Constructive Analysis (Grundlehren Der Mathematischen Wissenschaften)

Grundlehren der mathematischen Wissenschaften

279

A Series of Comprehensive Studies in Mathematics

Errett Bishop Douglas Bridges Constructive Analysis

Springer-Verlag Berlin Heidelberg NewYork Tokyo

Errett Bishop

Douglas Bridges

Constructive Analysis

Springer-Verlag Berlin Heidelberg NewYork Tokyo

Errett Bishop t Department of Mathematics University of California, San Diego, USA Douglas Bridges Department of Mathematics University of Buckingham Buckingham, MK18 1EG, England

Mathematics Subject Classification (1980)' 03F65, 46R05, 28C05, 28C10

ISBN 3-540-15066-8 Springer-Verlag Berlin Heidelberg New York Tokyo ISBN 0-387-15066-8 Springer-Verlag New York Heidelberg Berlin Tokyo

Library of Congress Cataloging in Publication Data Bishop, E rret t 1928-1983 Con· ,

structive analysis (Grundlehren der mathematischen Wissensch a ften

,

279) An outgrowth

of Foundations of constructive analysis I Errett Bisho p [1967] Bibliography p Inclu­ des index 1 Mathematical analysis - Foundations I 1945-

II

Bridges, D S (Douglas S ),

Bishop, Errett, 1928-1983 Foundations of constructive analysis

III Title IV Series QA299 8 B57

1985

515

85·2828

ISBN 0·387·15066·8 (U S)

This work is subject to copyright All rights are reserved, whether the whole or part of the material is concerned, specifically those of translation, reprinting, re·use of illustrations, broadcasting, reproduction by photocopying machine or similar means, and storage in data banks Under § 54 of the German Copyright Law where copies are made for other than private use a fee is payable to "Verwertungsge,ellschaft Wort", Munich © Springer· Verlag Berlin Heidelberg 1985

Printed in Germany Typesetting Universitatsdruckerei H Sturtz AG, Wiirzburg Printing GmbH Berlin Bookbinding Liideritz & Bauer·GmbH, Berlin ,

2141/3020-543210

Mercedes-Druck

This book is dedicated to the memory of

Errett Bishop in the hope that it will promote the achievement of a man remarkable as a person and as a mathematician. vivida vis animi pervicit, et extra processit longe Flammantia moenia mundi atque omne immensum Peragravit, mente animoque

.. the vital strength of his spirit won through, and he made his way far outside the flaming walls of the world and ranged over the measureless whole, in both mind and spirit

Preface

This work grew out of Errett Bishop's fundamental treatise' Foundations of Constructive Analysis' (FCA), which appeared in 1967 and which contained the bountiful harvest of a remarkably short period of research by its author. Truly, FCA was an exceptional book, not only because of the quantity of original material it contained, but also as a demonstration of the practicability of a program which most mathematicians believed impossible to carry out. Errett's book went out of print shortly after its publication, and no second edition was produced by its publishers. Some years later, 'by a set of curious chances', it was agreed that a new edition of FCA would be published by Springer Verlag, the revision being carried out by me under Errett's supervision; at the same time, Errett generously insisted that I become a joint author. The revision turned out to be much more substantial than we had anticipated, and took longer than we would have wished. Indeed, tragically, Errett died before the work was completed. The present book is the result of our efforts. Although substantially based on FCA, it contains so much new material, and such full revision and expansion of the old, that it is essentially a new book. For this reason, and also to preserve the integrity of the original, I decided to give our joint work a title of its own. Most of the new material outside Chapter 5 originated with Errett. In particular, there is a full and much improved account, in Chapter 6, of the Bishop-Cheng theory of integration; and there is a completely new approach to the theory of Banach algebras, in Chapter 9 Of special interest also is the last section of Chapter 5, in which are found necessary and sufficient conditions for the existence of a Riemann mapping function. One important part of FCA has been omitted from our new book: the work on martingales and ergodic theory. Errett had some ideas for improving that section, but unfortunately he had no time to put them on paper before he died. Following his advice, and not wishing to delay the publication of this book any longer, I decided to leave

VIII

Prefdce

out that material altogether Jane Bishop has suggested that I revise Errett's ergodic theory for publication separately in an expository paper, I intend to do this in the near future I have retained (with minimal changes) as a Prolog Errett's testamental preface to FCA; I have also retained his first chapter, 'A Constructivist Manifesto'. Errett saw, contributed to, and approved most of the main text of the book outside Chapter 7. In Chapter 7 he made substantial improvements to my original draft of the material on approximation theory (in Section 2) and the Radon-Nikodym theorem (in Section 3); he never saw the rest of the chapter. The other parts of the book which he did not see are the first section of Chapter 2; the discussion of the Jordan curve theorem preceding Lemma (7.9) of Chapter 5, and the proof of that lemma (which was based on a suggestion of his); Sections 8-10 of Chapter 6; and the footnotes at the end of each chapter. I do not know if Errett would have included a proof of the Jordan curve theorem. There is a strong case for its inclusion, in order to make the book self-contained. However, bearing in mind that the intuition underlying the Jordan curve theorem is clear, and that there is an excellent presentation of its proof in a paper by Julian et al. [5], I decided to follow the usual practice in texts on complex analysis, and omit its proof. The perceptive reader will notice that the literary style of this book is different from that of FCA. This is hardly surprising: most of the drafts, and the entire final manuscript, were prepared by me, and it seemed both natural and sensible to use my style, rather than adapt it to Errett's. In doing this, I was not implying any criticism of Errett's style; nor, incidentally, did he ever criticize mine. I take full responsibility for all the material in this book which Errett did not see, and for any errors and omissions in the final version of the text. Although in theory the prerequisites for understanding this book are few (some familiarity with elementary calculus, linear algebra, and the basic notions of abstract algebra), in practice it requires a level of mathematical maturity achieved by few undergraduates. Indeed, a better appreciation of the similarities and contrasts between the classical and constructive developments will be gained by the reader with some experience of classical functional analysis and measure theory. A word about internal references: within the text, a citation of the form (m n) refers to (Theorem, Proposition, or Lemma) n in Section m of the current chapter; a citation from another chapter will have the form' by (m.n) of Chapter ... '

Preface

IX

During the writing of this book, support was provided by the University of Buckingham, New Mexico State University, and Massey University. In addition, I enjoyed the hospitality of the Bishop family, of Bill and Nancy Julian, and of Edna and Bruce Mawson. Bill Julian, Ray Mines, Fred Richman, and Garth Dales read parts of the manuscript, and have given me good advice and much encouragement over the years. Several improvements to the book were brought about by the good offices of Michael Dummett. Julie Cakebread took out of my hands a large share of the burden of typing, which she carried with great patience and skill. Special thanks are due to my wife and children, who have borne with great fortitude my limited contribution to our family life over the past two years. Buckingham, May 1985

Douglas S. Bridges

Contents

Prolog

1

Chapter 1 A Constructivist Manifesto

4

1. The Descriptive Basis of Mathematics 2. The Idealistic Component of Mathematics 3. The Constructivization of Mathematics Notes

13

Chapter 2 Calculus and the Real Numbers

14

1. 2. 3. 4. 5. 6. 7.

14 18

Sets and Functions . . . . . . . . The Real Number System . . . . . Sequences and Series of Real Numbers Continuous Functions Differentiation . . . . . . Integration ...... . Certain Important Functions Problems Notes

Chapter 3

Set Theory

4

6 9

28 35 44'

5c1

55 62 64

67

1. Some Basic Notions of the Theory of Sets 2. Complemented Sets . . . . . . . . . . 3 Neighborhood Spaces and Function Spaces Problems Notes

67 72

Chapter 4

81

Metric Spaces

1. Fundamental Definitions and Constructions 2. Associated Structures . . . . . . . 3. Completeness ......... . 4. Total Boundedness and Compactness

75 78 79

81

87 89 94

Contents

Xl

5. Spaces of Functions 6. Locally Compact Spaces Problems Notes

100 109 121 125

Chapter 5 Complex Analysis

128

1. 2. 3. 4. 5. 6. 7.

128 130 134 142 152 163 180 210 213

The Complex Plane Derivatives Integration The Winding Number Estimates of Size, and the Location of Zeros Singularities and Picard's Theorem The Riemann Mapping Theorem Problems Notes

Chapter 6 Integration

215

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

216 222 232 236 245 252 259 265 277 282 292 297

Integration Spaces . . . . . . Complete Extension of an Integral Integrable Sets . . . . Profiles . . . . . . . . . . . Positive Measures on IR . . . . Approximation by Compact Sets Measurable Functions . . . . . Convergence of Functions and Integrals Product Integrals Measure Spaces Problems Notes

Chapter 7 Normed Linear Spaces

299

1. 2. 3. 4. 5. 6. 7. 8.

299 306 313 334 342 350 357 363 390 396

Definitions and Examples Finite-Dimensional Spaces . . The Lp Spaces and the Radon-Nikodym Theorem The Extension of Linear Functionals Quasinormed Linear Spaces; the Space Loo Dual Spaces . . . . . . . . . . . . Extreme Points . . . . . . . . . . . Hilbert Space and the Spectral Theorem Problems Notes . . . . . . . . . . . . . . .

XII

Contents

Chapter 8

Locally Compact Abelian Groups

399

1. Haar Measure Convolution Operators 3 The Character Group . 4. Duality and the Fourier Transform Problems Notes

419 424 434 447 449

Chapter 9

450

2.

Commutative Banach Algebras

399

1. Definitions and Examples . . . . . 2. Linear Equations in a Banach Algebra Problems Notes .

450

References

463

Symbols

467

Index

453 461 462

. 470

Prolog

Most mathematicians would find it hard to believe that there could be any serious controversy about the foundations of mathematics, any controversy whose outcome could significantly affect their own mathematical activity. Their attitude well represents the actual state of affairs: during a half-century of splendid mathematical progress there has been no deviation from the norm The voices of dissent, never much heeded, have long been silent Perhaps the times are not conducive to introspection Mathematics flourishes as never before, its scope is immense, its quality high. Mathematicians flourish as never before, their profession is respectable, their salaries good. Mathematical methods are more fashionable than ever before: witness the surge of interest in mathematical logic, mathematical biology, mathematical economics, mathematical psychology - in mathematical investigations of every sort. The extent to which many of these investigations are premature or unrealistic indicates the deep attraction mathematical exactitude holds for the contemporary mind. And yet there is dissatisfaction in the mathematical community. The pure mathematician is isolated from the world, which has little need of his brilliant creations. He suffers from an alienation which is seemingly inevitable. he has followed the gleam and it has led him out of this world. If every mathematician occasionally, perhaps only for an instant, feels an urge to move closer to reality, it is not because he believes that mathematics is lacking in meaning. He does not believe that mathematics consists in drawing brilliant conclusions from arbitrary axioms, of juggling concepts devoid of pragmatic content, of playing a meaningless game. On the other hand, many mathematical statements have a rather peculiar pragmatic content Consider the theorem that either every even integer greater than 2 is the sum of two primes, or else there exists an even integer greater than 2 that is not the sum of two primes. The pragmatic content of this theorem is not that if we go to the integers and observe, we shall see certain things happening

XII

Contents

Chapter 8 1. 2.

3 4.

Locally Compact Abelian Groups

Haar Measure Convolution Operators The Character Group . Duality and the Fourier Transform Problems Notes

Chapter 9

Commutative Banach Algebras

399 399

419 424 434 447 449 450

1. Definitions and Examples . . . . . 2. Linear Equations in a Banach Algebra Problems Notes .

450 453 461 462

References

463

Symbols

467

Index

. 470

Prolog

Most mathematicians would find it hard to believe that there could be any serious controversy about the foundations of mathematics, any controversy whose outcome could significantly affect their own mathematical activity. Their attitude well represents the actual state of affairs: during a half-century of splendid mathematical progress there has been no deviation from the norm The voices of dissent, never much heeded, have long been silent Perhaps the times are not conducive to introspection Mathematics flourishes as never before, its scope is immense, its quality high. Mathematicians flourish as never before, their profession is respectable, their salaries good. Mathematical methods are more fashionable than ever before: witness the surge of interest in mathematical logic, mathematical biology, mathematical economics, mathematical psychology - in mathematical investigations of every sort. The extent to which many of these investigations are premature or unrealistic indicates the deep attraction mathematical exactitude holds for the contemporary mind. And yet there is dissatisfaction in the mathematical community. The pure mathematician is isolated from the world, which has little need of his brilliant creations. He suffers from an alienation which is seemingly inevitable. he has followed the gleam and it has led him out of this world. If every mathematician occasionally, perhaps only for an instant, feels an urge to move closer to reality, it is not because he believes that mathematics is lacking in meaning. He does not believe that mathematics consists in drawing brilliant conclusions from arbitrary axioms, of juggling concepts devoid of pragmatic content, of playing a meaningless game. On the other hand, many mathematical statements have a rather peculiar pragmatic content Consider the theorem that either every even integer greater than 2 is the sum of two primes, or else there exists an even integer greater than 2 that is not the sum of two primes. The pragmatic content of this theorem is not that if we go to the integers and observe, we shall see certain things happening

2

Prolog

Rather, the pragmatic content of such a theorem, if it exists, lies in the circumstance that we are going to use it to help derive other theorems, themselves of peculiar pragmatic content, which in turn will be the basis for further developments. It appears then that there are certain mathematical statements that are merely evocative, that make assertions without empirical validity. There are also mathematical statements of immediate empirical validity, which say that certain performable operations will produce certain observable results: for instance, the theorem that every positive integer is the sum of four squares. Mathematics is a mixture of the real and the ideal, sometimes one, sometimes the other, often so presented that it is hard to tell which is which The realistic component of mathematics - the desire for pragmatic interpretation supplies the control which determines the course of development and keeps mathematics from lapsing into meaningless formalism. The idealistic component permits simplifications, and opens possibilities which would otherwise be closed. The methods of proof and the objects of investigation have been idealized to form a game, but the actual conduct of the game is ultimately motivated by pragmatic considerations. For 50 years now there have been no significant changes in the rules of this game. Mathematicians unanimously agree on how mathematics should be played. Accepted standards of performance suffice to regulate the course of mathematical activity, and there is no prospect that these standards will be changed in any significant respect by a revision of the idealistic code. In fact, no efforts are being made to impose such a revision. There have been, however, attempts to constructivize mathematics, to purge it completely of its idealistic content. The most sustained attempt was made by L.E.1. Brouwer, beginning in 1907. The movement he founded has long been dead, killed partly by compromises of Brouwer's disciples with the viewpoint of idealism, partly by extraneous peculiarities of Brouwer's system which made it vague and even ridiculous to practising mathematicians, but chiefly by the failure of Brouwer and his followers to convince the mathematical public that abandonment of the idealistic viewpoint would not sterilize or cripple the development of mathematics. Brouwer and other constructivists were much more successful in their criticisms of classical mathematics than in their efforts to replace it with something better. Many mathematicians familiar with Brouwer's objections to classical mathematics concede their validity but remain unconvinced that there is any satisfactory alternative. This book is a piece of constructivist propaganda, designed to

Prolog

3

show that there does exist a satisfactory alternative. To this end, we develop a large portion of abstract analysis within a constructive framework. This development is carried through with an absolute minimum of philosophical prejudice concerning the nature of constructive mathematics. There are no dogmas to which we must conform. Our program is simple' to give numerical meaning to as much as possible of classical abstract analysis. Our motivation is the well-known scandal, exposed by Brouwer (and others) in great detail, that classical mathematics is deficient in numerical meaning. Some familiarity with Brouwer's critique is essential. Following Brouwer, Chapter 1 is primarily devoted to an examination of the defects of classical mathematics, and a presentation of the thesis that all mathematics should have numerical meaning. Chapter 3 presents constructive versions of the fundamental concepts of sets and functions, and examines some of the obstacles to the constructivization of general topology The remaining chapters are primarily technical, and constitute a course in abstract analysis from the constructive point of view. Very little formal preparation is required of the reader, although a certain level of mathematical sophistication is probably indispensible. Every effort has been made to follow the classical development as closely as possible; digressions have been relegated to notes at the ends of the various chapters. The task of making analysis constructive is guided by three basic principles. First, to make every concept affirmative. (Even the concept of inequality is affirmative.) Second, to avoid definitions that are not relevant. (The concept of a pointwise continuous function is not relevant; a continuous function is one that is uniformly continuous on compact intervals.) Third, to avoid pseudogenerality. (Separability hypotheses are freely employed.) The book has a threefold purpose: to present the constructive point of view, to show that the constructive program can succeed, and to lay a foundation for further work. These immediate ends tend to an ultimate goal - to hasten the inevitable day when constructive mathematics will be the accepted norm Weare not contending that idealistic mathematics is worthless from the constructive point of view. This would be as silly as contending that unrigorous mathematics is worthless from the classical point of view. Every theorem proved with idealistic methods presents a challenge: to find a constructive version, and to give it a constructive proof. Errett Bishop

Chapter 1. A Constructivist Manifesto

1. The Descriptive BaSIS of Mathematics Mathematics is that portion of our intellectual actIvIty which transcends our biology and our environment The principles of biology as we know them may apply to life forms on other worlds, yet there is no necessity for this to be so. The principles of physics should be more universal, yet it is easy to imagine another universe governed by different physical laws. Mathematics, a creation of mind, is less arbitrary than biology or physics, creations of nature, the creatures we imagine inhabiting another world in another universe, with another biology and another physics, will develop a mathematics which in essence is the same as ours In believing this we may be falling into a trap. Mathematics being a creation of our mind, it is, of course, difficult to imagine how mathematics could be other than it is without our actually making it so, but perhaps we should not presume to predict the course of the mathematical activities of all possible types of intelligence. On the other hand, the pragmatic content of our belief in the transcendence of mathematics has nothing to do with alien forms of life. Rather, it serves to give a direction to mathematical investigation, resulting from the insistence that mathematics be born of an inner necessity The primary concern of mathematics IS number, and this means the positive integers. We feel about number the way Kant felt about space. The positive integers and their arithmetic are presupposed by the very nature of our intelligence and, we are tempted to believe, by the very nature of intelligence in general. The development of the theory of the positive integers from the primitive concept of the unit, the concept of adjoining a unit, and the process of mathematical induction carries complete conviction. In the words of Kronecker, the positive integers were created by God. Kronecker would have expressed it even better if he had said that the positive integers were created by God for the benefit of man (and other finite beings). Mathematics belongs to man, not to God. We are not interested in properties of the

1 The Descriptive Basis of Mathematics

5

posItIve integers that have no descriptive meaning for finite man. When a man proves a positive integer to exist, he should show how to find it. If God has mathematics of his own that needs to be done, let him do it himself. Almost equal in importance to number are the constructions by which we ascend from number to the higher levels of mathematical existence. These constructions involve the discovery of relationships among mathematical entities already constructed, in the process of which new mathematical entities are created. The relations which form the point of departure are the order and arithmetical relations of the positive integers From these we construct various rules for pairing integers with one another, for separating out certain integers from the rest, and for associating one integer with another Rules of this sort give rise to the notions of set and function. A set is not an entity which has an ideal existence. a set exists only when it has been defined To define a set we prescribe, at least implicitly, what we (the constructing intelligence) must do in order to construct an element of the set, and what we must do to show that two elements of the set are equal. A similar remark applies to the definition of a function: in order to define a function from a set A to a set B, we prescribe a finite routine which leads from an element of A to an element of B, and show that equal elements of A give rise to equal elements of B Building on the positive integers, weaving a web of ever more sets and more functions, we get the basic structures of mathematics: the rational number system, the real number system, the euclidean spaces, the complex number system, the algebraic number fields, Hilbert space, the classical groups, and so forth. Within the framework of these structures most mathematics is done. Everything attaches itself to number, and every mathematical statement ultimately expresses the fact that if we perform certain computations within the set of positive integers, we shall get certain results. Mathematics takes another leap, from the entity which is constructed in fact to the entity whose construction is hypothetical To some extent hypothetical entities are present from the start whenever we assert that every positive integer has a certain property, in essence we are considering a positive integer whose construction is hypothetical. But now we become bolder and consider a hypothetical set, endowed with hypothetical operations subject to certain axioms In this way we introduce such structures as topological spaces, groups, and manifolds. The motivation for doing this comes from the study of concretely constructed examples, and the justification comes from the possibility of applying the theory of the hypothetical structure to the

6

Chapter 1 A Constructivist Manifesto

study of more than one specific example Recently it has become fashionable to take another leap and study, as it were, a hypothetical hypothetical structure ~ a hypothetical structure qua hypothetical structure. Again the motivations and justifications attach themselves to particular examples, and the examples attach themselves to numbers in the ultimate analysis Thus even the most abstract mathematical statement has a computational basis The transcendence of mathematics demands that it should not be confined to computations that I can perform, or you can perform, or 100 men working 100 years with 100 digital computers can perform. Any computation that can be performed by a finite intelligence ~ any computation that has a finite number of steps ~ is permissible. This does not mean that no value is to be placed on the efficiency of a computation An applied mathematician will prize a computation for its efficiency above all else, whereas in formal mathematics much attention is paid to elegance and little to efficiency. Mathematics should and must concern itself with efficiency, perhaps to the detriment of elegance, but these matters will come to the fore only when realism has begun to prevail. Until then our first concern will be to put as much mathematics as possible on a realistic basis without close attention to questions of efficiency

2. The Idealistic Component of Mathematics Geometry was highly idealistic from the time of Euclid and the ancients until the time of Descartes, unfolding from axioms taken either to be self-evident or to reflect properties of the real world. Descartes reduced geometry to the theory of the real numbers, and in the nineteenth century Dedekind, Weierstrass, and others, by the arithmetization of the real number system, brought space into the concrete realm of objects constructed by pure thought Unfortunately, the promise held out to mathematics by the arithmetization of space was not fulfilled, largely due to the intervention, around the turn of the century, of the formalist program. The successful formalization of mathematics helped keep mathematics on a wrong course. The fact that space has been arithmetized loses much of its significance if space, number, and everything else are fitted into a matrix of idealism where even the positive integers have an ambiguous computational existence. Mathematics becomes the game of sets, which is a fine game as far as it goes, with rules that are admirably precise. The game becomes its own justification, and the

2 The Idealistic Component of Mathematics

7

fact that it represents a highly idealized version of mathematical existence is universally ignored. Of course, idealistic tendencies have been present, if not dominant, in mathematics since the Greeks, but it took the full flowering of formalism to kill the insight into the nature of mathematics which its arithmetization could have given To see how some of the most basic results of classical analysis lack computational meaning, take the assertion that every bounded nonvoid set A of real numbers has a least upper bound. (The real number b is the least upper bound of A if a ~ b for all a in A, and if there exist elements of A that are arbitrarily close to b.) To avoid unnecessary complications, we actually consider the somewhat less general assertion that every bounded sequence (x k ) of rational numbers has a least upper bound b (in the set of real numbers). If this assertion were constructively valid, we could compute b, in the sense of computing a rational number approximating b to within any desired accuracy; in fact, we could program a digital computer to compute the approximations for us. For instance, the computer could be programmed to produce, one by one, a sequence ((b k , m k )) of ordered pairs, where each b k is a rational number and each m k is a positive integer, such that Xj ~ bk + k - 1 for all positive integers j and k, and x mk ~ bk - k - 1 for all positive integers k. Unless there exists a general method M that produces such a computer program corresponding to each bounded, constructively given sequence (x k ) of rational numbers, we are not justified, by constructive standards, in asserting that each of the sequences (x k ) has a least upper bound. To see the scope such a method M would have, consider a constructively given sequence (n k ) of integers, each of which is either 0 or 1. Using the method M, we compute a rational number b 3 and a positive integer N == m3 such that (i) nj~b3 +! for all positive integers j, and (ii) nN~b3 -j. Either nN=O or nN = 1. If nN =0, then (i) and (ii) imply that

for all j. Since each nj is either 0 or 1, it follows that nj = 0 for all j Thus for each of the sequences (n k ) being considered, the method M either produces a proof that the n k are all equal to 0, or produces a positive integer N such that nN = 1. Of course, such a method M does not exist, and nobody expects that one will ever be found Such a method would solve most of the famous unsolved problems of mathematics - in particular, Fermat's last theorem, the Goldbach conjecture, and the Riemann hypothesis, since each of these problems can be reduced to finding, for a certain sequence (n k ) of the type being

8

Chapter 1 A Constructivist Manifesto

considered, either a proof that nk = 0 for all k or a proof that nk = 1 for some k. For another instance, consider the intuitively appealing theorem that every continuous real-valued function f on the closed interval [O,IJ, with f(O)
a ==

L3 k~

1

-k 11 k ,

There exists a unique constructively given continuous

function f on [0, 1J such that f(O) = -1, f (1) = 1, f(i) = f(~) = a, and f is linear on each of the intervals [0,1J, [1, nand [%,1]. If our theorem is valid, there exists a point Xo with f(x o) = 0 By computing a sufficiently close rational approximation to x o , we show that either Xo < ~ or Xo > i In the first case, a ~ 0, and therefore the first nonzero term of the sequence (n k ), if one exists, equals 1. Similarly, in the second case, the first nonzero term, if one exists, equals -1. Thus our theorem gives a method, which, applied to each of the sequences (n k ) being considered, either (i) proves that any term that equals 1 is preceded by a term that equals -1, or (ii) proves that any term that equals -1 is preceded by a term that equals 1. Nobody believes that such a method will ever be found. Brouwer fought the advance of formalism and undertook the disengagement of mathematics from logic He wanted to strengthen mathematics by associating with every theorem and every proof a pragmatically meaningful interpretation. His program failed to gain support He was an indifferent expositor and an inflexible advocate, contending against the great prestige of Hilbert and the undeniable fact that idealistic mathematics produced the most general results with the least effort More important, Brouwer's system itself had traces of idealism and, worse, of metaphysical speculation. There was a preoccupation with the philosophical aspects of constructivism at the expense of concrete mathematical activity A calculus of negation was developed which became a crutch to avoid the necessity of getting precise constructive results. It is not surprising that some of Brouwer's precepts were then formalized, giving rise to so-called intuitionistic

3 The Constructivization of Mathematics

9

number theory, and that the formal system so obtained turned out not to be of any constructive value. In fairness to Brouwer it should be said that he did not associate himself with these efforts to formalize reality, it is the fault of the logicians that many mathematicians who think they know something of the constructive point of view have in mind a dinky formal system or, just as bad, confuse constructivism with recursive function theory Brouwer became involved in metaphysical speculation by his desire to improve the theory of the continuum. A bugaboo of both Brouwer and the logicians has been compulsive speculation about the nature of the continuum In the case of the logicians this leads to contortions in which various formal systems, all detached from reality, are interpreted within one another in the hope that the nature of the continuum will somehow emerge. In Brouwer's case there seems to have been a nagging suspicion that unless he personally intervened to prevent it, the continuum would turn out to be discrete. He therefore introduced the method of free-choice sequences for constructing the continuum, as a consequence of which the continuum cannot be discrete because it is not well enough defined This makes mathematics so bizarre it becomes unpalatable to mathematicians, and foredooms the whole of Brouwer's program. This is a pity, because Brouwer had a remarkable insight into the defects of classical mathematics, and he made a heroic attempt to set things right.

3. The Constructivization of Mathematics A set is defined by describing exactly what must be done in order to construct an element of the set, and what must be done in order to show that two elements are equal. There is no guarantee that the description will be understood; it may be that an author thinks he has described a set with sufficient clarity but a reader does not understand. For an illustration, consider the set of all sequences (n k ) of integers. To construct such a sequence we must give a rule which associates an integer nk with each positive integer k in such a way that for each value of k the associated integer nk can be determined in a finite number of steps by an entirely routine process Now this definition could perhaps be interpreted to admit sequences (n k ) in which n k is constructed by a search, the proof that the search actually produces a value of nk after a finite number of steps being given in some formal system Of course, we do not have this interpretation in mind, but it is impossible to consider every possible interpretation of our definition

10

Chapter 1 A Constructivist Manifesto

and say whether that is what we have in mind. There is always ambiguity, but it becomes less and less as the reader continues to read and discovers more and more of the author's intent, modifying his interpretations if necessary to fit the intentions of the author as they continue to unfold. At any stage of the exposition the reader should be content if he can give a reasonable interpretation to account for everything the author has said. The expositor himself can never fully know all the possible ramifications of his definitions, and he is subject to the same necessity of modifying his interpretations, and sometimes his definitions as well, to conform to the dictates of experience The constructive interpretations of the mathematical connectives and quantifiers have been established by Brouwer. To prove the statement (P and Q) we must prove the statement P and prove the statement Q, just as in classical mathematics. To prove the statement (P or Q) we must either prove the statement P or prove the statement Q, whereas in classical mathematics it is possible to prove (P or Q) without proving either the statement P or the statement Q. The connective "implies" is more complicated. To prove (P implies Q) we must show that P necessarily entails Q, or that Q is true whenever P is true. The validity of the computational facts implicit in the statement P must ensure the validity of the computational facts implicit in the statement Q, but the way this actually happens can only be seen by looking at the proof of the statement (P implies Q). Statements formed with this connective - for example, statements of the type ((P implies Q) implies R) - have a less immediate meaning than the statements from which they are formed, although in actual practice this does not seem to lead to difficulties in interpretation. The negation (not P) of a statement P is the statement (P implies (0 = 1)) Classical mathematics makes no distinction between the content of the statements P and (not (not P)), whereas constructively the latter is a weaker statement. Brouwer's system makes essential use of negation in defining, for instance, inequality and set complementation. Thus two elements of a set A are unequal according to Brouwer if the assumption of their equality somehow allows us to compute that 0 = 1. It is natural to want to replace this negativistic definition by something more affirmative, phrased as much as possible in terms of specific computations leading to specific results. Brouwer himself does just this for the real number system, introducing an affirmative and stronger relation of inequality in addition to the negativistic relation already defined. Experience shows that it is not necessary to define inequality in terms of negation. For those cases in which an inequality relation in needed,

3 The Constructivization of Mathematics

11

it is better to introduce it affirmatively; the same remarks apply to set complementation. Van Dantzig and others have gone so far as to propose that negation could be entirely avoided in constructive mathematics Experience bears this out: in many cases where we seem to be using negation - for instance, in the assertion that either a given integer is even or it is not - we are really asserting that one of two finitely distinguishable alternatives obtains. Without intending to establish a dogma, we may continue to employ the language of negation but reserve it for situations of this sort (at least until experience changes our minds) and for counterexamples and purposes of motivation This will have the advantage of making mathematics more immediate, and in certain situations forcing us to sharpen our results Proofs by contradiction are constructively justified in finite situations. When we have proved that one of finitely many alternatives holds at a certain stage in the proof of a theorem, to finish the proof of the theorem it is enough to show that the theorem is a consequence of each of the alternatives. Should one of the alternatives lead to a contradiction - that is, imply (0 = 1) - either we may say that the alternative in question is ruled out and pass on to the consideration of the other alternatives, or we may be more meticulous and prove that the theorem is a consequence of the equality 0 = l. A universal statement, to the effect that every element of a certain set A has a certain property P, has the same meaning in constructive as in classical mathematics. To prove such a statement we must show by some general argument that if x is any element of A, then x has property P Constructive existence is much more restrictive than the ideal existence of classical mathematics. The only way to show that an object exists is to give a finite routine for finding it, whereas in classical mathematics other methods can be used. In fact, the following principle is valid in classical mathematics: Either all elements of A have property P or there exists an element of A with property (not P) This principle, which we shall call the principle of omniscience, lies at the root of most nonconstructivity in classical mathematics. This is already true of the principle of omniscience in its simplest form. if (n k ) is a sequence of integers, then either nk = 0 for some k or nk =1= 0 for all k. We shall call this the limited principle of omniscience. Theorem after theorem of classical mathematics depends in an essential way on the limited principle of omniscience, and is therefore not constructively valid. Some instances of this are: the theorem that a continuous realvalued function on a closed, bounded interval attains its maximum; the fixed-point theorem for a continuous map of a closed cell into

12

Chapter 1 A Constructivist Manifesto

itself, the ergodic theorem, and the Hahn-Banach theorem Nevertheless these theorems are not lost to constructive mathematics: each of these theorems P has a constructive substitute Q which is a constructively valid theorem Q implying P in the classical system by a more or less simple argument based on the limited principle of omniscience. For example, the statement "every continuous function from a closed cell in euclidean space into itself admits a fixed point" finds a constructive substitute in the theorem that such a function admits a point which is arbitrarily near to its image. The extent to which good constructive substitutes exist for the theorems of classical mathematics can be regarded as a demonstration that classical mathematics has a substantial underpinning of constructive truth When a classical mathematician claims he is constructivist, he probably means he avoids the axiom of choice. This axiom is unique in its ability to trouble the conscience of the classical mathematician, but in fact it is not a real source of nonconstructivity in classical mathematics. A choice function exists in constructive mathematics, because a choice is implied by the very meaning of existence. Applications of the axiom of choice in classical mathematics either are irrelevant or are combined with a sweeping use of the principle of omniscience. The axiom of choice is used to extract elements from equivalence classes where they should never have been put in the first place. For example, a real number should not be defined as an equivalence class of Cauchy sequences of rational numbers; there is no need to drag in the equivalence classes. The proof that the real numbers can be well ordered is an instance of a proof in which a sweeping use of the principle of omniscience is combined with an appeal to the axiom of choice. Such proofs offer little hope of constructivization: it is not likely that the theorem "the real numbers can be well ordered" will be given a constructive version consonant with the intuitive interpretation of the classical result. Almost every conceivable type of resistance has been offered to a straightforward realistic treatment of mathematics, even by constructivists. Brouwer, who has done more for constructive mathematics than anyone else, though it necessary to introduce a revolutionary, semimystical theory of the continuum Weyl, a great mathematician who in practice suppressed his constructivist convictions, expressed the opinion that idealistic mathematics finds its justification in its applications to physics. Hilbert, who insisted on constructivity in metamathematics but believed that the price of a constructive mathematics was too great, was willing to settle for consistency. Brouwer's disciples joined forces with the logicians in attempts to formalize

Notes

13

constructive mathematics Others seek constructive truth in the framework of recursive function theory Still others look for a short cut to reality, a point of vantage which will suddenly reveal classical mathematics in a constructive light. None of these substitutes for a straightforward realistIc approach has worked It is no exaggeration to say that a straightforward realistic approach to mathematics has yet to be tried It is time to make the attempt.

Notes Errett Bishop was never happy with the standard constructive interpretation of implication (the one given in Section 3). Among the alternatives he felt worthy of serious investigation is "Godel implication", as discussed in [9]. Bishop also worked on a deeper study of implication, but unfortunately he left only fragmentary notes on his ideas At first sight, Bishop' remark, "A choice function exists in constructive mathematics, because a choice is implied by the very meaning of existence", appears to be contradicted by counterexamples of the sort discussed in connection with the least-upper-bound principle. In fact, there is no contradiction here. To see this, consider a paraphrase of Bishop's remark: if to each x in a set A there corresponds an element y of a set B such that a given property P(x, y) holds, then it is implied by the very meaning of existence in constructive mathematics that there is a finite routine for computing an appropriate y in B from a given x in A; although this routine may not be a function relative to the given equality relation on A, it is a function relative to the equality relation of identity (intensional equality) on A, in which two elements are equal if and only if they are given as identically the same object

Chapter 2. Calculus and the Real Numbers

Section 1 establishes some conventions about sets and functions. The next three sections are devoted to constructing the real numbers as certain Cauchy sequences of rational numbers, and investigating their order and arithmetic. The rest of the chapter deals with the basic ideas of the calculus of one variable. Topics covered include continuity, the convergence of sequences and series of continuous functions, differentiation, integration, Taylor's theorem, and the basic properties of the exponential and trigonometric functions and their inverses Most of the material is a routine constructivization of the corresponding part of classical mathematics; for this reason it affords a good introduction to the constructive approach.

We assume that the reader is familiar with the order and arithmetic of the integers and the rational numbers. For us, a rational number will be an expression of the form plq, where p and q are integers with q*O. Two rational numbers plq and p'lq' are equal if pq'=p'q. The integer n is identified with the rational number nil. There are geometric magnitudes which are not represented by rational numbers, and which can only be described by a sequence of rational approximations. Certain such approximating sequences are called real numbers. In this chapter we construct the real numbers and study their basic properties. Then we develop the fundamental ideas of the calculus.

1. Sets and Functions Before constructing the real numbers, we introduce some notions which are basic to much of mathematics. The totality of all mathematical objects constructed in accordance with certain requirements is called a set. The requirements of the

1 Sets and Functions

15

construction, which vary with the set under consideration, determine the set. Thus the integers form a set, the rational numbers form a set, and (we anticipate here the formal definition of 'sequence') the collection of all sequences of integers is a set. Each set will be endowed with a binary relation = of equality. This relation is a matter of convention, except that it must be an equivalence relation, in other words, the following conditions must hold for all objects x, y, and z in the set: (11)

(i)

X=X

(ii) If X= y, then y=x (iii) If x = y and y = z, then x = z. The relation of equality given above for rational numbers is an equivalence relation. In this example there is a finite, mechanical procedure for deciding whether or not two given objects in the set are equal Such a procedure will not exist in general: there are instances in which we are unable to decide whether or not two given elements of a set are equal; such an instance, in the theory of real numbers, will be given later. We use the standard notation aEA to denote that a is an element, or member, of the set A, or that the construction defining a satisfies the requirements a construction must satisfy in order to define an object of A. We also use the notation {aI' a 2 , ... } for a set whose elements can be written in a (possibly finite) list. The dependence of one quantity on another is expressed by the basic notion of an operation An operation from a set A into a set B is a finite routine f which assigns an element f(a) of B to each given element a of A. This routine must afford an explicit, finite, mechanical reduction of the procedure for constructing f(a) to the procedure for constructing a. If it is clear from the context what the sets A and B are, we sometimes denote f by a 1-+ f(a), in order to bring out the form of f(a) for a given element a of A. The set A is called the domain of the operation, and is denoted by dmn! In the most important case, we have f(a}=f(a') whenever a,a'EA and a=a'; the operation f is then called a function, or a mapping of A into B, or a map of A into B For two functions f, g from A into B, f = g means that f(a) = g(a) for each element a of A. Taken with this equality relation, the collection of all functions from A into B becomes a set. The notation f: A -+ B indicates that f is a function from the set A to the set B. A function x whose domain is the set 7l+ of positive integers is called a sequence. The object x.=x(n) is called the nth term of the

16

Chapter 2

Calculus and the Real Numbers

sequence. The finite routine x can be given explicitly, or it can be left to inference: for example, by writing the terms of the sequence in order (X),X 2 ,· ) until the rule of their formation becomes clear. Different notations for the sequence whose nth term is Xn are: nt--->x n, (x),x 2 , ..• ), (xn):;O=), and (xJ Thus the sequence whose nth term is n 2 can be written nt--->n z, or (1,4,9, ... ), or (nZ):;o= I' or simply (nZ). A subsequence of a sequence (xn) consists of the sequence (xn) and a sequence (nk)r= I of positive integers such that n 1
of all ordered n-tuples (XI'···' xn) with Xl EX l' X z EX 2'.· , and XnEX n· Elements (x)' ... , xn) and (x~, ... , x~) of the cartesian product are equal if the coordinates (or components) Xk and x~ are equal elements of X k for each k. If X is a finite sequence of elements of a set B, then X can be identified with the element (x(l), ... , x(n)) of the cartesian product

where n is the length of x Returning to functions in general, we say that a function f: A ..... B maps A onto B if to each element b of B there corresponds an element a of A with f(a) = b. In other words, f maps A onto B if there is an operation g from B into A such that f(g(b)) = b for each b in B. A set A is countable if there exists a mapping of lL+ onto A, intuitively, this means that the elements of A can be arranged in a sequence with possible duplications. The elements of the cartesian product lL x lL of the set lL of integers with itself can be arranged in a sequence as follows. We order the elements (m, n) of lL x lL, first according to the value of Iml + Inl, then according to the value of m, and finally according to the value of

1 Sets and Functions

17

n. This produces the sequence (12)

«0,0), (-1,0), (0, -1), (0, 1), (1,0), (-2,0), (-1, -1), ... ),

in which each element of II x II occurs exactly once. In the sequence (1.2), omit every term (m, n) with n = 0, and replace each term (m, n) with n=l=O by min; this produces the sequence (1.3)

(0/ -1,0/1, -1/ -1, .. ),

in which every expression.p/q, with p and q integers and q =1= 0, occurs exactly once Keeping only the term 0/1 of (1.3), and those terms for which q > 0, p =1= 0, and p is relatively prime to q, we obtain a sequence (1 4)

(0/1, -1/1,1/1,

which has the property that for any given rational number r there exists exactly one term equal to r. F or each positive integer n, let lln be the set {O, 1, ... , n -l} If there is a mapping of lln onto the set A, then we say that A has at most n elements. A set with at most n elements for some n is said to be subfinite, or finitely enumerable. Note that every subfinite or countable set has at least one element. Before we introduce stronger notions than countability and sub· finiteness, we must discuss the composition of functions The composition of two functions f: A ---+ Band g: B ---+ C is the function g of: A ---+ C defined by (gof)(a)==g(f(a))

(aEA).

Composition is associative: h a (g oj) = (h a g) of

whenever the compositions are defined If f. A ---+ B, g: B ---+ A, and g(f(a)) = a for all a in A, then the function g is called a left inverse of f, and the function f is called a right inverse of g (Note that f has a left inverse if and only if it is one-one, in the sense that a = a' for all elements a, a' of A with f(a) = f(a')) When g is both a left and a right inverse of f, then it is simply called an inverse of j; f is then called a one-one correspondence, or a bijection, and the sets A and B are said to be in one-one correspondence with each other A set which is in one-one correspondence with the set ll+ of positive integers is said to be countably infinite. For example, let f be the sequence (1.4), and define a function g from the set
18

Chapter 2 Calculus and the Real Numbers

is count ably infinite. A similar proof using (1.2) shows that Z x Z is countably infinite A set which is in one-one correspondence with Z. is said to have n elements, and to be finite. Every finite set is countable. It is not true that every countable set is either countably infinite or subfinite. For example, let A consist of all positive integers n such that both nand n+2 are prime; then A is countable, but we do not know if it is either countably infinite or subfinite. This does not rule out the possibility that at some time in the future A will have become countably infinite or subfinite; it is possible that tomorrow someone will show that A is subfinite. This set A has the property that if it is subfinite, then it is finite. Not all sets have this property.

2. The Real Number System The following definition is basic to everything that follows. (2.1) Definition. A sequence (x.) of rational numbers is regular if (2.1.1)

A real number is a regular sequence of rational numbers. Two real numbers x == (x.) and y == (y.) are equal if (2.1.2)

The set of real numbers is denoted by JR.. (2.2) Proposition. Equality of real numbers is an equivalence relation. Proof: Parts (i) and (ii) of (Ll) are obvious. Part (iii) is a consequence of the following lemma. (2.3) Lemma. The real numbers x == (x.) and y == (y.) are equal if and only if for each positive integer j there exists a positive integer Nj such that (2.3.1) Proof: If x = y, then (2.3.1) holds with N; == 2j.

2. The Real Number System

19

Assume conversely that for each j in Z+ there exists ~ satisfying (2.3.1). Consider a positive integer n. If m and j are any positive integers with m~max V,N), then

Ixn - Ynl ~Ixn-xml +Ix m - Yml +Iym- Ynl ~(n-l

+m-l)+r 1+(n- 1 +m- 1 )<2n- 1 + 3r 1.

Since this holds for all j in Z +, (2.1.2) is valid.

0

Notice that the proof of Lemma(2.3) singles out a specific Nj satisfying (2.3.1). This situation is typical: every proof of a theorem which asserts the existence of an object must embody, at least implicitly, a finite routine for the construction of the object. The rational number Xn is called the nth rational approximation to the real number x (xJ Note that the operation from 1R. to CQ which takes the real number x into its nth rational approximation is not a function. For later use we wish to associate with each real number x=(x n) an integer K", such that

=

This is done by letting K", be the least integer which is greater than IXll +2. We call K", the canonical bound for x. The development of the arithmetic of the real numbers offers no surprises: we operate with real numbers by operating with their rational approximations. (2.4) Definition. Let x=(x n) and Y=(Yn) be real numbers with respective canonical bounds K", and Ky. Write k=max{Kx,K y}.

Let ex be any rational number. We define (a) x + Y

=(x

2n

+ Y2n):'= 1

(b) xy=(x 2kn Y2kn):'= 1

(c) max {x,y}

=(max {xn,Yn}):'=

1

(d) -x=( -xn):'= 1

(e) ex*=(ex,cx,ex, ... ). (2.5) Proposition. The sequences x + y, x y, max {x, Y}, - x, and ex* of Definition (2.4) are real numbers.

20

Chapter 2

Calculus and the Real Numbers

Proof (a) Write zn == x2n + Y2n Then x + Y == (zJ For all positive tegers m and n,

In-

IZm - znl:;:; IX 2m - xlnl + IYlm - Y2nl :;:; (2n)- 1+ (2m)- 1+ (2n)- 1+ (2m)- 1= n- 1+ m-I. Thus x + Y is a real number. (b) Write zn==xlknYZkn' Then xy==(zJ For all positive integers m and n, IZm - znl = IX lkm (Y2km - YZkn) + Y2kn(x lkm - X2kn )1 :;:; klYZkm - YZknl + klx 2km - X2kn l :;:; k«2km)- I + (2kn)-1 + (2km)- I + (2kn)- 1)= n- 1+ m- 1. Thus x Y is a real number. (c) Write zn==max{xn,Yn} Then max{x,y}==(z.). Consider positive integers m and n. For simplicity assume that Then IZm - znl = IX m- max {xn' Yn} I =xm -max {xn' Yn} :;:;x m-Xn :;:;n- 1+m- 1. Thus max {x,y} is a real number. (d) For all positive integers m and n, I-xm- (-xn)1

=

IX m -xnl :;:;m- I +n- I •

Thus - x is a real number (e) This is obvious D There is no trouble in proving that (x,y)~x+ y, (x,y)~xy, and are functions from lRxlR to lR, that x~-x is a function from lR to lR; and that rx~rx* is a function from
is therefore a function from lR to lR, and the operation (x, y)~min {x, y} == -max { -x, - y} is a function from lR x lR to lR The next proposition states that the real numbers obey the same rules of arithmetic as the rational numbers. (2.6) Proposition. For arbitrary real numbers x, y, and

numbers rx and {3, (a) x+y=y+x. xy=yx

Z

and rational

2 The Real Number System

21

(b) (x+y)+z=x+(y+z), x(yz)=(xy)z

(c) x(y+z)=xY+xz (d) O*+x=x, l*x=x (e) x-x=O* (f) IxYI = Ixllyl (g) (rx + (3)* = rx,* + [3*, (rx,[3)* = rx,* [3*, and (- rx,)* = - rx,*. We omit the simple proofs of these results. We shall use standard notations, such as x + y + z and max {x, y, z}, without further comment. There are three basic relations defined on the set of real numbers. The first of these, the equality relation, has already been defined. The remaining relations, which pertain to order, are best introduced in terms of certain subsets 1R + and 1R 0 + of 1R. (2.7) Definition. A real number x == (x n) is positive, or XE 1R +, if (2.7.1) for some n in '1.+. A real number x==(x n) is nonnegative, or xE1R o +, if (2.7.2) The following criteria are often useful (2.8) Lemma. A real number x == (xn) is positive exists a positive integer N such that

if and only if there

(2.8.1) A real number x == (xn) is nonnegative there exists Nn in '1.+ such that

if and only if for each n in

'1.+

(2.8.2) Proof: Assume that xE1R+. Then xn>n- 1 for some n in '1.+. Choose N in '1.+ with

Then Xm~Xn -Ix m -xnl ~xn _m- 1 _n- 1 ~xn-n-l-N-l

>N- 1

whenever m~ N. Therefore (2.8.1) is valid. Conversely, if (2.8.1) is valid, then (2.71) holds with n=N+l Therefore XE1R +.

22

Chapter 2 Calculus and the Real Numbers

Assume next that xElR o +. Then for each positive integer n, (m~n).

Therefore (2.8.2) is valid with Nn == n. Assume finally that (2.8.2) holds. Then if k, m, and n are positive integers with m ~ Nn , we have xk~xm-Ixm-xkl ~

_n- 1 _k- 1 _m- 1•

Since m and n are arbitrary, this gives x k ~ xElR o +. 0

-

k -1. Therefore

As a corollary of Lemma (2.8), we see that if x and yare equal real numbers, then x is positive if and only if Y is positive, and x is nonnegative if and only if y is nonnegative. It is not strictly correct to say that a real number (xn) is an element of IR +. An element of IR + consists of a real number (xn) and a positive integer n such that xn > n -1, because an element of IR + is not presented until both (xn) and n are given. One and the same real number (xn) can be associated with two distinct (but equal) elements of IR + Nevertheless we shall continue to refer loosely to a positive real number (xJ On those occasions when we need to refer to an n for which Xn > n -1, we shall take the position that it was there implicitly all along. The proof of the following proposition is now easy, and will be left to the reader. For convenience, IR * represents either IR + or IR 0 +. (2.9) Proposition. Let x and y be real numbers. Then

(a) x + YEIR * and XYEIR * whenever xEIR * and YEIR * (b) X+YEIR+ whenever xEIR+ and YElR o+ (c) IxiElR o + (d) max {X,Y}EIR* whenever xEIR* (e) min {X,Y}EIR* whenever XEIR* and YEIR*. We now define the order relations on IR. (2.10) Definition. Let x and Y be real numbers. We define x>y(ory<x)

ifx-YEIR+

X~y

if X-YElR o +.

and (or y;£x)

A real number x is negative if x < 0* - that is, if - x is positive.

2 The Real Number System

23

Consider real numbers x, x', y, and y' such that (i) x=x', y=y', and x>y We have X'-y'=X-YEJR+ and therefore (ii) x' > y'. We express the fact that (ii) holds whenever (i) is valid by saying that> is a relation on JR. More formally, a relation on a set X is a subset S of X x X such that if x, x', y, y' are elements of X with x=x', y=y', and (X,Y)ES, then (X',Y')ES. We express the fact that x> y if and only if y < x by saying that > and < are transposed relations. Similarly, ~ and ~ are transposed relations. If x < y or x = y, then x ~ y. The converse is not valid: as we shall see later, it is possible that we have x ~ y without being able to prove that x < y or x = y. For this reason it was necessary to define the relations < and ~ independently of each other. The following rules for manipulating inequalities are easily proved from Proposition (2.9). We omit the proofs. (2.11) Proposition. For all real numbers x, y, z, and t, (a) x
(c) x+ y~z+t

whenever

x~z

and

y~t

(d) x+y
(f) xyO* (h)

if x
(i)

max{x,y}~x

(g)

then -x> - y then -x~ - y

G) min {x, y} ~ x (k) if x~y and y~x, then

X=

y

(1) Ixl~O*

(m)

Ix+yl~lxl+lyl.

An important property of the relation <, of which we shall make no use, is the antisymmetry property, which states that at most one of the relations x < y and y < x is valid for given real numbers x and y. This negative statement has no place in the affirmative mathematics we are trying to develop, except as motivation. Its place is taken by the affirmative statement (k) of Proposition (2.11). As a general principle, negative statements are only for counterexamples and motivation; they are not to be used in subsequent work.

24

Chapter 2

Calculus and the Real Numbers

(2.12) Definition. For real numbers x and Y we write x =!= Y if and only if x < y or x > y. Inequality =!= is a relation because both < and > are relations. As motivation we have the negative statement that at most one of the relations x =!= y, x = y can hold for given real numbers x and y. In other words, at most one of the relations x < y, x> y, x = y can hold. This is clear from the definitions. The following proposition defines the inverse x -1 of a real number x =!= 0*, and derives the basic properties of the operation x f-+ X-I. (2.13) Proposition. Let x be a nonzero real number (so that IxIEIR. +). There exists a positive integer N with Ixml ~ N - 1 for m ~ N. Define Yn==(X N 3)-1

(n
and (n~N).

Then X-I == (Yn)~= 1 is a real number which is positive if x is positive, and negative if x is negative; also xx- 1 = 1 *. If t is any real number for which xt=1*, then t=X-I. The operation Xf-+X- 1 is afunction. If x =1=0 and y=!=O, then (xy)-1 =x- 1y-l. If IX =!= 0 is rational, then (1X*)-1 =(1X- 1)*. If x =!= 0, then (x- 1)-1 = x.

Proof: Our definitions guarantee that IYnl ~ N for all n. Consider positive integers m and n. Write j == max {m, N},

k

== max {n, N}.

Then IYm - Ynl = IYmllYnll x jN2 - X kN 21 ~ N 2«(jN 2)-1 +(kN 2)-I)=F 1 +k- 1 ~m-l + n- 1.

Therefore X-I is a real number. Assume now that x> 0*. Then by (2.8), Xn > 0 for all sufficiently large n. Hence Yn > K; 1 (where Kx is the canonical bound for x) for all sufficiently large n. It follows from (2.8) that X-I> 0*. A similar proof shows that X-I < 0* whenever x < 0*. Let k be the maxim urn of the canonical bounds for x and X-I. Write xx- 1 ==(z.). Then Therefore

IZn -1 *1 = IX 2nN kl- l1x 2nk - X2nN kl 2

2

~ IY2nkl«2nk)-1 +(2nN 2k)-l)~ n- 1

for n ~ N. It follows that xx- 1 = 1*.

2 The Real Number System

2S

If t is any real number with xt = 1*, then X-I =x-I(xt)=(x- I x)t=(xx-I)t=t.

If x= x', then x'x- l =xx- I =I*.

Therefore X-I = (X')-I. It follows that Xl->X- I is a function. If x~o and y~o, then (xy)x- I y-I =xx- I yy-I = 1*.

Therefore X-I y-I = (xy)-I. If IX ~ 0 is rational, then IX* == (IX, IX, ... ). Therefore (IX*)-I =(IX-t,IX- I , ... )=(IX- I )*.

For each x in lR +, X-I is in lR +, and thus (X-I)-I exists. Since x- I x=xx- I =I*, it follows that (X-I)-I=X. Similarly (X-I)-I=X if x is negative Therefore (x -1) - I = x whenever x ~ 0

o.

Of course, we often write x/y instead of xy-l when x and yare real numbers with y~o. As the previous propositions show, (IX f3)* = IX* f3*, (IX + f3)* = IX* + f3*, (-IX)*= -IX*, (IIXI)*=IIX*I, and (IX-I)*=(IX*)-I for all rational numbers IX and f3. Also 1X~f3 if and only if IX* ~f3*, where ~ stands for any of the relations =, <, >, and ~. This situation is expressed by saying that the map IXI->IX* is an order isomorphism from CQ into lR. This justifies identifying CQ with a subset of lR, as we previously identified Z with a subset of CQ. Henceforth we make no distinction between a rational number IX and the corresponding real number IX* The next lemma shows that the nIh rational approximation Xn to a real number x == (x n) actually approximates x to within n -I. (2.14) Lemma. For each real number x==(x n), we have Ix-xnl ~n-I

Proof: By (2.4) and the definition of tion to n- 1 -lx-xnl is

(nEZ+).

I I, the mlh rational approxima-

n-I-lx4m-xnl;;;;n-l_«4m)-I+n-I)= -(4m)-I> _m- 1 .

By (2.7), we have n- 1 -lx-x nIElR o+. Therefore Ix-xnl~n-I.

0

(2.15) Lemma. If x==(x n) and y==(Yn) are real numbers with x
26

Chapter 2 Calculus and the Real Numbers

Proof: By (2.4), we have Y-X=(Y2n-x2n):'=I. Since y-xe1R +, by (2.7) there exists n in Z+ with Y2n-X2n>n-l. Write

1X=!(X 2n + Y2n)·

Then Also, Y-IX~Y2n-1X -IY2n- yl ~!(Y2n-X2n)-(2n)-1 >0.

Therefore x < IX < y.

D

As a corollary, for each x in 1R and r in 1R + there exists with Ix -IXI < r. Here is another corollary. (2.16) Proposition. If

XI' ••. ' xn are real numbers with then x;>O for some i (1 ~i~n).

XI

IX

in
+ ... + xn > 0,

Proof. By (2.15), there exists a rational number IX with 0
Ix; -a;1 < (2n)-IIX. Then n

n

n

La;~ LX;- Llx;-a;!>!IX. ;=1

;=1

;=1

Therefore a; > (2n)-11X for some i. For this i it follows that x;~a;-lx;-a;I>O.

D

(2.17) Corollary. If x, y, and z are real numbers with Y < z, then either x y. Proof: Since z-x+x-y=z-y>O, either z-x>O or x-y>O, by

(2.16).

D

The next lemma gives an extremely useful method for proving inequalities of the form x ~ y. (2.18) Lemma. Let x and y be real numbers such that the assumption x> y implies that 0 = 1. Then x ~ y. Proof· Without loss of generality, we take y=O. For each n in Z+, either Xn ~ n -lor Xn > n -I. The case Xn > n - 1 is ruled out, since it implies that x> o. Therefore - Xn ~ - n - I for all n, and so - x ~ o. Thus x~O. D

2 The Real Number System

27

(2.19) Theorem. Let (an) be a sequence of real numbers, and let Xo and Yo be real numbers with Xo < Yo Then there exists a real number x such that xo~x~Yo and X =1= an for all n in 7l+.

Proof: We construct by induction sequences (xn) and (Yn) of rational numbers such that (i) xO~xn~xm
(m~n~l)

(ii) xn>a n or Ynx n- 1 or anx n_ 1 , let Xn be any rational number with xn_1<xn<min{an,Yn_l}' and let Yn be any rational number with xn
Similarly iYm-Yni Yn' then an> Y = x and so an =1= x. Thus x has the required properties. 0 Theorem (2.19) is the famous theorem of Cantor, that the real numbers are uncountable. The proof is essentially Cantor's" diagonal" proof. Both Cantor's theorem and his method of proof are of great importance. The time has come to consider some counterexamples. Let (n k ) be a sequence of integers, each of which is either 0 or 1, for which we are unable to prove either that nk = 1 for some k or that nk = 0 for all k. This corresponds to what Brouwer calls "a fugitive property of the natural numbers". For example, such a sequence can be defined as follows. Let nk be 0 if ul + Vi =1= Wi for all integers u, v, w, t with 0 < u, v, w ~ k and 3 ~ t ~ 2 + k. Otherwise let nk be 1. Then we are unable to prove nk = 1 for some k, because this would disprove Fermat's last theorem. We are unable to prove n k = 0 for all k, because this would prove Fermat's last theorem. Now define xk==O if nj=O for all j~k, and x k==2- m otherwise, where m is the least positive integer such that nm = 1. Then x == (x k ) is a

28

Chapter 2 Calculus and the Real Numbers

nonnegative real number, but we are unable to prove that x> 0 or x = O. Since nothing is true unless and until it has been proved, it is untrue that x> 0 or x = O. Of course, if Fermat's last theorem is proved tomorrow, we shall probably still be able to define a fugitive sequence (n k ) of integers. Thus it is unlikely that there will ever exist a constructive proof that for every real number x~O either x>O or x=O. We express this fact by saying that there exists a real number x ~ 0 such that it is not true that x>O or x=O. In much the same way we can construct a real number x such that it is not true that x~O or x~O.

3. Sequences and Series of Real Numbers We develop methods for defining a real number in terms of approximations by other real numbers. (3.1) Definition. A sequence (x.) of real numbers converges to a real number Xo if for each k in 7l+ there exists Nk in 7l+ with (3.1.1)

The real number Xo is then called a limit of the sequence (xJ To express the fact that (x.) converges to Xo we write lim X.=Xo .-00

or

as n-+oo or simply x. -+ xo' A sequence (x.) of real numbers is said to converge, or be convergent, if there exists a limit Xo of (xJ It is easily seen that if (x.) converges to both

Xo

and

x~,

then

Xo

=x~.

A convergent sequence is bounded: there exists r in IR. + such that n A convergent sequence of real numbers is not determined until the limit Xo and the sequence (Nk ) are given, as well as the sequence (x n ) itself. Even when they are not mentioned explicitly, these quantities are implicitly present. Similar comments apply to many subsequent definitions, including the following.

Ix.1 ~r for all

3 Sequences and Series of Real Numbers

29

(3.2) Definition. A sequence (x.) of real numbers is a Cauchy sequence if for each k in 7l+ there exists Mk in 7l+ such that (3.2.1)

IXm-x.l~k-l

(m,n~Mk)'

(3.3) Theorem. A sequence (x.) of real numbers converges

if and

only

if

it is a Cauchy sequence. Proof' Assume that (x.) converges to a real number xo' Let the sequence (Nk ) satisfy (3.1.1). Write M k ==N2k • Then

Ixm-x.1 ~ IX m-xol +Ix.-xol ~(2k)-1 +(2k)-1 =k- 1 for m, n~Mk' Therefore (x.) is a Cauchy sequence. Assume conversely that (x.) is a Cauchy sequence. Let the sequence (M k ) satisfy (3.2.1). Write Nk == max {3k,M 2k }. Then IXm-x.I~(2k)-1

(m,n~Nk)'

Let Yk be the (2kyh rational approximation to XNk ' For m~n, IYm-y.I~IYm-XNJ+lxNm -xNJ+lxN• -Y.I ~(2m)-1

+(2m)-1 +(2n)-1 +(2n)-1 =m- 1 +n- 1.

Therefore Y==(Y.) is a real number. To see that (xn) converges to y, we consider n ~ Nk and compute ly-x.1 ~ Iy- Y.I + Iy. -xN.1 + Ix N• -xnl ~n-l +(2n)-1 +(2k)-1 ~(3k)-1 +(6k)-1 +(2k)-1 =k- 1.

