integral
(ii) a primitive mapping
where g is real-valued and occurs in a single entry. If that entry is the ith entry, then the Jacobian matrix of @ is the identity matrix except in the i th row, where the entries are ... , Hence I det ~//(x) 1 = To prove Theorem 3.34 for a primitive mapping of this kind, it is enough to handle i = 1. If we write x = ( x l ,x') and y = ( y l ,x') with x' in Rn-l, Fubini's Theorem (Corollary 3.33) reduces matters to showing that
g, g.
I I.
under suitable hypotheses on g , and it is enough to prove that the inner integrals are equal for all x'. Theorem 1.34yields the equality of the inner integrals if g is a c1function for which g(xl,x') is defined for xl in an interval for any relevant x', and if x1 ( x l ,x') is everywhere positive at the points in question.
I
I
In the linear case a primitive mapping @ for which g(x) appears in the ith entry is given by a matrix that is the identity except in the ith row. For $' to be nonvanishing, the diagonal entry in the ith row must be nonzero. This kind of matrix is not always elementary but is the product of n elementary matrices. What needs to be proved for Theorem 3.34 is that apart from translations, any nonlinear rp as in Theorem 3.34 can be decomposed into the product of primitive transforn~ationsand flips, at least locally. The argumellt will peel primitive mappings from the right side of cp and flips from the left side. In that sense it will be a nonlinear version of column reduction with primitive mappings and row reduction with flips. The decomposition will be forced to be local because it uses the Inverse Function Theorem, which guarantees the existence of an inverse function only locally.
176
111. Theory o f Calculus in Several Real Variables
Lemma 3.35. Suppose that E is an open neighborhood of 0 in Rn and that + Rn is a C' function such that rp(0) = 0 and rp'(0)-' exists. Then there is a subneighborhood of 0 in Rn in which rp factors as rp : E
r p = p l O . . . O p n - l O $ n O . . . O ~ l ,
where each Pj is a flip or the identity and each ll'J is a primitive C' function in some open neighborhood of 0 such that Qj (0) = 0 and Qf(0)-l exists. PROOF.Let us set up an inductive procedure by assuming at the start that
with 1 5 io 5 n . We shall make use of the following formula for multiplying two matrices A and B when B has the property that it is equal to the identity matrix except possibly in row io . The formula is
It will be convenient to identify linear functions like rp'(x) with their matrices, so that the (i, j)th entry ~ ' ( x of ) ~rpf(x) ~ i s meaningful. Let j = jo be the least row index for which the ( j , io)t" entry of rp' (O) is nonzero. The index jo exists because rpf(0) is nonsingular, and jo is > io since the top io - 1 rows of rpl(x) match the corresponding rows of the identity matrix. Let identity function if jo = i n , Pi0 = flip of entries jo and io if jo > io. Then Pioo rp has the general form of (*) except that the iih and jAh entries have been interchanged. By inspection the Jacobian matrix at 0 of Pie o rp equals the identity matrix in rows 1 through io - 1 and has (io,io)t" entry nonzero. Thus if wc possibly incorporatc a composition with a flip into thc dcfinition of rp, we may assume that q1'(0)~,~, is nonzero. Put
{
10. Change o f Variables for the Riemann Integral
Then $ ' ( x ) is an n-by-n matrix with
where S i j i s the Krclnecker delta. Since det i/rl(0) = ~ I ' ( O ) ~ ,# ~ , 0, we can apply the Inverse Function Theorem (Theorem 3.17) to $ ,obtaining a C inverse function @-' that carries an open neighborhood of 0 onto an open subset of the domain of rp, has $-'(0) = 0 , and has derivative ( I / - ' ) ' ( y ) = I / ' ( x ) - ' , where x and y are related by y - $ ( x ) and x ( y ) . Using (**), we readily verify that if i # io, Ji j
'
I/-'
-(rp'(x)ioo)plrp'(x)ioj (~~(x)ioio)-l
if i = io # j, . . if1 = j = io.
Therefore J ij
if i # io,
# j, . . ifz = j = io.
- ( ' ( ) i o i ~ ) l ( ~ ) ~ ifj i = io (~'(x)i~i~)-'
Form q = rp o I/-'. By the chain rule (Theorem 3.10), we have ~ ' ( 0 )= ~ I ' ( o ) ( @ - ' ) ' ( 0 )and , this is nonsingular. Combining the formula for ((I/-' )ij with the chain rule and (**) gives
Since q f ( x ) i i , is 0 for i < io , the above formula shows that ~ ' ( x = ) ~Jij~for i < io. For i - io, the formula shows first that q ' ( ~ ) is ~0 , ~for j f io and then that qf(x)i, is 1 for j = io. Thus q f ( x ) i j = Jij for i 5 io. Consequently the i th cntry of ~ ( x is) xi ci if i 5 io , whcrc ci is a constant. Evaluating q at x = 0 , we see that ci = 0 . Thus q ( x ) has the same general shape as (*) except that the i t entry is now xi,. Following this argument inductively for i = 1, . . . , n - 1 leads us to a decom-
+
position q=p
n-lO...OplOqO$~lO...O$n-~l,
(t)
178
111. Theory o f Calculus in Several Real Variables
where each pj is a flip or the identity and where each llrJ is primitive. The function rj has ~ ( 0= ) 0 and ~ ' ( 0nonsingular, ) and v has the form
Therefore v is primitive. Solving the required form.
(I) for rp thus exhibits rp as decomposed into
PROOFOF THEOREM3.34. We are to prove that
whenever rp : U + q ( U ) is a c1function between open sets with a c1inverse and f : q ( U ) + R is continuous and has compact support lying in q ( U ) . In the argument we shall work with several functions in place of q , and the set U may be different for each. We have seen that (*) holds if q is a flip or an invertible primitive function. Let us observe also that (*) holds if rp is a translation rp(x) = x +xo for some xo in Rn; the reason is that (*) in this case can be reduced via successive uses of Fubini's Theorem (Corollary 3.33) to the 1-dimensional case, where we know it to be true by Theorem 1.34. If (.r) holds when q is either a : U + a ( U ) or : a ( U ) + B ( a ( U ) ) ,then (*) holds when rp is the composition y = p o a : U + j3( a ( U ) )because
the last two steps holding by the formula det(BA) = det B det A and the chain rule (Theorem 3.10). For any a in the given set U , Lemma 3.35 applies to the function pa carrying U - a to rp(U) - rp(a) and defined by rp,(x) = rp(x a ) - q ( a ) because pa ( 0 ) = 0 and v: ( 0 ) = ~ ' ( a )The . lemma produces an open neighborhood E, of
+
11. Problems
179
0 on which rp, factors as a composition of flips and invertible primitive functions. If t,, denotes the translation t,, (x) = x xo,then pa = tt_(,)o q o t, shows that rp = t,(,) o ip, o t-, . Therefore y factors on the open neighborhood E, a as the composition of translations, flips, and invertible primitive functions. From the previous paragraph we conclude for each a E U that (*) holds for rp if f is continuous and is compactly supported in the open neighborhood y ( E , a ) of y(a>. As a varies tllrougl~U ,the subsets V, = y (E, +a) of rp (U) f o i ~ ail n ope11cover of y (U) . Fix f continuous with compact support K in y(U) . By compactness a finite subfamily of the family {V,} forms an open cover of K . Applying Proposition 3.14, we obtain a finite family \V = {$} of continuous functions defined on q ( U ) and taking values in [0, 11 with the properties that
+
+
+
(i) each (ii)
+ is 0 outside of some compact set contained in some V,.
'&,,+ is identically 1 on K .
Then property (i) and the conclusion of the previous paragraph show that (*) holds @f = f on q ( U ) . Since there are only finitely for @f . From (ii), we have C@ many terms in the sum, we can interchange sum and integral and conclude that (*) holds for f . This completes the proof. One final remark is appropriate: Theorem 3.34 immediately extends from the scalar-valued case as stated to the case that f takes values in Rm or C m .
11. Problems 1 . Let IF be EX or @. Prove that the Hilbert-Schmidt norm satisfies (a) ITS1 5 IT1 IS1 if S is in L(Pn,I F m ) and T is in L(P", IFk), (b) 11 I = f i if n = m and 1 denotes the identity function on IFn 2.
Suppose that f : R" + Rm is a linear function with Jacobian matrix A . What is
f 3.
'(xo)?
Suppose that f : R2 -t R 1 has If (x)l 5 1x12 for all x . Prove that f is differentiable at x = 0.
4. L e t x = ( x l, . . . , x,) a n d u = ( u l, . . . , u,) b e i n EXn. For f : EXn + R differentiable at x , use the chain rule to derive a formula for -$f ( x + t u )
5. Compute exp rX from the definition for X and
( y i).
=
(A _; ), (A ) , ( _; A), (P h) ,
111. Theory o f Calculus in Several Real Variables
180
6.
It was observed in Section 6 in the context of polar coordinates that the Implicit Function Theorem implies the Inverse Function Theorem. Namely, the pair of polar-coordinate formulas (u, v ) = (r cos Q,r sinQ) was inverted by applying the Implicit Function Theorem to the system of equations
Using this example as a model, derive the Inverse Function Theorem in the general case from the Implicit Function Theorem in the general case. 7.
Define i l C O to mean limN+, disprove:
1 ; when the integrand is contmuous.
Prove or
Problems 8-9 use Fubini's Theorem to supplement the theory of Fourier series as given in Section 1.10.
8. Let f and g be continuous complex-valued periodic functions of period 2 n , and define their convolution to be the function
(a) Show that f * g is continuous periodic and that f * g = g * f . 00 (b) Let f ( x ) C00,_-,cneinx and g ( x ) C ,_-, dneinx. Prove that 00 (.f s > ( x ) cndneinx. (c) Prove that the Fourier series of f g converges uniformly.
*
-
-
*
9. Let f , g, and h be continuous complex-valued periodic functions of period 2n. Prove that f * (g * h ) = ( f * g ) * h. Problems 10-1 3 deal with homogeneous functions. I f f : Rn - {O} + R is a f~~nction not identically 0 such that f ( r x ) = rdf ( x )for all x in R n - { O } and all r z 0 , we say that f is homogeneous of degree d. For example, the function in the first problem below is homogeneous of degree 0 . 10. On lK2,define
Prove that $ and
5exist everywhere in R 2 and that f is not continuous at (0,O).
11. Problems
181
11. Let f : Rn - {O}+ R be smooth and homogeneous of degree d. (a) Prove that if d = 0,then f (x)is bounded on Rn - {O}and that f extends to be continuous at 0 only if it is constant. (b) Prove that if d > 0,then the definition f (0)= 0 makes f continuous for all x in Rn,while if d < 0, then no definition o f f (0)makes f continuous at 0. (c) Prove that a f is llonlogeneous of degree d - 1 unless it is identically 0. ax,
(d) If f is homogeneous of degree 1 and satisfies f (-x) = -f (x)and f (0)= 0,prove that each a f exists at 0 but that a f is not continuous at 0 unless it ax,
8%
is constant 12. On EX2,let f be the function homogeneous of degree 1 given by
if (x,y) = (o,o). ( 0 (a) Prove that f is continuous at (0,0). (b) Prove that and a f exist at (0,O)but are not continuous there. ay
(c) calculate
-$f ( t + t u ) 1 ,_,for x = 0 and u = (
)
Show that the formula
in Problem 4 fails, and conclude that f is not differentiable at ( O , 0 ) .
13. On R2,let f be the function homogeneous of degree 2 given by
lo (a) Prove that f ,
g,and
if (x,y) = (o,o).
are continuous on all of B".
(b) Prove that - and - exist at (0,0) but are not continuous there. (c) Prove that g ( 0 . O )= 1 and
$(0,O)=
-
1.
I
Problems 14-15 concern "harmonic functions" in {(x,y) E R2 I (x,y)I < I}, the open unit disk of the plane. A harmonic function is a complex-valued C2 function satisfying the Laplace equation ALL(.\-, y ) = 0,where A is the Laplacian A = aZ -
a2 + 3. 14. If (r,0) are regarded as polar coordinates, prove for all integers n that each ax2
function r n e i n eis a Cw function in the open unit disk and is harmonic there. Deduce that if {c,} is a doubly infinite sequence such that C,"=',c,rlneinO converges absolutely for each r with 0 5 r < 1, then the sum is a Cw function in the open unit disk and is harmonic there. 15. Prove that if u is harmonic in the unit disk, then so is the function u o R , where R is the rotation about the origin given by
(; ) u (z: )::'
(;).
182
111. Theory o f Calculus in Several Real Variables
Problems 16-20 illustrate the Inverse and Implicit Function Theorems.
+ y3 define a function from IK2 to R2 whose derivative at (1, 1 ) is given by the matrix ( ) . This
16. Verify that the equations u = x4y
+x
and v = x
::
matrix being invertible, the Inverse Function Theorem applies. ~ eth;t locally defined c1inverse function be given by x = F ( u , v ) and y = G ( u , v ) in an open neighborhood of ( u , v ) = ( 2 , 2 ) ,the point (2, 2) having the property that F ( 2 , 2 ) = 1 and G ( 2 , 2 ) = 1 . ~ i n $32, d 2). 17. Show that the equations
+ z2 = 0 , sin(uv) + 2z2 = 2,
x2 - y cos(uv) x2
+ y2
-
implicitly define x , y, z as c1functions of ( u , v ) near x = 1, y = 1 , u = n / 2 , v = 0 , and z = 0 , and find and for the function x ( u , v ) . Is the function x ( u , v ) of class Cm?
18. Regard the operation of squaring an n-by-n matrix as a function from IRn2 to IRn2, and show that this mapping is invertible on some open set of the domain that contains the identity matrix. 19. (Lagrange multipliers) Let f and g be real-valued C' functions defined on an open subset U of R n , and let S = ( x c u g ( x ) = 0 ) . prove that i f f has a local maximum or minimum at a point xo of S , then either gf(xo)= 0 or there exists a number h such that f ' ( x u ) k g f( x u )= 0 .
I
,1
+
20. (Arithmetic-geometricmean inequality) Using Lagrange multipliers, prove that any n real numbers a l , . . . , a, that are > 0 satisfy
CHAPTER IV Theory of Ordinary Differential Equations and Systems
Abstract. This chapter treats the theory of ordinary differential equations, both linear and nonlinear. Sections 1-4 establish existence and uniqueness theorems for ordinary differential equations. The first section gives some exa~rlplesof first-ordrr equations, mostly nonlinear, to illustrate cei.tain kinds of behavior of solutions. The second section shows, in the presence of continuity for a vectorvalued F satisfying a "Lipschitz condition," that the first-order system y' = F ( t , y) has a unique local solution satisfying an initial condition y (to) = yo. Since higher-order equations can always be reduced to first-order systems, these results address existence and uniqueness for nth-orderequations as a special case. Section 3 shows that the solutions to a system depend well on the initial condition and on any parameters that are present in F. Section 4 applies these results to existence of integral curves for a vector field and to construction of coordinate systems from families of integral curves. Sections 5-8 concern linear systems. Section 5 shows that local solutions of linear systems may be extended to global solutions and that in the homogeneous case the vector space of global solutions has dimension equal to the size of the system. The method of variation of parameters reduces the solution of any linear system to the solution of a homogeneous linear system. Sections 6-7 identify explicit solutions to nth-orderlinear equations and first-order linear systeins. The "Jordan canonical form" of a square matrix plays a role in the case of a system. Section 8 discusses power-series solutions to second-order homogeneous linear equations whose coefficients are given by convergent power series, as well as solutions that arise in the case of regular singular points. Two kinds of special functions are mentioned that result from this study -Legendre polynomials and Bessel functions.
1. Qualitative Features and Examples To introduce the subject of ordinary differential equations, this section gives examples of some qualitative features and complicated phenomena that can occur in such equations. Tf F is a complex-valiled filnction of n + 2 variables, a filnction y ( t ) is said to be a solution of the ordinary differential equation
F ( t , y , y', y" , . . . , y ( m ) ) = O of mth order on the open interval ( a , b) if
IV. Theory o f Ordinary Differential Equations and Systems
184
identically for a < t < b. The equation is "ordinary" in the sense that there is only one independent variable. The equation is said to be linear if it is of the form
+
a , ( t ) ~ ( ~ ) a,-i(t)~(~-')
+ . .. + al(t)yf + ao(t)y = q(t),
and it is homogeneous linear if in addition, q is the 0 function. A linear ordinary differential equation has constant coefficients if arn(t),. . . , ao(t) are all constant functions. Let us come to examples, which will point toward the enormous variety of phenomena that can occur. We stick to the first-order case, and all the examples will have F real-valued. Let us look only for real-valued solutions. Pictures indicating the qualitative behavior of the solutions of each of the examples are in Figure 4.1 . EXAMPLES.
(1) Simple equations can have relatively complicated solutions. This is already true for the equation y'
= l/t
on the interval (0, +a).
Intcgration shows that all solutions arc of thc form log t + c; on an intcrval of negative t's, the solutions are of the form log It 1 c. The c comes from a corollary of the Mean Value Theorem that says that a real-valued fullctioll on an open interval with 0 derivative everywhere is necessarily constant.' Another example, but with no singularity, is y' = ty. To solve this equation on intervals where y(t) # O,writeyl/y = t,sothatloglyl = i t 2 + a a n d l y l = ea et2/2. Thus y(t) = cet2/2,with c f 0 constant, on any interval where y(t) is nowhere 0. The function y(t) = 0 is a solution as well, and all real solutions on an interval are of the form y (t) = cet2l2with c real. See Figures 4.la and 4.lb. (2) Solutions may not be defined on obvious intervals. For the equation
+
ty'
+ y = sin t,
we can recognize the two sides as -$ (t y) and -$ (- cos t) . Therefore ty = c-cos t . Dividing by 1 , we obtain y ( 1 ) = -011 any interval that does not coritairi 0. What about intervals containing t = O? If we put t = 0 in the formula ty = c - cos t ,we see that c must be 1. In this case we can define y(0) = 0 there, and then y'(0) exists. We obtain the additional solution y(t> =
for t # 0,
( 0 fort = 0, on any open interval containing 0. Figure 4.lc shows graphs of some solutions. 'see Section A2 of the appendix for further information.
1. Qualitative Features and Examples
FIGURE4.1. Graphs of solutions of some first-order ordinary differential equations: (a) y' = lit, (b) y' = ty, (c) t y f + y = sint. (d) y' = y2 1, (e) y' = y2, (f) y' = y2I3.
+
(3) Even if the equation seems nice for all t ,the solutions may not exist for all t . An example occurs with y'=y
2
+l,
which we solve by the steps d (arctan y) = 1, arctan y = t I c, y = tan(t I c) . The solutions behave badly when t c is any odd multiple of n / 2 . Solutions are defined at lnost on intervals of length n.Figure 4.ld shows graphs of solne solutions for this example.
+
(4) Some solutions may look quite different from all the others. For example, with 2 Y=Y, 1 we solve by - l / y = t c for y # 0, so that y(t) = Also, y(t) = 0 is t+c
+
186
IV. Theory o f Ordinary Differential Equations and Systems
&
a solution. Here the solutions of the form y ( t ) = are not defined for all t , but the solution y ( t ) = 0 is defined for all t . We might think of y ( t ) = 0 as the limiting case with c tending to *co. Figure 4.le shows graphs of some of the solutions for this example.
(5) New solutions can sometimes be pieced together from old ones. For example, the equation is solved where y # 0 by the steps y-2/3y' = 1, 3y1I3 = t + c , and y ( t ) = & ( t + c ) ~ But . also y ( t ) = 0 is a solution. In fact, we can piece solutions of these types together. For example, the function $ ( t + ~ ) ~ fort<-1,
for - l ( t ( O , for0 it , is a solution on (-co,+co). Figure 4.lf shows graphs of some of the solutions for this example. One thing that stands out in the above examples is that the set of solutions seems to depend, more or less, on a single parameter c . The inference is that nothing much worse than the c occurs because somewhere an integration is taking place and the Mean value Theorem is controlling how many indefinite integrals there can be. One way of trying to quantify this statement about how the number of solutions is limited is to say that for any fixed t = to and given real number yo, there is only one solution y ( t ) near to with y (to) = yo. This statement is not quite accurate, however, as Example 5 shows. The uniqueness theorem in Section 2 will give a precise result. The data (to,yo) are called an initial condition. Something else that stands out, although perhaps not without the visual aid of the graphs of solutions as in Figure 4.1, is that the graphed solutions appear to fill the entire part of the plane corresponding to the t's under study. In the framework of thc prcvious paragraph, thc statcmcnt is that for any fixcd t = to and givcn real number yo, there exists a solution y ( t ) near to with y ( t ) = yo. The existence tlleorern in Section 2 will give a precise result.
WEAKVERSION OF EXISTENCE AND UNIQUENESS THEOREMS. Let D be a nonempty convex open set in R2, and let (to,yo) be in E . If F : D + R is a continuous function such that d ~ ( ty ), exists and is continuous in D, then for ay any sufficiently small open interval oft's containing to, the equation y' = F ( t , y ) has a unique solution y ( t ) with y (to) = yo such that the graph of t H y ( t ) lies in D .
2. Existence and Uniqueness
187
An improved theorem, together with a proof, will be given in Section 2. The proof of existence uses "Picard iterations," and the idea is as follows. First we convert the differential equation into an equivalent integral equation
Second we use the right side as input and the left side as output to define successive approximations to a solution:
yi(t) =
L
F ( s , yo(s)) ds
+ yo,
Third we use the Weierstrass M test to show that the series with partial sums N ( y n ( t )- y,-1 ( t ) ) is uniformly convergent. If the limiting yw(t) = yo function is denoted by y ( t ),we check that y ( t ) satisfies the integral equation from which we started. Hence y ( t ) is a solution of the differential equation.
+
2. Existence and Uniqueness In this section we state and prove the main existence and uniqueness theorems for solutions of ordinary differential equations. First let us establish an appropriate setting more general than the one in Section 1. The examples in Section 1 were all of the first order. They could all have been written in the form y = F ( t , y ) with F real-valued, and we considered real-valued solutions y ( t ) . From equations as simple as yf' + y f + y - 0, whose real-valued solutions are
we know that it can be easier to work, at least initially, with complex-valued solutions. In this particular case, it is easier as a first step to find all complexvalued solutions, namely
188
IV. Theory o f Ordinary Differential Equations and Systems
and then to extract the real-valued solutions from them. The solution method, which will be discussed in more detail in Section 6 below, involves finding all complex solutions of a certain polynomial equation with real coefficients, and the method is more natural if the coefficients of the polynomial equation are allowed to be complex. Thus right away, it is natural to consider first-order equations y' = F ( t , y ) with F complex-valued and to look for complex-valued solutions. The theory in Chapter I11 avoided working with functions of several variables in which some of the variables are complex, and we can update the theory of Chapter I11 here. The technique, which is to consider the complex variable y as two real variables Re y and Im y ,is again applicable. Thus we have only to think of F ( t , y ) as afunction of three real variables, even if we do not separate y into its two components in writing F ( t , y ) , and the theory of Chapter I11 applies directly. In adopting the point of view that y is actually two real variables, we need to apply the same consideration to y', and we are led to view y' = F ( t , y ) as a system of two simultaneous equations, namely Re y ' = Re F ( t , y ) and In1 y ' = Im F ( t , y ) . This viewpoint merely makes our functions conform to the prescriptions of Chapter 111. It is not necessary to work with the expanded notation; all we have to remember is that in this part of thc thcory wc ncvcr diffcrcntiatc a function with rcspcct to a complex variable. The utility of allowing y' = F ( t , y ) to represent a system of ordinary differential equations has, in any event, been thrust upon us. Let us consider the notion of a system a bit more With a little trick the second-order equation y" + y' + y = 0 can itself be transformed into a system, quite apart from the issue of real vs. complex variables. The trick is to introduce two unknown functions ul and u2 to play the roles of y and y'. Then ul and u2 satisfy u2 = u', and uh = uy = y" = -y' - y = -u2 - u l . In other words, ul and u2 satisfy the system u; = u2, u; = -u1
-
u2.
Conversely if u 1 ( t ) and u2( t ) satisfy this system of equations, then y ( t ) = ul ( t ) is a solution of y" + y' + y = 0. In this way, the given second-order equation is completely equivalent to a certain system of two first-order equations with two unknown functions. Let F be a function defined on an open set D of R x C k m and taking values in ck.A ck-valued functioll y ( t ) = ( y l ( t ) ,. . . , y k ( t ) ) is said to be a solutioil of the system F ( t , y , y', . . . , y(")) = 0 of k ordinary differential equations of order m in the open interval (a, b ) if F ( t , y ( t ) , y ' ( t ) , . . . , y ("1) = 0 identically fora < t < b . We saw that the single second-order equation y" y' y = 0 is equivalent to a certain first-order system of two equations, and the technique for exhibiting this
+ +
2. Existence and Uniqueness
189
equivalence works more generally: a system of k equations of order in that has been solved for the mth-orderderivatives is equivalent to a system of k m equations of first order. We shall consider first-order systems of the form y' = F ( t , y ) , where F is continuous on an open subset D of R x Cn and takes values in Cn. The example y' = y 2 / 3 in Section 1 fits these hypotheses, and we saw that the hoped-for uniqueness fails for this equation. In the weak theorem stated at the end of Section 1,an additional hypothesis was imposed in order to address this problem: for y' = F ( t , y ) with only real-valued solutions of interest, the hypothesis is that 3 P / 3 y exists and is continuous on the domain D of I;. Generalizing this condition presumably means saying something about partial derivatives in each of the directions yj for 1 < j < n . In addition, we must remember the injunction against differentiating with respect to complex variables. Thus we really expect a condition concerning 2n first-order derivatives. Fortunately there is an easily stated less-stringent condition that is nevertheless good enough. The condition is that F satisfy a Lipschitz condition in its y variable, i.e., that there exist a real number k such that IF(t, y1)
-
F ( t , y2)I 5 k l ~ l y2I
for all pairs of points ( t , y l ) and ( t , y2) in the domain D of F. If F is areal-valued continuous function of two real variables with a continuous partial derivative in the second variable, then the Mean Value Theorem gives
with 6 between yl and y2,provided the line segment from ( t , y l ) to ( t , y2) lies in the domain D of F . The partial derivative is bounded on any compact subset of D, and thus F satisfies, on any compact convex subset of D, a Lipschitz condition in the second variable.
Theorem 4.1 (Picard-Lindelof Existence Theorem). Let D be a nonempty open set in R' x C n ,let (to, yo) be in D , and suppose that F : D -f Cn is a continuous function such that F ( t , y ) satisfies a Lipschitz condition in the y variable and has I F ( t , y ) I < M on D. Let R be a compact set in R' x Cn of the form R = { ( t , y ) It - tol < a and ly - yo1 < b } , and suppose that K is contained in U . Put a' = min{a, b / M } . 'I'hen there exists a solution y ( t ) of the system Y' = F ( t , y ) on the open interval It - to1 < a' satisfying the initial condition
I
IV. Theory o f Ordinary Differential Equations and Systems
190
REMARKS.A variant of Theorem 4.1 takes D to be in R 1 x Cn but insists only on continuity of F, not on the Lipschitz condition. Then a local solution still exists for It - to I < a'. This better result, known as the "Cauchy-Peano Existence Theorem," appears in Problems 20-25 at the end of the chapter and is proved by an argument using Ascoli's Theorem. However, Example 5 in Section 1 shows that there is no corresponding uniqueness theorem, and within the text we omit the proof of the better existence theorem. Another variant of Theorem 4.1 assumes that the domain D of a given FRlies in E X 1 x EX'" FRtakes values in R'" and yo is in R n . Then y' = F R ( t , y ) has a solution y ( t ) such that ?(to) = yo and the range of y is R n . In fact, when FR satisfies a Lipschitz condition in the y variable, this variant is a consequence of Theorem 4.1 as stated. To derive this variant, one extends the given function FR from the subset of R 1 x R n to a subset of E X 1 x Cn by making it constant in Im y . Specifically the new system is y' = F ( t , y ) with F ( t , y ) = F R ( t , Re y ) , and the initial condition remains as y(to) = yo. The part of the system corresponding to equations for Im y' is just Im y' = 0, since F is real-valued, and therefore In1 y ( t ) is constant. Since yo is real, In1y ( t ) nust be 0. Thus Theorem 4.1 yields a solution y ( t ) with range W n under these special hypotheses.
PRoo~. Thc first stcp is to scc that thc sct of diffcrcntiablc functions t ++y ( t ) on It - tol < a' satisfying y' = F ( t , y ) and y(to) = yo is the same as the set of contir~uousfurictiorls 1 H y ( 1 ) 011 11 - 101 < a' satisfying the integral equatiori y(t) = F ( s , y(.r)) d.7 yo. If y is differentiable and satisfies the differential equation and the initial condition, then y is certainly continuous and hence s H F ( s , y ( s ) ) is continuous. Then ji: F ( s , y ( s ) )d s is differentiable by the Fundamental Theorem of Calculus (Theorem 1.32),and the differential equation shows that y ( t ) and F ( s , y ( s ) )d s have the same derivative for It - to I < a'. Thus they differ by a constant. The constant is checked by putting t = to, and indeed y satisfies the integral equation. Conversely if y is continuous and satisfies the integral equation, then s H F ( s , y ( s ) ) is continuous, and the Fundamental Theorem of Calculus shows that F ( s , y ( s ) ) d s is differentiable. This function equals y ( t ) - yo by the integral equation, and hence y is differentiable. Differentiating the two sides of the integral equation, we see that y satisfies the differential equation. Also, if we put t = to in the integral equation, we see that y satisfies the initial condition y(t0) = yo. Thus it is enough to prove existence for a continuous solution of the integral equation. For to - a' < t < to + a', define inductively
-(:
+
6
&
yo(t) = yo,
Y I ( t ) = yo
+
ft
F ( s , yo($))d s ,
2. Existence and Uniqueness
with the usual convention that of y n ( t ) lies in the set
A:
= - J;"' Let us see inductively that the graph
I
a'. The graph of yo(t) = yo is just { ( t , y o ) It - to1 i a ' ) , for It - tol and this lies in R'. The inductive hypothesis is that ( t , yn-1 ( t ) ) lies in R' for { ( t ,Y O ) It - tol 5 a'}. m e n
I
I&(')
-
nl
=
I
L F ( . ~ y,n - l ( L y ) d s 5 M l t
-
t o 5 Ma' 5 b,
and therefore ( t , y n ( t ) ) lies in R' for It - to1 5 a'. This completes the induction, and hence the graph of )i,%(t)lies in R' for It - to1 5 a'. Now write N
for N 0. We shall use the Weierstrass M test (Proposition 1.20),adapted to a series of functions with values in C n ,to prove uniform convergence of this series. Thus we are to bound 1 y, ( t ) - y,-1 ( t )1, and we shall do so inductively for n > 1. We start from the inequality I F ( t , y ) I 5 M on R' and the Lipschitz condition
Say that to 5 x 5 to
and
+ a' for definiteness. Then
IV. Theory o f Ordinary Differential Equations and Systems
kM(s -
-
to) ds
from the previous display
M k ( t - to)2 2!
Now wc carry out an induction. Thc basc casc is thc cstimatc carricd out abovc for I yl ( t )- yo ( t )1. The estimate for 1 y2 ( t )- yl ( t )1 suggests the inductive hypothesis,
Then we have
(S -
toy1
-
(?I -
I)!
ds
by inductive hypothesis
and the induction is complete. The argument when to - a' 5 t 5 to is completely similar, and the form of the estimate for the two cases combined is Iyn(t) - yn-1 ( t )I 5
Mkn-'It n!
-
toln
for It
-
tol 5 a'.
There is no harm in assuming that k is > 0 , and consequently
independently o f t . Since C z ( n ! ) - l k n ( a ' ) n= eka'is finite, the M test applies and shows that our series converges uniformly. Thus y ~ ( t converges ) uniformly for It - to 1 5 a', necessarily to a continuous function. We call this function y ( t ). For It - to I 5 a', we have
2. Existence and Uniqueness
193
On the right side, we have limN[ y N f l( t ) - yo] = y ( t ) - yo. Because of the Lipschitz condition the absolute value of the first term on the right side is
and this tends to O as n tends to infinity. Thus F ( s , v ( s ) )d s = ~
( f )
Yo.
and y ( t ) is a continuous solution of the integral equation.
Theorem 4.2 (uniqueness theorem). Let D be a nonempty open set in IR1 x Cn, let (to,yo) be in D , and suppose that F : D + Cn is a continuous function such that F ( t , y ) satisfies a Lipschitz condition in the y variable. For any a" > 0, there exists at most one solution y ( t ) to the system
on the open interval It
-
to 1 < a" satisfying the initial condition y(t0) = yo.
PROOF.As in the proof of Theorem 4.1, it is enough to prove uniqueness for the integral equation. Suppose that y ( t ) and z ( t ) are two solutions for It to 1 < a". Fix t > 0. Then I y ( t )- z ( t ) I is bounded by some constant C for It - to I 5 a" - t , and F is assumed to satisfy a Lipschitz condition I F ( t , y l ) - F ( t , y2) I 5 k 1 yl -y2 I on D . We argue as in the proof of Theorem 4.1, working first for to 5 t and starting from IY (t>- z(t>l 5 c and from
194
IV. Theory o f Ordinary Differential Equations and Systems
Inductively we suppose that
Then
and thus 1 y ( t ) - ~ ( tI 5 ) C(n!)-'kn(tfor t 5 to, and the combined estimate is
for all n . A similar estimate is valid
Since C (n!) -' kn 1 t - to I " converges, the individual terms tend to 0. Therefore y ( t ) = z ( t ) for It - toI < a"- t . Since t is arbitrary, y ( t ) = z ( t ) for It- to1 < a".
3. Dependence on Initial Conditions and Parameters In abstract settings where the existence and uniqueness theorems play a role, it is frequently of interest to know how the unique solution depends on the initial data (to, yo) such that y(to) = yo. To quantify this dependence, let us write the unique solution corresponding to y' = F ( t , y ) as y ( t , to, yo) rather than y ( t ). We continue to use y' to indicate the derivative in the t variable even though the differentiation is now actually a partial derivative.
Theorem 4.3. Let D be a nonempty open set in R 1 x C n , let ( t , y*) be in D , and suppose that F : D -r C n is a continuous function such that F ( t , y ) satisfies a Lipschitz condition in the y variable. Let R be a compact set in R1 x C n of the form R = { ( t , y ) It - t*I < a and ly - y*l < b } ,
I
suppose that R is contained in D, and let M be an upper bound for IF I on R . Put a' = min{a, b / M } . If Ito - t* 1 < a 1 / 2and 1 yo - y*l < b / 2 , then there exists a unique solution t H y ( t , to, yo) on the interval It - to I < a f / 2 to the system and initial data y' = F ( t , y ) and y(t0, to, yo) = yo, and the function ( t , to, yo) H y ( t , to, yo) is continuous on the open set
I
U = { ( t , to, yo) It - tol < a f / 2 , Ito - t*l < a f / 2 , lyo - y*l < b / 2 } . If F is smooth on D, then ( t , to, yo) H y ( t , to, yo) is smooth on U .
3. Dependence on Initial Conditions and Parameters
195
REMARK.It is customary to summarize the result about continuity qualitatively by saying that the unique solution depends continuously on the initial data.
PROOFOF CONTINUITY. Let us first check that there is indeed a unique solution for each pair (to,yo) in question and that its graph, as a function o f t , lies in R ' = { ( t , y ) I It-t*l 5 n'and ly
-
y*l 5 b }
For this purpose, fix to and yo with I to - t* 1 5 a f / 2 and lye - y* I 5 b / 2 . Use of the triangle inequality shows that the closed set with It - to1 < a f / 2 and 1 y - ynl < b / 2 lies within R . Thus IF I 5 M on this set. Theorem 4.1 shows that there exists a solution with graph in this smaller set for It - to I < a", where a" = min{af/2, ( b / 2 ) / M } .Now
and hence there exists a solution for It - to I < a f / 2 with graph in R . This solution y i t , to, yo) is unique by Theorem 4.2, and it is the result of the construction in the proof of Theorem 4.1 . The idea is to trace through the construction in the proof of Theorem 4.1 and to see that the function ( t , to, yo) H y i t , to, yo) is the uniform limit of explicit continuous functions on U . Imitating a part of the proof of Theorem 4.1, we define, for ( t , to, yo) in U , yo(t, to, yo) = yo, y i ( t , to, yo) = yo
+
y m ( t j to, yo) = yo
+
F ( s , yo(s, to, yo)) d s ,
1
F ( s , ym-i(s, to, yo)) ds.
We shall show by induction that y, i t , to, yo) is continuous on U . Certainly yo ( t , to, yo) is continuous on U . For the inductive step we need a preliminary calculation. Let Il be the closed interval between to and t , and let 12 be the closed interval between ti, and t'. Suppose we have two functions fi and f2 of a variable s such that (i) f l is defined for s between to and t with I fl I 5 M there, (ii) f 2 is defined for s between t; and t' with I f 2 5 M there, and (iii) I f ( s ) - f 2 ( s ) I 5 t on their common domain. If a' is > the maximum distance among to, t , th, t ' , let us show that
196
IV. Theory o f Ordinary Differential Equations and Systems
To show this for all possible order relations on the set { t o ,t;, t , t ' } , we observe that there is no loss of generality in assuming that to is the smallest member of the set. There are then six cases. Case 1. to < t; < t' < t , so that (iii) applies on [th, t ' ] . Then
and hence
Therefore (*) holds in this case. Case 2. to < t; < t < t ' , so that (iii) applies on
[th, t ] . Then
and hence
Therefore (*) holds in this case. Case 3. to < t < t' < t;. Then
1 i'
."")dsl
and
5
MI^; - t'13
so that (*) holds in this case. Case 4. to < t' < t; < t . Then
and
1f
f 2 ( s l d s l 5 M i l ' - ti1 = M(1t
so that (*) holds in this case.
-
t'l
-
it
-
t;~),
3. Dependence on Initial Conditions and Parameters
Case 5. to
and
( t ( ti, ( t'.
1
Then
L~' f 2 ( , ~ ) d ,5 ~ lM1t'- thl 5 Mlt'
-
ti,
so that (*) holds in this case. Case 6. to < t' < t < th. Then
and
t'l = M(lth
-
tl
+ It'
-
tl),
so that (*) holds in this case. With (*) proved we can now proceed with the inductive step to show that y,(t, to, yo) is continnons on [ I . Thus assume that y, l ( t , to, yo) is continnons on U . If ( t , to, yo) and (t', th, yh) are in U , then
where f i b ) = F ( s , y,-l(s, to, Y O ) ) and f2(s) = F ( s , yn-l(s, th, yh)). Thus (*I gives
if t is chosen such that 1 . f ~ ( s ) - .f2(s)I < t on the common domain of , f l and ,f2. Let t > O be given, and choose some 6 > O for uniform continuity of F on the set R . By uniform continuity of y,-l, choose 7 > 0 such that
Then I ( s , to, yo) - ( s , th, y;) 1 < rj implies fl ( s ) - f 2 ( s ) 1 < t on the common domain of fl and f2, and hence (**) holds. Therefore y, is continuous as a function on U . This completes the induction.
198
IV. Theory o f Ordinary Differential Equations and Systems
We know that y, ( t , to, yo) converges to a solution y ( t , to, yo) uniformly in t if (to,yo) is fixed. Let us see that the convergence is in fact uniform in ( t , to, y o ) . The proof of Theorem 4.1 yielded the estimate ly,(t, to, yo)
M kn(a'),
-
y,-l(t, to, yo11 5 k
3 -
n!
and this is independent of ( t , to, yo). Therefore the Weierstrass M test shows that y, ( t , to, yo) converges to y ( t , to, yo) uniformly on U . The uniform limit of continuous functions is continuous by Proposition 2.21, and hence y ( t , to, yo) is continuous. PROOF OF SMOOTHNESS. Under the assumption that F is smooth on D, we are to prove that y ( t , to, yo) is smooth on U . We return to the earlier proof of continuity of y ( t , to, yo) and show that each y,(t, to, yo) is smooth. This smoothness is trivial for n = 0, we assume inductively that y,-1 ( t , to, yo) is smooth, and we form ~ n ( tto, , yo) = yo
+
F ( s , y,-i(s, to, yo)) d s .
The function on the right side is the composition of ( t , to, yo) e ( t , to, to, yo) followed by ( t , to, so, yo) H S,: F ( s , y,-1 ( s , so, m)) d s . The chain rule (Theorem 3.10), the Fundamental Theorem of Calculus (Theorem 1.32), and Proposition 3.28 allow us to compute partial derivatives of this function, and another argument with (*) allows us to see that the partial derivatives are continuous. 'I'here is no difficulty in iterating this argument, and we conclude that y,(t, to, yo) is smooth. The same argument in the proof of Theorem 4.1 that enabled us to estimate the size of y, ( t , to, yo) - y,-1 ( t , to, yo) allows us to estimate any iterated partial dcrivativc of this diffcrcncc. Ncw constants cntcr thc cstimatc, but thc qualitativc result is the same, namely that any iterated partial derivative of y, ( t , to, yo) converges uniformly to that same iterated partial derivative of y ( t , to, yo). Applying Theorem 1.23, we see that y ( t , to, yo) is smooth. REMARK. Sometimes a given system y ' = F (t, y ) with initial CONCLUDING condition y (to) = yo involves parameters in the definition of F, so that effectively the system is y' = F ( t , y , h 1, . . . , h k ). A natural problern is to find conditions under which the dependence of the solution on the k parameters is continuous or smooth. The answer is that this problem can be reduced to the problem addressed by Theorem 4.3. We simply introduce k additional variables z,, one for each = 0 and new initial conditions parameter h j , together with new equations zj (to) = h j .
4. Integral Curves
199
4. Integral Curves If U is an open subset of Rn,then a vector field on U may be defined as a function X : U + Rn. The vector field is smooth if X is a smooth function. In classical notation, X is written X = a j ( x l ,. . . , x,) d , and the function carries ax, ( x l ,. . . , x,,) to ( a l (.ul, . . . , .up,),. . . , a,, (.ul, . . . , x,,)) . The traditional geometric interpretation of X is to attach to each point p of U the vector X ( p ) as an arrow based at p . This interpretation is appropriate, for example, if X represents the velocity vector at each point in space of a time-independent fluid flow. We have defined the term "path" ill a metric space to meal1 a coiltiiluous function from a closed bounded interval of EX1 into the metric space. The term curve in a metric space is used to refer to a continuous function from an open interval of R 1 into the metric space. A standard problem in connection with vector fields on an open subset U of R2 is to try to draw curves within U with the property that the tangent vector to the curve at any point rnatches the arrow for the vector Geld. An illustration occurs in Figure 4.2. This section abstracts and generalizes this kind of curve.
Ckl
FIGURE 4.2. Integral curve of a vector field. Let X : U + Rn be a smooth vector field on U . A curve c ( t ) is an integral curve for X if c is smooth and c f ( t ) = X ( c ( t ) ) for all t in the domain of c. Depending on one's interpretation of the informal wording in the previous paragraph, the present definition is perhaps more demanding than the definition given for R2 above: the expression c f ( t )involves both magnitude and direction, and the present definition insists that both ingredients match with X ( c ( t ) ) ,not just the direction.
Proposition 4.4. Let X : U + Rn be a smooth vector field on an open subset U of R n , and let p be in U . Then there exist an E > 0 and an integral curve c : (-e, E ) + U such that c ( 0 ) = p. Any two integral curves c and d for X having c ( 0 ) = d ( 0 ) = p coincide on the intersection of their domains.
200
IV. Theory o f Ordinary Differential Equations and Systems
PROOF.Apart from the smoothness the first conclusion is just a restatement of a special case of Theorem 4.1 in different notation. The conditions on c are that c be a solution of c' = X ( c ) and that c ( 0 ) = p. The existence of a solution is immediate from Theorem 4.1 if we put F = X , c = y , to = 0 , and yo = p . The way in which this application of Theorem 4.1 is a special case and not the general case is that F is independent o f t here The smoothness of c follows from Theorem 4.3, and the uniqueness follows from Theorem 4.2. The interest is not only in Proposition 4.4 in isolation but also in what happens to thc intcgral curvcs whcn X is part of a family of vcctor ficlds.
Proposition 4.5. Let x('),. . . , x ( ~be)smooth vector fields on an open subset U of IRn ,let p be in U , and let V be a bounded open neighborhood of 0 in Rm. For h in V ,put Xh = C?l h j X ( j ) .Then there exist an e > 0 and a system of integral curves c ( t , A), defined for t E ( - 6 , s ) and h E V , such that c( . , h ) is an integral curve for Xh with c(0, h ) = p . Each curve c ( t ,A ) is unique, and the function c : (-e, e ) x V + U is smooth. If m = n , if the vectors ~ ( ' ) ( p .).,. , x ( ~ ) are linearly independent, and if S is any positive number less than s , then the Jacobian matrix of h ++ c(S,h ) at h = 0 is nonsingular.
REMARK.In the final conclusion of this proposition, the open neighborhood of 0 within V is allowed to depend on 6. It follows from the final conclusioll that the Inverse Function Theorem (Theorem 3.17) and its corollary (Corollary 3.21) are applicable to the mapping h H c(6, h ) at h = 0 . These results produce a smooth inverse function carrying an open subneighborhood of 0 within V onto an open subneighborhood of p of U . In effect the inverse function assigns locally defined coordinates in h space to a neighborhood of U . ~ O O F We .
set up the system of equations c'
=
Xh o C , i.e.,
with initial condition c(0) = p. This is a smooth system of the kind considered in Theorem 4.3, and the h j with 1 5 j 5 m are parameters. The parameters are handled by the concluding remark in Section 3: we obtain unique solutions c ( t , h ) for t in some open interval (-e, e ), and ( t , h ) H ~ ( tl ,)is smooth. . .). ,,x ( " ) ( p )are linearly Now suppose that rn = n , that the vectors x ( ' ) ( ~ independent, and that 0 < S < s . The function c satisfies
(~)
5. Linear Equations and Systems, Wronskian
20 1
and we use this information to compute the Jacobian matrix of h H c ( S ,h ) at h = 0. The Fundamental Theorem of Calculus, Proposition 3.28, and (*) give
= 0 . Also, c ( t , 0 ) is constant Now ci ( 0 , h ) = pi for all h , and hence z ( 0 , A ) in t by (*), and the constant is c(0,O) = p. Finally when h is set equal to 0 in the term Z:=, hi& x ( i l ( c ( t , A ) ) d t , each hk becomes 0 , and thus the whole term becomes 0 . Thus the above equation specializes at h = 0 to
/,6
The vectors ~ ( j ) ( pare) by assumption linearly independent, and hence the determinant of the matrix [ x ( j ) ( p ) is ] not 0 . Consequently the Jacobian matrix h H c(6, h ) at h = 0 is nonsingular if 6 # 0.
5. Linear Equations and Systems, Wronskian
Recall from Section 1 that a linear ordinary differential equation is defined to be an equation of the type
with real or complex coefficients. The equation is homogeneous if q is the 0 function, inhomogeneous in general. In order for the existence and uniqueness theorems of Section 1 to apply, we need to be able to solve for y ( n ) and have all coefficients be continuous afterward. Thus we assume that a n ( t ) = 1 and that a,-1 ( t ), . . . , aO( t ) and q ( t ) are continuous on some open interval. Even in simple cases, the theory is helped by converting a single equation to a system of first-order equations. In Section 1 we saw an indication that a way to make this conversion is to put
202
IV. Theory o f Ordinary Differential Equations and Systems
and get yn-1 = Y (n-2) Yn = Y
yk-1 = yn
(n-1)
Y; = -ao(t)yl
-
+
.. . - a n - I Y ~ q(t).
If we change the meaning of the symbol y from a scalar-valued function to the vector-valued function y = ( y l , . . . , yn), then we arrive at the system
where A ( t ) is the n-by-n matrix of continuous functions given by
and Q ( t ) is the n-component column vector of continuous functions given by
In a general linear first-order system of the kind we shall study, A ( t ) can be any 12-by-12 matrix of continuous functions and Q ( t ) can be any column vector of continuous functions; thus the first-order system obtained by conversion of a single nth-order equation is of quite a special form among all first-order linear systems. For a system y' = A ( t ) y Q ( t ) as above, the Lipschitz condition for the function F ( t , y ) = A ( t )y I Q ( t ) is automatic, since
+
and since the function t H 11 A ( t ) 11 is bounded on any compact subinterval of our domain interval. By the uniqueness theorem (Theorem 4.2), a unique solution
5. Linear Equations and Systems, Wronskian
203
to the system is determined by data (to,yo), the local solution corresponding to (to,yo) being the one satisfying the initial condition that the vector y(to)equal the vector yo. If we track down what these data correspond to in the case of a single nth-order equation, we see that a unique solution to a single n"-order equation of the kind described above is determined by initial values at a point to for the scalar-valued solution and all its derivatives through order n - 1. First-order linear systems of size one can be solved explicitly in terms of known functions and integrations. Specifically the single homogeneous first-order equation y' = a ( t )y is solved by y ( t ) = c exp ( St a ( s ) d s ) , and the solution of a sirlgle irlhu111ugerle~uslirsl-urder equaliu11 car1 be reduced Lu llle hur~iuger~euus case by the variation-of-parameters formula that appears later in this section. However, there need not be such an elementary solution of a first-order linear system of size two, not even a system that comes from a single second-order equation. Elementary solutions exist when the coefficient matrix has constants as entries, and we shall address that case in the next two sections. Sometimes one can write down tidy power-series solutions when the coefficient matrix has nonconstant entries, and we shall take up that matter later in the chapter. For now, we develop some general theory about first-order linear systems, beginning with the homogeneous case. The linearity implies that the set of solutions to the system y' = A ( t ) y on an open interval is a vector space (of vector-valued functions) in the sense that it is closed under addition and scalar multiplication.
Theoreill 4.6. Let y' = A ( t ) y be a homogeneous linear first-order n-by-n system with A ( t ) continuous for a < t < b . Then (a) any solution on a subinterval (a', b') extends to a solution on the whole interval ( a ,b ) , (b) the dimension of the vector space of solutions on any subinterval (a', b') is exactly n , (c) if ul ( t ), . . . , u, ( t ) are solutions on an interval (a', b') and if to is in that interval, then ul, . . . , v, are linearly independent functions if and only if the column vectors vl ( t o ) ,. . . , vn [to) are linearly independent. PROOF.We begin by proving (c). If clul ( t ) + . . . + crur( t ) is identically 0 for constants cl , . . . , c, not all 0, then clvl (to)+ . . . + cr v, (to) = 0 for the same constants. Conversely suppose that cl vl (to) . . . c, v, (to) = 0 for constants not all 0. Put u ( t ) = clvl ( t ) + . . . + c,v, ( t ) . Then v ( t ) and the 0 function are solutions of the system satisfying the same initial conditions-that they are 0 at to. By the uniqueness theorem (Theorem 4.2), v ( t ) is the 0 function. This proves (c). The upper bound in (b) is immediate from (c) since the dimension of the space of n-component column vectors is n .
+ +
204
IV. Theory o f Ordinary Differential Equations and Systems
Let us prove that n is a lower bound for the dimension in (b) if the interval containing to is sufficiently small. By the existence theorem (Theorem 4.l), there exists a solution vJ( t ) on some interval It - to I < e, such that v, (to)= e, . The v, ( t ) are then solutions on It - to 1 < e with e = miniel, . . . , E,}, and they are linearly independent by (c). Hence the dimension of the space of solutions is at least n on the interval It - to I r F or on any subinterval containing to We are not completely done with proving (b), but let us now prove (a). Let u ( t ) be a solution on (a', b') . If we have a collection of solutions on different intervals containing (a', 6') and each pair of solutions is consistent on their common domain, then the union of the solutions is a solution. Consequently we may assume that u ( t ) does not extend to a solution on any larger interval. We are to prove that (a', b') = ( a ,b ) . Suppose on the contrary that b' < b. We use to = b' in the previous paragraph of the proof; the result is that on some interval It - b'I < e with e sufficiently small and at least small enough so that a' < b' - e , the space of solutions has dimension n with a basis { u l , .. . , v,}. By (c), the column vectors vl (b' - e ), . . . , v, (b' - e ) are linearly independent, and thus the restrictions of vl, . . . , v, to (b' - e, b') are linearly independent. The restriction of v ( t ) to the interval (b' - 6 , b') is a solution, and thus there exist constants C I , . . . , C , such that
+ +
But then the function equal to v ( t ) on (a', b') and equal to cl v l ( t ) . . . c,v, ( t ) on (b' - F, b' + e) extends v ( t ) to a solution on a larger interval and contradicts the rnaximality of the domain of u ( t ) . This proves that b' = b . Similarly we find that a' = a . This proves (a). We return to the unproved part of (b). Fix to in (a', b'). On a subinterval about to, the space of solutions has dimension n , as we have already proved. Let { v l ,. . . , v,} be a basis. By (a), we can extend v l , . . . , v, to solutions on (a', b'). Then the space of solutions on (a', b') has dimension at least n , and (b) is now completely proved. EXAMPLE. Let us illustrate the content of Theorem 4.6 by means of a single second-order equation, namely y" y = 0. We know that cl cost c2 sin t is a solution for every pair of constants cl and c2. TOconvert the equation to a system, we introduce yl = y and yz = y'. The system is then
+
+
( -:i) ,a matrix of constants. The scalar-valued
and hence the matrix is A(t) = solutions cost and sin t of y" + y \
,
= 0 correspond
to the vector-valued solutions
5. Linear Equations and Systems, Wronskian
(
cost) - sin t
and
205
( ) ,respectively; each of these has a scalar-valued solution in an t cos t
its first entry and the derivative in the second entry. In either case, both solutions are defined on the interval (-m, + X I ) .The theorem says that the restrictions of these two functions to any subinterval span the solutions on that subinterval. According to (c), the linear independence of the scalar-valued solutions cos t and sin t is reflected by the linear independence of the column vectors
)::(
( -_:)
and
for any lo in (-w, +w) . The latter independence we can see immediately
by observing that the matrix
cos to sin to
) has determinant equal to 1 and not 0.
The kind of matrix formed in the previous example is a useful tool when generalized to an arbitrary homogeneous linear system, and it has a customary name. Let v l (t), . . . , v,(t) be solutions of an n-by-n homogeneous linear system y' = A(t)y with A(t) continuous. The Wronskian matrix of vl,. . . , v, is the n-by-n matrix whose jth column is t v . If t l i , j denotes the ith entry of the jth solution, then
Since each column of W(t) is a solution, we obtain the matrix identity W1(t) = A(t) W(t).
EXAMPLE, CONTINUED. In the case of the single second-order equation y" y = 0, we listed two linearly independent scalar-valued solutions as cost and sint. When the equation is converted into a 2-by-2 homogeneous linear system, the Wronskian matrix is
+
w (t) =
-
cos t sint
sin t cost
For a general nth-order equation with vl, . . . , v, as scalar-valued solutions, the Wronskian matrix of the associated system is
Proposition 4.7. If v l (t), . . . , v,(t) are solutions on an interval of an n-by-n homogeneous linear system y' = A(t) y with A(t) continuous, then the following are equivalent: (a) vl,. . . , v, are linearly independent solutions,
IV. Theory o f Ordinary Differential Equations and Systems
206
(b) det W ( t ) is nowhere 0, (c) det W ( t ) is somewhere nonzero PROOF.By Theorem 4.6c, (a) here is equivalent to the linear independence of ul ( t o ) ., . . , u, ( t o ) ,no rnatter what to we choose, hence is equivalent to the condition det W ( t o )# 0, no matter what to we choose. The proposition follows. Wc shall usc thc Wronskian matrix of a homogcncous systcm to analyzc thc solutions of any corresponding inhomogeneous system.
Proposition 4.8. For an inhomogeneous linear system y' = A ( x ) y + Q ( t ) with A ( t ) and Q( t )continuous for a it ib , any solution y* ( t )on a subinterval (a', b') of ( a ,b) extends to be a solution on ( a ,b ),and the most general solution y ( t ) is of the form y ( t ) = h ( t ) + y* ( t ), where y* ( t ) is one solution of y' = A(t)y Q ( t ) and h ( t ) is an arbitrary solution of the homogeneous system y' = A(t)y.
+
PROOF.If y* and y** are two solutions of y' = A ( t ) y + Q ( t )on (a', b'), then
+
(y*"- y*)'(t)= ( A ( t ) y * * ( t+) Q
) - ( A ( t ) y * ( t ) Q) = A(t)(y*- y**)(t), and h = y** - y * solves y' = A(t)y on (a', b') . Conversely if h solves y' = A(t)y Q ( t ) on ( a f ,b') ,then
+
(y*
+ h)'(t)= y*'(t) + h f ( t ) +
+
+ h ) ( t )+ C_!(t),
= ( A ( t ) y " ( t ) C_!(t)) A ( t ) h ( t )= A(t)(y*
and y* + h is a solution of y' = A(t)y + Q ( t ) on ( a f ,b'). We are left with showing that any solution y * of y' = A ( t )y + Q ( t )on (a', b') cxtcnds to a solution on ( a ,b ) . As in thc proof of Thcorcm 4.6a, wc can form unions of functions and thereby assume that y* cannot be extended to be a solution on a larger interval. The claim is that (a', b') = ( a ,b ) . Assuming the contrary, suppose, for example, that b' < b. By the existence theorem (Theorem 4.l), there exists a solution y"*(t)of y' = A(t)y + Q ( t )for It - b'l < e if e is small enough. By the result of the previous paragraph, y*(t) = y ** ( t )+h ( t )on (b' - e , b') for a suitable choice of h that solves the homogeneous system y' = A(t)y on (b' - s , b') . Since y**(t)is given as a solution of y' = A(t)y I Q ( t )on (b' e , bf I e ) and since, by Theorem 4.6a, h ( t )extends to a solution of y' = A(t)y on (b' - E , b' e ),we see that y **(t)+h ( t )extends to a solutioil of y' = A(t)y + Q ( t )on (b' -e, b' +e) . Then the function equal to y * ( t )on (a', b') and to y**( t )+ h ( t )on (b' - e , b' +e ) extends y* ( t ) to a solution of y' = A(t)y + Q ( t ) on a larger interval, namely (a', b' e ) . We obtain a contradiction and conclude that b' must have equaled b. Similarly a' must equal a . Thus every solution of y' = A(t)y Q ( t ) on a subinterval extends to all of ( a ,b ) ,and the proof is complete.
+
+
+
5. Linear Equations and Systems, Wronskian
207
Theorem 4.9 (variation of parameters). For an inhomogeneous linear system y' = A ( x ) y Q ( t ) with A ( t ) and Q ( t ) continuous for a < t < b , let v l , . . . , v, be linearly independent solutions of y' = A ( t ) y on ( a ,b ) , and let W ( t )be their Wronskian matrix. Then a particular solution y* of y' = A ( t ) y Q ( t ) on ( a , b ) is given by
+
+
where
y*(t)=W(t)u(t),
W(t)uf(t)=Q(t).
That is,
REMARKS.Linearly independent solutions v l , . . . , v, as in the statement exist by Theorem 4.6. PROOF.For any differentiable vector-valued function u ( t ), y* ( t ) = W ( t ) u( t ) has (y*)' = W'u Wu' = AWu Wu' = Ay* Wu'.
+
+
+
+
Thus y* will have ( y*)' = Ay * Q if and only if Wu' = Q . Since Proposition 4.7 shows that w ( t ) - ' exists and is continuous, we can solve Wu' = Q for u . EXAMPLE, CONTINUED. Now consider the single second-order inhomogeneous y = tan t on the interval It 1 < n/2. We saw that we can linear equation y" take W ( t )= - s 'iOnS t1 cos 'Int We set up the system 1
(
+
)
-
cost sin t
s i n t ) ( cos t
( ) 0 tan t
of algebraic linear equations and solve for u/l and ub:
)j:(
=
(
cost sin t
-sint cos t
)(
A vector-valued function with derivative
sin t
-
0 tan t
(::)
1
=
(-=I
sin2 t sin t
for it 1 < n / 2 is
log(1 + sin t ) + log cos t - cos t
and we thus take y * ( t ) = (cos t ) u ( t ) + (sin t)u2( t ). The most general solution of the given inhomogeneous equation is therefore y* ( t ) cl cos t c2 sin t .
+
+
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IV. Theory o f Ordinary Differential Equations and Systems
6. Homogeneous Equations with Constant Coefficients In this section and the next, we discuss first-order homogeneous linear systems with constant coefficients. The system is of the form y' = Ay with A a matrix of constants. A single homogeneous nth-order linear equation with constant coefficients can be converted into such a first-order system and can therefore be handled by the method applicable to all first-order homogeneous linear systems with constant coefficients. But such an equation can be handled more simply in a direct fashion, and we therefore isolate in this section the case of a single nth-order equation. This section and the next will make use of material on polynomials from Section A8 of the appendix. The equation to be studied in this section is of the form
+
y(n) an-ly ( n - l ) + . . . + a l y ' + a o y = O with coefficients in C.Let us write this equation as L ( y ) = 0 for a suitable linear operator L defined on functions y of class C n :
The term a0 is understood to act as a0 times the identity operator. Since -$B t r e r t ,we immediately obtain ~ ( e '= ~ (rn )
=
+ an-lrn-l + . . . + alr + ao)ert.
Thc polynomial P ( h ) = hn+anPlhn-l
+...+alh+ao
is called the characteristic polynomial of the equation, and the formula L(ert) = P(r)ert shows that y ( t ) = ert is a solution of L ( y ) = 0 if and only if r is a root of the characteristic polynomial. From Section A8 of the appendix, we know that the polynomial P (A) factors into the product of linear factors h - r , the factors being unique apart from their order. Let us list the distinct roots, i.e., the distinct such complex numbers r , as rl , . . . , rk with k 5 n, and let us write m; for the number of times that h - r; occurs as a factor of P ( h ) , i.e.,the multiplicity of r; k as a root of P . Then we have 'j-j=l V Z ~= FZ and
Corresponding to this factorization of P is a factorization of L as
On the right side the individual factors commute with each other because differentiation commutes with itself and with multiplication by constants. The following lemma therefore produces n solutions of the given equation L ( y ) = 0 .
6. Homogeneous Equations with Constant Coefficients
209
Lemma 4.10. For m > 1 and r in C , all the functions e r t , t e r t , . . . , tm-'ert are solutions of the mth-orderdifferential equation
(2
PROOF. Direct computation gives - r ) " ( t k e r t ) = k ( k - 1 ) . . . (k - m 0 5 k 5 m - 1, and the lemma follows.
(2
-
r ) ( t k e r t )= k t k P 1 e r t ,and hence
+ l ) t k P m e Y t .The right side is 0 if
Lemma 4.11. Let r l , . . . , r ~ be distinct complex numbers, and let mj be N integers > 1. Then the z y = l mi functions
are linearly independent over C PROOF.Let k > 1 be an integer, let r be a complex number, and let P ( t ) be a polynomial of degree 5 k - 1. We allow P ( t ) to be the 0 polynomial. Then
+
L L ( ~p(tt))ertl ~ + = r ( t 9 ~ [ t ) ) e ' ~( ( k - l ) t k 1 dt
+ p'(t))ert,
from which it follows that
with Q ( t ) a polyiloillial of degree 5 k - 1 or the 0 polynomial. We shall prove by induction on N that if P I , . . . , PN are polynomials with complex coefficients such that Pj ( t ) e v is the 0 function, then all the Pj are 0 polynomials. For N = 1, if P ( t ) e r tis the 0 function, then P ( t ) is the0 function. Since a polynomial of degree k > 0 has at most k roots, we conclude that P has all coefficients 0. This disposes of the assertion for N = 1. Assume the result for N - 1, and suppose that we are given' ;that ="$ . c Pj (t)erjt+ PN ( t ) e ' ~is~the 0 function, where { r l ,. . . , r ~ - l r, ~are} distinct. Then
x?l
is the 0 function when qj = rj - r ~ for j 5 N - 1. If PN is the 0 polynomial, the inductive hypothesis shows that all Pj with j 5 N - 1 are 0 polynomials. Otherwise let PN have degree d , and differentiate (**) d 1 times. If Pj ( t ) for j 5 N - 1 is the sum of anjtnjplus lower-degree terms, then (*) shows that the result of the differentiation is that
+
IV. Theory o f Ordinary Differential Equations and Systems N-1
+
(a,,, (qj)d+ltnj lower-degree terms)eqjt j= 1 is the 0 function. By the inductive hypothesis each a,, has to be 0, and hence all coefficients of each Pj have to be 0 for j 5 N - 1. Then PN(t) is identically 0 and must be the 0 polynomial. This completes the induction. If we are given a linear combination of the functions in the statement of thc lcmma that cquals thc 0 function, thcn wc obtain a rclation of thc form 'j$l Pj(t)erjt = 0, and we have just seen that this rclation forces all Pj to be 0 polynomials. This completes the proof.
Proposition 4.12. Let the differential equation y(n) + anPly(n-l)
+ . . . + alyl+ aoy = 0,
with complex coefficients, have characteristic polynomial given by P(h) = n;=l (1- y i j m j with T I , . . . , rk distinct complex numbers and with the integers 0 such that x f = l mj
= n.
Then the n functions
form a basis over C of the space of solutions of the given equation on any interval. PROOF.Lemma 4.10 shows that the functions in question are solutions, Lemma 4.11 shows that they are linearly independent, and Theorem 4.6 shows that the dimension of the space of solutions on any interval is n . Since n linearly independent solutions have been exhibited, they must form a basis of the space of solutions. If the equation in Proposition 4.12 happens to have real coefficients, it is space of real-valued solutions. Since the meaningful to ask for a basis over R of the coefficients are real, we have L ( y ) = L(y) for all complex-valued functions y of class C n,and it follows that the complex conjugate of any complex-valued solution is again a solution. Thus the real and imaginary parts of any complex-valued solution are real-valued solutions. Meanwhile, the characteristic polynomial P of thc cquation has rcal cocfficicnts, and it follows that thc sct of roots of P is closed under complex conjugation. In addition, the multiplicity of a root equals the multiplicity of its complex conjugate. For any integer k > 0 and complex number a + bi with b 0, we have
+
~ t ~ e (+~( ~+ t ~~ ~e ()=~~r t-k~e a~tcos ) ~bt
+ r t k e a tsin bt
Thus tkeatcos bt and tkeatsinbt form a basis over C of the space spanned by tke(a+bi)t and tke(a-bi)t.The functions tkeatcos bt and tkea%inbt are real-valued,
7. Homogeneous Systems with Constant Coefficients
21 1
and thus we obtain a basis over C consisting of the real-valued solutions of the given equation if we retain the solutions tkert with r real and we replace any pair tke(afbi)t and tke(a-bi)tof solutions, b # 0, by the pair tkeatcosbt and tkeatsin bt. Let us see that these resulting functions form a basis over R of the real vector space of real-valued solutions. In fact, we know that they are linearly independent over R because they are linearly independent over C.To see that they span, we take any real-valued solution and expand it as a complex linear combination of these functions. The imaginary part of this expansion exhibits 0 as a linear combination of the given functions, and the coefficients must be 0 by linear independence. Thus the constructed functions form a basis over R of the space of real-valued solutions.
7. Homogeneous Systems with Constant Coefficients
Having discussed linear homogeneous equations with constant coefficients, let us pass to the more general case of first-order homogeneous linear systems with constant coefficients. We write the system as y' = Ay with A an n-by-n matrix of constants. In principle we can solve the system immediately. Namely, Proposition 3.13c tells us that $ (elA)= ~ eso that ~ each~ of the , n columns of etA is a solution of y' = Ay . At t = 0, etAreduces to the identity matrix, and thus these n solutions are linearly independent at t - 0. By Theorem 4.6 these n solutions form a basis of all solutions on any subinterval ( a , b) of (-00, fm). The solution satisfying thc initial condition y (to) = yo is y ( t ) = etAe-tnAyo,which is thc particular cjetAejof the columns of e" in which cj is the number linear combination Cj"=, cj = (ePbAyo) . In practice it is not so obvious how to compute etA except in special cases in which the exponential series can be summed entry by entry. Let us write down three model cases of this kind, and ultimately we shall see that we can handle general A by working suitably with these cases.
MODEL (1) Lct
212
IV. Theory o f Ordinary Differential Equations and Systems
be of size in-by-in with 0's below the main diagonal. Raising C to powers, we see that the (i, j)thentry of is 1 if j = i k and is 0 otherwise. Hence
+
with 0's below the main diagonal. (2) Let
A=
so that A = a1 + C with C as in the previous case. Since a1 and C commute, Proposition 3.13a shows that etA = eatetc. In other words, etA is obtained by multiplying every entry of the matrix etC in the previous case by eat. A matrix of this form A for some complex constant a and for some size m is said to be a Jordan block. Thus we know how to form e t Aif A is a Jordan block. (3) Let A be block diagonal with each block being a Jordan block:
(
block #1
A
block #2 block #k
Then
et block #I t, block #2
et hlock #k
Thus we know how to form etA if A is block diagonal with each block being a Jordan block. A matrix A of this kind is said to be in Jordan form.
7. Homogeneous Systems with Constant Coefficients
213
The theorem reduces any computation of a matrix etAto this case.
Theorem 4.13 (Jordan normal form). For any square matrix A with complex entries, there exists a nonsingular complex matrix B such that B - I A B = J is in Jordan form.
REMARKS.This theorem comes from linear algebra, but knowledge of it is beyond the algebra prerequisites for this book. The proof is long and is not in the spirit of this text, and we shall omit it; however, the interested reader can find a proof in many linear algebra books. As a practical matter, the proof will not give us any additional information, since we already know that etAyields the solutions to y' = Ay and the only remaining question is to convert the statement of the theorem into an explicit method of computation. Let us see what Theorem 4.13 accomplishes. The solution of y' = Ay with y (to)= yo is y ( t ) = e(t-tO)Ayo. Write B-I AB = J as in the proposition. Then Proposition 3.13d gives
If we can compute J , then Model Case 3 above tells us what e(t-to)Jis. If we can curnpule B alsu, lherl we recuvcr y ( t ) cxplicilly. The practical effect is that Theorem 4.13 gives us a method for calculating solutions. The idea behind the method is that the qualitative properties of B and J forced by the theorem are enough to lead us to explicit values of B and J . Let us go through the steps. A concrete example of J is a
1 0 a l O O a O
J
=
It is helpful to know the extent of uniqueness in Theorem 4.13. The matrix J is actually unique up to permuting the order of the Jordan blocks. The matrix B is not at all unique but results from finding bases of certain subspaces of Cn. The
214
IV. Theory of Ordinary Differential Equations and Systems
first step is to form the characteristic polynomial2 P ( k ) = det(h1 - A) of A. We have
= d e t ( ~ ) - ldet(h1 - A) det(B) = det(h1 - A),
and thus J has the same characteristic polynomial as A. The characteristic polynomial of J is just the product of expressions h - d as d runs through the diagonal entries of J . According to Section A8 of the appendix, the factorization of a polynomial with complex coefficients and with leading coefficient 1 into first-degree expressions h - c is unique up to order, and thus the factorization of P(h) tells us the diagonal entries of J . We still need to know the sizes of the individual Jordan blocks. The sizes of the Jordan blocks come from computing dimensions of various null spaces-or kernels, in the terminology of linear functions. If a occurs as a diagonal entry of J , think of forming J - a 1 and its powers, and consider the dimension of the kernel of each power. For example, with the explicit matrix J that is written above, we have
and dim ker(J - a l ) is the number of Jordan blocks of size > 1 with a on the . this case, diagonal, namely 4 in this case. Next we consider (J - a I ) ~In 0 0 1 0 0 0 0 0 0 a l )2 =
l an^
1 nonsingular
books write the characteristic polynomial as det(A - h 1) ,which is the same as the present polynomial if n is even but is its negative if n is odd. The present notation has the advantage that the notions of characteristic polynomial here and in the previous section coincide when an nth-order equation is converted into a first-order system.
7. Homogeneous Systems with Constant Coefficients
215
and dim ker(J - a ~ = ) ~7. This number arises as the sum of the previous number and the number of Jordan blocks of size > 2 with a on the diagonal. Thus dim ker(J - a 1)2- dim ker(J - a 1) in general is the number of Jordan blocks of size > 2 with a on the diagonal. Finally we consider ( J - a l ) 3 . In this case, the upper left part of ( J -a 1)3corresponding to diagonal entry a is all 0, and the lower = 8 This number arises as the right part is nonsingular; hence dim ker(J - a sum of the previous number and the number of Jordan blocks of size 2 3 with a on the diagonal. Thus in general, dimker(J - a1)3 - dimker(J - a1)2 is the number of Jordan blocks of size > 3 with a on the diagonal. In our example, the number dimker(J - a l ) k remains at 8 for all k 3 because 8 is the multiplicity of a as a root of P(L), and we are therefore done with diagonal entry a ; our computation has shown that the numbers of Jordan blocks of sizes 1 , 2 , 3 , 4 ,. . . , are 1 , 2 , 1, 0, . . . ,and a check on the computation is that l(1) + 2(2) + 3(1) = 8. Of course, we do not have J at our disposal for these calculations, but A yields the same numbers. In fact, we have B ( J - a l ) k ~ - = ' (A - a l ) k ,from which we see that x E ker(A - a l)k if and only if B-' x E ker(J - a 1)k. Hence
,
Since B is nonsingular, the dimension of the kernel of ( J dimension of the kernel of (A - a l ) k . Consequently dim ker(A
-
-
a l ) k equals the
a 1) = #{Jordan blocks of size > 1 with a on diagonal),
dim ker(A - u 1)2 - din1ker(A - u 1) = #{Jordan blocks of size
dimker(A
-
> 2 with a on diagonal},
a 113 - dim ker(A - a 112 = #{Jordan blocks of size 2 3 with a on diagonal}, etc
Repeating this argument with the other roots of P(h), we find that we can determine J completely. Calculating B requires working with vectors rather than dimensions. The columns of B are just Bel, . . . , Be,, and we seek a way of finding these. Fix attention on a root a of P (L). Consider an index i with 1 5 i 5 n , and suppose that the diagonal entry of J in column i is a . From the form of J , we see that either the i th column of J - a 1 is O or else it is ei-1. In the latter case, index i - 1 corresponds to the same Jordan block. Using the identity (A-a 1)B = B( J -a 1) , we see that either
216
IV. Theory o f Ordinary Differential Equations and Systems
and index i - 1 corresponds to the same Jordan block as index i in the latter case. Thus the vectors Be, corresponding to the columns with diagonal entry a and with smallest index for a Jordan block lie in ker(A - a 1). They are linearly independent since B is nonsingular, and the number of them is the number of Jordan blocks corresponding to diagonal entry a . We saw that this number equals dimker(A - 01) Hence the vectors Be, corresponding to the smallest indices going with each Jordan block form a basis of ker(A - a l ) . Similarly
and index i - 2 corresponds to the sane Jordan block as index i in the latter case. Thus the vectors Bei corresponding to the columns with diagonal entry a and with smallest or next smallest index for a Jordan block lie in ker(A - a 112. They are linearly independent since B is nonsingular, and the number of them is the sum of the previously computed number, namely dim ker(A - a 1), plus the number of Jordan blocks of size 2 that correspond to diagonal entry a . We saw that this sum equals dim ker(A - a I ) ~ Hence . the vectors Bei corresponding to the two smallest indices going with each Jordan block form a basis of ker(A - a l l 2 . The new vectors Bei are therefore vectors that we adjoin to a basis of ker(A - a l ) to obtain a basis of ker(A - a ~ ) ~ . In setting up these vectors properly, however, we have to correlate the indices studied at the previous step with those being studied now. The relevant formula is that the new indices i have the property (A - a 1)Bei = Bei-1. To obtain vectors with this consistency property, we would take a basis S1 of ker(A - a 1) , extend it to a basis S2 of ker(A - a 1)2,discard the members of S1, apply A - a 1 to the members of S2 - S1,and extend (A - a l)(S2 - S1) to a basis TI of ker(A - a 1). Then Sk = (S2 - S1) U TI is a new basis of ker(A - all2. We can continue the argument in this way. It is perhaps helpful to read the general discussion of the argument side by side with the explicit example that appears below. We continue to find that the construction of new basis vectors gets in the way of the necessary consistency property with the earlier basis vectors. Thus we really must start with the largest index k such that ker(A - a l ) k # ker(A - ~ 1 ) ~ We ~ 'extend . a basis Sk-1 uf ker(A - ~ 1 ) Lu~a basis ~ ' Sk uf ker(A - a 1)k ,and form
These vectors will be the columns of B corresponding to the largest Jordan blocks with diagonal entry a . The vectors in
7. Homogeneous Systems with Constant Coefficients
217
are linearly independent in ker(A - al)k-2; we extend this set to a basis SLP2of ker(A - a l l k p 2 ,and we extend SL-, U (A - a l ) ( S k - SkPl)to a basis SL-, of ker(A -a l ) k - l . The adjoined vectors, together with the result of applying powers of A - a 1 to them, will be the columns of B corresponding to the next largest Jordan blocks with diagonal entry a . The process continues until we obtain a basis of kcr(A a l l k with thc ncccssary consistcncy propcrty throughout. Thcn we repeat the process for the other roots of P(h) and assemble the result. -
EXAMPLE. Let
4
1 -1 -2 2). 2 -2 The characteristic polynomial is P (A) = det(h 1-A) = h3,whose factorization is . the kernel of A, we find that dim ker A = evidently P (A) = (h o ) ~Computing 2, so that there are 2 Jordan blocks. Also, A2 = 0, SO that dim ker A2 = 3 and the number of blocks of size > 2 is 3 - 2 = 1 . Thus A=(-;
-
We form a basis of ker A by solving A reduction gives
(i )
XI =
ker A consists of
-ix:,
i:: x2
= 0. The standard method of row
+ i x ? with x? and x3 arbitrary, so that a basis of
and
(i)
. We extend this to a basis of ker A2 =
by adjoining, for example, the vector u =
(
.
Then Aul =
(- )
c3
The
vector Avl is in ker A, and we extend it to a basis of ker A by adjoining, for example, v2 =
(-6)
4 . Then vl, Avl, v2 form a basis of ker
=
c3,and the
above general method asks that these vectors be listed in the order Aul, v l , The matrix B is obtained by lining these vectors up as columns: / 4 1 -l\
The result is easy to check. Computation shows that Bpl = and then one can carry out the multiplications to verify that B-' AB = J .
v2.
218
IV. Theory o f Ordinary Differential Equations and Systems
8. Series Solutions in the Second-Order Linear Case In this section we shall consider, in some detail, series solutions for two kinds of ordinary differential equations. The first kind is y" P(t)yf Q ( t ) y = 0,
+
+
where P ( t ) and Q ( t ) are given by convergent power-series expansions for It1 < R:
We seek power-series solutions of the form
The same methods and theorem that handle this first kind of equation apply also to nth-orderhomogeneous linear equations and to first-order homogeneous systems when the leading coefficient is 1 and the other coefficients are given by convergent power series. The second-order case, however, is by far the most important for applications and is sufficiently illustrative that we shall limit our attention to it. The idea in finding the solutions is to assume that we have a convergent powerseries solution y ( t ) as above, to substitute the series into the equation, and to sort out the conditions that are imposed on the unknown coefficients. Our theorems on power series in Section I 7 guarantee us that the operations of differentiation and multiplication of power series maintain convergence, and thus the result of substituting into the equation is that we obtain an equality of a convergent power series with 0. Corollary 1.39 then shows that all the coefficients of this last power series must be 0, and we obtain recursive equations for the unknown coefficients. There is one theorem about the equations under study, and it tells us that the power series for y ( t ) that we obtain by these manipulations is indeed convergent; we state and prove this theorem shortly. Let us go through the steps of finding the solutions. These steps turn out to be clearer when done in complete generality than when done for an example. Thus we shall first make the computation in complete generality, then state and prove the theorem, and finally consider an important example. The expansions of y ( t ) and its derivatives are
8. Series Solutions in the Second-Order Linear Case
Substituting all the series into the given equation yields
If the series for y ( t ) converges and if the left side is expanded out, then the coefficients of each power o f t must be 0. Thus
These equations tell us that co and cl are arbitrary and that c2, c3, . . . are each determined by the previous coefficients. Thus c2, cs, . . . may be computed inductively. Since co = y(0) and cl = y f ( 0 ) ,this degree of flexibility is consistent with the existence and uniqueness theorems.
Theorem 4.14. If P ( t ) and Q ( t ) are given by convergent power series for It 1 < R , then any formal power series that satisfies y" P ( t ) y f Q ( t )y = 0 converges for It I < R to a solution. Consequently every solution of this equation on the interval -R < t < R is given by a power series convergent for It 1 < R .
+
+
PROOF.Fix r with 0 < r < R , and choose some R1 with r < R1 < R . Let the notation for the power series of P , Q , and y be as above. Theorem 1.37 shows that the series with terms la,Rr 1 and I b,R; 1 are convergent, and hence the terms are bounded as functions of n . Thus there exists a real number C such that la, 1 5 C / R y and Ib, 1 5 C / R y for all n > 0 . We shall show that lc, 1 5 M / r n for a suitablc M and all n > 0. The constant M will be fixed so that a large initial number of terms have Ic,,I 5 M / r n , and then we shall see that all subsequent terms satisfy the same inequality. To find an M that works, we start from the formula computed above for c:,
220
IV. Theory o f Ordinary Differential Equations and Systems
If M works for 0 , 1 , . . . , n
-
1 , then
and therefore
For n sufficiently large, the factor in parentheses is 5 1. At that point we obtain Ic, I 5 M r P n if IckI 5 M r P kfork i n , and induction yields the asserted estimate. Thus C cntn converges for It1 < r . Since r can be arbitrarily close to R , C cntn converges for It 1 < R . Finally we saw above that co and cl are arbitrary and can therefore be matched to any initial data for y ( 0 ) and y f ( 0 ). Consequently the vector space of powerseries solutions convergent for It1 < K has dimension 2. By Theorem 4.6, all solutions on the interval -R < t < R are accounted for. This completes the proof. As a practical matter, the recursive expression for c, becomes increasingly complicated as n increases, and a closed-form expression need not be available. However, in certain cases, something special happens that yields a closed-form expression for c, . Here is an example.
EXAMPLE. Legendre's equation is
with p a complex constant. To apply the theorem literally, we should first divide the equation by (1 - t 2 ) ,and then the power-series expansions of the coefficients will be convergent for It 1 < 1 . The theorem says that we obtain two linearly independent power-series solutions of the equation for It I i1. To compute them, it is more convenient to work with the equation without making the preliminary division. Then the equation gives us
8. Series Solutions in the Second-Order Linear Case
which yields the following formulas for the coefficients:
n(n - l)c,
-
[ ( n- 2 ) ( n - 3 )
+ 2(n
-
2) - p(p
+ l)]cn-2 = 0.
Thus we can write c , explicitly as a product. We can verify convergence of C cntn directly by the ratio test: since
we have convergence for It1 < 1. Observe that the numerator in the fraction on the right is equal to
and this is 0 when p is an integer 10 and n - 2 = p . Therefore one of the solutions is a polynomial of degree p if p is an integer > 0. Such polynomials, when suitably normalized, are called Legendre polynomials. The second kind of ordinary differential equation for which we shall seek series solutions is t2yf tP(t)yf Q(t)y= 0,
+
+
where P ( t ) and Q ( t ) are given by convergent power-series expansions for It1 < R:
The existence and uniqueness theorems do not apply to this equation on an interval containing t = 0 unless t happens to divide P ( t ) and t 2 happens to divide Q ( t ) . When this divisibility does not occur, the above equation is said to have a regular singular point at t = 0. The treatment of the corresponding nth-order equation is no different, but we stick to the second-order case because of its relative importance in applications. For this kind of equation, the treatment of first-order systems is more complicated than the treatment of a single equation of nth order.
222
IV. Theory o f Ordinary Differential Equations and Systems
Actually, the second-order equation above need not have power series solutions. The prototype for the above equation is the equation
with P and Q constant. This equation is known as Euler's equation and can be solved ill teilils of elemeiita~yfunctions. 111 fact, we iiiake a cl~angeof variables by putting t = e x and x = log t for t > 0. Then we obtain dy dt
dy d x d x dt
1dy t dx
- --- -- -
and
and hence the equation becomes
This is an equation of the kind considered in Section 6. A solution is est ,where s is a root of the characteristic polynorrlial s2 + (P - 1)s + Q = 0. If the two roots of the characteristic polynomial are distinct, we obtain two linearly independent solutions for x E (-oo, +m), and these transform back to two solutions t S of the Euler equation for t > 0. If the characteristic equation has one root s of multiplicity 2 , then we obtain the two linearly independent solutions eSxand xeSx for x E (-oo,+ a ) , and these transform back to two solutions x S and x S log x forx > 0. In practice, the technique to solve the Euler equation tzy" + tPy' + Qy = 0 is to substitute y ( t ) = t S and obtain s ( s - l ) t s + s P t S + Q t s = 0. This equation holds if and only if s satisfies
which is called the indicia1 equation. In h e gerleral Lase uf a regular singular puinl, we pru~cedby arralvgy and are led to seek for t > 0 a series solution of the form
+ + czt2 + . . . )
y ( t ) = tS(co clt
with co # 0.
Suppose that the power-series part C c,tn is convergent. We substitute and obtain
8. Series Solutions in the Second-Order Linear Case
223
Dividing by t S and setting the coefficient of each power of t equal to 0 gives the equations
+ + cob0 = 0 , + ( ( s + 1)clao + scoal) + (clbo + cob11 = 0, C ~ ( S+ 2 ) ( s + 1) + ((s +2)c2ao + .. .) + (czbo + . . .) = 0, cos(s - 1 ) c l ( s 1)s
SCOQ
+
Since co is by assumption nonzero, we can divide the first equation by it, and we obtain s ( s - 1) aos bo = 0 ,
+
+
which is the indicial equation for t2y' + t P ( t )y' + Q ( t )y = 0. This determines the exponent s . Then co is arbitrary, and all subsequent c,'s can be found equation above is recursively, provided the coefficient of c, in the ( n + never 0 for n > 1, i.e., provided
+
In othcr words, wc can solvc rccursivcly for all c, in tcrms of co providcd s n does not satisfy the indicial equation for any n > 1. We summarize as follows.
Proposition 4.15. If P ( t ) and Q ( t ) are given by convergent power series for It1 iR, then the following can be said about formal series solutions of t2?" t P ( t ) y f Q ( t ) y = O of the type t S ( c o+ c l t c2t2 .. .) with co # 0:
+
+
+
+
(a) If the indicial equation has distinct roots not differing by an integer, then there are formal solutions of the type xS(co clt c2t2 . . . ) for each root .F of the indicial equation. (b) If the indicial equation has roots rl 5 r2 with r2 - rl equal to an integer, then there is a 1-parameter family of formal solutions of the type tT2(cO c l t c2t2 . . .) with co # 0. If r1 i~2 in addition, there may be formal solutions trl(co c l t c2t2 + . . . ) with cg # 0, as there are for an Euler equation.
+ +
+ +
+
+
+ +
Theorem 4.16. If P ( t ) and Q ( t ) are given by convergent power series for It 1 < R , then all formal series solutions of t2?" t P ( t )y' Q ( t ) y = 0 of the type t S ( c o clt c2t2 . . .) with co # 0 converge for 0 it < R to a function that is a solution for 0 < t < R.
+ +
+
+
+
224
IV. Theory o f Ordinary Differential Equations and Systems
PROOF.As in the proof of Theorem 4.14, fix r with 0 < r iR , and choose some R1 with r < R1 < R . Let the series expansions of P i t ) and Q ( t ) be as above, so that there is a number C with lan1 5 C / R ; and I bn1 5 CIR;". Choose N large enough so that
for n 2 N . Then choose M such that lcnl 5 M / r n for n 5 N . We shall prove by induction on n that lcnl 5 M / r n for all n . The base case of the induction is 12 = N , whcrc thc incquality holds by dcfinition of M . Supposc it holds for 1, . . . , n - 1. The formula for cn is
Our inductive hypothesis gives
Thus
the second inequality holding by (*), and the induction is complete. It follows that C cntn converges for It 1 < r . Since r can be arbitrarily close to R , C cntnconverges for It 1 < R . This completes the proof. EXAMPLE. Bessel's equation of order p with p 2 0. This is the equation
It has P ( t ) = 1 and Q ( t ) = t 2 - p 2 , both with infinite radius of convergence. The indicia1 equation in general is s ( s - 1 ) + aos + bo = 0 and hence is
in this case. Thus s = * p . Theorem 4.16 shows that there is a solution of the form
226
IV. Theory o f Ordinary Differential Equations and Systems
For the case that s = - p , we obtain
and the denominator gives a problem for n = 2 p and for no other value of n . If p is an integer, the problematic n is even and we must have c,-2 = 0, c,-4 = 0 , . . . , co = 0. The condition co = 0 is a contradiction, and we conclude that there is no solution of the form t-P times a nonzero power series; indeed, Problems 18-19 at the end of the chapter will identify a different kind of solution. If p is a half integer but not an integer, then the problematic n is odd, and we are led to conclude that 0 = cn-2 = . . . = cs = e l , with co and c2+, arbitrary. There is no contradiction, and we obtain a solution of the form t-P times a nonzero power series.
9. Problems 1. For the differential equation yy' - -t: (a) Solve the equation. (b) Find all points (to,yo) where the the existei~cetheorem and the ui~iqueness theorem of Section 2 do not apply. (c) For each point (to, yo) not in (b), give a solution y ( t ) with y (to) = yo. 2. Prove that the equation y' = t + y2 has a solution satisfying the initial condition y(0) = 0 and defined for It1 < 1/2.
& ++ &
. 3. In classicalnotation, a particular vector field in the plane is given by & Find a parametric realization of an integral curve for this vector field passing through ( 1 , 1).
5 . Find all solutions on
(-GO,
+GO) to
y"
-
3y'
+ 2y = 4.
6. (a) For each of these matrices A, find matrices B and J , with J in Jordan form, 0 0 -1 such that A = B J : (b) For each of the matrices A in (a), find a basis of solutions y ( t ) to the system of differential equations y' = Ay .
9. Problems
227
7 . The nth-orderequation y'n'+annlY'nnl'+. . .+soy leads to a linear system z f = Az with
Prove that det(h 1 A) by cofactors. -
8.
= hn
+a
= 0 with constant coefficients
+ +
, h n l . . . ag by expanding the determinant
(a) Let { f,} be a uniformly bounded sequence of Riemann integrable functions on [0, 11. Define F,(t) = $ f,(s) d s . Prove that {F,}is an equicontinuous family of functions on [0, 11. (b) Prove that the set of functions y ( t ) on [0, 11 with yf' y = f ( t ) and y (0) = y f ( 0 ) = 0 is equicontinuous as f varies over the set of continuous functions on [0, 11 with 0 5 f ( t ) 5 1 for all t . (c) Let u ( t ) be continuous on [ a ,b]. Prove that the set of functions y ( t ) on [a,b] with y" q ( t ) y = f ( t ) and y (0) = yf(0) = 0 is equicontinuous as f ( t )varies over the set of continuous functions on [0, 11 with 0 5 f ( t ) 5 1 for all t .
+
+
+
9. The differential equation t2yf' + (3t - l ) y f y = 0 has an irregular singular point at t = 0 . (a) Verily Lllal ~ ~ ~ o ( n is! )a tlurrrral " puwcr series sululivr~u l lllc eyualiur~ even though the power series has radius of convergence 0 . (b) Verify that y ( t ) = t-'p-'l?is a solution for t > 0. Problems 10-13 concern harmonic functions in the open unit disk, which were introduced in Problems 1 4 1 5 at the end of Chapter 111. The first objective here is to use ordinary differential equations and Fourier series to show that all these functions may be expressed in a relatively simple form. The second objective is to use convolution, as defined in Problem 8 at the end of Chapter 111, to relate this formula to the Poisson kernel, which was defined in Problems 27-29 at the end of Chapter I. Problems 10-12 here are an instance of the method of separation of variables, a beginning technique with partial differential equations; this topic is developed further in the companion volume Advanced Real Analysis. In all problems in this set, let u ( x , y ) be harmonic in the open unit disk.
10. Write u ( x , y ) in polar coordinates as u(r cos Q , r sin Q ) = u(r, Q ). Using Fourier series, show for 0 5 r < 1 and any S > 0 that v(r, Q ) is the sum of an absolutely convergent Fourier series c, (r)einOwith Ic,(r) 1 5 ~ / forn0 5~r 5 1 - S for some M depending on 6.
x,"=-,
IV. Theory o f Ordinary Differential Equations and Systems
228
1 1. Let Re be the rotation matrix defined in Problem 15 at the end of Chapter 111. That problem shows that ( u o R,) ( x , y ) = v ( r , Q p ) is harmonic for each p. Prove that J:nn ( u o R q )( x , y)epik, d p is harmonic and is given in polar coordinates by ck(r)eike.
+
&
12. By computing with the Laplacian in polar coordinates and showing that ck(r) is bounded as r J, 0 , prove that c k ( r ) = a k r k for some complex constant ak. Conclude that every harmonic function in the open unit disk is of the form v(r, 0) = CZpm c n r n e i n Bthe , sum being absolutely convergent for all r with O s r < 1. 13. Deduce from Problem 8 at the end of Chapter I11 that if v(r, 8 ) is as in the previous problem and if 0 < R < 1, then u (r, Q ) = f R ( p )PrIR( Q - p ) d p for 0 5 r < R , where P is the Poisson kernel and f R is the C m function f R ( 8 ) = Cr=pm cn~ ~ e ~ ~ ~ .
& I:"_
Problems 14-17 concern homogeneous linear differential equations. Except for the first of the problems, each works with a substitution in a second-order equation that simplifies the equation in some way. 14. If a ( t ) is continuous on an interval and A ( t ) is an indefinite integral, verify that all solutions of the single first-order linear homogeneous equation y' = a ( t ) y are of the form y ( t ) = ce A(" ).
15. (a) Suppose that u ( t ) is a nowhere vanishing solution of
on an interval, with P and Q assumed continuous. Look for a solution of the form u ( t ) v( t ) ,and derive the necessary and sufficient condition
(b) For y" - ty' solution.
+
-
y = 0 , one solution is e"/2. Find a linearly independent
+
16. Let yff P ( t ) y f Q ( t )y = 0 be given with P, P', and Q continuous on an interval. Write y ( t ) = u ( t ) v ( t ) ,substitute, regard u ( t ) as known, and obtain a second-order equation for v . Show how to choose u ( t ) to make the coefficient of u' be 0 , and thus reduce the given equation to an equation v" R (t)v = 0 with R continuous. Give a formula for R .
+
+
17. If L ( v ) = (pu')' - q v h r v , show that the substitution u = v f i changes L ( v ) = 0 into L o ( u ) = 0, where L o ( u ) = (p*uf)'- q*u hu with p* = p / r .
+
Problems 18-19 concern finding the form of the second solution to a second-order equation with a regular singular point. The first of the two problems amounts to a result in complex analysis but requires nothing beyond Chapter I of this book.
9. Problems
229
18. Suppose that CZoc,xn is a power series with co = 1. (a) Write down recursive formulas for the coefficients d, of a power series CEOdnxnwith dg = 1 such that ( C z Oc,xn) ( Cr=O d,xn) = 1 . (b) Prove, by induction on n , that if lc, 1 5 Mrn for all n > 0 , then Id, 1 5 M(M l ) n p ' r nfor all n > 1. (c) Prove that if f ( 0 ) f 0 and if f ( x ) is the sum of a convergent power series for Ix 1 < R for some R > 0 , then 1/ f ( x ) is the sum of a convergent power series for x 1 < s for sollle s > 0 .
+
19. Suppose that P ( t ) and Q ( t ) are given near t = 0 by power series with positive radii of convergence. Take for granted that if a ( t )is given by a power series with a positive radius of convergence, then so is ea(*) . Form the equation
let sl and s2 be the two roots of the indicia1 equation, and suppose that the differential equation has a solution given on some interval (0,s ) by f ( t ) = t S 1CEOcntnwith co # 0 . (a) Using Problem 15a, prove that the differential equation has a linearly independent solution given on some interval (0, s f ) by
g ( t ) = cf ( t )log t
+ tS2
00
with ko # 0.
kntn
n =O
(b) Prove that the coefficient c in g ( t ) is f 0 if sl = s2. (c) For Bessel's equation t"" ty' (t" -')y = 0 with p > 0 an integer and with sl = p and $2 = - p , show that the coefficient c in g ( t ) is # 0 . Thus there is a solution of the form J, ( t )log t tpP(power series) on some interval (0, s f ) .
+ +
+
Problems 20-25 prove the Cauchy-Peano Existence Theorem, that a local solution in Theorem 4.1 to y' = F ( t , y) and y(to) = yo exists if F is continuous even if F does not satisfy a Lipschitz condition. The idea is to construct a sequence of polygonal approximations to solutions, check that they form an equicontinuous family, apply Ascoli's Theorem (Theorem 2.56) to extract a uniformly convergent subsequence, and then see that the limit of the subsequence is a solution. A member 0 . With of the sequence of polygonal approximations depends on a number 6 notation as in the statement of Theorem 4.1, the construction for [to,to + a'] is as follows: Choose the S of uniform continuity for F and c on the set R . Fix a partition to < tl < . . . < t, = to + a f of [to,to + a t ] with maxk{tk- t k p l }< min(S,6/M). Define y ( t ) ,as a function of c , for t k 1 < t < t k inductively on k by y (to)= yo and
230
IV. Theory o f Ordinary Differential Equations and Systems
20. Check that the formula for y ( t ) when tk-1 < t 5 tk remains valid when t = tk-1, and conclude that y ( t ) is continuous. Then prove by induction on k that I y ( t ) - y(to)l 5 M ( t - to) < b for tk-1 < t < tk, and deduce that ( t , y ( t ) ) is in R' for to I t I to a'.
+
21. Provethat l y ( t )
-
y(tf)l
< M l t -t'l
i f t andt' arebothin[to,to + a f ] .
+
22. lhe function y f ( t ) is defined on [to, to a'] except at the points of the partition and is ;iven by y ' ( t ) = F(tk-1, y(tk-1)) if tk-1 < t < tk. Prove that y ( t ) = yo y f ( s ) d s for to < t < to a' and that l y f ( s ) - F ( s , y (s))l < c if tk-1 < S < tk.
+
+&
23. Writing y ( t ) = yo + J : [ F ( S , y ( s ) ) + [ y f ( s )- F ( s , y ( s ) ) ] ] d s andusing theresult of the previous problem, prove for all t in [to, to + a'] that
24. Let c, be a monotone decreasing sequence with limit 0, and let y, ( t ) be a function
+
a'] constructed as above for the number c,. Deduce from for t in [ t o , to Problem 21 that {y, ( t ) } is uniformly bounded and uniformly equicontinuous for t in [to, to a ' ] .
+
25. Apply Ascoli's Theorem to {y,}, and let y ( t ) be the uniform limit of a uniformly coilvergei~tsubsequei~ceof {y,}. Prove that y ( t ) is continuous, and use Problem 23 to prove that y ( t ) = yo $ F (s, y ( s ) ) ds . What modifications are needed to the argument to handle [to - a', to]'!
+
CHAPTER V Lebesgue Measure and Abstract Measure Theory
Abstract. This chapter develops the basic theory of measure and integration, including Lebesgue measure and Lebesgue integration for the line. Section 1 introduces mearures, including 1-dimensional 1.ehesgue measure as the primary example, and develops simple properties of them. Sections 2-4 introduce measurable functions and the Lebesgue integral and go on to establish some easy properties of integration and the fundamental theorems about how Lebesgue integration behaves under limit opemtions. Sections 5-6 concern the Extension Theorem announced in Section 1 and used as the final step in the construction of Lebesgue measure. The theorem allows o-finite measures to be extended from algebras of sets to a-algebras. The theorem is proved in Section 5, and the completion of a measure space is defined in Section 6 and related to the proof of the Extension Theorem. Section 7 treats Fubini's Theorem, which allows interchange of orde~.of integration unde~.i.ather general circumstances. This is a deep result. As part of the proof, product measure is constructed and important measurability conditions are established. This section mentions that Fubini's Theorem will be applicable to higher-dimensional Lebesgue measure, but the details are deferred to Chapter VI. Section 8 extends Lebesgue integration to complex-valued functions and to functions with values in Gniic-dimensional vecior spaces. Section 9 gives a careful definition of the spaces L 1 , L 2 , and Lm for any measure space, introduces the notion of a normed linear space, and verifies that these three spaces are examples. The main theorem of the section about L1, L 2 , and Lm is the completeness of these three spaces as metric spaces. In addition, the section proves a version of Alaoglu's Theorem concerning weak-star convcrgcncc.
1. Measures and Examples In the theory of the Riemann integral, as discussed in Chapter I for R1 and in Chapter I11 for Rn, we saw that Riemann integration is a powerful tool when applied to continuous functions. Riemann integration makes sense also when applied to certain kinds of discontinuous functions, but then the theory has some weaknesses. Without any change in the definitions, one of these is that the theory applies 1 only to bounded functions. Thus we can compute xP dx = [xpf l / ( p + I)]; = (p I)-' for p 2 0, but only the right side makes sense for - 1 < p < 0. More seriously we made calculations with trigonometric series in Section I.10 and found oo cosn0 oo sin n0 that ;log (-) 1 = En=l 7 and i(n - 8 ) = En=l 7 for0 < 8 < 2n.
+
Jb
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V Lebesgue Measure and Abstract Measure Theory
When we tried to explain these similar-looking identities with Fourier series, we were able to handle the second one because (n- 8) is a bounded function, but is unbounded. we were not able to handle the first one because log (-) Other weaknesses appeared in Chapters I-IV at certain times: when we always had to arrange for the set of integration to be bounded, when we had no clue which scqucnccs { c n }of Fouricr cocfficicnts occurrcd in thc beautiful formula givcn by Parseval's Theorem, when Fubini's Theorem turned out to be awkward to apply Lu discurilir~uuusfuncliu~is,arld wlicri llie clrangc-uf-variables furrnula did rivl immediately yield the desired identities even in simple cases like the change from Cartesian coordinates to polar coordinates. The Lebesgue integral will solve all these difficulties when formed with respect to "Lebesgue measure" in the setting of Rn. In addition, the Lebesgue integral will be meaningful in other settings. For example, the Lebesgue integral will be meaningful on the unit sphere in Euclidean space, while the Riemann integral would always require a choice of coordinates. The Lebesgue integral will be meaningful also in other situations where we can take advantage of some action by a group (such as a rotation group) that is difficult to handle when the setting has to be Euclidean. And the Lebesgue integral will enable us to provide a rigorous foundation for the theory of probability. There are five ingredients in Lebesgue integration, and these will be introduced in Sections 1-3 of this chapter: (i) an underlying nonempty set, such as R' in the case of Lebesgue integration on the line, (ii) a distinguished class of subsets, called the "measurable sets,'' which will form a "o-ringn or a "o -algebra," (iii) a measure, which attaches a member of [0, +oo] to each measurable set and which will be "length" in the case of Lebesgue measure on the line, (iv) the "measurable functions," those functions with values in R (or some more general space) that we try to integrate, (v) the integral of a measurable function over a measurable set. Let us write X for the underlying nonempty set. The important thing about whatever sets are measurable will be that certain simple set-theoretic operations lead from measurable sets to measurable sets. The two main definitions are those of an "algebra" of sets and a "o-algebra," but we shall refer also to the notions of a "ring" of sets and a "o--ringn in order to simplify certain technical problems in constructing measures. An algebra of sets A is a set of subsets of X containing 0 and X and closed under the operation of forming the union E U F of two sets and under taking the complement Ec of a set. An algebra is necessarily closed under intersection E f' F and difference E - F = E n Fc. Another operation under which A i s closed is symmetric difference,which is defined by
1. Measures and Examples
233
E A F = (E - F ) U ( F - E ) ; we shall make extensive use of this operation1 in Section 6 of this chapter. In practice, despite the effort often needed to define an interesting measure on the sets in an algebra, the closure properties2 of the algebra are insufficient to deal with questions about limits. For this reason one defines a o-algebra of subsets of X to be an algebra that is closed under countable unions (and hence also countable intersections). Typically a general foundational theorem (Theorem 5.5 below) is used to extend the constructed would-be measure from an algebra to a o-algebra. A ring R of subsets of X is a set of subsets closed under finite unions and under difference. Then R is closed also under the operations of finite intersections, difference, and symmetric differen~e.~ A o-ring of subsets of X is a ring of subsets that is closed under countable unions. (1) A = {m, X}. This is a o-algebra. (2) All subsets of X . This is a o-algebra. (3) All finite subsets of X . This is a ring. If the complements of such sets are included, the result is an algebra. (4) All finite and countably infinite subsets of X . This is a o-ring. If the complements of such sets are included, the result is a 0-algebra. (5) All elementary sets of R. These are all finite disjoint unions of bounded intervals in R with or without endpoints. This collection is a ring. To see the closure properties, we first verify that any finite union of bounded intervals is a finite disjoint union; in fact, if 11, . . . , I, are bounded intervals such that none contains any of the others, then Ik - uL1ll I, is an interval, and these intervals are disjoint as k varies; also these intervals have the same union as 11,. . . , I,. Now let E = U i ii and F = U j Jj be given. Since Ii fl Jj is an interval, the identity E n F = Ui,j(Ii n J j ) shows that E n F is a finite union of intervals. Since each Ii - Jj is an interval or the union of two intervals, the ( I i - J i ) then shows that E - F is a finite union of identity E - F = Ui intervals. (6) If C is an arbitrary class of subsets of X , then there is a unique smallest algebra A of subsets of X containing C. Similar statements apply to o-algebras,
nj
or
some properties of symmetric difference, see Problem 1 at the end of the chapter. ' ~ nalgebra of sets really is an algebra in the sense of the discussion of algebras with the Stone-Weierstrass Theorem (Theorem 2.58) The scalars replacing R or C are the members of the two-element field {O, I ] , addition is given by symmetric difference, and multiplication is given by intersection. The additive identity is 0, the multiplicative identity is X , and every element is its own negative. Multiplication is commutative. 3~ ring of sets really is a ring in the sense of modem algebra; addition is given by symmetric difference, and multiplication is given by intersection.
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V Lebesgue Measure and Abstract Measure Theory
rings, and o-rings. In fact, consider all algebras of subsets of X containing C. Example 2 shows that there is one. Let A be the intersection of all these algebras, i.e., the set of all subsets that occur in each of these algebras. If two sets occur in A, they occur in each such algebra, and their intersection is in each algebra. Hence their intersection is in A. Similarly A is closed under differences. If R is a ring of subsets of X , a set function is a function p : R + R*,where R*denotes the extended real-number system as in Section 1.1. The set function is nonnegative if its values are all in [O, fool, it is additive if p(M) = 0 and if p ( E U F ) = p (E) + p ( F ) whenever E and F are disjoint sets in R , and it is completely additive or countably additive if p (M) = 0 and if p(Ur==lEn) = C,"==, p(En) whenever the sets En are pairwise disjoint members of R with U,"=l E n in R. In the definitions of "additive" and "completely additive," it is taken as part of the definition that the sums in question are to be well defined in R*. Observe that completely additive implies additive, since p ( 9 ) = 0.
Proposition5.1. An additive set function p on aring R of sets has the following properties: (a) p ( U : En) = c=:~ p (En) if the sets En are pairwise disjoint and are in R. (b) p ( E I L J F ) + p ( E n F ) = p ( E ) p ( F ) if E and F are in 77.. (c) I f E and F a r e i n R a n d Ip(E)I i+m,then Ip(E n F ) I i+ m . (d) If E and F are in R and if Ip(E n F)I i +oo, then p ( E - F ) = P(E) - p ( E n F ) . (c) If p is nonncgativc and if E and F arc in R with E 5 F , thcn p (E) 5 p (F). ( f ) If p is nonnegative and if E , E l , . . . , E N are sets in R such that E 5 ~ f = En,then l P(E) i C;=l p(En). (g) If p is nonnegative and completely additive and if E , E l , E2, . . . are sets in R such that E 5 Ur=l E n ,then p (E) 5 C r = l p (En). PROOF.Part (a) follows by induction frorn the definition. In (b), we have E I L J F = ( E - F ) I L J ( E n F ) I L J ( F - E ) disjointly. Application of (a) gives p ( E U F ) = p ( E - F ) + p ( E n F ) + p ( F - E),with+ooand-mnotboth occurring. Adding p (E n F) to both sides, regrouping terms, and taking into accountthatp(E) = p ( E - F ) + p ( E n F ) a n d p ( F ) = p ( F - E ) + p ( E n F ) , wc obtain (b). Thc right sidc of thc idcntity p (E) = p ( E f' F) p (E - F) cannot be well defined if p ( E ) is finite and p ( E n F ) is infinite, and thus (c) follows. In the identity p (E) = p ( E n F ) + p ( E - F ) ,we can subtract p ( E n F ) from both sides and obtain (d) if p ( E n F ) is finite. For (e), the inclusion E 5 F forces F = ( F - E ) U E disjointly; then p ( F ) = p (F - E ) p (E) , and (e) follows. In (0, put Fn = En - U~I: E k . Then E = U:=l ( E f' Fn) disjointly, and (a) and
+
+
+
1 . Measures and Examples
(e) give p(E) = C:=l P(E f' Fn) i (g) is proved in the same way as (f).
N
235
p(F,) i Cy=l p(E,). Conclusion
Proposition 5.2. Let p be an additive set function on a ring R of sets. If p is completely additive, then p(E) = limp(E,) whenever {En}is an increasing sequence of members of R with union E in R Conversely if p (E) = lim p (En) for all such sequences, then p is completely additive. PROOF.First we prove the direct part of the proposition. For E and En as in the statement, let F1 = E l and Fn = En - EnP1for n ? 2. Then E n = Ui=, Fk disjointly, and p(E,) = C i = l p (Fk) by additivity. Also, E = Fk, and complete additivity gives p (E) = C,"==, p (Fk) = lim C:=, p (Fk) = lim p (En). The direct part of the proposition follows. For the converse let {F,} be a disjoint sequence in R with union F in R.Put En = Ui=l Fk.Then E, is an increasing sequence of sets in R with union F in R. We are given that p ( F ) = limp (E,), and we have p ( E n ) = xi=, p (Fk) by additivity and Proposition 5.la. Therefore p (F) = CEl p(Fk), and we conclude that p is completely additive.
UEl
Corollary 5 3 . Let p be an additive set furlctiorl on an algebra A of subsets of X such that I p (X) I < +oo . If p is completely additive, then p (E) = lim p (En) whenever {En}is a decreasing sequence of members of A with intersection E in A. Conversely if limp (E,,) = 0 whenever {E,,} is a decreasing sequence of members of A with intersection empty, then p is completely additive. PROOF. This follows from Proposition 5.2 by taking complements. A measure is a nonnegative completely additive set function on a o-ring of subsets of X. If no ambiguity is possible about the o-ring, we may refer to a "measure on X ." When we use measures to work with integrals, the o-ring will be taken to be a o-algebra; if integration were to be defined relative to a o-ring that is not a o-algebra, then nonzero constant functions would not be measurable. The assumption that our o-ring is a a-algebra for doing integration is no loss of generality. Even when the o-ring is not a o-algebra, there is a canonical way of extending a measure from a 0-ring to the smallest 0-algebra containing the o -ring. Proposition 5.37 at the end of Section 5 gives the details.
EXAMPLES. (1) For {M,X}, define p(X) = a 2 0. This is a measure. (2) For X equal to a countable set and with all subsets in the o-algebra, attach a weight 2 0 to each member of X . Define w ( E ) to be the sum of the weights for the members of E . This is a measure.
236
V Lebesgue Measure and Abstract Measure Theory
(3) For X arbitrary but nonempty, let p ( E ) be the number of points in E , a nonnegative integer or +oo. We refer to p as counting measure.
(4) Lebesgue measure m on the ring R of elementary sets of R. If E is a finite disjoint union of bounded intervals, we let m(E) be the sum of the lengths of the intervals. We need to see that this definition is unambiguous. Consider the special case that J = Il U . . . U I, disjointly with Ik extending from ak to b k , with or without endpoints. Then we can arrange the intervals in order so that bk = ak+l for k = 1 , .. . , r - 1. In this case, m ( J ) = b, - a1 and (bk ak) = b, a1 . Thus thc dcfinition is unambiguous in m(Ik) = this special case. If E = Il U . . . U I, = J1 U . . . U J,, then the special case gives m(Jk) = ,(Ij f' Jk) and hence x i = l m(Jk) = m(Ij f' Jk). Reversing the roles of the Ij's and the Jk's, we obtain x i = l m(Ij) = m(Ij f' Jk). Thus x i = 1 m(Jk) = x i = 1 m(Ij), and the definition of m on R is unambiguous. It is evident that m is nonnegative and additive. We shall prove that m is completely additive on R. Even so, m will not yet be a measure, since R is not a o-ring. That step will have to be carried out separately. Proving that m is completely additive on the ring R uses the fact that m is regular on R in the sense that if E is in R and if c > O is given, then there exist a cnmpact set K in 77. and an open set U in R such that K 5 E 5 U , m(K) ? m(E) - t , and m(U) 5 m(E) + t : In the special case that E is a single bounded interval with endpoints a and b, we can prove regularity by taking U = (a - t / 2 , b t/2) andbylcttingK = @ i f b - a 5 t o r K = [ a + t / 2 , b t / 2 ] i f b a > t . I n the general case that E is the union of n bounded intervals I j , choose Kj and Uj fur Ij and fur the nurniber t / n , and put K = Ujn=l li;. and U = Uj . Tlieri m(K) = xjn=l m(Kj) > ELl (m(lj) - t / n ) = m(E) - t , and Proposition 5.lf givcs m(U) 5 CYzlm(UJ) 5 CY=l (m(lj) t / n ) = m(E) t . -
-
Cj,k
xj,k
+
ULl
+
+
Proposition 5.4. Lebesgue measure m is completely additive on the ring R of elementary sets in R'. PROOF. Let {En}be a disjoint sequence in R with union E in R. Since m is nonnegative and additive, Proposition 5.1 gives m(E) > m(U:=l ~ k = ) x i = l m(Ek) for every n. Passing to the limit, we obtain m(E) 1 m(Ek). For the reverse ineq~lality,let t > 0 be given Choose by regularity a compact member K of R and open members Un of R such that K 5 E , Un 2 En for all n, m(K) 2 m(E) - t , and m(U,) 5 m(E,) ~ / 2 Then ~ . K 5 Ur=l Un, and Un for some N . Hence m(L) - t 5 the compactness implies that K 5 N N m ( K ) i E n = 1m(Un) i E n = 1(m(E,) t/2") 5 E r = l m(En) t . Since t is arbitrary, m (E) 5 C r = l m (En),and the proposition follows.
xEl
uL1 +
+
+
1 . Measures and Examples
237
The smallest o-ring containing the ring R of elementary sets in R1 is in fact a o-algebra, since R1 is the countable union of bounded intervals. For Lebesgue measure to be truly useful, it must be extended from R to this o -algebra, whose members are called the Borel sets of R1. Borel sets of R1 can be fairly complicated. Each open set is a Borel set because it is the countable union of bounded open intervals Each closed set is a Borel set, being the complement of an open set, and each compact set is a Borel set because compact subsets of R1 are closed. In addition, any countable set, such as the set Q of rationals, is a Borel set as the countable union of one-point sets. The extension is carried out by the general Extension Theorem that will be stated now and will be proved in Section 5. The theorem gives both existence and uniqueness for an extension, but not without an additional hypothesis. The need for an additional hypothesis to ensure uniqueness is closely related to the need to assume some finiteness condition on p in Corollary 5.3: even though each member of a decreasing sequence of sets has infinite measure, the intersection of the sets need not have infinite measure. To see what can go wrong for the Extension Theorem, consider the ring R'of subsets of R1 consisting of all finite unions of bounded intervals with rational endpoints; the individual intervals may or may not contain their endpoints. Tf a set function /L is defined on this ring by assigning to each set the number of elements in the set, then p is completely additive. Eachinterval in IW'can be obtained as the union of two sets-a countable union of intervals with rational endpoints and a countable intersection of intervals with rational cndpoints. It follows that thc smallcst o-ring containing R'is thc o-algebra of all Borel sets. The set function can be extended to the Borel sets in more than way. In fact, each one-point set consisting of a rational must get measure 1, but a one-point set consisting of an irrational can be assigned any measure. The additional hypothesis for the Extension Theorem is that the given nonnegative completely additive set function v on a ring of sets R be o-finite, i.e., that any member of R be contained in the countable union of members of R on which v is finite. An obvious sufficient condition for o-finiteness is that v ( E ) be finite for every set in R. This sufficient condition is satisfied by Lebesgue measure on the elementary sets, and thus the theorem proves that Lebesgue measure extends in a unique fashion to be a measure on the Borel sets. The condition of o-finiteness is less restrictive than a requirement that X be the countable union of sets in R of finite measure, another condition that is satisfied in the case of Lebesgue measure. The condition of o-finiteness on a ring allows for some very large measures when all the sets are in a sense generated by the sets of finite measure. For example, if R is the ring of finite subsets of an uncountable set and v is the counting measure, the o-finiteness condition is satisfied. In most areas of mathematics, these very large measures rarely arise.
V Lebesgue Measure and Abstract Measure Theory
238
Theorem 5.5 (Extension Theorem). Let R be a ring of subsets of a nonempty set X, and let v be a nonnegative completely additive set function on R that is o-finite on R. Then v extends uniquely to a measure F on the smallest o-ring containing R.
A measure space i s defined to be a triple ( X , A , / I , ) , where X i s a nonempty set, A is a o-algebra of subsets of X , and p is a measure on X. The measure space is finite if p ( X ) i+m;it is o-finite if X is the countable union of sets on which p is finite. The real line, together with the o-algebra of Bore1 sets and Lcbcsguc mcasurc, is a o-finitc mcasurc spacc.
2. Measurable Functions
In this section, X denotes a nonempty set, and A i s a o-algebra of subsets of X . The measurable sets are the members of A . We say that a function f : X + R* is measurable if (i) f -l ([-m,c ) ) is a measurable set for every real number c. Equivalently the measurability of f may be defined by any of the following conditions: (ii) f ([- m, c ] )is a measurable set for every real number c , (iii) f ( ( c ,+m])is a measurable set for every real number c, (iv) f -'([c, f o o l ) is a measurable set for every real number c. In fact, the implications (i) implies (ii), (ii) implies (iii), (iii) implies (iv), and (iv) implies (i) follow from the identities4
P' -'
EXAMPLES. (1) If A = {a, X } , then only the constant functions are measurable. (2) If A consists of all subsets of X , then every function from X to R* is measurable. 4~anipulationswith inverse images of sets are discussed in Section A1 of the appendix
2. Measurable Functions
239
(3) If X = R1and Aconsists of the Borel sets of R1, the measurable functions are often called Borel measurable. Every continuous function is Borel measurable by (i) because the inverse image of every open set is open. Any function that is 1 on an open or compact set and is 0 off that set is Borel measurable. It is shown in Problem 11 at the end of the chapter that not every Riemann integrable function (when set equal to 0 off some bounded interval) is Borel measurable However, let us verify that every function that is continuous except at countably many points is Borel measurable. In fact, let C be the exceptional countable set. The restriction of f to the metric space R - C is continuous, and hence the inverse image in R - C of any open set [-GO, c ) is open in R - C . IIence the inverse image is the countable union of sets ( a , b) - C, and these are Borel sets. The full inverse image in R of [ - X I , c) under f is the union of a countable set and this subset of R - C and hence is a Borel set.
(4) If X = R1and if A consists of the "Lebesgue measurable sets" in a sense to be defined in Section 5 , the measurable functions are often called Lebesgue measurable. Every Borel measurable function is Lebesgue measurable, and so is every Riemann integrable function (when set equal to 0 off some bounded interval).
The next proposition discusses, among other things, functions f +, f -, and = - min{f ( x ) , 01, and I f l(x) = I f ( x ) I . Then f = f + - f - a n d i f 1 = f f + f - .
I f 1 defined by f f (x) = max{f ( x ) , 01, f - ( x )
Proposition 5.6. (a) Constant functions are always measurable. (b) If f is measurable, then the inverse image of any interval is measurable. (c) If f is measurable, then the inverse image of any open set in R*is measurable. (d) If f is measurable, then the functions f +, f -, and I f 1 are measurable. PROOF.In (a), the inverse image of a set under a constant function is either M or X and in either case is measnrahle. Tn (h), the inverse image of an interval is the intersection of two sets of the kind described in (i) through (iv) above and hence is measurable. In (c) , any open set in R*is the countable union of open intervals, and the measurability of the inverse image follows from (b) and the closure of A undcr countable unions. In (d), (f +)-' ((c, +GO)) cquals f ((c, +GO)) if c > 0 and equals X if c < 0. The measurability of f - and I f 1 are handled similarly.
-'
Next we deal with measurability of sums and products, allowing for values f o o and -oo. Recall from Section 1.1 that multiplication is everywhere defined
in R*and that the product in R*of 0 with anything is 0.
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V Lebesgue Measure and Abstract Measure Theory
Proposition 5.7. Let f and g be measurable functions, and let a be in R.Then af and f g are measurable, and f +g is measurable provided the sum f (x) +g(x) is everywhere defined. PROOF.For f
+ g , with Q denoting the rationals,
If a = 0, then af = 0, and 0 is measurable. If a # 0, then
If f and g are measurable and are ? 0, then (fg)-'(c, + a 1 =
UrEQ,r>O f - I ( $ , +w] n g-l(r, + m i
z
if c 0, i f c < 0.
Hence f g is measurable in this special case. In the general case the formula ,fg = ,f +gf + f -g,f +g,f -gf exhibits f g as the everywhere-defined sum of measurable functions. -
-
Proposition 5 8 . If { f,} is a sequence of measurable functions, then the functions (a) sup, fn, (b) infn fn, (c) lim sup, f, , (d) lim inf .f, , are all measurable.
Uzl
PROOF. For (a) and (b), we have (sup f,)-'(c, +w] = ff,-'(c, +co] For (c) and (d), we have and (inf f,)-'([-a, c) = f;'[-m, c). lim sup, f, = inf, supk,,- fk and lim inf, f, = sup, infk,, fk .
Uzl
Corollary 5.9. The pointwise maximum and the pointwise minimum of a finite set of measurable functions are both measurable. PROOF.These are special cases of (a) and (b) in the proposition.
Corollary 5.10. If { f,} is a sequence of measurable functions and if f (x) = lim f, (x) exists in R*at every x , then f is measurable. PROOF. This is the special case of (c) and (d) in the proposition in which lim sup, f, = lim inf, f, .
3. Lebesgue Integral
24 1
The above results show that the set of measurable functions is closed under pointwise limits, as well as the arithmetic operations and max and min. Since the measurable functions will be the ones we attempt to integrate, we can hope for good limit theorems from Lebesgue integration, as well as the familiar results about arithmetic operations and ordering properties. If E is a subset of X , the indicator function5 Ic of E is the function that is 1 on E and is 0 elsewhere. The set ( I E ) - l ( c , +m] is @ or E or X , depending on the value of c. Therefore I E is a measurable function if and only if E is a measurable set. A simple function s : X R*is a function s with finite image coiltailled in R.Every simple function s has a unique representation as s = ~ fc,IE,, = where the c, are distinct real numbers and the En are disjoint nonempty sets with union X . In fact, the set of numbers c, equals the image of s , and En is the set where s takes the value c,. This expansion of s will be called the canonical expansion of s . The set s-' ( c , +a] is the union of the sets En such that c < c, , and it follows that s is a measurable function if and only if all of the sets En in the canonical expansion are measurable sets.
-
Proposition 5.11. For any function f : X + [0,+m] ,there exists a sequence of simple functions s, 10 with the property that for each x in X , { s , ( x ) } is a monotone increasing sequence in R with limit f (x) in R*. If f is measurable, then the simple functions s may be taken to be measurable. PROOF.For 1 5 n < m and 1 5 j 5 n2,,let
Then {s,} has the required properties. By convention from now on, simplefinnctions will always be understood to be measurable.
3. Lebesgue Integral Throughout this section, ( X , A, p ) denotes a measure space. The measurable sets continue to be those in A. Our objective in this section is to define the Lebesgue 5 ~ notedinchapter s II1,indicator functions are cal1ed"characteristic functions" by many authors, but the term "characteristic function" has another meaning in probability theory and is best avoided as a substitute for "indicator function" in any context where probability might play a role.
~
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V Lebesgue Measure and Abstract Measure Theory
integral. We defer any systematic discussion of properties of the integral to Section 4. Just as with the Riemann integral, the Lebesgue integral is defined by means of an approximation process. In the case of the Riemann integral, the process is to use upper sums and lower sums, which capture an approximate value of an integral by adding contributions influenced by proximity in the domain of the integrand. The process is qualitatively different for the Lebesgue integral, which captures an approximate value of an integral by adding contributions based on what happens in the image of the integrand. Let s be a simple function 0. I3y our convention at the end of the previous section, we have incorporated measurability into the definition of simple function. = be ~ the canonical expansion of Let E be a measurable set, and lets = ~ fc,IA, s . We define ZE(s)= ~ fc,p(A, = f'~E ) . This kind of object will be what we use as an approximation in the definition of the Lebesgue integral; the formula shows the sense in which ZE( s ) is built from the image of the integrand. If f > 0 is a measurable function and E is a measurable set, we define the Lebesgue integral of f on the set E with respect to the measure w to be
,
f (x) dp(x) = sup Zc(s) o i s i f, s simple
This is well-defined as a member of R* without restriction as long as E is a ~neasurableset and the measurable functioll f is 3 0 everywhere on X . It is evident in this case that f d p > 0 and that lE 0 d p = 0. For a general measurable function f ,not necessarily 2 0, the integral may or may not be defined. We write f = f + - f -. The functions f + and f - are > 0 and are measurable by Proposition S.hd, and consequently JE f + d p and f - d p are well-defined members of R*.If JE f + d p and JE f - d p are not both infinite, then we define
IE
1,
This definition is consistent with the definition in the special case f > 0, since such an f has f - = 0 and therefore f - d p = 0. We say that f is integrable if lE f + d p and JE f - d p are both finite. In this case the subsets of E where f is +oo and whcrc f is o o havc mcasurc 0. In fact, if S is thc subsct of E where f + is +oo,then the inequality f dw > ZE(C I S ) = Cw(S) for every C > 0 shows that p(S) 5 C-' f + d p for every C; hence p(S) = 0. A similar argument applies to the set where f - is +oo. We shall give some examples of integration after showing that the definition of lE f d p reduces to ZE(f ) if f is nonnegative and simple. The first lemma
IE
1,
+
3. Lebesgue Integral
243
below will make use of the additivity of p , and the second lemma will make use of the fact that p is nonnegative.
EL,
Lemma 5.12. Let s = cnlAabe the canonical expansion of a simple function 0, and let s = dnIBmbe another expansion in which the dm are - 0 and thc Bm arc disjoint and mcasurablc. Thcn Zb (s) = 1 dmp(Bmn E ) .
zL1
z
xE1
PROOF.Adjoin the term 0 . I((J_ to the second expansion, if necessary, to make B, = X . Without loss of generality, we may assume that no B, is empty. Then the fact that the sets B,,, are disjoint and nonempty with union X implies that the image of s is { d l ,. . . , d M } . Thus we can write dm = c,(,) for each m. Since A, = sP1( { c n } )we , see that Bm 5 A,(,). Since the B, are disjoint with union X , we obtain B,)C
uL1
&,
disjointly. The additivity of gives w(Ak n E) = I n(mj=k) w(B, f' E ) ,and thus c k p ( A kn E ) = C,, , , dmp(Bmn E ) . Summing on k , we obtain the conclusion of the lemma.
,=,
Lemma 5.13. If s and t are nonnegative simple functions and if t 5 s on E , then ZE ( t ) 5 ZE( s ).
q=,
PROOF. If s = cjIAJand t = z kK= , d k I ~are the canonical expansions of s and t , then Uj,,( A j n B k ) = X disjointly. Hence we can write s=
C cj1AjnBk
and
t=
C dkI~~nL7~.
Lemma 5.12 shows that
~ , ( S ) = C C ~ ~ (andA ~Z ~~ (~t )B= IC ,d ~~ ~~j E ~ j )n ~ ~ n E ) . j,k
j,k
We now have term-by-term inequality: either p ( A j n Bk n E ) = 0 for a term, or A J n B k n E # M a n d a n y x i n A j n B k n E h a s t ( x ) 5 s(x)andcxhibitsdk 5 c j .
Proposition 5.14. If s 10 is a simple function, then every measurable set E .
!E
s d p = Z E ( s ) for
PROOF. If t is a simple function with 0 5 t 5 s everywhere, then Lemma 5.13 gives Z E ( t ) 5 Z E ( s ) . Hence JE s d p = supo,,,, - - Z E ( t ) 5 Z E ( s ) . On , ~( ~~=)~jE s d p , and thus the other hand, we certainly have Z E ( s ) 5 S U -~ ~ Z. E s d p = ZE(S).
1,
V Lebesgue Measure and Abstract Measure Theory
244
EXAMPLES. (1) Let A = {m,X } and p ( X ) = 1. Only the constant f~~nctions are measurable, and J, c d p = 0 and J, c d p = c . (2) Let X be a nonempty countable set, let A consist of all subsets of X , and let p be defined by nonnegative finite weights w i attached ro each point i in X . If f = { f i } is a real-valued function, then the integral of f over X is C f i wi provided the integrals of f + and f - are not both infinite, i .e., provided every rearrangement of the series 2 ji wi converges in R*to the same sum. By contrast, f is integrable if and only if the series C fi wi is absolutely convergent. In the special case that all the weights w i are 1, the theory of the Lebesgue integral over X reduces to the theory of infinite series for which every rearrangement of the series converges in R* to the same sum. This is a very important special case for testing the validity of general assertions about Lebesgue integration. (3) Let ( X , A, p) be the real line R1 with A consisting of the Bore1 sets and with y equal to Lebesgue measure m . Recall that real-valued continuous functions on R1 are measurable. For such a function f , the assertion is that
iu,xlLX fdm=
f ( t )d t ,
the left side being a Lebesgue integral and the right side being a Riemann integral. Proving this assertion involves using some properties of the Lebesgue integral that will be proved in the next section. We give the argument now before these propelties have been established, in order to emphasize the importance of each of these properties: If h > 0, then 1 1 -
h
[
la,x+K)ia,xj f
dm
-
f dm] - f ( x ) = h
k,x+h) dm -
The absolute value of the left side is then
and the right side tends to 0 as h decreases to 0, by continuity of f at x . If h < 0, then the argument corresponding to the first display is
245
4. Properties o f the Integral
The absolute value of the left side is then 5 I f ( t ) - f ( x ) I , and this tends to 0 as h increases to 0, by continuity of f at x . We conclude that f d m is differentiable with derivative f . By the Fundamental Theorem of Calculus for the Riemann integral, together with a corollary of the Mean Value Theorem, j;LI,Xj f dm = f ( t )dt c for all x and some constant c. Putting x = a , we see that c = 0. Therefore the Riemann and Lebesgue integrals coincide for continuous functions on bounded intervals [ a ,b) .
+
1:
4. Properties of the Integral In this section, (X, A, p ) continues to denote a measure space. Our objective is to establish basic properties of the Lebesgue integral, including properties that indicate how Lebesgue integration interacts with passages to the limit. The properties that we establish will include all remaining properties needed to justify the argument in Example 3 at the end of the previous section.
Proposition 5.15. The Lebesgue integral has these four properties:
1,
(a) If f is a measurable function and p ( E ) = 0, then f d p = 0. (b) If E and F are measurable sets with F c E and i f f is a measurable func,f + d p and ,f - d p 5 JE ,f - d p . Consequently, if tion, then JF ,f + d p 5 f d p is defined, then so is dp. (c) If c is a constant function with its value in R*, then c dl1 = c p ( E ) . (d) If f d p is defined and if c is in R,then cf d p is defined and cf d p = c f d p . If f is intcgrablc on E , thcn so is cf .
IE
1,
SE
IE
IF
IF
IE
IE
I,
PROOF.In (a), it is enough to deal with f + and f - separately, and then it is enough to handle s > 0 simple. For such an s , Proposition 5.14 says that the , the definition shows that this is 0. In (b), Proposition integral equals Z E ( S ) and 5.14 makes it clear that the inequalities are valid for any simple function > 0, and then the general case follows by taking the supremum first for 0 5 s 5 f + and then for 0 5 s 5 f . In (c), if 0 5 c < +oo, then c is simple, and the integral equals Z E ( c ) - c p ( E ) by Proposition 5.14. If c - fm, then the case p ( E ) = 0 follows from (a) and the case p ( E ) > 0 is handled by the observations that jE c d p > Zb ( n ) = ~ z p ( Eand ) that thc right sidc tcnds to fcc as 12 tcnds to +oo. For c 5 0, we have c d p = - (-c) d p by definition, and then the result follows from the previous cases. In (d), we may assume, without loss of generality, that f 2 0 and c 2 0. Then cf d p = up^,,,,^ - ZE ( s ) = is proved. supo5ct5cfZ~(~t) = C SUPo5t5f Z E ( ~=) C JE f f P , and
1,
IE IE
Proposition 5.16. If f and g are measurable functions, if their integrals over E are defined, and i f f ( x ) 5 g ( x ) on E , then f d p 5 SE g d p .