D

A subsequence of a convergent sequence converges to the same limit If a sequence converges, then any sequence obtained from it by modifications (including, perhaps, insertions or deletions) which involve only a finite number of terms converges to the same limit If x == (x.) is a regular sequence of rational numbers, then (x:) converges to x, by (2.14). A sequence (x.) is increasing (respectively, strictly increasing) if x.+ 1 ~ x. (respectively, x.+ 1> x.) for each n. Decreasing and strictly decreasing sequences are defined analogously, in the obvious way. A theorem of classical mathematics states that every bounded increasing sequence of real numbers converges. A counterexample to this statement is given by any increasing sequence (x.) such that xn = 0 or x. = 1 for each n, but it is not known whether x. = 0 for all n It is useful to supplement Definition (3.1) by writing limx.= 00 .~oo

or

x. --+ 00

as

n --+ 00

30

Chapter 2 Calculus and the Real Numbers

to express the fact that for each k in 7l+ there exists Nk in 7l+ with xn>k for all n~Nk. We also define limxn= -

00

n~oo

or Xn-+ -

00

as

n---+ 00

to mean that lim -Xn= 00. n~oo

The next proposition shows that we may work with real numbers constructed as limits by working with their approximations. (34) Proposition. Assume that Xn-+XO as n-+oo, and Yn-+Yo as n-+oo, where Xo and Yo are real numbers. Then (a) xn+Yn-+xO+Yo as n-+oo (b) XnYn-+XoYo as n---+oo (c) max{x n, Yn} -+max{xo, Yo} as n---+oo (d) Xo = c whenever Xn = c for all n (e) if Xo =!= 0 and Xn =!= 0 for all n, then x; 1-+ xi) 1 as n-+ 00 (f)

if Xn~Yn

for all n, then xo~yo·

Proof: (a) For each k in 7l+ there exists Nk in 7l+ such that

Then Therefore xn+Yn-+xO+Yo as n---+oo. (b) Choose m in 7l+ such that IYol ~m and IXnl ~m for all n. For each k in 7l+ choose Nk in 7l+ with IXn-xol~(2mk)-1,

IYn-Yol~(2mk)-1

(n~Nk)·

Then for n ~ Nk , IX nYn-Xo Yol ~ IXn(Yn - Yo)1 + IYo(xn-xo)1 ~m(lYn - Yol + IXn-xol)~k-1. Therefore xnyn-+xoYo as n-+oo. (c) Since Imax{x n, Yn} -max{x o, Yo}1 ~max{lxn -xol, IYn - Yol}, it follows that max {xn,Yn}-+max{x o, Yo}

as n-+oo.

3 Sequences and Series of Real Numbers

31

(d) If Xn = c for all n, then (xn) converges to c. Therefore Xo = c. (e) Since IXol >0, IXnl ~ IXol-lxn-xol >1lxol whenever n is large enough, say for n ~ no' Let k and n be positive integers such that n~no and IXn-xol «2k)-llxoI2. Then Ix; I_XO II = IXnl-Ilxol-llxn -xol ~ 2lxol- 2(2k)-llxoI2 =k- I. Therefore x; 1-+ Xo I as n -+ 00. (f) We compute

Yo -Xo = lim Yn- lim xn= lim (Yn-xn)= lim IYn -xnl n~oo

.~oo

= lim max{Yn-xn, x n- Yn} =max{yo -X o, Xo - Yo} ~O, n~oo

by (a), (b), (c), and (d).

0

For each sequence (xn) of real numbers the number

is called the nth partial sum of (x.), and (sn) is called the sequence of partial sums of the sequence (xn)' A sum So of (xn) is a limit of the sequence (sn) of partial sums. We write

to indicate that So is a sum of (xn). A sequence which is meant to be summed is called a series. A series is said to converge to its sum. Thus the sequence (2 -n):."'= I converges to 0 as a sequence, but as a series it 00

converges to

L 2 - n= 1.

n= I A convergent series remains convergent, but not necessarily to the same sum, after modification of finitely many of its terms. 00

The series (x.) is often loosely referred to as the series 00

If the series 00

L

L xn converges, then x.-+O as n-+oo. 11=1

L x n. 11=

1

00

L

A series Xn is said to converge absolutely when the series IXnl converges. .= I n= I In classical analysis a series of nonnegative terms converges if the partial sums are bounded. This is not true in constructive analysis However, we have the following result.

32

Chapter 2

Calculus and the Real Numbers 00

L Yn

(3.5) Proposition. If

is a convergent series of nonnegative terms,

n= 1

and

if IXnl :-;:;; Yn for

00

L Xn converges.

each n, then

n= 1 00

Proof Since

L Yn

is convergent, the sequence of partial sums is a

n= 1

Cauchy sequence. Therefore for each kin '1.+ there exists an Nk in '1.+ with '" Yj:-;:;;k- 1

L

j=n+ 1

Then

00

Therefore the sequence of partial sums of the series sequence. By (3.3), the series converges. D

L Xn is

a Cauchy

n= 1

The criterion of Proposition (3.5) is known as the comparison test It follows from the comparison test that every absolutely convergent 00 series is convergent. The terms of an absolutely convergent series L Xn may be ren= 1

ordered without affecting the sum

So

of the series. More precisely, if A:

00

'1.+ -+'1.+ is a bijection, then

L x;'(n)

exists and equals so. This may

n= 1

00

L Xn is merely convergent.

not be true if the series

11=

1

A sequence (xn) is said to diverge if there exists E in 1R. + such that for each k in '1.+ there exist m and n in '1.+ with m, n~k and Ix", -xnl ~ E. The motivation for this definition is, of course, that a sequence cannot be both convergent and divergent. A series is said to diverge if the sequence of its partial sums diverges. 00

The series

L n-

1

diverges, because

n= 1

00

The series

L xn

diverges whenever there exists r in 1R. + such that

n= 1

IXnl ~r

for infinitely many values of n. 00

Let

L Xn "=1

00

and

L Yn

be series of nonnegative terms. The compari-

"=1

son test for divergence is that

00

L Xn 11=

00

diverges whenever

1

and there is a positive integer N with

L Yn 11=

Xn ~ Yn

for all n ~ N.

1

diverges

3 Sequences and Series of Real Numbers

33

The following very useful test for convergence and divergence is called the ratio test. 00

(3.6) Proposition. Let

L Xn be a series,

c a positive number, and N a

n= 1 00

positive integer. Then

L Xn converges if c < 1 and n= 1

(3.6.1)

and diverges

if c> 1 and

(3.6.2)

Proof. Assume that c< 1 and that (3.6.1) is valid. Then Ixnl ~cn-NlxNI 00

for n ~ N. By the comparison test,

L xn converges. n= 1

Next, assume that c> 1 and that (3.6.2) holds. Then IXnl~cn-N-1IxN+11~lxN+11

(n~N+l)

and 00

Therefore

L Xn diverges.

0

n= 1

A corollary of the ratio test is that if the limit L= limlxn+1x;11 n_oo 00

exists, then L Xn converges whenever L < 1 and diverges whenever L>1. n=l The ratio test says nothing in case L = 1. To handle this case, we introduce stronger tests based on Kummer's criterion. (3.7) Lemma. Let (an) and (xn) be sequences of positive numbers, c a 00

positive number, and N a positive integer. Then an Xn ---+ 0 as n -> 00 and (3.7.1) 00

while (3.7.2)

00

L xn diverges if L a;

1

diverges and

L Xn

converges

if

34

Chapter 2 Calculus and the Real Numbers

Proof: Assume that anxn-+O and that (3.7.1) is valid. Let e be an arbitrary positive number, and choose an integer v ~ N so that a k X k -ajxj~ce whenever j>k~v. For suchj and k we have j

L

j

L

xn~c-1

Xn(an_1Xn_1X,;-1-an) n=k+1 =c- 1(a k x k -ajxj)~ e.

n=k+1

Thus ttl Xn): 1 is a C:uch y sequence, and so Next assume that

L a,;-l ft=

diverges and that (3.7.2) holds. Then for

1

00

each n~N, Xn~aNxNa,;-l. Thus

L Xn

diverges, by comparison with

n= 1

00

D

n= 1

n~l xn converges.

(3.8) Lemma. Let (Yn) be a sequence of positive numbers, c a positive number, and N a positive integer such that n(YnY';-_,\ -1)~c

(n~N).

Then limYn=O. n~oo

Proof: For each n>N, YN y; 1 =(YN YN~ l)(YN+ 1YN~ 2)'" (Yn-1 Y';- 1) ~(1 +cN- 1) ... (1 +c(n-1)-1) n-1 ~1+c L k- 1. k=N

n-1

Given e>O, choose an integer v>N so that

L k- 1 >c- 1(e- 1YN-1)

k=N

for all n~v. Then for such n we have Yn<e. Hence Yn-+O as n--+oo.

D

The next convergence test is known as Raabe's test. 00

(3.9) Proposition. Let

L Xn ft=

1

be a series of positive numbers such that 00

n(Xnx';-;1-1) converges to a limit L. Then L xn converges n= 1 diverges if L< 1.

if

Proof: First note that n(n xn/(n + 1)xn+ 1 -1) =n(n+ 1)-1 (n(XnX';-; 1 -1) -1) --+L-1 as n-+ 00.

L>1, and

4. Continuous Functions nXn~O

If L>1, it follows from (3.8) that

35

as n-+oo. We then obtain the

00

convergence of

L Xn

an == n

by taking

(nEZ+) in Kummer's criterion.

n= 1

00

The same choice of an yields divergence of

L Xn in case L < 1.

0

n= 1

Important real numbers represented by series are 00

e=1+ L(n!)-l n= 1

and 00

n=4

L (-1)"(2n+ 1)-1.

n=O

The series for e converges by the ratio test. The convergence of the series for n is a consequence of the general result that a series 00

L (-1)" Xn converges whenever (i) Xn ~ 0 for all nand (ii) the sequence n= 1

is decreasing and converges to O. To see this, consider positive integers m and n with m ~ n. Then

(Xn)

O~(xn-xn+ 1)+(X n + 2 -Xn+3)+'"

+( _l)m+n Xm

m

=( -1)"

L (_1)kXk

=Xn-(X n + 1 -X n + 2 ) - .. ·

+( _1)m+n Xm~Xn'

It follows that the sequence of partial sums of the series is a Cauchy sequence. Therefore the series converges.

4. Continuous Functions A property P which is applicable to the elements of a set S is defined by a statement of the requirements that an element of S must satisfy in order to have property P. To construct an element of S with property P we must construct an element of S, perform certain additional constructions which depend on the property P, and prove that the entities constructed satisfy certain requirements that are characteristic of the property P. Each property P applicable to elements of a set S determines a subset of S which is denoted by {x: XES,

X

has property P}

{XES:

X

has property Pl.

or

36

Chapter 2

Calculus and the Real Numbers

When the context makes it clear which set S is under discussion, we also write simply {x: x has property P}. Properties applicable to elements of a set S, and subsets of S, are essentially the same things regarded from different points of view. Among the most important subsets of lR are the intervals. (4.1) Definition. For all real numbers a and b we define (a,b)=={x: xElR, a<x
Each of these sets in an interval, whose left and right end points are a and b, respectively. The interval (a, b) is open, [a, b] is closed, (a, b] is half-open on the left, and [a, b) is half-open on the right. If a
An interval I is nonvoid if we can construct a real number belonging to I. A nonvoid, closed, finite interval is called a compact interval. A nonvoid finite interval I with left and right end points a and b has length III == b - a If an interval I is a subset of an interval J (that is, if every element of I also belongs to J), then we say that I is a subinterval of J The rules for manipulating intervals, which we use in the sequel without mention or proof, are implicit in Proposition (2.11). (4.2) Definition. A nonvoid set A of real numbers is bounded above if there exists a real number b, called an upper bound of A, such that x:;:; b for all x in A. A real number b is called a supremum, or least upper bound, of A if it is an upper bound of A, and if for each e>O there exists x in A with x> b - e. We say that A is bounded below if there exists a real number b, called a lower bound of A, such that b:;:; x for all x in A. A real

4 Continuous Functions

37

number b is called an infimum, or greatest lower bound, of A if it is a lower bound of A, and if for each t: > 0 there exists x in A with x < b +t:.

The supremum (respectively, infimum) of A is unique, if it exists, and is written sup A (respectively, inf A). A classical theorem asserts that every nonvoid set of real numbers that is bounded above has a supremum. A counterexample to this is provided by the set {x n: nE71+} where xn =0 or xn = 1 for each n, but it is not known whether xn = 0 for all n. We now prove the constructive least-up per-bound principle. (4.3) Proposition. Let A be a nonvoid set of real numbers that is bounded above. Then sup A exists if and only if for all x, y in lR with x < y, either y is an upper bound of A or there exists a in A with x < a. Proof" If sup A exists and x < y, then either sup A < y or x < sup A; in the latter case we can find a in A with

supA-(supA -x) a l . We construct recursively a sequence (an) in A and a sequence (b n) of upper bounds of A such that for each n in 7l+, and (ii)

bn+l-an+l~i(bn-an)

Having found aI' ... ,an and b l , ... ,bn, if an+i(bn-a n) is an upper bound of A, we set bn+l=an+i(bn-an) and an+l=a n; while if there exists a in A with a>an+-i(bn-a n), we set an+l=a and bn+l=b n. This completes the recursive construction. By (i) and (ii), we have O~bn-an~(3/4)n-l(bl-al)

(nE71+).

Hence the sequences (an) and (b n) converge to a common limit t with an~t~bn for each n in 7l+. Since each b n is an upper bound of A, so is t. O~ the other hand, given t: > 0, we can choose n so that t ~ an > t -t:, where anEA. Hence t=supA. 0 In Proposition (4.3), if A is contained in some interval I, then in order to prove that sup A exists it is sufficient to consider arbitrary points x and y in I with x < y.

38

Chapter 2 Calculus and the Real Numbers

(4.4) Corollary. Let the subset A of IR. have the property that for each 1:>0 there exists a subfinite set {YI' ... ,Y.} of points of A such that for each x in A at least one of the numbers Ix-y11, ... ,lx-y.1 is less than 1:. (Such a set A is called totally bounded.) Then sup A and inf A exist.

Proof: It will suffice to prove that sup A exists. To this end, let x and Y be real numbers with xmax {ai' ... , aN} -

IX.

Either x < a. or a. < x + 2 IX. In the latter case, if aE A and we choose k with la-akl

so that y is an upper bound of A. Thus sup A exists, by (4.3).

D

Often when one real number depends on another the dependence is smooth, or continuous. An exact description of what this means is given in the following definition. (4.5) Definition. A real-valued function f defined on a compact interval 1 is continuous on 1 if for each I: >0 there exists w(l:) >0 such that If(x)-f(Y)I~1: whenever x,YEI and Ix-YI~w(I:). The operation I: f-+w(l:) is called a modulus of continuity for f A real-valued function f on an arbitrary interval J is continuous on J if it is continuous on every compact subinterval 1 of J. For example, when a and b are real numbers with a < b, then f is continuous on (a, b) if and only if it is continuous on [a + 15, b - 15] for each 15 with O
4 Continuous Functions

and

39

inff==inf{f(x): xE[a, b]}

(called, respectively, the supremum and the infimum of f on the interval [a, b]) exist. Proof. Consider any e > O. Choose real numbers a = ao ~ a1 ~ ... ~ a. =b such that ai+1-ai~w(e) (O~i~n-1), where w is a modulus of continuity for f Then for each x in [a, b] we have Ix - ad ~ w(e), and therefore If(x) - f(ai)1 ~e, for some i. Since e is arbitrary, it follows that the set {f(x): xE[a, b]} is totally bounded. Therefore supf and inf f exist, by (4.4). D

Let f: A -+ 1R and g: A -+ 1R be two functions defined on the same set A. We define the sum function f + g: A -+ 1R by (f + g) (x) == f(x) + g(x)

(XE A).

Such functions as f g, If I, and max {f, g} are defined similarly. If g(x) 0 for each x in A, then g -1: A -+ 1R is defined by

'*

g-l(X)==(g(X))-l

(xEA).

We also write 1/g for g-l, and fig for the quotient function fg-1. The proof of the following proposition, which resembles the proof of Proposition (3.4), is left to the reader. (4.7) Proposition. Let f and g be continuous real-valued functions defined on an interval I. Then the functions f + g, f g, and max {f, g} are continuous on I. If f is bounded away from 0 on every compact subinterval J of I - that is, if If(x)1 ~ c for all x in J and some c > 0 (depending on J) - then f- 1 is continuous on I.

Proposition (4.7) implies that the quotient of continuous functions is continuous, provided that the denominator is bounded away from 0 on every compact subinterval. It also implies that a polynomial function Xf->Co x· + C1 x·- 1 + ...

+ c.

is continuous on every interval, and that If I is continuous on each interval where f is continuous. The composition of continuous functions is continuous, in the sense that if f: I-+J and g: J-+1R are continuous, then gof is continuous, provided that f maps every compact subinterval of I into a compact subinterval of J. To prove this, it is sufficient to consider the case in which I and J are both compact. Let w be a modulus of

40

Chapter 2 Calculus and the Real Numbers

continuity for J, and a a modulus of continuity for g. Then if x, YEI, e > 0, and Ix - yl ~ w(a(e)), we have If(x) - f(y)1 ~ aCe). Therefore Ig(f(x)) - g(f(y))1 ~ e. It follows that go f is continuous, with modulus of continuity eHw(a(e)).

Classically, a continuous function maps an interval onto an interval. We now prove a weak version of this result, known as the intermediate value theorem. (4.8) Theorem. Let f be a continuous map defined on an interval I, and let a, b be points of I with f(a) 0, there exists x in [min {a, b}, max {a, b}] such that If(x)- yl <e.

Proof Since f is continuous, we must have a*b. We may assume that aO. Let m=:inf{lf(x)-yl'

a~x~b},

which exists by (4.6). Suppose that m>O. Then f(a)-y~ -m and feb) -y~m. Let w be a modulus of continuity for f on [a, b], and choose points a=xo~xl~ ... ~xn=b such that Xk+l-xk~w(m) for O~k~n -1. Then for such k we have If(Xk + 1) - Y - (f(Xk) - y)1 = If(Xk + 1) - f(Xk) I ~ m.

Since If(x) - yl ~ m for all x in [a, b], it follows that the quantities f(Xk) - y and f(xk+ 1) - yare either both positive or both negative. Therefore the quantities f(Xi) - Y (0 ~ i ~ n) are either all positive or all negative. Hence f(a) - y and feb) - yare either both positive or both negative. This contradiction ensures that the possibility m > 0 is ruled out; so that m < e, and the desired conclusion follows. 0 Under additional hypotheses satisfied by many of the common elementary functions of analysis, Theorem (4.8) can be strengthened to yield the conclusion that f(x)=y for some x in [min {a, b}, max {a, b}]. For example, this strong conclusion obtains whenever f is strictly increasing, in the sense that f(x) > f(x') for any two points x, x' of its domain with x> x'. For in that case, taking a < b we can construct sequences (an), (b n) in [a, b] such that for each n in Z+' ~ a2 ~

~ an ~ bn~

(i)

a = al

(ii)

f(an) ~ y ~ f(b n)

(iii)

bn+ 1 - an + 1 ~ (2/3) (b n- an).

"

...

~ b 2 ~ b1 = b

4 Continuous Functions

41

The sequences (an), (b n) then converge to a common limit x in [a, b] with I(x) = y. Just as sequences of real numbers can converge to real numbers, sequences of continuous functions can converge to continuous functions. In fact, most of the important functions of analysis are defined as limits of sequences of continuous functions. (4.9) DefinitiolL A sequence (!.) of continuous functions on a compact interval I converges on I to a continuous function I if for each e > 0 there exists N. in 7l + such that (4.9.1)

I/n{x) - f(x)1 ~ e

(xEI, n"?, N

E).

A sequence Un) of continuous functions on an arbitrary interval J converges on J to a continuous function I if it converges to I on every compact subinterval I of J; in that case, I is called the limit of the sequence Un). Definition (4.9) can be recast to bear a closer resemblance to Definition (3.1). To this end, we define the norm 11/111 of a continuous function I on a compact interval I to be the supremum of Ilion I. Then (In) converges to I on I if and only if for each k in 7l+ there exists N" in 7l+ with II!. - IIII~k-l (n"?,Nk) (4.1O) Definition. A sequence (In) of continuous functions on a compact interval I is a Cauchy sequence on I if for each e > 0 there exists ME in 7l+ such that (4.10.1)

1/",(x)-/n(x)l~e

(xEI; m,n"?,ME).

A sequence of continuous functions on an arbitrary interval J is a Cauchy sequence on J if it is a Cauchy sequence on every compact subinterval of J. The sequence (!.) is a Cauchy sequence on the compact interval I if and only if for every k in 7l+ there exists M" in 7l+ such that 11/",-/nIII~k-l

(n,n"?,M,,).

Notice that a sequence (c n) of real numbers converges if and only if the corresponding sequence of constant functions, which we also denote by (c n), converges on a given nonvoid interval I, and that a sequence of real numbers is a Cauchy sequence if and only if the corresponding sequence of constant functions is a Cauchy sequence on I. Because of these remarks, the following theorem is a generalization of Theorem (3.3).

42

Chapter 2 Calculus and the Real Numbers

(4.11) Theorem. A sequence (fn) of continuous functions on an interval 1 converges on 1 if and only if it is a Cauchy sequence on 1.

Proof. Assume that (fn) converges to f on 1. Let I be any compact subinterval of 1. For each 6>0 choose N. in Z+ satisfying (4.9.1), and write M.=N./ 2 . Then whenever m, n~M. and xEI, we have If". (x) - fn(x)1 ~ Ifm(x) - f(x)1

+ Ifn(x) -

f(x)1

~6/2+6/2 =6.

Therefore (fn) is a Cauchy sequence on I. It follows that (f,.) is a Cauchy sequence on 1. Assume conversely that (fn) is a Cauchy sequence on 1. Then for each x in 1, (fn(x)) is a Cauchy sequence of real numbers, whose limit we denote by f(x). We shall show that f: 1 ->JR. is a continuous function and that (fn) converges to f on 1. It is enough to show that f is continuous on each compact subinterval I of 1, and that (fn) converges to f on I. To this end, choose the positive integers M. such that (4.10.1) is valid, and for each n in Z+ let Wn be a modulus of continuity for fn on I. For each 6>0 write

where M=M./ 3 • Then whenever x,YEI and

Ix-YI~w(6),

we have

If(x) - f(Y)1 ~ If(x) - fM(X) I+ IfM(X) - fM(Y) I+ IfM(Y) - f(Y)1

= lim Ifn(x) - fM(X) I + IfM(X) - fM(Y) I + lim IfM(Y) - fn(Y) I 666 :=;;-+-+-=6. -3 3 3 Therefore f is continuous, with modulus of continuity w. Finally, if xEI, 6>0, and n~M .. then If,. (x) - f(x) I = lim Ifn(x) - fm(x)1 ~ 6. 111-00

0

Hence (fn) converges to f on I.

Notations to express the fact that (fn) converges to fare limfn= f and f,,-+f

We also write simply f,.-+f

as n->oo.

4. Continuous Functions

43

To each sequence (J.) of continuous functions on an interval 1 corresponds a sequence (gn) of partial sums, defined by n

gn==

Lk k~l

If (gn) converges to a continuous function g on 1, then g is the sum of co

L fn'

the series

co

n~l

co

and the series is said to converge to g on 1. If

n=1

00

then

L J. n~

L Ifni converges on 1,

is said to converge absolutely on 1. An absolutely con-

1

vergent series of functions converges. The comparison test and the ratio test for convergence carryover to 00

series of functions. The comparison test states that if

L gn

is a con-

n~l

vergent series of nonnegative continuous functions on an interval 1, 00

then the series

L fn

of continuous functions on 1 converges on 1

n~l

whenever Ifn(x)1 ~gn(x) for all n in Z+ and all x in 1. 00

The ratio test states that if

L fn is a series of continuous functions n~l

on an interval J such that for each compact subinterval 1 of J there exist a constant c[, O
then

L J. converges absolutely on J. n=1

00

A power series is a series of the form

L an(x -

x 0)", where an(x

n~O

- xo)n represents the function x 1-+ an (x - x 0)" and where ao(x - xo)O == ao for all x. The ratio test has the following corollary. 00

(4.12) Proposition. Let the power series

L an(x -xo)" have the property n~O

that there exist r>O and N in Z+ such that lan+ll~r-llanl for all n ~ N. Then the series converges absolutely on the interval (xo - r, Xo +r). Proof: If 1 is a compact subinterval of (xo - r, Xo + r), then there exists ro with 0 < ro < r such that Ix - xol ~ ro for all x in 1. Then

lan+l(x -xo)n+ll ~r-l ro lan{x -xo)"1 (n~ N, xE1). By the ratio test, the series therefore converges absolutely on 1.

D

44

Chapter 2 Calculus and the Real Numbers

5. Differentiation The rate at which a function is changing is a fundamental property of the function. Here is the precise definition of this concept. (5.1) Definition. Let f and g be continuous functions on a proper compact interval I, and let {) be an operation from JR. + to JR. + such that If(y) - f(x) - g(x) (y -

x)1 ~ ely - xl

whenever e> 0, x, YEI, and Iy- xl ~ {)(e). Then f is said to be differentiable on I, g is called a derivative of f on I, and {) is called a modulus of differentiability for f on I. If f and g are continuous functions on a proper interval J, then g is a derivative of f on J if it is a derivative of f on every proper compact subinterval of J; f is then said to be differentiable on J. To express that g is a derivative of f we write g=J,'

or

g=DJ,

or

df(x)

g(x)=~.

One way to interpret Definition (5.1) is that the difference quotient (f(y) - f(x)) (y _X)-1

approaches g(x) as y approaches x. In other words, g is the rate of change of f If f has two derivatives on I, then they are equal functions. (5.2) Theorem. Let f1 and f2 be differentiable functions on an interval I. Then It + f2 and f1 f2 are differentiable on I. In case f1 is bounded away from 0 on every compact subinterval of I, then 11- 1 is differentiable on I. The function Xl-+X is differentiable on JR.. For each c in JR. the function Xl-+C is differentiable on JR.. The derivatives in question are given by the following relations: (a)

D(f1 + f2)=Df1 +Df2

(b)

(c)

D(f1 f2)= f1 Df2 + f2 Df1- 1 = - It- 2 Df1

(d)

-=1

(e)

dc =0 dx .

dx dx

DIt

5. Differentiation

45

Proof" It is enough to consider the case in which I is compact. Let b1 and b2 be moduli of differentiability for ft and f2' respectively, on I, and W1 a modulus of continuity for ft on I. (a) Whenever x, YEI and Iy -xl ~b(B)=min{b1(B/2), b 2 (B/2n, we have If1(Y) + f2(Y) -(f1(X) + f2(X)) -(fleX) + f2(X))(Y -x)1 ~ Ifdy) - f1(X) - f{(x)(y -x)1 + If2(Y) - f2(X) - f2(X) (y-x)1

Thus f1 + f2 is differentable on I, with derivative fi + f2 and modulus of differentiability b. (b) Let M be a common bound for If11, If21, and If21 on the interval I. (For instance, define M=max{llftIII, Ilf211I' Ilf21II}.) Then whenever x, YEI and Iy - xl ~ b(B) = min {b 1«3 M)-l B), b 2 «3M)-1 B), W1 «3 M)-l Bn, we have Ifl(Y) f2(Y) - ft(x) f2(X) -(fleX) f2(X) + f2(X) flex)) (y -x)1 ~lfl(Y)llf2(Y)-f2(X)-f2(X)(y-x)1

+ Ifl(Y) - f1(X)llf2(X)lly -xl + If2(X)llft(y) - fleX) - f{(x)(y -x)1 ~3M(3M)-1 Bly-xl =Bly-xl. Therefore fl f2 is differentiable on I, with derivative ft f2 + f2 modulus of differentiability b. (c) For each B>O write

fl and

b(B) =min {b 1(! M- 2 B), W1(! M- 4 Bn

where M=max{llft- 11II,llfiIII}. Then whenever x, YEI and Iy-xl ~ b(B), we have If1- 1(y) - ft-l(X)+ ft-2(X) fleX) (y -x)1 = Ift-l (x) fl- l (y)llf1(Y) - fleX) - fl(Y) fl-l(X) fleX) (y -x)1 ~ M21fl (y) - ft (x) - fleX) (y - x)1 +M 2 If{(x)fl(X)-11Ifl(y)-f1(X)lly-xl ~M2(!M-2 B) Iy-xl +M 4 (!M- 4 B) Iy-xl =Bly-xl. Therefore ft- 1 is differentiable on I, with derivative - ft- 2 f{ and modulus of differentiability b.

46

Chapter 2 Calculus and the Real Numbers

(d) This is obvious. (e) This is obvious too.

D

(5.3) Corollary. For all positive integers n, dx" Tx=nx"-l.

(5.3.1)

Proof: The proof is by induction on n. When n = 1, (5.3.1) is just (d) of Theorem (5.2). If (5.3.1) is true for a given value of n, then dx"+l

d(x·x")

dx

dx

by (b) of Theorem (5.2). Therefore (5.3.1) is true for all n.

0

Theorem (5.2) and its corollary imply the formula D(f1 f2- 1) = f2- 2(f2 D f1

-it D f2)

for the derivative of a quotient, and the formula D

(± an_kxk) f. ka _k xk - 1 =

k~

0

n

k~

1

for the derivative of a polynomial. The next theorem is the so-called chain rule for the derivative of a composite function. Its intuitive meaning is that the rate of change of quantity C with respect to quantity A is the product of the rate of change of C with respect to some third quantity B by the rate of change of B with respect to A. (5.4) Theorem. Let f: I -+lR and g: J -+lR be differentiable functions such that f maps each compact subinterval of I into a compact subinterval of J. Then go f is differentiable, and

(5.4.1)

(go f), =(g' 0 f) J'.

Proof. It is no loss of generality to assume that I and J are compact. Let bf be a modulus of differentiability and wf a modulus of continuity for f on I. Let bg be a modulus of differentiability for g on J.

For each B>O write where

5. Differentiation

Then for x, y in 1 and Iy - xl ~ c5(e) we have If(y) - f(x)1

~ c5 g (IX),

~ IX If(y)

- f(x)l.

Ig(f(y» - g(f(x» - g'(f(x» (f(y) - f(x»1

47

so that

Also If(y) - f(x)1 ~ 111'111 Iy -xl + If(y) - f(x) - f'(x) (y -xli

and

If(y) - f(x) - f'(x) (y - x)1 ~ PIy - xl·

Using these inequalities and noting that IX 111'111 <e/2, we compute Ig(f(y» - g(f(x» - g'(f(x» f'(x) (y -xli ~

Ig(f(y)) - g(f(x» - g'(f(x» (f(y) - f(x»

I

+ Ig'(f(x»llf(y) - f(x) - f'(x) (y - x)1 ~ IX If(y)

- f(x)1 + Ilg'IIJ If(y) - f(x) - f'(x) (y -xli

~ IX 111'111 Iy -xl + (IX +

e

Ilg'IIJ) If(y) - f(x) - f'(x) (y -xli

e

<"2ly-xl +"2ly-xl=ely-xl. It follows that go f is differentiable on 1, with derivative (g' of) I' and modulus of differentiability c5. D

The next lemma is known as Rolle's theorem. (5.5) Lemma. Let f be differentiable on the interval [a, b], and let f(a)

=f(b). Then for each e>O there exists x in [a,b] with Proo!' Let c5 be a modulus of differentiability for

If'(x)l~e.

f on [a, b]. Let

m=:inf{lf'(x)l: xE[a,b]}, which exists, by (4.6). Suppose that m>O. We may assume that f'(a)~m. For each x in [a,b] we have f'(x)~m. For if f'(x)<m, then f'(x) ~ - m, so that, by the intermediate value theorem (4.8), there exists ~ in [a,b] with If'WI<m; this contradicts the definition of m. Now choose points a=xo~xl~ ... ~xn=b so that Xk+l-xk~c5(!m) (O~k~n -1). Then 0= feb) - f(a) n-l

=

L (f(Xk+l)- f(Xk»

k=O n-1

=

k= 0 n-l

~

n-1

L f'(Xk)(Xk+l -xd+ L (f(Xk+1)- f(Xk)- f'(Xk)(Xk+l -Xk»

L

k= 0

n-l

L

m(xk+l -Xk)-!m(Xk+l -Xk) k=O k=O =-!m(b-a»O.

48

Chapter 2 Calculus and the Real Numbers

This contradiction ensures that m=O. The desired conclusion follows immediately. 0 Rolle's theorem implies the mean value theorem, which gives a basic estimate for the difference of two values of a differentiable function. (5.6) Theorem. Let f be differentiable on the interval [a, b]. Then for each E > 0 there exists x in [a, b] with If(b) - f(a) - f'(x)(b -a)1 ~E. Proof Define the function h on [a, b] by h(x) == (x - aHf(b) - f(a)) - f(x)(b - a)

(XE [a,

b ]).

Then h(b)=h(a)= -f(a)(b-a). By (5.5), there exists x in [a,b] with E~

Ih'(x)1 = If(b) - f(a) -f'(x)(b -a)l.

0

A function f on strictly increasing) if x, YEI and x> y We decreasing) if - f is

a proper interval I is increasing (respectively, f(x) ~ f(y) (respectively, f(x) > f(y)) whenever say that f is decreasing (respectively, stricdy increasing (respectively, strictly increasing). It follows from Theorem (5.6) that if f: I --> 1R is differentiable on I and f'(x) ~ 0 (respectively, f'(x) ~ 0) for all x in I, then f is increasing (respectively, decreasing) on I.

(5.7) DefinitiolL Let f, Pll,f(2), ... , P"-l) be differentiable functions on a proper interval I such that Df=P 1 ), DP 1 )=P 2), . ,Df(n-2)=f(n-l),

and set pn)==Df("-l). Then PO) is called the nth derivative of f on I, and is also written D"f; f is then said to be n times differentiable on I. The function f itself may be written PO) or DO! A natural way to simplify a continuous function and set it up for computation is to replace it by a polynomial approximation. The basic result on polynomial approximation of differentiable functions is Taylor's theorem «5.10) below). To see the motivation for Taylor's theorem, consider an n times differentiable function f on an interval I, and a point a in I. It is natural to approximate f by a polynomial of degree n whose derivatives of orders 0,1, .. ,n at a have the same values as the corresponding derivatives of f at a. The unique such

5. Differentiation

49

polynomial is obviously (5.8)

This approximation is useful for a given value b of x only when there exists a good estimate for the remainder n j
Then there exists c with

min{a,b}~c~max{a,b},

IR -

such that

I

j
where R is given by (5.9). Proof: Let

M=.= 1 + max {f(ll(a)/l !,f(2l(a)/2!, ... ,j
IRI ~ If(b) - f(a)1 +

L lj
n

~!E+M

n

L bk<-h+M L (E/2nM)=E,

k=l k=l and so we can take c=.=b. In case O
1! 2! j
g'(x) = - f'(x) + f'(x) - f"(x)(b - x) + f"(x)(b - x)fIn + 1l(X) f (nl(x) + _ _ (b_X)n-1 (b- )n+R(b- )-1 ... (n-l)! n! x a

f (n+1 l(X)

:------:-,-(b _X)n +R(b -a)-l, n.

SO

Chapter 2 Calculus and the Real Numbers

since the terms cancel in pairs except for the last two. By Rolle's theorem, there exists c with min{a,b}~c~max{a,b} and 19'(c)l~e(b _a)-I. The required estimate for R now follows. D An important special case occurs when f is infinitely differentiable - that is, po) exists for all positive integers n - on an open interval I==(a-t,a+t). In that case, if r"f(" + 1) (S.11)

--=----:---+0

n!

then the Taylor series for

as n-+oo (O
f about a, 00

p")(a)

"~O

n.

L -,-(x -a)",

(S.12)

converges to f on I. Note that although condition (S.11) enables us to prove the convergence of (S.12) by the comparison test, Theorem (S.10) is needed to show that the sum of the Taylor series is f In fact, the series (S.12) converges to f under the weaker condition

r"po) ---+0

n!

as n-+oo (O
To prove this, we need a different estimate of the remainder term, which is derived using the theory of integration from Section 6 below. (See Problem 23.) Examples of (S.12) will occur later, when some of the common functions of analysis are expanded in Taylor series.

6. Integration Differentiation deals with the instantaneous behavior of a function. We now look at a process, called integration, which deals with the behavior of a function on the average; in fact, it is essentially a rigorous formulation of the averaging process. Differentiation and integration will turn out to be inverse processes in a certain precise sense. If a = ao ~ al ~ ... ~ a" = b are real numbers, the finite sequence P == (ao, ... , an) is called a partition of the interval [a, b], n is called the length of P, and mesh P==max {a;+1 -a;: O~i~n -I} is called the mesh of P. A partition Q==(ao, ... ,a:") of [a,b] refinement of P if for each i (0 ~ i ~ n) there exists j with aj = a;.

lS

a

6 Integration

51

For each partition P == (ao, ab ... , an) of [a, b] and each continuous function f on [a, b], S(f, P) will denote an arbitrary sum of the type n-1 L f(xi)(ai+1 -ail

(6.1)

i~O

where xiE[ai, ai+1] (0 ~ i ~ n -1). In particular, if ai = a + i(b -a)/n (O~i~n), the quantity

a

a)

bn- 1 ( bS(f,a,b,n)==- L f a + i -

(6.2)

n i~O n is one of the numbers S(f, Pl· Two partitions P == (ao, ... , an) and Q == (b o , ... , bm ) of [a, b] are separated if aO
(6.3.1)

converges to a limit, which is called the integral of f from a to b and is written b

Jf(x)dx.

(6.3.2)

a

Moreover, if w is a modulus of continuity for f, then for any e>O and any partition P of [a,b] with meshP~w(e), we have

(6.3.3)

jS(f, P) -

!

f(x) dxj

~ e(b -

a).

Proof: Let e and {) be arbitrary positive numbers. Consider partitions P==(ao, ... ,an) and Q==(bo, ... ,b",) of [a,b] with meshP~w(e) and meshQ~w({)). To begin with, assume that b-a>O and that P and Q are separated. Let R == (co, ... , cr ) be the common refinement of P and Q obtained by ordering the numbers ao, ... , an, bb ... , bm - 1 in a strictly increasing sequence. Let ZkE[Ck,Ck+1] (O~k~r-l). For each i in {O, .. ,n-l}, let denote summation over those indices k for which

L i

a(;;;;ck
IS(f, P) -S(f, R)I = =

j~t~f(Xi)(ai+l -ail - :t>(Zk)(Ck+l -Ck)j

[t~f(Xi) ~(Ck+1 -Ck) - ~t~ ~f(Zk)(Ck+1 -Ck)j "-1

"-1

~ L L If(Xi)- f(Zk) I (Ck+l-Ck)~ L Le(Ck+1- ck)=e(b-a). i=O i

i=O i

52

Chapter 2 Calculus and the Real Numbers

Similarly, IS(f, Q) - S(f, R)I ~ b(b - a)

It now follows that IS(f, P) -S(f, Q)I ~ (6 + b)(b -a).

(6.3.4)

Now consider the general case, and let ct be an arbitrary positive number. If b - a is small enough, then IS(f, P) - S(f, Q)I ~ (6 + b)(b - a) + ct.

(6.3.5)

If b - a > 0, then applying the first part of the proof to separated partitions which are sufficiently close approximations to P and Q, we again obtain (63.5). As ct>O is arbitrary, we now see that (6.3.4) holds in the general case also. In particular, IS(f, a, b,j) - S(f, a, b, k) I ~ 26(b - a)

for all positive integersj,k~w(6)-1(b-a). Therefore (6.3.1) is a Cauchy sequence, and hence converges. For each integer k C. W(b)-l (b - a), inequality (63 4) entails IS(f, P) - S(f, a, b, k)1 ~ (6 + b)(b - a).

Passing to the limit as k ---> rxJ and If

f

(j ---> 0

gives (6.3.3).

D

and g are real-valued functions on a set X such that for all x in X, then we write f~g.

f(x)~g(x)

(6.4) Proposition. Let f and g be continuous functions on a compact interval [a, b], and let ct, f3 be real numbers. Then b

b

a

Moreover,

if

f

a

~ g,

then b

(6.4.2)

b

S(cif(x) + f3 g(x)) dx = ct Sf(x) dx + f3 Sg(x) dx.

(6.4.l )

b

Sf(x) dx ~ Sg(x) dx.

Proof. Equation (64.1) follows from the fact that S(ctf + f3g, a, b, n) = rJ.S(f, a, b, n) + f3S(g, a, b, n)

for all n in Z+, and (6.4.2) follows from the inequalities S(f,a,b,n)~S(g,a,b,n)

(nEZ+).

As a corollary we have b

c1 (b -a)~S f(x)dx~C2(b -a)

0

6 Integration

whenever

Cl ~f(X)~C2

for all x in [a,b]. By setting

-Cl

53

=C2

= Ilfll[a,b], we obtain (6.5) (6.6) Proposition. If f is a continuous function on a compact interval [a,b], then b

(6.6.1)

b

c

Jf(x)dx=Jf(x)dx+Jf(x)dx a

(a~c~b)

a

Proof Let a~c~b. For arbitrary partitions P==(a, .. ,c) of [a,c] and Q==(c, ... ,b) of [c,b] we have S(j, P) + S(j, Q) = S(j, R)

with appropriate choices of the approximating sums, where R is the partition (a, .. ,c, . ,b) of [a, b] formed by combining the partitions P and Q. Since mesh R = max {mesh P, mesh Q}, equality (6.6.1) follows by passage to the limit as mesh P --> 0 and mesh Q --> O. 0 For arbitrary real numbers a and b and each continuous function

f on [min {a, b}, max {a, b}] we define b

(6.7)

b

a

;.

;.

Jf(x) dx == Jf(x) dx - Jf(x) dx,

where ). == min {a, b}. In case a ~ b this agrees with the definition already given. Equality (6.4.1) is valid for arbitrary real numbers a and b, provided that the integrals are defined (which means that f and g are continuous on the interval [min {a,b}, max {a,b}]). Equality (6.6.1) is valid whenever f is continuous on the interval [min {a, b, c}, max {a, b, c}]. In particular, setting b = a in (6.6.1) gives c

a

Jf(x) dx = - Jf(x)dx a

whenever the integrals are defined. For arbitrary real numbers a and b equality (6.5) holds with Ilfll[a,b] replaced by the norm of f on [min {a, b}, max {a, b}]. We now prove that differentiation and integration are inverse processes, a result whose imposing title is the fundamental theorem of calculus.

54

Chapter 2 Calculus and the Real Numbers

(6.8) Theorem. Let f be a continuous function on a proper interval 1, let a be a point of 1, and write x

J

g(x) == f(t) dt

(xE1).

a

Then g' = f Also, if go is any differentiable function on 1 with go = f, then the difference g - go is a constant function. Proof: There is no loss of generality in taking 1 to be compact. Let w be a modulus of continuity for f on 1. For each e>O and each pair of points x, y in 1 with Ix - yl ~ w(e), we have Ig(y) - g(x) - f(x)(y - x)1 =

II II

f(t) dt -

I

f(x) dtl

= (f(t) - f(x)) dtl ~ely -xl· Therefore g is differentiable on 1, with derivative f and modulus of differentiability w. If also go = f on 1, then (g - go)' = 0 on 1. By the mean-value theorem, (g-goHa)=(g-go)(b) whenever a, bEl and a
(!.) be a sequence of continuous functions converging to a function f on a nonvoid interval 1. Let a be any point of 1, and write

(6.9) Lemma. Let

x

J

g(x) == f(t) dt

(xE1)

a

and x

J

gn(x) == fn(t) dt

(xE1)

a

for each positive integer n. Then gn -+ g on 1 as n -+ 00. Proof: It is no loss of generality to assume that 1 is compact. Let r be the length of 1. Then for each x in 1 we have Ign(x) - g(x)1 =

II

(fn(t) - f(t)) dtl

~r II!.

- fllr-+O

Therefore (gn) converges to g on 1.

0

as n-+ 00.

7 Certain Important Functions

55

(6.10) Theorem. Let (fn) be a sequence of differentiable functions which converges to a continuous f on a proper interval 1. Assume that the sequence (f:) of derivatives converges to a continuous function g on 1. Then f' = g on 1. Proof. It is no loss of generality to assume that 1 is compact. Let a be

any point of 1. For each x in 1 write x

h(x) == Jg(t) dt a

and x

Ja

hn(x) == f:(t) dt.

By (6.8), we have h'=g and h~=f: (neZ+). Hence fn-hn is constant on 1. By (6.9), we have hn -+ h on 1 as n -+ 00. Therefore f - h = lim (fn - hn)

is constant on 1. It follows that f' = hi = g on 1, as was to be proved. 0

7. Certain Important Functions A good point of departure for the study of the exponential function is obtained from an analysis of its rate of change. Intuitively the exponential function is that differentiable function on JR, normalized to have the value 1 at 0, which increases at a rate equal to its size. In other words, it is the solution of the differential equation Df = f normalized by the condition f(O) = 1. Clearly, such a function f, it it exists, is infinitely differentiable, and j
°

(7.1)

For each interval [-c,c] with c>O, cnllfll[_c.c]/n! -+0 as n-+oo, and hence the series converges to f on [ - c, c]. Therefore if f exists, it is unique, and is given by (7.1). On the other hand, (7.1) converges on JR, by the ratio test. We therefore define the exponential function exp by means of the series (7.1): 00

(7.2)

exp (x) ==

xn

L -n!

n=O

(xeJR).

56

Chapter 2 Calculus and the Real Numbers

Clearly exp (0) = 1 Also, (7.3)

d exp(x) dx

00

nxn-1

co

xn-1

L ~,-= n~l(n-l). L --, =exp(x), n~O n.

by (6.10). The function exp has the desired properties, and is unique. (7.4) Proposition. The function exp satisfies the functional equation (7.4.1)

exp (x + y) = exp (x) exp (y)

(x, YEIR).

Proof: Let z be any positive real number. Then exp (z) =to 0, by (7.2). Define the differentiable function f: IR --+ IR by f(x) =(exp (Z))-l exp(x+z).

The derivative of f is df(x) d(x + z) ~=(exp(z))-l exp(x+z) dx

f(x),

by (5.4). Also, f(O) = 1. By the uniqueness of the exponential function, f(x) = exp (x) - that is, (7.4.2)

exp (x + z) = exp (x) exp (z).

Consider now arbitrary real numbers x and y. Choose z with z>o and z + y > O. Then, by (7.4.2), exp(x+ y) exp(z)= exp(x+ y+z) = exp (x) exp (y + z) = exp (x) exp (y) exp (z), which is equivalent to (7.4.1).

D

It can be shown that exp is the unique continuous function from IR to IR that has the value 1 at 0 and satisfies (7.4.1). Equality (7.4.1) implies that exp( -x) = (exp (X))-l. Also,

(7.5)

exp(x»O

(XEIR).

For, given x in IR and choosing z>o so that x+z>O, we have exp (z) ~ 1 and exp (x + z) ~ 1; whence exp (x) = exp (x + z)(exp (Z))-l >0. The theory of the exponential function can also be approached as follows. To construct a function f with f(O) = 1 and f' = f, notice that f' = f implies, by the mean-value theorem, that f(x +15) is approximately equal to f(x)+!5f'(x)=(1 +!5)f(x) when 15 is small. Therefore, if t is any real number and n is a large positive integer, f(t/n)

7 Certain Important Functions

57

1S approximately 1 + (tin), f(2 tin) = f(tln + tin) is approximately (1 + (tin)) 2, and so forth, and thus f(t)= f(n(tln)) is approximately (1 + (tln))n. This heuristic argument suggests that we define

exp(t)= lim n-oo

(1 +~)n. n

It is not hard to see that t n(n -1) (t)2 ( 1+-t)n =1+n-+ n n 2! n

+ ...

00

converges on 1R to

L t nIn!

as n --+ 00. This approach to the theory of

n=O

the exponential function would therefore lead to the same series representation. We want the logarithmic function to be the inverse function to the exponential function, but we shall not define it in that way. Arguing heuristically on the basis of the chain rule (5.4), for each x> 0 we want dx d d d 1 = dx = dx (exp (In (x)) = exp (In (x)) dx In(x)=x dx In (x).

This gives (7.6)

d

dx In(x)=x- 1

(x >0).

We therefore define In (x) to be the integral of X-I; specifically x

(7.7)

In (x) = St- 1 dt

(x >0).

1

By (6.8), In is differentiable on (0, (0) and (7.6) is valid. Let y be any positive real number. By (5.4.1), the derivative of the function (7.8)

x ~ In (x y) -In (y)

is

X-I. Also, (7.8) vanishes at 1. By the last statement of (6.8), In(x) =In(xy)-ln(y); in other words,

(7.9)

In(xy)=ln(x)+ln(y)

(x,y>O).

This is the functional equation for the logarithmic function, corresponding to the functional equation (7.4.1) for the exponential function.

58

Chapter 2 Calculus and the Real Numbers

By (7.5) and (5.4), the composite function In 0 exp exists and differentiable everywhere on IR, and d dx(ln(exp (x))=1

IS

(XEIR).

Since also In(exp(O»=ln(I)=O, we have In (exp(x»=x

(7.10)

(XEIR),

by the last part of (6.8). Consider x> 0 and write y == exp (In (x». Then In (y) = In (exp (In (x») = In (x), by (7.10). Thus y

0= In (y) -In (x) = St- 1 dt. x

If y>x, this gives O;;:;y-l(y-X), a contradiction. By (2.18), we therefore have y~x. Similarly, x~y; whence x=y. Thus exp(ln(x»=x for all x> o. It follows that the functions exp: IR --+ IR + and In: IR + --+ IR are inverse to each other. For any a> 0 we now define

aX == exp (x In (a»

(XEIR).

We leave the reader to confirm that the map X1-+ aX has the familiar properties. Note that for a fixed x> 0 the map t 1-+ t of IR + into IR + extends to a continuous map of IR 0 + into IR 0 + with OX = 0, this follows from the fact that t is arbitrarily small for all positive numbers t sufficiently close to 0 The trigonometric functions sin and cos also can be approached via an intuitive analysis of their rates of change. From this analysis we are led to believe that X

X

(7.11)

d

.

-COSX= -Slnx, dx

d . dx smx=cosx

(XEIR).

We therefore define these functions by power series constructed in such a way that (7.11) will hold. Remembering that cos (0) = 1 and sin (0) = 0, we are forced to define 00

(7.12)

cosx==

n=O 00

sinx==

x2n

L (_l)n _ _

(2n)! ' X2n+1

L (-I)n(2 n+. 1)1

n=O

Theorem (6.10) implies that (7.11) is valid.

(XEIR).

7 Certain Important Functions

59

(7.13) Proposition. The functions sin and cos satisfy the functional equations

(7.13.1)

sin (x + y) = sin x cos y + cos x sin y

(7.13.2)

cos (x + y) =cos x cos y -sinx sin y

for all x, y in 1R. Proof Consider both sides of (7.13.1) as functions of x, say f(x) and g(x). The functions f and g have Taylor series about x = 0 which

converge on 1R. The coefficients of these series are expressed in terms of the various derivatives of f and g at x = O. Therefore to show that f = g on 1R it is enough to show that f(n)(o) = g(n)(o) for all nonnegative integers n. Since J<2) = - f and g(2) = - g, it is enough to notice that f(O) = sin y = sinO cos y + cos 0 sin y= g(O)

and 1'(0) = cos y = cos 0 cos y + ( - sin 0) sin y = g'(O).

Therefore (7.13.1) is valid. Equation (7.13.2) follows from (7.13.1) by differentiation with respect to x. 0 Putting y= -x in (7.13.2) gives the useful identity cos 2 X+ sin 2 x = 1.

(7.14)

Further study of the trigonometric functions depends on properties of the number n, which we construct as twice the first positive zero of the cosine function. (7.15) Theorem. The sequence (X n) of real numbers defined inductively by the equations (i)

Xl == 1

(ii) IS

Xn+l ==xn+cosxn

(n~

1)

Increasing, and converges to a limit nl2 such that cos (nI2) = 0 and

cosx>O for all x with

0~x
Proof: To begin with, we prove that

(7.15.1) Consider first the case n = 1, when we have 0 ~ t ~ Xl = 1. Then

60

Chapter 2

Calculus and the Real Numbers

Next assume that (7.15.1) is true for a particular value of n; then Xn+l>X n. Consider a value of t with xnO. Hence t

cos t = cos Xn -

Jsin x dx > cos Xn -

(Xn+l - x.) =0.

Since cos is a continuous function and cos t > 0 whenever 0 ~ t ~ Xn or cost>O whenever O~t~Xn+l. By induction, (7.15.1) holds for all n. Since cos Xn > 0 for all n, the sequence (xn) is increasing. From the mean-value theorem and (7.15.1) we obtain

xn
sin Xn ~ sin 0 = 0

(7 15.2)

for all n, and sin t ~ sin 1 whenever 1 ~ t ~ Xn for some n Also by the mean-value theorem, for each n ~ 2 there exists t in [x n_ b xn] such that Xn+l -xn=xn+cosxn-(Xn_l +cosxn_d =Xn-Xn_l +cosxn-COSXn_l is arbitrarily near to Xn -Xn-l -(sin t)(xn -xn-d ~(xn -xn_d(l -sin 1). Therefore This gives Xn+l

-xn~(1-sin

l)n-l(X2 -xd

for all n ~ 1. Thus (xn) is a Cauchy sequence, whose limit we denote by n/2.

If 0 ~ x < n/2, then x < Xn for some n, and thus cos x > 0, by (7.15.1). Taking limits on both sides of (ii) as n -+ 00 we see that cos (n/2) =0. 0

By (7.15.2), sin(n/2)~0. Since cos (n/2) =0, it follows from (7.14) that sin (n/2) = 1, and hence from (7.13.1) that sin(x+n/2)=cosx for each x in IR.. Similarly, sin (n/2 -x) = cos x and cos (x + n/2) = -sin x. Hence sin (x + n) = cos (x + n/2) = - sin x, and thus sin (x + 2n) = sin x. Similarly, cos(x+n)= -cosx and cos(x+2n)=cosx. On the other hand, if a> 0 and sin (x + tI) = sin x for all x, then cos tI = sin (n/2 + a) = sin (n/2) = 1. Since (as the reader may prove) cos is a strictly decreasing function on [0, n] and a strictly increasing function on [n,2n], and since cos 0 = cos 2n = 1, it follows that a ~ 2n. Thus the function sin, and similarly the function cos, is periodic, with period 2n.

7 Certain Important Functions

61

Next we study the function arc sin, which is inverse to the function sin. For motivation we take the derivative of the equation sin (arc sin x) = x and get d

cos (arcsin x) -(arc sin x) = 1

dx

Since 1 = cos 2 (arc sin x) + sin 2(arc sin x) = cos 2(arc sin x) + x 2 , it follows that

With this motivation, we define

" 2 )-tdt arcsinx=J(1-t

(-1<x<1).

o

Then arc sin is differentiable on ( -1,1), and

~(arcsinx)=(1-x2)-t. dx

To discuss the range of arcsin, we first observe that the function sin is strictly increasing on [-nI2, n12]. Since sin ( - n12) = -1 and sin (nI2) = 1, it follows from the remarks after (4.8), and the continuity of sin, that for each x in (-1,1) there exists y in (-nI2, n12) with x =siny. For any y in (-nI2, n12) we have d d y arc sin (sin y) = (1 - sin 2 y) - t cos Y = 1.

Hence, as arc sin (sin 0) = 0, (7.16)

arc sin (sin y) = y

( -n12 < y < nI2).

It follows that the function arc sin maps the interval (-1,1) into the interval (-nI2, nI2). Moreover, if -1 < x < 1 and z = sin (arc sin x), then arc sin z = arc sin x, by (7.16); so that

Therefore x = z = sin (arc sin x). It follows that the functions sin: ( - n12, n12) ---+ ( - 1, 1) and arc sin: ( - 1, 1) ---+ ( - n12, n12) are inverses.

62

Chapter 2

Calculus and the Real Numbers

Problems The word "not" will be used throughout in the sense of the discussion at the end of Section 2. 1. Construct a set A such that x = y for all elements x and y of A, but A is not subfinite or void.

2. Construct a mapping f from a set A to a set B such that f is onto B, but there does not exist a mapping g: B --+ A with f(g(b» = b for all

bin B 3. Prove that (x, y) ~ x y is a function from IR x IR into IR. 4. Let x and y be real numbers such that x =1= y entails 0 = 1 Show that x=y. 5. Show that if Xl>. ,X n are real numbers such that Xl .. Xn < 0, then Xi < 0 for some i. 6. Call a pair (S, T) of nonvoid subsets of the set Q of rational numbers a Dedekind cut if s < t for all s in Sand t in T, and if for arbitrary rational numbers x,y with x
o.

9. Construct a real number that does not have a decimal expansion. 10. Construct a real number that is not rational and not irrational. (A real number x is irrational if x =1= r for each rational number r.) 11. Construct a continuous function f: [0, 1] --+ IR such that there does not exist a point x in [0,1] with f(x) equal to the supremum of f

Problems

63

12. Show that a continuous function on a compact interval admits a modulus of continuity which is a continuous function. 13. A mapping f of an interval I into 1R is weakly injective if x = y whenever f(x) = f(y), and is injective if f(x) f(y) whenever x y. Prove that the following statements are equivalent.

*

*

(i) Every weakly injective map f: [0,1] ~ 1R is injective. (ii) If x is a real number such that the equality x = is impossible, then x*o. (iii) For each sequence (x n) in {O, I}, if it is impossible that xn=O for all n, then there exists n with Xn = 1 (Markov's principle).

°

(There is no known proof or counterexample for Markov's principle.) 14. Let f: [0, 1] ~ 1R be a continuous function with f(O) < f(I). Show that there is a sequence (Yn) of elements of [/(0),/(1)] such that if f(O)~y~f(l) and if Y*Yn for each n, then there exists x in [0,1] with f(x)= y. 15. Let f: [0,1]~1R be a continuous function with f(O)
°

16. Let f: [0, 1] ~ 1R be a continuous function with f(O) < and f(I»O, such that there exists a positive integer n with If(x)I+If'(x)1 + ... +lpn)(x)I>O for all x in [0,1]. Show that there exists x in [0,1] with f(x)=O. 17. Let f: [0, 1] ~ 1R be a polynomial function with f(O) < f(I»O. Show that there exists x in [0,1] with f(x)=O.

°

and

18. Show that if f: [0, 1] ~ [0, 1] is a continuous function, then for each 8>0 there exists x in [0,1] with If(x)-xl<8. 19. Let f: [0, 1] ~ 1R be differentiable and locally nonconstant, in the sense that if X,YE[O, 1] and x
*

20. Construct a continuous function f: [0, 1] ~ 1R such that f' does not exist on any proper compact subinterval of [0, 1]. (In other words,

64

Chapter 2

Calculus and the Real Numbers

the assumption that to a contradiction.)

f' exists on a proper compact subinterval leads

21. Let f and g be continuous functions on the compact interval [a,b], such that g(x)~O for all x in [a,b]. Prove that for each E>O there exists c in [a, b] such that

II

f(x) g(x) dx - f(c) I g(x) dxl < E.

22. Prove that under the hypotheses of Taylor's theorem, b f(n+l)(t) R= J , (b-ttdt. a

n.

Hence prove that for each E> 0 there exists c with min {a, b} ~ c ~max{a,b}, such that

IR-

I

f(n+1)(c) (b-at+ 1 <E. (n+ I)!

23. Prove that if f is infinitely differentiable on the open interval I=(a-t,a+t), and if rnpn l ---+0

n!

then the Taylor series for

as n-+oo (O
f about a converges to f on

I.

Notes As the definitions of Section 1 indicate, a single concept of classical mathematics may split into two or more distinct concepts when looked at constructively. For example, the distinction between finite and subfinite sets does not exist classically; nor does the distinction between weakly injective and injective mappings (Problem 13). In [7] what we have called a map of A onto B is called a surjection. a function f: A -+ B is there said to map A onto B only if it has a right inverse. Although there is a meaningful distinction involved here (see Problem 2), we prefer to use the simple expression 'onto' for the more general situation. When a right inverse map exists we can say so explicitly if we need to Problem 2 indicates that the axiom of choice is not true in constructive mathematics In certain formal systems attempting to de-

Notes

65

scribe constructive mathematics it can be shown that the axiom of choice entails the principle of omniscience [42]. An alternative to Definition (2.1) would be to define real p.umbers by the method of Dedekind cuts (see Problem 6). Another possibility would be to define a real number as a sequence (xn) of rational numbers and a sequence (Nk ) of integers, such that Ix",-xnl~k-l whenever m, n ~ Nk • Notice that a real number is a regular sequence of rational numbers, not an equivalence class of regular sequences of rational numbers To define a real number to be such an equivalence class would be either pointless or incorrect: pointless (but correct) if the equivalence class is required to be specified by giving some particular regular sequence of rational numbers that belongs to it, and incorrect otherwise. Precisely what is meant by a subset (in particular, what it means for IR + and IR 0 + to be subsets of IR) will be discussed in Chapter 3. In classical mathematics, a proof like that of Lemma (2.18), in which we assume a certain hypothesis, derive a contradiction, and therefore conclude that the negation of our hypothesis is true, is called an indirect proof. It is necessary to exercise caution in thinking of Lemma (2.18) in this way Although x ~ y is equivalent to the negation of x> y, it is not true that x> y is equivalent to the negation of x ~ y There is a paradox growing out of Theorem (2 19) which the reader should resolve Since every regular sequence of rational numbers can presumably be described by a phrase in the English language, and since the phrases in the English language can be sequentially ordered, the regular sequences of rational numbers can be sequentially ordered, in contradiction to Theorem (2.19) The counterexamples following Theorem (2.19) are deceitful. The reader is asked not to form the impression that the purpose of constructive mathematics is to consider pathological numbers like the ones discussed here. The only reason for discussing such numbers is to show that certain statements (in this case, the statement that for each x in IR 0 + either x> 0 or x = 0) are not constructively valid. To appreciate this point better, the reader should examine the counterexamples given in Chapter 1, where the treatment is more detailed Sometimes a mathematician, when confronted with a counterexample of this sort, will say he is not interested in real numbers that have been artificially twisted out of shape. One reply to this is to point out that the set of real numbers x such that either x> 0 or x ~ 0 is not complete (nor is it closed under addition). The 'choice' involved in the proof of (b) of Proposition (3.4) is performed according to a definite rule.

66

Chapter 2

Calculus and the Real Numbers

Strictly speaking, a continuous function on a compact interval I is a pair (I, w) consisting of a function I: I ---+ IR and a modulus of continuity w for I on I; two continuous functions (f1' w 1) and (f2' w 2) are equal if 11 and 12 are equal functions. In practice, as in Definition (4.5), we call I itself a continuous function, in the spirit of the remarks preceding Proposition (2.9). Notice that divergence is a positive notion. As we remarked in Chapter 1, a non void bounded set of real numbers need not have a least upper bound. This, of course, makes some parts of constructive mathematics harder than the corresponding classical material, but it also makes some parts more interesting. The concept introduced in Definition (4.5) is classically called uniform continuity. We abbreviate this to continuity, since the classical version of continuity (that is, pointwise continuity) is not used anywhere in the book. Classically, one proves that on a compact interval, pointwise continuity implies uniform continuity. Constructively, there is no known proof of this result, and so we assume uniform continuity from the start. Nothing essential is lost by this assumption, and a certain simplicity is gained. Similar remarks apply to Definitions (4.9) and (5.1). It would be more correct to call Lemma(5.5) a constructive substitute for Rolle's theorem. the classical result states that under the hypotheses of Lemma (5.5), f'(x) = a for some x in (a, b). It follows from Problem 19 that as long as we are concerned with the familiar elementary functions of real analysis, the sharp classical form of Rolle's theorem is constructively valid. The proof of Theorem (5.10) illustrates the minor inconveniences caused by not being able to compare arbitrary real numbers. For Riemann integration of possibly discontinuous functions see Problem 10 of Chapter 6.

Chapter 3. Set Theory

The chapter begins with a fuller discussion of sets and functions, and includes the constructive meaning of such terms as subset, union, and inclusion. Certain classical laws of the algebra of sets carryover, and others do not. In Section 2 we introduce the basic notion of a complemented set, which is used in Chapter 6 to facilitate the development of the theory of measure and integration. The chapter closes with some remarks on general topology which support the conclusion that in most cases of interest it is inappropriate to define continuous functions in terms of the family of open sets.

Very little is left of general topology after that vehicle of classical mathematics has been taken apart and reassembled constructively. With some regret, plus a large measure of relief, we see this flamboyant engine collapse to constructive size. First, however, we come to terms with another old friend, the reverend theory of sets.

1. Some Basic Notions of the Theory of Sets As discussed previously, a set is defined by describing what must be done to construct an element of the set, and what must be done to show that two elements of the set are equal. A function f: A --+ B is a finite routine which, applied to any element a of a set A, produces an element b == f(a) of a set B, in such a way that f(a) = f(a') whenever a = a'. The set of all functions from a set A to a set B will be written F(A, B). When A is not countable, the set F(A, B) seems to have little practical interest, because to get a hold on its structure is too hard. For instance, it has been asserted by Brouwer that all functions in F(JR., JR.) are continuous, but no acceptable proof of this assertion is known.

68

Chapter 3 Set Theory

The development of Section 1 of Chapter 2 remains in force, and should be reviewed by the reader at this point. (11) Definition. A subset of a set B is a pair (A, i) consisting of a set A and a function i: A -4 B, called the inclusion map, such that for all a, a' in A, a = a' if and only if i(a) = i(a'). Although both the set A and the inclusion map i must be given for a subset (A, i) of B to be described, we shall speak somewhat loosely of 'a subset A of a set B', the inclusion map being understood. If A is a subset of B, it is convenient at times to identify an element a of A with its image i(a) in B. Also, we often say that an element a of A belongs to B, and that an element b of B belongs to A in case b = i(a) for some a in A. Sometimes the construction of an element of a set A implicitly entails the construction of an element of a set B, without this being an explicit part of the definition of A or B. In such a case we are at liberty to bring the inclusion map to the fore and realize A as a subset of B, or not, as we see fit, provided that the equality relation on A agrees with the equality relation on B. In this spirit we regard 7L as a subset of
1 Some Basic Notions of the Theory of Sets

69

iCc) == iA (a) Define two elements c, c' of An B to be equal if iCc) = iCc'). (In practice, we will identify the element c == (a, b) of An B with either the element a of A or the element b of B.)

(1.3.3) To show that A is included, or contained, in B, construct a functionf: A---+B such that iA=iBof In that case, we write AcB; we also write B::::J A. (1.3.4) To show that A=B, show that AcB and BcA.

The standard laws of the algebra of sets follows from Definition (1.3). These are the commutative laws AuB=BuA,

AnB=BnA,

the associative laws Au~u~=~u~u~

An~n~=~n~n~

the distributive laws An(Bu ~=(AnB)u(An~,

and the laws

AnBcA,

Au(Bn

~=(AuB)n(Au~,

AcAuB.

The void subset qJ of a set S is defined as follows To construct an element x of qJ, construct an element s of S and prove that 0 = 1; the inclusion map is defined by i(x) == s. The definition of qJ is negativistic, and we prefer to mention the void set as seldom as possible. Two subsets A and B of a set S are disjoint if An B is the void subset of S. A set S is nonvoid if we can construct an element of S (1 4) Definition. A family of subsets of a set S is a pair (T, A) == (A(t»tET consisting of an index set T and a finite routine A. which associates a subset ),(t) of S with each element t of T, such that A.(t) = A(t') whenever t, t' E T and t = t'. In case (T, A) has the property that for all t and t' in T, A(t) = A(t') if and only if t = t', we call (T, A) a set of subsets of S. The set-theoretic operations can be generalized so that they apply to an arbitrary family (T, A.) of subsets of a set S. For example, the subset (1.5)

U ),(t)=={s:

tET

sd(t) for some t in T}

of S is called the union of the family (T, A) More precisely, to construct an element u of A(t) we first construct an element t of T, and

U

rET

70

Chapter 3

Set Theory

then construct an element v of ).(t); the inclusion mapping is defined by taking i(u) == ir(v), where ir is the inclusion map of A(t). Elements u and u' of U A(t) are defined to be equal if i(u) = i(u') rET

As another example, in case T is nonvoid, the subset

n

A(t) == {s:

SES,

SE).(t) for all t in T}

rET

of S is called the intersection of the family (T, ).). More precisely, if 1" is chosen in T, then an element u of A(t) is a finite routine which

n

rET

associates an element Xr of A(t) with each element t of T, such that ir(xr)=ir·(x r·) whenever t, t'ET; the inclusion map of A(t) is defined

n

n

rET

by taking i(u) == it(x t). Elements u and u' of A(t) are defined to be equal if i(u) = i(u'). rET In case T== 71.+, a family (T, A) is just a sequence of subsets of S. We <Xl

then denote the union of (T, A) by

U A(n),

and the intersection by

nA(n). Analogous notations are used in similar situations, for exam"=1

00

"=1

pIe, if T== {l, ... , N}, then (T, A) is just a finite sequence of subsets of S, N

and we denote the union of (T, A) by

U A(n), or by A(I) u ... U A(N). "=1

If (T, A) is a family of subsets of a set S, then we define the corresponding (cartesian) product to be the set ).(t) consisting of all

TI

maps 1. T -->

rET

U A(t) such that l(t)E).(t) for all t in teT

T. In case T== 71. + we

00

denote the product by

TI A(n). "=1

An important family of subsets of a set S is the family of detachable subsets. The detachable subsets are indexed by the set T==F(S, {a, I}) of functions from S to {a, I}; for each 1 in T the set A(f) consists of all S in S with l(s) = 1. (Thus a subset A of S is detachable if and only if there exists a finite routine for deciding whether or not an arbitrary element of S belongs to A.) The detachable sets A(f) and A(g) are equal if and only if 1 =g; therefore the detachable sets form a set of subsets of S. For all 1 and g in F(S, {a, I}) the following algebraic laws are valid: A(f) n A(g) = ).(f g), A(f) U A(g) = A(f + g - 1 g). Each detachable set A(f) has a complement - A(f) == A(I - f). The operation A(f) 1-+ - A(f) of complementation obeys the following algebraic laws: -( -A(f» = A(f), -(A(f)UA(g»=( -A(f»n( -A(g», -(A(f)nA(g»=( -A(f)U( -A(g».

1. Some Basic Notions of the Theory of Sets

71

A set S is called discrete if D::={(s,s'): SES, S'ES, s=s'}

is a detachable subset of S x S. This means that there exists a finite routine for deciding whether or not two given elements of S are equal. A partial function from a set S to a set T is a mapping f: A -+ T whose domain A is a subset of S. Two partial functions f and g from S to T are equal if (i) their domains are equal subsets of S, and (ii) f(x) = g(x) for all x in dmn f If f and g are partial functions from S to lR, we define f + g and f g to be those partial functions with domain X::= dmn f n dmn g such that (f + g) (x) = f(x) + g(x) and (f g) (x) = f(x) g(x) for all x in X. We then say, for example, that f + g is the pointwise sum of f and g. If c is a real number, we write c for the constant mapping Xl-+C on any nonvoid subset of S. We also write - f for the function (-1) f, and f - g for f + ( - g). If g(x) =1= 0 for all x in dmn g, we define the function g -1 on dmn g by g-1 (x)::= (g(X))-1.

We often write fig instead of fg-1. If f is a mapping of S into lR, we define the supremum of f to be the real number supf::=sup{f(x): XES} when it exists; and we define the infimum of f to be the real number inf f::=inf{f(x): XES} when it exists. Let (J,.)~1 be a sequence of partial functions from a set S to lR. eo

Then the supremum of the sequence (J,.) is the partial function V J,. from S to lR defined by •~ 1 (1.6) for all s such that the right side of (1.6) exists as a real number. The infimum of (J,.) is the partial function eo

eo

/\ f.::= -

V ( - t.)·

n=1

11=1

72

Chapter 3 Set Theory

Clearly,

for all s in the domain of

1\

fn.

n=l

The supremum and infimum of a finite sequence (f1, .. , fN) of partial functions are defined in the obvious way, and are denoted by N

V fn

N

and

1\

fn, respectively. We also write N

max {f1, .. ·,iN} =f1 v ... V fN=

V f,. n=1

and

N

min {f1, .. , fN} =f1"·

"fN=

1\

fn

n=1

For each partial function f from S to IR we write If I = max U; - f}, f+ = max {f, OJ, and f- =max { - f, OJ. Thus If I = f+ + f- and f =f+-f-·

2. Complemented Sets It is often necessary to consider families of sets that are closed under

countable unions and countable intersections, and under a suitable operation of complementation. The operation of complementation entails difficulties: on the one hand, we do not wish to define complementation in terms of negation; but on the other hand, this seems to be the only general method available. The way out of this awkward position is to permit great flexibility in the definition of complementation. A very flexible notion is based on the concept of a set with an inequality. (2.1) Definition. An inequality on a set X that for all x, y, and z in X, (a)

if x = y and x =1= y, then 0 = 1

(b) (c)

if x =1= y, then y=l=x if x =1= y, then either x =1= z or y =1= z.

1S

a binary relation =1= such

If x =1= y, then x and yare said to be unequal, or distinct.

The standard inequality on IR is defined by x =1= y

if and only if

Ix - yl > o.

2 Complemented Sets

73

Note that this inequality has the additional property: if x =1= Y entails 0 = 1, then x = y. An inequality with this property is said to be tight. (2.2) Definition. Let X be a nonvoid set with an inequality =1=. A complemented set in X is an ordered pair A == (A 1, A 0) of subsets of X such that x =1= y whenever xEA 1 and YEA 0. The characteristic function of a complemented set A is the function XA: A 1 U A -+ {a, I} defined by if xEAt,

°

if xEAo The union (respectively, intersection) of a sequence

(An)~ 1

of com-

plemented sets in X is the complemented set n?l An (respectivelY, n01 An) with characteristic function nY/An (respectively, n0/An). The

union and intersection of a finite sequence (A1' ,AN) of complemented sets are defined similarly. We often write, for example, N

A1 v ...

V

AN instead of

V An n=l

If A and B are complemented sets in X, then the complement of B in A is the complemented set A - B which has characteristic function XA (1 - XB). The complement of B is the complemented set - B with characteristic function 1 - XB (Thus if B == (B\ BO), then - B = (BO, B1 ).) A complemented set A is a subset of a complemented set B if Ale B1 and BO c A 0; we then write A < B. The complemented sets A and B are equal if A1=B1 and AO=Bo.

Unless we state otherwise, we shall assume that if A is a complemented set, then A 1 is the first component of A, and A ° is the second. We shall also write xEA, and say that x belongs to A, when we mean that xEA1. The following algebraic laws are valid for complemented sets: -(-A)=A, <Xl

- V An= "=1

<Xl

/\ (-An), n=1

and <Xl

- /\ An= n=1

<Xl

V (-An). "=1

74

Chapter 3 Set Theory

In addition, all the usual finite algebraic laws which do not involve the operation of set complementation are satisfied. For instance, the distributive law v v (V A W) = (V v V) A (V V W) is valid for complemented sets V, V, and W. On the other hand, the law V A (V V - V) = V is not valid. The infinite distributive law 00

0()

(2.3)

Vv

1\

V.=

n=1

1\ (Vv V.) PI=1

also fails. To see this, take X == JR and let (nk)k..l be a sequence of integers, each of which is either 0 or 1, for which it is not known if nk = 0 for any k. Write Vk =(JR, qJ) if nk = 1, =(qJ,JR)

if nk=O

and

Then V v Vk = (JR, qJ) for each k, and thus the right side of (2.3) equals (JR, qJ); but we have no way of constructing any element of JR which 00

belongs to V v

1\

Vk. Hence (2.3) is not valid.

k~1

The invalid law (2.3) has the constructive substitute (2.4)

(V v - V)

A

(V

V

ZI

V.) =(V V

-

V)

A

.01

(V v v.),

whose proof is left to the reader. The dual law (2.5)

(V

A -

V)

V

(V

A

.Yl V.) =(V

A -

V)

V

.Yl (V

A

V.)

is also valid. If to each complemented set A we assign the set A I , then we obtain a map j from the set of complemented sets in X to the set of subsets of X; this map preserves all the above operations except complementation. Since (S, qJ) is a complemented set for each subset S of X, j is onto the set of subsets of X. Consider the special case X == JR., with the standard operations. Then (JR 0 +, JR -) is a complemented set, where JR.-is the set of negative numbers. Thus we can consider JR.-to be the complement of JR 0 +. A different result would be obtained by basing the notion of complementation on negation - that is, by defining x to be in the complement of JR 0 + precisely when the assumption xEJR 0 + leads to a

3 Neighborhood Spaces and Function Spaces

75

contradiction - since we have no way of showing that XElR - whenever xElR and the assumption XElR 0 + is contradictory. On the other hand, it is true that xElR 0 + if the assumption XElR - leads to a contradiction; this follows from Lemma (2.18) of Chapter 2. Thus if we were to base complementation on negation, the complement of lR would be lR 0 +, but the complement of lR 0 + would not be lR -. Hence a complemented set would not necessarily equal the complement of its complement.

3. Neighborhood Spaces and Function Spaces Sometimes a set X comes with a family of subsets that can be used to define a notion of proximity, for instance, the real numbers come with the family of all open intervals. Classically, this observation leads to the idea of a topological space, but constructively it seems more natural to introduce the concept of a neighborhood space, as follows. (3.1) Definition. A neighborhood space is a pair consisting of a set X and a family of subsets of X, called neighborhoods, such that if U and U' are neighborhoods and XE U n U', then there exists a neighborhood V with XEVC U n U'. When it is clear what the neighborhoods are, we usually refer to X itself as a neighborhood space. If x is an element of a neighborhood U, then U is called a neighborhood of x. A subset of X which is equal to the union of some family of neighborhoods is called an open set. Thus a set V is open if and only if each x in V has a neighborhood U with XE U c V. The intersection of finitely many open sets is open, and the union of an arbitrary family of open sets is open. The union of all neighborhoods which are included in a given subset Y of X is an open set called the interior of Y; it is the largest open subset of Y, and is written yo. A subset Y of X is closed if XE Y whenever x is a point of X such that each neighborhood of x contains a point of Y. The intersection of any family of closed sets is a closed set. The closure of a subset Y of X consists of all x in X each of whose neighborhoods contains a point of Y; it is the smallest closed set containing Y, and is written Y. A subset Y of X is dense in X if its closure equals X. Associated with each neighborhood space X is a subset Cont(X, lR) of F(X, lR) whose elements are called the weakly continuous functions on X. A function f: X -> lR belongs to Cont(X, lR) if

76

Chapter 3

Set Theory

and only if for each x in X and each e > 0 there exists a neighborhood U of x such that If(x) - f(x')1 ~ e (x' E U) In practice, the set Cont(X, IR) turns out to be of little interest, because it is not possible to get a good hold on its structure. To see this by an example, consider the set IR of real numbers and let F denote the set of continuous functions from IR to IR as defined in Chapter 2 It is easy to see that every continuous function is weakly continuous. Therefore Fe Cont(X, IR) c F(IR, IR).

As we remarked previously, Brouwer asserts that all functions in F(IR, IR) are continuous, so that these inclusions are actually equalities; he also contends that this assertion is a theorem We support the first claim and reject the second; while reflection makes it extremely plausible that F = F(IR, IR), to accept Brouwer's arguments as a proof would destroy the character of mathematics. It seems likely that we are in the tantalizing situation in which the equalities F = Cont(lR, IR) and Cont(lR, IR) = F(IR, IR) will never be proved and never be counterexampled. This situation is typical. For most sets X we shall be interested in a certain well-structured subset of F(X, IR) (rather than in F(X, IR) itself) which it is not possible to realize as the set of weakly continuous functions relative to any neighborhood structure on X. Therefore it makes sense to focus attention on subsets of F(X, IR) instead of on neighborhood structures For convenience these subsets are required to satisfy certain minimal restrictions. (3.2) DefinitiOlL A function space is a pair consisting of a set X and a subset F of F(X, IR), called the topology, such that the following conditions are satisfied. (a) F contains the constant functions. (b) The pointwise sum and product of two elements of F belong to F. (c) The composition hof of an element f of F and a continuous function h: IR --+ IR is in F. (d) Uniform limits of elements of F are in F; in other words, if cP maps X into IR and for each e > 0 there exists f in F such that IcP(x)-f(x)I~E for all x in X, then cPEF. The elements of F are then called the continuous functions of the topology.

3 Neighborhood Spaces and Function Spaces

When there is no confusion over the topology in usually refer to X itself as a function space. The set F(X, lR) is a topology on X. In particular, topology on lR. A more interesting topology on lR continuous functions as defined in Chapter 2. Let F be a topology on a set X. For all f and g in F,

77

question, we F(lR, lR) is a is the set of

since

max{1, g} = f +max{g - 1, O} = f +t(g - f + Ig- fl), max{1,g} belongs to F; similarly, min{1,g} belongs to F. Since fg=i((f + g)2 _ P _g2),

the requirement that products of elements of F belong to F is actually superfluous A topology F for a set X introduces a notion of proximity into X. As we remarked above, a notion of proximity is introduced into X classically not by giving a family of functions, but by giving a family of subsets, either open sets or neighborhoods. Classically, this is equivalent to giving a family of functions from X to {O, I}. Constructively, there is a vast difference: since functions are sharply defined, whereas most sets are fuzzy around the edges, only the all-toorare detachable subsets of X correspond to functions from X to {O, I}. The fuzziness of sets is another reason to focus attention on function spaces instead of on neighborhood spaces. The truth of the matter is that neither function spaces nor neighborhood spaces are as important as certain related structures - metric spaces and normed linear spaces - which will be studied 10 later chapters. One can pass from a neighborhood structure on a set X to a topology, or go the other way. (3.3) Proposition. The set Cont(X, lR) of weakly continuous functions on a neighborhood space X is a topology for x.

Proof The proof is routine, and is left to the reader.

D

Conversely, a topology F on a set X induces a neighborhood structure on X, obtained by taking neighborhoods to be sets of the form Vf=={XEX: f(x»O} where f is an element of F. To see this, consider elements 1, g of F and define h == min {1, g}. Then Vh = Vf n Vg , so that the intersection of neighborhoods is a neighborhood. Thus X is a neighborhood space.

78

Chapter 3 Set Theory

Any set X can be topologized: take F == F(X, IR). This is the finest topology, and the most natural; its lack of structure keeps it from being important. At the other extreme, X can be topologized by letting F consist of the constant functions and nothing else; this topology is too small to be useful. Any subset Fo of F(X, IR) can, of course, be used to generate a topology F for X; the topology in question is the smallest subset of F(X, IR) containing Fo and satisfying conditions (a)--{d) of Definition (3.2). In fact, F is just the subset of F(X, IR) obtained inductively from the following principles of construction. (i) (ii) (iii) (iv)

The elements of Fo and the constant functions are in F. The pointwise sum and product of two elements of F are in F. hof is in F whenever fEF and h: IR-+IR is continuous. A uniform limit of elements of F is in F.

Problems 1. Prove the algebraic laws stated after Definition (1.3).

2. A family of sets consists of an index set T, a finite routine A. that assigns to each element t of T a set A.(t), and a finite routine ¢J that assigns to each ordered pair (t', t) of elements of T with t = t' a function ¢J(t', t) from A.(t) to A.(t') such that (i) ¢J(t, t) is the identity map on A.(t) and (ii) ¢J(t", t')o¢J(t', t)=¢J(t", t) whenever t=t'=t". (This definition is due to Richman. The sets A.(t) are not necessarily subsets of some fixed set.) Define the exterior union (or disjoint union) of such a family. In the case where the sets A.(t) are all subsets of a fixed set S, relate the exterior union to the interior union defined at (1.5). 3. Show that the family of detachable subsets of a set X need not be closed with respect to countable unions. 4. Let S be a discrete set, and f an element of F(S x s, {O, I}) such that elements sand s' of S are equal if and only if f(s, 8') = 1. Show that the relation =1=, defined by setting s =1= 8' if and only if f(8, 8') = 0, is an inequality relation on S.

Notes

79

5. Let S be a discrete set with inequality defined as in Problem 4, and let (T, A) be the family of detachable subsets of S. Show that if a, bES and a*b, then there exists t in Twith aEA(t) and bE -A(t). 6. Call two elements a and b of a set A weakly unequal if there exists a function f. {a, b} -.lR with f(a) f(b). Show that a and b are weakly unequal if and only if a = b entails 0 = 1.

*

*

7. Let A and B be arbitrary sets, and let be an inequality relation on B. Show how to define an inequality relation on F(A, B). 8. Let A be a function from a set A into F(A, {O, I}). Show that there exists f in F(A, {O, I}) such that f *A(a) for all a in A, where is the inequality on F(A, {O, I}) defined in Problem 7. (This is a constructive version of Cantor's result that a set is always smaller than the set of its subsets.)

*

9. Prove (2.4) and (2.5). 10. Show that the union of two closed subsets of lR need not be closed. 11. A subset S of lR is convex if tx+(l-t)y belongs to S whenever x, YES and 0 ~ t ~ 1. A function space (X, F) is connected if for each f in F the closure of f(X) is a convex subset of lR. Show that a product X =Xl x .. X X N of connected spaces (Xl, Fd, ... , (X N, FN) is connected, where the topology for X is the topology generated by the set of all functions (Xl' ... , xN)t-+(fkEFk' 1'5;, k '5;, N). 12. Prove that a function space (X, F) is connected if and only if for arbitrary x, y in X, arbitrary bounded elements fl, ... , fn of F, and an arbitrary e > 0, there exist Xl> ... , XN in X such that Xl = X, XN = y, and

Notes Contrary to classical usage, the notion of equality is a convention. We define what it means for elements of a given set to be equal, but the notion of equality of elements of different sets is not defined. The only way in which elements of different sets A and B can be regarded as

80

Chapter 3 Set Theory

equal is by realizing A and B as subsets of a third set S. For this reason we define the operations of union and intersection only for sets which are given as subsets of a given set. (See, however, Problem 2.) An alternative approach to Definition (1.4) would be to define a family of subsets of a set S as a subset of the cartesian product S x T, where T is the index set for the family. The concept of a complemented set suffers from what appears to be a glaring deficiency· if X is a set with many elements, it seems unnatural to call (qJ, qJ) a complemented set in X, especially since (qJ,X) is also a complemented set. On the other hand, there is nothing to force us to consider complemented sets like (qJ, qJ) In the applications there will always be some condition that keeps the complemented sets A with which we are concerned from having the property that both Aland A 0 are ridiculously small. Brouwer's contention that all elements of F(1R,1R) are continuous seems to contradict claims of certain recursive function theorists, who give examples of elements of F(1R, 1R) that are not continuous In both instances, the claims are based on extramathematical considerations. Brouwer analyzes all possible techniques for constructing elements of F(1R,1R), and comes to the conclusion that all such elements are continuous. The recursive function theorists analyze the possibilities for constructing real numbers, and come to the conclusion that they all possess a certain property (they are recursive); in addition, they show how to construct a discontinuous function on the set of recursive real numbers. These two positions are, in fact, compatible; they do not contradict each other, because it is possible to believe both that all constructive real numbers are recursive, and that without making use of some unprovable hypothesis (such as the hypothesis that all constructive real numbers are recursive), the only elements of F(1R, 1R) that can be constructed are continuous. Extramathematical considerations of both types (especially the first) are useful in indicating that we should not try to do certain things constructively; but they have no place in the actual development of constructive mathematics. Definition (3.2) should not be taken too seriously. The purpose is merely to list a minimal number of properties that the set of all continuous functions in a topology should be expected to have. Other properties could be added, to find a complete list seems to be a nontrivial and interesting problem.

Chapter 4. Metric Spaces

The concept of a metric is defined, some examples are studied, and various techniques for constructing metrics are developed The neighborhood structure of a metric space is defined, and the notions of weakly continuous and uniformly continuous functions are introduced. Completeness is defined in Section 3, and the construction of the completion is carried through. Following Brouwer, we define a compact space to be a metric space that is complete and totally bounded. Compact and locally compaU spaces are studied in Sections 4-6. Constructivizations of various classical results, such as Ascoli's theorem, the Stone- Weierstrass theorem, and the Tietze extension theorem, are given. The concept of a located set, due to Brouwer, plays an important role. Crucial for later developments is Theorem (49), a partial substitute for the classical result that a closed subset of a compact space is compact.

The sets that occur in analysis are constructed from the set of real numbers by certain standard methods, for instance by the formation of subsets or cartesian products of sets already defined. They often carry a structure, such as a function space structure or a neighborhood structure, built up at the same time the sets themselves are constructed. Of special importance is the structure given by a metric, where a metric on a set X is a function on X x X satisfying certain properties that we intuitively associate with distance. A metric is yet another way of introducing a notion of proximity into X.

1. Fundamental Definitions and Constructions Guided by our knowledge of the real numbers and our intuition, we introduce the notion of distance as follows. (1.1) Definition. A metric on a set X is a function p: X x X such that for all x, y, and z in X

-+ 1R 0 +

82

Chapter 4

(1 1.1)

(1.1.2) ( 1.1.3)

Metnc Spaces

p(x,y)=O

if and only if x=y

p(x, y) = p(y, x) p(x, y) ~ p(x, z) + p(z, y)

(the triangle inequality).

A pair (X, p) where p is a metric on X is called a metric space. When there is no confusion over the metric, we often refer to X itself as a metric space. The standard inequality relation on a metric space (X, p) is defined by x =1= y if and only if p (x, y) > O. It often happens that in place of (1.1.1) the weaker condition p(x,y)=O

if X=y

is satisfied. In this case p is called a pseudometric, and (X, p) is called a pseudometric space. Associated with each pseudo metric space (X, p) is a metric space (X 0, p) obtained by taking X 0 to be the set X with the equality relation x = y modified to mean that p(x, y) =0. The theory of pseudometric spaces therefore is essentially no different from the theory of metric spaces. The most important example of a metric space is the set of real numbers, metrized by defining p(x, y) == Ix - yl

(x, YElR)

In order to define metrics for other sets, it is best to establish some general methods of metrization. (1.2) Definition. Let f: Y--+X be a function from a set Y to a pseudo metric space X The induced pseudometric p* on Y is defined by (1.2.1)

p* (y, i) == p(f(y), f(i»

(y, i

E

Y).

For p* to be a metric it is sufficient that p be a metric and that y = y' whenever fey) = fey')· As an example of the use of Definition (1.2), the inclusion map i: Y--+ X from a subset Y of a metric space X into X induces a metric on Y. In particular, the above metric on 1R induces metrics on CQ, 71., and the various intervals. As another example, if f: X --+1R is any real-valued function on a set X, the pseudometric p* is given by p* (x, y) = If(x) - f(y)l.

1 Fundamental Definitions and Constructions

83

(1.3) Definition. Let (Xl> pd, ... , (Xn' Pn) be metric spaces. The product metric P on the cartesian product space X == Xl X . • x X n is given by

•

p(X, y) == L Pi(Xi, Yi)

(1 3.1)

i=l

There are other ways of metrizing a finite product of metric spaces. In fact, JR" is usually given the metric d defined by d(x,y)== (tl(Xi-Yi)2r,

to correspond to the geometry of lR· as we know it from experience. This metric is so important that we digress at this point to give the proof that it actually is a metric. To this end, we establish two basic inequalities. The first is known as the Cauchy-Schwarz inequality. (1.4) Proposition. Let

Xl> ...

,x.,

Yl. ...

,Y. be real numbers. Then

(1.4.1) Proof: We compute

PI

PI

PI

PI

PI

PI

=iLxt LyJ+iLxJ LYt- LXiYi LXjYj i=l

j=l

j=l

i=l

i=l

j=l

= L i(xt yJ +xJ yt - 2XiYjXjYi) i.i=l n

= Li(XiYj-XjYi)2~O. i,j=l

This is equivalent to (1.4.1).

D

The next inequality is called Minkowski's inequality. (1.5) Proposition. Let (1.5.1)

Xl.

.,

X.,

Yl.

, Y.

be real numbers. Then

84

Chapter 4

Metnc Spaces

Proof Using (1.4.1), we compute n

PI

PI

PI

I(Xi-Yi)2= LX?+ Ly?-2L x iYi

i=l

i=1

~itIX?+it/?+2 (~x?)t (t/?r =(

(tl x?

This is equivalent to (1 5 1).

r (t/?rr· +

0

(1.6) Corollary. The function d is a metric on JR". Proof Clearly, d satisfies Minkowski's inequality, d(x,

conditions

and

(1.1.1)

z) =(tl «Xi - Yi) -(Zi - Yi))2)

(1.1.2).

By

t

~ (tl (Xi - Yi)2 r +(tl (Yi - zif r = d(x, y) + d(y, z)

X==(XIo

for all ,x"), Y==(YIo d also satisfies (1.1.3). 0

,Yo), and

Z==(ZIo ... ,z") in JR". Therefore

Related to the metric d on JR" is a certain metric on the set F of continuous, real-valued functions on a compact interval [a, b]. For arbitrary elements f. g of F we define d(f.

g) == (! If(x) - g(xW dx f·

Minkowski's inequality for integrals, d(f. g) ~ d(f. h) + d(h, g)

(f. g, hE F),

follows from (1.5.1) by approximation. There is also a Cauchy-Schwarz inequality for integrals,

IJ f(x) g(x) dxl ~ (J f(X)2 dx)t (J g(X)2 dx)"t

(f. gEF),

which follows from (1 4 1) by approximation. The Cauchy-Schwarz and Minkowski inequalities admit many other generalizations, as will be seen later. In the meantime, we end our digression, and return to the consideration of metric spaces in general

1. Fundamental Definitions and Constructions

85

A metric space (X, p) is called bounded if there exists a real number C > 0, called a bound for (X, p), such that p(x, y) ~ C for all x and Y in X. A subset Y of a non void metric space X is bounded if, for all (equivalently, some) x in X, the set Yu {x} with the induced metric p* is a bounded metric space. A subset Y of IR can be bounded as a metric space but not bounded as a subset of IR. A countable product of metric spaces can be metrized when each of the spaces is bounded. (1.7) Definition. Let

«Xn,Pn))~l

be a sequence of metric spaces, each

wi th bound 1 The product metric p on X ==-

TI X

n

is defined by

n=1

L2- npn(x n,Yn)

P«Xn),(Yn))==-

«xn), (Yn)EX).

11=1

Definition (1 7) is useful even when the .spaces (X n, Pn) are not necessarily bounded, because there is a standard method for transforming an arbitrary metric into a metric bounded by 1. (1.8) Proposition. Let (X,p) be a metric space. Let h IR o + -+IR o + satisfy the conditions

if and only if u=O h(u + v) ~ h(u) + h(v) (u, VE IR 0+).

(1.8.1)

h(u)=O

(1.8.2)

Then d ==- hap is a metric on X Proof Condition (1.1.1) is satisfied because of (1.81). Condition (1.1.2) is obviously satisfied To check (1.1.3) note that d(x, z) = h(p{x, z)) ~ h(p(x, y) + p(y, z)) ~ h(p(x,

for all x, y, z in X.

y)) + h(p(y, z)) = d(x, y) + dey, z)

0

(1.9) Corollary. If p is any metric on a set X, then the function p' defined by p'(x,y)==-min {p(x,y), I} is a bounded metric on X.

Proof. The proof is obvious.

0

(X,YEX)

86

Chapter 4

Metnc Spaces

This corollary leads us to define the product of an arbitrary sequence ((X n, Pn)) of metric spaces to be the product of the associated sequence ((X n, p~)). The definition of a continuous function on a compact interval generalizes in the context of a metric space, and leads to the important idea of a uniformly continuous function. (1.10) Definition. A function f: X ~ X' from a metric space (X, p) to a metric space (X', p') is uniformly continuous if there exists an operation w from IR+ to IR+ such that p'(f(x),f(Y))~E whenever X,YEX, E>O, and p(x, y) ~ wee). The operation w is called a modulus of continuity for f on X For example, if Xo is a point in an arbitrary metric space X, then the function XI--+p(x,xo) is uniformly continuous on X, with modulus of continuity EI--+E. To see this, consider points x and y in X We have p(x, xo) ~ p(x, y) + p(y, xo) and therefore p(x, xo) - p(y, xo) ~ p(x, y). Similarly, p(y, xo) - p(x, xo) ~ p(x, y) Therefore Ip(x, xo) - p(y, xo)1 ~ p(x, y)

It follows that Ip(x, xo) - p(y, xo)1 ~ E whenever p(x, y) ~ E. For each x==(x n ) in the product (X,p) of a sequence ((Xn,Pn)) of metric spaces (each with bound 1), and each k in Z+, write nk(x) == Xk. Then the projection nk: X ~Xk of X onto X k is uniformly continuous,

with modulus of continuity

EI--+2- k E.

(1.11) Definition. A sequence (1.) of functions from a set S to a metric space (X, p) converges uniformly to a function f: S ~ X if for each E > 0 there exists N. in Z + such that p(J.(X),f(X))~E

(n~N.,xES).

+ g and product fg of uniformly continuous functions f" X ~IR and g: X ~IR are uniformly continuous, and f- 1 is uniformly continuous if If(x)l~c for all x in X and some c>O. The composition f2 0 h of uniformly continuous functions f1: X 1 ~ X 2 and f2: X 2 ~ X 3 is uniformly continuous. The limit f: X 1 ~ X 2 of a uniformly convergent sequence (fn) of uniformly continuous functions from a metric space Xl to a metric space X 2 is uniformly continuous. (1.12) Proposition. The sum f

Proof. The proofs of the various statements are similar to certain proofs in Chapter 2, and are left to the reader. D

2 Associated Structures

87

A uniformly continuous function f from a metric space X I onto a metric space X 2 which has a uniformly continuous inverse is called a metric equivalence, and X I and X z are said to be equivalent metric spaces. Two metrics Pl and pz on the same set X for which the identity function from X to itself is a metric equivalence of (X, PI) with (X, pz) are called equivalent metrics on X. The product metric P on JR' is equivalent to the metric d. An arbitrary metric P is equivalent to the bounded metric P' of Corollary (1.9). If (X.,P.) and (Y",,o.) are equivalent metric spaces for each n in 7L+, then the products of the sequences «Xn,Pn» and «Y",,on» are equivalent.

2. Associated Structures Every metric space has a neighborhood structure and a topology. (2.1) Definition. The open sphere of radius r>O about a point x in a metric space X is the subset S(x, r)=:o {YEX: p(x,y)
of X. The point x is the center of S(x, r). The neighborhood structure of X is defined by taking the neighborhoods to be the open spheres. To see that the open spheres give a neighborhood structure, note first that if YES(x,r) and o
where t =:0 min {ro - p(x, xo), rl - p(x, xd}. Therefore the open spheres give a neighborhood structure. When applied to subsets of a metric space X, the notions of open set, closed set, dense set, interior, and closure refer to this neighborhood structure. A subset Y of X is closed if and only if it contains all points that are limits of sequences of points of Y, in the sense of (2.2) Definition. A sequence (x.) of elements 'of a metric space X converges to an element x of X if for each E > 0 there exists N, in 7L + such that

88

Chapter 4

Metric Spaces

In that case, x is called the limit of (x n ) in X, and we write either lim Xn=X or Xn-.X

as n-.oo.

In case X== IR., these notions of convergence and limit coincide with those introduced in Chapter 2. The closed sphere of radius r ~O about a point x (the center) in a metric space X is the set Sc(x,r)=={YEX· p(x,y)~r}.

It is a closed subset of X. We now introduce the notion of a located set, which will be of great importance in all our constructive analysis. (2.3) Definition. A subset A of a metric space X is located in X if the

distance

p(X, A) == inf {p(x, y): YEA}

from x to A exists for every x in X. There exist subsets of IR. that are not located (Problem 8). If A is located in a metric space X, then the closure A of A in X is located, and p(x, A) = p(x, A) for each x in X. Moreover, the distance function X 1-+ p(x, A) satisfies the inequality (24)

p(x, A) ~ p(x, y) + p(y, A)

(x, YEX),

because (2.4) is equivalent to the assertion that p(x, A) ~ p(x, y) + p(y, z) for all z in A; this holds because p(x,A)~p(x,z)~p(x,y)+p(y,z). The metric complement X -A (or just -A) of a located subset A of a metric space X is the set X -A=={XEX: p(x,A»O}.

Because of (24), this is an open set. Moreover, for each x in X and each e> 0, either p(x, A) > 0 or p(x, A) < e; whence Au - A is dense in X. The topology associated with the neighborhood structure of a metric space X is called the weak topology, and continuous functions in the weak topology are weakly continuous. A weakly continuous function f" X -.IR. has the property: if Xn -. x as n -. 00, then f(xn)-.f(x) as n-'oo. The weak topology is not of much use. More useful is the uniform topology, defined as follows. (2.5) Definition. The uniform topology F for a metric space X consists of all uniformly continuous functions f. X -.IR.

3 Completeness

89

Since every uniformly continuous function is weakly continuous, the weak topology is larger than the uniform topology. Other topologies for a metric space X are obtained by considering the set of all J. X --> IR which are uniformly continuous on each member of a given family of subsets of X. For example, this was done when we defined the continuous functions on an arbitrary interval in IR

3. Completeness The construction which led from
1S

The set X of all regular sequences of elements of X is called the completion of X Elements (x.) and (y.) of X are equal if p(x., Y.):;:; 2n- 1

The metric (3.1.1)

(nEZ+).

p on X is defined by p(x,Y)== lim p(xn,Yn)

for all x==(x.) and Y==(Y.) in X. The limit (3.1.1) exists because p(xm' Ym) - p(x., Y.) :;:; p(xon, x.) + p(x., Y.) + P(Y., Ym) - p(x., Yn)

= p(x,", x.) + P(Ym, Y.):;:; 2(m- 1 + n- 1 )

(m, nEZ+).

The operation p is a function from X x X to IR o +. For if x==(x.), x' == (x~), Y == (Y.), and y' == (y~) are elements of X with x = x' and Y = y', then p(x, y) = lim p(x., Y.) :;:; lim (p(x., x~) + p(x~, y~) + p(y~, Y.)) =p(x', y').

Similarly, p(x', y'):;:; p(x, y). Therefore p(x, y) = p(x', y') (3.2) Proposition. The function

p is a metric on X.

90

Chapter 4 Metnc Spaces

Proo!' Condition (1.1.2) is clearly satisfied. To check (1.1.3), consider points x==(x.), Y==(Y.), and z==(z.) in X. We have p(x, y) = lim p(x., Y.) ~

lim p(x., z.) + lim p(z., Y.)

= p(x, z) + p(z, y).

As to condition (1.1.1), it is obvious that p(x,y)=O if X= y. Conversely, if p(x, y) = 0, then for all n in 7r, p(x., Y.) ~ p(x., X.,) + p(x." y.,) + p(y." Y.) ~2(n-l

Therefore x = y.

+m- 1 )+ p(x." y.,) --+ 2n- 1

as m --+ 00.

0

The inclusion map i: X --+ X is defined for each x in X by taking i(x) to be the regular sequence each of whose terms is x. The map i realizes X as a subset of

X,

with preservation of metric:

p(x, y) = p(i(x),

The subset X is dense in

iCy»~

(x, YEX).

X.

(3.3) Definition. A Cauchy sequence (x.) of elements of a metric space X is a sequence such that for each e> 0 there exists a positive integer N. with p(xm,x.)~e

(m,n?;,NJ.

The metric space X is said to be complete if every Cauchy sequence in X converges. It is obvious that a convergent sequence in an arbitrary metric space is a Cauchy sequence.

(3.4) Theorem. The completion complete.

X

of an arbitrary metric space X is

Proof. The proof is similar to the second half of the proof of (3.3) of Chapter 2, and is therefore omitted. 0 (3.5) Proposition. A metric space X is complete

if and only if

X

=

X.

Proof. By (3.4), if X = X, then X is complete. Assume conversely that X is complete. Certainly, X eX. To show that X eX, consider an arbitrary element x == (x.) of X. Then (x.) is a Cauchy sequence of elements of X, and so converges to an element Xo of X. For each n in

3 Completeness 7/,+

91

we have m-oo

PJI-OO

Therefore x = i(xo), by definition of equality on X. It follows that XcX. D A closed subset Y of a complete metric space X is complete. For, any Cauchy sequence (Yn) of elements of Y converges to a point Yo in X which, because Y is closed, belongs to Y. Conversely, a complete subset Y of an arbitrary metric space X is closed. For if the point Xo in X is in the closure of Y, then Xo = lim Yn for some sequence (Yn) of points of Y. Since Y is complete, the Cauchy sequence (Yn) converges to a point Yo in Y. Thus Xo = Yo, and Xo is in Y. A complete metric space stays complete under an equivalent metric. Therefore the following result has intrinsic significance. (3.6) Theorem. The product (X,p) of a sequence «Xn,Pn)) of complete

metric spaces is complete. Proof By the above remark, there is no loss of generality in assuming that (X n' Pn) has bound 1 for all n. Let (Xk):== 1 be a Cauchy sequence of elements of X. For each k, Xk==(X~)~l is a sequence with X~EX. (nE7/,+). For each n the sequence (X~)k~l is a Cauchy sequence, because

Let x~ be the limit of (X~)k~l in X •. The sequence (Xk) converges to the point X O == (x~)~ 1 of X, since for all N in 7/,+ we have co

p(x\X O)=

L 2-nPn(x~,x~) .~1

N

~

L 2-nPn(x~, X~)+2-N .~1

~2-N+l

if k is sufficiently large.

0

The next lemma is useful in constructing uniformly continuous functions with values in a complete metric space. (37) Lemma. Let Y be a dense subset of a metric space X, and f: Y -+ Z a uniformly continuous function from Y to a complete metric space Z, with modulus of continuity w. Then there exists a uniformly

92

Chapter 4

Metric Spaces

continuous function g: X f (y) = g(y) for all Y in Y

---+

Z with modulus of continuity

tw

such that

Proof Since Y is dense in X, for each x in X there exists a sequence (Yn) of points of Y that converges to x If E is any positive number, then p(f(Ym).!(Yn» ~ E whenever P(Ynl> Yn) ~ W(E). Therefore (f(Yn)) is a Cauchy sequence of elements of Z, whose limit we denote by g(x). Now consider E> 0 and points x and x' of X with p(x, x') ~tW(E) Let (Yn) and (y~) be sequences of points of Y converging respectively to x and x' Then P(Yn,Y~)~W(E), and thus p(f(Yn).!(Y~»~E, for all sufficiently large n Therefore p(g(x), g(x')) = lim p(f(Yn), f(Y~)) ~ E Since E is arbitrary, we see from the case x = x' that g(x) = g(x') (whence g is a function) and that g(x) does not depend on the particular choice of the sequence (Yn) of points of Y converging to x. Clearly, g(y) = f(y) for all Y in Y For general x and x', the above argument shows that g is uniformly continuous, with modulus of continuity D

two

Our next lemma will be used later in the book, and illustrates a proof technique which is often effective in the context of a complete metric space. (3.8) Lemma. Let A be a complete, located subset of a metric space X, and x a point of X. Then there exists a point a in A such that

p(x, a»O entails p(x, A»O. Proof" Define recursively an increasing sequence ).: 7l.+ that for each n in 7l.+ ).(n)=O

=>

p(x,A)
).(n) = 1

=>

p(x, A) > (2n)-1

---+

{O, I} such

and With z any point of A, construct a sequence (an) in A such that: if ),(1) = 1, then an = z for all n; if ),(1) = 0 = },(n), then p(x, an) < n- 1 , while if },(1)=0 and },(n) = 1, then an=anl> where m is the unique positive integer with l(m) =0 and l(m + 1) = 1. Then (an) is a Cauchy sequence of elements of A: in fact, p(am,an)~2n-l whenever m~n. It therefore converges to a point a in A with p(a,an)~2n-l for each n. Suppose that p(x,a»O, and compute N in 7l.+ so that p(x,a»3N- 1 • Then

p(x, aN) ~ p(x, a) - p(a, aN) > N- 1 •

3 Completeness

93

Therefore A(N) cannot be 0, so that A(N)=l Hence p(x,A»(2N)-1 >0. 0 An immediate corollary of Lemma(3.S) is that p(x, A) >0 whenever A is a complete, located subset of a metric space X, and x is a point of X such that x =1= y for all y in A We now arrive at the Baire category theorem. (3.9) Theorem. Let (Un) be a sequence of dense open sets in a complete

n Un is also dense in X. 00

metric space X. Then the intersection U ==

n~l

Proof. Let S(xo, ro) be any open sphere in X. Since U1 is dense, there exists a closed sphere S c (x1,rd with 0 < r1 ~ 1 such that Sc(x1,rd c S(xo,ro)n U1·

Continuing inductively, we construct a sequence (S c(xn, rn)) of closed spheres such that 0 < rn ~ n- 1 and Sc(xn' rn)cS(xn_ 1, rn_ 1)n Un

for all n in Z+. Whenever m~n, Xm belongs to S(xn,rn), and therefore p(x nlO xn) < rn ~ n- 1. Hence (xn) is a Cauchy sequence. Since all except finitely many terms of this sequence lie in any given closed sphere S c(xn, rn), the limit x lies in each of these spheres Therefore cc

XES(Xo, ro)n nUn. n=l

It follows that U is dense in X.

0

Theorem (3.9) is one of the most useful versions of Cantor's diagonal technique, which was used in the proof of Theorem(2.19) of Chapter 2. In fact, the latter theorem is a special case of Corollary (3.10) below. A located subset Y of a metric space X is nowhere dense in X if the metric complement - Y of Y is dense. With this definition we reformulate Theorem (3.9) (3.10) Corollary. Let (y") be a sequence of nowhere dense subsets of a complete metric space X. Then every open sphere in X contains a point y whose distance to each Y" is positive. Proof" Let Sex, r) be an open sphere in X. By (3 9), there exists a point

n - Y". The distance from y to Y" is positive because yE 00

y in Sex, r) n

- Y".

0

n~l

94

Chapter 4 Metnc Spaces

4. Total Boundedness and Compactness A more significant concept for metric spaces than boundedness is the concept of total boundedness, defined as follows. (4.1) Definition. Let (X, p) be a metric space. An e approximation to X is a subset Y of X such that for each x in X there exists y in Y with p(x, y) < e. We say that X is totally bounded if for each e > 0 there exists a finite e approximation to X. The reader can show as an exercise that X is totally bounded if for each e > 0 there exists a subfinite e approximation {XI. ... , x n } to X. A subset of a metric space is totally bounded if and only if its closure is totally bounded The property of total boundedness is preserved under passage to an equivalent metric. The product of a sequence of totally bounded metric spaces is totally bounded. We call a metric space X separable if it contains a countable dense subset. The product of a sequence of separable metric spaces is separable. A totally bounded metric space X is separable. for if Sn is a <Xl

finite n- 1 approximation to X (nEZ+), then USn is a countable dense n=1 set in X. (4.2) Proposition. The image f(X) of a totally bounded metric space X under a uniformly continuous mapping f: X ~ Y is totally bounded.

Proof Let w be a modulus of continuity for f For each e>O let {XI. ... , x n } be an w(ie) approximation to X. Then if x is any point of X, we have p(x, Xi) <w(te) for some i with 1 ;;£i;;£n. For this i we have p(f(x), f(Xi»;;£te < e. It follows that {f(xd, ... , f(x n )} is a subfinite e approximation to f(X). 0 (4.3) Corollary. Let f: X ~ 1R. be a uniformly continuous function on a totally bounded metric space X. Then the supremum and infimum of f exist.

Proof. By (4.2), we see that f(X) is totally bounded. The result now follows from (4.4) of Chapter 2. 0 It follows that if X is totally bounded, then the diameter

diamX=sup{p(x,y): X,YEX} of X exists.

4. Total Boundedness and Compactness

95

(4.4) Proposition. A totally bounded subset Y of a metric space X is located. Proof. If x is any point of X, then yf--> p(x, y) is a uniformly continuous function on Y, as we have seen. By (4.3), its infimum p(x, Y) exists. Thus Y is located. 0 (4.5) Proposition. A located subset Y of a totally bounded metric space X is totally bounded.

Proof: Consider 8> 0, and let {XI, ... , xn} be an 8/3 approximation to X. For each i choose Yi in Y with P(Xi,y;) y) < 8/3. This gives p(y, Yj) ~ p(y, Xj) + p(Xj, Yj) < 8/3 + 8/3 + 8/3 = 8.

Thus the subfinite set {Yl> . . , Yn} is an 8 approximation to Y. Since arbitrary, it follows that Y is totally bounded. 0 (4.6) Proposition. A subset X of JR" is totally bounded

8

is

if and only if it

is located and bounded. Proof Proposition (4.4), together with the fact that a totally bounded metric space is bounded, implies that the conditions are necessary for X to be totally bounded. Conversely, assume that X is located and bounded. Because X is bounded, there exists c>O such that Xc [ - c, C ]n. The product space [ - c, cJ" is totally bounded. As a located subset of a totally bounded metric space, X is totally bounded, by (4.5).

0

One of the most important concepts of classical analysis is that of compactness, effective application of which cuts infinite sets down to finite, and therefore manageable, size. Regrettably, most such applications use either the 'open cover' or the 'sequential' forms of compactness, neither of which is appropriate in a constructive setting. For example, there is no known case of a nontrivial metric space for which there exists a routine method for extracting finite subcovers from arbitrary open covers. We are therefore relieved to find that the other common form of compactness of metric spaces is acceptable, fruitful, and hence adopted as definitive in constructive analysis. (4.7) Definition. A compact metric space, or simply a compact space, is a metric space which is complete and totally bounded.

96

Chapter 4

Metric Spaces

The compact intervals of real numbers are compact in the sense of the above definition. A subset of 1R.. is compact if and only if it is closed, located, and bounded. The product of a sequence of compact spaces is compact. Total boundedness is the more important of the two properties which occur in the definition of compactness, since a metric space which is totally bounded but not necessarily complete can always be compactified by embedding it in its completion and taking its closure. Let X be an arbitrary metric space. For each located subset B of X the function x I--> p(x, B) is uniformly continuous, by (2.4). Therefore the supremum m(A,B)=sup.{p(x,B): xEA}

exists for each compact set A c X. The function p defined by p(A,B)=max {m(A,B), m(B,A)}

is well defined on Cx C, where Cis the set of all compact subsets of X. We shall show that p is a metric on C. Clearly, peA, B) = pCB, A), and p(A,A)=O. Assume that p(A,B)=O. Then m(A,B)=O, so that p(x,B) = 0 for all x in A. Since B is closed, it follows that XE B whenever xEA, whence AcB. Similarly, BcA. Thus A=B whenever p(A,B)=O. It remains to check that p satisfies the triangle inequality To this end, consider arbitrary compact subsets A, B, and C of X, and an arbitrary x in X. By (2.4), p(x, C)~inf{p(x,y)+p(y, C):YEB} ~inf{p(x,y)

YEB} +m(B, C)~p(x,B)+p(B, C).

Therefore meA, C)~m(A,B)+p(B, C)~p(A,B)+p(B, C).

Similarly, m( C, A) ~ peA, B) + pCB, C).

Therefore peA, C)~p(A,B)+p(B, C),

and so p satisfies the triangle inequality. Turning now to the study of sufficient conditions for a subset of a compact space to be compact, we first prove that every compact space is a union of finitely many compact sets of arbitrarily small diameter. (48) Theorem. Let X be a compact space, and e a positive number. Then there exist finitely many compact subsets X b ... , X. of X, each with diameter at most E, whose union is X.

Proof. By the total boundedness of X, there exist subsets XL ... , X! of X, each consisting of one point, such that for each x in X at least one

4 Total Boundedness and Compactness

97

of the numbers p(x, XJ) (1 ~j~n) is less than 3- 2 e. We define inductively sequences (XD~ b ... , (X~)~ 1 of subfinite subsets of X such that for 1 ~j ~ nand i"?; 1, (i)

X} cXJc

(ii) (iii)

if XEX~+l, then p(x,X))<3- ie if p(x,X)<3- i- 1e, then p(x,X)+1)<3- i - 2 e

.

The sets X} (1 ~j ~ n) have already been defined. Assume therefore that xL . ,X~ have been defined and satisfy the relevant conditions. Let {Xb .. ,XN} be a 3- i - 2 e approximation to X. Write the set {(m,j): 1 ~ m ~ N, 1 ~j ~ n} as a union of two disjoint sets Sand T such that p(x m ,X)<3- ie if (m,j)ES, and p(x m,X)>i3- i e if (m,j)ET For 1 ~j ~ n construct X~ +1 by adjoining to Xj all those points X m , if any, for which (m,j)ES. Clearly, the sets Xj+1 are subfinite and satisfy (i) and (ii). To check (iii), consider x in X with p(x,xj)<3- i - 1e, and choose Xm with p(x,x m )<3- i - 2 e. We have P(Xm' Xj) ~ p(xm' x) + p(x, xj) < 3- i- 2 e+ 3- i - 1e
This completes the inductive construction of the sets Xj + 1 with the required properties. 00 For 1 ~j ~ n write 1) == Uxj, and let Xj be the closure of 1). Then i=l

Xj is closed, and therefore complete. To prove that Xj is totally bounded, consider any positive integer m and any point Y of 1). Choose i with yE Xj If i ~ m, then p(Y, Xj) = o. If i > m, apply (ii) to construct points Yi = y, Yi-1 EXj-\ ... , YmEXj such that

Then p(Y, Xj)~P(Yi,Ym)~

L

P(YbYk-1)

k~m+1

00

~

L

3- k + 1e=i3- m + 1e.

k~m+1

Thus Xj is a subfinite i3- m + 1 e approximation to 1). Since m is arbitrary, we see that 1), and therefore Xj' is totally bounded.

98

Chapter 4

Metnc Spaces

A simple induction argument using (ii) now shows that xj has i-I

diameter less than 2 L 3 - k E for each i ~ 2. Hence k~1 <Xl

diamXj~2

L 3- k E=E. k~1

To show that X = XI U ... u X., consider any x in X, and choose j (1~j~n) with p(x,X)<3- 2 E. It follows from (iii) by induction on i that p(x,Xj)<3- i - I E for all i in 71+. Hence XE~=Xj. 0 A compact space X will be given the uniform topology, so that a continuous function f: X -> 1R is one which is uniformly continuous. More generally, we define a continuous function from X to an arbitrary metric space to be one which is uniformly continuous. Each continuous function f: X -> 1R on a compact space X gives rise to an important class of compact subsets of X. Let P(O!) be a property of real numbers O!, and let (O!.) be a sequence of real numbers such that P(O!) is valid for each real number O! with O! O!. for all n Then P(O!) is said to hold for all but countably many real numbers, the sequence (O!.) is called the excluded sequence, and a real number O! with O! O!. for all n is said to be admissible for the property P(O!)

'*

'*

(4.9) Theorem. Let f" X

-> 1R be a continuous function on a compact space X. Then for all but countably many real numbers O! > a == inf {f(x): XEX} the set X,,== {XEX· f(x)~O!}

is compact. Moreover, if O! is admissible for this property, and E is any positive number, then there exists 15 > 0 such that p(Xt' X ,,) ~ E for all admissible t with IO! - t I <15, where p is the metric on the set of compact subsets of x. N(k)

Proof. For each k in 71+, represent X as a finite union X

=

U X~

of

j~1

compact sets, each of diameter less than k- 1 • Define the excluded sequence (O!.) to be an ordering of the real numbers cjk==inf{f(x): XEXn

(kE71+, 1~j~N(k)).

Let O! > a be admissible relative to (O!.). For each j and k with Cjk < O! choose xj in Xj with f(x~) < O!. We have (4.9.1) Consider any point x in X" and any positive integer k. Choose j with 1~j~N(k) and xEXj. Then Cjk~f(X)~o!, so that Cjk
4. Total Boundedness and Compactness

have

p(x,X~)
99

Hence {x~: 1 ~j~N(k), Cjk
is a subfinite k- 1 approximation to X". Therefore X" is totally bounded. As X" is closed, it is complete and therefore compact. Also, it follows from (4.9.1) that p(x,Xa)
(4.9.2)

(XEXj, Cjk
For an arbitrary e>O, choose k in Z+ with k- 1 <e. Write t5 == min {IIX -cjkl: 1 ~j ~ N(k)} ,

and consider any admissible t with IIX-tl
D

For an application of Theorem (4.9), we need a further definition. (4.10) Definition. Let f be a mapping of a metric space (X, p) into a metric space (X', p'). We say that f is injective, or an injection, if f(x) =!= fey) whenever x, y belong to X and x =!= y. We say that f is hyperinjective, or a hyperinjection, if for any two compact subsets A, B of X with inf{p(x,y): xEA, YEB}>O, there exists r>O such that p'(f(x), fey»~ ~ r for all x in A and y in B. (4.11) Proposition. Let f be a continuous, hyperinjective mapping of a

compact space (X, p) into a metric space (X', p'). Then the inverse map g: f(X) -+ X is uniformly continuous and hyperinjective on f(X), and f(X) is compact. N

Proof. Given e>O, represent X as a finite union X

=

U Xj of compact j=l

sets, each of diameter less than e/3. If sup {p(x, Xj). XEX} < e/3 for some j (1 ~j ~ N), then p(g(x), g(y»

~ diam

X <e

(x, yEf(X».

In proving g uniformly continuous, we may therefore assume that sup {p(x, Xj): XEX} >0 for each j; whence, by (4.9), there exists r such that 0 < r < e/3 and each of the sets Sj=={XEX·

p(x,X)~r}

={XEX:

-p(x,Xj)~

-r}

(l~j~N)

100

Chapter 4

Metric Spaces

is compact. Define and <5 == min {<5 b

,

<5 N }

Since O 0 Let x and Y be points of X with p'(f(x), fey)) < <5, and choose j so that YEX j • Then p'(f(x),j(Xj)) <<5j ; so that p(x, X j ):;:; r, and therefore p(x, z) < 2r for some z in Xj. It follows that p(x, y):;:; p(x, z) + diam Xj < 2r+ E/3 < E.

Hence g is uniformly continuous on f(X). The continuity of f ensures that g is hyperinjective. Finally, if (x n ) is a sequence in X such that (f(x n )) is a Cauchy sequence in f(X), then (x n ) is a Cauchy sequence, as g is uniformly continuous. Let x == lim x n • As f is continuous, (f(xn)) converges to f(x). Thus f(X) is complete. Since f(X) is totally bounded by (42), it is compact.

0

5. Spaces of Functions Some of the most important metric spaces of analysis are spaces of functions Here is a basic example. (5 1) Definition. For each compact space X and each metric space Y, we shall use C(X, Y) to denote the set of all (uniformly) continuous functions from X to Y For C(X, JR) we write simply C(X). The metric p on C(X, Y) is defined by p(f, g) == sup {p(f(x), g(x)): XEX}

(I, gE C(X, Y))

If Y is complete, then so is C(X, Y) (c.f (4.11) of Chapter 2). In case Y==JR, the metric p is related to the norm

Ilfll == Ilfllx==sup {If(x)1 XEX} on C(X) by the equality p(f,g)= Ilf -gil

(I, gE C(X)).

A subset S of C(X, Y) is equicontinuous if the functions in S have a common modulus of continuity. This concept is important because it affords a means of proving that certain subsets of C(X, Y) are totally bounded. The result in question is called Asco/i's theorem, or Arzeld's theorem.

5 Spaces of Functions

101

(5.2) Theorem. Let S be an equicontinuous subset of C(X, Y) such that for each e > 0 there exists an e approximation {X1o .. , xn} to X for which the subset A == {(f(xd, ... ,f(xn)): fES} of yn is totally bounded. Then S is totally bounded. Proof. Let w be a common modulus of continuity for the functions in S. Consider e > 0, and let {X1o ... , xn} be an w(e/3) approximation to X such that the above set A is totally bounded. Let ft, .. ,fm be elements of S such that the points u;==C[;(xd, ... ,j;(xn ))

(1 ~i~m)

form an e/4 approximation to A. Then for an arbitrary

f in S,

n

L p(!.(Xj),f(Xj)) < e/4 j=l

for some value of i, which we fix For an arbitrary x in X there exists Xj with p(x, Xj) < w(e/3). Then p(!.(x), f(x)) ~ p(!.(x), !.(Xj)) + p(!.(Xj), f(Xj)) + p(f(Xj), f(x)) ~ e/3 + e/4 + e/3.

Since x is an arbitrary point of X, it follows that p(j" f) < e. Thus {ft, ... ,fm} is an e approximation to S. Since e is arbitrary, S is totally bounded. 0 To get a good application of Theorem (5.2), we need three lemmas. (53) Lemma. A finite sequence ([a1o b1J, ... , [an, bnJ) of compact intervals has nonvoid intersection if and only if (5.3 1)

Proof The condition is clearly necessary, since if there exists x

In

n [ak,bkJ then a;~x~bj for all i andj. Assume conversely that (5.3.1) n

k=l

holds. Then min {b1o ... ,bn} -max {a1o ... , an} ~O. Therefore the point x == min {b 1 , intervals. D (5.4) Lemma. Let

.. ,

bn } belongs to each of the given

C10 .. , Cn be real numbers, and r10 ., rn nonnegative real numbers. Then there exists a real number x such that

(54.1)

102

Chapter 4 Metnc Spaces

if and only if (5.4.2)

Proof Condition (5.4.2) is obviously necessary for (5.4.1). Assume conversely that (5.4.2) is satisfied. Then Ci-ri;£Cj+rj for all i and j. The intervals [Ci -r;, Ci + ri] therefore have a common point x, by (5.3). Clearly, x satisfies (5.4.1) D (5.5) Lemma. Let K and rij (1;£ i, j;£ n) be positive real numbers Then

the set Y of points x == (Xl> Ixd;£K,

...

,x.) in JR' such that IXi-xjl;£rij

(l;£i,j;£n)

is totally bounded. Proof For each X==(XI, .. , x.) in JR', write cx==max{lxi -Xjl riil: 1;£ i,j;£n} and

f(x) == max {I, cx } - I x. Then f is a continuous function from the product space [- K, K]' into y, and is onto Y because f(x) = x for all x in Y. Since [ - K, KY is totally bounded, so is Y, by (4.2). D (5.6) Theorem. Let A: JRo+-+JR o + be a strictly increasing, continuous function such that },(O) = 0 and such that ).,(x + y);£ },(x) + },(y) for all x, y in JR 0 +. Let K be a positive number, and let S consist of all continuous functions f from a compact space X to JR such that II f II ;£ K and If(x) - f(y)l;£ },(p(x, y» (x, yeX).