IE
V Lebesgue Measure and Abstract Measure Theory
246
REMARK. Observe that the inequality f ( x ) 5 g ( x ) is assumed only on E , despite the definitions that take into account values of a function everywhere on X. This "localization" property of the integral is as one wants it to be. PROOF.First suppose that f > 0 and g > 0. If s is any simple function with 0 5 s 5 f , d e f i n e t t o e q u a l s o n E a n d t o e q u a l O o f f E . Then0 5 t 5 g,and Lemma 5.13 gives Z E ( S )= ZE( t ) 5 g d y . Hence f d p 5 lE g d p when f >Oandg>O. In the general case the inequality f ( x ) 5 g ( x ) on E implies that f + ( x ) 5 gf ( x ) on E and f - ( x ) > g - ( E ) on E . The special case gives JE f d p 5 g+ d p and f-d p 2 g - d p . Subtracting these inequalities, we obtain the desired result.
IE
SE
if
SE
1,
+
iE
Corollary 5.17. I f f and g are measurable functions that are equal on E and f d p is defined, then 1 , g d p is defined and JE f d p = JE g d w .
JE
PROOF.Apply Proposition 5.16 to the following inequalities on E , and then sort out the results: f + 5 g+, f + > gf , f - 5 g - , and f - > g-. Corollary 5.18. If f is a measurable function, then f is integrable on E if either (a) there is a function g integrable on E such that I f ( x ) I 5 g ( x ) on E , or (b) p ( E ) is finite and there is a real number c such that I f (x) I 5 c on E. ~ O O F For . (a), apply Proposition 5.16 to the inequalities f + 5 g and f - 5 g valid on E . For (b), use the formula for JE c dw in Proposition 5 . 1 5 and ~ apply (a).
Wc turn our attcntion now to propcrtics that indicatc how Lcbcsguc intcgration interacts with passages to the limit. These make essential use of the complete additivity of the measure p . We shall bring this hypothesis to bear initially through the following theorem.
ix
Theorem5.19. Let f be a fixed measurable function, and suppose that f d p is defined. Then the set function p ( E ) = f d p is completely additive.
SE
PROOF.We have p(M) = 0 by Proposition 5.15a, since p ( M ) = 0. We shall prove that if f > 0, then p is completely additive. The general case follows from this by applying the result to f + and f - separately and by using the fact that f + d y and Jx f - d p are not both infinite. Thus we are to show that if E = UT==l En disjointly and if f > 0, then p ( E ) = CEl p (E,). N For simple s > 0 with canonical expansion s = En=, cnIA, , the identity Z F ( s )= c n p ( A nn F ) and the complete additivity of p show that Z F ( s )is
ix
c:=~
247
4. Properties o f the Integral
a completely additive function of the set F. Thus for s simple with 0 5 s 5 f , we have
Hence We now prove the reverse inequality. By Proposition 5.15b, p(E) > p(E,) for every n , since f = f +. Hence if p(En)= +afor any n , the desired result is proved. Thus assume that p(E,) < +oo for all n . Let c > 0 be given, and choose simple functions t and u that are > 0 and are 5 f and have
Let s be the pointwise maximum s = max{t, u } . Then s is simple, and Lemma (s) > Zbl ( t ) and Zb2 (s)> ZL, (u). Consequently 5.13 gives ZLl
Since t is arbitrary, p (ElU E2) > p (El)+ p(E2). By induction, we obtain p(E1 U . . . U En) > p(E1)+ . . . + p(En)for every n , and thus p(E) 2 ,o(E1)+ . . . + p(E,) by another application of Proposition 5.15b. Therefore p (E) >
CT=l p ((E), and the reverse inequality has been proved.
We give five corollaries that are consequences of Corollary 5.17 and Theorem 5.19. Tlie first tliree iiiake use oiily of additivity, not of coiiiplete additivity.
Corollary 5.20. If IE f d p is defined, then I;, IE f d p is defined and equals [E
f
d ~ .
PROOF.It is sufficient to handle f and f separately. Then both integrals are defined, and lE f d p = lE IEf dp JEc 0 d ~ = JE I E d ~~ IE f dlJ = I E d~ ~ . +
+
+
~
E
C
248
V Lebesgue Measure and Abstract Measure Theory
Corollary 5.21. If JE f d p is defined, then integrable on E , so is If 1.
I
JE
PROOF. Let El = E f' f P 1 ( [ 0+m]) , and E2 use of the triangle inequality gives
If f is integrable on E , both
IElj'+ dw and
JE2
f dpl 5
=
JE
', f 1 d p . If f is
E f' f P 1 ( [ - m ,0 ) ) . Then
f - d p are finite. Their sum is
JE
If Id
JE
Corollary 5.22. If f is a measurable function and p(E A F ) = 0 , then f d p = JF f d p , provided one of the integrals exists.
~ .
PROOF.Without loss of generality, we may assume that f > 0. Then both integrals are defined. Since E A F = ( E - F ) U ( F - E ) ,we have w(E - F) = p(F - E ) = 0 . Then Theorem 5.19 and Proposition 5.15a give f dp = f ' P = J F - E f ' P + JE'EIiF f ' P = f J E r l F f d~ = O +
IE
I
Corollary 5.23. Iff is a measurable function and if the set A = {x f (x) # 0 ) has p ( A ) = 0 , then Jx f d p = 0. Conversely if f is measurable, is > 0 , and has 1, f d p = 0, then A = { x f ( x ) # 0 ) has w ( A ) = 0.
I
REMARKS.When a set where some condition fails to hold has measure 0 , one sometimes says that the condition holds almost everywhere, or a.e., or at almost every point. If there is any ambiguity about what measure is being referred to, one says "a.e. [ d p ]." Thus the conclusion in the converse half of the above proposition is that f is zero a.e. [ d p ]. PROOF.For the first statement, Corollary 5.20 gives f d p = IA f d p = f d p = 0. Conversely let A, = f ([:, +m]). This is a measurable set. Since f is > 0, A = Uz,A,, . Proposition 5 . l g and complete additivity of p give p ( A ) 5 C z l p(A,) . If p(A,) > 0 for some n , then f dp = > dl[ = 11 (A,%) > 0 , and we nbtain a contradiction f dl[ f dl1 We conclude that p(A,) = 0 for all n and hence that w ( A ) = 0.
-'
SAn
+IAL
IAni
i
1,
Corollary 5.24. If f > 0 is an integrable function on X , then for any t > 0, there exists a 6 > 0 such that JE f d p ( t for every measurable set E with p(E) 5 6.
4. Properties o f the Integral
249
PROOF.Let t- > 0 be given. If N > 0 is an integer, then the sets SN = { x E X f (x) 2 N ) form a decreasing sequence whose intersection is S = { x E X f ( x ) = f o o l . Since f is integrable, p ( S ) = 0 and therefore f d p = 0 . The finiteness of J, f d p , together with Corollary 5.3 and the complete additivity of E H lE f dw given in Theorem 5.19, implies that limN JsN f d p = 0 . Choose N large enough so that f d p 5 t / 2 , and then choose 6 = t l ( 2 N ) . If p ( E ) 5 6 , then
I
I
AN
and the proof is complete. In a number of the remaining results in the section, a sequence { f , } of measurable functions converges pointwise to a function f . Corollary 5.10 assures us that f is measurable. Suppose that jE f, d p exists for each n . Is it true that J, f d p exists, is it true that lim, JE fn d p exists, and if both exist, are they equal? Once again we encounter an interchange-of-limits problem, and there is no surprise from the general fact: all three answers can be "no" in particular cases. Examples of the failure of the limit of the integral to equal the integral of the limit are given below. After giving the examples, we shall discuss theorems that give "yes" answers under additional hypotheses. EXAMPLES. ( 1 )Let X be the set of positive integers, let A consist of all subsets of X , and let p be counting measure. A measurable function f is a sequence { f ( k ) }with values in R*. Define a sequence { f,} of measurable functions for n > 1 by taking lln
fn(k)= ( 0
ifk 5 n , i f k > n.
Then J;, f, d p = 1 for all n , lim f, = 0 pointwise, and
I*
lim f n d p < lim
fn d p .
( 2 )Let the measure space be X = R' with the Bore1 sets and Lebesgue measure m. Define
for 0 < x < l / n , 0 otherwise. Then the same phenomenon results, and everything of interest is taking place within [0, 11. So the difficulty in the previous example does not result from the fact that X has infinite measure. n
fn(x) =
250
V Lebesgue Measure and Abstract Measure Theory
Theorem 5.25 (Monotone Convergence Theorem). Let E be a measurable set, and suppose that { f,} is a sequence of measurable functions that satisfy 0 5 f i ( x ) 5 f2(x) 5
... 5 fn(x) 5 ...
for all x . Put f ( x ) = lim, f n ( x ) , the limit being taken in R*. Then lim,, f,, d p both exist, and
IEf
d p and
REMARKS.This theorem generalizes Corollary 1.14,which is the special case of the Monotone Convergence Theorem in which X is the set of positive integers, every subset is measurable, and p is counting measure. In the general setting of the Monotone Convergence Theorem, one of the by-products of the theorem is that we obtain an easier way of dealing with the definition of JE f d p for f 2 0. Instead of using the totality of simple functions between 0 and f , we may use a single increasing sequence with pointwise f , such as the one given by Proposition 5.11. The proof of Proposition 5.26 below will illustrate how we can take advantage of this fact. PROOF.Since f is the pointwise limit of measurable functions and is 2 0, f is measurable and jE f d p exists in R*. Since { f , ( x ) } is monotone increasing ill n , the same is time of { f, d p } . Therefore lim, f , d p exists in R*. Let us call this limit k . For each n , fn d p 5 f d p because f, 5 f . Therefore k 5 jE f d w , and the problem is to prove the reverse inequality. Let c be any real number with 0 < c < 1, to be regarded as close to 1, and let s be a simple function with 0 5 s 5 f . Define
iE
lE
lE
iE
These sets are measurable, and E l g E 2 g E3 g . . . g E . Let us see that E = E n . If f ( x ) = 0 for a particular x in E , then f,(x) = 0 for all n and also c s ( x ) = 0. Thus x is in every E n . If f (x) > 0, then the inequality f ( n ) s (x)forces f ( n ) > cs ( n ). Since f , ( x ) has increasing lirnit f ( x ), f , ( x ) must be > c s ( x ) eventually, and then x is in E n . In either case x is in U r = l E n . Thus E = Ur==l En. For every n ,we have
Uzl
Since, by Theorem 5.19, the integral is a completely additive set function, Proposition 5.2 shows that lim sdp = s d p . Therefore k 1c s d p . Since c is arbitrary with 0 < c < 1 , k > s d p . Taking the supremum over s with 0 5 s 5 f , we conclude that k > f dp.
lEn
lE lE iE
lE
25 1
4. Properties o f the Integral
Proposition 5.26. If f and g are measurable functions, if their sum Iz = f + g is everywhere defined, and if f dp g d p is defined, then h d p is defined and
IE
l h d w =
iE
+ 1,
L f dw+Lgdw.
REMARK.It iilay seeill suiyi-king that coiilplete additivity plays a i-ole ill the proof of this proposition, since it apparently played no role in the linearity of the Riemann integral. In fact, although complete additivity is used when f and g are unbounded, it can be avoided when f and g are bounded, as will be observed in Problems 4 2 4 3 at the end of the chapter. The distinction between the two cases is that the pointwise convergence in Proposition 5.11 is actually uniform if the given function is bounded, whereas it cannot be uniform for an unbounded function because the uniform limit of bounded functions is bounded. PROOF. The sum h is measurable by Proposition 5.7. For the conclusions about integration, first assume that f > 0 and g > 0. In the case of simple functions s = t + u with t > 0 and u > 0, we use Proposition 5.14 and Lemma 5.12. The proposition shows that we are to prove that Z E ( S )= Z E ( ~ ) Z E ( U ) , and the lemma shows that we can use expansions of t and u into sets on which t and u are both constant and the conclusion about Z E ( S )is evident. If f and g are 10 and are not necessarily simple, then we can use Proposition 5.11 to find increasing sequences {t,} and { u n }of simple functions > 0 with limits f and g. If s, = t, + u,, then s, is nonnegative simple, and {s,} increases to h . For each n , we have just proved that S, d p = tn d p u , d p , and therefore JE h dw = jE f dw jE g d p by the Monotone Convergence Theorem (Theorem 5.25). The next case is that f > 0, g 5 0, and h = f g > 0. Then f = h + (-g) with h > O and (-g) > O, so that j' d p = jE h d p jE(-g) d p . HenceiE h d p = f d p + g d p , provided the right side is defined. For a general h > 0, we decompose E into the disjoint union of three sets, one where f > 0 and g > 0,one where f 10 andg < 0,and one where f < 0 and g > 0. The additivity of the integral as a set function (Theorem 5.19), in combination with the cases that we have already proved, then gives the desired result. Finally for general h , we have only to write h = h+ - h- and consider h f and h- separately.
+
+
IE
IE
SE
SE
IE
+
+ iE +
Corollary 5.27. Let E be a measurable set, and let { f,} be a sequence of measurable functions > 0. Put F ( x ) = CZl f,(x) . Then JE F d p = jb ffil dw.
c,"==,
PROOF.Apply Proposition 5.26 to the nth partial sum of the series, and then use the Monotone Convergence Theorem (Theorem 5.25).
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V Lebesgue Measure and Abstract Measure Theory
The next corollary is given partly to illustrate a standard technique for passing from integration results about indicator functions to integration results about general functions. This technique is used again and again in measure theory. Corollary 5.28. If f > 0 is a measurable function and if v is the measure v(E) = JE ,f d p , then JE g dv = JE g,f d p for every measurable function g for which at least one side is defined.
REMARKS.Tlle set furictivrl v is a rrleasure by Tllevrerrl5.19. I11 the situatiori of this corollary, we shall write v = f d p . PROOF.By Corollary 5.20 it is enough to prove that
For g = I E , (*) is true by hypothesis. Proposition 5.26 shows that (*) extends to be valid for simple functions g > 0 . For general g > 0, Proposition 5.11produces an increasing sequence Isn}of simple functions > 0 with pointwise limit g . Then (*) for this g follows from the result for simple functions in combination with monotone convergence. For general g , write g = gf - g - , apply (*) for gf and g - , and subtract tlie results using Propositioii 5.26. Theorem 5.29 (Fatou's Lemma). If E is a measurable set and if { f,} is a sequence of nonnegative measurable functions, then
lirn inf f., d p 5 lirn inf
f., d p .
In particular, if f (x) = limn f, (x) exists for all x ,then
REMARK.Fatou's Lemma applies to both examples that precede the Monotone Convergence Theorem (Theorem 5.25), and strict inequality holds in both cases. PROOF.Set g,, (x) = infk,,, fk (x) . Then lirn,, g,, (x) = lim inf f,, (x) , and the Monotone Convergence Theorem (Theorem 5.25) gives
But g,(x) 5 f,(x) pointwise, so that
and the theorem follows.
IEg., d p 5 JE f., d p for all n . Thus
253
5. Proof o f the Extension Theorem
Theorem 5.30 (Dominated Convergence Theorem). Let E be a measurable set, and suppose that { f,} is a sequence of measurable functions such that for some integrable g , If, 1 5 g for all n. If f = lim f, exists pointwise, then limn JE f n dw exists, f is integrable on E, and
PROOF.The set on which g is infinite has measure 0, since g is integrable. If we redefine g , f, ,and f to be 0 on this set, we change no integrals and we affect the validity of neither the hypotheses nor the conclusion. By Corollary 5.18, f is integrable on E, and so is f, for every n. Applying g) d p 5 Fatou's Lemma (Theorem 5.29) to f, g 2 0, we obtain jE (f lim inf JE( f, + g ) d p . Since g is integrable and everywhere finite, this inequality becomes f dw 5 l i m i n f k f, d ~ .
+
+
l
A second application of Fatou's Lemma, this time to g ['(g - f ) d l 5~ lim inf S E ( g - f,) d11,. Thus
f d p 2 lim sup
and Therefore lim JE
fn
fn
-
f,
>
0, gives
d ~ .
dw exists and has the value asserted.
Corollary 5.31. Let E be a set of finite measure, let c 0 be in R, and suppose that { f,} is a sequence of measurable functions such that If, 1 5 c for all n . If f = lim f, exists pointwise, then lim f, d p exists, f is integrable on E , and
SE
PROOF.This is the special case g = c in Theorem 5.30.
5. Proof of the Extension Theorem
In this section we shall prove the Extension Theorem, Theorem 5.5. After the end of the proof, we shall fill in one further detail left from Section 1-to show
254
V Lebesgue Measure and Abstract Measure Theory
that a measure on a o-ring has a canonical extension to a measure on the smallest o -algebra containing the given o -ring. Most of this section will concern the proof of the Extension Theorem in the case that X is measurable and v ( X ) is finite. Thus, until further notice, let us assume that X is a nonempty set, A is an algebra of subsets of X , and v is a nonnegative completely additive set function defined on A such that v ( X ) i+m. In a way, the intuition for the proof is typical of that for many existenceuniqueness theorems in mathematics: to see how to prove existence, we assume existence and uniqueness outright, see what necessary conditions each of the assu~nptionsputs on the object to be constlucted, and then begin the proof. With the present theorem in the case that v ( X ) is finite, we shall assign to each subset E of X an upper bound p* ( E ) and a lower bound p , ( E ) for the value of the extended measure on the set E . If the existence half of the theorem is valid, we must have p , ( E ) 5 p* ( E ) for E in the smallest o-algebra containing A. In fact, we shall see that this inequality holds for all subsets E of X . On the other hand, if p , ( E ) < p * ( E ) for some E in the o-algebra of interest and if our upper and lower bounds are good estimates, we might expect that there is more than one way to define the extended measure on E , in contradiction to uniqueness. That thought suggests trying to prove that p, (E) = w * ( E ) for the sets of interest. One way of doing so is to try to prove that the class of subsets for which this equality holds is a 0-algehra containing A, and then the common valile of / I , and / I * is the desired extension. This procedure in fact works, and the only subtlety is in the definitions of p L( E ) and p* ( E ). We give these definitions after one preliminary lemma that will make p , and p* well defined. For orientation, think of the setting as the unit interval [0, 11, with Lebesgue measure to be extended from the elementary sets to the Bore1 sets. In this case the families U and K in the first lemma contain all the open sets and all the compact sets, respectively, and may be regarded as generalizations of these collections of sets. Lemma 5.32. Let U be the class of all countable unions of sets in A, and let K be the class of all countable intersections of sets in A. Then p* and p , are consistently defined on U and K , respectively, by letting p* ( U ) = lim v (A,)
and
p , ( K ) = lim v(C,)
whenever {A,} is an increasing sequence of sets in A with union U and {C,} is a decreasing sequence of sets in A with intersection K . Moreover, p* and p , have the following properties: (a) p* and p , agree with v on sets of A, (b) p* ( U ) 5 p* ( V ) whenever U is in U , V is in U , and U g V ,
5. Proof o f the Extension Theorem
255
(c) p , ( K ) i p * ( L ) w h e n e v e r K i s i n K , L i s i n K , a n d K 5 L , (d) lirn p*(U,) = p*(U) whenever {U,} is an increasing sequence of sets in U with union U . PROOF. If {B,} is another increasing sequence in A with union U , then Proposition 5.2 and Theorem 1.13 give lirn v(Arn)= lirn ( lim v (A, n
rn
in
n B,))
= lirn ( lirn v ( A , n
rn
n B,))
= lirn v (B,). n
Hence p* is consistently defined on U. Similarly if {D,} decreases to K , then Corollary 5.3 and Theorem 1.13 give
v ( X ) - lirn v (C,) = v ( X ) - lim ( lim v (C, n D,)) rn
rn
n
and hence lim, v(C,) = lim, v(D,) . Thus p , is consistently defined on K . The set functions w* and w, are defined on all of U and K because a set that is a countable union (or intersection) of sets in an algebra is a countable increasing union (or decreasing intersection). Of the four properties, (a) is clear, and (b) and (c) follow from the inequalities
( U )=
sup
v(A)5
A S U ,A E A
sup
v ( A )= p*(V)
ASV, A E A
and
inf
A2L, AtA
v ( A ) = p,(L).
In (d), U is in U , since the countable union of countable unions is again a countable union, and (b) shows that limp*(U,) 5 p* ( U ). For each n ,let {A:)} be an increasing sequence of sets from A with union U,. Arrange all the A:) in a sequence, and let Bk denote the union of the first k members of the sequence. Then {Bk}is an increasing sequence with union U . Let t > 0 be given, and choose M large enough so that p* (BM) 1p* ( U )- t . Since the sets U, increase, since BM is a finite union of sets A:), and since A:) 5 U,, we must have p* ( U N )> p* ( B M )for some N . But then
Since t is arbitrary, lirn p* (U,) 3 p* ( U ). For each subset E of X , we define inf
p*(U)
and
w,(E) =
sup
p*(K).
K S E , KEK:
Conclusions (b) and (c) of Lemma 5.32 show that the new definitions of p* and p , are consistent with the old ones. The set functions w* and w, on arbitrary subsets E of X may be called the outer measure and the inner measure associated to v .
V Lebesgue Measure and Abstract Measure Theory
256
Lemma 5.33. If A and B are subsets of X with A 5 B , then p* ( A ) 5 p * ( B ) and p , ( A ) 5 p , ( B ) . In addition, (a) if E 5 U,"=,E n , then p * ( E ) 5 C,"=, p*(E,), (b) if F and G are disjoint, then p , ( F ) p,(G) 5 p,(F U G )
+
PROOF.Since p * ( A ) is an infimum over a larger class of sets than y* (B) is, we have p * ( A ) 5 p* ( B ) . Similarly p, ( A ) 5 p , ( B ) . For (a), lct E 5 U,"==l En.In thc spccial casc in which En is in U for all n , let {F:)} be, for fixed n and varying m , an increasing sequence of sets in A with unionE,. F o r a n y N , w e t h e n h a ~ e l J , " = ~ ( ~ k ~ ) ~~. .k. ~~= E ) l )U . . . U E N . Hence 00
p * ( E ) j p*
( U E,)
= limp* N
n=l
= lim lim v ( ~ , k 'U ) N
m
5 lim lim N
m
(6
E.)
by Lemma 5.3Zd
n=l
. . . U FAN))
C v (F:))
by definition of p* on U by Proposition 5. lf
n=l
For general subsets En of X , choose Un in U with U, 2 En and p* (U,) 5 p* (En)+ t-/2". Then li 5 U , U,, and the special case applied to the U, shows that
Hence p* ( E ) 5 Enp*(E,) , and (a) is proved. For (b), let F and G be disjoint. In the special case in which F and G are in K, let IF,} and {G,} be decreasing sequences of sets in A with intersections F and G . Then by definition of p , on K
p , ( F U G ) = lim v (Fn U G,) = lim
(v(F,)
= p*(F)
+ v(G,)
-
v(F, f' G,))
by Proposition 5.lb
+y*(G),
the last step holding by Corollary 5.3, since F n G is empty. For general disjoint
257
5. Proof o f the Extension Theorem
subsets F and G in X , choose K and L in K with K 5 F , L 5 G , p , ( K ) p* ( F ) - t , and w* ( L ) > p* (GI - t . Then
>
the middle step holding by the special case. Hence p , ( F U G ) 1 p , ( F )+w, ( G ), and (b) is proved.
Lemma 5.34. For every subset E of X , p , (E) IAor IC.
p* ( E ). Equality holds if E
i s in
PROOF.The proof is in three steps. First we prove that if U is in U and K is in K , then w * ( U ) i w, ( U ) and p* ( K ) 5 p , ( K ) . In fact, choose C in A with C g U and p* ((U 5 v ( C ) t . Then p * ( U ) 5 v ( C ) + t- 5 p , ( U ) + t- by Lemma 5.33 since C c U . Hence p* ( U ) 5 p,(U). Similarly choose D in A with D 2 K and p , ( K ) > v ( D ) - t. Then w, ( K ) > v ( D ) - t > p* ( K ) - t , and hence w,(K) > w* ( K ) . Second we prove that if K is in K , then w * ( K ) = p , ( K ) . In fact, choose C in A with C 2 K and v ( C ) - p , ( K ) 5 t . Then C - K is in U , and
+
p * ( K ) 5 v ( C ) 5 P*(C - K )
5 ( p * (~K)
+ p*(K)
+p , ( ~ ) )
-
p,(K)
by Lemma 5.33a
+p*(K)
5 v ( c > p*(K)+ P*(K) 5 p*(K) t -
+
by the previous step by Lemma 5.33b by the choice of C .
Combining this inequality with the previous step, we see that p* ( K ) = p , ( K ) . Third we prove that p , ( E ) 5 p* ( E ) for every E . In fact, find K in K and U in U with K 5 E 5 U , p , ( K ) 2 p , ( E ) - c , and p* ( U ) 5 p* ( E ) t-. Then p* ( E ) F , ( K ) t = p * ( K ) t i p * ( U ) t i P* ( E ) 2 t , and the proof is complete.
+
+
+
+
+
Define a subset E of X to be measurable for purposes of this section if p* ( E ) = k*( E ) ,arid lcl B be llle class uf rrieasurable subsels uf X . Lcrrirrra 5.34 shows that U and K are both contained in B.
Lemma 5.35. If U is in U and K is in K with K 5 U , then
If E is measurable, then for any t > 0, there are sets K in K and U in U with KgEgUand p * ( E - K ) p*(U - K ) t.
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V Lebesgue Measure and Abstract Measure Theory
PROOF.For the first conclusion, U - K is in U and hence p*(U - K) = p , (U - K ) = p,(U) - p , ( K ) = p*(U) - p , ( K ) by Lemma 5.34, Lemma 5.33b, and Lemma 5.34 again. For the second conclusion choose K in K and U in U with K c E c U , p,(K) ? p , ( E ) , and p * ( E ) ? p * ( U ) Since P,(E) = P " ( E ) by the assumed measurability, we see that ~i,(K) + > ~ r (* U ) - hence that p * ( U ) - p,(K) 5 t . The result now follows from Lemma 5.33 and the first conclusion of the present lemma.
+5
5. 5
5,
Lemma 536. The class B of measurable sets is a o-algebra containing A, and the restriction of p* to I3 is a measure.
PROOF.Certainly B 2 A. The rest of the proof is in three steps. First we prove that the intersection of two measurable sets is measurable. In fact, let F and G be in B, and use Lemma 5.35 to choose K 5 F and L 5 G with p * ( F - K ) 5 t a n d p * ( G - L ) 5 t . SinceFnG g ( F - K ) U ( K n L ) U ( G - L ) , p*(F n G ) p* ( F - K ) + p* ( K f' L ) + p* ( G - L ) by Lemma 5.33a by definition of K and L < p*(K n L ) + 2 t by Lemma 5.34 = p , ( ~n L ) + 2t since K n L c F f' G . i W * ( Fn G ) + 2 t Second we prove that the complement of a measurable set is measurable. Let E be measurable. By Lemma 5.35 choose K in K and U in U with K 5 E 5 U andp*(U - K ) 5 t . SinceUC5 EC 5 K C a n d K C - U C =U - K,wehave
w* ( E c )5 w* ( K c - U c )+ p* ( U c ) = w*(U
5t
-
K ) + p,(UC)
by Lemma 5.3321 since U Cis in K
+ w*(EC).
Thus the complement of a measurable set is measurable, and B is an algebra of sets. Third we prove that the countable disjoint union of measurable sets is measurable, and p* is a measure on B. In fact, let [ E nJ be a sequence of disjoint sets in B. Application of Lemma 5.3321, Lemma 5.33b, and Lemma 5.34 gives
5. Proof o f the Extension Theorem
259
The end members of this chain of inequalities are equal, and thus equality must hold throughout: p , ( U , E n ) = p* (U, E n ) = C p* ( E n ) .Consequently U , En is measurable, and p* is completely additive. PROOF OF THEOREM5.5 UNDER THE SPECIAL HYPOTHESES. We continue to assume that the given ring of subsets of X is an algebra and that v ( X ) is finite. Define B to be the class of measurable sets in the previous construction. Then Lemma 5.36 shows that B is a o-algebra containing "4. Hence B contains the smallest o-algebra C containing A. Lemma 5.36 shows also that the restriction of p* to C is a measure extending v . This proves existence of the extension under the special hypotheses. For uniqueness, suppose that p' is an extension of v to C. Proposition 5.2 and Corollary 5.3 show that p' has to agree with p* on U and with p, on K. If K 5 E 5 U with K in K and U in U , then we have
Taking the supremum over K and the infimum over U gives p , ( E ) 5 p' ( E ) 5 y * ( E ) . Since E is in B, p , ( E ) = p * ( E ) , and we see that p f ( E ) = y * ( E ) . Thus w' coincides with the restriction of p* to C. This proves uniqueness of the extension under the special hypotheses. Now we return to the general hypotheses of Theorem 5.5-that R is a ring of subsets of X, that v is a nonnegative completely additive set function on R,and that v is o-finite-and we shall complete the proof that v extends uniquely to a measure on the smallest o-ring C containing R. PROOF OF THEOREM 5.5 IN THE GENERAL CASE. If S is an clcmcnt of R with v ( S ) finite, define S f' R = { S f' R R E 72). Then (S, Sf' R,vlsnn) is a set of data satisfying the special hypotheses of the Extension Theorem considered above. By the special case, if Cs denotes the smallest o-algebra of subsets of S containing S nR,then v SnR has a unique extension to a measure ps on Cs . The measures ps have a certain consistency property because of the uniqueness: if Sf 5 S , then PS s,nn = P S I . Now let (S,] be a sequence of sets in R with union S in C and with v ( S n ) finite for all n . Possibly replacing each set S, by the difference of S, and all previous Sk's, we may assume that the sequence is disjoint. We define ps on the o-algebra S n C of subsets S by p s ( E ) = Enps,, ( E n S,) for E in S n C. Let us check that p s is unambiguously defined and is completely additive. If {T,} is another sequence of sets in R with union S and with v(T,) finite for all in, then the corresponding definition of a set function on S f' C is p/s(E) = Em~ L . T (, E f' T,). The consistency property from the previous paragraph gives
I
I
1
260 US
V Lebesgue Measure and Abstract Measure Theory
ps,,( E
write
n Sn n T,)
_
= pl.
( E n Sn n Tm). Then Corollary 1.15 allows us to
and we see that ps is unambiguously defined. To check that ps is completely additive, let Fl , F2, . . . be a disjoint sequence of sets in S nC with union F. Then the complete additivity of pg ,in combination with Corollary 1.IS, gives
and thus ps is coinpletely additive. The measures ps are consistent on their common domains. To see the consistency, let us see that ps and p~ agree on subsets of S n T . Let S be the countable disjoint union of sets S, in R,and let T be the countable disjoint union of sets T,, in 'R. 'I'hen S f' 1' is the countable disjoint union of the sets Sn n 1;. If fi is in ( S n T ) nC, then Corollary 1.15 and the consistency property of the set functions p~ for R in R yield
Hence the measures ps are consistent on their com~nolldo~nai~ls. If M denotes the set of subsets of X that are contained in a countable union of members of R on which v is finite, then M is closed under countable unions and differences and is thus a o-ring containing R.It therefore contains C,and we conclude that every member of C is contained in a countable union of members of R on which v is finite. It follows that we can define p on all of C as follows: if E
5. Proof o f the Extension Theorem
26 1
is in C, then E is contained in some countable union S of members of R on which v is finite, and we define p ( E ) = p s ( E ) . We have seen that the measures ps are consistently defined, and hence p ( E ) is well defined. If a countable disjoint En of sets in C is given, then all the sets in question lie in a union E = U:=, single S , and we then have p ( E ) = p s ( E ) = CZl ps(E,) = p(E,). In other words, p is completely additive. This proves existence. For uniqueness let E be given in C, and suppose that S is a member of C containing E and equal to the countable disjoint union of sets S, in R with v(S,) finite for all n. We have seen that the value of p ( E f' Sn) = ps,, ( E f' S,) is determined by vlsn,,, hence by v on R. By complete additivity of p , p ( E ) is determined by the values of p ( E n S,) for all n. Therefore p on C is determined by v on R. This proves uniqueness. As was promised, we shall now fill in one further detail left from Section 1-to show that a measure on a o-ring has a canonical extension to a measure on the smallest a-algebra containing the given a-ring. Proposition 5.37. Let R be a o-ring of subsets of a nonempty set X , let R, be the set of complements in X of the members of R , and let A be the smallest o -algebra containing R. Then either (i) R = R, = A or (ii) R n R , = D a n d A = R U R , . In the latter case any measure p on R has a canonical extension to a measure p1 on A given by p l ( E ) = S U ~ { ~ ( FF) E R a n d F g E } for E in R,. This canonical extension has the property that any other extension ~2 satisfies P2 > P I .
I
PROOF.If X is in R , then R is closed under complements, since R is closed under differences; hence R = R, = A. If X is not in R , then R n R, = M because any set E in the intersection has Ec in the intersection and then also X = E U EC in the intersection. In this latter case it is plain that "4 2 R U R,. Thus (ii) will be the only alternative to (i) if it is proved that B = R U R, is a o-algebra. Certainly B is closed under complements. To see that B is closed under countable unions, we may assume, because R is a o-ring, that we are to check the union of countably many sets wit11 at least one in R,. Thus let { E n } be a sequence of sets in R , and let IFn} be a sequence of sets in R,. Then E = U,"=l En is in R and F = F," is in R , since R is a o-ring. The union of the sets En and Fn in question is E U Fc = ( F - E ) " , is exhibited as the complement of the difference of two sets in R , and is therefore in R,. Thus A is closed under countable unions and is a a-algebra.
V Lebesgue Measure and Abstract Measure Theory
262
In the case of (ii), let us see that p1 is a measure on A. If we are to check the measure of a disjoint sequence of sets in A, there is no problem if all the sets are in R , since p1 = p is completely additive. There cannot be as many as two of the sets in R, because no two sets F1 and F2 in R, are disjoint; in fact, F1 f' F2 = ( F t U F;)' exhibits the intersection as in R,, and the empty set is not a member of R, Thus we may assume that the disjoint sequence consists of a sequence {E,} of sets in R and a single set F in R,. If E = U:=l E n , p 1 ( E n ). So it is enough to see then p 1 ( E ) = p ( E ) = C r = l p(E,) = that w l ( E U F) = w ( E ) + w l ( F ) . If E' is a subset of F that is in R,then p1 ( E U F ) p ( E U E') - p ( E ) + p ( E f ). Taking the supremum over all such E' shows that pl ( E U F ) ) p ( E ) + p1 ( F ) . For the reverse inequality let S be a member of R contained in E U F. Then the sets E n S and F n S = S - F c are in R , and thus p ( S ) = p ( E n S ) p ( F n S ) ( p ( E ) p 1 ( F ). Taking the supremum over S gives w l ( E U F ) 5 w ( E ) p 1 ( F ) . Thus p1 is completely additive. If p2 is any otherextension, any set F in RChasp 2 ( F ) 1p 2 ( E ) = p ( E ) for all subsets E of F that are in R. Taking the supremum over E gives p 2 ( F ) > pl ( F ), and thus p2 > p1 as set functions on A.
IR
,
+
+
+
6. Completion of a Measure Space If ( X ,A, p ) is a measure space, we define the completion of this space to be the measure space (X, p) defined by
x,
3:
{EAZ
1
c
E is in A and Z Z' for some Z' r A with p ( Z ' ) = 0
Tt is necessary to verify that the result is in fact a measure space, and we shall carry out this step in the proposition below. In the case of Lebesgue measure m on the line, when initially defined on the o-algebra A of Bore1 sets, the sets in 0-algebra x a r e said to be Lebesgue measurable.
Proposition 5.38. If (X, A, w ) is a measure space, then the completion ( X ,3,A,) is a measure space. Specifically (a) 2is a o-algebra containing A, (b) the set function p is unambiguously defined on x , i.e., if El A Z 1 = E2 A Z2 as above, then w ( E l ) = p ( E 2 ) , (c) p is a measure on x , and p ( E ) = p ( E ) for all sets E in A.
6. Completion o f a Measure Space
263
In addition, (d) if E is anymeasure on x s u c h that E(E) = p ( E ) for all E in A, then E =ponA, (e) if p(X) < +oo and if for E 5 X , p,(E) and p * ( E ) are defined by p,(E)=
sup
P(A>
and
inf
= A?E, AEA
A C E , AEA
p(A),
then E is in x i f and only if p, (E) = p*(E) .
x
PROOF.For (a), certainly A 5 because we can use Z = Z' = M in the definition of 3. Since (E A Z)" = (E A Z) A X = ( E A X) A Z = E c A Z , 3 is closed under complements. To prove closure under countable unions, let us first prove that
Z= ( E U Z
1
E is in A and Z g Z' for some Z' E A with p ( Z f ) = 0
(*)
Thus let E U Z be given, with Z 5 Z'. Then E U Z = E A (Z A (E n Z)) with Z A (E f' Z) (I Z'. So E U Z is in 3.Conversely if E A Z is in we can write E A Z = ( E - Z f ) U ( ( E n Z ' ) - Z ) U ( Z - E ) ) with((EnZf)-Z)U(Z-E)) g Z', and thcn wc scc that E A Z is of thc form E" U Z" with E" in A and Z" 5 Z'. Returning to the proof of closure under countable unions, let E, U Z, be given i n x w i t h z , 5 ZI, andp(Zh) = 0. ThenU,(E,UZ,) = (U, E,)U(U,Z,) with Un Z, 5 U, ZI, and p( U, zI,) = 0. In view of (*), x i s therefore closed under countable unions. For (b), we take as given that E l A Z1 = E2 A Z2 with Z1 5 Z; , Z2 5 Z;, and p (Zi) = p (Zi) = 0. Then (E A E2) A (Zl A Z2) = m and hence E l A E2 = Z1 A Z 2 g ZiUZ;. Thereforep(E1-E2) 5 p(E1 A E2) 5 p(ZiUZ;) = Oand similarly p(E2 - E l ) = 0. It follows that p ( E l ) = p ( E l - Ez) + p ( E l f' E2) = p ( E 1 n E2) = p ( E 2 - El) p ( E l f' E2) = p(E2), and p is unambiguously defined. For (c), we see from (*) that can be defined equivalently byF(E UZ) = p ( E ) if Z 5 Z' and p ( Z f ) = 0. If a disjoint sequence En U Z, is given, then we find thatp(Un(~U n 2,)) = F((U, En) U ( Un z n ) ) = P( Un En) = C p(En) = C p(E, U Z,) ,and complete additivity is proved. Taking Z = M in the definition p ( E U Z) = p ( E ) , we obtain p ( E ) = p ( E ) for E ill A. For (d), we use (*) as the description of the sets in R Let E U Z be in 3 with E in A , Z 5 Z', and Z' in A with p(Z') = 0. Then Proposition 5.lf gives E ( E n Z) 5 E(Z) 5 E(Zf) = p ( Z f ) = 0, so that P ( E n Z) = P(Z) = 0. Meanwhile, Proposition 5.lb gives E ( E U Z) E(E f' Z) = E(E) E(Z). Hence E(E U Z) = E(E) = p ( E ) .
x,
+
+
+
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V Lebesgue Measure and Abstract Measure Theory
For (e), it is immediate that p,(E) 5 p * ( E ) for every subset E of X . Let E = C U Z be in with C in A, Z g Z ' , and Z' in A with p ( Z 1 ) = 0. Then p ( C ) i p * ( E ) i p * ( E ) i p(C U 2') i p ( C ) p ( Z f ) = p ( C ) . Since the expressions at the ends are equal, we must have equality throughout, and therefore p*(E) = p*(E). In the converse direction let p, (E) = p * ( E ) . We can find a sequence of sets A, E A contained in E with lim p(A,) = p , ( E ), and we may assume without loss of generality that {A,} is an increasing sequence. Similarly we can find a decreasing sequence of sets B, E A containing E with lim w(B,) = p * ( E ) . Let A = U, A, and B = 0, B, . When coillbined with the equality p , ( E ) = p* ( E ), Proposition 5.2 and Corollary 5.3 show that p ( A ) = p,(E) = p* ( E ) = p ( B ) . SinceA 5 E 5 B,wehavep(B-A) = p ( B ) - p ( A ) = Oand E = A U ( E - A ) with E - A g B - A a n d p ( B - A ) = O . By (*), E i s i n x .
+
A variant of Proposition 5.38e and its proof identifies the o-algebra on which the extended measure is constructed in the proof of the Extension Theorem (Theorem 5.5) in the special case we considered. In the special case of the Extension Theorem, the given ring of sets is an algebra A, and v ( X ) is finite. The set function v gets extended to a measure p on a o -algebra B that contains the smallest o-algebra C containing A. The sets of B are those for which p,(E) = p * ( E ) , where ~ L * ( E=)
inf
U>E, U € U
1
)
and
I
=
sup
~A,*(K),
KGE, KcK
K and U having been defined in terms of countable intersections and countable unions, respectively, from A. The variant of Proposition 5.38e is that a subset E of X has p,(E) = p * ( E ) if and only if E is of the form C U Z with C in C, Z g Z', and Z' in C with p ( Z f ) = 0. In other words, (X, B, p ) is the completion of ( X ,C, p ) . The proof is modeled on the proof of Proposition 5.38e. If E = C U Z is a set in C with C in C,Z 5 Z ' , and Z' in C with p ( Z 1 ) = 0, then p ( C ) 5 p, ( E ) 5 p* ( E ) ( p(C U 2') 5 p ( C ) p ( Z f ) = p ( C ) . We conclude that p*(E) = /J*(E). In the converse direction let p , ( E ) = p* ( E ). We can find an increasing sequence of sets A, E K g C contained in E with lim p(A,) = p , ( E ) , and we can find a decreasing sequence of sets B, E U 5 C containing E with limp(B,) = p * ( E ) . Let A = U, A, and B = 0, B,. Arguing as in the proof of Proposition 5.38e, we have p ( A ) = p , ( E ) = p * ( E ) = p ( B ) , p ( B - A ) = p ( B ) - p ( A ) =O,andE = AU(E-A)withE-A g B-Aandp(B-A) = O . Thus E = C U Z with C = A and Z = E - A. This calculation has the following interesting consequence.
+
7. Fubini's Theorem for the Lebesgue Integral
265
Proposition 5.39. In R1, the Lebesgue measurable sets of measure 0 are exactly the subsets E of R1 with the following property: for any t > 0, the set E can be covered by countably many intervals of total length less than t-. PROOF.Within a bounded interval [ a ,b], the above remarks apply and show that the Lebesgue measurable sets of measure 0 are the sets E with p * ( E ) = 0, where p * ( E ) = infu3E, u ~ pu* ( U ) . The sets U defining p * ( E ) are countable unions of intervals, and the proposition follows for subsets of any boundedinterval [ a ,bl. For general sets E in R1,if the covering condition holds, then Proposition 5 .lg shows that E has Lebesgue measure 0. Conversely if E is Lebesgue measurable of measure 0, then E n [- N , N] is a bounded set of measure 0 and can be covered by countably many intervals of arbitrarily small total length. Let us arrange that the total length is i2-N t. Taking the union of these sets of intervals as N varies, we obtain a cover of E by countably many intervals of total length less than t .
7. Fubini's Theorem for the Lebesgue Integral
Fubini's Theorem for the Lebesgue integral concerns the interchange of order of integration of functions of two variables, just as with the Riemann integral in Section 111.9. In the case of Euclidean space R n , we could have constructed Lebesgue measure in each dimension by a procedure sirnilar to the one we used for R' . Then Fubini's Theorem relates integration of a function of m +n variables over a set by either integrating in all variables at once or integrating in the first m variables first or integrating in the last n variables first. In the context of more general measure spaces, we need to develop the notion of the product of two measure spaces. This corresponds to knowing R7'hnd IRrhith their Lebesgue measures and to constructing Rmf with its Lebesgue measure. In the theorem as we shall state it, we are given two measures spaces (X, A, p ) and ( Y ,B, v), and we assume that both p and v are a-finite. We shall construct a product measure space(X x Y, A x B, p x v ) , and the formula in question will be
This formula will be valid for f ? 0 measurable with respect to A x I?. The technique of proof will be the standard one indicated in connection with proving Corollary 5.28. We start with indicator functions, extend the result to simple functions by linearity, and pass to the limit by the Monotone Convergence
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Theorem (Theorem 5.25). It is then apparent that the difficult step is the case that f is an indicator function. In fact, it is not even clear in this special case that the inside integral IE ( x , y) d v ( y ) is a measurable function of X , and this is the step that requires some work. We begin by describing A x B, the o-algebra of measurable sets for the product X x Y Recall from Section A1 of the appendix that X x Y is defined as a set of ordered pairs. If A g X and B g Y, then the set of ordered pairs that constitute A x B is a subset of X x Y, and we call A x B a rectangle6in X x Y . The sets A and B are called the sides of the rectangle.
Proposition 5.40. If A and t? are algebras of subsets of nonempty sets X and Y, then the class C of all finite disjoint unions of rectangles A x B with A in A and B in B is an algebra of sets in X x Y . In particular, a finite union of rectangles is a finite disjoint union. PROOF.The intersection of the rectangles R1 = A1 x B1 and R2 = A2 x B2 is the rectangle R = ( A 1f' A2) x ( B 1f' B2) because both R1 f' R2 and R coincide . with the set { ( x ,y ) E x x Y x E A l , x E A2, y E B 1 , y E ~ 2 )Therefore
I
and the right side is a disjoint union if both Ui ( A i x Bi) and U j (Cj x D j ) are disjoint unions. Moreover, the right side is in C if both unions on the left are in C. Therefore C is closed under finite intersections. Certainly M and X x Y are in C. The identity
exhibits the complement of a rectangle as a disjoint union of rectangles. Since the complement of a disjoint union is the intersection of the complements, C is closed under complementation. Thus C is an algebra of sets, and the proof is complete. If A and B are o-algebras in X and Y , then we denote the smallest o-algebra containing the algebra C of the above proposition by A x B. The set X x Y , together with the 0-algebra A x B, is called a product space. The ilieasurable sets of X x Y are the sets of A x B. 'The word "recta~gle"was used with a different meaning in Chaptes 111, but t11ei.e will be no possibility of confusion for now. Starting in Chapter VI, both kinds of rectangles will be in play; the ones in Chapter I11 can then be called "geometric rectangles" and the present ones can be called "abstract rectangles i'
7. Fubini's Theorem for the Lebesgue Integral
267
Let E be any set in X x Y. The section Ex of E determined by x in X is defined by Ex = { y ( x , y) is in E } .
I
Similarly the section EY determined by y in Y is
I
EY = { x ( x , y) is in E } . The section Ex is a subset of Y, and the section EY is a subset of X
Lemma 5.41. Let {E,} be a class of subsets of X x Y, and let x be a point of X. Then
(a) (U, Ea), = U , (Ea)x, (b) E , ) ~= (Ea)x, (c) ( E , - E B ) , = (E,), - (Ejj), and, in particular, ( E s ) , = Y
(n,
n,
(EB), . PROOF.These facts are special cases of the identities at the end of Section A1 of the appendix for inverse images of functions. In this case the function in question is given by f ( y ) = ( x , y ) . -
Proposition 5.42. Let A and B be o-algebras in X and Y, and let E be a measurable set in X x Y. Then every section Ex is a measurable set in Y, and every section EY is a measurable set in X . PROOF.We prove the result for sections E x , the proof for EY being completely analogous. Let E be the class of all subsets E of X x Y all of whose sections Ex are in B. Then E contains all rectangles with measurable sides, since a section of a rectangle is either the empty set or one of the sides. By Lemma 5.41a, I is closed under finite unions. Hence E contains the algebra C of finite disjoint unions of rectangles with measurable sides. By parts (a) and (c) of Lemma 5.41, E is closed under countable unions and complements. It is therefore a o-algebra containing C and thus contains A x B. A corollary of Proposition 5.42 is that a rectangle in X x Y is measurable if and only if its sides are measurable. The sufficiency follows from the fact that a rectangle with measurable sides is in C, and the necessity follows from the proposition. From now on, we shall adhere to the convention that a rectangle is always assumed to be measurable. We turn to the implementation of the sketch of proof of Fubini's Theorem given earlier in this section. The basic question will be the equality of the iterated integrals in either order when the integrand is an indicator function. If E is
V Lebesgue Measure and Abstract Measure Theory
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a measurable set in X x Y , then we know from Proposition 5.42 that Ex is a measurable subset of Y . In order to form the iterated integral
we compute tlie illside integral as v(E,), and we liave to be able to foim the outside integral, which is i, v(Ex)dp(x). That is, we need to know that v(Ex) is a measurable function on X . For the iterated integral in the other order, we need to know that p(EY) is measurable on Y. 'I'he proof of this measurability is the hard step, since the class of sets E for which v(E,) and p(EY) are both measurable does not appear to be necessarily a o-algebra, even when and v are finite measures. To deal with this difficulty, we introduce the following terminology: a class of sets is called a monotone class if it is closed under countable increasing unions and countable decreasing intersections. It is readily verified that the class of all subsets of a set is a monotone class and that the intersection of any nonempty family of monotone classes is a monotone class; hence there is a smallest monotone class containing any given class of sets. The proof of the lemma below introduces the notation f and J, to denote increasing countable union and decreasing countable intersection, respectively.
Lemma 5.43 (Monotone Class Lemma). The smallest monotone-class M containing an algebra A of sets is identical to the smallest o-algebra A containing A. PROOF.We have M c because is a monotone class containing A. To prove the reverse inclusion, it is sufficient to show that M is closed under the operations of finite union and complementation, since a countable union can be written as the increasing countable union of finite unions. The proof is in three steps. First we prove that if A is in A and M is in M, then A U M and A f' M are in M. For fixed A in A, let UAbe the class of all sets M in M such that A U M and A f' M are in h4. Then U A 2 A. If we show that U A is a monotone class, then it will follow that UA 2 M . For this purpose let
2
Un
.r U
and
2
Vn J, V
with
U, and V, in UA.
By definition of U A ,the sets Un U A , U, n A , V, U A , and Vn f' A are in M . But
U , U A T U U A and V , U A J , V U A and
U,nAfUnA, V,nAJ, V n A .
Therefore U and V are in U A ,and UA is a monotone class.
7. Fubini's Theorem for the Lebesgue Integral
269
Second we prove that M is closed under finite unions. For fixed N in M, let UN be the class of all sets M in M such that N U M and N f' M are in M. Then UN 2 A by the previous step. The same argument as in that step shows that UN is a monotone class, and hence UN = M. Third we prove that M is closed under complements. Let N b e the class of all sets in M whose complements are in M . Then N > A, and it is enough to show that N i s a monotone class. If C, T C
and
D, J, D
with
C , and D, in N,
then C and D are in M since C, and D, are in M . Now C: J. C c and
D: T DC,
and by definition of N , C i and Di are in M . Therefore C Cand DC are in M , and C and D must be in N. That is, N i s a monotone class.
Lemma 5.44. If (X, A, p ) and ( Y ,B, v ) are o-finite measure spaces, then v ( E x )and p(EY) are measurable functions for every E in A x B. PROOF IF p ( X ) < +oo AND v ( Y ) < +oo. Lcl h l be 1l1cclass uf all sets E in A x B for which v ( E x )and p(EY) are measurable. We shall show that M is a monotone class containing the algebra C of finite disjoint unions of rectangles. If R = A x B is a rectangle, then v(Rx)=v(B)IA
and
p(RY)=p(A)IB,
and so R is in M . If E and F are disjoint sets in M, then
for each x , and similarly for p for each y . By Proposition 5.7, v ( ( E U F),) and p ( ( E U F)Y) are measurable. Hence E U F is in M , and M contains C. If { E n } and IFn}are increasing and decreasing sequences of sets in M ,then the finiteness arid coniplete additivity of v irniply that
and and similarly for p . Since the limit of measurable functions is measurable (Corollary 5 .lo), we conclude that M is a monotone class. Therefore M contains A x B by the Monotone Class Lemma (Lemma 5.43).
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V Lebesgue Measure and Abstract Measure Theory
PROOF FOR o-FINITE p AND v . Write X = Uz=l Xm and Y = U z l Yn disjointly, with p ( X m ) < +cc and v(Y,) < +cc for all m and n . Define Amand B, by $I, = { A ~ X , I A i s i n A }
and
B n = {BnY,I BisinB},
and define pm and v, on Amand t?, by restriction from p and v . Then the triples ( X m, Am,p,) and (Y, , B, , v,) are finite measure spaces, and the previous case applies. If E is in A x B, then Em, = E n ( X , x Y,) is in Am x B,, and so v ( ( E m , ) , ) and p(IEm,)Y) are measurable with respect to Amand t?, ,hence with respect to A and B. Thus
exhibit v ( E x )and p(EY) as countable sums of nonnegative measurable functions. They are therefore measurable. The next proposition simultaneously constructs the product measure and establishes Fubini's Theorem for indicator functions.
Proposition 5.45. Let (X, A, p ) and ( Y ,B, v ) be o-finite measure spaces. Then there exists a unique measure p x v on A x t? such that
for every rectangle A x B. The measure p x v is o-finite, and
for every set E in A x B. PROOF.In view of the measurability of v ( E x ) given in Lerntna 5.44, we can define a set function p on A x t? by
Then p(M) = 0, and p is nonnegative. On a rectangle A x B , we have P ( Ax B ) = w ( A ) v ( B )
(*I
since v ( ( A x B),) = v ( B ) I A .We shall show that p is completely additive. If { E n }is a disjoint sequence in A x B, then
7. Fubini's Theorem for the Lebesgue Integral
p(
(Un E,,)
=
v ( ( u E,),) d p ( x )
=
v ( u (E,),) d p ( x )
=
=
by definition of p
n
[C y*
by Lemma 541a
v ( ( ~ , ) , ) ]d p ( r ) since the sets ( E n ) , are disjoint for each fixed x
vE
nd (x)
by Corollary 5.27
Now X x Y = Urn,,( X , x Y,). Since p has just been shown to be completely additive and since p and v are o-finite, (*) shows that p is o-finite. Also, (*) completely determines p on the algebra C of finite disjoint unions of rectangles. By the Extension Theorem (Theorem 5.5), p is completely determined on the smallest o-algebra A x B containing C. Defining o ( E ) = Jy w(EY) d v ( y ) and arguing in the same way, we see that o is a measure on A x B agreeing with p on rectangles and determined on A x B by its values on rectangles. Thus we have p = o on A x 13, and can define p x v = p = o to complete the proof.
Lemma 5.46. I f f is a measurable function defined on a product space X x Y , then for each x in X , y H f ( x , y ) is a measurable function on Y , and for each y in Y , x H f ( x , y ) is a measurable function on X . PROOF.For each fixed x , the formula
exhibits the set on the left as a section of a measurable set, which must be measurable according to Proposition 5.42. The result for fixed y is proved similarly.
Theorem 5.47 (Fubini's Theorem). Let ( X , A, p ) and ( Y , B, v ) be o-finite measure spaces, and let (X x Y, "4 x B, p x 11) be the product measure space. If f is a nonnegative measurable function on X x Y , then f ( x , y ) d v ( y ) and f ( x , y) d p ( x ) arc mcasurablc, and
V Lebesgue Measure and Abstract Measure Theory
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PROOF.Lemma 5.46 shows that f ( x , y ) is measurable in each variable separately and hence that the inside integrals in the conclusion are well defined. If f is the indicator function of a measurable subset E of X x Y, then the theorem reduces to Proposition 5.45. The result immediately extends to the case of a simple function f 10. Now let f be an arbitrary nonnegative measurable f~~nction Find by Proposition 5.11 an increasing sequence of simple functions s, > 0 with pointwise limit f . The sequence of functions & s n ( x , y ) d v ( y ) is an increasing sequence of nonnegative functions, and each is measurable by what we have already shown for simple functions. By the Monotone Convergence Theorem (Theorem 5.25),
Therefore J, f (x, y ) d v ( y ) is the pointwise limit of measurable functions and is measurable. Similarly Jx f ( x , y ) d p ( x ) is measurable. For every n , the result for simple functions gives
By a second application of monotone convergence,
By a third application of monotone convergence, lim
1 [l
sn(x. y ) d v ( y ) ]dw(*) =
[lip
X
l
s n ( x , Y ) dvO.)] d ~ ( x ) .
Putting our results together, we obtain
L x Y
f
d(w
X
v) =
[l
f i x , Y ) d v i y ) ] dw(*).
The other equality of the conclusion follows by interchanging the roles of X and Y . Fubini's Theorem arises surprisingly often in practice. In some applications the theorem is applied at least in part to prove that an integral with a parameter is finite or is 0 for almost every value of the parameter. Here is a general result concerning integral 0.
7. Fubini's Theorem for the Lebesgue Integral
273
Corollary 5.48. Suppose that ( X ,A, p) and (Y,B, v) are o-finite measure spaces, and suppose that E is a measurable subset of X x Y such that
I
for almost every x [dp].Then p({x (x,y)
E
E}) = 0 for almost every y [dv].
REMARKS.In words, if the x section of E has v measure 0 for almost every x in X ,then the y section of E has p measure 0 for almost every y in Y . For example, if one-point sets in X and Y have measure 0 and if every x section of E is a finite subset of Y , then for almost every y in Y, the y section of E has measure 0 in X . PROOF.Apply Fubini's Theorem to I E . The iterated integrals are equal, and the hypothesis makes one of them be 0. Then the other one must be 0, and the conclusion follows. When one tries to drop the hypothesis in Fubini's Theorem that the integrand is nonnegative, some finiteness condition is needed, and the result in the form of Theorem 5.47 is often used to establish this finiteness. Specifically suppose that ,f is measurable with respect to A x B but is not necessarily nonnegative. The assumption will be that one of the iterated integrals
is finite. Then the conclusions are that (a) f is integrable with respect to p x v ; f (x,y) d v (y)is defined for almost every x [dp]; if it is redefined to (b) be 0 on the exceptional set, then it is measurable and is in fact integrable
[&I; (c) a similar conclusion is valid for f (x,y) dp(x); (d) after the redefinitions in (b) and (c), the double integral equals each iterated integral, and the two iterated integrals are equal. These conclusions follow immediately by applying Fubini's Theorem to f + and f - separately and subtracting. The redefinitions in (b) and (c) are what make the subtractions of integrands everywhere defined. One final remark is in order: The completion of A x B is not necessarily the same as the product of the co~npletionsof A and B, and thus the statement of Fubini's Theorem requires some modification if completions of measure spaces are to be used. We shall see in the next chapter that Bore1 sets in Euclidean space behave well under the formation of product spaces, but Lebesgue measurable sets do not. Thus it simplifies matters to stick to integration of Borel-measurable sets in Euclidean space whenever possible.
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8. Integration of Complex-Valued and Vector-Valued Functions Fix a measure space (X, A, p ) . In this chapter we have worked so far with measurable functions on X whose values are in R* ,dividing them into two classes as far as integration is concerned. One class consists of measurable functions with valucs in [O, fool, and wc dcfincd thc intcgral of any such function as a mcmbcr of [0, +a]. The other class consists of general measurable functions with values in R*. Tlle irlkgral in h i s Lase Lan end up being any lllirlg in R*, and Illere are some such functions for which the integral is not defined. It is important in the theory to be able to integrate functions whose values are complex numbers or vectors in Rm or C m ,and it will not be productive to allow the same broad treatment of infinities as was done for general functions with values in R*. On the other hand, it is desirable to have the flexibility with nonnegative measurable functions of being able to treat infinite values and infinite integrals in the same way as finite values and finite integrals. In order to have two theories, rather than three, once we pass to vector-valued functions, we shall restrict somewhat the theory we have already developed for general functions with values in R* . Let us label these two theories of integration as the one for scalar-valued nonnegative measurable functions and the one for integrable vector-valued functions. Thc first of thcsc thcorics has alrcady bccn cstablishcd and nccds no changc. Thc second of these theories needs some definitions and comments that in part repeat skps Lake11 with Ricrrrarlr~irltcgralivrl in Sc~Liuns1.5, 111.3, and 111.7 and in part are new. In applications of this second theory later, if the term "vector-valued" is not included in a reference to a function either explicitly or by implication, the convention is that the function is scalar-valued. In the theory for vector-valued functions, we shall be assuming integrability, and the integrability will force the function to have meaningful finite values almost everywhere. Our convention will be that the values are finite everywhere. This will not be a serious restriction for any function that can be considered integrable, since we can redefine such a function on a certain set of measure 0 to be 0, and then the condition will be met without any changes in the values of integrals. Thus let a function f : X + Cm be given. Since the function can have its image contained in Rm, we will be handling Rm-valued functions at the same time. Since m can be 1, we will be handling complex-valued functions at the same time. Since the image can be in Rm and n z can be 1, we will at the same time be recasting our theory of real-valued functions whose values are not necessarily nonnegative. We impose the usual Hermitian inner product ( . , . ) and norm I . I on Cm. The function 7 : X + Cm is the composition of f followed by complex conjugation in each entry of Cm. We can write f = Re f + i Im f , where
8. Integration o f Complex-Valuedand Vector-ValuedFunctions
275
&
Re f = $ (f + f ) and Im f = (f - f ) , and then the functions Re f and Im f take values in Rm. Following the convention in Section A7 of the appendix, let {u1 , . . . , urn}be the standard basis of Rm . By a basic open set in C m ,we mean a set that is a product in R2" of bounded open intervals in each coordinate. In symbols, such a set is centered at some vo E C m ,and there are positive numbers Cj and qj such that the set is {v
t
Cm I I(Re(v - uo), uj)l < $, and I(Im(v - vo), uj) 1 < qj for 1 5 j 5 m}.
We say that f : X + Cm is measurable if the inverse image under f of each basic open set in Cm is measurable, i.e., lies in A.
Lemma 5.49. A function f : X + Cmis measurable if and only if the inverse image under f of each open set in Cm is in A. PROOF.If the stated condition holds, then the inverse image of any basic open set is in A, and hence f is measurable. Conversely suppose f is measurable, and let an open set U in Cm be given. Then U is the union of a sequence of hasic open sets [I,, and the measilrahility of f , in comhination with the formnla f (u) = U, f (u,) , shows that f (u) is in A.
Proposition 5.50. A function f : X + Cm is measurable if and only if Re f and Im f are measurable. ~ O O F In . view of Lemma 5.49, we can work with arbitrary open sets in place of basic open sets. If U and V are open sets in Rm ,then the product set U i V is open in Cm, and f (U + i V) = (Re f )-I (U) f' (Im f )-I (V) . It is immediate that measurability of Re f and Im f implies measurability o f f . Conversely if we specialize this formula to V = R m ,then we see that measurability of f implies measurability of Re f . Similarly if we specialize to U = R m ,then we see that measurability of f implies measurability of Im f .
+
Proposition 5.51. The following conditions on a function f : X + Cm are equivalent: (a) f is measurable, (b) (f, v ) is measurable for each v in Cm, (c) (f, uj) is measurable for 1 5 j 5 m . REMARKS. When infinite-dimensional ranges are used in more advanced texts, (a) is summarized by saying that f is "strongly measurable," and (b) is summarized by saying that f is "weakly measurable."
V Lebesgue Measure and Abstract Measure Theory
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PROOF.Suppose (a) holds. The function in (b) is the composition off followed by the continuous function ( . , v) from Cm to C . The inverse image of an open set in C is then open in C m ,and the inverse image of the latter open set under f is in A. This proves (b). Condition (b) trivially implies condition (c). If (c) holds, then Proposition 5.50 shows that (Re f , uj) and (Im f , uj) are measurable from X into R. Thus the inverse image of any open interval under any of these 2m functions on X is in A. The inverse image of a basic open set in Cmunder f is the intersection of 2m such sets in A and is therefore in A. Hence (a) holds.
Proposition 5.52. Measurability of vector-valued functions has the following properties: (a) If f : X + Cm and g : X + Cm are measurable, then so is f + g as a function from X to Cm. (b) If f : X + Cm is measurable and c is in C , then cf is measurable as a function from X to Cm. (c) If f : X + Cm is measurable, then so is f : X + C m . (d) If f : X + C and g : X + C are measurable, then so is f g : X + C . (e) If f : X + Cm is measurable, then I f 1 : X + [0, +oo) is measurable. ( f ) If { f,} is a sequence of measurable functions from X into Cm converging pointwise to a filnction f : X + C m ,then f is measnrahle. PROOF.Conclusions (a) through (e) may all be proved in the same way. It will be enough to illustrate the technique with (a). We can write the function x H f (x) g(x) as a composition of x H (f (x), g(x)) followed by addition (a, b) H a +b . Let an open set in Cmbe given. The inverse image under addition is open in Cm x Cm, since addition is continuous (Proposition 2.28). The inverse image of a product U x V of open sets in C"' x Cnhnder x H (f (x), g (x)) is f (U) n gP1( V ) , which is in A because f and g are measurable, and therefore the inverse image of any open set in Cm x Cm under x H (f (x), g(x)) is in A. This handles (a), and (b) through (e) are similar. For (f), we apply Proposition 5.50 to f , and then we apply the equivalence of (a) and (c) of Proposition 5.51 for Ref and Im f . In this way the result is reduced to the real-valued scalar case, which is known from Corollary 5 .lo.
+
Tf E is a measurable subset of X , we say that a function f : X + @ is integrable on E if Re f and Im f are integrable on E , and in this case we define 1, f d~ = iE',Ref d ~ + i [ ~ ' E mdf ~ .
Proposition 5.53. Let E be a measurable subset of X . Integrability on E of functions from X to C has the following properties:
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V Lebesgue Measure and Abstract Measure Theory
Proposition 5.54. Let E be a measurable subset of X. Integrability of vectorvalued functions on E satisfies the following properties: (a) If f and g are functions from X into Cm that are integrable on E , then
f
+ g is integrable on E , and Jx (f + g) d p = Jx f d p + fx g d p .
(b) If f is a function from X into Cm that is integrable on E , then cf is integrable on E , and JE cf d p = c f d p . (c) A function f : X + Cm is integrable on E if and only if Re f and Im f are integrable on E , and then f d p = Re f d p i & Im f d p . (d) If f is a function from X into Cm that is integrable on E and if v is a member of Cm,then x H (f (x) , v) is integrable on E and (f (x) , v) dp(x) = (JEf (x) d/J(x), v). (e) If f is a measurable function from X into C"'such that If 1 is integrable on E , then f is integrable on E , and f (x) dp(x) 5 If (x) I dp(x). ( f ) (Dominated convergence) Let fn be a sequence of measurable functions from X into Cm intcgrablc on E and converging pointwisc to f . If thcrc is a measurable function g : X + [O, fool that is integrable on E and has f, (x) I 5 g(x) for all x in E , then f is integrable on E , limn JE fn d p exists in C m ,and limn JEfn d~ = JEf d p .
IE
+
IE
I IE
I IE
PROOF.All of the relevant questions about measurability are addressed by Propositions 5.50 and 5.52. Conclusions (a), (b) , (c), and ( f ) about integrability are immediate from Proposition 5.53. For (d), let v = C cjuj with each cj in C . Since f is by assumption integrable, (f, v) = (f, C cjuj) = CjCj (f, uj) exhibits (f, v) as a linear combination of functions integrable on E. Therefore (f, v ) is integrable on E . To obtain the formula asserted in (d), we first consider v = ui. Then the definition of iEf d~ gives (JE f d ~u i ,) = (Cj( j E ( f j uj) d ~ ) u j u, i ) = iE(f; ui) d ~ . Multiplying by t i and adding, we obtain (IE f d p , v) = JE (f, v) d p. This proves (d). For (e), let f : X + Cm be integrable on E. The asserted inequality is trivial if f d p = 0. Otherwise, for every v in C m ,
IE
I(/Efd~,u)I=IJE'E(f'u)d~I
1 ( f v) 1 d 5 lvl jE1 f 1 dw 5
JE
by Proposition 5 . 5 3 ~ by Proposition 5.16 and the Schwarz inequality.
Takingv = i E f d w g i v e s I i E f d w 1 2 5 I J E f f d ~ l J E ~ f ~ S idn ~ c e. J E f d w has been assumed nonzero, we can divide by its magnitude, and then (e) follows.
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9. L 1 , L2, LCO,and Normed Linear Spaces Let ( X , A, p ) be a measure space. In this section we introduce the spaces L ( X ), L2( x ),and L" ( X ). Roughly speaking, these will be vector spaces of functions on X with suitable integrability properties. More precisely the actual vector spaces of functions will form pseudometric spaces, and the spaces L (x), L ~ ( x )and , L m ( X ) will be the corresponding metric spaces obtained from the construction of Proposition 2.12. They will all turn out to be vector spaces over R or @. It will matter little whether the scalars for these vector spaces are real or complex. When we need to refer to operations with scalars, we may use the symbol F to denote R or @, and we call F the field of scalars. We shall make explicit mention of R or @ in any situation in which it is necessary to insist on a particular one of R or @. The three spaces we will construct will all be obtained by introducing "pseudonorms" in vector spaces of measurable functions. A pseudonorm on a vector space V is a function 11 . 11 from V to [O, foe) such that7 (i) IlxII 'Oforallx E V , (ii) llcxII = IclIlxll for all scalars c and allx E V , (iii) (triangle inequality) Ilx y 11 5 Ilx 11 1 1 y 1 1 for all x and y in V . We encountered pseudonorms earlier in connection with pseudo inner-product spaces; ill Propositioii 2.3 we saw liow to form a pseudoiioim froiii a pseudo inner product. However, only the pseudonorm for L 2 ( X ) arises from a pseudo inner product in the construction of L' , L,, and L" . The definitions of the pseudonorms in these three instances are
+
Ilflll
Ilf
=J;,lfld~
112 =
11 f 1 1 ,
+
( I xl f
I ~ ~ P ) ~ ' ~
= "essential supremum" of f
( x ), for L~ ( x ), for L
for LCO(X)
Once we have defined "essential supremum," all the above expressions are meaningful for any measurable function f from X to the scalars, and the vector space V in each of the cases is the space of all measurable functions from X to the scalars such that the indicated pseudonorm is finite. In other words, V consists of the integrable functions on X in the case of L ( X ), the square-integrable functions on X in the case of L 2 ( x ) ,and the "essentially bounded" functions on X in the case of L m ( X ). We need to check that I I . I 1 , , I I . I I,, and I I . I 1 , are indeed pseudonorms and that the spaccs V arc vector spaccs in each case. 7 ~ h word e "seminorm" is a second name for a function with these propeaies and is generally used in the context of a family of such functions. We shall not use the word "seminorm" in this text.
V Lebesgue Measure and Abstract Measure Theory
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For L' (X), properties (i) and (ii) are immediate from the definition. For (iii), we have I f (x) g(x)l 5 I f (x)l Ig(x)l for all x and therefore Il f gill = If + g I d p 5 JX I f I d p + I x Igldp = llf Ill + llgIl1. For L 2 ( x ) , let V be the space of all square-integrable functions on X. The space V is certainly closed under scalar multiplication; let us see that it is closed under addition. If f and g are in V , then we have
I*
+
+
+
+
for every x in X. Integrating over X, we see that f g is in V if f and g are in V . Also, the left side is > 41 f (x) I Ig(x) 1, and it follows that f is integrable whenever f and g are in V . Then the definition (f, g), = f fg d p makes V into a pseudo inner product-space in the sense of Section 11.1. Hence Proposition 2.3 shows that the function 11 . 112 with Il f 112 = ( f , f ):I2 is a pseudonorm on V . For Lm(X),we say that f is essentially bounded if there is a real number M such that I f (x) 1 5 M almost everywhere [dp]. Let us call such an M an essential bound for If 1 . When f is essentially bounded, we define 11 f 1 , to be the infimum of all essential bounds for I f I . This infimum is itself an essential bound, since the countable union of sets of measure 0 is of measure 0. The infimumof the essential bounds is called the essential supremum of I f 1 . Certainly 11 . 1 , satisfies (i) and (ii). If If 1 is bounded a.e. by M and if IgI is bounded a.e. by N , then If + gl is bounded everywhere by I f I + Ig 1 , which is bounded a.e.by M +N . It follows that f + g is essentiallyboundedand If fgllw 5 Illf l + Iglll, 5 Ilf I , + llgll,. So (iii) holds for 11 . 1 ,. A real or complex vector space with a pseudonorm is a pseudo normed linear space. Such a space V becomes a pseudometric space by the definition d ( f ,g ) = 11 f - gll, according to the proof of Proposition 2.3. Proposition 2.12 shows that if we define two members f and g of V to be equivalent whenever d ( f , g) = 0, then the result is an equivalence relation and the function d descends to a welldefined metric on the set of equivalence classes. If we take into account the vector space structure on V , then we can see that the operations of addition and scalar multiplication descend to the set of equivalence classes, and the set of equivaleiice classes is tlieii also a vector space. Tlie argument for addition is tliat i f d ( f l , f2) =Oandd(gl,g2) =O,thend(fl + g l , f2+g2)isObecause
IE
The argument for scalar multiplication is similar, and one readily checks that the space of equivalence classes is a vector space.
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This construction is to be applied to the spaces V we formed in connection with integrability, square integrability, and essential boundedness. The spaces of equivalence classes in the respective cases are called L 1( X ),L 2 ( x ) ,and L m ( X ) . These spaces of equivalence classes are pseudo normed linear spaces with the additional property that 11 f 11 = 0 only for the 0 element of the vector space. ) If there is any possibility of confusion, we may write ~ l ( or~ L )1 ( X ,~ r or L' ( X , A, p ) in place of L ( X ), and similarly for L2 and L a . A pseudo normed linear space is called a normed linear space if 1 1 f 11 = 0 implies f is the 0 element of the vector space. Thus L' ( X ), L ~ ( x )and , LW(X) are normed linear spaces. In practice, in order to avoid clumsiness, one sometimes relaxes the terminology and works with the members of L ( X ), L2 ( x ), and L" ( X ) as if they were functions, saying, "Let the function f be in L' ( X ) " or "Let f be an L' function." There is little possibility of ambiguity in using such expressions. The 1-dimensional vector space consisting of the field of scalars IF with absolute value as norm is an example of a normed linear space. Apart from this and IFm, we have encountered one other important normed linear space thus far in the book. This is the space B ( S ) of bounded functions on a nonempty set S . It has various vcctor subspaccs of intcrcst, such as thc spacc C ( S ) of boundcd continuous functions in the case that S is a metric space. The norm for B ( S ) is the supremum norm or the uniform norm defined by
'
I l f l,
= sup I SES
f ($11
The corresponding metric is d ( f ,g )
=
Ilf
-
gll,,,
=
sup I f ( s ) - g(s)I, sES
and this agrees with the definition of the rnetric in the exarnple in Chapter 11. Proposition 2.44 shows that the metric space B ( S ) is complete. Any vector subspace of B ( S ) is a normed linear space under the restriction of the supremum norm to the subspace. In working with specific normed linear spaces, we shall often be interested in seeing whether a particular subset of the space is dense. In checking denseness, the following proposition about an arbitrary normed linear space is sometimes helpful. The illtersectioll of vector subspaces of X is a vector subspace, and the intersection of closed sets is closed. Therefore it makes sense to speak of the smallest closed vector subspace containing a given subset S of X .
Proposition 5.55. If X is a normed linear space with norm 11 . 11 and with IF as field of scalars, then (a) addition is a continuous function from X x X to X , (b) scalar multiplication is a continuous function from IF x X to X ,
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V Lebesgue Measure and Abstract Measure Theory
(c) the closure of any vector subspace of X is a vector subspace, (d) the set of all finite linear combinations of members of a subset S of X is dense in the smallest closed vector subspace containing S.
+
+
+
PROOF.The formula I l ( x y ) - (xo yo) ll 5 Ilx - xo ll I l y - yo ll shows continuity of addition because it says that if x is within distance t / 2 of xo and y is within distance t / 2 of yo, then x + y is within distance t of xo +yo. Similarly the formula Icx -coxoII i Ilc(x -xo)II II(c-co)xoII = IclIlx -xoII Ic-coIIxoII showsthat llcx-coxoII 5 6 ( ~ c o ~ + l ) + 6 assoonass ~ ~ x o ~ ~ 5 1,lc-coI 5 6,and Ilx-xoII 5 8. IftwithO it 5 lisgivenandifweset8 = ( 1 c ~ 1 + 1 + l l x ~ 1 l ) - ~ t , then we see that Ic - co i 6 and Ilx - xo 11 i 6 together imply llcx - cox0 II i t . Hence scalar multiplication is continuous. This proves (a) and (b). From (a) and (b) it follows that if x, + x and y, + y in X and c, + c in IF, then x, + y, + x + y and c,x, + c x . This proves (c). For (d), the smallest closed vector subspace Vl containing S certainly contains the closure V2of the set of all finite linear combinations of members of S. Part (c) shows that V2 is a closed vector subspace, and hence the definition of V1implies that Vl is contained in V2. Therefore Vl = V2,and (d) is proved.
+
+
Proposition 5.56. Let ( X , A, p ) be a measure space, and let p = 1 or p = 2. Then every indicator function of a set of finite measure is in L P ( X ) , and the smallest closed subspace of L P ( X ) containing all such indicator functions is LP ( X ) itsclf.
REMARK.Proposition 5.55d allows us to conclude from this that the the set of simple functions built from sets of finite measure lies in both L' ( X ) and L ~ ( x ) and is dense in each. It of course lies in L m ( X ) as well, but it is dense in L m ( X ) if and only if p ( X ) is finite.
ix
( 1 ~d p) = ~ p(E) PROOF.If E is a set of finite measure, then the equality shows that I E is in LP for p = 1 and p = 2. In the reverse direction let V be the smallest closed vector subspace uf LP containing all indicator functions of sets of finite measure. Suppose that s = CkckIEk is the canonical expansion of a simple function s 2 0 in LP and that ck > 0. The inequalities 0 5 ck ZEk (I s imply that ck ZEk is in LP . Hence ZEk is in LP, and w ( E k )is finite. Thus every nonnegative simple function in LP lies in V . Let f 3 0 be in LP, and let s, be an increasing sequence of sin~plefunctions 0 with pointwise limit f . Since 0 5 s, 5 f , each s, is in LP. Since If - snip has pointwise limit 0 and is dominated pointwise for every n by the integrable function If lP, dominated convergence gives lim Jx If - S, l P d p = 0. Hence s, tends to f in LP. Combining this conclusion with the result of the previous paragraph, we see that every nonnegative LP function is in V . Any LP function
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283
is a finite linear combination of nonnegative LP functions, and hence every LP function lies in V . Let us digress briefly once more from our study of L ' , L ~and , L" to obtain two more results about general normed linear spaces. A linear function between two normed linear spaces is often called a linear operator. A linear function whose range space is the field of scalars is called a linear functional. The following equivalence of properties is f~lndamentaland is often used without specific reference.
Proposition 5.57. Let X and Y be normed linear spaces that are both real or both complex, and let their respective norms be 11 . I l x and 11 . 11 y . Then the following conditions on a linear operator L : X + Y are equivalent: (a) L is uniformly continuous on X , (b) L is continuous on X , (c) L is continuous at 0 , (d) L is bounded in the sense that there exists a constant M such that
forallx in X . PROOF.If L is uniformly continuous on X, then L is certainly continuous on X. If L is continuous on X , then L is certainly continuous at 0 . Thus (a) implies (b), and (b) implies (c). If L is continuous at 0 , find 6 > 0 for t = 1 such that Ilx - Oilx 5 6 implies IIL(x) - L(0)lly 5 1 . Here L ( 0 ) = 0 . If a general x 0 is given, then IlxIlx # 0 , and the properties of the norm give I I ( S / ~ ~ X I ~=~ 8. ) X Thus I ~ ~ 11 L ((SlllxI x ) x )11 5 1 . By the linearity of L and the properties of the norm, ( 6 / 1 1 x 1 1 ~ ) 1 1 ~5( 1x.)Therefore ll~ IIL(x)ll, 5 6-I 11x1l~, and L is bounded with M = 8-l. Thus (c) implies (d). If L is bounded with constant M and if c > 0 is given, let S = c / M . Then llxl - x211x 5 6 implies
+
Thus (d) implies (a). If L : X + Y is a bounded linear operator, then the infimum of all constants M such that IIL(x)lly 5 Mllxllx forallx in X is again such aconstant,anditis called the operator norm 11 L 11 of L . Thus it in particular satisfies
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V Lebesgue Measure and Abstract Measure Theory
As a consequence of the way that L and the norms in X and Y interact with scalar multiplication, the operator norm is given by the formulas
except in the uninteresting case X = 0. It is easy to check that the bounded linear opcrators from X into Y form a vcctor spacc, and thc opcrator norm makcs this vector space into a normed linear space that we denote by B ( X , Y ) . When the domain and range are the same space X , we refer to the members of B ( X , X ) as bounded linear operators on X . The normed linear space B ( X , X ) has a multiplication operation given by composition. When Y is the field of scalars F , the space B ( X , F ) reduces to the space of continuous linear functionals on X . This is called the dual space of X and is denoted by X* . For example, if X = L (w) , then every member g of L" ( y ) defines a member x i of X* by x i ( f ) = f g d y for f in L' ( y ) ; the linear functional x i has Ilxi 1 1 5 Ilglloo. We shall be interested in two kinds of convergence in X * . One is norm convergence,in which a sequence { x ; } converges to an element x* in X* if Ilx,T - x* 1 1 tends to 0. The other is weak-star convergence, in which {x,"}converges to x* weak-star against X if limn x," ( x ) = x* ( x ) for each x in X .
'
Theorem 5.58 (Alaoglu's Theorem, preliminary form). If X is a separable normed linear space, then any sequence in X * that is bounded in norrrl has a subsequence that converges weak-star against X . R~MAKKS. In Chapter VI we shall see that L' and L 2 are separable in the case of Lebesgue measure on IK1 and in the case of many generalizations of Lebesgue measure to N-dimensional Euclidean space. PROOF.Let a sequence {x:}:=~ be given with Ilx; 1 1 5 M , and let { x k } be a countable dense set in X . For each k , we have Ix,"(xk)I i Ix,"IIllxkII i MllxkII, and hence the sequence { x i ( x ~ ) } : = ~of scalars is bounded for each fixed k . By the Bolzano-Weierstrass Theorem, {x,"( x k ) } z lhas a convergent subsequence. Since we can pass lu a cunvergcn~subsequence uf any subseque~~ce fur any parlicular k , we can use a diagonal process to pass to a single convergent subsequence { ~ , * ~ } f " = , such that liml x:/ ( x k ) exists for all k . Now let xo be arbitrary in X , let t > 0 be given, and choose xk in the dense set with llxk - xo 1 1 < t . Then
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Thus lim sup Ix:, (xo)- x:,, ( x o )I 5 2 M t . Since t is arbitrary, we conclude that l,l'+oo
{x,*,(xo)},",,is a Cauchy sequence of scalars. It is therefore convergent. Denote the limit by x * ( x o ) ,so that liml xi, ( x o ) = x * ( x o )for all xo in X . Since limits respect addition and multiplication of scalars, x* is a linear functional on X . The computation Ix*(xo)l = lim~x,",(xo)l= liml x,*,(xo)l i limsup, llxHllllxoll i M 11x0 11 shows that x* is bounded. Hence { x ~ , converges } ~ ~ to x* weak-star against X .
Now, as promised, we return to L l , L ~and , Lm . The completeness asserted in the next theorem will turn out to be one of the key advantages of Lebesgue integration over Riemann integration.
Theorem 5.59. Let ( X ,A, p) be any measure space, and let p be 1, 2, or oo. Any Cauchy sequence { f k } in LP has a subsequence { fkn} such that I l fkn - fk,,,l i p 5 C,,,i,(,,,) with EnC, < fm. A subsequence { fkn} with this property is necessarily Cauchy pointwise almost everywhere. If f denotes the almost-everywhere limit of { f,,}, then the original sequence { f k } converges to f in LP. Consequently these three spaces LP ,when regarded as metric spaces, are complete in the sense that every Cauchy sequence converges.
REMARKS.The broad sweep of the theorem is that the spaces L ' , L ~ and , are complete. Rut the detail is important, tan. First of all, the detail allows us to conclude that a sequence convergent in one of these spaces has an almost-everywhere convergent subsequence. Second of all, the detail allows us to conclude that if a sequence of functions is convergent in LP1 and in Lp2, then the limit functions in the two spaces are equal almost everywhere. I,"
PROOF.Let { f,} be a Cauchy sequence in LP . Inductively choose integers nk by defining no = 1 and taking nk to be any integer > nk-1 such that 11 f, - f,, 11, 5 2-k for m > nk; we can do so since the given sequence is Cauchy. Then the subsequence { f k n }has the property that 11 fm - f, l p 5 2-min{m,n) for all m > 1 and n 3 1. This proves the first conclusion of the theorem. Now suppose that we have a sequence {.f,} in LP such that Il .f, - .f, 1 1 , 5 Cminjm,,) with EnC, = C < +oo. We shall prove that { f,} is Cauchy pointwise almost everywhere and that if f is its almost-everywhere limit, then fn tends to f in LP. First suppose that p < oo.Let g, be the function from X to [0, +co] given by
V Lebesgue Measure and Abstract Measure Theory
286
and define g ( x ) = lim g, ( x ) pointwise. Then
(/x gn dp)liP = Ilgnll, 5
llfl
By monotone convergence, we deduce that g is finite a.e., and consequently the series
21
fk(x) - f k P l ( x )
lP
+C
llP
l l f k - fk-1
k=2
( 1, g d p )
= Ilg 1
,
is finite. Thus
converges in W for a.e. x [ d p ] .
k=2
(**I
By redefining the functions fk on a set of p measure 0, we may assume that the series (**) converges pointwise to a limit in R for every x . Consequently the series M
is absolutely convergent for all x and must be convergent for all x . The partial sums for the series without the absolute value signs are f, ( x )- fl ( x ) ,and hence f ( x ) = lim f, ( x ) exists in W for every x . For every n ,
and we have seen that gP is integrable. By dominated convergence, we conclude thatlim,i, I f - f,lPdp = J,lim, I f ( x ) - f,(x)IPdp(x) = O . Inotherwords, lim, 11 f - f,llp = 0. Therefore f, tends to f in L P ( p ) . Next suppose that p = oo.Let { f,} be any Cauchy sequence in LCO. For each m and n , let Em, be the subset of X where I , f , - ,f, I > 11 ,fm - ,f, 1 1 , , and put E = Urn,,Em,. This set has measure 0. Redefine all functions to be 0 on E . The redefined functions are then uniformly Cauchy, hence uniformly convergent to some function f , and then f, tends to f in L m ( X ) . For any p , we have shnwn that the original Cauchy sequence { f,%}has a convergent subsequence { f,,} in LP. Let f be the LP limit of the subsequence. Given t > 0, choose N such that n 2 m 2 N implies 11 f, - fm 11, 5 t ,and then choose K such that 11 f,, - flip 5 t fork 1K . Fix k ? K with nk ? N . Taking m = nk, we see that II f, - f 1 1 , 5 I I f, - f,, 11, I f , , - f 11, 5 2t- whenever n > nk. Thus { f,} converges to f . This completes the proof of the theorem.
+
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287
In Section 9 we introduced integration of functions with values in Rm or C m . The definitions of L 1 , L ~and , L" may be extended to include such functions, and we write L 1 ( x ,C m ) ,for example, to indicate that the functions in question take values in Cm. In the definitions any expression If ( x )I or If 1 that arises in the definition and refers to absolute value in the scalar-valued case is now to be understood as referring to the norm on the vector space where the functions take , Lm spaces are further normed linear their values. The vector-valued L 1 ,L ~and spaces, and one readily checks that Theorem 5.59 with the above proof applies to them because the range spaces are complete. The triangle inequality for a pseudo normed linear space says that the norm of the sum of two elements is less than or equal to the sum of the norms, and of course the inequality instantly extends to a sum of any finite number of elements. But what about an integral of elements? In the case that the linear space is one of the precursor spaces "V" for L ' , L2, or L m , the setting is that of functions of two variables. One of the variables corresponds to the measure space under study, and the other corresponds to the indexing set for the integral of the norms. Thus we could, if we wanted, force the situation into the mold of vector-valued functions whose values are in a space of functions. But it is not necessary to do so, and wc do not. Hcrc is thc thcorcm.
Theorem 5.60 (Minkowski's inequality for integrals). Let ( X ,A, w ) and (Y, B, v) be o-finite measure spaces, and put p = 1,2,or m. If f is measurable o n X x Y.then
in the following sense: The integrand on the right side is measurable. If the integral on the right is finite, then for almost every y [dv] the integral on the left is defined; when it is redefined to be 0 for the exceptional y 's, then the formula holds. REMARK.An extension of this theorem to values of p other than 1 , 2 , oo will be given in Chapter IX, and that result will have the same name. PROOF.Thc right sidc of thc intcgral formula is unchangcd if wc rcplacc f by If 1 , and thus we may assume that f 10 without loss of generality. If p = 1, then the formula for f > 0 reads
In fact, equality holds, and the result just amounts to Fubini's Theorem (Theorem 5.47).
V Lebesgue Measure and Abstract Measure Theory
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Let p
= 2.
We have 2 Ilf ( ~ ~)112,dv(y) 9 =
and this is measurable by Fubini's Theorem. Hence I l f (x, Y ) I I ~ , ~is, ,measur(~) able. The idea for proving the inequality in the statement of the theorem is to imitate the argument that derives the triangle inequality for L~ from the Schwarz inequality. That earlier argument is
The adapted argument is
the second and third lines following from Fubini's Theorem and the Schwarz inequality. Let p = cc.This is the hard case of the proof. We proceed in three steps. The first step is to prove the asserted measurability of 11 f (x, y) Il,,dviy), and we do so by first handling simple functions and then passing to the limit. If s = c, IE, is the canonical expansion of a simple function s > 0 on X x Y and if x is fixed, then Ils(x, y) I , ds(y) = max {c, v((E,),) > 0 ) . Inotherwords,ifk, is the indicator function of the set {x E X v ((En),) > 0}, then s = max{clkl, . . . , cNkN}. Each function c,k, is measurable by Lemma 5.44, and the pointwise maximum s is measurable by Corollary 5.9. Returning to our function f > 0, we use Proposition 5.11 to choose an increasing sequence {s,} of nonnegative simple functions with pointwise limit f . We prove that Ilsn(x, y) lloo,d,,(y) increases to II f (x,y) l,() for each x ,and then the measurability follows from Corollary 5 .lo. Since x is fixed in this step, let us drop it and consider an increasing sequence {s,} of nonnegative measurable functions on Y with limit f on Y; we are to show that 11 ,f lion = lim 11 sn1 ,. The numbers 11 sn11 , are monotone increasing and are 5 I I f 1 ,. Thus lim llsn 1 , 5 Il f 1 ,. Arguing by contradiction, suppose that equality fails and thatlimlls,II, 5 M < M + t < 11 f 1 ,. Then {y s,(y) M + E } hasmeasure0 for every n , and so does U, {y s,(y) > M c ) , by complete additivity. On the other hand, { y f (y) > M + t-} is a subset of this union, and it has positive measure since M + t < 11 f 1 ,. Thus we have a contradiction and conclude that lim IIsnII, = Ilf 1 ,. Consequently IIf (x, ~ ) l l , , ~ , , is ( ~measurable, ) as asserted.
xL1
I
I
I
I
+
I
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10. Problems
The second step is to prove that any measurable function F > 0 on Y has 11 F 1 , = sup, Fg d v where the supremum is taken over all g 2 0 with llglll 5 1. Certainly any such g has L ~ g d v l5 IIFII,[ygd~ = IIFII,, and therefore sup, F g dv 5 11 F 1 ,. For the reverse inequality, let IE be the indicator function of a set of finite positive measure, and put g = v ( E ) - l IE . Then F g d v = V(E)-' F d v 3 infE(F). If m is less than 11 FII,, then the set E where F is > m has positive measure, and the inequality reads m 5 Fg dv for the associated g . Hence m 5 sup, Fg d v . Taking the supremum of such m's, we obtain 11 F 1 , 5 sup, Fg d v and the reverse inequality is proved. The third step is to use the previous two steps to prove the inequality in the statement of the theorem for f > 0. Let g be any nonnegative function on Y with iyg dv 5 1. Thcn Fubini's Thcorcm, thc rcsult of thc first stcp abovc, and thc result in the easy direction of the second step above give
I& 1, I& I
Iy
I
iE
I d;
&
&
1,
Taking the supremum over g and using the result in the hard direction of the second step, we obtain the inequality in the statement of the theorem.
10. Problems 1. Let X be a finite set of n > 0 elements. (a) If A is an algebra of subsets, what are the possible numbers of sets in A? (b) Show that symmetric difference A A B = ( A - B ) U ( B - A ) is an abelian group operation on the set of all subsets of X and that every nontrivial element has order 2. (c) If B is a class of subsets containing 0 and X and closed under symmetric difference, what are the possible numbers of sets in B? (d) Prove or disprove: The class of sets in (c) is necessarily an algebra of sets. (e) Show that intersection and symmetric difference satisfy the distributive law A n (BAC) = ( A nB ) A ( A n C ) . 2. Exhibit a completely additive set function p on a a-algebra and two sets A and B such that p ( A ) < 0 and p ( B ) < 0 but p ( A U B ) > 0 .
3. Let { E n }be a sequence of subsets of X , and put 00
00
00
00
A = ~ U andE ~B = U ~ E ~
V Lebesgue Measure and Abstract Measure Theory
290
Prove that the indicator functions of E k , A , and B satisfy ZA = lim sup ZEn
and
I B = lim inf ZEn n
n
Suppose that p is a finite measure defined on a a-algebra and { E n }is a sequence of measurable sets with
n = l k-n
n = l k-n
Call the set on the two sides of this equation E . Prove that limn p(E,) exists and cquals p ( E ) . Let X be the set of rational numbers, and let R be the ring of all finite disjoint unions of bounded intervals in X , with or without endpoints. For each set E in R,let p ( E ) be its length. (a) Show that p is nonnegative additive. (b) Show that p is not completely additive. Prove that if E is aLebesgue measurable subset of [0, 11 of Lebesgue measure 0 , then the complement of E is dense in [0, 11. Let p be a measure defined on a a-algebra. Prove that if the complement of p ( A ) is finite every set of measure +cc is of finite measure, then sup,(,),+, and there is a set B with p ( B ) = SUP^(^)<+^ p ( A ) . If f is a measurable function, prove that f Rorel subset of the real line.
-'( E )is measurable whenever E is a
For the measure space ( X , A, p ) in which X is the positive integers, A consists of all subsets of X , and p is the counting measure, the theory of Lebesgue integration becomes a theory of infinite series. Restate Fatou's Lemma and the Dominated Convergence Theorem in this context. Suppose on a finite measure space that { f n } is a sequence of real-valued integrable functions tending uniformly to f . Prove that limn Jx fn d p = Jx f d p . This problem involves a Cantor set C in [0, 11 built using fractions rn as in Section 11.9. (a) Show that C has Lebesgue measure ( 1 - r,). (b) Prove that the indicator function Zc is discontinuous at every point of C and only there. Thus the set of discontinuities of Zc is not of measure 0 if ( 1 - rn) > 0. (c) Show that if the result of redefining Ic on a set of Lebesgue measure 0 is a function f , then the only possible points of continuity of f are those where f isO. (d) Conclude that there exists a Lebesgue measurable function on [0, 11 that is not Riemann integrable and cannot be redefined on a set of measure 0 so as to be Riemann integrable.
nzl
nzl
29 1
10. Problems
12. Let ( X , A, p ) be any measure space, and let ( X , A, 2 be its completion. Prove that if f is a function measurable with respect to 3,then f can be redefined on a set of jZ-measure 0 so as to be measurable with respect to A. 13. Let X be an uncountable set, and let A be the set of all countable subsets of X and their complements. Prove that the diagonal { ( x , x ) x E X } is not a member uT the a-algebra A x A, ~ I srrrallesl C a-algebra ~ u r ~ l a i r ~all i r ~r g e~~anglw ss i ~ ~ sides in A.
I
14. Let (R', B, m ) be the real line with Lebesgue measure on the Bore1 sets, and let ( X , A, p ) be a a-finite measure space. If f > 0 is a measurable function on X , prove that the "region under the graph of f ," defined by
R
=
I
{ ( x ,Y ) 0 5 Y <
f
(XI),
is a measurable subset of X x R1 and that its measure relative to p x m is f ( X I dP ( X I . 15. Let A b e aa-algebraof subsets of anonempty set X,let F : C n lx . . . xCnk + C N be continuous, and let f , : X + Cnj be measurable with respect to A for 1 < j < k . Prove that x H F (fl ( x ) , . . . , fk ( x ) ) is measurable with respect to A.
1,
16. This problem complements the proof in Theorem 5.59 that L' is a complete metric space. For a > 1, suppose that 0 < a, < 1 and C z la, = +oo. Find a measure space ( X , A, p ) and a sequence of functions f, with I f , 11 = a, and { fn (x) } convergent for no x . 17. (Egoroff's Theorem) Let ( X , A, p ) be a finite measure space. Suppose that f, and f are measurable functions with values in R such that lim f,(x) = f ( x ) pointwise. The objective of this problem is to prove that lim f, = f "almost uniformly." By considering the sets EM, = { x t X
I f n ( x )- f(x)I < l / M f o r n
> N}
for M fixed and N varying, prove that if c 0 is given, then there exists a measurable subset E of X with p ( E ) < such that lim f, ( x ) = f ( x )uniformly forx in EC. 18. (a) Derive the Dominated Convergence Theorem for a space of finite measure from Egoroff's Theorem (Problem 17) and Corollary 5.24. (b) Derive the Doiiiiiiated Colivergeiice Theorem for a space of iiifiiiite nieasme from the Dominated Convergence Theorem for a space of finite measure. Problems 1Y-21 use Egoroff's Theorem (Problem 17) to show how close pointwise convergence is to L' convergence on a measure space ( X , A, p ) of finite measure. Theorem 5.59 shows that if a sequence converges in L'(X), then a subsequence converges almost everywhere. These problems address the converse direction in a way different from Problem 16. Suppose that f, and f are integrable functions with values in R such that lim f, ( x ) = f ( x ) pointwise.
V Lebesgue Measure and Abstract Measure Theory
292
19. Suppose that f, lim, f, dp =
1,
> 0 for all n and that lim1,
f, d p = f d p for every measurable set E .
lE
1,
1, f d p .
Prove that
20. Suppose that f, > 0 for all n and that lim f, dp = Jx f dp . Use the previous problem and Egoroff's Theorem to prove that lim If, - f 1 d p = 0 .
1,
21. A sequence {g,} of nonnegative integrable functions is called uniformly integrable if for any c > 0 , there is an N such that f n ( x ) > N l gn d p .L c for all a . Suppose that the members of the given convergent sequence {,f,J are nonnegative. Using Egoroff's Theorem in one direction and the previous problem in the converse direction, prove that lim, f, d p = f dp if and only if the f, are uniformly integrable.
1,
1,
Problems 22-24 concern the extension of measures beyond what is given in Theorem 5.5 and Proposition 5.37. Let p be a finite measure on a a-algebra A of subsets of X , and define p, and p* on all subsets of X as in Lemma 5.32 and immediately after it. Let E be a subset of X that is not in A, and let B be the smallest a-algebra containing E and the members of A. 22. Show that there exist two sets K and U in A such that K g E g U , p * ( E ) = p ( K ), and p* (E) = p ( U ). Show that K and U have the further properties that U r g E r g K r ,p * ( E r )= p ( U r ) ,and p * ( E r )= p ( K r ) . 23. Show that the sets K and U of the previous problem satisfy p* ( A nE ) = p ( A nK ) and p * ( A n E ) = p ( A n U ) for every A in A. 24. Fix t in [0, 11. Show that the set function a defined for A and B in A by a [ ( An E ) U ( B n E C ) ]
is defined on all of 8, is a measure, agrees with p on A, and assigns measure t p * ( E ) ( 1 - t ) p * ( E )to the set E .
+
Problems 25-33 concern a construction by "transfinite induction" of all sets in the smallest a-algebra containing an algebra of sets. In particular, it describes how to obtain all Bore1 sets of the iiltei-val [O, 11 of the line fi-om the eleillentary sets ill that interval. Later problems in the set apply the construction in various ways. This set of problems makes use of partial orderings as described in Section A9 of the appendix, but they do not use Zom's Lemma. The set of countable ordinals is an uncountable partially ordered set Q, under a partial ordering 5 , with the following properties: (i) Q has the property that x ( y and y ( x together imply x - y , Q is "totally ordered" in the sense that any x and y in the set have either xiyoryix, (iii) C2 is "well ordered" in the sense that any nonempty subset has aleast element, (iv) for any x in Q, the set of elements 5 x is at most countable.
(ii)
Take as known that such a set Q exists.
10. Problems
293
25. Prove that any countable subset of Q has a least upper bound. 26. This problem asks for a proof of the validity of transfinite induction as applied to Q. Let 1 be the least element of Q, and let "<" mean "< but not =I' Suppose that some p(w) is specified for each w in Q. Suppose further that p ( 1 ) is true and that if for each w > 1, p ( w f ) is true for all w' < w , then p ( w ) is true. Prove Llral y ( w ) i b lrut: lur all w ~ I Q. I 7. Let X be a nonempty set, let A be an algebra of subsets of X, and let B be the smallest a-algebra containing A. This problem uses '2 to describe "constructively" B in terms of A. We define by transfinite induction two successively larger classes of sets U, and K , for each countable ordinal a > 1. Let U l be the set of all countable increasing unions of members of A, let K , for ac > 1 be the set of all countable decreasing intersections of members of U,, and let Ua for ac > 1 be the set of all countable increasing unions of members of previous Kg's. (a) Prove at each stage ac that 24, and K, are both closed under finite unions and finite intersections. (b) Prove that B is the union of all K , for ac in Q.
28. For the case that v(X) < t o o , prove the uniqueness half of the Extension Theorem (Theorem 5.5) by using the transfinite construction of Problem 27. [Educationalnote: It is not known how to prove the existence half of the Extension Theorem in this "constructive" way.] 29. Prove the Monotone Class Lemma (Lemma 5.43) by making use of the transfinite construction of Problem 27.
30. Devise a transfinite construction of all finite-valued Borel measurable functions on R1 that starts from continuous functions and alternately allows pointwise increasing limits and pointwise decreasing limits. The construction is to be in the spirit of Problem 27. Show that all finite-valued Borel measurable functions are obtained in this way if the indexing is done with '2. 3 1. This problem "counts" the number of Borel sets of the real line, using Problem 27. It uses the material on cardinality in Section A10 of the appendix. (a) Prove that (i) '2 has the same cardinality as some subset of R, (ii) the set of all sequences of members of R has the same cardinality as R, (iii) if A g B g C and if A and C have the same cardinality as R,then so does B, (iv) if a set A has the same cardinality as R and if for each a in A, B, is a set with the same cardinality as R,then UatAB, has the same cardinality as R. (b) Deduce that the set of all Borel sets of R has the same cardinality as R itself.
294
V Lebesgue Measure and Abstract Measure Theory
32. The standard Cantor set C in [0, 11, built using fractions r , = 1/3 as in Section 11.9, is a Borel set of Lebesgue measure 0 by Problem 11. Prove that C has the same cardinality as W.Conclude that the cardinality of the set of all Lebesgue measurable sets equals the cardinality of the set of all subsets of R. [Educational note: From this and Problem 3 1 it follows that there exists aLebesgue measurable set in [0, 11 that is not a Borel set.] 33. For the standard Cantor set C as in the previous problem, show that the indicator function Zc, of any subset C' of C is continuous on C". Conclude that the cardinality of the set of Riemann integrable functions on [0, 11 equals the cardinality of the set of all subsets of W.[Educational note: From this and Problems 30-3 1, it follows that there exists a Riemann integrable function on [0, 11 that is not Borel measurable .] Problems 3 4 4 1 show how to produce nontrivial nonnegative additive set functions on the set of all subsets of an infinite set from Zom's Lemma (Section A9 of the appendix). A filter F on a nonempty set X is a nonempty class of subsets of X such that (i) if E is in F and F 2 E, then F is in F , i.e., F i s closed under the operation of forming supersets, (ii) if E and F are in F , so is E f' F, (iii) 0 is not in F . An ultrafilter is a filter that is not properly contained in any larger filter.
34. Verify the following: (a) {X} is a filter. (b) Any filter is closed under finite intersections. (c) A one-point set and all of its supersets form an ultrafilter. (Such an ultrafilter is called a trivial ultrafi1ter.j (d) If X is infinite, then the set F of all subsets whose complements are finite sets is a filter. 35. Use Zorn's Lemma to show that every filter is contained in some ultrafilter. 36. Show that if C is a nonempty class of subsets of X, then there is a filter containing C if and only if no finite intersection of members of C is empty. 37. Prove that a filter F i s an ultrafilter if and only if A U B in F implies that either Ais in F o r B isin F. 38. Prove that a filter F is an ultrafilter if and only if for every A F o r Ac is in F .
c X , either A is in
39. Prove that the nonzero additive set functions defined on the set of all subsets of a set X and having image 10, l } stand in one-one correspondence with the ultrafilters on X, the correspondence being that the sets in the ultrafilter are exactly the sets on which the set function is 1. Prove that the set function is a measure if and only if the corresponding ultrafilter is closed under countable intersections.
10. Problems
295
40. Let X be any infinite set. Prove that X has a nontrivial ultrafilter, hence that X has a nonnegative additive set function p that assumes only the values 0 and 1 and is not a point mass. 41. Prove that the set Z+of positive integers has no nontrivial ultrafilter closed under countable intersections, i.e., that the set function p in the previous problem is not a measure. Problems 4 2 4 3 concern a theory of integration in which complete additivity is dropped as an assumption. An example is given in Problems 3 9 4 1 of a nonnegative additive set function on the set of all subsets of an infinite set that is not completely additive. For the present set of problems, let X be a nonempty set, let A be a a-algebra of subsets, and let p be a nonnegative additive set function on A such that p ( X ) < +oo. Imagine an integration theory for J, f d p with the definitions just as in the case that p is a measure. All the properties of the integral proved in the text before the Monotone Convergence Theorem would still be valid, except that the f d p as a function of E would be merely additive, rather than completely integral additive, and hence we would have to drop Corollary 5.24 and the converse half of Corollary 5.23.
1,
42. Let f be > 0, and let s, be the standard pointwise increasing sequence of simple functions with limit f , as in Proposition 5.1 1. Show that the convergence of s,, to f is uniform if f is bounded. 43. Use the result of the previous problem to show in this theory that JE f d p JE g d p if f and g are bounded and measurable.
+
lE( f +g) d p =
CHAPTER VI Measure Theory for Euclidean Space
Abstract. This chapter mines some of the powerful consequences of the basic measure theory in Chapter V. Sections 1-3 establish properties of Lebesgue measure and other Borel measures on Euclidean space and on open subsets of Euclidean space. The main general property is the regularity of all such measures -that the measure of any Borel set can be approximated by the measure of compact sets from within and open sets from without. Lebesgue measure in all of Euclidean space has an additional property, translation invariance, which allows for the notion of the convolution of two functions. Convolution gives a kind of moving average of the translates of one function weighted by the other function. Convolution with the dilates of a fixed integrable function provides a handy kind of approximate identity. Sc~Liul~ 4 givcb Lllc liual lu1111ul Lllc ~ u ~ ~ ~ p i iull i Lllc b u R~CIII~IIII ~~ alld Lcbcbguc i~~lcgialb, a preliminary form having been given in Chapter 111. Section 5 gives the final form of the change-of-variables theorem for integration, starting from the preliminary form of the theorem in Chapter I11 and taking advantage of the ease with which limits can be handled by the Lebesgue integral. Sard's 'l'heorem allows one to disregard sets of lower dimension in establishing such changes of variables, thereby giving results in their expected form rather than in a form dictated by technicalities. Section 6 concerns the Hardy-Littlewood Maximal Theorem in N dimensions. In dimension 1, this theorem implies that the derivative of a I -dimensional Lebesgue integral with respect to Lebesgue measure recovers the integrand almost everywhere. The theorem in the general case implies that certainaverages of afunction over small sets about apoint tend to the function almost everywhere. But the theorem can be regarded as saying also that a particular approximate identity formed by dilations applies to problems of almost-everywhere convergence, as well as to problems of norm convergence and uniform convergence. A corollary of the theorem is that many approximate identities formed by dilations yield almost-everywhere convergence theorems. Section 7 redevelops the beginnings of the subject of Fourier series using the Lebesgue integral, the theory having been developed with the Riemann integral in Section 1.10. With the Lebesgue integral and its accompanying tools, Fourier series are meaningful for more functions than before, Dml's test applies even to a wlder class of Klemann Integrable functions than before, and Fejer's Theorem and Parseval's Theorem become easier and more general than before. A completely new result with theLebesgue integral is theltiesz-FischerTheorem, which characterizes the trigonometric series that are Fourier series of square-integrable functions. Sections 8-10 deal with Stieltjes measures, which are Borel measures on the line, and their application to Fourier series. Such measures are characterized in terms of a class of monotone functions on the line, and they lead to a handy generalization of the integration-by-parts formula. This formula allows one to bound the size of theFourier coefficients of functions of bounded variation, which are differences of monotone functions. In combination with earlier results, this bound yields
1. Lebesgue Measure and Other Bore1 Measures
297
the Diiichlet-Jordan Theorem, which says that the Fourier series of a fuilctioil of bounded variatioil converges pointwise everywhere, the convergence being uniform on any compact set on which the function is continuous. Section 10 is a short section on computation of integrals.
1. Lebesgue Measure and Other Borel Measures Lebesgue measure on R1 was constructed in Section V.l on the ring of "elementary" sets -the finite disjoint unions of bounded intervals -and extended from there to the o-algebra of Borel sets by the Extension Theorem (Theorem 5 3 , which was proved in Section V.5. Fubini's Theorem (Theorem 5.47) would have allowed us to build Lebesgue measure in EXN as an iterated product of 1-dimensional Lebesgue measure, but we postponed the construction in IRN until the present chapter in order to show that it can be carried out in a fashion independent of how we group 1-dimensional factors. The Borel sets of IR1 are, by definition, the sets in the smallest o-algebra containing the elementary sets, and we saw readily that every set that is open or compact is a Borel set. We write Blfor this o-algebra. In fact, B1may be described as the smallest o-algebra containing the open sets of R1or as the s~nallest0-algebra containing the colnpact sets. The reason that the open sets generate Blis that every open interval is an open set, and every interval is a countable intersection of open intervals. Similarly the compact sets generate Dl because every closed bounded interval is a compact set, and every interval is the countable union of closed bounded intervals. Now let us turn our attention to IRN . We have already used the word "rectangle" in two different senses in connection with integration-in Chapter I11 to mean an N-fold product along coordinate directions of open or closed bounded intervals, and in Chaptcr V to mcan a product of mcasurablc scts. For clarity lct us rcfcr to any product of bounded intervals as a geometric rectangle and to any product of measurable sets as an abstract rectangle or an abstract rectangle in the sense of Fubini's Theorem. In RN,every geometric rectangle under our definition is an abstract rectangle, but not conversely. Define the Borel sets of IRN to be the members of the smallest o-algebra BN containing all compact sets in RN. It is equivalent to let BN be the smallest o-algebra containing all open sets. In fact, every open geometric rectangle is the countable union of compact geometric rectangles, and every open set in turn is the countable union of open geometric rectangles; thus the open sets are in the smallest o-algebra containing the compact sets. In the reverse direction every closed set is the complement of an open set, and every compact set is closed; thus the compact sets are in the smallest o-algebra containing the open sets. Functions on RN measurable with respect to BN are called Borel measurable functions or Borel functions. Any continuous real-valued function f on IRN
298
VI. Measure Theory for Euclidean Space
is Borel measurable because the inverse image f -'((c, f m ] ) of the open set ( c , f m ] has to be open and therefore has to be a Borel set. Proposition 6.1. If m and n are integers > 1, then Bm x B, the product set Rm x Rn = Rmf
=
Bm+, within
PROOF.If U is open is Rm and V is open in Rn,then U x V is open in Rm+n, and it follows that Urnx Un 5 Urn+,. For the reverse inclusion, let W be open in R m f n. Then W is the countable union of open geometric rectangles, and each of these is of the for111 U x V with U ope11 ill Rm and V open in Rn. Siilce each such U x V is in Bm x B,, so is W . Thus we obtain the reverse inclusion Bm+n 5 Bm x a n . Lebesgue measure on EXN will, at least initially, be a measure defined on the o -algebra BN. Proposition 6.1 tells us that the o -algebra on which the measure is to be defined is independent of the grouping of variables used in Fubini's Theorem. It will be quite believable that different constructions of Lebesgue measure by using different iterated product decompositions of R N ,such as ( I R 1 x R 1 )x IR1 and R1 x ( R 1x R 1 ) will , lead to the same measure, but we shall give two abstract characterizations of the result that will ensure uniqueness without any act of faith. These characterizations will take some moments to establish, but we shall obtain useful additional results along the way. The procedure will be to state the constructions of the measure via Fubini's Theorem, then to consider a wider class of measures on BN known as the "Borel measures," and finally to establish the two characterizations of Lebesgue measure among all Borel measures on R N . It is customary to write dx in place of dm(x)for Lebesgue measure on R 1 ,and we shall do so except when there is some special need for the symbol m . Then the notation for the measure normally becomes an expression like dx or dy instead of m. To construct Lebesgue measure dx on R N ,we can proceed inductively, adding one vanable at a time. Fubini's Theorem allows us to construct the product of Lebesgue measure on RN-I and Lebesgue measure on IR1,and Proposition 6.1 shows that the result is defined on the Rorel sets of TRN. T,et 11s take this particular construction as an inductive definition of Lebesgue measure on R N . It is apparent from the construction that the measure of a geometric rectangle is the product of the lengths of the sides. Altewatively, we could constluct Lebesgue measure on RN inductively by grouping EXN as some other Rm x RNPmand using the product measure from versions of the Lebesgue measures on Rm and IRN-". Again the result has the property that the measure of a geometric rectangle is the product of the lengths of the sides. It is believable that this condition determines completely the measure on R N ,and we shall give a proof of this uniqueness shortly.
300
VI. Measure Theory for Euclidean Space
arbitrary, E l U E2 is in A. Hence A is closed under finite unions, and A is an algebra of sets. The proof that A is closed under countable unions takes two steps. For the first step we let a sequence of sets E, in A be given with union E , and first assume that all E, lie in one of the sets FM in (i) above. Let r > 0 be given and choose, by (*) for each E,,, closed sets C,, and open sets U,, such that C,, g E , 5 U, and p(U, - C,) it / 2 , . Possibly by intersecting U, with F&+l, we may assumc that all U, lic in thc compact sct F M + ~Sct U = UzlU, and C = U,"==, C,. Then C g E c U with U open but C not necessarily closed. Nevertlieless, we have U - C 5 Ur=l (U, - C,), and Proposition 5.lg gives p(U - C ) 5 Czl p(U, - C,) it . The sets S, = U C , form a decreasing sequence within FM+i with intersection U - C . Since p(FM+i) is finite, Corollary 5.3 shows that p(S,) decreases to p(U - C ) , which is < t. Thus there is some m = mo with w(S,,) < t . The set C' = Uzll C , is closed, and we have C 1 5 E 5 U and p(U - C 1 )= p(S,,) < r . Therefore E is in A. For the second step we let the sets En be general members of A. Since A is an algebra, En n (F,+i - F,) is in A for every n and m . Applying the previous step, we see that El, = E n (F,+=, - F,) is in A for every m . The sets El, have union E ,and EA, is contained in F,+=, - F, . Changing notation, we may assume that the given sets E, all have En 5 F,+i - F,. If r > 0 is given, construct U, open and C, closed as in the previous paragraph except that U, is not constrained to lie in a particular F M . Again let U = Ur==,U, and C = UzlC,, so that C c E c U and p(U - C ) < t. The set U is open, and this time we can prove that the set C is closed. In fact, let {xk}be a sequence in C convergent to some limit point xo of C . Thc point xu is in somc FM sincc thc scts FM havc union thc wholc spacc. Since FM C F&+, and F&+, is open, the sequence is eventually in F;+, . The inclusion C , 5 En 5 F,+l - F, shows that C , n FM+i = M for n M + 1. Thus no term of the sequence after some point lies in CM+i,CM+2,. . . , i.e., all the terms of the sequence after some point lie in C,. This is a closed set, and the limit xo must lie in it. Therefore xo lies in C , and C is closed. This proves that E is in A. Hence A is a 0-algebra and must contain all Borel sets. From (*) for all Borel sets, it follows that every Borel set E satisfies p ( E ) = sup p ( C ) = inf p ( U )
uE,
CCE, C closed
U2E, U open
Proposition 5.2 shows that the sets F, of (i) have the property that p ( C ) = sup p ( C n F,) foreveryBore1set C . When C is closed, the sets C n F , are compact, and thus (**) implies the equality asserted in the statement of the theorem. This completes the proof. Recall from Section 111.10 that the support of a scalar-valued function on a metric space is the closure of the set where it is nonzero. Let C ~ ~ , ( I W ~ be ) the
30 1
1. Lebesgue Measure and Other Bore1 Measures
space of continuous scalar-valued functions on RN of compact support. If there is no special mention of the scalars, the scalars may be either real or complex. If K is a compact set and the open sets U, are as in (ii) before Theorem 6.2, Proposition 2.30e gives us continuous functions f, : R N + [O, 11 such that f, is 1 on K and is 0 on U,' . The support of the function f, is then contained in u:', which is compact. By replacing the functions f,,by g,, = mini f i , . . . , f,,}, we may assume that they are pointwise decreasing. Consequently (iii) tliere exists a decreasing sequence of real-valued r1lerrlbers of Ccom(RN) with pointwise limit the indicator function of K .
I,,
IRN
Corollary 6.3. If p and v are Borel measures on R N such that f dp = f dv for all continuous functions on RN of compact support, then p = v .
PROOF. Let K be a compact subset of R N ,and use (iii) to choose a decreasing sequence { f,} of real-valued members of c,,,(R~)with pointwise limit the indicator function IK . Since fl is integrable, dominated convergence allows us to deduce JR, I K d p = IK dv from the equality f, d p = f, dv for all n . Thus p ( K ) = v ( K ) for every compact set K . Applying Theorem 6.2, we obtain p ( E ) = v ( E ) for every Borel set E.
IRN
IRN
IRN
Corollary 6.4. Let p = 1 or p = 2. If w is a Borel measure on R N ,then (a) CCom(RN) is dense in LP(RN,p ) , (b) the smallest closed subspace of L P ( R ~p,) containing all indicator functions of compact sets in ktN is LP(ktN,w ) itself. as for REMARK. The scalars are assumed to be the same for CCom(RN) L ' ( R N ,w ) and L 2 ( R N ,w); the corollary is valid both for real scalars and for complex scalars.
PROOF.If E is a Borel set of finite p measure and if t is given, Theorem 6.2 E and p ( E - K ) < t. Then allows us to choose a compact set K with K IIE - IK IP dw = w ( E - K ) < t ,and consequently the closure in L p ( R N )of the set of all indicator functions of co~npactsets contains all indicator functions of Rorel sets of finite / I measure. Proposition 5.56 shows conseclilently that the smallest closed subspace of LP ( R N )containing all indicator functions of compact ) This proves (b). sets is L P ( R ~itself. For (a), let K be compact, and use (iii) to choose a decreasing sequence { f,} of rcal-valucd mcmbcrs of c , ~ , ( Rwith ~ ) pointwisc limit I K . Sincc f f is integrable, dominated convergence yields limn J, I f n - IK IP d p = 0. Hence the closure of c,,,(R~) in LP(RN)contains all indicator functions of compact sets. By Proposition 5.55d this closure contains the smallest closed subspace of LP(RN)containing all indicator functions of compact sets. Conclusion (b) shows that the latter subspace is L P ( R N )itself. This proves (a).
302
VI. Measure Theory for Euclidean Space
Fix an integer n > 0, and let ( a l , . . . , a N )be an n-tuple of integers. The diadic cube Q, ( a l ,. . . , a N )in IRN of side 2-" is defined to be the geometric rectangle Q n ( a l , .. . , a ~ =) { ( x l , .. . , X N ) 2-,aj < xj 5 2-,(aj
+ 1 ) for 1 5 j 5 N } .
Let Q , be the set of all diadic cubes of side 2-'\ m e members of Q , are disjoint and have union R N . Thus we can associate uniquely to each x in RN a sequence { Q,} of diadic cubes such that x is in Q , and Q , is in Q,. Since for each n , the members of Q,+l are obtained by subdividing each member of Q, into 2N disjoint smaller diadic cubes, the diadic cubes Q , associated to x must have the property that Q , 2 Q,+l for all n > 0.
Lemma 6.5. Any open set in R N is the countable disjoint union of diadic cubes. PROOF.Let an open set U be given. We may assume that U # R N ,so that U C# M.We describe which diadic cubes to include in a collection A so that A has the required properties. If x is in U ,then D ( x , U C )= d is positive since U C is closed and nonempty. Let {Q,} be the sequence of diadic cubes associated to x . The distance between any two points of Q, is 5 2 - , f l , and this is < d if n is sufficiently large. Hence Q , is contained in U for n sufficiently large. The cube in A that contains x is to be the Q , with n as small as possible so that Q , U. The construction has been arranged so that the union of the diadic cubes in A is exactly U . Suppose that Q and Q' are members of A obtained from respective points x and x' in U . If Q f' Q' # M,let x" be in the intersection. Then Q and Q' are two of the diadic cubes in the sequence associated to x", and one has to contain the other. Without loss of generality, suppose that Q 2 Q'. Then x' lies in Q as well as Q', and we should have selected Q for x' rather than Q' if Q Q'. We conclude that Q = Q', and thus the members of A are disjoint. Each collection Q , is countable, and therefore the collection A is countable.
+
Proposition 6.6. Any Borel measure on RN is determined by its values on all the diadic cubes. REMARK.We shall apply this result in the present section in connection with Lebesgue measure on RN and in Section 8 in connection with general Borel measures on R 1 . PROOF.The values on the diadic cubes determine the values on all open sets by Lemma 6.5, and the values on all open sets determine the values on all Borel sets by Theorem 6.2.
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Corollary 6.7. There exists a unique Borel measure on R N for which the measure of each geometric rectangle is the product of the lengths of the sides. The measure is the N-fold product of 1-dimensional Lebesgue measure. REMARKS. The uniqueness is immediate from Proposition 6.6. The first version of Lebesgue measure that we constructed has the property stated in the corollary and therefore proves existence. All the other versions of Lebesgue measure we constructed have the same property, and so all such versions are equal. The corollary therefore allows us to use Fubini's Theorem for any decomposition R N = Rm x Rn with m + n = N . As in the 1-dimensional case, we shall often write d x for Lebesgue measure.
Corollary 6.7 gives one characterization of Lebesgue measure. We shall use Proposition 6.6 to give a second characterization, which will be in terms of translation invariance.
Proposition 6 8 . Under a Borel function F : R N + R N ' , F - ' ( E ) is in BN whenever E is in B N , . In particular, this conclusion is valid if F is continuous. PROOF.The set of E's for which F - (E) ~ is in BN is a o-algebra, and the result will follow if this set of E's contains the open geometric rectangles of RN' . If Fj denotes the jth component of F , then Fj : RN + R1is Borel measurable and Proposition 5 . 6 ~ shows that FjP1( U j ) is a Borel set in R N if Uj is open in R 1 . Tllcn F-l(U1 x . . . x UN1)= F ~ - ' ( u ~is) a Burcl a d in R N .
Corollary 6.9. Any homeomorphism of RN carries BN to BN Corollary 6.9 is a special case of Proposition 6.8. The particular homeomorphisms of interest at the moment are translations and dilations. Translation by xo is the homeomorphism rx, ( x ) = x + xo . Its operation on a set E is given by rxo(E)= {txo ( x ) I x E E } = { x xo I x E E } = E xo, and its operation on a function f on IRN is given by r,, ( f )( x ) = f (rXi1( x ) ) = f ( x - xo). Its operation on an indicator function IE is r,, ( I c ) (.u) = IE ( x - .uo) = Ic+,, (.u) = I,% ( c )( x ). Because of Corollary 6.9, translations operate on measures, the formula being r x o ( p ) ( E )= p(rx;l ( E ) ) ;since homeomorphisms carry compact sets to compact sets, the right side is a Borel measure if is a Borel measure. The actions of r,, oil fuiictioiis aiid measures are related by iiitegratioii. Iff > 0 is a Borel fuiictioii, then so is r,, ( f ), and f d (r,, p ) = rx; ( f )d p ; this formula is verified by checking it for indicator functions and then passing to simple functions ? 0 by linearity and to Borel functions f > 0 by monotone convergence. Dilation 6, by a nonzero real c is given on members of RN by 6, ( x ) = cx ,and the operations on sets, functions, indicator functions, and measures are analogous
+
iRN
+
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VI. Measure Theory for Euclidean Space
to the corresponding operations for translations. Although dilations will play a recurring role in this book, the notation 6, will be used only in the present section.
Theorem 6.10. Lebesgue measure m on RN is translation invariant in the sense that rq ( m ) = m for every xo in R N . In fact, Lebesgue measure is the unique translation-invariant Borel measure on RN that assigns measure 1 to the diadic cube Qo(O,. . . , 0). The effect of dilations on Lebesgue measure is that Sc(m) = ~ c l - ~ i.e., m , JRN f ( C X ) d x = 1clpN JRN f ( x )d x for every nonnegative Borel function f . REMARKS. From one point of view, translation and dilation are examples of bounded linear operators on each L p ( R N ,d x ) , with translation preserving norms and with dilation multiplying norms by a constant depending on p and the particular dilation. From another point of view, translation and dilation are especially simple examples of changes of variables. Operationally the theorem allows us to write d y = dx when y = t,, ( x ) and d z = lclN d x when z = c x . These effects of translations and dilations on integration with respect to Lebesgue measure are special cases of the general change-of-variables formula to be proved in Section 5. ~ O O F For . any xo in RN, m and t,, ( m ) assign the product of the lengths of the sides as measure to any diadic cube. From Proposition 6.6 we conclude that m = tx,( m ). The assertion about the effect of dilations on Lebesgue measure is proved similarly. We still have to prove the uniqueness. Let p be a translation-invariant Borel measure. The members of Q, are translates of one another and hence have equal p measure. The members of Q,+l are obtained by partitioning each member of Q, into 2N members of Q,+l that are translates of one another. Thus the p measure of any member of Q,+l is 2TN times the p measure of any member of Q, . Consequently the p measure of any diadic cube is completely determined by the value of p on Qo(O,. . . , 0 ) , which is a member of g o . The uniqueness then follows by another application of Proposition 6.6.
For a continuous function on a closed bounded interval, it was shown at the end of Section V.3 that the Riemann integral equals the Lebesgue integral. The next proposition gives an N-dimensional analog. A general comparison of the Riemann and Lebesgue integrals will be given in Section 4.
Proposition6.11. For a continuous function on a compact geometric rectangle, the Riemann integral equals the Lebesgue integral. PROOF.The two are equal in the 1-dimensional case, and the N-dimensional cases of each may be computed by iterated 1-dimensional integrals- as a result of
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305
Corollary 3.33 in the case of the Riemann integral and as a result of the definition of Lebesgue measure as a product and the use of Fubini's Theorem (Theorem 5.47) in the case of the Lebesgue integral. So far, we have worked in this section only with Lebesgue measure on the Borel sets. The Lebesgue measurable sets are those sets that occur when Lebesgue measure is completed. The Lebesgue measurable sets of measure 0 are of particular interest. In Section 111.8 we defined an ostensibly different notion of measure 0 by saying that a set in RN is of measure 0 if for any t > 0, it can be covered by a countable set of open geometric rectangles of total volume less than t ,and Theorem 3.29 characterized the Riemann integrable functions on a compact geometric rectangle as those functions whose discontinuities form a set of measure 0 in this sense. Later, Proposition 5.39 showed for R1that a set has measure 0 in this sense if and only if it is Lebesgue measurable of Lebesgue measure 0. This equivalence extends to RN,as the next proposition shows.
Proposition 6.12. In RN,the Lebesgue measurable sets of measure 0 are exactly the subsets E of RN with the following property: for any t > 0, the set E can be covered by countably many geometric rectangles of total volume less than t . PROOF.Let m be Lebesgue measure on RN.If E has the stated property, let E , be the union of the given countable collection of geometric rectangles of total volume < l / n used to cover E . Proposition 5.lg shows that m ( E , ) < l / n , and hence the Borel set E' = Ek has m ( E ' ) < l / n for every n . Therefore m ( E f ) = 0. Since E c E', E is Lebesgue measurable and has Lebesgue measure 0. Conversely if E is Lebesgue measurable of Lebesgue measure 0 and if t > 0 is given, we are to find a union of open geometric rectangles containing E and having total volume < t. Find a set E' in BN with E 5 E' and m ( E 1 )= 0. It is enough to handle E'. Writing RN as the union of compact geometric cubes C , of side 2n centered at the origin and covering E' f' C, up to t/2", we see that we may assumc that E' is boundcd, bcing containcd in somc cubc C , . Within R' n [-n, n ] ,we know that the set of finite unions of intervals is an algebra A?) of sets such that B y ) = Bl n [-n, n] is the smallest o-algebra containing A?). Applying Proposition 5.40 inductively, we see that the set of finite disjoint unions of N-fold products of members of A?) is an algebra A$), and then Proposition 6.1 shows that the smallest o-algebra containing A$) is U:) = U N n C,. Proposition 5.38 shows that the measure rn on B:) is given by m*, where m * ( A ) is the infimum of countable unions of members of A$) that cover A. Consequently the subset E' of C , can be covered by countably many
0,
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VI. Measure Theory for Euclidean Space
geometric rectangles of total volume < t . Doubling these rectangles about their centers and discarding their edges, we obtain a covering of E' by open rectangles of total volume < 2Nt ,and we have the required covering. Borel measurable sets have two distinct advantages over Lebesgue measurable scts. Onc advantagc is that Borcl mcasurablc scts arc indcpcndcnt of thc particular Borel measure in question, whereas the sets in the completion of a a-algebra relative to a Borel measure very much depend on the particular measure. The other advantage is that Fubini's Theorem applies in a tidy fashion to Borel measurable functions as a consequence of the identity Bm x B, = Bm+, given in Proposition 6.1. By contrast, there are Lebesgue measurable sets for RN that are not in the product of the o-algebras of Lebesgue measurable sets from Rm and RN-". For example, take a set E in R' that is not Lebesgue measurable; such a set is produced in Problem 1 at the end of the present chapter. Then E x {0}in R2is a subset of the Borel set R1x {0},and hence it is Lebesgue measurable of measure 0. However, E x {0}is not in the product o-algebra, because a section of a function measurable with respect to the product has to be measurable with respect to the appropriate factor (Lemma 5.46). On the other hand, Lebesgue measurable functions are sometimes unavoidable. An example occurs with Riemann integrability: In view of Proposition 6.12, Theorem 3.29 says that the Riemann integrable functions on a compact geometric rectangle are exactly the functions whose discontinuities form a Lebesgue measurable set of Lebesgue measure 0, and Problems 3 1-33 at the end of Chapter V produced such a function in the 1-dimensional case that is not a Borel measurable function. The upshot is that a little care is needed when using Fubini's Theorem and Lebesgue measurable sets at the same time, and there are times when one wants to do so. The situation is a little messy but not intractable. Problem 12 at the end of Chapter V showed that a Lebesgue measurable function can be adjusted on a set of Lebesgue measure 0 so as to become Borel measurable. Using this fact, one can write down a form of Fubini's Theorem for Lebesgue measurable functions that is usable even if inelegant.
2. Convolution Convolution is an important operation available for functions on RN. On a formal level, the convolution f * g of two functions f and g is
2. Convolution
307
One place convolution arises is as a limit of a linear combination of translates: We shall see in Proposition 6.13 that the convolution at x may be written also as JRN f ( y ) g ( x - y ) d y . If f is fixed and if finite sets of translation operators rYgand of weights f ( y i ) are given, then the value at x of the linear combination f ( y i ) r y zapplied to g and evaluated at x is f ( y i ) g ( x - y i ) . Corollary 6.17 will show a sense in which we can think of f ( y ) g ( x - y ) d y as a limit of such expressions. To make mathematical sense out o f f * g ,let us begin with the case that f and g are nonnegative Borel functions on R N . The assertion is that f * g is meaningful as a Borel functioii 10. I11 fact, ( x , y ) H f ( x - y ) is the coiilpositioii of the continuous function F : R2N + RN given by F ( x , y ) = x - y , followed by the Borel function f : R N + [0, +m]. If U is open in [0, + X I ]then , f ( u ) is in BN ,and Proposition 6.8 shows that ( f o F)-' ( ( I ) = F - I ( f ( u ) ) is in B I N . Then the product ( x , y) H f ( x - y ) g ( y ) is a Borel function, and Fubini's Theorem (Theorem 5.47) and Proposition 6.1 combine to show that x H ( f * g ) (x) is a Borel function 10.
xi
xi
I,,
Proposition 6.13. For nonnegative Borel functions on R N , (a) (b)
f *s=s*f, f *(g*h) =( f *g)*h.
PROOF.We use Theorem 6.10 for both parts and also Fubini's Theorem for (b). For (a), the changes of variables y H y + x and then y H -y give f ( x - y > g ( y >d y = JRN f ( - Y ) ~ ( Y x ) d x = JRN f ( Y ) ~ ( -x Y ) d y . For (b), thc computation is
+
the change of variables y
H
y
+ z being used for the fourth equality.