Then S is compact Proof Since S is obviously closed in C(X), to show that S is compact it is enough to show that it is totally bounded. Since S is equicontinuous, by (5.2) it will suffice to show that for every finite sequence (Xl, ... , x.) of unequal points of X the subset A=={(f(xd, .. ,f(xn»: feS} of JRn is totally bounded. To this end, for 1;£ i, j ;£ n define

rij = A(p(X;, Xj»

=1

*

if i j, if i=j.

5 Spaces of Functions

103

Note that rij>A.(O)=O whenever i*j. By (5.5), the set B={U=(Ul, ... ,U.)EIRn. IUil:;:::;K, IUi-ujl:;:::;rij (1:;:::;i,j:;:::;n)}

is totally bounded. Therefore it is enough to show that A = B, or (since obviously AcB) that BcA. Since X is totally bounded, (Xl, ... ,X.) can be extended to a sequence (Xk)k= 1 of elements of X which is dense in X. Consider an element U of B. We continue the finite sequence (Ul, . . ,u.) to an infinite sequence (Uk) such that for all i,j in 7L+, (5.6.1) This is done inductively. Certainly, (5.6.1) holds for 1 :;:::;i,j:;:::;n. Assume that Ulo ., Urn have already been constructed to satisfy (5.6.1). Then, by (5.4), Urn+l can be constructed to satisfy (5.6.1) because the hypotheses on 2 ensure that

lUi -ujl:;:::; 2(P(Xi' Xj)):;:::; 2[P(Xi' xm+d + P(Xm+lo Xj)] :;:::;A.(P(Xi' xrn+d)+2(p(xjo Xm+l))

(1 :;:::;i,j:;:::;m).

Thus the sequence (Uk) is constructed inductively. Let Y be the dense subset {Xl, X2, ... } of X. Define h: Y -+JR. by h(xi)=ui (iEZ+). By (5.6.1), we see that h is uniformly continuous. It follows from (3.7) that h has a uniformly continuous extension f to X. By (5.6.1), the fact that Y is dense, and the continuity of the various functions involved, we see thatfES. Since u=(f(xd, .,f(x.)), U belongs to A. Thus BcA, as was to be proved. 0 (5.7) Corollary. Let X be compact, and let K, c, and (X be positive numbers with (X:;:::; 1. Let S be the set of all f in C( X) with II f II :;: :; K which satisfy the Lipschitz condition

If(x) - f(y)l:;:::; c p(x, y)~

(x, YEX)

Then S is compact. Proof· The result follows from (5.6) by taking 2(t)=ct~ for each t~O. (It is a simple exercise to show that (s + t)~ ~ s~ + t a whenever s;;::-: 0, t;;::-: 0, and O<(X~ 1.) D Our next theorem will be the Stone- Weierstrass theorem, the basic result in the theory of approximation by real-valued functions on compact spaces. Some groundwork is necessary.

104

Chapter 4 Metne Spaces

(5.8) Definition. A polynomial of degree N in n variables is a function p: lRn --+ lR of the form P(Xb ... , xn) =

ai, in x~l .. x~n.

L O~il+

+i"~N

If p(o) = 0, we say that the polynomial p is strict.

(5.9) Lemma. For each 8>0 there exists a strict polynomial p' lR--+lR such that Ilxl- p(x)l;;;;; 8 whenever -1;;;;; x;;;;; 1. Proof. Consider the function tl-+(1 - t)t == exp(t In (1 - t))

on the interval (-1,1). For the moment, fix t in (-1,1). For nth derivative of the above function at t is

n~1

the

We show that the Taylor series 00

(5.9 1)

1 - L (2n n!)-11.3.5 ... (2n - 3) t"

converges to (1-t)t. To this end, write N

R N==(1-t)t-1+ L(2"n!)-11.3.5 .(2n-3)t". "=1

By Taylor's theorem, there exists c such that min{O,t};;;;;c;;;;;max{O,t} and RN = _(2 N+1N !)-1 1.3.5 ... (2N -1)(1 _c)t- N(t _C)N t. Hence

Now, by the ratio test, for Ixl < 1 the series 00

L (2"+1 n!)-11.3.5 ... (2n-1)x" "=1

converges, and therefore (2"+l n !)-11.3.5. (2n-l)x"--+O as n--+oo. Since I(t -c)(1 -c)-ll < 1, it follows that RN--+O as N --+ 00. Hence the series (5.9.1) converges to (1-t)i 00 By Raabe's test, the series L (2" n !)-1 1.3.5 (2n - 3) converges. "=1

Hence, by the comparison test, the series (5.9.1) converges (in the sense

5 Spaces of Functions

105

of Definition (4.9) of Chapter 2) to a continuous function [ -1,1]. Since

f on

f(t)=(1-t)-!

(5.9.2)

for all t in ( -1, 1), it follows by continuity that (5.9.2) holds for all t in [ -1,1]. We therefore have 1(I-t)t-g(t)I~Ii/2

(tE[O, I])

for some partial sum g(t) of the series (5.9.1). Then Ilxl-g(l-x 2 )1=1(1-(I-x 2 ))t- g (l-x 2 )1 ~1i/2

for all x with - 1 ~ x ~ 1. The polynomial p defined by p(x) == g(1 _x 2 ) - g(l)

therefore satisfies our requirements.

0

(5.10) Definition. Let G be a set of real-valued functions on a set X Then d(G) is the set of all real-valued functions f on X of the form f==pog, where p: lR"--+lR is a strict polynomial and g: X --+lR" has the form g(X)==(gl(X), ... ,g"(x)) for some gl, .. ,g. in G. d(G) is the smallest set of real-valued functions on X containing G and closed with respect to the operations of addition, multiplication, and multiplication by real numbers. (5.11) Lemma. Let G be a set of continuous functions on a compact space X, and let f, g be functions belonging to the closure H of d(G) in C(X). Then If I, max {f, g}, and min {f, g} also belong to H.

Proof' Clearly, H is closed with respect to the operations of addition, multiplication, and multiplication by real numbers. Hence H = d(H). Choose c > so small that II cf II ~ 1. Let Ii> 0, and let p be the polynomial function of (5.9) Then

°

Ilcf(x)l- p(cf(x))1 ~ Ii

Since po(cf) belongs to H, and Therefore IfIEH, and so

Ii

(XE X).

is arbitrary, it follows that IcflEH

max {f,g} =t(f + g+ If - gI)EH. Similarly, min {f,g}EH.

0

106

Chapter 4 Metnc Spaces

(5.12) Lemma. Let G be a set of continuous functions on a compact space X, and let H be the closure of d(G) in C(X). Let h be a function in H with O
L

l-c 2 1Ihll- 2 < 1. Hence the series Ilhll- 2 h(l-A)n converges uniformly on X to the function n=O Ilhll- 2h(1-(1-A»-1 =h- l . Since each term of the series belongs to H, and H is closed, it follows that h- l belongs to H. 0 (5.13) Definition. Let X be a compact space. A subset G of C(X) is separating if there is an operation " from IR. + to IR. + such that whenever e>O, x and y belong to X, and p(x,y)~e, there exists g in G satisfying (5.13.1)

Ig(z)1 ~ e

(ZEX, p(x, z) ~ "(e»

and (5.13.2)

Ig(z)-11 ~e

(ZEX, p(y,z)~"(e»,

and such that, whenever e>O and YEX, there exists g in G satisfying (5.13.2). We now come to the Stone- Weierstrass theorem. (5.14) Theorem. If G is a separating set of continuous functions on a compact space X, then d(G) is dense in C(X). Proof' Let H be the closure of d( G) in C(X). Clearly, H = d(H). We must show that H = C(X). First we show that 1 E H. To this end, let {Xl' ... ,xn } be a ,,(!) approximation to X, where" is the operation of Definition (5.13). Then for each i (1 ~ i ~ n) there exists gi in G with gi(Z)~t whenever P(Xi,Z)~"(t) By (5.11), we have h=max{gb ... ,gn} EH. Now, for each Z in X we have P(Xi'Z)~"(!), and therefore gi(Z)~!, for at least one value of i. It follows that h ~!. By (5.12), we have h- l EH. Thus 1=hh- 1 EH. We now prove (5.14.1) For each y in X and each r in (O,i], there exists A in H such that 0~A~1, A(z)=1 whenever p(y,z)~"(r), and A(Z)=O whenever p(y,z)~3r.

5 Spaces of Functions

107

To this end, let {x1, ... ,x n } be a c=min{r,o(r)} approximation to X. The set {l, ... , n} is the union of finite or void sets Sand T, with p(xi,y»r for all i in S, and p(x i ,y)<2r for all i in T. If S is void, then p(y,z)<2r+c~3r for each Z in X, and we can take A=1 Thus we may assume that S is nonvoid. By (5 13), for each i in S there exists gi in G such that gi(z)~r~i whenever P(xi,z)~o(r), and gi(Z);;;; 1-r;;;;i whenever p(y,z)~o(r) Write hi =min{1,max{0,2g i -1}}

Then hiEH, 0~hi~1, hJz)=O whenever whenever p(y, z) ~ oCr). Write

(iES).

P(xi,z)~o(r),

and h i(z)=1

ieS

Then AEH, 0~A~1, and A(z)=1 whenever p(y,z)~o(r). Consider Z in X with p(y,z);;;;3r, and choose i with l~i~n and P(Xi,z)~c~r. Then

p(x i , y);;;; p(y, z)- P(Xi' z);;;; 3r - r= 2r. Hence i cannot belong to T, and so iES Since P(Xi,z)~(~o(r), we have hi(z) =0, and therefore A(Z)=O. Thus (5.14.1) is proved We are now prepared to show that every f in C(X) belongs to H. Let w be a modulus of continuity for f Let c be any positive number, and write r=min{i,iw(c)}. Let {Yl, ... ,Ym} be a oCr) approximation to X. By (5.14.1), for 1 ~j~m there exists Aj in H with O~Aj~ 1, Aiz)= 1 whenever p(Yj,z)~o(r), and Aiz)=O whenever p(Yj,z);;;;3r. Thus m

A=

I

Aj is in H; also, A;;;; 1, since for each

Z

in X we have

j~l

p(Yj,z)~o(r), and therefore A(z);;;;Aiz);;;;l, for some j. By (5.12), we have A-1EH. The function m

g= If(Y)A-1A j j~l

therefore belongs to H. Let Z be any point of X, and note that 3r<w(c). Since Aj(Z) vanishes whenever p(Yj' z);;;; 3r, and since If(z) -f(Y)I~c whenever p(Yj'z)~w(c), we have

If(z) - g(z)1 =

/Jl (f(z) - fey)~ A-l(Z) AiZ)/ m

~

I

If(z)- f(y)1 A-1(Z)A/Z)

j~l

m

~

Idj~l

1

(z)Aj(z)=c.

108

Chapter 4

Metnc Spaces

Since I: is an arbitrary positive number, it follows that j EH, as was to be proved. 0 (5.15) Corollary. Let «X n,Pn)):,"'=1 be a sequence of compact spaces, with product (X, pl. Let G be the set of all functions on X of the form fnonn, where nn: X -+Xn is the projection of X onto X n, and fn: X n -+ IR is continuous. Then the set H of all h in C(X) of the form h = hi + ... + hn, where each hi is a finite product of functions in G, is dense in C(X). Proof' We may assume that 1 is a bound for each of the spaces X n. Clearly, H = d(G) In view of (5.14), to finish the proof it is enough to show that G is separating. To this end, let I: > 0, choose N in 7l+ so 00

L

2- n
N

L 2- nPn(x n, Yn) > p(x, y) -

e/2 ~ e/2.

n=1

Therefore P.(x n,Yn»e/2 for some n with n~N. Write g=::fnonn, where the continuous function fn: X n -+ IR is defined by fn(zn)=::(Pn(x n,Yn))-1 Pn(xn,z.) If Z=::(Zk)EX and

p(x,z)~c5(e),

(znEXn)

then

Ig(z)1 = If.(zn)1 ~ 2e- 1P.(x n, zn) ~2e-12np(x,z)~2cI2N

Similarly, if

p(y,z)~c5(e),

Ig(z) -11

c5(e)=e.

then

= Ifn(zn) - fn(Y.) 1 ~ 2e- 1IPn(x n, z.) - Pn(x n, Yn)1 ~ 2e- 1Pn(Yn,

Therefore G is separating

zn) ~2c 12N c5(e) = e.

0

(5.16) Corollary. Let X be a compact space with positive diameter, and let G consist of all functions of the form Xf-+p(x,x o) with XoEX. Then d(G) is dense in C(X). Proof' Let d=::diamX and c=d/7. Let F be a finite c approximation to X Consider any x in X, and suppose that p(x,x')<2c for all x' in

6 Locally Compact Spaces

109

F. Let y and z be arbitrary points of X, and choose y' and z' in F such that p(y, y') < c and p(z, z') < c. Then p(y, z)~ p(y, y')+ p(y', x)+ p(x, z')+ p(z', z) < 6c. Hence d ~ 6c, a contradiction. It follows that p(x, x') > c for some x' in F Let e>O, and write b(e)=min{e 2 ,ce}. Consider any x and y in X with p(x,y)~e. Define g in d(G) by g(z)=p(X,y)-l p(x,z)

(ZEX)

If p(x, z) ~ b(e), then Ig(z)1 ~ e- 1 p(x, z) ~ e,

and if p(y, z)~b(e), then Ig(z) -11 = p(x, y)-llp(x, z) - p(x, y)1 ~C1

p(y, z)~e.

On the other hand, if YEX is arbitrary and we choose y' in F so that p(y,y'»c, then (5.13.2) holds with g the function Zf-+p(y,y')-l p(Z,y') in d(G). Hence d(G) is separating, and so, by (5.14), d(G)=d(d(G)) is dense in C(X). D The case n= 1 and X = [ -1,1] of our third corollary is the famous Weierstrass approximation theorem. (5.17) Corollary. Every continuous function on a compact set Xc lR" can be arbitrarily closely approximated on X by polynomial functions. Proof: First consider the case in which n = 1. By (5 9), the function xf-+lx-xol (with XoEX) can be arbitrarily closely approximated on X

by polynomials. The theorem then follows from (5 16). Next consider the case where X = [a, b]" for some compact interval [a, b]. The result then follows from (5.15) and the case just proved. Finally, consider the general case. Since X is bounded, there exists a compact interval [a, b] with X c [a, b]". By the case just considered, each of the functions Xf-+p(x,x o) (with xoE[a,b]") can be arbitrarily closely approximated by polynomials on [a, b ]". The result then follows from (5.16). D

6. Locally Compact Spaces Many important metric spaces, such as the euclidean spaces lR", that are not compact have a property almost as good.

110

Chapter 4

Metnc Spaces

(6.1) Definition. A nonvoid metric space X is locally compact if every bounded subset of X is contained in a compact subset. In that case, a function f: X -. Y from X to a metric space Y is continuous if it is uniformly continuous on every compact subset of X, or, equivalently, on every bounded subset of X. A continuous function g: X -.IR such that the set X,,={XEX:

g(x)~ex}

is bounded for all cx in IR is called a compactifier for X. If Xo is any point in the locally compact space X, then the function XI-+ p (x, xo) is a compactifier. The term 'compactifier' comes from the fact that if g is any compactifier, then Xu. is either void or compact for all except countably many real numbers cx. This follows from Theorem (4.9). Since a locally compact space is a countable union of compact spheres (by Theorem (4.9)), it is separable. Since any Cauchy sequence is bounded, a Cauchy sequence in a locally compact space is contained in a compact set, and therefore converges. Hence a locally compact space is complete. (6.2) Proposition. A locally compact subset Y of a metric space X is located. Proof: Let x be any point of X. Let YoEY, and choose c>2p(x,Yo) so

that

Y,,= {YE Y: p(y, yo)~c} is compact Since p(x, YcJ~p(x,yo)
so that YEY", and therefore p(x,y);;;;p(x,y"). It follows that p(x,Y) exists and equals p(x, y"). Hence Y is located. D (6.3) Proposition. A closed, located subset Y of a locally compact space X is locally compact. Proof: Let Yo be an arbitrary point of Y. For each ex in IR write

Xa= {XEX: p(x, yo)~ex}.

Let B be any bounded subset of Y, and choose c > 0 so that B c: Y n Xc and X 4c is compact. Note that if XEX and p(x, Yo) < 2c, then

6 Locally Compact Spaces

III

p(X, YnX 4c ) exists and equals p(x, Y): for in that case, if YEY and p(x, Y) < p(x, y) + 2c - p(x, Yo), then p(y,yo)~p(x,y)+p(x,Yo)
Y)+2c

~p(x,Yo)+2c<4c

and so YEYnX 4c ' Thus if M=sup{p(x, Yo). XEX 4c } <2c, then YnX 4c is located in X 4c ; in view of (4.5), it follows that YnX 4c is a compact subset of Y containing B. Thus we may assume that M> c; whence there exists r in (c, 2c) such that the set V= {XEX: r~p(x, Yo)~4c}

is compact. Then U= Vu(YnX 4 J is located in X 4c ' For, given x in X 4c ,we have either ror: for all x in V, we have V nX,,= YnX~::J YnXc::JB. Hence Y is locally compact.

D

(6.4) Proposition. The product X of finitely many locally compact spaces Xl' ... , X n is locally compact

Proo!, Let A be a bounded subset of X. By definition of the product metric (1.3),

where Ai is some bounded subset of Xi (1 ~i~n). Let Bi be any compact subset of Xi with AicB i. Then B=B l x B2 x .. X Bn IS a compact subset of X, and AcB. Therefore X is locally compact. D For maps f: X -+ Y between locally compact spaces, the notion of metric equivalence is too restrictive. More useful is the concept of a homeomorphism. (6.5) Definition. A continuous map f from a locally compact space X onto a locally compact space Y, with an inverse g: Y-+ X which is also continuous, is called a homeomorphism. In case p and p' are two metrics on the same set X, such that the spaces (X, p) and (X, p') are both locally compact and such that the identity map xt--+x is a

112

Chapter 4

Metnc Spaces

homeomorphism from (X, p) to (X, p'), the metrics p and p' are called homeomorphic metrics on X. Let f: X --+ Y be a continuous map on a locally compact space X. If B is any bounded subset of X, then there exists a compact set K with Be K. Since f is continuous on K, f(K) is totally bounded. Hence f(B) is bounded. Thus f takes bounded sets into bounded sets. In particular, the property of boundedness is intrinsic - that is, preserved under homeomorphisms Certain spaces which are not necessarily locally compact become locally compact under a new metric. For example, let g: X --+1R. be a function from a metric space X to 1R. such that (i) the set X 1% == {XE X: g(x)~oc} is compact for certain arbitrarily large values of oc, and (ii) g is uniformly continuous on each of the sets Xa. Then X may not be locally compact, because there may be a bounded subset of X that is not contained in any Xa; but X is locally compact under the new metric Po defined by Po(X, y)== p(x, y) + Ig(x) - g(y)1

(x, YEX),

and g is a compactifier for (X, Po). This construction can be used when X is the metric complement of a located subset A of a locally compact space Y. In this case the function g can be taken as xf-+h(x) +p(x,A)-t, where h is a compactifier for Y. We shall show that every locally compact space can be viewed as the metric complement of a one-point set in a compact space. (6.6) Definition. A one-point compactification of a locally compact space X is a triple consisting of a compact space (Y, d), a point w of Y, and a continuous injection i: X --+ Y such that (i)

i(X)=Y-{w}

(ii) d(w, i(K» > 0 for each compact set K c X (iii) the inverse j: i(X)--+X of the map i is uniformly continuous on each compact subset of i(X). The point w is called the point at infinity, and the map i the inclusion map, of the one-point compactification When no confusion is likely, we refer to the metric space Y itself as a one-point compactification of X. (6.7) Proposition. The inclusion map from a locally compact space X into a one-poirJ.t compactification Yof X is hyperinjective and uniformly continuous.

6 Locally Compact Spaces

113

Proof We use the notation of (6.6). Given E > 0, choose r in (0, E/2) so that S={YEY: d(y,w)~r} is a compact subset of i(X). Then j(S) is totally bounded, and therefore located in X. Choose c > 0 so that T= {XEX:

p(x,j(S»~

3c}

is compact. Let c5 be a modulus of continuity for i on T, and consider points x,x' of X with p(x,x')~min{c, c5(E)}. Either p(x,j(S»>c or p(x,j(S»<2c In the former case, choose Q( such that 0<0( ~min{r,d(i(x),w)} and K={YEY:

d(y,w)~Q(}

is compact. Then Kci(X), i(X)EK, and ScK. Since p(x,j(S»>O and j is uniformly continuous on K, we see that d(i(x), S) > 0, and hence that d(i(x), w)~r. Moreover, as p(x',j(S»~p(x,j(S»-

p(x, x'»O,

a similar argument shows that d(i(x'), w)~r. Hence d(i(x), i(x'»~2r <E. On the other hand, if p(x,j(S»<2c, then p(x', j(S» < 3c, both x and x' belong to T, and therefore d(i(x), i(X'»~E. Since e is arbitrary, it follows that i is uniformly continuous on X. Now let A and B be compact subsets of X such that O<m=inf{p(x, y): xEA, YEB} Choose a compact subset C of X with Au Be C. There exists t with O
is compact. If c5' is a modulus of continuity for j on L, then d(i(x), i(y»~c5'(mI2) for all x in A and y in B. Hence i is hyperinjective. D We have yet to establish that one-point compactifications exist. (6.8) Theorem. Every locally compact space has a one-point compactification. Proof· Let (X, p) be a locally compact space, and let (x.) be a dense

n [0, 1J with 00

sequence in X. Let Z be the compact product space

n=1

product metric d defined in the usual way, and let w be the sequence

114

Chapter 4 Metnc Spaces

with each term 1. For each n in Z+, define the bounded continuous function In on X by In(x)=min{l, p(x, xnn

(XEX).

Then X is a subset of Z, with the inclusion map i(x)=(fn(X))~=l. For all x and Y in X,

defined by

00

d(i(x), i(y)) =

L: 2 - nI/n(x) - In (y)1 .=1 00

=

L: 2- nlmin {I, p(x, Xn)} -

min {I, p(y, xn)}l

n=l 00

~

L:2- nlp(x,x n)-p(y,xn)1

n=l 00

~

L:2- np(x,y)=p(x,y) . • =1

Hence i is uniformly continuous. To show that i(X) u {w} is totally bounded in the metric d, for each N in Z+ we define SN= {XEX: p(x, X1)~OCN}

and TN={xeX: p(x,x1)~N-l},

where OC N is chosen in (N -1, N) to make SN compact. By (4.2), i(SN) is totally bounded. Let 6 be an arbitrary positive number. Since i(X) = i(S N) u i(TN), to prove that there exists an 6 approximation to i(X) it will suffice to show that d(i(x), W)<6 for some N and all x in T, To 00

this end, choose M in Z+ so that

L:

2- n <6, and let N be an

n=M+1

integer greater than 2+max{p(x1,Xn): l~n~M}. Consider an arbitrary point x in TN. For 1 ~n~ M we have p(x, xn)~P(x, x 1)- P(X1' x n) ~ N -1- p(xp x.) > 1

and therefore In (x) = 1. Hence 00

d(i(x), w)=

L:

2-' I/n(x)-11

n=M+1 00

~

L:

2- n <6,

n=M+1

as we wanted. Hence i(X)u{w} is totally bounded. Since Z is complete, the closure Y of i(X)u {w} in Z is compact.

6 Locally Compact Spaces

115

-

Consider any compact subset K of X. Since (x.) is dense in X, k

K e

U {XEX:

p(x, x.)
.~l

for some k in Z+; whence d(w,i(K))>2- k - 1 • It follows that i(X)eY -{w}. We now prove that the inversej: i(X)--+X of the mapping i is uniformly continuous on i(K). To do this, let e be an arbitrary positive number, and set t == min {4 -1, (2e)-1}. Choose a positive integer v such that v Ke

U {XEX:

p(x,xn)
"=1

Given points x, y of K with p (x, y) > e, choose n (1 ~ n ~ v) so that p(x, xn) < t e. Then p(x, xn) < (1- t) e < p(x, y)- p(x, xn)~ p(y, x n),

so that d(i(x),

i(y))~2-'(min{l, ~2-V(min{1,

p(y,

x.n -min{l, p(x, x n)})

(l-t) e} -tel

=2- V min{l-t e, (1-2t) e} >0. Hence j is uniformly continuous on i(K). Now consider any set SeX such that

d(i(x),w)~!5

for some 15>0

00

and all x in S. Choose N so that N

we have

L

2 -. <15. Then for each x in S

n~N+l

L2-'lfn(x)-11>0; so that 1f..(x)-II>O, and therefore n=1

p (x, x n) < 1, for some n with 1 ~ n ~ N. Thus S is bounded, and so contained in a compact subset of X. Hence j is uniformly continuous on i(S). It now follows that j is uniformly continuous on each compact set Le i(X): for since i(X)e Y- {w}, we have d(w, L) >0, by (3.8). It remains to show that Y-{w}ei(X). Consider therefore any point y of Y- {w}. Let 15 ==td(y, w), and choose a sequence (yn) in i(X)u{w} such that d(y,y.)<min{n-1,!5} for each n. Then d(y.,w)~!5 and so y. belongs to {i(x): XEX, d(i(x),

w)~!5}.

The uniform continuity of j on this set implies that (j(y.)) is a Cauchy sequence in X, so that (j(y.)) converges to a point x of X. As i is uniformly continuous, (Yo) converges to i(x). Hence y = i(X)Ei(X), as was to be proved. 0 All one-point compactifications of a given locally compact space are metrically equivalent (see Problem 18). The following result will be useful later.

116

Chapter 4

Metric Spaces

(6.9) Proposition. Let (X, p) be a compact space with positive diameter, and w a point of x. Then Xo=X -{w} is locally compact relative to the metric Po(x, y)=p(x, y)+ Ip(x, W)-l_ p(y, w)-ll, and X is a one-point compatification of Xo with point at infinity wand inclusion map i: Xo---+X given by i(x)=x. Proof As diam X> 0, X 0 is nonvoid. Since P ~ Po' the map i is uniformly continuous and injective on (Xo, Po)· If LcX-{w} is compact relative to p, then pew, L»O, by (3.8); whence, by (1.12), the mapping j inverse to i is uniformly continuous on L. It remains to prove that (X 0' Po) is locally compact, and that condition (ii) of (6.6) holds. To this end, consider a subset K of X 0 that is bounded relative to Po. Choose Xo in Xo and c>O such that Po(x,xo)~c for all x in K Then

(6.9.1) By (4.9), there exists a p-compact set L c X - {w} with K c L. Since the mappings i and j are uniformly continuous on L relative to Po and p, respectively, we see that L is po-compact. Hence (X 0' Po) is locally compact. Finally, if K is po-compact, then pew, i(K)) > 0, by (6.9.1). D Important examples of spaces of functions attach themselves to a locally compact space. (6.10) Definition. Let (X, p) be a locally compact space. A mapping f. X ---+ lR is said to vanish at infinity if for each r > 0 there exists a compact subset K of X such that If(x)1 ~r for all x in -K We use Coo(X, lR), or simply C",,(X), to denote the set of all continuous functions from X to lR which vanish at infinity For each f in Coo(X) the norm IIfll =sup{lf(x)l: XEX} of f exists To see this, consider arbitrary numbers IX, {3 with 0~1X<{3 Choose a compact set K c X such that If(x)1 ~ {3 for all x in - K Then either IlfIIK>1X or IlfIIK<{3. In the former case, If(x)I>1X for some x in K; in the latter, If(x)1 ~ {3 for all x in the dense subset K u - K of X, and hence (by continuity) for all x in X Thus Ilfll exists, by (4.3) of Chapter 2. The standard metric p on Coo(X) is given by p(f, g)=

Ilf -

gil

(f, gE Coo(X)).

6 Locally Compact Spaces

117

To illustrate the application of the one-point compactification, we shall prove that the metric space CC(,(X) is both complete and separable. (6.11) Lemma. Let (X, p) be a locally compact space; let a one-point compactification of X consist of the compact space (Y, d), the point at infinity w, and the inclusion map i: X ---> Y - {w}; and let f be an element of Coo (X). Then there exists a unique element f# of C(Y) such that

(611.1)

for all x in X,

f#(i(x»=f(x)

and

f#(w)=O.

Proof. In view of (3 7) and the fact that i(X) u {w} is dense in Y, it will suffice to prove that the map f#. i (X) u {w} ---> lR defined by condition (6111) is uniformly continuous To this end, let j: i(X)--->X be the inverse of the mapping i Given c > 0, choose a compact set K c X such that If(x)lO. Choose c with 0<4c
w)~c}

is a compact subset of i(X) containing i(K). Then j(S) is totally bounded, and so foj is uniformly continuous on S. Let b be a modulus of continuity for foj on S, and consider points x, y of i(X)u {w} with d(x, y)~min{c, b(c)}. It is enough to prove that

If#(x)- f#(y)1 ~c.

(6.11.2)

We may assume that x=j=w. If y=w, then

d(x, i(K»

~ d(w,

i(K)) - d(x, w) > 3 c,

so that as i is continuous, j(X)E X - K and therefore

If# (x) - f# (y)1 = If(j(x))1
If# (x) - f# (y)1 ~ IfU(x))1 + IfU(y))1 < c. In the second,

d(x,

w)~d(y,

w)-d(x, y»c,

both x and y belong to S, and so (6.11.2) clearly holds.

0

(6.12) Corollary. If X is a locally compact space, then each element of Coc(X) is uniformly continuous

118

Chapter 4

Metnc Spaces

Proof· Let f belong to Coo (X) In the notation of (611), if b1 is a modulus of continuity for i on X, and b z is a modulus of continuity for I'll on Y, then it is clear that e ~ b1 (b z(e)) is a modulus of continuity for f on X. D

Under the conditions of Lemma (6.11), it is routine to verify that the map f~f# is a norm-preserving algebra isomorphism which identifies Coo (X) with the metric space G of all elements g in C(Y) with g(w)=O. Since C(Y) is complete and G is a closed subset of C(Y), it follows that G is complete. Hence Coo(X) is complete To prove that it is separable, let (xn) be a dense sequence of points of Y. Let r consist of all functions on Y of the form po h, where p: 1R. n --> 1R. is a strict polynomial with rational coefficients, and h· Y --> 1R. n has the form h(x) == (p(x, x;J, ... , p(x, Xi)) for some i 1 , ••• , in in 7l+ Then r is a countable subset of C(Y) which, by a straightforward extension of (5.16), is dense in C(Y) Thus {g-g(w): gET} is a countable dense subset of G, so that G, and therefore Coo (X), is separable. (6.13) Definition. Let X be a locally compact space. A mapping f: X --> 1R. is said to have compact support if there exists a compact subset K of X such that f(x) = 0 for all X in - K. K is then called a support of f Extending the notation of Definition (5.1), we use C(X,1R.), or simply C(X), to denote the set of all continuous functions from X to 1R. with compact support The elements of C(X) are also called test functions. Clearly, C(X)e Coo(X). In fact, we can say more (6.14) Proposition. If X is a locally compact space, then C(X) is dense in Coo (X) Proof: Given fin Coo(X) and e>O, choose a compact set K eX such that If(x)1 < e/2 for each x in - K Choose positive numbers r, b so that the sets Kr== {XEX: p(x,

K)~r}

and Kr+
K)~r+b}

are compact. Define a continuous map g: X

--> 1R.

g(x)==f(x) max{O, I-b- 1 p(x, Kr)}

by (XEX).

6 Locally Compact Spaces

119

Then g(x) = 0 for each x in - Kr+ d' so that gE C(X). For each x in X we have either p(x,K)
Hence Ilf -gil ~E.

0

For future reference we remark that, once the metric space
E

a positive number Then there exist nonnegative test N

N

functions fl' ···,fN on X such that L fn~l, L fn(x) = 1 for all x in K, n~l

n~l

and each fn has a compact support of diameter less than

E.

N

Proof: By (4.8), we can represent K as a union K

=

U Kn of compact n=1

sets, each of diameter less than E/2. Choose rx. so that 0 < rx. < E/4 and each of the sets Sn= {XEX: p(x, Kn)~rx.}

is compact. Then for each n, diam Sn < E and Sn continuous mapping gn: X --+ [0, 1] defined by gn(x)=max{O, l-rx.- 1 p(x, Knn

IS

a support of the

(XEX). N

Given x in K, we have p(x,K;)=O for some i; whence Lgn(x)~1. Thus the functions n~ 1 (1 ~k~N)

satisfy our requirements.

0

The problem of extending a continuous function to a larger space arises often and in many forms. The following solution to a special case of this problem is the famous Tietze extension theorem, or rather as much of it as is constructively valid.

120

Chapter 4

Metric Spaces

(6.16) Theorem. Let Y be a locally compact subset of a metric space X, and Ie 1R a proper compact interval. Let j: Y--> I be continuous. Then

there exists a function h: X --> I which is uniformly continuous on bounded subsets of X, and which !>atisfies h(y)= fey) for all y in Y. Proof· Without loss of generality, we may assume that 1= [ -1,1]. Moreover, adjoining points to X if necessary, we may assume also that X - Y contains at least two distinct points x- and x+. This will serve to make certain sets which will be defined below non void. We first prove (6.16.1) Let N be a nonnegative integer, and let g be a continuous mapping of Y into the interval [- (3j4)N, (3j4)N]. Then there is a mapping go of X into [ -~(3j4t, ~(3j4)N] which is uniformly continuous on bounded subsets oj X, and which satisfies (6.16.2)

To this end, extend g to the locally compact subset Z=Yu{x-,x+} of X by writing g(x-)= -(3j4t, g(x+) = (3j4t· Then g is continuous on Z. Let ¢ be the mapping xr-+min{p(x,x-), p(x,x+)} on Z Then ¢ is a compactifier for Z with ¢(x-)=O=¢(x+). Hence there exists a sequence (tIn) of positive numbers with IX" --> 00 as n--> 00, such that the set {ZEZ: ¢(z);;=;;tI n} is compact for all n in 7l+ By (49), there exists IX in (.H3j4t, ~(3j4)N) such that the sets An={ZEZ: g(z);;=;;

-tI,

¢(Z);;=;;lX n }

and are compact for each n. Therefore the sets 00

U An

A={ZEZ: g(z);;=;; -IX}=

n=l

and B={ZEZ: g(z)~IX}=

U Bn n~l

are locally compact Thus p(x, A)+ p(x, B) exists for each x in X. Moreover, p(x, A) + p(x, B) is bounded away from 0 on each bounded subset S of X To see this, choose c > 0 such that p (x, x +) ~ c for all x in S, and let [) be a modulus of continuity for g on {XEZ: p(x, x+);;=;;2c} If XES and p(x, A)+ p(x, B)<min{c,t[)(IX}}, then there exist a in A and b in B with p(a,x+)~2c, p(b,x+);;=;;2c, and pea, b);;=;; [)(IX); whence Ig(a)- g(b)1 ~IX Since this is impossible, we conclude that p(X, A)+ p(x, B}~min{c,t[)(IX}} (XES).

Problems

121

Hence go(X)= lX(p(X, A) + p(x, B))-1 (p(x, A) - p(x, B))

(XEX)

defines a mapping go that is uniformly continuous on each bounded subset of X. Clearly, Igo(x)1 ;:::;~(3/4t for all x in X. Consider any y in Y. If YEA, then O~g(Y)+IX=g(y)-go(Y)~ -(3/4)N +H3/4t

and so Ig(y) - go(y)1 ;:::; (3/4)N + 1

By a similar argument, the last inequality holds if YEB. On the other hand, if -1 (3/4t ;:::; g(y) ;:::;1 (3/4)N, then Ig(y) - go(y)1 ;:::; Ig(y)1 + Igo (y)1 ;:::; i(3/4)N < (3/4)N+ 1.

Thus (6 16.2) holds, and the proof of (6.161) is complete. Now apply (6.16.1) repeatedly, to construct a sequence Un) of mappings of X into JR, each uniformly continuous on bounded subsets of X, such that for each n in 71.+, Ifn(z)1 ;:::;~(3/4)n-1

(XE X)

and jI(Y) Then

L fn

kt/k(y)i;:::; (3/4)"

(YE Y).

converges uniformly on X to a function F: X ---+JR such

that F(y) = f(y) for each y in Y. By (1.12), F is uniformly continuous on bounded subsets of X. The function h=min{1,max{ -l,F}} therefore has the desired properties.

0

Problems 1 A neighborhood space X satisfies the second axiom of countability if there exists a sequence (Un) of neighborhoods in X such that for each x in X and each neighborhood U with XE U, there exists n in 71.+ such that XE Un C U. Show that a metric space X is separable if and only if the associated neighborhood space satisfies the second axiom of countability.

122

Chapter 4

Metnc Spaces

2. Show that a located subset of a separable metric space is separable. 3. Show that a dense subset of a separable metric space is separable. 4 Let (Fn) be a sequence of located subsets of a complete metric space

n-F,. can be metrized to be a complete metric 00

X. Show that G ==

space.

n~l

5 Call a bounded sequence (xn) in a metric space fickle if for each 1'>0 there exists N in 71+ such that whenever n1
(i) (ii)

IR a is located in IR.

(iii)

IRa is a complete subset of IR.

9. Show that the set of all compact subsets of a compact space X is compact with respect to the metric defined in Section 4. 10. Let K1:=J K 2 :=J ... be a decreasing sequence of compact subsets of a metric space X such that p(Km,Kn)-+O as m,n-+oo (where P is the

nKn 00

metric on the set of compact subsets of X). Show that compact.

n=l

IS

Problems

123

11. Let h be a mapping of a metric space X into a compact space Y, such that foh is uniformly continuous for each continuous map f: Y--+ JR. Prove that h is uniformly continuous. 12. Let h be a mapping of a compact space X into a metric space Y, such that foh is continuous for each uniformly continuous map f: Y--+ JR. Prove that h is continuous if and only if heX) is totally bounded. Prove also that if Y is locally compact, then h is continuous. 13 Prove the converse of Ascoli's theorem. 14. Let S consist of all differentiable functions f on [0, 1] with Ilf I ~ Co and II!'II ~Cl' where Co and C l are positive constants. Show that S is a totally bounded subset of q[O, 1J). 15. Let S consist of all N times differentiable functions f on [0, 1] such that Ilf(n)ll~c" (O~n~N), where co, ... ,cN are positive constants. Show that S is a totally bounded subset of q[O, 1J). 16. Show that there exists a metric d on the set M of irrational numbers such that (i) a subset of M is locally compact with respect to d if and only if it is locally compact with respect to the metric p on JR, (ii) d and p are homeomorphic metrics on such sets, and (iii) if (xn) is any sequence of irrational numbers converging in the metric p to a rational number, then there exists e > 0 such that for infinitely many n there exists m>n with d(xm,xn)~e. 17. Show that if every continuous m]ection on a compact space is hyperinjective, then every continuous injection of a compact space into JR + has positive infimum. 18. For k = 1, 2 let X k be a locally compact space with one-point compactification (Yk , W k ' ik ). Show that f: Xl--+X 2 is a homeomorphism of Xl onto X 2 if and only if there exists a metric equivalence ¢ between l';. and Y2 such that ¢(W I )=W 2 and i 2 of=¢oi l • 19. Let X be a locally compact space with one-point compactification (Y, w, i), and let S be a nonvoid closed subset of X Prove that S is locally compact if and only if the closure T of i(S)u {w} in Y is compact; in that case, prove also that (T, w,j) is a one-point compactification of S, where j is the restriction of i to S.

124

Chapter 4

Metric Spaces

20 Let f. [O,l]--+lR be continuous, with f(O}=O and f(l}=1. Show that the set {x. f(x}=(J(} is compact for all but countably many tx in [0,1]. 21. Let the compact space X be locally connected, in the sense that for each E > 0, X is a union of finitely many compact sets, each connected in the sense of Problem 11 of Chapter 3, and each with diameter at most E. Prove that if f. X --+lR is continuous, then {XEX: f(x}=(J(} IS compact or void for all but countably many (J( in lR. 22. A uniform space consists of a set X and a set M of pseudometrics on X, such that x = x' if and only if P(x, x') = 0 for all P in M A function f: X --+ Y from a uniform space (X, M) to a uniform space (Y, N) is uniformly continuous if for each d in N and each E > 0 there exist PI' ... ' Pm in M and c'5 > 0 such that d(j (x},f(x'}) ~ I:; whenever Pi(X, x'} ~ c'5 (1 ~ i ~ m). If f has an inverse which is also uniformly continuous, then f is called a metric equivalence Show that if M is countable, then there exists a metric P on X such that the identity map i: X --+X is a metric equivalence of (X, M) with (X, {p}) 23. Call a uniform space (X, M) totally bounded if X is totally bounded with respect to the pseudo metric PI + .. + Pm whenever PI' ... , Pm belong to M Show that a uniformly continuous map takes totally bounded sets to totally bounded sets. 24. A filter in a set X is a collection F of nonvoid subsets of X such that (i) if U, VEF, then Un VEF, and (ii) if U c V and U EF, then VEF. If (X, M) is a uniform space, a filter F in X is a Cauchy filter if for each P in M and each E>O there exists U in F such that p(X,y}<E for all x, y in U A Cauchy filter F is said to converge to a point a in X if for each P in M and each 1:;>0 there exists U in F such that p(a,y}<E whenever yE U. If every Cauchy filter in X converges, X is said to be complete. Relate this definition to the definition of completeness for metric spaces. 25 Show that an arbitrary uniform space has a completion analogous to the completion of a metric space. 26. Let

«X., M.)):;C~

I

be a sequence of uniform spaces, and define the

nX. and letting M be 00

product uniform space (X, M) by taking X==-

•~ I

the set of all pseudometrics «x.), (y.))

I--->

p(x;, y;)

«x.), (Y.)EX)

Notes

125

with i in 7l+ and p in Mi Show that (X, M) is complete if each (X.,M.) is complete, and totally bounded if each (X.,M.) is totally bounded

Notes Within the main body of this text, we have only defined the product of a family of subsets of a given set. However, with the aid of Problem 2 of Chapter 3 we can define the product of an arbitrary sequence of sets. Definition (1.7) then applies to such a product In classical analysis two metrics are equivalent if they give rise to the same family of open sets. This notion of equivalence is of no use constructively. The definition given in Section 1 is especially suited to compact spaces; other notions are more suited to other spaces. (See, for example, Definition (6.5).) The definition of a located set is due to Brouwer. In classical analysis every nonvoid subset of a metric space is located This is not true constructively: see Problem 8. Proving certain sets located is an important, and often nontrivial, part of constructive analysis In this connection, the reader might consult the article 'Locating Metric Complements in JR.', on pages 240-249 of [74]. It is natural to conjecture that the remark following Lemma (3 8) can be strengthened, to the effect that inf{p(x, A): xEK}>O whenever K is a compact subset of a metric space X, A is a complete, located subset of X, and x =l= y for all x in K and y in A. No acceptable proof or counterexample is known for this conjecture (The conjecture holds within Brouwer's intuitionistic mathematics, but is false under the hypothesis that all real numbers are recursive. See [52J ) The definition of a nowhere dense set can be extended to cover sets that are not necessarily located: We say that a subset A of a metric space X is nowhere dense if there exists a dense open subset B of X such that x =l= y for all x in A and y in B (so that (A, B) is a complemented set in X) Proposition (4.6) would be false without the condition that X is located. Definition (4.7) is due to Brouwer. The classical definition of a compact space, as one for which every open cover has a finite subcover, would not do: there would be no nontrivial example of a compact space! (Note that whereas Brouwer contends that the interval [O,lJ is compact in the 'open cover' sense, for each 8>0 there exists a sequence (U.) of open intervals whose union contains all the

126

Chapter 4 Metnc Spaces

recursive points of [0,1], such that any finite collection of the Un has total length less than e.) Theorems (4.8) and (4.9) are partial substitutes for the classical result that a closed subset of a compact space is compact. In Proposition (4.11) the continuity of I is not needed to prove that g is uniformly continuous, but it is needed to prove that I(X) is compact. In fact, an injection between metric spaces is hyperinjective if and only if its inverse map is uniformly continuous. The extra hypothesis in Ascoli's theorem (Theorem (5.2)), that for each e > there exists an e approximation of the type described, unfortunately makes it much harder to apply than its classical counterpart. For example, Theorem (5.6) would be trivial if this condition did not have to be checked The proof of Theorem (5.14) is another instance of minor complications caused by not being able to compare arbitrary real numbers. We are forbidden to assume that either P(Xi' y) > r or P(Xi' y) ~ r. Definition (6.1) differs from the classical version, which states that a locally compact space is one in which every point has a compact neighborhood. The open interval (0,1) is locally compact in this sense, but not constructively. Of course, (0,1) can be given a new metric to make it locally compact constructively (as indicated in the discussion preceding Definition (6.6)). The one-point compactification Y of a locally compact space X is much simpler to define classically; in particular, condition (ii) of Definition (6.6) is classically superfluous Constructively, there is much more to Y than the union of X and a point at infinity. Problems 5, 6, and 7 indicate an attempt to beef up lR. Such attempts have not led to any significant results. (See the Appendix, and also Troelstra's papers [88], [89] ) Problem 12 is a partial substitute for the classical result that a pointwise continuous function on a compact space is uniformly continuous. In the last part of Problem 12 is h continuous even if Y is not locally compact? (This is an open question.) In connection with a possible link between Problems 20 and 21, see [49]. At first sight a uniform space appears to be a natural and fruitful concept for constructive mathematics, a promising substitute for the concept of a topological space. In fact, this is not the case. For instance, just to construct a compact uniform space X, such that the assumption that X is metrizable leads to a contradiction, seems to be a hard problem. Here is an attempt that fails. Let I be a proper compact interval, and S a set such that the assumption that S is countable leads to a contradiction. Let X consist of all functions from

°

Notes

127

S into I, and for all s in S and all 1, g in X write Ps(f, g) == If(s) - g(s)l. There is no known choice of S which makes X totally bounded (in the sense of Problem 23) relative to the uniform structure defined by the pseudometrics Ps' because the known sets S all have the property that there exist points sand t in S for which we are unable either to construct f in X with f(s) =1= f(t), or to prove that f(s) = f(t) for all f in X. This means that we are unable to show that S is totally bounded in the pseudometric Ps + Pt; thus we are unable to show that S is totally bounded relative to the given uniform structure. Of course, there are important constructively defined uniform spaces that are not necessarily metrizable. Examples of such spaces are found among the locally convex spaces discussed in the Problems and Notes at the end of Chapter 7

Chapter 5. Complex Analysis

The con~tructive development of elementary complex analysis, through Cauchy's integral formula, is a simple matter, the material seems to have a natural constructive cast. This is carried out in Sections 1, 2, 3, and 4. Next it is shown that under certain conditions it is possible to find the zeros of an analytic function. Although results of this type involve much more careful estimates than their classical counterparts, nothing basically new is required. The next section discusses singularities, and includes proofs of two constructively distinct versions of the Picard theorem on the range of a differentiable function in the neighborhood of a singularity, the proofs are based on Schottky's theorem, the classical proof of which readily adapts to a constructive one The last section provides a constructive version of the Riemann mapping theorem, following the classical Koebe approach as presented by Ostrowski and Warschawski. Some care must be taken in finding the right definitions, since the Riemann mapping theorem is not constructively valid without additional restrictions on the domain. The set of real numbers is too thin for certain purposes. Many beautiful phenomena become fully visible only when the complex numbers are brought to the fore.

1. The Complex Plane The complex numbers are obtained by adjoining a square root -1 to the real numbers.

of

(1.1) Definition. A complex number is an ordered pair z == (x, y) of real numbers. Two complex numbers Zl == (Xl' YI) and Z2 == (X2' Y2) are equal if Xl =x 2 and YI =Y2' Zl and Z2 are unequal, Zl qoZ2' if Xl qoX 2 or YI qo Y2' The sum of Zl and Z2 is Zl

+Z2==(XI +X 2'YI +Y2)'

I The Complex Plane

129

and the product is

Zl Z2 =(X1X2 - Y1 Y2' Xl Y2 +X2 Y1)· The norm, or modulus, of the complex number number Izi (x 2 + y2)-l:.

Z

=(x, y)

is the real

=

The set

(C

of complex numbers is called the complex plane.

Notice that the map (x,y)l->(x,y) is a bijection of that IZ1 -z21 =d((x 1, Y1)' (x 2, Y2))

(C

onto IR2, and

for all zl = (Xl' yd and Z2 = (X2' Y2) in (C, where d is the metric on IR 2 defined in Section 1 of Chapter 4. The complex numbers therefore form a locally compact metric space, with metric p defined by P(Zl' Z2)= IZ1 -z21

(Zl' Z2 E (C).

The function zl->lzl is a compactifier for (C. The operations of addition and multiplication are continuous functions from (C x (C to (C (that is, they are uniformly continuous on each bounded subset of (C x (C). These operations obey the usual rules of arithmetic, which we do not list here. The additive identity is (0,0), and the multiplicative identity is (1,0). The map XI-> (x, 0) from IR to (C preserves sums, products, and distances. We therefore identity IR with the subset {(x, 0): XEIR} of (C, by identifying each real number x with the complex number (x, 0). The element (0,1) of (C is written i, and satisfies i 2 = -1. Thus the complex number (x, y) may be, and usually is, written x + i y. We also call x the real part of z, written Re z, and y the imaginary part, written 1m z. With each complex number z=x+iy we associate a complex number z*=x-iy, called the conjugate of z. The map ZI->z* preserves sums, products, and distances, as can be verified by computation. Also, Consequently, and thus

IZ1z21=lz11Iz21, for all complex numbers Zl and Z2. The modulus function also satisfies a version of the triangle inequality,

IZ1 +z21~lz11+lz21, because of its relation to the metric.

130

Z-1

Chapter 5 Complex Analysis

Each complex number z =l= 0 has a multiplicative inverse given by == z* Izl- 2, since z(z* Izl- 2)= (z z*) Izl- 2 = Izl2 Izl- 2 = 1.

2. Derivatives The notion of the derivative is analogous to the notion of the derivative of a function of a real variable. (2.1) Definition. Let f and g be continuous complex-valued functions on a compact set K c CC, and c5 an operation from IR + to IR + such

that (2.1.1 )

If(w)- f(z)- g(z) (w-z)1 ~elw-zl

whenever z, WE K, e > 0, and 1w - zl ~ c5(e). Then g is called a derivative of f on K - written g=!" g=Df, or g(z)=df(z)/dz - and c5 is called a modulus of differentiability for f on K. The function f is then differentiable on K. In order to discuss differentiation on an open set, we need some preliminaries. (2.2) Definition. For each located set K c CC and each r > 0, we write

Kr== {ZECC: p(z, K)~r}. A totally bounded set K c CC is well contained in an open set U c CC if Krc U for some r>O. We then write Kcr::, U. If K c CC is totally bounded, then Kr is compact for each r > O. To see this, we note that Kr is then closed and bounded, and we refer to (63) of Chapter 4 and the following lemma. (2.3) Lemma. Let K be a located subset of CC, and r a positive number. Then Kr is located, and (2.3.1)

p(z, Kr)=max{O, p(z, K)-r}

for each z in CC. Proof Consider an arbitrary complex number z If Z1 EK and IZ21 ~ r, then

2. Derivatives

131

Therefore IZ-(Zl +z2)1 ~tx(z), where tx(z) is the right-hand side of (2.3.1). On the other hand, let I: be any positive number, and choose , in K with Iz-'I O. In the latter case, let " be that unique nonnegative multiple of z - , for which lei =min{r,lz-W. Then ,+" belongs to K., and Iz-(' +01 =max{lz-'I-r, O} <max{p(z, K)-r, O} +1:. Since I: > 0 tx(z). D

IS

arbitrary, it follows that p(z, Kr) exists and equals

(2.4) Definition. A function f: U -> (c defined on an open subset U of (c is continuous if it is continuous on every compact set K which is well contained in U. If f and g are continuous on U and I' = g on every compact set K ~ U, then g is called a derivative of f on U, and f is said to be differentiable on U.

The derivative of f on a compact or an open subset of (C is unique, if it exists. Many of the results obtained in Chapter 2 for differentiation in the real domain carryover to the complex numbers: for instance, the rules for differentiating sums, products, quotients, and polynomials. The chain rule, (2.5)

(go fY

= (g' of) 1',

is valid whenever f is a differentiable function from a compact subset K of (C into a compact subset L of (C and g is a differentiable function on L In case f is a differentiable function from an open subset U of (C into an open subset V of (C, and g is differentiable on V, equality (2.5) is valid under the extra hypothesis that f(K)~ V for each compact K~ U. (See Theorem (5.17) below for conditions under which this extra hypothesis holds.) It is easily seen that if f and g are complex-valued functions on an open set U c (C such that for each compact set K ~ U there exists an operation /j from lR+ to lR + with respect to which (2.1.1) holds, and such that either f is continuous or g is bounded, then f and g are continuous, and therefore g = 1'. There is another kind of derivative in the complex domain (2.6) Definition. Let f and g be continuous complex-valued functions on a compact set K c (C, and /j an operation from lR + to lR + such

132

Chapter 5 Complex Analysis

that li(xz+iy)- f(x l +iy)-g(x l +iY)(XZ-XI)I~Elx2-Xll

whenever Xl +iYEK, Xz+iYEK, E>O, and IXZ-XII~c5(E) Then g is called the partial derivative of f with respect to X on K - written g =Dx!' or g=fx, or g(z)=("1f(z)/(ix - and c5 is called a modulus of xdifferentiability for f on K We say that g = fx on an open set U if g = fy on every compact set K rt:, U. The partial derivative fy of f with respect to y is defined similarly. It is clear that a differentiable function f on a compact set K (respectively, on an open set U) has partial derivatives fx and fy on K (respectively, on U), and that

fx= 1',

1,.= if'.

The following converse to this simple remark is a basic criterion for differentiability. (2.7) Theorem. Let f be a continuous function on an open set U c
Proof Let Krt:, U be compact Choose r>O so that K,rt:, U, and let Zl=x1+iYI and z2=x z +iyz be any points of K with IZZ-zll~r. Then f(zz)- f(ZI)- fx(zj) (zz -Zj) = f(x z + i yz) - f(x j + i yz)+ i (Xl + i yz) - f(x l + i Yl) - fx(ZI) (x z - Xl) - iy(ZI) (yz - Yl) = f(x z + i yz)- f(x l + i Yz)- fx(xl + i yz) (x z -Xl) + f(x l + i Yz) - f(x 1 + i Yl) - fy(x 1 + i YI)(Yz - Yl) + (x z - Xl) (fx(X I + i Yz) - f,(x l

+ i Yl))·

If c5 denotes the minimum of r, the modulus of continuity for fx on Ky, and the moduli of x- and y-differentiability for f on Ky, then it follows that If(zz)- i(zj)- ix(Zl) (zz -zl)1 ~381z2 -zll

whenever Zl,ZzEK and

IZ2-Z11~c5(8).

Therefore f is differentiable on 0

K, and l' = ix. It follows that f is differentiable on U, with l' = fx

2 Derivatives

133

A function f:
t---+

L akZ

n - k•

Here are some other examples.

k~O

(2.8) Definition. The exponential function exp:
exp(z)=:0 exp(x)(cos Y + i sin y)

(z =:0 X+ i YE
where on the right-hand side, exp denotes the exponential function on lR

This function has partial derivatives expx(z) = exp(x) (cos y+ i sin y) and expy(z) = exp(x) (- sin Y + i cos y) By Theorem (2.7), it is differentiable, with derivative d exp(z) -d-=-z- = expx(z) = exp(z)

For all Zl =:0 Xl + i Yl and Z2 =:OX 2 + i Yz in
134

Chapter 5

Complex Analysis

if it converges to f uniformly on each compact set K ~ U, in the sense that for each E > 0 there exists N. in 7l+ such that If.(z)-f(z)I~E

(n~N.,zEK)

We leave the reader to make the appropriate definition of a Cauchy sequence of continuous mappings of U into
3. Integration We introduce the notion of a path, which integrations are perfurmed

IS

a domain over which

(3.1) Definition. A path 'I is a continuous map of a proper compact interval [a, bJ into
3 Integration

135

and if tE[b l , b l +b 2 -a 2]

Y3(t)=Y2(A,(t))

is called the sum of the paths YI and Y2, and is written YI +Yz. It is clear that path addition is associative. Every path is a finite sum of differentiable paths. The negative of a path Y is the path -y==yoA" where A. is the linear map of the parameter interval of Y onto itself which is everywhere decreasing. The end points of - yare the end points of Y reversed. For paths YI and Y2 we write YI -Y2 instead of YI +( -')'2)' Next we study a form, which is an object to be integrated along a path. (3.3) Definition. A form on a compact subset K (respectively, an open subset U) of
where f and g are continuous functions on K (respectively, on U). If y==rx.+i{3 is a path (where rx. and {3 are real-valued functions), and if OJ is a form on car y, then we define the integral of w along Y to be b

SW == S(f(y(t)) rx.'(t) + g(y(t)) {3'(t)) dt, a

where the integral on the right-hand side is to be interpreted as a sum of integrals over the intervals of differentiability of y. If the sum Y+ b of the paths Y and b is defined, and if w is a form on car (y + b), then S OJ = Sw + SOJ. y+b

b

If Y and b are equivalent paths, then SOJ = Sw whenever the integrals y

b

are defined. For any path y and any form OJ on car y,

S W=

-

SOJ.

-y

The magnitude of Sw can be estimated as follows. For each path y

y == a. + i {3 the length Iy I of y is defined by b

b

Iyl == Sly'(t)1 dt == SIrx.'(t) + i{3'(t)1 dt. a

a

Clearly IYI + yzi = IYII + IY21 whenever YI + Y2 exists. For a form w == f dx +gdy defined on a compact set K, the norm Ilwil K of w on K is

136

Chapter 5 Complex Analysis

defined by

IlwlI K== 11(lf1 Z+ IgIZ)tIIK =SUp {(If(xW + Ig(xW)t: XEK}.

Cauchy's inequality gives the estimate b

IS wi ~ S(If(y(t)W + Ig(y(t)W)t(a'(t)2 + f3'(t)2)t dt ~

IlwllKlyl

whenever y lies in K If the function f has partial derivatives on an open set U, then df will denote the form fxdx+ fydy on U. Thus we denote by dz the form dX+idy, and by dz* the form dx-idy. If h is a continuous function, and w == f dx + g dy is a form, both defined on the same set K, then hw will denote the form hfdx+hgdy Thus hdz is the form hdx+ihdy. For each path y in K we have the estimate

If h d zl

=i!

h(y(t))(a'(t) + i f3'(t)) dti b

~

IlhllKS la'(t)+if3'(t)1 dt= IlhllKlyl. a

In case y is a differentiable path and j is a differentiable function on car y, the chain rule for differentiating composite functions gives b

(3 4)

Sdf = S1'(z) dz == S1'(y(t)) y'(t) dt a

d = S-d f(y(t))dt= f(y(b))- f(y(a}}. b

a

t

This formula can be generalized to an arbitrary path y by breaking the parameter interval [a, bJ into the intervals of differentiability of y. In case y is a closed path, the right side of (3.4), and therefore the left side, vanishes. A path y == a + i f3 is called linear if both a and f3 are linear functions on the parameter interval. The unique linear path y == lin (Zl' zz) with parameter interval [0,1J and end points Zl and Z2 is given by We write Z2

Sw==Sw

3 Integration

137

for the integral of the form w along this path. The polygonal path Y== poly (ZI' ... , z.) whose vertices are the complex numbers ZI' .. , z. IS defined by poly (ZI' ... ,z.)==lin(zl,z2)+ ... +lin(z._I'z.), and the paths lin(zi,zi+l) (1~i~n-1) are called the linear pieces ofy. A subset S of
O(j~O

for

l~j~n,

and

0(1

+ ... +rx n =l}

It is easily seen to be compact. If y == poly (ZI' ... ' z.) is a polygonal path, we write span y == span (ZI' ... , z.). (3 5) Definition. A continuous function analytic on U if

J on an open set U c
IS

S f(z)dz=O

(3.5.1 )

whenever y == poly (z l' Z2' Z3' ZI) is a triangular path in U whose span is well contained in U Our goal is to show that J is analytic if and only if it is differentiable, and to extend (3.5.1) to a large class of closed paths y in U. (3.6) Proposition. A differentiable Junction J on an open set U in
Proof" Let Yl==polY(ZI,Z2,Z3,ZI) be any triangular path whose span is well contained in U, and let B be any positive number. We have

S f(z) dz = S J(z) dz + S J(z) dz + S f(z) dz + S J(z) dz ')11

')Ill

}'12

}'13

')114

where YU==POlY(ZI,t(ZI +Z2),t(ZI +Z3),ZI), Y12 == poly (Z2' t(Z 2 + Z3)' t(ZI + Z2)' Z2)' Y13 == poly (Z 3' HZl + Z3)' t(Z2 + z 3)' Z3)'

and YI4==poly(t(ZI +Z2),t(Z2+ Z 3),t(ZI +Z3),t(ZI +Z2))· It follows that for at least one value of i,

1S J(z) dzl >-1-1 S f(z) dzl- B/8. ')Ili

')It

138

Chapter 5 Complex Analysis

Hence IJ J(z) dz l<4IJ J(z)dzl+te, 11

where Y2=Yli. The length of Y2 is tlYll, and span Y2 cspan Yl. Continuing in this way, we construct a sequence (Yn) of triangular paths with (3.6.1)

span Yn+l cspan Yn

(3.6.2)

IYn+ll =t IYnl =2- nIYll

and 1n

Yn+ 1

for all n in Z+. Hence, by induction,

n-l

IJ J(z)dzl<4n-1IJ J(z)dzl+e

(3.6.3)

11

L 2-

k

(nEZ+).

k=1

1n

Let {) be a modulus of differentiability for J on span Yl. By (3.6.1) and (3.6.2), we can choose the positive integer N so that Iz-z'I~{)(e) whenever z,z'EK=spanYN. Let zoEK. Then

IJ(z)- J(zo)- f'(zo)(z-zo)l~elz-zol for all z in K. Now, by (3.4),

J(J(zo)+ f'(zo)(z-zo»dz=O 1N

because the integrand J(zo) + f'(zo)(z - zo) is the derivative of the function ZI-+J(zo)z+tf'(zo)(Z-zo)2. Hence, by (3.6.3) and (3.6.2),

IJJ(z)dzl <4N- 11J J(z)dzl +e 11

1N

1N

~4N-IIYNI ellz-zoIIK+e ~4N-IIYNI2e+e

=IYI1 2e+e. Since e is an arbitrary positive number, it follows that J f(z)dz=O. Therefore J is analytic on U. 0 11 (3.7) Lemma. Let J and g be continuous Junctions on an open set U in CC such that %2

(3.7.1 )

J

g(z 2) - g(ZI) = J(z) dz %1

3. Integration

holds whenever Zl,Z2EU and V, and g'=!

span(zl,z2)~

139

V Then g is differentiable on

Proof: Let K~ V be compact. Choose r>O with K,~ V, and let J be a modulus of continuity for f on K,. Then if e > 0, Zl ' Z2 E K, and IZ2 -zll ~min {r, J(e)}, we have

Ig(Z2) - g(Zl) - f( Zl)(Z2 - zl)1

II

= (f(z) - f(zl» dzi ~sup

{If(z) - f(zl)l: ZEK" Iz - zll ~ J(e)} IZ2 - zll

~elzz -zll·

Therefore g' = f on K. Since K is an arbitrary compact set well contained in V, we have g'= f on U. 0 (3.8) Lemma. Let V be a convex open set in
g(z)= S f(Od,

(3.8.1) is differentiable, and g' =

Zo

!

Proof: Let Zl and Zz be arbitrary points of U. Using the convexity of V, it is easy to establish that span(zO,zl,z2)~ U. By definition of 'analytic', ZI

Z2

Zo

S f(z)dz+ S f(z)dz+ S f(z)dz=O, Zo

ZI

Z2

so that Z2

g(zz) - g(Zl) = Sfez) dz.

Z,

The desired conclusion now follows from (3.7).

0

(3.9) Lemma. If f is an analytic function on a convex open set V in
°

Proof: Let Zo be any point of U. Define the function g on V by (3.8.1). Then by (3.8), we have g' = f on U. If y: [a, b] ---+
J f(z)dz=g(y(b»-g(y(a»=O. ¥

0

140

Chapter 5 Complex Analysis

(3.10) Lemma. Let U c ([ be open, K compact, and E>O, such that Kc~ U Let Yo and 1'1 be closed paths in K with common parameter interval [a, b], such that IYo(t) -YI(t)1 ~tE

(tE[a, b]).

Then Jor all analytic Junctions J on U,

j

j

J(z) dz =

J(z) dz.

}'t

)10

ProoJ Choose points to=a
Then for

0~i~n-1,

is a closed path whose carrier lies in the closed sphere of radius tE +tE = 3E/4 about the point Yo(t} Since the open sphere of radius E about Yo(t;) is convex and lies in U, by (3.9) we have

j J(z)dz=O

(3.10.1)

~.

for any analytic function J on U. Summation of (3.10.1) for gIVes

0~i~n-1

"-1

0=

L (j J(z)dz+ j J(z)dz1'i

i= 0 ai

j

= j J(z)dz- j J(z)dz+ j f(z)dz- j Yo

)11

aD

j J(z)dz) jlh f(z)dz

J(z)dz-

ai + 1

an

= j f(z)dz- j J(z)dz (since the paths

0"0

and

0""

are the same), as was to be proved.

0

(3.11) Definition. Closed paths Yo and 1'1 in a compact set K c ([ are homotopic in K if there exists a continuous function 0": [0,1] x [0,1]--+K such that for each t in [0,1], the function 0",: [0,1]--+([ defined by O",(x) == O"(t, x) is a closed path, with 0"0 equivalent to Yo and 0"1 to 1'1· The function 0" is called a homotopy of Yo and 1'1. In case U c ([ is open and K ~ U, Yo and 1'1 are said to be homotopic in U. In case 1'1 is a constant path, Yo is said to be null-homotopic. Our next result is the famous Cauchy integral theorem.

3 Integration

141

(3.12) Theorem. Let the closed paths Yo and 1'1 in the open set U c
Jf(z)dz= Jf(z)dz

(3.12.1)

for all analytic functions f on U. Proof: Let K ~ U be compact, and let u be a homotopy of Yo and 1'1 in K. Choose e>O so that K.~U. Choose points to=O
The desired equality (3.12.1) now follows. An immediate corollary is that

0

Jf(z) dz = 0 whenever

f is analytic

~

on an open set U and the closed path I' in U is null-homotopic in U. (3.13) Definition. An open set U c
For simply connected open sets, Theorem (3.12) takes the following form. (3.14) Corollary. If I' is any closed path in a simply connected open set Uc
Corollary (3.14) is the form in which Cauchy's integral theorem is usually stated Another version is the following. (3.15) Corollary. Let Yo and 1'1 be paths in a simply connected open set U c
J

f(z)dz=O,

}'O-'Yl

which holds by (3.14).

D

(3.16) Theorem. If f is analytic on the simply connected open set U c cr, then there exists a differentiable function g on U with g' = f, and g is unique to within an additive constant.

142

Chapter 5 Complex Analysis

Proof. Let Zo be any point of U. For each z in U let 'Yz be a path in U from Zo to z. Define the function g on U by g(z):: S f(Od(. y.

By (3.15), any other choice of 'Yz would give the same value for g(z). Also, g(Z2)-g(Zl)= Sf(Od(, y

where'Y is any path in U from Zl to Z2. In particular, Z2

g(Z2)-g(Zl)=

S f(Od( ZI

whenever span (Zl' Z2)11: U. From (3.7) it follows that g is differentiable on U and that g' = f Now let h be any function on U with h' = f, and consider any points Z1>Z2 in U and any path 'Y in U from Zl to Z2. We have g(Z2) - g(Zl)= Sg'(z)dz= Sf(z)dz y

= Sh'(z)dz=h(z2)-h(zl)' y

by (3.4). Therefore g - h is constant on U.

0

4. The Winding Number Further developments depend on the concept of how many times a closed path winds around a point. (4.1) Proposition. Let 'Y be a path in the metric complement
a point

Zo

in
(4.1.1)

where Zl and Z2 are the left and right end points of 'Y. Proof" Since every path is a finite sum of differentiable paths, it is enough to consider the case where 'Y is differentiable. For each u in the parameter interval [a,b] of'}' we write u

A(U):: J('Y(t) - ZO)-l 'Y'(t) dt, a

4. The Winding Number

143

so that .A.: [a,b] --+ CC is continuous, A(b)= J(Z-ZO)-I dz, and A, (a) =0. Also, A,'(u) = (y(u) - ZO)-I y'(u).

Therefore d du «y(u)-zo) exp( -A(U))) = (y(u) - zo)( - A.'(u)) exp ( - A(u)) + y'(u) exp ( - A(u)) =0.

Therefore the function ul-+(y(u)-zo)exp(-A,(u)) is constant on [a,b], and so equals its value at a, which is y(a)-zo. In particular, (y(b) - zo) exp (- A(b)) = y(a) - zo,

which is equivalent to (4.1.1).

0

A complex number w is a logarithm of a complex number z if exp(w)=z. Every complex number z*,O has at least one logarithm w, and the numbers (4.2)

(nEZ)

w+2nin

form the totality of logarithms of z. The fact that z has at least one logarithm follows from (4.1). Since exp (2 n i) = cos 2n + i sin 2n = 1, the numbers (4.2) are all logarithms of z. Conversely, if ( is any logarithm of z, then exp«(-w)=1, and thus e"(cosv+isinv)=1, where (-w=u+iv; hence u=O, v=2nn for some n in Z, and consequently ( has the form (4.2). If U is a simply connected open subset of CC - {OJ, Zo is any point of U, and a is any logarithm of Zo, then by (3.16) there exists a unique differentiable function g on U with g'(Z)=Z-1 and g(zo)=a. Then g(z) -a= S(-I de, where y is any path in U from Zo to z. By (4.1), exp(g(z))=z

(ZEU).

The function g is called a branch of the logarithmic function in U. Proposition (4.1) has the following corollary. (4.3) Proposition. If y is a closed path in CC-{zo}, then the quantity j(y,zo)=(2ni)-1 J(Z-ZO)-I dz y

is an integer, called the winding number of y with respect to zoo

144

Chapter 5

Complex Analysis

Proof: By (41),

exp (2 nij(y, zo)) = 1. Therefore j(y, zo) is an integer.

0

(4.4) Proposition. Let K c 0 such that j(y,zd=j(y,Z2) whenever Z1 and Z2 belong to K and IZ1 - z21 ~ b. Proof· Choose 15>0 so that with IZ1 -z21 ~b. Then

(cary)~~
Y1(t)=y(t)-(Z1 -Z2)

Let Z1, Z2 belong to K,

(O~t~

1)

defines a closed path 1'1 in (car y)~, and rjJ(t,~)=y(t)-(Z1-Z2)~

(O~t, ~~1)

defines a homotopy of I' and 1'1 in (car y)~ By Cauchy's integral theorem, j(y, z) = j(Y1' z) for each z in K. Hence 1

2n i j(y, Z1) = Jy'(t) dt/( y(t) - Z1) o 1

= Jy~(t)dt/(Y1(t) -Z2) o

= 2nij(Y1' Z2) = 2n ij(y, Z2). Thusj(y,z1)=j(y,z2).

D

ce, and let Z1' Z2 be two points of ce which can be joined by a closed path in - cary. Then j(y, Z1) =j(y,Z2)·

(4.5) Corollary. Let I' be a closed path in

Proof" Let 0": [0,1] -> 0 such that j(y, z)=j(y, z') whenever z and z' belong to car cr and Iz - z'l ~ b. Choose points 0 = to < t1 < ... < tn = 1 so that Icr(tk+ 1) - O"(tk)1 ~ 15 (0 ~ k ~ n -1). Then j(y, cr(t k )) = j(y, O"(tk+ 1)) (0~k~n-1), and so j(y, Z1)=j(y, Z2)· D

An important example is the path I' defined by y(t)=zo+rexp(int)

(0~t~2n),

where r>O. (In case n= 1, I' is called the circular path of radius r about the point zo' or the circle of radius r with center zo.) The winding number of I' is 2" j(y,zo)=(2ni)-1 Jr- 1 exp(-int)inrexp(int)dt=n. o

4 The Winding Number

145

This accords with the intuitive meaning of winding. By (4.5), j(y, z) = j(y, zo) = n whenever Iz - zol < r. As another example, let Zo be any point in
Let Y2 be any path from zo-x 2 to ZO+Xl which lies in U2 ==
Let 0( be the circular path of radius r ==~(XI + x 2 ) with center Zo +~(XI -x 2 ). Then 0( passes through the points ZO+XI and Zo-X 2 • The path 0( is the sum of the paths 0(1 and 0(2' where 0(1 is a semicircular path from Zo + XI to Zo - X 2 lying in UI , and 0(2 is a semicircular path from ZO-X 2 to ZO+XI lying in U2. By (3.15), j(YI +Y2,zo)=(2ni)-I(J (z-zo)-Idz+ J (z-zo)-Idz)

= j(O(, zo) = 1. In particular, j(y, zo) = 1 whenever Y is a triangular or rectangular path surrounding Zo and traversed in the positive (anticlockwise) direction.

(4.6) Lemma. Let f be a differentiable function on an open set U c
Z

be a point of U, and let the mapping h be defined by h(C)==(fW- f(z))(C -Z)-I

(CEU - {z}).

Then h extends to an analytic function on U. B==Sc(z,r)~ U. Let (j be a modulus of differentiability for f on B, with (j < 4r, and let e be a positive number. Let K be the compact set

Proof· Choose r>O so that

{CE
"1 - 2

max {lei -zl, le2 -zl} < (j(e) or

146

Chapter 5 Complex Analysis

In the former case,

Ih«(2) - h«(dl ~ Ih«(2) - f'(z) I+ Ih«(l) - f'(z) I

=

IJ«(2)(2-- J(z) Z

f'(z) 1 +

IJ«(l)(l-- J(z) - f'(z) Z

1

~e+e=2e.

In the second case, we may assume that 1(1 -zl>tb(e). Then 1(2 -zl>tb(e), both (1 and (2 belong to K, and so Ih«(d-h«(2)I~e. Since e>O is arbitrary, it follows that h is uniformly continuous on B- {z}. By (3.7) of Chapter 4, we see that h has a unique continuous extension to B. Hence, clearly, h has a continuous extension to U. To show that the extended function h is analytic on U, consider points Zl,Z2,Z3 in U with K=span(zl,z2,z3)~ U. Choose r>O so that K.~U. Either p(z,K»r/2 or p(z,K)
Ih«()d(=O,

where Y=POlY(Zl,Z2,Z3,Zl)' Assume therefore that p(z,K)
I h«) d( = I h«() d( + I h«() d( + I h«() d(, 11

where Y1=polY(Zl,Z2,Z,Zl)' Y2=poly(z2,z3,z,z2)' and Y3=poly(z3,zl' z, z3)' Thus we have reduced the proof of (4.6.1) to the case in which z is one of the vertices of Y, say z = Zl' In this case, given e > 0 and using the continuity of h on K, we can find ex> 0 so that (4.6.2)

II h«() d(l < e

whenever m=min{lz2-zl,lz3-zl}<ex. Either m<ex or m>O. In the latter case, choose a point w of lin (z, Z2) so that

p(z, span (w, Z2, Z3)) > 0 and

II h«() d( -

I h«() d(l < e, l

where A=poly(w,z2,z3' w). By (3.6), I h(Od(=O, whence (4.6.2) holds. l

Thus (4.6.2) holds in all cases. As e>O is arbitrary, (4.6.1) obtains, and so h is analytic on U. 0 The following result, embodying Cauchy's integral Jormula, uses the winding number to establish a representation of a differentiable function by means of an integral.

4 The Winding Number

147

(4.7) Theorem. Let f be a differentiable function on an open set U c ce, let Z be a point of U, and let y be a closed path in U - {z} which is null-homotopic in U. Then

j(y,z)f(z)=(2ni)-1 Jf(C)(C _Z)-l de. Proof: The function CI-+(f(C)- f(z))(C _Z)-l extends to an analytic function h on U, by (46). Since Jh(C)dC=O, 1

(2nO- 1 J f(C)(C _Z)-l dC=(2ni)-1 J f(z)(C _Z)-l dC=j(y,z)f(z).

D

Cauchy's integral formula is often used in conjunction with the following proposition (4.8) Proposition. Let h be a continuous function on the carrier of the

path y in by

ce, and

let n be a positive integer. Then the function f defined

is differentiable on the metric complement - car y of car y, and where g(z)=(2ni)-lnJ h(C)(C _z)-n-1 dC.

l' = g,

Proof: Let K be any compact set well contained in - car y. Write C=sUP{lh(C)lkt1IC-ZI-k: CEcary, ZEK}.

Let w be a common modulus of continuity for the continuous functions (C,z)I-+(C _Z)-k (1 ~k~n) on the product space (cary)xK. Consider e>O and Zl,Z2 in K with IZ1 - z21 ~ w(e). Then If(z2) - f(zl) - g(Zl)(Z2 - zl)1

= (2 n)-lIJ h(O((C - z2)-n - (C - zl)-n - n(C - Zl)-n-1(Z2 - Zl)) dCI =(2n)-1Iz2 -zl'ls h(O 1

(i (C -Zl)-k(C -Z2)-n-1+k k~l

- n(C - Zl)-n-1 )

dcl

148

Chapter 5 Complex Analysis

= (2 n)-llz 2 - Zlllf h(()ktl (( - ZI)-k((( - Z2)-n-1 +k _(( _ ZI)-n-l +k) dcl ~(2n)-llz2-ZIII')I1 CEo

lt follows that

l' = g on

K, and therefore on - cary.

0

(4.9) Lemma. Let U c O with Kr~ U, and let {C l , ... ,(n} be an r/2 approximation to K. Let b be a modulus of differentiability for f, and w a common modulus of continuity for f and g, on the union of the closed spheres Sc(Ci,r) (1~j~n) Consider points ZI and Z2 in K, and a positive number e, such that

Iz 2 - zil ~ min {r/2, beE), w(e)}.

We have IZ1-()
J (z2)1 ~ e,

Ig(zl) - g(z2)1 ~ e,

and If(Z2) - f(ZI) - g(ZI)(Z2 - zl)1 ~ elz 2 - z11

Therefore f and g are continuous on K, f is differentiable on K, and g= l' on K. Since K~ U is arbitrary, the result follows. 0 (4.10) Theorem. Let f be differentiable on an open set U c
has derivatives of all orders on U, and for any closed path 'Y in U which is null-homotopic in U, any Z in U -cary, and any integer n~O, (4.10.1)

Proof" To show that f has derivatives of all orders, it is enough to show that l' is differentiable, since by the same token, f" will be differentiable, and so on. Consider a closed sphere Sc(zo,r)~ U.

4 The Winding Number

149

Choose t > r with S c(zo, t)~ U. Let y be the circular path of radius t about Zoo By (4.7) and (4.8), we see that f'(z)=(2ni)-1 Sf(O(C-z)-2dC

whenever Iz - zol < t. Hence, again by (4.8), f' is differentiable on Sc(zo,r). It follows from (4.9) that f' is differentiable on U. The formula (4.10.1) now follows from (4.7) and (48) 0 (4.11) Corollary. A function f analytic on an open set U c
Recall the definition of convergence of a sequence of continuous functions, at the end of Section 2 above. (4.12) Corollary. Let (fn) be a sequence of differentiable functions on an open set U c
Therefore f is analytic, and consequently differentiable, on U.

0

(4.13) Corollary. If f is differentiable on Sc(zo, ro) c
Ipn)(zo)1 :;=;n! ro-nsup{lf(z)l: Iz-zol=ro}

(n~O).

Proof' For O
Then Ipn)(zo)1 = (2 n)-l n! IS f(C)(C - zo)-n-l del :;=;(2n)-1 n! Iyl r- n - 1 sup{lf(z)l: Iz-zol =r} =n'r-nsup{lf(z)l: Iz-zol=r}.

Letting r ---+ ro gives the result.

0

The inequalities (4.13 1) are known as Cauchy's inequalities.

150

Chapter 5 Complex Analysis

(4.14) Corollary. If f is differentiable on an open set U c (C, and Zo is any point of U, then the function zl-+(f(z)- f(zo» (Z-ZO)-l on U - {zo} extends to a differentiable function on U. Proof. The result follows from (4.6) and (4.11).

0

Power series are uniquely suited to the representation of differentiable functions in the complex domain. The complex variable version of Taylor's theorem reads as follows. (4.15) Theorem. Let f be differentiable on an open sphere S

=S(zo, r)

c (C. Then the series

(4.15.1)

if and only if the

converges to f on S

coefficients an are given by

(4.15.2) Proof· Consider a compact set K~S. Let T=Sc(zo,ro), where c T. Write t =1(r + ro), and let ')I be the circular path of radius t about ZOo If ZET and NEZ+, then for all C with Ie -zol =t we

0< ro < rand K have N

(C-Z)-l=

L

(z-zo)"(C-zo)-n-l

+ (C-Z)-l(Z-ZO)N+l(C-ZO)-N-l.

Hence, by (4.10), N

f(z)-

L (n!)-l pn)(zo)(z-zo)" n~O

=(2n

i)-I! f(0

(C _Z)-l_ nto (Z-ZO)n(c -zo)-n-l) dC

=(2n i)-l(Z-ZO)N+l Jf(C)(C -Z)-l(C -Zo)-N-ldC.

Since 1(2ni)-I(z-zot+ 1 Jf(C)(C -Z)-l(C -ZO)-N-1dCl 1

Ie -zol ~t} 2(r-rO)-l t -N-l =2rO(r-rO)-1(rO/tt sup{lf(C)I: Ie -zol ~t} ~(2n)-1 r~+ll')ll

-+0

sup{lf(C)I:

as N -+ 00,

the series (4.15.1) with coefficients given by (4.15.2) converges uniformly to f on T, and therefore on K.

\

4. The Winding . . . umber

151

If, conversely, (4.15.1) converges uniformly to f on T, then for each integer n ~ we have (n!)-l j
°

00

= Lak(2ni)-lS('-zot- n- 1d,.

'1-+('

k=O

y

For k=t=n, the map _zo)k-n-1 is the derivative of'l-+(k-n)-l(, - zo)k-n on car y. It follows from (3.4) that (n!)-l j
Hence, by the ratio test, a power series L an(z - zo)" converges abn=O

solutely and uniformly on every compact set KfCS(zo, r) if there exists N in Z+ such that lan+11~r-1Ianl whenever n~N. We end this section with a result needed in the proof of the Casorati-Weierstrass theorem ((6.10) below). 00

(4.16) Proposition. Suppose that the series L bnzn is uniformly conn=O

vergent on each compact set well contained in A == {ze R}, where R ~ 0. Then the series converges for all z in R, and let y be the circular path of radius r about 0. By (4.8),

,n) (' - Z)-l d,

f(z) ==(2n i)-l { (~o bn

defines a differentiable function on - cary. For each z in S(O, r) and each integer N ~ 0, since z"=(2ni)-1 S ,n(,_z)-ld,

(n~O),

we have

k(Z) - nto bnznl = 1(2n i)-1 ! C=~+l bn,n) (' - z)-l d'i ~(2n)-1Iyl (r-lz-z ol)-l sup

{I

~ n=N+1

bn,nl: ,ecar y}.

152

Chapter 5 Complex Analysis

Since

L bn ,n

00

00

converges uniformly on car y, it follows that

n~O

L bn zn n~O

converges to f(z) uniformly on each compact set well contained in 00

S(O, r) As r>R is arbitrary, we now see that function. 0

ZI---+

L bnzn is

an entire

n~ 0

5. Estimates of Size, and the Location of Zeros It is a famous principle of classical analysis, called the maximum principle, that an analytic function attains its maximum on the boundary. The next proposition is the constructive version of this result.

(5.1) Definition. If K is a compact subset of
(zEK),

then B is called a border for K. (5.2) Proposition. If B is a border for the compact set K c 0 so that K n B ~ is compact, and If(z)I;;:;;llfIIB+E whenever zEKnBd' Consider Zo in K with p(zo, B) > fJ/2. Let y be the circle of radius r about ZO' where r is chosen so that p(zo,B)-c5/4
Then Izl-zol=r, and p(zl,B);;:;;lzl-'I=lzo-'I- r
Let the path YI be the arc of the circle y extending in length an amount fJ/2 on either side of Zl' Then IYII=fJ, YI lies in B d , and If(z)l;;:;; IlfIIB+E for all z in caryl' Let Y2 be the complementary arc, so

5 Estimates of Size, and the Location of Zeros

153

that ')1=')11 +')12 Then

If(zo)1 = (2 n)-1 IS f(z) (z -

ZO)-1

dzl

~(2n)-11')111 (1IfIIB+e) r- 1+(2n)-11')1211IfIIK r - 1

= (2n r)-1 (c5(llfIIB + e) + (2n r -

c5) IlfII K).

This inequality also holds for Zo in K n B d , since the right-hand side is greater than or equal to IlfIIKnB •. It therefore holds for all Zo in K. Hence IlfilK ~(2n r)-1(c5(llfIIB+ e) + (2n r - c5) Ilfll K) and therefore Since this is true for all e > 0, the proposition is proved.

K

D

For each compact set K c
-->

We have the following corollary to Proposition (5.2). (5.3) Corollary. Let K c 0 or m(f, K) < m(f, B). In the first case, the function f- 1 is differentiable on K, and

Thus m(f, K) = m(f, B). In the second case, consider an arbitrary positive integer n. Either m(f,K)
r~O,

we write

r(z, r)= {CE
Ie -zl=r}.

In the special case where z = 0, we write r(r) instead of r(0, r). The case n = 1 and c5 = 0 of the next lemma is known as Schwarz's lemma.

154

Chapter 5 Complex Analysis

(5.4) Lemma. Let h be a differentiable function on a closed sphere S=Sc(w,r)c
and write

(ZES).

f(z)=h(z)(z-z1) .,. (z-z.) Then If(z)I~(lz-wl+c5)·(r-o)-·

Ilflls

(ZES).

Proof. Iflz-wl=r, then Iz-zkl~r-o (1~k~n). Thus

Ilhll s= IlhIIT(w,r) ~(r _0)-· Ilf IIT(w,r) = (r _15)-· Ilf lis, since r(w, r) is a border for S. Then If(z)I~lz-z11

for all

Z

in S.

.. ·lz-z.lllhlls~(lz-wl+o)·(r-o)-· Ilflls

D

(5.5) Lemma. Let r, 15, d, and 0( be positive numbers with r> 3£5, and let n be a positive integer. Let f be differentiable on a closed sphere Let Z1""'Z. be points of S with IZj-zkl~d (1 ~j
S=Sc(w,r)c
If(zk)1 ~O(,

Izk-wl ~o

(1 ~k~n).

J#ite T=Sc(w,r-30). Then

Ilfll T ~ Ilfll s(r - 215)" (r - 0)-· + 20( nCr + 0)"-1 d-·+ 1. Proof Let g be the polynomial of degree n -1 with the same values as f at the points Z1' ... , z.; so that n

g(z)=

I

k~1

where

f1

f(zk)

f1 «Zk- Z)-1(Z-Z)), k

means the product taken from j = 1 to j = n with the factor

k

corresponding to j = k left out. Then (5.5.1) since

Iz-Zjl ~Iz-wl +Izj-wl ~r+o

(ZES, 1 ~j~n).

The function f - g vanishes at Z1' ... , Z•. By (4.14), f(z)-g(z)=(z-z1)'" (z-zn) h(z)

(ZES)

for some differentiable function h on S. It follows from (5.4) that (5.5.2)

II! -gIIT~(r-20)·(r-o)-· II!- glls'

5 Estimates of Size, and the Location of Zeros

155

By (5.5.1) and (5.5.2), II!IIT~ Ilf -gIIT+ IlgIIT~(r-2t5)"(r-t5)-" Ilf -glls+ Ilglis ~(r-2t5)"(r-t5)-"(IIflls+ Ilglls)+ Ilglls ~ Ilflls(r - 215)" (r - 15)-" + 2ac nCr + 15)"-1 d-"+ 1.

0

A complex-valued function f is said to be nonzero if there exists z with f(z)=l=O. We say that f is nonvanishing on a set AcCC if f(z)=l=O for all z in A. (5.6) Lemma. Let r, 15, and I: be positive numbers with r> 315, and let f be a nonzero differentiable mapping on a closed sphere S=Sc(w, r)c CC. Then there exists s such that 15 - I: < S < 15 and m(f, r(w, s)) > O. Proof' To begin with, assume that IlfIIT>O, where T=Sc(w,r-3t5). Choose an integer N;:;: 2 so that

lI!IIs(r - 2t5t (r - t5)-N
Choose ac > 0 so that 2rxN(r+ t5t- 1 d- N+ 1
then there exist points l~j
Zk

with

IZk -

l~k~N}
wi = rk and If(zk) I< ac (1 ~ k ~ N). If

IZj-Zkl;:;: IZk-wl-lzj-wl = rk-rj;:;:d. It follows from (5.5) that

II !II T ~ Ilflls(r - 2t5)N (r - 15)- N + 2rxN(r + t5)N -1 d- N+1 iac>O. This completes the proof in case IlfIIT>O. In the general case, choose 15' so that 0 < 15' < min {r/3, r - 3t5} and IlfII Sc (w.r-3d'»0. By the foregoing, there exists t with O O. Hence II f II T > 0, and the result follows. 0 (5.7) Lemma. Let K c CC be compact, and let B be a border for K. Let f be a differentiable function on K with m(f, B) > m(f, K) = O. Then for

156

Chapter 5 Complex Analysi,

each t > 0 there exists a closed sphere S = Sc(w, to) c K of radius to ~ t such that m(f, r(w, to))>m(f, S)=O. Proof: Choose (in K with Ifml<m(f,B). Then p(CB»O, and so K'= {zEK: p(z, B)~IX}

is nonvoid whenever 0 < IX < p(C B). Choose IX in (0, t) so that K n B 2 • and K' are compact, and m(f,KnB2a»~m(f,B). Let {ZI' ""ZN} be a tlX approximation to K'. As m(f, B) > 0, f is nonzero on each of the spheres SC(Zj' p(Zj' B)). By (5.6), for each j (1 ~j ~ N) there exists tj in (1X/4, 1X/2) such that Choose (' in K with

If((,)1 <minam(f, B), c I '

... ,

c N}·

Then (' cannot belong to KnBza; so that ('EK', and therefore ('ES(Zk' t k ) for some k. Hence c k > If((')1 ~ m(f, SC(Zk' t k))

and so m(f, SC(Zk' tk))=O, by (5.3). Thus we can take W=Zk and to=tk' D

(5.8) Lemma. Under the hypotheses of Lemma (5.7), there exists with f(z)=O.

Z

in K

Proof Using (5.7), construct recursively a sequence (Sn);~)~l of closed spheres with K:;)SI:;)Sz:;)'" such that m(f,~»m(f,Sk)=O, where ~ is the boundary of Sk' and such that the radius of Sk is at most k- 1 . Then these spheres have a common point Z with f(z)=O. D (5.9) Definition. A polynomial z~ao Z· + a l zn-I least k if an_/oF 0 for some j ~ k.

+ ... + an

has degree at

We now prove the fundamental theorem of algebra. (5.10) Theorem. If the polynomial p(z)=aoz n+ ... +a. has degree at least k, then there exist complex numbers Zl' ... , Zk and a polynomial q such that p(Z)=(Z-ZI) '" (Z-Zk) q(z) (ZE
Proof. Choose j ~ k with an _ j

=1= 0,

and then choose r > 0 with

j-I

lan_)r j > Ilan_mlrm+lp(O)I. m~O

5 Estimates of Size, and the Location of Zeros

157

Then either an _ m =1= 0 for some m > j or else i-I n (5.10.1) lan_ilri > Ilan_mlrm + I lan_mlrm+lp(O)I. m~O

m~i+1

In the former case, we replace j with m and repeat the above construction. Eventually, this process leads to values of j and r for which n

(5.101) holds, where

follows that

I

lan_ml rm is taken as 0 if j=n. It then

m~i+1

inf{lp(z)l: Izl=r}>lp(O)I; whence, by (5.8), there exists a complex number ZI with P(Zl) = O. Using the process of polynomial division, we find a polynomial ql(z)==boz n - 1+ ... +b n _ 1 and a constant c in
(ZE
Since P(Zl)=O, we have c=O. Now if an_i=l=O, then either j=n, and hence b o = a o =1= 0, or j < n; in the latter case, bn _ i + ZI bn_i-I = an_i =1= 0, and so either bn_i=l=O or bn - i - I =1=0. Hence bn_1_s=l=O for some s?;,k -1, and so ql has degree at least k -1. Replacing p with ql and repeating the construction, we find that p(z)=(Z-Zl) (z-zz) qz(z)

(ZE
where q 2 has degree at least k - 2. By induction, we finally construct the desired numbers ZI' ... , Zk and the desired polynomial q == qk· 0 (5.11) Theorem. Let K

c m (f, K) = O. Then there exist finitely many points zl' .. 'Zn of K, and a differentiable function g on K, such that

I(z)=(Z-ZI) ... (z-z,) g(z)

(zEK)

and m(g,K»O. Proof· Let r be the diameter of K, which is posItIve since m(f,B»m(f,K). Choose b in (O,2r/3) so that KnB3b is compact and If(z)1 >lm(f, B) for all Z in K n B 3o . Choose a positive integer v so that (r- 3 b/2)V(r- b)- v IlfilK
158

Chapter 5

Complex Analysis

m(f, B»O, we have P(Zl' B»O. By (4.14), the function ZI-+ (Z-Zl)-l!(Z) on S(Zl,P(Zl,B))-{zd extends to a differentiable function on S(Zl' p(Z l' B)). Hence there is a differentiable function g 1 on K such that !(Z)=(Z-Zl)gl(Z) (zEK).

Since p(zl,B»O and m(f,B»O, we have m(gl,B»O. Thus either m(gl,K)=O or m(gl,K)=m(gl,B»O. In the latter case, we take g=gl and are finished. Otherwise, there exist Z2 in K and a differentiable function g2 on K with !(z)=(Z-Zl)(Z-Z2)g2(Z) (zEK) and m(g2,B»0 Repeat this process recursively, so that, unless the process stops at some previous stage, we obtain at the nth stage points Zl' ... 'Zn of K and a differentiable function gn on K with !(Z) = (z - Zl) ... (z - zn) gn(z)

(ZE K)

and m(gn' B) > O. If m(gn' K) > 0, then the process stops and we are through. Otherwise, m(gn' K) = 0, and we continue to the (n + l)th stage. Either the process proceeds for N stages or it terminates at some stage n < N. Consider the former case. By the choice of N, there exist v of the points Zl' ... ' ZN whose mutual distances are less than fJ. Call these points w1 , ... , w Then V •

(5.11.1)

f(Z)=(Z-W 1 )

•••

(z-w.) h(z)

(zEK)

for some differentiable function h on K. Since !(w1)=0 and 1!(z)1 >~m(f, B) for all Z in K nB36' we have ro=p(wl,B»3fJ.

Hence, by (2.3), there exists, in B36 with (5.112)

"-w11
Since B is a border for K, we have (5.11.3)

Since r;;;;ro>3fJ, we have (5.11.4)

(r - 3fJ/2) (r - fJ)-l - (ro - 3fJ/2) (ro _ fJ)-l =~fJ(r -

As Iwi-w11
(l~i~v),

fJ)-l (ro - fJ)-l (r - ro);;;; O.

it follows from (5.11.1)-(5.114) and (5.4) that

I!ml ~(" -wll+fJ)V(ro-fJ)-V II!IIK ~(ro-3fJ/2)V(ro-fJ)-V II!IIK ~(r- 3fJ/2)V (r- fJ)-V IlfIIK<~m(f, B).

5. Estimates of Size, and the Location of Zeros

159

This contradicts the fact that CEB 3d , and rules out the possibility that the process continuous for N stages. Therefore the process stops at some stage n < N, and the requirements of the theorem are satisfied by the function g == gn and the associated points Zl' ... , Zn· 0 (5.12) Lemma. Let f be a nonzero differentiable function on a connected open set U c
D

We now have an important corollary of Theorem (5.11). (5.13) Theorem. Let f be a nonzero differentiable function on a connected open set U c
e > o. Then either m(f, K) > 0 or there exist finitely many points Zl' ... , Zn of U and a differentiable function g on U such that

(5.13.1) for each

Z

in U, m(g,K»O, and p(zi,K)<efor eachj.

Proof: Choose r so that 0
m(f, r(Ci , s)) > 0 N

Let K' be the closure of the set USc(Ci,s). Then KcK', K' is a i=l N

compact subset of K,/3' and B' ==

U r(Ci , s)

is a border for K' with

i=l

m(f,B'»O. Either m(f,K'»O, in which case m(f,K»O, or m(f, K') < m(f, B') In the latter case, by (5.11), there exist points Zl' ... , Z. of K' and a differentiable function g on K' such that (5.13.1) holds for each Z in K', and meg, K'»O. By (4.14), g extends to a differentiable function on U such that (5.13.1) holds for all Z in U.

160

Since

Chapter 5

K cK',

p(zj,K)~r/3
Complex Analysis

meg, K»O. Finally, for eachj. 0

since

K' c K r/ 3 ,

we

have

In the above proof, the connectedness of U is used only to establish that f is nonzero on each of the spheres SC('k' r). It follows that the conclusion of Theorem (S.13) will hold even if U is not connected, provided that f does not vanish on K. (S.14) Proposition. Let f be a differentiable function on an open set U c 0, and f- 1 is differentiable on K.

Proof: By the above remark, either m(f, K) > 0 or there exist finitely many points Zl' ... , Z. of U and a differentiable function g on U such that fez) = (z - Zl) ... (z - zn) g(z) (ZE U) and OO for each follow. 0

Z

Z

in K

l~j~n}.

in K. The desired conclusions now

(S.1S) Corollary. If f is a nonvanishing differentiable function on an open set U c
Proof This is an immediate consequence of (S.14).

0

Corollary (S.1S) enables us to introduce an important type of differentiable function. Let U be a simply connected open subset of
Z

in U, g(z)= If'COf(O-ld'+a,

where y is any path in U joining Zo to z. The function g is called a branch of the logarithm of f on U. The following is another useful consequence of Theorem (S.13).

5 Estimates of Size, and the Location of Zeros

161

(5.16) Proposition. Let f be a nonzero differentiable function on an open sphere S == S(za, r) c o. Then there exists t' > 0 such that m(f, r(za, t')) > 0 and It - t'l < E. Proof. In view of (5.13), we may assume that there exist points Zl' ... , zn of S and a differentiable function g on S such that f(z) = (z - Zl) ... (z -zn) g(z)

(ZES)

and O
Let t' be any real number with t-E
(1~j~n)

n

If(z)1 ~ c

=c

TI (Iz -

j=l

zal-Izj- zal)

TI (t' -Izj- zal)·

j=l

Hence

n

m(f, r(za, t'))~c

TI (t' -lzj-zol»O.

D

j=l

A mapping f: A -+
Z

and z'

(5.17) Theorem. Let f be a nonconstant differentiable function on a connected open set U c
for each compact set K well contained in U.

Proof We first prove that f(K)~ f(U) whenever K == S c(za, r) is an arbitrary closed sphere well contained in U. Let ')I: [0,1] -+
f(')I(1)) - f(')I(O)) = Sf'(')I(t)) ')I'(t) dt,

°

f' is nonzero on U. It follows from (5.12) and (5 16) that m(f',r(za,t))>O for values of t arbitrarily close to r. Increasing r if

necessary, we may therefore assume that O
where B==r(zo,r). Choose R>r so that in (0, R - r) so that

Sc(za,R)~

U, and compute b

If(Z2) - f(Zl)- f'(Zl)(Z2 -zl)1 ~!CIZl - z21

162

Chapter 5 Complex Analysis

whenever ZlEB and IZ1-z21~!5 Let Cbe an arbitrary complex number with p(C,f(K)) < c!5/8. Either p(C feB)) > c!5/8 or p(C,f(B)) < c!5/4. In the former case we have m(f-C,B»m(f-C,K), so that f(z)-C=O for some z in K, and therefore CEf(U). In case p(C,f(B))
> c!5 -tc IZl -z21-c!5/4 = c!5/4. Hence m(f -

Cr(Zl' !5)) ~ c!5/4 > If(zl)- CI ~ m(f - CS C(Zl' !5)).

It follows that f(z)-C=O for some z in SC(Zl,!5), so that CEf(U) in this case also. As C is arbitrary, we see that (f(K))cb/8 c feU). Hence f(K)~f(U).

It now follows that feU) is open and that f(K)~ feU) for each compact set K~ U. To prove that feU) is connected, consider any two points Wi =: f(zl) and w2=: f(z2) in feU). Let '}I be a path in U joining Zl and Z2' Then by the foregoing, fa '}I is a path in feU) joining Wi and

w2 •

0

(5.18) Corollary. Let f be a nonconstant differentiable function on a connected open set U c IlfiIK' Proof By (5.17), there exists r>O such that wEf(U) whenever p(w,f(K)) IlfII K-r/4. Either IWolO. In the first case let w=:r/2, and in the second let w=:(1+irlw ol- 1)w o. Then in each case we have Iw-wol IlfiIK'

Choose C in U with f(C) = w. Then If(C)I> II !II K' By continuity, we also have CEU-K. D (5.19) Corollary. Let f be a nonconstant differentiable function on S(O,l)c II f II T(r)" By the maximum principle (5.2). we have IIflll(R)= IIfllsc(o.R)~ If(Ol > Ilfllr(r)'

D

6. Singulanties and Picard's Theorem

163

6. Singularities and Picard's Theorem In this section we prove two constructive versions of a famous theorem of Picard concerning the range of a differentiable function on the punctured disk {ZE
Zoo

Then

Proof: Consider such z, t 1, and t 2. By (4.14), there is a differentiable function g on A such that g(O=(f(O- f(z»«( _Z)-l

«(EA-{z}).

By Cauchy's integral theorem (3.12), we have 0=

S g(Od( - S g(Od( y(t2)

y(t,)

= S f(o(' _Z)-l d, -2nif(z)j(y(t2)'z) y(t2)

y(t,)

The result now follows becausej(y(t 1),z)=0 andj(y(t 2 ),z)=1.

0

Our next theorem generalizes Taylor's theorem. (6.2) Theorem. Let f be differentiable on an open annulus A==: {ZE
g(z) ==:

L an(z -

zo)", uniformly convergent on each compact set well con-

n=O

00

tained in S(zo,r2), and a power series h(z)==:

L bn(z-zo)-n in (Z-ZO)-l, n~l

uniformly convergent on each compact set well contained in {z: Iz-zol >r1}, such that f=g+h. Moreover, the power series g, h with these properties are unique, and for each r in (r1' r2), (6.2.1)

a n=(2ni)-1

S1(0«( -zo)-n-1 d( y(r)

164

Chapter 5 Complex Analysis

and bn=(2ni)-1 S f(O«( -ZO)"-I dC y(r)

(6.2.2)

where y(r) is the circular path of radius r about Zo Proof. Fix r in (r1' r2). In view of (6.1), to prove the existence of g and h it will suffice to prove that if r1 < tl < t2 < r2, then the series 00

(6.2.3)

g(z)=

L an(z-zo)" n=O

with coefficients an given by (62.1) converges to (2ni)-1 S f«()«( _Z)-I d( y(t2)

(6.2.4)

uniformly on each compact set well contained in S(zo, t 2 ); and that the series 00 (6 2 5) h(z)= L bn(z-zo)-n n=1

with coefficients bn given by (6.2.2), converges to -(2ni)-1 S f(O«(-Z)-ld( yet,)

(6.2.6)

uniformly on each compact set well contained in T= {ZE tl}' The first of these facts is established by applying (4.15) to the mapping zH(2ni)-1 S f(O«(-Z)-ldC y(t2) which, by (4.8) and (3.12), is differentiable on - car y(t 2 ) and has derivative at Zo given by (2ni)-1 n! S f«()« -zo)-n-l d(=(2ni)-1 n! S f«()«( -zo)-n-l d(. y(t2) y(r) On the other hand, iflz-zol>t1 and NE7L+, then N

«( _Z)-l =«( _Z)-l«( -ZO)N(Z-ZO)-N -

L (z-zo)-n«( -ZO)"-1

whenever I( - zol = t 1 ; so that N

I S f(0«( _Z)-1 d( + L ( S f(O«( -zo)"-l d()(z-zo)-nl yet,)

n= 1

y(tj)

=I(Z-Zo)-N S f(0«(-Z)-1«(-ZO)Nd(l

yet,) ~ 2n IIf IIr(zo.t,)t 1(lz - zol- t1)-1(tdlz - ZOI)N.

nth

6 Singulanties and Picard's Theorem

165

As N --> CXJ this last expression converges to 0 uniformly on each compact set K ~ T. It follows from (3.12) that the series (6.2.5), with coefficients bn given by (6.2.2), converges to the expression (62.6) uniformly on K Conversely, if co

co

converges uniformly to f on each compact set well contained in A, then for each integer m we have y(r) 00

00

=

L ak I ('-zot+m-1d,+ L bk I ('-zo)-k+m-1d,. k~O

k~1

y(r)

y(r)

For each nonzero integer j. the derivative of the map 'Hr1(,-Zo}' on -car'}'(r) is 'H('-zo)i-1.1t follows from (3.4) that

I f(0('-zot-

1

y(r)

d,={2ni)a_ m =(2ni)bm

if m~O, if m>O.

Hence the coefficients an' b nare given by (62.1) and (622) respectively

0

In Theorem (6.2) the series 00

ex,

n~O

n~1

L an(z-zot+ L bn(z-zo)-n, with coefficients given by (6.2.1) and (6.2.2), is called the Laurent series, or Laurent expansion, of f in A. It is convenient to write a_n=b n for each n in lL+, so that the Laurent series of f is 00

n= -

00

with coefficients an given by an={2ni)-1

I f{O(' -zo)-n-l d,

(nElL).

y(r)

Theorem (6.2) and these remarks apply with the obvious modifications when f is a differentiable function on an unbounded open annulus A={z: r1
166

Chapter 5 Complex Analysis

Choose r>O so that S(zo,r)c U, and let f have Laurent serIes 00

I

n= -

an(z-zot in the annulus A::{z: O
integer We say that f has a pole of order at most vat Zo if an=O for all n< -v; if also a_v =1=0, then we say that f has a pole of determinate order at zo' and that the order of this pole is v. On the other hand, if a_ n =1= 0 for infinitely many positive integers n, then we say that f has an essential singularity at ZOo Note that if v~O and f has a pole of order at most v at zO' then Zo is a removable singularity of f, by (6.2) and (412). (6.4) Lemma. Let U c
suppose that rand

rJ.

exist with the given properties. Let

-zo)" be the Laurent expansion of integer n and each t in (0, r), we have

f

I

n= -

an(z 00

in A. Then for each negative

lanl = 1(2ni)-1 Jf«()«( -zo)-n-1 d(l ~rJ.t-n, where y is the circular path of radius t about Zo. Letting t ~ 0, we see that an = 0 Thus f has a pole of order at most 0 at ZO. 0 We say that a map f: A f(z)=!= ()( for all z in A.

~


rJ.,

where ()(E
(6.5) Lemma. Let U c
Proof. Let

I

n= -

an(z - zo)n be the Laurent expansion of f in an an00

nulus A::{z: OO. It clearly suffices to prove that a_v=O or a_v=l=O. The map zt-+(z-zoY(f(z)-rJ.) extends

6 Singulanties and Picard's Theorem

167

to a differentiable mapping g on S(zo, r). Since g(zo +~rH= 0, we see from (513) that either m(g,Sc(zo,~r))>O, in which case a_v=g(zoH=O, or there exists, in S(zo, r) with g(O = O. In the latter case, were' =1= Zo we would have f(O=(' -zo)-V g(')+a=a, a contradiction. Hence '=zo, and a_v=g(zo)=O.

0

(6.6) Lemma. Let U c
Proof' Let

I

n= -

an(z-zo)" be the Laurent expansion of f in an an00

nulus A={z: OO, in which case f has a pole of order o at zo; or there exist points Z1' ""Zn of S(zo,r) and a differentiable function g on S(zo, r) such that f(z)=(Z-Z1)'" (z-zn)g(z)

(ZES(zo,r))

and m(g,Sc(zo,~r))>O. In the latter case, as f is nonvanishing in A, we must have Z1 = ... =zn=zO; whence f has a pole of determinate order -n at zoo 0 (6.7) Lemma. Let f be a nonvanishing differentiable function on an open annulus A={ZE
Proof By (66), f- 1 has a pole of determinate order at ZOo Hence there is an integer N and a nonvanishing differentiable function g on S(zo,r) such that f- 1(z)=(z-zot g(z) whenever zEA. Then g-1 is differentiable on S(zo, r), by (5.15), and 00

f(Z)=(Z-ZO)-N

I

bn(z-zo)"

(zEA),

00

where

I

bn(z-zot is the Taylor expansion of g-1 in S(zo,r). Since b o

n~O

=g(ZO)-1 =1=0, f has a pole of determinate order Nat ZOo Now /,(z)f(Z)-1 = - N(z -

ZO)-1

+ g'(z) g(Z)-1

(zEA).

168

Chapter 5 Complex Analysis

Since g' g-I is analytic on S(zo, r), it follows that j(foy,0)=(2ni)-1 Sf'(z)f(z)-Idz = -N(2ni)-1 S(z-zo)-Idz=-N andsoN=-j(foy,O)

D

In order to discuss the behaviour of a differentiable function in the neighborhood of an essential singularity, we need some information about entire functions. We say that an entire function g is of irifinite degree if the function Zf--+g(Z-I) has an essential singularity at O. This is equivalent to the condition that for any polynomial p, the function g - p IS nonzero. (6.8) Lemma. Let g be a nonconstant entire function, and p a polynomial. Then there exists' in
Proof Let p(z) =

L: Pnzn,

where N is a positive integer. Choose r > 0

n= 0

so that m(g,r(r))>O. Either Ilpllr(r)<m(g,r(r))-l, in which case we need only choose' so that I(I=r and IlpllT(r) O. Thus there exists j with Pj =1= O. If j = 0, choose' so that g(O =1= 0 and g(O=l=POI. Then either

jJ/n,nj < Ig(O-1 - Pol and therefore g(O-1 =l=p(O, or o<jnt/n,nj. It follows that we can take j~l.

Now let z be any zero of p. IfO<m(g,Sc(z,l)),then we can take ,=z. On the other hand, if m(g,Sc(z, l))
(6.9) Lemma. Let g be an entire function of irifinite degree, and let r, e be positive numbers Then there exists R > r such that either m(g,r(R))<e or g(z)=O for some z with r
Proof By (5.13), there exist an integer N ~ 0, a polynomial p of degree N, and an entire function h, such that all the zeros of p belong to S(O,r+l), m(h,Sc(O,r))>O, and g=ph Clearly, h is of infinite degree 00

Let

L bnz"

be the Taylor expansion of h- I in S(O, r). According to N

n~O

(6.8), we can find ( such that h(O=l=O and h«()-I

=1=

L bn(n. n~O

By (5.13)

6 Singulanties and Picard's Theorem

169

and (5.15), h- 1 is defined and differentiable on some connected open N

set A with (EA and S(O,r)cA, By (5.12), the mapz~h(z)-1- Ibnz n n~

0

of S(O, r) into N with bj =1= 0 Since p has degree N, we can choose R > r so that IlpllT(R)<eRjlb) and m(h,r(R))>O. Either m(h,Sc(O,R))>O or m(h, S dO, R)) < m(h,r(R)). In the former case, for all s in (0, R) we have Ibjl = 1(211: i)-1 Sh(Z)-1 z-j-1 dzl y

~ s- j

Ilh -111 T(s) = s- jm(h, r(S))-I,

where y is the circular path of radius s about O. By continuity, Ibjl~R-jm(h,r(R))-I. Thus m(g, r(R)) ~ Ilpll T(R) m(h, r(R)) ~ IlpIIT(R) Ib)-l R-j<e.

On the other hand, if m(h, S c(O, R)) < m(h,r(R)), then h has a zero z in S c(O, R). Clearly, r < Izl < R, and z is a zero of g. D We now arrive at the Casorati- Weierstrass theorem. (6.10) Theorem. Let f be differentiable in the open annulus A:: {ZEO. Then there exists z in
Proof' We may take both Zo and ( equal to O. Let

I H= -

Laurent expansion of f in A, and choose t > r- 1 so that

anz n be the 00

In~l anznl < el2

whenever Izl
I

g(z)::

a_nz n is an entire function. Clearly, g is of infinite degree. By

n~O

(6.9),

Iz I<

t-

there exists z such that Iz- 1 1>t and Ig(z-I)I<eI2. Thus 1 < rand

D A remarkable classical theorem of Picard states that under the hypotheses of Theorem (6.10), f takes every complex value, with at most one exception. Before proving constructive versions of Picard's theorem, we need a succession of auxiliary results, some of which are of considerable interest in their own right. The first of these yields an inequality known as Caratheodory's inequality.

170

Chapter 5 Complex Analysis

(6.11) Proposition. Let f be differentiable on the closed sphere Sc(O,R) c
A(r) =:= sup {Ref(z): Izl=r} Then

R+r 2r IlfIIT(r)~ R -r If(O)1 + R -r A(R)

(6.11.1) for each r in (0, R).

Proof; Applying the maximum principle (5.2) to the differentiable function z r-+exp (f(z)), we see that Ref(z)~A(r)

(lzl~r~R).

Consider first the case f(0) = O. Then A(R) ~ A(O) = O. Let E be an arbitrary positive number, and define a differentiable function g on S=:= Sc(O, R) by f(z)

g(z) =:= 2(A(R) + E) _ f(z)

(ZES).

Then g(O)=O, and Ig(z)1 < 1 for all z in S. By (4.14), there is a differentiable function ¢ on S such that g(z) = z¢(z) for all z in S. Hence, by Schwarz's lemma, if Izl=r
E

is arbitrary, it follows that

(6.11.2)

2r

Ilfllr(r)~ R-r A(R)

(O~r
when f(O)=O. In the general case, we apply (6.11.2) to the function f - f(O), to obtain 2r Ilfllr(r)~ R_r(A(R)+ 1f(O) I) + If(O)1 (O~r
0

1)

2X+ (6.12) Lemma. If x~ 1, then (4x 3)-1
(6.12.1) We have

00

exp«4x3)-1)~ 1

+

2: (4x n=1

3

)-n= 1 +(4x 3 _1)-1.

6 Singulanties and Picard's Theorem

171

Now the function g defined by g(t):=4t 3 -2t-1 has positive derivative throughout [1, (0). Hence g(x)~g(I)= 1, and therefore (4x 3 -l)-1«2x)-1. Inequality (6.12.1) now follows. 0 (6.13) Lemma. Let f be a non vanishing differentiable mapping on the open sphere S(O,l)c2. Suppose that lX-l~lf(O)I~1X and lX-l~lf(O)-n Let g be a branch of the logarithm of f on S(O, 1) such that Ilmg(O)1 ~n+ 1. Then lIn (lg(O)1)1 ~ 3 In (IX) + 2ln (2). Proof: First observe the following facts: 1-cosx~lxl

(6.13.1) (6.13.2)

(4x 3 )-1
(XEJR)

e~: 1 )
In(x)+n+ 1 ~4X3

(6.13.3)

(x~2).

Inequalities (6.13.1) and (6.13.3) are proved by elementary differential calculus; inequality (6.13.2) follows from (6.12) and properties of the logarithmic function. For convenience write,:= f(O). Then IX- 2 ~ I' _112 = 1'12 - 21'1 cos (1m g(O»+ 1

= (1- 1'1)2 + 21'1 (1- cos (1m g(O») ~(1-IW2+21X1Img(O)I,

by (6.13.1). Either (1-1'1)2 «21X2)-1 or (41X2)-1«1-IW2. In the former case, Ig(O)1 ~ IImg(O)1 ~ (21X)-1(1X- 2 _ (1- 1'1)2) > (41X 3 )-1. In the latter, 1(1-IWI>(21X)-1 and so either "I >(21X+ 1)/2 IX or "1< (21X-l)/21X. Hence, by (6.13.2), Ig(O)1

~ Iln(lml~min {In e~: 1), In (2:: 1 )}>(41X )-1. 3

Thus in both cases, (41X 3 )-1 ~ Ig(O)1 ~ lIn (IWI + 11m g(O)1 ~ln(lX) + ~41X3,

n+ 1

(since 1X- 1 ~ "I ~ IX)

172

Chapter 5 Complex Analysis

by (6.13.3). Hence

lIn (lg(O)1)1 ~ In (40(3) = 3In (0() + 2In (2)

0

(6.14) Lemma. Let cjJ be a continuous map of [0,1) into IR 0+, and let K

be a positive constant, such that Then (6.14.1)

for each r in [0, 1). Proof. First consider the case where cjJ is bounded by some constant M>O. Let O~r=ro
~

- ro)- 2(K cjJ(r2)t(r2 - r1)- 2)t

...

(6.14.2)

where n

0(=

~)2-i+l i~l

and

n

f3=

L2- i +

1•

i~l

Now n

0(=

L 2-i + i~l

n-1

1

+t

Lj2- i +

1

i~l

= f3 +t(X-tn2- n +1 and so (X=2f3-n2- n + 1 • Since f3-+2 as n-+oo, we see that (X-+4 as n -+ 00. We now obtain (6.14.1) by letting n -+ 00 in (6.14.2).

6 Singulanties and Picard's Theorem

173

Now consider the general case. Let t be an arbitrary number in (0,1), and define (O~r<

t/I(r)=¢(tr)

Then for

0~r
1).

we have

o~ t/I(r) ~ t- 2 K t/I(R)t(R -

r)- 2

and t/I(r)~sup{¢(x): O~x~t}.

By the first part of the proof, we have t/I(r)~28t-4

K 2 (1-r)-4

for each r in [0,1). We now obtain (6.14.1) by letting t -+ 1.

0

(6.15) Definition. A differentiable mapping which omits the values 0 and 1 on an open set U c
The omission of the values 0 and 1 by a differentiable function on S(O, 1) leads to a remarkable restriction on the growth of the function. (6.16) Lemma. Let oc > 2, and let f be a Picard function on S(O, 1) such that max {If(O)I, 11-f(O)l}oc- 1. Then

If(z)1 ~exp(218e18 oc 8(1_r)-4) whenever Izl ~ r < 1. Proof: Since both f and consider branches of their and gl be branches of the respectively, on S(O, 1) such (6.16.1)

1 - fare non vanishing, it is natural to logarithms on S(0,1). Accordingly, let go logarithm of f and the logarithm of 1 - f, that IImgk(O)1 ~n+ 1 for k=O, 1. Then

Igo(O)1 ~ lIn (If(O)1)1 + 11m go(O)1 ~ln (oc)+ n + 1

and similarly, (6.16.2)

For each differentiable function ¢ on S(O, 1) define A(¢,r)=sup{Re¢(z): Izl=r}

(0~r<1).

M(r)=max{llgoll rw Ilg11Ir(r)}

(O~r
Define also Let r
174

Chapter 5 Complex Analysis

Either 1 < A( - go, t) or A( - go' t) < 4n. In the former case, let e be an arbitrary number with 0<e<1, and compute C so that ICI=t and (6.16.4)

max {1, A( - go, t)-e} < Re (- go(C)= In (If(C)I- 1).

Then If(C)I<e- 1
- L k- 1 Zk. Hence if k=l

00

v == (2 ni)-l (gl (C) +

k~l k- 1 f(C)k) ,

then exp(2niv)=1, and so v is an integer. We have (6.16.5)

It follows from this and the maximum principle that

(6.16.6) Since 1- f omits the value 1, gl omits the value 2niv. Hence there exist branches of the logarithm of gl -2niv on S(O,1). Let h be such a branch with IImh(O)I~n+e. By Carathi:odory's inequality, (6.16.7)

2t 2R II hllr(t) ~- A(h, R) +-lh(O)I· R-t R-t

Now, by (6.16.5) and (6.16.4), we have (6.16.8)

Ilhll ret) ~ Ih(C)1 ~ In (lgl(C) - 2nivl- 1 )

~ln (l~llf(Olkrl) ~In«2If(C)I)-l) > A( - go, t)-e-In (2). On the other hand, we have (6.16.9)

A(h, R)= sup {In (lgl(Z)- 2nivl): Izl = R} ~In(llglllr(R) +2n Ivl)
by (6.16.6). Also, if v =1= 0, then

6. Singulanties and Picard's Theorem

175

and so, by (6.16.6), (6.16.10)

Ih(O)1 ~ In (Igl (0) - 2 nivl) + n + 8 ~In (Igl (0)1

+ 2nlvl)+ n + 8

~ In (Igl (0)1

+ 1 + IlgIIIT(R)) + n + 8 + lin (Igl (0)1)1·

This inequality also holds if v = 0, in which case Ih(O)1 ~ lin (Igl (0)1)1 + n + 8.

From (6.16.7)-(6.16.10) we now have A( -go,t)~ Il h llT(t)+8+ln(2)

2t

~ R _ tin (21Ig111T(R) + 1)

2R

+ R -t (In (Igl (0)1 + 1 + IlgIIIT(R))+ n+8

+ lin (Ig 1(0)1)1) + 8 + In (2) and so 4R

A( - go, t) ~ R -t (In(21Ig 111 T(R) + 1 + Ig 1(0)1) + n + 28 + In(2) + IIn(lg 1(0)1)1).

Clearly, this last inequality also holds in the case where A( -go' t) < 4 n. Since 8 is arbitrary, it now follows that in both cases, (6.16.11)

4R A ( - go' t) ~ R _ t (In (2 II gIll T(R) + 1 + Ig 1(0)1) + n

+ In (2) + lIn (Igl (0)1)1). From (6.16.3) and (6.16.11) we now obtain 8Rt Ilgo IIT(r) ~ (R _ t)(t -r) (In (211g IIIT(R) + 1 + Ig 1(0)1) + n

+ In (2) + lIn (Ig 1(0)1)1 + Igo(O)I) ~ 32(R _r)-2 (In (2M(R)+

1 + Igo(O)1 + Igl (0)1)+ n

+ In (2) + lIn (Ig o(O)!)I + lIn (Ig 1(0)1)1 + Ig o(0)1 + Ig 1(0)1) ~ 32(R _r)-2 (In (4M(R) + 4)+ n+ In (2) + lIn (lgo(O)1)1 + lIn (Ig 1(0)1)1 + Igo(O)1 + Ig 1(0)1), the last inequality following from the maximum principle. Interchanging the roles of f and 1 -fin the foregoing, and referring to (6.16.1),

176

Chapter 5 Complex Analysis

(6.16.2), and (6.13), we see that

M(r) ~ 32(R -r)- 2 (In (M(R) + 1) + 11: + 3ln (2) + lIn (lgo(O)I)1 + lIn (Ig 1(O)!)I + Igo(O)1 + Ig 1(0)1) ~ 32(R -

r)- 2(ln (M(R) + 1)+ 8ln (ex) + 7ln (2) + 311: + 2)

< 32(R - r)- 2(ln (M(R) + 1)+ 8ln (ex) + 17). Hence

M(r) + 1 < 32(R - r)- 2(ln (M(R) + 1)+ 8ln (ex) + 18) =32(R - r)- 2 In (e 18 ex 8 (M(R) + 1». Since In (x) ~ xt for all positive numbers x (the reader may prove this as a simple exercise in calculus), it follows that

M(r) + 1 <2 5 e 9 ex 4(R-r)-2(M(R)+ l)t

(O~r
1).

Hence, by (6 14),

°

M(r) < M(r) + 1 ~218e18(X8(I_r)-4

whenever ~ r < 1. For such r and all z in S c(O, r) it follows from the maximum principle that If(z)1 ~ II f II T(r) ~ eM(r) ~ exp (2 18 e 18 ex 8(1- r)- 4). (6.17) Lemma. Let

°<

0

ICI = r < 1, define h: S(O, 1)-+
h(z)=(z-cm*z-I)-l

(zES(O,I»,

and let t=2r/(1 +r2). Then h is a differentiable bijection of S(O, 1) onto itself with inverse h, and S c(O, r) c h(S c(O, t».

°

Proof: Since ICI < 1, the denominator of h(z) is bounded away from on S(O, 1), so that h is differentiable. A simple computation shows that h is a bijection of S(O, 1) onto itself with inverse h. By the maximum principle, Ih(z)I~llhllr(r) (ZESc(O,r». Write C=rei~, and consider an arbitrary z=re ifJ in r(r). We have Ih(z)12 = (z - O(z* - C*)(C* z _1)-l(C z* _1)-1

= (z z* + CC* - 2 Re C* z)(CC* z z* + 1- 2 Re C* Z)-l = (2r2 - 2r2 cos (ex - /3»(r4 + 1- 2r2 cos (ex - /3»-1. It is now a straightforward exercise in calculus to show that II h II T(r) = t. Hence

S c(O, r) = h(h(S c(O, r») c h(S c(O, t».

0

6 Singularities and Picard's Theorem

177

For the remainder of this section let tfJ be the continuous map of 1R + x [0,1) into 1R + defined by tfJ(oc, r) = 1 + exp (21B e lB (oc + 3)B (1 + r 2 )4(I-r)- B)

(6.18)

(oc>O,

O~r<

1).