In order to have a well-defined expressioa for f * g when f and g are not necessarily > 0, we need conditions under which the nonnegative case leads to something finite. The conditions we use ensure finiteness of ( I f 1 * Igl) ( x ) for almost every x . For real-valued f and g , we then define f * g ( x ) by subtraction at the points where ( I f 1 * Igl)(x) is finite, and we define it to be O elsewhere. For complex-valued f and g ,we define ( f * g ) ( x ) as a linear combination of the
VI. Measure Theory for Euclidean Space
308
appropriate parts where (If 1 * Igl) (x) is finite, and we define it to be 0 elsewhere. When we proceed this way, the commutativity and associativity properties in Proposition 6.13 will be valid even though f and g are not necessarily > 0.
Proposition 6.14. For nonnegative Bore1 functions f and g on R N , convolution is finite almost everywhere in the following cases, and then the indicated inequalities of norms are satisfied: (a) f o r f i n ~ ' ( R a~n )d g i n ~ ' ( R ~ ) , a n d t h eI l f n 5 IlflllIlglllr (b) for f in L1(RN)and g in L2(RN),and then I l f * gll, 5 I l f I l l llgl12, for f in L ~ ( R andg ~ ) in L ' ( R ~ ) ,and then 11 f * gl12 5 11 f l1211glll, (c) f o r f inL1(RN) andginL"(IfSN),andthen I l f *glloo i I I f lllIlglloo, f o r f inL"(RN) andginL1(RN),andthen I l f *gll, 5 Ilfll,llglll, (d) for f in L ~ ( R and ~ ) g in L ~ ( R ~and ) , then 11 f * gll, 5 11 f 11211gl12. Consequently f * g is defined in the above situations even if the scalar-valued functions f and g are not necessarily 3 0, and the estimates on the norm of f * g are still valid. PROOF.For (a) and the first conclusions in (b) and (c), let p be 1,2, or oo as appropriate. By Minkowski's inequality for integrals (Theorem 5.60),
llf
* gllp = 1 = &'mPN
IRN
f ( Y ) ~ (x Y ) ~ Y Ii I ~ , ~ llf ( Y ) ~ ( xY)II~,~~Y , ~ = [Rm' nhl If (Y>Illgllp d~ = Ilf l111811p9 If (y)l I l g ( ~ ~ ) l l pd~ IRN
the next-to-last equality following from the translation invariance of d x . The second conclusions in (b) and (c) require only notational changes. For (d) ,we have
5 supx I l f Il2llg(x - y)112,, = Ilf ll2llgll2, the inequality following from the Schwarz inequality and the last step following from translation invariance of dy and invariance under y H -y . Going over these arguments, we see that we may use them even if f and g are not necessarily 2 0. Then the last statement of the proposition follows.
Next let us relate the translation operators of Section 1 to convolution. The formula for thc cffcct of a translation opcrator on a function is tt( f ) (x) = f ( x t ) .
Proposition 6.1 5. Convolution commutes with translations in the sense that ~ t ( f* g > = ( t t f ) * g = f * r t g . PROOF.It is enough to treat functions 3 0. Then we have tt(f * g)(x) = (f *g)(x-t) = [ R ~ f ( ~ - t - y ) g ( ~ > d ~ , w h i c h e q u a l s ~ ~ ~ ( r t f > ( x - = ~)g(~)d~ ((rt f ) * g ) (x) on the one hand and, because of translation invariance of Lebesgue f (X - y)g (y - t) dy = (f * ttg) (x) on the other hand. measure, equals
IRN
2. Convolution
309
Proposition 6.16. If p = 1 or p = 2, then translation of a function is continuous in the translation parameter in L P ( R ~d ,x ) . In other words, if f is in LP relative to Lebesgue measure, then limh,o Ilrt+hf - rt f l i p = 0 for all t . REMARK.However, continuity fails on L". In this case, there is a substitute result, and we take that up in a moment.
PROOF. Let f be in LP. By translation invariance of Lebesgue measure, Ilrt+h f - rt f 1 1 , = 11th f - f 1 ,. If g is in c,,,IR~), thcn llrhg - gll; = SmpNIg(x - h ) - g ( x )IP d x ,and dominated convergence shows that this tends to 0 as h tends to 0 . Let c > 0 and f be given. By Corollary 6.4a, c,,,(R~) is dense in L P ( R N ,d x ) , and thus we can choose g in C,,,(RN) with I f - g 11, c t . Then
Corollary 6.17. Let p = 1 or p = 2, and let gl, . . . , g, be finitely many functions in L P ( R N ). If a positive number t and a function f in L ( R N )are given, then there exist finitely many members yj of R N , 1 5 j 5 n , and constants cj such that f * gk - x;=;=,cjr,gkIP c t for 1 5 k 5 r .
1
REMARK.In the case r = 1, the corollary says that any convolution f * g can be approximated in LP by a linear combination of translates of g . The result will be used in Chapter VIII with r > 1. PROOF.Let V be the set of functions f in L ' ( R ~ for ) which this kind of approximation is possible for every t > 0. The main step is to show that V contains the indicator functions of the compact sets in R N . Let K be compact, and let IK be its indicator function. Proposition 6.16 shows that the functions y H rygk are continuous from K into L P ( R N ) for 1 5 k 5 r , and therefore these functions are uniformly continuous. Fix c > 0, and let S > 0 be such that Ilrygk - rY1gkIp c t for all k whenever ly - yll c 6 and y and y1 are in K . For each y in K , form the open ball B(6; y) in R N . These balls cover K , and finitely many suffice; let their centers be yl, . . . , y,. Define sets S 1 , . . . , S, inductively as follows: Sj is the subset of K where 1 y - yj I < S but 1 y - yj I 2 S for i < j . Then K = Sj disjointly. By the choice of S, we have ll rygk - ryjgk l i p < t for all y in ,Sj and all k . 1~JsingMinkowski's ineclilality for integrals (Themem 5.60), and writing m for Lebesgue measure, we have
UGl
3 10
VI. Measure Theory for Euclidean Space
Summing over j gives
Since t is arbitrary, IK lies in V . If f l and f2 are in V and if g l , . . . , g, are given, then we may assume, by taking the union of the sets of members yj of RN and by setting any unnecessary constants cj equal to O, that the translates used for ji and j2 with the same E > 0 are the same. Thus we can write f l * gh - C r = l cjrYigk < c/2 and
I f2 * gk
-
x ; = l d j ryjgk
I
+
11,
1
11,
< r / 2 for suitable yj 's and c j ' s , and the triangle
1
+
inequality gives (f l f2) * gk - Z;=l ( c j d j ) r y j g k c r . Hence V is closed under addition. Similarly V is closed under scalar multiplication. If fi + f in L 1 with fi in V and if t > 0 is given, choose I large enough so that I f - f i l l "/(2max llgkl,). If I f i * g k - C;=l c j(1) \;)gkI, < c/2,then the inequality 11 f * gk - fi * gkllp 5 11 f - fillllgkllp and the triangle inequality
1 * gk
together give f
-
1,
E(il=, c j(1) r ~ ) g k YJ
it.
Hence f is in V ,and V is closed.
By Corollary 6.4b, V = L ( R N ) ,and thc proof is complctc. In some cases with L W ( R N ) ,results have more content when phrased in terms of the supremum norm 11 f I s , p = s u p X E W f ~(x)I defined in Section V.9. For a coiitiiiuous fuiictioii f , tlie two norms agree because the set where If (x) I > M is open and therefore has positive measure if it is nonempty. For a bounded function f , the condition limh,o lithf - f Il , = 0 is equivalent to uniform continuity of f , basically by definition. The functions f in L w for which limh,o 1 1 4 f - f 1 1 , = 0 are not much more general than the bounded uniformly continuous functions; we shall see shortly that they can be adjusted on a set of measure 0 so as to be bounded and uniformly continuous.
Proposition 6.18. In RN with Lebesgue measure, the convolution of L' with L m , or of L w with L 1 ,or of L~ with L 2 results in an everywhere-defined bounded uniformly continuous function, not just an L m function. Moreover,
I l f *gllsup 5 l l f
111
11~11W>
I l f *81ISUP5 I l f
11W11~111~or
I l f *glls"p 5 I l f
11211~112
in the various cases. PROOF.We give the proof when f is in L' and g is in La, the other cases being handled similarly. The bound follows from the computation I l f gll,, =
*
supx I i R N f ( ~ Y ) ~ ( Y ) ~5 Y ~
llgllwJR'mnN
-
y)ldy
=
Ilf
lllll&llm'
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311
For uniform continuity we use Proposition 6.15 and the bound 11 f I l f l l l g l l , to make the estimate
* gllSupi
and then we apply Proposition 6.16 to see that the right side tends to 0 as h tends to 0.
A corollary of Proposition 6.18 gives a first look at how differentiability interacts with convolution.
Corollary 6.19. Suppose that f is a compactly supported function of class Cn on RN and that g is in LP(RN,d x ) with p equal to 1 , 2 , or oo.Then f * g is of class C n ,and D( f * g ) = ( Df ) * g for any iterated partial derivative of order 5 n. PROOF.First suppose that n = 1. Fix j with 1 (: j (: N , and put D; = a/a.x;. The function ( D jf ) g is continuous b y Proposition 6.18. If we can prove that Dj ( f * g ) ( x ) exists and equals ( ( D jf ) * g ) ( x ) for each x ,then it will follow that Dj ( f * g ) is continuous. This fact for all j implies that f * g is of class C l , by Theorem 3.7, and the result for n = 1 will have been proved. The result for higher n can then be obtained by iterating the result for n = 1. Thus we are to prove that Dj ( f g ) ( x ) exists and equals ((D;f ) g ) ( x ) for each x . In the respective cases p = 1 , 2 , m, put p' = oo,2, 1. Let e; be the jth standard basis vcctor of RN and lct h bc rcal with I h 1 i 1. Proposition 6.15 givcs
*
*
*
Proposition 3.28a shows that h-' (rPhej f - f ) converges uniformly, as h + 0, to Dj f on any compact set; since the support is compact, h-' (tPhej f - f ) converges uniformly to D; f on RN. Hence the convergence occurs in L" , and dominated convergence shows that it occurs in L and L~ also. Combining Proposition 6.18 and (*), we see that
The right side tends to 0 as h equals ((D.jf ) * 8 )( x ) .
+ 0, and thus indeed D; ( f
* g ) ( x ) exists and
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Twice in Chapter I we made use of an "approximate identity" in R1,a system of functions peaking at the origin such that convolution by these functions acts more and more like the identity operator on some class of functions. The first occasion of this kind was in Section 1.9 in connection with the Weierstrass Approximation Theorem, where the functions in the system were p, (x) = cn(l - x ~ on) [- ~1, 11 with the constants c, chosen to make the total integral be 1 The polynomials p,, had the properties (i) Vn(x) 3 0, (ii) ~ n ( x ) d x= 1, (iii) for any 6 > 0, sup,,lxl,, rp,(x) tends to 0 as n tends to infinity, and the convolutions were with continuous functions f such that f (0) = f (1) = 0 and f vanishes outside [0, 11. The second occasion was in Section 1.10 in connection with FejCr's Theorem, where the functions in the system were trigonometric polynomials KN(x) such that
(9 KN(x) 1 0 , (ii) KN(x) dx = 1, (iii) for any 6 > 0 , s ~ p , , ~ , ~KN , ~ (x) tends to 0 as n tends to infinity. In this case the convolutions were with periodic functions of period 2 n over an interval of length 277, and the integrations involved dx instead of dx. Now we shall use the dilations of a single function in order to produce a more robust kind of approximate identity, this time on RN. One sense in which convolution by this system acts more and more like the identity appears in Theorem 6.20 below, and a sample application appears in Corollary 6.21. The corollary will illustrate how one can use an approximate identity to pass from conclusions about nice functions in some class to conclusions about all functions in the class.
& sfn
&
Theorem 620. Let p be in L' (RN,dx) ,not necessarily
0. Define
pF(x) = E - ~ ~ ( B - ~ x for ) B > 0,
I,,
and put c = rp(x) d x . Then the following hold: (a) if p = 1 or p = 2 and i f f is in Lp(RN,dx),then
(b) the conclusion in (a) is valid for p = oo if f is in L"(RN, dx) and limt+o l l t t f - f lloo = 0, (c) if f is bounded on RN and is continuous at x , then lime$o(cpe* f )(x) = c f (x) > (d) the convergence in (c) is uniform for any set E of x's such that f is uniformly continuous at the points of E .
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VI. Measure Theory for Euclidean Space
Corollary 6.21. I f f in L " ( R ~ ,dx) satisfies limt+o littf - f 1 , = 0, then f can be adjusted on a set of measure 0 so as to be uniformly continuous. PROOF.Let q be a member of c,,,(R~) such that JRN q(x) dx = 1. Fix a sequence {F,} decreasing to 0 in Kt1. Proposition 6.18 shows that each qEa* f is bounded and uniformly continuous for every n , and Theorem 6.20 shows that {qEn* f } is Cauchy in L". Since the L" and supremum norms coincide for continuous functions, {q,, * f } is uniformly Cauchy and must therefore be uniformly convergent. Let g be the limit function, which is necessarily bounded and uniformlycontinuous. Then 11 f -gII, i II f -qFn* f ll,+Ilq,n* f -gI,,and both terms on the right tend to 0 as n tends to infinity. Consequently 11 f -g 1, = 0 , and g is a bounded uniformly continuous function that differs from f only on a set of measure 0.
3. Borel Measures on Open Sets A number of results in Sections 1-2 about Borel measures on RN extend to suitably defined Borel measures on arbitrary nonempty open subsets V of R N , and we shall collect some of these results here in order to do two things: to prepare for the proof in Section 5 of the change-of-variables formula for the Lebesgue integral in RN and to provide motivation for the treatment in Chapter XI of Borel measures on locally compact Hausdorff spaces. Throughout this section, let V be a nonempty open subset of R N . We shall make use of the following lemma that generalizes to V three properties (i-iii) listed for RN before Theorem 6.2 and Corollary 6.3. Let Ccom(V) be the vector space of scalar-valued continuous functions on V of compact support in V. If nothing is said to the contrary, the scalars may be either real or complex.
Lemma 6.22. (a) There exists a sequence {F,},"==,of compact subsets of V with union V such that F, 5 Fl+l for all n . (b) For any compact subset K of V ,there exists a decreasing sequence of open sets U, with compact closure in V such that U, = K . (c) For any compact subset K of V, there exists a decreasing sequence of functions in Ccom(V)with values in [0, 11 and with pointwise limit the indicator function of K . PROOF.In (a), the case V = RN was handled by (i) before Theorem 6.2. For V # R N , we can take F, = {x E V D(x, VC) 2 l / n and 1x1 i n} as long as n is 2 some suitable no. We complete the definition of the F,'s by taking F1 = . . . = Fno-l = F,.
I
3. Bore1 Measures on Open Sets
315
In (b), the case V = RN was handled by (ii) before Theorem 6.2. For V # RN, every x in K has D ( x , V C ) > 0 since V C is closed and is disjoint from K . The function D ( . , V C )is continuous and therefore has a positive minimum on K . Choose no such that D ( x , V C )> l / n o for x in K , i.e., Ix - yl > l / n o for all x E K and y E V C . Then D ( y , K ) 1 l / n o if y is not in V . Let U,, = { y E RN D ( y , K ) (: l / n } for n no. This is an open set containing K , and its closure in RN is contained in the set where D ( y , K ) 5 l / n , which in turn is containcd in V . Thc sct whcrc D ( y , K) 5 l / n is closcd and boundcd in RN and hence is compact. Therefore u,C' is contained in a compact subset of V . We ~ u r ~ p l c lllie t : deliriiliur~u l ale U,'s by lellirlg 0'1, . . . , U,, all equal U,,+l. For (c), we argue as with (iii) before Corollary 6.3. Choose open sets U, as in (b) that decrease and have intersection K , and apply Proposition 2.30e to obtain continuous functions f, : RN + [O, 11 such that f, is 1 on K and is 0 on U,". The support of the function f, is then contained in u:', which is compact. By replacing the functions f, by g, = mini f l , . . . , f,}, we may assume that they are pointwise decreasing. Then (c) follows.
I
,
The Borel sets in the open set V are the sets in the o-algebra
of subsets of V . We can regard V as a metric space by restricting the distance and because V is open, the open sets of V are the open sets of function on RN, RN that are subsets of V . We shall prove the following proposition about these Borel sets after first proving a lemma.
Proposition 6.23. The a-algebra B N ( V ) is the smallest a-algebra for V containing the open sets of V , and it is the smallest o-algebra for V containing the compact sets of V .
Lemma 6.24. Let X be a nonempty set, let U be a family of subsets of X , let B be the smallest o-ring of subsets of X containing U , and let E be a member of B. Then B f' E is the smallest o-ring containing U n E. PROOF OF LEMMA6.24. Let
A be the smallest o -ring containing Uf' E ,and let
A' be the smallest o -ring containing Uf'E ' . Since B n E is a o -ring of subsets of X colltaillillg U n E , A i s colltailled in B n E. Silllilarly A' 5 B n Er . Thus the set of unions AU A' with A E A and A' E A' is contained in B, contains U ,and is closed under countable unions. To see that it is closed under differences, let A1 U A; and A2 U A; be such unions. Then ( A l U A;) - ( A 2U A;) = ( A 1- A2) U ( A ; - A;) exhibits the difference of the given sets as such a union. Hence the set of such unions is a a-ring and must equal B.
VI. Measure Theory for Euclidean Space
316
PROOF OF PROPOSITION 6.23. The statement about open sets follows from U to be the set of open sets in RN, and E to Lemma 6.24 by taking X to be RN, be V . The set U f' E is the set of open subsets of V , and the lemma says that the smallest o-ring containing U f' E is BN( V ) . This is a o-algebra of subsets of V since V itself is in U f' V . Let {F,,}be the sequence of compact subsets of V produced by Lemma 6 22a F,, V is a member of the smallest o-ring of subsets containing Since V = the compact subsets of V . If F is a relatively closed subset of V ,then each F n F, is compact, and the countable union F is therefore in this o-algebra. Taking complements, we see that every open subset of V is in the smallest o-algebra of subsets of V containing the compact sets. Therefore BN( V ) is contained in this o-algebra and must equal this o-algebra.
Uzl
A Borel function on V is a scalar-valued function measurable with respect to BN ( V ) . A Borel measure on V is a measure on BN( V ) that is finite on every compact set in V .
Theorem 625. Every Borel measure
w on the nonempty open subset V
of
RN is regular in the sense that the value of EL on any Borel set E in V is given by p(E)
=
sup BgE, K compact in V
p(K) =
:if,
~ ( u ) .
U open in V
REMARK.If p is a Borel measure on V and if we define v ( E ) = p ( E f' V ) for Borel sets E of RN, then v is a measure on the Borel sets of RN, but v need not be finite on compact sets. Thus Theorem 6.25 is not a special case of Theorem 6.2. PROOF. This is proved from parts (a) and (b) of Lemma 6.22 in exactly the same way that Theorem 6.2 is proved from items (i) and (ii) before the statement of that theorem.
Corollary 6.26. If p and v are Borel measures on V such that f dv for all j' in Ccom(V), then p = v.
1, f d p =
PROOF.This is proved from Theorem 6.25 and Lemma 6 . 2 2 ~ in the same way that Corollary 6.3 is proved from Theorem 6.2 and item (iii) before the statement of that corollary.
Corollary 6.27. Let p = 1 or p = 2. If p is a Borel measure on V , then (a) Ccom(V) is dense in L P ( V , p ) , (b) the smallest closed subspace of L P ( V , w) containing all indicator functions of compact subsets of V is L P ( V , p ) itself,
3. Bore1 Measures on Open Sets
317
(c) CcOm(V), as a normed linear space under the supremum norm, is separable, (d) LP(V, p ) is separable. PROOF.Conclusions (a) and (b) are proved from Lemma 6 . 2 2 ~ with the aid of Propositions 5.56 and 5.55d in the same way that Corollary 6.4 is proved from item (iii) before Corollary 6.3 with the aid of those propositions. For (c), Lemma 6.22a produces a sequence {Fn}r==lof compact subsets of V , cover V, they with union V such that Fn Fz+lfor all n. Since the open sets FO cover any compact subset K of V, and K must be contained in some F z . Let us put that observation aside for a moment. For each n , we can identify the vector subspace of Ccom(V)of functions supported in F,O with a vector subspace of C(Fn). The latter is separable by Corollary 2.59, and hence the vector subspace of CcO,(V) of functions supported in F,O is separable. If we form the union on n of these countable dense sets of certain vector subspaces and if we take into account that the functions supported in any compact subset of V have compact support within some F,", we see that CcOm(V)is separable. For (d), we apply (a) and (c) with V replaced by F," and take into account that y (F,) < oo. Let f be arbitrary in LP(F,O,y ) . If t > 0 is given, choose F," g in Ccom(F,O)with 11 f - g lip 5 t . Then choose h in the countable dense set Dn of CcO,(F,") such that lli: - hllsup i t . Since I l f - hllp i I l f - gllp llg - hIlp and Ilg - All; = Ig(x) = h(x)lPdw(x) i tPw(E,). We obtain 11 f hllp 5 c I c p ( ~ , ) ~ ' pHence . the closure in LP(V, p) of the countable set D = U,"==, D, contains f . In particular, D" is a vector subspace containing all indicator functions of compact subsets of V . By (b), D"' = LP(V, p ) .
1
SF;
+
Proposition 6.28. With V still open in RN,let V' be an open set in RN'. Under a continuous function or even a Borel measurable function F : V + V', F - I (E) is in BN(V) whenever E is in BN,(V') . PROOF.The set of E's for which F-'(E) is in BN (V) is a o-algebra, and this o-algebra contains the open geometric rectangles of RN' by the same argument as for Proposition 6.8. Thus it contains BNl(V') .
Corollary 6.29. If V' is a second nonempty open subset in RN besides V, then any homeomorphism of V onto V' carries BN (V) to -IJN (V') . If K is a nonempty compact subset of RN, it will be convenient to be able to speak of the Borel sets in K , just as we can speak of the Borel sets in an open subset V of RN. The theory for K is easier than the theory for V ,partly because Borel measures on K can all be obtained by restriction from Borel measures on RN.
318
VI. Measure Theory for Euclidean Space
The Borel sets in K are the sets in BN ( K ) = BN n K . Using Lemma 6.24, we readily see that B N ( K )is the smallest o-algebra for K containing the compact subsets of K ; the argument is simpler than the corresponding proof for Proposition 6.23 in that it is not necessary to produce some sequence of sets by means of Lemma 6.22. A Borel function on the compact set K is a scalar-valued function measurable with respect to BN ( K ). A Borel measure on K is a measure on BN ( K ) that is finite on compact subsets of K . In this situation regularity is a consequence of the regularity of Borel measures on R N ,and no separate argument is needed. In fact, if p is a Borel measure 011 K , we call define v ( E ) = p ( E n K ) for each Borel subset E on R N ,and then the finiteness of p ( K ) implies that v is a Borel measure on R N . Borel measures on RN are regular, and therefore we have
v(E)=
sup
v ( K f )=
v(U).
inf U X ,
K'GE,
U open in IRN
K' compact in RN
Replacing E by E f' K and substituting from the definition of v , we obtain the following proposition.
Proposition 630. Every Borel measure p on a compact nonempty set K in R N is regular in the sense that the value of p on any Borel set E in K is given by p(E) =
sup KGE,
K' compact in K
p ( K f )=
inf
p(U).
U>E,
U relatively open in K
4. Comparison of Riemann and Lebesgue Integrals
This section contains the definitive theorem about the relationship between the Riemann integral and the Lebesgue integral in R N . The Riemann integral is defined in Section 111.7, the Lebesgue integral is defined in Section V.3, and Lebesgue measure in R N is defined in Section 1 of the present chapter. In order to have a notational distinction between the Riemann and Lebesgue integrals, we write in this section R -/A f d.w for the Rie~nannintegral of a bounded realvalued function on a compact geometric rectangle A , and we write Ja f dx for thc Lcbcsguc integral.
Theorem 6.31. Suppose that f is a bounded real-valued function on a compact geometric rectangle A in R N . Then f is Riemann integrable on A if and only if f is continuous except on a Lebesgue measurable set of Lebesgue measure 0. In this case, f is Lebesgue measurable, and R f dx = f dx.
4. Comparison o f Riemann and Lebesgue Integrals
319
PROOF.Proposition 6.12 shows that a set in RN has "measure 0" in the sense of Chapter I11 if and only if it is Lebesgue measurable of measure 0, and Theorem 3.29 shows that f is Riemann integrable on A if and only if f is continuous except on a set of measure 0. This proves the first conclusion of the theorem. For the second conclusion, suppose that R f dx exists. Lemma 3.23 shows that there exists a sequence of partitions P ( ~of) A, each refining the previous one, such that the lower Riemann sums L ( P ( ~ )f,) increase to R f dx and the upper Rienlarin surrls U(P(k),f ) decrease to R Jk f d n . Fur each k , we define Burel functions Lk and Uk on A as follows: If x is an interior point of some component (closed) rectangle S of P(", we define Lk(x) = ms( f ) , where ms(f ) is the infimum of f on S ; otherwise we let Lk(x) = 0. If we write I SI for the volume of S , then the Lebesgue integral of Lk over S is given by J;, Lk(x)dx = m r ( f )1 SI . Consequently
iA
We define Uk(x) similarly, using the supremum Ms (f) of f on S instead of the infimum, and then
Let E be the subset of points x in A such that x is in the interior of a component rectangle of ~ ( for~ all1 k. The set A - E is a Borel set of Lebesgue measure 0. Since P ( ~'1+is a refinement of P ( ~for ) every k , we have Lk(x) 5 Lk+' (x) and Uk(x) > Uk+l(x) for all x in E and all k. Therefore L (x) = lim Lk(x) and U(x) =1imUk(x)existforxinE. SinceLB(x) ( f ( x ) 5 U k ( x ) f o r x i n E , w e see that L(x)if(x)iU(x) forallxinE. Define L(x) = U(x) = 0 on A - E . Then L and U are Borel functions with L(x) 5 U(x) everywhere on A. On E , we have dominated convergence, and thus Lk(r)d.x
and
l
U (.x) d.x = lip
l
Uk (.XId-x.
The set A - E has Lebesgue measure 0, and therefore these equations imply that Lx(x)dx
and
~~(x)dx=ii~J,u~(x)dx.
3 20
VI. Measure Theory for Euclidean Space
Consequently
Since L ( x ) 5 U ( x )on A, Corollary 5.23 shows that the set F where L ( x ) = U (x) is a Borel set such that A - F has Lebesgue measure 0. Since the inequalities L ( x ) 5 .f ( x ) 5 U ( x ) are valid for x in E , ,f ( x ) equals L ( x ) at least on E n F. The set E n F is a Borel set, and L is Borel measurable; hence the restriction o f f to E nF is Borel measurable. The set A - (E f' F) is a Borel set of measure 0, and the restriction of f to this set is Lebesgue measurable no matter what values f assumes on this set. Thus f is Lebesgue measurable. Then the Lebesgue integral f d x is defined, and we have
1,
L ( x )d x = R
S,
f dx.
5. Change of Variables for the Lebesgue Integral A general-looking change-of-variables formula for the Riernann integral was proved in Section 111.10. On closer examination of the theorem, we found that the result did not fully handle even as ostensibly simple a case as the change from Cartesian coordinates in &t2 to polar coordinates. Lebesgue integration gives us methods that deal with all the unpleasantness that was concealed by the earlier formula.
Theorem 6.32 (change-of-variables formula). Let cp be a one-one function of class C' from an open subset U of RN onto an open subset p ( U ) of RN such that det cpf(x)is nowhere 0. Then
for every nonnegative Borel function f defined on cp(U). REMARK.The o-algebra on p ( U ) is understood to be BN n p ( U ) , the set of intersections of Borel sets in RN with the open set cp(U). If f is extended from q ( U ) to RN by defining it to be 0 off q ( U ) ,then measurability of f with respect to this o-algebra is the same as measurability of the extended function with respect to BN.
5. Change o f Variables for the Lebesgue Integral
321
PROOF.Theorem 3.34 gives us the change-of-variables formula, as an equality of Riemann integrals, for every f in Cco,(rp(U)). In this case the integrands on both sides, when extended to be 0 outside the regions of integration, are continuous on all of RN,and the integrations can be viewed as involving continuous functions on compact geometric rectangles. Proposition 6.11 (or Theorem 6.31 if one prefers) allows us to reinterpret the equality as an equality of Lebesgue integrals In the extension of this identity to all nonnegative Bore1 functions, measurability will not be an issue. The function f is to be measurable with respect to BN(rp(U)),and Corollary 6.29 shows that such f's correspond exactly to functions f o q measurable with respect to B N ( U ). Using Theorem 5.19, define a measure p on BN ( U ) by
Corollary 5.28 implies that p satisfies
for every nonnegative g on U measurable with respect to B N ( U ) . Next define another set function v on B N ( U ) by
where rn is Lebesgue measure. It is immediate that v is a measure, and we have Passing /,(,j IE ( P - ' ( Y ) ) d y = luioj I ~ ( Ed~) = m ( ~ ( E )=) v ( E ) = IE to simple functions > 0 and then using monotone convergence, we obtain
So
for every nonnegative g on U measurable with respect to BN ( U ). If in (**) and (*) we take g = f o y, with f in Ccom(y,(U))and we substitute into the change-of-variables formula as it is given for f in Ccom(y,(U)), we obtain the identity
for all g in C,,,(U). From Corollary 6.26 we conclude that p = v . Hence (7) holds for every nonnegative g on U measurable with respect to BN (U) . We unwind (7) using (**) and (*) with g = f o y, but now taking f to be any nonnegative function on rp(U) measurable with respect to BN(rp(U)),and we obtain the conclusion of the theorem.
VI. Measure Theory for Euclidean Space
3 22
Let us return to the example of polar coordinates in Kt2, first considered in Section 111.10. The data in the theorem are
(z) and we have Since det p' (
a)
r cos 0 (rsine)'
=y(i)=
w = Kt2- {(i)I x
? o}
= r , Theorem 6.32 gives
S
f ( r c o s 8 , r sin8)r d r dtl
O
for every nonnegative Borel function f on y ( U ). The set of integration y ( U ) on the left side is not quite the whole plane; it omits the part of the x axis where x > 0. But this is a harmless defect: this subset of the x axis is contained in the entire x axis, which is an abstract rectangle in the sense of Fubini's Theorem and has measilre 0. Thils the formilla can he changed to read
for every nonnegative Borel function f on R2.Here is an application of this formula that we shall use in proving the Fourier Inversion Formula in Chapter VIII. 00
Proposition 633.
p-nx2
dx = 1 .
REMARK.Since we now know from Theorem 6.31 that there is no discrepancy between the Riemann integral and the Lebesgue integral with respect to Lebesgue measure, there will be no harm in the future in writing limits of integration in the usual way for integrals with respect to l-dimensional Lebesgue measure. PROOF.We use polar coordinates and Fubini's Theorem to compute the square of the integral in question:
Since the integral in question is certainly z 0, the proposition follows.
5. Change o f Variables for the Lebesgue Integral
323
Proposition 6.33 is closely related to properties of the "gamma function," a certain function of a complex variable that reduces essentially to the factorial function on the positive integers. The definition of the gamma function makes use of the expression t S defined for 0 < t < +oo and s in C by t S = eS'Ogt. Fix s E C with Res > 0. The function t H ts-lePt is continuous on (0, foe) and hence Bore1 measurable Let us see that it is integrable with respect to Lebesgue measure. Since itS-'e-7 = t R e S - l e - t ,we may assume that s is real (and positive) in showing the integrability. Integrability on (0, I] is no problem, since we know that jt t"' dt < m for s > 0 To handle [I, + m ) , let n be an integer s - 1 Then ts-I 5 tn = 2"n! 5 2"n! ZxU, = 2nn!et/2.Hence ts-'ePt < 2"n ! e - t / 2 ,and the integrability on [1, + m ) follows. With this integrability in place, we define the gamma function by
(5(ijn)
r(s) =
6"
ts-1 e -t dt
h(i)L
forRes>O.
Proposition 634. The gamma function has the properties that (a) l-(s + 1) = s r ( s ) forRes > 0, (b) l-(1) = 1 and r(n + 1) = n! for integers n > 0, (c) l- (;) = f i . PROOF.Part (a) follows from integration by parts, which needs to be done on an interval [s, 11/11 and followed by passages to the limit s + 0 and M + m. In ePtdt = 1, (b), the formula r ( 1 ) = 1just amounts to the elementary integral and then the formula T ( n 1 ) = n ! for integers n > 0 follows by iterating (a). For (c), the change of variables t = n x 7 gives
JbOO
+
JbOO d x = 1 j"" dx, Proposition 6.33 allows us to conclude that 2 Jb" eCzx2dx = 1 . Hence l- (i)= f i . Since
e-"x2
2
e-"x2
-00
It is often true in applications of the change-of-variables formula that the set p(U) does not exhaust the set that one might hope to have as region of integration. For polar coordinates the exceptional set was the part of the x axis with x 2 0, and an easy argument showed that the exceptional set had measure 0. In a more complicated example, that easy argument will not ordinarily apply, but still the exceptional set has a certain "lower-dimensional" quality to it. A general result saying that certain lower-dimensional sets have measure 0 will be given as a corollary of Sard's Theorem, which we prove now. Let $I : V + RN be a smooth map defined on an open subset V of RN.A critical point x of $I is a point where I/Tf(x)has rank < N . In this case, I/T(x) is called a critical value.
3 24
VI. Measure Theory for Euclidean Space
Theorem 6.35 (Sard's Theorem). If @ : V + RN is a smooth map defined on an open subset V of IRN,then the set of critical values of @ is a Borel set of Lebesgue measure 0 in RN. PROOF.The set where @'(I) has rank 5 N - 1 is relatively closed in V and hence is the union of countably many compact sets. The set of critical values is then the union of the compact images of these sets and consequently is a Borel set. Let us see that this Borel set has Lebesgue measure 0. Since V is the countable union of compact geometric rectangles and since the countable union of sets of measure 0 is of measure 0, it is enough to prove the theorem for the restriction of @ to a compact geometric rectangle R . For points x = (xl, . . . , xN) and x' = (xi, . . . , x h ) in R , the Mean Value Theorem gives
where Z- is a point on the line segment from x to I ' Since the on R , we see as a consequence that
are bounded
with a independent of x and x' . Let L, (x') = (Lx31 (x') , . . . , L,J (x')) be the best first-order approximation to @ about x , namely
Subtracting this equation from (*), we obtain
Since
is smooth and r i - x I 5 Ix'
-
x I, we deduce that
with b independent of x and x'. If x is a critical point, let us bound the image of the set of x' with Ix' - x I 5 c. The determinant of the linear part of L, is 0, and hence L, has image in a hyperplane, not necessarily a coordinate hyperplane. By (t), @(xf)has distance
5. Change o f Variables for the Lebesgue Integral
5 bc2 from this hyperplane. In each of the N
-
325
1 perpendicular directions,
(**) shows that @ ( x f )and @ ( x ) are at distance 5 ac from each other. Thus
$(x') is contained in a box1 centered at $ ( x ) with volume 2 N ( a c ) N - 1 ( b ~= 2) 2NaN-lbCN+l We subdivide R into M N smaller compact geometric rectangles whose dimensions are l / M times those of R If d is the diameter of R and if one of these smaller geometric rectangles R' contains a critical point x ,then any point x' in R' has Ix' - x I 5 d / M . By the result of the previous paragraph, I/J of R' is contained in a box of volume 2 N a N - 1b ( d / ~ ) ~ +The l . union of these boxes, taken over all of the smaller geometric rectangles containing critical points, contains the critical values. Since there are at most M N of the smaller geometric rectangles, the outer measure m* of the set of critical values, where m refers to Lebesgue measure, is 5 2NaN-1bdN+1M-1.This estimate is valid for all M , and hence the set S of critical values has m* ( S ) = 0. Therefore the Borel set S has Lebesgue measure 0. Corollary 6.36. If I/J : V + RN is a smooth map defined on an open subset V of RM with M < N , then the image of @ is a Borel set of Lebesgue measure 0 in R N . PROOF. Sard's Theorem (Theorem 6.35) applies to the composition of the projection R N + R M followed by @. Every point of the domain is a critical point, and hence every point of the image is a critical value. The result follows.
We define a lower-dimensional set in R N to be any set contained in the countable union of smooth images of open sets in Euclidean spaces of dimension < N . The following result is immediate from Corollary 6.36. Corollary 6.37. Any lower-dimensional set in R N is Lebesgue measurable of Lebesgue measure 0. The N-dimensional generalization of polar coordinates in R2 is spherical coordinates in R N . In the notation of Theorem 6.32, we have
Y
cos 01
and r smO1... s ~ n t l ~ - ~ c o s O ~ - ~ r sin01 ... s i n 0 ~ - 2s i n B ~ - l
his box need not have its faces parallel to the coordinate hyperplanes.
VI. Measure Theory for Euclidean Space
3 26
Problem 2 at the end of the chapter asks for three things to be checked: (i) the determinant factor in the change-of-variables formula is given by
I det q1'1
= rN-' sinNp2O1 sinNp302 . . .
(ii) rp is one-one on U , (iii) h e ~urriplerncntof y ( U ) in RN is a luwer-clir~iensiurlalscl. Then it follows that the change-of-variables formula applies and that the integration over p ( U ) can be extended over RN. We can write the result as
x rN-I sinNp281 . . . sin
doNpl . . .dol dr.
The expression sinNp2Q1 . . . doNpl . . . dQl we abbreviate as dw. Geometrically it is the contribution to Lebesgue measure on R N from the sphere sN-l of radius 1 centered at the origin. In Chapter XI we shall speak of Borel sets in any compact metric space. The sphere sN-lis a compact metric space, and we shall note that dco refers to a rotation-invariant Borel measure on sN-l.We write
This constant is evaluated in Problem 12 at the for the "area" of the sphere sN-l. end of the present chapter with the aid of Proposition 6.33. In terms of dw, the change-of-variables formula for spherical coordinates is
LN
f (x) dx =
Sn r=O
f ( r ~rNP1 ) dodr. w€SN-'
This formula allows us quickly to check the integrability of powers of 1x1 near the origin and near oo. In fact, we have
and frorri wllicli we see that 1x Iq is integrable near
0
forq > -N,
rn
forq < -N.
6. Hardy-Littlewood Maximal Theorem
327
6. Hardy-Littlewood Maximal Theorem This section takes a first look at the theory of almost-everywhere convergence. The theory developed historically out of Lebesgue's work on an extension of the Fundamental Theorem of Calculus to general integrable functions on intervals of the line, work that we address largely in the next chapter. We shall see gradually that the theory applies to a broader range of problems than the ones immediately generalizing Lebesgue's work, and one can make a case that nowadays the theory in this section is of considerably greater significance in real analysis than one might expect from Lebesgue's work on the Fundamental Theorem. The theory brings together two threads. The first thread is the observation that an effort to differentiate integrals of general integrable functions on an interval of the line can be reinterpreted as a problem of almost-everywhere convergence in connection with an approximate identity of the kind in Theorem 6.20. In explaining this assertion, let us denote Lebesgue measure by m as necessary. To differentiate F ( x ) = j""' f ( t )d t , one forms the usual difference quotient hP1[ F ( x h ) - F ( x ) ], which can be written for h > 0 as
+
or as f * rph(x),where rp(y) = m ( [ - 1, 0])-'l~-l,o1( y ) .Here rp has integral 1,and ph is the normalized dilated function defined in Section 2 by rph ( y ) = h rp (h y ) in the 1-dimensional case. Theorem 6.20 says for p = 1 and p = 2 that as h decreases to 0, f * rph converges to f in LP i f f is in LP. Also, f * rph converges uniformly to f if f is bounded and uniformly continuous, and f * rph ( x )converges to f (x) at the point x if f is bounded and is continuous at x . The problem about differentiation of integrals asks about convergence almost everywhere. We shall want to have a theorem in RN,and for this purpose an N-dimensional does not seem attractive for generalizing. Instead, let us generalversion of I[-l,ol ize from 11,taking the N-dimensional problem to involve a ball B of radius 1 centered at the origin; there is some flexibility in choosing the set B, and a cube centered at the origin would work as well. We write r B for the set of dilates of the members of B by the scalar r . Thus we investigate
'
as r decreases to 0; equivalently we investigate
'
328
VI. Measure Theory for Euclidean Space
The second thread comes from making a simple observation and then trying to prove the converse in specific settings, as improbable as it sounds. The observation is that if some sequence of nonnegative functions indexed by n is to converge almost everywhere, its supremum on n must be finite almost everywhere. A converse would say that a finite supremum almost everywhere implies convergence almost everywhere Banach succeeded in proving an abstract such converse in a 1926 paper, making use of the completeness of the space of functions he was studying. In a celebrated 1930 paper, Hardy and Littlewood proved a concrete such converse in connection with differentiation of integrals; they obtained a quantitative estimate about the supremum, and then the almost everywhere convergence followed from that estimate and from the fact that the convergence certainly takes place for nice functions. Here is an N-dimensional version of the basic theorem in that direction.
Theorem 6.38 (Hardy-Littlewood Maximal Theorem). If f is in L 1(RN) , then
for every 6 > 0, where
Before examining the statement of the theorem more closely and then proving the theorem, let us see how to derive a corresponding N-dimensional convergence result from it, and let us see how the first part of Lebesgue's version of the Fundamental Theorem of Calculus, the part about differentiation of integrals, follows as well.
Corollary 6.39. If f is integrable on every bounded subset of R N , then
~ O O F Since . the convergence for a particular x depends on the behavior of the function only near x ,we may assume that f is identically 0 off some bounded set. The effect of this assilmption for onr pilrposes is that f then has to he in L' (RN). Define P
bearing in mind that f "(x) = sup,,, T,(l f I)(x). If g is continuous of compact support, then limTJoT,g (x) = g(x) everywhere by Theorem 6 . 2 0 ~Let . t > 0 be
6. Hardy-Littlewood Maximal Theorem
given, and choose by Corollary 6.4 a function g in c,,,(R~) with 11 f Then
329 -
gill
it
.
If the left side is > e , at least one of the terms on the right side is > e / 2 . Hence
I
By Theorem6.38 and the inequality a m { x I F ( x ) I > a } 5 llFll1, the Lebesgue measure of the right side is
I
Since t is arbitrary, S ( 5 ) = { x lim sup 11; f ( x ) - j'(x)I > 6 ) has measure O . Letting 6 tend to 0 through the values 1i n , we see that S ( 0 ) has measure 0 , i.e., that limrJo T, f ( x ) = f ( x ) almost everywhere.
Corollary 6.40 (first part of Lcbcsguc's form of thc Fundamental Thcorcm of Calculus). If f is integrable on every bounded subset of Kt1, then JaX f ( y ) d y is differentiable almost everywhere and
~ O O F For .
f in L (IR1),let f * be as in Theorem 6.38, and define
If(x+t)ldt
and
f:*(x) =
h>O
Then
If(x+t)Idt
=2f*(x),
VI. Measure Theory for Euclidean Space
3 30
and similarly f;* ( x ) 5 2 f * ( x ). From Theorem 6.38 it follows that
I
and
m { x fi**(x> > 6 ) 5 l 0 1 1 IfI I / <
'I'he same argument as for Corollary 6.39 allows us to conclude, for any f integrable on every bounded subset of R', that limILJo f ( x + t )dt = f ( x )
$1;
,IPh + 1
0
a.e. and limhJo f ( x t ) d t = f ( x ) a.e. Hence almost everywhere for such f .
2 laxf ( t ) d t
= f (x)
Let us return to Theorem 6.38. The function f * ( x ) is called the HardyLittlewood maximal function of f . It is measurable because the supremum over rational r gives the same value of f * ( x ) for each x . If we let 6 tend to cc in the inequality m { x f * ( x ) > 6 } 5 5 N f / 6 , we see immediately that f * ( x ) is finite almost everywhere, i.e., that the supremum in question is actually finite almost everywhere. The inequality is a quantitative version of that qualitative conclusion. For any situation in which it is desired to prove an almost-everywhere convergence theorem, there is an associated maximal function in modem terminology, which can be taken as the supremum of the absolute value of the quantity for which one is trying to prove almost-everywhere convergence. In the above case we used the supremum for I f 1 instead, which in principle could be larger. There is no hope that the Hardy-Littlewood maximal function f * is actually in L' if f is not a.e. the 0 function because the occurrence of large values of r in the supremum already rules out L behavior: in fact, f * ( x )is necessarily 1a positive multiple of IxlPN for large 1x1, and thus f * cannot be integrable. On the other hand, f * is close to integrable: We shall see in Section 10 that the integral of any nonnegative function g can be computed in terms of the function m { x g ( x ) > $ ) of 6 , the formula being g ( x ) d x = 1; m { x g ( x ) > 6 ) d $ . Theorem 6.38 shows that the integrand in the case of f * is 5 a multiple of 116, and 116 is close to being integrable on ( 0 , +a)This . is a better qualitative conclusion than merely finiteness almost everywhere, and Theorem 6.38 is a quantitative version of just how close f * is to being integrable. The particular property of f * that is isolated in Theorem 6.38 arises fairly often. If g 10 is integrable and S is the set where g > 6 , then g 1 61s everywhere; hence llglll 6 m ( S ) and m ( S ) 5 llglll/$. A function g is said to be in weak L' if m { x Is(x)l > i c/c
I
1 1
I
IRN
I
'
I
c}
for some constant C and for all $ > 0. Theorem 6.38 says that the nonlinear operator f H f * carries L' to weak L with C bounded by a multiple of the L'
'
6. Hardy-Littlewood Maximal Theorem
331
norm of f , and an operator of this kind that satisfies also a certain sublinearity property is said to be of weak type (1, 1). We retum to this matter, with the definition in a clearer context, in Chapter IX. Now let us prove Theorem 6.38. One modem proof uses the following covering lemma, which takes into account the geometry of RN in a surprisingly subtle way. Once the lemma is in hand, the rest is easy.
Lemma 6.41 (Wiener's Covering Lemma). Let E g RN be a Bore1 set, and suppose that to each x in E there is associated some ball B ( r ; x ) with r perhaps depending on x . If the radii r = r ( x ) are bounded, then there is a finite or countable disjoint collection of these balls, say B ( r l ; X I ) , B(r2;x 2 ) ,. . . , such that either the collection is infinite and infl,j,, rj # 0 or
In either case.
REMARK. The shape of the sets of B ( r ; x ) is not very important. What is important is that there be some neighborhood B of the origin that is closed under the operation of multiplying all its members by - 1 and by any positive number r 5 1. The other sets are obtained from B by dilation and translation. PROOF. Let
A1 = {all sets B ( r ; x ) in question} and
I
R1 = sup { r B ( r ; x ) is in A1for some x }
By hypothesis, R1 is finite. Pick some B ( r l ; x l ) with rl 2 ~ R Iand , let A2 =
{members of A1disjoint from B ( r l ;x l )}
If PI2 is empty, let all further Kj's be O and let all further B ( r j ;x j ) ' s be empty. If is nonempty, let
I
R2 = sup { r ~ ( rX); is in A2 for some x } .
Pick B(r2;x2) in Az with r2
>
~ 2Let .
A3 = {members of A2 disjoint from B(r2;x2)},
332
VI. Measure Theory for Euclidean Space
and proceed inductively to construct R j , B ( r j ;x j ) , A,etc. The numbers Rj are monotone decreasing. We may assume that lim Rj = 0, since otherwise infj r j # 0 and m ( B ( r j ;x j ) ) = +oo. Let Vj = union of all sets in Aj - Aj+1
for j ? 1
Vo = union of all sets in A1.
and
Then Vo = U E l V, ; in fact, if B(r; x ) is in A 1 , then the equality lim Rj = 0 forces there to be a last index j such that B(r; x ) is in A j , and this j has the property that B (r ; x ) is in Aj and not Aj+l. Since E c U,,,B(r; x ) = Vo = U E l V j ,the proof will be complete if we show that Vj B(5rj;x j ) . (*) Thus let B ( r ; x ) be in Aj - Aj+1. Then r 5 Rj ,
and rj 2 + R ~Consequently . r 5 2rj and
This condition means that there is some p in B(rj;0 ) with Ix q is any member of B(r; x ) , then
-
xj
-
pl < r . If
Thus q is in B(2r +rj; x j ) C B(5rj;x j ) , and (*) follows. PROOFOF THEOREM6.38. Let E = { x f * ( x ) z 61. If x is in E , then m ( B ( r ;0))-' 1 f (yjl d y > 6 for some r > 0. Associate this r to x in applying Lemma 6.41. Since
lB(r;xj
we see that r N 5 t - l r n ( ~ ( 10))-' ; 11 f 11 1. Hence the numbers r are bounded. Thus the lemma applies, and we obtain
the last inequality holding because of the disjointness of the sets B(rj;x j ) .
6. Hardy-Littlewood Maximal Theorem
333
Let us return to the theme of almost-everywhere convergence in connection with approximate identities. Theorem 6.38 has the following consequence of just that kind. Corollary 6.42. Let rp > 0 be a continuous integrable function on RN of the form rp(x) = rpo(lxI), where rpo is a decreasing C function on [O, a), and define p, ( x ) = C Np(e-'x) for e > 0. Then there is a constant C , such that sup I(% &>O
* f)(x)I
5 C,~+(X)
for all x in RN and for all f in L' ( R N.)Consequently if
JRM p ( x ) d x =
1, then
almost everywhere for each f in L ( R N.) PROOF.Put + ( r ) = -rpA(r) > 0, so that rpo(r) - rpo(R) = J ; + ~ ( s ) d s . The integrability of p and the fact that po is decreasing force limR,, po(R) = 0, and we obtain rpo(r) = ITw @ ( s )d s and p ( x ) = ( r )dr . Meanwhile, the integrability of rp, together with the formula for integrating in spherical coordinates, shows that po ( r )rN-I dr = C < +m. Integrating by parts on the interval [0, MI gives
hx";+
Jb"
C
> J:
po(r) r N p ldr = + [ r n ( r )I N ] :
and thus j$
JbOO @ ( Y )r N dr 5 C <
+ k J: 1
~ ( rr N) d r ,
00.
The form of rp implies that rpF(x)= cN + ( I . ) dr. If, as in the statement of Theorem 6.38, we let B be the ball of radius 1 centered at the origin, we obtain
5 m ( R ) [ J F z o?Ir(r)r N d r ] f " ( X I . The right side is 5 C , f " ( x ) with C , = C N m ( B ) . Applying Theorem 6.38, I ( p s * f ) ( x ) is of weak type (1 ,I). Since we see that the operator f H p, * f converges pointwise (and in fact uniformly) to f when f is in C,,,(RN) , the same argument as for Corollary 6.39 shows that limElo(pE* f ) ( x ) = f ( x ) almost everywhere for each f in L ( R N.)
'
VI. Measure Theory for Euclidean Space
3 34
1
EXAMPLE. An example of a function p as in Corollary 6.42is P ( x ) = 1 +x2 in R1. We shall see in Chapter VIII that the function h ( x , y ) = ( P y * f ) ( x ) for this q is the natural function on the half plane y > 0 in IE2 that is harmonic, i.e., a2h a2h has -+ - = 0, and has boundary value f . Corollary 6.42says that h ( x , y ) ax2 a y 2 has f ( x ) as boundary values almost everywhere i f f is in L' (IR1)
7. Fourier Series and the Riesz-Fischer Theorem As mentioned at the beginning of Chapter V, the use of the Riemann integral imposes some limitations on the subject of Fourier series that no longer apply when one uses the Lebesgue integral. In this section we shall redo the elementary theory of Fourier series of Section 1.10 with the Lebesgue integral in place, with particular attention to the improved theorems that we obtain. It will be assumed that the reader knows the theory of that section. n]with the o-algebra of Bore1 The underlying measure space with be [-n, sets and with the measure d x , where d x is 1-dimensional Lebesgue measure. The complex-valued functions under consideration will be periodic of period 2n, thus assuming the same value at n as at -n . The spaces L l , L 2 , and L m will refer to this measure space when no other parameters are given. Since the measure - L1. of the whole space is finite, these spaces satisfy the inclusions L m 5 L~ C The functions in L m being essentially bounded, they are certainly integrable and square integrable. The inclusion L2 5 L' follows from the Schwarz inequality:
&
1
J-"n I f
11 dx 5 (& J?" l f l 2 d x ) l I 2 ( & J?" 1 d x ) l I 2 . There is another way of viewing this measure space that will be especially helpful in relating convolution to the theory. Namely, a periodic function on the line of period 2n may be viewed as a function on the unit circle of C with the angle as parameter. In fact, convolution is a construction that combines group theory with measure theory when the measure is invariant under the group, and that is n].The limits why convolution appears more natural on the circle than on [-n, uf irllegraliv~ldu riul have Lu be wrillcm diffcrerilly lrurn Llie way llicy are wrillcri on the line, but we must remember that functions are to be extended periodically when we interpret integrands. The factor in front of the measure means that all convolutions of functions are to contain this factor. Thus the definition of convolution for nonnegative f and g is
&
Convolution is commutative and associative on the circle just as in Proposition 6.13, and the various norm estimates of Section 2 are valid in the setting of the
7. Fourier Series and the Riesz-Fischer Theorem
335
circle. The use of dilations has no analog for the circle, and thus the circle has no approximate identities of the form p, . If f is in L ' , the trigonometric series
is called the Fourier series of f . This time we regard the integral as a Lebesgue integral. We write
A Fourier series can be written also with cosines and sines, and the coefficients a, and b, are unchanged from Section 1.10.
Theorem 6.43. Let f be in L ~ Among . all choices of d P N , .. . , d N , the expression
is minimized uniquely by choosing d,, for all n with In 1 5 N , to be the Fourier f (x)ePin.' dx. The minimum value is coefficient c, =
& in,
PROOF.The proof is the same as for Theorem 1.53.
Corollary 6.44 (Bessel's inequality). Iff is in L2 with f ( x ) then
In parlicular,
Cz-, lc, l 2
Cz-,
&nx
,
is Gnile.
PROOF.The proof is the same as for Corollary 1.54. Corollary 6.45 (Riemann-Lebesgue Lemma). If f is in L' and has Fourier cn = 0. coefficients {c,}E-,, then limlnl+oo REMARK.Since L2 is properly contained in L' ,this corollary is not a special case of Corollary 6.44, unlike the situation with the Riemann integral.
VI. Measure Theory for Euclidean Space
336
PROOF.The result is immediate from Corollary 6.44 in the case of L~ functions and in particular in the case of continuous functions. Write c, (h) for the nth Fourier coefficient of any function h . Let t > 0 be given. Choose by Corollary 6.27a a continuous g with I l f - g I l l i t/2. Then choose N such that Inl 1 N implies Icn(g)I 5 t / 2 . Then Ic,(f)I 5 Icn(f - g)I Icn(g)l 5 t since If - glle-inxl d-x = Ilf - g l l ~i ~ 1 2 . Ic,,(f - g>l i
+
-&
Theorern 6.46 (Di~li'sLesl). Let f be i11 L' ,and fix x in [-n, n]. If Illere is a constant S > 0 such that
then limNs, ( f ; x ) = f ( x ). REMARK.The condition in the corresponding result for the Riemann integral, namelyTheorem1.57,wasthat~f(x+t) f ( x ) I M l t l f o r I t l <Sandsome constant M . The condition in the present theorem is satisfied by f ( x ) = at x = 0, and the condition in the earlier theorem is not. PROOF. The proof is the same as for Theorem 1.57 except that we need to appeal to the improved version of the Riemann-1,ehesgne 1,emma in Corollary 6.45. Now we work toward a proof of Parseval's Theorem for all of L ~ We . need to know about Fourier coefficients of convolutions.
-
00 Proposition 6.47. Iff ( x ) C,=-, ( f * g ) ( x ) -- C,"=-,cndneinx.
cneinx and g ( x ) --
2,=-, 00
dneinx,then
PROOF.This is a consequence of Fubini's Theorem and the translation invariance of Lebesgue measure:
7. Fourier Series and the Riesz-Fischer Theorem
337
The proof of the version of Parseval's Theorem for all of L2 will make use of the FejCr kernel KN(t) introduced in Section 1.10. We do not need to recall the exact formula for KN,only the fact that it is a trigonometric polynomial of degree N with the following three properties: (9 K,(x) > 0, KN(x) dx = 1, (ii) (iii) for any 6 > 0, sup,,lxl,, KN(x) tends to 0 as n tends to infinity.
&
These three properties identified KN as an approximate identity in the setting of periodic functions, and FejCr's Theorem in the form of Theorem 1.59 gave the consequence for convergence at points of continuity of f . With the Lebesgue integral, we get also results about norm convergence in L and L2.
Theorem 6.48 (FejCr's Theorem). Let f be in L l . Then (a) l i m ~l l KN * f - f 11 1 = 0 with no additional hypotheses on f , (b) limNllKN * f - f l 1 2 = 0 i f f isalsoin L2, (c) l i m (KN ~ * f ) (XO)= f (x0) if f is bounded on [-n, n ] and is continuous at xo > (d) thc convcrgcncc in (c) is uniform for xo in E if f is boundcd on [-n, n ] and is uniformly continuous at the points of E, (e) l i r n ~ ( K* ~f ) (no) = (f (no+) f (no-)) if f is bounded on [-n, n ] and has right and left limits f (xO+) and f (xo-) at xo.
+
PROOF.For (a) and (b), let p llKn*f
-
=
1 or p
=2
as appropriate. Then
f llp KN(t)[f (X 1
-
t)
-
f (XI]d t
by (ii) by (i) and Theorem 5.60
Givcn r > 0, choosc S by Proposition 6.16 to makc thc first tcrm of thc final bound be < e / 2 , and then choose No by (iii) to make the second term of the final bound be < r / 2 for N 1No. Then the final bound is < t- for N 1No. Parts (c) and (d) are proved exactly as in Theorem 1.59. For (e), we may assume without loss of generality that xo = 0 because convolution commutes with translations. If we can prove (e) for a single function g with a jump discontinuity
338
VI. Measure Theory for Euclidean Space
at x = 0 equal to the jump for f , then we can apply (c) to f - g and deduce (e) for f . Let us see that such a function g may be taken as a multiple of
In fact, a computation at the beginning of Section 1.10 shows explicitly that the series C,"=, (sinnx)/n converges to h(x) for x # 0, but we do not need this fact. (sin nx)/n is the Fourier series of h , a All that we need is that the series C,"=, fact that we can readily check from the definition. The sum of this series at x = 0 is manifestly 0, and this sum matches the average of the jumps + The CesBro sums of the series C,"=, (sin nx)/n must have the same limit 0, according to Theorem 1.47, and (e) is proved.
(5 2).
Theorem 6.49 (Parseval's Theorem). If f is a function in L2 with f (x) a3 Cn=-, cneinx, then
-
and
PROOF.Fromthe first conclusion of Theorem6.43, we obtain0 5 11 f -sN 11; i and we know from Theorem 6.48b that 11 f - (KN* f ) 1 ; tends to 0. This proves the first formula, and the second formula follows by passing to the limit in the second conclusion of Theorem 6.43.
11 f - (KN * f ) ll;,
Corollary 6.50 (uniqueness theorem). If f is in L' and has all Fourier coefficients 0, then f is the 0 element in L l . PROOF.Proposition 6.47 shows that the Fourier coefficients of KN * f are cn(KN* F) = c,(KN)cn(f ) , and this is 0 for all n. By Proposition 6.18, KN * f is continuous, and thus KN * f = 0 by Corollary 1.60. Since KN * f tends to f in L' according to Theorem 6.48a, we conclude that f is the 0 element in L' . Now we come to the Riesz-Fischer Theorem, which historically was a great triumph for the Lebesgue integral over the Riemann integral. The result uses the completeness of L2 and has no counterpart with Riemann integration.
Theorem 6.51 (Riesz-Fischer Theorem). If {cn}is a given doubly infinite sequence of complex numbers with C,"=-, lcn12 < oo,then there exists an f in L2 whose Fourier series is zr=-, cneinx.
8. Stieltjes Measures on the Line
339
PROOF.Define FN ( x ) = '&n15N cneinx.For M > N , Parseval's Theorem lcn12,and the right side tends (Theorem 6.49) gives 11 f M - fN 11; = to 0 as M and N tend to infinity because of the convergence of C,"=-,lcnl 2 . Thus { f N } is a Cauchy sequence in L2. By Theorem 5.59, L2 is complete as a metric space, and thus { f N } converges in L2. Let f be (a function representing) the limit element in L ~ The . inner product in L2 is a continuous function of the L2 function in the first variable, and therefore the Fourier coefficients o f f satisfy c n ( f ) = ( f , einX)= l i p ((f". einx). As soon as N gets to be > In 1, (f N , einx)equals c, . Thus c, ( f ) = c, for all n , and f has the required properties.
8. Stieltjes Measures on the Line A Stieltjes measure2 is a Borel measure on R'. Lebesgue measure d x is an example, as is any measure f ( x ) d x in which f is nonnegative and Borel measurable and is integrable on every bounded interval. A completely different kind of Stieltjes measure is one that attaches nonnegative weights to countably many points in such a way that the sum of the weights in any bounded interval is finite. In this section we shall see that the Stieltjes measures stand in one-one correspondence with a class of monotone functions on the line that we describe shortly. We shall also obtain an integration-by-parts formula in which a Stieltjes measure plays the role of the derivative of its corresponding monotone function. If a Stieltjes measure p is given, we associate to p the function F : R1 + R 1
The function F is called the distribution function of w . It has the following
proper tie^:^ (i) F is nondecreasing ,i.e., is monotone increasing, (ii) P' is continuous from the right in the sense that P'(xo) = limx~x, b'(x) for every xo in R 1 , i.e., limn F(x,) = F ( x o ) whenever {xnIn->lis a sequence tending to xo such that xn > xo for all n > 1, (iii) F (0) = 0.
an^
books, this one included, take Stieltjes measures by definition to occur on the line. However, there is a theory, albeit a somewhat unsatisfactory one, of "Stieltjes measures" in higherdimensional Euclidean space. It is of interest chiefly in probability theory. 3 ~ alternative n definition says F ( x ) equals -p[x, 0) and p[0, x) in the two cases, and then property (ii) says that F is continuousfrom the left. The choice made here between these alternatives is governed by keeping technicalities to a minimum in Section 10.
340
VI. Measure Theory for Euclidean Space
Properties (i) and (iii) are immediate from the definition. With (ii), there are two cases according as the limit xo is 5 0 or > 0, and both cases are settled by the complete additivity of p . The measure is completely determined by its distribution function F. In fact, the definition of F forces p ( ( a , b ] ) = F ( b ) - F ( a ) , and Proposition 6.6 implies that p is determined as a Bore1 measure by this formula.
Theorem 6.52. The Stieltjes measures p stand in one-one correspondence with the functions F : R1 + R1 satisfying (i), (ii), and (iii), the correspondence being that F is the distribution function of p . PROOF.We have seen that each p leads to an F and that F uniquely determines p . We need to see that every F satisfying (i), (ii), and (iii) arises from some p . If such a function F is given, we define a set function p on bounded intervals by v ( ( n ,bl) = F ( b ) - F ( n ) , p ( ( a , b ) ) = limF(b -
i)F ( a ) , -
~ ( [ ub ], ) = F ( b ) - l i m F ( u I2
p ( [ a ,b ) ) = l i m F ( b n
A),
i)- l i m F ( a
-
n
);1
We extend / I to the ring 'R,of elementary silhsets of R',i .e., the ring of all finite disjoint unions of bounded intervals, by setting p of a finite disjoint union of bounded intervals equal to the sum of the values of p on each of the intervals, just as with Lebesgue measure in Example 4 at the end of Section V.1. To see that p is unambiguously defined and is additive on R,we readily reduce matters, just as with Lebesgue measure, to showing that if an interval is decomposed into the union of two smaller intervals, then p of the union is the sum of p of the components. Thus let a 5 b 5 c , and let an interval I from a to c be the union of an interval from a to b and an interval from b to c. If the interval I from a to c is (a, c ),then the two possible cases are handled by ,I
( ( a ,b ) ) + [I ( [ b ,c ) ) = lim F ( b - ): I1
and p((a,b])
+ p ( ( b ,c ) ) = F ( b )
-
F(a)
-
F(n)
+ lim F (c A) -
I1
+ limF(c
-
-
-
lim F ( b I1
A)
F ( b ) = ~ ( ( ac ),) .
If the interval I from a to c has one or both endpoints present, then the computation is the same except that F ( a ) is replaced by limn F ( a if a is in I and
i)
8. Stieltjes Measures on the Line
341
i)
limn F ( c - is replaced by F(c) if c is in I . Thus p is unambiguously defined, and it is nonnegative and additive. The next step is to prove, just as with Lebesgue measure in Section V.1, that p is regular on R in the sense that for each E in R and t > 0 , we can find a c o m p a c t K i n R a n d a n o p e n U i n R s u c h t h a t K 5 E 5 U,m(K) m ( E ) - r , and m(U) 5 m ( E ) + t As with Lebesgue measure, the proof comes down to the case that E is a single interval, and this time there are four subcases. Choose n large enough so that < t, and then for[a,b),
takeK = [ a , b - :] and U = ( a - , , b1 ) ,
for[a,b],
takeK = [ a , b ] and U = ( a - : , b + ; ) ,
1
+ A, b] and U = (a, b + A), take K = [a + i, b - A] and U = (a, b).
for (a, b ] , take K = [a for (a, b),
An exception occurs in the definition of K if the listed left endpoint of K exceeds the listed right endpoint of K , and then K is defined to be empty. Each of these definitions contains a parameter n ; if we write K, and U, for the corresponding sets K and U, then we can check from the definitions and property (ii) of the function F that limn p(Kn) = p ( E ) and limn p(Un) = p ( E ) . The regularity condition for fi follows from these limit relations. The next step is to prove that p is completely additive on R by imitating the proof for Lebesgue measure. In fact, the proof of Proposition 5.4 applies word-for-word except that m has to be changed to p throughout and the word "proposition" in the last sentence of the proof should be changed to "complete additivity." Then p extends to a measure on the Bore1 sets by Theorem 5.5. The extended measure is o-finite on IR1 because R' is the countable union of bounded intervals and p is finite on every bounded interval. Finally we need to show that the distribution function G of p is equal to F. Our definitions make G(x) =
-w((x,O])=-(F(0)-F(x))=F(x)
ifxiO,
~ ( ( 0XI) ,
ifx
=
F(x)
F(O)
=
F(x)
> 0.
Thus G = F
(1) Let F be any continuous distribution function that has a continuous derivative f except possibly at finitely many points. If x is a point of continuity of f , then the Fundamental Theorem of Calculus (Theorem 1.32) gives
342
VI. Measure Theory for Euclidean Space
2 Jb" f ( t ) d t = f(x). Put -I;Of(t)dt
G(x) =
{ j:
j (t) d t
ifxio, ifx
> 0.
Then G is a continuous distribution function, and the formula for the derivative of the integral shows that Ft(x) = Gf(x) except at finitely many points. Recursive application of the Mean Value Theorem, starting from x = 0, to F - G on intervals having F' - G' = 0 in their interiors, shows that F = G everywhere. The Stieltjes measure p associated to F, by the uniqueness in Theorem 6.52, is given by
The special case with F(x) identically equal to x has f identically equal to 1, and the measure is just Lebesgue measure. (2) The function F with F (x) =
0 -1
forx>O, forx(0,
has the three properties of a distribution function, and the associated measure p is a point mass assigning weight 1 to x = 0. The measure p takes the value 1 on every Borel set containing 0 and takes the value 0 on every Borel set not containing O . This measure is sometimes called the delta measure at O or "delta mass" at 0. Whenever a Stieltjes measure v has v({p}) > 0 for some p in R1, we say that I? contains a point mass at p of weight ~ ( { p } )Then . v is the sum of a point mass at p of weight v({p}) and a Stieltjes measure containing no point mass at p . (3) Let {x,} be a sequence in R.For example, {x,} could be an enumeration of the rationals. Let {w,} be a sequence of positive numbers with C w, < oo, and define ( w, for x 0,
Then F satisfies (i), (ii), and (iii), and hence F is the distribution function of some Sticltjcs mcasurc p . Thc mcasurc is givcn by
It is a countable sum of point masses. The function F, though monotone increasing, is discontinuous at every x, , and this set is allowed to be dense.
8. Stieltjes Measures on the Line
343
(4) This example will be a nonzero Stieltjes measure that is carried on a Bore1 set of Lebesgue measure 0 and yet has a continuous distribution function. We start from the standard Cantor set C in [0, 11described in detail in Section 11.9. This set is compact and is obtained as the intersection of a sequence {C,} of sets with each C, consisting of the finite union of closed bounded intervals. The set Co is [0, 11, and C,,+l is obtained from C,, by removing the open middle third of each of the
FIGURE 6.1.
Construction of a Cantor function F. Graphs of approximations Fl , F2, F3, F4to F .
constituent closed intervals of C,. The Lebesgue measure of C, is (2/3)", and thus C has Lebesgue measure limn(2/3), = 0. The measure p we construct will have p(C) = 1 and p ( C 9 = 0, yet it will assign 0 measure to every one-point set. The properties that are needed of the corresponding distribution function F so that p has these properties are that F is continuous, F is 0 for .\- 5 0, F is 1 for x 2 1, and F is constant on every open interval I of [0, 11 - C ,i.e., on every open interval of every [0, 11 C,. This condition will make p ( I ) = 0 for all such I . Since the metric space [0, 11 - C has a countable base, it is the union of countably many suc11 open illtelvals I, and thus p([0, 11 - C) = 0. Since F is constant for x 5 0 and for x 2 1, p is 0 on (-cm, 0) and (1, +cm) as well, and thus p(Cc) = 0. To obtain the distribution function F, we construct a sequence of approximating functions F, and show that the sequence is uniformly Cauchy. The set C," f' [O, 11 is the union of 2" - 1 disjoint open intervals. On the kth such interval we define F, to be k2-,. We let F, (x) = 0 for x 5 0 and F (x) = 1
344
VI. Measure Theory for Euclidean Space
for x > 1. On the complementary closed intervals, define F, in any fashion that makes F, monotone increasing and continuous. Graphs of F1 through F4 are shown in Figure 6.1 with the interpolation in the graphs done by straight lines. The result is that IFn(x>- Fn+k(x)I 5 zpn for all x . Hence {F,} is uniformly Cauchy and therefore uniformly convergent. Let F be the limit function. The function F continuous by Theorem 1.21, and it is monotone increasing, satisfies F ( 0 ) = 0 , and is constant on every open interval contained in C C . According to Problem 15 at the end of the chapter, it is independent of the method of interpolation used in constructing the F,'s. The function F is called the Cantor function corresponding to the standard Cantor set. The most general monotone increasing function F on R 1 is not far from being the distribution function of some Stieltjes measure. In the first place the monotonicity of F implies that F has left and right limits at every point, and consequently its only discontinuities are jumps. There can be only countably many such jumps: in fact, if thcrc wcrc uncountably many jump discontinuitics, there would be uncountably many in some bounded interval, and that interval would contain uncountably many of magnitude at least l / n for some integer n ; hence F would have to be unbounded on that bounded interval. Let us define a function F1 by F l ( r ) = lim,$, F ( t ) . This is well defined, since F has right limits at every point, and we have F ( x ) = Fl ( x ) except on a countable set. If we define F2(x) = Fl (x) - Fl ( 0 ), then F2 satisfies the three defining properties (i), (ii), (iii) of a distribution function. If is the Stieltjes measure corresponding to F2under Theorem 6.52, we call p the associated Stieltjes measure for F .
Theorem 6.53 (integration by parts). Let a < b, let F be a monotone increasing function on R' that is continuous from the right at a and b, and let p be the associated Stieltjes measure. If G is a C' complex-valued function on [ a ,b] with derivative g ,then
PROOF.Without loss of generality, we may assume that G is real-valued. Let F2be the distribution function of p. By construction of F2, there is a constant c such that F - F2 = c except possibly at points of discontinuity of F, and the set S of such points S within [ a ,b] is countable. This exceptional countable set S does not contain a or b, since F and F2 are continuous from the right at a and b.
8. Stieltjes Measures on the Line
We have
and also
Thus
Comparing this formula with the formula in the statement of the theorem, we see that if the theorem holds for F2, then it holds for F. Changing notation, we may therefore assume that F is the distribution function of p. Let P be a partition a = xo < xl < . . . < x,-1 < x, = b of [ a ,b ] with mesh to be specified. For 1 5 i 5 n , we use the Mean Value Theorem to choose ti c ( x i P 1x, i ) with G ( x i )- G ( x i P l )= g ( t i ) ( x i- x i - l ) . We can do so since we have assumed that G is real valued. Then we have
and
We shall show for small enough mesh that Cr=lF ( x i ) g ( t i ) ( x i x i p l ) is close to lab F ( x ) g ( x )d x and that ELl G ( x i P l ) ( F ( x i-) F ( s - 1 ) ) isclose to /ia,,l G d p . Let M be an upper bound for Ig 1 and IF I on [ a ,b ] ,and let t > 0 be given. Choose a number 6 > 0 by uniform continuity of G and g such that Ix - x'l < 6 implies IG(x) - G(xf)I < t / M and Ig(x) - g(xf)I < t / ( M ( b - a ) ) , as well as another condition to be specified. If the mesh of the partition is i6 , then
346
VI. Measure Theory for Euclidean Space
n
C
i=l
Ax,-l,XJ
( c l M ) d~
( f I M ) ( F i b )- F ( a ) ) since IF1 5 M . 5 2e =
Also,
+(F(b) 5 c +2 ~ ~
=c
-
F (n))MS
by monotonicity of F
8 .
Thus if 6 satisfies the additional condition that 6 < t / ( 2 ~ ~then ) , the absolute value of the difference of the two sides in the formula of the theorem is < 2 t + 2 t = 4 t . This completes the proof.
9. Fourier Series and the Dirichlet-Jordan Theorem A real-valued function f on a bounded interval [a,b] is said to be of bounded variation on [ a ,b] if there is a constant M such that every partition
has
n
9. Fourier Series and the Djrichlet-Jordan Theorem
347
Let us write 11 f l l B v for the least M such that this inequality holds. The set of functions f of bounded variation on [a, b ] is a pseudo normed linear space in the sense of Section V.9, the pseudonorm being 11 . l l B v . The only functions f with 11 f l B V = 0 are the constants. Examples of functions of bounded variation are furnished by arbitrary bounded monotone functions and by any function with a continuous derivative. In fact, if f is monotone increasing, then f is of bounded variation with 11 f 11 B v = f ( b ) f ( a ) . If f has f' continuous, then the Mean Value Theorem gives
and we see that f is of bounded variation with 11 f 11 B v 5 11 f 'llsUp(b - a). Let us associate two functions on [a,b] to f i f f is of bounded variation. For a real number r ,define r f = max{r, 0) and r - = - min{r, 0}, so that r = r f - r and Ir I = r f + r - . The functions are the positive and negative variations of f , given by V f ( f ) ( l - )=
sup
C ( f (xi,
P with xo=a i = l and x, =x
-
f
(xi-1))'.
n
v-tr,cx,=
sup
C(f(xi)-f(xi-l))-,
P with xo=a i=1 and x, =x
the supremum in each case being taken over all partitions of [ a ,x ] .
Proposition 6.54. If f is of bounded variation on [ a ,b ] , then V f ( f ) and V - ( f ) are monotone increasing functions such that
for all x in [ a ,b ] . In particular, f is the difference of two monotone increasing functions. REMARK.Since monotone functions have left and right limits at each point, it follows that every function f of bounded variation has left and right limits at each point. We denote these by f ( x - ) and f ( x + ) , respectively. The function f is continuous from the left at x if and only if f ( x - ) = f ( x ), and it is continuous from the right at x if and only if f ( x ) = f (x+).
VI. Measure Theory for Euclidean Space
348
PROOF.It is evident from the definitions that V f ( f ) and V - ( f ) are monotone increasing. Fix x ,and let P be a partition of [ a ,x ] . Then
Hence
C (f (xi) i=l
-
f
( x i - I ) ) +=
C ( f (xi) i=l
-
f ( x i - I ) )+ ( f ( x ) - f ( a ) )
and V f ( f ) ( x >5 v - ( f ) ( x )
+ ( f (x) f -
(a)).
Also,
and V - ( f ) ( x ) 5 V f ( f ) ( x )- ( f ( x ) - f ( a ) ) .
Therefore
f
( x )- f ( a ) = V f ( f ) ( x > - V-(f>(x>,
and the proof is complete
Theorem 6.55 (Dirichlet-Jordan Theorem). If f is a function of bounded variation on [-n, n ] ,then the Fourier series of f converges at each point to ( f ( x - ) + f (x+)) and it convcrgcs uniformly to f ( x ) on any compact sct on which the periodic extension of f is continuous. By way of preparation, it will be convenient to extend the definition of Fourier series to allow integrable functions to be replaced by more general Borel measures. If p is a Borel measure on [-n, n ] ,we want to be able to regard p as periodic. One way to proceed would be to insist that really be a measure on the circle group, hence be defined on (-n, n ] . Alternatively, we could insist that any point
9. Fourier Series and the Djrichlet-Jordan Theorem
349
mass contributing to p at -n be matched by an equal point mass for p at n . A way of avoiding point masses contributing at the endpoints is to change the interval [-n, n ]to a suitable [c- n, c +n ] ;we can find a number c with no point masses at the ends because only countably many point masses can contribute to p and still have p be a finite measure. In any event, we define the Fourier-Stieltjes series of ~r to be the series
E
n=-oo
cneinx
with
c, =
e-lnX d p ( x ) . (-n,nl
& &
The usual factor of is dropped because we identify an integrable function f 0 with the measure f d x when making the generalization. From the definition of the Fourier-Stieltjes coefficients, we see immediately that Icn I 5 p ( ( - n , n ] ) ; hence the coefficients are bounded. PROOFOF THEOREM 6.55. We take the given function f to be periodic of period 2 n . On some closed interval [ a ,b ] containing [-n, n] in its interior, let us decompose f according to Proposition 6.54 as f = f ( a ) V + (f ) - V - ( f ) . It is then enough to prove the theorem for the monotone increasing functions f ( a ) V + (f ) and V - ( f ) separately. These functions need to be extended to i ~be constant to the all of R', and we nlay nlake that extensioii by taking t l ~ e i to left of [ a ,b] and to the right of [ a ,b ]. Changing notation in the theorem, we may assume from the outset that f is monotone and bounded, though no longer periodic. Neither the Fourier coefficients of j'nor the hoped-for values of the sum of the Fourier series are changed if we adjust f on a subset of the countable set where f is discontinuous. Thus we may assume without loss of generality that f is continuous from the right at every point. Let f ( x ) cneinxbe its Fourier series. Let p be the Stieltjes measure associated to f . Applying integration by parts (Theorem6.53) on theinterval [-n, n ]with G ( x ) = ePi" andg(x) = -inePinx, we obtain
+
+
- xr=-,
The left side is -2nincn, and the right side is the sum of two bounded terms and the negative of a Fourier-Stieltjes coefficient of p . These Fourier-Stieltjes coefficients are bounded, and hence I c, I 5 C / 1 n 1 for some constant C . Let S N ( X ) = ~ f = cneinX - ~ be the N ' ~partial sum of the Fourier series of f , and let o N ( x ) = Z f = OQ ( x ) be the N" Cesjro sum. We know that ON ( x ) = ( K N * f ) ( x ), where KN is the FejCr kernel. FejCr's Theorem (Theorem 6.48) shows that limN o N ( x ) = ( f ( X - ) + f (x+)) for all x and
&
350
VI. Measure Theory for Euclidean Space
that limN o N ( x ) = f ( x ) uniformly on any compact set of points where f is continuous. The Tauberian theorem stated as Proposition 1.50 allows us to conclude that s~ ( x ) converges and has the same limit as o N ( x )if it is shown that the sequence n(cneinx c-ne-inx) is bounded for n > 0 . But this boundedness is immediate from the estimate I c, I 5 C / l nl for the Fourier coefficients of f .
+
10. Distribution Functions This section concerns the computation of integrals. A measure space ( X , "4,p) will be fixed throughout. A need to estimate integrals arises in two quite distinct situations, and the emphasis is different for the two situations. One is in connection with problems in Fourier analysis and differential equations, and the underlying measure space typically has X equal to R N ,A equal to the o-algebra of Borel sets, and p equal to Lebesgue measure. The other is in connection with probability theory, and the underlying measure space is typically a complicated space with p ( X ) = 1. Although the word "distribution" acquires multiple meanings in the process, the theory can begin at the same point in the two cases. Let f : X + R be a measurable function. We define a measure p f on the Borel sets of R and a function h f : ( 0 , +m)+ [0, +m]by p f ( b )= p ( f hf(U = ~
I f ( x ) E b }) for each Borel set b , +m)>> = ~ ( { x x I I f (x>l > $1)
( b ) )= p ({x E X
( l fl - l ( ( c ,
E
Proposition 6.56. If ,f : X + IR is a measurable function, then (a) Jx @ (f ( x ) )dp ( x ) = JR @ ( t )d p f ( t ) for every nonnegative Borel measurable function @ : R + R ,
Jbm
(b) J;: @ ( I f ( x )) d p ( x ) = I Z f ( ( ) q ( e )d( whenever q ( ( ) d( is a Stieltjes measure on R 1 and @ is its distribution function.
PROOF.In (a), when @ is an indicator function I E , the two sides of the identity are p( f P1 ( E ) ) and p f ( E ) , and these are equal by definition of p f . We can pass to nonnegative simple functions by linearity and then to general nonnegative Borel measurable functions by monotone convergence. 111 (b), wlieii f is a iioiiiiegative siiiiple fuiictioii s, let s = C:=l ck IEk be tlie canonical expansion of s as a linear combination of indicator functions, with the cj's arranged so that cl > c2 > . . . > c, 3 0 . Put c,+l = 0 . Then we have
10. Distribution Functions
35 1
=
c y = l c : = j p ( E j ) [ @ ( c k ) - @(ck+l)l
=
Cj"=, p ( E j ) @ ( c j )
= z = l JEj
since @(O) = 0
d~(x)
= J;, @ ( s ( x ) )& ( X I .
This proves (b) for nonnegative simple functions f . For a general measurable If 1 on X , choose an increasing sequence of nonnegative simple functions s, with pointwise limit If 1 . The definition of 4, in terms of rp makes 4, monotone increasing and continuous, and thus @ (s, ( x ) ) increases to @ ( 1 f ( x ) 1 ) . Also, the set { x E X If ( x )I > for each fixed 6 , is the increasing union of the sets { x E X s , ( x ) > 61, and thus hSn( 6 ) = p ( { x E X s , ( x ) > 6 ) ) increases to h f ( 6 ) = p ( { x E X If ( x )1 > for each 6 . Hence we can pass to the limit in the identity for each s, and obtain the identity for If I by monotone convergence. This proves (b) for a general measurable If 1 .
I
I
0,
I
I
(1)
For applications to Fourier analysis and differential equations, it is (b) that is important, and the function @ of most interest is @ ( t ) = tP with 0 < p < +oo. The formula in this case is
Somewhat unfortunately, the function h f is called the distribution function of f ; the term does not conflict with the notion of the "distribution function" of a Stieltjes measure as long as one does not make any associations between functions and measures. A special case of the displayed formula is that X is RN, p is Lebesgue measure, If ( x ) I d x = hf d6, a and p is 1. In this case the formula simplifies to formula that was mentioned after the statement of the Hardy-Littlewood Maximal Theorem (Theorem 6.38). The displayed formula shows that If IP d p can be computed from the function h f , and it is apparent that the integral cannot be finite if h f ( 6 ) is everywhere > some positive multiple of 6 - p . This observation can be improved upon without the aid of Proposition 6.56 in the following way. We have If (x)lP d p ( x ) > If ( x ) l P d p ( x )Z t p p ( { x E X If ( ~ 1 1> $ 1 . Thus we obtain
i,,
hxEX
I
an inequality that goes under the name Chebyshev's inequality.
(e)
352
VI. Measure Theory for Euclidean Space
11. Problems 1. Let S' be the unit circle of @, let T be the subgroup of elements of finite order, and let E be a subset of S' that contains exactly one element of each coset in s ' / T . (Such a set E exists by the Axiom of Choice.) Prove that E is not a Lebesgue measurable subset of the circle and therefore that the corresponding subset of (-n, n ]is not Lebesgue measurable on R'.
2. Let rp be the mapping given explicitly in Section 5 that allows one to substitute in an expression in Cartesian coordinates and obtain an expression in spherical coordinates. Let U be the domain of rp . Prove that (a) the determinant factor in the change-of-variables formula is given by
(b) y, is one-one on U , (c) the complement of p ( U ) in RN is a lower-dimensional set.
3. Let L be a nonsingular N-by-N real matrix. Prove that
for every nonnegative Borel measurable function f .
4. Let MN denote the N'-dimensional Euclidean space of all real N-by-N matrices, and let d x refer to its Lebesgue measure. Prove that
for each nonsingular matrix y and Borel measurable function f formula, yx is the matrix product of y and x .
> 0.
In the
5 . Fix a with 0 < a < 1. Suppose f : R + @ is periodic of period 237, is smooth except at multiples of 2 n , and satisfies the inequalities If ( x )I < C Ix l a , I fl(x)I I ~ 1 x 1 ~ and - ' , I fl'(x)I I ~ l x l for ~ -1x1~I 1. (a) By breaking the integral at Ix 1 = 1/In 1, prove that the Fourier coefficients c, o f f satisfy lc,l < ~ / l n l ' + ~ . (b) How can one conclude from (a) that the Fourier series of f converges uniformly? Why is the limit equal to f ? (c) Prove or disprove: The real and imaginary parts of the function f are of bounded variation on every bounded interval.
6. Let p be a nonzero measure on the a-algebra of all subsets of R' assigning to each set either measure 0 or measure 1. Prove that p is a point mass.
7 . Determine all Stieltjes measures v f 0 on the line with
for all continuous nonnegative functions f and g. Problems 8-10 make use of Fubini's Theorem in unexpected ways.
8.
(a) Show that the complement of any Lebesgue measurable set of Lebesgue mcasurc 0 in RN is dcnsc. (b) Let p be a Stieltjes measure on the line, and let E be a Borel set in I W ~ with Lebesgue measure 0. Prove that p ( E t ) = 0 for almost every t with respect to Lebesgue measure. (c) Suppose that a Stieltjes measure p on the line satisfies lim,,o p ( E t ) = p ( E ) for each bounded Borel set E in R1. Prove that p ( E ) = 0 for every Bore1 set E of Lebesgue measure 0.
+
+
9. In potential theory apositive charge onR3 is by definition any finite Borel measure p , and its potential h is the function h(x) = w .Prove that the potential W3 12-YI is finite almost everywhere with respect to Lebesgue measure.
l
10. Let P ( x l , . . . , x,) be a real-valued polyno~llialon R" that is not idelltically 0. Prove by induction that the set in R" where P = 0 has Lebesgue measure 0. Problems 11-14 concern the gamma function and some associated changes of variables. 11. Prove that
+
by starting from the product of T ( x y ) and the left side, substituting for r (x+ y), making a change of variables, using Fubini's Theorem, and making another change of variables.
SEN
e p n x 2dx first in Cartesian coordinates by means 12. By evaluating the integral of Proposition 6.33 and then in spherical coordinates by means of the changeof-variables formula for multiple integrals, obtain an expression for the area QNP1 = JSN-l d m of the sphere sNpl.Express the answer in terms of a value of the gamma function.
13. Let I be the "cube" of all u = ( u l , . . . , u,) in R" with0 < ui < 1 for all i , and let S be the "simplex" of all x = ( x l , . . . , x,) in Rn with xi > 0 for all i and Cy=lxi < 1. Define x = p(u) by
VI. Measure Theory for Euclidean Space
ny=J
(a) Prove that C:=,,xi - 1 1 - z ~j.i (b) Prove that rp maps I one-one onto S , with inverse given by
14. Using Problems 11 and 13, prove for the simplex S in Problem 13 that
when aj > 0 for all j Problems 15-17 concern the Cantor function for the standard Cantor set.
15. Prove that the values of the Cantor function F for the standard Cantor set are independent of the method of defining the approximating functions Fn on the complementary closed intervals as long as F, is monotone increasing and continuous. 16. Compute J: F ( x j dx if F is the Cantor function for the standard Cantor set.
17. The Stieltjes measure p corresponding to the Cantor function for the standard Cantor set C is called the Cantor measure. The set C consists of the members of [0, 11 that can be expanded in the digits 0, 1, 2 of base 3 without using any 1's. Show, for each n-tuple of 0 ' s and 2 ' s , that p attaches measure 2-" to the subset of C whose base 3 expansion begins with that n-tuple. Problems 18-20 introduce the Poisson integral formula for the unit disk in IX2.The Poisson kernel was the subject of Problems 27-29 at the end of Chapter I and is given by 00 1-r2 Pr (Q) = yInein0 = 1 -2rcosQ + r 2 ' n=-00 Harmonic functions in the unit disk were the subject of Problems 14-15 at the end of Chapter I11 and also Problems 10-13 at the end of Chapter IV. The present set of problems begins to relate the Poisson kernel to harmonic functions via convolution.
11. Problems
355
&
18. If f is in L ( [ - n , n ] , d o ) , then the Poisson integral of f is the function in the unit disk defined in polar coordinates by
If c, is the nth Fourier coefficient of f , prove that u ( r , Q ) = Cr==_, c,r " e i n O , and conclude that u is harmonic in the open unit disk.
&
19. If p equals 1 or 2 and if f is in L P ( [ - n , n ] , d o ) , prove that the Poisson integral u(r, 8 ) o f f has the properties that Ilu(r, .)Ilp 5 11 flip for 0 5 r < 1 and that u (r, . ) tends to f in LP in the sense that lim,? 1 11 u (r, . ) - flip = 0 .
&
20. Suppose that f is in L"([-n, n ] , d 8 ) and that u(r, 8 ) is the Poisson integral off. (a) Prove that lim,? 1 u (r, 0 ) = f ( 0 ) uniformly on any set of 0's where f is uniformly continuous. (b) For f of class c2, prove that the Poisson integral o f f is the only har~llonic function u(r, Q )in the disk such that limrTlu(r, Q ) = f ( Q )uniformly in Q . (c) Prove that u(r, . ) tends to f weak-star in L m relative to L 1 in the sense that limYtlSTnnu ( r , t))g(t))dt) = STnnf (t))g(t))dH for all g in L' . (Weak-star convergence was defined in Section V.9.) Problems 21-25 concern functions of bounded variation. For such a function f , the positive and negative variations o f f were defined in Section 9 , and their values at x were denoted by V + (f ) ( x ) and V p (f ) ( x ). 21. Prove that the product of two f~~nctions of bounded variation on [ a ,h ] is of bounded variation. 22. This problem concerns a certain minimality property of the decomposition f ( x ) = f ( a ) V f ( f )( x ) - V ( f )( x ) of a function f of bounded variation on [ a ,61. Prove that if gl and g2 are any two nonnegative monotone increasing functions such that f ( x ) = f ( a ) gl ( x ) - g 2 ( x ) for all x , then V + (f ) ( x ) 5 gl ( x ) and V p( f )(I)1 g2(x).
+
+
27. Prove that i f f is of bounded variation on [ a ,b ] and is continuous at a point x in ( a , b ) , then both V+(f ) and V ( f ) are continuous at x . 24. If f is of bounded variation on La, b j , dehne the total variation of f as the function given by n
the supremum being taken over all partitions of [ a ,x ] . Prove that V ( f ) ( x ) = V + ( f ) ( x ) V p ( f ) ( x )forallx.
+
356
VI. Measure Theory for Euclidean Space
25. Prove that the function f on [- 1, 11 given by for x # 0, for x = 0, is not of bounded variation. Prove or disprove that the function g on [-I, 11 given by g(n) =
is of bounded variation.
{
?;
sin(l/x)
for x # 0. for x = 0,
CHAPTER VII Differentiation of Lebesgue Integrals on the Line
Abstract. This chapter concerns the Fundamental Theorem of Calculus for the Lebesgue integral, viewed from Lebesgue's perspective but slightly updated. Sectioil 1 coiltaiils Lebesgue's illaiil tool, a theorem saying that illoilotoile fuilctioils on the line are differentiable almost everywhere. A relatively easy consequence is Fubini's theorem that an absolutely convergent series of monotone increasing functions may be differentiated term by term. The result that the indefinite integral :J f ( t ) dt of a locally integrable function f is differentiable alillost evelywhere with derivative f follows readily. Section 2 addresses the converse question of what functions F have the property for a particular f that the integral ~ , fb ( t ) dt can be evaluated as F (b) - F ( a )for all a and b. The development involves a decomposition theorem for monotone increasing functions and a corresponding decomposition theorem for Stieltjes measures. The answer to the converse question when f 3 0 and F' - f almost evelywhere is that F is "absolutely continuous" in a sense defined in the section.
1. Differentiation of Monotone Functions The generalization of the Fundamental Theorem of Calculus to the Lebesgue integral was the crowning achievement of Lebesgue's book. We have already stated and proved a particular result in that direction as Corollary 6.40, using a more recent method that is of continual applicability in analysis. The statement f ( t ) d t is of the part of the Fundamental Theorem in that corollary is that differentiable almost everywhere with derivative f (x) i f f is a Bore1 function on the line that is integrable on every bounded interval. In this chapter we shall develop that and allied results using something closer to Lebesgue's original method. These allied results are chiefly of historical interest, no longer being of great importance as analytic tools. However, their beauty is undeniable and by itself justifies their inclusion in this book. In addition, these allied results motivate some results in Chapter IX,particularly the RadonNikodym Theorem, that might seem strange indeed if the historical background were omitted. The starting point is the almost-everywhere differentiability of monotone functions on the line, given in Theorem 7.2 below. Since monotone functions include the distribution functions of Stieltjes measures, this differentiability shows at
lox
358
VII. Differentiation o f Lebesgue Integrals on the Line
once that functions of the form JaX f ( t )d t with f > 0 are differentiable almost everywhere, and then we are well on our way toward a more traditional proof of the Fundamental Theorem for the Lebesgue integral. The advantage of starting with all monotone functions is that one can address at the same time the differentiability of all distribution functions of Stieltjes measures, not just those of measures f ( t ) d t From this fact one can attack the question of how close the derivative f ( t )is to determining the function of which it is the derivative almost everywhere. This is the second aspect of the traditional Fundamental Theorem of Calculus as in Theorem 1.32: for the case of continuous f , any two functions with derivative f everywhere on an interval differ by a constant. There is a certain formal similarity between the theory of differentiation of monotone functions and the theory of the Hardy-Littlewood Maximal Theorem as in Chapter VI. Wiener's Covering Lemma captured the geometric core of the theorem in Chapter VI, and another covering lemma captures the geometric core here. This is the Rising Sun Lemma, which will be given as Lemma 7.1. By way of preliminaries, any open subset U of R' is uniquely the union of countably many disjoint open intervals, the open interval containing a point x in U being the union of all connected subsets of U containing x . These sets give thc rcquircd dccomposition of U by Propositions 2.48 and 2.51. An opcn subsct of an interval ( a , b ) is necessarily open in R1, and hence it too is uniquely the countable union of disjoint open intervals.
Lemma 7.1 (Rising Sun emm ma).' Let g : [ a ,b ] + R be continuous, and define
I
E = { x E ( a , b) there exists $
E
( a , b) with $ > x and g ( $ ) > g ( x ) } .
The set E is open in ( a , b ) . If E is written as the disjoint union of open intervals with endpoints ak and b k ,then g ( a k ) 5 g ( b k ) for each k .
FIGURE7.1. Rising Sun Lemma. Graph showing three open intervals produced by the lemma. 'some authors call this result Riesa's Lemma
1. Differentiation of Monotone Functions
359
REMARK.The Rising Sun Lemma is so named because of the situation in Figure 7.1. The sun rises in the east, viewed as the direction of the positive x axis. It casts shadows within the graph of g , and the content of the lemma is the nature of those shadows. Although the conclusion of the lemma is that g ( a k ) 5 g ( b k ) for all k , the reader can observe in the figure that g ( a k ) = g ( b k ) for the open intervals that are shown. This observation is valid in general except possibly when ak = a , but the observation is not needed in the proof of Theorem 7.2 below. PROOF.Ifxo E E and 6 E ( a , b) have 6 > xo a n d g ( 6 ) > g ( x o ) ,then every x in ( a , 6 ) with Ig(x) - g(xo)l < i ( g ( 6 ) - g ( x o ) ) lies in E . Hence E is open. Let E be the disjoint union of intervals ( a k ,b k ) . Fix attention on one such interval ( a k ,b k ). We make critical use of the fact that the point bk is not in E. If xo satisfies ak < xo < b k , we prove that g ( x o ) 5 g ( b k ). Once we do so, we can let xo decrease to ak and use continuity to obtain the assertion g ( a k ) 5 g ( b k ) of the lemma. Arguing by contradiction, suppose that g ( x o ) g ( b k ) . Since xo is in E , there exists xl > xo with g ( x l ) > g ( x o ) . If xl > b k , then the inequality g ( x l ) > g ( x o ) > g ( b k ) forces bk to be in E. Since bk is not in E , we conclude that xl 5 b k . The set of all x with xl 5 x 5 bk and g ( x ) > g ( x l ) is closed, bounded, and nonenlpty, arid we let n2 be its largest element, so that n2 5 bk . Since g ( x 2 ) > g ( x l ) > g ( x o ) > g ( b k ) , we must have x2 < b k ; in fact, x2 = bk would yield the contradiction g ( b k ) > g ( b k ) . From ak < xo < x2 < bk and ( a k ,b k ) C E , we see that x2 is in E . Hence there is some 6 > x2 with g ( 6 ) > g ( x 2 ) . Then the conditions g ( 6 ) > g(bk) and bk 6 E together force 6 to be 5 bk. So xz < 6 5 bk with g ( 6 ) g ( x l ) , in contradiction to the maximality of x 2 . This contradiction allows us to conclude that g ( x o ) 5 g ( b k ) ,and the proof is complete.
Theorem 7.2 (Lebesgue). If F is a monotone increasing function on an interval, then F is differentiable almost everywhere in this sense: the set where F fails to be differentiable is a Lebesgue measurable set of Lebesgue measure 0. In addition, if the definition of F' is extended so that F ' ( x ) = 0 at every point where F is not differentiable, then F' is Lebesgue measurable. REMARK. Recall that any monotone increasing function F can have only countably many discontinuities, and these are all given by jumps. In other words, F has, at each point x , left and right limits F ( x -) and F ( x +), and the only possible discontinuities occur when one or both of the equalities F ( x - ) = F ( x ) and F ( x ) = F ( x + ) fail. PROOF.The second statement is a consequence of the first. In fact, if E is the Lebesgue measurable set of measure 0 where F is nondifferentiable and if B is a Borel set of measure 0 containing E , then the sequence of Borel functions
3 60
VII. Differentiation o f Lebesgue Integrals on the Line
+
G,(x) = & ( F ( x l / n ) - F i x ) ) converges everywhere on Bc to a function G. If G is extended to the domain of F by defining it to be 0 on B , then G is a Bore1 function that equals F' except on a subset of B, and hence F' is Lebesgue measurable. Let us come to the conclusion about differentiability. Possibly by taking the union of countably many sets, we may assume that the domain of F is a bounded interval [ a ,b ]. For a < x < b ,define U, ( x ) = lim sup ( F ( x
+h)
-
F (x))
h $0
and
+
L, ( x ) = lim inf ( ~ ( x h ) - F i x ) ) , h $0
Ul( x ) = lirn sup ( F ( x + h ) - F ( x ) ) hfO
and
LL( x ) = lim inf ( F ( x htO
+h)
-
F(x)).
We shall prove that and almost everywhere. If the latter inequality is applied to
-
F (-x) ,we obtain also
almost everywhere. Putting these inequalities together, we have Ul( x ) 5 L, ( x ) 5 U, ( x ) 5 Ll ( x ) 5 Ul ( x ) almost everywhere, and equality must hold throughout, almost everywhere. The points where equality holds throughout and also U, ( x ) < foo are the points where F is differentiable, and hence the two inequalities U, ( x ) < +oo and U, ( x ) 5 LL( x ) prove the theorem. For most of the proof, we shall assume that F is continuous. At the end we return and show how to modify the proof to handle discontinuous F . First we consider the inequality U, ( x ) < foe. The subset E of ( a , b ) where this inequality fails is, for cach positivc intcgcr 1 2 , containcd in thc sct whcrc U, ( x ) > 12. If - F ( x ) > n for some 6 > x . That is, g ( $ ) > g ( x ) for 6-x the continuous function g ( x ) = F ( x ) - n x . In the notation of Lemma 7.1, E is covered by a systeiii of disjoint open iiitervals ( a k ,b k ) such that g ( a k ) 5 g ( b k ) for each such interval. Thus n(bk - a k ) 5 F ( b k ) - F ( a k ) for each. Summing on k gives n Ck(bk - a k ) 5 Ck( F ( b k )- F ( a k ) ) 5 F i b ) - F ( a ). Thus the exceptional set E can be covered by a system of open intervals of total measure 5 (F( b )-F ( a ) ) .Since n is arbitrary, Proposition 5.39 shows that E is Lebesgue measurable of Lebesgue measure 0.
U, ( x ) > n , then ' ( ' )
1. Differentiation of Monotone Functions
361
Next we prove that Ur( x ) 5 Ll ( x ) almost everywhere on ( a ,b ) . If 0 5 p < q are rational numbers, we prove that the set E,, where
has Lebesgue measure 0. The countable union of such sets is the exceptional set in question, and thus we will have proved that the exceptional set has measure 0. F ( 0 - F(x) If L l ( x ) < p , then there exists $ E ( a ,b ) with $ < x and < P, &-x hence with p( - F ( ( ) < px - F ( x ) . Define g(z) = pz F (-z) for z in [-b, -a]. If y = -x and rl = -6, then p q + F(-7) > p y + F ( - y ) and hence g(7) > g ( y ) with > y . Applying Lemma 7.1 to g on the interval [-b, -a], we obtain a disjoint system of open intervals (-bi, -ai) covering the set of y's where Ll(-y) < p and having g(-bi) 5 g(-ai) in each case. Thus -phi F ( b i )i -pai F (ai).In other words, the set of x's where Ll ( x ) < p is covered by a disjoint system of open intervals (ai,b i ) such that
+
+
+
fur each such interval. Applying the lerrlrrla tu gp( A ) = F ( n )- q n 011 the interval [ai,bi], we obtain a disjoint system of open intervals ( a i j ,b i j )indexed by j and having gp(aij)5 gp(bij). Thus (*) and
hold in each case. Summing (**) over j, we obtain
Summing this inequality over i and dividing by q gives
If we repeat this argument with [aij,bij]in place of [ a ,b ] ,we obtain intervals (aijuv,bij,,) and an inequality
Iteration gives m(E,,) 5 ( p / q ) " ( b- a ) for every n , and therefore m(E,,) = 0. This completes the proof in the case that F is continuous.
362
VII. Differentiation o f Lebesgue Integrals on the Line
If F is possibly discontinuous, we modify Lemma 7.1 and the present proof as follows. Each function g that arises has right and left limits g(x+) and g ( x - ) at each point x , and we let G ( x ) be the largest of g ( x - ) , g ( x ) , and g(x+). A modified Lemma 7.1 says that the set of x in ( a , b ) for which there is some 6 E ( a , b ) with 6 > x and g ( 6 ) > G ( x ) is an open set whose component intervals (al,,b k ) have g(ak+) 5 G ( b k )for each k . Going over the proof of Lemma 7.1 carefully and changing g to G as necessary, we obtain a proof of the modified Lemma 7.1. The modifications necessary to the present proof are as follows. In the proof that U,(x) i+aalmost everywhere, the set E is to be taken to be the set where F is continuous and this inequality fails. The inequality that results from applying the modified Lemma 7.1 is n (bk - a k ) 5 F (bk+) - F (ak+), and this inequality can be summed on k without any further change. Similarly in the proof that U,(x) 5 L L ( x )almost everywhere, the set E,, is to be taken to be the set where F is continuous and L L ( x )< p < q < U F ( x ) .Inequality (*) becomes F(bi -) - F (ai+) 5 y (bi - ai) . When we consider the interval [ a i ,b i ],the value of F (bi+) is not relevant, and the value of F ( b i ) can be adjusted to equal F(bi -) for purposes of understanding F between ai and bi. With that understanding, incquality (**) bccomcs q ( b i j - a i j ) 5 F(bij+) - F ( a i j + ) , and stcp (7) is replaced by
The two inequalities at the ends have come about from (*) and (**), and the critical observation is that the convention F ( b i ) = F(bi-) makes the middle inequality hold. The rest of the argument proceeds as in the case that F is continuous, and then the theorem is completely proved.
Theorem 7.3 (Fubini's theorem on the differentiation of series of monotone functions). If F = C F, is an absolutely convergent sequence of monotone increasing functions on [ a ,b ] ,then F f ( x )= CT==l FA(x) almost everywhere. PROOF.Without loss of generality, we may assume that F, ( a ) = 0 for all n . Then F, ( x ) > 0 for all n and x . Possibly by lumping terms, we may assume also that F ( b ) - CE=l F k ( b ) 5 2-". Since F ( x ) F k ( x ) is a lnonotone increasing function that is 0 for x = a , we have 0 5 F (x)-
C Fk ( x ) 5 2-, k=l
+
for a i x i b. The decomposition F ( x ) = Fk(x) ( CEnflF ~ ( x ) ) exhibits F as the sum of n + 1 monotone increasing functions, and thus we have
1. Differentiation of Monotone Functions
363
C:=l FL(x) 5 F f ( x ) at all points where all the derivatives exist. In view of Theorem 7.2, this inequality holds almost everywhere. Consequently
almost everywhere. Now consider the series
Then
Thus G satisfies the same kind of inequality that F did in (*), and we can conclude that G satisfies the analog of (w),namely
The right side is finite almost everywhere by Theorem 7.2, and thus the individual terms F f ( x ) F,/(x) of the series tend to 0 almost everywhere. This complctcs thc proof. From Theorems 7.2 and 7.3, we can derive the first part of Lebesgue's form of the Fundamental Theorem of Calculus. This same result was stated as Corollary 6.40 and was proved in Chapter VI by using the Hardy-Littlewood Maximal Theorem.
Corollary 7.4 (first part of Lebesgue's form of the Fundamental Theorem of Calculus). If f is integrable on every bounded subset of IK1, then f ( y ) d y is differentiable almost everywhere and
Jz
PROOF.It is enough to prove the theorem for functions vanishing off an interval [ a ,b]. Let A be the set of all Bore1 sets E [ a ,b] such that I E ( t )d t = I E ( x ) almost everywhere. Then A contains the elementary sets within [a,b ] , and A is closed under complements within [ a ,b ] . If { E n }is an increasing
&
3 64
VII. Differentiation o f Lebesgue Integrals on the Line
sequence of sets in A with Eo = @ and with union E , let us write Ib = ((IE, - IE,-l). This is a series of nonnegative functions. Putting F ( x ) = SaxIE ( t )d t and F, ( x ) = Sax(IEn( t ) - I,+, ( t ) )d t and applying Corollary 5.27, we obtain F ( x ) = C z l F,(x). ThenTheorem7.3 gives F f ( x ) = C r = l FA(x) = N lim~ x:=Z/=, FA ( x ) = l i m E ~ n = , ( I E ,( X I - IE,-~( X I ) = l i m ~ ( X I = IE (4 almost everywhere. Thus E is in A, and A is closed under increasing countable unions. Since A is closed under complements as well, A is closed under decreasing countable intersections. Then the Monotone Class Lemma (Lemma 5.43) shows that A contains all Borel sets. Now consider the set F of all integrable Borel functions f for which the f ( t ) d t = f ( x ) holds. We have just seen that almost-everywhere equality lax F contains all indicator functions of Borel subsets of [ a ,b]. By linearity, F contains all nonnegative simple functions vanishing off [ a ,b]. Let f > 0 be an integrable function on [ a ,b], and let {s,} be an increasing sequence of nonnegative simple functions with pointwise limit f . The functions s, are in F.Put so = 0, and let F ( x ) = Jax f ( t )d t and Fn(x) = JaX ( s n ( t )- s n - ~ ( t ) )d t . Since s, ? sn-1, each F, is monotone increasing. Corollary 5.27 shows that F ( x ) = C z l F , ( x ) , N and Theorem 7.3 then shows that F 1 ( x )= C Z l FA(x) = limN En==, FA(x) = N limN En=,(s,(x) - snPl( x ) ) = limNs,(x) = f ( x ) almost everywhere. Thus F contains all nonnegative integrable Borel functions, and by linearity it contains all integrable Borel functions.
x;==,
&
2. Absolute Continuity, Singular Measures, and Lebesgue Decomposition In this section we address questions about the Lebesgue integral raised by the sccond part of thc Fundamcntal Thcorcm of Calculus in Thcorcm 1.32. For continuous integrands f , the result is a kind of uniqueness statement, asserting that any function with derivative f differs from Saxf ( t ) d t by a constant function. From a practical point of view, this is the really important part of the theorem for calculus, since it provides a technique for evaluating definite integrals: find any function whose derivative is the given function, evaluate it at the endpoints, and subtract the results. With the Lebesgue integral and equality of derivatives only almost cvcrywhcrc, thc uniqucncss rcsult is not as sharp. Thc practical aspcct of a uniqueness theorem is largely lost, and the resulting theory ends up having to be appreciated only as an end in itself. We begin at the following point.
Proposition 7.5. Every monotone increasing function on IK1 is uniquely the sum of an indefinite integral G ( x ) = f ( t )d t , where f > 0 is integrable on every bounded interval, and a monotone increasing function H such that H' ( x ) = 0 almost everywhere.
Jb"
2. Absolute Continuity, Singular Measures, and Lebesgue Decomposition
365
PROOF. Let F be a given monotone increasing function on IW'. If F = G H with G as in the statement of the proposition and with H f ( x ) = 0 almost everywhere, Corollary 7.4 shows that we must have f = F'. This proves uniqueness. For existence we take f = F'. Regard h as a positive number tending to 0 through some sequence, so that hP1 ( F (t + h ) - F ( t ) ) tends to f' ( t ) for almost every t . If a < b , then
+
=
;Lbfh
F ( t ) dt
-
;lath
-
F ( t )dl.
The right side tends to F ( b ) - F ( a ) if a and b are points of continuity of F. By Fatou's Lemma (Theorem 5.29), j;lb f ( t ) d t 5 F ( b ) - F ( a ) if a and b are points of continuity of F. The points of continuity of F are dense, and thus for general a and b , we can find sequences of points of continuity decreasing to a and increasing to b. Passing to the limit, we obtain
lb
f ( t )dt 5 F i b ) - F(a+) 5 F(b) - F ( a )
(*I
for all a and b. Hence f is integrable. With G ( x ) as in the statement of the proposition, (*) gives G (b)- G ( a ) 5 F ( b )- F ( a ) .Equivalently, F ( a )- G ( a ) 5 F ( b ) - G (b). Thus the function H (x) = F ( x ) - G ( x ) is monotone increasing with F = G + H . Since F and G have derivative f almost everywhere, H has derivative 0 almost everywhere. Thus we want to identify all monotone increasing functions with derivative zero almost everywhere. The first step is to see that the question of discontinuities of a monotone function can be completely eliminated from the problem.
Proposition 7.6. Let c be a real number. If {xn}is a sequence in [a,b] and if {c,} and {d,} are sequences of positive real numbers with C c, finite and C d, finite, then the function
n with x n 5x
n w~th X, <X
is a monotone increasing function on [a,b] with F f ( x )= 0 almost everywhere.
366
VII. Differentiation o f Lebesgue Integrals on the Line
PROOF.The function F is certainly monotone increasing. It is the convergent sum of the constant function c and monotone increasing functions of the form for x < x,, for x = xI1,
+
for x > x,, C, d, and the function F, has derivative 0 except at the point x,. Thus the proposition follows immediately from Theorem 7.3.
A monotone increasing function on the line whose restriction to every closed bounded interval is of the form in Proposition 7.6 is called a saltus function; the name comes from the Latin word for "jump." Since R' is the countable union of closed bounded intervals, it follows from Proposition 7.6 that every saltus function has derivative 0 almost everywhere.
Proposition 7.7. Any monotone increasing function F on R1 is uniquely the sum F = G + S of a continuous monotone increasing function G with G(0) = 0 and a saltus function S . PROOF. For existence, it is enough to obtain the decomposition without insisting on the normalization G(O) = 0, since the sum of a saltus function and a constant is a saltus function. Let xo be a point of continuity of F , and enumerate the points of discontinuity of F as x, , n > 1. For each n > 1, define c, = F(x,) - F(x,-) and d, = F (x,+) - F(x,). Let S be the saltus function
and put G = F - S . Then G is continuous everywhere. To see that G is monotone increasing, let a < b be points of continuity of F and S . We start from the equality S(x,+) - S(x,-) = F (x,+) - F (x,-) and sum for x, with a < x, < b to obtain
Hence F ( u ) - S ( u ) 5 F ( b ) - S ( b ) , and we conclude that G ( u ) 5 G ( b ) at all points of continuity a < b of F and S . These points are dense, and G is continuous everywhere. Hence G ( a ) 5 G ( b ) whenever a < b, and G is monotone increasing. This proves existence. Uniqueness follows from the fact that S(b-) - S(a+) = Ca<x,
2. Absolute Continuity, Singular Measures, and Lebesgue Decomposition
367
Consequently we need to understand the continuous monotone increasing functions F on the line with F f ( x )= 0 almost everywhere. The Cantor function for the standard Cantor set, constructed as in Section VI.8, is an example. For such a function, F - F ( 0 ) satisfies the defining properties of the distribution function of a Stieltjes measure p on R1. The continuity of F is equivalent to the fact that p contains no point masses. The following property isolates the meaning of having derivative zero almost everywhere.
Proposition 7.8. Suppose that p is a Stieltjes measure with no point masses. If thc distribution function F of p has F f ( x ) = 0 at cvcry point of a Borcl sct E , then (E) = 0 . REMARK.The proof will use the Rising Sun Lemma (Lemma 7.1). Problem 3 at the end of the chapter asks for an alternative proof by means of Wiener's Covering Lemma (Lemma 6.41). A proof using Wiener's Covering Lemma does not make use of the continuity of F, and therefore it is not necessary to assume in the proposition that p has no point masses. PROOF.We may confine our attention to an interval [ a ,b ] ,taking E to be a subset of [ a ,b] . Since p has no point masses, we may discard a and b from E . Fix a positive integer n . For every point x in E , we have F 1 ( x )< Therefore to each such x , we can associate some $ > x with 6 in ( a ,b ) such that
A.
A
This inequality says that 6 - F ( 6 ) > x - F ( x ) ,hence that the continuous function g with g ( x ) = x - F ( x ) has g ( 6 ) > g ( x ) . The Rising Sun Lemma (Lemma 7.1) applies and shows that E is covered by countably many disjoint open intervals ( a k ,bk) with g ( a k ) I g(bk). n u s ak - F ( a k ) I $ bk - ~ ( b k ) for each k . Adding, we obtain p ( E ) 5 E p ( ( a i , bx)) = k
Since 12 is arbitrary, p ( E )
E F(bi) k
-
F(ai))5
$ E (bk
-
ax) 5 $ ( b - a ) .
k
= 0.
Again consider a colltinuous nlollotone fullction F with derivative zero almost everywhere. The function F - F ( 0 ) is the distribution function of some Stieltjes measure p with no point masses, and Proposition 7.8 shows that there is a Borel set E such that p ( E ) = 0 and m ( E c ) = 0, where m is Lebesgue measure. In other words, is concentrated completely on the set EC of Lebesgue measure 0. A Stieltjes measure p for which there is a Borel set F with p ( F C ) = 0 and
368
VII. Differentiation o f Lebesgue Integrals on the Line
m ( F ) = 0 is called a singular Stieltjes measure or a "Stieltjes measure singular with respect to Lebesgue measure." If also it contains no point masses, it is said to be continuous singular. The Stieltjes measure associated to the Cantor function for the standard Cantor set is an example. We can summarize matters either in terms of decompositions of monotone functions or in terms of decompositions of Stieltjes measures The result in the case of monotone f~lnctionsis a first answer to the question of uniqueness in the Fundamental Theorem of Calculus for the Lebesgue integral; the result in the case of Stieltjes measures gives the Lebesgue decomposition of Stieltjes measures.
Theorem 7.9. Every monotone increasing function F on IR1 decomposes uniquely as the sum F = G + H + S , where G is the indefinite integral G ( x ) = f ( t )d t of a function f > 0 integrable on every bounded interval, H is the distribution function of a continuous singular measure, and S is a saltus function. The function f is the derivative of F.
1;
PROOF.Proposition 7.7 allows us to write F = P + S uniquely, where S is a saltus function and P is continuous and monotone increasing with P ( 0 ) = 0. Proposition 7.5 says that P = G + H uniquely, where G is an indefinite f ( t )d t and H is monotone increasing with H ' ( x ) = 0 integral G ( x ) = almost everywhere. The function f is the derivative of F. The function H has H (0) = 0 and is continuous because P and G have these properties, and therefore H is the distribution function of a Stieltjes measure p containing no point masses. Since H ' ( x ) = 0 almost everywhere, Proposition 7.8 shows that p is singular.
Id(
Corollary 7.10 (Lebesgue decomposition). Every Stieltjes measure p decomposes uniquely as the sum = f d x w, p d , where f > 0 is a function integrable on every bounded interval, p, is a continuous singular measure, and p d is a countable sum of point masses such that the sum of the weights on any bounded interval is finite.
+ +
PROOF.This follows by applying Theorem 7.9 to the distribution function of p.
The final question that we address in this section is how to recognize the f ( t ) d t from among all the monotone particular monotone function G ( x ) = fi~ncticlnsF = G + H + ,C described in Theorem 7.9.
JbX
Proposition 7.11. With m denoting Lebesgue measure, the following conditions on a Stieltjes measure p, are equivalent: (a) pa is of the form wa = f d x for some function f on every bounded interval,
> 0 that is integrable
2. Absolute Continuity, Singular Measures, and Lebesgue Decomposition
369
(b) for each bounded interval [ a ,b ] and number t > 0, there exists a number 6 > 0 such that p a ( E ) < t whenever E is a Borel subset of [ a ,b ] with m ( E ) < 8, (c) p,(E) = 0 whenever E is a Borel subset of R' with m ( E ) = 0. REMARK.A Stieltjes measure pa satisfying the equivalent conditions in this proposition is said to be absolutely continuous or "absolutely continuous with respect to Lebesgue measure." From any of these defining conditions, we see right away that an absolutely continuous measure contains no point masses. PROOF. Corollary 5.24 shows immediately that (a) implies (b). To see that (b) implies (c), let E be a Borel set E in IK1 with m ( E ) = 0. Applying (b) to E f' [ a ,b] gives w,(E f' [ a ,b ] ) < t for every positive t , and hence p,(E n [ a ,b ] ) = 0 . Since [ a ,b ] is arbitrary and pa is completely additive, pa (E) = 0 . To see that (c) implies (a), we appeal to Corollary 7.10 to decompose p, according to the Lebesgue decomposition as
where p, is continuous singular and w d is discrete. The measures p, and pd have the property that there is a Borel set E with m ( E ) = 0 such that p s ( E c ) = /ld(EC= ) 0. Condition (c) shows that p,(E) = 0. Evaluating (*) at E , we obtain 0 = p, ( E ) = 0 + p , ( E ) + p d ( E ) . Therefore p , ( E ) = p d ( E ) = 0 . Since p s ( E C )= p d ( E C )= 0 also, we must have ps = pd = 0, and then (.r) shows that w, = f dx. In Chapter IX the implication (c) implies (a) will be generalized to a result in abstract measure theory known as the Radon-Nikodym Theorem. Meanwhile, it is conditions (b) and (c) that we can translate into a condition on the corresponding distribution function, and then we shall have our second and final answer to the question of uniqueness in the Fundamental Theorem of Calculus for the Lebesgue integral. A monotone increasing function F on the line is said to be absolutely continuous if for each bounded interval [ a ,b] and number t > 0, there exists a 6 > 0 such that on any countable disjoint union U k( a k ,b k ) of intervals within [ a ,b ] having total length < S , the variation C , ( F ( b k )- F ( o k ) )of F on that union of intervals is < t-.
Proposition 7.12. A Stieltjes measure is absolutely continuous if and only if its distribution function is absolutely continuous. PROOF. Let p be a Stieltjes measure with distribution function F. Suppose that p is absolutely continuous. Fix an interval [ a ,b], let t > 0 be given, and choose S > 0 by (b) in Proposition 7.1 1 such that m ( E ) < S implies p ( E ) < t .
3 70
VII. Differentiation o f Lebesgue Integrals on the Line
If the set A = U k( a k ,b k ) is a countable disjoint union of intervals within [ a ,b ] having total length < 6 , then m ( A ) < 6 , and hence p ( A ) < t . Therefore ( F i b k )- F ( a k ) )= p ( ( b k - a k ) ) = p ( A ) < t, and we conclude that F is absolutely continuous. Conversely suppose that F is absolutely continuous, and suppose that E is a Bore1 set with m (E) = 0 Fix an interval [ a ,b ] , and let t 0 be given By absolute continuity of F , there exists a S > 0 such that on any countable disjoint union U k( a k ,b k )of intervals within [ a ,b] having total length < 8 , the variation Ck( F i b k )- F ( a k ) )of F on that union of intervals is < t . With 6 defined in this way, we can find a countable disjoint union of intervals U k ( a k ,bk) covering E n [ a ,b ] and having total length < 8 . Then p ( E f' [ a ,b ] ) 5 p ( U k ( a k ,bk)) = p ( ( a k ,bk)) = ( ( F ( b k ) - F ( a k ) )< t . Since t is arbitrary, p ( E n [ a , b ] ) = 0. Since [ a ,b ] is arbitrary, p ( E ) = 0. Therefore p satisfies (c) in Proposition 7.1 1 and is absolutely continuous.
xk
xk
xk
xk
Corollary 7.13 (second part of Lebesgue's form of the Fundamental Theorem of Calculus). Let F be a monotone increasing function on IE1, and let f be its almost-everywhere derivative. Then J~~ f ( t )d t = F ( b ) - F ( u ) whenever a < b if and only if F is absolutely continuous.
JbX
+
+
PROOF.By Theorem 7.9 we can write F i x ) = f ( t ) d t H ( x ) S ( x ), where H is the distribution function of a continuous singular measure and S is a saltus function. For a < b , we then have
Ib lab
F(b) - Fia) =
f ( t )d t
+ (H(b)
-
His))
+ (S(b)
-
S(a)).
If F ( b ) - F ( a ) = f ( t )d t whenever a < b , then the monotonicity of H and S forces H and S to be constant functions, say with H ( 0 ) S ( 0 ) = c . Substituting, f ( t )d t +c for all x . The function f ( t ) d t is absolutely we see that F (x) = continuous by Proposition 7.12, and the additive constant c does not hurt matters. Thus F is absolutely continuous. Conversely if F is absolutely continuous, then it is continuous, and its monotonicity forces F - F ( 0 ) to be a distribution function of some Stieltjes measure p . Proposition 7.12 shows that the measure p is absolutely continuous, and Proposition 7.1 1 shows that p is of the form p = g d x . Therefore F ( x )- F ( 0 ) = g ( t ) d t . By Corollary 7.4, g = F f = f . Hence F ( b ) - F ( a ) = f ( t ) dt whenever a < b .
+
1;
1;
Jb"
1:
3. Problems 1. In the Rising Sun Lemma (Lemma 7 .I), show that g ( a k ) = g(bk) if ak # a . Give an example of a continuous g for which one of the intervals ( a k ,b k ) has ak = a and g ( a k )< g ( b k ) .
3. Problems
371
2. Let 112 be Lebesgue measure. Does there exist aLebesgue measurable set E such that m ( E n I) = i m ( 1 ) for every bounded interval I? Why or why not?
3. Prove Proposition 7.8 using Wiener's Covering Lemma (Lemma 6.41) instead of the Rising Sun Lemma (Lemma 7.1). 4.
Find all continuous monotone increasing functions on R' with derivative 0 at all but countably many points.
5. Cantor scts within [0, 11 wcrc introduccd in Scction 11.9. Each is associated to a sequence {r,},,l of numbers with 0 < r , < 1, the standard Cantor set being obtained when r , = 1/3 for every ?z. Section V1.8 showed how to associate a distribution function to the standard Cantor set, and in similar fashion one can associate a distribution function to any Cantor set. Let C be a Cantor set, let F be the associated distribution function, and let p be the associated Stieltjes measure. The Lebesgue measure of C is the number P = ( 1 - r , ) . Prove that (a) p is singular if P = 0, (b) p is absolutely continuous if P > 0, being of the form P ' Ic ( x )d x .
nzl
Problems 6-7 concern the Lebesgue set of an integrable function f on an interval [ a ,61. This is the set where If ( t )- f ( x )I dt exists and equals 0 . Many almosteverywhere convergence results involving f are valid at every point of the Lebesgue set. Such results may be regarded as relatively straightforward consequences of Corollary 7.4. Conversely an almost-everywhere convergence theorem that fails to hold at somc point of thc Lcbcsguc sct might wcll bc cxpcctcd to involvc somc ncw idea.
& 1'
6. For f integrable on [ a ,61, prove that almost every point of ( a , b) is in the Lebesgue set o f f by showing that the Lebesgue set of f is the same as the set where If ( t ) - r I d t f If (x) - r 1 for some rational r .
& 1;
The Fejkr kernel, which was defined in Section 1.10 and studied further in Section VI.7, is the periodic function defined for t in [-n, n] by K N ( t ) = 1 lLcos(N+l)t L . et f be integrable on [-n, n], regard f as periodic, and let x N+l be in the Lebesgue set of f . Prove that limN( K N* f ) ( x ) = f ( x ) by following these steps: (a) Check that the estimates K N ( t ) < N 1 and K N ( t ) < c / ( N ~are ~ )valid for all N and for It1 < n. (b) Check that the problem is to show that K N ( t ) I . f ( x - t ) - f (x)I dt tends to 0 as N tends to infinity. (c) Break the integral in (b) into pieces where It 1 < 1 / N , where 2kL1/ N < and where 1 / N 1 l 4 5 It1 5 n.Using It1 5 2 9 for~ 1 5 k 5 the better of the bounds in (a) in each piece, prove the statement that (b) says needs to be shown.
+
h,,,n
VII. Differentiation o f Lebesgue Integrals on the Line
372
Problems 8-12 concern singular Stieltjes measures, which for notational convenience we assume are continuous singular. In all these problems it is assumed that p is a continuous singular measure and m is Lebesgue measure. Among other things these problems prove that the indefinite integral of p has derivative 0 almost everywhere with respect to Lebesgue measure, i.e., d p ( t ) = 0 a.e. [ d x ] ,with the tools of Chapter VI and without Theorem 7.2.
& l;
8. If c
0 is given, prove by considering m such that p ( U ) ( E and 172 ( U C )= 0.
IW'
+ p that there exists an open set U in
9. If U is an open subset of JR1 and v is a Stieltjes measure with v ( U ) = 0, prove that limh$o( 2 h ) - ' v ( ( x - h , x + h ) ) = 0 for all x in U . 10. Let v be any finite Stieltjes measure, and define v y x ) = sup ( 2 h ) - l v ( ( x - h , x
+ h)).
h 10
I
Prove for each 6 > 0 that m { x v * ( x ) > 6 ) 5 5v(JR1)/ 0 be given, and choose an open set U as in Problem 8. Define Stieltjes measures p1 and p2 by p 1 ( A ) = p ( A n U ) and p2 ( A ) = p ( A - U ). Use Problem 9 to prove that limhJo( 2 h ) - l p 2 ( ( x- h , x h ) ) = 0 a.e. [ d x ] ,and use Problem 10 to prove for a11 f > 0 that
+
I
m { x limsup ( 2 h ) - l P 1 ( ( x- h , x
+ h ) ) > () < 5c/(
hJ.0
12. Deduce from Problem 11 that limhJo( 2 h ) - ' p ( ( x - h , x
+
h ) ) = 0 a.e. [ d x ] .By reviewing the proof of Corollary 6.40, show how the argument in Problems 8-1 1 d p ( t ) = 0 a.e. [ d x ] . can be adjusted to yield the better conclusion that
& loX
CHAPTER VIII Fourier Transform in Euclidean Space
Abstract. This chapter develops some of the theory of the LQN Fourier transform as an operator that carries certain spaces of complex-valued functions on IEN to other spaces of such functions. Sections 1-3 give the indispensable parts of the theory, beginning in Section 1 with the definition, the fact that integrable functions are mapped to bounded continuous functions, and various transformation rules. In Section 2 the main results concern L' , chiefly the vanishing of the Fourier transforms of integrable functions at infinity, the fact that the Fourier transform is one-one, and the all-important Fourier inversion formula. The third section builds on these results to establish a theory for L'. The Fourier transform carries functions in L1 n L' to functions in L 2 ,preserving the L2 norm; this is the Plancherel formula. The Fourier transform therefore extends by continuity to all of L', and the Riesz-Fischer Theorem says that this extended mapping is onto L'. These results allow one to construct bounded linear operators on L' commuting with translations by multiplying by La functions on the Fourier transform side and then using Fourier inversion; a converse theorem is proved in the next section. Section4 discusses the Fourier transform on the Schwartz space, the subspace of L 1 consisting of smooth functions with the property that the product of any iterated partial derivative of the function with any polynomial is bounded. The Fourier transform carries the Schwartz space in one-one fashion onto itself, and this fact leads to the proof of the converse theorem mentioned above. Section 5 applies the Schwartz space in R' to obtain the Poisson Summation Formula, which relates Fourier series and the Fourier transform. A particular instance of this formula allows one to prove the functional equation of the Riemann zeta function. Section 6 develops the Poisson integral formula, which transforms functions on LQN into harmonic functions on a half space in LQN+l. A function on LQN can be recovered as boundary values of its Poisson integral in various ways. Section 7 specializes the theory of the previous section to m f l , where one can associate a "conjugate" harmonic function to any harmonic function in the upper half plane. There is an associated conjugate Poisson kernel that maps a boundary function to a harmonic function conjugate to the Poisson integral. The boundary values of the harmonic function and its conjugate are related by the Hilbert transform, which implements a "90" phase shift" on functions. The Hilbert transform is a bounded linear operator on L' and is of weak type (1, 1).
1. Elementary Properties Although the Fourier transform in the one-variable case dates from the early nineteenth century, it was not until the introduction of the Lebesgue integral early in the twentieth century that the theory could advance very far. Fourier
3 74
VIII. Fourier Transform in Euclidean Space
series in one variable have a standard physical interpretation as representing a resolution into component frequencies of a periodic signal that is given as a function of time. In the presence of the Riesz-Fischer Theorem, they are especially handy at analyzing time-independent operators on signals, such as those given by filters. An operator of this kind takes a function f with Fourier series f (x) -- Cz-, c,einx into the expression Cz-, m,,c,,einx, where the constants mn depend only on the filter. If the original function f is in L~ and if thc constants inn arc boundcd, thc Ricsz-Fischcr Thcorcm allows onc to intcrprct the new series as the Fourier series of a new L~ function T ( f ) ,and thus the effect uf llle filler is lu carry f lu T ( f ) . If one imagines that the period is allowed to increase without limit, one can hope to obtain convergence of some sort to a transform that handles aperiodic signals, and this was once a common attitude about how to view the Fourier transform. In the twentieth century the Fourier transform began to be developed as an object in its own right, and soon the theory was extended from one variable to several variables. The Fourier transform in Euclidean space RN is a mapping of suitable kinds of functions on RN to other functions on RN.The functions will in all cases now be assumed to be complex valued. The underlying RN is usually regarded as space, rather than time, and the Fourier transform is of great importance in studying operators that commute with translations, i.e., spatially homogeneous operators. One example of such an operator is a linear partial differential operator with constant coefficients, and another is convolution with a fixed function. In the latter case if F denotes the Fourier transform and h is a fixed function, the f ) , the product on the right side being relevant formula is T(h* f ) = T(h)T( the pointwise product of two functions. Thus convolution can be understood in terms of the simpler operation of pointwise multiplication if we understand what Fdoes and we understand how to invert F. In the actual definition of the Fourier transform, factors of 2 n invariably pop up here and there, and there is no universally accepted place to put these factors. This ambiguity is not unlike the distinction between radians and cycles in connection with frequencies in physics; again the distinction is a factor of 2 n . The definition that we shall use occurs quite commonly these days, namely
with x .y referring to the dot product and with the 2 n in the exponent. The formula for F-' will turn out to be similar looking, except that the minus sign is changed to plus in the exponent. Some authors drop the 2n from the exponent, and then a factor of ( 2 ~ is )needed ~ ~in the inversion formula. Other authors who drop
1. Elementary Properties
375
the 2n from the exponent also include a factor of ( 2 ~ in )front ~ of~ the integral; then the inversion formula requires no such factor. Still other authors who drop the 2n from the exponent insert a factor of ( 2 ~ ) in~ the ~ formula ' ~ for both F and its inverse. In all cases, there is a certain utility in adjusting the definition of convolution by an appropriate power of 2 n so that the Fourier transform of a convolution is indeed the pointwise product of the Fourier transforms. The relationships among these alternative formulas are examined in Problem 1 at the end of the chapter. At any rate, in this book we take the boxed formula above as the definition of Llle Fvurier ~ramslvnnvf a f u ~ ~ d i uf nin L (RN, dx) . C U I ~ V U ~was U ~deG11eCI ~ U I ~i11 Section VI.2. Although there are many elementary functions for which one can compute the Fourier transform explicitly, there are precious few for which one can make the pair of calculations that compute the Fourier transform and verify the inversion formula. One example is e-nlx12,which will be examined in the next section. Recall from Section V1.1 that the translate r,, f is defined by r,, f (x) = f (x - xo).