Note that for each r in [0,1) the map oct-+tfJ(oc,r) is strictly increasing on 1R+; and that for each oc>O the map rt-+cI>(oc,r) is strictly increasing on [0, 1). The following result is known as Schottky's theorem. (6.19) Theorem. Let f be a Picard function on S(O, 1), and oc a positive number with If(O)1 ~ oc. Then (6.19.1)

If(z)1 ~ tfJ(oc, Izl)

(zES(O,I)).

Proof Either min{lf(0)1,ll-f(0)1}>(oc+3)-1 or min {If(O)I, 11-f(0)1}
(6.19.2)

min {II f II F(r» 111 - f II r(r)} < cI>(oc, r) -1.

Clearly, we may assume that min {II f II F(r» 111- f II F(rJ > 2. To begin with, take 1f(0)1 If(O)1 or m(f, r(r))
(wES(O,l))

defines a differentiable map of S(O,I) onto itself with h(O) = ,. Thus F = f h is a Picard function on S(O,I) with max {IF(O)I, 11- F(O)I} < 3 and min {IF(O)I, 11-F(0)1}>1/3. Write t=2r/(1+r2). Then by (6.17) and (6.16), 0

II f II F(r) ~ sup {If(z)l: zEh(S c(O, t))}

= IIFIIr(t) ~cI>(oc,r)-1.

Thus (6.19.2) holds On the other hand, if 11- f(O)1
=

178

Chapter 5 Complex Analysis

takes at least one of the values 0, 1 in the annulus B r/2 < Izi < 3r/2}.

={z E
Proof' For each t>O define "1(t)=cP(2t,1) and "n+ 1(t)= cP("n(t),1)

(nEZ+).

"n

Then each is an increasing continuous map of IR + into IR +. Let Let O
"="52'

f is nonconstant, and so 1 - f is nonzero. Consider first the case r = 1. By (5.13), there exist r1, r 2 such that 1/4 O.

(6.20.1)

Choose' so that 1'1 =1 and If(')1 < 2m(f, r(1)), and define zn=,exp(nni/26)

(n=0,1, ... ,52).

Since (6.20.2)

IZn+ 1 - znl < n/52 < 1/16

(n= 0,1, ... ,51),

we have 51

(6.20.3)

r(1) c

U0 S(Zn' 1/16).

n~

For 0~n~51 the map wl-+f(zn+w/8) is a Picard function gn on S(0,1), because S(z.,1/8)cK and we are assuming (6.20.1). Thus if IX> 0, If(zn)1 ~ IX, and ZES c(zn' 1/16), then If(z)1 = Ig.(8(z - z.))I ~ cP(lX, 81z - znl) ~ cP(IX,1)' Since If(zo)1 < 2m(f, r(1)) and (6.20.2) holds, it readily follows by induction that (6.20.4)

As

If(z)1 ~ "n+ 1(m(f, r(1)))

"n ~ ".+

(ZES c(zn' 1/16), n =0,1, ... ,51).

for each n, it follows from (6.20.3) and (6.20.4) that This contradiction shows that (6.20.1) is ruled out. Hence either 1

Ilfllrtt)~"(m(f,r(1))).

6 Singularities and Picard's Theorem

179

or m(l-f, K) < min {m(l -f, r(r1)), m(l-f, r(r2 ))}·

Thus by (5.3) and (5.11), I takes at least one of the values 0, 1 in K. Since K c B, the desired conclusion follows. In case r is an arbitrary number in (O,t], the conclusion follows by applying the foregoing argument to the function Z 1-+1(2r z) on A. 0 We now prove a first constructive version of Picard's theorem. (6.21) Theorem. II I is a Picard lunction on an open annulus A == {ZE
00

Consider first the case v ~ O. Let

L

anz n be the Laurent expansion

n= -00

of I in A. Let n < - v be an integer, and suppose that an =1=0. For each s in (0, t] we have la nl=I(2n:i)-1

Jl(z)z-n- 1 dzl y.

:;;; s-n I flins):;;; s-n ~(m(f, res))), where

~

is the mapping introduced in (6.20). Since n < 0, it follows that ~(m(f,r(s)))~snlanl--+oo

as s--+O.

Choose So in (O,t] so that ~(m(f,r(s))»~(l) for all s in (O,so)' For such s, since ~ is an increasing function, we have m(f, res)) ~ 1, and therefore I I -111 nS):;;; 1. Hence II -1(z)1 :;;; 1 whenever 0 < Izl :;;; so' By (6.4), 1-1 has a pole of order at most 0 at O. Hence, by (6.7), I has a pole of determinate order v at O. Since n < - v, we must have an = 0, a contradiction. Thus, in fact, an = 0 for all n < - v, and so I has a pole at O. By (6.6), this pole is of determinate order. Now consider the case v
180

Chapter 5

Complex Analysis

(6.22) Theorem. Let f be differentiable on an open annulus A=: {ZE
'I.

Proof· We may take zo=O and r= 1. Moreover, replacing f by the

map Zf-+(,'-O-l(f(z)-O, 00

we may assume that' =0 and "= 1. Let

L n= -

anz n be the Laurent 00

expansion of f in A, and choose N ~ - 1 with aN =1= O. By the CasoratiWeierstrass theorem, f is nonzero, and there exists s such that

o<s < min H, (b(l)-llaNI)-l/N} and m(f,r(s))O With "Is the circular path of radius s about 0, we have laNI= 1(2ni)-1

Jf(Z)Z-N-l dzl ~S-N Ilfllr(s).

y.

Thus, as b is increasing and N < 0, II f II 1(s);;;; SN laNI > b(l);;;; b(m(f, r(s))).

We now see from (6.20) that f takes at least one of the values 0, 1 in the annulus {ZE
7. The Riemann Mapping Theorem In this section we prove the beautiful theorem of Riemann, that under very general conditions any two simply connected open sets in
f(z)=(2n)-1

.

Jo f(zo+Re'
(R2-r2)d¢ 2 R (8 ¢)

r

cos

-

+r

2·

Proof: It is enough to consider the case Zo = 0 and R = 1. Let "I be the circular path of radius 1 about the point O. By Cauchy's integral

7 The Riemann Mapping Theorem

formula (4.7) and a simple continuity argument, for

181

Izl=r
f(z)=(271:0- 1 Jf(O«( _Z)-1 d(

(7.1.1 )

=(271:)-1 Jf«()(( _z)-1(i()-1 de· Similarly, by Cauchy's integral theorem (3.12), O=(2ni)-1 Jf(()z*(1-z*()-1 d(

(7.1.2)

=(2n)-1 Jf«()z*«(* _z*)-1(i()-1 de. y

Addition of (7.1.1) and (7.1.2) gives f(z)=(2n)-1 {f«()

t =::; -z(!

Writing (= e i 4> and z = re i9 , where

+zz*) (i()-1 de.

e, ¢EIR, we get

which is the required formula in the case

Zo

= 0 and R = 1.

0

A consequence of Proposition (7.1) is another bound for an analytic function inside a circle. This should be compared with the bound given by Caratheodory's inequality (6.11). (7.2) Proposition. If f is differentiable on a closed sphere S=Sc(zo,r), and if Iz-zol=r
R-r

If(z)1 ~lf(zo)1 R+r + Ilflls R+r

(7.2.1)

Proof. There is no loss of generality in taking z in the form z=re i9 , and let and Then

2"

By (7.1), (2n)-1

J Ad¢=1. Also, o

Zo

= 0 and R = 1. Write 1-r l+r

B=--.

182

Chapter 5 Complex Analysis

2 n If(z)1 =

IYf(e itP )A dcfJl

~ IY f(eitP)B dcfJl + Ir f(e itP )(A - B) dcfJl ~2n If(O)IB +2n Ilflls(1-B)

This is just (7.2.1) in the case

Zo =

0 and R = 1.

(by (4.7)). 0

A weak form of Proposition (7.2) carries over to arbitrary connected open sets. (7.3) Corollary. Let U be a connected open subset of 0 so that K 2e c U, and let {Zj, ... ,zm} be an 8 approximation to K. By the above, there exists Co in (0,1) such that If(Zk)1 ~ Co (1 ~ k ~m) whenever f satisfies our conditions. Consider such a function f and any point Z of K. Choose k with Iz - zkl < 8. Since S C(Zk' 28) C U, (7.2) and elementary calculus show that

Thus we may take c ==~co +~.

0

(7.4) Proposition. Let 0<8< 1, and let f be a differentiable function on a closed sphere S == S dO, R), such that f(O) > 0 and 1- e~ If(z)1 ~ 1 for all Z in S. Then whenever Izl=r
Iw-112 = l-w-w* +ww* ~2-w-w*

(Iwl ~ 1)

7 The Riemann Mapping Theorem

183

we see from (4.7) that 2n

2n

JIf(ei9)-112de~ J(2-f(e i9 )-f(e i9 )*)de o

0

=2n(2- f(O)- f(0)*)~4ne.

The Cauchy-Schwarz inequality for integrals implies that

rlf(ei9)-1Ide~ C{lf(ei9)-Wder C{der ~2n(2e)t.

It follows from (4.7) that for Izl=r<1,

If(z) -11 = (2n)-1

Ir

(f(e i9 ) _1)(e i9 - Z)-l e i9 del 2n

~(2n)-1(1-r)-1

JIf(e i9 )-11 de o

~ (1- r)-1(2e)t.

D

(7.5) Corollary. Let U and V be open subsets of CC with S(O, R) c Un V with f(O)=g(O)=O and f'(0) >0, such that go f: U -+ U and fog: V -+ V are the identity maps. Then

(7.5.1) whenever Izl = r < R2. Proof: By (4.14), (5.4), and the continuity of f and g, we have If(w)1 ~R-llwl and Ig(w)1 ~R-llwl whenever Iwl
Izl = Ig(f(z))1 ~ R -llf(z)l· Thus R2~IRz-lf(z)I~1

(0
The mapping zI-+Rz-1f(z) on U-{O} extends to a differentiable map h on U, by (4.14). Applying (7.4) to this map, and using a simple continuity argument, we have IRz- 1f(z)-11

~ (1-r R -

2)-1(2(I_R2))t

whenever 0
184

Chapter 5 Complex Analysis

The extension of this ineq uali ty to the case Iz I= r < R 2 follows by continuity. 0 We are interested in whether two connected open subsets of
h.(z) == (z - a)(1- a* Z)-l If Irxl=1, then rxh. is an equivalence of {z: a*z=l=1} with {w. a*w=l= - rx}, the inverse equivalence being rx* h_ a •. indeed, given w == rxh.(z) with a* z =1= 1, we have

z = (w + rxa)(rx + a* W)-l = rx*(w+ rxa)(1 + rx*a*w)-l = rx* h_a.(w).

(7.7) Proposition. A mapping f: S(O, 1) -+
Proof: It is routine to verify that a mapping of the stated type is an equivalence of S(O, 1) with itself. Conversely, consider an arbitrary equivalence f of S(0,1) with itself, and write b == f(O) Then h == hb a f is an equivalence of S(O, 1) with itself, and h(O)=O. Choose the complex number rx so that Irxl = 1 and rx* h'(O) > O. (Note that h'(O) g'(h(O)) = 1, where g is the inverse equivalence of h; so that h'(O) =1=0.) Then ¢==rx*hbof is an equivalence of S(0,1) with itself, ¢(O)=O, and ¢'(O) >0. By (75), ¢ is the identity map on S(O, 1). Therefore f is the restriction to S(0,1) of the inverse of rx*h b, so that f(z)=rxh.(z) for all z in S(0,1), where Irxl=1, a== -rx*b, and lal<1 D (7.8) Definition. An equivalence malized if f(O) = 0 and f'(0) = 1.

f of S(0,1) with an open set is nor-

7 The Riemann Mapping Theorem

185

Our next aim is to show that the range of a normalized equivalence on S(O,l) always contains the open sphere S(O,t). In order to achieve this aim, we first need estimates of integrals over certain paths. A Jordan curve is a map of the form f y, where y is a circular path in
If J is a Jordan curve, then for any two points a, b of - car J either (i) a and b can be joined by a polygonal path bounded away from J, or (ii) the winding numbers of J with respect to a and b differ by 1. Moreover, if Z is any point of car J, then there exist points a, b of - car J which are arbitrarily close to z, such that j(J, a) t= j(J, b).

Since j(J, 0 = 0 for all , sufficiently far from car J, we see that z I-+j(J,z) either maps -carJ onto {O,l} or maps -carJ onto {O, -I}. The set of points z in - car J with j(J, z) t= 0 is called the interior of J, and the set of points z withj(J,z)=O is called the exterior of J. The Jordan curve theorem can be proved without reference to any of the other results in this section (see [5J). However, such a proof would take us too far afield, and we shall simply accept the Jordan curve theorem on trust. A consequence of the Jordan curve theorem (which we shall likewise take on trust) is that if J is a piecewise differentiable Jordan curve, and if j(J,z)=l for some complex number z, then S(xdy J

-ydx)~O

Intuitively, if j(J,z)=l, then J winds round z in the positive (anticlockwise) direction, and so tsexdy-ydx) represents the J

area enclosed by J, which cannot be negative. This result is needed in the proof of our next lemma. (7.9) Lemma. Let r be a positive number, and f a nonvanishing equivalence of A={z: Izl>r} with an open set, such that f has a Laurent expansion of the form 00

f(z)=cz+

L a_nzn=1

n

(zEA)

186

Chapter 5 Complex Analysis

with c =t= O. Then for each R > r,

1m Sf*(z) f'(z) dz ~ 0, where

y is the circular path of radius R about o.

Proof: Writing f=u+iv with u and v real-valued functions, we have

1m Sf*(z) J'(z)dz = S(u(vxdx + uxdy) - v(uxdx -

Vx

dy))

y

= S(u(vxdx + vydy)- v(uxdx + uydy)), y

by the Cauchy-Riemann equations. Hence

ImS f*(z)J'(z)dz = S(udv -vdu)= S(xdy-ydx), J

y

where J is the differentiable Jordan curve f 0 y. Thus we need only prove that j(J, 0) = 1. (Note that j(J, 0) exists, since f is nonvanishing.) 00

Writing 1:=

I

a_ n z- n -

J'(z)lf(z)=z-l

1,

for Izl sufficiently large we have

(l-c-ln~l na_nz- n- 1) (1 +c- 11:)-1

f:

na_nz- nn=l =z-1(1+terms in Z-2).

=Z-l (1-c- 1

1)

(1-c- 1 1: +c- 2 1: 2 _

•.. )

By the uniqueness of the Laurent expansion of J'lf in A, the last expression must be that expansion for all z in A. Hence j(J,0)=(2ni)-1 S(J'(z)lf(z))dz y

=(2ni)-1 S(Z-l + terms in z-3)dz =(2ni)-1(2ni+0)= 1.

D

The next result is known as the area theorem. (7.10) Proposition. Let r be a positive number, and f a nonvanishing equivalence of A={z: Izl>r} with an open set, such that f has a Laurent expansion of the form 00

f(z)=cz+

L a_nz- n

(zEA)

n= 1 N

with c=t=O. Then

L nla_nI2r-2n~lcI2r2 for all N n= 1

in 7l+.

7 The Riemann Mapping Theorem

187

Proof: Consider any R > r, and let y be the circular path of radius R about O. For Izl=R we have ZZ*=R2 and therefore

I: a~m(z*)-m) (c - n=1I: na_ nz- n- 1)

f*(z) j'(z) = (c* z* +

m=1

00

00

L a~mR-2mzm_c* L na_ nR 2z- n- 2

=lcI 2R 2z- 1+C

m= 1

n= 1 00

-

"i..J na - na* R-2mZm-n-1. -m m.n= 1

Hence 00

Jf*(z)j'(z)dz=\C1 2R2 JZ-1 dz+c L a~mR-2m Jzmdz m=1

y

00

-C*

L na_ nR2 Jz-n-Zdz n= 1

y

00

- L

na_na~mR-2mJZm-n-1dz

m,n= 1

=2ni

(lc lZ R Z+0-0-

y

I: nla_ nl2R -zn).

n= 1

It follows from (7.9) that 00

L nla_nI2R-2n~lcI2R2. n= 1

Hence for each N in Z+ we have N

L nla_nI2R-2n~lcI2R2. 11=

1

Letting R--+r, we now obtain the required conclusion.

0

For a good application of the area theorem we need two more lemmas. (7.11) Lemma. Let f

be a differentiable map on an open sphere S,=S(wo,r), such that f(wo)=wo=l=O and If(w)I=lwl for each w in S. Then f is the identity map on S.

Proof: Choose t so that 0 < t < rand w =l= 0 for each w in T '= S(wo, t). Then h(w) '= w- 1f(w) defines a differentiable function h on T, and Ih(w)1 = 1 for each w in T. By (5.18), h(w) = h(w o) = 1, and so f(w) = w, for all w in T. It follows from (5.13) that f(w)=w for all w in S. 0

188

Chapter 5 Complex Analysis

(7.12) Lemma. Let f be a normalized equivalence of S(O,I) with an open set U, and let h be the unique differentiable function on S(O, 1) with h(O)=1 and h(z)=z- l f(z) for O1} by F(z) == z exp ( -t),(z- Z)) (Izl > 1). Then F is an equivalence of A with an open set. Proof. For each z in A we have

(7.12.1) Let Z10 Zz be points of A with F(zd=F(zz). Then f(zlZ)=f(ZZZ), so that zf=z~. Suppose that ZI=1=ZZ' Then ZI= -Zz, and so F(zd= -F(zz). Hence F(zd= -F(zd; so that O=F(zd=ZI exp( -t},(zl Z))

and therefore ZI=O. This is absurd, since ZIEA. Thus ZI=Z2' It follows that there is a mapping G: F(A) -+ A such that FoG: F(A) -+ F(A) and Go F: A -+ A are the identity maps. Also, by (5.17), F(A) is open. Now, from (7.12.1) we see that g(F(z)-Z)=z-z =1=0

(zEA),

where g: U -+ S(O, 1) is the inverse equivalence of f Thus by (5.15), ¢(w)==g(W- Z)-1

(wEF(A))

defines a nonvanishing differentiable map ¢ on F(A). Consider an arbitrary open sphere S==S(wo,r) contained in F(A). Let zo=G(w o) and let rJ. be any logarithm of zoo Let L be a differentiable branch of the logarithm of ¢ on S with L(wo) = 2rJ.. Define a differentiable function Gs: S -+
(WES).

(7.12.2) For each w in S we have (7.12.3) Hence Gs(w)EA. By (7.12.1) and (7.12.3), F(G s(W))2 = f(G s(w)-Z)-1 = f(g(w- 2))-1= WZ.

7 The Riemann Mapping Theorem

189

Hence (7.12.4)

IF(Gs(w))l =

Iwl

(WES).

It follows from (7.12 2), (7.12.4), and (7.11) that for each w in S,

F(Gs(w)) = w = F(G(w)) and therefore G(w) = Gs(w). Thus G is differentiable on S. Since S is arbitrary, it follows from (4.9) that G is differentiable on F(A). Hence F is an equivalence of A with F(A). 0 (7.13) Proposition. If f is a normalized equivalence of S(O, 1) with an open set, then 11"(0)1 ~ 4.

Proof. Let h, )" and F be as in (7.12), so that F is a nonvanishing equivalence of A == {z· Izl > I} with an open set. Let

be the Taylor expansion of h in S(O, 1) is

f in S(O, 1). Then the Taylor expansion of 00

h(z)=I+ L anz nn~

1•

Z

Thus X(O) = h'(O)/h(O) = az, and so the Taylor expansion of A in S(O, 1) is of the form 00

A(z)=azz+ Lbnz n. n~Z

Thus for

Izl > 1 we have F(z)=zexp =Z

(-t (azz-

2

+ Jzbnz- Zn ))

(l- t (azz- 2 + n~2bnZ-Zn) + terms in Z-4),

where the final expression converges to F(z) uniformly on each compact set well contained in A. Hence the Laurent expansion of F in A is of the form F(z) = z -taz Z-l + terms in z- 3. By (7.10), we have H-a21~1 and therefore II"(O)I=2Ia21~4. We now prove the Koebe covering theorem.

0

190

Chapter 5 Complex Analysis

(7.14) Theorem. If f is a normalized equivalence of S(O, 1) with an open set, then S(O, i) c f(S(O, 1)). Proof Consider an arbitrary complex number w with Iwl 0. If Iwl < m(f - w, r(r)), then m(f - w, S c(O, r)) ~ If(O) - wi = Iwl < m(f - w, r(r)),

and so, by (5.11), there exists z in S(O, r) with f(z) = w. So we may assume that Iwl>O. Suppose that m(f -w,Sc(O,r))>O. Then elementary calculations show that g(z) == f(r z)/r(1 - w- 1 f(r z))

defines a normalized equivalence g on S(0,1) with g"(O) = r(f"(O) +2w- 1 ). By (7.13), we have rlf"(0)+2w-ll~4 and If"(0)1~4. Thus Iwl- 1 ~H4r- 1 + 11"(0)1) ~ 2(r- 1 + 1), and so Iwl ~ r/2(1 + r), a contradiction. Hence m(f - w, S c(O, r)) < m(f - w, r(r)), so that there exists z in S (0, r) with f (z) = w 0 The Riemann mapping theorem will provide conditions under which there exists an equivalence of a simply connected open set U c
cc.

3

d(x, x') == C~l (Xk -

X~)2 )

t

for all x==(X 1 ,X 2,X 3 ) and x'==(x~,x~,x'3) in (C. We call (C the extended complex plane, or the Riemann sphere. Denote the point (0,0,1) of (C by 00, and define the inclusion map i:
(ZE
Then for all z, z' in
and therefore (7.15)

d(e (z), e (z')) ~ 2p(z, z').

The reader should verify the following statements.

7 The Riemann Mapping Theorem

191

(7.16) The map e is a uniformly continuous bijection of CC onto (C - {oo}, with inverse j given by j(X 1,X 2,X 3)==(X l +ix 2 )(1-x 3 )-1.

(7.17) The triple (cC, oo,e) is a one-point compactification of

cc.

(7.18) A subset U of CC is open if and only if e(U) is open in (C. If U c CC is open, then a subset K of U is compact and well contained in U if and only if e(K) is compact, and well contained in e(U) in the sense that {XE(C: d(x,e(K))~r} ce(U) for some r > 0.

From now on we shall identify CC with the dense subset CC - { oo} of (C. With this identification, CC receives the metric d given by (7 19)

d(z, z') == 2(1 + IzI2)- t(l

+ Iz'1 2)-t Iz -

z'l

(z, z' ECC).

By (7.18), the metrics p and d give rise to the same open sets and the same compact sets. However, they are not equivalent metrics, because (CC,p) is complete and (CC,d) is not. We shall continue to write S(z,r) and Sc(z,r) for open and closed spheres in (CC, p). To avoid confusion, we shall write S<Xl(z, r) == {z' ECC: d(z, z') < r}

and Sc<Xl(z,r)=={z'ECC:

d(z,z')~r}

for open and closed spheres in (CC, d). Many notions (such as that of being well contained in an open set) which are defined relative to one of the metrics p and d make sense also when defined relative to the other. Whenever this is the case, we shall assume that the definition applies when the metric (p or d) is replaced by its alternative. We shall make the same assumption for results which are stated in terms of one metric, but which hold equally with respect to the other. The geometrical interpretation of the map e. CC -+ (C - { oo} is of some interest. The reader may verify that if z == (x, Y)ECC, then ,,·(z) is the unique point, other than N==(O,O, 1), in which (C is met by the line through N and (x, y, 0) in IR 3. Also, if Zo and Zl are distinct points of CC, then the plane containing Po==e(zo), p"==e(zl)' and N intersects (C in a circle; the arc of this circle that is bounded away from N is the image of the line segment span (zo, z d under e; the arc joining Po and N, but excluding N, is the image under e of the set {tzo+(l-t)Zl·

192

Chapter 5 Complex Analysis

t ~ I}; and the arc joining II and N, but excluding N, is the image under i of {tz 1+(I-t)zo: t~ I}. As the proof of the next lemma shows, reference to this interpretation enables us to establish certain facts without recourse to tedious analytic estimation.

(7.20) Lemma. Let zo, Z1 be complex numbers, and a>O Then there exists a path y: [0, 1] ---+
sup {d(zo,ZO+t(Z1 -zo)).

O~t~

I}

~d(zo,zd+a,

in which case we may take y==lin(zo,zd; or, as we now assume, Zo =l=z1. Let Po==t(zo), ll==t(zd, and N==(O,O, 1). To begin with, suppose that d(Po, P1 ) < 2. Then the plane containing Po,~, and N intersects
(O~t~l),

since Po~ is the longest chord with end points on the arc 11. So in this case we may take y == lin (zo, z d. In case N lies on 11, consider an arbitrary positive number R>max{lzol,lz11}. The line through Zo and Z1 in
(O~t~

1)

Then YO+Y<Xl+Y1 is a path in
Hence, by continuity, (7.20.1) holds if we take Y==Yo+Y<Xl +Y1·

7. The Riemann Mapping Theorem

193

We now remove the restriction that d(Po, Pd < 2. Choose ( in
t

in [0,1],

d(zo,zo+t«( -zo»~2p(zo,zo+t«( -zo» ~2tl( -zol


Thus for each tin dmn 1'10 d(zo, YI(t» ~ d(zo, () + d«(, Yl(t» ~ d(zo,

() +d«(, zd + B/2

~ 2d(zo,

() + d(zo, ZI) + B/2

< 8/2 + d(zo, zd + B/2 =d(zo,zd+B. It follows by continuity that (7.20.1) holds with y=lin(zo, ()+YI·

D

We now return to the consideration of equivalences of S(O, 1) with open subsets of 0 Hence, by the maximum principle (5.2), 1 = h(O)-1 ~ Ilh - IIIF(r) = r/m(f, r(r»

and therefore m(f, r(r»

~ r.

D

(7.22) Lemma. Let f be a normalized equivalence on S(O,l), let 8>0, and let 0 < c < 1. Then there exists R in (c, 1) such that d(f(z),J(r(r))) ~ B whenever Izl=R and Ri(l+r) for eachj in 7L+. It follows from (7.21), and (4.9) of Chapter 4, that there exists fJ in (4B- I, 8c I) such that Aj={zEr(r):

If(z)l~fJ}

is compact for each j in 7L +. Let Mj=sup {1f'(z)l: ZEAj}

(jE7L+).

194

Chapter 5 Complex Analysis

There exists a positive integer N such that if Wb ... , WN are complex numbers with IW j - wkl ~ 2 - 8 e whenever 1 ~j < k ~ N, then there exists k with l~k~N and IWkl>8c 1: in fact, if {Cl' ... 'C} is a 2- g e approximation to S(D, 9 c 1), then we may take N == 1 + v. Suppose that M j > 2- 5 e(l-rj)-l

(7.22.1)

~j ~ N).

(1

For eachj with l~j~N choose Zj in Aj with 1f'(zj)I>2- 5 e(l-rj)-1, and define a normalized equivalence hj on S(D, 1) by h-(z)_f(zj+(rj + 1 -rJz)-!(Zj) J f'(zj)(rj+ 1 -rj)

(zES(D,I».

We show that (7.22.2) Suppose, on the contrary, that p(f(Zj).!(r(rj+ 1»)
C- f(Zj) 1<1 If'(zJ(rj+ -rJ

'4.

1

By the Koebe covering theorem (7.14), there exists Z in S(D, 1) with h (z) J

C- !(Zj) f'(zj)(rj+ 1 -rj)

and therefore C=!(zj+(rj+1-r)z). But IZj+(rj+l-r)zl
p(f(Zj), !(r(rj+

d»

> 2- 8 e.

It follows that m(f -w,Sc(D,rj+d)<2- g e<m(f -w,r(rj+d)

for each W in S(f(zj),2- g e). Hence S(f(Zj),2- g e)c!(Sc(D, rj+ d), by (5.11). Now let 1 ~j < k ~ N, and suppose that If(zj) - !(Zk) I< 2- 8 e; so that S(!(zj),2- g e)nS(f(zk),2- g e) is nonvoid. Then the path lin (f(Zj), !(Zk» lies in !(S(D,I» The image of this path under the inverse of the equivalence! is a path joining Zj and Zk, and so comes arbitrarily close to r(rj+ 1)' by (4.8) of Chapter 2. So lin (f(Zj).!(Zk» comes arbitrarily close to f(r(rj+ d). In particular, we can find t in (D, 1) and z' in r(rj + d such that I!(z) + t(!(Zk)-!(z)- !(z')1 <2- 8 e-I!(z)- f(z&

7 The Riemann Mapping Theorem

195

Hence p(f(z),f(r(rj + l)))~ If(z) - f(z') I <2- S e-(I-t) If(zj)- f(zk) I <2- se,

which contradicts (7.22.3). Thus If(z) - f(Zk) I ~ 2 - S e whenever 1 ~j < k ~ N. By our choice of N, there exists k with 1 ~ k ~ Nand If(zk)I>8e- l . This is impossible, as zjEAj; so condition (7.22.1) is ruled out. Thus there exists J with I~J~N and M J <2- 4 e(l-rJ)-1. Let R=rJ> R4c 1 or If(z)I If(z)l. (This is possible by (5.19).) Then d(f(z), fez')) ~ 4If(z')1 (1 + If(z)1 2 )-t(1 + If(z'W)-t < 4If(z)I- 1 < e.

Hence d(f(z).J(r(r))) < e. In case If(z)1 < p, suppose that p(f(z), f(r(r))) > e12. Then for each w in S(f(z), e14) we have m(f - w, S c(O, r)) < el4 < m(f - w, r(r)).

Hence S(f(z),e/4)cf(S(O,r)), by (5.11). Let g:f(S(O,I))--.S(O, I) be the inverse of f Then A.(w) 4(g(!ew + fez)) - z) 'I' eg'(f(z)) (wES(O,I)) defines a normalized equivalence ¢ on S(O, 1). As zEAJ> we have Ig'(f(z))I- 1= If'(z) I < 2- 4 e(I-R)-1 =2- 4 eIR- 1 z _ZI-1 and therefore 14(R -1 z - z)/eg'(f(z)) I
By (7.14), there exists w in S(O, 1) such that ¢(w) = 4(R -1 z - z)/eg'(f(z)).

Hence g(!Bw+f(z))=R-1z; so that IR- 1zl
If y is a path in
196

Chapter 5 Complex Analysis

U c CC, defined by ~

U

=:0

{ZECC: d(z, z') >0 for all z' in U}.

We write X ~ U instead of X n( - U). The reader should prove that if U c CC is open, then ~ U consists of those complex numbers z for which it is contradictory that ZE U. We now introduce the condition of mappability of an open set U c CC. As we shall see, this condition is both necessary and sufficient for U to be equivalent to S(O, 1). (7.23) Definition. Let U be a nonvoid subset of CC, Zo a point of U, and e a positive number An e-border of U relative to Zo is a subfinite subset B of ~ U such that y lies in U whenever y is a path in CC with left end point Zo and with d*(y, B) ~ E. We say that U is mappable if it is open and simply connected, and if there is a distinguished point Zo of U relative to which there exist E- borders of U for each E > O. If U is mappable, then any point of U will act as a distinguished point (Problem 20).

(7.24) Lemma. Let U c CC be mappable, with distinguished point zoo Let E

be a positive number, and B an E-border of U relative to zoo Then for all z in CC with d(zo,z)
ZEU

Proof. Consider any z in CC with d(zo,z)
tJ =:od(zo, B) -d(zo, z) -E. By (7.20), there exists a path y. [0, 1]-+CC with left end point right end point z, such that d(zo, yet)) ~ d(zo, z) + tJ

Zo

and

(0 ~ t ~ 1)

For each t in [0,1] we have d(y(t), B) ~ d(zo, B) -d(zo, yet)) ~ d(zo,

Hence y lies in U, and so ZE U.

B) -d(zo, z) -tJ = E.

D

(7.25) Lemma. Let f be a differentiable function on S(O,l), let 0 < r < 1, and let y be a path in CC with left end point f(O), such that d*(y,f(r(r))) >0. Then y lies in the open set f(S(O,r)). Proof: Note that since d(f(O),f(r(r))) >0, f is nonconstant. Thus f(S(O,r)) is open, by (517). Write E=:od*(y,f(r(r))) Since
7 The Riemann Mapping Theorem

197

point compactification of
Choose M>O so that

Izl ~M whenever d(z, oo)~c/2. Let ex=min{c/2,8/(2+M 2)}.

We first show that S"'(w, ex) c f(S(O, r)) whenever wEcary nf(S(O, r)). To this end, consider such a point w and any in S<>"(w, ex). We have

e

m(f -e, rer)) = p(C, f(rer))) ~ td(e, f(rer))) ~ 1(d(w, f(rer))) -dew, C)) ~t(8-ex)

On the other hand, since wEf(S(O, r)) and d(C, (0) ~ d(w, (0) -dew, 0> c/2,

we have m(f -e,Sc(O,r))~1C -wi ~Hl

+M2)d(C, w)<1(1 +M2)ex

=1(2 + M2)ex

-tcx ~1(8 -ex).

It follows from (5 11) that there exists z in S(O, r) with fez) = C. Hence S<>"(w, ex) c f(S(O, r)). Now let () be a modulus of continuity for y on its domain [a, b J, relative to the metric d on
y(to) = f(0) E car y nf(S(O, r)),

it follows by induction that for 0 ~ k ~ n -1, y([ tk , tk + 1J) c S (OO(y(t k ), ex/2) c SOO(y(t k ), ex) c f(S(O, r)).

Hence car y ~ f(S(O, r))

D

(7.26) Theorem. If f is an equivalence of S(0,1) with an open set U, then U is mappable.

Proof: Replacing f by the map fr---+ 1'(0)-1 (f(z) - f(O)), if necessary, we may assume that f is normalized. Since U is nonvoid and simply connected, we need only prove that for each 8>0 there is an 8-border of U relative to O. We may assume that 0 < 8 < 1. Using (7 22), construct a strictly increasing sequence (rj) in (0, 1) with limit 1, such that d(f(z), f(rer))) < 2- j -

38

198

Chapter 5 Complex Analysis

whenever jE71+, zEr(rj), and rj
d(f(zd, 0;£

L d(f(zj),f(Zj+ d);£ E/8. j= 1

Either ,*00, in which case ,E
d(f(zd, oo);£d(f(zd, O+d(" (0)<3E/16.

Now by (7.21), there exists Z in r(r1 ) with If(z)1 < 1 and therefore (7.26.2) It follows from (7.26.1), (7.26.2), and (4.8) of Chapter 2 that there exists Z;. in r(rd with E/8
d(C (0) ~ d(f(z1), (0) -d(J(z;'), 0> 0,

" belongs to
<E/8 + E/16 + 3E/16 + E/8 =E/2. SO, replacing' by,' and Zj by z} (]> 1) in the case d(C (0) <E/16, we may assume that 'E
j ---+00. But ZjEr(rj) for each j, and so is open, it follows that' E - U.

Izi = 1, which is absurd. Since U

To construct an E-border of U we now let {WI> ... , WN} be an E/4 approximation to J(r(rd) in the metric d. By the foregoing, for each k (1 ;£ k;£ N) there exists 'k in ... , 'N}) ~ E. Let ZE r(rd, and choose k with d(J(z), Wk) < E/4. Then d(f(z), car y) ~ d('k, car y) - d('k, f(z)) ~ E -d('k, Wk) -d(Wk' J(z))

>E -E/2 -E/4=E/4.

7 The Riemann Mapping Theorem

199

Hence d*(y,f(r(rd»'i;;6/4. By (725), y lies in U. Thus {ell .. ,eN} is an 6-border of U relative to O. 0 The converse of Theorem (7.26) is the Riemann mapping theorem. In order to prove the latter, we first investigate a new property of open sets. (7.27) Definition. Let U be a nonvoid subset of f,l(z). In that case, U is open, f,l is a function on U, and the positive number f,l(z) is called the maximal extent of U about z. The next two results show that mappability and the maximal extent property are equivalent properties of a simply connected open set. (7.28) Proposition. Let U be a mappable set. Then U has the maximal extent property. Moreover, if ZoEU, 6>0, Be is an 6-border of U relative to zo, and f,l is the maximal extent of U about zo, then (7.28.1)

Proof: Consider any point Zo in U, and recall that Zo will serve as the distinguished point of U. For each 6> 0 let Be be any 6-border of U relative to zoo Then for all positive band 6 we have (7.28.2)

e

To see this, suppose that d(zo,Be»d(zo,B~)+6, and choose in B~ with d(zo,B e»d(zo,e)+6. Then eEU, by (7.24). But this is absurd, as eEB~c -U. Thus (7.28.2) holds. Hence

Id(zo, Be) -d(zo, B~)I ~ max {b, 6}

(b, 6 > 0).

Since lR. is complete, it follows that there exists a nonnegative number f,l such that (7.28.1) holds for each 6>0. Consider any r>f,l. By (7.28.1), we have d(zo,Be)0. Hence SOO(zo,r)nB., and therefore SOO(zo, r)- U, is nonvoid. It follows that if t > 0 is chosen with SOO(zo, t) c U, then we must have f,l'i;; t>O. Finally, consider any z in SOO(zo,f,l). Choosing b>O so that d(z,zo)
mal extent property is mappable.

200

Chapter 5

Complex Analysis

Proof Let U be a simply connected open set with the maximal extent property, and for each Z in U let J.L(z) be the maximal extent of U about z. Let Zo be any point of U, and let e be an arbitrary positive number less than 8 J.L(zo) Construct recursively a sequence (Bn)~ 0 of subfinite subsets of U, and for each n and each Z in Bn a finite subset B(z) of U, such that (i) Bo={zo} (ii) for each Z in Bn, B(z) is a finite 5(2- 5 e) approximation to the set Soo(z, J.L(z»

(iii) Bn + 1 =

U B(z) zEBn
Let S consist of

Zo

together with all points z of

U Bn such

that there

n=O

is a finite sequence (zo, z l ' ... , zn) with zk+ 1 E B(Zk) for each k (0 ~ k ~ n -1), zn=z, and min {d(Zi' Z). 1 ~i<j~n} >2- 6 e. Since (
U Bn for some N,

bounded, S c

and therefore S is subfinite Note that

n= 0

if Z'ES and zEB(z'), then either ZES or d(z,S)<2- 5 E, in other words, d(z,S)<2- 5 e. For each Z in S choose a point ,(z) in SOO(z,J.L(z)+2-4e)~U. We shall prove that the subfinite set

is an e-border of U relative to zo0 To this end, let y: [0,1] -+(y(tk)' 2- 5e)

Let Wk==y(tk)

(O~k~m)

(0 ~ k ~m -1).

and

Sk==SOO(Wk, 2d(Wb Wk+ d + 2- 4 e)

(O~k ~m -1).

We prove by induction that there exist Zo, , Zm-I in S such that Sk c Soo(Zb J.L(Zk» for each k. As e < 8 J.L(zo), we have So = Soo(zo, 2d(zo, wd + 2- 4 e) c Soo(zo, 2- 3 e) c Soo(zo, J.l(zo».

Now assume that we have found Zo, , Zk in S with Sj c Soo(Zj, J.L(z) for O~j~k<m-l. Either d(Wk+I,Zk)<J.L(Zk)-2- 3 e or d(Wk+I,Zk) >J.L(zk)-3(2- 4 e) In the former case, Sk+ 1 C Soo (Wk+ 1,2- 3 e) c S"'(Zk' J.L(Zk»'

and we need only set Zk+ 1 == Zk In the other case, since d(w k, Zk) < J.L(Zk)

7 The Riemann Mapping Theorem

201

(by our induction hypothesis), it follows from (ii) that d(Wk+ l' B(Zk)) ~ d(w k, B(Zk)) + d(Wk' Wk+ 1) <5(2- 5 1;)+2- 5 I; <2- 2 1;

Choose Z in B(zk) so that d(W k+ 1 ,z)<2- 2 e, and then choose Zk+l in S with d(z, Zk+ 1) <2- 5 1;. We have Il(Zk+ 1) > d(C(Zk+ 1)' Zk+ 1) - 2 - 4I; ~d(C(Zk+ 1)' Wk+ 1) -d(Wk+ l' z) -d(z, Zk+ 1) - 2 -41; >d*(y, B) - 2- 2 I; - 2 - 5 I; - 2 -4 e

>1;/2.

Thus if Cbelongs to Sk+ b we have d(Zk+

b

0 ~ d(Zk+ b

z) + d(z, Wk+

d + 2d(Wk+ b

Wk+ 2) + 2- 4 1;

< 2- 5 I; +2- 2 1;+2- 4 " +2- 4 e < e/2 < Il(Zk + d·

Hence Sk+l cS OO (Zk+l,ll(zk+d), and our induction is complete It now follows that m-I

cary~

U Sk c

U.

k=O

Thus B is an e-border of U relative to

Zo,

and so U is mappable.

0

In view of Proposition (729) it is clear that S(O, 1) is mappable. If U is a non void subset of 0 there exists an e-border B of U relative to Zo with B c S(O, 1).

Clearly, a sequestered set is mappable. (7.31) Lemma. Let U be a sequestered set, lal < 1, and lexl = 1. Then V == exha( U) is a sequestered set, and exha (restricted to U) is an equiva-

202

Chapter 5 Complex Analysis

lence of V with V. Moreover,

if

ZoEV, e>O,

(7.31.1)

and BcS(0,1) is an e-border of (V,p) relative to Zo, then och.(B) is an e'-border of (V,p) relative to och.(zo), with och.(B)cS(O, 1). Proof. By (77) and (5.17), V is an open subset of S(O,I), and och. (restricted to V) is an equivalence of V with V. Consider any positive numbers e,e' satisfying (7.31.1). Choose r>1 so that Sc(O,r)cdmnh. and e' >e(1 + lal)(l-rlal)-l. Choose also R>1 so that if Iwl~R, then Ihb(w)l~r, where b==-oca. Then if ZES(O, 1) and Iz'l ~r, we have (7.31.2)

Ih.(z) - h.{z')1 = 1(1 -a* a)(z - z')(1 - a* Z)-l (1 -a* z')-ll ~ (1-laI 2 )(1 -lal)-l(1 - r lal)-llz - z'l

= (1 + lal)(l-rlal)-llz -z'l. With Zo and B as above, let K==och.(B). Then KcS{O, 1), by (7.7). Let y:[O,I]-+([; be a path with left end point och.(zo), such that p*(y,K)~e'. Consider any t in [0,1] such that ly(t)I~R. With Z'==OC*hb(y(t)), we see from (7.31.2) that for each z in B, loc* hb(y(t)) - zl = Iz' - zl

+ lal)-l(1-rlal) loch.(z) -och.{z')1 = (1 + lal)- 1(1 -rial) loch.(z) - y{t)l.

~ (1

Since OCh.(Z)EK, it follows that loc* hb(y(t)) - zl ~ (1 + lal)- 1(1 -r lal)e' > e

(zEB).

As B is an e-border of (V,p), it follows that if t>O and y([O,t]) c S c{O, R), then the restriction of the path oc* hb 0 I' to [0, t] lies in V. Now, either 111'11 1. In the latter case, choose {3 in (I,R) so that A == {tE[O, 1]: ly(t)1 ~ {3} is compact, and let .==infA. Then y{[O,.])cSc(O,{3)cSc(O,R); also, as 11'(0)1 < 1 and 11'(.)1 ~{3> 1, the continuity of I' ensures that .>0. Hence the restriction of oc* hb 0 I' to [0,.] lies in V; so that 1'(.) = och.(oc* hb(y(.))) E V c S(O, 1),

which is absurd. Thus the case Ilyll > 1 is ruled out, and so 111'11
7 The Riemann Mapping Theorem

203

=rxhao(rx*hboy) lies in V, and so K is an e'-border of (V,p) relative to rxha(zo). Since e, e' are arbitrary, V is sequestered. D

(7.32) Lemma. Let V be a mappable set such that OE- V, and let s be a branch of the square root function on V: s(z) =.: exp (t log z)

(ZE V),

where log is any branch of the logarithm on V. Then Vo =.: s(V) is mappable, and s: V -+ Vo is an equivalence. Moreover, if 'EV, e>O, and B is a te 2-border of (V, d) relative to " then Bo =.: {w: W2 EB} is an e-border of (Vo, d) relative to

sm.

Proof First note that Vo is non void, open, and simply connected. Clearly, s is an equivalence, with inverse so: Vo -> V given by so(w)=.:w 2. With "e, B, and Bo as above, observe that since Vo is open and Bc-V, we have Boc~Vo. Let y: [O,I]->CC be a path with left end point such that d*(y, Bo) ~ e. Then So ° Y is a path with left end point ,. Consider any t in [0,1] and any Z in B. Construct w in CC with w2 = z. Then

sm,

2Iy(t)2 -w 2 1

d(so(y(t)), z) = (1

+ II' (tW)t(1 + IW14)t

1 2Iy(t)-wl 2Iy(t)+wl >-------~~----~ ------~~----~ =2 (1 + ly(tW)t(1 + Iwl2)t (1 + ly(tW)t(1 + Iwl2)t ~ td*(y, BO)2 ~ te 2

since both wand - w belong to Bo. Thus d* (so ° y, B) ~ te 2, so that So ° Y lies in V. Hence y=so(sooy) lies in V o , by (517). Thus Bo is an eborder of Vo relative to s(Q. Since e is arbitrary, Vo is mappable. D It is left to the reader to prove.

(7.33) Lemma. With V, s, and Vo as in Lemma (732), suppose also that V is sequestered. Then Vo is sequestered. Moreover, if'EV, e>O, and B is an E2-border of (V,p) relative to " then Bo=':{W:W 2 EB} is an Eborder of (Vo,p) relative to sm; and if BcS(O, 1), then BocS(O, 1).

(7.34) Definition. Let V be a sequestered set containing 0, and let J1 be the maximal extent of (V, p) about (which exists, by (7.28)) Note that J1<1, since S(O, 1) contains points of -V. Choose a in S(O,l)-V with J1
°

204

Chapter 5 Complex Analysis

Let s be that branch of the square root function on V with O
¢u=a*hbosoaha on U is called the canonical map of U. (7.35) Lemma. Let U be a sequestered set containing 0, and let J1 be the maximal extent of U about O. Then the canonical map ¢u is an equivalence of U with a sequestered set ut such that ¢u(O) = 0, ¢~(O) > 1 + l2(1- J1)2,

and J1~J1t, where J1t is the maximal extent of O
ut

about O. Moreover,

if

Proof. Lemmas (7.31)--{7.33) ensure that ¢u is an equivalence of U with ut = ¢u(U), and that U t is a sequestered set. Now in the notation of Definition (7.34), we have ¢u(O) = a* hb(s(lal)) = a* hb(b) = 0 Also, ah~(0)=a(1-laI2),

s'(lal) =! lal-!, and

a*hi,(b)=a*(I-lal)-l. Therefore, by (2.5),

¢u(O)=!lal-!(1 + lal) = 1 +!lal-!(I-laI 1)2 > 1 +t(1-lal!)2 > 1 + t(1 - [t(1 + J1)]t)2 ;?; 1+t(1-[l+tJ1W = 1 +ir(1-J1)2.

To prove that J1~J1t, first note that the inverse of ¢u is the restriction to U t of the composition tf; of the inverse of a* h b , the map ZI-+Z2, and the inverse of aha. Each of these mappings carries S(O,I) into itself; also, tf;(0)=0. By (414) and (5.4), we have 1tf;(z)l~r-llzl whenever Izl~r
1tf;(z)1 ~ Izl

(zES(0,1))

Consider any 8 > 0, and define lb l )2 (1- la l). "=82(1l+lbl l+lal

7 The Riemann Mapping Theorem

Construct a b-border BcS(O, 1) of (V,p) relative to (7.33), for any e' > e the set

205

° By (7.31) and

Bt == {ct*hb(w)' w 2EcthaCB)}

is an e' -border of (vt, p) relative to 0, and Bt c S(O, 1). Since B = t/I (Bt), we see from (7.35.1) that p(O, B);;::; p(O, Bt). It follows from (7.28), with reference to the remarks preceding (7 30), that fl;;::; fl t + E. Since e is arbitrary, we therefore have fl;;::;/1t. Now let O
S c(O, r) = cPu(t/I(S c(O, r») c cPu(S c(O, r».

0

(7.36) Lemma. Let Vo be a sequestered set containing 0, and let cPo be the canonical map of Vo onto V 1 == cPo(Vo). Continuing in this way, define a sequence (Vn)~ 0 of sequestered sets with OE Vn (nEZ+), and for each n a canonical map cPn of Vn onto Vn+ 1 Let /1n be the maximal extent of Vn about 0. Then /10;;::;/11;;::; ., lim fln= 1, and (7.36.1) for each n. Proof. By (7.35), we have flo;;::; fl1 ;;::; .... For each nonnegative integer n write

By (4.13), IcPo.n + 1(0)1;;::; flo 1.

On the other hand, by (7.35), IcPo.n+1(0)1=lcP~(0)

.. cPo(O) 1 > (1 + ti(1 - fln)2) ... (1 +:h( 1 - flo)2) ~(1 + i2(1-fl.)2)"+ 1.

This yields (7 36.1), which in turn implies that fl.

~

1 as n ~ 00

0

(7.37) Lemma. Let f be an equivalence of an open set V, and let S c(zo, r) c V. Let s, c be positive numbers with s < r, such that (7.37.1)

If(z) - f(z')1

~

clz -z'l

whenever Iz -zol;;::;s and Iz' -zol =r. Then (7.37.1) holds for all z in S c(zo,s) and z' in S c(zo, r). Proof. Fix z in Sc(zo,s). By (4.14), there is a differentiable function F on V such that F(z') = (f(z') - /(z»(z'

-Z)-l

(z' E V - {z}).

206

Chapter 5 Complex Analysis

We have IF(z')I~c whenever Iz'-zol=r. Suppose that m(F,Sc(zo,r)) 0 such that If(z)-f(z')I~clz-z'l for all z, z' in K. Proof Since f is an equivalence, the differentiable function I' is nonvanishing on V. Hence m(f',K»O, by (5.14). We may assume that diam K > O. Thus if w is a modulus of differentiability for f on K, there exists r with 0 < r < w(tm(f', K)) such that L== {(z,z')EK x K: Iz -z'l ~r}

is compact. Since f is an equivalence, there exists IX> 0 such that If(z) - f(z') I~ IX diam K whenever (z, Z')EL. Let c == min {IX, tm(l', K)}, and consider any z, z' in K. We have either Iz-z'l<w(tm(I',K)) or (z, Z')EL. In the first case,

If(z) - f(z')1 ~ II'(z')llz -z'l-tm(l', K)lz -z'l ~tm(l', K)lz

-z'l ~ clz -z'l.

In the second case, we have If(z)-f(z')I~lXdiamK~clz-z'l.

0

(7.39) Proposition. Let Vo be a sequestered set containing 0, define the sequences (Vn), (rPn), and (I1n) as in Lemma (7.36), and let rPO.n==rPn-lo ... 0rPo. VO---->Vn

(nEZ+).

Then the sequence (rPO,n):'~ 1 converges on Vo to an equivalence rP of Vo with S(O, 1). Proof For all integers m, n with 0 ~ m < n define

Let K ~ Vo be a compact set containing O. By (7.3), there exists c in (0, 1) such that II rPO,n II K ~ c for all n in Z +. Given E > 0, choose R > 0 so that c
7 The Riemann Mapping Theorem

207

By (7.36), there exists v in '1.+ such that J.l..?;,R for all n?;,v. If n>m?;,v, then by (7.35) and (75), for all z in K we have IrPo,.(z) - rPo,m(z)1 = I rPm,.(rPO,m(z)) - rPo,m(z)1 ~3(R2 -lrPo,m(z)I)-1(1-R)t ~3(R2-C)-1(1-R)t<8

Since 8 is arbitrary, it follows that (rPo,.):~ 1 converges uniformly on K to a continuous map of K into S c(O, c). Hence, as K is arbitrary, (rPO,.)~l converges on Uo to a continuous function rP: Uo--+S(O, 1). By (4.12), rP is differentiable. To construct the inverse !/I: S(O, 1) --+ Uo of rP, for all integers m, n with O~mr for all n?;,N. For all such n, !/In,o is defined on Sc(O,r). By (5.17), L=!/IN,O(Sc(O,r)) is a compact set well contained in Uo. Since rPo,N(L)=Sc(O,r), it follows from (7.35) that (7.39.1)

rPo,n(L) = rPN,n(rPO,N(L))::::l S c(O, r)

(n?;, N)

By (7.3), there exists s in (0,1) such that IlrPo,nllL~S for all n in '1.+. By (7.36), there exists an integer No?;, N such that S c(O, t(1 + s)) C Un for all n?;,No. Given 8>0, choose a positive number R so that t(1+s)
For all sufficiently large m, n with m
IrPm,n(z)-zl <8

(ZESC(O,t(1 +S))).

In particular, there exists N1 ?;, No such that if n > m?;, Nb then (7.39.2) holds with 8=!(1-s). Thus if n>m?;,Nb Izl ~s, and Iz'l =t(1 +s), then IrPm,.(z) - rPm,n(z')I?;' Iz -z'I-lrPm,n(z) - zl-lrPrn,n(z') -z'l ?;, Iz - z'l-i(1- s) ?;, i(l-s)?;, at:lz - z'l,

where at:=(1-s)/2(1 +3s). Hence, by (7.37), IrPm, n(z) - rPm, n(z')I?;, at: Iz -z'l

whenever n>m?;,Nb ZESC(O,S), and Z'ESC(O,t(1 +s)). Now since rPO,Nl is an equivalence, (7.38) shows that there exists c > such that

°

208

Chapter 5 Complex Analysis

Thus for all z, z' in L and all n > N 1, we have (7.39.3)

IcPo.n(z) - cPO.n(z')1 = IcPN,.n(cPO.N,(Z)) - cPN ,.n(cPo.N,(z'))1 ~CtlcPO.N,(Z)-cPO.N,(Z')1 ~Ctc Iz-z'l

By (7.39.1) and our choice of s, Sc(O,r)cSc(O,!(1 +s)). Thus if 13>0, then by the argument leading to (7.39.2), we can find an integer N2 > Nl such that IcPmn(z)-zlm~N2 and zeSc(O,r). For such m, n, and z, both t/lm.O(z) and t/ln.O(z) belong to L, by (7 39.1); so that, by (7393), It/lm.o(z) -t/ln. o(z)1 ~Ct-l c- 1 IcPo.n(t/lm.O(z)) -cPO.n(t/ln.o(z))1

= Ct- 1 c- 1 IcPm.n(z) -zl < B. It follows that the sequence (t/ln.O)~=N converges uniformly on Sc(O, r) to a continuous mapping. Moreover, as t/ln.O(Sc(O,r))cL for each n~N, lim t/ln.O carries Sc(O,r) into Vo. Since r is arbitrary, it follows n -+ 00

that there is a continuous map t/I. S(O, 1) -+ Vo such that for each compact set K~S(O, 1), t/ln.O is defined on K for all sufficiently large n, and t/ln.O-+t/I uniformly on K as n-+oo. By (4.12), t/I is differentiable on S(O,I).

To show that cP: Vo -+ S(O, 1) is an equivalence, it remains to prove that t/locP: Vo-+Vo and cP0t/l: S(0,1)-+S(0,1) are the identity maps. Let z be an arbitrary point of Vo. Since (cPo ..(z)) converges to cP(z), there exists a compact set J ~ S(O, 1) such that cPo.n(z)eJ for each n in 7l +. Given 13 > 0, choose v in 7l + so that

Taking n = k, we have Iz -t/l(cPo.k(z))1 ~B

(k~

v).

Letting k-+oo, we find that Iz-t/l(cP(z))I~B. Since 13 and z are arbitrary, t/I 0 cP is the identity map on Vo. Similarly, cP 0 t/I is the identity map on S(0,1). 0 In order to extend Proposition (7.39) to the Riemann mapping theorem, we need one more lemma. (7.40) Lemma. Let Vo be a mappable set, woeVo, and S(',r)c -Vo. Then for each 13 > there exists an B-border B of (Vo, d) relative to wo, such that Iw - " > r/4 for all w in B.

°

7 The Riemann Mapping Theorem

209

Proof Consider any e>O. We may assume that e is so small that d(w,w'»e whenever Iw-Clr/4 for all W in S, and Iw-Cl r/4 for all w in B, and B c ~ Va. Consider any path y: [0, 1] -+
(740.1)

Let O~t~l, and suppose that Iy(t)-Cl
(7.41.1)

S(-a,r)c~Va·

To this end, consider any w in S(-a,r). We have l-w-al=lw(-a)l
-r

6(s(za)+a)

.

210

Chapter 5

Then

Complex Analysis

r t(w)=b+ 6(w+a)

(WE-{ -a})

defines an equivalence of -{ -a} with -{b}, and maps the set A={wE
into S c(O, 5/6). Since Uo c A (by (7.41 1)), the restriction of t to Uo is an equivalence of Uo with a nonvoid, open, simply connected subset V of S c(O, 5/6), also, 0 = t(S(Zo))E V Given E> 0, it remains to construct an Eborder of V in S(O, 1) relative to O. To this end, first note that (7412) . .

It(w)-t(w')I=

rlw-w'l <8I w - w'l 6lw+allw' +al = 3r

( , A) w, WE.

We may assume that O<E
d(z,b)~E}

is compact. Let q: - {b} ---+ - { -a} be the inverse of the equivalence t. Then q(K) is compact, by (5.17) In view of this and (7.41.2), we can find 15 > 0 such that d(t(w), t(w')) < E whenever w, w' belong to Au q(K) and d(w,w')1 In the former case, for each tin [0,1] we have y(t)EK; so that if d(q(y(t)), Bo) < 15, then

d*(y,B)~E.

d*(y,B)~d(y(t),

t(Bo))<E

- a contradiction Thus d*(qoy,Bo)~t5. Since q(y(O))=s(zo), it follows that q 0"1 lies in Uo , and hence, by (5.17), that "I = t 0 (q 0 "I) lies in V. To complete the proof that B is an E-border of V, it remains to rule out the possibility 11"1 II > 1. If 11"1 II > 1, then there exists ae in [0, 1] such that ly(ae)I>1 and y([0,ae])cSc(0,2). By the argument just used, we see that the restriction of "I to [0, rx] lies in V, which is absurd because V cS(O,l) and ly(rx)I>1. Thus the case 11"111>1 is ruled out, and the proof is complete. 0

Problems 1. Prove that every complex number has a square root. 2. Is homotopy an equivalence relation on the set of all closed paths lying in a given compact set K?

Problems

211

3. Show that a closed path in an open set V c
need not be closed.

5. Prove that homotopic closed paths in a compact set K the same winding number with respect to each point of - K.

c


6. Let f and g be complex-valued functions on an open set V c
whenever z,wEK, e>O, and continuous (so that f' = g).

Iw-zl~b(e).

Show that

f and

g are

7 Let G be the set of all analytic functions in V=={ZE
p(j,g)==

L 2- n min{1,sup{lf(z)-g(z)I'

Izl~rn}}

n= 1

For each sequence O
If(z)l~cn

whenever

Izl~rn

(n=1,2,3,

.)}

is compact. 9. Show that the set of differentiable functions on D=={ZE
Izl~l}

is

10. Call open sets Vi and V 2 in
11. A sequence (Yn) of complex numbers in an open set V c
212

Chapter 5 Complex Analysis

numbers P(Yn, K; are positive. Let the set U - {(Ynn consist of all z in U with z =1= Yn for all n. A continuous function f: U - {(y.n --> (C is called meromorphic in U if for each compact set K ~ U there exist a differentiable function h on K and a polynomial p with at least one nonzero coefficient, such that fp = h on K n (U - {(Yn)}). Show that such a function f is differentiable on U - {(Yn)}. 12. Show that sums and products of meromorphic functions in an open set U are meromorphic in U Give conditions for the reciprocal l/f of a meromorphic function f to be merom orphic. 13. Let f: U - {(Yn)} --> (C be meromorphic on U Call a finite set A c71+ detached if P(Ym,Yn»O whenever nEA and mE71+-A. Let y be any closed path in U such that P(Yn, car y) > 0 for all n, and such that y is null-homotopic in U. Show that the set A(m, y) =:0 {nE71+ . j(y, Yn) = m} 00

is detached for each m in 7l+, and that

U A(m, y) is finite m~l

14. Continuing Problem 13, for each detached set Ac71+ define the residue r(f, A) of f on A in such a way that

S1(z) dz =2ni L mr(f, A(m, y)) for each closed path y with homotopic in U.

cary~

U - {(Yn)}, such that y is null-

15 Show that (z -a)(z -b) =(z -min {a, b})(z -max {a, b}) for all real numbers a and b, and all z in (C Conclude that the linear factorization of a polynomial over (C is not unique. 16. Show that an entire function g has infinite degree if and only if g - p is nonzero for each polynomial function p. 17. Prove that If f is a nonconstant entire function, and C (' are distinct complex numbers, then f attains at least one of the values C('.

18. Construct an analytic function f on A=:o{z: O
Notes

213

19. Show that the map Zf--+ Z- 1 of CC - {OJ onto itself extends to a metric equivalence of the Riemann sphere (C with itself. 20. Prove that if V is a mappable set, then for each e > 0 there exists an e-border of V relative to z.