'
'
Proposition 8.1. The Fourier transform on L (RN) has these properties: (a) f in L implies that f i s bounded and uniformly continuous with 11
.fils.,
5
Ilf Ill> (b) f in L' implies that the translate r,, f and the product f (x)e-2nix'Y~have (r,, f ) (y) = e-2nix0.yf^((y) and F ( f ( x ) e 2 n i x . ~ ~ )= ( y(r,,,f^)((y), ) (c) f and g in L' implies f * g = f g^, (d) f in L' implies f * = f , where f * (x) = f (-x) , (e) (multiplicationformula) f and rp in L1 implies JRN f q d x = JRN fhrp dx , A
A
A
(0 f
A
in L1 and 2nixj f in L1 implies that
everywhere and satisfies (g) f in L' and
."
exists in the ordinary sense
= F(-2niq f ) ,
5 existing in the L' sense, i.e., limh+o hP'(r-h,
existing in L , implies
'
F(z)
f
-
f)
(y) = 2niyj f(y). This formula holds also
when f is in L' nC ,the ordinary
is in L1,and f vanishes at infinity.
PROOF. All the integrals will be over l K N , and we drop ktN from the notation.
~or(a),wehaveI ~ ^ ( ( Y5) J" ) I I f (x)I dx =
I l f lll,andhence llfil,,, 5 I f I l l . Also,
376
VIII. Fourier Transform in Euclidean Space
On the right side the second factor of the integrand is bounded by 2 and tends to 0 for each x as yl - y2 tends to 0. Thus the right side tends to 0 by dominated convergence at a r g e depending only on yl - yz . dX = For (b), ( % o f ) ( y ) = J f ( x - xo)e -2nix.y d x = f (x)e-2ni(x+x~j.~ e-2nixo-y f ( Y ) and
1
A
For (c), we use Fubini's Theorem. The standard technique for verifying the theorem's applicability was mentioned near the end of Section V.7. Let us see the technique in context this once. The procedure is to write out the computation, blindly making the interchange, and then to check the validity of the interchange by imagining that absolute value signs have been put in place. What needs to be verified is that the double or iterated integrals with the absolute value signs in place are finite. The computation here is
The steps with absolute value signs in place around the integrands are
11 If ( x
-
11 If ( X t ) g ( t ) e - 2 n i x .d~x~d t = 11 If ( ~ ) g ( t ) e - ~ " ~ ( ~ d+x~d)t..v l
t ) g ( t ) e - 2 " i x . d~ t~d x =
-
The first interchange is valid, but the first and second integrals are not so clearly finite. What is clear, because f and g are integrable, is that we have finiteness for the third integral, and the second and third integrals are equal by a translation in the inner integration. Thus the computation o f f * g ( y ) is justified. For (d), wc havc f * ( y ) = l f ( - X ) ~ - ~ " d~ x~ = ' Yl f d x = f (y) . For (el, we use Fubini's Theorem, justifying the details in the same way as in (c). We obtain A
A
A
1 f F d x = 11 f ( x ) c P ( y ) e - 2 " i ~d. xy d x =
j"l f
( x ) ' p ( y ) e 2niy.x d x d y = J f^q d y ,
and the interchange is valid because f and q have been assumed integrable. For (f), we apply (b) and obtain 12-l
(T(y
+hej)
-
T ( y ) ) --
F(f (x) h-'
(e-"""j'-"
-
l))(y).
2. Fourier Transform on L 1 , Inversion Formula
377
Application of the Mean Value Theorem to the real and imaginary parts of h-1(e-2"ihej.x- 1) shows for Ihl 5 1 that
and
I Re(h-l I
(e-2nihej.x -
Im(h-l
1))
(e-2nihei.x -
hence that
1 = 1hp1(l
-
cos 2nhxj)l i 2n lxj 1
l ) ) l = lh-' sin2nhxjl 5 2nlxjl,
I h - l ( e - 2 ~ i h e j . ~-
111 i 4nIxjI.
Since xi f ( x ) is assumed integrable, we have dominated convergence in the computation of the limit of 3(f ( x ) h-l (e-2"ihej.x - 1 ) ) ( y ) as h tends to 0 ,
a7
and we get 3(-2 n i x j f ) ( y ) = - ( y ) . ayj
For the first part of (g), the assumptions and (a) give
The left side equals
I By)(h- (e 1
2nihej.y -
I
1 ) ) - 3(%) ( y ) by ( b ) ,and this tends A
to f ? y ) 2 n i i - F ( % ) ( y ) l .~ e n c 3(?)(~7) e = 2 n i y jf ( y ) . xi For the second part of (g), let xj denote the tuple of the N - 1 variables other than x j . Then integration by parts in the variable xj gives
2(x)e-2nin.y d x j dxj
T ( ~ ) ( Y= )J R N - l f z axj limn j': -
LN-1
=-
1
- [RN-l
1
LX( x ) e - 2 n i x . ~d x j d x j
n axj
[ f ( 1-2nix.y ]xj=-n dxj limn
[r, f (x)(-2niyj)e-2"ix.Yd x j dxj
=0+2ni~~f^(y),
as asserted.
2. Fourier Transform on L l , Inversion Formula
The maill tlleol-em of this section is the Fourier inversion for~nulafor L ' ( R N.) The Fourier transform for R1is the analog for the line of the mapping that carries a function f on the circle to its doubly infinite sequence { c k }of Fourier coefficients. The inversion problem for the circle amounts to recovering f from the ck's. We know that the procedure is to form the partial sums s, ( x ) = C:=-,ckeikxand to look for a sense in which Isn} converges to f . There is no problem for the case
378
VIII. Fourier Transform in Euclidean Space
that f is itself a trigonometric polynomial; then s, will be equal to f for large enough n , and no passage to the limit is necessary. The situation with the Fourier transform is different. There is no readily available nonzero integrable function on the line analogous to an exponential on the circle for which we know an inversion formula with all constants in place. In order to obtain such an inversion formula for the Fourier transform on L' , it is necessary to be able to invert the Fourier transform of some particular nonzero function explicitly. This step is carried out in Proposition 8.2 below, and then in general. The analog for the we can address the inversion problem of L' (RN) circle of what we shall prove for the line is a rather modest result: It would say that if C lckl is finite, then the sequence of partial sums converges uniformly to a function that equals f almost everywhere. The uniform convergence is a relatively trivial conclusion, being an immediate consequence of the Weierstrass M test; but the conclusion that we recover f lies deeper and incorporates a version of the uniqueness theorem.
Proposition 8 2 .
= e-"I~1~.
~((e-"1~1~)
REMARKS. Readers who know about the Cauchy Integral Theorem from complex-variable theory or else Green's Theorem in the theory of line integrals will recognize that the calculation below amounts to an application of one or the other of these theorems to the function ePnz2 over a long thin geometric rectangle next to the x axis in C. However, the present application of either of these theorems is so simple that we can without difficulty substitute a proof of one of these theorems in the special case of interest, and hence neither of these other theorems needs to be assumed. As the proof below will show, matters come down to the Fundamental Theorem of Calculus in its traditional form (Theorem 1.32). PROOF.The question is whether
and the integral on the left is the product of N integrals in one variable. Thus the
J -00
We start by observing that
Write
2. Fourier Transform on L 1 , Inversion Formula
Direct calculation gives1 au av -- and ax ay Regard n as positive and large. Then
I:n u ( s , O ) d s =
-
au
av
ay
ax
--
I:nu(s, y ) d s
f , , f f $(s, t ) d t d s
by Theorem 1.32
=+I:,# $ ( s , t ) d t d s = 6I:'-", e ( s ,t ) ds d t =
v(a,t )dt
-
by (**) by Fubini's Theorem
JbY v(-n, t ) d t
by Theorem 1.32.
With y fixed we let n tend to infinity. Then v ( n , t ) and v (-n, t ) tend to 0 uniformly for t between 0 and y by inspection of v , and hence the right side of our expression tends to 0. Thus 00 U ( S , 0 )ds = ~ ( 3y ,) d s ,
I-oo
I-",
which says that
Sirrlilarly we calculate
J:n
U ( S , 0 )ds -
I:n U ( S , y ) d s
I:n # % ( S , t ) d t d s = -I:n fl %(S, t ) d t d s = Jb'J:n g(s,t ) d s d t = JbY u ( n , t ) d t + J : u(-n, - -
by (**)
-
t )dt.
Again we can see that the right side tends to 0 , and thus
I-cov ( s , 0 ) d s = I-", 00
~ ( 3y ,)
ds,
which says that
Taking (*) into account and combining (7) and (It),we obtain 00
lco 00
e-z(x2+2ixy)d x = e - ~ y 2
e-n(x+i~!2 d x = e-rry2
and the proposition follows from the formula sition 6.33.
l-",e-""'
e-zx2 d x ,
dx = 1 given in Propo-
'The equations (**) are called the Cauchy-Riemann equations. They occur again in Section 7.
VIII. Fourier Transform in Euclidean Space
3 80
We shall use dilations to create an approximate identity out of e-nlxlzin the style of Section VI.2. Put y(x) = e-nlXIZand define y,(x) = E - ~ ~ ( E -for '~) F > 0. Whenever rp in an integrable function and rp, is formed in this way, we have
the next-to-last equality following from the change of variables F-'x H x . For the particular function p(x) = e-nlxIZ,this calculation shows that G(?) = e-""z~12.In particular, is > 0 and vanishes at oo for each fixed E > 0. As e decreases to 0, 6 increases pointwise to the constant function 1. The constant c in Theorem 6.20 for this p is c = JR, q(x) dx = 1 by Proposition 6.33. That theorem gives various convergence results for y, * f , one of which is that y, * f converges to f in L1 if f is in L1.
Theorem 8 3 (Riemann-Lebesgue Lemma). If f is in L1(RN),then the continuous function Tvanishes at infinity. PROOF.The continuity of T i s by Proposition 8.la. Put ~ ( x ) e-nlxlz and form y, . Then parts (c) and (a) of Proposition 8.1 give
and Theorem 6.20 shows that the right side tends to 0 as e decreases to 0. Hence ( y ) tends to f i y ) uniformly in y . Since T i s bounded (Proposition
e-ns2y12 f
A
8 .la), e-""21y12 f^(?) vanishes at infinity. The uniform limit of functions vanishing at infinity vanishes at infinity, and the theorem follows.
Theorem 8.4 (Fourier inversion formula). If f is in L ' ( I w ~ and ) f^is in L'(RN), then f can be redefined on a set of measure 0 so as to be continuous. After this adjustment, f (x)
=
lN
f^(y)e2nixvdy
PROOF.By way of preliminaries, recall from Proposition 8.le that the multiplidx whenever f and g are both integrable. cation formula gives J" f g d x = J" With e fixed for the moment, let us apply this formula with g(x) = e-n"lx12. The remarks before Theorem 8.3 about how the Fourier transform interacts with dilations show that ?(?) = e-Ne-n"-zlylz. Tn other words, if we take y(x) = e-nlxlz, then
Tg
3. Fourier Transform on L', Plancherel Formula
381
To prove the theorem, consider first the special case that f is bounded and continuous. If we let e decrease to 0 in (*), the left side tends to f (0) by Theorem 6.20c, and the right side tends to f i y ) d y by dominated convergence since f^is assumed integrable. Thus f (0) = f*(y)d y . Applying this conclusion to the translate r-, f and using Proposition 8 .lb, we obtain
LN
IRN
as required. Without the special assumption on f ,we adjust the above argument a little. Using the equality y, (- y ) = p,(y), we apply (*) to the translate r-, f of f to get
1fiy)e2nix.ye-ns21y12d y = 1f ( x + y)rp, ( y )d y = S f ( x -Y)v,(Y)~Y = (v,*f)(x).
1
As e decreases to 0, the left side tends pointwise to f ^ ( y ) e 2 n i xdy , ~ by dominated convergence, and the result is a continuous function of x , by a version of Proposition 8.la. The right side tends to f in L 1 by Theorem 6.20, and hence Theorem 5.59 shows that a subsequence of rp, * j' tends to j' almost everywhere. 'I'hus f ( x ) = lPLN f ^ ( y ) e 2 s i xd' ~y almost everywhere, with the right side continuous.
Corollary 8.5. The Fourier transform is one-one on L 1( R N ). PROOF. If f is in L 1 and f^is identically 0, then f^is in L' , and the inversion formula (Thcorcm 8.4) applics. Hcncc f is 0 almost cvcrywhcrc.
3. Fourier Transform on L 2 , Plancherel Formula We mentioned in Section 1 that the Fourier transform is of great importance in analyzing operators that commute with translations. The initial analysis of such operators is done on L 2 ( R N ) and , this section describes some of how that analysis comes about. The first result is the theorem for R N that is the analog of Parseval's Theorem for the circle.
Theorem 8.6 (Plancherel formula). I f f is in L 1( R N )f' L 2 ( R N )then ,
Ilf
II f ^ l 1 2
112.
REMARKS.There is a formal computation that is almost a proof, namely
=
VIII. Fourier Transform in Euclidean Space
3 82
the middle equality using the Fourier inversion formula (Theorem 8.4). What is needed in order to make this computation into a proof is a verification that the Fourier inversion formula actually applies. We know that f * * f is in L since f * and f are in L 1 ,and we know from Proposition 6.18 that f * * f is continuous, being in L2 * L2. But it is not immediately obvious that the Fourier transform to which the inversion formula is to be applied, namely f * * f = 1 f f2, is in L l . We handle this question by proving a lemma that is a little more general than necessary.
'
A
A
Lemma 8.7. Suppose f is in L ( R N ), is bounded on RN, and is continuous at 0 . If f^(y) > 0 for all y , then f^ is in L (RN). rp,
PROOF. Put ( P ( x )= epnIxl2and rp,(x) = epNrp(eplx).Then the function * f is continuous by Proposition 6.18 since rp, is in Lm and f is in L 1 ,and
*
by Theorem 6 . 2 0 ~The . function is in L 1 ,and f^is bounded. Hence rp, f^ is iii L . By the Fourier iiiversioii fol-mula (Tlieorem 8.4),
w^,
'
f =
Letting e decrease to 0 and taking into account the monotone convergence, we obtain f ( 0 ) = JR, f i y ) d y . Therefore f^is integrable. 8.6. The remarks after the statement of the theorem prove PROOFOF THEOREM everything except that the Fourier transform f * * f = I f^12 is in L 1 ,and this step is carried out by Lemma 8.7. A
Abstract linear operators between normed linear spaces were introduced in Section V.9, and Proposition 5.57 showed that boundedness is equivalent to uniform continuity. Let us make use of such operators now. Theorem 8.6 allows us to extend the Fourier transform for RN from L1 fl L2 to L ~ In. fact, Proposition 5.56 shows that L1 f' L2 is dense in L 2 . The conclusion of Theorem 8.6 i~npliesthat the linear operator F is bounded relative to the L2 norms on domain and range, and hence it is uniformly continuous. Since the range space L2 is complete (Theorem 5.59), Proposition 2.47 shows that F extends to a continuous map F : L2 + L2. This extended map, also called F, is readily checked to be linear and then is a bounded linear operator satisfying IlFf l 2 = Il f 112 for all f in L2.
3. Fourier Transform on L', Plancherel Formula
383
If f is in L2(RN),we can use any approximating sequence from L1 n L2 to obtain a formula for Ff . One such is f IB(R;O), as R increases to infinity through some sequence. Thus
3f(y)
=
lim
f
(x)e-2nix.~dx.
R+oo
Corollary 8.8. If f is in L1(RN) and g is in L2(lXN),then F ( f F ( f > F ( s >""dk*) = F(g).
* g)
=
PROOF.Set g, = gIB(,,o),so that g, is in L1 n L2 for all n and g, + g in L2.Then f * g n + f *ginL2sinceIlf * g n - f *gl12= I l f * ( g , - g ) I 1 2 i llfllIgn -g1I2. ThereforeF(f)F(g,) = F ( f *g,) + F ( f * g ) in L2. Since 3 ( f ) is a bounded function and F(g,) + F(g) in L2, we see that F ( f )F((g,) + 3 ( f )3(g) in L ~ Hence . 3 (f * g) = 3 ( f )3(g). The identity 3(g*) = is proved similarly.
Theorem 8.9 (Riesz-Fischer Theorem). The Fourier transform operator carries L2(RN)onto L2(RN).
F
IR,
PROOF.The operator F is built from the integral f (x)e-2nix'Yd x . In a similar fashion, build an operator Z from f (x)e2nix.~ dx, or equivalently define Zf (y) = Ff (-y). Then IIZf 112 = 11 f 112 for f in L2. It is sufficient to prove that 3 Z = 1 on L2, since for any f in L2, the equation F Z = 1 implies that Zf is a rnernber of L2 carried to L2 by 3. Moreover, 3 Z is cuntinuous, being bounded. It is therefore enough to prove that 3 Z f = f for f in a dense subspace of L2. We shall do so for the dense subspace L n L2. For a function f in L1f' L2 with the additional property that f^is in L (and also L2 by Theorem 8.6), Theorem 8.4 for Z (or Theorem 8.4 applied to the function f (-x)) shows that 3 Z f = f . For a general f in L~ n L ~ form , 9, * f , where ~ ( x )= e-nlx12. Then p, * f = 6 f is in L1 f' L2; in fact, it is in L~ by Proposition 6.14 and Theorem 8.6, and it is in L1 because f^is bounded and 6 is in L1. By the special case just proved, FZ(p, .r f ) = p, * f . Since 3 Z i s continuous and cp, .r f + f in L~ by Theorem 6.20a, F Z f = f . Thus F Zf = f on the dense subspace L1 f' L2, and the proof is complete.
lRN
-
A
We shall be interested especially in bounded linear operators A on L2(lXN) that commute with translations, i.e., that satisfy A(tx f ) = tx(Af) for all x in RN and all f in L ~Recall . that the operator norm 11 A 11 of a bounded linear operator on L2 is the least C such that llAf 112 5 Cll f 112 for all f in L2.
VIII. Fourier Transform in Euclidean Space
3 84
EXAMPLES. (1) The translation t,, is an example of a bounded linear operator on L~ that commutes with translations; the commutativity in question follows from the commutativity of RN as an additive group, and the equality Iltno,f112 = 1l.f 112 shows that t,, is bounded with Iltxo 11 = 1. In terms of Fourier transforms, Proposition 8.1 shows that (rxof ) (y) = e-7"i"0.y&). (2) Another example of a bounded linear operator on L~ that commutes with translations is the operator Ag = f * g for fixed f in L'. This commutes with translations by Proposition 6.15, and it is bounded with 11 A1 5 11 f I l l by Proposition 6.14. Proposition 8.1 shows that Ag = f g. (3) Let M(y) be any LDOfunction on RN, and for f in L2, define Af by = Mf? The function f^is in L2 by the Plancherel Theorem, ~ f ^ i in s L2 since M is essentially bounded, and ~ f ^ i the s Fourier transform of some L2 function by the Riesz-Fischer Theorem. We take this L2 function to be Af . The ( M F f ) . From the inequalities 11 Af 112 = 11 M F f 112 5 brief formula is Af = FP1 IIMII,IIFf 112 = IIMIDOIl f l12,wesee that Aisboundedwith IIAll 5 IIMII,. The bounded linear operator A commutes with translations, since A
A
A
3
F(A(txf))(y) and
= F ( F - ~ M F ~f)(y) ~ =
M F t xf (y)
=
~ ( ~ ) e - ~ (y) ~ ~ ~ ' ~ F f
F((t, (Af )) (y) = e-2"ix'yF(Af )(y) = e-2"ix'yM ( y ) F f (Y)
One speaks of the function M as a multiplier on L? The previous two examand ples are instances of this construction. Example 1 has M(y) = e-2nixo.~, Example 2 has M(y) = &). We shall see in Theorem 8.14 in the next section that every bounded linear operator A on L2 commuting with translations arises from some such essentially bounded M and that 11 A 11 = I l M 11;, for this reason a bounded linear operator on L2 that commutes with translations is often called a "multiplier operator" or a "bounded multiplier operator" on L2.
4. Schwartz Space This section introduces the space S(RN) of Schwartz functions on R N . This space is a vector subspace of L1(RN),so that the Fourier transform is given on it by the usual concrete formula; S(RN)will turn out to be another space besides L7 that is carried onto itself by the Fourier transform. Working with S(RN)provides a convenient way for using the Fourier transform and derivatives together, as becomes clearer when one studies partial differential equations. If Q is a complex-valued polynomial on IRN , define Q (D) to be the partial differential operator with constant coefficients obtained by substituting, for each
4. Schwartz Space
385
-&
j with 1 j j 5 N , the operator Dj = for q . A Schwartz function on RN is a smooth function such that P (x) Q ( D )f is bounded for each pair of polynomials P and Q . An example is the function e-"1x12, since its iterated partial derivatives 1 ~some polynomial R . The Schwartz space are all of the form ~ ( x ) e - " l ~for
S = S ( R N )is the set of all Schwartz functions. The Schwartz space S is evidently a vector space, and it is closed under partial differentiation and under multiplication by polynomials. Closure under partial differentiation is in effect built into the definition. To see closure under multiplication by polynomials, it is enough to check closure under multiplication by each monomial x,. This closure follows readily from the formula Q ( D ) ( x jf ) = Q " ( D )f + x j Q ( D )f , where Q" is 0 or is a polynomial having degree strictly lower than Q has. If f is a Schwartz function, then P ( x )Q ( D )f is actually integrable, as well as bounded, for each pair of polynomials P and Q . In fact, (l+Ix 12)N ~ ( xQ )( D )f is bounded, and therefore P ( x )Q ( D )f is 5 a multiple of the integrable function (1 + 1x 12)-N. In particular, S is contained in L 1 ,L 2 , and L m . Finally the Fourier transform 3carries S into itself. In fact, parts (f) and (g) of Proposition 8.1 give ~ ( xQ )( D ) ~ = P ( x ) 3 ( Q ( - 2 n i x ) f ) = 3 ( ~ ( ( 2 n i ) -Dl ) Q ( - 2 n i x ) f ) ,
and the right side is the Fourier transform of an L 1 function and therefore is bounded. Proposition 8.10. The Fourier transform 3 is one-one from S ( R N ) onto S ( R N ), and the Fourier inversion formula holds on S ( R N ) .
PROOF.Since S g L l , 3 is one-one on S as a consequence of Corollary 8.5. Since 3 ( S ) g S g L 1 , Theorem 8.4 shows that the Fourier inversion formula holds on S . Let (Zf ) ( x ) = ( F f ) ( - x ) for f in L l . Then Z(S) 5 S . The Riesz-Fischer Theorem (Theorem 8.9) shows that FZ = 1 on L' f' L 2 , and hence FZ = 1 on S as well. Therefore if f is in S , then g = Zf is a member of S such that 3 g = f , and we conclude that Fcarries S onto S. To make effective use of Proposition 8 .lo,we need to know that S ( R N is ) quite large, large enough so that we call shape fuiictioiis suitably when we need tlieni. For U open in R N , let C&,(U) denote the vector space of smooth complexvalued functions on U whose support is a compact subset of U . It is apparent that C&(U) is closed under pointwise multiplication and that every member of Cg,,,(U) extends to a member of Cg,,,(RN) when set equal to 0 off U . But it is not apparent that C&(U) contains nonzero functions. We shall construct some.
3 86
VIII. Fourier Transform in Euclidean Space
Lemma 8.11. If S 1 and S2 are given positive numbers with S 1 < S2,then there exists I) in cEm(RN) such that @ ( x )depends only on Ix 1, I) is nonincreasing in 1x1, $ takes values in [0, 11, $ ( x ) = 1 for 1x1 5 S l , and $ ( x ) = 0 for 1x1 > S2. ~ O O F We . begin from the statement in SectionI.7 that the function f : R + R with f ( t )equal to e-'lt2 for t > 0 and equal to 0 for t 5 0 is smooth everywhere, including at t = 0. (The verification that f is smooth occurs in Problems 20-22 at the end of Chapter I.) If S > 0, then it follows that the function g 6 ( t ) = f (8 t )f (8 - t ) is smooth. Consequently the function ha( t ) = ga ( s )d s is smooth, is nondecreasing, is 0 for t i -6, is some positive constant for t > 6 , and takes only values between 0 and that positive constant. Forming the function ha,, ( t ) = ha(Y + t ) h g(Y - t ) with r at least 38 and dilating it suitably, we obtain a smooth even function @J, ( t ) with values in [O, c ],the function being identically 0 for t 1 > S2 and being identically c for It 1 5 S 1 . Putting @ ( x )= c-' $r0(lxI), we obtain the desired function.
foe
+
Proposition 8.12. If K and U are subsets of RN with K compact, U open, and K G U , then there exists rp E C g m ( U ) with values in [0, 11 such that rp is identically 1 on K . PROOF.There is no loss of generality in assuming that K is nonempty and U is bounded. The continuous distance function D ( x , U C )is everywhere positive on the compact set K and hence assumes a positive minimum c. Define K' to be the set { x E RN I U ( x , K ) i i c } . 'I'his set is compact, contains K , and has nonempty interior. Since the interior is nonempty, K' has positive Lebesgue measure I K' 1 . Applying Lemma 8.11, let h be a nonnegative smooth function that vanishes identically for Ix 1 ? i c and has total integral 1. Define rp = Iz * I K / ,where I K / is the indicator function of K ' . Corollary 6.19 shows that rp is smooth. The function rp is > 0 and has sup Irpl i IlhII 1 IIIKt I , = 1. We have rp ( x ) = h ( x - y ) IKj (y) d y . If x is in K and h ( x - y ) is nonzero, then Ix - yl 5 i c . Then D ( y , K ) 5 Ix - yl 5 i c , and y is in K'. Hence I K ' ( y ) = l , a n d r p ( x ) = J R M h ( x - y ) d y = 1. Next, suppose D ( x, U c ) i i c and h ( x - y ) is nonzero, so that again Ix - y 1 i i r . The claim is that y is not in K ' , i.e., that D ( y , K) > i r . Assuming the contrary, we can find, because of the compactness of K , some k E K with ly - kl i i c . Then every uc E U c satisfies c i luc - kl i luc - xi+ Ix-yl+ly-kl 5 l u C - x ~ + ~ c + ~ c , a n d w e o b t a i n ~ u ~1 - ic.Takingthe xl infimum over u C ,we obtain D ( x , U C )> k c , and this is a contradiction. Thus y is not in K', and the integrand is identically O whenever D ( x , U L )5 i c . Hence rp ( x ) = 0 if D ( x , U C )i i c , and the support of rp is a compact subset of U . This completes the proof.
IRN
4. Schwartz Space
387
Every function in c & ( R N ) is the Fourier transform of some Schwartz function by Proposition 8.10, and there are many such functions by Proposition 8.12. With this fact in hand, we can prove the theorem about operators commuting with translations that was promised in the previous section. We begin with a lemma.
Lemma 8.13, If A is a bounded linear operator on L 2 ( R N )commuting with translations, then A commutes with convolution by any L function.
'
PROOF.
We are to show that A( j'*g) = j'*(Ag) if j' is in L and g is in L2. Let
> 0 be given. Corollary 6.17, with gl = g and g2 = Ag , shows that there exist yl, . . . , yn in R N and constants c l , . . . , cn such that f * g - Cy=l cjryJg < t and f ~g - Ej"=, cjrYjAAgI2 < t-. Then we have t
1
1 *
1
112
x7=l
The first term on the right side is i llAll f * g cjtyjgll i t-IlAll, the sccond tcrm is 0 sincc A commutcs with translations, And thc third tcrm is < t by construction.
Theorem 8.14. If A is a bounded linear operator on L 2 ( R N )commuting with U C lhal ~ ~ A f = F-' ( M F f ) fur all f in L2. As amemberofLDO,Misuniqueandsatisfies IIMII, = IIAIl. lranslaliuns, lllcrl Lllcre cxisls am Lm fun~liunM S
REMARKS.The idea of the proof comes from the corresponding result for L2 of the circle, where it is easy to define M. Call the operator T in the case of the circle. Each function eikx is in L2, and the given operator 1' satisfies tx,(l'(eikx))= T ( ~ ( e i k x ) ) = ~ ( ~ i k k ( x - x o= ) ) , - i k x o ~ eikx ). If we write f for the L , function ( xo T (eikx)and form the Fourier series expansion f ( x ) C cneinx,then r,, f has Fourier series r,, f ( x ) C c,e-inxoeinx by linearity and boundedness of rxo. Since we have just seen that rxof = ePikxof ,we conclude that C c,e-inxoeinx = Cne-ikxoeinx . If c, # 0 for some n unequal to k , then we do not have the term-by-term match required by the uniqueness theorem. Hence only ck can be nonzero, and we have l'(eik")= ckeik". The number ck is the value of the multiplier M at the integer k. In the actual setting of the theorem, the circle is replaced by R N , and individual exponential functions are not in L ~ Thus . this easy process for obtaining M is not available, and we are led to construct M by successive approximations.
-
-
c
PROOF.Choose, by Proposition 8.12, functions Qk E c & ( R N ) with (i) 0 5 Q k ( y )5 1 forall y ,
388
VIII. Fourier Transform in Euclidean Space
(ii) Q k ( y )= 0 for lyl > k + 1, (iii) a k ( y )= 1 for lyl 5 k . Then = @m,nO,kj if j # k , and Qk is in c,&(RN) and hence in the Schwartz space S ( W N ) .Put rpk = F-l (Qk). Proposition 8.10 shows that rpk is in S,and therefore pk is in L 1 f' L 2 . Since the Fourier transform carries convolution into pointwise product, we have p, * pk = ymln(,,kjif j # k. Define
as an L' funcliun. Lcrrlrrra 8.13 sllvws lllal A c u ~ r l ~ n u will1 ~ c s cvrlvvlulivrl by an L' function, and thus pk * ~ p k + l= A ( ~*I pk+l) ~ = ~p~ = ~ ( r *p pk+2) ~ = pk Apk+'. Consequently
*
@kMk+l = @kMk+2 and
Mk+l ( y ) = Mk+2(y) for lyl i k .
This equation shows that if we put
then M is consistently defined and is locally in L ~ . Let So = F - ~ ~ C & ( R ~CQS ( W N ) . If a member f of So has f ? y ) = 0 for I y 1 > k , then f @k+l = f and hence f * pk+l = f . Application of A gives A f = A( f * pk+l) = f * A Q + ~ .If we take the L2 Fourier transform of both sides and use Corollary 8.8, we obtain F ( Af ) = Mk+1 The right side equals M Tsince f ^ ( y )= 0 for 1 y 1 > k , and thus
71
whenever f is in So and f ^ ( y )= 0 for 1 y 1 > k . The subspace c Z m ( W N )of L~ is dense by Corollary 6.19 and Theorern 6.20a. Since the T , Fonrier ~ transform carries 1,2onto T , and ~ preserves norms (Theorems 8.6 and 8.9), So is dense in L'. Let a general f in L2 be given, and choose a sequence { f,} in So with f , a f in L ~ Then . F(Af,) + F ( Af ) in L2. By Theorem 5.59 we can pass to a subsequence, still written as { f,}, so that f , + f and F ( Af,) + F ( Af ) almost cvcrywhcrc. Conscqucntly
almost everywhere.
5. Poisson Summation Formula
389
To see that M is in L W , suppose that I M ( y )1 > C occurs at least on a set E of positive finite measure. Then IE is in L ~ If. we put f = 3 - I ( I E ) ,then we have I I A l l l l f 112 > I I A f 112- IIF(Af112= I I M 3 ( f ) l 1 2 = I I M z ~ 1 1 2 ? C11I~12= CIlf 112) andhence lAll C. Therefore IlAll IIMII,. Inparticular,M isin L w . In the reverse direction we have IIAfl12 = I13(Af)l12 = IlM3(f)l12 5 IIMII,II3(f)ll2 = IIMll,llf ll2 for all f in L 2 >and thus ll All 5 IIMII,. We conclude that 11 M 1 1 , = 11 All . This completes the proof of existence. If we liave two candidates for tlie iiiultiplier, say M aid M l , tlieii subtraction of the equations 3 ( Af ) = M 3 ( f ) and 3 ( A f ) = M 1 3 (f ) shows that 0 = ( M - M l ) F ( f ) for all f in L ~ Therefore . M = M l almost everywhere. This proves uniqueness.
'
'
5. Poisson Summation Formula
The Poisson Summation Formula is a result combining Fourier series and the Fourier transform in a way that has remarkable applications, both pure and applied. Nowadays the formula is expressed as a result about Schwartz functions and therefore fits at this particular spot in the discussion of the Fourier transform. Part of the power of the formula comes about because it applies to more settings than originally envisioned. The Euclidean version applies to the additive group RN,the discrete subgroup of points with integer coordinates, and the quotient group equal to the product of circle groups. In this section we shall take N = 1 simply because a theory of Fourier series has been developed in this book only in one variable. We begin by stating and proving the 1-dimensional version of the theorem.
Theorem 8.15 (Poisson Summation Formula). If f is in the Schwartz space s(IR'), then
PKOOP.Define F ( x ) = x,"=-, f ( x + n ) . From the definition of S,it is easy to check that this series is uniformly convergent on any bounded interval and also the series of kt" derivatives is uniformly convergent on any bounded interval for each k . Consequently the function F is well defined and smooth, and it is periodic of period one. We form its Fourier series, taking into consideration that the period is 1 rather than 2n ; the relevant formulas for Fourier series when the period is L rather than 2n are f(x)
- n=-w 2
Cnr2"'"/~
/
L
with cn = If (t)e-2"'nt/L d t . 2 L -L
3 90
VIII. Fourier Transform in Euclidean Space
A smooth periodic function is the sum of its Fourier series, and thus
The Fourier coefficient in parentheses in (*) is
and the theorem follows by substituting this equality into (*).
Corollary 8.16. C,"=-,
e-"'-2n2
00
-
r En=-,
e-nr2n2
foranyr > 0.
PROOF.The remarks before Theorem 8.3 show that it we define q ( x ) = ePnx2 and rp,(x) - ~ - ' ~ ( e - ' xthen ) , g ( y )- $ ( i ( e y ) . If we put f ( x ) - rrp,(x) -nr-'xz ,then it follows that = re-nrzxz. Applying Theorem 8.15 to this f
T(y)
and setting x = 0 gives the asserted equality. In one especially significant application of the 1-dimensionalEuclidean version of the Poisson Summation Formula to pure mathematics, the remarkable identity in Corollary 8.16 can be combined with some complex-variable theory to obtain a functional equation for the Riemann zeta function, which is initially defined for cornplcx s with Rc s > 1 by O0 1 {(s)=c-= n=l nS
<
n pprime
(I--)
1
-1
.
PS
<
<
The functional equation relates ( s )to ( 1 - s ) . More precisely the function ( s ) extends to be defined in a natural way2 for all s in C - { I } ,and the functional equation is A(1 - S ) = A(s),
where A ( s ) = < ( s ) r ( i s ) n - 1 ' .
This implication is just the beginning of a deep theory in which Fourier analysis, complex-variable theory, algebraic number theory, and algebraic geometry come
he natural way is as an analytic function in C - {I]. The function has a simple pole at s = 1.
5. Poisson Summation Formula
391
together to yield a vast array of surprising results about prime numbers. The derivation of the functional equation of the Riemann zeta function uses some complex-variable theory, and we shall not give it. In real-world applications the 1-dimensional Fourier transform is of great significance because of its interpretation in signal processing. A given function f ( t ) on $8' is interpreted as the voltage of some signal, written as a function of time, and the Fourier transform f^((w) gives the components of the signal at each frequency w . Tlle Plaricllerel forrrlula states the corrlforting fact that energy can be computed either by summing the contributions over time or by summing the contributions over frequency, and the result is the same. Convolutions are of special significance in the theory because they represent the effects of timeindependent operations on the signal- such as the passage of the signal through a filter. To make numerical computations, one takes some discretized version of the signal, obtained for example by rapid sampling over a long interval of time. The discrete signal, which may well be obtained at 2n points for some n , is then regarded as periodic. In other words, the signal is really a function on a cyclic group of order 2". Computing a convolution involves multiplying each translate of one function by the other function at Zn points, adding, and assembling the results. The number of steps is on the order of 22n. Alternatively, a convolution can be computed using Fourier transforms: One computes the Fourier transform of each function, does a pointwise multiplication of the new functions, and then computes an inverse Fourier transform. The pointwise multiplication involves only 2n steps, which is relatively trivial compared with 22n steps. How many steps are involved in the computation of a Fourier transform? Naively it would seem that an exponential depending on y has to be multiplied by the value of the function at each point x and the results added, hence 22n steps. However, the mechanism of the Poisson Summation Formula contains a better way of carrying out the computation of the Fourier transform that involves only about n2" steps. The algorithm in question is known as the fast Fourier transform and is discussed in more detail in Problems 13-18 at the end of the chapter. The upshot is that the Poisson Summation Formula leads to a practical device that cuts down enormously on the cost of analyzing signals mathematically. Although the Poisson Summation Formula as stated in this section relates the real line, the subgroup of integers, and the quotient circle group, the fast Fourier transfornl iterates versions of the for~nulafor settings that are different fro111 this. The groups in question are cyclic of order 2k with k 5 n. We can take the subgroup to have order 2 , and the quotient group then has order 2k-1. A still more general version of the Poisson Summation Formula applies to any "locally compact" abelian group with a discrete subgroup having compact quotient. This more general version of the formula is used in the full-fledged application to pure
VIII. Fourier Transform in Euclidean Space
392
mathematics that combines Fourier analysis, complex-variable theory, algebraic number theory, and algebraic geometry.
6. Poisson Integral Formula
I
Let R y f l be the open half space { ( x ,t ) x E R N and t > 0 ) . We view the boundary { ( x ,0) x E R N }as R N . F01- a function f in L p ( R N )for y equal to 1, 2, or m, we consider the problem of finding u ( x , t ) that is defined on R:", has f' as boundary value in a suitable sense, and is harmonic in the sense of being a c2function satisfying the Laplace equation Au = 0, where A is the Laplacian
I
We studied the corresponding problem for the unit disk in a sequence of problems at the ends of Chapters I, 111, IV, and VI. In that situation the open disk played the role of the open half space, and the circle played the role of the Euclidcan-spacc boundary. Wc wcrc ablc to scc that thc uniquc possiblc answcr, at least if f is of class c2,is given by the Poisson integral formula for the unit disk: u(~,O)=-
2n
S"-,
f(O-v)Pr(p)dq,
The situation with R y f l is different. One complication is that the boundary is not compact, and a discrete sum can no longer be expected. Another is that the harmonic function with given boundary values need not be unique; in fact, the function u ( x , t ) = t is a nonzero harmonic function with boundary values given by f = 0, and thus we cannot expect to get a unique solution to a boundary-value problem unless we impose some further condition on u . In effect, the condition we impose will amount to a growth condition on the behavior of m at infinity. A partial compensation for these two complications is that the boundary is now the Euclidean space R N ,and dilations are available as a tool. Let us make a heuristic calculation to look for a harmonic function with given bouiidaiy values. Suppose u ( x , t ) is tlie solutioii we seek that coi-responds to f . Then we expect that the translate t,,u(x, t ) is the solution corresponding to rx, f ( x ) . We might further expect that the mapping f H u( . , t ) is bounded on L ~ ( R ~By) Theorem . 8.14 we would therefore have
6. Poisson lntegral Formula
393
for some multiplier lnt ( y ) ;the Fourier transform is to be understood as occurring in thex variable only. If tl > 0 is fixed, then u ( x , t+tl) is harmonic with boundary value u ( x , t l ),and so i2'(y, t t l ) = m , ( y ) Z ( p ,t l ) = m , (y)m,, ( y ) f ^ ( y ) .The left side equals m,+,, ( y ) f ^ ( y ) and , therefore
+
Since this is only a heuristic calculation anyway, we might as well assume that m is jointly measurable. Then we deduce that
for some L" function g . To compute g , we use the condition A u = 0 more explicitly. Formally, as a result of the Fourier inversion formula, u (x , t ) is given as
Without regard to the validity of the interchange of limits, we differentiate under the integral sign to obtain
and Summing the derivatives and taking into account that f ^ ( y ) is rather arbitrary, we conclude that g ( y ) 2 = 4 n 2 l y 12. Since n z t ( y ) is an Lm function, we expect that the negative square root is to be used for all y . Thus g ( y ) = -2n 1 y 1. Therefore our guess for the multiplier is
This is an L' function, and we begin our investigation of the validity of this answer by computing its "inverse Fourier transform," to see what to expect for the form of the bounded linear operator f H u ( . , t ) .
Lemma 8.17. For t > 0,
3 94
VIII. Fourier Transform in Euclidean Space
REMARK.The idea is to handle t = 1 first and then to derive the formula for other t's by taking dilations into account. For t = 1 , we express e-2nl~Ias an integral of dilates of e-"Iy1', and then the integral in question will be computable in terms of the known inverse Fourier transforms of dilates of e-"lylz. PROOF.In one dimension, direct calculation using calculus methods on the intervals [O, +m) and (-oo,0] separately gives
+
Since ( 1 v2)-I is integrable, the Fourier inversion formula in R1 (Theorem 8.4) then shows that
Putting u = ( y (with y in RN yields
Jr
CO Any /3 > 0 has /3-l = e-pS ds, and hence ( 1 + v2)-' = n Jb e-(l+"jns ds. Substitution for ( 1 v2)-I in (*), interchange of integrals by Fubini's Theorem, and use of the formula in R1 for the inverse Fourier transform of a dilate of e-""' gives
+
and this is our formula for e-2nI~Ias an integral of dilates of e-nl~12. We multiply both sides by e 2 n i x . integrate, ~, interchange the order of integration, use the formula in RN for the inverse Fourier transform of a dilate of e-nl~12, and make a change of variables n s ( 1 + 1 x 1 ~+ ) s. The result is
6. Poisson lntegral Formula
395
The proof is completed by making use of the effect of the Fourier transform on dilations. We have just seen that the function p ( x ) = (1 is integrable with Fourier transform C ~ ' F ( ~ = ) cN -1e-2"lyl. Then q, ( x ) = t-Nrp(t-'x) = t(t2+1x l 2 ) - ; ( ~ + h a s ~ o u r i e r t r a n s f o r m c ~ ~=~cN (-1t ~e)-2ntlyl
+
,
for t 0, with C N as in Lemma 8.17, to be the Poisson kernel for R:". The Poisson integral formula for R?+' is u ( x , t ) = ( P , * f ) ( x ), and the function u is called the Poisson integral o f f .
Proposition 8.18. The Poisson kernel for R T f l has the following properties: (a) P,(x) = t - " ~ ~ ( t - ' x ) , (b) P, is integrable with P, ( y ) = e-2"tlyI, (c) Pt 2 0 and P,(x) d x = 1 , ( 4 Pt * Ptl = Pt+tl, (e) P ( x , t ) is llar~nonicin N + 1 variables.
IRN
PROOF. Conclusion (a) is by inspection. For (b), the formula for P, shows as y tends to infinity. that P,,for fixed t , is continuous and is of order 1 Therefore P, is integrable. The formula for then follows from Lemma 8.17 and the Fourier inversion formula (Theorem 8.4). In (c), the first conclusion is by inspection of the formilla, and the second conclilsion follows from (h) hy setting y = 0. Conclusion (d) follows from the corresponding formula on the Fourier transform side, namely P, P,' = P,+,, , and conclusion (e) may be verified by a routine computation.
P?
A -
h
Theorem 8.19. Let p be 1 , 2 , or oo,let f be in L P ( R N ), and let u ( x , t ) = (P, f ) ( x ) be the Poisson integral of f . Then (a) u is harmonic in N 1 variables, (b) Ilu(. t > I p5 l l f lip> (c) u ( . , t ) converges to f in LP as t decreases to 0 provided p < oo, (d) u ( x , t ) converges to f ( x )uniformly for x in E as t decreases to 0 provided f is in L" and f is uniformly continuous at the points of E , (e) the maximal function f ** ( x ) = sup,,o I ( P ,* f ) ( x )1 satisfies an inequality i c 11 f 11 /$ with C independent of f and $ , m ( { x f ** ((x) ( f ) (Fatou's Theorem) u ( x , t ) converges to f ( x ) for almost every x in R N .
*
+
9
I
$1)
REMARKS.The theorem says that u is harmonic and has boundary value f in various senses. The hypothesis for ( f ) is really that f is the sum of an L' function
396
VIII. Fourier Transform in Euclidean Space
and an Loo function, and every L~ function has this property, as will be observed in the proof below.
PROOF.Let us leave aside (a) for the moment. Conclusion (b) is immediate from Proposition 6.14 and parts (a) and (c) of Proposition 8.18. Conclusions (c) and (d) follow fromparts (a) and (c) of Theorem 6.20. Conclusion (e) follows from Corollary 6.42 and the Hardy-Littlewood Maximal Theorem (Theorem 6.38), and conclusion (f) for L' functions f is part of Corollary 6.42. Now suppose that f is an L" function. Fix a bounded interval [ a ,b] and write f = f ; f2 with f ; equal to 0 off [ a ,b] and f2 equal to 0 on [ a ,b]. Then Pt * fl converges almost everywhere to fl since fl is integrable, and P, * f2 converges to 0 everywhere on ( a , b) by (d). Hence Pt * f converges almost everywhere on ( a , b);since ( a , b) is arbitrary, P, * f converges almost everywhere. This proves (f). Now we return to (a). Since P ( x , t ) is harmonic, conclusion (a) represents an interchange of differentiation and convolution. The prototype of the tool we need is Corollary 6.19, but that result does not apply here because P, does not have compact support. If we break a function f into two pieces, one where If I is > 1 and one where If 1 is 5 1, we see that any L~ function is the sum of an L1 function and an L" function. Thus it is enough to prove (a) when f is in L 1 or Loo. Let p be P or one of P's iterated partial derivatives of some order, let 1 5 j 5 N + 1, and define Dj Lu be a / a x j if j 5 N ur a/a t if j = N + 1. I1 is sufficient to check that
+
tends to 0 pointwise as h tends to 0. Taking Proposition 6.15 into account, we see that we are to check that
tends to 0 as h tends to 0. Proposition 6.18 shows that it is enough to have
tend to 0 in Loo of the x variable for each fixed t if f is in L 1 ,or in L' of the x variable for fixed t if f is in L". The Mean Value Theorem shows that this expression is equal to
for some h' between 0 and h , with h' depending on x and t , and a second application of the Mean Value Theorem shows that the expression is equal to
7. Hilbert Transform
397
We are to show that this tends to 0, for fixed t, uniformly in x and in L1 of the x variable as h tends to 0. It is enough to show for each fixed t that
is dominated in absolute value by a fixed bounded function of x and a fixed L' function of x when h satisfies Ih 5 minil, t}. An easy induction on the degree shows that any dth-orderpartial derivative of P(x, t) is of the form Q (x, t) (t2 + lx 1 ~ ) - 1' I p(d ~, where ~ Q (x, t) is a homogeneous polynomial in (x, t) of degree d 1 . Since any monomial of degree 1 is bounded by a multiple of (t2+ 1x 12)'I2, the dth-orderpartial derivative is bounded by a multiple of (t2 + I x 1 2 ) - ~ ( ~ + l ) - f ( d - l ) (**I
+
Thus the desired properties of the expression (*) will follow if it is shown that (**) has these properties. This is a routine matter for d > 1, and the proof of (a) is complete.
7. Hilbert Transform This section concerns the Hilhert transform, the honnded linear operator H m L2(IKN)given by F ( H f ( Y ) = -i (sgny)(Ff )(y). Formally this operator has the effect, for y z 0, of mapping exponentials by
and hence of mapping cosines and sines by cos(2nx . y)
H
sin(2nx . y)
and
sin(2nx . y)
H - cos(2nx
. y).
For this reason, engineers sometimes call the Hilbert transform a "90" phase shift." The notion is of such importance that there is even a piece of hardware called a "Hilbert transformer" that takes an input signal and produces some kind of approximation to the Hilbert transform of the signal .3 We shall do some Fourier analysis in order to identify H more directly. To get an idea what H is, we begin by computing the effect on L~ of composing the Hilbert transform and convolution with the Poisson kernel P, (x) . Then we examine what happens when e decreases to 0. 3The delay in time that a Hilbert transformer requires in producing its output imposes a built-in theoretical limit for how good the approximation to the Hilbert transform can be. An exact result would require an infinite time delay.
VIII. Fourier Transform in Euclidean Space
398
i sgn t ) e - 2 n ~ l t l e 2nix.t d t = - -1 .X n E ~ + x ~ '
Lemma 8.20. For s > 0,
PROOF. This result follows by direct calculation, using calculus methods on the intervals 10, foo) and (-oo,01 separately. 1 x If we define Q ( x ) = - -for x in IR', then Q , ( x ) = E - l Q ( e - ' x ) = n l+x2 1 x -is the function in the statement of Lemma 8.20. We define n € 2 + x2
for E > 0, to be the conjugate Poisson kernel on R?. The function Q , is not in L' (Kt1).However, it is in L ~ ( R ' ) and , therefore the convolution of Q, and any L2 function is a well-defined bounded uniformly continuous function. For f in L ~the , function v ( x , e ) = ( Q , * f ) ( x ) is called the conjugate Poisson integral off.
Proposition 8.21. The conjugate Poisson kernel for Kt: has the following properties : (a) the function v ( x , y ) = Q , ( x ) is harmonic in R$ ,and the pair of functions u and v with u ( x , y ) = Py ( x ) satisfies the Cauchy-Riemann equations au ax
av ay
- --
and
au ay
--
au
--
ax'
(b) the L~ Fourier transform F ( Q , ) ( y ) equals -i (sgn y)e-2n"lYI, (c) Q s * = Qs+sl. REMARKS.A fundamental result of complex-variable theory is that if u and v are C' functions on an open subset of C satisfying the Cauchy-Riemann equations, then f'(z) = u ( x , y ) i v ( x , y ) is an "analytic" function in the sense that in any open disk about any point in the open set, f (z) equals the sum of a power series convergent in that disk. We shall not make use of this fact, but the analyticity of u i v is the starting point for a great deal of analysis that will not be treated in this book. In the special case of the Poisson kernel and the conjugate Poisson kernel, the function f is f ( z ) = i / ( n z ) .
+
+
PROOF. Part (a) is a routine calculation. For (b), we know that Q , is in L 2 and has an L2 Fourier transform g = Fie,). The inverse Fourier transform F-l on L2 satisfies F - ' ( ~ = ) Q,, and ( b )will
399
7. Hilbert Transform
follow if we show that F-l ( f ) = Q, ,where f (t) = -i (sgn t)e-2n"ltl. For each integern>O,letf,(t)be f(t)forItl (nandOforIt1 > n . T h e n f,+ fin^^ by dominated convergence, and hence F-l (f,) + F-l ( f ) in L2. By Theorem 5.59 a subsequence of F-'(f,) converges almost everywhere to F-' ( f ) . Since f is in L1, Lemma 8.20 shows that F-l (f,)(t) = f (t)e2"'It dt converges pointwise to Q, (x) ,and therefore F-l (f) = Q,. For (c), Corollary 8.8 shows that F(Q, * P,t) = F(Q,)F(P,j). In combination with Proposition 8.18b, conclusion (b) of the present proposition gives F(Q,)(y)F(P,,)(y) = -i(sgn y)e-2n("f"')lyl a.e., and this is F(Q,+,I)(~)a.e. by a second application of (b).
I:,
Theorem 8.22. Let f be in L~(IR'),and let u(x, y) = (P, * f)(x) and V(X,y) = (Qy * f ) (x) be the Poisson integral and conjugate Poisson integral of f . Then (a) the function v is harmonic in R:,
and the pair of functions u and v satisfies the Cauchy-Riemann equations, (b) the function Q, * f is in L2(R1) for every E > 0, and its L2 Fourier transform is F(Q, * f ) (y) = -i (sgn y)e-"El)'I.F( f ) (y) , > 0, (c> 11 QE * f 112 = llpE * f 112 5 I l f 112 for (d) Q, * f + H ( f ) in L~ as e decreases to 0.
'
PROOF. Conclusion (a) is handled just like Theorem 8.19a. In the proof of Theorem 8.19a, the integrability of P, did not play a role; it was the integrability of the iterated partial derivatives of P, (i.e., the case d > 0) that was important. The estimates involving such derivatives here are the same as in that case. For (b), put g = F(Q,).F(f). This is an L~ function since F(Q,) is in L" by inspection and since F ( f ) is in L~ by the Plancherel formula. Define fn = IB(n,O) f , SO that each fn is in L1 f' L2 and also f, + f in L2. Since F(Q,) is in La, the Plancherel formula shows that g,, = F(Q,)F( f,,) is in L~ for each n and converges to g in L2. Since f, is in L1 and Q, is in L ~Corollary , 8.8 gives F(Q, * fn) = F(Q,)F( fn) = gn for all n. Thus Q, * f , = F-l (g,) . We now let n tend to infinity. We know that 11 Q , * f, - Q , * f llSuP < 11 Q , 11211 f, - f l12. Since Q, is in L2 and f, + f in L ~Q,, * f, converges uniformly to Q, * f . On the other hand, F-' (gn) converges to F - I (g) in L2, and Theorem 5.59 shows that a subsequence converges almost everywhere. Therefore F-' (g) = Q, * f . Consequently F(Q, * f ) = g = F(Q,)F( f ) , and we obtain F ( Q , * f ) (y) = -i (sgn y)e-2""1yl.F(f )(y). Conclusions (c) and (d) follow by taking L~ Fourier transforms and using (b), Proposition 8.18b, and the Plancherel Theorem. This completes the proof.
VIII. Fourier Transform in Euclidean Space
400
To get a more direct formula for the Hilbert transform, we introduce the functions
for 1x1 < 1, and Let $(x) = Q(x)
-
h(x), so that $,(x) = e-'$(e-'x)
=
Q,(x)
-
h,(x).
Lemma 8.23. The function $ on R1 is integrable, and f",$(x) dx
= 0.
PROOF.For 1x1 < 1, we have I / T ( ~=) Q(x) = n-'x/(l + x 2 ) . This is a continuous odd function, and therefore it is integrable on [- 1, 11 with integral 0 . For 1x1 r> 1, we have ~ ( x = ) n-'(& - ) ,1 = -n-' (-). This is an integrable function for Ix I > 1; since it is an odd function, its integral is 0. Theorem 8.24. Let Iz, be defined as above. If f is in L2(IR1),then /z, * f is in L2(R1)for every e > 0 , and h, * f + H ( f ) in L2 as e decreases to 0. REMARKS.More concretely the limit relation in the theorem is that Hf (x) =
lim & $0
The integrand on the right side is the product of two L2 functions on the set where It 1 3 e , and it is integrable by the Schwarz inequality.
PROOF.We have h, * f = Q, * f - $, * f . The term Q, * f is in L2 by Theorem 8.22b, and the term Ilr, * f is in L2 by Lemma 8.23 and Proposition 6.14. As e decreases to 0, Q, f tends to Hf in L2 by Theorem 8.22d, and @, ~i. f tends to 0 in L2 by Theorem 6.20a. This completes the proof.
*
Now that we have the concrete formula of Theorem 8.24 for the Hilbert transform on L2 functions, we can ask whether the Hilbert transform is meaningful on other kinds of functioiis. For exaiilple, we could ask, If we have soiile otlier vector space V of functions and V f' L2(R1) is dense in V , can we extend H to V ? The , answer for V = L1( R 2 i s unfortunately negative. In fact, i f f is in L' n L ~then the Fourier transform f will be continuous and the Fourier transform of Hf will have to be -i(sgn y)f? If F(0)is nonzero, then -i(sgn y) f is not continuous and cannot be the Fourier transform of an L function. A
'
7. Hilbert Transform
40 1
However, in Chapter M we shall introduce LP spaces for 1 5 p 5 oo,thereby extending the definitions we have already made for p equal to 1,2,and m. Toward the end of the chapter, we shall see that the Hilbert transform makes sense as a bounded linear operator on L p ( R 1 )for 1 < p < oo. This boundedness is an indication that the Hilbert transform is not a completely wild transformation, and the result in question will be used in the problems at the end of Chapter IX to prove that the partial sums of the Fourier series of an LP function on the circle converge to the function in LP as long as 1 ip < oo. Actually, this boundedness on LP will be a consequence of a substitute result about L' that we shall prove now. Although the IIilbert transform is not a bounded linear operator on L 1 ,its approximations in the statement of Theorem 8.24 are of weak type (1, I), in the same sense that the passage from a function to its Hardy-Littlewood maximal function in Chapter VI was of weak type (1, 1).
Theorem 8.25. Let h l be the function on R' equal to l / ( n x ) for 1x1 equal to 0 for Ix i1. For f in L' ( I R 1 ) + L2(IR1),define
> 1 and
as the convolution of the fixed function h l in L~ with a function f that is the sum of an L 1 function and an L2 function. Then
with the constant A independent of f ,and
for every 6 > 0, with the constant C independent of 6 and f REMARK.This result about the approximation H1 to H on L1 and L2 will be enough for now. The result for L1 is much more difficult than the result for L2. Tn the next chapter we shall derive from Theorem 8.25 a boundedness theorem for all the other approximations HE = h, * ( . ) on LP(Ki1)for 1 ip im, with a bound independent of e , and then it will be an easy matter to get the boundedness of the Hilbert transform H itself on LP for these values of p. PROOF. A preliminary fact is needed that involves a computation with the function h 1. We need to know that
402
VIII. Fourier Transform in Euclidean Space
whenever 0 < Ir'l 5 r . To see this, we break the region of integration into four sets-one where 1x1 > 2 r , 1x1 > 1, and Ix +r'l > 1; a second where 1x1 > 2 r , 1x1 < l,andIx+rfl > 1;athirdwhereIxI > 2r,IxI > l,andIx+rfl < 1;anda fourth where Ix I > 2 r , Ix I < 1,and Ix +r '1 < 1. For the fourth piece the integrand is 0. For the second and third pieces, the integrand is 5 1 in absolute value, and the set has measure 5 2; hence each of these pieces contributes at most 2 For the first piece the absolute value of the integrand is ~ r ' l / l x ( x rf)l 5 2 r / x 2 ; thus thc absolutc valuc of thc intcgral is 5 >2r 2 r / x 2 d x = 2. This provcs (+). It is an easy matter to prove that H1 is a bounded linear operator on L ~ In . , is in L by Lemma 8.23. Thus Theorem 8 . 2 2 ~gives fact, h 1 = Q 1 - ~ / rand IIHlf l l , 5 IIQl * f l l , Il+ * f l l , 5 I l f 112+ I I ~ 1 I 1 I I f 112. Inotherwords,H1 is boundedon L~ with 11 H111 5 1 + II1/rIIl. Put A = 1 + Il1/r11~. The heart of the proof is the observation that if F is in L l , vanishes off a bounded interval I with center yo and double4 I * , and has total integral F ( y ) d y equal to 0, then I 1 H l I~LI(R-I*) 5 611 Fll 1 (**I
hxl
+
+ +
To see this, we use the fact that the total integral of F is 0 to write
'lhking the absolute value of both sides and integrating over R - 1*, we obtain
the last step holding by (*). This proves (**). Let the L 1 function f be given. Fix 6 > 0. We shall decompose the L 1 function f into the sum f = g + b of a "good" function g and a "bad" function b, in a manner dependent on $ . The good function will be in Lm and hence will be ; effect of applying HI to it will be controlled by the bound in L f' L" g L ~the of HI on L ~ The . bad function will be nonzero on a set of rather small measure, and we shall be able to control the effect of HI on it by means of (**). We begin by constructing adisjoint countable systemof bounded open intervals Ik such that (9 C k m ( I k ) 5 511f 111/6> (ii) If ( x )1 5 6 almost everywhere off U, I k , (iii) I f ( y )I d y 5 26 for each 12.
& lIk
4 ~ h "double" e of a bounded interval I is an interval of twice the length of I and the same center.
7. Hilbert Transform
403
1 Namely, let f * ( x ) = sup,,, 5 &x-,,x+,l If ( t )I d t be the Hardy-Littlewood maximal function of f , and let E be the set where f * ( x ) > 6. The set E is open. In fact, if f * ( x ) > 6, then 1 f ( t )1 d t 3 4' c for some t > 0. For S > 0, the inequality &xph,x+h+2Sl -f ( t )I d t > t shows that
&
+
& hx-h,x+hl
+
<+
f * ( x +S) > ($ t ). Hence f * ( x +S) > 4' for S sufficiently small. Similarly f * ( x - b') > $ for b' sufficiently small. Since E is open, E is uniquely the disjoint union of countably many open intervals, and these intervals will be the sets Ik.The disjointness of the Ik's and the Hardy-Littlewood Maximal Theorem (Theorem 6.38) together give
and this proves (i) and the boundedness of the intervals. The a.e. differentiability of integrals (Corollary 6.39) shows that If ( x )1 5 f * ( x ) a.e., and therefore If (x)I 5 4' a.e. off E = Uk Ik.This proves (ii). If I = ( a , b ) is one of the Ik's, then a is not in E , and we must have 1 If (t)l d t 5 f ' ( a ) 5 4'. Therefore &a,bl If ( t )1 d t 5 2$. This proves (iii). With the open intervals Ikin hand, we define the decomposition f = g b by
i[b-2(b-aj,bl
&
+
for x
( 0
I
Since { x I H l f ( x > l > 6} enough to prove
I 1
C {x1
6
uk~ k .
I
IHlg(x)l > $/2} U { x IHlb(x)l > 6/2},itis
r r ~ ( { x IH1g(x)l 6/21) 5 ~ ' l f l l l l / 4 ' and m ( { x IH1b(x)l > $12)) 5 ~ ' l l f ill/$ for some constant C' independent of 4' and f . The definition of g shows that J, lg(x)l d x i JIk If (x)l d x for all k and that Ig(x)l = I f ( x ) l for x @ U k I k ; therefore I , l g ( x ) d x 5 ~ , I f ( x ) I d x . Also, properties (ii) and (iii) of the Ik's show that Ig ( x )1 5 24' a.e. These two inequalities, together with the bound I I H l g ll 5 All g l l 2 ,give
VIII. Fourier Transform in Euclidean Space
404
I
Combining this result with Chebyshev's inequality m({x IF (x) 1 > B } ) 5 pP2 m'f F (x) l 2 d x for the function F = Hlg and the number = (12, we obtain
iR
This proves the bulleted item about g . For b , let bk be the product of b with the indicator function of I k . Then we have b = Ckbk with the sum convergent in L' . Since H1 is convolution by the L2 function h l , H 1 b = CkH1bk with the sum convergent in L 2 . Lumping terms via Theorem 5.59 if necessary, we may assume that the convergence takes place a.e. Therefore I Hl b(x) 1 5 CkI Hlbk (x) I a.e. Using monotone convergence and (**), we conclude that
I
~ i n c e m ( { xE U kI ; } ) 5 511 f l l l / $ by(i),weobtainm({x IHlb(x)I > $121) 5 1711 f 11,/6, and the bulleted item about b follows.
8. Problems 1. For each of the following alternative definitions of the Fourier transform in RN, find a constant a such that the Fourier inversion formula is as indicated, and find a constant ,i3 such that when convolution is defined by
then the Fourier transform of the convolution is the product of the Fourier transforms. f (x)epix.Yd y and inverse Fourier transform (a) Fourier transform f*(y) = f ( x ) = a JRN f*((y)eix.~ dy. (b) Fourier transform f ( y ) = ( 2 ~ ) f (x)epix.y ~ ~ d y and inverse Fourier transform f (x) = a JRN f*(y)eix.yd y . (c) Fourier transform f i y ) = f (x)epix'Yd y and inverse Fourier transform f ( x ) = w lRN f (y)eix.Yd y .
I,,
IRN
A
A
I,,
405
8. Problems
2.
SEN
Let ( u , = u(x)U(X)dx if u and v are in L 2 ( R N ) ,and let .F denote the Fourier transform on L 2 ( R N ) .Prove for every pair of functions f and g in L2 that ( f , g)2 = (.F(f ) , 8 g ) ) 2 .
3. Prove that the Poisson kernel P and the conjugate Poisson kernel Q for satisfy the identity Q , * Q,, = P,+,, .
4. This problem is an analog for the Fourier transform of Problem 20c of Chapter VI concerning Fourier series and weak-star convergence. Weak-star convergence was defined in Section V.9. (a) If f is in La ( R N )and Pt is the Poisson kernel, prove that Pt f converges to f weak-star against L' ( R N )as t decreases to 0. In other words, prove (P, * f ) ( x ) g ( x )d x = f ( x ) g ( x )d x for every g in L 1 . that limtJo (b) Theorem 8.19b shows that IIP, * f ,1 5 11 f lloo i f f is in L'O(W~).Prove
*
l,,
that lim,Jo llPt
l,,
*f
lla
=
Ilf
Iloo.
5. Let M + ( R N ) be the space of finite Borel measures on R N . This problem introduces convolution and the Poisson integral formula for M ( R N ) . Each finite Borel measure on R N defines, by means of integration, a bounded linear functional on the normed linear space Cc,,(RN) equipped with the supremum norm, and thus it is meaningful to speak of weak-star convergence of such measures against C C o m ( R N ) . (a) The convolution of a finite Borel measure p on R N with an integrable function f is defined by ( f * p) ( x ) = f ( X - y ) d p ( y ) . Define the convolution ~i = ~i1 * ~ 1 2of two members of M f ( R N )by ~i( E ) = JRN pl ( E - x ) d p 2 ( x ) for all Borel sets E . Check that the result is a Borel measure and that the definition for f d x * p , i.e., for the situation in which p1 and p2 are specialized so that p1 = f d x and p2 = p , is consistent with the definition in the special case. (b) With convolution of finite Borel measures on R N defined as in (a), prove that S R N g PI * ~ 2=) JRN JR.v ~ ( x + Yd) ~( xl )d ~ 2 (forevery ~ ) bounded Borel function g on R N . (c) Verify that 11 Pt * p 11, 5 ( R N )if p is in m + ( R N ) .Prove the limit formula
SRN
limtJ.0 II Pt * p II = p ( R N > . (d) If p is in M + ( R N ) ,prove that the measures (P, p ) ( x ) d x converge to p weak-star against C , ~ , ~ , ( Ras~t) decreases to 0. In other words, prove that limt$0 JRN ( P , * p ) ( x ) g ( x )d x = JRN g ( x ) ~ P ( xfor ) every g in C c m ( R N ) .
*
Problems 6-12 examine the Fourier transform of a measure in M + ( I R N ) ,ultimately proving "Bochner's theorem" characterizing the "positive definite functions" on R N . They take for granted the Helly-Bray Theorem, i.e., the statement that if { p , } is a sequence in M + ( R N )with p , ( R N ) bounded, then there is a subsequence {p,,} convergent to some member p of M f ( R N )weak-star against Cc0,(RN). The HellyBray Theorem will be proved in something like this form in Chapter XI.
VIII. Fourier Transform in Euclidean Space
406
6. If p is in M f ( R N ) ,the Fourier transform of p is defined to be the function p ( y ) = JRN ep2nix.yd p ( x ) . (a) Prove that is bounded and continuous. (b) Prove that the Fourier transform of the delta measure at 0 does not vanish at infinity. (c) Prove that p 1 * pz - Ez when convolution of finite measures is defined as in Problem 5 . (d) By forming p, * p , prove that @ can equal 0 for some p in M f ( R N )only A
A
ifp=o.
A continuous function F : RN + C is called positive definite if for each finite set of points x l , . . . , xk in EXN and corresponding system of complex F ( x i - x j ) & 5 > 0 holds. Prove that numbers t l , . . . , t k , the inequality the continuous function F is positive definite if and only if the inequality lRN lRN F ( X - y ) g ( x ) g ( y )d x d y > 0 holds for each member g of c,,,(JR~). 8. Prove that the Fourier transform of any member p of M f ( J R N ) is a positive definite function.
7.
9. Using sets of one and then two elements x, in the definition of positive definite, prove that a positive definite fuilctioil F illust have F (0) > 0 a i d IF (x)1 5 F (0) for all x . 10. Supposc that F is positivc dcfinitc, that cp
2 0 is in L ' ( I w ~ and ) , that @ ( x ) =
lRN e2nzxy p ( y ) d y . Prove that F ( x ) @( x ) is positive definite. 11. Suppose that F is positive definite. Let s > 0, and let cp be as in Problem 10, so that p ( x ) = ~ ~ and @ ( x ) = ~ ep""21x12. e ~ ~
(a) The function Fo ( x ) = F ( x )@ ( x ) is positive definite by Problem 10. Prove that it is integrable. (b) Using Problem 2 and the alternative definition of positive definite in Problem 7,prove that ( y )l g ( y )l 2 d y > 0 for every function g in C c o m ( R N.) (c) Deduce from (b) that the function fo = is > 0. (d) Conclude from (c) that fo is integrable with fo d y = F ( 0 ) ,hence that f o ( y ) d y is a finite Borel measure.
lRN 6
6
l,,
12. (Bochner's Theorem) By combining the results of the previous problems with
the Helly-Bray Theorem, prove that each positive definite function on R N is the Fourier transform of a finite Borel measure. Problems 13-1 8 concern a version of the Fourier transform for finite abelian groups, along with the Poisson Summation Formula in that setting. They show for a cyclic group of order m = pq that the use of the idea behind the Poisson Summation Formula makes it possible to compute a Fourier transform in about p q ( p q ) steps rather than the expected m2 = p2q2 steps. This savings may be iterated in the case of a cyclic group of order 2n so that the Fourier transform is computed in about ~ 2 2 ~
+
~
-
~
407
8. Problems
steps rather than the expected 22n steps. An organized algorithm to implement this method of computation is known as the fast Fourier transform. 13. Let G be a finite abelian group. A multiplicative character x of G is a homomorphism of G into the circle group { e i e } .If f and g are two complex-valued functions on G , their L2 inner product is defined to be Cttcf (t)g(t). (a) Prove that the set of multiplicative characters of G forms an abelian group under pointwise multiplication, the identity e l e m z t being the constant function 1 and the inverse of x being z. This group G is called the dual group of G . (b) Prove that distinct multiplicative characters are orthogonal and hence that the members of GI form a linearly independent set. (c) Let J, be the cyclic group (0, 1 , 2 , . . . , m - 1 ) of integers modulo m under addition, and let = e2nilm.Fork in J, define amultiplicative character X , of J, by X, (k) = (c:)~. Prove that the resulting rn multiplicative characters exhaust J, and that x,x,, = xn+,,. Therefore J, is isomorphic to J,. For Problems 16-1 8 below, it will be convenient to identify X, with X , ( 1 ) = < .; (d) If G is a direct sum of cyclic groups of orders m 1 , . . . , m, ,use (c) to exhibit mj distinct members of GI. Using (b) and the theorem that every finite abelian group is the dkect sun1 of cyclicgoups, c2nclude for any finite abelian group G that these members of G exhaust G and form a basis of L"G).
<,
n;=,
14. Let G be a finite abelian group, and let GI be its dual group. The Fourier transform of a function f in L ~ ( Gis) the function T o n GI given by = CttGf ( t ) ) o . Prove that the Fourier transform mapping carries L ~ ( Gone) one onto L ~ ( G I ) and that the correct analog of the Fourier inversion formula is f(t) = I G ' ~ ~ , ~ f * ( ~ ) ~ ( t ) , w IGIh istheorderof ere G.
T(x)
15. Let G be a finite abelian group, let H be a subgroup, and let G / H be the quotient ) define group. If t is in G , write t for the coset of t in G / H . Let f be in L ~ ( Gand F ( t ) = ChtH f ( t h ) as a function on G / H . Suppose that x is a member of GI that is identically 1 on H , so that x descends to a member j( of G / H . By imitating steps in the proof of Theorem 8.15, prove that f ( x ) = F*(i).
-
+
A
16. Now suppose that G = .I, with rn = pq; here p and q need not be relatively prime. Let H = 10, q , 257, . . . , ( p - l ) q } be the subgroup of G isomorphic to J,, so that G / H = (0, 1, 2 , . . . , g-1) is isomorphic to J,. Prove that the characters x of G identified as in Problem 13c with <,:
52,
<:,
A
VIII. Fourier Transform in Euclidean Space
408
F*
is computed from the definition of Fourier transform in Problem 14, then the number of steps involved in its computation is about q 2 , apart from %constant factor. Show therefore that the total number of steps in computing f at these special values of x is therefore on the order of q 2 p q .
+
,;<mp+/<,
17. In the previous problem show for each k with 0 5 k 5 p-1 that the value of T a t
,
+
CHAPTER IX L~ Spaces
Abstract. This chapter extends the theory of the spaces L ~L , ~and , LCO to include a whole family of spaces L P , 1 5 p 5 00, in order to be able to capture finer quantitative facts about the size of measurable functions and the effect of linear operators on such functions. Sections 1-2 give the basics about LP. For general measure spaces these consist of Holder's inequality, Minkowski's inequality, a completeness theorem, and related results. For Euclidean space they include also facts about convolution. Sections 3 4 develop some tools that at first may seem quite unrelated to L P spaces but play a significant role in Section 5. These are the Radon-Nikodym Theorem and two decomposition theorems for additive set functions. The Radon-Nikodym Theorem gives a sufficient condition for writing a measure as a function times another measure. Section 5 identifies the space of continuous linear functionals on L P for 1 5 p < 00 when the underlying measure is o-finite. For one thing this identification makes Alaoglu's Theorem in Chapter V concrete enough so as to be quite useful. Section 6 discusses the Marcinkiewicz Interpolation l'heorem, which allows one to reinterpret suitably bounded operators between two pairs of LP spaces as bounded between intermediate pairs of LP spaccs as wcll. Thc thcorcm has immcdiatc corollarics for thc Hardy-Littlcwood maximal function and an approximation to the Hilbert transform, and Section 6 goes on to use each of these corollaries to derive interesting consequences.
1. Inequalities and Conlpleteness In the context of any measure space, we introduced in Section V.9 the spaces L l , L 2 , and L". Since then, we have used these three spaces to capture quantitative facts about the size of measurable functions. The construction in each case involved introducing a certain pseudonorm in a vector space of functions, thereby making the vector space into a pseudo normed linear space and in particular a pseudometric space. The corresponding metric space obtained from the construction of Proposition 2.12 was L ' , L ~or, Lm in the respective cases. For each of the three, the vector-space stlucture for the pseudometric space yielded a vectorspace structure for the metric space, and L ' , L ~ and , Lm were normed linear spaces. As was true in Chapters V and VI, it continues in the present chapter to be largely a matter of indifference whether the functions in question are real valued or complex valued, hence whether the scalars for these vector spaces are real or complex.
lX.LP Spaces
410
Now we shall enlarge the family consisting of L 1 ,L ~L", to a family LP for 1 5 p 5 cc in order to be able to capture finer quantitative facts about the size of measurable functions. Enlarging the family in this way makes it possible to get better insight into the behavior of specific operators and to make more helpful estimates with partial differential equations. Let ( X , A, p ) be a measure space. We have already dealt with p = oo. For 1 5 p < oo,we consider the set V = Vpof measurable functions f on X such that If l P d p is finite. This integral is well defined; in fact, f measurable implies If I measurable,andalso,forc > O , ( If lp)-'(c, +m) = i f 1-'(cl'p, +oo). The sel V is in he1 a veclur space uf funcliuns. 11 is ccrlainly elused under s ~ a l a r multiplication; let us see that it is closed under addition. I f f and g are in V ,then we have
(If
(x)I
+ I g ( ~ ) li) ~( m a x I l f ( x ) l ,lg(x)Il +maxIIf ( x ) I ,l g ( x ) ~ l ) ~ = 2 P m a x I l f(x)lP3Ig(x)IPli Z P I f
(,)IP
+2PIg(x)lP
for every x in X . Integrating over X , we see that f + g is in V if f and g are in V . Following the construction of the prototypes L' and L2 in Section V.9, we irrlruducr llie erpreaaiurl 11 f I p = If I P L Z ~ ) ' ' ~fur f iri V P We wuuld like 11 . l i p to be a pseudonorm in the sense of satisfying
(Ix
'
(i) Ilxllp 0 for all x E V , (ii) llcxllp = ~ c ~ ~ ~ x ~ ~ p f o r a l l s c a l a r s c acnVd a, l l x (iii) Ilx y l i p i Ilx l i p l l y l i p for all x and y in V . Properties (i) and (ii) are certainly satisfied, but a little argument is needed to verify (iii). We return to this matter in a moment. Once the function 11 . 11, on the vector space V p is known to be a pseudonorm, Vp meets the conditions of being a pseudo normed linear space in the sense of Section V.9. We can pass to the set of equivalence classes just as in that section, and this set is defined to be LP or LP(X) or LP(X, w ) . The equivalence class of 0 is again the set of all functions vanishing almost everywhere. The function 1 1 . l i p is well defined on LP, and LP is a normed linear space. In particular, it has the structure of a metric space. This handles 1 5 p < oo,and the space L" was constructed in Section V.9. As is true with L ' , L 2 , and L", one sometimes relaxes the terminology and works with the men~bersof L p ( X ) as if they were functions, saying, "Let the function f be in LP(X)" or "Let f be an LP function." There is little possibility of ambiguity in using such expressions. Let us return to property (iii) above. This will be proved as Minkowski's inequality below. But first we prove a numerical lemma and then "Holder's inequality," which is a version for LP of the Schwarz inequality for L ~ Holder's .
+
+
1. Inequalities and Completeness
41 1
1 1 inequality makes use of the dual index p' to p ,defined by the equality -+- = 1. P P' The dual index to 1 is oo,and vice versa. The index 2 is its own dual.
Lemma 9.1. If s , t , a , and j3 are real numbers
> 0 with a + p = 1, then
PROOF.If any of s, t , a , p is 0, the result is certainly true. If all are nonzero, corlsider the furlctiorl ,f ( x ) = axN-I
+ (1
-
a)xZ,
defined for x > 0. The derivative f f ( x ) = (1 - o l ) o l ~ " - ~ (x 1 ) is < 0 for 0 < x < l , i s = Oforx = 1,andis > Oforx > 1. Therefore f ( x ) assumesits absolute minimum value for x = 1. Since f (1) = 1, we have for all x > 0. The lemma follows by putting x multiplying both sides by sZtD.
= t/s
in this inequality and by
REMARK.Alternatively, this lemma can be proved by Lagrange multipliers in the same way that Problem 20 at the end of Chapter 111 suggested using for the arithmetic-geometric mean inequality.
Theorem 9 2 (Holder's inequality). Let (X, A, p) be any measure space, let 1 5 p 5 oo,and let p' be the dual index to p . If f is in LP and g is in LP',then f g is in L ' , and
5 1l.f l l p l l ~ l l p ~ . REMARK.The inequality holds trivially if I1 f l i p = +m or Ilgllp, = +oo ll.fgIl1
PROOF.We already know the result if y = 1 and p' = oo or the other way around. Thus suppose that p > 1 and p' > 1. We may assume that neither f nor g is 0 almost everywhere. Then we can apply Lemma 9.1 with a = p - l ,
p
= p'-', S
=
If P ) l P , lXIf l P d~
and
t =
lg(x)lP'
lXl g l p ' d ~ '
getting
If (x)g(x>l If ( x ) I P 5 I I ~ I I ~ I I ~ I PI ~Jx If l p d Integrating, we obtain
and the conclusions of the theorem follow.
+
lg(x)l~'
~ pfJx I g I p ' d ~ ~
lX.LP Spaces
412
Theorem 9.3 (Minkowski's inequality). Let (X, A, p) be any measure space, a n d l e t l i p 5 oo.If f andgareinLP,then f + g i s i n L P a n d
REMARK.The theorem assumes the usual convention that f I g is made to be 0 at any point x where f (x) + g(x) is not defined. The set where this change occuls is of measure 0 since f and g have to be finite almost evelywhere to be in LP. PROOF.We have already seen that f + g is in LP, and we know the inequality for p = 1 and p = oo from Section V.9. For 1 ip ioo,let p' be the dual index. We apply Holder's inequality (Theorem 9.2) to f and I f + g l ~ - ' and to g and I f + lp-' to obtain
+
the last step holding because ( p - 1)p' = p . If 11 f g I p = 0, the inequality of the theorem is certainly true. Otherwise the inequality of the theorem follows after dividing the inequality of the display by ( J X 1 f g P da)l'p', which we know to be finite, and using the fact that 1 - 1- 1.
+
P'
Thus LP is a nonned linear space for 1 properties.
p
-
P
oo.Let us derive some of its
Proposition 9.4. Let ( X , A, p ) be a measure space, and let 1 5 p < m. Then every indicator function of a set of finite measure is in LP(X), and the smallest closed subspace of LP(X) containing all such indicator functions is LP(X) itself. Consequently (a) the set of simple functions built from sets of finite measure lies in every Lp(X) for 1 5 p 5 oo and is dense in Lp(X) if 1 5 y < oo, (b) 1 5 p1 5 p 5 p2 5 oo and p ioo together imply that LPl (X)nLPZ(X) is dense in LP(X) .
In addition, (c) 1 5 p1 5 p 5 p2
oo implies that LP(X) 5 LP1(X) + LPZ(X).
1. Inequalities and Completeness
413
PROOF.The conclusion in the second sentence of the proposition is proved by the same argument as for Proposition 5.56. Part (a) then follows from Proposition 5.55d. Part (b) follows by combining these two results once it is known that LP1( X ) n LP2( X ) C L P ( X ) . For this inclusion let f be in LP1( X ) n LP2( X ). We may assume that p < oo.If p2 < oo,then
and hence f is in LP ( X ). If pz is in LPl and pl < oo.Thus
=
oo,then { I f
: ,1}
has finite iileasure since f
I* I f l P d p = ~ l f l > l ) I f l P d w + J ; l f 1 5 1 ) I f l P d P s l l f lP, P ( I l f l
> 11) + J ; l f 1 5 1 }
I f lP1
d P < +W,
and again f is in L P ( X ) . This completes the proof of (b). For (c), let f be in LP , and write f = fl f2, where
+
otherwise
otherwise
iIf
Then
~ l . f i l " d ~ =l . f~l ' "~d ~~5 ~
I.flpdp<
shows that fl is in LP1( X ) . It is apparent that f2 is in L" ( X ), and thus f2 is certainly in LP2( X ) if p2 = oo.If p2 < oo , then
shows that f2 is in L" ( X ). This proves (c) . Holder's inequality allows us to prove the following supplement to the conclusions of Proposition 9.4. Proposition 9.5. Let ( X , A, /J) be any measure space. Let 1 i pl < p < pz, a n d d e f i n e t w i t h o s t 5 l b y L = - + 4 . a hen P
Pl
P2
PFXI~F.First suppose that p2 < oo. Since 1 > E , we can find b with P P1 1 1 < b < +oo such that - = E.If b' denotes the dual index, then 1= bp Pl b'p
lX.LP Spaces
414 1 P
(P
1 1 b p - P -
1-t PI
$)$ = P 2 ( ;
-
P2 -
. Define a by the equation a b = pl . Then ( p - a ) b f =
6)=
1 P2(;
-
Y )= p2.
We wnte If l P = If la If P - a . Application of Holder's inequality with index b and dual index b f gives / If l p d p 5 ( / 1 flab d w ) l I b ( / i f ~ ( p - ~ ) ~ ' d p and )~'~', hence f 11, 5 ( i1 f l a b d p ) l " b p ) ( / 1 f I ~ P - dp)l'(b'p), ~)~' We have seen that a b = pl , l / ( b p ) = ( 1 - t ) / p l ,( p - a ) b f = p2, and l / ( b f p )= t l p 2 . Thus the inequality reads I I f 11, 5 11 f f f11i2,and the proof is complete wheii p2 ioo. When p2 = oo,we write 1 f l P = If lP1 If I P - P l . Replacing If lp-pl by its essential supremum gives J" If l p d p 5 I f IIP,-pl / I f lP1 d p and hence 11 f 1 1 , is
II;;"~
This completes the proof when p2 = oo. We have already made serious use of the completeness of LP for p equal to 1 , 2, and oo as proved in Theorem 5.58. As might be expected, this result extends to be valid for the other values of p .
Theorem 9.6. Let (X, A, p ) be any measure space, and let 1 5 p 5 oo. Any Cauchy sequence { f k } in LP has a subsequence { fk,} such that I l fk, - fk, 1 1 , 5 C,,,lnz,n~ with CrlC , < +m. A silhserlnence {hn} with this property is necessarily Cauchy pointwise almost everywhere. If f denotes the almosteverywhere limit of { f,,}, then the original sequence { f k } converges to f in LP . Consequently the space LP ,when regarded as a metric space, is complete in the sense that every Cauchy sequence converges. REMARK.As in the case with p equal to 1,2, and oo,the detail is important. The detailed statement of the theorem allows us to conclude, among other things, that if a sequence of functions is convergent in LP1 and in LP2, then the limit functions in the two spaces are equal almost everywhere. PROOF.We rnay assurrle that p < cx,,tlie case p = oo liaving been handled in Theorem 5.58. The argument for 1 5 p < oo is word-for-word the same as in the proof for p = 1 and p = 2 of Theorem 5.58.
+
+
In Section V.9 the inequality 11 f g l i p 5 11 flip Ilg 11, for p equal to 1 , 2 , or oo says in words that "the norm of a sum is 5 the sum of the norms." In that section we obtained a generalization for those values of p , saying that "the norm of an integral is 5 the integral of the norms." The generalization continues to be valid for the other p's under study; the proof amounts to a direct derivation from Holder's inequality.
1. Inequalities and Completeness
415
Theorem 9.7 (Minkowski's inequality for integrals). Let ( X , A, p) and (Y,B, v) be o-finite measure spaces, and let 1 5 p 5 oo.If f is measurable on X x Y , then
in the following sense: The integrand on the right side is measurable. If the integral on the right is finite, then for almost every y [dv]the integral on the left is defined; when it is redefined to be 0 for the exceptional y's, then the formula holds. PROOF.Theorem 5.60 handles p = 1 and p = oo,and we may assume that 1 < p < oo.The measurability question is handled for 1 < p < oo in the same way as in Theorem 5.60 for p = 2. In proving the inequality, we may assume without loss of generality that f 2 0. The generalization of the computation in the proof of Theorem 9.3 makes use of Fubini's Theorem and proceeds as follows:
The next-to-last step uses Holder's inequality (Theorem 9.2), and the last step uses the fact that ( p - l)pf= p . In order to complete the proof, we need to be able to divide by the factor (SY jXf (xl,y) dp(xl) dv (Y))l'". There is no problem with the theorem if this factor is 0, since then the left side of the inequality of the theorem is 0. A problem occurs if this factor is infinite. Instead of trying to prove directly that this factor is finite (and hence the division is allowable), let us retreat to the special case that f is bounded and is equal to 0 off an abstract rectangle of finite p x v ~neasure.Then the factor in question is certainly finite, the division is allowable, and we obtain the inequality of the theorem. To handle general measurable f 0 , we do not attempt to justify this division. Instead, we observe that the validity of the inequality in the theorem when f is bounded and is equal to 0 off a set of finite x v measure implies the validity of the inequality in general, by a routine application of monotone convergence. This completes the proof.
I
1'
lX.LP Spaces
416
The last basic fact about LP spaces is the identification of continuous linear functionals on LP, at least when p is finite. Deriving the necessary tools for this analysis will require a digression, and we shall return to this topic in Section 5. Meanwhile, we can easily obtain one part of the identification of continuous linear functionals, as in Proposition 9.8 below. It amounts to a combination of Holder's ineq~lalityand a converse, and it gives a way of computing LP norms by starting with computations that are linear.
Proposition 9.8. Let (X, A, p ) be any measure space, let 1 5 p i oo, and let p' be the dual index. If p ioo,then
and this equality remains valid for p
= m if
is o-finite.
REMARK.The equality can fail when p = m and p is not a -finite. Problem 4 at the end of the chapter gives an example. PROOF.
With 1 5 p 5 oo,ifg is in LP' with IgIp, 5 1, thcn Holdcr's inequality
1
gives IJ'fgdd 5 IfgI d p 5 I f llpIgllp, 5 I f I l p . Taking the supremumover g with lgll,, i 1 shows that sup, J ' f g d p l i I l f 1 ,. For the reverse inequality we may assume that 11 f 1 , # 0. First suppose that 1 < p < cm.Define g(x) by
I
(P-1)~'
ThenSlg(x)lp'dw=Ilflli S~f(x)l(~-~)~'dw=~lflli~J'If(x)I~d~= 1 For this g , we have f g d p = 11 f , ( p - l1 ) f Pj d p = 1 f lip. Thus the supremum over the relevant g's of .[ f g d p l is ? II f II ,. Nextsupposethatp = 1. Ifwedefineg(x)tobe f (x)/I f(x)l when f ( x ) # 0 andtobeowhen f ( x ) = O , t h e n I g l l , = landI/'fgdpl= J'1f1~/lfldp= 11 f ~ , , a n d t h e s u ~ r e m u m o v e r ~ o ff g d p l i s ? I l f l l , . Finally suppose that p = oo.Let r > 0 be given with r 5 11 f II,, and let E be the set where If (x)I 11 f 1 , - t Since p is o-finite, there must exist a subset of E with nonzero finite measure. If F is such a subset and if g(x) is defined to be p ( ~ ) - ' f o / 1 f (x) 1 when x is in F and to be 0 when x is in FC,then llg 11 = 1 and fgdpl = p ( ~ ) - ' J ~ l f f d 1 p IIfII, - r . Thus thesupremumover g of & f g d p is > 11 f 11, - t . Since t is arbitrary, the supremum over g of J;, f g d ~ isl > Ilf 1 ,.
11
I
11
'
IA
I
I
1
417
2. Convolution Involving LP
2. Convolution Involving LP In this section we collect results about LP spaces that extend facts proved about L' , L ~and , L" in the first three sections of Chapter VI.
Proposition 9.9. If p is a Borel measure on a nonempty open set V in RN and i f 1 5 p < co,then (a) C,,,(V) is densein I,P(V, / I ) , (b) the smallest closed subspace of LP(V, p) containing all indicator functions of compact subsets of V is LP(V, p ) itself, (c) LP(V, w ) is separable. PROOF. Parts (a) and (b) are proved from Lemma 6.22c, the regularity of p (Theorem 6.25), Proposition 9.4, and Proposition 5.56 by the same kind of
argument as for Corollary 6.4. Part (c) is obtained as a consequence in the same way that Corollary 6.27d follows from the other parts of that corollary. The remaining results in this section concern Lebesgue measure in R N , and the LP spaces are understood to be LP(RN, {Borel sets}, dx) .
Proposition 9.10. Let 1 ip icc, and let p' be the dual index. Convolution is defined in the following additional cases beyond those listed in Proposition 6.14, and the indicated inequalities hold: (e) forf i n ~ ~ ( ~ ~ , d x ) a n d ~ i n L P ( R ~ , d x ) , a n d tfugllpiII h e n I I f IIIIgllp, forf inLP(RN,dx) a n d g i n L l ( ~dx),andthen ~, I l f * g l l p i I l f lpIlgII1, ( f ) for f' in Lp(RN,dx) and g in L ~ ' ( Rdx), ~ , and then I l f' * g l , i
llf l l p l l ~ l l p ~ > for f in L P ' ( R ~dx) , and g in L P ( R ~dx), , and then I l f
* g 1,
i
llf l l p ~ l l ~ l l p . PROOF.The two conclusions in (e) follow from Minkowski's inequality for integrals (Theorem 9.7) in the same way that the special case of p = 2 was proved in Proposition 6.14 from Theorem 5.60. The two conclusions in (f) follow from Holder's inequality (Theorem 9.2) in the same way that the special casc p = p' = 2 was provcd in Proposition 6.14 from thc Schwarz inequality.
Proposition 9.11. If 1 5 p < cc, then translation of a function is continuous in the translation parameter in LP(RN, dx). In other words, iff is in LP(RN,dx) , then limh,o 11 t t + h f - tt 1 , = 0 for all t . PROOF.This follows from the denseness of C,,,(RN) in LP(RN,dx) (Proposition 9.9a) and is proved in the same way that Proposition 6.16 is derived from Corollary 6.421.
lX.LP Spaces
418
Proposition 9.12. Let 1 5 p 5 oo,and let p' be the dual index. Then the convolution of an LP function with an LP' function results in an everywhere-defined bounded uniformly continuous function, not just an Lm function. Moreover,
I l f * gllsup5 I l f l l P l l ~ l l P ~ . PROOF.This extends Proposition 6.18 and is derived for 1 < p < oo from Propositions 9.10 and 9.11 in the same way that Proposition 6.18 is derived for p = 2 from Propositions 6.14 and 6.16. Theorem 9.13. Let cp be in L' ( I R N , d x ) ,define ( x ) =e N ( x )
lim llpE* f
-
cf
for E > 0:
lip
= 0.
& $0
PROOF.This is derived from Minkowski's inequality for integrals (Theorem 9.7) and the continuity of translation in LP (Proposition 9.1 1) in the same way that Theorem 6.20a is derived for p = 2 from Theorem 5.60 and Proposition 6.16.
3. Jordan and Hahn Decompositions Now we digress before returning in Section 5 to the subject of continuous linear functionals on LP spaces. The subject of the present section is decompositions of additive and completely additive real-valued set functions into positive and negative parts. This material will be applied in Section 4 to obtain the Radon-Nikodym Theorem, an abstract generalization of some consequences of Lebesgue's theory of differentiation of integrals. In turn, we shall use the Radon-Nikodym Theorem in Section 5 to address the subject of continuous linear functionals on LP spaces. A real-valued additive set function v on an algebra of sets is said to be bounded if Iv(E)I 2 C for all E in the algebra. A real-valued completely additive set function on a o -algebra of sets is said to be a signed measure.
Theorem 9.14 (Jordan decomposition). Let v be a bounded additive set function on an algebra A of sets, and define set functions v f and v- on A by v + ( E ) = sup v ( F ) and v - ( E ) = - inf v ( F ) FGE, F EA
FGE, FEA
419
3. Jordan and Hahn Decompositions
Then vf and v- are nonnegative bounded additive set functions on A such that v = vf - v-. They are completely additive if v is completely additive. In any event, the decomposition v = vf - v- is minimal in the sense that an equality v = pf - p- in which pf and p- are nonnegative bounded additive set functions must have vf 5 pf and v- 5 p-.
PROOF.First let us see that vf is additive always. In fact, let El and E2 be disjoint members of A. If F c ElU E2,then the additivity of v implies that v(F)= v(F n El)+ v(F f' E2)5 vf (El) vf(E2).Hence
+
+
On the other hand, if F15 E and F2 5 Ez, then v (F1) v (Fz) = v (Fl U Fz)i v +(ElU E2).Taking the supremum over Fland then over F2gives
+
vf (El) vf (E2)5 vf (ElU E2). Thus vf is additive. Second let us see that vf is completely additive if v is completely additive. Let Enbe a disjoint sequence of sets in A whose union E is in A.If F 5 E ,then the complete additivity of v implies that v(F)= v(F f' En)5 vf(E,). Hence vf (E)5 vf(En).On the other hand, the fact that vf is nonnegative additive implies for every N that vf (En)= vf(ElU . . . U EN)5 vf (E). Thus vf (En)5 vf(E).Therefore vf is completely additive. Third let us see that v = vf - v-. This equality will imply also that vis additive and that v- is completely additive if v is completely additive. Form v(E)+vP(E)= v(E)+sup,,,{-v(F)}; we are to show that this equals vf (E). For any F g E , we have V(E) (-v(F))= v(E - F) i vf (E). Taking the supremum over F gives v(E) + v-(E) 5 vf (E). In the reverse direction, F 5 E implies that v(F) = v(E)- v(E - F) 5 v(E)+ supKe{-v(G)} = v(E)+ vP(E).Taking the supremum over F gives vf(E)5 v(E)+ vP(E). This proves the decomposition v = vf - v-. Finally we prove the minimality of the decomposition. Let v = pf - pwith ic + and ic- nonnegative additive. If F E , then we can write v(F) = pf (F)- p-(F) 5 pf (F) 5 pf(E). Taking the supremum over F gives vf (E)(: pf(E).Similarly v- (: p-.
C,,
Czl
C,"==,
C,,
Cf=l
+
Theorem 9.15 (Hahn decomposition). If v is a bounded signed measure on a o-algebra A of subsets of X , then there exist disjoint measurable sets P and N in A with X = P U N such that v(E)> 0 for all sets E g P and v(E) 5 0 for all sets E 5 N.
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420
PROOF.Write v = v f - v- as in Theorem 9.14. If in A with v ( A ) > v f ( X ) - t . Then
t
> 0 is given, choose A
+t 5 t
v - ( A ) = v f ( A )- v ( A ) 5 v f ( A ) - v f ( x )
and v f ( A " ) = v f ( x ) v f ( A ) 5 v ( A ) + t v + ( ~5) t . By taking Po = A and No = A C , we see that for any c z 0 we can write X = Po U No disjointly with v f ( N o ) 5 t and v P ( P o )5 t . For n > 1, write X = P, U N , disjoiiitly wit11 v f ( N , ) 5 2-, and v P ( P n )5 2-,. Define -
-
nz==n
P = Ur==l P, and N = PC = U:=, N,. These sets are in A since A is a o-algebra. Theorem 9.14 shows that v- is co~npletelyadditive, and hence v- (P) 5 Cr==l v- ( n:', P,) . The right side is 0 since v - ( n;=, P,) 5 v P ( P n f k )5 2 - ( n f k )for all k > 0, and therefore v - ( P ) = 0. In addition, every n has v f ( N ) 5 v f ( UE==,v f ( N , ) ) 5 00 Ern==, v f ( N , ) 5 C;=, 2-m = 2-"+ l , and thcrcforc v f ( N ) = 0.
4. Radon-Nikodym Theorem The Lebesgue decomposition of Chapter VII says that any Stieltjes measure p on f d x + p , with p , concentrated on a Bore1 the line decomposes as p ( E ) = set of Lebesgue measure 0. The function f is obtained in that chapter as the derivative almost everywhere of the distribution function of w , hence as the limit of p ( I ) / m ( l ) as intervals I shrink to a point; here m is Lebesgue measure. In this formulation of the result, the geometry of the line plays an essential role, and attempts to generalize to abstract settings the construction of f from limits of p ( I ) / m ( I ) havc not bccn fruitful. Nevertheless, the Lebesgue decomposition itself turns out to be a general measure-theory theorem, valid for any two measures in place of p and dx, as long as suitable finiteness conditions are satisfied. For a reinterpretation of the results of Chapter VII, the heart of the matter is that one can tell in advance which p's have p ( E ) = f d.u with the singular term p , absent. The answer is given by the equivalent conditions of Proposition 7.11,which are taken in that chapter as a definition of "absolute continuity" of p with respect to dx . The remarkable fact is that those conditions continue to be equivalent when any two finite measures replace p and dx. This is the content of the Radon-Nikodym Theorem, which we shall prove in this section, and then a version of the Lebesgue decomposition will follow as a consequence. Let X be a nonempty set, and let A be a o-algebra of subsets of X . If p and v are measures defined on A, we say that v is absolutely continuous with respect to p , written v << p , if v ( E ) = 0 whenever p ( E ) = 0.
IE
1,
4. Radon-Nikodym Theorem
421
Theorem 9.16 (Radon-Nikodym Theorem). Let ( X , A, p ) be a o-finite measure space, and let v be a o-finite measure on A with v << p . Then there exists a measurable f 2 0 such that v ( E ) = f d p for all E in A, and f is unique up to a set of w measure 0.
iE
The Radon-Nikodym Theorem has two chief initial applications. One is to the identification of continuous linear functionals on LP for 1 5 p < m, and the other is to the construction of "conditional expectation" in probability theory. The application to LP will be given in Section 5, and the application to conditional expectation appears in Problems 23-26 at the end of the chapter. In both applications one needs a version of the theorem in which the completely additive set function v is complex-valued but not necessarily > 0. We take up this extension of the theorem later in this section. Most of the effort in the proof goes into showing existence when p and v are both finite measures, as we shall see. In this setting we can quickly use the Hahn decomposition (Theorem 9.15 ) to get an idea how to construct f : Imagine that v ( E ) = f d p for all E . Fix c and d , and let S be the set of x's where c 5 f (x) < d . On any subset E of S , we then have c p ( E ) 5 v ( E ) 5 d p ( E ) . In other words, the bounded signed measure v - c p is 2 0 on every subset of S , and the bounded signed measure v - d p is 5 O on every subset of S. Let X = PC U Nc and X = Pd U Nd be Hahn decompositions of v - c p and v - d p with respect to p . Then it is reasonable to expect S to be PCn Nd . In particular, c is a good lower bound for the values of f on S. It is easy to imagine that we can use this process repeatedly to obtain a monotone sequence of functions f, > 0 tending to the desired function f . Actually, this argument can be pushed through, but handling the details is a good deal more complicated than one might at first suppose. The reason is that a Hahn decomposition is not necessarily unique. Sets of measure 0 account for the nonuniqueness, and the particular measures yielding these sets of measure 0 are constantly changing. The complication is that one has to adjust all the Hahn decompositions to satisfy various compatibility conditions. We shall not pursue this idea because a simpler proof is available.
iE
PROOF OF UNIQUENESS IN THEOREM9.16. Suppose that f and g are nonnegative measurable functions with f dp g d p for every measurable E. If F is a set where the equal integrals f dw and lF g dw are finite, then ( f - g ) d p = 0 for every measurable subset E n F of F. If E is taken as the set where f > g , then Corollary 5.23 shows that f = g a.e. on E f' F. Similarly f = g a.e. on the set Ec n F, where f 5 g . Thus f = g a.e. on F. By o-finiteness of p and v , we can write X = X, disjointly with p(X,) and v (X,) finite for all n . Taking F equal to each X, in turn, we see that f = g a.e. on each X,, and we conclude that f = g a.e. on X .
iE
SEnF
iF
iE
UZl
422
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PROOF OF EXISTENCE IN THEOREM 9.16 WHEN p AND V ARE FINITE. Let F(v) be the set of all f > 0 in L ' ( x ,p ) such that JE f d p 5 v ( E ) for all sets E in A. The zero function is in F ( v ),and thus it makes sense to define
C = sup
1
f dp.
f € F ( T ( ~X )
Let { f,} be a sequence in F ( v ) with limn Jxf, d p = C . Let us show that there is no loss of generality in assuming that the f, satisfy fl < f? < . . . . To show this, it is enough to show that g and h in F ( v ) implies that max{g, h } is in F ( v ). We have
and hence max{g, h } is indeed in F ( v ) . With the f a ' s now increasing with n , put f (x) = limn f ( x ) . Monotone convergence shows that f is in F ( v ) and Jx f d p = C . Define
Then vo is a measure, vo << p , and the class F(v0) for vo consists of 0 alone. We shall complete this part of the proof by showing that vo = 0. If vo # 0, choose n large enough so that v o ( X ) - p ( X ) > 0 , and put v; = vo p . Let X = P U N be a Hahn decomposition for v; as in Theorem 9.15, and define g = I F . Then the calculation
A
shows that g is in F(v0). Hence g = 0 a.e. [ d w ] ,and w ( P ) = 0. Since vo << p , we obtain v o ( P ) = 0 and therefore also vh(P) = 0. Then vt, 5 0, and we must have v o ( X ) - $(x) 5 0. This contradicts the choice of n , and the proof of existence is complete when p and v are finite. PROOF OF EXISTENCE IN THEOREM 9.16 WHEN p AND v ARE a -FINITE. Write X as the countable disjoint union of sets X , such that p(X,) and v(X,) are both finite. Ifwe put p,(E) = p ( E n X , ) and v,(E) = v ( E n X , ) , then p, and v, are finite measures such that v, << w, ,and the above special case produces functions f, ? 0 such that v,(E) = JE f, d p , for all E . Since v,(Xi) = 0, we may assume that f,(x) = 0 for x $ X,. Let f > 0 be the measurable function that equals f, on X , for each n. Then our formula reads v ( E n X,) = f dp for all n and for all E . Summing on n , we obtain v ( E ) = JE f d p for all E in A.
4. Radon-Nikodym Theorem
423
Corollary 9.17. Let (X,A, p) be a finite measure space, and let v be a (realvalued) bounded signed measure on A with v << p in the sense that p(E) = 0 implies v(E) = 0. Then there exists a function f in L' (X,p) such that v(E)= f dp for all E in A, and f is unique up to a set of p measure 0.
1,
PROOF.Let v = v+ - v- be the Jordan decomposition of v as in Theorem 9.14, and let X = P U N be a Hahn decomposition of v as in Theorem 9.15. Suppose p (E)= 0. Since p is nonnegative, we obtain p (En P) = 0 and p (En N ) = 0, and the assumption v << p forces
and
0 = V(E n N ) = -v-(E
n N) = -v-(E).
Therefore v+ << p and v- << p,and the corollary follows by applying Theorem 9.16 to vf and v- separately. Corollary 9.18. Let (X,A, p) be a o-finite measure space, and let v be a o-finite measure on A. Then there exist a measurable f 2 0 and a set S in A with p(S) - 0 such that v - f dp + v,,where us(,?) - v(E f' S).The measure v, is unique, and the function f is unique up to a set of p measure 0.
REMARK.The measure v,,being carried on a set of p measure 0, is said to be sing~~lar with respect to / I . Themeasure f d / is, ~ ofconrse,ahsoIiltelycontinnons with respect to p.The decomposition of v into the sum of an absolutely continuous part and a singular part is called the Lebesgue decomposition of v with respect to p.The corollary asserts that this decomposition of measures exists and is unique. PROOF.As in the proof of Theorem 9.16, we can reduce matters to the case that v and p are both finite, and it is therefore enough to handle this special case. Among all sets E in A with p(E) = 0, let C be the supremum of v(E). The number C is finite, being 5 v(X).Choose a sequence of sets En in A with p(E,) = 0 and v(E,)increasing to C . Without loss of generality, we may assume Proposition 5.2shows that p(S) = 0 and that El c E2 c . . . . Put S = U, E,,. v (S)= C .Define v, (E)= v(E f' SC)and v,(E)= v (E n S). Then v, and v, are measures, and v = v, + v,. Certainly v, is singular with respect to p, being carried on the set S of p lneasure 0. Let us see that v, is absolutely continuous. Thus suppose that p(E) = 0. Then p(S U E) 5 p(S) + p(E) = 0,and the construction of C shows that v(SU E) 5 C = v(S).Therefore v(SU E)- v(S)5 0 and v(SU E)- v(S)= 0. Hence 0 = v(S U E)- v(S)= v(E - S) = v(E f' Sc)= v, (E), and v, is indeed absolutely continuous. Applying the Radon-Nikodym Theorem (Theorem 9.16), we obtain v = v, + v, = f dp + v,.This proves existence.
424
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For uniqueness, suppose that we have v = f d p + v, = f ' d p + v,# with v, and v,#carried on respective sets S and S' of p measure 0. The functions f and f are integrable with respect to p , and we have f - f') d p = v,#(E)- v , ( E ) . Taking E to be any subset T in A of S U s', we see that 0 = u,#(T)- v , ( T ) . Therefore v,#(T) = v,(T) whenever T 5 S U s". On ( S U s')", we have u,#((s U s')') = v,((S U s')') = 0. Thcrcforc v,# = us. Thc uniqucncss of the function part follows from the uniqueness in the Radon-Nikodym Theorem, wllicli is part uf the staterrlent of that tlleurerri (Theureni 9.16).
'
IE(
5. Continuous Linear Functionals on LP We return to the question of identifying the continuous linear functionals on LP spaces. Let ( X , A, p) be a fixed o-finite measure space. The space L P ( X , p ) is a normed linear space and, as such, is both a vector space and a metric space. The scalars may be real or complex. Recall from Section V.9 that a linear functional on LP (X, p ) is a linear function from L P ( X , p ) into the scalars. Proposition 5.57 shows that a linear functional x" is continuous if and only if it is bounded in the sense that Ix" ( f )I 5 C 1 1 flip for some constant C and all f in LP. The inequality Ix*( f ) I i C 11 f l i p holds for all f in LP if and only if it holds for all f with 11 flip 5 1, if and only if it holds for all f with 11 flip = 1. If there is such a constant C , then the finite number
is the least such constant C and is called the norm of x * . Since Ilx* 11 is one such constant C , we have Ix*if > I 5 Ilxll*llf lip. Let p be the dual index to p , defined by 1P
+ 4P
= 1. Each member g
of L P ' ( X ,b) provides an example of a continuous linear functional on LP by the formula x* ( f ) = f g d p . The linear functional x* is bounded, hence continuous, as a consequence of Holder's inequality: J; f g d p 5 llg I , I f 1, . This inequality shows that Ilx* 1 1 5 ilglp,. Proposition 9.8 shows that equality Ilx*II = Ilgllp, holds if p is o-finite. Theorem 9.19 gives a converse when 1 5 y < oo,saying that there are no other examples of continuous linear functionals if p is o -finite. By contrast, there can be other examples in the case of L W ( X ,p). For example, for the situation in which X is the set of positive integers and A consists of all subsets of X and p is the counting measure, Problems 3 9 4 3 at the end of Chapter V show how to construct a bounded additive set function on A that is not completely additive,
I
1
5. Continuous Linear Functionals on LP
425
and they show how this set function leads to a notion of integration (hence a linear functional) on this L" space; this linear functional is not given by an L function.
Theorem 9.19 (Riesz Representation Theorem for LP). Let ( X ,A, p ) be a o -finite measure space, let 1 5 p < m, and let p' be the dual index to p . If x* is
a continuous linear functional on LP(X, p ) ,then there exists a unique member g of L P ' ( X ,p ) such that r
for all f in LP. For this function g , Ilx*ll
=
llgllp,.
REMARKS.For 1 5 p < oo,Proposition 9.9 shows that LP ( V , p ) is separable if ~1 is a Bore1 measure on an open subset of RN.For this or any other setting in which any of these LP spaces is separable, Alaoglu's Theorem (Theorem 5.58) says that any bounded sequence in LP(V, p)* has a weak-star convergent subsequence. Because of Theorem 9.19 we know what the members of the dual space are. Thus any bounded sequence in LP' has a subsequence that is convergent weak-star against LP. In effect we obtain a nonconstructive way of producing members of LP' . Problem 8 at the end of the chapter will illustrate the usefulness of this technique.
Ur=l
PROOF OF UNIQUENESS. Write X = X n disjointly with p(X,) finite for all n . If Jx f g d p = 0 for all f in LP, then J, IAnx,g d p = 0 for every measurable subset A of X . Taking A successively to be each of the sets where Re g or Img is 2 0 or is 5 0 and applying Corollary 5.23, we see that g is 0 almost everywhere on X , for each n . Hence g is 0 almost everywhere. PROOF OF EXISTENCE IF p ( X ) IS FINITE. Temporarily let us suppose that the underlying scalars are real. Define a set function v on Jt by v ( E ) = x * ( I E ;) v is well defined because every IE is in LP, and v is additive because x* is linear. If En is an increasing sequence of measurable sets with union E ,then limn IE, = IE pointwise, and hence limn I IE - IE, IP = 0 pointwise. By dominated convergence, limn 11 IE - IE, 1 1 , = 0. Thus
Iv(E) - v(E,)I
=
I ~ * ( l a- lan)l 5 Ilx*llllIa
-
Ia,llp,
and the right side has limit 0. By Proposition 5.2, v is completely additive. The set function v is bounded because Iv(E)I = Ix*(IE)I 5 I I X * I I I I / ~=~ ~ ~ I ~ X * I I ( ~ . I ( E ) ) ' I5 P ~lx* ~ l ( j ~ ( ~ ) )and l l Pit, satisfies v << 11, because if jl,(E) = 0, then IE is the 0 function of LP and thus v ( E ) = x * ( I E ) = x*(O) = 0. By the Radon-Nikodym Theorem in the form of Corollary 9.17, there exists an integrable real-valued function g such that v ( E ) = jE g d p for all E , i.e., x* ( 1 ) =
1
for every measurable set P .
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426
By linearity, this equality extends to show that x* (s) = J, sg d p for every simple function s . Let f > 0 be in LP, and choose an increasing sequence {s,} of simple functions > 0 with pointwise limit f . We shall show that f g is integrable and x*(f ) = J f g dw. In fact, let A be the set where g(x) > 0. Then limn 1 f lA- snIAIP= 0 pointwise, and hence limn 11 f lA- snlAllP= 0 by dominated convergence. Since
and since the right side tends to 0, the set { x W ( s n I ~of) }numbers is bounded. Thus the set { JX snIAgd p } of equal numbers is bounded. Since g > 0 on A, the functions .snIAg increase to f IAg, and thus f lAgd p is finite by monotone convergence. In other words, f gf is integrable. Similarly f g- is integrable, and thus f g is integrable. Since limn x * (s, IA) = X * (f IA) and limn J;i s, IAgd p = f IAgd p and since a similar result holds for g - , we conclude that
1,
1,
This conclusion, now proved for f > 0, immediately extends by linearity to all f in LP and completes the verification that x*(f ) = f g d p in the case that the scalars are real. If the scalars are complex, we apply the above argument to the restrictions of Re x* and Imx* to the real-valued functions in LP ,obtaining real-valued functions gl andg2inLp1withRex*(f) = fgldpandImx*(f) fg2dpforall real-valued f . Then x*(f ) = J, f (gl ig2) d p for all real-valued f , and it follows that this same equality is valid for all complex-valued f . Since gl and g2 are in LP, SO is g1 ig2. This completes the verification that x*(f ) = fg d p for a suitable g in the case that the scalars are complex. Finally Proposition 9.8 shows that Ilx* 11 = Ilg lip, and completes the proof of the theorem under the assumption that p(X) is finite.
IX
+
+
=A
1,
PROOFOF EXISTENCE IF p(X) IS o-FINITE. Again we temporarily suppose that the underlying scalars are real. Since /c is 0-finite, we can write X as the increasing union of sets En of finite measure. Let L{ be the set of members of LP that vanish off E n ,and let x; be the restriction of x* to L{ . Find, by the special case just completed, a function g, for each n such that x,"(f,) = JEn fngnd p for all fn in L: . The already proved uniqueness result implies that the restriction of g,+l to En equals gn almost everywhere [dp]. Let g be the measurable function equal to gl on El and equal to g, on En - EnP1if n > 2. Let A be the set where g(x) > 0, and let f > 0 be in LP. Then f IEnnA increases to f I A ,and dominated
6. Marcinkiewicz Interpolation Theorem
convergence implies that limn 11 f IbnnA - f I A l i p pointwise to f I A g ,monotone convergence gives
= 0.
427
Since f IbnnAg increases
the last equality holding since I f IEnnA- f IAll, tends to 0. Hence f g + is integrable. By proceeding similarly with the set where g ( x ) i0 and by writing a general f as f = f + - f -, we conclude that f g is integrable for every f in LP and x r ( f ) = Jx f g d p , provided the scalars are real. Again there is no difficulty in extending the argument to the case that the scalars are complex, and Proposition 9.8 shows that Ilx* 11 = llg lip,.
6. Marcinkiewicz Interpolation Theorem This section concerns linear functions and some almost-linear functions between LP spaces. We saw evidence in Proposition 9.5 that the LP spaces behave collcctivcly likc a wcll-bchavcd family of spaccs. That rcsult spccifically gavc an upper bound for 11 flip in terms of 11 flip, and 11 flip, when pl 5 p 5 p p It turns out that linear functions between pairs of LP spaces satisfy inequalities of a similar sort. There are two classes of results in this direction. Results of the first kind use methods of complex analysis, address bounded linear operators only, and give estimates for a one-parameter family of operators that are sharp at the ends. The l Theorem," whose precise general main result of this kind is the " ~ i e s z Convexity statement and proof we omit. The thrust of the theorem is that if a linear operator T satisfies the two estimates IlT(f)ll,, 5 M l l l f l l p l and IIT(f)ll,, 5 M2Ilf ll,,, then T satisfies also an estimate I T ( f )1 1 , i M 11 f 11, for all pairs (p, q) such that (P1' 1 )lies on the line segment in the (L, I) plane from (1, I)to &). The 9 P 9 PI 91 conclusion gives also some specific information about M. The existence of some M in the Riesz theorem can be obtained in most cases of interest by a corresponding real-analysis result known as the "Marcinkiewicz Interpolation Theorem." We include below a statement of the Marcinkiewicz theorem in general and the proof in a special case of exceptional interest. The Marcinkicwicz thcorcm imposcs somc restrictions on thc pairs ( p l ,q l ) and ( p 2 ,92) that are not needed in the Riesz theorem, but situations that do not
(i,
'The pei.son in question h e i ~ is Marcel Riesz, whose name is associated also with collvergellce of the partial sums of the Fourier series of an LP function in LP for 1 < p < oo. The other mentions of the name "Riesz" in this book, namely in connection with the Rising Sun Lemmaof Section VI1.I and various results known as the Riesz Representation Theorem, refer to Frigyes Riesz.
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428
satisfy these restrictions are of comparatively little interest in applications. In any event, in the situations where the Marcinkiewicz theorem applies, it is only the specific information about M in the Riesz theorem that does not come out of the real-analysis proof of the Marcinkiewicz theorem. Let us mention without proof two consequences of the Riesz Convexity Theorem- the Hausdorff-Young Theorem and Young's inequality. The linear operator T in the Hausdorff-Young Theorem is the Fourier transform F , and the instances of the theorem that we knew previously are when ( p , p') equals (1, m) or ( 2 , 2 ) . The numerology that allows the Riesz Convexity Theorem to apply is that 1
-
P
-
1-t
+ -t
-
1
2
and
1 p'
1-t
+ -2t
- --
00
for the same t : HAUSDORFF-YOUNG THEOREM.If 1 p 2 and if p' is the dual index, then the Fourier transform F , initially defined on the dense subspace L 1( R N )n L 2 ( R N )of L p ( R N ), satisfies
for such f and therefore extends to all of L P ( R N )in such a way that this same inequality holds. If one tries to derive the Hausdorff-Young Theorem from the Marcinkiewicz InterpolationTheorem, one gets only the conclusion IIF( f ) I I p , i M II f l i p without the improvement on the bound: M 5 1. The linear operator T in Young's inequality can be taken to be g H f * g with f fixed in LP . The instances of the inequality that we knew previously are when (q, r) equals (1, p) or ( p ' , m) . The relevant numerology is that
YOUNG'S INEQUALITY. Let p ,q ,and r be three indices > 1 and 5 oo such that $ = 1+ 1- - 1. Then convolution f * g is well defined for P 4 f in L P ( R N )and g in L ~ ( R ~ and ) it,satisfies
Ilf
*gll, i l l f
llpll&ll,.
6. Marcinkiewicz Interpolation Theorem
429
By way of preparation for the statement of the Marcinkiewicz theorem, let (X, A, p) and (Y, B, v) be o-finite measure spaces, and let T be a function from a vector subspace of measurable functions on X, modulo sets2 of p measure 0 , into measurable functions on Y, modulo sets of v measure 0. We say that T is a sublinear operator if I T (f +g) 1 5 I T ( f ) 1 + I T (g) I for all f and g in the domain of T . The two examples of T to keep in mind are the sublinear operator f H f * in RN of passing to the Hardy-Littlewood maximal function, as in Section VI.6, and the linear operator f H Hl f in R1 of forming a certain approximation HI to the IIilbert transform, as in Section VIII.7. More specifically the IIardy-Littlewood maximal function of a locally integrable function f on RN is defined as f*(x)=
sup r n ( ~ , ) - ~ l ~ f ( x - y ) l d y , w h e r e B r = B ( r : 0 ) i n R N , O
and the sublinear operator T is Tf = f * . The approximation HI to the Hilbert transform is defined for f in L' L~ by
+
as the convolution with a fixed L~ function. Let 1 5 p , q 5 oo. We generalize the notion of boundedness of a linear operator between LP(X) and L q Y ) so that we can work with sublinear operators as well as linear ones. A sublinear operator T is said to be of type (p, q) or strong type (p, q) if 11 Tf 11, 5 M 11 f 1 , with M finite and independent o f f . The least M for which this inequality holds is called the norm or operator norm of T . If q < oo,then Chebyshev's inequality from Section VI.10 gives
and for any M such that ITf
1Iq
5 MI f
l i p for all f ,it follows that
If q < oo,a sublinear operator T is said to be of weak type ( p , q ) if it satisfies
his condition means that the domain of T is to be regarded as a vector subspace of measurable functions, except that two functions are identified if they differ only on a set of measure 0.
lX.LP Spaces
430
for some M. In this case the least such M is called the weak-type norm of T . We already encountered the definition of weak type (1, 1) in Section VI.6. If q = oo, the convention is that weak type ( p , cm) is the same as strong type ( p , cm). Consider our two examples. The operation T ( f ) = f * of passing to the Hardy-Littlewood maximal function in RN is of weak type ( 1 , 1 ) by the HardyLittlewood Maximal Theorem (Theorem 6-38),and the evident inequality
shows that f H f " is of type (cm, cm) as well. The linear operator T(f ) = H I f of passing to the approximation H1 to the Hilbert transform in R1 is of weak type (1, 1) and type (2,2)by Theorem 8.25.
Theorem 9.20 (Marcinkiewicz Interpolation Theorem). Let ( X , A, p ) and ( Y , B, v ) be o-finite measure spaces, and let ( p l ,q l ) and ( p z ,q z ) be two pairs of indices between 1 and oo. Suppose that 1 5 pl 5 ql 5 co,1 5 p2 5 q2 5 oo, and pl f p2. Let T be a sublinear operator from LP1( X , p ) LP2(X,p ) to the space of measurable functions on Y modulo sets of v measure 0, and suppose that T is of wcak typcs ( p l , q l ) and ( p 2 , q 2 ) with rcspcctivc wcak-typc norms M I and M2. Fix t with 0 < t < 1, and define ( p , q ) by
+
1
1-t
P
PI
t +-
-- -
P2
and
1
-
4
-
1-t
+ -.q2t
-
41
Then T is of strong type (p, q ) with
with the constant C depending only on t , M I , M2, p l , q l , p2, q2 and with C bounded as a function of t as long as t is bounded away from 0 and 1. Before discussing the proof, let us apply the theorem to our two examples, the Hardy-Littlewood maximal function and the approximation H I to the Hilbert transform. Then let us draw some consequences of these applications. As was said before the statement of Theorem 9.20, the sublinear operator f H f * is of wcak typc ( 1 , 1 ) and strong typc ( 2 , 2 ) . Thc thcorcm immcdiatcly givcs thc following corollary.
Corollary 9.21. If 1 ip 5 oo,then there exists a constant Ap such that the Hardy-Littlewood maximal function satisfies
for all f in L p ( R N ) .
6. Marcinkiewicz Interpolation Theorem
43 1
The case of this result in one dimension implies something in N dimensions that we have not obtained earlier. If f is locally integrable on RN, one says that strong differentiation holds for f at x if lim centered at x
A consequence of Corollary 9.21 is that strong differentiation holds almost evelywhere for each f in LP(RN) for p > 1. The proof is outlined in Proble~ns 13-15 at the end of the chapter. By contrast, it is known that there are functions in L1(RN) for which strong differentiation fails everywhere. In the second example the operator H I that approximates the Hilbert transform is of weak type (1, 1) and strong type (2,2), and Theorem 9.20 allows us to conclude that it is of strong type ( p , p) for 1 < p 5 2. But we can do better. The operator H1 is convolution by the function hl with hl (x) = l / ( n x ) for 1x1 2 1 and hl(x) = 0 for 1x1 < 1. The function hl is in LP for all p > 1, and Proposition 9.10f shows that h 1 * f is well defined as a bounded continuous function whenever f is in some L-ith 1 < q < oo. Thus H1 is defined on all LP classes for 1 < p < m , and a general result that we prove below as T,emma 9.22 shows that an ineclilality 11 H1f l i p i ApII f Il .' for all f in T,P implies 11 HlgIP, 5 A, lldl,, for all g in LP', provided p' is the dual index to p and 1 < y < w. Thus the bouridedrless result for H1 on LP extends to 1 < y < oo. Next, we define the dilate h,(x) = sP1hl(x) in the usual way and put HEf = h, * f . We shall see for every E > 0 that II HEf 1 , i Apll f 1 , with the same constant Ap, and finally we shall see that we can let e decrease to 0 and obtain the Hilbert transform H as a well-defined linear operator on all LP classes for 1 < p < oo;the estimate is 11 Hf 11, 5 Ap 11 f l i p , again for the same Ap. Problems 20-22 at the end of the chapter indicate how to use this boundedness to prove that the Fourier series of any LP function on [-n, n ] converges to the function inLPifl
lX.LP Spaces
432
PROOF.For any function F on R N , define ~ ' ( x = ) F ( - x ) and observe that = 11 F 11, for 1 5 r 5 oo. If g is an integrable simple function, then ( h # * g ) ( x ) = h ( y - x ) g ( y ) d y = h ( - y - x ) g # ( y ) d y = ( h * g # ) ( - x ) . Thus this g and an integrable simple function f together satisfy
11 ~ ' 1 1 ,
and
1
1
*
I@#* g ) ( - y ) f ( y ) d y = j"S h # ( - y x ) g ( x ) f ( y )d x d y = 11h ( x + y ) g ( x ) f ( y ) d x d y .
S ( h g U ) ( y ) f( y ) d y =
-
Because f and g are in every LY class, the right sides of these two displays are finite when absolute value signs are inserted in the integrands. Thus Fubini's Theorem applies and shows that the two right sides are equal. Combining this fact with Holder's inequality and the hypothesis about h , we obtain
whenever f and g are integrable simple functions. If a general fo in LP is given, wc can find a scqucncc f, ofintcgrablc simplc functions such that 11 f,- fo 11, + 0 , and we apply this inequality to each f , . Then the left side of the inequality tends to j"(h * R') (y),fo( y ) d y and the right side tends to A, I I .fo 11, llgll,, . Taking the supremum over all fo with 11 fo 1 1 , 5 1 and applying Proposition 9.8, we find that
I
llh * g'll,,
1,
5 Apllgll,,
=
~ ~ l l g ' l l , , .In other words,
for every integrable simple function g,, . For a general g in LP' ,choose a sequence of integrable simple functions g, with 11 g, - g lip, + 0. Since h is in LP ,it follows from Proposition 9.10f that h * g, converges to h * g uniformly. On the other hand, the inequality I l h * ( g , - g,) Il, i A p llg, - gn Il, shows that { h * g,} is Cauchy in Lp'. By Theorem 5.58, { h* g,} converges to some function in Lp' and has an almost-everywhere convergent subsequence to this function. Since h * g, converges uniformly to h * g , we conclude that h * g, converges to h * g in L P ' . Therefore Il h * g lip, 5 A p llgIp,, and the proof is complete. Again let h l be the function on R1 equal to l / ( n x ) for 1x1 > 1 and equal to 0 for Ix < 1. This is in L r ( R 1 )for every r > 1. Our operator giving an
6. Marcinkiewicz Interpolation Theorem
433
approximation to the Hilbert transform is Hl f = I z l * f . Using our results from Chapter VIII along with the Marcinkiewicz Interpolation Theorem, we saw earlier in this section that H1 satisfies 11 Hl f l i p i A p 11 flip for 1 < p i 2 and all f in Lp ( R 1 ) .Lemma 9.22 shows that this inequality remains valid for 1 < p < cm. From this result we can extend the Hilbert transform to L p ( R 1 ) for all p with 1 < p < m , a s follows. Theorem 923. Let 1 ip
cm, let
i
and define H, f = h , * f for f in LP and e > 0. Then (a) there exists a constant Ap independent of e such that 11 H, f 11, 5 Ap 11 flip for all f in L P , (b) the limit H f ( x ) = lim exists in LP for every f in LP, (c) the operator H satisfies 11 Hf 11, 5 A p 11 flip for every f in Lp. PROOF.Convolution with h , is well defined on LP because h , is in L P ' , p' being the dual index for p . The three computations H, f ( x ) = ( f
* h,)(x) = 1 f
= J e-l f,-1 (e-lx
and
l lg,-1 ( x )l P
-
( x - y)e-'hl(e-'y) d y = l f ( x - ~ Y ) ~ I (dYy ) y)hl ( y )d y = € - ' ( H I f,-i)(eP1x),
1
d x = eP Ig(ex) l P d x = eP1+pl Ig(x) l P d x
allow us to write
This proves (a), the constant A p being any constant that works for H I . In Lemma 9.24 below we show by a direct computation that (b) holds for the dense subset of C functions f of compact support. Let us deduce (b) for general f in LP from this fact and (a). In fact, if we are given f , we choose a sequence f, in the dense set with fn + f in LP. Then
lX.LP Spaces
434
Choose n to make the first and third terms small on the right, and then choose e and e' sufficiently close to 0 so that the second term on the right is small. The result is that HE,,f is Cauchy in LP along any sequence en tending to 0. This proves (b), apart from the direct computation for the dense subset. In (b), we proved that H,f 3 H f in LP. Then (a) gives llHfllp = lim,&*11 H, f l i p 5 lim supEJoA p 11 f l i p = A p 11 f 11 P' This proves (c) and completes the proof of Theorem 9.23 except for the following lemma.
Lemma 9.24. If f is a c function of compact support on R1,then lim exists uniformly and in LP for every p > 1. PROOF.Let 11 . 11 denote the supremum norm or the LP norm. By the Cauchy criterion it is enough to show that
tends to 0 for the above interpretations of 11 . 11 as el and e2 tend to 0. Since If '(u) I 5 M , use of the Mean Value Theorem on Re f and Im f shows that I f ( x - t ) - f ( x ) I 5 2 M l t I . Supposethat0 ( € 1 5 e 2 5 1. IfEisacompactset containing the sum of any member of the support of f and any x with Ix I 5 1 , then it follows that
The right side tends to 0 as complete.
el
and
e2
tend to 0, and the proof of the lerrlrrla is
Having now completely proved Theorem 9.23, let us return to a discussion of the proof of the Marcinkiewicz theorem, Theorem 9.20. The proof is considerably simplified by assuming that ql = pl and q 2 = p2, which happens to be the special
6. Marcinkiewicz Interpolation Theorem
435
case of most interest to us, and we shall give a proof only under this additional hypothesis. The idea in the special case will be to estimate integrals of powers of functions by using Proposition 6.56b to reduce the estimates to facts about distribution functions. The proof in general has the same flavor as the argument we give, but it involves also a subtler decomposition of f into two parts, a nonobvious application of Holder's inequality, and a clever use of Proposition 9.8. PROOF OF THEOREM 9.20 WHEN pl = ql < p2 = 42.We divide matters into two cascs, thc first whcn p2 ioo and thc sccond whcn p2 = oo. We begin with the case with p2 < oo. Let
be the distribution function of T f as in Section VI.lO. Proposition 6.56b shows that 00 I I T ~ I=I P ;
1
C ~ - ~ A ( 16 C) =2 ' ~
I"
dl.
(*I
With $ > O fixed, we shall estimate h ( 2 $ ) . We decompose f as f = fl with f2(x) =
{;(XI
otherwise
+ f2
if I f ( X I I i 8 otherwise
}
Jilst as in the proof of Proposition 9.4c, fl is in T,pl ( X , ! I ) and f2 is in T Y 2 ( X ,1 1 ) Because f = fl fi, sublinearity of T gives IT f I 5 I T f l I IT f2 . If h l and h z are the distribution functions of T f l and T f 2 and if a > 0 is given, then
+
+
because IT f 1 can be > 2a only if at least one of I Tfl I and IT f2 1 is > a . For every a > 0 , the assumption that T is of weak types ( p l , p l ) and ( p 2 ,p2) gives US
)
Mlllf1llp1 h 1 ( ~ ) 5 ( For a
=
6, we therefore obtain
and
hl(u)c(
M211f211p,
). P2
IX.LP Spaces
436
With the estimate for h ( 2 6 ) in hand, we can now let 6 vary and estimate 11 T f From (*) and (**) we obtain 11 Tf 11; 5 Il 12, where
+
1,
h =~PPM?
and
6p-p2-1
/
11; .
i f ( x ) l P 2d p ( x ) d l .
(1.fcxil55)
Fubini's Theorem gives
Similarly
and thus I I T11; ~5 C P I f~ 11; as required. The remaining case to handle has p2 - GO. The general line of the argument is the same as above, but there are small differences. With 6 fixed, the definitions of f l and f2 are adjusted to be
f1(x) = and f i = f
-
otherwise,
f l . Then I I fill, 5 ~ / I I T I I , ll~fiII, , 5 6,andh2(6)
Hence
and then the proof can proceed along the lines above.
7. Problems 1. For a measure space of finite measure, prove that LP More particularly prove, for the case that the total m whenever p > q > 1.
C L4 whenever p > q > 1. e is 1,that I f 11, 5 11 f 1 1 ,
2. Let p , q , r be real numbers in [ I , +oo] with 1+ 14 + P
+ $ = 1 and Holder's inequality,prove thatsX I f g h l P
=
1. Using the equality
d p 5 1 f II,Igll,llhll,.
437
7. Problems
3. For a measure space of finite measure, let { f , } be a sequence of measurable functions converging pointwise to f . Suppose that 1 5 q < p < oo,and suppose that the sequence of numbers { l l f l i p } is bounded. Using Egoroff's Theorem (Problem 17, Chapter V) or uniform integrability (Problem 21, Chapter V), prove that f, + f in Lq . 4.