Z

in V and each

21 Show that a set is mappable relative to one of the metrics p, d on CC if and only if it is mappable relative to the other. 22. Call a simply connected open set V c CC strongly mappable if (i) V is bounded, and (ii) there exist a compact set K and a point Zo of CC - K such that V consists of all points of CC - K which can be joined to Zo by a path in CC - K. Construct a bounded mappable set that is not strongly mappable. 23. Let V be a mappable set containing 0, and let ¢ be an equivalence of V with S(O, 1) such that ¢(O)=O. Prove that I¢'(O)I is the maximum of 11'(0)1 as f ranges over the differentiable mappings from V into S(O, 1) with }(O)=O

Notes It is not hard to show that the following statements are equivalent· (i) K~S(O,l) for every compact set KcS(O,I); (ii) inf{f(x): O~x~l}>O for every continuous function f. [0,1] -+ 1R +. No proof or counterex-

ample is known for either statement. (Brouwer claims that (ii) is true; but (ii) is false under the hypothesis that all real numbers are recursive.) A simple modification of the proof of Proposition (3.6) shows that if f is pointwise differentiable (in the classical sense) on an open set V c CC, then f is analytic on u. It follows from Corollary (4.11) that a pointwise differentiable function on an open set V c CC is differentiable, in striking contrast to the real case. The notion of a border (Definition (5.1)) is a substitute for the classical notion of a boundary Certain compact sets have a border but do not have a compact boundary. Classically, the proof of Lemma (5.8) is quite simple if there were no such z, then f- 1 would be differentiable on K, contradicting the assumption that m(!, K) = 0 Under the hypotheses of Theorem (5.13), we cannot expect to locate precisely the zeros of f which lie in K to see this, take U == CC,

214

Chapter 5 Complex Analysis

== S c(O, 1), and J(z) == z - (, where it is

not true that either' > 1 or 1. Theorems (6.21) and (6.22) are equivalent classically, but provide totally different constructive information. Problem 18 shows that a further classically equivalent formulation of Picard's theorem does not hold. (The material in Section 6 is based on joint work of Bridges, Calder, Julian, Mines, and Richman. An earlier constructive presentation of the Picard theorems was given by Belinfante [4].) Note that the Picard theorem on entire functions (Problem 17) does not follow constructively from Theorem (6.22). It can be proved by an argument analogous to that used for Theorem (6.22). Details of the generalization of the notion of winding number, and a particularly elegant proof of the Jordan curve theorem, are found in K

,~

[5]. The notion of mappability (Definition (7.23)) replaces Bishop's original notion, which is not a necessary condition for the existence of an equivalence with S(O, 1). See Problem 22, where the original notion is called strong mappability. The work in Section 7 amplifies and corrects the material in [29]. Not every simply connected, bounded, open set U is equivalent to S(O, 1). To see this, consider any sequence (nJ:'= 1 in {a, I} such that we 00

do not know if there exists k with nk = 0, and take U == U Uk' where Uk==S(O, 1) if nk=l=O, and Uk==S(I, 1) if nk=O k=l The reader is invited to give a routine for computing the value of the mapping function ¢ of a given mappable set U at a given point z to within a given accuracy E. He may also give a routine for computing a modulus of continuity for ¢ on a given compact set K ~ U The mapping function ¢ on a mappable set U containing is obtained classically as an analytic function from U into S(O, 1) which vanishes at and has the maximum possible value of I¢'(O)I. Constructively, this approach does not work, because it is not a priori evident that such a function exists. Once the mapping function is known to exist, it can easily be shown to be such a function (Problem 23).

°

°

Chapter 6. Integration

An integration space consists of a set X with an inequality relation, a set L of partial functions from X to JR, and a function I: L ---+ JR, called an integral, which has certain properties classically equivalent to those of a Daniell integral. Integration spaces are introduced in Section 1, and several examples are given. The most important example occurs when X is a locally compact space, L is the set of test functions on X, and the integral is a positive measure on X - that is, a linear map Ji.: L ---+ JR such that Ji.(f) ~ 0 whenever f is a nonnegative test function. The goal is to define the class Ll of integrable functions, construct the integral, and investigate its properties. The construction of the integral is carried out in Section 2. Integrable sets are discussed in Sections 3 and 4; an integrable set is a complemented set whose characteristic function is integrable The classical theorem that every compact set in a locally compact space X is integrable with respect to a positive measure on X fails in the constructive setting. However, Theorem (4.11) says that there are sufficiently many integrable sets for our needs, this theorem is proved using the properties of profiles, which are introduced in Section 4 In Section 5 we show that every positive measure on JR is induced by a monotone function. In Section 6 we consider a positive measure on a locally compact space, and show that integrable sets can be approximated arbitrarily closely from within by compact integrable sets Measurable functions and simple functions are discussed in Section 7. In the next section we prove various results about the convergence of sequences of integrable functions .. in particular, we prove the monotone and dominated convergence theorems associated with the name of Lebesgue. In Section 9 we construct product integrals and prove Fubini's theorem. The final section of the chapter deals with measure spaces, in which the attention is focussed on certain complemented sets with properties abstracted from those of the integrable sets in an integration space, every integration space gives rise to a measure space, and vice versa. Any constructive approach to mathematics will find a crucial test in its ability to assimilate the intricate body of mathematical thought

216

Chapter 6 Integration

called measure theory, or the theory of integration. This subject was initiated in 1904 with a book by Lebesgue, and now pervades abstract analysis in a way so essential that there is little point in trying to give meaning to that branch of mathematics without laying a constructive foundation for measure theory. It was recognised by Lebesgue, Borel, and other pioneers in abstract function theory that the mathematics they were creating relied, in a way almost unique at that time, on set-theoretic methods, and led to results whose constructive content was problematical. Today it is true more than ever that much of abstract analysis has no ready constructive interpretation. It is to be expected that such work will be seen in a proper constructive light only as the result of an investigation which recreates analysis from the very beginning in accord with constructive principles. We are at the stage in such an investigation in which we can properly take up the study of integration, basing our work on the theory of complemented sets from Chapter 3 and the metric space theory of Chapter 4. In two respects this study is crucial. First, as already indicated, measure theory and its techniques underlie many parts of modern analysis. Second, measure theory provides the proper framework for a discussion of discontinuous functions, which in most cases seem to be realized best in a measure-theoretic framework, as functions defined on a full set with respect to the appropriate integral.

1. Integration Spaces In this section we introduce the notion of an integration space, and give some examples of this fundamental concept. Throughout our discussion of integration theory, when we refer to a function f: X -+ Y between sets with inequality relations, we shall assume that f has the property of strong extensionality for all x and x' in X, if f(xHf(x'), then x =1= x'. We define .9'(X, Y) to be the set of all strongly extensional partial functions from X to Y, elements f and g of .9'(X, Y) being equal if they are equal as partial functions. In the special case where Y is IR., we write .9'(X) == .9'(X, IR.) If L1 is any of the symbols 1:, U, V, or /\, then fj, will denote 00

n,

n

fj, n= 1

Another convention that we shall adopt in the context of integration theory is that if (fn) is a sequence of real-valued partial functions on a set X, and A is any nonvoid subset of dmnfn' then Lfn is the

n n

n

1 Integration Spaces

217

function defined on A by

provided that the series on the right converges for each x in A This usage of the notation I fn differs from that introduced in Section 4 of n

Chapter 2; however, the intended interpretation should be clear from the context (1 1) Definition. A triple (X, L, 1) is an integration space if X is a nonvoid set with an inequality =1=, L is a subset of ~(X), and I is a mapping of L into IR such that the following properties hold. (111) If f,gEL and rx,/3EIR, then rxf +/3g, If I, and f /\ 1 belong to L, and [(rxf +/3g)=rx/(!)+/3I(g). (1.1.2) If f ELand Un) is a sequence of nonnegative functions in L such that I I (fn) converges and I I (fn) < I (f), then there exists x in X

•

such that If.(x) converges and If.(x)
to be a weaker condition than (1 1 2). In stating (1.1.4), we note that f /\ r belongs to L for each r > 0, because f /\r=r(r-1f /\ 1). Before we give any examples of integration spaces, it is convenient to derive some simple consequences of properties (11 1)-(1.1.4). First, if f belongs to L, then so does f - f, and l(f - !) = l(f) - I(!) = O. Next, if j and g belong to L, then, by (1.1.1), so do the functions fvg=g+~(f-g+lf-gl) and f /\g= -(-fv -g). In particular, fEL if and only if f+ = f v Of and f- = (-!) v Of belong to L; in that case, since f = f+ - f-, it follows from (1.1.1) that I(!) = l(f+) - l(f-). (1.2) Lemma. Let f ELand let (fn) be a sequence of nonnegative elements of L such that l(fn) converges and I(!) + I(f.) > O. Then there "

L

L

218

Chapter 6 Integration

exists x in dmnf n(n dmnfn) such that

I

fn(x) converges and f(x)

+Ifn(x»O 00

Proof" Choose N in 7l+ so that

I

IUn)<~(lU)+ II(fn))' Then

00

By (1.1.2), there exists x in dmnfn(ndmnfn) such that N

00

I

converges and

fn(x)
n=N+1

the result follows

I

I

fn(x)

n

fn(x). Since fn(x)~O for each n,

n=l

D

(1.3) Lemma. If fEL and 1(1»0, then f(x»O for some x.

Proof. Choose p in L with I(P) = 1. It suffices to take fn == p - p (n ~ 1)

in (1.2).

D

(1.4) Proposition. If fEL is nonnegative, then

IU)~O.

Proof. This is an immediate consequence of (1 3).

(1.5) Proposition. If f

E L,

D

then IIU)I ~ I(lfl)·

Proof" This follows from (1.1.1) and (1.4), since If 1- f and are both nonnegative. D

Ifl-( -

I)

We can now show that the domain of an element of Lis nonvoid; in fact, we can show more. (1.6) Proposition. If Un) is a sequence of elements of L, then is nonvoid.

ndmn fn n

Proof: Choose p in L with I(P) = 1. Since the series

I

[Un - fn) con-

n

verges to 0, we have I(P)+ IJUn-fn»O. Hence, by (1.2), there exists x in

ndmnUn- fn)= ndm~fn' n

D

n

It is time we gave some examples of integration spaces. Accordingly, let X be a locally compact space, and recall that C(X) is the space of all continuous functions f: X --+ IR with compact support; we

1 Integration Spaces

219

call the elements of C(X) test functions. A mapping /1: C(X) -+ IR. is linear if /1(rxf + Pg) = rx/1(f) + p/1 (g)

for all f, g in C(X) and all rx, P in IR.. (1.7) Definition. A positive measure on a locally compact space X is a nonzero linear map /1 of C(X) into IR. such that /1(j)~0 for each nonnegative test function f on X. For such /1 we commonly write Sf d/1 instead of /1(f). The set of positive measures on X is denoted by M+(X). (1 8) Lemma. Let /1 be a positive measure on a locally compact space

X, f a test function, and (f.) a sequence of nonnegative test functions such that S fn d/1 converges and S f. d/1 < S f d/1. Then for each E > 0

'L

'L

n

n

there exists a nonnegative test function g with support of diameter less than E, such that 'LS fngd/1 converges and 'LS fngd/1<J fgd/1. n

n

Proof· Let K be a compact support of

f

By (615) of Chapter 4, there N

exist nonnegative test functions g l' ... , gN such that

N

'L gk ~ 1, 'L gk(X) k~l

k~l

= 1 for each x in K, and each gk has a compact support of diameter N

less than

E.

Thus

'L 'L Jfngk d/1 converges, and k~

1 n

N

N

'L 'LJ fngkd/1~'LJ fn d /1<J f d/1= 'L k~

S fgk d /1.

1 •

Hence there exists k such that

'L S fngk d/1 converges to a sum less than

Jf gk d/1. It only remains to set g"=: gk for this value of k.

0

(1.9) Lemma. Let /1 be a positive measure on a locally compact space X, f a test function with Jf d/1 > 0, and K a compact support of f Then f(x»O for some x in K. Proof. It is enough to prove that II f II > O. Define a test function g by g(x)=:(l-p(x,KW

(XEX).

Then Ilfllg- f~O, so that

J

J

O~J
Hence Ilfll Jgd/1~J f d/1 >0, and therefore Ilfll >0.

0

220

Chapter 6 Integration

(110) Theorem. If fJ. is a positive measure on a locally compact space X, then (X, C(X), fJ.) is an integration space. Proof It is clear that (1.1.1) and (1 1 3) hold, and that lim S U n~

1\

n)dfJ.

00

= S f dfJ. for each f in C(X). If f E C(X) and K is a compact support of f, define an element g of C(X) by (1.1 0.1)

g(X)=(1-p(x,K))+

(XEX).

Then for each n in 7l+ we have 0;:::;lfll\n-1;:::;n-lg. Since fJ. is positive, it follows that O;:::;S (If I 1\ n-1)dfJ.;:::;n- 1 gdfJ.;

J

whence lim J(lfll\n-1)dfJ.=0. Thus (1.1.4) holds To complete the proof, it remains to verify (1 1.2). Accordingly, consider f in C(X) and a sequence Un) of nonnegative test functions such that Jfn dfJ. converges to a sum less than Jf dfJ.. Let K be a

L n

compact support of f Applying (1 8) recursively, construct a sequence (gk) of nonnegative test functions such that gk has a compact support of diameter less than k- 1 , and n

By (1.9), for each k in 7l+ there exists x k in X with k

0;:::;

L Ungl"

gk)(Xk)«fgl···gk)(X k)·

For 1 ;:::;j;:::;k we have giXk»O; thus Xj and Xk belong to each compact support of gj' and therefore p(xj,Xk);:::;j-l. Hence (x k) is a Cauchy sequence in X; it therefore converges to a point x of X. Since the j

functions fn and gi are nonnegative, we have 1 ;:::;j;:::;k, so that

L fn(x k) < f(x k)

for

n= 1

j

L In(x);:::; f(x) n= 1

Also, f(xk»O and therefore XkEK (kE71+); so that xEK Define g as in (1101) Choose oc>O so that

L JfndfJ.+ oc + oc JgdfJ.< Jf

dfl·

Choose also a strictly increasing sequence (N(k));;'~ 1 of positive tegers such that L in d fl<2- 2k rx (kE71+). 00

n~N(k)

J

In-

1 Integration Spaces

221

Applying the argument of the preceding paragraph with (In) replaced by the sequence

we obtain a point x of K with N(j+ I)

ot:g(x) + II (x) + ... + Jj(x) + 2i L

In(x) ~ f(x)

UE7I.+)

n ~ N(j)

N(j+ I)

Thus

L

In(x)~2-il(x)

for each j; so that Lln(x) converges, and

n~N(j)

therefore ot:g(x) + L In(x) ~ I(x). Since ot:g(x)=ot:>O, it follows that Lln(x) < f(x). This completes the proof. D Here are some important examples of positive measures. Every point x in a locally compact space X gives rise to a positive measure bx , the point mass at x, defined by Sldbx=/(x)

(fEC(X)).

The set 71. of integers is locally compact, and C(71.) consists of all functions I: 71. -.1R which vanish outside a finite subset of 71.. Every nonnegative function ot:: 71. ~ 1R determines a positive measure Jla on 71., defined by 00

SldJl7=

L

ot:(n)f(n)

(fEC(71.)),

11= - 0 0

where, of course, the sum is actually finite. Conversely, if positive measure on 71., then Jl = Jl. for a unique ot:, in fact, ot:(n)

=S andJl

)J.

is a

(nE7I.),

where an is the element of C(71.) whose value at n is 1, and whose value elsewhere is O. In case ot:(n) = 1 for all n in 71., the measure Jla. is called counting measure on 71.. b Lebesgue measure Jl on 1R is defined by S I d)J.= S I(x)dx, where a

[a, b] is any compact interval supporting the test function f Test functions and positive measures have some useful properties relative to certain maps. To see this, let X and Y be locally compact spaces, and ).: X -. Y a continuous map. Then for each I in C(Y) the mapping 10). is continuous on X. Suppose that). is proper, in the

222

Chapter 6 Integration

sense that

A:: {XEX: A(X)EB}

is a bounded subset of X for each bounded subset B of Y. Then f A belongs to C(X): for if B is a compact support of f, then any compact subset of X containing A is a support of f 0 A. Thus A induces a map A*: C(Y)~ C(X) given by 0

A*(f)::foA

In turn, A* induces a map A*.

(fEC(Y)).

M+(X)~M+(Y),

Jf dA*(Ji.):: p*(f) dJi.

given by

(Ji.EM+(X),fEC(Y)).

A positive measure Ji. on X can be multiplied by a nonzero continuous function h: X ~ IR 0 +, to give a positive measure h Ji. defined by

Jf

d(hJi.):: Jfh dJi.

(f E C(X)).

A positive measure Ji. on X is said to be supported by a locally compact subset Y of X if f dJi. = 0 whenever f E C(X) and f(y) = 0 for all y in Y. (For example, the measure hJi. is supported by Y if h vanishes on - Y.) If this is the case, then consider any function g in C(Y). By the Tietze extension theorem, g has a continuous extension g* to X with compact support. Moreover, if g is nonnegative, then g* can be chosen nonnegative. Since Ji. is supported by Y, the integral Jg * d Ji. does not depend on the choice of the extension g *. Therefore the equation Jgdv::Jg*dJi.

J

defines a positive measure v on Y, called the restriction of Ji. to Y.

2. Complete Extension of an Integral Let (X, L, I) be an integration space. In order to develop the theory of integration, we need to extend the domain of the integral I. (2.1) Definition. An element f of jii"(X) is an integrable function if there exists a sequence (fn) of functions in L such that I(lfnD con-

L n

verges, and f(x) = Ifn(x) whenever

Ilfn(x)1 converges. The sequence n

(fn) is then called a representation of f

by elements of L, or a representation of f in L We write Ll for the class of integrable functions.

2 Complete Extension of an Integral

223

Under these conditions, since 11(1.)1 ~ 1(11.1) for each n, the series L I(f.) converges absolutely to a real number. Moreover, if (g.) is n

another representation of I by elements of L, then LI(g.)= LI(f.). To see this, suppose that

L I(g.) oF L 1(1.). 00

•

Then

•

.~N+

<Xl

L .~

•

L

•

•

Choose in turn

tx,

N so that

00

1(11.1) < tx,

and

L .~N+

1

1(11. - g./) < tx. I

<Xl

1(11.1)+

N+ I

L .~

I(II.-g./)<2tx
N+ I

•


~I(I.tl (f. -g·)I)· Hence there exists converge and •

Thus

x

in X such that L

•

II.(x)1

and L II.(x) -g.(x)1

•

~ ~+ III.(x)1 + n~ ~+ III. (x) - g.(x)1 < I.tl (I. (x) - g.(X))I·

L Ig.(x)1 converges, and •

0< I.tl (I.(x)- g.(X))I- .~~+ III.(x) - g.(x)1 ~ IL(f.(x) -

•

g.(x))I·

This contradicts the fact that L I.(x) = I(x) = L g.(x). It follows that •

L 1(1.) = L I(g.). •

•

In view of this and the fact that an element c/J of L has a representation (c/J, 0 c/J, 0 c/J, ... ) in L, we can extend the domain of I to L I by defining the integral of the function I in LI to be I(!) == L 1(1.), where

(I.)

•

is any representation of I by elements of L. The extension of I to LI is called the complete extension of the original mapping I: L --> JR. Since, as we shall see shortly, the integral cannot be extended any further in this way, the integral I on Ll is also called a completely extended integral on X.

224

Chapter 6

Integration

Let f and g be integrable functions, with respective representations (fn) and (gn) in L, and let rx be any real number. Then the functions f + g, rxf, If I, and j /\ 1 are integrable, with respective representations (f1' g 1 ,fz , g Z, ... ), (rxf1' rxfz, rxf3' ... ),

(In,f1' - f1' IfI + f21-lfll, j20 - f2' IfI + f2

+ f 31-lfl + f21,

),

and (fl /\ 1,f1' - fl' (f1

+ f2) /\ 1 -

fl /\ 1,f2' - f2' .. )

The apparently redundant terms in the last two representations are included to ensure that each point where the sum of the terms of the representation is absolutely convergent belongs to the domain of f It is clear that the integral is linear - that is, l(rxf + f3g)=rx1(f)+ f31(g) for all integrable functions f, g and all rx, f3 in JR.. It is also clear that (2.2) whenever (fn) is a representation of f in L. It follows from (2.2) that (2.3)

11(f)1 ;£1(lfl)·

For if 11(f)I>I(lfl), then for some n in 7l+,

which contradicts Proposition (1.5). To complete the description of LI as a set, we must introduce an appropriate notion of equality of its elements. This involves a discussion of sets containing the domain of an integrable function. (2.4) Definition. A subset F of X is called a full set if there exists an integrable function f with dmn f c F. (2.5) Proposition. In order that F c X be a full set, it is necessary and sufficient that there exist a sequence (fn) of integrable functions with dmnfn cF.

n n

Proof: The necessity of the condition is clear. To prove the sufficiency, suppose that such a sequence (fn) exists. For each n in 7l+ choose a

2 Complete Extension of an Integral (fn,k)';'~ 1

representation

225

of fn in L, and set 00

IXn == 1 + L 1(lfn,kl), k~

1

Arrange the terms

2- nIX; 1 fn,k

(n, kEZ+)

into a single sequence (¢n)':~ l ' Then the series that the function ¢ == L ¢n with domain {XEX.

L 1(I¢nl) converges,

so

L l¢n(x)1 converges} 00

is integrable. For each x in dmn ¢ and each n, the series L Ifn,k(X)1 converges Hence k~ 1 dmn¢cn dmnf.cF n

and so F is a full set.

0

(2.6) Corollary. A countable intersection of full sets is a full set. Proof: This follows immediately from (2.4) and (2.5).

0

In constructive mathematics, full sets playa role analogous to that of the complements of sets of measure zero in classical measure theory; roughly, what happens outside a full set may be ignored. To make the last remark more precise, we need a lemma. (2.7) Lemma. Let (f.) be a sequence of elements of L such that L 1(lfnl) n

converges, and L fn(x) ~ 0 whenever L Ifn(x)1 converges. Then L 1(fn) ~ O. n

n

Proof: Since L 1(lf.l) converges, the series L 1(f/) and L 1(1.-) both

•

n

converge, and L1(f.)= L1(f/)- L1(fn-) n n n

Also, Llfn(x)1 converges if and only if both Lf/(x) and Ifn-(x) n

converge, in which case Lfn+(x)- I f..-(x) = Ifn(x)~O. n

n

226

Chapter 6 Integration

Now suppose that

I

l(fn) <0. Choose a>O so that

n

00

Choose also N in 7L+ with

I

l(fn-)
00

I

I/(f/)+ n~

n

N

00

l(fn-)< I/(fn-)+2 +I

n~

n~

1

I

l(fn-)-2a

N+ I

N

< I/(fn-)· n~

1

Hence there exists x such that If/(x), Ifn-(x), and therefore If.(x)l, are convergent and •

I

n

N

If/(x)< n

It follows that

I

I

ft=

fn-(x)~If.-(x). 1

n

fn(x) < 0, a contradiction. Hence

I

l(fn) ~ O.

0

• If A c X and f, g are elements of ff(X) such that f(x) = g(x) for all x in A, we say that f and g are equal on A. We then write: f = g on A. The intended meaning of similar expressions, such as "f ~ g on A", should now be clear. (2.8) Proposition. If f and g are integrable functions such that f a full set, then I(f) ~ I(g).

~g

on

Proof: Choose a full set F such that f(x)~g(x) for all x in F. Let h be an integrable function with domain H contained in F. Then fl == f + h - hand g I == g + h - h are integrable functions, each with domain H; 1(f1)=/(f); and l(gl)=/(g). Since gl - fl ~O on H, we can apply (2.7) to a representation of g 1 - fl' to show that l(fl) ~ I(g 1). Hence 1(f)~/(g). 0 (2.9) Proposition. Let fEff(X), let g be integrable, and suppose that f = g on a full set. Then f is integrable, and l(f) = leg)·

Proof" Let h be an integrable function whose domain H is contained in a full set on which f = g. Then the function g + h - h with domain H is integrable, and f = g + h - h on H. Any representation of g + h - h is therefore a representation of f The result now follows. 0

2 Complete Extension of an Integral

227

Although we shall not use the following lemma for some time, it is convenient and appropriate to place it in the context of our discussion of full sets

(2.10) Lemma. Let f be an integrable function. Then there exist nonnegative integrable functions ¢ and 1/1, each having a representation by elements of L that are everywhere nonnegative, such that f = ¢ -1/1 on a full set Proof: Choose a representation Un) of f in L. Define the integrable functions ¢ = L Ifni and IjJ = L (Ifnl- fn)' each with domain n

n

F={XEX' Llfn(x)1 converges}. n

Then ¢ and IjJ are nonnegative integrable functions, with respective representations (Ifni) and (Ifnl-!") by elements of L that are everywhere nonnegative; and f = ¢ -1jJ on the full set F. 0 We say that two integrable functions f and g are equal if f(x) = g(x) for all x in a full set This notion of equality completes our construction of Ll as a set. Let !::::. stand for any of the relations =, <, >, ~, and ~ For arbitrary functions f and g we write f !::::.g if f(x) !::::.g(x) for all x in a full set. From now on, when we describe a partial function f on X as nonnegative, we mean that f ~ 0 on a full set; and when we describe a partial function f as bounded, we mean that If I ~ c on a full set for some c>O.

(2.11) Definition. The norm of an element f of Ll is the real number

Ilflll =1(lfl). (2.12) Proposition. Let f be an element of L l · Then f=O

if

and only

if

Ilflll =0. Proof' It follows from (2.8) that if f = 0, then I fill = O. Conversely, assume that I fill = O. Let Un) be a representation of f in L, and construct a strictly increasing sequence (N(k))r'= 1 of positive integers such that

228

Chapter 6

(This that

Integration

~s ~~~sible

in view of (2.2)) Then

k~1 I(\:t>n\)

converges, so

k~l\n~1 fn\ converges on a full set F. For each x in the full set Fn{xEX: Llfn(x)1 converges}

we have

!~~ \:t>n(X)\ =0 and hence f(x)= Lfn(x)=O.

0

It readily follows from (2.12) that I is a mapping of the set LI into 1R It is now clear that

(2.13) defines a metric p on the set L I. When we describe LI as a metric space, it will normally be the metric (2 l3) that we have in mind It is tempting to imagine that we could continue extending our notion of integral by a procedure similar to the one we used to extend the map 1 from L to L I . The next lemma enables us to show that no further extension of this sort is possible. (2.14) Lemma. If fELl and e>O, then there exists a representation Un) of f with L 1(lfni):;:; 1(lfl) + I: n

Proof. Let (gn) be a representation of f in L, and N a positive integer N

~

L

such that

I(lgnl)<e/2. Set fl

n~N+I

==

L gk'

and fn==gN+n-1 (n~2).

k~1

Then Un) is a representation of have

f in

L. Moreover, for each m> N we

~I (Ifni):;:; 1(\ktl gk\) + E/2 = I (Iktl gkl) + 1(Iktl gkl-Iktl gkl) + 1:/2 :;:;1

(Iktl gkl) + 1(lk~ t+ gkl) + e/2

:;:;I(lktl

I

gkl) + k~t+

I

1(lgkl)+e/2
gkl) +e

The result now follows, since I(lktl gkl)---+/(lfl) as m-+oo.

0

2 Complete Extension of an Integral

229

We now arrive at the constructive version of Lebesgue's .\eries theorem.

(2 15) Theorem. Let (fn) be a sequence of integrable functions such that Then

L 1(lfnl) converges

F=={XEX: Llfn(x)1 converges} is a full set,' the function f == L in with domain F is integrable, and

Proof' For each positive integer n choose a representation (f".k)k"~ fn in L such that

1

of

LI(lfn kl)
L L1(1f",ki) converges; so that

Then

n

k

A == {XE X: L L Ifn,k(X)1 converges} k

n

is a full set, and the function

L L fn.k

with domain A is integrable and

k

n

has a representation given by any enumeration of the terms fn,k (n, k ~ 1) as a single sequence. Moreover, for each x in A and each n~ 1 we have fn(x)

= Lfn.k(X); whence L Ifn(x)1 converges and Lfn(x) k

=

n

"

L L fn.k(X) Thus A c F, F is a full set, and f = L L fn,k on A. By n

(2.9),

k

11

f is an integrable function.

k

I

Now let N E71+, and let (gn) be a representation of f Then B==A n {XEX: L Ign(x)1 converges}

N

L fn

I

n~ 1

is a full set, and oc I N I "~~+ 1 ~ 1f",k(X)I- ~ gn(X) ~ n~ ~+ I lfn (x)l- f(x) - n~l fn(x) ~ 0 00

for each x in B. Hence, by (2.7), l(jr - nt/nl) =

~1(gn);£ n~~+ ~1(lfn.kl) 1

00

;£

L n~N+

(l(1fnl) +2-"). 1

in L.

230

Chapter 6

Integration

Letting N ---> 00, we see that this last expression, and hence

tends to 0

0

(2.16) Corollary. L is dense in LI relative to the metric arising from the norm II III' Proof· If f is any element of LI and Un) is a representation of f by elements of L, then N ---> 00, by (2.15). 0

I.

fn EL for each N; and

n~ I

Ilf - I. fnll n~ I

--->0 as I

(2.17) Corollary. LI is complete with respect to the metric arising from the norm II III' Proof: Let Un) be a Cauchy sequence in L I , and construct a strictly increasing sequence (n(k»~~ I of positive integers such that II fn(j) - fn(k) I I <2- k Then that

L 1(lfn(k+

I) -

U~ k ~ 1).

fn(k)l) converges. By (2.15), we can find f in

LI

such

k

as N ---> 00. Since Un) is a Cauchy sequence, it follows that II f - fn III ---> 0 as n--->oo. Hence LI is complete. 0 (2.18) Theorem. (X,LI,I) is an integration space. Proof" We have already seen that (X,LI,I) satisfies (1.1.1). To verify (1.1.2), let Un) be a sequence of nonnegative elements of L I , and f an arbitrary element of L I , such that L IUn) converges to a sum less than n

IU)· To begin with, assume that f ~ O. Choose

0(

> 0 so that

L IUn) n

+30«IU). By (2.14), for each n in 7l+ there exists a representation Un,k)~= I of fn in L such that

L 1(lfn)

< IUn} + 2 - n 0(.

k

On the other hand, if (cPk) is a representation of f in L, then we can find

N

so that

IU)
k~~+/(lcPkl)
Then

2 Complete Extension of an Integral

231

I: I: I(lf.,kl) converges, and n

k

00

I:I:/(lf.,ki)+ •

I: l(lcPki)
k

+1

•

Since (X, L, I) is an integration space, there exists a point x such that

I: I: If.,k(x)1 and I: IcPk(x)1 are convergent, and •

k

k

~ ~ If.,k(x)1 + k~~+

1

IcPk (x)1 < Ikt1 cPk(X)I·

I:lf.(x)1 converges, and • I: f.(x) = I: I: f.,k(X) ~ I: I: If.,k(x)1

Hence xEdmnfn(ndmnf.),

•

"

n

n

k

< Ikt1

k

cPk(X)I-lk~ ~+

1 cPk(X)

I

~ II: cPk(X) I= If(x)1 = f(x). k

This verifies (1.1.2) in the case this, in view of the inequality

f

~ O.

The general case follows from

I(lfl) + I: I(f.) < 1(lfl + f) .

• Since (1.1.3) holds in L, it clearly holds in L 1 • It therefore remains to verify (1.1.4). Accordingly, consider an arbitrary f in L1 and an arbitrary e > O. By (2.16), there exists g in L with I f - gill < e/3 Choose N in Z+ so that I(g I\. N) > I(g) - e/3. For each b > 0 we have 1/(f I\. b) - I(g I\. b)1 ~ 1(lf I\. b - g I\. bl) ~ 1(lf - gl) <

e/3.

Hence for all n ~ N, 1(f)~/(f I\.n)~/(f

I\.N)

> l(g I\. N) - e/3 > leg) - 2e/3 >/(f)-e.

Thus l(f I\. n) -+ I(f) as n -+ 00. On the other hand, choosing m in Z+ so that 1(lgll\.m- 1)<e/2, for each n~m we have O~/(lfll\.n-1)~/(lfll\.m-1)


232

Chapter 6

Integration

Hence I(lfl An - 1) -> 0 as n -> 00. This completes the verification of (U.4). D We call (X, L 1 ,!) the complete extension of the original integration space (X, L,!), and refer to (X, L 1 ,!) as a completely extended integration space

3. Integrable Sets In this section we shall assign a value to certain complemented sets in X, where (X, L 1 , 1) is a completely extended integration space. Recall that the characteristic function of a complemented set A == (A 1, A 0) in X is the map XA: Al u A °-> {O, I} defined by XA(X) = 1

if xEA1,

=0

if xEAo

(3.1) Definition. A complemented set A in X is said to be integrable, or an integrable set, if XA is an integrable function. We then define the measure of A to be IleA) == I(XA) If A == (A 1, A 0) is integrable, then Al u A ° is a full set. Our first lemma is a constructive expression of the classical statement that the complement of a set of measure 0 is a full set. (3.2) Lemma. If A==(Al,AO) is an integrable set with /l(A)=O, then AO is a full set.

Proof: Since L I(lnxAI) = L n/l(A) = 0, L nXA is an integrable function, n

with domain F=={XEX: LnXA converges}. n

Clearly, FcAo. Hence AO is a full set.

D

(3.3) Lemma. If A is a complemented set such that AO is full, then A is integrable, and IleA) = O. Proof' Choose an integrable function f with domain contained in AO. Then Y 1 = f - f on the full ~et dmn rand is integrable Hence, by (2.9), XA is integrable, and /l(A)=I(j -f)=0. 0

r- r

3 Integrable Sets

233

(3 4) Proposition. If F is a full set and A is an integrable set with positive measure, then F n A I is nonvoid.

Proof Choose an integrable function f with dmn f e F. Then g == J1(A)(l

+ I(lfl))-llfl

is a nonnegative integrable function, and I(g) < J1(A); so that there exists x in dmn f with

o~ J1(A)(l + I(lfl))- Ilf(x)1 < xix). Hence XA(x)=l and therefore xEFnAI

0

(3.5) Proposition. Let A and B be integrable sets. Then A /\ B and A vB are integrable, and

J1(A) + J1(B) = J1(A v B) + J1(A /\ B).

Proof We have

and on the set F=(A I uAO)n(BI uBO)

Since F is the intersection of two full sets, it is a full set. The result now follows from (2 9). 0 (3 6) Proposition. Let A and B be complemented sets such that A and A /\ B are integrable Then A - B is integrable, and J1(A) = J1(A /\ B) + J1(A - B).

Proof' This follows from (2.9) and the fact that XA-B=XA-XAI\B

on the full set (A I u A 0) n (BI u BO).

0

(3.7) Proposition. If A and B are integrable sets with Al eBI, then J1(A) ~ J1(B).

Proof. This follows from (2.8) and the fact that XA ~ XB on the full set (A I uAO)n(Bl uBO). 0

It follows from Proposition (3 5) that a finite union or intersection of integrable sets is integrable. Additional numerical information is

234

Chapter 6 Integration

needed before we can assert the integrability of the UnIon or tersection of a sequence of integrable sets.

In-

(3.8) Proposition. If (A.) is a sequence of integrable sets such that

;~~ ILlY1A.)

(respectivelY,

V A. (respectively, /\ •

;~~ ILl~lA.))

exists and equals A., then

A.) is integrable and has measure A-

•

;~~ ILlY1A.)

Proof· Suppose that

exists and equals A- For each

k

V A. and Xk == XBk· Then Bk and B k+1

positive integer k define Bk ==

.~

- Bk are integrable, 0 ~ Xk+ and I(lXll)+

1-

1

Xk ~ 1 on the full set dmn Xk n dmn Xk+ l'

L I(lx.+ 1 -

X.I)= I(xd+

•

L I(x.+ 1 •

X.)

= lim IL(B k ) = Ak~oo

Thus by (2.15),

• is an integrable function with domain F== {XEX:

(X.(x)):~ 1

converges},

and I(x) = A.. Given x in F, we have either X(x»O or X(x)<1. In the former case, X.(x) = 1 for all sufficiently large n, and so X(x)= li"m X.(x) = 1 =

V X.(x). n

11-00

In case X(x)< 1 we have X.(x)=O for all n, and therefore X(x)=O = V X.(x). Thus X equals the characteristic function of V A. on the

•

full set F. Hence, by (2.9),

• V A. is integrable and has measure A. .

. A similar argument disposes of the case where lim IL ( A. A.) eXIsts. D k~oo \,.~ 1 (3.9) Lemma. Let (A.) be a sequence of integrable sets. If A ~ ~ Ai ~ ... and A.== lim IL(A.) exists, then /\ A. is integrable, and IL(/\ A.) = A.. If 11-00

A~ c Ai c...

J.I.(V An)=A.

11

and A. == lim IL(A.) exists, then

V A.

n

is integrable, and

11-00

• Proof: We prove only the first part, since the proof of the second is similar. Accordingly, assume that A ~ ~ Ai ~ ... and that A. == lim IL(A.)

3 Integrable Sets

235

exists. In view of (3 8) it is enough to prove that

/ll~IAn)=/l(Ak) For k each J.L

k

we

clearly

l~\ Ak)
have

(kEZ+).

/ll~l An) ~ /leAk)·

there exists x in

A~

such that

Suppose

xEA~

that

for some

n with 1 ~ n ~ k. Since Ai c A! (1 ~ n ~ k), this is impossible. The result

now follows.

D

(3.10) Proposition. If (An) is a sequence of integrable sets such that ~>(An) converges, then V An is integrable and /l(V An) ~ ~:>(An)· 11

11

11

PI

Proof: For arbitrary positive integers j, k with k > j, it follows from (3.5)-(3.7) that

Hence

(/llY1An)):1 is a Cauchy sequence, and so converges to a V An is integrable, and

limit in 1R.. By (3.8),

n

(3.11) Proposition. If A is an integrable set and f is an integrable function, then the functions XAf and x-Af are integrable. Proof: Since f = f+ - f-, we may assume that f ~ O. Choose a strictly increasing sequence (nk)f~ 1 of positive integers such that O~I(f - fA n k )<2- k

(k~ 1).

Taking no == 0, define gk == min {(nk -nk-1)XA,f -

fA

n k_ 1}

(k~ 1).

236

Chapter 6 Integration

Then gk ~ 0 for each k, and that the series

L J(gk)

L gk defines an

converges. It follows from (2 15)

k

integrable function on the set of points

k

where it converges. Since XA! = L gk on the full set dmn XA ( l dmn f, we k

see from (2.9) that XA! is integrable. It now follows that LA! = ! - XA! is integrable. D

4. Profiles Further developments in integration theory depend on the construction of an abundant supply of integrable sets. The notion of a profile is introduced in order to make this construction possible.

(4.1) Definition. Let J be a proper, although not necessarily finite, interval in lR. Let C be a set of continuous mappings of J into [0,1] with the profile property. if Xo and x I are interior points of J with Xo < x I' then there exists an element h of C such that h(x)=O whenever XEJ and x~xo, and h(x)=1 whenever XEJ and X~XI' Let the map A' C --+ lR be increasing, in the sense that A(¢) ~ A(t/J) whenever ¢, t/J E C and ¢ ~ t/J on J. Then the pair (C, A) is called a profile for J. The most important example of a profile is associated with the integration space (X, L l , J), which we shall continue to study in this section Define functions h(xo, Xl' .). lR 0+ --+ [0,1] by h(xO,xl,x)==(x l -xo)- l(min {x,x l } -min {x,x o})

(O<XO<XI' x~O), and let C== {h(xo, XI")' 0 <xo <xd.

Let! be a nonnegative integrable function. Since! 1\ r = r(r- I! 1\ 1) is integrable for each r>O, we see that h(xo,xl,.)o! is integrable whenever 0< Xo < XI Hence A(h(xo, Xl' .» == J(h(xo, Xl' .) of)

(0 <XO < Xl)

defines an increasing map A. C--+lR, and the pair (C,A) is a profile the (I,f)-profile for lR °+.

4 Profiles

237

(4.2) Definition. Let (t&', A) be a profile for the interval J, let [a, b] be a compact subinterval of J, and let E>O We say that [a,b] has profile lower than E relative to (t&', A) if there exist points a', b' of J with a'
(4.2.1)

~E

relative to (t&',A).

When it is clear which profile is under consideration, we shall abbreviate (42.1) to [a, b] ~e. If [a,b]~E relative to (t&',A) for each E>O, we say that [a,b] has arbitrarily low profile relative to (t&', A). There now follows a succession of lemmas which will enable us to prove the fundamental theorem in the theory of profiles: relative to a given profile on an interval J, {x} has arbitrarily low profile for all but countably many points x of J. (4.3) Lemma. Let (t&', A) be a profile for an interval J, E a positive number, and a, b, c points of J such that a;£bi£c, [a,b]~E, and [b,C]~E.

Then

[a,c]~2E.

Proof: Since [a, b] ~ e, there exist points 0(, p of J with 0( < a;£ b < p, and functions ¢I' t/ll in t&', such that ¢1(x)=l whenever XEJ and x~O(, t/ll(X)=O whenever XEJ and x;£p, and A(¢I)-A(t/ll)<E Likewise, there exist points ,}" fJ of J with'}' < b i£ c < fJ, and functions ¢2' t/l2 in t&', such that ¢2(X) = 1 whenever XEJ and x~'}', t/lz{x)=O whenever XEJ and x;£fJ, and A(¢2)-A(t/l2)<E. It will suffice to prove that A(¢l) -A(t/l2)<2E Since 0;£¢2' t/l1;£1 and ¢2(x)=1 whenever t/l1(X»O, we see that ¢2 ~ t/ll on J Hence A(¢l) - A(t/l2) = A(¢l)- A(t/ll) + A(t/ll) - A(t/l2)

< E + A(¢2) - A(t/l2) < 2E.

0

(4.4) Lemma. Let (t&', A) be a profile for an interval J, let E and fJ be positive numbers, and let [a, b] be a compact subinterval of J such that O
Proof' Let N

~

2 be an integer with [a, b]

~E-

3(N _1)-1 E.

238

Chapter 6 Integration

Construct real numbers Xo < Xl < ... < X N + 2 in (a, b) so that b - Xo < lJ/2 and XN+ 2 -a < {)j2. By the profile property, for each n (0 ~ n ~ N + 1) there exists hn in rf such that hn(x)=O whenever XEJ and x~xn' and hn(x) = 1 whenever XEJ and x;;;; Xn+ 1 On the other hand, there exist points a', b' of J with a'
L (.,1(h n)- .,1(hn+ 3))= .,1(h o) -

.,1(hN_ 1) + .,1(h1)- .,1(hN)+ .,1(h2) - .,1(hN+ 1)

n~O

~

3(.,1(c/J)- .,1(t/J» < 3E.

Thus there exists k (0 ~ k ~ N - 2) such that .,1(hk)- .,1(h k+3) < 3(N _1)-1 E

and therefore (.,1(c/J)- .,1(hk+ 3»

+ (.,1(h k)- .,1(t/J)) <E.

Hence there exist positive numbers E 1 , E2 with E1 + E2 < E, .,1(c/J) -.,1(hk+3)<E1' and .,1(h k )-.,1(t/J)<E 2· Since a'
0

(4.6) Lemma. Let (rf, A) be a profile for an interval J, M a nonnegative integer, E a positive number, and [a, b] a proper compact subinterval of J such that [a, b] ~(M + 1)E Then there exist nonnegative integers A and B, and a point x in (a,b), such that max{b-x, x-a}
Proof· By (4.4), there exist positive numbers E1 and E2 , and a point x in (a, b), such that max{b-x, x-a}
4 Profiles

239

and 1:1 + 1:2 < (M + 1)1:. By (4.5), there exist integers A, B such that A+B=M, I: I C 1 O. Then there exist points Xl"",XM of [a,b] such that {x}~1: whenever xE[a,b] and x=i=x k for all k. Proof Let e represent either a void sequence ( ), or a finite sequence (e l , ... , en) of integers each of which is either 0 or 1. If e is ( ), write lel=O, e*O = (0), and e*I=(I); if e=(el, ... ,en), write lel=n, e*O=(el, ... ,en,O), and e*I=(e l , ... ,en,I). Since [a, b] c]D, we can find a positive integer M such that [a, b] ~ (M + 1)1:. With each e we shall associate a nonnegative integer M" and a proper compact interval I" c [a, b] satisfying the following conditions.

(i) If lel=O, then M,,=M and 1,,= [a,b]. (ii) For each positive integer n, L M" = M. (iii) I.~(M.+l)l:. (iv) 11,,1 < (3/4)le l -l(b -a).

{"

I"I~ n)

(v) I"'ocI e, 1"1 cI., and M.-o+M.*1 =M•. (vi) If lel=le'l and e=i=e', then the intervals I. and I., do not overlap, that is, I. n Ie' has void interior. (vii) U {I.: lel=n} is dense in [a,b] for each n in '1.+. We proceed by induction on lei. The construction for lei =0 is carried out according to the prescription (i). Assume then that for all e with Iel ~ N, I. and M. have been constructed so that the appropriate parts of (i)-(vii) hold. Consider any e with lei = N. By (4.6), we can find nonnegative integers A and B, and a point x of I. [a., bJ, such that

=

max {b. -x, x-aJ <~II.I < (3/4)N(b -a), [a.,x]~(A+l)l:, [x,bJ~(B+l)l:,

and A+B=M•. Set M •• o=A, M._! =B, I •• o=[a.,x], and Ie.!=[x,bJ. This procedure defines M. and I. for all e with lel=N+1. Moreover, it follows immediately from this definition and our induction hypothesis that the appropriate parts of (i)-(vii) hold for all e with lei = N + 1. Thus our inductive construction is complete.

240

Chapter 6

Integration

For each e let x. be the midpoint of I. By induction on n we now define finitely many sequences Sk == (Xn,k)~~ 0 (1 ~ k ~ M) of points of [a, b], with the following properties. For each pair (n, k) there exists e (perforce unique) with lei = nand Xn,k=X,. (II) For each e with lei = n there are exactly M. values of k for which Xn,k = X" (III) If Xn,k=X., then either xn+ 1,k=X•• O or xn+ l,k=X.*l· (I)

We begin the induction by setting XO,1=X O,2=",=X O,M==X() Now assume that Xn k has been defined and satisfies the appropriate parts of (I)-(III) for O~' n ~ Nand 1 ~ k ~ M. Then for each e with lei = N there are exactly M. values of k for which XN,k=X •. For M •• o of these values (it doesn't matter which) we define XN+ 1,k==X.*O; for the remaining M"l values we define XN+ 1,k==X.*1' Taken with our induction hypothesis, this construction ensures that the appropriate parts of (1)(III) hold for 0 ~ n ~ N + 1 and 1 ~ k ~ M. This completes the definition of the sequences Sk' Since Xn,k and Xn+1,k both belong to the same I. with lel=n, we have IXn,k - xn+ 1,kl ~ 11.1 < (3/4)n- l(b -a). Hence for all m> n, 00

IXn,k - xm,kl <

L (3/4)j-1(b -

a)= 4(3/4)n-1(b - a).

j=n

Thus Sk is a Cauchy sequence. Let Xk be its limit. Then XkE [a, b], since I. c [a, b] for each e; and

Now consider any point x in [a, b] with x =1= x k (1 ~ k ~ M). Choose a positive integer n so that Ix-xkl>5(3/4)n-1(b-a) for 1~k~M. Then for such k we have Ix-x n,kl>(3/4)"-1(b-a). Since the points Xn,k (1~k~M) are the midpoints of those intervals I. with lel=n and M. > 0, it follows from (iv) above that x has a positive distance from all such intervals. Therefore, by (vii), x has a distance 0 from the union of those intervals I. with lel=n and M.=O. Since I.~(M.+1)e=e for each such e, it follows that {x} ~e. 0 Here, at last, is the result for which we have been preparing (4.8) Theorem. If (tC, A) is a profile for an interval J, then for all but

countably many points x of J, the set {x} has arbitrarily low profile.

4 Profiles

Proof' Write 1" ==

r.

"" i., U

241

where each i. is a proper compact interval

"=1

contained in For each pair (n, k) of positive integers there are finitely many points x~~L x~~L ... ,x~~Z of i. such that {X}~k-I whenever xEi. and x =!= X~)k (1 ~j ~ N). Let (x.) be a rearrangement of all the points x~L together with those end points (if any) of i that belong to i, in a single sequence. Then {x} has arbitrarily low profile whenever XEi and x =!= x. for all n D Note that if {x} has arbitrarily low profile, then x is an interior point of i; this follows from Definition (42). (49) Definition. Let (C, A) be a profile for an interval i. A point x of i is smooth for (C, A) if there exists a real number c with the property: for each e>O there exists 15>0 such that 1..1(4))- CI<e whenever 4>EC, 4>(t)=O for all t in i with t~x-c5, and 4>(t)=1 for all t in i with t~x+c5.

Taken with Theorem (48), the following theorem shows that all but countably many points of i are smooth. (4.10) Theorem. If (C, A) is a profile for an interval i, then any point of i with arbitrarily low profile is smooth for (C, A) Proof: Let x be a point of i with arbitrarily low profile. Chose r > 0 so that x - r and x + r are interior points of i Construct a strictly decreasing sequence (IX.) of numbers in (0, r) converging to 0, and sequences (4).), (1/1.) of elements of C, such that for each n, 4>.(t) = 1 whenever tEi and t~X-IX., 1/I.(t)=O whenever tEi and t~X+IX., and A(4).)-A(I/I.)
=1

if tEl and t ~ x -IX., if tEi and t~X-IX.+ I'

and g.(t)=O

=1

if tEi and t~ x +IX.+ I' if tEi and t~x+ IX.

o~ min {A(f.) -

A(fm)' A(gm) - A(g.)}

~ A(f.)-A(g.)~ ..1(4).) - ..1(1/1.)
I .

It follows that (A(f.)) and (A (g.)) are Cauchy sequences in JR., and that

they have a common limit c.

242

Chapter 6

Integration

Now consider an arbitrary E>O. Choose N in '1.+ so that N-1<E, and set c5=Ot: N + 1 . Let ¢ be any element of tC such that ¢(t)=O for all t in J with t~x-c5, and ¢(t)=l for all t in J with t~x+c5. Then fN~¢~gN on J. Since A(fN)~c~A(gN)' it follows that IA(¢)-cl =max {A(¢)-c, c- A(¢)} ~ A(fN) - A(gN) < N- 1 <

Hence {x} is smooth for (tC, A).

E.

D

We now return to our integration space (X, L 1 , I). For each realvalued function f with dmn f c X, and each real number t, we define the complemented sets (f

~t)=({XEX: f(x)~t},

{XEX: f(x)
and (f >t)=({XEX: f(x»t}, {XEX: f(x)~t}).

We also write (f~t)=

-(f>t)

(f
-(f~t).

and Note that these are indeed complemented sets, in view of the convention about functions that we established at the beginning of this chapter. This brings us to the theorem on integration for which the theory of profiles was originally developed. (4.11) Theorem. Iff is an integrable function, then for all but countably many positive numbers t, the complemented sets (f ~ t) and (f > t) are integrable and have the same measure. Moreover, for each admissible t > 0 and each E> 0, there exists c5 > 0 such that (4.11.1)

IJl((f ~ t)) - Jl((f ~ t'))1 < E

whenever t' > 0 is admissible and

It - t' I< c5.

Proof: Replacing f by f+, we may assume that f~O. By (4.8) and (4.10), there is a sequence (xn) of positive numbers such that if t>O and t=\=xn for all n, then {t} is smooth relative to the (I,f)-profile for IR 0 +. Call such t admissible relative to the excluded sequence (xn). Consider an arbitrary admissible t > O. Choose a positive integer N with t>2- N • For each integer n~N and each x in X, define ¢n(x) =h(t- 2- n, t- 2- n-

1,

f(x))

4 Profiles

243

and t/ln(x)=h(t+2- n-

1,

t+2- n,f(x)).

Then ¢n?;, ¢n+ 1 ?;, t/I n+ 1?;, t/I n on dmn f Also, since t is smooth, the sequences (I(¢n))';:N and (l(t/ln)':~N converge to a common limit c in 1R Thus ()()

()()

Hence ()()

X=¢N-

I

(¢n-¢n+l)

defines an integrable function on the full set F= {XEX:

n~}¢n(X)-¢n+ l(X)) converges},

and I(X)=c. For each x in F, either X(x»O or X(x)< 1. In the former case we have ¢n(x»O, and therefore f(x)?;,t-2- n, for all sufficiently large n; so that f(x)?;,t, ¢n(X) = 1 for all n?;,N, and therefore X(x)=1. On the other hand, if X(x) < 1, we can choose m with ¢m(x) < 1. We then have f(x)~t-2-m-l; so that f(x)
x'=t/lN+

I

(t/ln+ 1 -t/ln)

n~N

is an integrable function defined and equal to XU>t) wherever the ()()

series

L (t/ln+l-t/ln)

converges, and that I(X')=c. Hence (f>t) is an

n~N

integrable set, also with measure c. Now let 1'>0 be arbitrary. Choose k?;,N so that I(¢k)-I(t/lk)<E, and write c5=2- k - 1 . Consider any admissible point t' with It-t' I 0 and the complemented set (f?;, t) is integrable, then f ~ O.

244

Chapter 6

Integration

Proof By (411), there is a decreasing sequence (tn) of positive numbers converging to 0 such that (f ~ tn) is integrable for each n Since

L J.l((f~tn))=O,

we see from (3.10) that A=

n=1

V (f~tn) is integrable n

and has measure 0 By (3.2), A 0 is a full set. Clearly,

f

~0

on A 0

D

(413) Proposition. If f is a nonnegative integrable function, and A an integrable set of positive measure such that f > 0 on AI, then I(j) > O. Proof: It is enough to prove that I(XA f) > O. Hence, replacing f by XAf, we may assume that A = (f > 0). Choose a decreasing sequence (tn) of positive numbers converging to 0 such that An=(f~tn)

is an integrable set for each n It will suffice to find N with J.l(A N) > 0: for then we will have I(f)~ I(XANf) ~ tNJ.l(AN) >

Accordingly, construct an increasing map A: 7l+ A(n)=O

=>

J.l(An)
A(n)= 1

=>

J.l(An»O.

o. -->

{O, 1} such that

and We may assume that J.(1)=0. If J.(n)=O, set Bn=An. If A(n)= 1, set Bn=Am, where m is the unique integer with A(m)=O and J.(m+1)=1. Then each Bn is integrable, and B: c B~ c ... Also, 1J.l(B;)- J.l(B) I <2F 1

(i~j~ 1);

so that (J.l(B n)) is a Cauchy sequence and therefore converges in 1R By (3.9), B V Bn is an integrable set, and J.l(B) = lim J.l(B.). Either

=

n

n-oo

J.l(B»O or J.l(B) <J.l(A). In the former case, choose N in 7l+ with J.l(BN»N- I . Were J.(N)=O we would have BN=AN' and therefore J.l(B N) < N- 1 , a contradiction. Thus )_(N) = 1, and therefore J.l(A N) > 0 In case J.l(B) <J.l(A), choose x with XA(X) = 1 and XB(X)=O. Then f(x»O, and so xEA1 for some N. Were A(N)=O we would have A1 =B1cBI, and therefore XB(X) = 1, a contradiction Thus J.(N)=l and J.l(AN»O. This completes the proof. 0

In order to prove the converse of Proposition (4.13) we need two useful lemmas. (4.14) Lemma. Let f be a nonnegative integrable junction, and (tn) a decreasing sequence of positive numbers converging to 0 such that the

5 Positive Measures on IR

245

complemented set (f ~ t n ) is integrable for each n. Then

lim I(XC!
"-00

Proof: For each n we have

o~ I(X(!
0

(415) Lemma. If f i5 an integrable (unction, and c a positive number, then there exists 15>0 such that II(XAf)l
N)+ I(XA(f A N))


whence II(XAf)I <e.

D

(416) Proposition. If f is an integrable function with I(f»O, then there exist an integrable set A with positive measure, and a positive number r, such that f ~ r on A 1. Proof· With c:=-!I(f), choose 15 as in (4.15). By (411) and (4.14), there exists r>O such that A:=(f~r) is integrable and I(LAf) tI(f), so that, by our choice of 15, Il(A) ~ J D

(4.17) Corollary. There exists an integrable set with positive measure. Proof. This follows from (1.1 3) and (4.16).

D

5. Positive Measures on 1R Theorem (411) enables us to characterize the positive measures on 1R in terms of a generalization of the Riemann integral First, we introduce a convention. If A is a located subset of a 10cally compact space X, then we identify A with the complemented set (A, {XEX: p(x,y»O for all y in A}).

246

Chapter 6 Integration

If A is closed, then this complemented set is just (A, - A), by (3.8) of Chapter 4; in that case we also identify the metric complement of A with the complemented set (- A, A). It should be clear from the context whether or not such identifications are being made Let f be a continuous function on a proper compact interval J == [a, b J in JR. Let a be an increasing function defined on a dense subset A of J, where A contains the end points of J. The Riemannb

Stieltjes integral S f(x) da(x) is introduced by a definition analogous to a

that of the Riemann integral in Chapter 2. Here is a sketch of how this is done; the details are left to the reader. We consider only partitions P == (a o , '" ,an) formed from the points of A, with a o = a and an = b. With each such P we associate sums of the form n

S(f, P) ==

L f(xJ(a(x;) -

rx(x i _ 1))

i= 1

where a i _

1

~ Xi ~ a i .

There exists a unique constant b

L

== S f(x) drx(x)

such that IS(f, P) - LI is arbitrarily small when the mesh of P is sufficiently small. The value of L is not changed if A is replaced by a subset of A which is dense in J and contains the end points of J. The integral L depends linearly on the test function f, and L?; 0 whenever f?; 0 on J. Therefore the equation b

(5.1)

b

Sf dll a == S f(x)drx(x) a

(fE C([a,bJ))

a

defines a positive measure Ila on [a, b J A similar construction associates a positive measure 11(1. on JR with an increasing function rx defined on a dense subset of JR: for each f in C(JR) choose a proper compact interval [a, bJ which supports f and b

whose end points are in the domain of rx, and define Sf dlla.== S f dll", b

a

where Sf dll. is given by (5.1). a

We shall show that every positive measure on JR is of the form 11" for some increasing function rx defined for all but countably many real numbers (5.2) Lemma. Let 11 be a positive measure on JR. Then there exists a set S consisting of all but countably many real numbers, such that for all x, y in S,

5. Positive Measures on IR

247

(i) {x} is integrable and Jl({x}) =0 (ii)

if x
measure (iii) for each e > 0 there exists c'5 x (e) > 0 such that if f is integrable, o~ f ~ 1, and f(t) = 0 for all t in dmn f with Ix - tl ~ c'5 x (e), then Sf dJl<e. Proof. First note that every compact interval J in IR is contained in an integrable proper compact interval K Indeed, if g(x)o:=max{O,1-p(x,J)}

(XEIR),

then gE C(IR) and we can take K to be any integrable set of the form (g ~ r) with 0 < r < 1. It follows that IR is the union of an increasing sequence of integrable proper compact intervals. Hence it will suffice to prove that for each integrable compact interval J 0:= [a, b] with a < b, there exists a sequence (x.) of real numbers such that (i)-(iii) hold whenever x and y belong to (a, b) and are distinct from each x •. To this end, construct h in C(IR) so that h(x)=x for all x in J. Applying (4.11) to the integrable functions (b-h)xJ and (h-a)xJ, we can find a sequence (x.) in IR such that if a < x < b and x =t= x. for each n, then the sets [a,x] and [a,x) are integrable and have the same measure; the sets [x,b] and (x,b] are integrable and have the same measure; and for each e > 0 there exists wx(e) > 0 such that

IJl([a, x])- Jl([a, x'])1 < e and IJl([x, b]) - Jl([x', b])1 < e

whenever a<x'
= X[x,b]- X(y,b] =

X[x,b]- X[y,b]

= X[x,y)

on a full set - namely, the intersection of the full set - {y} and the domains of these characteristic functions. Hence [x, y] and [x, y) are integrable, with measure Jl([x, b]) - Jl([y, b]). This establishes (ii).

248

Chapter 6

Integration

Now let E>O be arbitrary, and define c5==bAE)==tmin{w x (E/2), x-a, b-x} Consider any integrable function f such that 0 ~ f ~ 1, and f(t) = 0 for all t in dmn f with Ix - tl ~ b. Choose an admissible point x' with x-w x(E/2)<x' <x -b Then X(x,b] ~ fX[a,x] + X(x,b] ~ X[x' ,x] + X(x,b] = X[x' ,b]

on a full set - namely, the intersection of - {x} and the domains of these functions. Thus we have j.t((x, b]) ~ I h[a,x] dj.t + j.t((X, b]) ~

j.t([X', b ])<j.t([x, b]) + E/2,

by our choice of x'. Hence O~I h[a,x] dj.t < E/2.

A similar argument shows that O~I h[x,b]dj.t<E/2

Since x is admissible and a < x - b < x + b < b, we have f = fX[a,x] + fX[x,b] on a full set, and therefore O~I f dj.t<E

Thus (iii) holds, and the proof is complete

D

if x, YES and xO there exists b==b(x,Y,E) with O
(5.3) Lemma. Under the conditions of Lemma (5.2),

(5 3.1)

II f

dj.t- j.t([x,y])1 <E

whenever f is integrable, 0 ~ f ~ 1, f(t) = 0 for all t in dmn f with either t~x-b or t~y+b, and f(t)=l for all t in dmnf with x+b~t~y-b

Proof: Let E be an arbitrary positive number. In the notation of (5.2), let b == b(x, y, E) == min {i(y - x), tb x (E/2), tbiE/2)}.

Choose points a, x', y', and b of S so that a<x-b <x+ b <x' <x+2b

and y-2b
5 Positive Measures on IR

249

Consider an arbitrary integrable function f satisfying the properties in the statement of the lemma. By (5.2), we have }L([X', y']L~ Jf d}L

(5.3.2)

J

J

= !x[a.x·] d}L + !x[x'.y'] d}L + < e/2 + }L([x', y']) +e/2

J

!x[Y'.b]

d}L

= }L([x', y']) + e. Also, by (5 2), }L([x', y']) ~ }L([x, y])

(5.3.3)

= }L([x, x']) + }L([x', y']) + }L([y', y])

< e/2 + }L([x', y']) + e/2

(by our choice of b)

= }L([x', y']) + e. Inequality (5.3.1) follows from (5.3.2) and (5.3.3).

D

(5.4) Lemma. Under the conditions of Lemma (5.2), fix Xo in S, and define a map ot: S - {xo} --+ 1R. by ot(X) = - }L([x, xo]) =}L([xo,x]) Then

ot

if XES and x<x o , if XES and x>xo'

extends to an increasing map of S into 1R. such that

(5.4.1 )

ot (y) - ot (x) = }L ([ x, y])

whenever X,YES and x
(5.4.2)

lot(y) - ot(x) - }L([x, y])1 < 3e,

where e > 0 is arbitrary. To this end, we refer to (5.2) and assume first that Xo is so close to x that Xo < y, lot(x)1 < e, and there exists x' in S with x' <min {x,x o}, }L([x',x])<e, and }L([x',xo])<e. Then, by (5.2), }L([x, y]) ~ }L([x', y])

= }L([x', x]) + }L([x, y]) < }L([x, y]) + e. Similarly, }L([X o' y]) ~ }L([x', y]) < }L([x o , y]) + e.