This problem produces an example of a measure space in which two distinct members of Lm act as the same linear functional on L ' . The measure space ( X , A, p ) has X co~~sisti~lg of a single point y , A = 10, X } , and p ( X ) = +o. (a) Show that dim L (X) = 0 and dim Lo' ( X ) = 1. (b) Proposition 9.8 assumed a-finiteness to ensure its conclusion when p = oo. Show that the conclusion of Proposition 9.8 fails for p = ca in this example.
5. If f is 1-eal-valuedand integi-able on the measme space ( X , A, p ) , what ai-e all the Hahn decompositions for the signed measure v ( E ) = f dp?
SE
6. Provide examples of each of the following. Each example can be produced on one of the following three algebras of subsets of a set X : the finite subsets of a X and their complements, all subsets of a countable set X, the Bore1 sets of x = [O, 11. (a) An additive set function v on an algebra of sets with Iv(X)I < cc but with sup, Iv(E)I = ca. (b) A counterexample to the Hahn decomposition if the assumption "a-algebra" is relaxed to "algebra" but the other assumptions are left in place. (c) A finite measure 1 1 and anon a -finite measure p , both defined on a a -algebra, such that v << p but v is not given by an integral with respect to p . Problems 7-8 concern harmonic functions and the Poisson integral formula for the unit disk in JR2. These matters were the subject of Problems 27-29 at the end of Chapter I, Problems 14-15 at the end of Chapter 111, Problems 10-13 at the end of Chapter IV, and Problems 18-20 at the end of Chapter VI. Problem 7 updates the results from Chapter VI so that they apply for 1 5 p < ca,and Problem 8 uses weak-star convergence to establish a converse result.
&
7.
If 1 5 p < oo and if f is in L" (1-n, nj, d o ) , prove that the Poisson integral u(r, 0 ) of f has the properties that 11 u(r, . ) l i p 5 11 flip for 0 5 r i1 and that u(r, . ) tends to f in LP in the sense that limrTlIlu(r, .) - f 11, = 0.
8.
Suppose that 1 < p' 5 oo and that u(r, Q ) is a harmonic function on the open unit disk such that up^,^.,^ Ilu(r, .)IIP,is finite. By using Problem 13 at the end of Chapter IV and taking a weak-star limit of a suitable sequence of functions u(r,, 8 ) with {r,} increasing to 1, prove that u ( r , 0) is the Poisson integral of a function in LP' ([-n,XI, d ~. )
&
Problems 9-12 concern decomposing any bounded nonnegative additive set function on an algebra into acompletely additive part and a "purely finitely additive" part. They
IX.LP Spaces
438
make use of Zorn's Lemma (Section A9 of the appendix). A bounded nonnegative additive set function p will be called purely finitely additive if there is no nonzero completely additive set function v such that 0 5 v ( E ) 5 p ( E ) for all E . 9. Suppose that p is an additive set function on the cr-algebra of all subsets of the integers such that p has image 10, l } and p ( { n } ) = 0 for every integer n . Prove that ~c is purely finitely additive. (Such a ~c was constnlcted by means of a nontrivial ultrafilter in Problems 3 9 4 1 at the end of Chapter V.)
10. Ube Zurrl'b Lsrrirrra Lu bhuw Llral any buur~dsdnur~ncgalivt:addilivt: bsl ~ U I I L L ~ U I I is the sum of anonnegative completely additive set function and a purely finitely additive set function.
1 1. Prove that if v is a bounded nonnegative completely additive set function and if p is bounded nonnegative and purely finitely additive with 0 5 p ( E ) 5 v ( E ) for all E , then p = 0. 12. Deduce from the previous problem and the Jordan Decomposition Theorem that the decomposition of Problem 10 is unique. Problems 13-15 prove the theorem, for the case of JR2, of Jessen-MarcinkiewiczZygmund concerning strong differentiation of integrals of LP functions almost everywhere when p > 1. Strong differentiation holds at (x, y) for the locally integrable function f on IFt2 if lim
diam(R)+O, m(R) R =eeometric rectan~le
J f ( u , .Id u d s
= f ( x , y)
Let f ** be the associated maximal function, given by
f * * ( x ,Y ) =
SUP diam(R)+O. R=geometric rectangle centered at (x,y)
I l f[ (u,u)dudu. m(R) R
13. Let f l ( x , y) be the value of the one-dimensional Hardy-Littlewood maximal function of y H f ( x , y ) , and let f 2 ( x , y) be the value of the one-dimensional Hardy-Littlewood maximal function of x H fl ( x , y). Prove that f ** ( x , y ) 5 f2(x, Y ) . 14. Using Corollary 9.21 and the previous problem, prove that I I f ** 11, 5 A; I I f l l , if 1
15. Conclude that strong differentiation holds almost everywhere for each f in L P ( R ~i f) 1 < p 5 GO. Problems 16-19 concern the Hilbert transform H defined in Section VI11.7 and Theorem 9.23. The operator H is defined on L P ( R ' ) for 1 < p < GO. Recall the functions h,, Q,, and $, on R' satisfying Q, = h, $s. Let f be in LP, and let f * be the Hardy-Littlewood maximal function of f .
+
439
7. Problems
16. Prove that there exists a continuous integrable function @ > 0 on JR1 of the form Q ( x ) = Qo(lxl),where Qo is adecreasing function on [0,G O )such , that the function $, for s = 1 satisfies I $rl 1 5 a.
c1
17. Deduce from the previous problem and Corollary 6.42 that I (1Cr,* f ) ( x )1 5 Cf * ( x ). How does it follow that limEJo($, * f ) ( x ) = 0 almost everywhere for all f in LP, 1 ( p ( GO?
1 8. Prove that Q, * f = PC * ( H f ) for f P ( x , s ) is the Poisson kernel.
F
LP with I < 11 < m, where P,(x) =
19. Deduce from the previous two problems that the limit in the equality H f ( x ) = lim-
f(x
-
t)dt
of Theorem 9.23 may be interpreted as an almost-everywhere limit if f is in L P ( R ' ) and 1 < p < GO. Problems 20-22 prove the theorem of M. Riesz that the partial sums of the Fourier series of a function in LP([-n, n])converge to the function in LP if 1 i p ioo. Recall from Sections 1.10 and V1.7 that if f is integrable on [ - n , n ] , then the nth partial sum of the Fourier series of f is given by ( S , f ) ( x ) = ( D , f ) ( x ), where sin(,++)t D, is the Dirichlet kernel D, ( t ) = and the convolution is taken relative to
*
sln ?t
20. Suppose it can be proved that llStlf l i p i Ap 11 flip for 1 < p < 'LXI with Ap independent of n and f . Prove that S, f + f in LP for all f in LP, provided 1ipioo. 2sin(n+' t 1 21. Define E,(t) = for i t l 5 n and E,(t) = 0 for It < =. Then extend E,(t) periodically. Show that D, - En = p, is integrable on [ - n , n]with IIp, I l l 5 C independently of n , and say why it is therefore enough to prove that the operators T, with T, f = En f satisfy 11 T, f l i p ( Bp 11 f 11 for 1 < p < oo with Bp independent of n and f .
*
22. In E, ( t ) , write sin(n + 4 ) t as a linear combination of two exponentials eikt, rewrite each exponential as epik(xp")eikx , and decompose the operator T, as the corresponding sum of two operators. By relating these two operators separately to the operators H, in Theorem 9.23, prove that the T,'s satisfy the desired estimate IIT,f l l , 5 Bpllf 11,. Problems 23-26 develop a kind of function-valued integration known as conditional expectation in probability theory. They make use of the Radon-Nikodym Theorem (Theorem 9.16). Let ( X , A, p ) be a measure space with p ( X ) = 1.
IX.LP Spaces
440
23. If f is integrable and if B is a a-algebra contained in A, prove that there exists a function E [f IB] that
(i) is measurable with respect to B and (ii) has f d p = E [f IB] d p for all B in B. Show further that E [f IB] is unique in this sense: any two functions satisfying
1,
1,
(i) arid (ii) diClcr vr~lyWII a be1 ir~8 u l p rricaburc 0.
24. Suppose that X is a co~~ntable disjoint union of sets X, in A and that B consists of all possible unions of the X , 's . Give an explicit formula for E [f 1 B] . 25. Show that if B
= A, then
E [f IB] = f almost everywhere.
26. Let 8 and C be a-algebras with C C B C A. Prove the following: almost everywhere. (a) E [ E [f B ] I C] = E [ f (b) If f and g are integrable and everywhere finite, then E [ f +g 1 Bl = EE[f PI+ E[glBl almost everywhere. (c) If g is measurable with respect to B and if f and fg are integrable, then E [ f g 1 B] = g E [ f lB] almost everywhere. (d) I f f a n d g a r e i n ~ ~ ( ~ , A , p ) , t h e nf SE, [ g l B I d p = S , E [ f l @ g d p .
CHAPTER X Topological Spaces
Abstract. This chapter extends considerably the framework for discussing convergence, limits, and continuity that was developed in Chapter 11: topological spaces replace metric spaces. Section 1 makes various definitions, including definitions for the terms topology, open set, closed set, continuous function, base for a topology, separable, and subspace. It introduces two general kinds of constructions useful in analysis and other fields for forming new topological spaces out of old ones-weak topologies and quotient topologies. The section gives several examples of each. Sections 2-3 develop standard facts, mostly elementary, about how certain combinations of properties of topological spaces imply others. Examples show some limitations to such implications. Properties that are studied include Hausdorff, regular, normal, dense, compact, locally compact, Lindelof, and a-compact. Section 4 discusses product topologies on arbitrary product spaces, an example of a weak topology. The main theorem, the Tychonoff Product Theorem, says that the product of compact spaces is compact. Section 5 introduces nets, a geneialization of sequences. Sequences by themselves are inadequate for detecting convergence in general topological spaces, and nets are a substitute. The use of nets in many cases provides an easier way of establishing properties of subsets of a topological space than direct arguments with open and closed sets. Section 6 elaborates on quotient topologies as introduced in Section 1. Conditions under which a quotient space is Hausdorff are of particular interest. Sections 7-8 prove and apply Urysohn's Lemma, which says that any two disjoint closed sets in a normal topological space may be separated by a real-valued continuous function. This result is fundamental to serious uses of topological spaces in analysis. One application is to showing that every separable Hausdorff regular topology arises from a metric. Section 9 extends Ascoli's Theorem and the Stone-Weierstrass Theorem from their settings in compact metric spaces in Chapter I1 to the wider setting of compact Hausdorff spaces.
1. Open Sets and Constructions of Topologies In applications involving metric spaces, we have seen several times that the explicit form of a metric may not at all be one of objects of interest for the space. Instead, we may be interested in the open sets, or in convergence, or in continuity, or in some other aspect of the space. The same open sets, convergence, and
442
X . Topological Spaces
continuity may come from two different metrics, and we have even encountered notions of convergence that are not associated with any metric at all. We saw in Section 11.5,for example, that we could associate three different natural-looking metrics to the product X x Y of two metric spaces, and the three metrics led to the same open sets, the same convergence of sequences, and the same continuous functions On the other hand, the notions in Chapter V of pointwise convergence, convergence almost everywhere, and weak-star convergence were defined without reference to a metric, and depending on the details of the situation, there need not be metrics yielding these notions of convergence. We have brushed against further, more subtle situations with one or the other of these phenomena-no special distinguished metric or no metric at all-but there is no need to produce a complete list. The present chapter introduces and studies an abstract generalization of the notion of a metric space, namely a "topology," that makes it unnecessary to have the kind of explicit formula demanded by the definition of metric space. The framework for a "topological space" consists of a nonenlpty set and a collection of "open sets" satisfying the conditions of Proposition 2.5. Thus let X be a nonempty set. A set 7 of subsets of X is called a topology for X if (i) X and 0 are in I; (ii) any union of members of 7 is a member of 7, (iii) any finite intersection of members of 7 is a member of 7 The members of Tare called open sets, and (X, I)is called a topological space. When there is no chance for ambiguity, we may refer to X itself as a topological space. Every metric space furnishes an example of a topological space by virtue of Proposition 2.5; we refer to the topology in question as the metric topology for thc spacc. Two othcr cxamplcs of gcncral constructions lcading to topological spaces will be given later in this section, and some specific examples of other kinds will be given in Section 2. Neighborhoods, open neighborhoods, interior, closed sets, limit points, and closure may be defined in the same way as in Section 11.2. As remarked after Corollary 2.1 1, the proofs of certain results relating these notions depended only on the definitions and the three properties of open sets listed above. These results are Proposition 2.6 and Corollary 2.7 characterizing interior, Proposition 2.8 giving properties of the family of all closed sets, Proposition 2.9 relating closed sets to limit points, and Proposition 2.10 and Corollary 2.1 1 characterizing closure. Thus we may take all those results as known for general topological spaces, and it is not necessary to repeat their statements here. The notion of continuity extends to topological spaces in straightforward fashion. Specifically the definition of continuity at a point is extracted from the statement of Proposition 2.13: if X and Y are topological spaces, a function
I . Open Sets and Constructions of Topologies
443
X + Y is continuous at a point x E X if for any open neighborhood V of f ( x ) in Y , there is a neighborhood U of x such that f ( U ) g V . Then Corollary 2.14 is immediately available, saying that if f : X + Y is continuous at x and g : Y + Z is continuous at f ( x ) , then the composition g o f is continuous at x . Proposition 2.15 and its proof are available also, saying that the function f . X + Y is continuous at every point of X if and only if the inverse image under f of every open set in Y is open in X ,if and only if the inverse image under f of every closed set in Y is closed in X . We say that f : X ; . Y is continuous if these equivalent conditions are satisfied. The function f : X + Y is said to be :Y X a homeomorphism i f f is continuous, f is one-one and onto, and f is continuous. The relation "is homeomorphic to" is an equivalence relation. Now let us come to the two general constructions of topological spaces, known as "weak topologies" and "quotient topologies." Both of these have many applications in real analysis. The notion of "weak topology" starts from the fact that the intersection of a nonempty collection of topologies for a set is a topology; this fact is evident from the very definition. The prototype of a weak topology is the "product topology" for the product of a nonempty set of topological spaces. In the terminology of Scction A1 of thc appcndix, if S is a noncmpty sct and if X, is a noncmpty sct for each s in S , then the Cartesian product X = X s,sX, is the set of all functions f from S into Us,, X , such that f (s) is in X , for all s E S. Now suppose that each X , is a topological space, and let p, : X + X , be the sth coordinate function, given by p , ( f ) - f ( s ). If X is given the discrete topology 23,in which every subset of X is open, then each p , is continuous; in fact, the inverse image of an open set in X , is some subset of X , and every subset of X is in D. Form the collection of all topologies I, on X such that each p , : X + X, is continuous relative to I,.The collection is nonempty since D is one. Let 7 b e their intersection. The inverse image of any open set in X , under p , lies in 7, for each a and hence lies in 7 Therefore each p , is continuous relative to 7 We speak of T a s the "weakest topology" on X such that all p , are continuous, and this topology for X is called the product topology for X . We shall study product topologies in more detail in Section 4. More generally let X be a nonempty set, let S be a nonempty set, let X , be a topological space for each s in S , and suppose that we are given a function f , : X + X , for each s in S . If X is given the discrete topology, then every f , is continuous. Arguing as in the previous paragraph, we see that there exists a smallest topology for X making all the functions f , continuous. This is called the weak topology for X determined by { fs}s,S.
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EXAMPLES. (1) Let ( X , d ) be a metric space. Then the weak topology for X determined by all functions x H d ( x , y ) as y varies through X is the usual metric topology on X , as we readily check from the definitions.
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(2) Let X be a normed linear space with field of scalars IF, such as an LP space for 1 5 p 5 oo,and let X* be the vector space of continuous linear functionals on X, as introduced in Section V.9. (For X = LP with 1 5 p < 00 and with the assumption that the underlying measure is o-finite, Theorem 9.19 identified X* explicitly as LP',where p' is the dual index to p.) Each member x of X defines a function f, : X* + IF by the formula f, (x*) = x*(x). The weak topology on X* determined by X is called the weak-star topology on X* relative to X. The words "relative to X" are included in the terminology because two normed linear spaces X might have the same set X* of continuous linear functionals. In Section V.9 we introduced a notion of weak-star convergence but no iiietric associated to it. In problems at the ends of Chapters VI, VIII, and IX, this kind of convergence became a powerful tool for working with harmonic functions, Poisson integrals, and positive definite functions. Later in the present chapter we shall relate topologies to convergence of sequences,' and it will be apparent that weak-star convergence as defined in Section V.9 is the appropriate notion of convergence for the newly defined weak-star topology. (3) The construction in Example 2 can be transposed to other situations in which a topology is to be imposed on a vector space. For example, let X be a normed linear space with field of scalars IF equal to R or @, and let X* be the vector space of continuous linear functionals on X. Then X* indexes a set of functions x* : X + IF. The weak topology on X determined by X* is known as the weak topology on X. This topology arises in some advanced situations, but we shall not have occasion to make use of it in the present volume.
(4) We have encountered three vector spaces of scalar-valued smooth functions on open sets of Euclidean space-in Section 111.2the space C a 3 ( U )of all smooth functions on U ,in Section VIIIA the space Cz,,, ( U ) of all smooth functions on U with compact support contained in U , and in Section VIIIA the space S ( R N )of Schwartz functions defined on RN.The subject of partial differential equations makes extensive use of functions of all three of these kinds, and it is necessary to be able to discuss convergence for them. The easiest convergence to describe is for C m ( U ) ,where convergence is to mean uniform convergence of the function and all of its partial dcrivativcs on cach compact subsct of U . Uniform convcrgcncc by itself is captured by the supremum norm, and somehow we want to work here will1 llle suprcrriumi nonns uf llle furlclium arld each u l ils parlial dcrivalivcs on each compact subset. The appropriate topology turns out to be the weak topology determined by all the functions f H 11 f - g 11 , where 11 . 11 is the supremum of some iterated partial derivative on some compact subset of U . This construction is carried out in detail in the companion volume Advanced Real Analysis. A topology for the Schwartz space S ( R N )is obtained in a qualitatively similar way. And to "nets :' which are a generalization of sequences
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A topology for C & ( U ) is more subtle, and it too is constructed in the companion volume.
The second general construction of topological spaces is the "quotient topology" for the set of equivalence classes on X when X is a topological space and sorrle equivalence relation2 llas been specified 011 X.If the relation is written as -, the set of equivalence classes may be written as X/-, and the quotient map, i.e., passage from each member of X to its equivalence class, is a well-defined function q : X + X/-. With a topology in place on X ,define a subset U of X/ to be open if q - l ( ~ is ) open. Since inverse images of functions preserve set-theoretic operations, it is immediate that the resulting collection of open subsets of X/ is a topology for X/ and that this topology makes q continuous. This topology is called the quotient topology for X / -. In any other topology 7'on X / -, any subset V of X/ that is open in I'but not open in the quotient topology must have the property that q - l ( ~ is ) not open; this condition implies that q is not continuous when 7'is the topology on X/ -. Therefore the quotient topology is the finest topology on X/-- that makes the quotient map continuous-in the sense that it contains all topologies making q continuous.
-
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-
-
(1) Let (X, d) be a pseudometric space such as the set of all integrable functions on some measure space (S, A , / I ) with d ( g , h ) = js /;.lg- hl d j ~ The . pseudometric on X gives X a topology. For x and y in X , define x y if d ( x , y ) = 0 . The result is an equivalence relation, and we know from Proposition 2.12 that the pseudometric d descends to be a metric on the set X/-- of equivalence classes. The quotient topology on X/ coincides with the topology defined by this metric.
-
-
(2) Let X be the interval [-n, n]with its usual topology from the metric on R, let S 1 be the unit circle in @ with its usual topology from the metric on @, and let q : X + S' be given by q ( x ) = eix. We can consider S' as the set of equivalence classes of X under the relation that lets n and n be the only nontrivial pair of elements of X that are equivalent. The function q is continuous, and it carries compact sets to compact sets. In Problem 11 at the end of the chapter, we shall see that q exhibits S' as having the quotient topology.
(3)Let X be the line R with its usual metric, let S 1 be the unit circle as in the previous example, and let q : X + S' be given by q (x) = eix. The domain X is a group, and the function q identifies S 1 set-theoretically as the quotient group R/2nZ,where Z is the subgroup of integers. This example illustrates the natural 'Equivalence relations and their connection with equivalence classes are discussed in Section A6 of the appendix.
446
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topology to impose on any quotient of a group when the group has a topology for which all translations are homeornoq~hisms.~ In many situations the problem of describing what sets are to be open sets in a topological space is simplified by the notion of a base for a topology. By a base B for thc topology 7on X is mcant a subfamily of mcmbcrs of 7such that cvcry member of 7 i s a union of sets in B. In Chapter I1 the topology for a metric space was really introduced by specifying that the farrlily of all open balls is to be a base. Arguing as with Proposition 2.31, we obtain the following result. Proposition 10.1. A family B of subsets of a nonempty set X is a base for some topology 7 on X if and only if (a) X = U B E B R and (b) whenever U and V are in B and x is in U f' V ,then there is a B in B such thatxisinBandB G U n V.
In this case the topology 7 is necessarily the set of all unions of members of B, and hence 7 is determined by B. A family B of subsets of X is a base for a on X if and only if (a) holds and particular given topology lo (b') for each x E X and member U of 3 containing x ,there is some member B of I? such that x is in B and B is contained in U. REMARK.Condition (b) is satisfied if B is closed under finite intersections. Thus any family of subsets of X that is closed under finite intersections and has union X is a base for some topology on X . A topological space ( X , 7 ) is said to be separable if 7has a base consisting of only countably many sets? A separable metric space has a countable base consisting entirely of open balls. As with metric spaces, there is a natural definition of subspaces for general topological spaces. If ( X , 7 ) is a topological space and if A is a nonempty subset of X , then the relative topology for A is the family of all sets U n A with U in 7 We can write I n A for this family. It is a simple matter to check that I n A is indeed a topology for A , and we say that ( A , If' A ) is a topological subspace of ( X , 7 ) . If there is no possibility of confusion and if the relative topology is understood, we may say that "A is a subspace of X ." 3 ~ h definition e of "topological group," which is given in the companion volume Advanced Real Arxalysis, imposes further conditions beyond the fact that every translation is a homeomorphism. 4 ~ o m eauthors use the word "separable" to mean that X has a countable dense set, but the meaning in the text here is becoming more and more common. The existence of a countable dense set is not a particularly useful property for a general topological space.
2. Properties o f Topological Spaces
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Proposition 10.2. If A and B are subspaces of a topological space X with B g A g X , then the relative topology of B considered as a subspace of X is identical to the relative topology of B considered as a subspace of A. PROOF.The relative topology of B considered as a subspace of X consists of all sets U n B with U open in X , and the relative topology of B considered as a subspace of A consists of all sets (U n A) n B with U open in X . Thus the result follows from the identity (U n A ) n B = U n ( A n B ) = U n B . The next two propositions are proved in the same way as Proposition 2.26 and Corollary 2.27.
Proposition 10.3. If A is a subspace of a topological space X , then the closed sets of A are all sets F n A , where F is closed in X. Consequently B is closed in A if and only if B = B"' n A . Proposition 10.4. If X and Y are topological spaces and f : X + Y is continuous at a point a of a subspace A of X , then the restriction f : A + Y is continuous at a . Also, f is continuous at a if and only if thc function fo : X + f ( X ) obtained by redefining the range to be the image is continuous a1 a .
,1
2. Properties of Topological Spaces Proposition 2.30 listed certain properties of metric spaces as "separation properties." These properties are not shared by all topological spaces, and instead we list them in this section as definitions. After giving the definitions, we shall examine implications among them and some roles that they play. The disproofs of certain implications provide an opportunity to introduce some further examples of topological spaces beyond those obtained from the constructions in Section 1. Let (X, 7 )be a topological space. We say that (i) X is a TI space if every one-point set in X is closed, (ii) X is Hausdorff if for any two distinct points x and y of X , there are disjoint open sets U and V with x E U and y E V , (iii) X is regular if for any point x E X and any closed set F g X with x $ F, there are disjoint open sets U and V with x E U and F g V , (iv) X is normal if for any two disjoint closed subsets E and F of X , there are disjoint open sets U and V such that E U and F 5 V .
448
X . Topological Spaces
Proposition 2.30 listed one further property of an arbitrary metric space X, namely that any two disjoint closed sets can be separated by a continuous function from X into [0, 11. Urysohn's Lemma in Section 7 will establish this property for any normal topological space.
Proposition 10.5. If (X, 7 )is a topological space, then (a) X is TI if and only if for any pair of distinct points x and y, there are opensets U and V suchthatx E U , y @ U , x @ V,andy E V, (b) X is regular if and only if for any point x and any closed set F with x $ F , there is an open set U such that x E U and UC1n F = M, (c) X is normal if and only if for any pair of disjoint closed sets E and F , there is an open set U such that E g U and uC'n F = a. PROOF. If X is T1 and if x and y are given, we can choose U = {ylCand V = {x}". In the reverse direction, if x is given, choose, for each y # x , an open set Vy such that x @ V, and y E V,; then {x}" = U, Vy is open, and hence {x} is closed. If X is regular and if x and F are given, we can choose disjoint open sets U and Vwithx E U a n d F 5 V. ThentheclosedsetVChasVc2 U a n d V c n F = 0; therefore also Vc 2 u"' and u"' n F = 0.In the reverse direction, suppose that x and F are given and that U is an open set with x c U and UC' n F = 0; choosing V = ( u ~ ' ) we ~ , see that x E U , F 5 V , and U n V = 0. If X is normal and if E and F are given, we call choose disjoint open sets U and VwithE g U a n d F g V.ThentheclosedsetVChasVC2 U a n d V C n F = M; therefore also Vc ; ,u"' and u"' n F = 0. In the reverse direction, suppose that E and F are given and that U is an open set with E 5 U and UC' n F = 0; choosing V = (u"')", we see that E g U , F g V ,and U f' V = M.
Proposition 10.6. If (X, 7 )is a topological space and (a) if X is T1 and normal, then X is regular, (b) if X is T1 and regular, then X is Hausdorff, (c) if X is Hausdorff, then X is T 1 . ~ O O F In . (a), if x and a disjoint closed set F are given, then [xJ is closed, and the fact that X is normal implies that we can separate the closed sets {x} and F by disjoint open sets. I11 (b), if x and y are distinct points in X, then {y}is closed and the fact that X is regular implies that we can separate the point x and the disjoint closed set {y}by disjoint open sets. In (c), the fact that X is Hausdorff means that for any two distinct points x and y, there are disjoint open sets U and V with x E U and y E V . Then X satisfies the condition in Proposition 10.5a that was shown to be equivalent to the TI property.
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(1) A space that is not T I , regular, or normal. Let X = { a ,b, c } , and let { a } , { a ,b}, { a ,cl, { a ,b, ell. (2) A space that is T I but not Hausdorff. Let X be an infinite set, and let 7 consist of the empty set and all complements of finite sets.
I= ID,
(3) A Hausdorff space that is not regular. Let X be the real line. A subset U of X is to be in 7 if for each point x of U ,there is an open interval I, containing x such that every rational number in I, is in U . Then every open interval is in 7,and hence X is certainly Hausdorff. On the other hand, the set of rationals is open in this topology, and therefore the set of irrationals is closed. The set of irrationals cannot be separated from the point 0 by disjoint open sets.
(4) A Hausdorff regular space that is not normal. Let X be the closed upper half plane {Imz > 0) in C. A base for Tconsists of all open disks in X that do not meet the x axis, together with all open disks in X that are tangent to the x axis; the latter sets are to include the point of tangency. It is easy to see that X is Hausdorff, but a little argument is needed to see that X is regular. To begin with, every open set in the usual metric topology for X is in I,and hence every closed set in the usual metric topology for X is closed relative to 7 Let p be a point in X , and let F be a 7closed subset of X not containing p. There is no difficulty in separating p and F by disjoint open sets if p has y coordinate positive, and we therefore assume that p lies on the x axis. Since F is closed, Proposition 10.1 produces a basic open set U tangent to the x axis at p such that U n F = @ . If D denotes a strictly smaller basic open set tangent to the x axis at p, then '' is p itself. Thus the only point of the ordinary boundary of U that lies in D F n D"' = M, and it follows that D and (D"')" are disjoint open sets separating p and F. Consequently X is regular. We postpone the argument that X is not normal until Section 7, when Urysohn's Lemma will be available.
(5) A normal space that is not regular. Let X { a } ,and la, b}.
=
{ a ,b } ,and let Tconsist of @,
We shall see in Section 5 that the Hausdorff property is exactly the right condition to make limits be unique, hence to allow a reasonable notion of convergence. Also, in the construction of a quotient space, it is often a subtle matter to decide whcthcr thc quoticnt spacc is Hausdorff; wc shall obtain sufficicnt conditions in Section 6. The property of regularity makes possible a generalization of the passage from a pseudometric space of points to a metric space of equivalence classes. The point of departure is the following proposition; we shall examine the resulting quotient space further in Section 6.
X. Topological Spaces
450
Proposition 10.7. Let X be a regular topological space. For points x and y in X, define x y if x is in {y}'. Then is an equivalence relation.
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PROOF.Certainly x lies in {x}" , and if x lies in {y}C' and y lies in {z}",then x lies in {z}"'.For the symmetry property, we argue by contradiction and use the regularity of X . Suppose that x lies in {y}C'but y does not lie in {x}C'.Regularity allows us to find disjoint open sets U and V such that y E U and {x}"' g V . Then the closed set V C contains y and hence also {y}''. Since x lies in {y}'', x lies in V C .But this relationship contradicts the fact that x lies in V . We conclude that is symmetric and is therefore an equivalence relation.
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Subspaces of topological spaces inherit certain properties if the original space has them. Among these are T I , Hausdorff, and separable. A subspace of a normal space need not be normal, as is seen by taking X = {a,b , c , d},and I= (0, {a},{a,b},{a,c},{a,b , c,d}},the subspace being {a,b,c}and the relatively closed subsets of interest being { b }and {c}.Let us state the result for regularity as a proposition.
Proposition 10.8. A subspace of a regular topological space is regular. PROOF.Within a subspace A of X , let F be a relatively closed set, and let x he a point of A not in F. Ry Proposition 10.3we have F = F" n A , the closnre being taken in X . Since x is in A but not F ,x is not in FC'. Since X is regular, we can find disjoint open sets U and V in X with x E U and F" (I V . Then U n A and V f' A are disjoint relatively open sets containing x and F . As with metric spaces, a subset D of a topological space X is dense in A if
D"' 2 A; D is dense if D is dense in X . A set D is dense if and only if there is some point of D in each nonempty open set of X . If X is separable, then X has a countable dense set; we have only to select one point from each nonempty member of the base. The properties of bases of a topological space X become more transparent with the aid of the notion of a local base. A set U,of open neighborhoods of x is a local base at x if each open set containing x contains some member of Ux. If B is a base, then the members of B containing x form a local base at x . Conversely if U,is a local base for each x ,then the union of all the Ux's is a base. We say that X has a countable local base at each point5 if a countable such U,can be chosen for each x in X . Metric spaces have this property; the open balls of rational radii centered at a point form a local base at the point. 5 ~ o mauthors e say instead that "X satisfies the first axiom of countability" or "X is first countable" if this condition holds. In the same kind of terminology, one says that "X satisfies the second axiom of countability" or "X is second countable" if X is separable in the sense of Section 1.
3. Compactness and Local Compactness
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EXAMPLE 4, CONTINUED. A space that has a countable dense set and has a countable local base at each point and yet is not separable. As in Example 4 earlier in this section, let X be the closed upper half plane {Imz > 0} in C. A base for 7 consists of all open disks in X that do not meet the x axis, together with all open disks in X that are tangent to the x axis; the latter sets are to include the point of tangency. For a point p on the x axis, the open disks of rational radii with point of tangency p form a countable local base, and for a point p off the x axis, the open disks within the open upper half plane having center p and rational radius form a countable local base. A countable dense set consists of all points with rational coordinates and with y coordinate positive. We shall see in Corollary 10.10 in the next section that a separable regular space has to be normal, and this X is not normal, according to the statement in Example 4 and the proof to be given in Section 7. Thus X cannot be separable.
3. Compactness and Local Compactness Let X be a topological space. In this section we carry over to a general topological space X some definitions made in Section 11.7 for metric spaces. A collection Ld of open sets is an open cover of X if its union is X. An open subcover of U is a subsct of U that is itsclf an opcn covcr. We begin with a new term, saying that the topological space X is a Lindelof space if every open cover of X has a countable subcover. Proposition 2.32 showed that a metric space X is separable if and only if X is a Lindelof space. For general topological spaces it is still true that any separable X is a Lindelof space, by the same argument as for the implication that condition (a) implies condition (b) in Proposition 2.32. In fact, every subspace of a separable space is separable, and hence every subspace of a separable space is Lindelof. However, aLindelof space need not be separable, as the following example shows rather emphatically. EXAMPLE.We construct a topological space (X, I)that is Hausdorff and normal, has a countable dense set, has a countable local base at each point, is Lindelof, yet is not separable. Take X as a set to be the real line. The intersection of any two bounded intervals of the form [ a ,b) is an interval of the same kind, and the union of all such intervals is the whole line. Hence the bounded intervals [ a ,b) for111 a base for solne topology on the line, and this topology we take to be 7 It is called the half-open interval topology for the real line. Since every ordinary open interval of the line is the union of intervals [ a ,b) , any open set in the usual metric topology is open in the half-open interval topology. Any two distinct points of X may be separated by ordinary disjoint open intervals, and therefore X is Hausdorff. To see that X is regular, let a point x and a closed set
X . Topological Spaces
452
F with x not in F be given. Since x is in the open set FC,some [x,x + t) is disjoint from F. Then U = [x,x t) and V = (-00, x) U [x t,+m)are disjoint open sets separating x and F, and we conclude that X is regular. Once we prove that X is Lindelof, it will follow from Proposition 10.9 below that X is normal. The rationals form a countable dense subset of X , and the set of all intervals [x, x + is a countable local base at x. The space X is not separable. In fact, if B is any base, we can find, for each x , some open neighborhood B, of x that is in U and is contained in [x,x + 1). If x < y , then x cannot lie in By and hence B, # B y ; therefore U has to be uncountable. Finally let us see that X is Lindelof. Let an open cover U of X be given, and fix a negative real number xu. Consider the set S(xo) of all real numbers x such that some countable collection of members of U covers [xo,x]. Since xo is covered by some member of U , the set S(xo) contains xo. If the set contains an element xl , then the member of the countable collection that covers xl must contain [xl, xl t) for some t > 0. Thus xl + is in S(xo), and S(xo) contains no largest element. We shall show that S(xo) = [xo,+m). If the contrary is true, then S(xo) nust be bounded. In this case, let c be the least upper bound. For large enough n , c - is in S(xo). Taking the union of the countable collections that cover [xo,c - ,together with one more set to cover c, we obtain a countdble collection that covers [xo,c], and we see that c is in S(xo). Thus c is in S(xo), and we have a contradiction to the fact that S(.uo) contains no largest element. We conclude that some countable no matter what xo is. Taking the union of subcollection of U covers [xo,+m), the countable subcollections corresponding to each negative integer, we obtain a countable subcollection of U covering (-m, +m). Thus X is Lindelof.
+
+
i)
+
5
i]
It is not always so obvious when a topological space is normal. The next result provides one sufficient condition.
Proposition 10.9 (Tychonoff's Lemma). Every regular Lindelof space is normal. F % ~ o F . T,et X he regular and T,indeliif, and let disjoint closed silhsets E and F of X be given. By regularity and Proposition 10.5b each point of E has an open neighborhood whose closure is disjoint from F. Therefore the class L4 of open sets with closures disjoint from F covers E . Similarly the class V of open sets with closurcs disjoint from E covcrs F. Thus U U V U {X - (EU F)} is an opcn cover of X. Since X is Lindelof, there exist sequences of sets U, in U and V, in V such that E 5 U,and F 5 U r = l Vn. Put
Uzl
UL = (/, -
U v': ksn
and
VA = Vn -
u
ksn
U;'
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453
When m 5 n , we have Vm 5 Uk5,vkC'.Then UA n Vm = M,and hence the smaller set U i n VA is empty. Reversing the roles of the U's and the V's shows that UA n VA is empty for rn > n . Therefore UA f' VA = M for all n and rn . Define 00
U= Then U f' V = U,
UUA
00
and
V=
U VA
(UL n Vk) = M . Also,
the last equality holding since {U,} covers E. The right side here equals E since vkC'5 X - E for all k , and therefore E 5 U . Similarly F 5 V . The proof is complete.
Corollary 10.10. Every regular separable space is normal PROOF.A separable space is automatically Lindelof, and thus the corollary follows from Proposition 10.9. Let us return to the concluding example in Section 2, in which X as a set is the closed upper half plane {Imz > 01 but in which the topology is nonstandard near the real axis. It was shown in Section 2 that this particular X is regular, and it was stated that Urysohn's Lemma would be used in Section 7 to show that X is not normal. By Corollary 10.10, X cannot be separable. This completes the argument that X has a countable dense set and has a countable local base at each point yet is not separable. We can now proceed with carrying over some definitions from Section 11.7, valid there for metric spaces, to a general topological space X . We call X compact if every open cover of X has a finite subcover. A subset E of X is compact if it is compact as a subspace of X , i.e.,if every collection of open sets in X whose union contains E has a finite subcollection whose union contains E. It is immediate from the definition that the union of two compact subsets is compact. This definition generalizes the property of closed bounded sets of Rn given by the Heine-Bore1 Theorem. We shall see that the Heine-Bore1 property, rather than the Bolzano-Weierstrass property for sequences, is the useful property to carry over to more general situations in real analysis. In fact, in several places in this book, we have combined an iterated application of the Bolzano-Weierstrass property with the Cantor diagonal process to obtain some conclusion. This construction is tantamount to proving that the product of countably many compact
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metric spaces, which is a metric space essentially by Proposition 10.28 below, is compact. There will be situations for which we want to consider an uncountable product of compact metric spaces, and then arguments with sequences are not decisive. Instead, it is the Heine-Bore1 property that is relevant. The Tychonoff Product Theorem of Section 4 will be the substitute for the Cantor diagonal process, and the use of nets, considered in Section 5, will be analogous to the use of sequences. A number of the simpler results in Section 11.7generalize easily from compact metric spaces to all compact topological spaces or at least to all compact Hausdorff spaces. We list those now. A consequence of Proposition 10.12 below is that compactness is preserved under homeomorphisms. A set of subsets of a nonempty set is said to have the finite-intersection property if each intersection of finitely many of the subsets is nonempty.
Proposition 10.11. A topological space X is compact if and only if each set of closed subsets of X with the finite-intersection property has nonempty intersection. PROOF.Closed sets with the finite-intersection property have complements that are open sets, no finite subcollection of which is an open cover.
Proposition 10.12. Let X and Y be topological spaces with X compact. If + Y is continuous, then f ( X ) is a compact subset of Y.
f :X
-'
PROOF.If {U,} is an open cover of f ( X ) ,then { f (U,)} is an open cover of X . Let { f - ' ( u ~ ) } ; = ~be a finite subcover. Then {Uj}jn=lis a finite subcover of
f(W. Corollary 10.13. Let X be a compact topological space, and let f : X + R be a continuous function. Then f attains its maximum and minimum values. PROOF.By Proposition 10.12, f ( X ) is a compact subset of R. Arguing as in the proof of Corollary 2.39, we see that f ( X ) has a finite supremum and a finite infimum and that both of these must lie in f (X) .
Proposition 10.14. A closed subset of a compact topological space is compact. PKOOP.Let E be a closed subset of the compact space X , and let U be an open cover of E . Then U U { E C }is an open cover of X. Passing to a finite subcover and discarding E", we obtain a finite subcover of E . 'I'hus E is compact.
Lemma 10.15. Let K and E be subsets of a topological space X , and let K be compact. Suppose that to each point x of K there are disjoint open sets U, and V, such that x is in U, and E g V,. Then there exist disjoint open sets U and V such that K g U and E 5 V .
3. Compactness and Local Compactness
455
PROOF.As x varies through K , the open sets Ux form an open cover of K . By compactness, a finite subcollection of the Ux'sis a cover, say Ux,, . . . , UXn. Put U= Uxk and V = Vxk.Then K 5 U and E 5 V. Also, U n V = ( u i = l UXk)n (n;=l 'Xk) = U:=l (UXkn(ny=l 'Xl)) c U:=l('Xkn X' k) = @, and thus U and V have the required properties.
Proposition 10.16. Every compact Hausdorff space is regular and normal. PROOF.Let X be compact Hausdorff. If apoint x and a closed set F with x 6 F are given, we observe by Proposition 10.14 that F is compact. The Hausdorff property of X allows us to take E = {x} and K = F in Lemma 10.15, and we obtain disjoint open sets U and V such that x is in V and F c U. Thus X is regular. If disjoint closed sets E and F are given, then F is compact by Proposition 10.14. The fact that X has been shown to be regular allows us to take K = F in Lemma 10.15, and we obtain disjoint open sets U and V such that E 5 V and F 5 U. Thus X is normal.
Proposition 10.17. In a Hausdorff space every compact set is closed. PROOF.Let X be a Hausdorff space, and let K be a compact subset of X . Fix x in K C .The Hausdorff property of X allows us to take E = {x} in Lemma 10.15, and we obtain disjoint open sets Ux and V, such that x is in Vx and K c Ux. Letting x now vary, we see that K C = UxtKcV,. Hence K C is open and K is closed.
Corollary 10.18. Let X and Y be topological spaces with X compact and with Y Hausdorff. If f : X 3 Y is continuous, one-one, and onto, then f is a homeomorphism.
-'
PROOF.We are to show that f :Y X is continuous. Let E be a closed subset of X , and consider ( f -')-' (E) = f (E) . The set E is compact in X by Proposition 10.14, f (E)is compact by Proposition 10.12, and f (E)is closed by Proposition 10.17. Since the inverse image under f of any closed set is closed, f is continuous.
-'
-'
A topological space is locally compact if every point has a compact neighborliood. Coiiipact spaces are locally coiilpact, but tlie real line wit11 its usual topology is locally compact and not compact. In a sense to be made precise in the next two propositions, locally compact Hausdorff spaces are just one point away from being compact Hausdorff. Let ( X , I)be an arbitrary topological space. Define a new set X* by X* = X U {cc},where cc is not already a member of X , and define T*to be the union
456
X . Topological Spaces
of 7 and the set of all complements in X * of closed compact subsets of X . We shall verify in Proposition 10.19 that I*is a topology for X * . The topological space ( X * , I * ) is called the one-point compactification of ( X , I ) . By way of examples, the one-point compactification of R may be visualized as a circle and the one-point compactification of R2may be visualized as a sphere.
Proposition 10.19. If ( X , 7 ) is a topological space, then ( X * , I * ) is a compact topological space, X is an open subset of X * , and the relative topology for X in X* is 7 PROOF.To see that I * is a topology, we observe first that and X* are in I * . If U and V are in I * , there are three cases in checking that U n V is in T * : If U and V are both in I;then U n V is in 7since 7 i s closed under finite intersections. If U is in I and V is not, then VC is closed compact in X , and X - VC is thus open in X ; since I i s closed under finite intersections, U n V = U n ( X - Vc) is in I If U and V are not in 7,then the complements U C and V C in X * are closed compact subsets of X ; so is their union ( U n V)', and hence U n V is in I * . We still have to check closure of I*under arbitrary unions. Suppose that U, is in I for a in an index set A and Vg has closed compact complement for Vp' is a in an index set B . Then Ua,, U, is in I and if B is nonempty, closed subset of one V i and hence is compact; in this case, (UBEB v ~ is) closed compact in X , and hence UgEBVB is in T * . Thus we have only to check that U U V is in 7 * if U is in I and Vc is closed compact in X . As the intersection of two closed sets, one of which is compact, ( X - U ) n Vc = ( X - U ) n ( X - V ) is closed and compact in X , and thus U 1-1 V = ( ( X - U ) n VC)' is in I * . Thus I*is a topology. To see that X * is compact, let U be an open cover of X * . Find some V in U containing the point m. The members of U n I cover the compact subset VC of X , and there is a finite subcollection V that covers VC. Then V U {V} is a finite subcollection of U that covers X * . The set X is in I and is therefore in I * . Thus X is open in X * . To complete the proof, we are to show that I*n X = I We know that I*n X 2 I If V is a member of 'P that does not lie in 'Z then Vc is closed compact in X , and its complement X - V C = V n X in X is open in X. Hence V n X is in I
no:,
Proposition 10.20. If X * is the one-point compactification of a topological space X , then X * is Hausdorff if and only if X is both locally compact and Hausdorff. PROOF.Suppose that X is locally compact and Hausdorff. Since X is Hausdorff, any two points of X can be separated by disjoint open sets in X , and these sets will be open in X * . To separate a point x in X from m, let C be a compact
3. Compactness and Local Compactness
457
neighborhood of x in X . Since X is Hausdorff, C is closed in X . Thus C Cis in I*.Then C0 and Cc are disjoint open sets in X* such that x is in C Oand oo is in C C,and X * is Hausdorff. Conversely suppose that X* is Hausdorff. Proposition 10.19 shows that X is a subspace of X * . Since any subspace of a Hausdorff space is Hausdorff, X is Hausdorff. To see that X is locally compact, let x be in X , and find disjoint open sets U and V in X* such that x is in U and rn is in V. Then U must be in I; and V" must bc closcd compact in X . Sincc U n V = M,U 5 V" . This inclusion exhibits Vc as a compact neighborhood of x , and thus X is locally compact. Corollary 10.21. Every locally compact Hausdorff space is regular. PROOF.If X is locally compact Hausdorff, Propositions 10.19 and 10.20 show that the one-point compactification X * is compact Hausdorff and allow us to regard X as a subspace of X * . Proposition 10.16 shows that X * is regular, and Proposition 10.8 shows that X is therefore regular. A locally compact Hausdorff space need not be normal; an example is given in Problem 5 at the end of the chapter. The remainder of this section concerns senses in which a locally compact Hausdorff space is almost normal.
Corollary 10.22. If K and F are disjoint closed sets in a locally compact Hausdorff space and if K is compact, then there exist disjoint open sets U and V such that K 5 U and F 5 V. PROOF.This is immediate from Lemma 10.15 and Corollary 10.21 Corollary 10.23. If K is a compact set in a locally compact Hausdorff space, then there is a compact set L such that K 5 Lo. PROOF.Let X be locally compact Hausdorff, and form the one-point compactification X * . Since X * is compact Hausdorff by Proposition 10.20, Proposition 10.17 shows that K is closed in X* and Proposition 10.16 shows that X * is regular. Thus Proposition 10.5b shows that we can find an open set U in X * such that m is in U and u"' f' K = M. Then K 5 X * - u"' 5 X * - U . By definition of the topology of X * , the set L = X * - U is compact in X . Its subset X * - UC'is open and is therefore contained in Lo. Thus K 5 Lo 5 L with L compact.
A topological space is called o-compact if there is a sequence of compact sets with union the whole space. The real line with its usual topology is o-compact. For that matter, so is the subspace of rationals since each finite subset is compact.
X . Topological Spaces
458
Proposition 10.24. A locally compact topological space is o-compact if and only if it is Lindelof. Consequently every o-compact locally compact Hausdorff space is normal. PROOF.If X is o-compact, write X = U,"=,K, with K, compact. If U is an open cover of X ,then Uis an open cover of each K,, and there is a finite subcover U , of K,. Then U,"=l U, is a countable subcover of U , and X is Lindelof. Conversely if X is locally compact and Lindelof, choose, for each x in X ,a compact neighborhood K, of x , and let Ux be the interior of Kx . As x varies, the U, form an open cover of X.Since X is Lindelof, there is a countable subcover { U x n } z l Since . we have U,,, 5 K,,, for all n , {Kxn}r=lis a sequence of compact sets with union X . Hence X is o-compact. Finally if X is locally compact Hausdorff and o-compact, hence also Lindelof, then Corollary 10.21 shows that X is regular, and Tychonoff's Lemma (Proposition 10.9) shows that X is normal.
Proposition 10.25. In a o-compact locally compact Hausdorff space, there exists an increasing sequence {K,} of compact sets with union the whole space and with K, 5 K:++, for all n .
PROOF.Let X be a locally compact Hausdorff space such that X = U,"=,L , with L, compact. Replacing L , by the union of the previous members of the sequence, we may assume that L , 5 L,+l for all n 3 1. Put Lo = KO = @. Use Corollary 10.23 to cl~ooseK1 coiilpact with L 1 5 K:. Inductively suppose that n > 0 and that for all k with 0 < k 5 n ,a compact set Kk has been defined such that Lk U Kk-1 5 K i . Applying Corollary 10.23, we can find a compact set K,+l such that the compact set L,+, U K, is contained in K;+,. Then KI,+l K{ for all k > 1 as required, and X = Ur=l K, since K, g L, and L, = X .
UZl
4. Product Spaces and the Tychonoff Product Theorem The product topology for the product of topological spaces was discussed briefly in Section 1. If S is a nonempty set and if X, is a topological space for each s in S , then the Cartesian product X = X s,sX,, as a set, is the set of all functions f from S into USES X, such that f (s) is in X, for all s E S . Thc topology that is imposed on X is, by definition, the weakest topology that makes the sthcoordinate function p, : X + X, be continuous for every s . Let us investigate what sets have to be open in this topology, and then we can look at examples and see better what the topology is. If Us is any open subset of X,,then p;l (Us)has to be open in X since p, is continuous. For example,
4. Product Spaces and the Tychonoff Product Theorem
459
if S = { I , 21, we are considering X = X 1 x X2. A set py1 ( U 1 )is of the form U1 x X 2 , and a set p;' (U2) is of the form X 1 x U2. These have to be open if U1 is open in X I and U2 is open in X 2 . The intersection of any two such sets, which is of the form U 1 x U 2 ,has to be open in X , as well. We do not need to intersect these sets further, since p r l ( U 1 )n p , l ( ~ l ) = (U1 n V 1 ) .By the remark with Proposition 10.1, the sets p l l ( U 1 )f' p y 1 ( U 2 )with U1 open in X 1 and U2 open in X2 form a base for some topology on X = X I x X2. These sets have to be open in the product topology, and pl and p2 are indeed continuous in this topology. Therefore the product topology on X = X 1 x X2 has
prl
{p;'
( ~ 1n ) py1(u2)
I U 1open in X I , U2 open in ~
2
}
as a base. More generally the product topology on X = X 1 x . . . x X, has
( i)p*' k=l
I
( U k ) Uk open in Xk for each k
as a base. When the index set S is the set of positive integers, the product X = X A E S X n , as a set, is the set of sequences { f ( n ) j n E SAgain . any set p i 1 (U,) with U , open in X , has to be open in X . Hence any finite intersection of such sets as n varies has to be open. But there is no need for infinite intersections of such sets to be opcn, and a basc for thc product topology in fact consists of alljifilzite intcrscctions of sets (U,) with U , open in X,. The use of linile inlcrsecliums, arld rlvl inlinile i~llcrsccliuns,persisls for all S and gives us a description of a base for the product topology in general. When S = [O, 11 and all X, are [O, 11, the description of the product topology has a helpful geometric interpretation. The set X consists of all functions from the closed unit interval to itself, and we can visualize these in terms of their graphs. A basic open set of such functions imposes restrictions at finitely many values of s, i.e., at finitely many points of the domain. At such values of s,the graph of a function in the basic open set is to pass through a certain window Us depending on s . At all othcr valucs of s,thc function is unrcstrictcd.
Proposition 10.26. The topological product of Hausdorff topological spaces is Hausdorff. PROOF.Let a product X = X ,,,X, be given, let p, : X + X , be the sth coordiiiate function, and let two distinct menibers f and g of X be given. Members of X are functions of a certain kind, and these two functions, being distinct, have f ( s ) # g ( s ) for some s E S . Since X , is Hausdorff, we can choose disjoint open sets Us and V , in X , such that f (s) is in Us and g ( s ) is in V, . Then p ; l ( ~ , ) and p;' (V,) are disjoint open sets in X such that f is in p;'(US) and g is in p ; ' ( ~ , ) .
460
X . Topological Spaces
Theorem 1027 (Tychonoff Product Theorem). The topological product of compact topological spaces is compact. REMARKS.This theorem is a fundamental tool in real analysis. We shall give the proof and then discuss how the theorem can be regarded as a generalization of the Cantor diagonal process used in the proofs earlier of the fact that any totally bounded complete metric space is compact (Theorem 2.46), the Helly Selection Principle (Problem 10 at the end of Chapter I), Ascoli's Theorem (Theorems 1.22 and 2.56), and, by implication, the Cauchy-Peano Existence Theorem for differential equations (Problems 24-29 at the end of Chapter IV). The proof will make use of Zorn's Lemma (Section A9 of the appendix), which is one formulation of the Axiom of Choice. Actually, the Axiom of Choice arises in two rnore transparent ways in the proof as well. One is simply in the statement that the topological product is a topological space; for this to be the case, the product has to be nonempty, and that is the content of the Axiom of Choice. The other is the construction of a particular element x in the product that occurs near the beginning of the proof below. PROOF.Let X = X s,sX, be given witheach X , compact, andlet p, : X + X , bc thc sth coordinate function. Wc arc to provc that any opcn covcr of X has a finite subcover, and we begin by proving a special case. Let Sbe the family of all sets y;l ( U s ) as U s varies through all ope11sets of X , arid as s varies. We know that finite intersections of members of S form a base for the product topology on X . For the special case let U be an open cover of X by members of S, we shall produce a finite subcover. For each s, let B, be the family of all open sets U s in X , such that p;' ( U s )is in U. We may assume for each s that no finite subfamily of B, covers X , since otherwise the corresponding finitely many sets p;l ( U s ) would cover X . By compactness of X,, B, does not cover X,; say that x, is not covered. The point x of X whose sth coordinate is x , then belongs to no member of U , and U cannot be a cover. This contradiction shows that the special U has a finite subcover. Now let U be any open cover of X , and suppose that no finite subfamily of U covers X . Let C be the system of all open covers V of X such that 24 c V and such that no finite subfamily of V covers X . The set C is partially ordered by inclusion upward and is nonempty, having U as a member. If (VaJ is a chain in C, then we shall show that V = U, V, is in C and hence is an upper bound in C for the chain {V,}. I11 fact, V is certainly an open cover. If it has a finite subcover, then each member of the finite subcover lies in one of the covers, say Val. Since {V,} is a chain, all members of the finite subcover lie in the largest of those V,, 's. Thus one of the Val's fails to be in C,and we arrive at a contradiction. We conclude that every chain in C has an upper bound in C. By Zorn's Lemma let U * be a maximal cover from C of X .
4. Product Spaces and the Tychonoff Product Theorem
46 1
The family S n U*of all members of U*that are in the family S of the first paragraph of the proof has the property that no finite subfamily is a cover of X . By the result of the first paragraph, S n U * cannot be a cover of X. Hence we shall have arrived at a contradiction if we show that the union of the members of U * is contained in the union of the members of S n U * . Let U be a member of [A*, and fix a point x in U . Since finite intersections of members of S form a base, Proposition 10.1 shows that there are members S1 n . . . n S, of S such that x i s i n S 1 n . . . n S , andS1 n . . . n S , g U . Weshallshowthatoneofthesets Sj is in U *,hence in U * n S,and then the proof will be complete. If S1 is in U*, we are finished. Otherwise, by the maximality of U*, there are finitelymanyopensetsC1, . . . , Ck o f U * suchthatX = S1 U C 1 U . . . U Ck . Again by the maximality, no open set containing S1 can belong to U*, since the union of that set with C1 U . . . U Ck would be X. Proceeding inductively, suppose we have shown that no open set containing S1 n . . . n Si is in U * and that there are open sets D l , . . . , Dm in U*with
If, as we may assume, Si+1is not in U *, then by maximality of U * ,there are open setsH1, . . . , E, inU* suchthat X - Si+1 U E 1 U . . . U E , . Then
and
si+l
(sl n . .. n s ~ + u ~ )( s ~ +n~( D ~ u . . . u D,)) g ( s l n . . . n s , + ~ )u ( D ~u . . . u D,).
=
Hence
That is,
Therefore, once again by maximality of U * ,no open set containing S1n . . .f' Si+1 can be in U *, and the induction is complete. In particular, U , which is an open set coiitaiiiiiig S1n . . . n S, ,is not ill U * . This coiitradictioii coiicludes tlie proof. As announced above, the Tychonoff Product Theorem is a generalization of the Cantor diagonal process. In fact, let us see how that diagonal process may be used to show directly that the product of a sequence of copies of [0, 11 is compact. Denote the product as a set by X = X [0, 11. A member of X is a sequence {x,} with terms x, . Let us impose on X the Hilbert-cube metric of Example 11 in Section 11.1:
z1
X . Topological Spaces
462
We show below in Corollary 10.29 that this metric on X yields the product topology. By Theorern 2.36 the space X will then be compact if every sequence in X has a convergent subsequence. A sequence in X means a system { x i r n ) }in which the nth term of the mth sequence is ximi").Convergence is term-by-term convergence. To produce a convergent subsequence of sequences, we iterate use of the Bolzano-Weierstrass property of [0, 11. Remembering that m tells which sequence we are dealing with, we find first a subcollection mk of the indices m such that we have convergence along the mk's for n = 1, then a subcollection inkl of that such that we have convergence along the mk,'s for n = 2, and so on. Since the intersection of all these sequences may be empty, we instead obtain a convergent subsequence of our sequences by requiring that the kth term of the desired subsequence be the kth term of the kth subsequence. This "diagonal process" thus shows that any sequence in X has a convergent subsequence. Hence X, being a metric space, is compact. The general Tychonoff Product Theorem may thus be viewed as a topological generalization of the diagonal process to product spaces with an uncountable number of factors. Here is one way in which the Tychonoff Product Theorem is used in real analysis. For the situation in which we have a set Y and a system of functions f, : Y C for s in some set S, the first section of this chapter introduced the weak topology for Y determined by { fs},,s. This is the weakest topology making all the functions f , continuous. Often in analysis a set Y and a system of functions f , of this kind arise in a construction, and then this weak topology is imposed on Y. Tn favorable cases it turns out that each function f, is hounded on Y . In this case if there are enough functions f, to separate points of Y (i.e., enough so that for each x and y there is some s with f , ( x ) f f , ( y ) ) , then Y is a candidate for a compact Hausdorff space. To see what is needed for compactness, let X, be a compact subset of C containing the image of f , , and let X = X s,sX,. Define a function F : Y + X by " F ( y ) is the function whose sth coordinate is f , ( y )." It is readily verified that F is a homeomorphism of Y onto a subspace of the compact Hausdorff space X. Thus Y is compact if and only if F ( Y ) is closed in X . Checking that a set is closed is much easier than checking compactness directly, and it is especially easy if one uses "nets," which are the objects introduced in the next section as a useful generalization of sequences. To complete our discussion, we still need to prove that the Hilbert-cube metric 00 on X = X n=, [O, 11 yields the product topology. It will be helpful to prove the following more general result and to obtain the statement about the Hilbert cube as a special case.
-
5. Sequences and Nets
463
Proposition 10.28. Suppose that X is a nonempty set and {d,},?~ is a sequence of pseudometrics on X such that d, ( x , y) 5 1 for all n and for all x and y in X . Then d ( x , y) = C z l 2-,dn(x, y ) is a pseudometric. If the open balls relative to d, are denoted by B,(r; x ) and the open balls relative to d are denoted by B ( r ; x ) , then the B,'s and B's are related as follows: (a) wllcrlcvcr surne B,(r,; x ) is give11 will1 r, > 0, Illere cxisls surne B ( r ; x)
with r > 0 such that B ( r ; x ) 5 B, (r,; x ) , (b) whenever B ( r ; x ) is given with r > 0, there exist finitely many r, > 0, say for n 5 K , such that K B,(r,; x ) 5 B ( r ; x ) . PROOF.For (a), choose r = 2-"r, . If d ( x , y) < r , then 2-"d, ( x , y ) < r for all m and in particular d, ( x , y ) < 2,r = r , . For (b), choose K large enough so that 2-K < 712, and put r , = r / 2 for n 5 K . If y is in n f = l B , ( r , ; x ) , t h e n d , ( x , y ) < r, = r / 2 f o r n 5 K . Hence d ( x , y ) 5 ~ f 2-'dn(x, = ~ y) Cr=K+l2-, < C L 1Zpny/2 2-K < r / 2 + r / 2 = r . Thcrcforc y is in B(r ; x ) .
+
Corollary 10.29. The Hilbert-cube metric on X = X product topology.
+
[O, 11 yields the
PROOF.Proposition 10.28a implies that any basic open neighborhood of x in the product topology contains a basic open neighborhood in the Hilbert-cube metric topology. Proposition 10.28b shows that any basic open neighborhood of x in the Hilbert-cube metric topology contains a basic open neighborhood in the product topology.
5. Sequences and Nets Sequences are of limited interest in general topological spaces. Nets, which are generalized sequences of a certain kind, are a useful substitute, and we introduce them in this section. Using nets, we shall be able to see that product topologies are appropriate for detecting pointwise convergence in the same way that the metric topology obtained from the supremum norm is appropriate for detecting uniform convergence. We begin with two exan~plesthat illustrate some of the difficulties with using sequences in general topological spaces. We use the natural definition suggested by Section 11.4-that a sequence {x,} in X converges to xo if for each neighborhood of xo, there is some N depending on the neighborhood such that x, is in the neighborhood for n > N . We say that the sequence is eventually in the neighborhood. The point xo is a limit of the sequence.
464
X . Topological Spaces
(1) Let X be the set of positive integers, and let a topology for X consist of the empty set and all sets whose complements are finite. If x, = 2n, then the sequence {x,) converges to every point of X and hence does not have a unique limit. The space X is TI and has a countable local base at each point, but X is not Hausdorff.
(2) Lct X bc thc sct of points (712, 7%) in thc planc with 112 and n intcgcrs > 0 . Define a topology for X as follows. Any set not containing (0,O)is to be open. If a set U contains (0, 0 ) ,then U is defined to be open if tliere are only finitely rrlany columns C , = {(m, n ) I n = 0 , 1 , 2 . . . } such that C , - (U n C,) is infinite. Enumerate X , and define x , to be the nth point in the enumeration. It is easy to check that the image of the sequence {x,} has (0,O) as a limit point and that no subsequence of {x,} converges to (0,O). The space X is Hausdorff but does not have a countable local base at (0,O). Thus the elementary results in Section 11.4 do not generalize to all topological spaces. But Proposition 2.20 (the uniqueness of the limit of any sequence) is still valid if X is Hausdorff, and Proposition 2.22 and Corollary 2.23 (the characterization of limit points and of closed sets in terms of sequences) are still valid if X has a countable local base at each point. Nets will cure the problem about characterizing limit points and closed sets without countable local bases but not the problem about nonuniqueness of limits, and thus we shall be able to work well with nets in all Hausdorff spaces. In particular we shall be able to use nets in uncountable products of Hausdorff spaces, which arise frequently in real analysis and tend not to have a countable local base at each point. Before defining nets, let us give one positive result whose statement mixes topological spaces and metric spaces. If S is any nonempty set, we have made B ( S ),the vector space of all bounded scalar-valued functions on S , into a normed linear space-and hence a metric space-by means of the supremum norm. If S is a topological space, let C ( S ) be the subset of continuous members of B ( S ) ; this is a vector subspace and hence is itself a normed linear space.
Proposition 10.30. If S is a topological space and { f,} is a sequence of scalarvalued functions continuous at s~and converging uniformly to a function f ,then f is continuous at xo. Consequently the subspace C ( S ) of B ( S ) is a closed subspace, and C ( S ) is complete as a metric space. PROOF.Given t > 0 , choose N such that n For any s, we then have
> N implies Il f,
-
f, ,ll
< t.
5. Sequences and Nets
465
Since ,fN is continuous at SO, there exists a neighborhood of so such that the right side is < 3t for s in that neighborhood. Thus f is continuous at so. If { f,} is a sequence in C ( S ) converging uniformly to f in B ( S ) , then f is in C ( S ) ,by the result of the previous paragraph. Since convergence of sequences in B ( S ) is the same as uniform convergence, Corollary 2.23 shows that C ( S ) is a closed subset of B ( S ) . Propositions 2.43 and 2.44 then show that C ( S ) is complete as a metric space. Now we turn our attention to nets. In the indexing for a net, the set of positive integers is replaced by a "directed set," which we define first. Let D be a partially ordcrcd sct in thc scnsc of Scction A9 of thc appcndix, thc partial ordcring bcing denoted by 5 . We say that ( D , 5 ) is a directed set if for any a and p in D , there issomeyinDwithaiyand~iy.
(1) Take D to be the set of positive integers, and let 5 have the usual ~neaning. (2) Let S be a nonempty set, take D to be the set of all finite subsets of S , and let a 5 B mean that the inclusion a holds.
(3) Let X be a topological space, let x be a point in X , take D to be the set of all neighborhoods of x , and let a i 8 mean that a 2 8 . let
(4) Let ( D l , 51) and (D2, 5 2 ) be two directed sets, take D to be Dl x D2, and (011,012) i ( P I ,8 2 ) mean that 011 i1 PI and 012 5 2 82.
If X is a nonempty set, a net in X is a function from a directed set D into X . If D needs to be specified to avoid confusion, we speak of a "net from D to X ." The function will often be written a H x, or {x,}. If E is a subset of X , the net is eventually in E if there is some a0 in D such that a0 5 a implies that x, is in E. The net is frequently in E if for any a in D , there is a p in D with a 5 p such that xa is in E. It is important to observe that the negation of "the net is eventually in E" is that "the net is frequently in the complement of E ." The directedness of the set D plays an important role in the theory by allowing us to work simultaneously with finitely many conditions on a net. For example, if {x,} is eventually in E l and eventually in E 2 , then it is eventually in E l n E2. In fact, the given conditions say that there are members a1 and 012 of D such that x, is in E l for a1 5 a and x, in E z for a2 5 a . The directedness implies that
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a1 5 ao and 1x2 5 ao for some ao in D. Then {x,} is in El n E2 for ao 5 a . This kind of argument will be used often without mention of the details. If X is a topological space, a net { x a }in X converges to xo in X if {x,} is eventually in each neighborhood of xo. In this case we write x, + xo, and we say that xo is a limit of { x a } . Because of the availability of Examples 3 and 4 above, it is an easy matter to characterize the terms "Hausdorff," "limit point," "closed set," and "continuous at a point" in terms of convergence of nets.
Proposition 10.31. A topological space X is Hausdorff if and only if every convergent net in X has only one limit. PROOF. Suppose that X is Hausdorff and that x, + xo and x , + yo with xo f y o Choose disjoint open sets U and V with xo in U and yo in V . By the assumed convergence, { x a }is in U eventually and is in V eventually. Then it is in U n V = m eventually, and we have a contradiction. Suppose that X is not Hausdorff. Find distinct points xo and yo such that every pair of neighborhoods U of xo and V of yo has nonempty intersection. For any such pair ( U , V ) , define xu," to be some point in the intersection. Combining Examples 3 and 4 above, we see that ( U , V) ++ x ~v is , a net in X converging to both xo and yo.
Proposition 10.32. If X is a topological space, then (a) for any subset A of X and limit point xo of A , there exists a net in A - { x o } converging to xo , (b) any convergent net {x,} in X with limit xo in X either has xo as a limit point of the image of the net or else is eventually constantly equal to xo. PROOF.For (a), the definition of limit point implies that for each neighborhood U ofxo, the set U n ( A - { x o } ) is nonempty. Ifxu denotes apointin the intersection, then U H X U is a net in A - {xo}converging to xo. For (b), suppose that xo is not a limit point of the image of the net. Then there exists a neighborhood U of xo such that U - { x o }is disjoint from the image of the net. Since the convergence implies that the net is eventually in U , it must be true that x, = xo eventually.
Corollary 10.33. If X is a topological space, then a subset F of X is closed if and only if every convergent net in F has its limit in F. PROOF.Suppose that F is closed and that {x,} is a convergent net in F with limit xo. By Proposition 10.32b,either xo is in the image of the net or xo is a limit point of the image of the net. In the latter case, xo is a limit point of the larger set F. In either case, xo is in F ; thus the limit of any convergent net in F is in F.
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Conversely suppose every convergent net in F has its limit in F. If xo is a limit point of F, then Proposition 10.32a produces a net in F - {xo}converging to xo. By assumption, the limit xo is in F .Therefore F contains all its limit points and is closed.
Proposition 10.34. Lct f : X + Y bc a function bctwccn topological spaces. Then f is continuous at apoint xo in X if and only if whenever {x,} is a convergent net in X witli limit no, tlierl { f (A,)} is corlvergerlt in Y witli lirrlit f (no). REMARKS.This result needs to be used with caution if Y is not known to be Hausdorff. For example, let X and Y both be the set {a,b}. Let the topology for X be discrete and the topology for Y be indiscrete, consisting only of @ and the whole space. Every function f : X + Y is continuous. Suppose that f (a) = f (b) = a. Take xo = b and x, = b for all a. Then { f (x,)} converges to both a and b. Hence we cannot evaluate f (xo)as just any limit of { f (x,)}; we have to pick the right limit. PROOF.Suppose that f is continuous at xo and that {x,} is a convergent net in X with limit xo. Let V be any open neighborhood o f f (xo).By continuity, there cxists an opcn ncighborhood U of xo such that f ( U ) 5 V . Sincc x, + xu, thc members x, of the net are eventually in U .Then f (x,) is in f ( U ) g V for the same u's, lierlce eventually. Therefore { f (A,)} coriverges to f (no). Conversely suppose that x, + xo always implies f (x,) + f (xo). We are to show that f is continuous. If V is an arbitrary open neighborhood of f (x0),we seek some open neighborhood of xo that maps into V under f . Assuming that there is no such neighborhood for some V , we can find, for each neighborhood U of xo, some x u in U such that f (xu) is not in V . Then x u + xo, but f (xu) does not have limit f (xo)because f (xu) is never in V . This is a contradiction, and we conclude that some U maps into V under f ;thus f is continuous.
Proposition 1035. Let X = X s,sX, be the product of topological spaces X,, and let p, : X + X , be the sthcoordinate function. Then a net {x,} in X converges to soiile xo ill X if and oiily if the net {p,(x,)} ill X , converges to p, (xo) for each s in S . REMARK.This is the sense in which the product topology is the topology of pointwise convergence. In combination with Corollary 10.33, this proposition simplifies the problem of deciding when a subset of a product space is closed in the product topology. PROOF.If {x,} converges to xo, then Proposition 10.34and the continuity of p, together imply that {p,(x,)} converges to p, (xo).
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Conversely suppose that { p ,(x,) } converges to p,(xo) for all s . Fix s . If Usis an open neighborhood of p, (xo)in X,,then {p,(x,) } is eventually in Us. Hence there is some a0 such that p,(x,) is in Uswhenever a0 5 a.For the same values Thus {x,} is eventually in ps-' (Us). of a,{x,} is in p;' (Us). Any neighborhood N of xo in X contains some basic open neighborhood of the form U = (Us,) n . .. n (USn). It follows from the result of the previous paragraph that {x,} is eventually in each p;' ( U s ) hence , is eventually in the intersection U ,and hence is eventually in N . Therefore {x,} converges to xu. One can express also the notion of compactness in terms of nets, the idea being that compactness of X is equivalent to the fact that every net in X has a convergent subnet, for an appropriate definition of "subnet." The remainder of this section will deal with this question. Carrying out the details of this equivalence is harder than what we have done so far with nets. Actually, the main benefit of the equivalence is the resulting simplification to proofs of compactness, especially to the proof of the Tychonoff Product Theorem. Since we have already proved the Tychonoff Product Theorem without nets, the material in the remainder of this section will be used only in minor ways in the rest of the book.6 Lct D and E bc dircctcd scts. A function from E to D , writtcn p H a,, is cofinal if for any in D, there is a v in E such that 5 a, whenever v 5 p . If p H aa,is cofinal and if a H x, is a net from D to X , then the composition p H x , ~is a net from E to X and is called a subnet of the net a H x,. The prototype of a subnet is a subsequence. In this case, D and E are both the set of positive integers, and the function from E to D is k H nk. If the sequence is {a,}, then the subnet/subsequence is {an,}.For a general subnet one might expect that it would suffice always to take E to be a subset of D and to let the function from E to D be inclusion. However, this definition of subnet is insufficient to prove the desired characterization of compactness in terms of nets and subnets. A net from a directed set D to a nonempty set X is called universal if for any subset A of X ,the net is eventually in A or eventually in A". It of course cannot be eventually in both, since otherwise it would eventually be in the intersection, namely the empty set.
Proposition 10.36. Each net in a nonempty set X has a universal subnet. REMARK. The proof will use Zorn's Lemma. Apart from this one use, the only other uses of the Axiom of Choice in the remainder of this section are transparent ones. 6 ~ e tplay s a more significant role in the companion volume Advanced Real Analysis. 7 ~ h i definition s is not the standard one given in Kelley's General Topology, but it leads to the standard definition of "subnet."
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PROOF.Let D be a directed set, and let a H x, be a net from D to X . Consider all families Cp of subsets of X that are closed under finite intersections and have the property, for each A in Cg ,that the net is frequently in A. There exists such a family, the singleton family {X}being one. Partially order the set of such families by inclusion upward, saying that Cg 5 Cg, when Cp g Cp,. In any chain of Cp's, let C, be the union of the sets in the various members of the chain. Since closure under intersection depends only on two sets at a time and since the other property of a Cg depends only on one set at a time, C,, is again a family of this kind. By Zorn's Lemma let C be a maximal such family. Let us prove for each subset A of X that either A or AC is in C . In fact, if for every B in C, the net is frequently in A n B , then C U { A }is a family containing C and satisfying the two defining properties of one of our families. By maximality, C U { A }= C. Hence A is in C. Assuming that A is not in C, we obtain a set B in C such that the net fails to be frequently in A f' B . Then B is a member of C such that the net is eventually in ( A n B)'. Sinlilarly if we assume that A' is not in C, we obtain a set B' in C such that the net is eventually in ( A c f' BB')'.If neither A nor Ac is in C, then the net is eventually in ( A n B)" n (A" n B')"
=
(ncu B") n ( A u B'")
=
(A' n ( A u B ' ~ ) u) ( B n~( A u B"))
= (A'
n B ' ~ u) ( B n~( A u B ' ~ ) )
g B ' ~u B~ = ( B n B ' ) ~ , and it cannot be frequently in B n B'. This contradicts the fact that B
f' B' is in C because C is closed under finite intersections. This completes the proof that either A or AC has to be in C. The members of C form a directed set under inclusion downward, i.e., with partial ordering A 5 B if A 2 B. Form C x D as a directed set under the definition in Example 4 at the beginning of this section. We construct a subnet as follows. For each ordered pair ( A , B ) in C x D, let a ( A , p ) be an element of D with B 5 a ( ~ , p and j with x,,,,,, in A; this choice is possible since D is directed and the given net is frequently in A . The function ( A ,p ) H a ( ~ , pis) cofinal because for any p E D, the domain value ( A , p ) has p 5 a ( ~ ), ,whenever ( A , p ) 5 ( B , y ) . Thus ( A ,p ) H X,(A,D) is a subnet. To complete the proof, we show that this subnet is universal. For any subset A of X , we have seen that either A or AChas to be in C. Without loss of generality, assume that A is in C. For any fixed p , the inequality ( A ,p ) 5 ( B , y ) implies that x , , , , ~ ,is in the subset B of A , and hence the subnet is eventually in A.
Proposition 1037. The following three statements about a topological space X are equivalent:
X . Topological Spaces
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(a) X is compact, (b) every universal net in X is convergent, (c) every net in X has a convergent subnet. PROOF.To provc that (a) implics (b), lct {x,} bc a univcrsal nct in X , and suppose that {x,} is not convergent. For each x in X , there is then an open neighborhood Ux of x such that {x,} is not eventually in U,. Since the net is universal, it is eventually in (U,)" for each x . The open sets Uxcover X . By compactness, let {U,, , . . . , U x n }be a finite subcover. The net is eventually in each (UYJ)"and hence is eventually in their intersection. But their intersection is empty since X = Ux, . We have arrived at a contradiction, and thus {x,} must be convergent. Statement (b) implies statement (c) since every net has a universal subnet, by Proposition 10.36. To prove that (c) implies (a), suppose that X is noncompact. We shall produce a net with no convergent suhnet. If Uis an open cover of X with no finite subcover, we shall use Uto define a directed set. Let F b e the set of all finite subcollections of members of U. This is directed under inclusion upward: a 5 p if a 5 p . For each a in F , the set X - Uu,, U is not empty since Uhas no finite subcover, and we let x, be an element of X - U,,, U . 'I'hen a H x , is a net. Suppose that {x,} has a convergent subnet, with some xo as limit. For any neighborhood N of xo, {x,} is frequently in N . Since U is a covering, there is some U in U with xo in U . By construction, {x,} is not in U as soon as a has { U } 5 a . We conclude that no subnet of {x,} converges.
Ukl
Proposition 10.37 gives the statement about general topological spaces that extends the equivalence of the Bolzano-Weierstrass property and the HeineBore1 property of closed bounded subsets of Euclidean space. To illustrate the power of nets, we can now use them to give a second proof of the Tychonoff Product Theorem (Theorem 10.27). SECONDPROOF OF TYCHONOFF PRODUCTTHEOREM.Let X = )( s E S X Sbe given with each X , compact, let p, . X + X , be the sthcoordinate function, and let {x,} be a universal net in X . Fix s, and let A, be any subset of X,. Since the net is universal, it is eventually in p;'(A,) or in (p;'(A,))". Since (p;' (A,))' = p ; l ( ( ~ , ) C )the , net {p,(x,)} is eventually in A, or in (A,)". Thus {p,(x,)} is a universal net ill X,. By Proposition 10.37 and the coiiipactiiess of X, , { p ,(x,)} converges to some member x, of X , . Now let s vary. Forming the member x of X with p, ( x ) = x, for all s and applying Proposition 10.35,we see that x, + x . By Proposition 10.37, X is compact.
47 1
6. Quotient Spaces
6. Quotient Spaces
--
If X is a topological space and is an equivalence relation on X , then we saw in Section 1 that the set X / of equivalence classes inherits a natural topology known as the "quotient topology." If q : X + X / is the quotient map, then a subset U of X / - ~ .is defined to be open in the quotient topology if q - l ( U ) is open in X . The quotient topology is then the finest topology on X / that makes the quotient map continuous. Without some assumption that relates the equivalence relation to the topology of X , we cannot expect much from general quotient spaces. In this section we shall investigate situations in which the quotient space does have reasonable properties. Ultimately our interest will be in four situations, some of which are hinted at in Section 1:
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(i) the passage from a regular topological space to the quotient when the equivalence relation is that x y if x is in I ~ ) C(Proposition ' 10.7), (ii) the passage from a compact Hausdorff space X to the quotient when the equivalence relation is closed as a subset of X x X (to be discussed in Problem 11 at the end of the chapter), (iii) the passage frorri a "topological vector space" or "topological group" to a coset space (to be discussed in the companion volume Advanced Real Analysis), (iv) the piecing together of a "manifold," or a "vector bundle," or a "covering space" from local data (to be discussed in the companion volume Advanced Real Analysis).
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We begin with some general facts. The first is a kind of "universal mapping property" for all quotient spaces. Its corollary describes a situation in which we can recognize a given space as a quotient even if it was not constructed that way: we say that a function F : X + Y is open if F carries open sets to open sets.
Proposition 10.38. (a) Let F : X + Y be a continuous function between topological spaces, let be an equivalence relation on X , and let q : X + X / be the quotient map. Suppose that F has the property that F ( x l ) = F ( x 2 ) whenever xl x2, so that + Y such that F = f o q . Then there exists a well-defined function f : X / f is continuous. (b) The quotient X / is characterized by the property in (a) in the following sense. Suppose that q' : X + Z is any continuous function of X onto a topological space Z such that (i) xl x2 implies q f ( x l )= q f ( x 2 ) , (ii) whenever F : X + Y is a continuous function such that xl x2
-
-
-
-
-
-
-
X . Topological Spaces
472
implies F ( x l ) = F ( x 2 ) ,there exists a continuous function f' : Z + Y with F = f ' o q'. Then Z is canonically homeomorphic to X / -.
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PROOF.In (a), we want to know that f - ' ( U ) is open in X / whenever U is open in Y . By definition of the quotient topology, f ( U ) is open in X / if and only if q-l ( f (u))is open in X . This set is F - l ( u ) , which is open since F is assumed continuous. In (b), suppose Z and q' are such that q' : X + Z has the stated properties. We apply the result of (a) with F taken to be q : X + X / -, and the property of Z gives us a continuous function f' : Z + X / such that q = f' o q'. Then we apply the result of (a) with F taken to be q' : X + Z , and (a) shows that the function f : X / + Z with q' = f o q is continuous. Combining these two equations gives us q = f' o f o q and q' = f o f' o q'. Thus f' o f is the identity on the image of q , and f o f' is the identity on the image of q'. Since q is onto X / and q' is onto Z , f : X / + Z is a homeomorphism.
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Corollary 1039. Let F : X + Y be a continuous function from one topological space onto another, and define xl x2 if F ( x l ) = F ( x z ) . Let q : X + X / be the quotient map, and let f : X / + Y be the continuous map such that F = f o q . If F is open, then f is a homeomorphism and hence Y can be regarded as a quotient of X .
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REMARK.The continuity of f is the conclusion of Proposition 10.38a.
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PROOF.The function f : X / + Y is continuous, one-one, and onto. To see that f is open and hence is a homeomorphism, let an open set U in X / be given. Then F ( ~ - '(u))is open because q is continuous and F is open. Since F ( ~ - ' ( u ) )= f ( q ( q - ' ( ~ ) )=) f ( U ) ,we see that f ( U ) is open. Hence f is open.
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EXAMPLE. Let X = X ,,,X, be a product of topological spaces, fix s in S , and let p, : X + X , be the sthcoordinate function. We shall show that p, is x2 open, so that X , can be regarded as the quotient of X by the relation that xl if p , ~( A ' ) = p , ~( n z )for all s' # s. If U is an open set in X and n is in U ,then we ( U 1 )n . . . n (U,) about x that is contained can find a basic open set Vx = in U . Then p,(Vx) equals Uj if s = sj , and it equals X , if s is not equal to any sj . In either case, p, (V,) is open. Thus p, ( U ) contains a neighborhood of each of its points and must be an open set. So p, is open.
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A key desirable property of a quotient space is that it is Hausdorff. The Hausdorff property is what makes limits unique, after all, and it therefore paves the way to doing some analysis with the space. The next proposition gives a useful necessary condition and a useful sufficient condition.
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Proposition 10.40. Let X be a topological space, let be an equivalence relation on X , and let R be the subset { ( x l x, 2 ) I X I x 2 } of X x X . If the quotient topology on X / is Hausdorff, then R is a closed subset of X x X . Conversely if R is a closed subset of X x X and if the quotient map q : X + X / is open, then X / is Hausdorff.
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PROOF.Suppose that X / is Hausdorff. If ( x , y ) is not in R , then q ( x ) and q ( y ) are distinct points in X / -. Find disjoint open sets U and V in X / -- such ) open sets in that q ( x ) is in U and q ( y ) is in V . Then q-l (u)and q - l ( ~ are X with the property that no member of q-' ( U ) is equivalent to any member of q-l ( V ) . Thus q - l ( U ) x q-l ( v )is an open neighborhood of ( x , y ) that does not meet R. Hence R is closed. Conversely if R is closed and ( x , y ) is not in R , then there exists a basic open set U x V of X x X containing ( x , y ) that does not meet R . The sets q ( U ) and q ( V ) are open in X / since q is open, they are disjoint since no member of U is equivalent to a member of V , and they are neighborhoods of q ( x ) and q ( y ), respectively. Thus X / -- is Hausdorff.
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A special case is the situation with a pseudometric space in which the equivalence relation is that x y if x and y are at distance 0 from one another. A generalization of this relation was given in Proposition 10.7, which said that in y if x is in {y}"' is an equivalence a regular topological space the relation x relation. The corollary to follow gives properties of the quotient space when this equivalence relation is used.
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Corollary 10.41. Let X be aregulartopological space, let -- be the equivalence relation defined by saying that x y if x is in {y}"', and let q : X + X / -- be the quotient map. Then
-
(a) (b) (c) (d)
q is open, and every open set in X is the union of equivalence classes, X/ is regular and Hausdorff, X normal implies X / -- normal, X separable implies X / separable.
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~ O O F First . we show that every open set is a union of equivalence classes. y . If y were not in U , then y Suppose that x is in an open set U in X . Let x would be in the closed set U r and hence {y}" would be contained in U r . Since x y , x is in { y ) ~ and ' , we are led to the contradiction that x would be in U C , hence in U n U c = 0 . So U is a union of equivalence classes. Then it follows that q-' ( q ( U ) ) = U , and the set q ( U ) has the property that its inverse image is open in X . By definition of the quotient topology, q ( U ) is open. Therefore q is an open map. This proves (a).
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To prove the Hausdorffproperty in (b), we shall apply Proposition 10.40. Since (a) shows that q is open, it is enough to show that the subset R = { ( x ,y ) I x y} of X x X is closed. If ( x , y) is not in R , then x is not in {y}C'.By regularity of X , choose disjoint open sets U and V in X such that x is in U and {y}c' g V . Since U and V are unions of equivalence classes and are disjoint, no member of U is eq~livalentto any member of V Therefore (U x V ) f' R = M, and every point of RChas an open neighborhood lying in RC.Hence R is closed. As a result of (a), the open sets in X are in one-one correspondence via q with the open sets in X / -, and the same thing is true for the closed sets. Under this correspondence disjoint sets correspond to disjoint sets. Then regularity in (b), as well as conclusions (c) and (d), follow immediately.
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7. Urysohn's Lemma
According to Proposition 10.31, a Hausdorff topological space has unique limits for convergent sequences and nets. Corollary 10.41 shows that regularity of a space makes it possible to pass to a natural quotient space that is regular and Hausdorff. The following theorem exhibits a special role for the condition that a space be normal.
Theorem 10.42. (Urysol~n'sLemma). If E and F are disjoint closed sets in a normal topological space X , then there exists a continuous function f from X into [0, 11 thatis0 on E andis 1 on F. PROOF. Proposition 1 0 . 5 shows ~ in a normal space that between a closed set and a larger open set we can always interpolate an open set and its closure. Starting from E g FC,we find an open set U1/2with E
C U1/2 C i ~ 1 / 2 ) ~C' F C
Then we can find open sets U1/4and U3/4with Proceeding inductively on n ,we obtain, for each diadic rational number r = m/Zn with 0 < r < 1, an open set U, between E and FC such that r < s implies (u,)" 5 U s . Put U I = X . For each x in X , define f (x) to be the greatest lower bound of all r such that x is in U, . Then f is 0 on E , is 1 on F, and has values in [0, 11. To see that f is continuous, let x be given, let r and s be diadic rationals in (0, 1) with r < f ( x ) < s, and choose diadic rationals r' and s f with r < r' < f ( x ) < sf < s . ( I f f ( x ) = 0, we omit r and r'; if f ( x ) = 1, we omit s and s f.) We are to produce an open neighborhood U of x with f ( U ) C (r,s ) . If U = U,, - (u,,)"', then U is open with r' 5 f ( U ) 5 s f . Thus r < f ( U ) < s as required. We conclude that f is continuous.
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EXAMPLE.In Example 4 of Section 2, we produced a certain Hausdorff regular space X that is not normal, but we deferred the proof that X is not normal until we had Urysohn's Lemma in hand. We can now give that missing proof. As a set, X is the closed upper half plane {Imz > 0) in C. A base for the topology in question consists of all open disks in X that do not meet the x axis, together with all open disks in X that are tangent to the x axis; the latter sets are to include the point of tangency. For a point p on the x axis, the open disks of rational radii with point of tangency p form a countable local base. Arguing by contradiction, suppose that X is normal. Any subset of the x axis in X is closed in X , and we take E to be the set of rationals on the axis and I; to be the set of irrationals on the axis. Urysohn's Lemma (Theorem 10.42) supplies a continuous function f : X + [O, 11 such that f (E)= 0 and f (F) = 1. Define a sequence of functions f , : R + [O, 11 by f n ( x ) = f (x, the notation (x,y) indicating a point in the (x, y) plane. The functions f, are continuous in the ordinary topology on R since the topology on X is the ordinary topology of the half plane as long as we stay away from the x axis. At any point (x , 0) of the x axis, the sets
i),
Urn = { x 301 u
~ ( i(x,;i))
i)
form a local base at (x,U), and (x, is in Urn for n > m . 'I'he continuity of j therefore yields limn f (x , = f (x,0). In other words, limn fn exists pointwise on R and equals the indicator function of the set of irrationals. The sequence { f n } is therefore a sequence of continuous real-valued functions on R whose pointwise limit is everywhere discontinuous. However, Theorem 2.54 implies that the set of discontini~itiesof the limit filnction is of first category in R,and the Raire Category Theorem (Theorem 2.53) implies that R is not of first category in itself. Thus we have a contradiction, and we conclude that X cannot be normal.
i)
Corollary 10.43. If E and F are disjoint closed sets in a compact Hausdorff space X , then there exists a continuous function f : X + [O, 11 that is 0 on E andis 1 on F. PROOF.This follows by combining Proposition 10.16 and Theorem 10.42.
Corollary 10.44. If K and F are disjoint closed sets in a locally compact Hausdorff space X and if K is compact, then there exists a continuous function f : X + 10, 11 that is 1 on K , is 0 on F, and has compact support. PROOF.Using Proposition 10.19, regard X as an open subset of the one-point compactification X*. Proposition 10.20 shows that the compact space X * is Hausdorff. Choose disjoint open sets U and V in X by Corollary 10.22 such that K G U and F G V . Choose L compact in X by Corollary 10.23 such that K 5 Lo. Then M = L f' ( X - V ) is compact in X by Proposition 10.14,
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and K c Lo n U 5 L o n ( X - V)O c ( L f' ( X - V))' = M o . Hence K and X* - M O are disjoint compact sets in X * . Corollary 10.43 produces a continuous g : X* + [O, 11 such that g is 1 on K and is 0 on X* - M O . Since FG v (x-L)UV=X-(L~(x-V))=X-M~X-MO~X*-M~, the function f = g has the required properties.
c
1
8. Metrization in the Separable Case A problem about topological spaces, now completely solved, is to characterize those topologies that arise from metric spaces. Such a space is said to be metrizable. We consider only the separable case and prove the following theorem. Theorem 10.45 (Urysohn Metrization Theorem). Any separable regular Hausdorff space X is homeomorphic to a subspace of the Hilbert cube C = X [O, 11 and is therefore metrizable. PROOF.The Hilbert cube C is seen as a metric space in Example 11 in Section 11.1,Corollary 10.29 identifies it as a product space, and the Tychonoff Product Theorem (Theorem 10.27) shows that it is compact. Let pn : X > [O, 11 be the nth coordinate function. By Corollary 10.10, X is 1101-mal. Fix a countable base B for the open sets. Enumerate the countable set of pairs ( U , V ) of members of I? such that u"' 5 V . To the nth pair, associate by Urysohn's Lemma (Theorem 10.42) a continuous function fn : X + [O, 11 such that f , is 1 on uC'and is 0 on V C. Let F : X + C be defined by "F(x) is the sequence whose nL" term is f, (x) ." We are to show that F is continuous, is one-one, and is open as a function onto F ( X ) . The continuity of p, o F = f , for each n means that F - l p i l of any open set in [0, 11 is open in C . Since F-I of a basic open set in C is the finite intersection of the various ~ - ' ~ ; l ' sof open sets, F is continuous. To see that F is one-one, let x and y be distinct points of x . By Proposition 10.6c, X Hausdorff implies that { y } is closed and hence that {y}' is an open neighhorhood of .x. Choose a hasic open set V containing .x and contained in { y } ' . By Proposition 10.5b and the regularity of X , choose a basic open set U containing x such that UC' (I V . Then ( U , V ) is one of our pairs, and the corresponding function f, has f , ( x ) = 1 and f n ( y ) = 0. Hence F ( x ) # F ( y ) , and F is onc-onc. To see that F carries open sets of X to open sets in F ( X ) , let W be open in X , and fix x in W . Arguing as in the previous paragraph, we can find basic open sets U and V such that x is in U and Uc' c V 5 W . The corresponding f, then has f,(x) = 1 and f n ( V c ) = 0. Hence f n ( W c ) = 0. The set N, of y's such that f, ( y ) > 0 is open in X and contains x . The product (0, 11, x ( X [O, Ilk) is
9. Ascoli-Arzeli and Stone-Weierstrass Theorems
477
open in C , and its intersection with F ( X ) is the same as F (N,) f' F ( X ) . Thus F ( N , ) f' F ( X ) is relatively open in F ( X ). Then F ( x ) lies in this relatively open set, which in turn lies in F ( W ) , and it follows that F ( W ) is a relatively open neighborhood of each of its members.
Corollary 10.46. Every separable compact Hausdorff space is metrizable. PROOF.This is immediate from Proposition 10.16 and Theorem 10.45.
In Section 11.10 we studied Ascoli's Theorem (Theorem 2.56) and the StoneWeierstrass Theorem (Theorem 2.58) as tools for working with continuous functions on compact metric spaces. In turn, these theorems were illuminating generalizations of results about continuous functions on closed bounded intervals of the line, particularly the classical version of Ascoli's Theorem (Theorem 1.22) and the Weierstrass Approximation Theorem (Theorem 1.52). In this section we shall extend these results to the setting of continuous functions on compact Hausdorff spaces. The proof of the extended Ascoli theorem will be our first example of how the Cantor diagonal process gets replaced by an application of the Tychonoff Product Theorem (Theorem 10.27) when one is dealing with an uncountable number of limiting situations at once. The Stone-Weierstrass Theorem in the more general setting becomes in part a tool for dealing with large abstract compact Hausdorff spaces that arise in functional analysis. The starting point for this investigation is the general form of Alaoglu's Theorem: which says that the closed unit ball in the dual X * of a normed linear space X is compact in the weak-star topology; closed subsets of this space play a foundational role in the theory of Banach algebras. We work in this section with a compact Hausdorff space X and with the algebra C ( X ) of bounded continuous scalar-valued functions on X . The scalars may be real or complex. Corollary 10.13 shows that if f is a continuous scalar-valued function on X , then If 1 attains its maximum value on X . The set C ( X ) is a subspace of the normed linear space B ( X ) of bounded scalar-valued functions on X , the norm being 11 f I , = supxExIf ( x ) 1 . Convergence in B ( X ) is uniform convergence. Proposition 10.30 shows that C ( X ) is a closed subspace of B ( X ) and is complete as a metric space. We begin with the extended Ascoli theorem. Let F = { f a } be a set of scalar-valued functions on the compact Hausdorff space X . We say that F is cquicontinuous at x in X if for each t > 0, there is an open neighborhood (I,,, 'A preliminary form of this theorem was given as Theorem 5.58. The general form appears in the companion volume Advanced Real Analysis.
478
X . Topological Spaces
of x such that I f a ( y ) - f a ( x ) l < t for all y in U,,, and all fa in F. We say that F is equicontinuous if it is equicontinuous at each point. Not having a metric to compare different points of X ,we no longer define a notion of "uniform equicontinuity." It is immediate from the definition that any subset of an equicontinuous family is equicontinuous The definition of equicontinuity at x reduces to the definition of continuity if F has just one member, and therefore every member of an equicontinuous family is continuous. As in Section 11.10 the set F i s uniformly bounded on X if it is pointwise bounded at each x E X and if the bound for the values If ( x )I with f E F c a n be taken independent of x .
Lemma 10.47. If F = { f a } is equicontinuous at x in X , then the closure F"' of F i n the product topology on CX is equicontinuous at x . REMARK.Consequently every member of Fc' is continuous at x t
PROOF.Let U,,, be as in the definition of equicontinuity of F at x . For each > 0, the set of functions f E CX such that
for a particular y in X is a closed subset of C X . Thus the set of functions f E CX ,being an intersection of closed sets, such that this inequality holds for all y in U,,, is closcd, and it contains F. In turn, thc intcrscction G of thcsc scts takcn ovcr all t > 0 is closed in CXand contains F. For each t > 0, each g in this closure G salislics 1l1e imcyualily I g ( y ) - g ( x ) I < 2t wllcr~every is i11 U,,, . Therelure G is equicontinuous at x , and so is its subset F'.
Theorem 10.48 (Ascoli-Arzelh Theorem). If { f,} is an equicontinuous family of scalar-valued functions defined on a compact Hausdorff space X and if { f,} has the property that { f, ( x ) } is bounded for each x , then { f,} has a uniformly convergent subsequence. PROOF.We may assume that there are infinitely many distinct functions f,, , since otherwise the assertion is trivial. Let I f n ( x ) I 5 ex for all n , and form the 5 ex]. The space C is compact by the product space C = X [z E C Tychonoff Product Theorem (Theorem 10.27), and we are now assuming that there are infinitely lnany ~ n e ~ l ~ bofe rthe s sequence { f,} in the space. Let S be the image of the sequence as a subset of C. If S were to contain all its limit points, then each f , would have an open neighborhood in C disjoint from the rest of S ; these open sets and Sc would form an open cover of C with no finite subcover, in contradiction to compactness of C . Thus S has a limit point f not in S. By Lemma 10.47 and the remarks before it, the family S U { f } is equicontinuous.
I
9. Ascoli-Arzeli and Stone-Weierstrass Theorems
Let
t
479
> 0. We shall complete the proof by producing an fN in S such that (x) I < t for all x . By equicontinuity find an open neighborhood Ux
I f N (x) - f
Uximplies
for each x such that y
E
and
I fn(y) - fn(x) I < ~ / 3 f ( Y ) - f @)I < 6/3.
for all n
The open sets U, cover X, and finitely many of them suffice to cover, by the compactness of X. Thus there are finitely many points xl , . . . , xk in X with the property that for each y in X , there is some xj with 1 5 j 5 k such that
for all n . Since f is a limit point of S , choose N such that
Then for every y in X, there is an x; such that
Thus f N is within distance t of f ,as asserted.
Corollary 10.49. If X is a compact Hausdorff space, then a subset F = { f a } of C (X) is compact if and only if (a) F is closed in C (X) , (b) the set { f a } is pointwise bounded at each point in X , and (c) F is equicontinuous. In this case, F i s uniformly bounded. PROOF.Suppose that the three conditions hold. Being a subset of C(X), F i s a metric space under the restriction of the metric. By Theorem 2.36, F will be compact if we prove that every sequence has a convergent subsequence. Because of (b) and (c), Theorem 10.48 shows that every sequence in F has a uniformly Cauchy subsequence. By (a) and the completeness of C(X) given in Proposition 10.30, F i s complete as a metric space. Hence the Cauchy subsequence converges to an element of F. Coiiversely suppose that F is conipact. Property (a) follows since coiiipact sets are closed in any metric space. For (b) and the stronger conclusion that F is uniformly bounded, the function f H 11 f l , is a continuous function on the compact set F , and Corollary 10.13 shows that it is bounded. For the equicontinuity in (c), let t > 0 and x be given. Theorem 2.46 shows that F is totally bounded as a metric space. Hence we can find a finite set f i , . . . , fi
480
X . Topological Spaces
F such that each member f of F has supyGsI f ( y ) - f ,( y ) 1 < t for some j. By continuity of each fi , choose an open neighborhood U,,, of x such that I f i ( x ) - f , ( y ) l < t for 1 5 i 5 I for all y in U,,,. I f f is some member of F and if f j is the member of the finite set associated with f ,then y E U,,, implies
in
Hence F i s equicontinuous at each x in X . Now we come to the extended Stone-Weierstrass Theorem. We are interested in showing that certain subalgebras of the algebra C ( X ) of continuous scalarvalued functions on a compact Hausdorff space X are dense in C ( X ) . Except for the dropping of the assumption that X is metric, the assumptions and notation are the same as in Section 11.10. In particular the scalars for the subalgebra and for C ( X ) may be real or complex, and the statement of the theorem is slightly different in the two cases.
Theorem 10.50 (Stone-Weierstrass Theorem). Let X be a compact Hausdorff space. (a) If A is a rcal subalgcbra of rcal-valucd mcmbcrs of C ( X ) that scparatcs points and contains the constant functions, then A is dense in the algebra of real-valued members of C (X) in the uniform metric. (b) If A is a complex subalgebra of C ( X ) that separates points, contains the constant functions, and is closed under complex conjugation, then A is dense in C ( X ) in the uniform metric. REMARKS.Curiously, Urysohn's Lemma (Corollary 10.43) does not play a role in the proof. Instead, the role of Urysohn's Lemma is to ensure that C ( X ) is large in applications, and then the present theorem has serious content. Tlle actual proof of Theorem 10.50 is word-for-word the same as for Theorem 2.58, and there is no need to repeat it.
10. Problems 1 . T.et f and g he contin~lo~l~ filnction~fi-om a topological ?pace intn a Ha~xdoi-ff
space Y . (a) Prove that the set of all points x in X for which f ( x ) = g ( x ) is closed. (b) Prove that if f ( x ) = g ( x ) for all x in a dense subset of X , then f = g . 2. (Dini's Theorem) Let X be a compact Hausdorff space. Suppose that the function f, : X + R is continuous, that f l 5 f2 5 f 3 5 . . . , and that f (x) = lim f,(x) is continuous and is nowhere +oo. Use the defining property of compactness to prove that { f,} converges to f uniformly on X .
481
10. Problems
3.
(Baire Category Theorem) Prove that a locally compact Hausdorff space cannot be the countable union of closed nowhere dense sets.
4.
Prove that a locally compact dense subset of a Hausdorff space is open.
5.
This problem produces a locally compact Hausdorff space that is not normal. Verify the details of the construction. Let X be a countably infinite discrete space, and let Y be an uncountable discrete space. Let X* and Y * be their one-point compactifications, with the added points denoted by x , and y,. The locally compact Hausdorff space is Z = X* x Y * - { ( x , , y,)} with the relative topology. Two closed subsets that cannot be separated by disjoint open sets are A = ( { x , } x Y * )- { ( x , , y,)l and B = ( X * x { y , } ) - { ( x , , y,)}. If X is compact, prove that each infinite subset of X has a limit point.
6. 7.
Let U be the family of subsets of R consisting of all sets { x E R I x < a } , together with 0 and R. (a) Prove that U is a topology for R and that it is not Hausdorff. (It is called the upper topology of R.) (b) If {t,},,D is a net in R,define limsup, t, to be the infimum over n of supmED, ? ,., Prove that a net {t,},,D in R converges to t relative to Z4 if and only if lim sup, t, 2 t .
8. Let ( X , I)be a topological space, and let Ube the upper topology of R as in the f : X + R is said to be upper semicontinuous previous problem. A f~~nction if it is continuous with respect to l a n d U. (a) Prove that upper semicontinuity o f f : X + R is equivalent to the condition that lim sup f ( x , ) 5 f ( x ) whenever x, + x in X . (b) Prove that the function f : R + R that is 1 at x = 0 and is 0 elsewhere is upper semicontinuous. (c) Prove that if f and g are upper semicontinuous functions on X and if c is nonnegative real, then f g and cf are upper semicontinuous. (d) Prove that if { fs}sts is a nonempty set of upper semicontinuous functions on X such that infScs f ( x ) > -cc for all x E X , then infScs f, is upper semicontinuous. (e) Prove that if f is a bounded real-valued function on X , then there exists a unique smallest upper semicontinuous function f with f ( x ) > f ( x ) for all x .
+
-
-
9. Let ( X , 7 ) be a topological space. A function f : X + R is lower semicontinuous if - f is upper seinicoiitiiiuous. 111 this case if f is bounded, let f _= -(- f ) - ,with the right side defined as in the previous problem. Let the oscillation Q of f be defined by Q ( x ) = f ( x ) - f- ( x ) for x in X . -
(a) Why is Qf upper semicontinuous? (b) Prove that this definition agrees with the one in Section 11.9. (c) Prove that f is continuous if and only if Qf is identically 0.
X . Topological Spaces
482
-
10. Let X be a Hausdorff topological space in which there are two disjoint nonempty closed sets A and B. Let be the equivalence relation that identifies all elements of A with each other, identifies all elements of B with each other, and otherwise identifies no distinct points of X. (a) Prove that the subset of pairs ( x , y ) in X x X with x y is closed. (h) Give an example of this kind in which X / is not Hausdorff. 11. Let X be a compact Hausdorff space, and let be an equivalence relation on X such that thc subsct R g X x X of pairs ( x , y) with x -, y is closcd. Lct q : X + X/-- be the quotient map. (a) Pruvt: lur cad1 x E X ll~alq p l q ( x )is a ~ l u s e dsubssL u l X. (b) If U g X is open, prove that V = { x E X I q p l q ( x ) g U } is open by first proving that V c = p2((Uc x X) f' R), where p2 : X x X + X is the projection to the second coordinate. (c) Prove that the compact quotient X/ -- is Hausdorff. (d) Prove that the quotient map is closed, i.e., that closed sets map to closed sets. (e) Is the quotient map necessarily open'! (f) As in one of the examples in Section 1, let X be the interval [-n, n ] , and let S1 be the unit circle in C. Let -- be the equivalence relation that lets -n and n be the only nontrivial pair of elements of X that are equivalent, and form X/ -. Prove that X/ -- is homeomorphic to S 1 and that under this identification the quotient map may be taken to be the function p : X + S' given by p ( x ) = eix.
-
-
-
Problems 12-15 concern connectedness and connected components. Most of the definitions and proofs in the first three are rather similar to those in Chapter I1 ($11.8 and Problems 11-13) for the special case of metric spaces. A topological space X is connected if X cannot be written as X = U U V with U and V open, disjoint, and nonempty. A subset E of X is connected if E is connected as a subspace of X , i.e., if E cannot be written as a disjoint union (E n U) U (E n V) with U and V open in X and with E n U and E n V both nonempty. 12. (a) Prove that a continuous function between topological spaces carries connected sets to connected sets. (b) A path in a topological space X is a continuous function from a closed bounded interval [ a ,b ] into X. Why is the image of a path necessarily comiected? 13. (a) If X is a topological space and {E,} is a system of connected subsets of X with a point xg in common, prove that U, E, is connected. (b) If X is a topological space and E is a connected subset of X, prove that the closure E'' is connected. 14. (a) A topological space X is pathwise connected if for any two points xl and x2 in X, there is some continuous p : [ a ,b ] + X with p ( a ) = xl and p ( b ) = x2. Why is a pathwise-connected space X necessarily connected?
10. Problems
483
(b) A topological space X is called locally pathwise connected if each point has arbitrarily small open neighborhoods that are pathwise connected. Prove that if X is connected and locally pathwise connected, then it is pathwise connected. 15. In a topological space X, define two points to be equivalent if they lie in a connected subset of X. (a) Show that this notion of equivalence is indeed an equivalence relation. The equivalence classes are called the connected components of X . (b) Prove that the connected components of X are closed sets. (c) Prove that the connected components of X are open sets if X is locally connected, i.e., if each point has arbitrarily small connected neighborhoods. Problems 16-17 concern partitions of unity, which were introduced in Section 111.5. An open cover U of a topological space is said to be locally finite if each point of x has a neighborhood that lies in only finitely many members of U. 16. Suppose that U is a locally finite open cover of a normal space X. By applying Zom's Lemma to the class of all functions F defined on subfamilies of U such that F ( U ) ,for each U in the domain of F, is an open set with F (u)"g U and
prove that it is possible to select, for each U in U , an open set VU such that V;' g l J and such that {VU I l J E U }i s an open cover of X . 17. Prove that if Uis a locally finite open cover of a normal space X , then it is possible to select, for each U in U, a continuous function fU : X + [O, 11 such that fU is 0 outside U and such that f u ( x ) = 1 for all x E X.
xu,u
Problems 18-20 establish the Tietze Extension Theorem. Let X be a normal topological space, and let C be a closed subset of X. Suppose that f is a bounded real-valued continuous function defined on C. The theorem is that there exists a continuous function F : X + R such that F = f and sup,,, IF ( x )I = sup,,c If ( x )1 .
1,
18. Let go = f , co = sup,,, Igo(x) I , Po = { x E C I go ( x ) 2 c o / 3 } , and NO = { x E C I g o ( x ) 5 -co/3}. Show that there is a coiitiiiuous fuiictioii f i from X into [-co/3, c o / 3 ]that is co/3 on Po and -co/3 on No. 19. In the previous problem, put gl = go - Fo on C , and let cl = Igl ( x )I . Show that cl 5 $ 0 . When the result of the previous problem is applied to gl in order to produce a function Fl , what properties does F1 have? 20. Show that iteration of the above results produces a sequence of continuous functions F, : X + R such that the series F , ( x ) is uniformly convergent on X and such that the sum F ( x ) = CZoF , ( x ) is continuous. Show also that F has F = f and satisfies IF ( x )1 = If ( x )I .
xZo
1,
X . Topological Spaces
484
Problems 21-28 concern order topologies. Suppose that X is a set with at least two elements and having a total ordering, i.e., a partial ordering < such that
(i) x < y and y < x together imply x = y, (ii) any x and y in the set have either x < y or y
< x.
Define x < y to mean that x < y and x # y. The order topology on X is the topology lor whi~11a base ~ u r ~ s i sor l s all sets { x I x < b } , { x I u < x } , arld { x I u < x < b } . For a nonempty subset Y of X , the terms "lower bound," "upper bound," "greatest lower bound," and "least upper bound" are defined in the expected way. Examples are given by the real line R with its usual topology, the set Q of countable ordinals (as defined in Problems 25-33 at the end of Chapter V) with its order topology, and other examples given below.
21. Prove that every open interval { x I a < x < b} in X is open and every closed interval { x I a < x < b} is closed. 22. Prove that X is Hausdorff and regular in its order topology. 23. Prove that every nonempty subset with an upper bound has a least upper bound if and only if every every nonempty subset with a lower bound has a greatest lower bound. In this case, X is said to be order complete.
24. Suppose that X is order co~nplete. (a) Prove that a nonempty subset Y of X is compact if and only if Y is closed and has a lower bound and an upper bound. (b) Prove that X is locally compact.
25. (a) Prove that if there exist a and b in X with a < b and with no c such that a < c < 6 , then X is not connected, in the sense of Problems 12-15. Let us say that X has a gap when such a and b exist. (b) Prove that if X is order complete and has no gaps, then X is connected.
26. The set X
= [0, 1) U [2,3) is totally ordered. Prove that this X is connected in its order topology, and conclude that the order topology is different from the relative topology for X as a subspace of R.
27. The set X
= [0, 1) U (1,2]is totally ordered. Prove that this X is not connected in its order topology but has no gaps.
28. Let X and Y be two totally ordered sets with at least two elements apiece. Define the lexicographic ordering on X x Y to be the total ordering given by (xl,yl) ( (x2,y2) if xl < x2 or else xl = x2 and yl 5 y2. (a) Prove that the lexicographic ordering on [0, 11 x [O, 11 makes the space compact connected but not separable. (b) The long line is defined to be the product Q x [0, 1) with the lexicographic ordering, where Q is the set of countable ordinals as defined in Problems 25-33 at the end of Chapter V. Prove that the long line is locally compact and connected but not separable.
CHAPTER XI Integration on Locally Compact Spaces
Abstract. This chapter deals with the special features of measure theory when the setting is a locally compact Hausdorff space and when the measurable sets are the Borel sets, those generated by the compact sets. Sections 1-2 establish the basic theorem, the Riesz RepresentationTheorem, which says that any positive linear functional on the space Ccom(X)of continuous scalar-valued functions of compact support on the underlying space X is given by integration with respect to a unique Borel measure having a property called regularity. The steps in the construction of the measure run completely parallel to those for Lebesgue measure if one regards the geometric information about lengths of intervals as being encoded in the Riemann integral. The Extension Theorem of Chapter V is the main technical tool. Section 3 studies more closely the nature of regularity of Borel measures. One direct generalization of a Euclidean theorem is that the space of continuous functions of compact support in an open set is dense in every LP space on that open set for 1 5 p < oo.A new result is the Helly-Bray Theorem-that any sequence of Borel measures of bounded total measure in a locally compact separable metric space has a weak-star convergent subsequence whose limit is a Borel measure. Section 4 regads Ccom(X) as a nonned linear space under the suprenmm noun and ide~ltificsthe space of continuous linear functionals, with its norm, as a space of signed or complex Borel measures with a regularity property, the norm being the total-variation norm for the signed or complex Borel measure.
1. Setting This chapter brings together the measure theory of Chapters V-VI and the theory of topological spaces of Chapter X in a way that takes many of our earlier most interesting examples into account. Specifically we shall study the special features of measure theory when the underlying space is a locally compact Hausdorff space. Our primary example from earlier is that of Lebesgue measure, first on IR1 and then in I R N . In R' we considered also the class of all Stieltjes measures and showed how they are classified by monotone functions satisfying certain properties. We introduced Borel measures in EXN but did not attempt to classify them. Along the way we saw glimpses of some other examples: The unit circle of C can be regarded as [-n, n] if we identify -n and n , and we obtained Lebesgue measure on the circle. As we saw, any open set or any compact set in IRN has
XI. Integration on Locally Compact Spaces
486
a theory of Borel measures associated with it. Most of our concrete examples of such measures when N > 1 came about as a consequence of the changeof-variables formula for multiple integrals. Of particular interest is what we anticipated in Section VI.5 would ultimately come to be regarded as a "rotationinvariant measure on the sphere," the sphere sN-' being a compact metric space. This measure corresponds to the expression d w when Lebesgue measure dx on IRN is written in spherical coordinates and the factor rN-l dr is dropped. In the concrete case of IR3, in which r is the radius, O1 is the latitude from the north pole, and 02 is the longitude, Lebesgue measure is given by d x = r 2 sin 81 do2 dol dr and we have d w - sin Ol doz do1. The change-of-variables formula in the N variable case then reads
LN
f ( x )d x =
Sm1 r=O
f ( r w )r N 1 d w dr
w€SN-l
for every Borel measurable function f 1 0 on IXN. We shall be making sense of d w as a genuine measure on sNP1 in the course of the present chapter. In the opposite direction it is important not to get the idea that all important measure-theoretic examples in mathematics arise from locally compact Hausdorff spaces. Examples that arise from probability theory need not fit this pattern. This fact becomes clearer after one encounters some specific measure spaces that arise in the theory.l Let us turn to the setting of this chapter, a locally compact Hausdorff space X . In order that the measure theory have some connection with the topological-space structure, we shall build our o -algebra out of topologically significant sets. There will be a choice for how to do so, and we come to that point in a moment. We shall follow as much as possible the pattern of the development of Lebesgue measure on an interval of IR1 or on all of IR1, as occurred in Chapter V, in order to construct measures on X . The thing that is missing for general X occurs right at the start: it is the kind of geometric information that goes into regarding the length of an interval as a quantity worthy of study. That is where an ingenious idea comes into play, that of studying linear functionals on the vector space C,,,(X) of continuous scalar-valued functions on X that vanish off a compact subset of X . As in earlier chapters, it will not be important whether the scalars for C,,,(X) are real or complex, and the reader may hx attention on either of these. On an interval [ a , b ] , we thus consider the space C ( [ a ,b ] ) of scalar-valued continuous functions on the interval. The particular linear functional of interest f ( x )d x ,the notation with the R being as is the Riemann integral ( f ) = R lab in Section VI.4. This kind of integral is a fairly simple object analytically; it was ' ~ h measure-theoretic e foundations of probability theory are discussed in the companion volume Advanced Real Analysis.
1. Setting
487
quickly shown to make sense in Theorem 1.26. Our point of view will be that the Riemann integral encodes information about the lengths of all intervals. Why might one consider linear functionals? In the subject of linear algebra, linear functionals play an important role. Two important ways of realizing subsets of Euclidean space are parametric form and implicit form. In the case of a vector subspace of R n , the idea of parametric form leads us to represent the subspace as all linear combinations of members of a spanning set. If we use implicit form instead, the subspace is realized as all vectors satisfying a set of homogeneous linear equations, thus as the kernel of some linear function. The most primitive case of the latter is that there is just one nontrivial equation. Then the linear function has range the scalars, and the linear function is a linear functional. When there are several equations, the subspace is in effect described as the intersection of the kernels of several linear functionals. Thus linear functionals in linear algebra arise in describing vector subspaces, specifically in describing subspaces by limiting their size from the outside. In analysis we have occasionally needed this kind of control of a subspace in proving theorems by an approximation argument. Two nontrivial examples were the proofs in Chapter VI of differentiation of integrals and the proof in Chapter M of the boundedness of the Hilbert transform. In each case we proved a theorem for "nice" functions, and we obtained some estimate for all functions of interest. To connect the one conclusion with the other, we needed to know that the subspace of "nice" functions is dense. Corollary 6.4 was a result of this kind, saying that c,,,(R~) is deiise ill L' ( R N )and in L ~ ( R ~The ) .proof given for Corollary 6.4 was more like an argument using spanning sets, showing that we can pass from c,,,(R~) to simple functions and then recalling that simple functions are dense as a consequence of basic properties of the Lebesgue integral. However, we can visualize another argument of this kind, one with continuous linear functionals. If one could prove, for any proper closed vector subspace of our total space of functions (L' or L2 or something else), that there is a nonzero continuous linear functional on the total space vanishing on the closed subspace, then we could test whether a given vector subspace is dense by examining the effect of continuous linear functionals when restricted to the subspace. Historically this idea began to be applied in analysis in the early part of the twentieth century at about the same time that people began thinking frequently about spaces of functions and not just individual functions. The key general existence tool for such continuous linear functionals was the Hahn-Banach Theorem, which we shall take up in Chapter XII. In any event, out of this confluence of ideas arose the idea of considering continuous linear functionals on Ccom(X)as capturing enough information about X to make measure theory possible. The continuity of a linear functional will actually be somewhat concealed in what we do for most of this chapter, and
488
XI. Integration on Locally Compact Spaces
instead we impose on the linear functional the natural condition that it needs to satisfy in order to provide a notion of integration-that it be > 0 on functions > 0. Let us be more precise about the definitions. Let X be a locally compact Hausdorff space, and let C C o m ( X )be the vector space of scalar-valued functions on X that vanish outside some compact set For a specific ft~nctionf ,the support off is the closure of the set where f is not zero. The members of CCo,(X) are then the continuous scalar-valued functions on X having compact support. A linear functional & on C,,,(X) is said to be positive if & ( f ) 2 0 whenever f 2 0. The Riesz Representation Theorem, to be stated formally in Section 2 with all details in place, will say that to any such & corresponds a measure p on a certain o-algebra of "topologically significant" sets such that f(f)=
jxf d~
for all f
E
c,,,(x).
The "topologically significant" sets have to include the sets necessary to make each f in C C o m ( X )measurable. At first glance it might seem that the smallest o-algebra containing the open sets is the right object. But in fact this o-algebra is unnecessarily large. In an uncountable discrete space, we do not need to have every subset measurable in order to have all the functions of compact support be measurable. Accordingly we define the a-algebra B ( X ) of Bore1 sets of X to be the smallest u-algehra containing all compact silhsets of X . The plan of attack now follows the steps in the construction of Lebesgue measure. We take the compact subsets of X to be the analog of the bounded intervals in IR1, and we thus define the "elementary sets" in X to be the sets in the smallcst ring K ( X ) containing all thc compact scts. In thc casc of EX1,cvcry sct in the ring generated by the bounded intervals is a finite disjoint union of sets that are the difference of two bounded intervals. We shall prove for X in Section 2 that every member of K ( X ) is a finite disjoint union of sets that are the difference of two compact sets. For Kt1,we defined the measure of the difference of two bounded intervals to be the difference of their lengths as soon as the second interval is contained in the first; this was no loss of generality because the intersection of two bounded intervals is a bounded interval. The measure of a finite disjoint union was defined as the sum of the measures. We showed that this was well defined, and then we had a finite-valued nonnegative additive set function on a ring of sets. For X , we define the measure of a compact set K by the natural formula
where IK as usual is the indicator function of K . The intersection of two compact sets is compact, and thus we can define the measure of K 1 - K2 for K 1 and K2
1. Setting
489
compact, to be p ( K 1 )- p ( K 1 n K z ) . We define the measure of the disjoint union of such sets K 1 - K2 to be the sum of the measures. We have to prove that this is well defined, and then we have a finite-valued nonnegative additive set function p on the ring K ( X ). The next step for IR1 was to prove complete additivity on the ring generated by the bounded intervals. With X , the problem is the same; we are to prove complete additivity on the ring K ( X ) . Suppose that this has been done. Since p is everywhere finite-valued on K ( X ) , we can apply the Extension Theorem (Theorem 5.5) to extend p to the generated o-ring. Either this o-ring is already the generated a-algebra B ( X ),or Proposition 5.37 supplies a canonical extension to a measure on the generated o-algebra B ( X ) . This completes the construction of the measure p on B ( X ) . It is then a fairly easy matter to see that C ( f ) is recovered as the integral of f if f is in Cco,(X): In the case of IR1, we carried out this step by first establishing the Fundamental Theorem of Calculus for the Lebesgue integral of a continuous function; the argument appears at the end of Section V.3. A more direct argument would have been possible, and that direct argument works for general X . Thus the problem comes down to proving that the set function, as defined on the ring of sets, is actually completely additive on that ring. In the case of IR1,that complete additivity was an easy consequence of "regularity" of Lebesgue measure on the ring generated by the bounded intervals; in other words, the measure of any set in the ring could be approximated from within by the measure of compact sets in the ring and from without by the measure of open sets in the ring. Exactly the same approach works for general X , but the regularity has to be established. Quantitatively the construction of the measure comes down to defining w ( K ) for K compact as above and then proving three identities: (i) p ( K 1 ) + p ( K 2 )= p ( K 1 U K 2 ) + p ( K 1 n K 2 ifK1 ) andK2arecompact, (ii) sup l ( f ) = F ( K ) - p ( K - U ) if U is any open set contained in f ~CCOrn(X), Oifi l u
(iii)
some compact set K , sup p ( K ) = sup KCU, K compact
C ( f ) if U is open and has compact closure.
f ECcom(X), O 5 f 51u
Identity (i) and an elementary but lengthy computation in elementary set theory together allow us to prove that w is well defined on the ring K ( X ) under the definitions above. Once p has been so extended, the right side of (ii) is just p ( U ) if U is open with compact closure. Thus (iii) says that p ( U ) is the supremum of p ( K ) over compact sets K contained in U ,provided U is open and has compact closure. Since p ( U ) is trivially the infimum of p ( V ) for open sets V in K ( X ) containing U , this is the regularity conclusion for U . It is easy to see that the subclass of K ( X ) for which regularity holds is a ring and contains the compact
XI. Integration on Locally Compact Spaces
490
sets, and hence regularity is established for K ( X ). When the locally compact Hausdorff space X is a metric space, the three identities above are fairly easy to prove. When X is metric, any indicator function IK for K compact is the pointwise decreasing limit of members of Cco,(X) that are > 0. In fact, if D( . , K ) is the distance to K ,then the sequence { f,} with f,(x) = max{O, 1-n D ( x , K ) }has the required properties A little trick proves in this case that p ( K ) = limn l (f,). To prove (i), we choose such sequences { f,} and {g,} for K1 and K2. If q~ is a member of Ccom(X)that is identically 1on the union of the supports of f i and gl ,then fn gn = minIfn gn, cpl (maxIfn gn, cpl - cp) decomposes f, g, into the sum of such sequences for K 1 U K2 and K 1 f' K 2 , and identity (i) follows from linearity of l and a passage to the limit. Identities (ii) and (iii) follow from equally simple arguments. The difficulty for a general locally compact Hausdorff space X is that the indicator function of a compact set need not be a pointwise decreasing limit of a sequence of continuous functions. The technicalities introduced by this fact have the effect of making the proofs of (i) ,(ii), and (iii) be more complicated, but these complications need not obscure the line of argument that is so clear in the metric case.
+
+
+
+
+
2. Riesz Representation Theorem Throughout this section we fix the locally compact Hausdorff space X . We continue to let Cco,(X) be the space of continuous functions of compact support, K ( X ) be the ring of elementary sets, and B ( X ) be the a-algebra of Borel sets. A subset E of X is said to be bounded if it is contained in a compact set, hence if E"' is compact; it is o-bounded if it is contained in the countable union of compact sets. The class of all o-bounded Borel sets is a o-ring containing K ( X ) , and it is therefore the smallest o-ring containing K ( X ) . A measure on the Borel sets of X is called a Borel measure if it is finite on every compact set. A Borel measure is said to be regular if it satisfies p(E) =
sup
/1(K)
for every set E in B ( X )
KCE, K compact
w(E)=
w(U)
inf
for every o -bounded set E in U ( X ) .
W E , U open o-bounded
Theorem 11.1 (Riesz Representation Theorem). If l is a positive linear func, there exists a unique regular Borel measure p on X such tional on C c o m ( X )then that for all f E c,,,(x). l t f )= dl*
lxf P
2. Riesz Representation Theorem
49 1
EXAMPLES. (1) If X is the line EX1 and l is given by Riemann integration l ( f ) = 72 lab f ( x )d x whenever [ a , b] contains the support of f , then C is a positive linear functional on C,,,(IW~) and the corresponding is Lebesgue measure. (2) If X = s2is the unit sphere in EX3,parametrized by latitude Q1 from 0 to n and by longitude H2 from 0 to 237, then l (f ) = 72 f ( H I , &) sin H1 do2 dB1 is a positivc lincar functional on c ( s 2 ) ,and thc corrcsponding mcasurc, which is written dw in the same way that Lebesgue measure is written as dx ,is a rotationinvariant measure on the sphere such that iR3 F ( x )d x = Js2 F ( Y < , ) ) Y ~d<,)d~ for every nonnegative Borel function on RN . The proof of this identity and of the rotation invariance will be indicated in Problem 5 at the end of the chapter. (3) If X is general and if p is a regular Borel measure on X , then C ( f ) = 1 , f d p is a positive linear functional on C,,,(X). The proof of Theorem 1 1 . 1 will occupy the remainder of this section. We begin with some lemmas clarifying the nature of the ring K ( X ) , the linear functional C , and general compact and open subsets of X . Then we recall the definition of y ( K ) for compact sets and establish the identities (i), (ii), and (iii) in Section 1. Finally we give the details of how the three identities imply the theorem. We begin with information about the ring IC(X).
Lemma 11.2. The members of the ring K ( X ) are exactly all finite disjoint unions of subsets of V of the form K - L with K and L compact and L g K g V . The ring K ( X ) may be characterized also as the smallest ring containing all bounded open subsets of X. PROOF.If K 1 - L1 and K2 - L2 are two sets of the same kind as V in the statement of the lemma, then the identity ( K 1 - L1) U (K2 - L2) = ((K1U K2) - (L1 U L2)) U ( ( K 2 nL1) - (L1 n L 2 ) ) U ((K1n L2) - (L1 n L2)) shows that a union of two such sets is a disjoint union, and the identity ( K i - L i ) - (K2 - L2) = ( ( K i n L2) - ( L i n L2)) U ( K i - ( L i U ( K i n K2.1)) shows that the difference of two such sets is such a set. Therefore the collection of all such sets is a ring of subsets of X . This ring contains all compact sets because any compact set K is of the for~nK - D , and hence this ring equals K ( X ). Any open bounded set U is the difference of the compact sets UC'and UC'- U , and hence it lies in K ( X ). In the reverse direction Corollary 10.23 shows that any compact set K is contained in the interior Lo of some compact set L . Thus K is the difference of the bounded open sets Lo and Lo - K , and K ( X ) is contained in the smallest ring containing all bounded open sets.
492
XI. Integration on Locally Compact Spaces
Next we observe some properties of the linear functional l . It is to be understood throughout the section that & is a positive linear functional on Cc,,(X) . The positivity implies that l ( f - g ) > 0 if f - g > 0 ; the linearity therefore gives &( f ) 2 & ( g )for f g . The linear functional has a kind of continuity property, according to the following lemma.
Lemma 11.3. Let K be a compact set, and let { f,} be a sequence in C c o m ( X ) converging uniformly to a member f of Ccom( X ) in such a way that support( f,,) 5 K for all n . Then limn L ( f , ) exists and equals l ( f ). PROOF. Corollaries 10.23 and 10.44 show that there exists a function F in Cc,,(X) such that F takes values in [0, 11 and is 1 on K . Since f , - f < I f , - f I and-(fn-f)
whcrc c , = 1 1 f , - f llsup Thc assumcd uniform convcrgcncc mcans that c , tends to 0. Since L ( F ) is some fixed constant, the asserted convergence of & (f,) follows.
Lemma 11.4 (Dini's Theorem). If { f,} is a sequence of functions in C c o m ( X ) decreasing pointwise to 0, then { f,} converges uniformly to 0. PROOF. Because of the pointwise decrease to 0, all the functions ,f, have support contained in the compact set K = support(f l ) . Let r > 0 be given, and let U , be the open set where the continuous function f , is < t. The pointwise decrease implies that the Un are increasing with n , and the limit of 0 implies that each x in K is in some U,. Thus the open sets U , form an open cover of K . By compactness, there is a finite subcover. Since the sets IT, are increasing, some particular U N covers K . Then 11 fnllsup < t for n > N . The final step of preparation is to observe some properties of compact and open sets. A bounded subset of X is said to be a Gg if it is the countable intersection of bounded open sets. It is said to be an F, if it is the countable union of compact sets. We shall be especially interested in compact Gs's and in open bounded F, 's.
Lemma 11.5. Let f be a member of Cc,,(X) with values in [0, 11. If r > 0, then the set where f is > r is a compact Gs. If r > 0, then the set where f is > r is a bounded open F, . PROOF.The set where f is > r is closed because of continuity, and this closed set is a subset of the compact support. Hence the set is compact. Similarly the set where f is > r is open because of continuity, and this open set is a subset of the compact support. Hence the set is bounded.
2. Riesz Representation Theorem
493
When r 10, the set where f is > r is the union, for n > 1, of the sets where f is > r F o r r > 0 when N is large enough so that r - $ > 0, the set where f is 1u is the intersection, for n > N , of the sets where f is > r The lemma follows.
+ i.
i.
Lemma 11.6. (a) If K is a compact Gg, then there exists a decreasing sequence of bounded open sets U, such that U, 2 u,":, for all n and r-),"==,U,= K . (b) If U is a bounded open F,, then there exists an increasing sequence of compact sets K, such that K, c K,",, for all n and U,"=, K, = U. PROOF.For (a), let {V,} be a sequence of bounded open sets with intersection K . This is possible since K is a Gg. Without loss of generality we may assume that the V, decrease with n . We define the sequence {U,} inductively on n . Put U1 = Vl. If U, has been constructed, use Corollary 10.22 to find an open set v,' such that K g V; and v;"' g U,, and then define U,l+l = V; n V,+I. Then the sets U, have the required properties. For (b), let {L,} be a sequence of compact sets with union U . This is possible since U is an F,. Without loss of generality we may assume that the L, increase with n. We define the sequence {K,} inductively on n. h t K1 = L 1. If K, has been constructed, use Corollary 10.23 to find an open set V,' such that U 2 v;"' and V,' 2 K,. The compact set Lk = v,'" has (Lk)" 2 V,'. If we define K,+l = Lk U L,+l ,then the sets K, have the required properties.
Lemma 11.7. (a) If K is a compact G g ,then there exists a decreasing sequence of functions f, in Cc,,(X) with values in [O, 11 such that each f, is 1 on some neighborhood of K and lim f, = IK pointwise. (b) If U is a bounded open F,, then there exists an increasing sequence of functions f, in Cc,,(X) with values in [0, 11 such that each f, has compact support contained in U and lim f, = Iu pointwise. PROOF.For (a), apply Lemma 11.6a to choose a sequence of bounded open sets U, with intersection K such that U, 2 u,c\~ for all n. Using Corollary 10.44,let g, be a member of Cc,,(X) with values in [0, I] such that g, is 1 on uC,: and is 0 off U,, and put f, = min{gl, . . . , g,}. Then the functions f, have the required properties. For (b), apply Lemma 11.6b to choose a sequence of compact sets K, with union U such that K, 5 Kl+l for all n . Using Corollary 10.44, let g, be a member of Cc,,(X) with values in [O, 11 such that g, is 1 on K, and is 0 off K,"+, , and put f, = max{gl, . . . , g,}. Then the functions f, have the required properties.
494
XI. Integration on Locally Compact Spaces
Now we begin the proofs of the three identities in Section 1. If K is compact, let
P(K) = inf C(f),
the infimum being taken over all f in Ccom(X)such that f > IK. Since [(mini f , 1)) 5 C( f ) , there is no harm in considering only those f ' s taking values in [0, 11. It is immediate from this definition and the positivity of e that p is nonnegative and monotone in the sense that K' 5 K implies p(Kf) 5 p ( K ) . Thc ncxt lcmma is thc kcy to bcing ablc to provc thc thrcc idcntitics in Scction 1.
Lemma 11.8. If K is a compact subset of X, then the infimum of C(f) over all f in Cco,(X) such that f > IK equals the infimum of & ( f ) over all f in Cc,,(X) with values in [O,1] such that f 1IN for some neighborhood N of K depending on f . REMARK.In particular, p(K) can be computed by using only functions f 1IK that are equal to 1 in some neighborhood of K .
PROOF.The problem is to show that the first infimum II is not less than the second infimum 1 2 . Let c > 0 be given. Choose f in Cco,(X) with values in [O, 11 such that f > IKand t ( f ) 5 Il + c , and let L be the set where f is > 1. Lemma 11.5 shows that L is a compact Gg, and Lemma 11.7a produces a decreasing sequence of functions .f, in C,,,(X) with values in 10, 11 such that each f, is 1 on some neighborhood of L and lim f, = ILpointwise. Then the sequence {max{f,, f }} is pointwise decreasing with limit m a x { l ~f, } = f , and hence {maxif,, f } - f } is a pointwise decreasing sequence in C,,,(X) with limit 0. By Dini's Theorem (Lemma 11.4), the sequence {max{f,,, f } - f } converges uniformly to 0, and hence C (max{f, , f }) decreases to C( f ) . For some sufficiently large no, we therefore have l(max{f,,, f }) 5 Il + 2 c . The function maxi f,,, f } is one of the functions that figures into 12, and thus I2 5 Il 2 t . Since t is arbitrary, I25 Il.
+
Lemma 11.8 puts us in aposition to prove identity (i) in Section 1 and to deduce that p extends in a well-defined fashion to a nonnegative additive set function on K(X). We make use of the formula a + b = min{a, b} + max{a, b}, from which it follows that a = min{a, b} (max{a, b} - b).
+
Lemma 11.9. If K1 and K2 are any two compact subsets of X, then
REMARK.The argument in Lemma 11.8 adapts to give a quick proof of the present lemma when X is a metric space. In the metric case we can find a decreasing sequence { f,} of functions 5 1 in C,,,(X) with pointwise limit IK,. If
2. Riesz Representation Theorem
495
f 1I K 1 ,then the proof of Lemma 11.8 shows that f, f converges uniformly to f and hence L( f, f ) decreases to L( f ) . It follows that L( f,) decreases to p(K1) whenever fn decreases to I K l . If we similarly choose {g,} decreasing to I K ~and choose, by Corollary 10.44, a function rp E C,,,(X) with values in [O, 11 that is identically 1 on the support of f i + gl, then the formula stated just above shows that f,, g,, = minIf,, g,,, (n} (maxi f,, g,,, (n} - (n). The , the second term first term on the right side decreases pointwise to I K l U K 2and decreases to I K l n K 2 .Thus a passage to the limit in the formula l(f,) + l(gn) = L(min{f, g,, rp}) L((max{f, g,, rp} - rp)) immediately yields the result of the present lemma.
+
+
+
+
+
+
+
PROOF.Let f and g be functions in CCom(X)with values in [0, 11 such that f > I K 1 and g > I K z , and choose, by Corollary 10.44, rp E C,,, (X) with values in [0, 11 that is identically 1 on the support of f g . Then we have f g = mini f g , rp} (max{f g , rp} - rp). The first term on the right side K z the , second term is > I K ln ~ 2Therefore . is > l K I U and
+
+
+
+
+
Taking the infimum over f and then over g , we obtain
For the reverse direction let F be a member of Ccom(X)with values in [0, 11 that is > I K I U Kand z is equal to 1 at least on some open set [J containing K1 U K2. Similarly let G be a member of Cc,(X) with values in [0, 11 that is > I K , " ~ , and is equal to 1 at least on some open set V containing K1 n K2. Lemma 11.8 shows that F and G are the most general functions of a kind needed for the computation of p(K1 U K2) and F ( K ~f' K2). The sets U and V have compact closure in X since they are subsets of the supports of F and G . Choose, by Corollary 10.44, q~ E Ccom(X)with values in [0, 11 that is identically 1 on the support of F G . Let Vo be an open set with K1 f' K2 C VO C vOC1 C V. Then (K2 - VO)n Kl = K2 n V," n Kl g Vo n V," = D. So there exists an open set W such that K2 - Vo g W g w"' g K,". We define f and g to be members of C,,,(X) having compact support contained in U and having with values in [0, 11 such that
+
and
XI. Integration on Locally Compact Spaces
496
The functions f and g exist by Corollary 10.44 if it is shown that the closed sets K1 and w"' are disjoint and the closed sets K2 and support( f ) - V are disjoint. The sets K1 and w"' are disjoint since w"' 5 KF. For K2 and support(f) - V, we observe that support(f) C ((WC')C)C'C (WC)"' = WC C (K2 - VO)' = Vo U K," 5 V U K,". Therefore
We conclude that f and g exist. By inspection, f 1I K l and g 1I K L , from which f g 1I K l I K 2 . Then min{f + g , q} is 1 on K1 U K2 and is 0 off U . Since F is 1 on U , we obtain
+
Since f equals f
+
+ g > I K ] + I.y2 = I K ~ U K+~IKl"K2, the function maxi f + g , rp} +g
-
-
rp
1 on K1 U K2, and this in turn is 5 1 everywhere. Let us see that
everywhere. The only points x at which (**) could possibly fail are those where G(x) < 1, hence points of VC. At such points the definition of g shows that f (x) + g(x) 5 1. If also x is in U , then q(x) = 1 and we compute that maxi f (x) + g(x), q(x)} - q(x) = 1 - 1 = 0. Thus (**) holds at points of U n Vc. At points of Uc n Vc, the equality f (x) = g(x) = 0 implies that max{f (x) + g(x), rp(x)} - rp(x) = rp(x) - rp(x) = 0. Thus again (**) holds, and hence (**) holds at every point of VC,therefore everywhere. Addition of (*) and (**) gives f g 5 F G everywhere. Therefore
+
+
Taking the infimum over F and then over G gives p(K1 U K2) p(K1) p(Kz) and completes the proof of the lemma.
+
+ p(K1 n K2) 1
Lemma 11.9 yields by iteration a corresponding formula with the sum of n terms on each side. This extension of Lemma 11.9 is a computation in Boolean algebra involving no analysis at all-only the fact that the collection of compact sets is closed under finite unions and intersections. The details are carried out in the next lemma.
2. Riesz Representation Theorem
Lemma 11.10. If K 1 . . . , Kn are compact subsets of X , then
PROOF.The argument is by induction on n , the base case of the induction being the case n = 2 that was settled by Lemma 11.9. Thus let n > 2, and assume the identity for the case n - 1. The inductive hypothesis gives
We shall prove by induction on r
> 1 that
the base case of this induction being r = 1,where this identity reduces to (*). The proof for the case r = n will complete the inductive step for the outer induction and thereby will complete the proof of the lemma. To pass from r to r 1 in the inner induction, the question is wlletller
+
The union of the two sets on the left here is the first set on the right side. In view of Lemma 11.9, this formula will follow if it is shown that the second set on the right side is the intersection of the two sets on the left. The intersection of the two sets on the left side is equal to
+
A term in the union in this expression is an intersection of at least r 1 of the sets K1, . . . , K,, the last of which is K n , namely the ones corresponding to indices i i , . . . , ii and i, = n . Every intersection of exactly r 1 of the sets K1, . . . , K, occurs if the last one is Kn because we can take i l = i i , . . . , i,-1 = i i P l . Any intersection of more than r 1 sets is contained in one with exactly r 1 sets,
+
+
and thus
(**I
(n;+jK,)
equals Uliili...ii,+l=n
+
, as asserted.
498
XI. Integration on Locally Compact Spaces
A further formality is the derivation from these results that p extends in a welldefined fashion to a nonnegative additive set function on the ring K ( X ) . Again no analysis is involved, only the one additional fact that the intersection of two sets of the form K - L with K and L compact is again of this form, specifically that (K - L) n (K' - L') = (K n K') - (L U L').
Lemma 11.11. The set function y extends in a well-defined fashion to a nonnegative additive set function on K(X) under the definition
whenever Kj and Lj are compact with Lj Kj for each j with 1 5 j 5 n and the sets K1 - L1, . . . , K, - Ln are painvise disjoint. REMARKS.Lemma 11.2assures us that every member of K (X) is of the form in this lemma. The subtlety of the lemma arises from the fact that the sets Kj need not be disjoint. PROOF.First let us see that y is well defined in the case j = 1, i.e., that K' - L' = K - L with L' 5 K' and L 5 K implies y ( K f ) - y ( L f ) = p ( K ) - //,(I,).We are to show that p(K') + p(1,) = p ( K ) + j~(l.'),and 1,emma 11.9 shows that it is enough to show that K' U L = K U L' and K' f' L = K n L'. Supposex isin K ' U L . I f x i s i n L , t h e n x isin K,henceinK UL'. I f x i s i n K' instead, then either x has to be in L' in the case that x is not in K' - L' or x has to bc in K in thc casc that x is in K' - L' = K - L. So K' U L 5 K U L'. If x is in K' n L ,then x is not in K - L and must be in L' in order to avoid being in K' - L'. S o x is in L U L' 5 K U L'. Reversing the roles of K' - L' and K - L , we see that K' U L = K U L' and K' n L = K n L'. Next suppose that K' - L' = Uj"=, (Kj - Lj) with L' 5 K', Lj 5 Kj for each j, and the sets Kj - Lj disjoint. We are to show that y ( K f ) - y(Lf) = C,"=1 (p(Kj) - p(Lj)),i.e.,thatp(K1)+ CJn_lp-L(Lj)= p(L') + CJn=l p(Kj). The argument will generalize that in the previous paragraph: The set K' - L' has complement L' U K f C and , therefore the given condition of disjointness means that n
X = (L' U KfC)U
U(Kj
-
Lj)
j=1 disjointly. h t L,+l = K' and K,+l = L', so that we are asking whether
(*I
2. Riesz Representation Theorem
In view of Lemma 11.lo, it would be enough to show that
for 1 < k < n + 1. The left side is the set of x lying in at least k of the sets L i , and the right side is the corresponding set for the Ki's. Thus it is enough to prove that the set of x lying in exactly r sets Ki is contained in the set of x lying inexactlyr sets L i , f o r l < r < n + 1. We check this condition separately for the three cases x E L', x $ K', and x E K' - L'. From (*) we see that x in L' U KfCimplies that x is not in any Kj - Lj for 1 5 j 5 n. Hence for the first two cases, x is in Lj with 1 5 j 5 n if and only if x is in Kj. C a s e l . x E L ' . F o r x t o b e i n r o f t h e ~ e t s K .~.,. , K,+l,xmustbeinr - 1 of the sets K1, . . . , Kn,hence in r - 1 of the sets L1, . . . , L n . Since x is in L', it Therefore x is in r of the sets L 1 , . . . , Ln+1. is in K' = Case2. x $ K'. Forx t o b e i n r o f t h e s e t s K 1 , . . . , K,+l,x m u s t b e i n r o f the sets K1, . . . , K,, hence in r of the sets L1, . . . , L,. Since x is not in K', it is not in L,+l. Therefore x is in r of the sets L1, . . . , L,+l. Case 3. x E K' L'. Since x is not in L' U K", (*) shows that x is in exactly o n e K j - L j w i t h l < j < n . F o r x t o b e i n r o f t h e ~ e t s K.~. ., , Kn+l,xmust be in r. of the sets K1, . . . , K,, hence in r. - 1 of the sets L1, . . . , L,. Since x is in K' = L,+l, it is in r of the sets L1, . . . , L,+l. For the general case, suppose that U,"=, (Kj/ - Li) = Uy=?=, (Kj - Lj). Intersecting both sides with Ki - Li , we obtain
The case just proved shows that
and hence
Similarly
Therefore Cr=l I"(K~- Li)
=
(Kj - Lj), and the proof is complete.
XI. Integration on Locally Compact Spaces
500
In short order, we can now prove identities (ii) and (iii). Lemma 11.12 will prove (iii) ,and Lemma 11.13 will prove (ii) .
Lemma 11.12. If U is any bounded open subset of X , then sup
L(g)=
ggCcom(X), 05g5lu,
sup p ( K ) = KGU, K compact
sup
L(f).
f gCcom(X), O5f 5 l u
support g L U
PROOF.Let S 1 ,S2, S3 be the three suprema in question. We first check that S1 < S2 5 S3. If g contributes to S l , then g < Isupport < IU.If h E Ccom(X) has Isupport ( h , then g ( h and hence L(g) ( L ( h ). Taking the infimum over all such h , we obtain L(g) 5 support g ) ( S2. Taking the supremum over all g therefore gives S1 5 S2. Next if K is compact with K c U , Corollary 10.44 allows us to find f E CcO,(X) with values in [0, 11 such that f is equal to 1 on K and equal to 0 on U c . Then IK f I U . The definitions of p ( K ) and S3 yield p ( K ) 5 t (f ) ( S3. Taking the supremum over all K therefore gives S2 5 S3. To complete the proof, we show that S1 2 S3. Let t > 0 be given. Choose f in Cc,,(X) such that 0 5 f 5 IUand l (f ) 1S j - t, and lct V bc thc sct where f is > 0. Lemma 11.5 shows that V is a bounded open F, , and Lemma 11.7b produces an increasing sequence of functions fnin C C o m ( Xwith ) values in [0, 11, each with support some compact subset of V , such that lim f n= Iv pointwise. Then the sequence {minifiz, f }} is pointwise increasing with limit minil", f }. If x is a point where Iv( x ) < f ( x ) , then f ( x ) > 0, x is in V , and Iv( x ) = 1, contradiction. So there is no such point, and min{Iv, f } = f . Therefore the sequence { f - mini f,, f }} is a pointwise decreasing sequence in Ccom(X)with limit 0. By Dini's Theorem (Lemma 11.4), the sequence { f - min{f, , f }} converges uniformly to 0, and hence L (mini f, , f }) increases to L ( f ). For some sufficiently large no, we therefore have L(min{f,, , f }) > S3 - 2 t . The function min{f,,, f } is one of the functions that figures into S 1 , and thus S1 2 .!?(min{f,,, f )) 2 S3 - 2 t . Since t is arbitrary, S1 2 S3.
Lemma 11.13. Let p be extended to a nonnegative additive set function on K ( X ) as in Lemma 1 1 . 1 1 . Tf I/ i s a bounded open subset of X , then / I ( [ / )= SUPKGU, K compact p ( K ) . PROOF.For the bounded open set U ,let S 1 , S2, S3 be the three equal suprema of Lemma 11.12. By definition, p ( U ) = p ( L ) - p ( L - U ) for any compact set L containing U , and we are to prove that p ( U ) = S2. If K is a compact subset of U , then K U ( L - U ) is a disjoint union contained in L , and we have p ( K ) p ( L - U ) = p ( K U ( L - U ) ) 5 p ( L ) . Taking the supremum over all such K , we obtain S2 p ( L - U ) 5 p ( L ) , i.e., S2 5 p ( U ) .
+
+
2. Riesz Representation Theorem
50 1
Let Iz be any member of C,,,(X) with values in [0, 11 such that Iz > ILPU and such that h is 1 on an open neighborhood N of L - U . Then L g N U U. For each point x of U, find an open neighborhood Ux of x with u:' 5 U . Then N and the U,'s form an open cover of L , and there is a finite subcover. Let us say that L 5 N U Ux, U . . . U Uxn.The set K = u,":U . . . U u : is a compact subset of U , and L g N 1-1 K . Choose, by Corollary 10.44, a function f E C,,,(X) with values in [0, 11 such that f is 1 on K and is 0 off U. This function has 0 5 f 5 I u . Since f i s l o n K a n d h i s l o n N , h + f i s ? I o n L . Hence p ( L ) i l ( h f ) = t ( h ) l ( f i l ( h ) S3. Thus p ( L ) i p ( L - U) S3 and p ( U ) S3. Since S3 - S2 by Lemma 11.12, p ( U ) - Sz as required.
+
+
+
+
PROOFOF EXISTENCE IN THEOREM11.1. If K is compact, we define p (K) ,just as we did earlier in this section, to be the inlirnurn of .t(f ) over all f in C,,,(X) such that f 2 IK . Lemma 11.11 shows that p extends, necessarily in a unique fashion, to a well-defined nonnegative additive set function on K(X). Consider the set C of all members E of K(X) satisfying the following regularity property: for each t- > 0 , there exist compact K and open bounded U with K 5 E 5 U and p ( U - K) < t . Lemma 11.13 shows that every open bounded set is in C. We show closure of C under finite unions. If E l and E2 are in C, then we can choose K1 and K2 compact and Ul and U2 bounded open such that K1 g E l 5 U1, K2 5 E2 5 Uz, p(Ul - K1) < t / 2 , and p(Uz - Kz) < t / 2 . Then KlUK2 C ElUE2 c UlUU2and(UlUU2)-(KlUK2) C (Ul-Kl)U(Uz-K2). It follows that ~ ( ( UUI U2) - (KI U K2)) 5 ~ ( ( U-I KI)) p((U2 - K2)) < t , and C is closed under finite unions. We show closure of C under differences. If E l and E2 are in C,then we again choose K1 and K2 compact and U1 and U2 bounded open such that K1 5 E l 5 U1, K2 C E2 C U2, w(U1 - K1) < t / 2 , and p(U2 - K2) < t / 2 . Then K1 - U2 C E l - E2 5 U1 - K2, and (U1 - K2) - (K1 - U2) 5 (U1 - K1) U (U2 - K2). Hence p((U1 - K2) - (K1 - U2)) i p(U1 - K1) p(U2 - K2) < t, and C is closed under differences. By Lemma 11.2, C equals K(X). Thus every set in K(X) satisfies the regularity property. Next let us see that p i s completely additive on C. Let En be a disjoint sequence of sets in K(X) with union E in K(X). For every N , we have c= :~ p(En) = p ( E l U ... U EN) 5 p ( E ) . Hence CEl p ( E n ) 5 p ( E ) . For the reverse inequality, let t > 0 be given. Choose, by the regularity property, K compact and U, open bounded with K 5 E ,E, 5 Un,p (E- K ) < t ,and p(U,- En) < ~ / 2 ~ . Then K C E = U,"=, E n C U,"=, Un . In other words, the sets U, form an open cover of the compact set K . Some finite subcollection is a cover, and thus K C U1 U . . . U U N f o r s o m e N . Thenwehave
+
+
p(E) = p ( E
-
K)
+ p ( K ) i + p(Ul U . .. U UN) t
502
XI. Integration on Locally Compact Spaces
Since t is arbitrary, p ( E ) i C z l p(E,). Therefore p ( E ) = Czl p(E,), and p is completely additive on K(X) . The Extension Theorem (Theorem 5.5) shows that p extends uniquely to a measure on the smallest o-ring containing K(X), i.e., the o-ring of o-bounded Borel sets. Proposition 5.37 shows further that p extends canonically to a measure on the 0-algebra of all Borel sets under the definition p(E) =
sup
p(F).
FCE, FcR(X), F o-bounded
This defines p on B(X). We are left with showing that p is regular and that e (f = J;, f d/J for every f E Ccom(X). In showing that .t(f ) = &. f d p for every f E Cc,,(X), it is enough to handle an arbitrary f 10. Fix t- > 0, and fix an integer N such that 11 f l S up < N r For 0 5 n 5 N , define f, = mini f , nt}. Each f, is in CcO,(X), the function fa is 0, and the function f N is f . For 0 5 n < N , define g, = f,+l - f,. We can recover f from the g,'s as f = g,. For n 1, define K, = {X I f (x) > nc}, and let KO = support( f ) . All the sets K, are compact, and they decrease in size with n . In this notation the formula for g, is
>
xfil
ifx $ Kn, ifx~K,-K,+l, if x
E
K,,+l.
CG,+~ i gn i ~ I K , .
Conscqucntly
(4)
Integration therefore gives
The inequality given as IK,+l i t c l g , in (*) implies that W(K,+~)i t-lt(g,). The other inequality E - ' ~ , 5 IKn in (*) says that any h r Cc,,(X) with IK, 5 h has t-'g,, 5 h . Taking the infimum over h yields t-'&(g,,) 5 p(K,,). Thus we have cp(Kn+l) 5 l(gn) 5 c ~ ( K n ) . (77) Subtracting (7) and ( t t ) , we obtain
Since f =
N-1
g,, summing from n = 0 to n = N
-
1 gives
2. Riesz Representation Theorem
503
I
Since t is arbitrary, Jx f d p - . t ( f ) l = 0 . Thus l ( f ) = f d p . Fix a compact subset KO of X , form the o-ring B ( X ) n K O ,and let A ( K o ) be the collection of members E of B ( X ) n KO such that p ( E ) is the supremum of w ( K ) over all compact subsets K of E and w ( E ) is the infimum of w ( U ) over all bounded open sets in X that contain E ; the open sets in question need not lie within KO Since the sets in A ( K o ) all have finite measure, the regularity condition on E is that there exist, for each t 0 , K compact and U bounded open with K 5 E g U and p(U - K ) it-. The same arguments as at the beginning of the present proof show that A ( K 0 ) is closed under finite unions and differences. To see closure under countable disjoint unions, let { E n }be a disjoint sequence in A ( K o ) with union E , let t be given, and choose K, compact and U, bounded open with K, 5 En 5 U, and p(U, - K,) < t / 2 " . Applying Corollary 10.23, let L be a compact subset of X with KO 5 Lo. The sets Kn are disjoint, and thus C,"=, w(K,) converges. Choose N such that C,"=,+, w(K,) < t . Define N
U = Lo f' U,"=,U,, K = U,=, K,, K , = U,"=,K,, and F = U,"=,+, K,. Then K is compact, U is bounded open, and K c E c U . Since K , = K U F , we have
Thus A ( K o ) is closed under countable disjoint unions and is a o-ring. Since the compact subsets of KO are in A ( K o ) , we conclude that &KO) = B(Ko). This proves regularity for all bounded sets. If E is o-bounded, we can choose an increasing sequence {L,} of compact sets whose union contains E . Put En = E n L,. Given t > 0, we apply the previous step to choose Kn compact and Un bounded open such that K, 5 En 5 Un and p(Un - K,) < t/2,. Taking w U = U,=, U, and K , = U,"=,K, ,we have K , c E c U and p(U- K,) < t . Thus p ( U ) 5 p ( E ) + t, and p ( E ) 5 p(K,) + t . The first of these inequalities, being possible for any t , shows that 11 ( E )is the infimum of the measures of open o-bounded sets containing E . Since p(K,) = limN p ( ~ f K,) = ~by complete additivity, the second of these inequalities, being possible for any c , shows that p ( E ) is the supremum of the measures of compact sets contained in E. This proves regularity for all 0-bounded sets. If E is a Borel set that is not o-bounded, we know that p ( E ) is the supremum of the measures of p ( F ) for o-bounded Borel subsets F of E , and we know that p ( F ) is the supremum of the measures of p ( K ) for compact subsets K of F. Therefore p ( E ) is the supremum of the measures of w ( K ) for compact subsets K of E. This completes the proof of regularity of p .
504
XI. Integration on Locally Compact Spaces
PROOF OF UNIQUENESS W THEOREM11.1. Let p be the constructed measure, and let v be a second measure satisfying the properties of the theorem. The assumed regularity of v implies that it is enough to prove that v ( K ) = p ( K ) for every compact subset K of X . Fix K , and let a be the infimum defining w ( K ) , namely the infimum of L( f ) over all f E Cc,,(X) with values in [0, 11 such that IK 5 f . Integrating this ineq~lalitywith respect to I ) ,we see that 11 ( K ) 5 f dl) and therefore v ( K ) 5 a . Suppose that v ( K ) < a . By Corollary 10.23 and the assumed regularity of v , we can find a bounded open set U with U 2 K and v ( U ) < a . By Corollary 10.44 we can find a function g E C,,,(X) with values in [0, 11 such that g is 1 on K and is 0 off U . Then IK 5 g 5 I u . IIence Iu d v = v ( U ) < a 5 L(g), and we obtain a gdv 5 l(g) = & g d p = contradiction. We conclude that v ( K ) = a = p ( K ),and the uniqueness follows.
3. Regular Borel Measures The fact that compact sets for a general locally compact Hausdorff X need not be countable intersections of open sets suggests a look at the ring of sets generated by thc compact scts that arc indccd such intcrscctions, as wcll as thc associatcd o-algebra. The sets in this o-algebra are known as "Baire sets,'' and it turns out that the rnernibers of Cc,,(X) are measurable wit11 respect to this o -algebra. The a-algebra of Baire sets can be strictly smaller than the o-algebra of Borel sets, and thus one can make a case for limiting oneself to Baire sets all along. This would be a fine point, one not worth pursuing here, but for one fact: the o -algebra of Baire sets for X x Y is a correct o -algebra to use in Fubini's Theorem for changing iterated integrals over X and Y to a double integral-and this may not be true when Borel sets are used. This fact about Fubini's Theorem might seem to be a telling argument for replacing Borel sets by Baire sets everywhere in the theory. The difficulty is that it is a little tedious to check constantly whether sets are Baire sets-for example, whether one-point sets are Baire sets. Thus the normal practice is to work with Borel sets and to resort to Baire sets only when Fubini's Theorem comes into play in a way that makes the distinction important. The most frequent case that arises in applications of Fubini's Theorem in this theory is that a function on X x Y is continuous with compact support, in which case only Baire sets are involved anyway. Thus let X be a locally compact Hausdorff space. The sets in the smallest o-algebra B ( X ) containing the compact sets are the Borel sets, and the sets in the smallest o-algebra B o ( X ) containing the compact G8's are the Baire sets. Measurable functions in the first case will be called Borel measurable functions or Borel functions, and measurable functions in the second case will be called
3. RegularBorel Measures
505
Baire measurable functions or Baire functions. We shall observe in Corollary 11.16 below that every member of C,,,(X) is a Baire function. If the locally compact Hausdorff space X is a metric space, then any closed set F is the intersection of the sets Un = {x I D ( x , F) < where D(. , F) is the distance to the set F. Consequently every compact subset of X is a Gs, and every Bore1 set is a Baire set.
i},
Proposition 11.14. If K and U arc subscts of X with K compact, U opcn, and K 5 U , then there exist a compact Gs, say KO,and an open bounded F, , say 110,such that K g lJo g KO g 11. PROOF. Choose by Corollary 10.44 a member f of C,,,(X) with values in [O, 11 such that f is 1 on K and is 0 on U c . If KO is the set where f is > and Uo is the set where f is > then Lemma 11.5 shows that KO and Uo have the required properties.
i,
Corollary 11.15. Any 0-compact open subset of X is a Baire set PROOF. If U = Ur=l Kn is open with each K, compact, we can apply Proposition 11.14 to the inclusion K , g U and find a set (K,)o that is a compact Gs and has K, g (K,)o g U . Then U = Ur=l(K,)o exhibits U as the countable union of compact Gg's, hence as a Baire set. Corollary 11.16. Every member of C,,,(X)
is a Baire function.
PROOF.This is immediate from Lemma 11.5 and Corollary 11.15. Proposition 11.17. If X and Y are 0-compact, then the product a-algebra for X x Y obtained from the Baire sets of X and Y is the 0-algebra of Baire sets of X x Y. PROOF.If K x and K y are compact Gs's in X and Y, then K x x K y is a compact Gs in X x Y , and it follows that Bo ( X ) x Bo ( Y ) 5 Bo ( X x Y ) . For the reverse Un with inclusion let K be a compact Gg in X x Y , and write K as K = each U, open. We construct open sets S, in B o ( X ) x Bo(Y) with K 5 S, 5 U , , and then it follows that K = S,, and K is a Baire set. To do so, it is enough to show that if K 5 W with W open, then there is an open set S in Bo (X) x B o ( Y )with K 5 S 5 W . For each (x, y) in K ,find open neighborhoods Ux of x and Vy of y such that Ux x Vy g W. Proposition 11.14, applied to the inclusioll {x} 5 Ux and tl~eato the inclusioll {y} g Vy , shows that we may assume that Ux and V, are open F, 's. In view of Corollary 11.15, they are then Baire sets. Hence Ux x Vy is in Bo ( X ) x Bo ( Y ) . As ( x , y ) varies, the sets Ux x V, form an open cover of K , and there is a finite subcover. We can take S to be the union of the elements in the finite subcover, and then S has the required properties.
(-)El
XI. Integration on Locally Compact Spaces
506
Now we turn our attention to measures. A Baire measure on X is a measure on the Baire sets that is finite on every compact Gs. The restriction of a Borel measure to the Baire sets is a Baire measure. We are going to prove that Baire measures are automatically regular in the same sense that Borel measures in IRN are automatically regular.
Proposition 11.18. Every Baire measure p is regular in the following sense: p(E) =
sup
p(K)
for every set E in Uo(X),
KCE, K compact Gs
p(E) =
inf
p(U)
for every o -bounded set E in B0 ( X ) .
U2E, U open F ,
REMARK. Since Baire sets and Borel sets are the same in a metric space, this proposition generalizes the known regularity of Borel measures on any open subset of Rn, as given in Theorem 6.25. PROOF.If L is a compact G a ,then p ( L ) is certainly the supremum of p ( K ) for the compact Gs's contained in L . Suppose that U is o-bounded open with L g U . Proposition 11.14produces a bounded open set Uo that is an Fa and has L 5 Uo 5 U . Collsequelltly p ( L ) is the illfimum of p(UO)fol the open F,'s containing L . Thus every compact Gs satisfies the stated regularity condition. The remainder of the proof runs parallel to the proof of regularity at the end of the proof of existence for Theorem 11.1, and we shall be brief. Fix a compact Gg in X , say KO.Form the o-ring B o ( X )f' K O ,and let & ( K O ) be the collection of members E of & ( X ) n KO such that p ( E ) is the supremum of p ( K ) over all compact subsets K of E that are Gsls and p ( E ) is the infimum of p ( U ) over all open supersets U of E that are F, 's; the open sets in question need not lie within KO.Sincc thc scts in ( K O )all havc finitc mcasurc, thc regularity condition on E is that there exist, for each t > 0, K compact and U open of the correct kind with K 5 E 5 U and p(U - K ) < t. The same arguments as earlier show that & ( K O ) is closed first under finite unions and differences, then under countable disjoint unions. Thus & ( K O ) is a o-ring containing all compact Gs's, and we conclude that A ( K o ) = B ( K o ) . This proves regularity for all bounded Baire sets. If the Baire set E is o-bounded, we can choose an increasing sequence ( L , ] of compact Ga's whose union contains E . Put En = E n L,. Then the same argument as earlier, using the sets E n , shows that the regularity collditioll holds for E . Finally if E is a Baire set that is not o-bounded, we know that p ( E ) is the supremum of the measures of p ( F ) for o-bounded Baire subsets F of E , and we know that p ( F ) is the supremum of the measures of p ( K ) for compact subsets K of F that are Gs's. Therefore p ( E ) is the supremum of the measures of w ( K ) for compact subsets K of E that are Gs's.
3. RegularBorel Measures
507
Proposition 11.19. If v is a Baire measure on X , then there is one and only one regular Borel measure p on X whose restriction to the Baire sets is p .
PROOF.Since the members of C,,,(X) are Baire functions (Corollary 11.16), we can define a positive linear functional t on C,,,ll(X) by t (f ) = f d v . The uniqueness of the extending p follows from the uniqueness part of Theorem 11.1. For existence we take p to be the regular Borel measure given by the existence part of Theorem 1 1 . l . We are to prove that / I and v agree on Baire sets. The measures p and v agree on compact Gs's by Lemma 11.7a and dominated convergence. By regularity of Baire measures (Proposition 11.IS), p and v agree on all Baire sets.
I;,
Proposition 11.20. Suppose that X is compact and that y and v are Borel measures on X with w regular. If v is absolutely continuous with respect to p , then v is regular.
PROOF.Let t > 0 be given. The Radon-Nikodym Theorem (Theorem 9.16) and Corollary 5.24 together show that there exists 6 > 0 such that any Borel set A with p ( A ) < S has v ( A ) < c. Let E be a Borel set to be tested for regularity under v . Since p is regular, we can choose K compact and U open with K 5 E 5 U and p(U - K ) < 8 . Then v(U - K ) < c , and it follows that v ( E ) is approximated within t by v ( K ) and v ( U ) . Proposition 11.21. Tf / I is a regular Rorel measure on X and if 1 5 p < oo, then (a) C,,,(X) is dense in LP(X, p ) , (b) the smallest closed subspace of LP ( X , p ) containing all indicator functions of compact Gs's in X is LP(X, p ) itself.
REMARK.This generalizes conclusions (a) and (b) of Proposition 9.9 from open subsets of RN to all locally compact Hausdorff spaces. PROOF.If E is a Bvrel set of finite p measure and if t is given, the regularity E and kr(E - K ) (: c of ~i allows us to choose a compact set K with K Then we can find a bounded open set U with K 5 U and p(U - K ) < c , and Proposition 11.14gives us a compact Gs set KOsuch that K 5 KO 5 U . We have iRN I I E - I K I ~ ~ / J -w= ( E - K ) < t,JRN I I u - I K ~ ~ ~=/ J/ J- - ( U - K ) < t, and 1 Iu - IKoIP d p = p(U - KO)< r . Consequelltly we see in succession that the closure in LP(X, p ) of the set of all indicator functions of compact sets contains all indicator functions of Borel sets of finite p measure, the closure in LP(X, p ) of the set of all indicator functions of bounded open sets contains all indicator functions of Borel sets of finite p measure, and the closure in LP(X, p ) of the set of all indicator functions of compact Gs 's contains all indicator functions
IRN
508
XI. Integration on Locally Compact Spaces
of Borel sets of finite p measure. Proposition 5.56 shows consequently that the smallest closed subspace of LP(X, p ) containing all indicator functions of compact Baire sets is LP(X, p ) itself. This proves (b). For (a), let KObe a compact Gg, and use Lemma 11.7a to choose a decreasing sequence { f,} of real-valued members of Cco,(RN) with pointwise limit IK,. Since f / is integrable, dominated convergence yields lim,, If, - IKuIP d p = 0. Hence the closure of C,,,(X) in LP(X, p ) contains all indicator functions of compact Gs's. By Proposition 5.55d this closure contains the smallest closed subspace of LP(X, p ) containing all indicator functions of compact Gg's. Conclusion (b) shows that the latter subspace is LP(X, p ) itself. This proves (a).
INN
Corollary 11.22. Suppose that X is a locally compact separable metric space. If p is a Borel measure on X and if 1 5 p < cm,then (a) C,,,(X), as a normed linear space under the supremum norm, is separable, (b) LP(X, p ) is separable. ~ open REMARK.This generalizes Corollary 6 . 2 7 ~and Proposition 9 . 9 from subsets of R N to all locally compact separable metric spaces. The measure p is automatically regular by Proposition 11.8 since Baire measures and Borel measures coincide in any locally compact metric space. PROOF.Part (a) is proved by the same argument as for Corollary 6 . 2 7 ~What . is rcquircd is a substitutc for Lcmma 6.22a in ordcr to obtain a scqucncc {F,Jr==l of compact subsets of X with union X such that F, Fz+, for all n . It was observed at the begirlrlirlg of Sectiori X.3 that separable irnplies Lindelijf, arid it follows from Proposition 10.24 that X is consequently o -compact. Application of Proposition 10.25 then gives the sequence {F,},"_, . Corollary 2.59 is still to be applied to C (F,) ; since F, is a compact metric space, the corollary shows that C (F,) is separable, and the argument goes through. Part (b) follows from (a) and Proposition 11.21a in the same way that Corollary 6.27d follows from parts (a) and (c) of that corollary. The sequence { F , , } g l of the previous paragraph is to be used in the argument.
Theorem 11.23 (Helly-Bray Theorem). Let X be a locally compact separable metric space. If { p , } is a sequence of Borel measures on X with { p ,( X )} bounded, say by M, then there exist a Borel measure on X and a subsequence {p,,c}such that p ( X ) 5 M and limn f dp,, = f d p for all f in Cco,(X) . REMARKS.In the terminology of Section V.9, the measures p, are continuous linear functionals on the normed linear space Cco,(X), and the norm of the linear functional corresponding to p, is p, ( X ) . The convergence is weak-star convergence, and the limiting linear functional is given by a Borel measure p
4. Dual to Space o f Finite Signed Measures
509
with p ( X ) 5 M. The theorem amounts to an application of the preliminary form of Alaoglu's Theorem (Theorem 5.58) and the identification of the limit as a measure. PROOF. The proof consists of filling in the details in the remarks above. We regard Y = Cco,(X) as a normed linear space with the supremum norm. Any Borel measure v on X defines by integration a linear functional on Y with norm given by Ilv 11 = sup,fEc,,,(x), II ixf du The right side is certainly ( 11 f llsupv(X). In the reverse direction, let {K,} be an increasing sequence of compact subsets of X with union X , so that limn v ( K n ) = v ( X ) . Choose functions fn : X + [O, 11 in Ccom(X)by Corollary 10.44 such that fn is 1 on K,. Then I l frills,, 5 1 for all n , and Jx f,dv > JKn dv = v(K,). Hence 1 1 v 1 1 > lim sup, v(K,) = v ( X ), and we conclude that 11 v 1 1 = v ( X ) . Thus the given sequence {p,} corresponds to a sequence in Y * with lip, 11 M for all n . Corollary 11.22 shows that Y is separable. Theorem 5.58 therefore applies and yields a subsequence {p,,} and a member l of Y* with Il11 5 M such that limk Jx f dpnk= l (f ) for all f in Ccom(X).If f > 0, limk Jx f dpnk is certainly > 0, and thus l is a positive linear functional on Cco,(X) . The Riesz Representation Theorem (Theorem 11.1) produces a Borel measure p on X with f d p for all f i n Cco,(X). Since Il11 5 M , we have p ( X ) 5 M. l ( f )= Il,f
1
1.
4. Dual to Space of Finite Signed Measures
We continue in this section with X as a locally compact Hausdorff space. We now change the point of view a little and regard Cco,(X) as a normed linear space = supxExIf (x) 1. The problem is to identify under the supremum norm 11 f l , all continuous linear functionals on this normed linear space. We shall see shortly that it is enough to handle the case that X is compact. If X* is the one-point compactification of X , then two spaces to be considered in conjunction with C,,,(X) are C ( X * ) ,the space of continuous scalar-valued functions on X * ,and C o ( X ) ,the space of continuous scalar-valued functions on X that "vanish at infinity." When applied to a function f , the term vanishes at infinity means that for any c > 0 , there is some compact set with the property that If ( x )I 5 c outside that set. It is equivalent to say that f extends to a member of C ( X * )that is 0 at oo. The three spaces Cc,,(X), C o ( X ) ,and C ( X * )are related. 111 the first place, Ccom(X)is dense in C o ( X ). In fact, if f is in C o ( X )and if c > 0 is given, we find K compact with If ( x )I 5 t outside K . Corollary 10.44 supplies a member g of Cco,(X) with values in [0, 11 that is 1 on K . Then the product f g is in C,,,(X), and I l f - f g llsup i t . Thus Ccom(X)is dense in C o ( X ). Any continuous linear functional on Cco,(X) is uniformly continuous by Proposition 5.57, and
5 10
XI. Integration on Locally Compact Spaces
Proposition 2.47 shows that it extends uniquely to a continuous linear functional on C o ( X ). Thus the continuous linear functionals on C o ( X ) and C,, ( X ) are in one-one correspondence by restriction. If we identify C o ( X ) as the subspace of C ( X * ) of functions equal to 0 at o o , then every continuous linear functional on C ( X * ) restricts to a continuous linear functional on C o ( X ) In the reverse direction every continuous linear functional on C o ( X ) extends (nonuniquely) to a continuous linear functional on C ( X * ) . In fact, let lobe a continuous linear functional on Co( X ) , and fix a member fo of C ( X * ) with fo ( o o ) = 1. If f is any member of C ( X * ),then f - f ( m )fo is in C o ( X ) and it makes sense to define & ( f ) - lo( f - f ( m )f o ) . Since
f (m>fo>l 5 Illoll l l f - f ( ~ ) f o l l s , p 5 ll~oll(llfllsup + I f ~ ~ ) l l l f o l l s u5p )ll~oIl(1+ llfollsup~llf
Il(f>l = Ilo(f
-
llsup9
t is bounded on C ( X * ) and is therefore continuous. Thus the study of continuous linear functionals on C,,(X) reduces to the case that X is compact. The first result below shows that any continuous linear functional on C ( X ) with X compact is a finite linear combination of positive linear functionals. In view of Theorem 11.l ,it is therefore given as a finite linear combination of integrations with respect to regular Bore1 measures. The remainder of the section will be devoted to making this result look tidier and seeing what happens to various norms under the correspondence. Proposition 11.24. Let X be a compact Hausdorff space, and let & be a continuous linear functional on C ( X ) . If & takes real values on real-valued functions, define, for f > 0 in C ( X ), l + ( f ) = sup l ( g )
and
lP(f)=t+(f)-~(f);
05gif
then l f and l - extend to positive linearfunctionals on C ( X ) such that l = l+-l- . If & does not necessarily take real values on real-valued functions, then & is a complex linear combination of positive linear functionals on C ( X ). FROOF. The fimctions f andg in this argument will all he in C ( X ) .For general -
t not necessarily taking real values on real-valued functions, define &(f ) = L( f ) . We readily clieck tliat & is a coiitiiiuous linear fuiictioiial oil C ( X ) , tliat l R=
.e>
+ and LI = &(L - &) are continuous linear functionals on C ( X ) taking real values on real-valued functions, and that & = l R+ i l l exhibits L as a complex linear combination of continuous linear functionals taking real values on real-valued functions. This reduces the proposition to the case that L takes real values on real-valued functions. 1 ?(&
4. Dual to Space o f Finite Signed Measures
511
In this case, for f > 0, inspection gives the following: l ( f ) = l f ( f )-l - (f ) , &+(O) = &-(0)= 0 , l f ( c f ) = c L f ( f ) f o r c > 0 , a n d L P ( c f ) = c L P ( f ) f o r c > 0. In addition, l f ( f ) > 0 for f > 0 because
and !-( f ) ? O for f ? O because
To complete the proof, all that we have to do is show that .tf (f l + f2) l + ( f l ) l + ( f2) whenever f l > 0 and f2 > 0. The argument for > is that
+
f
+
( f + ~ f2) =
sup f ( g ) ? oigifi+f2
=
sup .e(gl + g2) 8l,82,
0581 5 f 1 , 05825f2 -
sup l ( g l ) +
0581 sf1
sup l ( g 2 ) = . e f ( f l ) + e + ( f
2)
05825f2
+
For the reverse direction, let g be arbitrary with 0 5 g 5 f l f i , and set gl = min{g, f l } and g2 = g - g l . Certainly 0 5 gl 5 f l . Let us show that 0 5 g2 5 f 2 . In fact,
Thus g2 is certainly ? 0 . In addition, the computation
shows that g i is 5 f 2 . Thus any g with 0 i g i fl rlecnmposition
Taking the supremum over g ,we obtain &+( f l proof is complete.
+ f2 gives us a corresponding
+ f2) i &+( f l ) + &+( f 2 ) , and the
512
XI. Integration on Locally Compact Spaces
Let us reinterpret matters in terms of Borel measures. We begin with the realvalued case. Recall from Section IX.3 that a real-valued completely additive set function p on a o -algebra is called a signed measure. It is bounded if I p (E)I 5 C for all E in the algebra. In this case Theorem 9.14 shows that it has a Jordan decomposition p = pf - p - , where ,of and p- are uniquely determined finite measures such that any decomposition p = I)+ - 11- as the difference of finite measures has ,of 5 v f and p- 5 v - . We say that a bounded signed measure p on the Borel sets of the compact Hausdorff space X is a regular Borel signed measure if its Jordan decomposition is into regular Borel measures. If p = v f - v - is any decomposition of a bounded signed measure p on the Borel sets as the difference of regular Borel measures, then the equalities pf 5 v f and p- 5 v- that compare the decomposition with the Jordan decomposition force pf and p- to be regular, in view of Proposition 11.20. Hence p is a regular Borel signed measure. The regular Borel signed measures form a real vector space M ( X , R ) . To see closure under vector space operations, we observe from the definition of regularity that the sum of two (nonnegative) regular Borel measures is a regular Borel measure. From this fact we can see that the sum of two regular Borel signed mcasurcs is rcgular and hcncc that M ( X , R ) is closcd undcr addition: in fact, if p = pf - p- and o = of - o - are given in their Jordan decompositions, then the formula ( p + o ) + - ( p +o)- = ( p f + o f ) - ( p - f o p ) shows that p + o is the difference of two regular Borel measures and hence is regular. Thus M ( X , R) is a real vector space. Proposition 11.25. The real vector space M ( X , R ) becomes a real normed linear space under the definition Ilp 11 = pf ( X ) p P ( X ) , where p = pf - p- is the Jordan decomposition of p .
+
'
PROOF.Certainly lip 1 0 with equality if and only if p = 0. Also, if p has the Jordan decomposition p = pf - p - , then - p = p- - pf is the Jordan decomposition of - p , and it follows that I l cp 11 = Ic 1 I l p 11 for any real scalar c . Finally consider 11 p + o 11. If p = pf - p- and o = o+ - o- are Jordan decompositions, then the formula ( p + o ) + - ( p + o ) - - (P+ + o f ) - (p- + o - ) shows that ( p + o ) + 5 pf + o f and hence ( p + o ) + ( x ) 5 p f ( x ) + o f ( x ) . Similarly ( p + 0 ) - ( X ) 5 p - X ) + o P ( X ) . Adding thcsc incqualitics, wc obtain l l p + o l l 5 Ipll + Iloll. Returning to the statement of Proposition 11.24, let us write C ( X , R ) or C ( X , C) for the space of continuous scalar-valued functions when the field of scalars is important, reserving the expression C ( X ) for situations in which the scalars do not matter. Suppose that L is a continuous linear function on C ( X ) that takes real values on real-valued functions. The proposition shows that .t is the
4. Dual to Space o f Finite Signed Measures
513
difference of two positive linear functionals. By Theorem 11. l , l operates as the difference of two integrations: L( f ) = Jx f d v f - J;, f d v - , where v f and vare the regular Borel measures corresponding to l f and l - . Then l corresponds f d p, to a regular Borel signed measure p and is given by integration: L ( f ) = the integral with respect to the signed measure being interpreted as the difference of two integrals with respect to measures. Conversely any regular Borel signed measure p yields a continuous linear functional l on C ( X ) by the definition l ( f )=J;,fd p . In particular the passage to integration gives us a real-linear mapping of M ( X , R) oilto the space C ( X , R ) * of contiiluous linear fuilctioiials 011 the real vector space C ( X , R ) . Both of these spaces are normed linear spaces, and the theorem is that the map is one-one and that the norms match.
Theorem 11.26. The real-linear map of M ( X , R ) onto C ( X , R)" given by p H L with L( f ) = jx f d p is one-one and norm preserving. R~MAKK AS . in Section V.9 the norm Ill 11 of l is the least constant C such that . The constant C equals the supremum of I L( f ) 1 over all f with I l f llsup 5 1.
I L( f ) 1 5 C 11 f 11 ,,,for all f
I;,
PROOF.To see that the map is one-one, suppose that f' dp = O for all f' in f dpf = f d p - , and the uniqueness part of Theorem 11.1 C ( X , R ) . Then shows that p f = p - . Hence p = p f - p- = 0. Now suppose that l and p correspond. Then we have
1,
I l ( f >=
i
IJ;,f d p f -
J x f dp-1
I;,I f 1 d p f +J;i
+p-(X)llf
5 pf(X)llf l, Taking the supremum over all f with 11 f
I f 1 dp-
I,
ll,,,.
i 1, we obtain
For the inequality in the reverse direction, let t > 0 be given, and let X = P U N be a Hahn decomposition (Theorem 9.15) for p . By regularity of ,of on P and p- on N , choose compact subsets K p and K N with K p 5 P , K N 5 N , p f ( P K p ) < t, and p - ( N K N ) < t . Sincc p f ( N ) = 0 and p - ( P ) = 0, -
-
pf (X
-
Kp)
and
p-(X
-
K N ) < t.
(*)
By Urysohn's Lemma (Corollary 10.43), we can find a continuous function f : X + [-I, 11 such that f is 1 on K p andis -1 on K N . Then
5 14
XI. Integration on Locally Compact Spaces
By (*) the first two terms on the right side are each it-. Since p + ( K ; n K ; ) = p f ( P - K p ) < t a n d p p ( K g n K ; ) = p - ( N - K N ) < t,andsince I f I , 5 1, the third term on the right side is 5 2 t . Therefore IL( f ) - I l p 11 < 4 t , and our function f has the property that lL( f ) I ? (llpll - 4t-Ill f ll,,. In other words, llLll 2 lip 11 - 4 t . Since t is arbitrary, llLll lip 11. This completes the proof.
1
'
Now let us consider the case in which the values are complex. A regular Borel complex measure on the compact Hausdorff space X is an expression p = p ~ + i p I in which p ~ and pl are regular Borel signed measures. In other words, it is a complex-valued set function whose real and imaginary parts are regular Borel signed measures. The space M ( X , C ) of these is a complex vector space, and we shall make it into a normed linear space shortly. Meanwhile, the space C ( X , C ) * of continuous linear functionals on C ( X , C ) is a complex f d p to handle members of normed linear space. Extending the definition of M ( X , C ) ,we see from Proposition 11.24that the complex-linear map of M ( X , C ) into C ( X , C ) " given by p H L with L( f ) = Jx (;, d p is one-one and onto. To have a theorem in this case that parallels Theorem 11.26, we need to define the norm on M ( X , C ) . Doing so on an element p is not just a matter of combining the norms of the real and imaginary parts of p any more than writing the norm of a complex-valued L' function can be done in terms of the L' norms of the real and imaginary parts. A more subtle definition is needed. We define the total variation I p 1 of a member p of M ( X , C ) to be the nonnegative set functioii whose value oil a Borel set E is tlie supreilium of all finite sums Cy=l I p ( E j )1 with E = E j disjointly. The total-variation norm of the member p of M ( X , C ) is defined to be Ilp 11 = Ip 1 ( X ) . It is a simple matter to verify that the total-variation norm is indeed a norm.
Proposition 11.27. The total variation Ipl of a member p of M ( X , C) is a regular Borel measure, there exists a Borel function h with Ilhll,,, 5 1 such that p = h d 1 p 1 , and the total-variation norm on M ( X , C ) makes M ( X , C) into a norrned linear space in such a way that & f d p 5 1 1 p 1 1 11 f llSup for every bounded Borel function f . Moreover, I p 1 equals p f p- if p is real valued and has p = p+ - p- as its Jordan decomposition.
I
1 +
REMARK. It follows that if p is real valued and if X = P U N is a Hahn decomposition (Theorem 9.15) for p , then the corresponding function h may be taken to be +1 on P and - N on N . PROOF.To see that Ipl is additive, let E and F be disjoint Borel sets. If E = Uy=n=l E i disjointly and F = F j disjointly, then E U F = (u:=' E ~ U) (US=, F ~ disjointly, ) and hence C:=l Ip(Ei)l C;=l I P ( F j )I 5 I p l ( E U F ) . Taking the supremum over systems { E i } and then over systems
UEl
+
4. Dual to Space o f Finite Signed Measures
515
[ F j } ,we obtain Ip l ( E ) + Ip 1 ( F ) 5 Ipl ( E U F ) . In the reverse direction let E U F = U:='=,G k disjointly. Then E = U:='=,(En G k ) disjointly, and F = U L 1 ( Fn G,) disjointly. Hence
and this is 5 Ipl(E) + Ipl(F). Taking the supremum over systems [ G k } ,we obtain Ipl(E U F ) 5 Ipl(E) + Ipl(F). Thus Ipl is additive. To prove that pi is completely additive, let E = UrZlEn disjointly. For every N , C f = l Ipl(En) = Ipl(E1 U . . . U E N ) i Ipl(E1, andhence C,"==, Ipl(E,) i Ipl ( E ) . For the reverse inequality let [ G ~ } ~ Pbe, ~a finite collection of disjoint Borel sets with union E . Then En = ( E n f' G k ) disjointly, and hence
ULl
Thus I p 1 ( E ) 5 C Z l I p 1 (E,,), and I p 1 is completely additive. The measure I p 1 is certainly finite on X and hence on all compact sets. To see regularity, we write p = p~ ipl = p i - p i + ipl+ - ip;. Writing a set E as the disjoint union of n sets Ei and writing out p ( E i ) according to this expansion of p , we see that I p 1 ( E ) 5 p i p: + p;) ( E ). Each measure on the right side is regular, and Proposition 11.20 therefore shows that I p 1 is regular. For the existence of h , let us write p in terms of its real and imaginary parts as p = p~ i p l . If E is a Borel set, then the definitions give Ipl ( E ) 1 Ip(E)I > I ~ R ( E ) Iand similarly Ipl(E) > Ipz(E)I. Hence p~ << Ipl and pl << 1 p 1. By the Radon-Nikodym Theorem (Corollary 9.17), there exist functions h~ and hz integrable [d lp I ] such that pR = h R d lp 1 and pz = h z d lp 1 . Thus the I p 1 integrable complex-valued function h = h R + i h z has p = h d 1 p 1 . Wc shall show that 1%has 11%(x) I 5 1 a.c. [d1 p I]. If thc contrary wcrc thc casc, thcn there would exist a constant c with Icl = 1 and an t > 0 such that Re(ch) > 1 t un a scl E uf pusilivc I p I rrlcasurc and we wuuld lravc
+
+ +
+
+
a contradiction. Thus h exists as asserted.
XI. Integration on Locally Compact Spaces
516
I
The inequality J,; f dpI 5 Ilp 1 1 1 f IlSup
follows from the existence of Iz since
= I J;r f h d l p l l i I I ~ ~ I I , , J i ; , I ~I III,,,IPI(X> P ~ I = I I II,,IIPII. ~ I J;,Finally f if p is real valued, then any Borel set E satisfies Ip(E)I =
Ipf (E) - p-(E)I 5 pf (E) + p P ( E ) . If E is the disjoint union of Borel sets E l , . . . , E n , we consequently have
Taking the supremum over all decompositions of E of this kind gives I p 1 (E) 5 pf (E) + p-(E) . For the reverse inequality let X = P U N be a Hahn decomposition (Theorem 9.15) for p , so that ,of (E) = p ( P f' E ) and p-(E) = -p (N nE ) . Then E is the disjoint union of E f' P and E n N , and thus ,of (E) + p-(E) = I p ( E n P ) I + I p ( E n N ) I 5 IpI(E). Inotherwords,IpI = p++pPasasserted.
Theorem 11.28. The one-one complex-linear map of M(X, C) onto C (X, @)* given by p
H 2!.
with l (f ) = j;, f dp is norm preserving.
I
I
F'ROOF. If f is in C (X), then Proposition 11.27 gives It( f ) 1 = & f dp i Ilpll 11 f I l s up. Taking the supremum over all f with 11 f l s up 5 1, we obtain Il.tll 5 llpll. For the reverse inequality let t > 0 be given, and choose a finite disjoint collection of Borel sets E l , . . . , En with union X such that C:=l lp(Ei) 1 3 llpll - c . Since 101 is regular, we can find compact sets Ki g Ei such that Ipl(Ei - Ki)I 5 t / n foreachi. We shall define disjoint open sets Ui with K, c Ui for all i . First we find disjoint open sets U1 and Vl containing K1 and Kz U . . .U Kn. Having inductively chosen disjoint open sets U1, . . . , Uj and Vj such that Ki c Ui for i 5 j and K.t+1 U . . . U Kn Vi , we use Corollary 10.22 to choose disjoint open subsets Ui+1 and Vi+1 of Vi containing Ki+1 and Kif2 U . . . U K n . In this way we obtain the disjoint open sets U1, . . . , Un with Ki c Ui for all i . For 1 5 i 5 n , choose j; E C(X) with values in [0, 11 such that j; is 1 on Ki and is 0 off Ui. Choose ci E C for each i such that cip (Ei) = Ip (Ei)l, and define fo = C:=l ci fi . The function fo has 11 fo l s up = 1 since the sets Ui are disjoint. Then
Hence
and
Therefore
Ilfll
=
Ilfll
lfoII,,,
1If(fo>l 1IlPll - I t ( f o > - l l ~ l l 1 l IlPll - 3 ~ .
Since t is arbitrary, Iltll > l p l l .
5. Problems In all problems for this chapter, X is assumed to be a locally compact IIausdorff space. Sometimes additional hypotheses are imposed on X .
1. (a) Prove that if X is a-compact, then the a-algebra of Borel subsets of X coincides with the a-algebra of intersections of X with the Borel subsets of the one-point compactification X* . (b) Prove that if X is an uncountable discrete space, then the a-algebra of Borel subsets of X is strictly smaller than the a-algebra of intersections of X with the Borel subsets of the one-point compactification X * .
2. Prove that if X is u-compact and f : X + C is continuous, then f is a Borel function. 3.
Suppose that X is a-compact. Prove that if p is a regular Borel measure on X and if f is Borel measurable, then there exists a Baire measurable function g such that f = g except on a Borel set of p measure 0.
4.
(Lusin's Theorem) Let X be compact, let p be a regular Borel measure on X, let f be a Borel function on X , and let c > 0 be given. By first considering simple functions and then passing to the limit via Egoroff's Theorem, prove that there exists a compact subset K of X with p (KC)< c such that f is continuous.
1
5. This problem establishes the rotation invariance of the Borel measure dw on the sphere s2 C IR3 obtained fromRiemann integration with respect to sin O1 d B 1 dQ2, where Q1 and Q2 are latitude and longitude with 0 < Q1 < n and 0 < Q2 < 2n. The measure dw was constructed by means of the Riesz Representation Theorem as one of the examples in Section 2.
518
XI. Integration on Locally Compact Spaces
(a) A rotation in JR3 is the linear function L determined by a matrix A with AAtr = 1 and det A = 1. For 0 < a < 1 < b < m, let Sab be the subset of IR3 given in spherical coordinates by a < r < b,O 5 Q1 5 n,0 5 & 5 2 n . Show that Sab is carried to itself by any such rotation L . (b) For any bounded Borel function F : Sab + C, let ( L F ) ( x ) = F ( L - ' ~ )if .r is in Sab and L is a rotation. Prove that LF dx = F dx. (c) Let f : S2 + @ be any continuous function, and define ( Lf ) (o)= f ( ~ ~ ' w Extend ) . f to a function F defined on Sab by the definition F ( r w ) = f (wj . Prove that Fdx = r 2 d r ) ( js, f ( w )d w ) and deduce that Lf dw = f dw. (d) Deduce from (c) that d w ( L ( E ) )= d w ( E ) for every Borel subset E of s 2 .
lAob
Isab
lS2
lx2
lAob
(lab
6. Let X be compact. (a) Let {K,} be a collection of compact subsets of X closed under finite intersections, and let K = K,. Prove that every regular Borel measure p on X has the property that p ( K ) = inf, p ( K , ) . (b) If p is a nonzero regular Borel measure on X assuming only the values 0 and 1, prove that p is a point mass. (c) If p is a nonzero regular Borel measure on X with
n,
for all f and g in C ( X ) ,prove that p is a point mass. (d) If l is a positive linear functional on C (X) that is multiplicative in the sense that l (f g ) = l (f ) l ( g ) for all f and g in C ( X ), prove that l is zero or l is evaluation at some point of X .
7 . This problem continues the investigation of harmonic functions and Poisson integrals in the unit disk of R ~ following , up on Problems 7-8 at the end of Chapter IX. Problem 8 in that series provides orientation. The new ingredient for the present problem is weak-star convergence of sequences in M ( s ' , @) against C ( S 1 ) ,where S' is the unit circle. (a) State and prove a characterization of the harmonic functions u ( r , 8 ) on the open unit disk such that s ~ p ~ , , ,Ilu(r, ~ . ) 11 is finite. (b) (Herglotz's Theorem) prove that if u(r, Q ) is a nonnegative harmonic function on the open unit disk, then there is a Borel measure p on the circle such that u ( r , 0 ) = &-n,nl P,. (0 - p) d p ( p ) . Problems 8-10 construct a Borel measure p on a compact space such that p is not regular. The totally ordered set Q of countable ordinals was introduced in Problems 25-33 at the end of Chapter V. Let Q* = Q U {m},totally ordered so that every element of Q is less than {m}.Give Q* the order topology, as discussed in Problems 25-32 at the end of Chapter X.
5. Problems
519
8. Prove that Q* is compact Hausdorff.
9. Prove that the class of all relatively closed uncountable subsets of Q is closed under the formation of countable intersections. 10. Define p on the Borel sets of Q* to be 1 on those sets E such that E - {oo}contains a relatively closed uncountable subset of Q , and put v ( E ) = 0 otherwise. Prove that p is a Bore1 measure that is not regular.
Problems 11-14 concern decomposing any finite Borel measure on a compact X into a regular Borel measure and a "purely irregular" Borel measure. They make use of Zorn's Lemma (Section A9 of the appendix). A Borel measure p will be called purely irregular if there is no nonzero regular Borel measure u such that 0 5 v ( E ) 5 p ( E ) for every Borel set E. 11. Use Zom's Lemma to show that any Borel measure on X is the sum of a regular Borel measure and a purely irregular Borel measure. 12. Prove that i f v is aregular Borelmeasure, if p is purely irregular, andif0 5 p 5 v , then p = 0. 13. Deduce from the Jordan decomposition (Theorem 9.14) that the decomposition of Problem 11 is unique. 14. Prove that the irregular Borel measure constructed in Problem 10 is purely irregular. Problems 15-19 concern extension of measures from finite products of compact metric spaces to countably infinite such products. Let X be a compact metric space, N and for each integer n > 1, let X, be a copy of X. Define = X n,~Xn, 00 arid lsl Q = X n=lX,. Ea~11ul and ib givsr~~ 1prudud s Lupulugy. I l E is a Borel subset of d N ) ,we can regard E as a subset of Q by identifying E with E x (X x"). In this way any Borel measure on Q ( N )can be regarded as a measure on a certain a-subalgebra .En of B ( Q ) . 15. Prove that Ur=lF, = F i s an algebra.
zN+l
16. Let v, be a (regular) Borel measure on Q(,) with v ( Q ( " ) )= 1, and regard v, as defined on E,, . Suppose for each n that u, agrees with v,+l on &. Define v ( E ) for E in F to be the common value of v,(E) for n large. Prove that v is nonnegative additive, and prove that in a suitable sense v is regular on F . 17. Using the kind of regularity established in the previous problem, prove that u is completely additive on F . 18. In view of Problems 16 and 17, v extends to a measure on the smallest a-algebra for Q containing F . Prove that this a-algebra is B ( Q ) . 19. Let X be a 2-point space, and let v, be 2-" on each one-point subset of a ( " ) . Exhibit a homeomorphism of onto the standard Cantor set in [0, 11 that carries u to the Cantor measure defined in Problems 17-20 at the end of Chapter VI.
CHAPTER XI1 Hilbert and Banach Spaces
Abstract. This chapter develops the beginnings of abstract functional analysis, a subject designed to study properties of functions by treating the functions as the members of a space and formulating the properties as properties of the space. Section 1 defines Banach spaces as completenormed linear spaces andgives anumber of examples of these. The space of bounded linear operators from one normed linear space to another is a normed linear space, and it is a Banach space if the range is a Banach space. Sections 2-3 concern Hilbert spaces. These are Banach spaces whose norms are induced by inner products. Section 2 shows that closed vector subspaces of such a space have orthogonal complements, and it shows the role of orthonormal bases for such a space. Section 3 concentrates on bounded linear operators from a Hilbert space to itself and constructs the adjoint of each such operator. Sections 4-6 prove the three main abstract theorems about the norm topology of general normed linear spaces-the Hahn-Banach Theorem, the Uniform Boundedness or Banach-Steinhaus Theorem, and the Interior Mapping Principle. A number of consequences of these theorems are given. The second and third of the theorems require some hypothesis of completeness.
1. Definitions and Examples Functional analysis puts into practicc an idca from thc carly twcnticth ccntury, that sometimes properties of functions become clearer when the functions are regarded as the members of a space and the properties are formulated as properties of the space. We encountered some simple examples of this situation already in Chapter I1in the examples of metric spaces. Uniform convergence was encoded in the metric on spaces of functions, and other kinds of convergence were captured by other metrics. In Chapter V we introduced the spaces L 1( X ), L 2 ( x ) ,and L m ( X ) of functions (or really equivalence classes of functions), all of which were proved to be complete. The property of completeness was a useful property of the space as a whole that led, for one thing, to the Riesz-Fischer Theorem in Chapter VI. More complicated properties led us to various kinds of differentiability of integrals in Rn in Chapters VI and IX and to boundedness of the Hilbert transform in Chapter IX. The development of measure theory on locally compact Hausdorff spaces in Chapter XI rested on an analysis of positive linear functionals on the space of continuous functions of compact support.
1. Definitions and Examples
521
The different spaces-of functions, measures, and whatever else-that arise in this way have some properties in common, and we study them in this chapter in a setting that emphasizes these common properties. We shall work with normed linear spaces, which were defined in Section V.9. With such spaces the field of scalars IF can be either R or C. Recall then that a normed linear space X is a vector space over IF with a norm, i e , a function 11 . 11 from X to [0, +m) S U C ~ that IlxII > Owithequality if andonly ifx = 0, 1~x11= IcIIlxII ifcis ascalar,and Ilx +yII 5 IlxII + IIyII. Thenorm yields ametricd(x, y) = Ilx - yI1,andwecan then speak of the norm topology on X . Proposition 5.55 showed that addition and scalar multiplication are continuous, that the closure of any vector subspace of X is a vector subspace, and that the set of all finite linear combinations of members of a subset S of X is dense in the smallest closed subspace containing S . Completeness plays an increasingly important role as one studies such spaces, and it is customary to introduce a definition to incorporate this notion: a normed linear space X is a Banach space if X is complete as a metric space. The metricspace completion of a norrned linear space is automatically a norrned linear space that is complete, hence is a Banach space. Let us consider some examples of normed linear spaces, some old and some new. Except as indicated, they will all be Banach spaces.
(1) Euclidean space IfS" and complex Euclidean space C n ,written briefly as = (al, . . . , a,) with llall equal
IF'" The space consists of n-tuples of scalars a
xi=,
lakI)l i 2 . It was to the Euclidean norm n 1 of Section 11.1, namely l a I = ( remarked in Section 11.7that these spaces are complete, hence are Banach spaces. (2) Finite-dimensional normed linear spaces. It can be shown that each finitedimensional normed linear space X is complete. In fact, any linear map carrying a vector-space basis of X to a vector-space basis of some IFn, normed as in the previous example, can be shown to be uniformly continuous with a uniformly continuous inverse, and the completeness of X follows.
(3) B ( S ) , the space of bounded scalar-valued functions on a nonempty set S with the supremum norm, defined in Section 11.1. Proposition 2.44 establishes the completeness.
(4)C ( S ),the space of bounded continuous scalar-valued functions on a metric space or topological space S , defined in Section 11.4in the metric case and Section X.5 in general. The norm is the supremum norm. Corollary 2.45 and Proposition 10.30 establish the completeness of C ( S ) . When S is locally compact Hausdorff, we defined C o ( S ) in Section XI.4 to be the subspace of C ( S ) of all members vanishing at infinity. This is complete. However, the subspace C,,,(S) of continuous scalar-valued functions of compact support is usually not complete.
522
XII. Hilbert and Banach Spaces
(5) LP(S, A, p ) , the space of equivalence classes of pth-power integrable functions on a measure space ( S , A, p ) . This is a normed linear space for 1 5 p < oo with norm I flip = f ( s ) ~ d ~ ( ~ ) These ) " ~ .spaces were introduced in Section V.9 for p = 1 and p = 2 and in Section IX.l for general p . Theorem 5.59 established the completeness for p = 1 and p = 2, and Theorem 9.6 established the completeness for general p. (6) L m ( S , A, p ) , the space of equivalence classes of essentially bounded functions on a measure space ( S ,A, p ) . This is a normed linear space with norm the essential supremum norm. This space was introduced in Section V.9 and was proved to be complete in Theorem 5.59. (7) Sequence spaces c , co, and &$ and l p for 1 5 p ( oo. These are is LP(S, A, p) when special cases of various exanlples above. The space S = {I, 2, . . . , n}, A is the set of all subsets, and p is counting measure, the norm being I ( a l , .. . , a , ) = (C:=llaklp)"" if p < oo and being Il(al,. . . , a,)ll = maxl,k,, lakl if p = oo. The space l g specializes to IFn when p = 2. The spaceplF is the version of l[ when S is the set of positive integers; the members of this space are thus all sequences for which the norm is finite. The sequence spaces c and co can be regarded as subspaces of C ( S )when S is the set of positive integers. The space c consists of all convergent sequences, and co is the space of sequences vanishing at infinity; in both cases the norm is the supremum norm. All these examples are Banach spaces. They tend to be useful in testing guesses about properties of normed linear spaces. We shall not need them explicitly, and this traditional notation for them will not recur after the end of this section. (8) M ( S ) , S being a compact Hausdorff space. This is the space of regular Bore1 signed or complex measures on S , introduced as M ( S ,R) or M ( S , C) in Section XI.4. The norm is the total-variation norm. Theorems 11.26 and 11.28 identify these spaces with duals of spaces of continuous functions, and Proposition 12.1 below will show that they are complete as a consequence. (9) C N( [ a ,b]), the space of scalar-valued functions on a bounded interval [ a ,b] with N bounded derivatives, the norm being
(Is
lt
It is shown in Problem 2 at the end of the chapter that this space is complete. This space is an indication of how normed linear spaces can carry information about derivatives. Indeed, normed linear spaces carrying information about derivatives play a significant role in the subject of partial differential equations.'
his is one of the themes of the companion volume Advanced Real Analysis
1. Definitions and Examples
523
(10) H m ( D ) , the space of bounded functions in the open unit disk D = {lzl < 1) in C such that the function is given by a convergent power series. The norm is the supremum norm. It is shown in Problem 3 at the end of the chapter that this space is complete. (11j A ( D ),the space of bounded continuous functions on the closed unit disk whose restriction to the open unit disk is given by a convergent power series. The norm is the supremum norm. It is shown in Problem 3 at the end of the chapter that this space is complete.
Two further kinds of nornled linear spaces are worth mentioning now. One is that any real or complex inner-product space X in the sense of Section 11.1 gives an example of a normed linear space. Recall that an inner product on X is a function ( . , . ) from X x X to IF that is linear in the first variable, is conjugate linear in the second variable, is symmetric if IF = R or Hermitian symmetric if IF = C,and has ( x ,x ) 10 for all x with equality if and only if x = 0. Such an inner product satisfies the Schwarz inequality I ( x , y) I i (x, x) l / ' ( y , y) ' I 2 , according to Lemma 2.2, and then the definition Ilx 11 = (x, x)lI2 makes X into a normed linear space, according to Proposition 2.3. As a normed linear space, an inner-product space may or may not be complete. , p ) , with ( f , g) = J;, f g d p , is an example in which the Any space L ~ ( sA, associated nonned linear space is complete. An inner-product space whose associated normed linear space is complete is called a Hilbert space. The other kind of normed linear space worth mentioning now involves bounded linear operators. Recall from Section V.9 that a linear function L : X + Y bctwccn two normcd lincar spaccs with rcspcctivc norms 11 . I l x and 11 . I l y is often called a linear operator. Proposition 5.57 showed that a linear operator L is continuous at a point if and only if it is continuous everywhere, if and only if it is uniformly continuous, if and only if it is bounded in the sense that 11 L (x) 11, i M llx I l x for some constant M and all x in X . The least such constant M is called the operator norm of L , written 11 L 11. We can define addition and scalar multiplication on bounded linear operators from X to Y by addition and scalar multiplication of their values:
+
Then L 1 L2 and c L are linear operators by the elementary theory of vector spaces, and the inequalities
and
524
XII. Hilbert and Banach Spaces
+
+
+
show that L1 Lz and cL are bounded with llLl Lzll 5 IILIII llLzll and llcLII 5 IclIILII. Applying the latterconclusion to c-'whenc f Ogives IILII = I ~ C - ' ( C L5 ) I II C ~ - ' ~ ~ C L5I I ~ c l - l l c l l l= ~ ~IILII, ~ and we conclude that llcLll = I cl 11 L 11 . Since it is plain that 11 L 11 > 0 with equality if and only if L = 0, the set of bounded linear operators from X to Y, with the operator norm, is a normed linear space. We denote this normed linear space by B(X, Y).
Proposition 12.1. If X and Y are norrlled linear spaces and if Y is conlplete, then the normed linear space B(X, Y) is a Banach space. REMARKS.In the special case in which Y is the set IF of scalars, the linear operators are called linear functionals,in terminology we have used repeatedly. The normed linear space IF = IF1 is complete, and therefore the normed linear space of bounded linear functionals on X is a Banach space. The space of bounded linear functionals is called the dual space of X and is denoted by X*. More explicitly the norm of an element x* of X* is2
Proposition 12.1is implicitly saying that X* is always complete. PROOF.Let {L,} be a Cauchy sequence in B(X, Y) . Since in any metric space the members of a Cauchy sequence are at a bounded distance from any particular element, the sequence {I1L, II } is bounded. Let C = sup, II L, 1 1 . If x is in X , then {L, (x)} is a Cauchy sequence since / I L, (x) - L, (x) 11 5 11 L,- L,II Ix I I x . Bycompletenessof Y, L(x) = limn L,(x) exists. Continuityof addition and scalar multiplication in X implies that L (x +xf) = limn L, (x +xf) = lim,(L,(x) L,(xf)) = limn L,(x) limn L,(xf) = L(x) L(xf) and that L(cx) = limn L,(cx) = lim,(cL,(x)) = c limn L,(x) = cL(x). Therefore L is a linear operator. For boundedness of L, we have IIL,(x)lly 5 IIL,II IlxIlx 5 CIIxIlx for all n. Hence continuity of the norm function implies that II L (x) 11, = II lim L, (x) 11, 5 liminf, IIL,(x)ll, 5 CllxIlx, and L is bounded with IILII 5 C . To complete the proof, we show that 11 L,, - L I + 0 Assurning the contrary, we can pass to a subsequence and then change notation so that 11 L, - L 11 ? t for some t > 0 for all n . Then for each n,we can find x, in X with llxnllx = 1
+
+
+
'A superscript * has also been used in this book to indicate a one-point compactification, but t11ei.e ~leedIICVM be ally c o ~ ~ f u s iabout o ~ l this lota at ion. One-pint compactifications arise in pactice only for locally compact Hausdorff spaces, and one can show that a normed linear space is locally compact only if it is finite dimensional, For finite-dimensional normed linear spaces it is always clear from the context whether * refers to the dual space or to the one-point compactification.
1 . Definitions and Examples
525
such that 11 Ln(x,) - L(xn)11 > t / 2 . Choose and fix N so that in > N implies 11 LN - L, 11 5 t/4. Whenever m > N , the triangle inequality gives
in contradiction to the fact that lim, Lm(xN)= L(xN) EXAMPLESOF DUAL SPACES. (1) Lp(S, A, w)* E L ~ ' ( s ,A, w) if 1 i p < oo, is o-finite, and p' is the dual index with 1 + 4P = 1, according to the Riesz Representation Theorem P (Theorem 9.19). Specifically to each x* in LP(S, A, p)* corresponds a unique g in LP'(S, A, p ) with x*( f ) = f g d p for all f in LP(S, A, p ) , and this g has Ilx*ll = Ilgll,,. It can be shown that the hypothesis of o-finiteness of p can be dropped if 1 < p < oo,but Problem 4 at the end of Chapter IX shows that the hypothesis cannot be completely dropped for p = 1.
IS
(2) (&{)* E litand (lp)* E lp' for 1 5 p < oo if p' is the dual index. This is a special case of Example 1. In particular, the first of these duality results for p = 2 says that [Rn)*g Rn and (cCn)*g cCn. (3) C (S)* E M(S) if S is a compact Hausdorff space, according to Theorems 11.26 and 11.28. Specifically to each x" in C (S)* corresponds a unique p in M(S)withx*(f) = J;, f dpforall f inC(S),andthisphas IIx*I = IlpII. Since M ( S ) is in this way identified as the dual space of some normed linear space, it follows from Proposition 12.1 that M(S) is a Banach space. (4) ( l r ) * E and (co)* E l l .The isomorphism ( l r ) * E l: is the special case of Example 3 in which S = 11, . . . , n } . To see the isomorphism (co)* r" 11, we take S to be the set of positive integers and form the one-point compactification S*. The continuous scalar-valued functions on S*, with their supremum norm, can be identified with the normed linear space c of convergent sequences. Thus M(S*). The members of co are the Example 3 in this setting says that c* members of c that vanish at oo,and any point mass at oo in a member of M(S*) has no effect on the subspace co. It readily follows that the dual of co consists of the members of M(S*) with no point mass at oo,and these elements, with their norm, may be identified with l l .
LA
From one point of view, Hilbelt spaces are palticularly simple Banach spaces, and we shall study them first. The geometry of Hilbert space will be the topic of the next section, and the section after that will give a brief introduction to bounded linear operators from a Hilbert space to itself.
526
XII. Hilbert and Banach Spaces
2. Geometry of Hilbert Space Hilbert spaces were defined in Section 1 as complete normed linear spaces whose norms arise from an inner product. Euclidean space Rn and complex Euclidean space Cn are examples, and every space L2(S, A, w ) with ( f , g ) = f fg d w is a Hilbert space. We shall see in this section that every Hilbert space shares many geometric facts in common with the finite-dimensional examples Rn and 6 " . The expansion of square integrable functions on [-n, n]in Fourier series will be seen to be an example of expansion of all members of a Hilbert space in terms of an "orthonormal basis." Let H be a real or complex Hilbert space with inner product ( . , . ) and with norm 11 . 11 given by llln 11 = (11, u ) ' / ~ .Lemma 2.2 shows that H satisfies the Schwarz inequality
The Schwarz inequality implies the estimate
from which it follows that the inner product is a continuous function of two variables. We shall make frequent use of the formula
which is what one combines with the Schwarz inequality to prove the triangle inequality for the norm. With the additional hypothesis that ( u , v) = 0, this formula reduces to the Pythagorean Theorem
Direct expansion of the norms squared in terms of the inner product shows that H satisfies the parallelogram law
+ + Ilu
~ l u v12
-
v112 = 211~11~ +211~11~
for all u and v in H.
Actually, there is a converse to this formula, due to Jordan and von Neumann, whose details are left to Problems 19-24 at the end of the chapter: a Banach space
2. Geometry of Hilbert Space
527
is a Hilbert space if its norm satisfies the parallelogram law. The idea is that the inner product in a Hilbert space can be computed from the identity
where the sum extends for k E {0,2} if the scalars are real and extends for k E {0, 1 , 2,3} if the scalars are complex. This identity goes under the name polarization. For the result of Jordan and von Neumann, one dejines ( u , v) by this formula, shows that the result is an inner product, and proves that 11 u 112 = ( u , u ) . 'I'he following lemma, which makes use of the completeness, is the key to all the geometry.
Lemma 12.2. If M is a closed vector subspace of the Hilbert space H and if u is in H , then there is a vector v in M with vll = inf Ilu
-
w€M
REMARK.Examination of the proof will show that we do not make full use of the assumption that M is closed under addition and scalar multiplication, only llral M is clused ur~dcrpassage Lu cvrivex curnbirra~iu~~s, i.e., lliat x arid y in M imply that t x + (1 - t ) y is in M for all t with 0 5 t 5 1. Thus it is enough to assume that M is a closed convex set, not necessarily a closed vector subspace. Ilu
PROOF.Let d = infWGntIlu - lull, and choose a sequence {w,} in M with - w, 11 + d . By the parallelogram law,
Since i ( w ,
+ w,)
is in M ,
We conclude that 11 w , - w,112 + 0 , and {w,} is Cauchy. By completeness of H , {w,} is convergent. If v = lim w , , then v is in M since M is topologically closed. Since 11 11 - w, 11 + d , continuity of the norm gives 11 LL - v I - d . Two vectors m and I) in H are said to be orthogonal if ( Z Av,) = 0. The set of all vectors orthogonal to a subset M of H is denoted by M 1 . In symbols, M~
= {u E H
I
( u , v ) = 0 for all
v
E
M}.
We see by inspection that M1 is a closed vector subspace. Moreover, ~ f hrll ' =0 since any u in M f'M1 must have ( u , u ) = 0. The subspace M 1 will be of greatest interest when M is a closed vector subspace, as a consequence of the following proposition.
528
XII. Hilbert and Banach Spaces
Proposition 12.3 (Projection Theorem). If M is a closed vector subspace of the Hilbert space H , then every u in H decomposes uniquely as u = v w with v in M and w in MI.
+
REMARKS.One writes H = M @ MIto express this unique decomposition of vector spaces. Because of this proposition, M1 is often called the orthogonal complement of the closed vector subspace M . It is essential that M be closed in this proposition. In fact, consider the vector subspace M of polynomials in L ~ ( [ o 11). , This is dense as a consequence of the Weierstrass Approximation Theorem, and consequently no L2 function other than 0 can be in MI.Thus not every member of L~ is the sum of a member of M and a member of M1.
PROOF.Uniqueness follows from the fact that M f' M I = 0. For existence let IIu - wII. u bein H,andchoose v in M by Lemma 12.2 with Ilu - vII = infWEM If m is any member of M with llm 11 = 1, then the vector v + (u - v, m)m is in - 2Re(x, y) + llY112 gives M and the formula Ilx - Y112 = 11x11~
Hence (u - v , m) = O. Since every nonzero member of M is a scalar multiple of a member with Imll = 1, u - v is in M1.
Corollary 12.4. If M is a closed vector subspace of the Hilbert space H , then M1l = M. PROOF.From the definition we see that M 5 ML1. If u is in ML1, write u = m+ml withm E M and rnl E M1 by ~ r o ~ o s i t i o12.3. n Then0 = ml+(rn - u) with m1 E M~ and m - u E MI1. By the uniqueness in the decomposition H = M1 @ MI1 of Proposition 12.3, m1 = 0 and m - u = 0. Therefore u = m is in M, and MI1 = M.
'l'heorem 125 (Riesz Representation 'I'heorem). H' & is a continuous linear functional on the Hilbert space H , then there exists a unique v in H with L(u) = (u, v) for all u in H. This vector u has the property that llLll = Ilvll. REM~,ICS.It is instructive to comparc this rcsult with thc vcrsion of thc Riesz Representation Theorem in Theorem 9.19, which applies to LP(S, A, p ) for 1 5 p < oo and in particular to L2(s, A, p ) . That theorem associates to a continuous linear functional L on this L2 space a member g of the space such that E( f ) = Js f g d p for all f in the space. The present theorem, applied with H = L2(s, A, p ) , instead yields a member v of the space such that E( f ) = f G d p
5 30
XII. Hilbert and Banach Spaces
up to a scalar, and Pn ( t ) is a suitably normalized polynomial solution, necessarily of degree n . These can be shown to be orthogonal3 in L ~ ( [ -1, 11, d t ) . Another example was constructed from the Bessel function
which was defined in Section 1V.8. There are infinitely many distinct positive real numbers k, such that Jo(k,) = 0, and it can be shown that the functions x h JO(k,x) are o r t l ~ o ~ o i iin a l L'([o, ~ 11, x d x ) . If an ordered set of n linearly independent vectors in H is given, the GramSchmidt orthogonalization process, which appears in Problem 6 at the end of the present chapter, gives an algorithm for replacing the set with an orthonormal set having the same linear span. Let M be a closed vector subspace of H , so that H = M @ M' by Proposition 12.3. Tlle linear projection operator E of H on M along M L ,given by the identity on M and the 0 operator on M', is called the orthogonal projection of H on M . The linear operator E is bounded with 11 E 11 5 1 because if u E H decomposes as u = in + m', the Pythagorean Theorem gives
We are going to derive a forrnula for E in terms of orthonorrnal sets.
Lemma 12.6. If { u j } is an orthonormal sequence in the Hilbert space H and if { c j } is a sequence of scalars, then CE1c j u j converges if and only if CE I cj < 00, and in this case
When the series converges, the sum CTlcj u j is independent of the order of the terms.
PROOF.For m
> n, we have
x:=,
x,"=,
This shows that the sequence { c j u j } is Cauchy in H if and only if Icj l 2 is convergent, and the first conclusion follows since H is complete. When 3 ~ h verification e appears in the problems in the companion volume Advanced Real Analysis. 4 ~ g a i the n verification appears in the problems in the companion volume Advanced RealAnalysis.
531
2. Geometry of Hilbert Space
{ CC1c j u j } is convergent, we denote its limit by CF1c j u j , and continuity of the norm yields 11 CEl cjuj 11 = limp I Cf=,cjuj 11. Since we have seen that the second conclusion of the lemma follows. 1 EL1cjuj I = (ZY=lIcj 1 2 )
Let u = C jc j u j ,and let Ckcj,ujk be arearrangement, necessarily convergent by what has already been proved. Suppose that the rearrangement has sum u'. The Ici l 2 = llu'112 since rearrangements equality just proved shows that llu 112 = of series of nonnegative reals have the same sums. Continuity of the inner product, together with the same computation as made above, gives
xrl
( u , u') = lim (
ci ui ,
cjkujk)= lim
P 4
p,q
C
liiip, i = j ~with k s q
lei 12.
The limit on the right is C r l Ici l 2 since Cklcjkl 2 is a rearrangement of X i Ici 1 2 , and hence ( u , u ' ) = C r l 1cil2 = llu112 = 1lu'l2. Therefore Ilu - u'112 = ( u , u ) - 2 R e ( u , u') + ( u ' , u') = 1lul2 - 211u112 + llu112 = 0 , and u' = u .
Proposition 12.7. Let S be an orthonormal set in the Hilbert space H , and let M be lllc srrrallcs1 ~ l u s c dvcclvr subspact: of H ~unbainingS . Fur cad1 u in H , there are at most countably many members v, of S such that ( u , v,) # 0 , and thus the series
has only countably many nonzcro tcrms. Thc scrics convcrgcs indcpcndcntly of the order of the nonzero terms, E is the orthogonal projection of H on M , and E satisfies
REMARK.The final inequality of the proposition is Bessel's inequality.
PROOF.Let v,,, . . . , van be a finite subset of S , and form the vector u' =
C;=,( u , v a j ) v a j .Taking the inner product of both sides with u gives
and Lemma 12.6 gives n
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XII. Hilbert and Banach Spaces
Therefore0 5 1lu-u'1~= l l ~ 1 1 ~ - 2 ~ e (~u' ), + l l u ' 1= ~ llu112 - llu'112,and weobtain
=
no matter what finite subset v,, , . . . , van of S we use. The sum of uncountably many positive real numbers is infinite, since otherwise there could be only finitely many greater than l / n for each n . Since 11 u 112 < oo, (**) implies that there can be only countably many a's with (u, u,) l 2 nonzero. This proves the first conclusion. If we enumerate those a's and apply Lemma (u, ua)ua to a sum independent of the 12.6, we obtain the convergence of order of the terms. It is evident from the formula that E is linear and that E(u) = 0 if u is in M ~ Inequality . (**) shows that the partial sums u' of E(u) have Ilu'll 5 Ilu 1 1 , and the continuity of the norm therefore implies that IIE(u) 11 5 llull for all u . Hence E is continuous. Since E(u,) = u, for all a , E is the identity on all finite linear combinations of members of S. The continuity of E thus implies that E is the identity on all of M. Hence E is the orthogonal projection as asserted. The final assertion of the proposition follows from Lemma 12.6 and the inequality IIE(u)ll 5 IluII, which we have already proved.
x,,ES
Corollary 12.8. If S is an orthonormal set in the Hilbert space H , then the following are equivalent: (a) S is maximal among orthonormal subsets of H , (b) u = ~ v , E S ( u , v , ) v , f o r a l l u i n H , (c) ~ l u l l ' = C
V,
(d) ( u , u ) = E
ES
V,
I(u,u,)12
ES
foralluinH,
( u , v , ) ( u , ~ ) foralluanduinH
REMARKS.Condition (b) is summarized by saying that the orthonormal set S is an orthoilor~l~al basis of H . If H is infinite-dimensio~lal,an orthonor~llalbasis is not a basis in the ordinary linear-algebra sense; a passage to the limit is usually needed to expand vectors in terms of the basis. Condition (c), or sometimes condition (d), is called Parseval's equality. Thus the corollary says that the orthonormal set S is maximal if and only if it is an orthonormal basis, if and only if Parseval's equality holds.
2. Geometry of Hilbert Space
533
PROOF.Let M be the smallest closed vector subspace of H containing S. Then S is maximal if and only if M' = 0, and we replace (a) by this condition. If M' = 0, then E is the identity operator in Proposition 12.7, and the proposition shows that (b) holds. If (b) holds, Proposition 12.7 says that (c) holds. On the other hand, if (c) holds, then Proposition 12.7 says that Ilu 11 = 11 E ( u )11 for all u . For a vector m in M', which must have E ( M )= 0, this says that lmII = 0. Thus M' = 0, and (a) holds. Hence (a), (b), and (c) are equivalent. Finally (c) and (d) are equivalent by polarization. In the context of Fourier series, Parseval's equality ((c) in Corollary 12.8) was proved as Theorem 6.49, and that theorem showed also that any member of L 2 ( [ - n , n ] , d x ) is the sum of its Fourier series in the sense of convergence in L2. This conclusion was (b) in the corollary. The corollary is showing that the equivalence of (b) and (c) is just a result in abstract Hilbert-space theory. The extra content of Theorem 6.49 is that these conditions are actually satisfied by the system of exponential functions. One can show that the other two examples we gave in this section of orthogonal sets give orthonormal bases when normalized- the Legendre polynomials P, ( t ) on [ - I , 11 with respect to dt and the functions Jo(k,,t) on [0, 11 with respect to t dt.
&
Proposition 12.9. Let (X, p ) and ( Y , v ) be o-finite measure spaces, and suppose that L 2 ( X ,p ) has a countable orthonormal basis { u i }and L 2 ( y , v ) has a countable orthonormal basis { v j } .Then { ( x ,y) H ui (x) vj ( y ) }is an orthonormal basis of L 2 ( x x Y, p x v). PROOF.The functions u, (x)v,(y) are orthonormal, and Corollary 12.8 shows that it is enough to prove that this orthonormal set is maximal. Suppose that w ( x , y) is an L2 function on X x Y orthogonal to all of them. Then
for all i and j . Since { u ,} is an orthonormal basis of L2( x ,p ) , x H ( w ( x , .) , v,) is the 0 function in L 2 ( X ,y ) for each j . In other words, ( w ( x , . ), v,) = 0 for a.e. x [ d y ]for that j . Since the number of j's is countable, ( w( x , .) , u,) = 0 for all j f o r a e x [ d p ] Any such x has0 = I(ul(x, .), t5)l2= lul(x, y)12dl)(y) Integrating in x ,we see that w is the 0 function in L 2 ( x x Y, x v) .
'&
Proposition 12.10. Any orthonormal set in a closed vector subspace M of a Hilbert space H can be extended to an orthonormal basis of M . In particular any closed vector subspace M of H has an orthonormal basis.
5 34
XII. Hilbert and Banach Spaces
PROOF.As a closed subset of a complete space, M is complete, and therefore M is a Hilbert space in its own right. Order by inclusion all orthonormal subsets of M containing the given set. The given set is one such, and the union of the members of a chain is an orthonormal set forming an upper bound for the chain. By Zorn's Lemma we can find a maximal orthonormal set S in M containing the given one This satisfies (a) in Corollary 12 8 and hence is an orthonormal basis This proves the first conclusion, and the second conclusion follows from the first by taking the given orthonormal set in M to be empty.
Proposition 12.11. Any two orthonormal bases of a Hilbert space have the same cardinality. REMARKS.Cardinality is discussed in Section A10 of the appendix. The "same cardinality" whose existence is proved in the proposition is called the Hilbert space dimension of the Hilbert space. Problem 7 at the end of the chapter shows that two Hilbert spaces are isomorphic as Hilbert spaces if and only if they have the same Hilbert space dimension. Despite the apparent definitive sound of this result, one must not attach too much significance to the proposition. Hilbert spaces that arise in practice tend to have some additional structure, and an isomorphism of this kind need not preserve the additional structure. PROOF.Fix two orthonormal bases U = {u,} and V = { u p } of a Hilbert space H. We define two members u, and u , ~of U to be equivalent if there exists a sequence
with !la, = M, and M,,, = M,, ,with each M,] in U and each r)pJin V, and with each consecutive pair having nonzero inner product. Define an equivalence relation in V similarly. Each equivalence class is countable. In fact, consider the class of u,, , and consider sequences of a fixed length. Proposition 12.7 shows that only countably many members of V can have nonzero inner product with u,, , only countahly many members of U can have nonzero inner product with that, and so on. Thus there are only countably many sequences of any particular length. The countable union of these countable sets is countable, and thus there are only finitely many sequences connecting u,, to anything. Hence u,, can be equivalent to only countably many members of U . Let U1 and Vl be equivalence classes in U and V ,respectively, and suppose that u,, and up, are members of U1 and Vl with nonzero inner product. Expand u,, in terms of V as u, = Cp(u,, , u p )u p ,retaining only the terms with (u,, , u p ) # 0. One of the terms making a contribution is the one with up = up,, and it follows
3. Bounded Linear Operators on Hilbert Spaces
535
that any other term with (u,,, v g ~ # ) 0 has vg~equivalent to vg. Hence we have
ua0 =
x
( u , ~ ,vp)vp
and similarly
up, =
x
(uporu,)u,.
If u,;, is another member of UI and we expand it in terms of V, retaining only the nonzero terms, then the up's that occur have to be equivalent to one another. So we have u,;, = CVPEV2 ( u W ,uB)uBfor solne equivalence class V2 within V. If we form a sequence (*) connecting u,, and u,;, ,we see that at least one member of V2 is connected to at least one member of Vl . Thus Vl = V2. Consequently every member of U1 lies in the smallest closed vector subspace containing V1, and every member of Vl lies in the smallest closed subspace containing U1. In other words, U1 and V1 are orthonormal bases for the same closed vector subspace of H. If U1 is finite, then linear algebra shows that Vl is finite and has the same number of elements. Since U1 and Vl are countable, the only way that either can be infinite is if both are countably infinite. In any event, U1 and V1 have the same cardinality. Thus we have a one-one function carrying U1 onto Vl . Repeating this process for each equivalence class within U , we obtain a one-one function carrying U onto V.
3. Bounded Linear Operators on Hilbert Spaces In this section we briefly study bounded linear operators from a Hilbert space H to itself. In the finite-dimensional case we often make a correspondence between matrices and linear operators by using the standard basis of the space of column vectors. If {ei is this basis, then the correspondence between a matrix A = [ A i j ]and a linear operator L is given by Aij = ( L ( e j ) ,ei). If u = CJ. uJ. eJ . and v = Ci uiei are column vectors, then L ( u ) = C ju j L ( e j )and hence ( L ( u ) ,v) = ujCi ( L ( e j ) ,e i ) = GiAijuj. We could extend these formulas to the case of a general Hilbert space, not necessarily finite-dimensional, by using a particular orthonormal basis as the generalization of { e i } . But no particular such basis recommends itself, and we work without any choice of basis as much as possible, except for purposes of motivation. Instead, we may think of the function ( u , v) H ( L ( u ) ,v ) as a more appropriate -and canonical -analog of the matrix of L . Just as the operator norm of L is given by a formula that views L as an operator, namely
xi,
xij
536
XII. Hilbert and Banach Spaces
so there is a formula for computing the norm in terms of the function of two variables, namely IlLll = sup I(L(u), v)l. Ilull51, ll~llll
To verify this formula, fix u and let v have norm 5 1. Application of the Schwarz inequality gives I(L(u), v)l i IIL(u)llllvll i IIL(u)ll. On the other hand, if L(u) # 0, we take v = l l ~ ( u ) l l - ' ~ ( u )this ; v has llvll = 1, and we obtain I(L(u), v)I = lIL(u)Il-'(L(u), L(u)) = IIL(u>ll. Hence suPllvl151 I(L(u), v)ll = 11 L (m) 11. Taking the supremum over 11 m I 1 shows that the two expressions for 11 L 11 are equal. We shall work with the "adjoint" L* of a bounded linear operator L . In terms of matrices in the finite-dimensional case, the matrix of L* is to be the conjugate transpose of the matrix of L . In other words, the (i, j)thentry (L*(ej), e i ) ) of the matrix for L* is to be (L (e,), ej) = (ej , L (ei)). Passing to our functions of two variables, we want to arrange that (L*(u), v) = (u, L (v)) for all u and v. Let us prove existence and uniqueness of such a bounded linear operator.
Proposition 12.12. Let L : H 3 H be a bounded linear operator on the Hilbert space H. For each u in H , there exists a unique vector L*((u in H such that (L*(u),v)=(u,L(v)) forallvinH. As u varies, this formula defines L* as a bounded linear operator on H , and llL*11 = llLll. PROOF.The function v H (L(v), u) is a linear functional on H satisfying I(L(v),u)I i IILllllllullllvll,hence having norm i IILllllllull. Being bounded, the linear functional is given by (L(v), u) = (v,w) for some unique w in H , according to Theorem 12.5. We define L*(u) = w ,and then we have (L*(u) , v) = (u, L(v)). This formula shows that L* is a linear operator, and the computation llL*11 = sup I(L*(u), v)l = sup I(u, L(v))l = sup I(L(v), u)l = IILII Ilull51> 11~1151
Il~ll51> 11~1151
Il~ll51> Ilvll51
shows that 11 L*11 = 11 L 11. The bounded linear operator L* in the proposition is called the acljoiilt of L. The mapping L H L* is conjugate linear. We shall be especially interested in the case that L* = L, in which case we say that L is self adjoint. An example of a self-adjoint operator is the orthogonal projection E on a closed vector subspace M as defined before Lemma 3.6. In fact, if u in H decomposes according to H = M @ M' as u = u' + u", then the computation (1 - E)(u) =
4. Hahn-Banach Theorem
537
u - u' = u" shows that 1 - E is the orthogonal projection on M'. Hence ( E ( u ) ,( 1 - E ) ( v ) )= 0 for all u and v in H , and also ( ( 1 - E ) ( u ) ,E ( v ) ) = 0 . The first of these says that ( E ( u ), v ) = ( E ( u ) ,E ( v ) ), and the second says that ( E ( u ), E ( v ) ) = ( u , E ( v ) ). Combining these, we obtain ( E ( u ), v ) = ( u , E ( v ) ). Comparison of this formula with the formula in Proposition 12.12 shows that E = E*. The Banach space B ( H , H) is closed under composition. In fact, if L and M are in U ( H , H) ,then linear algebra shows L M to be linear, and the computation II(LM)(u)II = IIL(M(u))ll i IILllllM(u)ll i IILII IIMII Ilull h w s that
Hence LM is in U ( H , H ) if L and M are. Within U ( H , H ) , we have ( L M ) * = M* L*.
4. Hahn-Banach Theorem We return now to the setting of general normed linear spaces or Banach spaces. 'I'here are three main theorems concerning the norm topology of such spaces -the Hahn-Banach Theorem, the Uniform Boundedness Theorem, and the Interior Mapping Principle. These three theorems are the main subject matter of the remainder of this chapter. We shall often use symbols x , y , . . . for members of a normed linear space and symbols x * , y*, . . . for linear functionals. This notation has the advantage of allowing us to use symbols like x** for linear functionals on a space of linear functionals, an important notion as we shall see. We begin with the Hahn-Banach Theorem, which ensures the existence of many continuous linear functionals on a normed linear space. The theorem has applications even in situations in which one has a concrete realization of the dual space, because it shows that any closed vector subspace is characterized by the continuous linear functionals that vanish on the subspace.
Theorem 12.13 (Hahn-Banach Theorem). If Y is a vector subspace of a rlvnried lirlear space X arid if y* is a cv~~liriuuus lirlear fu~~cliurial v11 Y , llleri Illere exists a continuous linear functional x* on X with Ilx* 11 = 11 y* 11 such that x*(y)= y*(y)
for all y
E
Y.
The theorem as stated is derived from the following lemma, which itself goes under the name "Hahn-Banach Theorem" and has other applications quite distinct from Theorem 12.13 that are beyond the scope of this book.
538
XII. Hilbert and Banach Spaces
Lemma 12.14. Let X be areal vector space, and let p be areal-valued function on X with p(x
+ x') i p ( x ) + p ( x f )
and
p(tx) = tp(x)
for all x and x' in X and all real t > 0. If f is a linear functional on a vector subspace Y of X with f ( y ) 5 p ( y ) for all y in Y , then there exists a linear functional F on X with F ( y ) = f ( y ) for all y E Y and F ( x ) 5 p ( x ) for all x E X.
PROOF.Form the collection of all linear functionals on vector subspaces of X that extend f and that are dominated by p , and partially order the collection by saying that one is 5 another if the second is an extension of the first. If we have a chain of such extensions, then we can obtain an upper bound for the chain by taking the union of the domains and using the common value of the linear functionals on an element of this domain as the value of the linear functional forming the upper hound. The result is linear because any two members of the domain must lie in the domain of a single member of the chain. By Zorn's Lemma let fo , with domain Y o ,be a maximal extension. We shall prove that Yo = X . In fact, suppose that yl is a vector in X but not Yo. Every vector in the vector subspace Y1 spanned by yl and Yo has a unique representation as y c y l , where y is in Yo and c is in R. Define fl on Y l by
+
where k is a real number to be specified. For a suitable choice of k , fl will be bounded by p and will contradict the maximality of (fo, Y o ) . Let y and y' be in Yo. Then
and hence
-P(-yl
-
Y ) - fo(y) 5 P(Y'
+ yl)
-
fo(yf)-
Take the supremum of the left side over y and the infimum of the right side over y', let k be any real number in between, and define fl on Y 1by (*). 'I'o complete the proof, we are to check that ji ( x ) 5 p ( x ) for all x in Y 1 .'I'hus suppose that x = y cyl is arbitrary in Y l . If c = 0, then fl ( x ) 5 p ( x ) by the assumption on Yo. If c > 0, then
+
If c < 0, then
4. Hahn-Banach Theorem
539
PROOFOF THEOREM12.13. If the field of scalars is R,then Theorem 12.13 follows immediately from Lemma 12.14with p(x) = 11 y* 11 Ilx 11 and f = y* . If the field of scalars is C, if y* is given, and if, as we may, we regard X as a real normed linear space, then Re y* defined by (Re y*)(y) = Re(y*(y)) is a real linear functional on Y with
By what has already been proved, we can extend Re y* without an increase in norm to a real linear functional F defined on all of X . Define
We show that x* has the required properties. Certainly x* (xfx') = x* (x)+x* (x') and x*(cx) = cx*(x) for c real. Furthermore
Thus x* is complex linear. On Y, we have (Re y*)(iy)
+ i(Imy*)(iy) = y*(iy) = iy*(y) = -(Im?*)(?)
+i(Re?*)(y),
and thus (Re y *) (iy) = -(h y*) (y) . Substituting this identity into the definition of X* , we obtain
for y in Y. Thus x* is an extension of y*. Finally if x*(x) = reie for r and 8 real and r > 0, then
since the nonnegative numberx*((ePisx)has 0 imaginary part. Thus Ilx* 11 5 11 y* 1 1 . The reverse inequality follows because x* is an extension of y* , and the proof is complete.
Corollary 12.15. If Y is a closed vector subspace of a normed linear space X and if xo is a vector of X not in Y , then there exists an x* in the dual X* with x*(y) = 0 and
for all y
E
Y
x*(x0) = 1.
The norm of x* can be taken to be the reciprocal of the distance from xo to Y.
540
XII. Hilbert and Banach Spaces
PROOF.Let d > 0 be the distance from xo to Y, and let Z be the linear span of xo and Y. Every x in Z has a unique expansion as x = y cxo for some scalar c and some y in Y. For such an x ,let Z* ( x ) = c . Let us see that the linear function
+
z* on Z satisfies 1
Iz*II = d - .
First we check that Iz*(x)I 5 d-'IlxII: if c
(*)
+ 0, then
while if c = 0, then z * ( x ) = 0. Thus I z * ( x ) 1 5 d-lllx 11 for all x , and we obtain llz* l l 5 d-' . For the reverse inequality, let {y,} be a sequence in Y, not necessarily convergent, with limn llxo yn I = d . Then -
and hence 112" 11 1 d - l . This proves (*). Applying Theorem 12.13 to z * , we obtain the corollary. EXAMPLE. To illustrate Corollary 12.15, we re-prove the result of Proposition 11.21a that C ( S ) is dense in L P ( S , p ) if S is a compact Hausdorff space, p is a regular Borel measure on S , and p satisfies 1 5 p < cm.For definiteness let us suppose that the underlying scalars are real. If C ( S ) were not dense, then the corollary would produce a continuous linear functional C on L P ( S , p ) that vanishes on C ( S ) but is not identically 0 on L P ( S , p ) . Theorem 9.19 says that l has to be given by integration with some member g of L P ' ( S , p ) , where p' is the dual index: L( f ) = J;, f g d p for all f in L P ( S , p ) . Since L vanishes on C ( S ) , we have J;, f g d p = 0 for all f E C ( S ) . Thus Js f g + d p = Js f g - d p for all f E C ( S ) . Here g+ d p and g - d p are Borel measures on S , regular by Proposition 11.20, and they yield the same positive linear functional on C ( S ) . Applying the uniqueness in the Riesz Representation Theorem (Theorem 11.I), we obtain g+ d p = g - d p and therefore g+ = g - almost everywhere. Since g+ and g - are nowhere both nonzero, gf = g- = 0 almost everywhere. Hence g is the 0 function, and l = 0 , contradiction.
Corollary 12.16. If X is a normed linear space and if xo llicri Illere is an x* in X* will1 IIx*
11
=1
and
# 0 is a vector in X ,
x * ( x o ) = llxoII.
PROOF.Apply Corollary 12.15 with Y = 0 and multiply by llxo I the linear functional that is produced by that corollary.
4. Hahn-Banach Theorem
541
Corollary 12.16, when applied to xo = x - x', shows that there are enough continuous linear functionals on a normed linear space X to separate points. Also, it implies that the only vector xo in X with x*(xo) = 0 for all x * in X* is xo = 0. The third corollary we have already seen for LP spaces with 1 i p < oo in Proposition 9.8, at least when the measure space is o-finite.
Corollary 12.17. If X is a normed linear space and xo is in X , then
PROOF. If Ilx*ll i 1, then Ix*(xo)l i Ilx*llllx~ll i Ilxoll, and therefore SUplln~I151 Ix*(xo)I 5 llxO1 1 . Thc lincar functional of Corollary 12.16 shows that equality holds. We have seen for o-finite measure spaces that X = L (S, p ) may be identified with Lm(S, p) via integration. In turn every member of L ~ ( s ,p ) then acts as a continuous linear functional on Lm(S, p) via integration. This change of point of view amounts to the implementation of a certain canonically defined linear rrrappirlg vf X irilv X"", wl~icliwe rlvw define fur ge~lcralrlvnried lirlear spaces. Let X be a normed linear space, and let X** be the dual of X*. We define a linear operator i : X + X** by
and we call L the canonical map of X into X**
Corollary 12.18. If X is a normed linear space, then the canonical map + X** has Ili (x) 11 = llx 11 for all x and in particular is one-one. Consequently if X is complete, then i (X) is a closed vector subspace of X**. I
: X
PROOF.We have Ill(~)ll= sup I(1(x))(x*)l = sup Ix*(x)l = llxll, Ilx* lli 1
Ilx*lli 1
the last step holding by Corollary 12.17. This proves the first conclusion. Because I p~eservesrivmis, X cvrriplete ir~ipliesthat i(X) is a cor~ipletesubset of the complete space X** and is therefore closed, by Corollary 2.43. A Banach space X is said to be reflexive if the canonical map carries X onto X**. Warning: This is a more restrictive condition than to say that there is some norm-preserving linear mapping of X onto X**.
XII. Hilbert and Banach Spaces
542
Finite-dimensional normed linear spaces are reflexive since linear functionals in this case are automatically continuous and since the vector-space dual of a finite-dimensional vector space has the same dimension as the space itself. Hilbert spaces are reflexive as a consequence of the Riesz Representation Theorem in its form in Theorem 12.5. The spaces LP(S, p ) for a o-finite measure space, when 1 r p r on,are reflexive as a consequence of the Riesz Representation he or em' in its form in Theorem 9.19. However, L (s,p ) and La3(S,p ) are often not reflexive, as is shown below in Proposition 12.19 and Corollary 12.21.
Proposition 12.19. If ( S , y) is a o-finite measure space with infinitely many disjoint sets of positive measure, then L' ( S , p ) is not reflexive. PROOF.Theorem 9.19 shows that the Banach space X = L ' ( s , p ) has X* E L w ( S , p ) ,the isomorphism being given by integration. Therefore it is enough to produce a continuous linear functional on L a 3 ( S ,p ) that is not given by integration with an L function. Thus let { E n }be a sequence of disjoint sets of positive measure, and let Y be the vector subspace of functions in L w ( S , p) that are constant on each En and have values on the En's tending to a finite limit as n tends to infinity. Let y* of such a function be the limit Then y* is a linear functional on Y of norm 1 By the Hahn-Banach Theorem (Theorem 12.13), there exists a linear functional x* defined on all of L a 3 ( S ,p ) , having norm 1, and restricting to y* on Y. Suppose that there is some g in L 1( S , p ) with x* ( f ) = f g d p for all f in Y , quite apart from all f in Lw ( S , p ) . I f f is 1 on En and is 0 elsewhere, then x*(f ) - 0 , and hence jEng d p = 0 In other words, g d p = 0 for every n. If we next take f to be 1 on U,"==, En and to be O elsewhere, then x* (j ) = 1. On the other hand, this f has
'
I;,
IEn
and we have a contradiction.
Proposition 12.20. If X is a Banach space and its dual X* is reflexive, then X is reflexive.
PROOF. Let 1 : X + X** and L* : X* + X*** be the canonical maps. Arguing by contradiction, suppose that X is not reflexive. Since L ( X )is a closed proper vector subspace of X**, Corollary 12.15 produces a nonzero member x*** of X*** such that x * * * ( l ( X ) )= 0. Since X* is reflexive by assumption, there exists x* in X* with x*** = l * ( x * ) . If x is in X , then we have 0 = x***(i(x))= ( L * ( x * ) ) ( L ( x= ) ) ( ~ ( x ()x)* )= x * ( x ) ,and hence x* = 0. But then x*** * - 1 ( x * ) = 0, and we have a contradiction. 5 ~ c t u a l l ythe , o-finiteness is not needed for 1 < p < oo
5. Uniform Boundedness Theorem
543
Corollary 12.21. If ( S , p ) is a o-finite measure space with infinitely many disjoint sets of positive measure, then L W ( S ,p) is not reflexive. PROOF.Theorem 9.19 shows that the Banach space X = L ' ( s , p ) has X* E L" ( S , p ) ,the isomorphism being given by integration. If X* were reflexive, then X would have to be reflexive by Proposition 12.20, in contradiction to Proposition 12.19. 5. Uniform Boundedness Theorem
The second main theorem about the norm topology of normed linear spaces is the Uniform Boundedness Theorem, also known as the Banach-Steinhaus Theorem. This result involves a parametrized family of linear operators from one normed linear space into another, and it is assumed that the domain is complete. Two kinds of boundedness as a function of one variable are assumed-boundedness of each linear operator as a function on (the unit ball of) the domain and boundedness in the parameter for each fixed member of the domain. The conclusion is boundedness in the two variables jointly.
Theorem 12.22 (Uniform Boundedness Theorem). If {L,} is a set of bounded linear operators from a Banach space X into a normed linear space Y such that
then there is a constant C independent of x such that 1 1 L ,
11 5 C for all a .
PROOF.For each positive integer n,the set
is closed in X , being the intersection of inverse images of closed sets in Y under continuous functions, and F, = X by assumption. By the Baire Category Theorem (Theorem 2.53b), one of the sets, say F N , contains a nonempty open subset B of X . Then 11 L,(x) 11 5 N for all a and for all x in B. If B contains the open ball in X of radius 2r > 0 and center b , then Ilx 1 1 5 r implies that x + b is in B and that
independently of a . Hence Ilx 11 i 1 implies
Inotherwords, IIL,II 5 r-'(N + C b ) .
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XII. Hilbert and Banach Spaces
EXAMPLE.Let us use the theorem to give a proof that the Fourier series of a continuous periodic function need not converge at some point. Consider the Banach space X of all continuous periodic functions f on [-n, n ] with the supremum norm. Let D, be the Dirichlet kernel as in Section 1.10, given by sin((n I + ) t )
n
e"' =
~ , ( t )=
1
sin 2 t
k=-n
The nth partial sum of the Fourier series of f is
Define linear functionals l , on X by
+
+
Each of these is bounded; specifically Il, 11 5 2n 1 because I I D, I , 5 2n 1. If thc Fouricr scrics of cach continuous function f wcrc to convcrgc at 0, thcn lim, l , ( f ) would exist for each f , and hence we would have Il, ( f ) 1 5 C f for a ~.u~lskaril Cf irldeperideril of n . The U~iiforrnBoundcd~iessTlieurern would say that Ill, 11 5 C for some constant C independent of n . The norm equality of Theorem 11.26 or 11.28 would then allow us to conclude that I D,(t) I d t is bounded. In fact, the numbers I Dn(t)I d t are unbounded, according to the following proposition, and thus there exists a continuous periodic function whose Fourier series diverges at x = 0.
12
STn
Proposition 12.23. The numbers
have the property that L,
= 4nP2logn
+ 0(1),
where O(1) denotes an expression bounded as a function of n . Hence L, is unbounded with n . REMARK.The numbers L, are sometimes called Lebesgue constants.
6. Interior Mapping Principle
PROOF.By writing sin((n + i ) t ) = sin nt cos i t 1 D,(t) = sinnt cot 2t
+ cos nt = 2t-'
where h,(t) is bounded in the pair ( n , t ) for It1 expression bounded as a function of n , then
545
+ cos nt sin i t , we see that sinnt
+ h,(t),
n . If we let 0 ( 1 ) denote an
The first term on the right side is bounded, and the sum in brackets lies between -
n ln(l+;+...+&)
and
-
n I n ( ; + . . . + ; ) ,1
fl
which are upper and lower Riemann sums for n - l n t-l dt and have difference n P 1 n ( lThus the sum in brackets is equal to n P 1 n ( l o g n+ O ( 1 ) ) . The integral of sin nt over [0, n l n ] is 2 / n , and the result follows.
i).
6. Interior Mapping Principle The third main theorem about the norm topology of normed linear spaces is the Intcrior Mapping Principle. This rcsult involvcs a singlc boundcd lincar opcrator from one normed linear space into another, and it is assumed that the domain and the range are both complete. The theorem is that if the operator is onto the range, then it carries open sets to open sets.
Theorem 12.24 (Interior Mapping Principle). If L is a continuous linear operator from a Banach space X onto a Banach space Y, then L carries open subsets of X to open subsets of Y .
546
XII. Hilbert and Banach Spaces
PROOF.Let B, be the closed ball in X with center 0 and radius r , and let Us be the open ball in Y with center 0 and radius s. The proof is in three steps. The first step is to show that ( L ( B ~ ) )contains ~' an open neighborhood of 0 in Y. To do so, we use the fact that L is onto Y to write
Thus Y = U ~ l ( ~ ( ~ , and ) ) cthe' ,Baire Category Theorem (Theorem 2.53b) shows that one of the sets ( L ( B , ) ) ~ contains ' a nonempty open set. Since L is linear and since multiplication by 2n is a homeomorphism of Y , (L(B,))" = ( L(2nBl12))"'= ( 2 n( ~B ~ , ~ )=) ~(2n) ' ( L( B ~ , ~ ) and ) ~ ' we , see that ( L ( B ~ , ~ ) ) ~ ' contains some nonempty open subset V of Y. Tf v and v' are in V , they are in ( L ( B ~ , ~ )and ) ~ there ' exist sequences {v,} and {vk}in L ( B 1 p )with v, + v and u; + u'. By linearity, v, vl, is in L ( B 1 ) ,and passage to the limit shows that v - v' is in L ( B ~ ) ~The ' . set V - V of such differences v - v' is the union over u' E V of V - u', hence is the union of open sets and is open. Since 0 is in V - V , the set V - V is an open neighborhood of 0 lying in L ( B ~ ) ~ ' . The second step is to show that the image of any neighborhood of 0 in X is a neighborhood of 0 in Y. The previous step shows that (L(B1))'' 2 Us for some s > 0, and we show for every c > 0 that L(B,) 2 Us,,2. For t > 0, rrlultiplicativrl of the irlclusivrl (L(B1))" 2 Us by 1 shows that -
since multiplication by t is a homeomorphism of Y and L is linear. If y is in Usc,2, we are to produce x in Bc with L ( x ) = y , and we do so by successive approximations. Specifically we construct inductively the terms x, of a convergent series in X with sum x , as follows: Condition (*) with t = c / 2 allows us . x l , . . . , x,-1 have to choose a member xl of Bc,2 with 11 y - L ( x l )11 < 2 - 2 s ~ If been constructed with each xj in B2-j, and with
then y - L ( x l + . . . +x,-l) is in U2-t9c.Condition (*) with t we can find x, in B2pc with
= 2-,c
We now have Ily - L(x1
+ . . . + x,-1 + x,) II < 2-(n+1)sc.
shows that
6. Interior Mapping Principle
547
This completes the inductive construction of the x,'s, and we shall prove that the series C xn is convergent in X . Since X is complete, it is enough to show that the partial sums of C x , are Cauchy. If q > p , then The right side is 5 2-PC, and the partial sums of C x, are indeed Cauchy. Let x = C;==, x,. Taking p = 0 and using the continuity of the norm, we see that llxll 5 c. By continuity of L , we have y = limn L ( x l + .. . + x,) = L ( x ) . Consequently the member y of U s c pis of the form L ( x ) for some x in B,, as was asserted. The third step is to show that each open set of X is mapped to an open set of Y by L . Let U be open in X , let x be in U , and let N be an open neighborhood of 0 in X such that x + N g U . The previous step shows that there is some open neighborhood V of 0 in Y such that V g L (N) . Then L ( x ) + V is an open neighborhood in Y of L ( x ) with Therefore L ( U ) contains a neighborhood about each of its points and must be open.
Corollary 12.25. A one-one continuous linear operator L of a Banach space X onto a Banach space Y has a continuous linear inverse. PROOF.Since L is one-one onto, L-' exists. For L-' to be continuous, the inverse image under L-' of each open set is to be open. In other words, the direct image under L of any open set is to be open. But this is just the conclusion of Theorem 12.24. EXAMPLE. Let 3 b e the Fourier coefficient mapping, which carries functions in L' (& d x ) to doubly infinite sequences { c n }vanishing at infinity. The linear operator F has norm 1 when the space of doubly infinite sequences is given the supremum norm 11 {c,} l , = sup, Ic, 1. Corollary 6.50 shows that F i s one-one. Let us see that there is some doubly infinite sequence vanishing at infinity that is not the sequence of Fourier coefficients of some L' function. If this were not so, then Corollary 12.25 would say that F-' is bounded. We can obtain a contradiction if we produce a sequence { f,} of 1,' filnctions with 1 1 f, I l l = 1 for all n and with limn IlF(f,) llSup = 0. Form the Dirichlet kernel D, as defined in Section 1.10 and reproduced in the previous section. Its Fourier coefficients ck are 1 for Ikl 5 n and are 0 for Ikl > n, and thus IIF(D,)IIsUp = 1. Put f n = ~ n / l l D n l l l . Then l l f n l l l = 1 for all n , and IIF(fn)llsup = 1/11Dnlll. Proposition 12.23 shows that in fact limn 1/ 11 D, 11 1 = 0, and we obtain the desired contradiction. The conclusion is that the image of 3 o n L' fails to include some doubly infinite sequence { c n }vanishing at infinity.
XII. Hilbert and Banach Spaces
548
If f : X + Y is a function between Hausdorff spaces, the graph of f is the subset G = {(x, f (x)) I x E X} of X x Y. If f is continuous, then G is a closed set, as we see immediately by using nets. The converse fails because f : [O, 11 + R with f (0) = 0 and f (x) = l l x for x > 0 is a discontinuous function with closed graph. We shall be interested in the converse under the additional condition that our function f is linear. Our spaces being metric spaces, the condition that the graph be closed is that whenever {(xn,f (x,))} converges to some (x, y), then x is in the domain of f and f (x) = y. Linearity by itself is not enough to get an affirmative result. In fact, let X C([O, I]), let Xo be the vector subspace of functions with a continuous derivative, and let L : Xo + X be the derivative operator F H F'. If lim, F, = F in X and lim, FA = H , then Theorem 1.23 shows that F' exists and equals H. Hence the linear operator L : Xo + X has closed graph. However, L is unbounded since the function x H xn has norm 1 and its derivative has norm n .
Corollary 12.26 (Closed Graph Theorem). If L : X + Y is a linear operator from a Banach space X into a Banach space Y such that the graph of L is a closed subset of X x Y , then L is a bounded linear operator. PROOF.M a k e X e Y intoaBanachspacebydefining II(x, y)II = IlxIlx+IlyIlr. The graph G = {(x, L(x)) I x E X} of L is a vector subspace of X @ Y since L is linear, and it is closed by hypothesis. Thus G is a Banach space. The limear operalor P : G + X giver1 by P ((x, L (x)) = x is uric-one arld onlo, and Corollary 12.25 shows that the linear operator P-l : X + G given by P-l (x) = [x , L [x)) is continuous. If E denotes the projection of X @ Y to the Y coordinate, then E is bounded with norm 5 1, and hence the restriction E : G + Y is bounded with norm 5 1. Therefore the composition (E G ) o P-I : X + Y is bounded. But (E G ) ( ~ -(x)) l = E (x, L(x)) = L(x), and thus L is bounded.
I
I
I
EXAMPLE. Suppose that a Banach space X is the vector-space direct sum of two closed vector subspaces: X = Y @ Y ' . Let E : X + Y be the projection of X on Y given by E(y + y') = y. Corollary 12.26 implies that E is bounded. In fact, let x, = y, + yl, define a sequence in X, so that (x,, y,) defines a sequence in the graph of E . Suppose that limn(x,, y,) = (xo, yo) in X x X , i.e., that limn xn = xo and limn yn = yo. Here xo is in X, and yo is in Y since Y is closed. Then yi, - limn yl, - limn x, - lim, y, - xo - yo, and this is in Y' since Y' is closed. The equality xo = yo yh shows that E (xo) = yo, and therefore (xo, yo) is in the graph of E. In other words, the graph of E is closed. We conclude from Corollary 12.26 that E is bounded.
+
7. Problems 1. Let X be a normed linear space. (a) Prove that the closure of the open ball of radius r and center xo is the closed ball of radius r and center xo. (b) If X is complete, prove that any decreasing sequence of closed balls has nonempty intersection.
2. The normed linear space C ( N ) ( [ ab ,] ) was defined in Section 1. Prove that it is complete.
3. The normed linear space H" (D) and its vector subspace A ( D ) were defined in Section 1. Prove that H o a ( D )is complete and that A ( D ) is a closed subspace, hence complete.
4. Let X be aBanach space, let Y be aclosedvector subspace, anddefine Ilx+ Y 11 = infy,y I I x y 11 for x Y in the quotient vector space X / Y . (a) Show that 11 . +Y 11 is a norm for X/ Y . (b) By replacing a Cauchy sequence { x , Y} in X / Y by a subsequence such that Ilx,, - x,,,, YII 5 2Zk, show that the subsequence can be lifted to a Cauclly sequence in X and deduce that X / Y is a Banacll space.
+
+
+
+
5. Let v l , . . . , v, be vectors in an inner-product space. Their Gram matrix is the Herlllitiall lnatrix of iililer products given by G (vl , . . . , v,) = [ ( v ,, u J ) ] ,and det G ( v l ,. . . , v,) is called their Gram determinant. (a) If c l , . . . , c, are in C, let c =
()
. Prove that c t r G ( u . . . , v,)? =
c.
(b) Making use of the finite-dimensional Spectral Theorem, prove that there exists a unitary matrix u such that the matrix u Z 1G ( v l ,. . . , v,)u is diagonal with diagonal entries > 0. (c) Prove that det G ( v l ,. . . , v,) 2 0 with equality if and only if v l , . . . , v, are linearly dependent. (This generalizes the Schwarz inequality.)
6. (Gram-Schmidt orthogonalization process) Let ( u l , . . . , u,) be a linearly independent ordered set in an inner-product space, and inductively define u/l = v = u l v = ~~~:(u.~q,andv~=~~v;ll~'uk.Provethat the vectors v l , . . . , u, are well defined, that v l , . . . , u, are orthonormal, and that for cach k with 1 5 k 5 12, span{vl, . . . , v k }= span{u1, . . . , u k } . 7 . Let H1 and Hz be Hilbert spaces with respective orthonormal bases {u,} and
{ v g } .If there is a one-one function carrying the one orthonormal basis onto the other, prove that there is a bounded linear operator F : H1 + H2 carrying H1 onto H2 and preserving distances. Deduce that HI and Hz are isomorphic as Hilbert spaces if and only if they have the same Hilbert space dimension.
XII. Hilbert and Banach Spaces
550
8. Let ( S , p ) be a a-finite measure space, and let f be in L m ( S , p). (a) Show that multiplication by f is a bounded linear operator on L ~ ( sp, ) , and find the norm of this operator. (b) Find the adjoint of the operator in (a).
9. Suppose that X is a normed linear space and that its dual X* is separable in its norm topology, with {x:} as a countable dense set. For each n , choose x, in X with llx, 11 < 1 and IxX(x,)l 2 llx; 1 1 . Prove that {x,} is dense in X , so that X * separable implies X separable.
3
10. By considering the discontinuous indicator function I[,), where so is a limit point of S , prove that the Banach space C ( S ) is not reflexive if S is compact Hausdorff and infinite.
11. Without using the Baire Category Theorem,prove that the Uniform Boundedness Theorem for linear functionals implies the same theorem for linear operators. 12. Suppose for each n that L, : X + X f is a bounded linear operator from anormed linear space X to a Banach space X f such that 11 L, 11 5 C with C independent of n . Suppose in addition that {L, ( y ) }converges for each y in a dense subset Y of X . Prove that L ( x ) = limn L,(x) exists for all x in X and that the resulting fuiictioii L : X + X f is a bounded linear operator with 11 L 11 5 C . 13. Let X be anormed linear space, andlet {x,} be a subset of X . If sup, Ix*(x,)l < oo for each x* in X * , prove that sup, Ilx, 11 < oo. 14. Let X be a Banacli space. A subset E of X is coilvex if it coiltaiils all poiilts ( 1 - t ) x ty with 0 < t < 1 whenever it contains x and y. (a) Show that any closed ball { y ly - xl < r } is convex. (b) Give an example of a decreasing sequence of nonempty bounded closed convex sets in a Banach space with empty intersection.
+
I
15. Let X and Y be Banach spaces, and let L be a bounded linear operator from X onto Y. Suppose that {y,} is a convergent sequence in Y with limit yo. Prove that there exists a constant M and a sequence {x,} in X such that Ilx, 11 5 M 11 y, 11 for all n , L (x,) = y, for all n , and {x,} is convergent. Problems 16-1 8 introduce "Banach limits," a kind of universal summability method. Let X be the real Banach space of real-valued bounded sequences s = {s,}:=~ with the supremum norm. 16. Let Xo be the smallest closed vector subspace of X containing all sequences with terms sl , s2 - s2, s3 - s2, . . . such that {s,} is in X . Prove that the sequence e with all tei-111s 1 is not ill X o .
17. A Banach limit is defined to be any member x* of X* with Ilx* 11 = 1 , x * ( e ) = 1 , and x* ( x o )= 0 for all xo in Xo . Prove that a Banach limit exists.
55 1
7. Problems
18. Let LIM,,, s, denote the value of a Banach limit when applied to the member {s,} of X. Prove that this satisfies (a) LIM,,, s, > 0 if s, > 0 for all n. (b) LIM,, , s,+ 1 = LIM,, , s, for every {s,} in X . (c) LIM,,, s, = 0 if all terms s, are 0 for n sufficiently large. (d) lim inf, s, 5 LIM,,, s, 5 lim sup, s, for all {s,} in X. (e) LIM,,, s, = c if {s,} is convergent with limit c. Problems 19-24 establish the Jordan and von Neumann Theorem that a normed linear space satisfying the parallelogram law acquires its norm from an inner product, the definition of the inner product being (x, y ) = Ck$ llx i k y112, where the sum extends for k E 10, 2} if the scalars are real and extends for k E (0, 1 , 2 , 3} if the scalars are coillplex. The ilol-111 is 1-ecovei-edfi-om the iimei- product by the usual formula (x, x) = 11x112. Thus let X be a normed linear space with norm 11 . 11 such that the parallelogram law holds.
+
19. Check from the definition of (x, y ) that (x, x) = 1lx 112, that (x, x) 2 0 with equality if and only if x = 0, and that (x, y ) = ( y , x). 20. Prove the identity
for all x, y , z in X . 21. Derive the formula (xl previous problem.
+ x2, y)
= (xl, y )
+ (x2,y ) from the identity in the +
22. Let D be the set of rationals if the scalars are real, or the set of all a bi with a and b rational if the scalars are complex. Using the definition of (x, y ) and the result of the previous problem, prove that (rx, y) = r (x, y ) if r is in D. 23. By considering Ilx - r y 11 for r in D with r tending to (x, y ) / 11 y 11 2, prove that ( . , . ) satisfies the Schwarz inequality. 24. By estimating Ir (x, y) - (cx, y) I with the Schwarz inequality when c is a scalar and r is a member of D tending to c, prove that c(x, y ) = (cx, y ), thereby completing the proof that ( . , . ) is an inner product.
APPENDIX
Abstract. This appendix treats some topics that are likely to be well known by some readers and less well known by others. Section A1 deals with set theory and with functions: it discusses the role of fo~malset theo~y,it works in a simplified framework that avoids too nmch for~nalismand the standard pitfalls, it establishes notation, and it mentions some formulas. Some emphasis is put on distinguishing the image and the range of a function, as this distinction is important in algebra and algebraic topology and thcrcforc plays a role whcn real analysis bcgins to intcract scriously with algebra. Sections A2 and A3 assume knowledge of Section 1.1 and discuss topics that occur logically between the end of Section 1.1 and the beginning of Section 1.2. The first of these establishes the Mean Value Thcorcm and its standard corollaries and thcn gocs on to dcfinc thc notion of a continuous derivative for a function on a closed interval. The other section gives a careful treatment of the differentiability of an inverse function in one-variable calculus. Section A4 is a quick review of complex numbers, real and imaginary parts, complex conjugation, and absolute value. Complex-valued functions appear in the book beginning in Section 1.5. Section A5 states and proves the classical Schwarz inequality, which is used in Chapter I1 to establish the triangle inequality for certain metrics but is needed before that in Chapter I in the context of Fourier series. Sections A6 and A7 are not needed until Chapter 11.The first of these defines equivalence relations and establishes the basic fact that they lead to a partitioning of the underlying set into equivalence classes. The other section discusses the connection between linear functions and matrices in the subject of linear algebra and summarizes the basic properties of determinants. Section AS, which is not needed until Chapter IV, establishes unique factorization for polynomials with real or complex coefficients and defines "multiplicity" for roots of complex polynomials. Sections A9 and A10 return to set theory. Section A9 defines p&ial orderings and includes Zom's Lemma, which is a powerful version of the Axiom of Choice, while Section A10 concerns cardinality. The material in these sections first appears in problems in Chapter V, it does not appear in the text until Chapter X in the case of Section A9 and until Chapter XI1 in the case of Section A10.
A l . Sets and Functions Real analysis typically makes use of an informal notion of set theory and notation for it in which sets are described by propelties of their ele~l~e~lts and by operations on sets. This informal set theory, if allowed to be too informal, runs into certain paradoxes, such as the Russell paradox: "If S is the set of all sets that do not contain themselves as elements, is S a member of S or is it not?" The conclusion of the Russell paradox is that the "set" of all sets that do not contain themselves as elements is not in fact a set.
554
Appendix
Mathematicians' experience is that such pitfalls can be avoided completely by working within some formal axiom system for sets, of which there are several that are well established. A basic one is "Zermelo-Fraenkel set theory," and the remarks in this section refer specifically to it but refer to the others at least to some extent.' The standard logical paradoxes are avoided by having sets, elements (or "entities"), and a membership relation E such that a E S is a meaningful statement, true or false, if and only if a is an element and S is a set. The terms set, element, and E are taken to be primitive terms of the theory that are in effect defined by a system of axioms. The axioms ensure the existence of many sets, including infinite sets, and operations on sets that lead to other sets. To make full use of this axiom system, one has to regard it as occurring in the context of certain rules of logic that tell the forms of basic statements (namely, a = b, a E S , and "S is a set"), the connectives for creating complicated statements from simple ones (''Or,,>"and,,' "not," and "if. . . then"), and the way that quantifiers work ("there exists" and "for all"). Working rigorously with such a system would likely make the development of mathematics unwieldy, and it might well obscure important patterns and dircctions. In practicc, thcrcforc, onc comprorniscs bctwccn using a formal axiom system and working totally informally; let us say that one works "informally but carefully." The logical problems are avoided not by rigid use of an axiom system, but by taking care that sets do not become too "large": one limits the sets that one uses to those obtained from other sets by set-theoretic operations and by passage to subsets.' A feature of the axiom system that one takes advantage of in working informally but carefully is that the axiom system does not preclude the existence of additional sets beyond those forced to exist by the axioms. Thus, for example, in the subject of coin-tossing within probability, it is normal to work with the set of possible outcomes as S = {heads,tails} even though it is not apparent that requiring this S to be a set does not introduce some contradiction. It is worth emphasizing that the points of the theory at which one takes particular care vary somewhat from subject to subject within mathematics. For example, it is sometimes of interest in calculus of several variables to distinguish between the range of a function and its image in a way that will be mentioned below, but it is usually not too important. In homological algebra, however, the distinction is l~athematicianshave no proof that this technique avoids problems completely. Such a proof would be a proof of the consistency of a version of mathematics in which one can construct the integers, and it is known that this much of mathematics cannot be proved to be consistent unless it is in fact inconsistent. ' ~ o tevery set so obtained is to be regarded as "constructed." The Axiom of Choice, which we come to shortly, is an existence statement for elements in products of sets, and the result of applying the axiom is a set that can hardly be viewed as "constrncted."
A l . Sets and Functions
555
extremely important, and the subject loses a great deal of its impact if one blurs the notions of range and image. Some references for set theory that are appropriate for reading once are Halmos's Naive Set Theory, Hayden-Kennison's Zermelo-Fraenkel Set Theory, and Chapter 0 and the appendix of Kelley's General Topology. The Kelley book is one that uses the word "class" as a primitive term more general than "set"; it develops von Neumann set theory. All that being said, let us now introduce the familiar terms, constructions, and notation that one associates with set theory. To cut down on repetition, one allows some alternative words for "set," such as family and collection. The word "class" is used by some authors as a synonym for "set," but the word class is used in some set-theory axiom systems to refer to a more general notion than "set," and it will be useful to preserve this possibility. Thus a class can be a set, but we allow ourselves to speak, for example, of the class of all groups even though this class is too large to be a set. Alternative terms for "element" are member and point; we shall not use the term "entity." Instead of writing E systematically, we allow ourselves to write "in." Generally, we do not use E in sentences of text as an abbreviation for an expression like "is in" that contains a verb. If A and B are two sets, some familiar operations on them are the union A U B , the intersection A n B ,and the difference A - B ,all defined in the usual way in terms of the elements they contain. Notation for the difference of sets varies from author to author; somc othcr authors writc A \ B or A B for diffcrcncc, but this book uses A - B. If one is thinking of A as a universe, one may abbreviate A - B as B C ,the conlplenlent of B in A. Tlle ernpty set M is a set, and so is the set of all subsets of a set A , which is sometimes denoted by 2 A . Inclusion of a subset A in a set B is written A 5 B or B 2 A. Inclusion that does not permit A; in this case one says that A is a proper equality is denoted by A B or B subset of B or that A is properly contained in B . If A is a set, the singleton { A }is a set with just the one member A. Another operation is unordered pair, whose formal definition is { A , B } = { A }U { B }and whose informal meaning is a set of two elements in which we cannot distinguish either element over the other. Still another operation is ordered pair, whose formal definition is ( A , B ) = { { A } {, A ,B } } . It is customary to think of an ordered pair as a set with two elements in which one of the elements can be distinguished as coming first.3 Let A and B be two sets. The set of all ordered pairs of an element of A and
--
5
2
3 ~ n f o ~ ~ t u ~ ~a a"sequence" tely as in Chaptw I gets de~~oted by {xl,xz,. . . ] or {x,}:=~. If its notation were really consistent with the above definitions, we might infer, inaccurately, that the order of the terms of the sequence does not matter. The notation for unordered pairs, ordered pairs, and sequences is, however, traditional, and it will not be changed here.
556
Appendix
an element of B is a set denoted by A x B; it is called the product of A and B or the Cartesian product. A relation between a set A and a set B is a subset of A x B. Functions, which are to be defined in a moment, provide examples. Two examples of relations that are usually not functions are "equivalence relations," which are discussed in Section A6, and "partial orderings," which are discussed in Section A9. If A and B are sets, a relation f between A and B is said to be a function, written f : A -f B , if for each x E A , whenever y E B and z E B are such that (x, y) and (x, 2) are in f , then y = 2. If (x, y) is in f , we write f (x) = y . In this informal but careful definition of function, the function consists of more than just a set of ordered pairs; it consists of the set of ordered pairs regarded as a subset of A x B. This careful definition makes it meaningful to say that the set A is the domain, the set B is the range, and the subset of y E B such that y = f (x) for some x E A is the image of f . The image is also denoted by f (A). Sometimes a function f is described in terms of what happens to typical elements, and then the notation is x H f (x) or x H y , possibly with y given by some formula or by some description in words about how it is obtained from x . Sometimes a function f is written as f ( .), with a dot indicating the placement of the variable; this notation is especially helpful in working with restrictions of functions, which we come to in a moment, and with functions of two variables when one of the variables is held fixed. This notation is useful also for functions that involve unusual symbols, such as the absolute value function x ++ Ix 1 , which in this notation becomes I . I . The word map or mapping is sometimes used for "function" and for the operation of a function, particularly when a geometric context for the function is of importance. Often mathematicians are not so careful with the definition of function. Depending on the degree of informality that is allowed, one may occasionally refer to a function as f (x) when it should be called f or x H f (x) . If any confusion is possible, it is wise to use the more rigorous notation. Another habit of informality is to regard a function f : A -f B as simply a set of ordered pairs. Thus two functions f i : A + B and f2 : A + C become the same if f i (a) = f 2 (a) for all a in A. With the less careful definition, the notion of the range of a function is not really well defined. The less careful definition can lead to trouble in algebra, but it does not often lead to trouble in real analysis until one gets to a level where algebra and analysis merge somewhat. The set of all functions from a set A to a set B is a set. It is sometimes denoted by B A . The special case 2A that arose with subsets comes by regarding 2 as a set { 1 , 2 }and identifying a function f from A into { 1 , 2 }with the subset of all elements x of A for which f (x) = 1. If a subset B of a set A may be described by some distinguishing property P of its elements, we may write this relationship as B = {x E A I P } . For
A l . Sets and Functions
557
example, the function f in the previous paragraph is identified with the subset { x E A I f ( x ) = 1). Another example is the image of a general function f : A + B , namely f ( A ) = { y E B I y = f ( x ) forsomex E A}. Still more generally along these lines, if E is any subset of A , then f ( E ) denotes the set { y E B I y = f ( x ) for some x E E}. Some authors use a colon instead of a vertical line in this notation. A X ,an inThis book frequently uses sets denoted by expressions like UxES dexed union, where S is a set that is usually nonempty. If S is the set { l ,21, this reduces to A1 U A2. In the general case it is understood that we have an unnamed function, say f , given by x ++ A,, having domain S and range an unnamed set T , and UxEsA, is the set of all y E T such that y is in A, for some x E S . When S is understood, we may write U, A, instead of UxEsA,. Indexed intersections Ax are defined similarly, and this time it is essential to disallow S empty because otherwise the intersection cannot be a set in any useful set theory. that specializes in the There is also an indexed Cartesian product )( case that S = {1,2}to A1 x A2. Usually S is assumed nonempty. This Cartesian product is the set of all functions f from S into UxEsA, such that f ( x ) is in A, for all x E S . In the special case that S is 11, . . . , n}, the Cartesian product is thc sct of ordcrcd n-tuplcs from n scts A l , . . . , A, and may bc dcnotcd by A1 x . . . x A,; its members may be denoted by ( a l ,. . . , a,) with aj E Aj for 1 5 j 5 n . When the factors of a Cartesian product have some additional algebraic structure, the notation for the Cartesian product is sometimes altered; for example, the Cartesian product of groups A, is denoted by A,. It is completely normal in real analysis, and it is the practice in this book, to take the following axiom as part of one's set theory; the axiom is normally used without specific mention.
n,,,
nxts
Axiom of Choice. The Cartesian product of nonempty sets is nonempty. If the index set is finite, then the Axiom of Choice reduces to a theorem of set theory. The axiom is often used quite innocently with a countably infinite index set. For example, Proposition 1 . 7 asserts ~ that any sequence in R*has a subsequence converging to lim sup a,, ,and the proof constructs one member of the sequence at a time. When these members have some flexibility in their definitions, as is the case with the proof as it is written for Proposition 1.7c, the Axiom of Choice is being invoked. When the members instead have specific definitions, such as "the term LC, such that n is the s~llallestinteger satisfying such-and-such properties," the axiom is not being invoked. The proof in the text of Proposition 1 . 7 can ~ be rewritten with specific definitions and thereby can avoid invoking the axiom, but there is no point in undertaking this rewriting. In Chapter I1 the axiom is invoked in situations in which the index set is uncountable; uses of compactness provide a number of examples.
558
Appendix
From the Axiom of Choice, one can deduce a powerful tool known as Zorn's Lemma, whose use it is normal to acknowledge. Zorn's Lemma appears in Section A9 and is used in problems beginning in Chapter V and in the text beginning in Chapter X. If f : A + B is a function and B is a subset of B', then f can be regarded as a function with range B' in a natural way. Namely, the set of ordered pairs is unchanged but is to be regarded as a subset of A x B' rather than A x B . Let f : A + B and g : B + C be two fuiictioiis such that the range of f equals the domain of g . The composition g o f : A + C is the function with ( g o f ) ( x ) = g( f ( x ) ) for all x . Because of the construction in the previous paragraph, it is meaningful to define the composition more generally when the range of f is merely a subset of the domain of g . A function f : A + B is said to be one-one if f ( x l ) f f ( x 2 )whenever xl and x2 are distinct members of A. The function is said to be onto, or often "onto B," if its image equals its range. The terminology "onto B" avoids confusion: it specifies the image and thereby guards against the use of the less careful definition of function mentioned above. A mathematical audience often contains some people who use the careful definition of function and some people who use the less careful definition. For the latter kind of person, a function is always onto something, namely its image, and a statement that a particular function is onto might be regarded as a tautology. When a function f : A + B is one-one and is onto B ,there exists a function g : B + A such that g o f is the identity function on A arid f o g is the identity function on B . The function g is unique, and it is defined by the condition, for y E B , that g ( y ) is the unique x E A with f ( x ) = y . The function g is called the inverse function of f and is often denoted by f -'. Conversely if f : A + B has an inverse function, then f is one-one and is onto B. The reason is that a composition g o f can be one-one only if f is one-one, and in addition, that a composition f o g can be onto the range of f only if f is onto its range. If f : A + B is a function and E is a subset of A , the restriction of f to E , denoted by f is the function f : E + R consisting of all ordered pairs ( x , f ( x ) )with x E E, this set being regarded as a subset of E x B , not of A x B . One especially common example of a restriction is restriction to one of the variables of a function of two variables, and then the idea of using a dot in place of a variablc can bc hclpful notationally. Thus thc function of two variables might be indicated by f or ( x , y ) H f ( x , y ) , and the restriction to the first variable, for fixed value of the second variable, would be f ( . , y ) or x H f ( x , y ) . We conclude this section with a discussion of direct and inverse images of sets under functions. If f : A + B is a function and E is a subset of A , we have defined f ( E ) = { y E B I y = f ( x ) for some x E E}. This is the same
,I
A2. Mean Value Theorem and Some Consequences
559
,1
as the image of f and is frequently called the image or direct image of E under f . The notion of direct image does not behave well with respect to some set-theoretic operations: it respects unions but not intersections. In the case of unions, we have
( , )
the inclusion 2 follows since f U,, E , 2 f ( E , ) for each s ,and the inclusion 5 follows because any member of the left side is f of a member of some E,. In the case of intersections, the question f ( E f' F ) 2 f ( E )n f ( F ) can easily have a negative answer, the correct general statement being f ( E nF ) g f ( E )n f ( F ). An example with equality failing occurs when A = { I ,2, 31, B = { I , 21, f (1) = f (3) = 1, f (2) = 2, E = {1,2}and F = {2,3}because f ( E n F ) = {2} and f ( E l n f iF) = 11221. If f : A + B is a function and E is a subset of B, the inverse image of E under f is the set f - ' ( E ) = {x E A I f (x) E E } . This is well defined even i f f does not have an inverse function. (If f does have an inverse function f - ' , then the inverse image of E under f coincides with the direct image of E under f .) Unlike direct images, inverse images behave well under set-theoretic operations. If f : A + B is a function and { E , I s E S} is a set of subsets of B, then
-'
In the third of these identities, the complement on the left side is taken within B, and the complement on the right side is taken within A . To prove the first identity, we observe that f E.) f -'(E.) for each s E S and hence f ( E,) g f (E,) . For the reverse inclusion, if x is in f ( E , ) , lllerl x is in f ( E , ) fur cad1 s and Illus f ( x ) is in E, fur cad1 s . Hence f ( x ) is in E , , and x is in f ( E,) . This proves the reverse inclusion. 'I'he second and third identities are proved similarly.
-' (nsEs ' nsEs n,,, -' nSEs-' -' nsEs -' nsEs -'
A2. Mean Value Theorem and Some Consequences This section states and proves the Mean Value Theorem and two standard corollaries, and then it discusses the notion of a function with a continuous derivative on a closed interval. It makes use of results in Section 1.1 of the text.
560
Appendix
Lemma. Let [a,b ] be a nontrivial closed interval, and let f : [ a ,b] + R be a continuous function that is differentiable on ( a ,b ) and has f ( a ) = f (b) = 0 . Then the derivative f' satisfies f ' ( c ) = 0 for some c with a < c < b. PROOF.We divide matters into three cases. If f ( x ) > 0 for some x , let c be a member of [a,b] where f attains its maximum (existence by Theorem 1.11). Since f ( x ) > 0 somewhere, we must have a < c < b. Thus f f ( c ) exists. If f f ( c ) > 0, then the inequality limh+o h P 1 ( f(c h ) - f ( c ) ) > 0 forces f (c h ) > f ( c ) for h positive and sufficiently small, in contradiction to the fact that f attains its maximum at c. Similarly if f f ( c ) < 0, then we find that f (c h ) > f ( c ) for h positive and sufficiently small, and again we have a contradiction. We conclude that f ' ( c ) = 0. If f ( x ) 5 0 for all x and f ( x ) < 0 for some x , let c instead be a member of [ a ,b] where f attains its minimum. Arguing in the same way as in the previous paragraph, we find that f'(c) = 0. Finally if f ( x ) = 0 for all x , then f ' ( x ) = 0 for a < x < b , and f f ( c ) = 0 for c = 31 (a b ),for example.
+
+
-
+
Mean Value Theorem. Let [a,bl be a nontrivial closed interval. f : [ a , b] + R is a continuous function that is differentiable on ( a ,b) , then f'(c> =
f
If
(b) - f ( a ) b-a
for some c with a < c < b . PROOF.Apply the lemma to the function
which has g(a) = g(b) = 0 and g'(x) = f ' ( x ) -
F.
Corollary 1. A differentiable function f : ( a ,b) + R whose derivative is 0 everywhere on ( a ,b ) is a constant function. PROOF. If f (a') # f (b'), then the Mean Value Theorem produces some c between a' and b' where f ' ( c ) 0 .
+
Corollary 2. A differentiable function f : ( a ,b ) + 0 everywhere on ( a ,b) is strictly increasing on ( a ,b ) .
R whose derivative is
PROOF.If a' < b' and f (a') > f (b'),then the Mean Value Theorem produces some c with a' < c < b' where f ' ( c ) 5 0.
A3. Inverse Function Theorem in One Variable
561
In the setting of the Mean Value Theorem, it can happen that ff(x) has a finite limit C as x decreases to a (or as x increases to b). This terminology means that for any t > 0, there exists some S > 0 such that I ff(x) - CI < t whenever a < x < a 6 . In this case, f can be extended to a function F defined and continuous on ( - a , b], differentiable on (-oo, b), in such a way that F' is continuous at a . In fact, the extended definition is fora 5 x 5 b, F(x) = (x) f(a)+C(x-a) for-m<x(a. To see that F' (a) exists for the extended function F, let t > 0 be given and choose S >Osuchthata < x < a+SimpliesIfl(x)-CI < t - . I f a < x < a + S , t h e n the Mean Value Theorem gives
+
{
F(x) - F ( a ) = F ' ( ~ ) x-a witha < c < x < a+S,andhence
-cI 1-
< t. I f a - S < x < a,then
+
F(x) - F ( a ) ( f (a) C(x - a)) - f (a) - C = o . x-a x -a Thus F'(a) exists and equals C . The definitions make lim,,, F'(x) = F'(a), and hence F' is continuous at a .
I
As a consequence of this construction, it makes sense to say that a continuous function f : [a, b] + IR with a derivative on (a, b ) has a continuous derivative at one or both endpoints. This phrasing means that f ' has a finite limit at the endpoint in question, and it is equivalent to say that f extends to a larger set so as to be differentiable in an open interval about the endpoint and to have its derivative be continuous at the endpoint.
A3. Inverse Function Theorem in One Variable This section addresses one of the "further topics" mentioned at the end of Section 1.1 and assumes knowledge of Section 1.1 and some additional facts about continuity and differentiability of functions of a real variable. The topic is that of differentiability of inverse functions, the nub of the matter being continuity of the inverse function. The topic is one that is sometimes skipped in calculus courses and slighted in courses in real variable theory. Yet it is necessary for the development of one of the two functions exp and log, of one of the two functions sin and arcsin, and of one of the two functions tan and arctan unless actual constructions of both members of a pair are given. In principle the matter arises also with differentiation of the function xl/q on (0, m ) ,but the proposition of this section can be readily avoided in that case by explicit calculations.
562
Appendix
Proposition. Let ( a , b ) be an open interval in R, possibly infinite, and let f : ( a , b) + R be a function with a continuous everywhere-positive derivative. Then f is strictly increasing and has an interval ( c , d ) , possibly infinite, as its image. The inverse function g : ( c , d ) + ( a , b ) exists and has a continuous derivative given by g ' ( y ) = l / f ' ( g ( y ) ). PROOF.The function f is strictly increasing as a corollary of the Mean Value Theorem, and its image is an interval ( c , d ) because of the Intermediate Value Theorem (Theorem 1.12). Being one-one and onto, f has an inverse function g , according to Section A l . Fix yo E ( c , d ) , fix c'and d' such that c < c' < yo < d' < d , and consider y # yo in (c', d') . Put x = g ( y ) , xo = g(yo), a' = g(c') , and b' = g ( d ' ) . Then a < a' < xo < b' < b since f is strictly increasing. By Theorem 1.1 1, there exist real numbers m and M such that 0 < m 5 f ' ( t ) 5 M for all t E [a', b']. The Mean Value Theorem produces 6 between xo and x such that
and hence Ix - xo 1 5 m-' ly - yo 1. Since g is one-one, we have x # xo. Also, f (x) = y # yo = f ( x o ) . Thus it makes sense to form
f ' ( x o ) # 0, we have
Let r > 0 be given. Since lim,+, lim t+xo
t
f ' t ~ ~ ~-~ "
-
xo
f ( t ) - f (xo)
-
1
f '(~0) '
Choose y > 0 such that
as long as It - xo 1 < y with t # xo and t E [a', b']. Then put 6 = ym. If ly - yol < 6,then Ix -xoI 5 m - l l y - yoI < mP16 = y. Since t = x satisfies the condition It - xn I < q with t # xn and t E [a', b'l, it follows that
whenever ly - yol < 6 . Since r is arbitrary, the conclusion is that g'(yo) = l / f ' ( g ( y o ) ) . Since g is differentiable, g is continuous and also the composition f ' o g is continuous. Because f ' o g is nowhere zero, g' = 1 / ( f ' o g ) is continuous. This completes the proof.
A5. Classical Schwarz Inequality
563
A4. Complex Numbers Complex numbers are taken as known, and this section reviews their notation and basic properties. Briefly, the system C of complex numbers is a two-dimensional vector space over R with a distinguished basis (1, i } and a multiplication defined initially by 11 = 1, l i = i 1 = i , and i i = - 1. Elements may then be written as a + bi or a + i b with a and b in R; here a is an abbreviation for a1 . The multiplication is extended to all of C so that the distributive laws hold, i.e., so that (a bi) (c di) can bc cxpandcd in thc cxpcctcd way. Thc multiplication is associativc and commutative, the element 1 acts as a multiplicative identity, and every nonzero b element has a multiplicative inverse: (a + bi) ("-a2fb2 - i aZfb2) = Complex conjugation is indicated by a bar: the conjugate of a + bi is a - bi ifaandbarerea1,andwewriteafbi = a - b i . T h e n w e h a v e z + w = Z + G , E = r j i f r i s r e a l , a n d E = ZG. The real and imaginary parts of z = a + bi are Re z = a and Irn z = b. These may be computed as Re z = (z Z) and Im z = - $ (z - Z). The absolute value function of z = a bi is given by lzl = d m ,and this satisfies 1z12 = zZ. It has the simple properties that IZI = zl , I Re zl 5 121, and I Im zl 5 lzl . In addition, it satisfies
+
+
because
+
+
2
lzwl = zwzw = ZWZG= zjww = lz121w12,
and it satisfies the triangle inequality
because
lz + w12 = (z + w)(z + w)
= zZ
+ zG + WZ + wG
+ 2Re(zG) + lw12 i 1z12+ 21zGl + lw12 = 1z12 + 2121 1wl + 1w12 = (121 + lw112. = 1z12
A5. Classical Schwarz Inequality The inequality in question is as
follow^.^
4 ~ the n classical setting below, the inequality is often called the "Cauchy-Schwarz inequality" and may have other people's names attached to it as well. However, generalizations tend to be called simply the "Schwarz inequality," and this book therefore drops all names but Schwarz.
564
Appendix
Schwarz inequality. Let (al, . . . , a,) and (bl , . . . , b,) be n-tuples of complex numbers. Then
PROOF.We add n-tuples of complex numbers entry by entry, and we multiply such an 12-tuple by a complex scalar by multiplying each entry of the 12-tuple by that scalar. For any n-tuples of complex numbers a = (al, . . . , a,) and 1'2 ~bkl~)~'~,and b = (bl, . . . , b,,),define la1 = (xi=l lak12) ,PI = (a, b) = akbk. The Schwarz inequality says that 0 5 0 if b = (0,. . . , O), and thus we may assume that b is something else. In this case, Ibl 0. Then
+
and the asserted inequality follows.
A6. Equivalence Relations An equivalence relation on a set S is a relation between S and itself, i.e., is a subset of S x S, satisfying three properties. We define the expression a E 6 , written "a is equivalent to b ," to mean that the ordered pair (a, b) is a member of the relation, and we say that "E" is the equivalence relation. The properties are (i) a a for all a in S,i.e.,- is reflexive, (ii) a E b implies b E a if a and b are in S, i.e., E is symmetric. (iii) a E b and b E c together imply a E c if a , b, and c are in S , i.e., is transitive. An example occurs with S equal to the set Zof integers with a E b meaning that the difference a - b is even. The properties hold because (i) 0 is even, (ii) the negative of an even integer is even, and (iii) the sum of two even integers is even. There is one fundamental result about abstract equivalence relations. The equivalence class of a , written [a]for now, is the set of all members b of S such that a E b.
-.
-.
Proposition. If E is an equivalence relation on a set S , then any two equivalence classes are disjoint or equal, and S is the union of all the equivalence classes.
A7. Linear Transformations, Matrices, and Determinants
565
PROOF.Let [a]and [b]be the equivalence classes of members a and b of S . If [a]n [b] f m,choose c in the intersection. Then a c and b c. By (ii), c E b , and then by (iii), a E b. I f d is any member of [ b ] ,then b E d. From b and b d together imply a E d. Thus [b] [ a ] . Reversing the (iii), a roles of a and b , we see that [a] 5 [b]also, whence [a]= [ b ] .This proves the first conclusion. The second conclusion follows from (i), which ensures that rr is in [ a ] hence , that every member of S lies in some equivalence class.
-.
-.
-.
-.
-.
EXAMPLE.With the equivalence relation on Z that a b if a - b is even, there are two equivalence classes-the subset of even integers and the subset of odd integers. The first two examples of equivalence relations in this book arise in Chapter 11. The first example, which is in Section 11.2 and concerns a passage from "pseudometric spaces" to "metric spaces," yields equivalence classes exactly as above. The second example, which is in Section 11.3, is a relation "is homeomorphic to" and implicitly is defined on the class of all metric spaces. This class is not a set, and Section A1 of this appendix suggested avoiding using classes that are not sets in order to avoid the logical paradoxes mentioned at the beginning of the appendix. 'I'here is not much problem with using general classes in this particular situation, but there is a simple approach in this situation for eliminating classes that are not sets and thereby following the suggestion of Section A1 without making an exception. The approach is to work with any subclass of metric spaces that is a set. The equivalence relation is well defined on the set of metric spaces in question, and the proposition yields equivalence classes within that set. This set can be an arbitrary subclass of the class of all metric spaces that happens to be a set, and the practical effect is the same as if the equivalence relation had been defined on the class of all metric spaces. A7. Linear Transformations, Matrices, and Determinants A certain amount of linear algebra, done with real or complex scalars, is taken as known. The topics of vectors, vector spaces, operations on matrices, row reduction of matrices, spanning, linear independence, bases, and dimension will not be reviewed here. This section will concentrate on the correspondence beand nlatrices in the finite-dimensional case, and on tween linear t~-ansfor~~~ations the elementary properties of determinants. So as to be able to handle real and complex scalars simultaneously, we denote by IF either R or C. The linear transformations in question will be functions with domain IFn and range IFm. As is emphasized for the case IF = R in Section 11.1,the members of these spaces are to be regarded as column vectors with entries in IF even if, in order
566
Appendix
to save space, one occasionally writes them horizontally with commas between entries. This is an important convention, since it makes matrix operations and operations with linear transformations correspond to each other in the same order without the need to transpose any matrix. The standard bases for IFn and IFm are often denoted by { e l ,. . . , e n }and { u l ,. . . , u,}, respectively, in this book, where
are n-entry column vectors and
are rn-entry column vectors. A function T : IFn + I P is a linear function if it satisfies T ( x y ) = T ( x ) + T ( y ) and T ( c x ) = cT ( x ) for all x and y in IFn and all elements c of IF. Thc tcrms "lincar transformation" and "lincar map" arc uscd also. An example is obtained from any m-by-n matrix A with entries in IF, namely T ( x ) = A x , the right side being a matrix product. The size of A needs emphasis: the number of rows equals the dimension of the range, and the number of columns equals the dimension of the domain. P is a linear function, then there is a unique Conversely if T : IFn + I such matrix A such that T ( x ) = Ax for all x in IFn: the jth column of A is T ( e j ) for 1 5 j 5 n. For example, if T : R2 + R2 is the rotation about
+
the origin counterclockwise through an angle 8 , then T
(y) (
)
(
cos0 - sin0
()=
(z:)and
1
= cos s . consequently A = sin cos . Sometimes it is necessary to have a notation for the entries of a matrix A , and this text uses Aij to indicate the entry of A in the ith row and j" column. If a matrix is defined entry by entry, the entries being M i j , the text will occasionally refer to the whole matrix as [ M i j ] .This convention is especially handy if Mij is given by some nontrivial expression like aui / a x j that involves i and j . We can give a tidy formula for the correspondence T ct A if we define a dot product in IFm by
T
(al, . . . , a,) . ( b l , . . . , b,) = a l b l + . . . + a , b ,
with no complex conjugations involved. The correspondence of a linear function T in L (IFn, IFm) to a matrix A with entries in IF is then given by
A7. Linear Transformations, Matrices, and Determinants
567
The correspondence T ct A of linear functions to matrices carries certain vector spaces associated to T to vector spaces associated with A. The kernel of T , namely the set of vectors x with T (x) = 0, corresponds to the null space of A, the set of column vectors with Ax = 0. The image of T , as defined in Section A1 ,corresponds to the column space of A, the linear span of the columns of A. The method of row reduction of matrices shows that #{cc>lilmnsof A} = dirn(nill1 space c>f A)
+ dirn(span of rows c>fA),
while a little argument with bases shows that dim(domain of T) = dim(keme1 of T)
+ dim(image of T).
In these two equations the left sides are equal, and the first terms on the two right sides are equal. Therefore the second terms on the two right sides are equal, and we obtain dim(span of rows ofA) = dim(span of columns ofA) The common value of the two sides of this equation is called the rank of A or of T . Under this correspondence of linear functions between column-vector spaces with matrices of the appropriate size, composition of linear functions corresponds to matrix product in the same written order. In other words, suppose that T : IFn + IFm corresponds to A of size m-by-n and that U : Fm + Fk corresponds to B of size Iz-by-m. Then U o T : IFn + Fk corresponds to B A of size k-by-n. The determinant function A H det A has domain the set of all square matrices over IF and has range IF. It is uniquely defined by the three properties (i) det A is linear in each row of A if the other rows are held fixed, (ii) det A = 0 if two rows of A are equal, (iii) det I = 1 if I denotes the identity matrix of any size. These properties enable one to calculate det A by row reducing the matrix A. Specifically replacement of a row by the sum of it and a multiple of another row leaves det A unchanged, multiplication of a row by a constant to make the diagonal entry be one means pulling out the diagonal entry as a scalar factor multiplying the determinant, and interchanging two rows multiplies the determinant by -1. After the row reduction is complete for a square matrix, either the reduced rowechelon form is the identity matrix and (iii) says that the determinant is 1 or else the reduced row-echelon form has a row of O's, and (i) and (ii) imply that the determinant is 0. The determinant function has the following additional properties, which may be regarded as consequences of (i), (ii), and (iii) above:
568
Appendix
det A # 0 if and only if A is invertible, det A = det At', where At' is the transpose of A, det(A B) = (det A) (det B) , det A = C , (sgn O ) A ~ , , ( .~.). A,,,(,) if A is n-by-n with entries Ai,j; the sum is taken over all permutations o of 11, . . . , n}, with sgno denoting the sign of 0 , (viii) (expansion by cofactors) for n > 1 if Tij denotes the (n - 1)-by-(n - 1) matrix obtained by deleting the it" row and jt" column from the n-by-n matrix A. then det A = x;=, (-l)'+JAij det for all i and det A = Cy=l (-l)i+JAij det for all j, (ix) (Crarner's rule) if det A # 0, if v is in Rn, and if Ai denotes the matrix obtained by replacing the jthcolumn of A by v , then the jthentry of the unique solution x E Rn of Ax = v is xj = det A,/ det A. (iv) (v) (vi) (vii)
zij
xj
AS. Factorization and Roots of Polynomials The first objective of this section is to prove unique factorization of real and complex polynomials. Let IF denote either the reals R or the complex numbers
c. We work with polynomials with coefficients in IF. These are expressions P(X) = a n X n + . . . + a l X + a o w i t h a n , . . . , a1,aoinIF. Althoughitistempting to think of P(X) as a function with independent variable X, it is better to identify P with the sequence (ao,a l , . . . , a,, 0, 0, . . . ) of coefficients. For this setting, a polynomial (in one "indeterminate") may be defined as a sequence of members of IF such that all terms of the sequence are 0 from some point on. The indexing of the sequence is to begin with 0. Addition, scalar multiplication, and polynomial multiplication are then defined in the expected way so as to match the operations on functions. The usual associative, commutative, and distributive laws are then valid. Nevertheless, it is still convenient to use the notation X in writing explicit polynomials. If r is in IF,we can evaluate P(X) = anXn + . . . + a l X + a0 at r , and the result is the number P (r) = a n r n 1 . . . I a l r I a*. We say that r is a root of P if P (r) = 0. The degree of a polynomial P , denoted by deg P , is the largest integer n such that the coefficient of Xn is nonzero; the ~lotio~l of "degree" is left undefined for the 0 polynomial, i .e., the polynomial all of whose coefficients are 0. A factor of a polynomial A(X) is a polynomial B(X) such that A(X) = B(X)Q(X) for some polynomial Q(X); we say also that B(X) and Q(X) divide A(X). In this case, if B and Q are not 0, then A is not 0 and degA = deg B deg Q.
+
AS. Factorization and Roots o f Polynomials
569
Division Algorithm. If A(X) and B(X) are polynomials with coefficients in
IF and if B (X) is not the 0 polynomial, then there exist unique polynomials Q (X) and R(X) such that
+
(a) 4x1 = B(X) Q(X) R(X) and (b) either R(X) is the 0 polynomial or deg R < deg B. REMARK. This result codifies the usual method of dividing polynomials in high-scl~oolalgebra. That iilethod writes A(X)/B(X) = Q(X) + R(X)/B(X), and then one obtains the above result by multiplying by B (X) . The polynomial Q is the quotient in the division, and R(X) is the remainder. PROOF OF UNIQUENESS. If A = B e l + R1 also, then B(Q - e l ) = R1- R . Without loss of generality, R1 - R is not the 0 polynomial since otherwise Q Ql = 0 also. Then -
deg B
+ deg(Q
-
Q1) = deg(R1
-
R) 5 max{deg R , deg R1} ideg B ,
and we have a contradiction. PROOFOF EXISTENCE. If A = 0 or dcg A idcg B, wc takc Q = 0 and R = A, and we are done. Otherwise we induct on deg A. Assume the result lor degree 5 n - 1, arld lcl deg A = n. Wrile A = a n X n+ Al will1 Al = 0 or deg Al < deg A. Let B = b k x k+ B1 with B1 = 0 or deg B1 < deg B. Put Q = anbkl x " - ~Then .
with the right side equal to 0 or of degree < deg A. Then the right side, by induction, is of the form BQ2 R , and A = B(Q1 + Q2) + R is the required decomposition.
+
Corollary 1 (Factor Theorem). If r is in IF and P is a polynomial, then X - r divides P if and only if P ( r ) = 0. PROOF. If P = (X - r ) Q , then P ( r ) = (r - r ) Q(r) = 0. Conversely let P ( r ) = 0. Taking B(X) = X - r in the Division Algorithm, we obtain P = (X - r ) + R with R = 0 or deg R < deg(X - r ) = 1. In either event we have 0 = P (r) = (r - r ) Q(r) R(r), and thus R(r) = 0. Of the two choices, wemusthave R = O , a n d t h e n P = ( X - r ) Q .
+
Proposition. If P is a nonzero polynomial with coefficients in IF and if deg P = n , then P has at most n distinct roots.
570
Appendix
PROOF. Let r l , . . . , r,+l be distinct roots of P(X). By the Factor Theorem, X - rl is a factor of P ( X ) . We prove inductively on k that the product (X - rl)(X - r2) . . . (X - rk) is a factor of P ( X ) . Assume that this assertion holds for k, so that P (X) = (X - rl) . . . (X - rk) Q (X) and
Since the r,'s are distinct, we must have Q(rktl) = 0. By the Factor Theorem, we can write Q(X) = (X - rk+1)R(X) for some polynomial R(X) . Substitution givesP(X) = (X-rl)...[X-~k)(X-~k+l)R(X),and(X-rl)...(X-~k+l) is exhibited as a factor of P (X) . This completes the induction. Consequently
for some polynomial S(X). Comparing the degrees of the two sides, we find that deg S = - 1, and we have a contradiction. A greatest common divisor of polynomials A and B with B # 0 is any polynomial D of maximum degree such that D divides A and D divides B. The Euclidean algorithm is the iterative process that makes use of the Division Algorithm in the form
+
A = B Q i Ri, B=Ki422+K2, RI = R 2 Q 3 + R 3 ,
R,-2
= R,-lQ,
+ R,,
R1 = 0 or deg R1 < deg B, K 2 = O o r degK2 (degK1, R3 = 0 or deg R3 < deg R2?
R,
= Oor
degR, < degRnP1,
Rn-I = Rn &,+I. In the above computation the integer n is defined by the conditions that R, # 0 and that R,+l = 0. Such an n must exist since deg B > deg R1 > . . . 2 0 .
Theorem. Let A and B be polynomials with coefficients in IF and with B # 0 , and let R1, . . . , R, be the remainders generated by the Euclidean algorithm when applied to A and B . Then (a) R, is a greatest common divisor of A and B, (b) the greatest common divisor D of A and B is unique up to scalar multiplication, (c) any Dl that divides both A and B necessarily divides D, (d) there exist polynomials P and Q with A P B Q = D .
+
AS. Factorization and Roots o f Polynomials
571
+
PROOF. Let Dl divide A and B . From A = BQl R 1 , we see that Dl divides R 1 . From B = Rl Q2 R2, we see that Dl divides R2. Continuing in this way through RnPz = RnPlQ, + R, , we see that Dl divides R, . In particular any greatest common divisor D of A and B divides R, and therefore has deg D 5 deg R,. In the reverse direction, RnP1 = R,Q,+l shows that R,, divides R,,-l. From RnP2 = RlzPlQ,, + R,,, we see that R,, divides RlzP2. Continuing in this way through B = Rl Q2 + R2, we see that R, divides B. Finally A = B Q 1 + R1 shows that R, divides A and B . Thus R, is a divisor of both A and B , and we have seen that its degree is maximal. This proves (a). If D is a greatest common divisor of A and B, it follows that D divides R, and deg D = deg R,. This proves (b). We have seen that any Dl that divides A and B necessarily divides R, , and then (c) follows from the uniqueness of the greatest common divisor up to scalar multiplication. Put R,+l = 0, Ro = B , and R-1 = A. We prove by induction downward that there are polynomials Sk and Tk such that RkSk + Rk+1Tk = D. The base case of the induction is k = n , where we have R, 1 + R,+10 = D. Suppose that RkSk Rk+1Tk = D with k > 0. We rewrite R k P l = RkQk+1 Rk+1 as Rk+1 = RkPl - Rk Qk+1 and substitute to obtain
+
+
+
D
=
RkSk
+ Rk+lTk = RkSk + Rk-1 Tk
-
RkQk+l
In other words, we can take = Tk and Tk = Sk - Qk+1,and our inductive assertion is proved for k - 1. The assertion for - 1 proves (d). A nonzero polynomial P with coefficients in IF is prime if the only factors of P are the scalar multiples of 1 and the scalar multiples of P.
Lemma. If A and B are nonzero polynomials with coefficients in IF and if P is a prime polynomial such that P divides A B ,then P divides A or P divides B . PROOF.Suppose that P does not divide A. Then 1 is a greatest common divisor of A and P , and part (d) of the above theorem produces polynomials S and T such that AS + P T = 1. Multiplication by B gives AB S + P T B = B . Then P divides ABS because it divides AB, and P divides P T B because it divides P . Hence P divides B .
Theorem (unique factorization). Every polynomial of degree 3 1 with coefficients in IF is a product of primes. This factorization is unique up to order and to scalar multiplication of the prime factors. PROOF.If A is given and is not prime, decompose A = BC with deg B < deg A and deg C < deg A . For each factor that is not prime, write the factor as the product of two polynomials of lower degree. This process, when continued in
572
Appendix
this fashion, must stop since the degrees strictly decrease with any factorization. This proves existence. For uniqueness, assume the contrary and choose m > 1 as small as possible so that some polynomial has two distinct factorizations P1 . . . P, = Q 1 . . . Q , into primes, apart from order and scalar factors. Adjusting scalar multiples, we may assume that each PI and Q n has leading coefficient 1 and that there is a global coefficient multiplying each side. These global coefficients must be equal, being the coefficients of the largest power of X on each side. Thus we may cancel them and assume that each P, and Q k has leading coefficient 1. By the lemma, the fact that Q 1 is prime means that Q 1 must divide one of Pl , . . . , P,. Reordering the factors, we may assume that Q l divides P I . Since PI is prime, PI is a scalar multiple of Q 1 . Since P1 and Q 1 both have leading coefficient 1, P1 = Q 1 . Then we can cancel P1 and Q 1 from both of our factorizations, obtaining distinct factorizations with fewer than m factors on one side. By the minimality of m , either we have arrived at a contradiction or we now have the polynomial 1 left on one side. Then the other side is 1, and the two sides match. If IF is R,then x2+ 1 is prime. But x2+ 1 is not prime when IF = C since x2+ 1 = ( X + i ) ( X - i ) . The Fundamental Theorem of Algebra, stated below, implies that every prime polynomial over C is of degree 1. It is possible to prove the Fundamental Theorem of Algebra within complex analysis as a consequence of Liouville's Theorem or within modern algebra as a consequence of Galois theory and the Sylow theorems. This text gives a proof of the result in Section 11.7using the Heine-Bore1 Theorem and other facts about compactness.
Fundamental Theorem of Algebra. Any polynomial with coefficients in C and with degree 2 1 has at least one root. Corollary. Let P be a nonzero polynomial of degree n with coefficients in C , and let r l , . . . , rk be the roots. Then there exist unique integers mj > 0 such that P ( X ) is a multiple of ( X - grn]. The numbers mj have X jk = , wq = n .
E=,
PROOF.We may assume that deg P > 0. We apply unique factorization to P ( X ) . It follows from the Fundamental Theorem of Algebra and the Factor Theorem that each prime polynomial with coefficients in C has degree 1. Thus the unique factorization of P ( X ) has to be of the form c n,"=, (X - zl) for some complex numbers that are unique up to order. The zl's are roots, and every root is a 21, by the Factor Theorem. Grouping like factors proves the desired factorization and its uniqueness. The numbers mi have 'j-F=l mi = n by a count of degrees. The integers mj in the corollary are called the multiplicities of the roots of the polynomial P ( X ).
A9. Partial Orderings and Zorn's Lemma
573
A9. Partial Orderings and Zorn's Lemma A partial ordering on a set S is a relation between S and itself, i.e., a subset of S x S , satisfying two properties. We define the expression a 5 b to mean that the ordered pair ( a , b) is a member of the relation, and we say that "5" is the partial ordering. The properties are (i) a 5 a for all a in S , i.e., 5 is reflexive, (ii) a 5 b and b 5 c together imply a 5 c whellevel-a , b, and c are in S , i .e., 5 is transitive. An example of such an S is any set of subsets of a set X , with 5 taken to be inclusion 5.This particular partial ordering has a third property of interest, namely (iii) a 5 b and b 5 a with a and b in S imply a = b. However, the validity of (iii) has no bearing on Zorn's Lemma below. A partial ordering is said to be a total ordering or simple ordering if (iii) holds and also (iv) any a and b in S have either a 5 b or b 5 a . For the sake of a result to be proved at the end of the section, let us interpolate one further definition: a totally ordered set is said to be well ordered if every nonempty subset has a least element, i.e., if each nonempty subset contains an element a such that a 5 b for all b in the subset. A chain in a partially ordered set S is a totally ordered subset. An upper bound for a chaia T is an elenlent u in S such that c 5 u for all c in T . A maximal element in S is an element m such that m 5 a for some a in S implies a 5 m . (If (iii) holds, we can then conclude that m = a .)
Zorn's Lemma. If S is a nonempty partially ordered set in which every chain has an upper bound, then S has a maximal element. REMARKS.Zorn's Lemma will be proved below using the Axiom of Choice, which was stated in Section A l . It is an easy exercise to see, conversely, that Zorn's Lemma implies the Axiom of Choice. It is customary with many mathematical writers to mention Zorn's Lemma each time it is invoked, even though most writers nowadays do not ordinarily acknowledge uses of the Axiom of Choice. Before coming to the proof, we give an exa~npleof how Zow's Lemma is used. EXAMPLE.Zorn's Lemma gives a quick proof that any real vector space V has a basis. In fact, let S be the set of all linearly independent subsets of V , and order S by inclusion upward as in the example above of a partial ordering. The set S is nonempty because M is a linearly independent subset of V . Let T be a chain in S , and let u be the union of the members of T . If t is in T , we certainly
574
Appendix
have t 5 u . Let us see that u is linearly independent. For u to be dependent would mean that there are vectors x l , . . . , x, in u with rlxl . . . r,x, = 0 for some system of real numbers not all 0. Let xj be in the member tj of the chain T . Since tl g t2 or t2 g t l , x1 and x2 are both in tl or both in t 2 . To keep the notation neutral, say they are both in t i . Since ti 5 t3 or t3 5 t i , all of x l , x z , x3 are in ti or they are all in t 3 . Say they are both in t i . Continuing in this way, we arrive at one of the sets t l , . . . , t,, say t;, such that all of x l , . . . , x, are all in t;. The members of tl, are linearly independent by assumption, and we obtain the contradiction rl = . . . = r, = 0. We conclude that the chain T has an upper bound in S. By Zorn's Lemma, S has a maximal element, say i n . If in is not a basis, it fails to span. If a vector x is not in its span, it is routine to see that m U { x }is linearly independent and properly contains m , in contradiction to the maximality of m . We conclude that m is a basis.
+ +
We now begin the proof of Zorn's Lemma. If T is a chain in a partially ordered set S, then an upper bound uo for T is a least upper bound for T if uo 5 u for all upper bounds of T . If (iii) holds in S, then there can be at most one least upper bound for T . In fact, if uo and ub are least upper bounds, then uo 5 ub since uo is a least upper bound, and L L ~5 LLO since 1.16 is a least upper bound; by (iii), uo = u;.
Lemma. Let X be a nonempty partially ordered set such that (iii) holds, and write 5 for the partial ordering. Suppose that X has the additional property that each nonempty chain in X has a least upper bound in X. If f : X + X is a function such that x 5 f ( x ) for all x in X , then there exists an xo in X with f (xo) = xo.
PROOF.A nonempty subset E of X will be called admissible for purposes of this proof if f (E) 5 E and if the least upper bound of each nonempty chain in E , which exists in X by assumption, actually lies in E. By assumption, X is an admissible subset of X . If x is in X , then the intersection of admissible subsets of X containing x is admissihle. Thus the intersection A, of all admissihle silhsets containing x is an admissible subset containing x . The set of all y in X with x 5 y is an admissible subset of X containing x , and it follows that x 5 y for all y in A,. By hypothesis, X is nonempty. Fix an element u in X , and let A = A,. The main step is to prove that A is a chain. Once that is established, we argue as follows: Since A is a chain, its least upper bound xo lies in X , and since A is an admissible subset, xo lies in A . By admissibility, f ( A ) 5 A. Hence f ( x o )is in A. Since xo is an upper bound of A , f ( x o ) 5 xo . On the other hand, xo i f ( x o ) by the assumed property of f . Therefore f (xo) = xo by (iii).
A9. Partial Orderings and Zorn's Lemma
575
To prove that A is a chain, consider the subset C of members x of A with the property that there is a nonempty chain C, in A containing a and x such that a 5 y 5 x for all y in C, , .f(C, - { X I ) c Cx, and the least upper bound of any nonempty subchain of C, is in C,. The element a is in C because we can take C, = { a } . If x is in C , so that C, exists, let us use the bulleted properties to see that
We have A 2 C, by definition; also A n A, is an admissible set containing x and hence containing A,, and thus A 2 A,. Therefore A 2 A, U C, . For the reverse inclusion it is enough to prove that A, UC, is an admissible subset of X containing a . The element a is in C, and thus is in A, U C, . For the admissibility we have to show that f ( A , U C,) c A, U C, and that the least upper bound of any nonempty chain in A, U C, lies in A, U C,. Since x lies in A,, A, U C, = A, U (C, - { x } ) and f ( A , U C,) = f (A,) U f (C, - {x}) 5 A, U C,, the inclusion following from the admissibility of A, and the second bulleted property of C,. To complete the proof of (*), take a nonempty chain in A, U C, , and let u be its least upper bound in X; it is enough to show that u is in A, U C, . The element u is necessarily in A since A is admissible. Observe that y ix
and x i z
whenever y is in C, and z is in A,.
(**)
If the chain has at least one member in A,, then (**) implies that x 5 u , and hence the set of members of the chain that lie in A, forms a nonempty chain in A, with least upper bound u . Since A, is admissible, u is in A,. Otherwise the chain has all its members in C, , and then u is in C, by the third bulleted property of C, . This completes the proof of (*). Although we do not need the fact, let us observe that combining (**) and (*) yields A, n C, = { x }whenever C, exists. Thus C, = ( A - A,) U {x}if C, exists. In particular the defining properties of C, determine C, completely. Recalling that C is the subset of members of A such that C, exists, we shall show that C is an admissible set containing a . If we can do so, then it follows that C 2 A, = A aiid l~eiiceC = A. This fact, ill coilibiiiatioii wit11 (*) and (**), proves that A is a chain: if x and y are in A, (*) shows that y is in A, or C,, and (**) shows that x 5 y in the first case and y 5 x in the second case. Thus the proof will be complete if we show that C is admissible and contains a . We already observed that it contains a . We need to see that f ( C ) C and that the least upper bound of any nonempty chain of C lies in C . For the
576
Appendix
first of these conclusions, let us see that if x is in C , then CfcXjexists and can be taken to be C, U { f ( x ) } . Property (*) proves that C, U { f ( x ) }satisfies the first bulleted property of Cf(,), and the second bulleted property follows from that same property of C, and the fact that x 5 f ( x ) . Any nonempty chain in C, U { f (x)} either lies in C, and has its least upper bound in C, or else contains f ( x ) as a member and then has f ( x ) as its least upper bound Thus C, 1-1 { f ( x ) } satisfies the third bulleted property of Cf(,) and can be taken to be Cf(,). Hence f ( x ) isinC,and f(C) g C. Finally take any nonempty chain {x,} in C , and let u be its least upper bound, which is necessarily in A . Form the set (U, c,~) U { u } . It is immediate that this set satisfies the three bulleted properties of C, and therefore can be taken to be C, . Hence u is in C , and C contains the least upper bound of any nonempty chain in C . Then C is indeed an admissible set containing a . PROOFOF ZORN'SLEMMA. Let S be a partially ordered set, with partial ordering 5 , in which every chain has an upper bound. Let X be the partially ordered system, ordered by inclusion upward c, of nonempty chains5 in S. The partially ordered system X , being given by ordinary inclusion, satisfies property (iii). A nonempty chain C in X is a nested system of chains r, d ,C, and Uur , is a chain in S that is a least upper bound for C . The lemma is therefore applicable to any function f : X a X such that c g f ( c ) for all c in X . We use the lemma to produce a maximal chain in X . Arguing by contradiction, suppose that no chain within S is maximal under inclusion. For each nonempty chain c within S, let f ( c )be a chain with c g f ( c ) and c # f ( c ) . (This choice of f ( c ) for each c is where we use the Axiom of Choice.) The result is a function f : X + X of the required kind, the lemma says that f (c) = c for some c in X , and we arrive at a contradiction. We conclude that there is some maximal chain co within S. By assumption in Zorn's lemma, every nonempty chain within S has an upper bound. Let uo be an upper bound for the maximal chain co . If u is a member of S with uo 5 u , then co U { u }is a chain and maximality implies that co U {u} = co. Therefore u is in cn, and u 5 un. This is the condition that un is a maximal element of S.
Corollary (Zermelo's well-ordering theorem). Every set has a well ordering. ~ O O F Let . S be a set, and let E be the family of all pairs (E, iE)such that E is a subset of S and iEis a well-ordering of E . The family E is nonempty since
ere
a chain is simply a certain kind of subset of S, and no element of S can occur more than once in it even if (iii) fails for the partial ordering. Thus if S = {x,y } with x 5 y and y 5 x , then (x,y ] is in X and in fact is maximal in X.
A10. Cardinality
577
( M , a ) is a member of it. We partially order E by a notion of "inclusion as an initial segment," saying that ( E , i E i) ( F ,i F if )
(0 E c F , (ii) a and b in E with a 5 Eb implies a 5 Fb, (iii) a in E and b in F but not E together imply a S F b. In preparation for applying Zorn's Lemma, let C = {(E,, 5,)) be a chain in E, with the a's running through some set I . Define Eo = U, E, and define 50as follows: If el and e2 are in E o , let el be in E,, with a1 in I , and let e2 be in E,, with 012 in I . Since C is a chain, we may assume without loss of generality that (E,, , s,,) 5 (E,,, s,,), so that E,, c E,, in particular. Then el and ez are both in E,, and we define el 50 e2 if el i,,e2, or e2 50 el if e2 i,,e l . Because of (i) and (ii) above, the result is well defined independently of the choice of a1 and 012. Similar reasoning shows that ( 0 is a total ordering of Eo. If we can prove that 50is a well ordering, then ( E o ,10) is evidently an upper bound in E for the chain C,and Zom's Lemma is applicable. Now suppose that F is a nonempty subset of Eo. Pick an element of F, and let E,, be a set in the chain that contains it. Since (E,,, 5,) is well ordered and F nE,, is nonempty, F f' E,, contains a least element fo relative to i,,. We show that fo 50 f fur all f in F. In fact, if f is given, there are two possibilities. One is that f is in E,,; in this case, the consistency of 50with forces fo 50 f . The other is that f is not in E,, but is in some E,, . Since C is a chain and E,, c E,, fails, we must have (E,,, 5,") 5 ( E , , , (,,). Then f is in E,, but not E,,, and property (iii) above says that f0 fb,, j'.By the consistency of the orderings, fo 50 f . Hence fo is a least element in F, and Eo is well ordered. Application of Zorn's Lemma produces a maximal element ( E , i E of) E. If E were a proper subset of S , we could adjoin to E a member s of S not in E and define every element e of E to be 5 s. The result would contradict maximality. Therefore E = S , and S has been well ordered.
(,,
A10. Cardinality
Two sets A and B are said to have the same cardinality, written card A = card B, if there exists a one-one function from A onto B . On any set A of sets, "having the same cardinality" is plainly an equivalence relation and therefore partitions A into disjoint equivalellceclasses, the sets in each class having the same cardinality. The question of what constitutes cardinality (or a "cardinal number") in its own right is one that is addressed in set theory but that we do not need to address carefully here; the idea is that each equivalence class under "having the same cardinality" has a distinguished representative, and the cardinal number is defined to be that representative. We write card A for the cardinal number of a set A .
578
Appendix
Having addressed equality, we now introduce a partial ordering, saying that card A 5 card B if there is a one-one function from A into B . The first result below is that card A 5 card B and card B 5 card A together imply card A = card B.
Proposition (Schroeder-Bernstein Theorem). If A and B are sets such that there exist one-one functions f : A 4 B and g : B 4 A, then A and B have the same cardinality. PROOF.Define the function g-' : imageg + A by g-' (g(a)) = a ; this definition makes sense since g is one-one. Write (g o f)(") for the composition of g o f with itself n times, and define (f o g)(n) similarly. Define subsets A, and A; of A and subsets B, and BI, for n > 0 by A, = image((g
o
f )(n))- image((g
A;
= image((g o
f)(n) o g)
B,
= image ((f o
g)(n))- image(( f
BL = image(( f
o
g)(") o f )
-
-
o
f)(n) g),
image((g o f)(n+')), o g)(n) f ) ,
image(( f o g)("+'I),
and let 00
A,
=
n image(( f o g)(n)). n=O 00
0 image((g o f )I,()
and
B,
and
B
=
n=O
Then we have 00
A = A,U
00
U AnU U A;
n=U
n=O
=
B,U
02
00
n=O
n=O
U BnU U BI,,
with both unions disjoint. Let us prove that f carries A, one-one onto Bi. If a is in A,, then a = ( g o f)(n)(n) for sorne n E A and a is not of the forrn ( g o f)(n)(g(y)) with JJ E R . Applying f , we obtain f (a) = (f o ((g o f)@))(x) = (f o g)(n)(f (x)), so that f ( a ) is in the image of ((f o g)(n) o f ) . Meanwhile, if f (a) is in the image of (f o s)("+'), then f (a) = (f o s)("+') (Y) = f ((s f ) ( n ) ( ~ ( ~ for ))) some y E B . Since f is one-one, we can cancel the f on the outside and obtain a = (g o f)in)(g(y)), in contradiction to thc fact that a is in A,. Thus f carrics A, into BA, and it is certainly one-one. To see that f (A,) contains all of BA, let b E BI, be given. Then b = (f o g)(n)(f (x)) for some x E A and b is not of the form (f o g) ("+I) (y) with y E B. Henceb = f ( ( g o f)in)(x)),i.e.,b = f ( a ) with a = (g o f )(")(x). If this element a were in the image of (g o f o g, we could write a = (g o f)(n)(g(y))for some y E B, and then we would have
A10. Cardinality
579
b = f ( a ) = f ((g o f )(n'(g(y))) = (f o g)in+l'(y), contradiction. Thus a is in A,, and f carries A, one-one onto BA . Similarly g carries B, one-one onto A;. Since A; is in the image of g , we can apply g-' to it and see that g-' carries A; one-one onto B,. The same kind of reasoning as above shows that f carries A, one-one onto B,. In summary, f carries each A,, one-one onto BI, and carries A, one-one onto B, , while g-l carries each A; one-one onto B, . Then the function on A, and each A,, on each A;, carries A one-one onto B. Next we show that any two sets A and B have comparable cardinalities in the sense that either card A 5 card B or card B 5 card A.
Proposition. If A and B are two sets, then either there is a one-one function from A into B or there is a one-one function from B into A. ~ O O F Consider . the set S of all one-one functions f : E + B with E 5 A, the empty function with E = M being one such. Each such function is a certain subset of A x B. If we order S by inclusion upward, then the union of the members of any chain is an upper bound for the chain. By Zorn's Lemma let G : Eo + B be a maximal one-one function of this kind, and let Fo be the image of G . If Eo = A , then G is a one-one function from A into B . If Fo = B , then G-l is a one-one function from B into A. If neither of these things happens, then there exist xo E A - Eo and yo in B - Fo, and the function G equal to G on Eo and having G(xo) = yo extends G and is still one-one; thus it contradicts the maximality of G .
-
Cantor's proof that there exist uncountable sets, done with a diagonal argument, in fact showed how to start from any set A and construct a set with strictly larger cardinality. Proposition (Cantor). Tf A is a set and 2A denotes the set of all silhsets of A , then card 2A is strictly larger than card A.
PROOF.The map A H {A} is a one-one function from A into 2A. If we are given a one-one function F : A + 2 A ,let E be the set of all x in A such that x is notin F(x). If F(xo) = E,thenxo E E impliesxo 6 F(xo) = E,whilexo 6 E implies x E F (xo) = E . We have a contradiction in any case, and hence E is not in the image of F. We conclude that F cannot be onto 2A.
HINTS FOR SOLUTIONS OF PROBLEMS
Chapter I 1 . Let E be a nonempty set that is bounded above. Start with a member sl of E. Choose if possible an s2 in E with s2 - sl > 1 . Continue with s3 - s2 > 1 , s4 - s3 > 1 , etc .,until this is no longer possible; the existence of an upper bound forces the process to stop at some stage. Suppose that sk has been constructed at this stage. Define sk+, inductively for n > 1 to be a member of E with sk+, - sk+,-1 > 2-, if possible; otherwise define sk+, = sk+,-1. Then { s , } is bounded and monotone increasing. To complete the problem, one has only to show that limn s, is the least upper bound of E. 2 . Show that xl > ,6 and that ,,6 < x,+l < x, for n > 1. Then limn x, = c exists by Corollary 1.6, and c must satisfy c = $ ( c 2 a ) / c .
+
i(1
3. Write out a few cases and guess that the pattern is a2, = - 2-("-')) for n > 1 and a2,+l = 1 - 2Zn for n > 0. Prove each of these statements by induction. Since 4,+ and q n + l + 1 and since these two subsequences use all the terrns of the sequence, the only subsequential limits of { a k }are and 1 . Therefore lim sup ak = 1 and liminf ak - $.
4
+
4. The argument without paying attention to finiteness is that an+ bn 5 a, sup,?, 6, for n > k , then that sup,,,(a, 6,) < sup,,, a, sup,?, 6, for all r , and then that the limit of the sum is th;sum of the limits.
+
-
+
5. Only (ii) converges uniformly, the reason being that 0 5 x n / a 5 l / n and that lim l / n = 0 . There is uniform convergence in (i) on [O, 1 - E] because 0 < x n 5 ( 1 - c ) , , and there is uniform convergence in (iii) on [0, 1 - c ] because the Weierstrass M test applies with 1xkl/k < ( 1 - €1, and C k ( l - c), < +GO. 6 . The uniform convergence of CZoa , ( x ) follows from Corollary 1.18, and the pointwise convergence of CZola,(x) 1 follows because ( 1 x ) CZoxn = 1 for 0 < x < 1 and because every a , ( x ) is 0 for x = 1. The convergence of C,"==, la,(x) I cannot bc uniform bccausc thc sum is discontinuous and Thcorcm 1.21 says that it would have to be continuous.
7. Put g, = f - f,, so that g, is continuous and decreases pointwise to the 0 function. Let x = x, be a point where g, ( x ) is a maximum, and let Mn = g, ( x , ) . We are to prove that M, tends to 0 . Suppose it does not. If k > n , then Mk = gk(xk) > gk(x,) > g,(x,) = M,. So Mn decreases to some M > 0. Passing to a
5 82
Hints for Solutions of Problems
subsequence if necessary, we may assume by the Bolzano-Weierstrass Theorem that limn x, = x'. For k > n, we have gk(xn) > gn(xn) = M, > M. Letting n tend to infinity gives gk (x') 2 M since gk is continuous. This inequality for all k contradicts the assumption that limk gk (x') = 0.
8. The idea is to prove the four inequalities
x x 2m
+
(- l)kx2k+1/(2k I)! > sinx,
k =O
2m+l (-l)kx2k+1/(2k
+ I)! < sinx,
x x
2m+l
( - 1 1 ~ x ~ ~ / ( 2 k<) !cos X ,
k=O 2m+2 (-l)kx2k/(2k)! > cosx
together by an induction. They are to be proved in order for rn = 0, then in order for rn = 1, and so on. In each case of the inductive step, the left side minus the right side is 0 at x = 0 and has derivative equal to the previous left side minus right side. The Mean Value Theorem says that each left side minus right side at x > 0 equals the product of x and the left side minus right side at with 0 < < x . Substituting the previously proved inequality at ( then gives the result. In other words, everything comes down to proving the first inequality, namely x > sinx for x > 0. Arguing in thesamestyle,wehavex-sinx = 1-cos(with0 < ( < x . Soatleastx-sinx 2 0 . For 0 < x 5 n , we actually obtain x - sin x > 0. Since (x - sin x) 0, we have x-sinx>n-sinnforn<x.Thusx-sinx>Oforallx>0.
<
<
&
9. The thing to prove is that the remainder term (X - t)" -f ("+')(t) d t tends to 0 for eachx as n tends to GO. If x > 0, the absolute value is 5 ( n ! ) ~ ' (x - t)" d t = xn+' /(n I)!, which tends to 0 for any fixed x . If x 5 0, one argues in a similar fashion.
+
l;
10. By a diagonal process we can find a subsequence {F,,} convergent for each rational x . Let F be the resulting limit function, carrying the rationals in [- 1, 11 into [O, 11. If r and s are rationals with r 5 s ,then F (r) = limk F,, (r) 5 limk F,, (s) = F (s) . Thus F is nondecreasing on the rationals. For each real x with - 1 < x i1, define F ( x p ) to be the limit of F ( r ) with r rational as r increases to 1, and define F (x+) to be the limit of F (r) with r rational as r decreases to 1. Then F (xp) 5 F (xf ) for each x , and F (x+) 5 F (yp) if x < y. For each N > 0, it follows that there can be only finitely many x's for which F (x+) - F(xp) > 1 / N , and hence there can be at most countably many x's for which F ( x p ) # F ( x + ) . Let this exceptional set be denoted by C. For x not in C, define F (x) = F (x+) = F (xp) . For x not in C , let us show that limk Fnk(x) exists and equals F(x). lt r < x is rational, we have F (r) = lim infk F,, (r) 5 lim infk F,, (x); taking the supremum over r gives F (x) = F (xp) 5 lim infk F,, (x) . Arguing similarly with s rational and x < s , we have lim supk Fnk (x) 5 lim supk Fnk (s) = F (s), and hence lim supk Fnk(x) 5 F(x+) = F (x). Combining these two conclusions, we see that liminfk F,,(x) = lim supk Fnk (x) and that the common value of these limits is F (x) .
583
Chapter I
Thus {Fnk(x)}converges except possibly for x in C. At each point of C, the sequence is bounded. Since C is countable, another use of adiagonal process produces a subsequence of Fnkthat converges at every point of C, hence at every point of [- 1, 11. 11. If Ix 1 z l / lim sup ;/i;l;lT,then ;/i;l;lT > l / lx 1 for infinitely many n. Thus la,xnl > 1 for infinitely many n, and the terms of the series do not tend to 0. Hence the series cannot converge. In the reverse direction we want to see that the implies convergence of the series. We rewrite inequality 1x1 < l/ lim sup i lllxl. Choose anumberr withlimsup i r il/lxl. this as limsup Then 5 r for all sufficiently large n, 1x1 5 r 1x1 < 1 for all n sufficiently large, and la,xn I 5 (r Ix jn for all n sufficiently large. Thus C lanxnI is dominated term-by-term (from some point on) by the geometric series C s n , where s = r lx 1. Since s < 1, the geometric series converges, and hence so does C la,xn 1.
+
+
12. 1/(1 - xj2 = C z o ( n l)xn, log(1 - x) = - Czlx n / n , 1/(1 x2) = C r = o ( - ljnx2,, and arctanx = C z O ( - l ) n ~ 2 n /(2n +1 1). All these series have radius of convergence 1.
+
13. The proof of existence of arccosx uses the proposition in Section A3 of the appendix. The result of the calculation of the derivative is that $ arccos x = - 114for Ix I < 1. Then arcsinx arccos x has derivative 0 on (O,n/2) and hence is constant. The constant is evaluated by putting .\-= 0, and the result is that arcsin x arccos x = n / 2 on (0, n/2j.
+
+
14. The uniform version of Abel's Theorem is this: Let {a, (x)},,~ be a sequence of complex-valued functions with a, (x) converging uniformly to the limit s (x) . Then lim,? 1 C n > oa, (x)rn = s (x) uniformly in x. The proof is just a matter of seeing that the estimates in the proof of Theorem 1.48 can be made uniform in x under the stated assumptions. The result about CesBro sums is handled similarly. -
CEO
15. Write cosn0 = i(einB+ epi"') and sinno = $(ein'
- e-ing
). Then
1 - e-f(N+l)8 +T ,-,-i~ . Each numerator is bounded by 2, and each denominator gets close to 0 only as0 tends to a multiple of 2 n . This proves the estimate for the cosines, and the estimate for the sines works in the same way. 17. For (a), the relevant result is that when all a, are 0, Czl lb, l 2 equals 00 0 ° 1 1.f (,\-)12dx. Here Ib,12 is ( 4 / ~ ) ~ and If(,\-)12d,\(2n - 1j2 ' 3 7 -Z n=l n=l N
COS~@ =
EL1eine+ 31
N
e-in8
-
- 2
x
1 - ei(N+l)\ -
-
i
In
2n 0 ° 1 n - is just - = 2. ~ e n c e n (2n - 1j2 8 n=l
'
l;
18. Wehave F(x)f(y) = f ( t ) f ( y ) d t = ?;; f ( t + y ) d t = lYxY f(tjdt = F (x+y) - F (y). If F (x) # 0 for some x , we can divide and use the Fundamental Theorem of Calculus to see that f (y) has a continuous derivative everywhere. (If F (x) = 0 for all x , then differentiation gives f (x) = 0 for all x.) Differentiating the original
5 84
Hints for Solutions of Problems
+
identity in x gives f ' ( x )f ( y ) = f ( x y). When x = 0 , we obtain f ' ( 0 )f ( y ) = f ' ( y ) . Then $ ( f ( y ) e p f ' ( 0 )= ~ ) f '(y)epf'(')y+ f ( y ) ( - f ' ( ~ ) e p f ' ( ~= ) y0) , and hence f (y)epf'(0)yis constant. Thus f ( y ) = aefi(')y. In the original identity f ( x )f ( y ) = f ( x y ) , if we put x = 0 and choose y such that f ( y ) # 0, then we see that f (0) = 1. Hence f ( y ) = eY(')y if f is not identically 0 .
+
19. We may assume that f is not identically 0 . As in Problem 18, we have , f ( 0 ) = 1. By continuity of f , choose xo such that 1.f ( x ) - 1 I 5 when Ix l 5 1x0I. Then R e f (xo) > 0 , and we can choose a unique c with I Im(cxo)l < n / 2 such 2 that ecxO= f ( x 0 ) The equation for f shows that f ( $ x o ) = f ( x o ) ,and hence 1 f ( i x o ) equals eCXo12 or -ecx0I2. From i f ( i x o )- 1 I 5 m, we have Re f ( i x o ) > 0 . Since I Im(cxo/2)I < n / 2 , eCx0I2is the choice of square root of ecxowith positive real part, and we conclude that f ( i x o ) = eCX012. Iterating this argument, we obtain f (2-,xo) = ec2-"x0 for all n 2 0. The equation for f shows that f ( k x ) = f ( x l k for all integers k > 0 , and thus f (qxo) = eCqx0for every rational q of the form k/2" with k an integer 2 0 . From f ( x )f ( - x ) = f (0) = 1, we have f ( x p l )= f ( x ) ~ 'and , thus f (qxo) = eCqX0 for every rational number of the form k / 2 , with k any integer. Using continuity and passing to the limit, we obtain f ( r ) = eCrfor all real r .
&
21. This uses the discussion at the end of Section A2 of the appendix. For x # 0 , we compute that gf( x ) = ( R(x)/,Y(x))ep'lx2 for polynomials R and S with S not the 0 polynomial. Then limx,o g l ( x ) = 0 by Problem 20, and the appendix shows that g f ( 0 )exists and equals 0 . 22. Use Problem 2 1 and induction. 23. Since {s,} is convergent, it is bounded. Say lsnl 5 K for all a . Let t z 0 be given, and choose N such that n z N implies s , - sl < c / 2 . Write t, - s = CJMnJsJ- s = CJMnj(sJ - s ) by (i). A second application of (i) gives
Since N is fixed, (ii) shows that 2K those n , It, - sl < c.
Mnj < t / 2 for n sufficiently large. For
24. For Ceskro summability the ith row, for i > 1, has its first i entries equal to 1 / i and its remaining entries equal to 0 . For Abel summability the row going with ri has jth entry ( 1 ri)(ri)jfor j > 0. -
25. Certainly Mij > 0 for all i and j. The power series in Problem 12a shows that C jMij = 1foralli,and(ii)holdsbecauselim,~l(j+l)r~(l-r)2 = (j+1).1.0 = 0 .
Chapter I
585
26. Check that M as in the previous problem transforms the Cesaro sums into the Abel sums, and apply Problem 23. 27. This is handled by the same kind of computation as with the FejCr kernel.
28. The formula for P, ( 0 )comes from summing the two geometric series forn 2 0 and n < 0 and then adding the results. Properties (i) and (iii) are then immediate by irlhpc~Liun.Fur prupcrly (ii) we uhe Lhc herich expar~hiur~ ul P, ( 0 ) . Tl~curcrri1.31 allows the integration to be done term by term, and the result follows. 29. This is proved in the same way as FejCr's Theorem (Theorem 1.59). 30. Corollary 1.38 shows that f,'(xj = CZoen knxnpl and that f,"(xj = c,,kn(n - 1)xnp2for 1x1 < R. The point is to show that { fL(x)} is uniformly bounded and uniformly equicontinuous for 1x1 5 r , and then Ascoli's Theorem produces the required subsequence. For proving the equicontinuity, it is enough to prove that { fk'/(x)}is uniformly bounded for Ix 1 5 r . Fix r < R , and choose rl with r < rl < R . Since lim fk ( x ) = f ( x )uniformly for 1x1 5 r l , there is an M such that I fk(rl)l 5 M for all k. Thus I Enc , , ~ ~ F I 5 M for all k . Since cn,k > 0 for all n and k, cn,k 5 M r l n for all n and k. Since r < r l , choose N such that n 1 N implies n ( r / ~ l ) ~ -5' 1 and n ( n - l ) ( r / ~ i ) 5 ~ -1~for n > N. Since cn,k > 0 for all n and k , c,,kn 1x1"-l 5 c,,knrnpl 5 (c,,kr;-l)(n(r/rl)n-'> 5 c,,kr;-' for n z N and I X I 5 r . Summing on n z N and taking Corollary 1.38 into account, we see that
for 1x1 5 r . Thus 1x1 5 r implies that I fk/(x)l is
and { f ; ( x ) } is uniformly bounded for 1x1 5 r . A similar argument with fk/' shows that N-1
rz-0
and we find similarly that { f k / ' ( ~ is ) }uniformly bounded for Ix 1 5 r . This completes the pi-nnf. 31. Theorem 1.23 shows that the limit of the subsequence of first derivatives is the first derivative of the limit, the liiiiit being differentiable. 111 other wolds, f is differentiable for 1x1 < r ,and the subsequence converges to f f ( x ) there. Since r < R is arbitrary, f is differentiable for 1x1 < R . Now we can induct, replacing f and the sequence fk in Problem 30 by f ' and a subsequence of f; on a smaller disk, then passing to f 'I,and so on. The result is that f is infinitely differentiable for 1x1 < R . 32. This is proved in the same way as in Problem 9.
5 86
Hints for Solutions of Problems
5 rNtk if IzI 5 r , and C E O r N + k= r /(1 - r). Thus 1$zN + 1~ z+N l+ ' . . . I tends uniformly to 0 for z l < r . Since t H exp(t) is continuous at t = 0, the required convergence follows. 34. Corollary 1.38 shows from the behavior for z real that all c, are 0. 35. Write 1 2 1 3 exp (z 2.z 7z ...) = (nFk";exp($zk)) l exp ($zN &zN+' ...). Problem 7 1 shows that the left side is the uniform limit of exp(izk) for I z I < r if r < 1. Each factor of the finite product is given by a convergent power series with nonnegative coefficients, and Theorem 1.40 shows that the finite product is given by a convergent power series with nonnegative coefficients. By Problem 32, exp ( z + hz2 + 5z3 + . . . ) is given by a convergent power series for lzl < 1. Hence 1 2 1 3 exp ( z + ~z + ?z + . . .) - 1/ (1 - z) is given by a convergent power series for lzl < 1. For z = x real with 1x1 < 1, the series expansion of Problem 12b shows that our expression is exp ( - log(1 - x)) - 1/(1 - x) = 0. Thus our power series sums to 0 on the real axis. By Problem 34, it sums to 0 everywhere. 33. 1&zkl
+
+
+
+
+ nfil
+
Chapter I1
+
1. Let us c o n ~ p x ed(n, y) with d(n, z) d(z, y). If j contributes to d ( n , y), then xj # yj. Hence xj # zj or zJ # yj. Thus j contributes to at least one of d(x, z) and d ( z , y). 111ulllcr wurds, the ~ur~lribuliurl u l j Lu d(x, y) is 5 ~ l ~ur~lribuliur~ c u l j Lo d ( x , z) d(z, y). Summing on j gives the desired result.
+
2. Let (X, d) be the given separable metric space, define E to be the subset of members x of X such that every neighborhood of x is uncountable, and let F be the complement of E. If x is in F, we can associate to x some open neighborhood N , containing at most countably many elements, and N, is entirely contained in F. As x varies in F, the sets N, form an open cover of F. By Proposition 2.32b, some subcollection of the N, that is at most countable covers F. The union of these sets is open and is at most countable, and it equals F.
3. Let f (x) = l / x for 0 < x < 1, and let f (0) = 0. 4. Suppose that x is in U. Since A is dense, the set A f' B ( l / n ; x) is nonempty for each n > 1. Let x, be a member of it. Since U is open, B ( l / n ; x ) is contained in U if n is > N for a suitable N . Thus x, is in A n U for n > N and converges to x . By Proposition 2.22b, cithcr x, = x infinitcly oftcn, in which casc x is in A n U , or x is a limit point of A n U. In either case, U c (A n u)". 5. For (a), the sets E n are compact by the Heine-Bore1 Theorem. Then each En- U is compact. Their intersection is (Enf' UC)= ( E,) f' uCg U f' UC = 0. By Proposition 2.35 the system { E n- U }does not have the finite-intersection property. N Thus (En - U) = 0 for some N . Since E l 2 E 2 2 . . . , we find that E N - U = 0.Therefore E N C U. For (b), let U be empty, and let En = Q n + l/n].
nzl
nzl
[a,
Chapter 11
587
6 . Inboth parts o f the problem, let the metrics be d x , d y , dz. For (a),use continuity o f F to choose for each (xo,y ) some S 1 , , > 0 and S2,, > 0 such that the two inequalities dx ( x ,xo) < 6 1 ,and ~ d y(y', y) < 62,ytogether imply dz(F ( x , y'), F (xo,y ) ) < c/2. As y varies, the open balls B(62,,; y ) cover Y . Since Y is compact, a finite number o f them suffice to cover Y , say B(82,yl; y l ) , . . . , B(82,y,; y n ) Put 61 = min{61,,,, . . . , 61,,n}. Suppose now that d x ( x , xo) < 61 and that y' is in Y . Then y' is in some B(S2,,);y j ) . Hence we have d x ( x , x o ) < S 1 < S1,,) and d ( y ' , y,) < ST,,, , and we therefore obtain d y ( F ( x , y ' ) , F (xn,y,)) < t / 2 . Since also dx(xo,xo) = 0 andd(yf,y J ) < S2,y,, weobtain alsodZ(F(xo,y'), Fixa, y,)) < t / 2 . Combining these two results gives dZ ( F ( x . y'), F (xo.y')) < c. For (b),consider d z ( F ( x , y ) , F(xo,yo)),and let t > 0 be given. By uniform convergence, choose S 1 > 0 such that dx ( x ,xo) < S 1 implies d Z ( F( x , y), F (xo,y ) ) < t / 2 for all y. Proposition 2.21 gives us continuity o f Fixo, .), and thus there exists S2 > 0 such that d y ( y ,yo) < 62 implies d Z ( F ( x oy, ) , F(xo,yo)) < c / 2 . Then d x ( x , xo) < 61 and dr ( Y , yo) < 62 together imply d z ( F ( x ,y ) , F(x0, yo)) 5 d z ( F i x , Y ) , F(x0, y ) ) + dziF(x0, Y ) , F(x0, yo)) < ~ / +2 t / 2 = 6 . 7 . Let f : (0, 1 ) + R be defined by f ( x ) = l / x . Then the Cauchy sequence { l / n }is carried to a sequence that is not Cauchy in R. 8. Define inductively f ( O ) to be the identity and f ( k ) = f o f ( k p l ) for k > 0 . For existence we see inductively that d ( f (4 ( x ) ,f ( " ( y ) ) < r b ( x , y ) for all x and y . I f n 1 m and i f x is arbitrary but fixed, we then have d ( f (") ( x ) ,f (m)( x ) ) 5 d ( f tkil)(x),f t k ) ( x ) )< r k d (f ( x ) ,x ) < rmd(f ( x ) ,x ) / ( l - r ) . Hence the sequence ( f ( n )( x ) ]is Cauchy. Let x' be its limit. Since d ( f ( f ( n )( x ) ) ,f (") ( x ) )= d ( f ("+')(x),f cn)(x))5 r n d ( f ( x ) ,x ) / ( l - r ) and since d and f are continuous, d ( f ( x f ) , x ' )5 l i m s u p n r n d ( f ( x ) , x ) / (-l r ) = 0 . Thus f ( x f ) = x f . For uniqueness, let f (x") = x" also. Then d (x", x') = d ( f (x"),f ( x ' ) )since f fixes x' and x", and d ( f (x"), f (x')) < rd(xf',x') by the contraction property. Then ( 1 r)d(xf',x') 5 0 and we conclude that d(xf',x') = 0. Thus x" = x'. 9. I f no point is isolated, each one-point set is closed nowhere dense. The countable union o f these sets is the whole space, in contradictionto the Baire Category Theorem. An alternative argument is to appeal to Problem 2. 10. The set is closed and bounded, hence compact, and it is pathwise connected, hence connected. It is not, however, locally connected. Take, for example, the point p = [c,1/21 in X , where c is in C . The open ball o f radius 114 around p has the property that no open subiieighborhood o f p is connected. 11. Fix xo in X , and let U be the set o f all points in X that can be connected to xo by paths. 'l'he set U is nonempty, and we prove that it is open and closed. Being connected, it must then be all o f X . It is open because the local pathwise connectedness means that any x in U can be connected to every point in some neighborhood o f x by a path; hence U contains a neighborhood o f each o f its points and is open. To see that U is closed, let y be a limit point o f U . I f V is a pathwise connected open neighborhood o f y , the set U n V is nonempty because y is a limit point o f U . Let z
c;:;
-
c;:;
588
Hints for Solutions of Problems
be in U n V. Then xo can be connected to z by a path because of the defining property of U , and z can be connected to y by apath because V is pathwise connected. Hence xo can be connected to y by a path, and y is in U . 12. Any open subset of Rn is locally pathwise connected. So the desired conclusion follows from the previous problem. 13. Let the open set be U . For each x in U , let U, be the union of all connected subsets of U containing x . It was shown in Section 8 that this is connected. For x and y in U , either U, = Uy or U, n U y = @ for the same reason. Then U is the disjoint union of its subsets U,, which are connected. These are intervals, being connected, and they must be open in order not to be contained in larger connected subsets of U . 14. Same as for Proposition 2.21. 15. Suppose { f , } is totally bounded. Let c > 0 be given. Find, by total boundedness, real numbers t l , . . . , t, such that for any t , there is an index j = j ( t ) with 11 f , - hJ11 I c . Put L / 2 = max{ltl 1, . . . , It, I}. If we are given an interval of length > L , take t to be its center, so that the interval contains [t - L / 2 , t L / 2 ] . Choose j by total boundedness with 11 f , - hJ} < t . Then 11 Apt]- fo 11 < t . So t - tj is an e almost period, and this lies in [t - L / 2 , t L / 2 ] . Thus the Bohr condition holds. Conversely suppose that the Bohr condition holds and f is uniformly continuous. Let c > 0 be given, and find L as in the Bohr condition for c / 2 almost periods. Also, find some S for uniform continuity of f and the number 612. Choose tl, . . . , t, in I = [ - L / 2 , L / 2 ] such that any point in I is within 6 of one of t l , . . . , t,. Let us see that the open balls of radius c around f,,, . . . , J n together cover the set If,} of all translates. If t is given, find an L / 2 alrrlost period t - s in [t - L / 2 , t L / 2 ] . Here Is1 < L / 2 , so that 11 f,-, - fo 11 < c / 2 and 11 f , - f, 11 < 612. Since 1s - tj I < 8 for some j , we have Il,f, - ,hJ11 < c/2 by uniform continuity. Thus 11 ,f, - ,ftJ 11 < c. 16. Let Tf be the closure of the set of translates of f . This is complete by Problem 14. Theorem 2.36 shows that Tfis compact if and only if every sequence in it has a convergent subsequence, and this is the definition of Bochner almost periodicity. Theorem 1.46 shows that Tfis compact if and only if it is totally bounded, and this is equivalent to Bohr almost periodicity by Problem 15. 17. This is easier with the Bochner definition. For an example of closure under the various operations, consider closure under multiplication. Suppose that f and g are given and that we want a coi~vergentsubsequei~cefro~llthe sequeilce of trai~slates ( f g ) t n . First choose a subsequence of {t,} such that those translates of f converge uniformly, and the11choose a subsequence of that such that the kanslates of g converge uniformly. These sequences of translates o f f and g will be uniformly bounded, and then it follows that the sequence of products converges uniformly. For closure under uniform limits, we argue similarly with translates of each of the functions { f,} when lim f, = f uniformly. A Cantor diagonal process is used to extract the sequence of translates to use for f . 18. If c > 0 is given, let U, be the set where I f ( x ) - f,(x) I < c. This is open by the assumed continuity, and U z l U, = X by the assumed convergence. Since
+
+
+
589
Chapter 11
X is compact, some finite collection of U,'s suffices. Since the fa's are pointwise increasing with n , the Un7sare increasing, and thus X = U N for some N. For that N, If ( x ) - fN(x)l < t . Then If ( x ) - fn(x)I < t for n > N since the fn7s are pointwise increasing. 19. If 0 ( P,(x) ( ,,6 ( 1, then x > P , ( x ) ~and the recursion shows that prL+l( x ) > prL(-x).Also, prL+l(-x)= prL(x) ;(& prL(x))(,,6- prL(-x))5 P,(x> i ( l I ) ( , / - P,(x)) = &. 20. By Problem 19, P,(x) increases pointwise to some f ( x ) . Passing to the limit in the recursion gives f ( x ) = f ( x ) + $ ( x - f (x12),and thus f (x12 = x and f ( x ) = A. Since is continuous and [O, 11 is compact, Dini's Theorem (Problem 18) shows that the convergence is uniform.
+
+
+
+
21. If x and y are given, then we are given three relevant functions in A, possibly not all distinct. They are hl with h l ( x ) # h l ( y ) ,h2 with h 2 ( x ) # 0, and h3 with h 3 ( y )f 0 . If hl ( x ) or h l ( y ) is 0 , we can add a multiple of h2 or h3 to hl to obtain an h4 with h4(x) f h4(y),h4(x) # 0 , and h 4 ( y ) # 0 . The restrictions of h4 and hi to the two-element set { x , y} are linearly independent and therefore form a basis for the 2-dimensional space of restrictions. Hence some linear combination of h4 and hi equals the given f at x and y . 22. Let f be in C R ( S ) with f (so) = 0 . Since BC' = C a ( S ) , there exists a sequence {g,} in B with lim g, = f uniformly. Then lim g,(so) = f (so) = 0 in particular. Put f, ( s ) = g, ( s ) - g, (so). Then f,(so) = 0 . The inequality I f n ( ~ > -f (s)I = Ign(s) - f ( s )-gn(so)I 5 lgn(s)- f (s)l+ Ign(s0)I shows that If,} converges uniformly to f . The members of A are the members of B that vanish at so. The functions f, have this property, and thus { f,} is a sequence in A converging ui~ifor~llly to f .
24. For (a), we identify Co([O,+GO), E X ) with the subalgebra of C([O,+GO], E X ) of continuous functions equal to 0 at t o o . The function epx separates points on [O, fool. Apply Problem 22 to the algebra it generates, namely the algebra of all finite linear combinations of epnxfor n a positive integer. For (b), let t > 0 be given, and choose g ( x ) = C c , e p n x by (a) such that b up^^^<+^ If ( x ) - g(x)I 5 t. The hypothesis forces f ( x ) g ( x )d x = 0 , and this f(x12 d x f ( x ) ( f ( x )- g ( x ) )d x . ~ h u s is
So
lob
lob
Sob
o lb
f (x12dx 5 t If ( x )I d x . Since t is arbitrary, f (x12d x = 0. Therefore So f = 0. 25. Isorrietries are uniforrrily continuous. Applying Proposition 2.47 to the uniformly continuous function q2 o ( q c l ) of the dense subset q1 ( X )of XT into X z , v1 ( X I we obtain an isometry : XT + X; extending q2 o ( q c l Reversing the roles
I
1
Vl(X)).
Hints for Solutions of Problems
590
of X l and Xg ,we obtain an isometry @ : Xg + XT extending q1 o (q?
I
Iv2(,) .
Then
IWl
@ 0 is a continuous extension of the composition q l o (q;' q2,)) o q 2 0 (ql-I , which is the identity map on q1 (X) . Hence Q, o \I, is the identity on XT . Similarly \I, 0 Q, is the identity on X;. Thus \V is onto. This proves existence. For uniqueness let V and V*be two such maps. Then V p l o V*is a continuous extension of the identity map on the dense subset q l ( X ) of XT, and hence it is the identity. Therefore \V = q*. 26. 'l'heorem 2.60 says that X is dense in X * . 'I'hen X = X * if and only if X is closed, and this happens if and only if X is complete, by Proposition 2.43. 27. The only one of these that requires explanation is (iv). We may assume that none of r , s , and r s is 0. Write r = r n p k / n and s = u p z / v with p not dividing any of r , s , u , v . Without loss of generality, we may assume k < 1, so that rnax{lrI,, IsI,} = IrI, = p p k . Wehave
+
r+s
=NIP
k
/n+up I/v =pk(?+
=pk(+).
The denominator nu is not divisible by p . The part of the numerator within the parentheses is an integer, and we factor out any factors of p from it as pa with a > 0. Then we have Ir s 1, = p ~ ( ~ +and " ) this is < p p k as required. 28. For the triangle inequality, let r , s , t be given. Then Problem 27 gives d ( r , t ) = I r t l p = I(r s ) + ( s t ) I p <max{lr s l p , I s t l p } 5 I r s l p + I s t l p = d ( r ,s ) d ( s , t ) . 29. Part (a) will be illustrated by the more difficult (b) and (c). Multiplication by a member r of Q is auniformly continuous function from Q into QD; in fact, the equality Ir ( S - so) 1 , = r 1 , s - SO I p shows that if c is given, then the 6 of uniform continuity can be taken as Ir 1p1t. Proposition 2.47 then tells us how to form products r s for r in Q and s in 0,. For fixed s ,the result is a uniformly continuous map of Q into Q , since I . 1 , extends continuously to Q, and we have I (r - ro)s1, = Ir - ro 1 , 1s 1,. A second application of Proposition 2.47 extends the operation to a mapping of Q , x Q , into Q , that is uniformly continuous in each variable when the other variable is held fixed. In fact, it is continuous in both variables since Irs -rosoI, = I(r -rojs +ro(s - s o ) , < Ir - roIplslp + I ' O l p l ~ -sol 5 Ir - rolpls - ~ o l p+ Ir - r0lplsoIp + Irolpls -sol. For (c), take a shell Akn = { r E Qp p p k < lrlp < p n } . This is a closed subset of Q,, hence complete. Reciprocal is a mapping from Ank n Q into Akn that is uniformly continuous because r and s in Allk n Q implies I T - l - s p l 1, = 1 1 I(s - r ) / r s l p = 1s - rlplrl; Isl; 5 p2"ls - r 1,. Hence reciprocal extends to a uniformly continuous mapping from Ank to A k n These mappings are consistent as n and k tend to infinity, and thus reciprocal is a well-defined function from Q," to itself. It is continuous because the same computation as just given shows that I T - l - r t 1 1, = Ir -roIpIr 1 ; 1 lrOlp1.If we write IrIp - Ir -rolpI and require
+
+
I
ire
I,, then that r -ro 1, < of reciprocal at ro follows.
I I ~ ~ I ~
17-l -7;'
1,
= r -rol,(i
lro,)-' lro 1 ; ' .
and continuity
591
Chapter III
The abelian group axioms in (c) are associativity, commutativity, existence of the two-sided identity 1, and existence of two-sided reciprocals. To complete (c), we need associativity and commutativity. We can regard associativity as asserting the equality of two continuous functions from Q, x Q, x Q, to 0,. These are equal on Q x Q x Q , and this subset is dense. Hence the two functions are equal everywhere. Commutativity is proved similarly. The distributive law in (d) is proved by the same technique used for associativity in (c). Thus Qpis a field.
I
30. For (a), it is enough to prove that S = {t E Q Itl, < 1) is totally bounded. For x in Q , let C(S; x) = {t E Q It - xip 5 S}. It is enough to show for each
I
1
integer 1 > 0 that S U,","; C(ppZ;r ) . If t is given in S, t is of the form t = rnln with rn and n in Z and n nondivisible by p. Let npl denote the integer from 0 to p Z- 1 such that nnpl 1 mod p Z ,and let r denote the integer from 0 to p1 - 1 such that nplm = r mod pz. Then rn - nr = 0 mod p Z ,and so Irn - nr lp < ppZ. Since Inl, = 1, - r i p < ppZ.Thust i s i n ~ ( p ~ ' ; r ) . For (b), compact sets are closed and bounded by Proposition 2 . 3 4 ~Conversely let E be closed and bounded. The set T = {t E Q, Itl, < 1) is certainly closed. Since Qp is complete, T is complete. Part (a) shows that T is totally bounded. By Theorem 2.46, T is compact. The given set E is contained in some set T, = { t E Qp It Ip 5 f } . Multiplication by the member pp" of Q, carries T continuously onto T,, and T, is compact by Proposition 2.38. Since E is a closed subset of the compact set T,, Proposition 2.34b shows that E is compact.
-
1;
I
I
31. The first two assertions are routine coilsequelices of (ii), (iii), and (iv). Let us consider the quotient Z,/ P. We show that P is a maximal ideal. In fact, if I is ly P, then I co~ltai~ls sonle elenlent 1 with 11 lp = 1. an ideal in Zp p ~ o p e ~containing Then (iii) shows that t p l has 1tp1lp = 1 and lies in Z,. Since t is in I and t p l is in Z, , their product 1 is in I. Thus I = Z, . In other words, P is a maximal ideal. Hence Z p / P is a field. To complete the argument, we show that Z P has exactly p elements. Given x in Z,, choose rn/n in Q with x < pp','by denseness of
Q in 0,. Here
I "I
;Ip
< 1, and we may assume that n is nondivisible by p . Arguing
n p -
as in Problem 30a, we can find r in 10, 1, . . . , p
l} such that
1;
r
1
5 ppl. Then Is - r 1, 5 max { x , -r 5 p p l by the ultrametric inequality. So n P x = ( x - r ) + r w i t h x - r i n P . Thus{O, 1, . . . , p - 1}representsallcosetsofZ,/P. Finally no two distinct elements r and r' in 10, 1, . . . , p - l} have Ir - 7'1, < p p l because this inequality would entail having r - r ' divisible by p .
"1
1;
I,}
-
-
P
Chapter I11 1. For (a), I T S I = ~ Cj lTs(ej)12 = Cj I Ci(S(ej), ei)T(ei)12. Use of the triangle inequality and then the Schwarz inequality shows that this expression is 5
Hints for Solutions of Problems
592
G ( C l ( s ( e j ) .e i ) l ~ ( e i )2l ) 5 E j ( ( X i l ( s ( e j ) ,e i ) l 2 ) ' I 2 ( E il ~ ( e i ) l ~ ) "=~ ) ~ =
Cj
1
~ 17.1~. 1 Part ~ (b) is routine.
2. The member of L ( R n ,Rm)with matrix A. 3. limsuphiO (1hlp1lf (1%) 0 -01) 5 l i m s ~ p ~( ,1~1 ~ 1 ~ ~ 1= 1 ~0 1. ~ ) -
2
+
4. The formula is f ( x t u ) ( x ). The argument is written out = C j uj within the proof of Theorem 3.11. 5. et '0" sin'), (ytisint), ( c y h t s i n h t )
(:
o,),
(i;), (
s~ntcost
slnht cosh t
zslntcost
'
7. The equality is false because the left side is positive and the right side is 1 negative. In fact, the left side is So [ l i m ~ : ( e -~ 2epxY) ~ ~ d x ] d y , which equals N lim [-epXy / y + e p 2 x ~ / y ]d, y = [epy -ep2*] d y ; since epy > ep2y on ( 0 , 1))
1; $
1;
+
1
[ - e p x Y / x e ~ ~ " ~d x/ =x ] ~ the left side is > 0 . Meanwhile, the right side is [ep2" - e p x ]d x ; since ep2" < e p x on (1, GO),the right side is < 0 . 8. Define I . 112 as in Section 1.10, and let fx ( t ) = f (x - t ) ;the latter definition is not the one used earlier in the book. For (a), the Schwarz inequality gives
and the right side tends to 0 as x tends to xg by uniform continuity of f . This proves that f * g is continuous. The periodicity is evident. The proof that f * g - g * f is the same as the proof in Section 1.10 that f DN = DN f . For (b), ail application of Fubini's Theorem (Corollary 3.33) and a change of I 2 n variables gives f * g (x)epinxd x = ( E ) Jn f ( x - t)g(t)epinxdt d x =
*
& 1: (k)'s,"?;= f ( x t ) g ( t ) e p i n xd x d t
(&)'x ?;=
-
=
*
( k 2)J:
1:
Jz
f(x)g(t)epin(x+"d x d t
f ( x ) g ( t ) e p i n x ~ idnxt dt = cadn. For (c), we apply the Weierstrass M test. It is enough to prove that +GO,and the Schwarz and Bessel inequalities together do this:
C ~cndnl5 (Clcnl ) 112 (C1411 ) 112 < I l f ll2llgll2 2
2
-
EnIcndn1
=
<
+KJ.
9. Write out each side as an iterated integral, and apply Fubini's Theorem (Corollary 3.33). 10. For the partial derivatives, E ( O , 0 ) = f (&) IX=, = 0 and g ( 0 , O ) = 0 similarly. The fact that f is not continuous at (0,O) is a special case of Problem 1la. 11. For (a), the homogeneity says in particular that f ( r x ) = f ( x ) for r > 0 and 1x1 = 1. Then supYfoIf ( y )I = suplXl=l If ( x )1, and the right side is finite, being the maximum value of a continuous function on a compact set. If f ( y ) is continuous at
593
Chapter III
y = 0, then f (0) = limrLof (rx) = f (x) for every x with 1x1 = 1 and so f must be constantly equal to f (0). For (b), lim up,^,^ I f (rx)l = lim s ~ p , ~rd, 1~f (x)l = 0 if d > 0 since f (x) is bounded for Ix = 1. Thus f is continuous at 0 if d > 0 and f (0) = 0. If d < 0 , then lim s ~ p , ~r d, I ~f (x)I = +m if d i0 and f (x) f 0. For (c), we have f ( r x ) = r d f ( x ) for any x = ( x l , .. . , x,,) # 0. Put g = f o m, , where m, refers to multiplication by r . The homogeneity gives g = r d f , and thus (x) - rdaf (x). On the other hand, the chain rule gives (x) -
Cy=l~
ax,
ax,
ax,
( r x ) ~ (=xr X ) (rx). S O T ~ X ( X =)r X ( r x ) , and (c) follows. ax,
ax,
8x1
For (d), the given conditions say that f (tx)
+
ax,
= tf
(x) for all real t. Then X ( 0 ) ax,
=
limt,o tpl(f (0 te,) 0) = limtio tpltf (el) = f (eJ). On the other hand, (c) says that af/axJ is homogeneous of degree 0, and (a) says that af/axJ cannot be continuous at 0 unless it is constant. d dt
-
+
= 12. Part (a) follows from Problem 1lb. In (b), g ( 0 ) = $ f (0 t (1,O)) d af = 1 and 5(0) = f (0 ty)It=o= ;l;O1t,O = 0. The failure of continuity
2
+
is by parts (a) and (c) of Problem 11. For (c), we have $ f (0 tu)lt=, = $t cos3 tiable at x = 0 , the chain i-ule would give -$f (0
+
~ l =~ cos3= 0.~ If f were differen+ tu)It=, = u l g ( 0 ) + u2$(0)
=
cos Q. Since cos3 Q is not identically equal to cos Q , f is not differentiable at 0. 13. Part (a) follows from (a), (b), and (c) of Problem 11. About 0, the function f' is even in x and even in y, and hence the first partial derivatives are odd ahoilt 0. Then part (b) follows from Problem 1ld. To calculate the results for (c), we need to compute af(.\-,0) for .\-# 0 and g ( O , y) for y # 0. The first of these is .\-,and the ay second is -y. The formulas for the second partial derivatives follow. 14. Forn > O,rneLnO = ( ~ + i y ) ~ i s o f c l a s s C ~ ~ s, aonids ~ ~ e= ~ (' ~~ -' i y ) ~For . the first of these functions, (x +iy)" = n(n - l)(x + i y ) " ~ while ~ , <(x +iy)" =
+
$
ay
+
i2n(n - l)(x i y ) " ~ ~Hence . A(x iy)" = 0. The result for (x - iy)" follows by taking complex conjugates. The final conclusion is a routine consequence of Theorem 1.37, the complex-valued version of Theorem 1.23, and the fact that each term is harmonic. 15. This follows by direct calculation 16. In the notation of Theorem 3.17, p(x, y) is ( 2 . 2 ) One checks that p t ( l , 1) = near (2.2) has ft(2, 2) = pt(1, I)-' entry of this, namely 3/14. 17.
(<:f
:),
a is (1, 1). and b is
(: :). The locally defined inverse function f =
3/14 1 / 1 4 14), and E ( 2 , 2) is the upper left
All 6 derivatives of possible interest are given by the matrix product
Hints for Solutions of Problems
594
(; 1 ;) (:I 2
1 0
()
0 n / 2
I
) =
Then
2
0) = 0 and
( r / 20 ) =
-n/12. he f&ction'x(u, v ) is of class Cm by Corollary 3.21 18. The map in question is X H x2 and is the composition of X H ( X , X ) followed by ( U , V ) H U V . Here we can write U V = L ( U ) V = R ( V ) U , where L ( U ) is the linear function "left multiplication by U" on matrix space and R ( V ) is the linear function "right multiplication by V." The derivative of ( U , V ) H U V is then ( R ( V ) L ( U ) ) by Problem 2. Hence the derivative of X H x 2 , by the chain rule. is
+
At X = 1, this is R ( l ) L ( 1 ) , which is "multiplication by 2" and is invertible. The Inverse Function Theorem thus applies.
19. We may assume that gf(xo) # 0 , thus that $(xo) # 0 for some i. We take this i to be i = n; the other cases involve only notational changes. Write x = (x', x n ) with x' t and write xo = ( a ,b ) similarly. Then the Implicit F~~nction Theorem produces a real-val~~ed c1function h ( x f )defined on an open set V about the point a in w"-' such that h ( a ) = b , g(x', h ( x f ) )= 0 for all x' in V , and & ( a ) = - ( % ( a , b ) ) ~(*(a, ' b ) ) for 1 5 j < n . Let H ( x ) = (x', h ( x f ) ) . 8% ax, Form f o H, which has a local maximum or minimum at x' = a in V . All the first partial derivatives of this function must be 0 at x' = a . Thus, for 1 ( j ( n - 1 , 0= ( x o ) G ( a ) .Since Hi ( x ) = xi for i < n , all the terms of ( a ) = x:=l 8x1
8x1
this sum are0 except possibly forthe jth and the nth. Thus 0 =
*(xo) ax,
-
(C)(xo)(g( a ,b ) ) ~(*(a, ' 8%
=
*(xo)+ 8x1
( x o ) E( a )
b ) ) for j < n. The right side is 0 trivially
for j = n , and thus the result follows with A = -
(E( a ,b ) ) - l .
20. The difficulty in handling this inequality as a maximum-minimum problem is the question of existence. Lagrange multipliers can constrain matters to a compact set, and then existence is no longer an obstacle. The domain U initially will be the set where a1 > 0, . . . , a, > 0. Fix a number c , and let g ( a l , . . . , a,) = i1( a l . . . a,) c and f ( a l ,. . . , a,) = g a l a , . The subset of D where g(a1, . . . , a,) = 0 is compact, and f must have an absolute maximum on it. This maximum cannot occur where any aj equals 0 since f is 0 at such points. So it is at a point in thc sct U whcrc all a] arc > 0. Apply Lagrangc multiplicrs on U . The resulting equations are (al . . .a n ) ' l n / a j = l / n for 1 < j < n , as well as the constraint equation (al . . . a,) = c. The first n equations show that all aj's must be equal, and the constraint equation shows that they must equal c. The desired inequality is true in this case and hence is true in all cases.
+
+
-
+ +
595
Chapter IV
Chapter IV 1. For (a), y ( t )=
+-.
iy2= - i t 2 +
+
Adjusting c, we have y2 = -t2 c. Then For (b), the exceptional points are (to,0 ) . For (c), a solution with
r (to) = yo is Y ( t ) = S
C.
g n ( y o ) \ I m .
2. In Theorem 4.1, take a = 1 and b = 1. Then M = 2 and a' = therefore gives a solution for It 1 < 112.
i. The theorem
3. To be an integral curve, ( x ( t ) ,y ( t ) ) must satisfy x f ( t )= ,,5 and y f ( t )= 112. Then 2 m = t cl and y(t) = i t c2. At some unspecified time to, the curve is to pass through ( 1 , 1). Then x ( t O )= 1 and y (to) = 1 ; these force 2 = to cl and 1 = it0 c2. SO ( ~ ( t y) (,t ) ) = ( i ( t - to 212, i ( t - to 2)). If to = 0 , for example, the curve is ( x ( t ) ,y ( t ) )= (t 212, (t 2 ) ) .
+
+
+
(i +
+
+
+
+
4. This uses the multivariable chain rule, Proposition 3.28b, and the Fundamental Theorem of Calculus. The derivative in question is
6. For (a), J = O
i
(i ((:1 + (k ))
and B
e3'
( :)
=
t
and B =
-9
(1 y )
for the first, and l =
0
'
for the second. For (b), the bases are e3'
for the first, and
(i ) ( ) ,eit
,e i '
( b)
0
(:)
-i and
for h e second.
Part (b) can be solved directly without solving part (a) first. Consider the 2-by-2 example. The only root of the characteristic polynomial iq '3, and it haq milltiplicity 2. We solve ( A- 3 . l)ko = 0 and get ko = ( ; r ) . Then we solve ( A- 3 . l)lo = and get lo = ( c
ko =
(1).
+d2d).
a i d lo =
Choose any c $ 0 and any d , say c = 1 and d = 0. Then
( ),
and we obtain the solutions in the form given above.
For more complicated examples, the choice of these constants can get tricky, but this method works quickly for easy examples. 7. For n = 1, det(h
-
(-ao)) = h + q.Assume the result for n
-
1, and expand
596
Hints for Solutions of Problems
the nth-order determinant by cofactors about the first column. Then
the next-to-last equality following by induction. 8.1n(a),letIfn(t) 5 Mforalltandn.ThenIF,(t)-Fn(tl)l = f,(s)dsI 5 M 1 t - t'l . Thus equicolltilluity holds with S = c/M. In (b), we solve the equation explicitly, using variation of parameters. The solutions of the homogeneous equation are cl cost c2 sin t, and computation shows that the unique solution of the inhomogeneous equation with the given initial condition is y* (t) = -(cos t) (sin ,s)f.(,s)d,s (sin t) (cos s) f (,s)d,s. Each integral is equicontinuous by the same argument as in (a), and the operations of multiplication by a bounded continuous function and addition preserve the equicontinuity. In (c), we do not know explicit formulas for the solutions of the homogeneous equation, but the same argument as in (b) with variation of parameters will work anyway.
+
1;
+
1;
10. For any c2periodic function f , the nth Fourier coefficient cn of f has IcnI 5 np2 sup I f If 1. The function u(r, B ) , being acomposition of two c2functions, is c2for 0 < r < land101 < n,andhencesupI$Iisboundedbysome~forO < r < 1-6. Then we obtain Icn(r)I 5 ~ / n ~ .
11. The function (uoR,) (x, y ) e p i kis ~ of class c2jointly inx, y , p. By Proposition 3.28 we can pass the second derivatives with respect to x and y under the given integral sign with respect to p. The integrand is harmonic in (x, y) for each q ,and therefore the integral itself is harmonic. The integral itself is given by
The series in the integrand is uniformly convergent as afunction of p,by the estimate in Problem 10 andby the Weierstrass M-test. Theorem 1.31 says that wecaninterchange sum and integral, and then the right side above collapses to ck(r)eine.
597
Chapter IV
12. Starting from v(r, 0 ) = u(r cos0, r sinOj, we compute rule and obtain
$ and
b y the chain
au a r = c o s ~ ~ + s i n O and~ ~ = - r s i n ~ ~ + r c o s ~ & ay
9and 2 in terms o f the partial derivatives o f
Using the same technique, we form u , and we find that -
'
a2. a?
pI
i a2u
rar
72882.
Substituting u(r, 8 ) = ck(r)eikeand taking into account that ALL= 0 , we obtain
+
Thus r2ct rc; - k2ck = 0 . This is an Euler equation. The solutions are c k ( r ) = a k r k bkrCIk i f k # 0 and are ao bo logr i f k = 0. Taking into account that c k ( r )is differentiable at r = 0, we obtain e k ( r )= akr I k for all k . Substitution gives ~ ( r , @ j = x , " = ~ , c ,n rein8 .
+
+
xzCM
13. Since f R ( 8 ) = c,Rnemneand P r I R ( @= ) x E C M ( r / R ) l n e i n the e, result follows immediately from Problem 8b at the end o f Chapter 111. 15. For (a),substitute y = u v , y' = ufv+uv', and y" = uf'v +2u'v'+uv"into the equation for y , take into account that ZL"+ PLL'+QLL= 0 , and get ~ Z L ' U ' + L L U " + PLLV'= 0 . Put w = u'. W e can rewrite our equation as w' = ( - P - 2 u f / u ) w since u is assumed nonvanishing. Then Problem 14 gives w ( t ) = ceCJ ' p d t - 2 Pdtelog(ln12i= C u ( t ) C 2 e C j - P('idt
ce.r
J'(lr'/a) dt -
For (b), the formula in (a) gives v l ( t ) = ceC"l2, and hence y ( t ) = u ( t ) v( t ) = e"/2 J';e - ~ 2 / 2d s .
16. The substitution leads to uv"
+
+ (2u' + Pu)vf + (u" + Pu' + Qu)v = 0. Thus
the condition is 2u' Pu = 0. By Problem 14, u ( t ) is a multiple o f e C ~ ( P 1 2 )The dt. computation o f R ( t ) is then routine. 17. Substitution o f v indicated form.
=
urC1/2shows that L ( v )
=
r 1 / 2 ~ o ( with u ) Lo o f the
18. For (a),the formula is dn = - x i = l ckdnCk,with do = 1 . For (b), we have dl = e l & = e l , so that Idl = Icll 5 M T ' . Thus Idnl 5 M ( M I l ) n C 1 r nfor n = 1. Assume that ldkl < Mrk for 1 < k < I I . Then Idnl < lcnCklldkl< ( M ~ " ~ " ( M (+M1)"'r" ) Mrn + M2rn ( M + l l k C 1 This . is Icn I
+
xi:;
598
Hints for Solutions of Problems
For (c), we may assume that f ( 0 ) = 1. Write f ( x ) = CEO c n x n ,and define dn as in the answer to (a). The estimate in (b) shows that the power series g ( x ) = CEO dnxn has positive radius of convergence, and Theorem 1.40 shows that f ( x ) g ( x )= 1 on the common region of convergence. Then g ( x ) = 1/f (x) , and 1/f ( x ) is exhibited as the sum of a convergent power series.
+
+
19. The iiidicial equation is s ( s - 1) aos Do = 0 , where q = P ( 0 ) aiid bo = Q(0). Thussl +s2 = 1 -ao. In (a), we apply Problem 1% with u ( t ) = t S 1CZocnt? The expression P ( t ) in that problem has become t-' ~ ( there, ) and we obtain v f ( t )= ~ ( t ) - ~ e ~ .~In' ~ ' ~ " ) ~ ~ the integrand of the exponent, we separate the term -ao/t from the power series, and ~ ' ~ gseries h = t - q ~ ( t ) -x~power series, the we see that v'(t) = ~ ( t ) - ~ e - ~ power power series having nonzero constant term since exponentials are nowhere vanishing. This is of the form t-2Sl-a0 x power series as a consequence of Problem 18 and Theorem 1.40, the power series having nonzero constant term. When this expression is integrated to form u ( t ) ,the t p l produces a logarithm, and the rest produces powers o f t . Thus u ( t ) equals clog t t-2sl-q+1 x power series; here the power series has nonzero constant term. Then u ( t ) u ( t ) = c u ( t ) log t tS1t-2sl-q+1 x power series; once again the power series has nonzero constant term. The exponent of t in the 1 - a0 = -sl ($1 f a ) = $2, and (a) is done. second term is -sl In (b), we know that there is only one solution beginning with t S l , and thus we must have c # 0 in (a). Another way to see this conclusion is to recognize that the exponent of t-2sl-ao in v f ( t )is just - 1 since 2sl = sl s2. Thus the coefficient of t-' in integrating to form v ( t ) is not 0 , and the logarithm occurs. In (c), we know from a computation in the text that no series solution begins with t-P except when p = 0 , and thus the first argument for (b) applies.
+
+
+
+ +
+
20. When t = tk-1 is substituted into the formula valid for tk-1 < t < tk, we get y ( t ) = y i t k p l ) ;SO the formula is valid also at tk-1. We induct on k. For k = 0, y(to) = yo. Assume inductively for k > 0 that I y(tk-1) - y (to)l 5 Mltk-1 - to 1 5 Ma' 5 6 . For tk-1 5 t 5 tk, the displayed formula in the problem implies ly ( t )- y (tkPl)l= IF i t k p l ,y (tkPl))lIt - tk-1 I . Since (tk-1, y(tk-1)) lies ill R', IF1 is 5 M on it. Thus l y ( t ) - y(tkPl)l 5 Mlt - tk-11 5 Ma' < 6. If tl-1 5 t < tl, then adding such inequalities gives ly ( t ) - y (to)1 < Mlti - to1 + . . . Mltl-1 - t1-21 Mlt - tlPll = Mlt - tOl as required. Since It - to1 5 a', we have Mlt - to1 5 Ma' 5 h. Thus ( t , y ( t ) ) is in R'.
+
+
21. We may assume that t' 5 t . If t' and t lie in the same interval [ t k P l t, k ]of the partition, then y ( t ) - y ( t f )= F i t k p l ,y ( t k P l ) ) ( t t f ) . Taking absolute values gives l y ( t ) - y(t')l 5 Mlt - t'l. Otherwise let t' < tl < tk-1 < t . Then each pair of points (t', tl),(tl,tl+1), . . . ,(tkP2,t k P l ) , ( t k p lt,) lies in a single interval of the partition. Adding the estimates for each and taking into account that each difference of t values is > 0 , we obtain l y ( t ) - y(t')l i Mlt - t'l.
599
Chapter V k-1
+ h:-l
t,
y f ( s )d s = y f ( s )d s = C j = l S,i - 1 y f ( s )d s ( ~ ( t l-) Y ( ~ o > > ' ' ' f ( ~ ( t k - 1 )- ~ ( t k - 2 ) ) ( ~ ( t-) ~ ( t k - 1 ) )= y ( t ) - Y ( ~ o ) , by an application of the Fundamental Theorem of Calculus on each interval. If tk-1 < s < tk,wehave l y f ( s )- F ( s , y(s))l = IF(tk-l, y(tk-11) - F ( s , y ( s ) ) l . Here 1s - tk-11 5 Itk - tk-11 5 6 by the choice of the partition. Again by the choice of the partition, 1 y (s) - y (tk-l)l I MIS - tk-I 1 i M ( 6 / M ) = 6 . By the definition of 6 in terms of c and the uniform continuity of F , we conclude that 1 y f ( s )- F ( s , y ( s ) )1 < c . 22. Let tk-I 5 t 5 tk. Then
+
+
&
I
23. Wc have l y ( t ) - (yo + F ( s , y ( s ) ) d s ) I = J : [ y f ( s ) - F ( s , y ( s ) ) l d s I 5 Iyf(s) - F ( s , y(s))I d s i c d s i clt - to1 i ca'. 24. The statement of Problem 21 proves uniform equicontinuity with 6 = c / M . If we specialize to t' = to, it implies uniform boundedness. 25. Let y ( t ) = limy,, ( t ) uniformly. The functions y,, ( t ) are continuous, and the uniform limit of continuous functions is continuous. Hence y ( t ) is continuous. By Problem 23 we have y,, ( t ) - (yo F ( s , y,, ( s ) )d s ) 5 c,,af for each k . We take the limsup of this expression as k tends to infinity. We know that y,, ( t ) tends uniformly to y ( t ) . Then y,, ( s ) tends uniformly to y ( s ) uniformly for to ( s 5 t . By uniform continuity of F , F ( s , y,, ( s ) ) tends uniformly to F ( s , y (s)). By Theorem 1.31,L: F ( s , y,,(s))ds tends t o < F ( s , y ( s ) ) d s .
6
1
+&
I
Chapter V 1. For (a) aiid (c), the answer is 2k for 0 5 k 5 T I . However, the assertioii iii (d) is false; for a counterexample, take X = {1,2, 3,4}, and let 8 consist of all sets with an even ~lunlberof elements. For (b), the associativity is proved by observing that A A B AC is the set of all elements that lie in an odd number of the sets A , B, C. 2. Let X = ( 1 , 2, 31 with the 0-algebra consisting of all subsets. Take p ( { l } ) = ~ ( ( 3 1= ) +2, p({2}) = -3, A = {l,2}, and B = {2,3}. 4. This can be worked out carefully, but it is easier to use Problem 3 and apply dominated convergence to see that the measure of the left side is lim sup p (E,), and the measure of the right side is lim inf p(E,). 5. Part (a) is proved the same way as for Lebesgue measure. In (b), the interval I of rationals from 0 to 1 has p ( 1 ) = 1, and it is a countable union of one-point sets {p},each of which has p ( { p } ) = 0.
6. Argue by contradiction. If Ec is not dense, then there is a nonempty open interval U in [ 0 , 11 with U n Ec = O and hence U g E. Since p ( U ) > 0 , we lnust have p ( E ) > 0 . 7. As soon as sup p ( A ) is known to be finite, B can be constructed as the union of a sequence of sets whose measures increase to the supremum. Thus assume that the supremum of p ( A ) over all sets of finite measure is infinite. Then we can choose a disjoint sequence of sets A, with each p ( A , ) finite and with C p(A,) = +GO.A
600
Hints for Solutions of Problems
little argument allows us to partition the terms of the series into two subsets, with the series obtained from each subset divergent. Say the terms of one subset are p ( B i ) and the terms of the other are p ( C j ) . Since C p ( B i ) = +oo, the hypothesis makes p ( ( U i B,)') finite. A contradiction arises because ( U i ~ i 2 )U j~C j and U j C j has infinite measure. 8. Consider the set ,4 of all Borel sets E such that f - ' ( E ) is measurable. The set A is closed under complements and countable unions, and it contains all intervals. So it is a a-algebra containing all intervals and must consist of all Borel sets. 10. This problem can be done via dominated convergence, but let us do it from scratch in order to be able to quote it in solving Problem 18 and other problems. We have
and the right side tends to 0 by the uniform convergence. Thus limx fn d p = the limit existing.
lxf
d p,
11. In (a) the approximating sets are finite unions of intervals, and we can add ( 1 - r,). Then apply Corollary 5.3. For (b), the set C C their lengths to obtain is open, and every point of C Chas an open interval about it where Ic is identically 0; this proves the continuity at points of C c . To have continuity of Ic at a point xo of C , we would need Ic > 1 /2 on some interval about xo , and this would mean that Ic equals 1 on that interval and hence that the interval is contained in C . But C contains no intervals of positive length. Part (c) is handled by the same argument as (b). For (d), part (c) says that Ic cannot be redefined on a Lebesgue measurable set of measure 0 so as to be continuous except on a set of measure 0. Theorem 3.29 says that no f obtained by redefining Ic on a set of Lebesgue measure 0 can be Riemann integrable. On the other hand, Ic is measurable, being the indicator function of a compact set, and hence it is Lebesgue integrable.
nLl
12. Argue for indicator functions and then simple functions. Then pass to the limit to handle nonnegative functions. 13. Let D be the diagonal. Let B be the set of all subsets of A x A containing only countably many members of D. This is a 0-ring, and it contains all rectangles of the form A x A', A x B , and B x A with A , A', and BC in A. If Bc denotes the set of complements of members of B, then Proposition 5.37 shows that C = B U Bc is a a-algebra, and certainly C contains all rectangles AC x A'' with A and A' in A. Therefore C = A x A. If the set D is in A x A, either D or Dc must be in 8, and neither is the case. 14. Toprove that R is measurable, one first proves the assertion for simple functions
1 0 and then passes to the limit. For the rest Fubini's Theorem gives
Chapter V
60 1
15. This is proved in the same way as Proposition 5.52a. 16. The measure space is the unit interval with Lebesgue measure, and each f, is an indicator function. The set of which f, is the indicator function is the subset of EX between a k and a k written modulo 1, i.e., the set of fractional parts of each of these rational numbers. The divergence of the series forces these sets to cycle through the unit interval infinitely often, and thus f,(x) is 1 infinitely often and 0 infinitely often. 17. Frv~rlLllc clchrlilivr~ul EMN,we see UNEMN = X arid ERN = 0.The sets E M N are increasing as a function of N , and their complements are decreasing with empty intersection. Corollary 5.3 produces an integer C(M) such that p(E&,C(M))i~ / 2 Put ~ .E = U ME&,C(M).Then p ( E ) ic by Proposition 5.lg. If c' > 0 is given, we are to produce K such that I fk(x) - f (x)l < c' for all k > K and all x in E c . Choose Mo with l/Mo < c'. The integer K will be EM,,-(M),x is in EMO,C(Mo) in particular. Then C(Mo). Since x is in Ec = I fk(x) f (x)I < l/Mo < t' fork 1 C(Mo). 18. In (a), we may take the set of integration to be X. Let S be the set of measure 0 on which any of f, and f is infinite, and redefine all the functions to be 0 on S. Given c 0 , choose S 0 by Corollary 5.24 such that p ( F ) < S implies f, g d p < t . Let E be as in Egoroff's Theorem for the number 6 . Problem 10 shows that limJE, f, d p = JEC f d p , the limit existing. Also, f, dpI 5 If, 1 d p 5 J, g d p < F for all n, and similarly for f . Hence lirn sup,, J, f, d p - J, f d ~ r5 26. Since c is arbitrary, the result follows. In (h), consider the measure g d p and the serlilence of functions {h,} with h , ( x ) = fn(x)/g(x) when g(x) > 0 , h,(x) = 0 when g(x) = 0. After checking that h, is meas~u-able,use Corollary 5.28 and apply (a). The constant that bounds the sequence is 1. 19. By Fatou's Lemma, J,, f d p 5 liminf, JE, f, d p . Subtracting this from f d p = 1imSx f, d p gives JE f d p > limsup, f, d p . Another application of Fatou's Lemma gives liminf, lE fn d p > lE f d p , and we conclude that lim inf, f, d p = lim sup, f, d p = f d p , from which the result follows. 20. Let t 0 be given. Choose S > 0 by Corollary 5.24 such that p ( F ) 5 S implies f d p 5 t. Then choose E with p ( E ) < S such that f, converges to f uniformly off E. Problem 10 shows that there is an N such that If, - f 1 d p < c for n N , and Problem 19 shows that there is an N' such that I,f, - f 1 d p 5 f n d p + J E f d p 5 2JE f d p + ~ f o r n1 N'. Sincep(E) < S,2JE f d p + t 5 3.5. Then n > max{N, N'} implies J, I f , - f 1 d p < 46.
xi=l
nN
n,
-
I IE I
Sx
IE
I
SE
SE
IE
SE
SF
SE, lE
SE
21. Suppose that limJ, f, d p = J, f d p . Given c > 0 , choose S > 0 by Corollary 5.24 such that p ( E ) < S implies JE f d p < t . Then choose N such that N-' f d p t ) < 6 . For any n, the convergence of f, d p to Jx f d p implies that NP({x I f n ( ~ )1 N}) 5 fn(x)>Nl fn d p 5 JSX fn d p 5 JSX f d p c if n is sufficiently large. Hence p({x I f,(x) > N}) 5 N-' f d p t) iS for large
(Ix +
1,
(1,
+
+
Hints for Solutions of Problems
602
hx
1,
n , and therefore N} f d p it . Problem 20 shows that If, - f 1 d p 5 E if n is large enough, and then also n(xj'Ni I f, - f 1 d p 5 E . SO we have
hx1
hX
hxI
Ax
f,(xj>N} fn d p 5 1 f,(x)>N} If, - f 1 d p + f,(xj>N} f d p 5 E + c = 26 for n large, say n > N'. By increasing N and taking the integrability of f 1 , . . . , fWtpl into account, we can achieve the inequality fn(xj'Ni fn d p .L 22c for all n. Conversely suppose that { f,} is unifornlly integrable. Given t > 0 , find the N of uniform integrability, put S = t / N , and choose Eo by Egoroff's Theorem such that p(E0) < S and f, converges u ~ ~ i f o r ~off n l yEo. Then lit11J f,d p = f d p by
hx
Problem 10. Fatou's Lemma gives
SEof
.
Eo
1Eo.
d p .L liminf JEo f, d p , and we have
Ax
The first term on the right side is 5 fn(xj>Nl f, d p , which is 5 E by uniform integrability, and the second term on the right side is 5 NS = 6 because p ( E O )< S and f, ( x ) 5 N on the set of integration. Thus lim sup JEo f, d p 5 26, and we obtain
I
SE0 d ~ I
limsupn JEo f, d~ f 5 4c. 22. In the notation of Section 5, K = 24 = A since A is now assumed to be a 0-algebra. Thus p , ( E ) = SUPK,*, K c E p ( K ) and p * ( E ) = infutn, U Z E p ( U ) . Take a scyucrue o l scls K, irl A will1 lirrl (K,) = p, ( E ) ;willloul loss or gcrlcralily, the sets K, may be assumed increasing. Then we may take K to be the union of the K,. The construction of U is similar. The set K is any member of A such that p ( K ) is the supremum of p ( S ) for all S in A with S & E . Then p ( K c ) is the infimum of all p ( S C )= p ( X ) - p ( S ) for all SCin A with SC 2 E C . A similar argument applies to U and U C .The result is that U c c Ec & K c , p * ( E C j= p ( U C j ,and p"ECj = p ( K C j . 23. Lemma 5.33 gives p ( A n K ) 5 p* ( A n E ) , p ( A Cn K ) 5 p , ( A Cn E ) , and p * ( E ) = p ( K ) = p ( A n K ) + p ( A C n K ) i p * ( A n E ) + p , ( A C n E )i p,(E),from which we obtain p, ( An E ) = p ( A n K ) . The argument that p* ( An E ) = p ( A n U ) is similar. 24. The right side of the definition of cr depends only on A n E and B n E C ,and hence a is well defined. The formulas
and [ ( An E ) U ( B n EC)IC= ( A Cn E ) U ( B Cn E C )show that the sets in question form a a-algebraC. 'l'aking A = B shows that A C, and taking A = X and B = 0 shows that E is in C. Therefore B & C, and 0 is defined on all of B. The complete additivity of a results from the complete additivity of each of the four terms in the definition of a . Specifically let a disjoint sequence ( A , n E ) U ( B , n E ) be given, and let A = U , A, and B = U , B,. We have p,(A, n E ) = p ( A , n K ) , and the sets A, n K are disjoint; thus p,(An n E ) = p * ( A n E ) . The next term
Chapter V
603
is p*(A, n E ) = p(An n U), and the sets A, n U may not be disjoint. However, p * ( ~ ,n E ) p * ( ~ n , E ) = p ( ~ ,n U ) P(A, n U ) = p ( ~ ,n A, n E)+ p((Al u A2) n E ) , and p(A, n A, n U) = p*(A, n A, n E) = p*( 0 ) = 0. Thus the term with p*(A, n E ) behaves in additive fashion. Consequently p*(A n E ) > p ' ( ( U i = l ~ ~E ) )= n p'(Akn E ) . Letting n tend to infinity gives p'(An E) > EEl p*(Akn E ) . The reverse inequality follows from Lemma 5.33a, and thus the term p* (A,nE) is completely additive. The terms with the B,'s are handled similarly, and a is completely additive. Taking A = X and B = 0 , we see immediately that the formula for a (E) is as asserted. To prove that a ( A ) = p(A) for A in A, we take A = B. Then we see that U ( A ) = t p ( n~ K ) (1 - t ) p ( n ~ U) t p ( n~ K ~ ) (1 - t ) p ( n ~ uC)= tp(A) + (1 - t)p(A) = ~ ( ' 4 ) .
+
+
+
+
+
25. Each member of the countable set has only countably many ordinals less than it, and the countable union of countable sets is countable. Therefore some member of Q is not accounted for and is an upper bound for the countable set. Application of (iii) completes the argument. 27. For(a),ifU, UandV, f V,thenU,UV, f U U V a n d U , n V , f U n V . Similar remarks apply to K,. Then the assertion follows by transfinite induction. For (b), we know that K, is closed under finite unions and intersections, and we readily see that the complement of any set occurs at most one step later. Now let an increasing sequence of sets in various K,'s be given. Say that U, is in Ken. Problem 25 shows that there is a countable ordinal a 0 that is > all the a,, and then all the U, are in K,, . The union is then in U,o+l and necessarily in K,,+l. Hence the union is in the union of the K,'s. So the union of the K,'s is a a-algebra and must contain B. All the set-theoretic operations take place within B, and thus the union must actually equal B. 28. Proposition 5.2 and Corollary 5.3 show that the value of the measure is determined on all the new sets that are constructed in terms of the values on the previous sets. Problem 27 shows that all members of B are obtained by the construction, and hence p is completely determined on B. 29. Same argument as for Problem 27b. 30. At every stage of taking limits, we have closure under addition and scalar multiplication. Pointwise decreasing limits produce the indicator functions of finite unions of closed intervals, and pointwise increasing limits of them produce the indicator functions of arbitrary finite unions of intervals. Since the constants are present as continuous functions, we have the indicator function of every elementary set and its complement. These sets form an algebra. Going through the construction of Problem 27, we obtain the indicator function of every Borel set. Since we have closure under addition and scalar multiplication at each step, we obtain all simple functions. One increasing limit gives us all nonnegative Borel measurable functions,
604
Hints for Solutions of Problems
and a subtraction (allowable without another passage to the limit) gives us all Bore1 measurable functions. 32. To see that C has the same cardinality as R,we can make an identification of the disjoint union of R and a countable set. To do so, we write C as the members of [O, 11 whose base-3 expansions involve no 1's. For each such infinite sequence of 0's and 2's, we change all the 2's to 1's and regard the result as the base-2 expansion of some real number. This identification is onto [0, 11, and it is one-one if we discard from C all the sequences of 0's and 2's that end in all 2's. The standard Cantor set has Lebesgue measure 0, and thus any subset of it is Lebesgue measurable of measure 0. The cardinality of this set of subsets is the same as the cardinality of the set of subsets of R. In Section A.10 of the appendix, it is shown for any set S that the cardinality of S is less than the cardinality of the set of all subsets of S. So the cardinality of the set of Lebesgue measurable sets is at least that of the set of all subsets of R. 33. Since C c is open, any member x of C c has the property that Ici is 0 on some open interval about x. Thus Ic1 is continuous at x. Since C has Lebesgue measure 0 , Ic, is continuous except on a Lebesgue measurable set of measure 0. Theorem 3.29 shows that Ic, is Riemann integrable. Hence the cardinality of the set of Riemann integrable functions is at least that of the set of all subsets of R. 35. If F is the given filter, form the partially ordered set consisting of all filters on X containing F , with inclusion as the partial ordering. The union of the members of a chain is readily verified to be an upper bound for the chain, and Zorn's Lemma produces a maximal element. This maximal element is readily seen to be an ultrafilter.
36. The filter in question consists of all supersets of finite intersections of members of C. 37-38. Suppose that F i s an ultrafilter, A U B is in F , A is not in F , and B is not in F . Let Ffconsist of all sets in F a n d all sets B n F with F in F . Since B is not in F , Ffproperly contains F . Since F i s an ultrafilter, F' must fail to be a filter. On the other hand, by inspection, Ffsatisfies properties (i) and (ii) in the definition of filter. We conclude that 0 is in F f ,hence that there is a set F in Fwith B n F = 0 . Since .F satisfies (ii), the set ( A U B ) n F = ( A n F ) U ( B n F ) = A n F is in F . By (i), A i s in .F, contradiction. Conversely suppose that F is a filter such that either A or AC is in F for each subset A of X. If F i s not rnaxirnal, let B be a set that lies in sorrle filter Ffproperly containing Fwhile B is not itself in F . By hypothesis, Bc is in F a n d hence is in F f . But then B f' BC = 0 lies in F f ,in contradiction to (iii).
39. If an ultrafilter F is given, define p ( E ) = 1 if E is in F and define p (E) = 0 otherwise. Then p is defined on all subsets, and we have p ( 0 ) = 0 and p ( X ) = 1. If E and E f are disjoint, we are to show that
Chapter VI
605
If E U E f is not in F , then all terms in the displayed equation are 0 since F i s closed under supersets. If E U E f is in F , then Problem 37 shows that E or E f is in F, on the other hand, they cannot both be in F because F is closed under finite intersections and the empty set is not in F . Thus exactly one term on the left side of the displayed equation is 1, and the right side is 1. This proves additivity. Conversely if an additive set function p is given on all subsets of X that takes only the values 0 and 1 and is not the 0 set function, let Fconsist of the sets E for which p ( E ) = 1. It is immediate that (i) and (iii) hold in the definition of filter. For (ii), let E andEfbeinF. ThenEUEfisinF. Hencep(EnEf)+l = p(E)+p(Ef) = 1+1, and p ( E n E') = 1. Hence F i s closed under finite intersections and (ii) holds. Thus F i s a filter. If A is given, we have 1 = p(X) = p(A) p(AC),and hence exactly one of the sets A and A" is in F.By Problem 38, F i s an ultrafilter. The statement that complete additivity is equivalent to closure of the ultrafilter under countable intersections is a routine consequence of Corollary 5.3.
+
40. This follows from Problems 34d and 35. 41. Let S, be the set of all integers > n. Since S1 = X, S1 is in the ultrafilter. Since the ultrafilter is not trivial, {n} is not in it, and thus Problem 37 shows that S, is in it if Snpl is in it. Hence S, is in the ultrafilter for all n. The countable intersection S, is empty, and the empty set is not in any filter. Hence the ultrafilter is not closed under countable intersections. Corollary 5.3 shows that the corresponding set function is not completely additive.
n,
43. The proof of Proposition 5.26 shows that the result holds for simple functions
> 0. I f f > 0 and g > 0, choose the standard sequences t, and u, of simple functions increasing to f and g . These converge uniformly. Hence so does the sum s, = t, +u, . The same argument as for Problem 10 shows that lim .Y, d p = 1, (f g) d p , lim 1, t, d p = f d p , and liml, u, d p = g d p . Thus the result holds for bounded nonnegative f and g. The passage to general bounded f and g is achieved as in Proposition 5.26.
1,
lE
1,
+
Chapter VI
+
1. In additive notation, the sets E t for t in T are disjoint, and their countable union is s'. Since Lebesgue measure is translation invariant, these sets all have the samc mcasurc c . 'I'hcn complctc additivity givcs c oo = 2n,which is impossiblc.
2. Parts (b) and (c) are easy. For (a), expand the Jacobian determinant J ( N ) in cofactors about the first row, obtaining two terms-one each ti-om the first two entries of the first row. The first term is cos Q1 times a determinant of size N - 1 whose first column has a common factor of r cos Q1 and whose second column has a common factor of sin Q1,the remaining part of the determinant being J ( N - 1); thus the first term gives (r cos2 Ql sin 01) J ( N - 1). The second term is -(-r sin Q1) times a determinant of size N - 1 whose first column has a common factor of sin 01
606
Hints for Solutions of Problems
and whose second column has a common factor of r sin 01, the remaining part of the determinant being J ( N - 1);thus the second term gives (r sin3 Q 1J) ( N - 1). Adding the two terms gives J ( N ) = (r sin Q1) J ( N - I ) , and an induction readily proves the formula.
3. Replace f in Theorem 6.32 by f o L , and use p = L-'. Since p f ( x ) = L-' for each x , the result follows.
4. In the result of Problem 3, use L (x) = yx and replace f ( z ) by f ( z ) / I det z l N . Then the left side in Problem 3 is f ( y x ) / l det(yx)lNd x , while the right side is I d e t ~ ~ ~ ~ S ~ ~ f ( x ) / ~ d e t x lThus ~ d x .IdetylcNjMNf(yx)/Idet(x)lNdx =
lMN
I det L Icl
JMN f ( x ) / I detx I d x , and the problem reduces to showing that det L = (det Y ) ~One . way of doing this is to verify that this formula is true if y is the matrix of an elementary row operation and then to multiply the results. But a faster way is to let x l , . . . , x , be the columns of x , so that L ( x l , . . . , x,) = ( y x l , . . . , yx,). Then L as a matrix is given in block diagonal form by a copy of y in each block. Hence det L = (det Y ) ~In. a little more detail, the matrix of L is being formed relative to the following basis of M N : if Eij is the N-by-N matrix with 1 in the (i, j)thentry and 0 elsewhere, the basis is E l l , E21, . . . , E N 1 ,El2, . . . , E N N .
5. For (a), we have, for n # 0,
Let us call these terms I and 1 1 . Since If (x) I 5 CIXI" for 1x1 5 1,
For I I , we use integration by parts and take into account that the terms at n and -n cancel by periodicity:
Let us call the terms on the right I I I and I V . Since f ( x ) 5 C l x 1" for Ix 1 5 1,
The derivation of the formula for I I , when applied to following value for I V:
IV = -
4n {ff(l)ecinllnl n
-
f f (
-
i)e+inllnl
1
-
f' instead of f , gives the
1 dh51x15z
f
(x)ecinx
dx.
Chapter VI
Let us call the terms on the right V and V I . Since
607
If
'(x) 1 5
Since f "(x) is bounded for 1 5 Ix 1 5 n , we can write 0 < Ix I 5 n ,in view of the assumption on f ". Therefore
+
+
If
c lx la-'
for Ix 5 1,
"(x)l 5 ~ ' l 1"-"or x
+
Since2nIcnI 5 III IIIII IVI IVIl,weobtain Ic,I 5 ~ / l n l l + " . For (b),the uniform convergence follows by applying the Weierstrass M-test, and the limit is f as a consequence of the uniqueness theorem. In (c), aproof is called for. The crux of the matter is to show, under the assumption that f is real valued, that the variation V, off on [s,11,which gets larger as sdecreases to 0, is bounded. If xo < . . . ix, is a partition P of [s,11, then
with ni-1 < < ni. With s fixed, the right side is a Riemalllln suln for the boullded function xapl on [E, 11 and is 5 the corresponding upper sum U (P,xa-' ASwe insert points into the partition, the left sides increase and the right sides decrease to the 1 limit J, xapl d x = a p l(1 - s a ) . Hence V, 5 C a p l ( l - sa), and sup,,o V, 5 C / a .
I
6. The distribution function F of p must have F ( 6 )- F ( a )equal to 0 or 1 for all a and 6 . If c is the supremum of the x's for which there exists y > x with F (x) < F ( y ) , then F has to be k on (-00, c) and k 1 on [c, +oo) for the value of k that makes F(0) = 0. Hence p is a point mass at c with p({c}) = 1.
+
7. Let K be compact, and let f and g both be equal to the members of a sequence { f,} of continuous functions of compact support decreasing to the indicator function IKof K . Applying the identity to f, and passing to the limit, we obtain v(K) = v ( K ) ~Thus . v(K) is 0 or 1 for each compact set. By regularity v takes on only the values 0 and 1 on Bore1 sets. Then the argument (but not the statement) of Problem 6 applies, and there is some c with v equal to a point mass at c with v({c]) = 1.
8. In (a), if the complement of the set in question is not dense, it omits an open set. However, nonempty open sets have positive measure. IE(x - t) dt] dp(x). The inner integral equals the Lebesgue In (b), form measure of E for every x since Lebesgue measure is invariant under translations and the map t tt t . Hence the iterated integral is 0. The integral in the other order is O = JR1 'E(X - t ) d ~ . L ( x ) ] d = t JRl ' E + ~ ( ~ ) ~ P . L dt ( X= ) ] l R l piE -tt) d t , and Corollary 5.23 shows that p ( E t) is 0 almost everywhere. In (c), the same computation applies, and p ( E t ) is 0 almost everywhere. Under the assumption that limt,o p ( E t) exists, the limit must be 0, by (a).
lRl [lR,
[IR,
+
+
[IEl +
Hints for Solutions of Problems
608
+
9. Write l / l x l as a sum F1 F,, where F1 is l / l x l for 1x1 i1 and is 0 for 1x1 2 1 . Then F,(x - y ) d p ( y ) is bounded by p(JR3), and it is enough to handle the contribution from F1. For that we have F1 ( X - y ) d p ( y ) ]d x =
I,,
IR3[lR3
Ax,,1
[SR3 ~ i ( x - y d ) x ] d p ( y ) = I R 3 [ S R 3 FI ( x )d x ] d p ( y ) = p ( R 3 ) IXI-' d x , and this is finite in R 3 . Hence the inner integral Fl ( x - y ) d p ( y ) is finite almost everywhere. lR3
I,,
10. We proceed by induction on n , the case n = 1 following since finite sets have Lebesgue measure 0. Assume the result i n n - 1 variables, and let P ( x l , . . . , x,) $ 0 be given. Let E be the set where P = 0. This is closed, hence Bore1 measurable in R'" Fix ( x i , . . . , x:) with P ( x ; , . . . , x:) # 0 . The polynomial in one variable R ( x ) = P ( x i , . . . , xAPl,x ) is not identically 0, being nonzero at x = x: , and hence it vanishes only finitely often, say for x in the finite set F. Fix x' $ F . Then the polynomial Q ( x l , . . . , x n P l ) = P ( x l , . . . , xn-l, x') in n - 1 variables is not identically 0, being nonzero at ( x i , . . . , xAPl),and its set Ex, of zeros has measure 0 by inductive hypothesis. If m , denotes n-dimensional Lebesgue measure, then Fubini's Theoreill applied to IE gives
On the right side the first term is 0 since the 1-dimensional measure of F is 0, while the second term is 0 since the integrand is 0. Thus m ( E ) = 0.
12. In Cartesian coordinates we obtain l N ,hence 1. In spherical coordinates we obtain Q N P 1 r N p 1 e p n rd2r . Putting n r 2 = s shows that l F r N - 1 e - n r 2dr = [ ; ( ~ / n ) ( ~ - ~ ) /I 2n ~ ed s- ~= ~2 X - ~ / ~ F ( N Hence / ~ )Q. N P 1= 2 n N l 2 / r ( N / 2 ) .
zLl
13. Part (a) is carried out by showing by induction on k that xi = 1 - n:=, ( 1 - u i ) . The case k = n is the desired result. In (b), let 0 < ui < 1 for all i . Then xi r 0 for all i , and (a) makes it clear that Cy="=lj < 1. Therefore (o carries I into S. Define u = F ( x ) by the formula in (b). If all xi > 0 and C;=,xi < 1, then certainly ui > 0. Also, x , < 1 implies
/
~j=,
zi~f
xi < 1 x j , so that ui = xi (1 \-,.) < 1. Therefore F carries S into I. To complete the proof, we show that G o p is the identity on I and (o o F i s the identity on S. For @ o q ,we pass from u to x to v . Thus we start with vi, substitute the x's, use the inductive version of (a) to substitute the ZL'S, and then sort matters out to see that vi = u i . For q o @, we pass from x to u to y . Then we start with yi and substitute the u's to obtain yi = ( 1 - u i ) ) u i . To substitute for the u's in terms of the x's, i-1 we use the inductive version of (a) in the form yl = 1 - n l = l (1 - u l ) . This (1 - u i ) ) u i = (1 - ~fii~ ~ ~ ) x-~ / ( lx l ) . Then an induction on i gives shows that yi = x i , and hence (o o P i s the identity on S.
(ni~:
(nji:
zii:
Chapter VI
609
In (c), routine computation shows that p t ( u ) is lower triangular with diagonal entries 1, (1 - u l ) , ( 1 - u l ) ( l - u 2 ) ,. . . , ( 1 - u l ) . . . (1 - u n P l ) ,andhence the determinant is the product of these diagonal entries. Similarly G t ( x )is lower triangular withdiagonalentries 1, (1 - x l ) - ' , (1 - x l -x2)-I,. . . , (1 - x l - x 2 - . . . -x,-l)-l, and its determinant is the product of these diagonal entries.
14. The change of variables in Problem 13 gives
The right side is the product of 1-dimensional integrals of the kind treated in Problem 11. Substitution of the values from that problem leads to the desired result.
15. The ~llo~lotoilicity makes possible the estinlate of unifor~lcoilvergence, and the continuity then makes the limit continuous. A continuous function is determined by its values 011 a dense set, and C c is dense. 1
1
16. Foreachn,Fn(x) = 1 - F n ( l - x ) . Thuslo F n ( x j d x = 1-lo F n ( l - x j d x = 1 1 1 - So F, ( x ) d x and loFn ( x )d x = Passing to the limit and using uniform or 1 1 dominated convergence, we obtain So F ( x ) d x = 2..
i.
18. Use Proposition 6.47. Then u is harmonic by Problem 14 at the end of Chapter 111. 19. Since Pr has L' norm 1, the inequality Ilu(r, . ) l i p .5 I f 11, follows from Minkowski's inequality for integrals. For the limiting behavior as r increases to 1, wc cxtcnd f pcriodically and writc
the second step following since for integrals, we obtain
l l ~ ( '1 ~, f
llp
&z :I
Pr d q = 1. Applying Minkowski's inequality
&I :,
I
Pr(v>I l f (0 -
-
f ( 0 )Ilp,e
610
Hints for Solutions of Problems
since Pr
> 0 . The integration on the right is broken into two sets, S1 = ( - 6 , s )
and
S2 = [ - n , -61 U [ S , n ] ,and the integral is
5 sup vtS1
Ilf
(Q - P ) -
f
(Q)II,,,
+2 I f II,
sup Pr(P). vtsz
Let c > 0 be given. If 6 is sufficiently small, Proposition 6.16 shows that the first term is < c . With 6 fixed, we can then choose r close enough to 1 to make the second term ic.
20. For (a), we argue as in Problem 19, taking S1 and S2 to be as in that solution. Then
and the uniform convergence follows. car Inleino,where the c, For (b), the Poisson integral of f is of the form Czpm are the Fourier coefficients of f . Any other harmonic function in the disk is of the form 00 cLr lneine.Suppose this tends uniformly to f as r increases to 1. Then the difference is a series CL',d , r " e i " O that converges uniformly to O. 'I'hen the integral of the product of this series and epiketends to 0 . Interchanging integral and sum, we see that d k r k tends to 0 for each k. Therefore dk = 0 for each k. In (c) since Pr is even,
xn=p,
and thus J::, ( P , * f ) ( O ) g ( Q )dQ =
I::, ( p r * g)(O)f ( Q ) dB. Therefore
By the previous problem the right side tends to 0 as r increases to 1, and the weak-star convergence follows.
Chapter VI
61 1
21. Let Mf and M, be upper bounds for If 1 and Igl on [a,61. Then
+
22. Let us rewrite the given equation f ( x ) = f ( a ) g l ( x ) - g2(x) as gz(x) f ( x )- f ( a ) - gl ( x ) . If xi 3 xi-1, then subtraction of the values at x - xi and at x = xi-1 gives g2(xi) - g2(xi-l) f (xi) - f ( x i p l )= g l ( x i )- g l ( ~ i - l ) . If f (xi) - f (xi-1) 1 0, then f (xi) - f (xi-1) i gl(xi) - gl(xi-I) because g2 is monotone; if f (xi) - f (xi-1) < 0, then 0 5 g l ( x i ) - g l ( x i P l )because gl is + monotone. Therefore ( f (xi) - f ( x i P l ) ) 5 gl ( x i ) - g1 ( x i P l ) .Summing on i for + a partition of [a,x ] gives ( f (xi) - f ( x i p l ) ) 5 g l ( x ) - gl(a). If we take the supremum of the left side and recall that gl(a) > 0, we obtain V + ( f ) ( x )i gl ( x ) - gl ( a ) < gl ( x ) . Starting similarly from gl ( x ) - f ( x ) f ( a ) = g z ( x )and arguing in the same way, we obtain V (f ) ( x ) < gz ( x ) - g2 ( a ) 5 g2 ( x ) .
+
+
xy=l
+
23. Suppose that V +( f ) and V - ( f ) are both discontinuous at some x. Then V + ( f ) ( x - ) c < V + ( f ) ( x + and ) V P ( f ) ( x - ) c < V P ( f ) ( x +for ) some c > 0. Define for y < x ,
+
+
for y
V + ( f > ( y->6
= x,
fory > x,
and define g2(y) similarly except that V replaces V f . Then gl and g2 are both nonnegative, and gl - g2 = V + ( f )- V ( f )= f - f ( a ) . If gl and g2 are shown to be monotone, Then Problem 22 leads to the contradiction gl ( y ) < V +( f )( y ) for y > x , and we conclude that V + (f ) and V (f ) could not have been discontinuous. In proving monotonicity for gl ,it is necessary to make comparisons only of x with otherpoints y. Let h > 0. Forpoints y > x,we havegl(x h ) = V + ( f ) ( x +h ) - t > V + ( f ) ( x +-) c > V + ( f ) ( x p=) g l ( x ) . Forpoints y < x , we have g l ( x - h ) = V + (f ) ( x - I z ) 5 V + (f ) ( x - ) - gl ( x ) . Monotonicity for gz is proved in the same way.
+
24. The proof is similar in spirit to the proof of Proposition 6.54.
+ i)-lr-l
25. For f , let y, = (n + i ) - l r p l , so that f (y,,) is +(n if n is even and is -(n + i)-'n-l if n is odd. Compute the sum of the absolute values of the difference of values of f at yN, yN-l, . . . , y1 and see that this is unbounded as a function of N. The function g has a bounded derivative (even though the derivative is discontinuous), and this is enough to imply bounded variation.
Hints for Solutions of Problems
612
Chapter VII 1. If g ( a k )= g ( b k ) ,then ak would have to be in E. For the second part an example is g ( x ) = x on [0, 11; there is only one interval ( a k ,b k ) ,and it is (0, 1 ) . 2. No. Corollary 7.4 applied to ZE shows for almost all x that the quotient m ( E f' ( x - h , x h ) ) / m ( ( x- h , x h ) ) has to tend to 0 or 1 as h decreases to 0
+
+
3. We may work on a bounded interval I. Let c > 0 be given. If x is in E , then ( F ( x h ) - F ( x )1 5 c wheiiever I h 1 5 6, for some 6, depeiidiiig on x. For each such x , fix a positive number r, with r, < ;s,. Associate the set B(r,; x ) to x . Then
+
I~L-'
Applying Wiener's Covering Lemma, we can find disjoint sets B(r,; x i ) with E UE B (5rXz; xi). Then
Since I is fixed and c is arbitrary, p ( E ) = 0.
4. If F is the function in question, F - F (0) is the distribution function of some Stieltjes measure p containing no point masses. Proposition 7.8 shows that p ( E C )= 0 for some countable set E. Since ~ ( { p = } )0 for each point p , p ( E ) = 0 by complete additivity. 'l'hus /L = 0, and P must be constant. 5. For (a), the construction shows that F f ( x )= 0 for all x E C C .Then Proposition 7.8 allows us to conclude that p is singular. For (b), let F, be the nth constructed approximation to F (using straight-line interpolations), and let f, be its derivative (defined except on a finite set and put equal to 0 there). The function f, is a multiple c, of the indicator function of the subset C, of [0, 11 that remains after the first n steps of the construction, and also m ( C n )= ( 1 - r k ) . Since F,(x) = Sox f, ( t )dt for all x , we have 1 = F n ( l ) =
1,
1
n;_,
n;=,
(n;=,
c, Zcn ( t )dt = c, ( 1 - r k ) .Therefore f, = ( 1 - rk))-'zCn. Put f = P-' Zc . The functions f, converge pointwise to f , and they are uniformly bounded by the constant function P - l . By dominated convergence, F ( x ) = f ( t ) d t for 0 < x < 1 . Therefore F is the distribution function of the measure f ( t ) d t . 6 . Let E be the second described set. The complement of E has measure 0 by Corollary 7.4. Fix x in E , and let 6 > 0 be given. Choose a rational r suc,h that Ir - f(x)I < t . Forh > 0 ,
1;
The second term on the right side equals Ir - f (x)l < c , and the first term tends to If ( x ) - r ) I ic since x is in E . A similar argument applies if h < 0 .
Chapter VII
613
7 . Part (a) is routine, and part (b) follows by adapting part o f the argument for Theorem 6.48. In (c), the assumption that x is in the Lebesgue set implies that 0 , where cx( . ) is a function that If ( x - t ) - f ( x )I dt 5 hcx ( h ) for h tends to 0 as h decreases to 0. For each o f the described pieces o f the integral Kn(t)lf ( x - t ) - f ( x ) d t , we use one o f the two estimates in (a), specifically the estimate K N ( ~<) N 1 for the piece with It1 < 1/N and the estimate K N ( ~5) c / ( N t 2 ) for all the other pieces. The piece for 1/N then contributes 5 (N 1) If ( x - t ) - f (x)l dt 5 2 c , ( l / N ) , the piece for 2 k - 1 / ~5 It1 < 2 k l contributes ~ I5 i 2 k - 1 1 ~ ) -~22 k - l l N 5 1 t 1 1 2 k l N I f (X - t ) - f ix)I dt < 5 (2" ' / N ) 2 ( 2 " / ~ ) c x ( 2 k /=~ )4 . 2 " c X ( 2 " / N ) ,and finally the piece for Np114 5 It 5 n contributes 5 5 N ~ ~ ~ ~ ~ - If~ ( X/ ~- ,t )~-, f~ ( x ) ,I , dt 5 If ( x ) l ) .The sum o f the estimates is 5 N'I22n(ll f
htI,,
+
+
~,,,,,, -
+
and this tends to 0 as h decreases to 0 . (The use o f the shells with 2~~ is a device that appears frequently in Zygmund's Trigonometric Series and may be regarded as a kind o f manual integration by parts.) 8. Since p is singular, find a Bore1 set E with p ( E ) = 0 and m ( E c ) = 0. Let E > 0 be given. By regularity o f m + p , choose an open set U containing E such t h a t ( m + p ) ( U - E ) < E . T h e n p ( U ) ( p ( U - E ) + p ( E ) = p ( U - E ) < t,and m ( U c ) < m ( E C )= 0 . 9. About each x in U , there is some Six) such that ( x - h , x + h ) U for h 5 Six). Then v ( ( x - h , x + h ) ) = 0 for h 5 S i x ) , and the limit o f this is 0 as h decreases to 0 . 1 1 . Since U is open and p2(U) = 0 , Problem 9 gives lim ( 2 h ) - l p 2 ( ( x- h , x h&O
+h))= 0 +
for all x in U . Since m ( U c ) = 0 , litnhIn(2h)-'p~j(x- h , x h ) ) = o for almost every x in R1. The measure p1 has p1 ( R 1 )= p ( U ) < E ,and Problem 10 shows that m { x l i r n s u p p ~ ( ( x - h , x + h ) )> 6 ) h&O
I
5 m { x suppl ( ( x - h , x h >O
+ h ) ) > 5 ) < 5 ~ 1 ( ~ ' 1
12. It is enough to handle the case that p vanishes outside some interval and hence has p ( R 1 )finite. Combining the estimates for p1 and p2 gives
614
Hints for Solutions of Problems
I
+
Since t is arbitrary, nz { x lim suphJop ( ( x - h , x I z ) ) > t }= 0. Taking the union for = l l n , we conclude that the set where limsuphJop ( ( x - h , x h ) ) > 0 has measure 0. To get the better conclusion, the main step is to obtain a bound 10c/< for the maximal function formed from the supremum of v ( ( x ,x h ) ) or v ( ( x h, x ) ) . The proof of Corollary 6.40 shows how to derive this from Problem 10.
<
+
+
-
Chapter VIII 1. Let F be the Fourier transform as defined in the text. In each part of the problem, a can be computed by relating matters to the known facts about F , and /I can be corniputed directly fro111tlie definitions and Fubini's Theorern. In (a), we have f ( y ) f (x)epix.yd y f (x)ep2nix.(y/(2n)) dy = F f (yl(2n)). To obtain f ( x ) = ac f*(y)eix.yd y , we want f ( x ) = ac F f (y/(2n))eix.Yd y = ( 2 7 ~ ).Ff~ ( y~f ) e i x ' ( 2 "d~y' )f = ( 2 ~ ) ~( xa) .f With f * g ( x ) = P f ( x - t ) g ( t ) d t , wehave f T g ( y ) = 3j f (x-t)g(t)epiX.yd t d x = j3 JJ f (x-t)g(t)epiX.yd x d t = /3 f (x)g(t)ep"'"+"'.~ d x d t = j 3 f * ( ( y ) ~ ( yThus ) . ac = ( 2 ~and) j3 ~= 1~. In (b), we find similarly that f ( y ) = ( 2 n ) ~ ~( F y / f( 2 n ) ) ,and we are led to ( 2 ~( 2 )n )~~ ~ = a c1. So ac = 1 . Also, B ( 2 ~=)( ~2 ~and) ~= (~ 2 ~ ) ~ . In (c), we find similarly that ac = ( 2 n ) p N / 2and 3j = ( 2 ~ ) ~ This / ~normalization . has the property that ac and /3 are both 1 if d x is replaced by d ~ / ( 2 n ) ~ ~ % r o u ~ h o u t . A
I
=S
=l
1
l
SJ
A
2 . Thiq iq an operation called "polari7ationn in linear algebra, and it will he cg, explained further in Chapter XII. Application of the Plancherel formula to f f , and cg gives I l f + csll: = IlF(f>+ cF(g)ll:, l l f 11: = l l - ~ ( f ) l l :and , llcsll: = ~ ~ C . F ( ~We ) I :expand . the first one in terms of the inner product and subtract the other two to obtain
+
+
Then c ( f , + ~ ( fgl2, = c(.F( f ) , F ( g ) I 2 c ( . F ( f ) , .F(g))2. Taking c = 1 gives 2 R e ( f , g ) 2 = 2 R e ( F ( f ) ,F ( T ( ~ )whereas )~, taking c = i gives 2 I m ( f , g)2 = 2 Im(F( f ) , F ( T ( ~ )The ) ~ .result follows. 3. For any f in L ' , we have Q , * (Q,, * f ) = P,+,, * f because the Fourier transforms are equal. Also, ( Q , Q,,) f = Q , (Q,, f ) since we have finiteness when the functions are replaced by their absolute values. Moreover, the functions Q,* Q,, and P,+,I are bounded andcontinuous. Letting f run through an approximate identity formed with respect to dilations and applying Theorem 6.20c, we see that Q , * Q,, ( x ) = PC+,, ( x ) for all x .
*
*
*
4 . Since P, is even, J,, (P,* f ) ( x ) g ( x )d x = JRN Pt ( x - y ) f ( y ) g ( x )d x d y = JR.v JRN Pt (Y
SRN
*
-
Pt ( x - y ) f ( y ) g ( x )d y d x = x ) f ( y ) g ( x )d x d y , and thus
Chapter VIII
615
By Theorem 8 . 1 9 the ~ right side tends to 0 as t decreases to 0, and (a) follows. For (b), part (a) shows for each g with llglll 5 1 that f (x)g(x) dxl = lim,~o JRN(t: * f)(x)g(x)dxI. Since 4 * f (x)g(x)dxl 5 llt: * f ll,Ilglll 5 IIPt f ll,,wehave
*
I
I SEN
I lRN
whenever Ilg 11 5 1. For any c > 0 with 11 f 1 , - 6 > 0 , let S, be the set where I f 1 is 1 IIfI1,-t.Thenm(S,) >O.TakeEtobeanysubsetofS,withO< m(E) < +ca, and let g (x) be m (Ej-'fO/1 f (x) 1 on E and zero elsewhere. This function has llglll 5 1- hen f g d x l = JRN f g d x = m ( ~ 1 - lJE If dx 2 IIfII, - t. fgdxI 5liminf,~oIIP,*fII,. Sincecisarbitrary,IfII, 5 HenceIIfII,-c 5 liminftJo 11 Pt f 11,. On the other hand, Theorem 8.19b shows that 11 Pt f 1 , 5 I I f 1 ,. So wehave I l f I , 5 liminft~ollPt * f l , 5 l i m s ~ l ~IIPt~ *~ fo l , 5 Ilf 1 ,. Equality must hold throughout, and (b) is thereby proved.
I h.v
*
*
5. In (a), the set function is a measure by Corollary 5.27. It has p(JRN) equal to p (JRNjW2(JRN)and is therefore a Bore1 measure. If p 1 = f dx and ~ 1 . 2= p , then
In (b), we start with an indicator function and compute that
Then we pass to simple functions 2 0, use monotone convergence, and finally take lirlear ~urrlbirlaliurlsLu gel JRN g(x y) d p l (x) dp2(y) = JRN g d(p1 * ~1.2). In (c), we actually have 11 P, * p I = P ( J R ~ for ) every t 0 by Fubini's Theorem. Part (d) is handled in the same way as Problem 4a. First one shows that (P, p)(x)g(x) dx = JRN (P, g)(x) dp(x) for g in c,,,(JR~). The resulting estimate is JRN[(Pt g)(x) - g(x)l ~ P ( x ) I5 IIPt g - gllsUpw(JRN),and then (a) follows from Theorem 8.19d.
+
SRN
S,,
*
*
*
*
Hints for Solutions of Problems
616
6. Part (a) follows from the same argument as for Proposition 8.la. In (b), the measure S with S({O})= 1 and S ( R N - {O}) = 0 has T i y ) = 1 for all y. In (c), we use the result of Problem 5b with g ( x ) = e-2nix'hnd get J e-2"ix'"(pl * p 2 ) ( x ) = JJ e-2ni(x+y).t d p1 ( x ) d p 2 ( y ) = z ( t ) Z ( t ) . In (d), let q ( x ) = PI ( x ) . Then =0 implies @ @ *, = O for every s > O. Since p, * p is a function, Corollary 8.5 gives q, p = 0 for every s > 0. By Problem 5d, q, p converges weak-star to p against Cc,,(RN). Therefore J,, g d p = 0 for every g in Cc,,(RN), and Corollary 6.3 shows that p = 0.
*
*
7. This is the same kind of approximation argument as was done in Corollary 6.17. ~-X 8. We calculate that b(xi - xj)tiF = J e- ~ , L ~ I .j () tXi t j dp.L(t) = 2 ( C,,(e-2nit.xz(i)(e-2nit.xj t j ) ) d p ( t ) = j" e-2nit.xj(j d p ( t ) > 0.
xi,
J
I xj
,
Z J
xi,j
I
9. For the set {O},the condition is that F ( 0 ) l t ll2 1 0 for all tl;thus F ( 0 ) 1 0. For theset{x,O},theconditioni~thatF(O)1~~1~+~(x)~1~+~(-x)~2~+~(0)1~~~~ 10. Taking t1= t2= 1 shows that F i x ) + F ( - x ) is real; taking t1= i and t2= 1 shows that i( F (x) - F ( - x ) ) is real. Therefore F (x) + F ( - x ) = F (x) + F (-x) and F ( x ) - F ( - x ) = -F ( x ) + F ( - x ) . Adding we obtain F ( - x ) = F ( x ) . Hence -- ~ i x ) < 1 2- F ( x ) t1<2 < F ( O ) ( I < I I ~1<212). If F i x ) # 0 , we put q0(. t ) 11. Part (a) follows from the boundedness of F obtained in Problem 9. In (b), every g in Cc,,(RN) satisfies 0 5 JJ Fu(x - y ) g ( x ) g ( y )d x d y =
+
xi,J xi,]
-
xi,J
A
-
J ( F O * g ) i x ) g i x ) = J Fo * giy)?i;(y)dy=J F?liy)?i;(y)rndy=J F?l(y)l?iy)l2dy. For (c), if f is in L 2 , we can approximate f as closely as we like by a member g of CC,,iRN). Then fo1a2 = fol.Vf)12 2foRe(.T(f>i?- F ( f ) ) + fo Ig-F( f ) 12. We integrate and use the resulting f o r m ~ ~tol acompare fo d y with fo IF(f ) l 2 d y . By the Schwarz inequality and the Plancherel formula, the absolute value of the difference of these is 5 211foIlsupIlfl1211g - f l 2 + IIfollsupIg - f 11;. Since j" fo1a2 d y is > 0, it follows that j" fo IF(f)12 d y > 0 for all f in L2. Since F(f ) is an arbitrary L' function and fo is continuous, we conclude that fo is z 0. The integrability in (d) is immediate from Lemma 8.7, and the f o r m ~ ~ l afo d y = F ( 0 ) follows from the Fourier inversion formula.
+
la2
1
1
12. Lel 8 , be a scyuer1L.e clscreasir~gtu 0 , Is1 @ i r ~Prublsrri 11 be tl~e~ U I I L L ~ U I I , and write F, for the function F @ . Then Problem 1l d shows that p, = ( y ) d y is a finite Borel measure with p, ( R N )= F, (0,) = F ( 0 ) . The Helly-Bray Theorem applies and produces a subsequence of { p , } convergent to a finite Borel measure p weak-star against Cc,,(RN). We shall prove that F i x ) = e2"ix'?d p ( y ) , i.e., that v with v ( E ) = p ( - E ) is the desired measure. For each n , the Fourier inversion formula gives F, ( x ) = 1e2nix.yE( y ) d y = [ e2nix.?d p , ( y ) . Since F, ( x ) + F ( x )pointwise, the result would follow if we could e-n&.'x l 2
1
Chapter VIII
617
1
1
say that the weak-star convergence implies that e2nix.yd p n ( y ) + e2xix.yd p ( y ) . However, e2"ix'~is not compactly supported, and an additional argument is needed. First we extend the weak-star convergence so that it applies to continuous functions vanishing at infinity. I f f is such a function, we can find a sequence { f k } in C,,,(RN) converging to f uniformly. 'l'hen
Choose k to make 11 fk f llsup small. With k fixed, choose n to make the middle term small. Then the right side-is small since the numbers p , ( R N ) are bounded. This is not quite good enough by itself because e 2 " i x ' ~does not vanish at infinity. However, averages of it by L' functions (i.e., Fourier transforms of L' functions) vanish at infinity, and that will be enough for us. Define F # ( x ) = S e 2 " i x - ~ d p ( y )We . prove that F # ( x ) = F ( x ) for all x . It is enough to prove that F#$ d x = F $ d x for all $ in L' . Define lCrV ( y ) = e 2 n i x ' y $ ( ~d)x . The multiplication formula (for (.)" instead of (.)-) and the Riemann-Lebesgue Lemma give -
1
The right side equals all y.
1
1
1@ F d y by dominated convergence since IFn (y) 1 5 IF ( y )1 for
13. Part (a) is easy. In (b), if x is a character, then x ( x ) = x ( g x ) = x ( g )Z, x ( x ). Thus C, x ( x ) = 0 if there is some g with x (g) # 1, i.e., if x is not trivial. If x and X' are distinct characters, then is not trivial, and therefore x (x)X'(x) = 0. The orthogonality implies the linear independence. In (c), the element 1 of J, has order rn under the group operation of addition. Thus each character x of Jm must have x ( 1 ) equal to an mth root of unity. Since 1 generates Jm , x ( 1 ) determines x . Thus the listed characters are the only ones. In (d), any tuple ( n l , . . . , n,) with 0 < n, < rn, for 1 < j < r defines a character r 11 by ( k l , . . . , k,) H n , = , ( ~ ~ : There ) ~ j .are mJ distinct characters in this list, and they are linearly independent by (b). Since dim L 2 ( ~=) m,, these characters form a vector-space basis.
x,
x,
x,
X7
n;_l
n(rZl
) a consequence of Problem 13d, 14. Since the characters form a basis of L ~ ( G as we have f ( t ) = Ex,c X xf l ( t ) for some constants c,!. Multiply by and sum over
3
t to get T ( x ) = C,,
xrcxixl(t)x(t>. The orthogonality in Problem 13b shows that A
this equation simplifies to f ( x ) = cx Ct I ( t )l2 = I G Icx.
Hints for Solutions of Problems
618
15. f
!XI
=
CtGc f ( t ) x ( t )= CiFGIH ChGHf
!t+h)X
=
C;,,;, ~ ( i (;I )i
A
=F(i).
:<
16. The characters of G are the ones with ~ ~ (=1 ) for 0 5 n < m. Such a character is trivial on H if and only if x n ( q ) = 1, i.e., if and only if
<:
xy=jl
xy=jl(
17. fi<$+k) = f (i)
(i)
+
+
+
+
+
Chapter IX 1. Let r- = q / p , and let r-' be the dual index. Regxd If 1" as a product If IP . 1,and apply Holder's inequality with 1 f l p to be raised to the r power and 1 to be raised to the r f power. Compare with Problem 3 below, which is a more complicated version of the same thing. 2. The inequality is routine if any of the indices is ca. Otherwise, we have
3. Let us say that 11 ,fillip 5 C . Let t > 0 be given. By Egoroff's Theorem, find E with p ( E ) < such that fn tends to f uniformly on E C .Application of Holder's inequality with the exponent r = p / q and dual index r f = p / ( p - q ) to J, I ,fn14. 1 d p ( P P ~ ) / ( P ~< ) C~(E)(P-~)/< (P~) gives I I ~ ~ I E I I ~ ~ f n ~ ~ ( ~ ' ~l 'd d~ ~) ) ~ ~-~ ( / ~ C E ( P - ~ ) / ( P ~ ) Meanwhile, . we have
(IE
Chapter IX
619
The first term on the right is 5 ~ t ' ~ ~ q ) / ' J and ' q ) ,so is the third term, by Fatou's Lemma. The middle term tends to 0 as n tends to infinity because of the uniform convergence. Thus limsup, 11 f, - f l l q 5 ~ c c ( P ~ ~ ) / (Since P ~ ) c. is arbitrary, limsup,
Ilf,
-
f ll,
= 0.
4. L 1 is 0 , and Lm consists of the constant functions. All the constant functions give Lhc barrlc linear lurl~tiurlal"11 L1 be~aubetl~eirllcgral u l ~ l prudud c u l any constant function and the 0 function is 0. 5. Put P f = { f ( x ) > 0 } , N f = { f(x) < 01, and Z' = { f ( x ) = O}. If E is any measurable subset of Z f ,then X = P U N with P = P f U E and N = N f U (Z' - E ) is a Hahn decomposition. All other Hahn decompositions are obtained by adjusting P and N by taking the symmetric difference of P and of N with any set of p measure 0. 6. In (a), let X be the positive integers, and let the algebra consist of all finite subsets and their complements; let v of a finite set be the number of elements in the set, and let v of the complement of a finite set F be -v(F). In (b),use the same X and algebra, define v ( { 2 k } ) = 2 ~ ~and v({2k - 1 ) ) = - 2 ~ and ~ , extend v to be completely additive. In (c), let X = [0, 11,let the 0-algebra consist of the Bore1 sets, and take v to be Lebesgue measure and y to be counting measure. 7. Since P, has L norm 1, the inequality Il u (r, . ) 11, II f 11, follows from Minkowski's inequality for integrals. For the limiting behavior as r increases to 1, we extend f periodically and write
the second step following since for integrals, we obtain I l ~ ( r '1 , -f
llp
& I:n
Pr d q = 1. Applying Minkowski's inequality
% I::, Pr (PIIlf (Q
I
-
P ) - f ( Q )Ilp,e
since P, ? 0. The integration on the right is broken into two sets, S1 = (-8, 6) and S2 = [ - n , -61 U [S, n ] , and the integral is
5 sup vF~TI
Ilf (0 - v ) - f (Q)IIp,,+ 2 I f II,
sup P,(p). 9F ,T2
Let t > 0 be given. If S is sufficiently small, Proposition 9.1 1 shows that the first term is < c . With S fixed, we can then choose r close enough to 1 to make the second term < t .
8. Let p be the dual index to pf . Put r / R = r f in Problem 13 at the end of Chapter IV, so that
Hints for Solutions of Problems
620
for r f i1. Take a sequence of R's increasing to 1, and let {R,} be a subsequence such that { fR,} converges weak-star in LP' relative to LP. Let the limit be f . For each 0 andrf, P,,(Q . ) is in LP, and the equality u ( r f R n Q, ) = J:nn fR, (q)Pr,(Q -q) d q thus gives u ( r f ,8 ) = [:nnf (q)P,i (0 - p) d q , which is the desired result. -
&
&
9. If v is a measure with 0 5 v 5 p , then v ( { n } )= 0 for every n , and hence ~({integers})= 0. So v = 0. 10. Let p be given on the space X, and consider the set S of all completely additive v with 0 5 v 5 p. This contains 0 and hence is nonempty. Order S by saying that vl 5 v2 if vl ( E ) 5 v2 ( E )for all E . If we are given a chain {v,}, let C = sup, v, ( X ) . This is 5 p ( X ) and hence is finite. Choose a sequence {v,,} from the chain with v, ( X ) monotone increasing with limit C. If rn < n , let us see that v , 5 van. Since the v,'s form a chain, the only way this can fail is to have v , ( E ) > van( E ) for some E and also vam( E c ) ) van( E c ) . But then vam( X ) v,,, ( X ) by additivity, and this contradicts the fact that v , ( X ) is monotone increasing. So rn in implies vam 5 van. Define vo ( E ) = limk v,, ( E ) . Corollary 1.14 shows that vo is completely additive, and certainly vo < p. So vo is an upper bound for the chain. Zom's Lemma therefore shows that S has a maximal element v . Write a = p - v . This is bounded nonnegative additive as a result of the construction. If there were a completely additive h such that 0 5 h 5 a , then v h would contradict the construction of v from Zorn's Lemma. Thus a is purely finitely additive. 11. It is enough to prove that p is completely additive. If the contrary is the case, then there exists an increasing sequence of sets En with union E in the algebra such that the monotone increasing sequence [ p ( E , ) ] does not have limit p ( E ) . Since p is nonnegative additive, p(E,) < p ( E ) for all n. Thus lim, p(E,) ip ( E ) . Since v - p is nonnegative additive, v - p similarly has limn( v - p ) ( E n )5 ( v - p ) ( E ). Adding, we obtain limn v ( E n ) iv ( E ) ,in contradiction to the complete additivity of v .
+
+
+
12. Suppose p is nonnegative bounded additive. Let p = vl pl = v2 p2 with vl and v2 nonnegative completely additive and with pl and p2 nonnegative purely finitely additive. Then vl - v2 = p2 - pl. Let v+ - v be the Jordan decomposition of vl - ~ 2 Since . vl - v2 is completely additive, so are v+ and v p . The equality v+ - v = p2 - p1 and the minimality of the Jordan decomposition together imply that 0 5 v+ 5 p2 and 0 5 v 5 pl. Problem 11 then shows that v+ = v = 0. Hence vl - = 0, vl = v2, and pl = p2. 13. Let R 1
=
I x J be centered at ( x , y ) . Then & J I R I f ( ~ , ~ ) l d = ~d~
m ~ l [ & [ r l f ( ~ ' ~ ) ~ d u ] 5d u& J 1 f l ! u , y ) d u supremum over R gives f * * ( x y, ) 5 f 2 ( x , y).
A;
=
f2(x.y).
Tking the
14. ~ J I f * * ( x , y ) l P d x d
[l
Chapter IX
Corollary 9.21 a second time, we see that this is 5 A?
f@llf
621
1 [ I If ( x ,y)IPdy] d x
=
11; 15. This is done in the style o f Corollary 6.39.
16. Let Q 1 2 0 be adecreasing C 1 fu~lctio~ion [0, 11 with Q i ( 0 ) = 0 , Q 1 ( l )= 1, and @; ( 1 ) = - 1. Define @o ( x ) on [0, 11 to be @ l ( x ) / ( n ( l+ x 2 ) )on [O, 11 and to (lx1 ) has the required property. be l / ( n x ( l x 2 ) )on [ I ,+GO).Then @ ( x ) =
+
17. suP,,o I($, * f )(x)I 5 S U P , > O ( ~ $ ~I * If l>(x>5 SUP,>^(@, * If l ) ( x ) ,and then sup,,^ I ($, * f ) ( x )1 < Cf * ( x ) by Corollary 6.42. Since l,, $ ( x ) dx = 0 , the last part o f the proof o f Corollary 6.42 shows that lim,,o($, * f ) ( x ) = 0 a.e. for f in L (IR1).I f f is in LW(IR1)and a bounded interval is specified, we can write f as the sum o f an L' function carried on that interval and an Lm function vanishing on that interval. The L' part is handled by the previous case, and the L" part is handled on that bounded interval by Theorem 6 . 2 0 ~ .
+
18. We use the fact that Q, = h, $,, where $ is integrable with integral 0 . Since h , * f and $, * f are in LP, SO is Q, * f . Convolution by an L' function such as P, is continuous on LP by Proposition 9.10. With all limits being taken in LP as s f J 0 , we have P, * ( H f ) = P, * (lim(h,, * f )) = lim P, * (h,, * f ) = lim P, * (Q,! * f * f ) = lim P, * (Q,, * f ) - (lim P, * @,) * f . The second term on the right side is 0 . I f we think o f P, as in L' and Q,, as in LP',then we have P C * ( Q c f * f ) = (PC * Q c f ) * f = Qcf+e* f = (Pet * QE>*f = PC! * (QE * f ) . Thus lim P, * (Q,, * f ) = lim P,, * ( Q , * f ) = Q, * f , and we conclude that Pe*(Hf)=Q,*f. $,I
19. suP,>o I(he * f ) ( x > l 5 suP,,o ~ ( Q E * f ) ( x ) I + SUP,>o I(@, * f ) ( x ) I 5 ) C f * ( x ) ,the last inequality sup,,o I(P, * H f ) ( x ) I C f *(x) 5 C f ( H f ) * ( x + following from Corollary 6.42 for P,. Let 1 < p < GO. Then it follows from Corollary 9.21 that I sup,,o Ih, * f 1 1 1 , i Cp(llHf 1 1 , Il f lip),and we conclude from Theorem 9 . 2 3 ~that 11 sup,,^ Ih, * f 1 11, 5 Dp 11 f 1 ,. Lemma 9.24 shows that limEJo(h,* f ) ( x ) = f ( x ) everywhere i f f is in a certain dense subspace o f LP, and it follows as in Problem 15 that lim,$o(h, * f )(x) = f ( x ) almost everywhere i f f is arbitrary in LP.
+
+
20. Imitating the proof o f parts (a) and (b) o f FejCr's Theorem (Theorem 6.48), we readily prove that K , * f + f in LP, where K , is the FejCr kernel. Therefore finite linear combinations o f the exponentials are dense in LP([-n, n ] ) .For each such linear combination f o f exponentials, we have S, f = f for all sufficientlylarge n , and hence S, f + f in LP for a dense subset o f LP. Using the given estimate on IIS, f l i p and the convergence o f S, f on the dense set, we argue as in the proof o f Theorem 9.23b to deduce convergence for all f in LP. 2sin(n I
21. Let F,(t) = for 0 < It 1 < n , and extend F, periodically. Then $ F,(t) = sin(n i ) t = (sin i t ) ~ , ( t ) .Since ( t / 2 ) / sin i t = 1 t $ ( t ) with $ ( t ) bounded above and below by positive constants on [-n, n ] ,we see that
+
;)I
+
622
Hints for Solutions of Problems
D, (t)
-
Fn(t) =
[
-
1 F (t) = 2 (t) s i n
+t
. Then the functions
+ i ) t have D,
- Fn = @, and I $, 11 bounded. By inspection, 2sin(n+i)t 1 F, -En equals the function that is for It 1 < and is 0 for 5 It 5 n . These functions are 5 2(a i j for It I < and are 0 otherwise; so their L' norms are bounded. This proves that Dn - En = q, with IIqnII 5 C for some C. If ITnf llp 5 BpIIf Ilp,thenwehave II&f l l p = IIDn*f lip = IIEn*f +%*f llp I
@,(t) = 2$(t) sin(n
&
+
IIEn
* f ll, + I l ~ n* f l l ,
I BpIIf
Ilp
+ II%IIl Ilf
+i
llp,andwecantake Ap = Bp + C .
22. We have 2i sin(n jt = ei(n+i)" ee(n+i)
+
If we regard f as continued periodically to the interval [-3n, 3x1 and we put f equal to 0 outside that interval, then
where g(y) = -inf (y)eei(nf$ ) ( Y ) on [-3n, 3x1. With A, as the constant from Theorem 9.23, Theorem 9.23 gives
f , and the desired estimate for T, f follows. We get a similar estimate for T,(~)
1
23. Define a signed measure v on 8 by v(B) = f d p . Then v is absolutely continuous with respect to the restriction of p to B, and the Radon-Nikodym Theorem yields a function 8 measurable with respect to B such that v(B) = 8 d p for all B in B. This function g is E [ f IB]. Uniqueness is built into the uniqueness aspect of the Radon-Nikodym Theorem.
l,
24. For those n's such that p ( X , ) # 0, E[f lB] may be defined to be equal everywhere on X, to the constant p ( x n j e l f d p . For definiteness, E[f IB] may be defined to be 0 on each X, with ~c (X,) = 0.
Sx,
25. The function f satisfies the defining properties (i) and (ii) of E [ f 1 A].
26. In (a), we identify E[E[f 1 C] as E[f IC]. It is measurable withrespect toC and hence satisfies (i) toward being E [ f IC]. Any C E C has J , E [ E [f B ] I C] d p = E [f B ] d p . In turn this equals f d p since C is in B. Hence E [ E [ f I ] I C] satisfies (ii) toward being E [f l q .
1,
-/"
Chapter X
623
In (b), we identify E [ f IB] + E [ g as E [f +g 1 B ] . It is measurable with respect to B and hence satisfies (i). For (ii), each B in 8 has [,(E[ f lB] E [ g l B ] )d p = E [ f lB1 ' P + JB E[glBl ' P = JB f ' P + JB g d p = J B ( f + g ) d ~ . In (c), it is enough to handle f > 0, and then it is enough to handle g > 0. If g = IB with B E B , then we shall identify l B E 1f lB] as E l f l B I W j . Certainly l B E lf lBj satisfies (i). For (ii), each B' in B has J,, I R E [f If31 d y = E [f lB] d y = l,,,, f d y = l,, I , f d p . This handles g equal to an indicator function. Part (b) allows us to handle g equal to a sinlple function, and ~llonotoneconvergence allows us to handle g equal to any nonnegative integrable function. (For this last conclusion oneneeds to use that f > O implies E [ f In] > 0, hut this is built into the constnlction via the Radon-Nikodym Theorem.) In (d), the important thing is that X is a set in B . Then (ii) and (c) successively give Jx f E [ s l B l d~ = Jx E [ f E[glBl I 1 1 d P = E [ f lBlE[gl@ d ~ The . right side is symmetric in f and g, and hence the left side is also.
+
SB
J,
Chapter X 1. For (a), the diagonal A = { ( y ,y ) E Y x Y} is a closed subset of Y x Y since Y is Hausdorff, and the functioii I; : X + Y x Y given by I;( x ) = ( f ( x ) ,g ( x ) )is continuous. Therefore F - ' ( A ) is closed. 2. The argument is the same as for Problem 18 in Chapter 11. 3. We argue as in the proof of Theorem 2.53. Taking complements, we see that it is enough to prove that the intersection of countably many open dense sets is nonempty. Suppose that U, is open and dense for n > 1. Let xl be in U 1 . Since U1 is open, local compactness and regularity together allow us to find an open neighborhood B1 of xl with B:' compact and BY' U 1 . We construct inductively points x, and open neighborhoods B, of them such that B, G U 1n . . .nU, and B,C' G B n l . Suppose B, with n > 1 has been constructed. Since U,+l is dense and B, is nonempty and open, Un+l n B, is not empty. Let x,+l be a point in U,+1 n B,. Since U,+1 n B, is open, we can find an open neighborhood B,+l of x,+l in U,+l such that B : ~ G Un+1n B,. Then B,+l has the required properties, and the inductive construction is complete. The sets B,C' have the finite-intersection property, and they are closed subsets of B Y , which is compact. By Proposition 10.11 their intersection is nonempty. Let x be in the intersection. For any integer N , the inequality n > N implies that x, is in BN+1. Thus x is in B:+~ c BN C U 1 n . . . n U M .Since N is arbitrary, x is in U,.
c
nZ1
4. Let Y be a locally compact dense subset of the Hausdorff space X . If y is in Y , let N be a relatively open neighborhood of y such that N G K with K compact in Y. Since N is relatively open, N = U n Y for some open U in X. It will be proved that N = U , so that each point of Y has an X open neighborhood, and then Y will be open. The set K is compact in X and must be closed since X is Hausdorff. The points of U n K are in Y since K G Y, and hence U n K G U n Y = N. Consider
624
Hints for Solutions of Problems
a point x of the open set U - K . Suppose x is not in Y . Then x is a limit point of Y since Y is dense. Hence the open neighborhood U - K of y contains apoint y' of Y. Then y' is in U n Y = N g K and cannot be in U - K , contradiction. We conclude thatxisinY.ThenxisinUnY=N,andU=N.
5. First consider any continuous function f : Y* + [O, 11 with f (y,) = 0. The set of y's with f ( y ) > 1/12 is open and contains y,, thus is a compact subset of Y and must be finite. Hence the set of y's with f ( y ) = 0 has a countable complement. If Z is normal, apply Urysohn's Lemma to A and B, obtaining a continuous F : Z + [O, 11 with f ( A ) = 1 and f (B) = 0 . Enumerate the members of X as x l , x2, . . . . For fixed rL, f ( y ) = F ( x n ,y ) is coi~tii~uous fro111 Y * to [0, 11 and is 0 at y,. Thus F (x,, y ) > 0 only on a countable set S, of y 's, and F (x,, y) > 0 for some a at most on the countable set S = Ur=lSn. If yo is not in S , then x H F ( x , yo) is continuous from X * to [0, 11, is 0 for every x other than x,, and is 1 at x,. This contradicts the continuity, and we conclude that Z is not normal. 6. If E is an infinite set with no limit point, then E is closed and each x in E is relatively open. Hence each x has an open set U, in X with U, n E = { x } . These open sets and Ec cover X , and there is no finite subcover. Thus X compact implies that each infinite subset has a limit point.
-'
8. Part (a) follows from Problem 7b and Proposition 10.34. For (b), f (-00, a ) is 0 if a < 0, is R - {0} if 0 < a < 1, and is R if a > 1; hence it is open in every case. Part (d) follows from (a). For (e), there exists an upper semicontinuous function > f ( x ) ,namely the constant function everywhere equal to sup If ( x )1 . Then (d) shows that thc pointwisc infimum ovcr all uppcr scmicontinuous functions > f ( x ) meets the conditions on f .
+
9. For (a), we have Q ( x ) = f ( x ) (- f ) - ( x ) . Both terms on the right are upper semicontinuous, and the sum is upper semicontinuous by Problem 8c. For (c), f - ( x ) 5 f ( x ) 5 f P ( x ) = Q f ( x ) f - ( x ) . If Qf = 0, then f- = f = f shows that f is continuous with respect to all sets { x < b} and all sets { x > a } . Hence f ( a , b ) is open for every a and b , and f is continuous with respect to the metric = f and topology. Conversely if f is continuous, then the definition makes f ( - f ) = - f . Therefore f = f- = f and Qf = f - f- = O .
+
-'
10. In (a), that subset of pairs is ( A x A) U ( B x B ) U {(.w,.w) I x E X } , which is the union of three closed sets and hence is closed. In (b), let X be a Hausdorff space that is not normal, and take A and B to be disjoint closed sets that cannot be separated by open sets. 11. In (a), q p l q ( x ) = p2(({x} x X j n R ) , where p2 is the projection to the second coordinate of X x X . Since { x }is closed and X is compact and R is closed, ( { x } x X ) f' R is compact. Then q p ' q ( x ) is compact, hence closed, being the continuous image of a compact set. In (b), we have p2((UCx X ) n R ) = { y E X I (x,y ) E R for some x E U c }= { y E X I q p l q ( y )n U c # 01 = { y E X I q p l q ( y ) g U l C= V C .Since U is open,
Chapter X
625
the left side is closed, by the same considerations as in (a). Thus V Cis closed, and V is open. In (c), let q ( x ) and q ( y ) be distinct points of X / -. By (a), the disjoint subsets 4 p 1 4 ( ~and ) q p l q ( y )are closed. Since X isnormal, find disjoint open sets U1 and U2 containing q p l q( x ) and q p l q ( y ) ,respectively. Let Vl = { z E X I q p l q ( z ) U 1 } and V2 = {z E X I q p l q ( z ) U2}. These are disjoint sets, and they are open by because qp1q(V1)= V l is open, and similarly (b). Then q(V1) is open in X / q (V?)is open. The sets q ( V I) and q (V?)are disjoint because q p l q ()~=~ V I and q p 1 q ( V 2 )= V2 are disjoint. Thus q ( V l ) and q(V2) are the required open sets separating q ( x ) and q ( y ) . For (d), part (c) shows that X / is Hausdorff, and therefore its compact subsets are closed. The image of any closed set is X is the image of a compact set, hence is compact and must be closed. For (e), the answer is "no," and part (f) supplies a counterexample. For (f), the function p : X + S' is continuous, and Proposition 10.38a produces a continuous function po : X / + S 1 such that p = po o q , where q is the quotient map. Then po is continuous and one-one from a compact space onto a Hausdorff space and must be a homeomorphism.
'
-
-
-
12-13. The proofs are the same as in Section 11.8. 14. This is proved in the same way as in Problems 13 and 1 1 in Chapter IT.
15. For (a), call the relation -. This is certainly reflexive and symmetric. For transitivity let x y and y z. Then x and y lie in a connected set E, and y and z lie in a connected set F . The sets E and F have y in common, and Problem 13a shows that E U F is connected. Thus x 2 . Part (b) is immediate from Problem 13b. For (c), let x be given, and let U be a connected neighborhood of x . Then U lies in the component of x . Thus the component of x is a neighborhood of each of its points and is therefore open.
-
-
-
16. Form the class C of all functions F as described, including the empty function, and order the class by inclusion; for the purposes of the ordering, each function is to be regarded as a set of ordered pairs. The class C is nonempty since the empty function is in it. If we have a chain in C, we form the union F of the functions in the chain. We show that F is an upper bound for the chain. To do so, we need to see that the indicated sets cover X. Thus let x E X be given. Only finitely many sets U in Z4 contain n, by assunlptiun. Say these ale U1, . . . , U,. If one of these fails to be in the domain of F, then x lies in UvCu,vgdomain(F)V , and x is covered. Thus all u l U1, . . . , U,, rriay be assurricd Lo be i r ~Lllc dvrrrair~or F . Ea~11Uj is i r ~h e dvrrrair~ of some function Fj in the chain, and all of them are in the domain of the largest of the Fj's, say Fo. Since x is not in UvCu,vgdomain(F)V , it is not in the larger union Uvtu,v+domain(Fo)V . Thus it must be in UUCaomainFo(U). Since FO(U)" g U for each U , x must lie in some Fo(U,). Then x lies in F (U,) , and F is an upper bound for the chain. By Zorn's Lemma let F be a maximal element in C. To complete the argument, we show that every set inZ4lies in domain(F). Suppose that Uo is a set in Uthat is not
Hints for Solutions of Problems
626
in domain(F). Let Uf be the union of all F ( U ) for U in domain(F) and all V other than Uo that are not in domain(F). Since F is in C, Uf U Uo = X. Hence Ufc is a closed subset of the open set Uo. Since X is normal, we can find an open set W such that Ufc & W & wC'& U. If we define F ( U ) = W, then we succeed in enlarging the domain of F, in contradiction to the maximality of F. Hence every member of U lies in domain(F), as asserted. 17. Form the open sets Vu as in the previous problem. For each U in U , apply Urysohn's Lemma to find a continuous function gu : X -. [O, 11 with gu equal to 1 on Vu and equal to 0 on Uc. The open cover {Vu} is locally finite since Z4 is locally finite. Therefore g = CUtLlgu is a continuous function on X. Since gu is positive on VU and the sets VU cover X , g is everywhere positive. Therefore the functions f U = g u / g have the required properties.
18. If co = 0, take Fo = 0. If co # 0 , apply Urysohn's Lemma to obtain a continuous function h with values in [0, 11 that is 1 on Po and is 0 on No, and then put Fo = g2 ~ O-h ~1 C O . 19. On Po n C , go is > co/3 and Fo is co/3. Therefore go - Fo is > 0 and Similarly on No f' C , go - Fo is < 0 and > -2co/3. Elsewhere on C , go and Fo are both between -co/3 and co/3, and hence Igo - Fol 5 2co/3. Thus 1x0 - FoI ( 2co/3 everywhere on C. The function FI is continuous from X into R, has I F1I ( (fcO),and takes a value cl ( 5(ico) on {x E C I g l (x) 1 c1/3} and the value -cl on {x E C I gl (x) < -cl/3}.
< 2co/3.
5
4
20. Iteration produces continuous functions Fn : X + R with I Fn(x)l 5 (+)nco for all x in X and (x) ~ i ( x ) l5 ($)nco for all x in C. Let F ( x ) = Fn(x). The series converges uniformly on X by the estimate on Fn(x) and the Weierstrass M test, and Proposition 10.30 shows that F is continuous on X. If we let a tend to infinity in the estimate on f (x) Fi (x), we see that F and f agree on C . Finally for x in X ,
~ ~ i i
If
~yid
IF(x)l I
x n =O
x
IFn(x)I I $(;)"co = co = sup I f (y)l n =O Yt C
Thus IF I and I f 1 have the same supremum. 21. Every open interval is in the base and hence is open. The closed interval {a ( x ( b} is the complement of the open set {x ia} U {b ix} and is therefore closed. 22. Let a < b be given. If there exists a c with a < c < b, then the open sets {x < c } and {c < x} separate a and 6; otherwise the open sets {x < 6) and { a < x} separate them. Hence X is Hausdorff. Let a and a closed set F be given with a not in F. Since Fc is a neighborhood of a, there exists a basic open set B containing a that is disjoint from F . If B has some element larger than a, let d be such an element; otherwise let d be undefined.
Chapter X
627
If B has some element smaller than a , let c be such an element; otherwise let c be undefined. If c and d are both defined, then F {x < c } U {d < x}, while a is in { c < x < d}. If c is not defined but d is defined, then F {x < a} U {d < x}, while a is in B n {x < d}. If d is not defined but c is defined, we argue symmetrically. If neither c nor d is defined, then B = {a} is open and closed; hence Bc and B are the required open sets separating F and a .
23. Suppose that any nonempty set with an upper bound has a least upper bound, and let B be a set with a lower bound. We are to produce a greatest lower bound. Let F be the set of all lower bounds for E. This is nonempty, and all elements of F are 5 e , where e is an elenlent of E . So F has an upper bound. Let c be a least upper bound. We show that c is a greatest lower bound for E. If c is not a lower bound for E , then E has some e with e 5 c , e # c , i.e., with e < c. All f in F have f 5 e < c. So e is a smaller upper bound for F, contradiction. Thus c is a lower bound for E . If there is some greater lower bound, say d , then c < d 5 e for all e in E. This implies that d is in F, and hence c is not an upper bound for F.
24. In (a), suppose that Y is nonempty closed and has an upper bound and a lower bound. We are to prove that Y is compact. It is enough to handle a set Y = [a, b]. Let an open cover U of Y be given, and suppose there is no finite subcover. Let E be the set of all x in [a, b] such that some finite subcollection fromUcovers [a, x]. Then a is in E. Since E is nonempty and has b as an upper bound, the order completeness shows that E has a least upper bound c. Since we are assuming that U has no finite suhcover of [n, h], E" f' [ n , h] is nonempty. This set has a lower hound, namely n , and therefore it has a greatest lower bound d. I f e i s i n E a n d f i s i n E C n [ a , b ] , t h e n e5 f . S o e 5 d,andthenc 5 d . Suppose c < d. Then c must be in E . Any x with c < x < d cannot be in E or E c , and hence there is no such x. Then a finite subclass of U that covers [a, c], together with a member of U that contains d , is a finite open subcover for [a, dl and contradicts the fact that d is not in E. Thus c = d . Now suppose that c is in E c n [a, 61. Since c = d , E has no largest element. Choose a member U of U containing c , and find a basic open neighborhood B of c contained in U . Then B f' E must contain some c' with c' < c. A finite subclass of U covers [a, c'],and U covers [c', c]. Tl~usc is in E , and we have a contradiction. We conclude that c is in E . Since c = d , Ec n [a, b] has no smallest element. Choose a member U of U containing c , and find a basic open neighborhood B of c contained in U . Then B f' (ECf' [a, b]) must contain an element c' with c < c', and then there must be some c" with c < C" < c'. A finite subclass of U that covers [a, c], together with the set U , then covers [a, c"] and shows that c" is in E . This contradicts the fact that c is an upper bound of E . In (b), let x be given in X. If a < x < b for some a and b , then [a, b] is the required compact neighborhood of x. If x is a lower bound for X and there exists b with x < 6 , then [x, b] is the required compact neighborhood. If x is an upper bound
628
Hints for Solutions of Problems
for X and there exists a with a < x , then [a, x] is the required compact neighborhood. Since X has at least two members, there are no other possibilities. So X is locally compact. 25. In (a), the sets {x < b } and {a < x} are open and disjoint, contain a and b respectively, and have union X. Thus X is disconnected. In (b), suppose that X is order complete and has no gaps. Assume, on the contrary, that U and V are disjoint nonempty open sets with union X. Say that u < v for some u in U and v in V. It will be convenient to assume that u is not the smallest element in X and u is not the largest; when this assumption is not in place, the same line of proof works except that one may below have to use basic open sets of the form (r ix ] and {x < s}, as well as {r < x < s}. Form the set S of all x E X with x 5 u and (x, u] g V. This set has u as a lower bound, and we let b be the greatest lower bound. Then u 5 b 5 v . First suppose that b is in V. Choose a basic open set (r, s) V with r < b < s; this is possible by our temporary assumption because V is open. Then (max{u, r }, u] c V. If max{u, r} < b, then max{u, r } is in S and h is not a lower hound for S ; thus b 5 maxju, r }, i.e., b = u. This is impossible since b is assumed to be in V. We conclude that b is in U . Choose a basic open set (r, s) g U with r < b < s ; again this is possible by our temporary assumption because U is open. Since there are no gaps, we can find sf with b < s' < s . Then min{u, sf}is a lower bound for S , and b cannot be the greatest lower bound unless min{v, sf} 5 b , i.e., b = v. This is impossible since b is assumed to be in U , and we have arrived at a contradiction. 26. As an ordered set, X is the same as R, and hence its order topology is the same as for R,which is connected. In its relative topology, X is disconnected, being the disjoint union of the open sets [0, 1) and [2,3). 27. The subset [0, 1) is closed, being the intersection of all sets {x I x 5 y} for y E (1, 21. Similarly (1,2] is closed. Hence they are both open, and X is disconnected. It follows immediately from the definition that there are no gaps. 28. If a nonempty subset of points (x, y) is given, let xo be the least upper bound of the x's. If no (xo, y) is in the set, then (xo, 0) is the least upper bound for the set. If some (xo, y) is in the set, let yo be the least upper bound of the y's. Then (xo, yo) is the least upper bound of the set. We conclude that X is order complete. Problem 24a then shows that X is compact. This proves the compactness in (a). There are no gaps, and Problem 25b thus proves the connectedness. For each x E [O, 11, the set {(x, y) I 0 < y < 1) is open. Thus we have an uncountable disjoint union of open sets, and X cannot be separable. Part (b) is handled in the same way.
Chapter XI 1. In (a), every compact subset of X is compact when viewed as in X*, and this gives inclusion in one direction. In the reverse direction it is enough to show that
Chapter XI
629
when U is open in X * , then U - {oo}is a Borel set in X. Since X is a-compact, we can choose an increasing sequence of compact sets K, with K, Kt+, and K, = X. Then U n K,O+l is open and bounded, hence is a Borel subset of X . The countable union of these sets is U, and hence U is a Borel set. In (b), the Borel sets of X are the countable sets and their complements. However, every subset U of X is open in X and therefore open in X * . Its complement in X * is compact and is a Borel set in X*. Thus U is a Borel set in X*.
Uzl
2. Part (a) of the previous problem shows that every open subset of X is a Borel set, and hence every continuous function is a Borel function. 3. Use the regularity to show that the conclusion holds for indicator functions and hence simple functions. Then pass to the limit. 4. Let ZE be an indicator function. Given c > 0, find by regularity a compact set L and an open set U with L c E c U and p ( U - L) < c . The compact set K will be K = (U - L)" = L n Uc. Thus consider the restriction of IE to the compact set K . L e t x b e i n K . I f x i s i n E , t h e n x i s i n L. Theset U n K = Lisarelatively open neighborhood of x , and IE is identically 1 on this. Hence the restriction of I E to K is continuous at the points of E . Similarly if x is in E C ,then x is in UC.The set LCn K = Uc is a relatively open neighborhood of x , and we argue similarly. This handles indicator functions, and the result for simple functions follows immediately. Next suppose that f is a real-valued Borel function > 0. Choose an increasing scqucncc of simplc functions s, > 0 with limit f . Lct t > 0 bc givcn, and find, by Egoroff's Theorem, a Borel set E with p ( E C ) < t such that lims,(x) = f (x) uniformly for x in E . Next find, for each T I ,a compact subset K, of X with p(K,C) -: ~ / 2 such , that s, is continuous. The set F = E n K,) has complement of measure < 26, and the restriction of every s, to F is continuous. Since {s,} converges to f uniformly on E , the restriction of f to F is continuous. Using regularity once more, we can find a compact subset KO of F such that p ( F - KO) < C . Then p (KOC)< 3 ~and , the restriction of f to KOis continuous.
IKn
5. In (a), any rotation preserves Euclidean distances and fixes the origin. Since Sab is exactly the set of points whose distance d from the origin has a < d < b , Sab is mapped to itself. Part (b) follows from the change-of-variables formula (Theorem 6.32). The determinant that enters the formula is the determinant of the matrix of the rotation and is 1. The first conclusion of (c) is what the change-ofvariables formula gives for the transformation to spherical coordinates when applied to the set Sab if we take Fubini's Theorem into account. It yields L F dx =
SSnb
(Ir2
(Sab
lab
( J. ~'d ~~ ) Lf d m ) = T~ d r ) ( I s 2 Lf d m ) . Since r 2 d~ is not zero, we can divide by it and obtain the second conclusion of (c). Part (d) is proved by setting it up to be a special case of the uniqueness in Theorem 11.l. 6. In (a), monotonicity of p gives p ( K ) < inf, p(K,). Suppose that < holds. Choose by regularity an open set U containing K such that p ( U ) < inf, p(K,). The sets K i form an open cover of the compact set UC,and there is a finite subcover. The
630
Hints for Solutions of Problems
intersection of the complements is one of the sets K,,, and it has the property that K,, U. Monotonicity then gives p(K,,) < p ( U ) , and thus inf, p(K,) < p ( U ) , contradiction. For (b), consider all compact subsets K of X for which p (K) = 1. The intersection of any two of these is again one by Lemma 11.9. If KOis the intersection of all of them, then KOis compact, and (a) shows that p (KO)= 1. If KOcontains two distinct points x and y ,find disjoint open neighborhoods U, and Uy. Then KO = (KO- U,)U (KO- Uy) exhibits Kn as the union of two proper compact sets. At least one of them must have measure 1, and then KOis shown not to be the intersection of all compact subsets of measure 1. In (c) let KObe any compact G s , and choose a decreasing sequence { f,} in C(X) 2 with limit IKO Passing to the limit from the formula f: d p = (J, f, d p ) , we obtain p(Ko) = ~ ( K o ) Thus ~ . p(Ko) is 0 or 1. By regularity, p takes only the values 0 and 1, and (b) shows that p is a point mass. For (d), apply Theorem 11.1 and obtain the regular Borel measure p corresponding to L . Then [I, has the property in (c) and must be a point mass.
l,
7. The statement for (a) is that u(r, 8 ) is the Poisson integral of a signed or complex Borel measure on the circle if and only if s ~ p ~ , , ,IIu(r, ~ 8 ) 11 is finite. The necessity is proved in the same way as in Problem 7 at the end of Chapter IX.The sufficiency is proved in the same way as in Problem 8 in that group, except that the weak-star convergence is in M(circ1e) relative to C(circ1e). For (b), expand u(r, 0) in series as in Problem 13 at the end of Chapter IV. Since u is nonnegative, the L' norm over any circle centered at the origin is just the integral, and the result of integrating in 0 is that the n = 0 term is picked out. Thus Ilu(r, @)I1 = co for every r . The condition in (a) is satisfied, and u is therefore the Poisson integral of a Borel complex measure. Examination of the proof of (a) shows that the complex measure is a measure. 8. Order topologies are always Hausdorff. Since Q* has a smallest element and a largest element, Problems 23 and 24 of Chapter X show that L2* is compact if every nonempty subset has a least upper bound. Since the ordering for Q* has the property that every nonempty subset has a least element, the existence of least upper bounds is satisfied.
9. First we prove that the intersection of any two uncountable relatively closed sets C and D is uncountable. Assume the contrary. Since C fl D is countable and the countable union of countable sets is countable, there is some countable ordinal w greater than all members of C n D. Since C and D are uncountable, we can find a sequence w < acl < p1 < ac2 < p2 < . . . such that each acj is in C and each aj is in D . 'l'he least ordinal y greater or equal to all members of the sequence is a countable ordinal and has to be a limit point of both C and D. Since C and D are closed, y is in C f' D. But C n D was supposed to have no ordinals greater than w . This contradiction shows that C n D is uncountable, and of course it is relatively closed also. Now let a sequence of uncountable relatively closed sets C, be given. By the
Chapter XI
63 1
nr=l
previous step we may assume that they are decreasing with n. If C, = C is countable, then there is some countable ordinal w greater than all members of C. Replacing C, by C , n { x > w } we may assume that the C, have empty intersection. Let a, be the least member of C,. The result is a monotone increasing sequence since the C, are decreasing. If a is the least ordinal 2 all a,, then a is a countable ordinal. It is a limit point of each C,, hence lies in each C,. The existence of a contradicts the fact that the C, have been adjusted to have empty intersection. This contradiction shows that C , is uncountable. 10. For additivity the question is whether the union of two sets that fail to meet the condition of the previous problem can meet the condition. The answer is no because the previous problem shows that the intersection of any two sets meeting the condition again meets the condition. The complete additivity is then a consequence of Corollary 5.3 and the result of the previous problem. The measure p takes on only the values 0 and 1 and yet is not a point mass because one-point sets do not satisfy the defining property for measure 1. Problem 6b therefore allows us to conclude that p is not regular. 11. Let p be a Borel measure on X , and let S be the set of all regular Borel measures v with v 5 p . This contains 0 and hence is nonempty. Order S by saying that vl 5 v2 if v , ( E ) 5 vz ( E ) for all E . If we are given a chain {v,}, let C = sup, v, (X). This is 5 p ( X ) and hence is finite. Choose a sequence {v,} from the chain with v,,(X) monotone increasing with limit C . Then v,,(E) is monotone increasing for every Borel set E , and we define v ( E ) to be its limit. The complete additivity of v follows from Corollary 1.14, and it is easy to check that v, 5 v 5 p for all a . We have to check that v is regular. Let E > 0 be given, and choose v,, with v, ( X j > v ( X ) - E . If E is given, find K and U with K c E c U , K compact, U open, and v, (U - K ) < t . Then
nzl
u,, (U
-
K)
+ v((U
-
K)')
+
6
> u,,
=v,,(X)+c
+
(U
-
K)
+ v,,((U
> v(X)=v(U
-
-
K)')
+
6
K)+v((U
-
K)'),
and hence v,(U K ) t > v(U K ) . Since v,,(U K ) < t, we obtain v(U - K ) < 26. Thus v is regular. The decomposition readily follows. 12. This follows immediately from Proposition 11.20. 13. Let p - p, pp - v, v p with pr and v, regular and with pp and vp purely irregular. Write a = p, - v, = v p - p p in terms of its Jordan decomposition as 0. = O.+ - o . Then o 5 p, and o - 5 v, , and hei~ceo-+ and o are regular by Proposition 11.20. Also, cr+ 5 up and c r 5 p,, and the definition of "purely irregular" l u r ~ c cr+ s arid a Lu bc 0. T11cr1p, = v, and pp = vp. 14. Let p be as in Problem 10, and suppose that v is a regular Borel measure with v 5 p. Since ~ ( { G O } )= 0, Problem 6a shows that lim,?, v ( { x > a } ) = 0. For each n , let a , be the least ordinal such that v ( { x > a,}) 5 1In. The least ordinal > all a , is a countable ordinal B , and v ( { x > p } ) = 0. Since { x < p } is a countable set, p ( { x ip}) = 0. Therefore v ( { x < p } ) = 0, and we conclude that v = 0. -
+
-
+
+
-
Hints for Solutions of Problems
632
-
16. For the regularity any set in F i s in some F,.The sets in F, are of the form = E x ( X :,+,X,) with E 5 62"") and v ( E ) = v,(E). Given c > 0, choose K compact and U open in Q(") with K E g U and v,(U - K) < c . In Q, k is compact, 6 is open, KS c E c E , and v ( E - E ) < c.
Uzl
17. Let E = En disjointly in F. Since v is nonnegative additive, we have C z l v(E,) 5 v ( E ) . For the reverse inequality let c > 0 be given. Choose K compactandU,openwithK g E , E , g U,,v(U,-En) < c / 2 " , a n d v ( E - K ) < c . Then K c UT=lU,, and the compactness of K forces K U, for some N . N N Thenv(E) 5 v ( K ) + c 5 v ( u L 1u,)+E5 Cn=l v(U,)+c 5 C,=lv(Enj+2c 5 00 v(E,) 26. Since c is arbitrary, v ( E ) 5 C r = l v(E,).
c uL1
+
18. The key is that !2 is a separable metric space. Every open set is therefore the countable union of basic open sets, which are in the various .Fa's.
Chapter XI1 1. In (a), the closed ball is closed and contains the open ball; also every point of the closed ball is a limit point of the open ball since 1x1 - xo 11 = r implies that Il[(l-;)1(x1-xo)+xoI-xoIl = (1-$)llxl-xoll c r andlim,[(l-$)(XI-xo)+xol = x1. For (b), let the closed balls be B(r,; x,)". If rn > n , then Ilxm - x, 11 5 m since B(rm; xm)" g B (r,; x,)". Let r = lirn, r,. If r = 0, then {x,} is Cauclly and hence is convergent. In this case if x = limx,, then Ilx - x, 11 5 r, for all n , and hence x is in B(r,; x,)" for all n. If r > 0, fix no large enough so that r,, 5 3712. It is enough to show that x,, is in B(r,; x,)" for n > no. We may assume that x,, x,. The members of B (r,; x,) are the vectors of the form x, v with 11 u 11 5 r,, and these are assurrled to lie in B (r,, ; x,,). Therefore llx, - xno u 11 5 r,, for all such u . Take u = r,;lm(x, - x,,) Then r,, 2 llxn - x,, vll = ll(1 r&'mj(x, - x,,>ll = ( 1+r;'r,) ~ l x -x,, , 11. ere r&'r, > ( 3Z r ) p l r= SO llx, -xno 11 5 ( 1 5 ) ' r , , = +r,, 5 $r < v 5 r,, as required.
+
+
+ +
5.
+
+
+
2. Reduce to the real-valued case, and there use Theorem 1.23 and the remarks at the end of Section A3 of the appendix. 3. Convergence in either case is uniform convergence. For HCO(D),suppose therefore that { CEO c f ) z k }is a Cauchy sequence in H" (D) indexed by n. Write z = rei\ multiply by epim\ and integrate in 0 from -n to n . The result is that { c i ) r m }is Cauchy in n for each r < 1 and each rn. Then lim, c 2 ) r m = cmrm exists for each r and rn. Taking r = 112, we see that limn c i ) = cm exists for each rn. Arguing as in the proof of Theorem 1.37, we see that f ( z ) = CEOckzk is convergent for lzl < 1 and that the sequence of functions f,(z) = CEOc f ) z k converges to it pointwise. Since { f,} is uniformly Cauchy and pointwise convergent
ChapterXII
633
to f , it converges uniformly to f . For the vector subspace A ( D ) , we have A ( D ) = H m ( D ) f' c(D"'). Hence A ( D ) is a closed subspace of H W ( D ) .
+ +
4 . In (a), let us check the triangle inequality. For y E Y , we have Ila b y 11 5 Ila yfII Ilb ( y - yf)II for all y f E Y . Comparing the definition of Ila + b + Y 11 with the left side, weobtain l a b YII 5 Ila y'II lib+ ( y - yf)II for ally and yf in Y . Thus Ila b YII 5 Ila yfII 16 + yffII for all yf and yf' in Y . Taking the infimum over yf and yf' gives the desired conclusion. In (b), let a Cauchy sequence in XI Y be given. It is enough to prove that some subsequence in convergent. Thus it is enough to prove that if {x,} is a sequence in X with Ilx, - x,+l Y I/ 5 ZZ", then { x , + Y } is convergent in X I Y . We define a sequence in X with ?, = x, - y, and y, in Y such that I ?, 11 5 2 . 2-,. It is then easy to check that {Fa} is Cauchy in X and that if x f is its limit, then { x , + Y } tends to x f Y . To define the y,'s, we proceed inductively, starting with yl = 0. If yl, . . . , y, have been defined such that Il& - F k f l 11 5 2 . 2 Z k for k < n , choose y,+l in Y such that 11% - x,+l y,+lll 5 Ilx, - x,+l + Y 11 + 2Zn 5 2 . 2Zn. Then x,+l = x,+l y,+l has 11% x",+lll 5 2 . 2Zn, and the induction is complete.
+ + + + +
+ + + +
+ +
+
{z,} +
+
-
-
-
5. In (a), we have ctrG( v l , . . . , v,)F
xi
1 xi
1
=
C . . ci ( v i , vj)Fj = xi, (civ i , c j v j ) = 1,
I
2
( ci vi, C jcj v ! ) = ci vi . In (b), G ( u l , . . . , I!,) is Hermitian, and thus the finite-dimens~oi~al Spectral Theore~llsays that there exists a unitary matrix u = [ u i j ]with u Z 1G ( u l , . . . , u,)u diagonal, say = diag(dl, . . . , d,). Then d j = eJ k u l ~ ( q. . ,. , v,)uej, and this, by (a), equals ci ui with F = uej . Hence di 1 0 . In (c), we have detG(v1,. .. , v,) = d e t ( u Z ' G ( v l ,.. . , v,)u) = d1d2...dn 0 with equality if and only if some d j is 0. If dj = 0 , then civi = 0 for F = u e j , and hence v l , . . . , v, is dependent. Conversely if v l , . . . , v, is deci vi = 0 for some nonzero tuple ( c l , . . . , c,), and therefore pendent, then 0= ci vi, v j ) = ci (vi, v j ) for all j ; this equality shows that anontrivial linear conibination of the rows of G (ul , . . . , u,) is 0 , and hence det G (ul , . . . , u,) = 0 .
1 xi
1
xi
(xi
xi
xi
6. A single induction immediately shows the following: span{v;, . . . , vk} = span{ul, . . . , u k } ,uk is # 0 , and uk is defined. Then each uk has norm 1. If k < 1, then (vj, vk) = (ul (ul, v,)u,, vk) = ( u l , uk) - ( u l ,v k ) = 0. This proves the orthogonality.
7 . Define F on each u , to be the vector up given in the statement of the problem, and extend F linearly to a mapping defined on the linear span V of {u,}. Corollary 1 2 . 8 ~shows that 11 F ( u ) llH2 = lull for u in V. Corollary 12.8b shows that V is dense. Proposition 2.47 shows that F extends to a bounded linear operator from H1 into H2 satisfying 11 F ( u )11 = 11 u 11 H I for u in H I . Arguing in the same way with F-' proves that F is onto HZ. The second conclusion follows by using Proposition 12.1 1.
8. In (a), the boundedness is elementary, and the operator norm is 11 f 1,. the adjoint is multiplication by the complex conjugate o f f .
In (b),
634
Hints for Solutions of Problems
9. The linear span V of {x,} is a separable vector subspace. Suppose that it is not dense. Choose by Corollary 12.15 a member x* # 0 of X* with x * ( V ) = 0 . Since {x;} is dense, choose a subsequence { x , * ~with } xlk + x*. Then
Since tlle left side tends to 0 , so does the right side. Thus n;k tends to 0 , and n* = 0 , contradiction. 10. The dual of C ( X ) is M ( X ) . Define a linear functional x* on M ( X ) by x * ( p ) = p({so}).Then Ilx*II = 1, so that x* is in M ( S ) * . Let S, denote apoint mass at s . If x* were given by integration with a continuous function f , then we would have I{,)(s) = S,({so}) = x*(S,) = f dS, = f ( s ) . Thus the only possibility would be f = I{,), and this is discontinuous. 11. Let X and Y be normed linear with X complete, and let {L,} be a family of bounded linear operators L, : X + Y such that I L n ( x )11 5 C, for each x in X. For each y* in Y* with 11 y* 11 5 1, the linear functional y* o L, on X is bounded and has 1 y* ( L , ( x ) )1 5 C, . Since X is complete, the Uniform Boundedness Theorem for linear functionals shows that 1 y* ( L , ( x ) )1 5 C llx 11 for all x . Taking the supremum over y* and applying Corollary 12.17, we obtain 11 L, ( x )11 5 C I I x 11, as required. 12. Forx i n X and y in Y,we have
Ss
Given x E X and t > 0 , choose y in Y to make the first term < E , and then choose n and rn large enough to make the second term < c . It follows that { L , ( x ) } is Cauchy for each x . Since X' is complete, L ( x ) = limn L,(x) exists for all x . Continuity of addition and scalar multiplication implies that L is linear. Then IIL(x)ll = lim IILn(x)Il 5 liminf, IIL,IIIlxll 5 Cllxll. Hence IILII 5 C. 13. Proposition 12.1 shows that X* is a Banach space. We identify the elements x, in X with their images i(x,) under the canonical map i : X + X**. Corollary 12.18 shows that the element i(x,) of X** has IIi(x,)ll = IIx,ll. The hypothesis shows for each x* that I(i (x,))(x*)1 = x * (x,) 1 5 C,. for a constant C,* independent of a . Since X* is complete, the Uniform Boundedness Theorem (Theorem 12.22) shows that 111 (x,) 11 5 C for aconstant C independent of a . Applying Corollary 12.18 a secoiid time, we coiiclude that Ilx, 11 5 C indepeiideiitly of a . 14. For (a), let u and v have 11 u x 11 5 r and 11 v x 11 5 r . Then the estimate II(1-t)u+tu-xII = II(1-t)(u-x)+t(u-x)ll I II(1-t)(u-x)II+IIt(v-x)II = ( 1 - t ) 11 u - x 11 t 11 v - x 11 5 ( 1 - t)r tr = r proves the convexity. For (b),let X be the space of sequences s = {s,} with 11s 11 = C, Is,l Let Ek be the set of sequences with all s, > 0 , with 11s 11 = 1, and with sj = 0 for j 5 k. If s and t are two sequences with terms > 0 , then 11s t 11 = 11s 11 11 t 1 1 . The convexity follows, and everything else is easy. -
+
-
+
+
+
ChapterXII
635
15. Denote open balls in X by Bx and open balls in Y by By. The Interior Mapping Theorem says that L(Bx(1; 0 ) ) is open. Hence it contains a ball By(c; 0). Put C = c p l . By linearity, L(Bx(Cr; 0)) 2 By(r; 0) for every r 2 0. Since L is onto Y, we can choose xo in X with L (xo) = yo. Linearity gives L (Bx (Cr; xo)) 2 By (r ; yo). For each y,, we can take r = 211 y, -yo 11 and choose x, in Bx(C211 y, -yo 11; xo) with L(xn) = Yn. since.^, + .Yoyo,xn + xo. Also,wehave Ix, -xoII 5 2Cll.yn -.~oll. In this construction if yo = 0 , we could choose xo = 0 , and then the result follows with M = 2C. If yo f 0 , then llynll + ll yoll f 0 says that Ily, I 5 llyoll only finitely often. For these exceptional n's, we can adjust x, when y, = 0 so that x, = 0 , and then we have Ilx, 11 5 M 11 y, 11 for a suitable M and the exceptional n's. For the remaining n's, an inequality llx, 11 5 Mil y, 11 is valid as soon as {x,} is bounded, and {x,} has to be bounded since it is convergent. 16. It will be proved that the distance from e to Xo is > 1. The set Xoo of all sequences s l , $2 - $2, s3 - $2, . . . such that {s,} is in X is closed under addition and scalar multiplication. Hence it is a dense vector subspace of Xo, and it is enough to prove that Ile - sll 11 for all s in Xoo. L e t s be in Xoo, and let c = e - s . Adding t h e f i r s t n e n t r i e s g i v e s c l + ~ ~ ~ +=cn, - s , . Hence I c l + . . . + c , I 1 n - IlsII. If, by way of contradiction, llcll = 1 - c with t > 0, then IcjI 5 1 - c for all j, and we have Icl . . . c,I 5 n - n t . Thus n - llsll 5 n - n t , and wt: gcl n t 5 Ilsll, irl contradiction to the finiteness of 11s 11.
+ +
17. This is immediate from Corollary 12.15 and the previous problem.
18. For (a), let s 2 0 have Ilsll = 1. Then Ile - sll 5 1, and so Ix*(e - s)l 5 1. Since x y e ) = 1, this says that I 1 - x*(s) I 5 1. On the other hand, Ix*(s) 1 5 1 since ll.sII 5 1. Thus 0 5 x*(.s) 5 1 . We can scale this inequality to handle general .s. For (b), the two sequences differ by a member of Xo, on which the Banach limit vanishes identically; then (c) follows by iterated application of (b) since the Banach limit of the 0 sequence is 0 . In (d), let c 0 be given. By applying (c), we see that we may adjust the sequence so that sup, s, - inf, s, 5 c and so that the Banach limit is unchanged. By (a), Banach limits preserve order. Since (infs,)e 5 s 5 (sups,)e, we have inf s, 5 LIM,,, s, 5 sup s,. Since sups, = (sup s, - lim sup s,) lim sup s, 5 (sup, - inf,) lim sup, r lim sup, +c, we obtain LIM,,, s, r lim sup, +c. Since c is arbitrary, LIM,,, s, 5 lim sup,. Similarly lim inf s, 5 LIM,,, s,. Conclusion (e) is immediate from (d).
+
+
20. The parallelogram law gives
If we set z = 0 in this identity and then set y = 0 in it, we get two relations, one involving an expression for Ilx 2y 112 and the other involving an expression for IIx 2211~.If we substitute these relations into the displayed equation for the
+
+
Hints for Solutions of Problems
636
+
+
+ +
+
terms ~ l x 2y 112 and llx 2z112, we obtain the formula llx y z112 II y - z12 = 1lx+y11~+lx+z112- 11x112+ lly112+ 11~11~. ~ubstitutionof211y11~+211~11~lly+z112 for Ily - z112 in this formula gives the desired identity. 21. We have
Each term of the first line on the right is 0 because side simplifies to (xl, y) (x2, y), as required.
+
x,i k / 4
= 0,
and thus the right
22. Induction with the result of the previous problem gives (nx, y) = nix, y) 1 for every integer n 10. Replacing nx by z, we obtain ;(z, y) = (; 1 z, y). Hence ( r x , y) = r (x, y) for every rational r > 0. It follows from the definition of ( . , . ) that (-x, y) = -(x, y) and that if the scalars are complex, (ix, y) = i(x, y). Consequently (rx, y) = r (x, y) if r is in the set D. 23. We are to prove that I(x, y)l 5 IlxII IIy 11, and we may assume that y f O. If r is in D, we have
Letting r tend to (x, y)/lly
112
through members of D, we obtain
and it follows that I(x, y) I 5 Ilx 11 11 y 11
24. The Schwarz inequality gives
As r tends to c through D, the right side tends to 0, and the left side tends to I c P , Y) (CX,y)I. Hence cix, y) = (cx, y).
SELECTED REFERENCES
Adams, M., and V. Guillemin, Measure Theory and Probability, Wadsworth & Brooks/Cole Advanced Books and Software, Monterey, CA 1986; corrected reprint of the original, Birkhauser Boston, Inc., Boston, 1996. Bartle, R. G., The Elements of Real Analysis, John Wiley & Sons, New York, 1964; second edition, 1976. Bartle, R. G., and D. R. Sherbert, Introduction to Real Analysis, John Wiley & Sons, Inc., New York, 1982; second edition, 1992. Coddington, E. A., and N. Levinson, Theory of Ordinary Differential Equations, McGraw-Hill Book Company, New York, 1955. Doob, J. L., Measure Theory, Springer-Verlag, New York, 1994. Duistermaat, J. J., and J. A. C. Kvlk, Multidimensional Real Analysis, Vvl. I, Cambridge University Press, Cambridge, 2004. Duistermaat, J. J., and J. A. C. Kolk, Mz~ltidimensionalReal Analysis, Vol. 11, Cambridge University Press, Cambridge, 2004. Dunford, N., and J. T. Schwartz, Lirzear Operators, Part I, Interscience Publishers, Inc., New York, 1958. Halmos, P. R., Measure Theory, D. Van Nostrand Company, Inc., Princeton, 1950; reprinted, Springer-Verlag, New York, 1974. Halmos, P. R., Naive Set Theory, D. Van Nostrand Company, Inc., Princeton, 1960; reprinted, Springer-Verlag, New York, 1974. Hayden, S., and J. F. Kennison, Zermelo-Fraenkel Set Theory, Charles E. Merrill Publishing Company, Columbus, 1968. Ince, E. L., Ordinary Differential Equations, Longmans, Green and Co., London, 1927; reprinted Dover Publications, New York, 1944. Jackson, D., Fourier Series and Orthogonal Polynomials, Carus Mathematical Monographs, vol. 6, Mathematical Association of America, no city listed, 1941. Kaplan, W., Ordinary Differential Equations, Addison-Wesley Publishing Co., Reading, MA, 1958. Kelley, J. L., General Topology, D. Van Nostrand Company, Inc., Princeton, 1955; reprinted, Springer-Verlag,New York, 1975. Lebesgue, H., L e ~ o n ssur l'lntegration et la Recherche des Fonctions Primitives, Gauthiers-Villars, Paris, 1904; second edition, 1928; second edition reprinted by ~ d i t i o n Jacques s Gabay, Sceaux, 1989.
638
Selected References
Munroe, M. E., Introduction to Measure and Integration, Addison-Wesley Publishing Company, Inc., Cambridge, MA, 1953; second edition, Measure andlntegration, Addison-Wesley Publishing Company, Reading, MA, 1971. Riesz, F., and Bkla Sz.-Nagy, Functional Analysis, Frederick Ungar Publishing Co., New York, 1955. Royden, H. L., Real Anal,ysis, The Macmillan Company, New York, 1963; second edition, Macmillan Publishing Co., 1968; third edition, 1988. Rudin, W., Principles of Mathematical Analysis, McGraw-Hill Book Company, New York, 1953; second edition, 1964; third edition 1976. Rudin, W., Real & Complex Analysis, McGraw-Hill Book Company, New York, 1966; second edition 1974; third edition 1987. Saks, S., Theory of the Integral, second revised edition, G. E. Stechart and Company, New York, 1937; reprinted, Hafner Publishing Company, New York, no date, and Dover Publications, New York, 1964. Spivak, M., Calculus on Manifolds, BenjaminiCummings Publishing Company, Menlo Park, CA, 1965. Stein, E. M., Singular Integrals and Dzfferentiability Properties of Functions, Princeton University Press, Princeton, 1970. Stein, E. M., and G. Weiss, Introduction to Fourier Analysis on Euclidean Spaces, Princeton University Press, Princeton, 1971. Stone, M. H., "A generalized Weierstrass approximation theorem," Studies in Modern Analysis, R. C. Buck (ed.), Mathematical Association of America, distributed by Prentice-Hall, Inc., Englewood Cliffs, NJ, 1962. Williamson, R. E., and H. F. Trotter, M~~ltivariable Mathematics, Prentice-Hall, Inc., Englewood Cliffs, NJ, 1974; second edition 1979; third edition 1996. Zygmund, A,, Trigonometric Series, Vol. I, second edition, Cambridge University Press, Cambridge, 1959. Zygmund, A., Trigonometric Series, Vol. 11, second edition, Cambridge University Press, Cambridge, 1959.
INDEX OF NOTATION
See also the list of Standard Notation on page xxi. I11 the list below, ite~ns are alphabetized according to their key symbols. For letters the order is italic lower case, Roman lower case, italic upper case, Roman upper case, script, and blackboard bold. Next come items whose key symbol is Greek, and then come items whose key symbol is a nonletter. The last of these are grouped by type.
a,, 61 arccos, 79 arcsin, 51 arctan, 51 A ( D ) ,523 b n , 61 B(r; x ) , 83 B ( S ) , 87, 281, 521 B ( S , C ) , 87 B ( S , R ) , 87 B,, 297 B N , 297 h'(X), 488 a ( x , Y ) , 284, 524 Bo(X>,504 B N ( K > ,318 B N ( V ) , 315 c , 522 co, 522 c,, 61, 335 card, 577 A ~ ' ,93 cos, 47 C m , 142 C k , 142 C ( S ) , 100, 281, 464, 521 C ( S ,C ) , 100,512 C ( S , R ) , 100, 512
C o ( X ) , 509, 521 C m ( E ) , 142 c ~ ( E ) 142 , Ccom(X),300, 314, 486, 521 C N ( [ a bl), , 522 d , 41 d2, 75 d ( x , y ) , 83 d x , 298 dm, 326, 491, 517 D , 384 D,, 385 D N , 69 e , 47 e A , 148 e j , 566 exp, 47, 148 E x , 267 Ey, 267 E r f l Q 3 440 F,, 492 F,374 IF, 135 IF", 565 Gs, 492 H, 397 H I , 401 H W ( D ) ,523
640
Index of Notation
inf, 3, 6 I E , 171, 241 Z E ( ~ )242 , J p ( t ) , 225 K N , 71 K ( X ) , 490 12, 87 B!, B P , 522 lim, 5, 98 liminf, 7 lim sup, 7 log, 49 L ' , 89, 281 L ~ 89, , 281 Loo, 281 L,, 544 L ( Y ) , 208 L ( P , f ) , 27, 162 L P ( X ) , L P ( X ,A,w), L P ( X ,w ) , 281, 522 m , 236, 298 n z R ( f > , 162 M ( X ) , 522 M ( X , C),514 M ( X , R), 512 M R ( ~ )162 , A", 92 oscg(xo),119, 167 p', 411 X = P U N , 419 P ( h ) , 208 P ( x , t ) , P,(x), 395 P,(Q), 392 Q ( D ) , 384 Q ( x , € ) , Q e ( x ) , 398 Qn(a1,.. . , a ~ ) 302 , 134 Q,, 133 Q;, 134 R [ a ,bl , 27 R ( A ) , 162
ox,
R I Af d x , 318 R*,6, 85 SN ( f ; x ) , 62, 335 sin, 47 sup, 2, 5 Sn, 126, 326 s ( P , {ti},. f ) ,39 S ( P , I ~ R ) ,f ) , 163 S , 385 S(EtN), 385 Tr X , 149 ?; 442 T*, 455 u j , 566 u ( p , f 1, 27,162 V ( f 1 , 355 V * ( f ) ,V - ( f > , 347 Z,, 134
Greek T(s), 323 A , 181, 392 Axi, 27 E A F , 233 T ( s ) , 390 1 , 541 h f , 350 p*, 254,255 p,, 254, 255 p f , 350 w ( P ) , 27, 161 [ d ~ l248 , v = f d p , 252 an, 53 0 7 , 54 t,, 303 V E , 312 Q , 292 Q N , 326
Index of Notation
Isolated symbols and signs E, 554, 555 00,5 c, 2 , 555 +on, -00, 6 r f , r - , 347 f f , f - , 239 v f , v-, 418 --, 62, 335 <<, 420 @, 528
Specific functions ( . , .)2, 75 II . I, 75, 279 I . 1, 84, 136, 138, 161, 239, 514 x . y , 84 ( . , .), 89, 523 11 . 11, 90, 136, 279, 514, 521, 523 I . , I 133 II . 1 1 1 3 279 II . , I 279 I1 . llsup3 281 II . I," 347 II . ,I 410
Subscripts and superscripts X * , 128, 444, 455, 524 f * ( x ) , 328 M', 527 L*, 536 x * , 537 2A, 555
Intervals ( a ,b ) , 3 [a,bl, 3 [ a , b > ,3 ( a ,b l , 3
Operations on sets A x B, 266 X / w , 445, 471 f, 4,268 B ~ 556 ,
Other
UxESA X , Ox,, Ax, X x c , A x j
557
[ M i j ] , 566 ( X , 77, 442 a, + a , 5, 98 daxj' f i 139 x 7 = 1 aj(x1, - - . ( X , A, P I , 238 (D,i),465 {x,}, 465 a H x a , 465
9
Operations on functions f - ' , 558 f * g , 180, 306, 334 p x v, 270 f? 374 -b ~ ; 1f n . ~ ,~ ~ f n . . 27 , - a
-
SAf d x , n f d*, hbf d x , 27, 30
J a f d", f SE f JE
162
162 fdP3 242 f x > d p ( x ) , 242 9
~
n
a
)
199 ~
,
INDEX
a . ~ .248 , Abel summable, 54 Abel sums, 54 Abel's Theorem, 54 Abelian theorem, 55 absolute value, 563 absolutely continuous measure, 420 monotone function, 369 Stieltjes measure, 369 abstract rectangle, 266,297 additivc sct function, 234 bounded, 41 8 adjoint, 536 Alaoglu's Theorem prelimlnruy form, 284,509 algebra, 124 algebra of sets, 232 almost every, 248 almost everywhere, 248 almost period, 131 almost periodic, 131 Bochner, 131 Bohr, 131 alternating series test, 19 approximate identity, 58,60,3 12 arccosine, 79 archimedean property, 4 arcsine, 5 1 arctangent, 5 1 area of sphere, 326 arithmetic operations, 104 arithmetic-geometric mean inequality, 182 Ascoli's theorem, 22,121,229 Ascoli-Arzeli Theorem, 478 associated Stieltjes measure, 344 Axiom of Choice, 557 axiom of countability, 450 Baire Categoly Theorem, 118,120,481
Raire filnction, 50.5 Baire measurable function, 505 Baire measure, 506 Baire set, 504 ball, open, 83 Banach limit, 550 Banach space, 521 quotient, 549 reflexive, 541 Banach-Steinhaus Theorem, 543 base countable local, 450 for metric space, 106 for topology, 446 local, 450 basis of vector space, 573 standard, 566 Reqsel filnction, 22.5 Bessel's equation, 224 Bessel's inequality, 67,335,531 binomial coefficient, xxi binomial series, 52,56 Bochner almost periodic, 131 Bochner's Theorem, 406 Bohr almost periodic, 131 Bolzano-Weierstrass property, 108,109 Bolzano-Weierstrass Theorem, 9,108 Borel complex measure, regular, 5 14 Borel function, 297,316,318,504 Borel measurable, 239,297 Borel measurable function, 504 Borel measure 299,316,318,490 purely irregular, 5 19 regular,299,316,318,490 Borel set,237,297,315,318,488,504 Borel signed measure, regular, 512 bound greatest lower, 3
least upper, 2 lower, 3 upper, 2 bounded, 478 bounded addilive set lunclion, 418 bounded, essentially, 280 bounded hnct~on,87 bounded interval, 3 bounded linear operator, 283,523 bounded operator, 283 ho~~ndrd,pointwiur, 22,121,478 bounded set, 3,490 bounded, un~formly,22,121,478 bounded variation, 346,355
ckfunction, 142 Cm function, 142 canonical expansion of simple function, 241 canonical extension of measure, 261 canonical map, 541 Cantor 579 Cantor diagonal proccss, 24,461 Cantor function, 344 Cantor measure, 354,519 Cantor set, 118,2YU, 343,519 standard, 118 cardinal number, 577 cardinality, 577 Cartesian prndl~ct,556,557 Cauchy criterion, 10 uniform, 18 Cauchy Integral Theorem, 378 Cauchy sequence, 9,112 uniformly, 18 Cauchy-Peano Existence Theorem, 229 Cauchy-Riemann equations, 379,398 Cauchy-Schwarz inequality, 563 Cesiro summable, 53 Cesiro sums, 53,371 chain, 573 chain rule, 144 change-of-variables formula, 37,172,320 character, multiplicative, 407 characteristic function, 171,241 characteristic polynomial, 208,214 Chebyshev's inequality, 351 class, 555 equivalence, 564 class ck,142
class Cm, 142 closed geometric rectangle, 161 Closed Graph Theorem, 548 closed interval, 3,6 closed map 482 closed rectangle, 161 closed set, 3,92,442 closed subspace, 102 closed unit disk, 126 closed vector subspace, 104 cloullrr, 93,442 cofactors, 568 cofind, 468 collection, 555 compact metric space, 108 compact set, 108,453 compact topological space, 453 compactification, one-point, 456 complement, 555 complete metric space, 113 completely additive set function, 234 a-finitc, 237 completeness of L P , 285,414 completion of measure space, 262,291 completion ot metnc space, 128, 133 complex conjugation, 563 complex Euclidean space, 85 complex measure, regular Borel, 5 14 component, 483 component rectangles, 161 components of a function, 139 composition, 558 conditional expectation, 439 conjugate Poisson integral, 398 conjugate Poisson kernel, 398 conjugation, complex, 563 connected, 482 locally, 131 locally pathwise, 131 connected component, 483 connected metric space, 115 connected set, 115 constant coefficients, 208,211 content 0, 167 continuity at a point, 95 continuity, uniform, 111 continuous derivative on a closed interval, 561 continuous from the left, 339 continuous from the right, 339
continuous function, 10,96,443 uniformly, 11 continuous linear functional on L P ,425 continuous periodic, 67 cunlinuous singular Slielljes mcasure, 368 contraction mapping, 131 contraction mapplng prlnclple, 131 converge, 463,466 convergence norm, 284 nniform, 17 weak-star, 284,355,405,437 convergent sequence, 5,97 convolution, 58,180,306,334,417 of measures, 405 cosine, 47 countable, xxi, 106 countable local base, 450 countable ordinals, 292 countably additive set function, 234 counting measure, 236 covcr, 106,151 Cramer's rule, 568 critical point, 323 crltlcal value, 323 curve, 199 integral, 199 cut, 2 defined implicitly, 153 definite, 89 degree of polynomial, 568 delta mass, 342 delta measure, 342 dense, 106 dense set, 450 derivative, I40 at endpoint of intexval, 561 on a closed interval, 561 partial, 139 determinant, 567 Wronskian, 205 diadic cube, 302 diagonal process, 24 diffeomorphism, 160 difference, 555 differentiable, 139 infinitely, 46 periodic, 67
differential equation existence theorem for ordinary system, 189, 229 first-order o r d i n q linear, 228 ordinary, 183 ordinary homogeneous linear, 184,201 ordnary lnhomogeneous Ilnear, 201 ordinary linear, 184,201 system of ordinary, 188 uniqueness theorem for o r d i n q system, 193 differential, total, 1.57 differentiation, implicit, 153 differentiation of integrals, 35,329,363,370 strong,431,438 differentiation of monotone functions, 359 differentiation of series of monotone functions, 362 dilation, 303 Dini's test, 70, 336 Dini's Theorem, 78,132,480,492 direct image, 559 dircctcd sct, 465 Dirichlet kernel, 69 Dirichlet-Jordan Theorem, 348 d~scretemetnc, 87 discrete topology, 443 disk, 126,181 distance, 41,75,83 to a ~ e t96 , distribution function, 339, 35 1 divide, 568 Division Algorithm, 569 divisor, 568 greatest common, 570 domain, 556 Dominated Convergence Theorem, 253 dot product, 84,566 donbly infinite sequence, 5 dual group of finite abelian group, 407 dual index, 41 1,525 dual of normed linear space, weak-star topology, 444 dual space, 284,524 Egoroff's Theorem, 291,292,s 17 element, 554,555 elementary matrix, 174 elementary set, 233 entity, 554
equicontinuous, 22,121,227,230,477,478 uniformly, 22, 121 equivalence class, 564 equivalence relalion, 564 essential bound, 280 essential supremum, 280 essentially bounded, 280 Euclidean algorithm, 570 Euclidean norm, 84,85 Enclidean rpace, 84,521 Euler's equation, 222 eventually, 97,463,465 everywhere dense, 106 existence theorem integral curves, 199 system of ordinary differential equations, 189,229 expansion by cofactors, 568 expectation, conditional, 439 exponential, 47 of a matrix, 148 extended real number, 6 , 8 5 Extension Theorem, 238,253,271,489,502 F, ,492 factor of polynomial, 568 Factor Theorem, 569 family, 5.55 fast Fourier transform, 39 1,407 Fatou's Lemma, 252 Fatou's Theorem, 334,395 FejCr kernel, 71 FejCr's Theorem, 72,337,371 field of scalars, 279 filter, 294 finest topology, 445 finite abelian group, Fourier analysis on, 407 finite interval, 3 finite limit for derivative at endpoint, 561 finite measure space, 238 finite-intersection property, 109,454 first axiom of countability, 450 first category, 119 first countable, 450 first-order linear ordinary differential equation, 228 flip, 174 Fourier coefficient, 62,335
Fourier inversion formula, 380 Fourier series almost everywhere convergence ol Ccsho sums, 371 Bessel's inequality, 67,335 convergence In L P , 439 convolution, 180 Dini's test, 70, 336 Dirichlet-Jordan Theorem, 348 divergence for a cnntinnonr fnnction, 544 failure of some sequence vanishing at infinity to occur, 547 FejCr's Theorem, 72,337 localization, 71 Parseval's Theorem, 74,338 Riemann-Lebesgue Lemma, 67,335 Riesz-Fischer Theorem, 338 uniqueness theorem, 73,77,338 with harmonic functions, 227 with Lebesgue integral, 335 with Ricmann integral, 62 Fourier transform, 374 fast, 391,407 for hnlte abellan group, 407 of measure, 406 Fourier-Stieltjes coefficient, 349 Fourier-Stieltjes series, 349 freqnently, 465 Fubini's Theorem, 15,169,170,271,273 Fubini's theorem on differentiation of series of monotone functions, 362 function, 556 functional, linear, 283 Fundamental Theorem of Algebra, 112,572 Fundamental Theorem of Calculus, 35,329, 363,370
Gs, 492 gamma hnction, 323,353 gap, 484 geometric rectangle, 161,266,297 closed, 161 Gram determinant, 549 Gram matrix, 549 Gram-Schmidt orthogonalization process, 530,549 graph, 291 greatest common divisor, 570
greatest lower bound, 003 Green's Theorem, 378 Hahn decomposition, 419 Hahn-Banach Thzurcm, 537 half space Polsson lntegral formula, 395 Poisson kernel, 395 half-open interval, 3,6,45 1 Hardy-Littlewood maximal function, 330. 403,438 Hardy-Littlewood Maximal Theorem, 328, 333,430 harmonic function, 181,334,354,392,437 unit disk, 227,518 Hausdorff, 105,447 Hausdorff-Young Theorem, 428 hedgehog space, 88 Heine-Bore1 Theorem, 108,110 Helly's Selection Principle, 78 Helly-Bray Theorem, 405,508 Hcrglotz's Thcorcm, 5 18 Hermitian inner product, 85,89 Hermitian symmetric, 89 Hilbert cube, X X Hilbert space, 523 dimension, 534 Hilbert transform, 397,433 exiutenre almout everywhere, 438 Hilbea-Schmidt norm, 138 Holder's inequality, 41 1 homeomorphism, 96,443 homogeneous function, 180 homogeneous linear o r d i n q differential equation, 184,20 1,228 image, 556,567 direct, 559 inverse, 559 imaginary part, 563 implicit differentiation, 153 Implicit Function Theorem, 155 indicator function, 171,241 indicia1 equation, 222,223 indiscrete space, 87 infimum, 3 infinite Taylor series, 44 infinitely differentiable, 46 infinity, 5 , 6
vanish at, 509 inhomogeneous linear ordinary differential equation, 201 initial condition, 186 inner measurc, 255 inner product, 89,523 ~nner-productspace, XY pseudo, 90 integrable, 242,276 Lebesgue, 242 Riernann,27,42,162 uniformly, 292 vector-valued function, 274,277 integral curve, 199 integral, Lebesgue, 242 integral, Riemann, 42 integration by parts, 36,67,344 interchange of limits, 13 interior, 92,442 Interior Mapping Principle, 545 Intermediate Value Theorem, 12,116 intcrscction, 555 interval, 3, 6 inverse function, 558 Inverse Functlon 'l'heorem, 156,562 inverse image, 559 irregular singular point, 227 isometric, 127 i uometry, 127 Jacobian matrix, 140 Jessen-Marcinkiewicz-Zygmund, 438 Jordan and von Neumann Theorem, 526,55 1 Jordan block, 212 Jordan decomposition, 418 Jordan form, 212 Jordan normal form, 213 kernel Dirichlet, 69 FejLr, 71 of linear function. 567 LP completeness, 285,414 LP dual, 425 L P norm, 410 LP translation, 309,417 Lagrange multipliers, 182,411 Laplace equation, I8 1,392
Laplacian, 181,392 least upper bound, 2,574 Lebesgue constant, 544 Lebesgue decomposition, 368,423 Lebesgue inlegral, 242 Lebesgue measurable, 239,262,265,305 Lebesgue measure, 11X,236,298,303 Lebesgue set, 371 Lebesgue's theorem on differentiation of monotone functions, 359 T .rgrndrr polynomial, 221 Legendre's equation, 220 Leibniz test, 19 length, 118 lexicographic ordering, 484 limit, 5,463,466 Banach, 550 of a sequence, 98 point, 3,93,442 uniform, 17 limits, interchange of, 13 Lindclof spacc, 45 1 linear function, 566 linear functional, 283,525 norm, 424 positive, 488 linear map, 566 linear operator, 283,523 hol~ndrd,283,523 continuous, 283 linear ordinary differential equation, 184,201 constant coefficients, 208 homogeneous, 228 linear transformation, 566 Lipschitz condition, 189 local base, 450 countable, 450 localization of Fourier series, 71 localization of Lebesgue integral, 246 locally compact, 455 locally connected, 131,483 locally finite open cover, 483 locally pathwise connected, 131,483 logarithm, 49 long line, 484 lower bound, 3 greatest, 3 lower Riemann integral, 27, 162 lower Riemann sum, 27,162
lower semicontinuous, 481 lower-dimensional set, 325 Lusin's Theorem, 5 17 map, 556 linear, 566 mapping, 556 Marcinkiewicz Interpolation Theorem, 430 matrix, 566 elementary, 174 Jacohian, 140 Jordan form, 21 3 Wronskian, 205 maximal element, 573 maximal function, 330 Hardy-Littlewood, 330 Mean Value Theorem, 560 measurable, Borel, 239,297 measurable function, 238 Baire, 505 Borel, 504 mcasurablc, Lcbcsguc, 239,262,265,305 measurable set, 238,241,257,266 measurable vector-valued function, 275 measure, 235 absolutely continuous, 420 absolutely continuous Stieltjes, 369 associated Stieltjes, 344 Rairr, SO6 Bore1,299,316,318,490 Cantor, 354,519 counting, 236 delta, 342 inner, 255 Lebesgue, 236,298,303 outer, 255 product, 270 purely irregular Borel, 519 regular Borel, 299,316,318,490 regular Borel complex, 5 14 regular Borel signed, 5 12 signed, 418 singular, 423 singular Stieltjes, 368 Stieltjes, 339 measure 0,166,265 measure space, 238 completion, 262,29 1 finite. 238
a-finite, 238 member, 555 mesh,27,161 metric, 83 discrclc, 87 hedgehog, 88 product, 103 uniform, 87 metric space, 83 complete, 113 completion, 128,173 connected, 115 metric subspace, 102,104 metric topology, 442 metrizable, 476 Minkow ski's inequality, 41 2 for integrals, 287,415 monotone class, 268 Monotone Class Lemma, 268,269 Monotone Convergence Theorem, 15,250,251 monotone increasing, 339 monotonc scqucncc, 5 monotone set function, 494 multiplication formula, 375 multlpl~cat~ve character, 407 multiplicity, 27 of a root, 572 multiplier, 384 m111tiplirr operatnr, 384 negative, xxi negative variation, 347 neighborhood, 442 of a point, 92 of a subset, 92 net, 465 universal, 468 Newton's method, 78 nonnegative set function, 234 norm, 90,429,521 convergence, 284 essential supremum, 280 Euclidean, 84,85 Hilbert-Schmidt, 138 LP,410 of linear functional, 424 operator, 136,283,429,523 supremum, 28 1 total-variation, 5 14
uniform, 28 1 weak-type, 430 normal, 105,447 normed linear space, 281,521 Gnilc-dimensional, 521 pseudo, 280 weak topology, 444 weak-star topology on dual, 444 nowhere dense, 118 null space, 567 nl~rnher extended real, 6 real, 2 one-one, 558 one-point compactification, 456 onto, 558 open ball, 83 open cover, 106,45 1 locally finite, 483 open interval, 3,6 opcn mapping, 47 1 open neighborhood, 92,442 open rectangle, 161 open set, 3,X3,442 open subcover, 106,451 open subspace, 102 open unit disk, 181 operator bounded linear, 283 continuous linear, 283 linear, 283,523 multiplier, 384 norm, 136,283,429,523 self-adjoint, 536 sublinear, 429 order, 183,188 order complete, 484 order topology, 484 ordered pair, 555 ordering lexicographic, 484 partial, 573 simple, 573 total, 484,573 ordinals, countable, 292 ordinary differential equation, 183 constant coefficients, 208 existence theorem, 189,229
first-order linear, 228 homogeneous linear, 184,20 1,228 inhomogeneous linear, 20 1 linear, 184,20 1 uniqueness heorem, 193 ordinary differential equations system, 188 wlth constant coefficients, 21 1 orthogonal, 89,527 orthogonal complement, 528 orthogonal projectlon, 530 orthonormal, 529 orthonormal basis, 532 orthonormal set, maximal, 532 oscillation, 119,167,481 outer measure. 255 p-adic numbers, 133 pair ordered, 555 unordered, 555 parallelogram law, 526 paramctcrs, 198 Parseval's equality, 532 Parseval's Theorem, 74,338 partla1 denvatlve, 131) partial ordering, 573 partition, 26,161 partition of unity, 151,483 path, 116,199,482 pathwise connected, 116,482 locally, 131 periodic, 63 Picard iteration, 187 Picard-Lindelof Existence Theorem, 189 Plancherel formula, 38 1 point, 555 point mass, 342 pointwise hounded, 22,121,478 Poisson integral, 355 conjugate, 398 formula for half space, 395 formula for unit disk, 354,392,437 Poisson kernel conjugate, 398 for half space, 395 for unit disk, 392 Poisson Summation Formula, 389 polar coordinates, 173,322 polarization, 527
polynomial characteristic, 208,214 in one indeterminate, 568 Legendre, 22 1 prime, 571 positive, xxi posltlve deiinlte function, 406 positive linear functional, 488 positive variation, 347 potential theory, 353 power series, 44,46 prime polynomial, 571 primitive mapping, 175 probability, 171,241,339,439,486 product Cartesian, 556,557 dot, 84,566 inner, 89,523 measure, 270 metric, 103 topology, 443,458 product of mctric spaccs, 103 product of sets, 556 product of u-algebras, 266 projectlon, orthogonal, 530 Projection Theorem, 528 pseudo ~nner-productspace, 90 pseudo normed h e a r space, 280 psrl~domrtrir,87 pseudometric space, 83 pseudonorm, 90,279 purely finitely additive, 438 purely irregular Bore1 measure, 5 19 Pythagorean Theorem, 526 quotient map, 445,471 quotient space of a Banach space, 549 quotient topology, 443,445,471 radius of convergence, 45,78 Radon-Nikodym Theorem, 421,425,439,507, 515 range, 556 rank, 567 real number, 2 real part, 563 rearrangement, 12 rectangle, 161,267,266,297 abstract, 266,297
closed, 161 closed geometric, 161 geometric, 266,297 open, 161 reduclion u l order, 228 refinement, 162 of partition, 28 reflexive, 564,573 reflexive Danach space, 541 region under a graph, 291 reglllar, 105,236,447 regular Borel complex measure, 5 14 regular Borel measure, 299,3 16,318,490 regular Borel signed measure, 5 12 regular singular point, 221 relation, 556 equivalence, 564 function, 556 set, 473 relative topology, 446 relatively dense, 131 rcstriction, 558 Riemann integrable, 27,42, 162 Riemann integral, 27,42,162 Kiemann sum, 27,39,42,162,163 Riemann zeta function, 390 Riemann-Lebesgue Lemma, 67,335,380 Riesz Convexity Theorem, 427 Rirq7,Frigyeq and Marcel, 427 Riesz Representation Theorem, 425,490,528 Riesz's Lemma, 358 Riesz-Fischer Theorem, 338,383 ring of sets, 233 Rising Sun Lemma, 358 root, 568 Russell paradox, 553 saltus function, 366 Sard's Theorem, 324 sawtooth function, 65 scalar, 87,279 scalar-valued nonnegative function, 274 Schroeder-Bernstein Theorem, 578 Schwartz function, 385 Schwartz space, 385 Schwarz inequality, 75,84,90,526,564 second axiom of countability, 450 second countable, 450 section, 267
self-adjoint, 536 semidefinite, 90 seminorm, 279 separable, 107,446 separalt: poinls, 124,462 separation of variables, 227 separation properties, 105 sequence, 4,97 Cauchy,9,112 convergent, 5 donhly infinite, 5 monotone, 5 space, 87 series binomial, 52,56 Fourier (see Fourier series) Fourier-Stieltjes, 349 power, 44,46 Taylor, 46 trigonometric, 6 1 set, 554 sct function, 234 additive, 234 bounded additive, 41 8 completely addltlve, 234 countably additive, 234 monotone, 494 nonnegative, 234 p~ltplyfinitely additive,438 o-finite completely additive, 237 set theory, 553 Zermelo-Fraenkel, 554 side, 266 a-algebra of sets, 233 a-bounded set, 490 o-compact, 457 a-finite completely additive set function, 237 o-finite measure space, 238 a-ring of sets, 233 signed measure, 41 8 regular Borel, 5 12 Silverman-Toeplitz summability method, 80 simple function, 241 canonical expansion, 241 simple ordering, 573 sine, 47 singleton, 555 singular measure, 423 singular point
irregular, 227 regular, 221 singular Stieltjes measure, 368 smallest closed vector subspace, 281 smoolh lunclion, I42 smooth vector field, 199 solution, 183,188 sphere, 126 area, 326 spherical coordinates, 325 rtandard h a r k , 566 standard Cantor set, 11 8 Stieltjes measure, 339 absolutely continuous, 369 associated, 344 singular, 368 Stone-Weierstrass Theorem, 124,132,480 strong differentiation, 431,438 strong type, 429 subalgebra, 124 subcover, 451 subdivision point, 26 sublinear operator, 429 subnet, 468 subordinate to a cover, 151 subsequence, 5 subspace, 102,104 closed, I02 metric, 102,104 open, 102 smallest closed vector, 281 topological, 446 vector, I04 summability, 53 Abel, 54 Cesiro, 53 Silverman-Toeplitz, 80 support, 172,300,488 supremum, 2 essential, 280 norm, 281 symmetric, 89,564 symmetric difference, 232 system of ordinary differential equations, 188 constant coefficients, 211 existence theorem, 189,229 uniqueness theorem, 193
Tauberian theorem, 55 Taylor series, 46 infinite, 44 Taylor's Theorem, 43,146 Lends, 5 Tietze Extension theorem, 483 topological space, 442 topological subspace, 446 topology, 442 discrete, 443 finert, 44.5 half-open interval, 451 metric, 442 order, 484 product, 443,458 quotient, 443,445,471 relative, 446 upper, 481 weak, 443 total differential, 153 total ordering, 484,573 total variation, 355,514 total-variation norm, 514 totally bounded, 113 trace of matnx, 149 transfinite induction, 293 transitive, 564,573 translation, 303 in I.P,709,417 triangle inequality,41,76,83,90,279,563 trigonometric series, 61 trivial ultrafilter, 294 Tychonoff Product Theorem, 460,470 Tychonoff's Lemma, 452 type, 429 strong, 429 weak. 429 ultrafilter, 294 trivial, 294 ultrametric inequality, 133 unbounded interval, 3 Uniform Boundedness Theorem, 543 uniform Cauchy criterion, 18 uniform continuity, 11 1 uniform convergence, 17,99 uniform limit, 17 uniform metric, 87 uniform norm, 281
uniformly bounded, 22,99,121,478 uniformly Cauchy, 18 uniformly continuous function, 11 uniformly equicontinuous, 22,121 uniformly integrable, 292 union, 555 unique factorization, 571 uniqueness theorem Fourier series, 73,77,338 integral curves, 199 system of ordinary differential equations, 193 unit disk, 126, 181 harmonic function, 227,5 1X Poisson integral formula, 354,392,437 Poisson kernel, 392 unit sphere, 126 universal net, 468 unordered pair, 555 upper bound, 2,573 least, 2 upper Ricmann inlegral, 27,162 upper Riemann sum, 27,162 upper semicontinuous, 48 1 upper topology, 48 1 Urysohn Metrization Theorem, 476 Urysohn's Lemma, 474
Young's inequality, 428
vanish at infinity, 509 variation bounded, 346,355 negative, 347
Zermelo's well-ordering theorem, 576 Zermelo-Fraenkel set theory, 554 zeta function, 390 Zom's lemma, 573
of parameters, 207 positive, 347 total, 355,514 vector field, 199 vector subspace, 104 smallest closed, 281 vector-valued function integrable, 274,277 measurable, 275 weak~',330 weak topology, 443 determined by functions, 443,462 on normed linear space, 444 weak type, 331,429 weak-star convergence, 284,355,405,437 weak-star topology of dual of normed linear space, 444 weak-type norm, 430 Weierstrass Approximation Theorem, 60 Weierstrass M test, 20 well ordercd, 573,576 Wiener's Covering Lemma, 33 1 Wronskian determinant, 205 Wronskian matrix, 205
Corrections to Basic Real Analysis Page 12, line −7 of proof. Change “ 0 ≤ k ≤ 2−n ” to “ 0 ≤ k ≤ 2n ”. Page 12, line −2 of proof. Change “p2−n = b ” to “p2n = b ”. Page 18, lines 10–12. Change the two sentences “By (b), choose j1 ≥ j0 such that |bij − Lj | < 6ε whenever i ∈ {1, . . . , i1 −1} and j ≥ j1 . Then j ≥ j1 implies |bij − Lj | < 6ε for all i whenever j ≥ j1 ” to “By (ii), choose i2 ≥ i1 such that |bij − Lj | < 6ε whenever j ∈ {1, . . . , j0 −1} and i ≥ i2 . Then i ≥ i2 implies |bij − Lj | < 6ε for all j whenever i ≥ i2 ”. Page 23, line 1 of first display. Change “ |f (x) − f (y)| ” to “ |fn (x) − fn (y)| ”. Page 23, second paragraph, line 3. Change “|xk −yk | < δnk () ” to “|xk −yk | < 1/k ”, and change “|fk (xk ) − fk (yk )| ≥ ” to “|fnk (xk ) − fnk (yk )| ≥ ”. Page 23, second paragraph, lines 10–11. Change “fkl ” to “fnkl ” in 6 places. Page 23, second paragraph, line 12. Change “fk ” to “fnk ” in 2 places. Page 23, line −4. Insert the following sentence after “uniformly Cauchy.” “Redefining our indices, we may assume that nk = k for all k ”. Page 31, line 3 of Lemma 1.28. Change “[a, b]− ” to [a, b]∩ ”. Page 33, lines 12–16. Change “If c ≥ 0, then . . . from Lemma 1.25e ” to the following: “If c ≥ 0, then Mi = cMi and mi = cmi , so that U (P, cf ) = cU (P, f ) and L(P, cf ) = cL(P, f ). If c ≤ 0, then Mi = −cmi and mi = −cMi , so that U (P, cf ) = cL(P, f ) and L(P, cf ) = cU (P, f ). In either case, U (P, cf ) − L(P, cf ) = |c|(U (P, f ) − L(P, f )), and (b) follows from Lemma 1.25e ”. Page 34, line 4 of example. Change “ = 0 = limn fn dx ” to “ = 0 = limn fn dx ”. Page 37, lines −10 and −8. Change “ξ ” to “ξi ”. Page 37, lines −6 and −1. Change “Mi Mi∗ ∆yi ” to “
Mi Mi∗ ∆yi ”.
Page 42, line 12. Change “c = p + iq ” to c = r + is ”. Page 48, line 3. Change “
∞ n=0
” to “
N
”.
n=0
Page 49, line −8. Change “[0, +∞) ” to “[1, +∞) ”. Page 51, line 9 of proof of Corollary 1.45. Change “exactly” to “at least” in order to allow for the fact that two of the complex numbers in (x + iy) with 0 ≤ n ≤ 3 have real and imaginary parts ≥ 0 if (x, y) = (1, 0) or (x, y) = (0, 1). Page 51, line −6. Change “(sin x)/ cos x) ” to “(sin x)/(cos x) ”. Page 52, line 2 of Example. Change “of f by ” to “of F by ”. Page 55, line 2 of Example. Change “log(1 + t) ” to “log(1 + x) ”.
Page 56, line 1. Change “complex ” to “numerical ”. Page 56, line 3. Change “C ” to “R ”. Page 56, line 9 of Example. Change “sk =
∞ k=1
” to “sn =
n k=1
”.
Page 58, line 4 of Section 9. Change “on a bounded interval ” to “on a bounded interval [a, b] ”. 1/√n 1/√n Page 59, line 2 of proof. Change “ −1/√n (1 − x2 ) dx ” to “ −1/√n (1 − x2 )n dx ”. 1 Page 59, line 1 of caption of Figure 1.1. Change “cn −1 (1 − x2 )n dx ” to “cn (1 − x2 )n ”. Page 52, line 2, change “even ” to “odd ”. Page 52, line 3, change “odd ” to “even ”. n N Page 65, line −1. Change “ n=1 (sin nx)/n ” to “ k=1 (sin kx)/k ”. Page 70, line 5 of Theorem 1.57. Change “sn ” to “sN ”. Page 72, condition (iii). Change “as n tends to infinity ” to “as N tends to infinity ”. Page 73, line −5 of proof of Theorem 1.59. Change “use of (i) shows ” to “use of (i) and (ii) shows ”.
Page 75, line before Lemma 1.63. Change “ M 2π ” to “M ”.
Page 76, line −9. Change “|f + g| ” to |f − g| ”. Page 76, line −8. Change “
sup
|f (t) + g(t)| ” to “
t∈[xi−1 ,xi ]
sup
|f (t) − g(t)| ”.
t∈[xi−1 ,xi ]
Page 77, line −8. Change “sin x ” to “sin nx ”, and change “cos x ” to “cos nx ”. Page 78, Problem 1. This problem needs an extra step so as to be manageable as the first problem in the book. Therefore change it to read as follows: “(a) Derive the archimedean property (Corollary 1.3) from the convergence of bounded monotone increasing sequences (Corollary 1.6). “(b) Using (a), derive the least-upper bound property (Theorem 1.1) from the convergence of bounded monotone sequences (Corollary 1.6).” x x Page 79, Problem 12b. Change “ 1 ” to “ 0 ”. Page 79, Problem 13. Change “0 < x < π ” to “−1 < x < 1 ”, and change “0 < x < π/2 ” to “−1 < x < 1 ”. Page 79, Problem 19, line 4. Change “for all c ” to “for all x ”. Page 80, Problem 21, last line. Change “g (x) = 0 ” to “g (0) = 0 ”. Page 80, last line. Change “for k ≥ 0 ” to “for k ≥ 1 ”. Page 81, Problem 27, line 1. Change “sn (f ; x) ” to “sn (f ; θ) ”. Page 84, line −1. Change “b = y − z ” to “b = z − y ”.
2 Page 90, line 2 of proof of Lemma 2.2. Change “0 ≤ x − y −2 (x, y)y ” to 2 “0 ≤ x − y −2 (x, y)y ”.
Page 93, line 9. Change “limit point of X ” to “limit point of A ”. Page 96, line −2. Change “for fixed x is continuous in y ” to “for fixed y is continuous in x ”. Page 99, line 10. Change displayed line to read |fn (s)| ≤ |fn (s) − f (s)| + |f (s) − fN (s)| + |fN (s)| ≤ 2 + |fN (s)| Page 99, line 11. Change “+1 ” to “+2 ”. Page 110, proof of Corollary 2.37, line 5. Change “convergent subsequence ” to “convergent subsequence in Rn . By Corollary 2.23 the limit is in C ”. Page 112, line −12. Change “|Q(z)| ≤ ” to “|Q(reiθ )| ≤ ”. Page 114, line −6 of proof of Theorem 2.46. Change “2−m ” to “1/m ”. Page 114, line −5 of proof of Theorem 2.46. Change “2−m ” to “1/m ”. Page 115, line −1. Change “[c, u) ” to “[c, x) ”. Page 116, line 1 of proof of Proposition 2.49. Change “be metric spaces ” to “be metric spaces with X connected ”. Page 121, 6 lines before Theorem 2.56. Change “t ∈ [a, b] ” to “t ∈ S ”. Page 122, displayed definition of δn (). Change “|f (x) − f (y)| ” to “|fn (x) − fn (y)| ”. Page 122, paragraph “Thus we are to prove”, line 3. Change “δnk () ” to “1/k ”, and change “fk ” to “fnk ” twice. Page 122, lines −9 and −8. Change “fkl ” to “fnkl ” six times. Page 122, line −7. Change “fk ” to “fnk ” twice. Page 123, line 1. Insert the following sentence after “uniformly Cauchy.” “Redefining our indices, we may assume that nk = k for all k ”. Page 126, line 11. Change “dense in AR ” to “C(S, C) ”. Page 126, Example 3, line 3. Change “S m−1 ” to “S n−1 ”. Page 128, line 6. Change “relation ∼ on X ” to “relation ∼ on Cauchy(X) ”. Page 130, Problem 2. Change “an open set that is at most countable ” to “a countable open set ”. (This change is not logically necessary but may help avoid confusion, since “countable” and “at most countable” mean the same thing in this book.) Page 132, Problem 21, line 5. Change “Acl ” to “A ”. N N Page 234, statement of Proposition 5.1a. Change “ n−1 En ” to “ n=1 En ”. Page 239, line 5. Change “Problem 11 at the end ” to “Problem 33 at the end ”. Page 290, line −3. Change “exists a Lebesgue ” to “exists a bounded Lebesgue ”.
Page 366, second display on page. Change conditions at right side and one summation index on second line so that the display reads: if x ≥ x0 , x0 ≤xn ≤x cn + x ≤x <x dn “S(x) = 0 n − x<xn ≤x0 cn − x≤xn ≤x0 dn if x ≤ x0 , ”. ∂2 ” on the right side. ∂t2 m≥n ” to “supm∈D, m≥n tm ”.
Page 392, first display on page. Include the term “ + Page 481, Problem 7b, line 2. Change “supm∈D,
Page 481, Problem 7b, line 3. Change “lim supn tn ≥ t ” to “lim supn tn ≤ t ”. Page 556, lines 8–9. Change “whenever y ∈ B and z ∈ B are such that (x, y) and (y, z) are in f , then y = z ” to “there is exactly one y ∈ B such that (x, y) is in f ”. Page 557, lines 10–11. Change “range an unnamed set T , and x∈S Ax is the set of all y ∈ T such that y is in Ax for some x ∈ S ” to “range the set of all subsets of an unnamed set T , and x∈S Ax is the set of all y ∈ T such that y is in Ax for some x ∈ S ”. Page 581, number 1, line 1. Change “Let E ” to “The derivation for (a) is similar to the proof of Corollary 1.3. For (b), let E ”. Page 581, number 1, line 8. Add at the end: “Doing so makes uses of (a).” Page 589, number 21, line 1. Change “If x and y are given ” to “If x and y are given with x = y ”. Page 600, number 13. Replace with the following: “13. Let B be the set of all subsets E of X × X such that there exists a set SE in A with Ex = SE for all but countably many x in X. Every rectangle in A × A is in B. In fact, there are two kinds of sets to check, sets E = A × B with A countable, in which case Ex is empty except for x in the countable set A, and sets Ac × B with A countable, in which case Ex = B except for x in A. Also B is a σ-algebra.
In fact, let sets En in B be given with associated sets SEn . Then En x = ((En )x ) = SEn except when x is in the countable exceptional set for some n; also if E and SE are given, then (E c )x = (Ex )c = (SE )c except when x lies in the exceptional set for E. Finally the diagonal D is not in B and therefore cannot be in A × A. In fact, Dx = {x} for each x, and there can be at most one x with Dx = SD , whatever SD is.” 9/5/07