250

Chapter 6 Integration

Hence

la(y) -a(x) -IL([X, y])1 ~ 11L([x o, y]) -IL([X,

y])1 + IIX(x)1

~ 11L([x o , y]) -IL([X', y])1

+ 11L([x', y]) -IL([X, y])1 + B


consisting of all but countably many real numbers, and an increasing function IX: S -> JR, such that IL = IL". Proof: Construct S as in (5.2). Fix Xo in S, and let IX: S -+ JR be the map constructed in (5.4). Let f be an arbitrary test function. We must show that If dlL= I f dlLa= I f(x)dlX(x). Accordingly, let J == [a, b] be a proper compact interval supporting 1. such that a, bES and f(x) = 0 for all x sufficiently close to a or b. Let w be a modulus of continuity for 1. and B any positive number. Choose a partition P == (a o, a 1 , ••• , an) of J consisting of distinct points of S, such that mesh(P)<~w(B), ao=a, an=b, f(al)=O, and (5.5.1)

II f(x)dlX(x)-S(1.P)1
where n

S(1.P)==

L f(a)(IX(aj)-IX(aj _ »· l

j~

1

In the notation of (5.3), let t5==min{a 1 -a, b-an_ l , min t5(a k _ 1 ,ak ,n- 1 B)}. l~k~1I

Then

0< t5 ~ t5(a o, al' n- l B) ~~(al -ao) ~i W(B). Define a continuous map h: JR -+ JR by hex) == min {1, (215- 1 p(x, J) -1)+}

For each j (1 ~j ~ n), define

(XEJR).

t/I j in C(JR) by

t/I j(x) == (1- t5- 1 p(x, [a j _ 1 , aj ]»+

(XEJR).

5 Positive Measures on IR

251

n

Then h+

L IjJk is bounded away from 0 on JR, so that k~1

fj ==

tl

(h + k

IjJ k) - 1 IjJ j n

I

is a well defined element of C(JR). We have: O~fj~l,

j~

fix)=O if Ix-a)~Q)(B), or if x>b, and, by (5.3) and (54),

IS fj d,u -

2~j~n

and x
fj=l on J; 1

l~j~n-l

and

(IX (a) -1X(a j _ 1))1 < n - 1B.

Define a test function

n

g==

I

f(a)fj. 1

j~

Since f(a 1 )=O=f(a n), and fj vanishes outside [a,b] for has support [a,b]. We have

2~j~n-l,

g

n

~

Ilfll LIS fjd,u-(IX(a)-IX(aj_I))1 j~

1

n

~

Ilfll

L n-

1

B=

IlfiIB.

Since f and g have support [a, b], to estimate If(x) to consider the case xEJ. For each j we have

o~ fj(x) If(x) if either

Ix-ajl~Q)(B)

or

If(x) -

f(aj)1

Ix-a)~Q)(B).

g(x)1 =

Ijtl

~

g(x)1 it is enough

fj(X)B

Hence

fj(x)(f(x) - f(aj))1

n

~ L fj(X)B=B. j~

Thus that (5.5.3)

Ilf-gll~B.

Since both

1

f and g are supported by [a,b], it follows

IS f d,u - S gd,ul ~Slf - gl XJ d,u~ B,u([a, b]).

From (5.5.1)-(5.5.3) we obtain

IS f d,u- Sf(x)dlX(x)1 ~B(1 + III I + ,u([a, bJ)). Since B is arbitrary, SI d,u= SI(x)dlX(x), as was to be proved. 0

252

Chapter 6 Integration

6. Approximation by Compact Sets We shall show that, relative to a positive measure on a locally compact space, any integrable set can be approximated from within by a compact integrable set. Call a complemented set A == (A 1, A 0) in a locally compact space X compact if A 1 is compact, and closed if Al is closed. (6.1) Definition. A compact set K in a locally compact space X is to be strongly integrable with respect to a positive measure Jl on there exists c > 0 such that for each E > 0, there exists c5 > 0 such ISfdJl-cl<E whenever fEC(X), O~f~l, f(x)=l for all x in K, f(x)=O for all x with p(x,K)~c5.

said X if that and

Under the conditions of Definition (6.1), the constant c is uniquely determined by the strongly integrable set K. This follows from our next result. (6.2) Proposition. Let K be a strongly integrable set with respect to a

positive measure Jl on a locally compact space X. Then K is integrable and, in the notation of Definition (6.1), Jl(K)=c Proof" There exists a decreasing sequence (c5 n) of positive numbers converging to 0 such that ISfdJl-cl<2-n-1 whenever fEC(X), O~f ~ 1, f(x) = 1 for all x in K, and f(x)=O for all x with p(x, K) ~ c5 n • Construct test functions fl ~ f2 ~ f3 ~ ... such that for each n, O~fn~ 1, fn(x) = 1 for all x in K, and fn(x)=O for all x with p(x, K)~c5n' Then S Ifn+ 1 - fnl dJl = S Un - fn+ 1) dJl ~IS fndJl-cl+IS fn+ 1 dJl-CI <2- n- 1+2- n- 2 <2- n,

and so the series L S Ifn+ 1 - fnl dJl converges Hence n

F=={XEX: Llfn+l(X)-fn(x)1 converges}

is a full set, and the function

with domain F is integrable For each x in F, either x(x) > 0 or X(x)0, and therefore p(x,K)~c5n' for all sufficiently large n; so that x belongs to the closed set K, fn(x) = 1 for

6 ApprOlomation by Compact Sets

253

all n, and therefore X(x) = 1. In case X(x) < 1, fn(x) < 1 for some n; so that XE -K, fn(x)=O for all sufficiently large n, and therefore X(x)=O. Thus X= XK on the full set F; so that K is integrable, and Jl(K)= SXdJl= lim SfndJl=c.

0

n~oo

(6.3) Proposition. Let Jl be a positive measure on a locally compact space X, and ¢ a test function such that O<M=:sup{¢(x): XEX} Then for all but countably many real numbers r with 0 < r < M, the set X(¢,r)=:{xEX: ¢(x)~r}

is strongly integrable. Proof. Let K be a compact support of ¢. Then for each r > 0, X(¢,r)= {xEK: ¢(x)~r}.

By (4.9) of Chapter 4 and (4.11) above, for all but countably many real numbers r with O 0, r' is admissible relative to the excluded sequence in question, O,r) = X(gr) on the full set dmnX(gr)=X(¢,r)u{xEX: ¢(x)
Hence X(¢, r) is integrable and Jl(X(¢, r)) = Jl((¢ ~ r)). Let w be a modulus of continuity for ¢ on X. Let r be admissible, O,r) on X(¢,r)u -X(¢,r) (a full set). On the other hand, if f(x»O, then p(x,X(¢,r))<w(t15 r (8)), so that ¢(x)?;.r-t15 r (8»r', and therefore XX(cJ>,r')(x) = 1. Hence f ;;;;;XX(cJ>,r') on the full set dmn XX(cJ>,r')' Thus Jl(X(¢, r));;;;; Sf dJl ;;;;; Jl(X(¢, r')) < Jl(X(¢, r)) + 8,

and so IS f dJl- Jl(X(¢, r))l < 8. Since 8 is arbitrary, it follows that X( ¢, r) is strongly integrable. 0

254

Chapter 6 Integration

(6.4) Definition. Let C be a complemented set that is integrable with respect to a positive measure J1. on a locally compact space X, and let e be a positive number. We say that C is e-strongly integrable with respect to J1. if there exists a strongly integrable set S such that C 1 cS and J1.(S) - J1.( C) < e. (6.5) Lemma. Let J1. be a positive measure on a locally compact space X, A an integrable set with J.L(A) >0, and e a positive number. Then there exist a closed e-strongly integrable set C, a finite set HeX, and a positive number r < e, such that C 1 cAt, J.L(A) - J.L( C) < e, S c(y, r) is integrable and J.L( C " S c (y, r)) > for each y in H, and p(x, H) < e for each x in C 1 •

°

Proof: We may assume that e<min H, J1.(A)}. Let (fn) be a representation of XA by elements of C(X). The proof hinges on the idea that for a sufficiently large k and for a suitable c in (0,1), the complemented set

(tl fn ~ c) will approximate A as closely as we wish. By (2.15), we

can choose integers 1
L

JlfnldJ1.<2-2j-6e2

(j~I).

n= .(jJ

Define a sequence (gn) of test functions by .(1)-1

gl ==

L

fn

n= 1

and .(jJ- 1

gj==

L

Ifni

U~2).

n= .(j-l)

Since e < J1.(A), it follows from (6.5.1) that there exists x such that IXA(x)- gl(x)1 e. By (6.3) and (4.11), we can choose c in (e/2,e) so that B1=={XEX:

gl(X)~C}

is strongly integrable, and the complemented sets - B j (j ~ 2) are integrable, where

6 ApproXImation by Compact Sets

For each j

~2

255

we then have 2- j - 2cJ1.( -Bj)~S gjdJ1. 00

~ n=

I vU-

Slfnl d J1. 1)

<2- 2j - 4 t;2,

so that 00

I

Hence the series

J1.( -B) converges. Thus

j= 2 00

V

B=B 1 -

00

-Bj = /\ Bj

j= 2

j= 1

is integrable, and

J1.(B)~J1(B1) -J1.(Y2 -Bj)

(6.5.2)

00

~J1.(B1)-

I

J1.( -Bj)~J1.(B1)-t;/4.

j=2

Also, ifXA_ B1(x)=1, then XA(x)=1 and

gl(X)~C;

whence

(1- C)xA-B, (x)~ 1- g 1 (x)= IXA(X) - g 1 (x)l.

Hence (1-e)(J1(A)-J1.(B1»~(1-c)J1.(A-B1)

~SIXA -glldJ1.<e/8,

so that

J1(A) - J1(B 1) < !(1- e)-l e < e/4.

It follows from (6.5.2) that

J1.(A)- J1.(B) = J1.(A)- J1.(B 1) + J1.(B 1)- J1.(B) < e/2.

Note that J1.(B) >0, since e<J1.(A). The required e-strongly integrable set C will be obtained by cutting down B. Before we do this, we show that B l cA1. If xeB 1 , then xeB I cdmng l and vU)-l

L

Ifn(x)1 =gj(x)~2-j-2c

U~2).

n=vU-I) 00

Hence

I

Ifn(x)1 converges; so that XA(X) is defined, and

n= I 00

XA(X)= Lfn(x)~gl(X)n

I

n= 2

gn(x)~c-c/8>0.

256

Chapter 6 Integration

n 00

Since each BJ is closed, so is B1 =

BJ. Let {Xl' ... ' x N } be an 1'./4

j= 1

approximation to B 1. Using (6.3), choose r in (1'./4, 1'./2) so that S C(Xk' r) = {XEX: (1- p(x, x k ))+

~

1-r} N

is strongly integrable for each k. Then B1 c B 1 C map A: {1, ... , N} ---+ {O, 1} so that

US C(Xk' r)

Define a

k= 1

A(k) = 0 => Il(B " S C(Xk' r)) > 0

and A(k) = 1 => Il(B "S C(Xk' r)) < e/2N.

Let Since ktt(B "S C(Xk' r))

~ IllY1 (B "S C(Xk' r))) ~ Il(B) >0,

we may assume that H is nonvoid. By (6.3), there exists 0( in (r,e) such that S= {xEB 1: p(x, H)~O(} = {xEB 1. (1- p(x, H))+ ~ 1- O(} is strongly integrable. Then the complemented set

C=B"S is integrable, C 1 is closed, C 1 cB1 cAl, and p(x,H)~O«e for each in C 1 . Also, if YEH, then

X

Il( C "S c(y, r)) = Il(B "S c(y, r)) > o. N

Now, if XB_dx)=1, then xEB 1c U Sc(xk,r) and p(x,H)~O(>r, so that XE U S C(Xk' r). Thus k= 1 }'(k) = 1

(6.5.3)

Il(B)-Il(C)~Il(B

,,(

V }'(k)

~

L }'(k)

=

SC(Xk,r))) 1

Il(B" SC(Xk' r)) < Ne/2N= e/2

=1

and so Il(A) - Il( C) = Il(A) - 11 (B) + 11 (B) - Il( C) < 1'..

Finally, C 1 c S, S is strongly integrable, and, by (6.5.2) and (6.5.3), Il(S) - Il( C) ~ Il(B 1) - 11 (B) + 11 (B) - Il( C) <1'./4+1'./2<1'..

Hence C is e-strongly integrable.

0

6 Approximation by Compact Sets

257

Recall that a complemented set A is said to be a subset of a complemented set B if Al c:B I and BOc:Ao; we then write A
(6.7)

Theorem. Let 11 be a positive measure on a locally compact space X, and let A be an integrable set with Il(A) > O. Then for each e > 0 there exists a strongly integrable set K < A with Il(A - K) < e.

Proof Consider any e with O<e< 1. Using (6.5), we can construct recursively a sequence (C n ) of integrable sets, a decreasing sequence (rn) of positive numbers converging to 0, a decreasing sequence (en) of positive numbers, and a sequence (Hn) of finite sets, such that (i)

AI

::J

C~

::J q : : J ..

(ii) 11 (A) -Il( C I) < e/2 (iii) for each n, C n is closed and 2- n e-strongly integrable, and Il(C n)-Il(Cn+ 1 )<min {2- n -

I B,

2- nen}

(iv) for each n and each y in H n , Sc(y, rn) is integrable, and Il( C n /\ S c(y, rn» > 2Bn

(v) for each n and each x in C~, p(x, H n )<2- n e. By (i), Il(A) ~ Il( C I) ~ Il( C 2) ~ .... It follows from (iii) that the series converges. Since

L(11 ( Cn) -Il( C n + I»

m-I

Il(C m )=Il(C I ) -

L (Il(C n)-Il(Cn +

n~

I

we see that lim Il( Cm) exists; whence, by (3.9),

is integrable, and (6.7.1)

1»

(m~2),

258

Chapter 6 Integration

For m > n ~ 1 and each Y in H n , we have

m-1

m-1

>2en -

I

m-1

2- ke k>2e n - 2: 2- k en>en·

k~n

k~n

It follows from (6.7.1) that J.l(J "Sc(y,rn))~en' so that J "Sc(y,rn) is nonvoid. Thus there exists a point ¢(y) in J1 such that p(y, ¢(y)) ~ rn. Since p(x,H n)<2- ne for each x in J1, it follows that {¢(Y):YEHn} is a subfinite (rn+2-ne) approximation to J 1. Since rn--+O as n--+oo, we now see that J1 is totally bounded. Also, each Cn is closed, so that J1 is

closed and therefore compact. Write K=.J 1. Since KcA 1, we have KO such that

whenever g is a test function, 0 ~g ~ 1, g(x) = 1 for all x in L, and g(x) = 0 for all x with p(x, L) ~ f3. Fix such a test function g. Let f be a test function such that O~f~l, f(x) = 1 for all x in K, and f(x)=O for all x with p(x,K)~f3. Since K=J 1 cL, we have O~SfdJ.l-J.l(J) ~S(f v g)dJ.l- J.l(L) + J.l(L)- J.l(C m ) + J.l(C m)- J.l(J)

< 0(/3 + 0(/3 + 0(/3 = 0(. Since f and 0( are arbitrary, it follows from (6.1) and (6.2) that K is strongly integrable, with J.l(K) = J.l(J). Finally,

n

>J.l(A)-e/2- 2:2-n-1e =J.l(A)-e,

so that J.l(A-K)=J.l(A)-J.l(K)<e.

0

7 Measurable Functions

259

7. Measurable Functions Let I be a completely extended integral on a set X. We shall introduce an important class of functions which can be approximated by integrable functions. (7.1) Definition. A real-valued function f defined on a full set is measurable if for each integrable set A and each E > 0, there exist an integrable set B and an integrable function g such that Bl c A I, Il(A -B)<E, and If -gl <E on Bl. A measurable set is a complemented set whose characteristic function is measurable. (7.2) Lemma. Let f be a measurable function, and A an integrable set. Then there exists an integrable set B such that Bl c AI, Il(A - B) is arbitrarily small, and f is bounded on Bl.

Proof: Consider first the case where f is integrable. Choose any r > 0 so that the complemented set S == (If I ~ r) is integrable, and write B == A-S. Then B is integrable, B l cAl, and If(x)l 0, we can find an integrable function g and an integrable set K such that KI cAl, Il(A-K)<Ej2, and If-gl<1 on Kl. By the first part of the proof, there exist an integrable set B and a positive constant c such that BI cKl, Il(K -B)<Ej2, and Igl ~c on BI. Then BI cA 1, Il(A -B)<E, and If(x)1 ~ 1 +c for each x in Bl. D n

(7.3) Definition. A function of the form

L akXk,

where a 1 , ••• , an are

k~1

real numbers and Xl"'" Xn are characteristic functions of integrable sets, is called a simple function. Simple functions are clearly integrable. Sums, products, and constant mUltiples of simple functions are also simple functions. (7.4) Lemma. Let A be an integrable set, f a nonnegative, bounded, integrable function which vanishes throughout - A, and E a positive number Then there exists a simple function X such that 0 ~ X ~ f ~ X+BXA'

Proof: Choose c > 0 so that 0 ~f ~ c on a full set F. Compute real numbers ao==O, a 1 , •.. ,an with an>c, such that for each k (l~k~n), 0< a k - ak _ I < e and Bk == (f~ aJ is integrable. Then the complemented

260

Chapter 6 Integration

sets Ao==A-BI and Ak==Bk-Bk+l(l~k~n-l) are integrable. Write Xk == XA. (0 ~ k ~ n - 1), and define the simple function X by n-I X== L akXk· k~O

Let x belong to the full set S==Fn(AI uAO)ndmnxo ndmnXI n ... ndmnXn_l.

n-I

V

If

Xk(x)=l, then there is a unique i (O~i~n-l) with Xi(x)=l; so

k~O

that

ai~f(x)
I, x(x)=a i, and therefore O~f(x)-x(x)
Since f vanishes on AO, it follows that O~X(x)~f(x)~X(X)+exA(X). n-I On the other hand, if V Xk(X)=O and XA(X)= 1, then xEB~ for each k~O

k (0 ~ k ~ n), and hence f(x) ~ an > c, which is impossible. Thus if n-I V Xk(X)=O, then xEAo, and therefore f(x)= X(x)=O. Hence k~O

O~x~f;£x+eXA

on the full set S.

0

The following corollary will be useful later. (7.5) Corollary. If f is a nonnegative integrable function, then there exists a simple function X such that O~x~f and I(f-x) is arbitrarily

small. Proof: Let e be any positive number Using (4.14), choose an integrable set A such that I(X-Af)<e. Choose also N in 7l+ such that I(xAf - XAf 1\ N) < e By (7.4), there exists a simple function X such that O;£x~xAf 1\ N~X+(1

+ J.L(A))-I eXA

on a full set F. Then 0 ~ X~f on F, and I(f- X)=I(X_Af)+I(XAf - XAf 1\ N) +I(xAf 1\ N - X) <e+e+(1+J.L(A))-l eJ.L(A)<3e. 0 (7.6) Lemma. Let f be a measurable function, K an integrable set, and c a positive constant such that If I ~ c. Then for each e > 0 there exist a simple function X and an integrable set B such that BI c KI,

J.L(K -B)<e,

Ixi <2c, and If -xl XB<e.

Proof. We may assume that O<e
7 Measurable Functions

261

such that BlcKI, Jl(K-B)<e, and If-¢I<e/2 on BI. Applying (7.4) to the functions XB¢+ and XB¢-' construct a simple function X such that l¢xB-xl~2-leXB on a full set F. Then X vanishes on FnBo. For each x in FnGnBI we have If(x)-x(x)l<e, and therefore Ix(x)I<2c. Thus If- xl XB<e and Ixi <2c on the full set F n Gn(BI uBO). D (7.7) Proposition. If f and g are measurable functions and c is a real number, then the functions f + g, cf, fg, max {f, g}, min {f, g}, and If I are all measurable.

Proof. We prove only the measurability of fg If g is the characteristic function of an integrable set, then, in view of (3.11), it is trivial to prove that fg is measurable. In the general case, let A be any integrable set, and e any positive number. By (7.2), we can find an integrable set K and a constant c>O such that KlcAt, Jl(A-K)<1.:/2, Ifl~c on K 1, and g is bounded on K I. Applying (76) to the measurable functions fXK and gXK' construct an integrable set B and simple functions XI' X2' such that BI cKt, Jl(K -B) <e/2, If -x I IXB<e/2, and Ig-X 2IxB«2c)-l e. Then BlcAI, Jl(A-B)<e, X==XIX2 is a simple function, and Ifg -xlxB~ Ifllg -x 2IxB+ If -x l lx2xB ~ C(2C)-1 e+e/2=e. Hence fg is measurable D Two measurable sets A and B are disjoint if XA XB = O. Two subsets of a set are distinct if there is an element of one of them that does not belong to the other. Further progress depends on a simple, but crucial, combinatorial lemma. (7.8) Lemma. Let A(l), ... ,A(n) be integrable sets, and aI' ... ,an real numbers. Let P consist of all the detachable subsets of {l, ... , n}, and for each S in P let

A(S) ==U\ A(i)) t\ ( ieS

1\ -A(i)) ie-S

(an integrable set). Then A(S) and A(T) are disjoint when Sand Tare distinct elements of P. Moreover, if P(i)== {SEP: iES}, then XA(i)= LXA(S)

(i=l, ... ,n)

SEP(i)

and n

La;XA(i)= L(La;)XA(S) i= 1

n(A(i)1 uA(i)O). n

on the full set

i= 1

SeP ieS

262

Chapter 6 Integration

Proof: If Sand T are distinct elements of P, then A(S)l eA(i)l and A(T)l eA(i)O for some i, so that A(S) and A(T) are disjoint. Let x be any point of the full set

n n

F=

(A(i)l uA(i)O).

i~

Then for each

1

(l~i~n), xEA(i)l

xEA(i)° if and only if XE

U A(S)t,

if and only if XE

n A(S)o. Therefore XA(i)=

SEP(i)

the sets A (S) are disjoint, it follows that XA(i) = x in the full set F we have

L

and

SEP(i)

V

XA(S)' Since

SEP(i)

XA(S)' Also, for each

SEP(i)

n

L ai XA(i)(X) = L ai L i~

1

i~

1

XA(S)(X)

SEP(i)

(7.9) Proposition. If f is a measurable function and ¢: IR ~ IR is continuous, then ¢ of is measurable. Proof: Consider any integrable set A and any e > O. Choose an integrable set K such that K leA t, J.L(A - K) < e, and If I is bounded on Kl by some constant c > O. Let 1J be a modulus of continuity for ¢ on the interval [-2c,2cJ. Using (7.6), choose an integrable set B and a simple function X such that Bl eKl, J.L(K -B)<e, Ixi <2c, and n

If-xIXB<1J(e). In view of (7.8), we may assume that

x= L akXk' where k~

1

al' ... , an are real numbers and Xl' ... ' Xn are characteristic functions of disjoint integrable sets. Then BleAt, J.L(A-B)<2e, n

n

XB(¢oX)=¢(O)xBIl(1-xk)+ k~

is a simple function, and measurable. D

1

L ¢(ak)XkXB k~

1

l¢of-xB(¢oX)IXB~e.

Thus ¢of is

(7.10) Corollary. If f is a measurable function that is bounded away from 0 on a full set, then f - 1 is measurable.

Proof" Choose c>O so that If I ~c on a full set, and let ¢: 1R~1R be the unique continuous function such that ¢(x)=x- l if x< -c or x>c, and ¢ is linear on the interval [- c, c]. Then ¢ of is measurable, by (7.9), and ¢of=f-l on a full set. Thus f- 1 is measurable. 0

7 Measurable Functions

263

The next theorem provides us with one of the most important methods of proving that certain measurable functions are integrable. (7.11) Theorem. If f is a measurable function, and g is an integrable function such that If I ~g, then f is integrable.

Proof: Choose a decreasing sequence (rn) of positive numbers converging to 0 such that each of the complemented sets A(n)==(g~rn)

is integrable. We shall construct an integrable set B(n) approximating A(n) from within, and an integrable function fn approximating f on B(n)1, such that (i) if xEB(n)1 for all sufficiently large n, then limfn(x)=f(x) ,,~oo

n

(iii)lim Ifn(x)1 XB(n)(x)~2g(x) for all x in a full set S1 (iv) lim (XA(n) (x) - XB(n) (x)) = 0 for all x in a full set S2. To this end, first observe that, by (4.14), O~I(g) -

as N -+ 00.

I(XA(N)g) =1(X-A(N)g)-+O

Hence Ll«XA(n+ 1) - XA(n») g) = I (g) - I(xA(1)g)·

"

Define En

== min {rn' 2-"(1 + J.l(A(n)))- 1}

(nEZ+).

Since f is measurable, it follows from (4.15) that for each n in Z+ we can construct an integrable set B(n) and an integrable function f" such that B(n)lcA(n)l, J.l(A(n)-B(n))<2- n, I(XA(n)-B(,,)g) <2- n, and If - f,,1 < E" on B(n)1. This last property ensures that (i) holds. Clearly, the series LJ.l(A(n)-B(n)), Ll(xA(n)-B(n)g), and LE"J.l(A,,) all converge. n

n

n

Also, If I ~g and

Ifni XB(")~E,,XB(")+ If I ~EnXA(n)+ g ~ En r,,- 1 g + g ~ 2 g

on a full set F,.. Hence l(1fn+ 1 XB(,,+ 1) - fnXB(n)!) ~ 1(lfn+ 1 - fnl XB(,,+ 1) 1\ B(n») + l(1f,,+ 11 XB(n+ 1)-B(n»)

+ 1(lf,,1 XB(II)-B(n+ 1»)

264

Chapter 6

Integration

~ (En +

En+ I) IL(A(n + 1) /\ A (n)) + 2I(XA(n+ 1)- B(n)g) +2I(XA(n+ I)-B(H I)g)

~ En IL(A(n)) +

En+ I IL(A(n + 1)) + 21 (XA(n+ 1)- A(n)g) + 2I(XA(n)_ B(n)g) + 2I (XA(n+ 1)- B(n+ I)g)·

It follows that (ii) holds; so that

¢ ==fl XB(I)+ L(fn+ I XB(n+ 1)-fnXB(n»)

" is an integrable function defined on a full set contained in (B(n)1 u B(n)O). Clearly, lim Ifn(x)1 XB(n)(x) ~ 2g(x) for all x in the full

n n

n-oo

set

SI ==(nF,,)ndmn¢, n

so that (iii) holds On the other hand, since LIL(A (n) - B(n)) converges n

and XB(k)~XA(k) (k~I), L(XA(n)-XB(n») converges on a full set S2' Conn

dition (iv) follows immediately. Now consider any point x in the full set F==SI nS 2 , and any E>O Either If(x)-¢(x)I<E

(7.11.1)

or f(xH= ¢(x). In the latter case, choose N in 7l+ so that If(x) -¢(x)I~3rN' Then 3rN ~ If(x)1 + 1¢(x)1 ~ 3g(x)

and so g(x)~rN' Hence xEA~ for all n~N; so that xEB(n)l for all sufficiently large n, by (iv). Thus, by (i) and the definition of ¢, ¢(x) = f(x). Therefore (7.11.1) holds in both cases. Since x and E are arbitrary, it follows that f = ¢ on F; whence f is integrable, by (29). 0 (7.12) Corollary. If f is an integrable function and A is a measurable

set, then XAf and X- Af are integrable functions. Proof: Since IXAfl ~ If I on the full set (A I u A 0) n dmnf, XA f is integrable, by (7.7) and (711). Thus LAf =f- XAf is integrable. 0 (7 13) Proposition. Let f

be a measurable function, A an integrable set, and c a positive number. Then the function - c v (c /\ XAf) is integrable.

8 Convergence of Functions and Integrals

265

By (7.7), the function - c v (C A XA f) is measurable. Since on the full set (AluAO)ndmnf, the result follows from (7 11). D Proof

I-CV(CAXAf)I~cXA

8. Convergence of Functions and Integrals Let I be a completely extended integral on a set X. Proximity in the class of measurable functions is most conveniently described by a notion of convergence. Here are three such notions. (8.1) Definition. Let (J..) be a sequence of measurable functions, and f a real-valued function defined on a full set. The sequence Un) converges to f in measure if to each integrable set A and each e> 0 there corresponds N in '1.+ such that for each n ~ N, there exists an integrable set B with BleAt, Jl(A-B)<e, and If-J..I<e on BI. The sequence Un) converges to f almost everywhere if to each integrable set A and each E > 0 there correspond N in '1.+ and an integrable set B such that BleAI, Jl(A-B)<e, and If-fnl<e on BI for all n~N. The sequence Un) converges to f almost uniformly if to each integrable set A and each e > 0 there corresponds an integrable set B such that BI cAl, Jl(A-B)<e, and (fn) converges to f uniformly on BI. Clearly, almost-uniform convergence entails convergence almost everywhere, and convergence almost everywhere entails convergence in measure. We first prove the measurability of the limit function f in the case of convergence in measure (and therefore in the cases of convergence almost everywhere and almost uniform convergence). (8.2) Proposition. If a sequence Un) of measurable functions converges in measure to a function f, then f is measurable.

Proof. Let A be any integrable set, and e any posItIve number. By (8.1), we can find n in '1.+ and an integrable set C such that C l e A I, Jl(A - C) < E/2, and 1 f(x) - fn(x)1 < el2 for all x in C 1 . Since J.. is measurable, there exist an integrable function g and an integrable set B with BleC I , Jl(C-B)<eI2, and Ifn(x)-g(x)I<E/2 for all x in BI. Hence BI eA 1, Jl(A - B) <e, and If(x) - g(x)1 <E for all x in BI. Therefore f is measurable. D

266

Chapter 6 Integration

(8.3) Proposition. If (fn) is a sequence of measurable functions converg-

ing in measure to each of two integrable functions f and g, then f on a full set.

=

g

Proof Consider any r > 0 for which A=(lf-gl~r)

is an integrable set. Given e in (0, r), choose a positive integer Nand integrable sets B, C such that Bl cAl, f.1(A-B)<e/2, and If-fNI<e/2 on Bl; and C 1 c A 1, f.1(A - C) < e/2, and Ig - fNI < e/2 on C 1 • Then If-gl<e
function such that Ifni ~g for all n. Then If I ~g on a full set. Proof. Let F be a full set on which Ifnl ~ g for all n. Consider any r > 0 for which A =(Ifl-g~r) is an integrable set. Let e be any positive number. Then there exist a positive integer N and an integrable set B such that Bl cAl, f.1(A -B) <e, and If - fNI
IfN(X)I- g(x) = If(x)l- g(x) + IfN(X)I-lf(x)1 ~ r -If(x) -

fN(X)1 > r/2.

Since this contradicts our choice of F, we must have XB=O on the full set F n(BI uBO). Hence f.1(B) = 0, and therefore f.1(A)<e. Since e is arbitrary, it follows that f.1(A) =0. Thus If I ~g on a full set, by (4.12). D (8.5) Lemma. Let fl ~ f2 ~ . .. be an increasing sequence of integrable

functions converging in measure to an integrable function f. Then fn ~ f for all n. Proof The proof is similar to the proofs of the last two results, and is left to the reader. D We say that a sequence (fn) of functions converges to a function f pointwise on a set F if (fn(x)) converges to f(x) for each x in F. We now come to some results about the interchange of integrals and limits. The first of these is Lebesgue's monotone convergence theorem

8 Convergence of Functions and Integrals

267

(8.6) Theorem. Let (i.) be an increasing sequence of integrable func-

tions. Then (i.) converges in measure to some integrable function f if and only if t == lim 1 (f.) exists, in which case (f.) converges almost everywhere, and pointwise on a full set, to f, and 1 (i) = Proof Assume first that

t.

t exists. Then •

is an integrable function defined on the full set on which the right side converges, l(i)=t, f(x)= limf.(x) for all x in the domain of f, and .~oo

f. ~f

for each n. Let A be an integrable set, and Choose N in 7l+ so that

E

a positive number.

Choose also tJ in (E/2, E) so that the complemented set (i - fN ~ tJ) is integrable. Let B == A - (f - fN ~ tJ) Then B is an integrable set, Bl cAl, Il(A-B)~tJ-l l(f-fN)<E,

and on a full set, O~(i-f.) xB~(i-fN)xB< tJ

<E

(n~

N).

Thus (i.) converges almost everywhere to f Conversely, suppose that (i.) converges in measure to an integrable function f Replacing f. by f.-fl' if necessary, we may assume that f. ~ 0 for each n. By (8.5), f. ~ f on a full set for all n. Given E > 0, choose an integrable set A such that I(LAf)<E By (8.1) and (4.15), there exist a positive integer N and an integrable set B such that Bl cAl, I(XA_Bf)<E, and If-fNlxB«1+Il(A))-lE. For each n~N we have

o~ I(f) -

1 (f.) ~ I(f) - 1 (fN)

~I(x_ Af) + I(XA_Bf) + 1 (xB(i-fN))

<E +E + Il(B)(1 + Il(A))-l E< 3E. Hence 1 (f) = lim 1 (i.). It follows from the first part of the proof, and (8.3), that (i.) converges to full set. D

f almost everywhere, and pointwise on a

Minor adaptations of an argument used in the first part of the above proof enable one to show that if f and i. (n ~ 1) are integrable functions such that lim 1(lf-f.I)=O, then (fn) converges to f in measure. .~oo

268

Chapter 6 Integration

The next result gives a good criterion for the convergence of an integral to O. (8.7) Proposition. Let Un) be a sequence of nonnegative integrable functions converging in measure to O. Suppose that for each E > 0 there exist an integrable set A and a positive integer N such that I(XBfn)<E whenever n ~ Nand" B is a measurable set with Jl(A 1\ B) < N- 1. Then I Un)-O as n--> 00. Proof. Given E > 0, choose A and N as described above. There exists no ~ N such that to each n ~ no there corresponds an integrable set K with K 1 cA1, Jl(A-K)
o~ I Un) = I (XBfn) + I (XKfn) < E + Jl(K)(l + Jl(A))- 1 E < 2E whenever

n~no

Therefore IUn)-O as n-->oo.

0

We now arrive at Lebesgue's dominated convergence theorem. (8.8) Theorem. Let (fn) be a sequence of integrable functions converging in measure to an integrable function f, and let g be an integrable function such that Ifn I ~ g for all n. Then lim I (I f - fn I) = o. Proof" By (8.7), it is enough to show that for each E > 0 there exist an integrable set A and a positive integer N such that I(XBlf-fnl)<E whenever n~N and B is a measurable set with Jl(AI\B) 0, choose an integrable set A with I(LAg)<Ej4. By (4.l5), there exists N in 7l+ such that I(XKg)<Ej4 for each integrable set K with Jl(K) < N - 1. Consider any measurable set B with Jl(A 1\ B) < N- 1. Since XB= XAAB+ XB-A~XAAB+ X-A on a full set, for any integer n ~ 1 we have I(XB If-fnl)~I(XA"B If-fnl)+I(LA If-fnl} ~2I(XA" Bg) + 2I(LAg) < E Since

E

is arbitrary, the proof is complete.

0

Before going any further, we introduce two important types of integration space.

g Convergence of Functions and Integrals

269

(8.9) Definition. The integration space (X, L 1 , I), or just the integral I itself, is finite if the constant function 1 is integrable. The integration space, or just I, is a-finite if there exists a sequence (Kn) of integrable sets such that K~cKic ... , UK~ is a full set, and (XK) converges in

•

measure to 1; in that case, the sequence (Kn) is called an I -basis of X. Clearly, if I is finite, then it is a-finite. The reader is invited to verify that the complete extension of a positive measure on a locally compact space is a-finite. (8.10) Proposition. Suppose that I is a-finite, let (Kn) be an I-basis of X, and let f be a nonnegative measurable function. Then f is integrable if and only if fxKn is integrable for each nand t == lim I (j XKJ exists .. in n~a: which case I(f)=t. Proof: It readily follows from the appropriate definitions that the sequence (fxK) converges to f in measure, and pointwise on the full set

The result now follows from the monotone convergence theorem, (3 11), and (2.9) 0 A special case of Proposition (8.10) occurs when f is the characteristic function of a measurable set A. In that case, A is an integrable set if and only if each of the complemented sets A A K. is integrable and lim J1(A A Kn) exists. n~oo

(8.11) Lemma. Let I be a-finite, and let (A.) be a sequence of measurable sets such that J1(E) = lim J1(E A An) for each integrable set E. Then UA~ is a full set. n~oo n

Proof: First observe that E A An and E - A. are integrable for any n and any integrable set E, by (7.12). Let (Kn) be an I-basis of X. Write S== V A •. For all j, kin '1.+ we have

•

J1l~1 (K j -

A.»)

~J1(Kj- A

k

)= J1(K) - J1(K j A Ak)·

Since J1(K j AA k ) ..... J1(K j) as k ..... oo, it follows that

!~~ J1C~1(Kj-A.»)

=0. By (3.8), Kj-S= /\ (Kj-A.) is integrable and has measure O. It n

270

Chapter 6

Integration

follows from (8.10) that - S is integrable, with J1( -S)= lim J1(K j -S)=O. j-oo

Hence (_S)O is a full set. Since (-S)o=SleUA~, the result follows. 0 (8.12) Lemma. Let I be (i-finite, and let (in) be a sequence of measurable functions converging in measure to each of two measurable functions f and g. Then f = g on a full set. Proof. Let (Kn) be an I-basis of X. For each pair (n,j) of positive integers choose an integrable set A(n,j) such that A(n,N e K~, J1(K n - A(n,j» <2- j n-t, and If -gl
to a sum less than n- l , we see from (3.10) that is integrable and has measure less than n - 1 Thus An= /\ A(n,j)=K nj

V (Kn-A(n,j» j

V (Kn-A(n,j» j

is integrable, and J1(K n-A n)
by (8.11). Since

f = g on

A~

for each n, the result follows.

D

For each of the notions of convergence introduced in Definition (8.1) there is a corresponding notion of Cauchyness. (813) Definition. Let (in) be a sequence of measurable functions. We say that (in) is Cauchy in measure if to each integrable set A and each e > 0 there corresponds N in '1.+ such that for all m, n ~ N, there exists an integrable set B with Bl cAl, J1(A-B)<e, and Ifm-fnl <e on Bl. The sequence (fn) is Cauchy almost everywhere if to each integrable set A and each e>O there correspond N in '1.+ and an integrable set B such that BleAt, J1(A-B)<e, and Ifm-fnl<e on Bl for all m, n~N. The sequence (in) is Cauchy almost uniformly if to each integrable set A and each e > 0 there corresponds an integrable set B with BI e A I and J1(A - B) < e, such that for each (; > 0 there exists N in 'I. + with Ifm-fnl<(; on BI for all m, n~N. It is obvious that a sequence which is Cauchy almost uniformly is Cauchy almost everywhere, and that a sequence which is Cauchy

8 Convergence of Functions and Integrals

271

almost everywhere is Cauchy in measure. It is also obvious that a sequence of measurable functions which converges in one of the three possible senses is Cauchy in the corresponding sense. In order to prove a converse of this last remark for a-finite spaces, we need two lemmas. (8.14) Lemma. Suppose that I is a-finite, and let (A.) be a sequence of

measurable sets such that Il(E) = lim Il(E

1\

An) for each integrable set E.

Let (f.) be a sequence of measurable functions converging uniformly to a function f: UA! ---+ 1R on each A; Then (f.) converges almost uniform• ly to f Proof Note that f is defined on a full set, by (8.11). For any integrable set E, Il(E-EI\A N ) can be made arbitrarily small by choosing N large enough. Since (f.) converges uniformly to f on E1 n A1, the result follows. D (8.15) Lemma. Let (f.) be a sequence of measurable functions which is

Cauchy in measure, and let (f.(k»);;'~ 1 be a subsequence of (f.) which converges almost uniformly to a function f Then (f.) converges to f in measure. Proof: Consider any integrable set A and any e > O. Choose an integrable set K and a positive integer N1 such that K1 cAl, Il(A - K) < e/2, and If - f.(k) I< e/2 on K 1 for all k ~ N 1 . Then choose an integer N ~ N1 such that for all j, k ~ N there exists an integrable set BU,k) with BU,k)lcKl, Il(K-BU,k))<e/2, and Ifj-fkl<e/2 on BU,k)l. For each k~N the set B(k,n(k)) is integrable, B(k,n(k))lcAl, and Il(A-B(k,n(k)))<e. Also, if xEB(k,n(k))l, then If(x) - fk(x)1 ~ If(x) - f.(k)(x)1

+ If..(k) (x) -

fk(x)1

<e/2+e/2=e.

Hence (f.) converges to

f in measure. D

(8.16) Theorem. Suppose that I is a-finite, and let (f.) be a sequence of

measurable functions. If (f.) is Cauchy almost everywhere, then (f.) converges almost uniformly, and pointwise on a full set, to a measurable function. If (f.) is Cauchy in measure, then there is a measurable function f such that (f.) converges to f in measure, and some subsequence of (f.) converges to f almost uniformly, and pointwise on a full set.

272

Chapter 6

Integration

Proof' Let (Kn) be an I-basis of X. Consider first a sequence (fn) which is Cauchy almost everywhere For arbitrary positive integers m and n, choose an integrable set B(m,n) and a positive integer N(m,n) such that B(m,n)lcK~, J1(K n-B(m,n))<2- mn- l , and Ifi-fjIXB(m,n) <2- m for all i, j~N(m,n). By (310), V(Kn-B(m,n)) is integrable and has measure less than n -1 ; so that m An=

A B(m, n)=Kn- V(Kn-B(m,n)) m

m

is integrable, and )1(Kn-An)
UA~

is a full set, by (8.11). On the other hand, the completeness of 1R

n

ensures that (in) converges uniformly, and hence pointwise, to a function i UA~---+1R on each of the sets A~. By (8.14), (fn) converges n

i

almost uniformly to

Consider next a sequence (fn) which is Cauchy in measure We construct subsequences (fn,k)'t~ I of (in)' and integrable sets An' such that for each n in 7l+, (fn,S;'~ 1 is a subsequence of (fn- 1,k);;"~ 1; (fn,k);;"~1 converges uniformly on A~, A~cK~; and J1(K n -A n)
(310),

Kn-An=V(Kn-Bk(N(k),N(k+l)))

IS

integrable,

and

k

J1(K n - An) < n- 1; hence An is integrable To complete the induction, we

observe that, since 1R is complete, (In,k)':'~ 1 converges uniformly on A~. For each m, all but at most a finite number of terms of the sequence (fn,.)':~ 1 are also terms of (fm,k)':'~ 1 Hence (fn,.)':'~ 1 converges uniformly, and therefore pointwise, to a function i' U A~ ---+ 1R on each n

of the sets A~. Since (Kn) is an I-basis, and J1(K n -A n)
UA~

is a full set Also, (fn,n) converges almost uniformly to i,

n

by (8.14). Since (fn.n) is a subsequence of (fn), and the latter sequence is Cauchy in measure, (fn) converges to i in measure, by (8.15). D

8 Convergence of Functions and Integrals

273

It follows from Theorem (816) and Lemma (8.12) that, relative to a u-finite integral, almost-uniform convergence and convergence almost everywhere are equivalent properties.

(8.17) Lemma. Suppose that I i~ finite. Let f be an integrable function, and c a positive number such that If I ~ c. Then there is a sequence (Xn) of simple functions converging to f almost uniformly, and pointwise on a full set, such that c ~ Xl ~ X2 ~ ... Prooj By (7.5), there exists an increasing sequence (gn) of nonnegative simple functions such that gn ~ C - f for each n in 7L+, and I (c - f - gn)--+O as n--+ 00. By the monotone convergence theorem, (gn) converges almost everywhere to an integrable function g such that I (g) = I(c -1). It follows from (8 16) and (8 12) that (gn) converges to g almost uniformly, and pointwise on a full set. Hence g ~ c - f; so that Ilc-f-glll=I(c-f-g)=O, and therefore g=c-f on a full set. Thus (gn) converges to c - j almost uniformly, and pointwise on a full set. Writing Xn == c - gn (n ~ 1), we see that Xn is a simple function (since I is finite); c ~ Xl ~ X2 ~ ... on a full set; and (Xn) converges to f almost uniformly, and pointwise on a full set. 0

Observe that if B is an integrable set of positive measure, then fl-+ I (XBf) is an integral on L l . Let I B denote the complete extension of this integral, and /lB(E) the measure of a complemented set E that is integrable with respect to lB. (8.18) Lemma. Suppose that I is u-finite, and let B be an integrable set with positive measure. Then the integral IBis jinite, and I B(l) = /l(B). Proof Let (K (n));.""~ 1 be an I-basis of X Since I B(XK(n») = /l(B

1\

K (n))--+ /l(B)

as n--+ 00,

the monotone convergence theorem shows that, relative to I B' 1 converges pointwise on a full set to an integrable function f with IBU)=/l(B). But, by (8.9), (816), and (812), (XK(n») converges pointwise to 1 on a set that is full relative to I Since every I -full set is clearly an I B-full set, the result now follows. 0 (XK(n»);."'~

(8.19) Lemma. Let B be an integrable set with positive measure, and E a complemented set that is integrable with respect to lB. Then A == (E l nBl, EOuBO) is integrable with respect to I, and I1(A)=I1B(E)

274

Chapter 6 Integration

Proof: Choose a sequence (fn) of functions integrable with respect to I such that IIB(lfnl) converges, and Ifn(x) = XE(X) whenever L lfn(x)l n

n

n

converges. Then II(lXBfnl) converges; so that, relative to I,

n

is a full set contained in Bl u BO, the function LXBfn with domain F is integrable, and n n

Since

=0

if xEF (lBo,

we see that LXBfn= XA on F. The result now follows.

D

n

Using Lemma (8.19), the reader may show that if f is a measurable function relative to I, then f is measurable relative to I B for any integrable set B with positive measure. It is very convenient to have at hand a measurable set on which a given measurable function is less than a given real number. The theorem which makes available such sets bears a strong resemblance to the first part of Theorem (4.11) above. (8.20) Theorem. If I is a-finite and f is a measurable function, then for all but countably many real numbers r:x there exists a measurable set A such that f < r:x on A 1 and f ~ r:x on A 0 .

Proof" It is enough to prove the assertion for all but countably many in the interior of each interval of the form J == [ - c, c], with c > O. For this purpose, we may replace f by -cv(CI\f), so that lfl~c. To begin with, assume that I is finite; so that f is integrable, by (7.11). Construct a full set F and a sequence (Xn) of simple functions converging almost uniformly to f, such that for each x in F, c ~ X1 (x) ~ X2 (x) ~ ... and lim Xn(x) = f(x). In view of (7.8) we may assume that n--> 00 r:x

N(n)

Xn =

I i= 1

a(n, k) XA(n,kP

8 Convergence of Functions and Integrals

275

where a(n,I), ... ,a(n,N(n)) are real numbers, and A(n,I), ... ,A(n,N(n» are disjoint integrable sets. For each h in C(J) and each integrable function g with igi ~ c, the function hog is measurable (by (7.9)) and bounded; since 1 is finite, it follows from (7.11) that hog is integrable Hence Jhdv=l(hof)

(hEC(J))

defines a posItIve, and clearly finite, measure v on C(J). By (4.11), there is a set S consisting of all but countably many points of ( -c, c), such that the following properties hold for each a in S.

*

*

(i) a a(n, k) for all nand k, and a 0 (ii) The complemented set [ -c, a] is integrable with respect to v. (iii)For each a>O there exists b>O such that if a'ES and ia-a'iO, the complemented set (c+Xn~c+r) is integrable, and for each k (1~k~N(n» either a(n,k)
If xEAl, then xEA(n)l nF, and therefore f(x)~Xn(x)
Xl~X2~""

V XA(k)=XA(n)

we have

on the full set Fn(n(A(n)l n

k=l

A (n)O». It follows from (3.8) that in order to prove that A is integrable, it will suffice to show that U

(8.20.1)

lim Jl(A(n»=v([ -c,a]).

To this end, let a be any positive number, and choose b as in (iii) above. We may assume that -c~a-b
~

Xn) - a

(since g is decreasing)

I(XA(n)-a= Jl(A(n»-a.

276

Chapter 6

Integration

Since this holds for arbitrary nand

(8.20.2)

E,

we conclude that

v([ -c,O(])~Jl(A(n))

(nEZ+).

With E and 11 as before, next construct a decreasing function h in C(J) with O~h~l, h(x)=l whenever -c~x~0(-11/2, and h(x)=O whenever 0( ~ x ~ c. Choose 0(' in S so that 0( - 11 < 0(' < 0( - 11/2 Then h~X[-c,a'l and so Shdv~ v([ -c, O(']»v([ -c, o(])-E.

Now, since h is continuous, (h a Xn) converges almost uniformly to h of; also, h a Xn ~ h of for each n. By the dominated convergence theorem, l(hoXn)---+/(hof) as n---+oo. Therefore we can choose N so that l(hoX.)~/(hof)-E for all n~N. For such n, since hOX,~XA(') on the full set A(n)luA(n)D, we have Jl(A(n))~v([ -c,O(])

(by (8.20.2))

< Shdv + e = I(h a f)+ e ~ I(h a X.) + 2e ~ I(XA(.)) +28 = Jl(A(n)) + 28 Since e is arbitrary, (8.20.1) now follows. This completes the proof in the case where I is finite. Next consider the general case. Choose an I-basis (K.) of X, and write Bl==Kl' B.==K.-K._l (n~2). Let K be any integrable set with positive measure. By (8 18) and the case just considered, there is a set S consisting of all but countably many real numbers, such that if 0( E S and n E Z+, then there exists a complemented set E., integrable with respect to I K v Bn' such that XBJ < 0( on E~ and XBJ ~ 0( on E~. By (8 19), A.==(E~ n(K v BY, E~ u(K v B.)O)

is integrable with respect to I; clearly, B~ nA~. Let

..

f < 0(

1\

Bn

on A~ and

f

~ 0( on

.

F ==(U K~) n(n (K~ uK~)) n(n (A! uA~)) and

A

== V (A! nF, A~ nF) •

Then F is a full set, and for each x in F we have x E Bt for a unique value of k. It is easy to show that N

XA(X)=

L H=

1

XAn(x)

(xEFnK1, NEZ+)

9 Product Integrals

277

For each integrable set E and each N in '1.+ we therefore have (XA -

n

tl

XAn) XE A KN = 0

on the full set (EI uEO)n(K~uK~)nF.

Since It(E-E/\KN)-+O as N-+oo, it follows that

(tIXAn):~1

con-

verges to XA in measure. Hence A is measurable. Clearly, f < Q( on A I. On the other hand, if x E A 0, choose N so that x E B~. Since XAN(X);:::;XA(X) =0, we have xEB~nA~ and therefore f(x)~Q(. D

9. Product Integrals We shall construct the product of two integration spaces, and study the resulting product integral. The first step is to define the product of two complemented sets. (9.1) Definition. Let X and Y be two sets with inequality relations, A a complemented set in X, and B a complemented set in Y. The product of A and B is the complemented set

Clearly, A x B has characteristic function (x, y)t-+ XA (X)xB(Y)' with domain (A I u A 0) X (BI U BO). For the remainder of this section, let (X, L(I),I) and (Y, L(J), J) be integration spaces Write Il(A) for the measure of a complemented set A which is integrable relative to I, and v(B) for the measure of a complemented set B which is integrable relative to J. Let L be the subset of ff(X x Y) consisting of all functions of the form n

X ==

L ci XA(i)

x

B(i»

i= 1

where C l ' ... 'C n are real numbers; AI' ... ,A n are integrable sets relative to I; B I' ... ,Bn are integrable sets relative to J; and the domain of X is

n«A (i)l uA(i)o) x (B(i)l uB(i)O)). n

i= 1

278

Chapter 6 Integration

Consider

X == L Ci XA(i) x B(i)

any

L.

In

For

each

x

In

n(A (i)l uA(i)O) the function YHX(X,y) defined on n(B(i)l u B(i)O) is i= 1

n

n

i= 1

i= 1

simple relative to J, and has integral n

J(X)(X) == L ci v(B(i» XA(i) (x).

,=

1

n(A (i)l uA(i)O), n

This defines a simple function J(X) on the full set

i= 1

n

with I(J(X»= LC,J.L(A(i»v(B(i). Note that if X is everywhere non-

,=

1

negative, then so is J(X). n

(9.2) Definition. We say that an element L c, XA(,) x B(,) of L is a simple

,=

1

expression if whenever i =t= j, the integrable sets A (i) and A U) are either disjoint or equal; the integrable sets B(i) and BU) are either disjoint or equal; and A(i) and ACJ) are disjoint if B(i)=BU). n

Clearly,

L c, XA(,) x

,=

is a simple expression if the first two con-

B(,)

1

ditions of Definition (9.2) hold, and if also B(i) and BU) are disjoint whenever A (i) = A U) and i =t= j. (9.3) Lemma. Let the complemented sets A(l), ... ,A(n) be integrable with respect to I, let the complemented sets B(l), ... ,B(n) be integrable with respect to J, and let c l , •.. , C n be real numbers. Let P consist of all detachable subsets of {l, ... ,n}; for each S in P let A(S)==U\A(i»"( /\ -A(i), iES

iE-S

B(S) == (/\ B(i) " (/\ - B(i); ieS

and let

iE-S

n«A(i)l uA(i)O) x (B(i)l uB(i)°». n

H==

i= 1

For each i write P(i)=={SEP: iES}. Then (i) A(S) and A(T) (respectively, B(S) and B(T» are disjoint when Sand T are distinct elements of P (ii) for each i, XA(,) x B(,) = L XA(S) x B(T) on H SnTeP(,)

(iii) the function

L ( L c,) XA(S) x B(T) is a simple expression, defined S,TeP ,eSnT n

and equal to L

,=

n

C,

XA(,) x B(,) throughout H

1

(iv) L c,J.L(A(i)v(B(i»= L (

,=

1

L

S,TeP ,eSnT

c,) J.L(A(S» v(B(T».

9. Product Integrals

279

Proof: The proof is similar to that of Lemma (7.8), and is left to the reader. D

(9.4) Lemma. Let c l ' ... 'C n and d l , ... ,dm be real numbers; let A(l), ... ,A(n) and E(l), ... ,E(m) be integrable sets relative to I; and let B(l), ... ,B(n) and F(l), ... ,F(m) be integrable sets relative to J. Define elements f and g of L by n

f=.

L1

CiXA(i)xB(i)

i~

and m

g=.

L1

djXE(j)

X

F(j)·

j~

Suppose that f = g on dmnf n dmng. Then n

m

L i~

Ci

JL(A(i)) v(B(i)) =

1

Ld j~

j

JL(EU)) v (FU)).

1

Proof: Introducing coefficients equal to 0, if necessary, we may assume that m=n, and that A(i)=E(i) and B(i)=F(i) for 1 ~i~n. Define the sets P, A(S), B(T) as in Lemma (9.3). Write n

(X

L C JL(A(i)) v (B(i))

=.

i

i~

1

and n

/3 =. L di JL(A(i)) v(B(i)). i~

Let

1

n(A(i)1 uA(i)O) n

F=.

i~

1

and

n(B(i)1 uB(i)O). n

G=.

i~

1

Then F is full relative to I, and G is full relative to J. Suppose that (X"* /3. Then, by part (iv) of (9.3), there exist Sand T in P such that

(L

Ci -

ieS" T

L

d;)JL(A(S))v(B(T)HO.

ieSn T

Since v((B(T)1 n G, B(T)O n G)) = v(B(T)), there exists y in B(T)l n G such that

(L ieSnT

c;-

L ieSnT

di)JL(A(S));h(T)(yHO.

280

Chapter 6 Integration

Since ,u«A(S)l (\ F, A (S)O (\ F)) = ,u(A(S)), it follows that there exists x in A (S)l (\ F such that

(L

L

Ci -

iESr"IT

dJ XA(S)(X)xB(T)(yH O.

ieSnT

By parts (i) and (iii) of (9.3), we have

f(x,y)-g(x,y)=(

L

L d;) XA(S) (X)xB(T) (yH O.

Ci -

ieSnT

leSnT

This contradicts the hypothesis that f = g on dmnf (\ dmng. Hence, in fact, IX = {J. 0 It follows from Lemma (94) that (9.5)

(I x J)(tl ci XA(i) x B(i») ==

itl i

c ,u(A(i)) v(B(i))

defines a mapping I x J of L into JR. (9.6) Theorem. (X x Y, L, I x J) is an integration space.

Proof. Consider elements f, g of L, and real numbers IX, {J. It is trivial to verify that IXf+f3g and ifi belong to L, and that I(r:xf+fJg)=IXI(f) N

+ {J I (g). By (9.3), there exists a simple expression X ==

L ci XA(i) i~

x

B(i)

1

with the same domain as f, such that f = X on dmnf Clearly, f!\ 1 = X !\ 1 belongs to L. Also, N

(I x J)(f

!\

L (c i !\ n) ,u(A(i)) v(B(i)),

n) =

i~

1

so that N

lim (I x J)(f

!\

n)= L

Ci ,u(A(i))

v(B(i))= (I x J)(f)

n~oo

This proves the first part of (1.1.4); the second part is proved similar! y. On the other hand, if A and B are integrable sets with positive measure relative to I and J, respectively, then XA XB ELand (I x J)(XAXB)=,u(A)v(B»O; thus (1.13) holds. It remains to establish (1.1.2). Accordingly, let fEL and let (fn) be a sequence of nonnegative functions in L such that L(I x J)(fn) conn

verges to a sum less than (I x J)(f). Using the notation established immediately before (9 2), we see that

'iJ(J(fn)) = L(I x J)(fn)<(1 x J)(f) = I(J(f))

9 Product Integrals

281

Since (X, L(1), I) is an integration space, there exists Xo in X such that LJ(fn)(xo) converges to a sum less than J(f)(xo). Since (Y,L(J),J) is n

an integration space, it follows that there exists Yo in Y such that Lfn(x o, Yo) converges to a sum less than f(x o, Yo). This establishes n

(1.1 2)

0

We call the complete extension of the integration space (X x Y, L, I x J) the product of the integration spaces (X, L(1), I) and (Y, L(J), J); we call the completely extended integral I x J the product of the integrals I and J Extending the notation introduced before Definition (9.2), if the function yH f(x, y) is integrable with respect to J, we denote its integral by J(f)(x); if the function XH f(x, y) is integrable with respect to I, we denote its integral by I (f) (y). We now prove Fubini's theorem, which shows that an integration with respect to a product integral is equivalent to iterated integrations. (9.7) Theorem. Let f be integrable with respect to the product integral I x J on X x Y. Then there exists a full subset F of X such that the function yH f(x, y) is integrable with respect to J for each x in F, the function J(f) with domain F is integrable with respect to I, and I(J(f) = (1 x J)(f). Also, there exists a full subset G of Y such that the function XH f(x, y) is integrable with respect to I for each y in G, the function 1(f) with domain G is integrable with respect to J, and J(1(f) =(1 x J)(f). Proof: In view of (2.10), we may assume that f~ 0 and that there exists a representation (fn) of f by elements of L that are everywhere nonnegative. For each n in '1.+ the nonnegative function J(fn) is integrable with respect to I, and I(J(fn»=(1 x J)(f.). Thus L1(J(f..»

converges; so that LJ(fn) converges on a full subset F of X, LJ(fn) is n

n

integrable with respect to I, and I (LJ(fn» = L1(J(fn» = L(1 x J)(fn) = (I x J)(f) n

Consider any x in F. Since

n

J(fn)(x)~O

for each n, and LJ(fn)(x) n

converges, the series Lfn(x, y) converges for all y in a full subset S of

282

Chapter 6 Integration

Y, the function g" defined by g,,(Y)

=Lf.(x,y)

(YES)

is integrable with respect to J, and J(g,,) = LJ(f.)(x). Since

• Lf.(x, y) = f(x, y)

• whenever the left side converges, we see that g,,(Y) = f(x, y) for all y in s. Hence the function yH f(x, y) is integrable with respect to J, and J(f)(x) = J(g,,) = LJ (f.) (x).

Thus J(f)= LJ(f.) on the full set F. Therefore J(f) is integrable with

•

respect to I, and I(J(f))=(/ x J)(f). The second part of the theorem is proved similarly.

D

10. Measure Spaces Abstracting from the properties of the measure of an integrable set, we are led to the concept of a measure space. As we shall see, not only does every integration space give rise to a measure space, but the converse holds as well. (10.1) Definition. A measure space is a triple (X, M, Jl) consisting of a nonvoid set X with an inequality ,*, a set M of complemented sets in X, and a mapping Jl of Minto IR o +, such that the following properties hold. (10.1.1) If A and B belong to M, then so do A v B and A" B, and Jl(A) + Jl(B) = Jl(A v B) + Jl(A "B). (10.1.2) If A and A" B belong to M, then so does A - B, and Jl(A)

= Jl(A "B) + Jl(A -

B).

(10.1.3) There exists A in M such that Jl(A) >0. (10.1.4) If (A.) lim Jl ( k-oc

is

a

sequence

of elements

of M

AA.) exists and is positive, then nA; is nonvoid.

n= 1

n

such

that

10. Measure Spaces

283

We then call Il the measure, and the elements of M the integrable sets, of the measure space (X, M, Il). For each A in M the nonnegative number Il(A) is called the measure of A. An integration space (X, L, I) gives rise to a natural measure space (X, M, Il), in which M is the set of integrable sets relative to I, and Il is

the map AI-+I(XA) on M. This measure space is said to be induced by (X,L,I). Consider an arbitrary measure space (X, M, Il). The following sequence of lemmas will enable us to show that (X, M, Il) gives rise to an integration space. (10.2) Lemma. If A is an element of M such that A 1 is void, then Il(A) =0.

Proof· Suppose that Il(A) > O. Write A. == A for each positive integer n. Then, by (10.1.4), A 1 = A~ is nonvoid. This contradiction ensures that Il(A)=O. D •

n

n•

(10.3) Lemma. If K 1 , ... ,K. are elements of M, and F== (Kt uK?), then «(/), F) also belongs to M. ;~ 1 Proof: By (10.1.2), for each i we have

«(/),Kf uK?)=K;-K;EM. Hence (g),F) =

V(0,K;1 uK?) belongs to M, by (10.1.1).

D

i= 1

(10.4) Lemma.

•

Let A, K l '

n (K;l uK?). Then (A

1

... ,

K. be elements of M, and F ==

nF, AO nF) belongs to M and has measure equal

;= 1

to Il(A). Proof. By (10.3), B == (O, F) belongs to M. It follows from (10.1) that A 1\ B and A - B belong to M. Moreover, Il(A)=Il(A I\B)+Il(A-B) =1l«(/),(A 1 uAO)nF)+Il(A 1 nF,Ao nF) =1l(A 1 nF,Ao nFl,

by (10.2).

D

(10.5) Lemma. If A, B, and K1,.··,K n belong to M, and

•

n(Kt uK?), then Il(A)~Il{B).

i= 1

if

XA~XB

on

284

Chapter 6 Integration n

Proof. Write F== n(KfuK?). By (10.4), A'==(AlnF,AOnF) and i~1

B'==(B I nF,BOnF) belong to M, J.l(A')=J.l(A), and J.l(B')=J.l(B). Since XA ~ XB on F, we have A' A B' = A' and therefore (by (10 1.2)) J.l(A) = J.l(A') = J.l(B') - J.l(B' - A') ~ J.l(B') = J.l(B).

0

The notion of a simple function extends in a natural way from the context of an integration space to the present one: a simple function is n

L ai XAW

a function of the form X ==

where ai' .. , a. are real numbers,

i= I

are

A(l), ... ,A(n)

n •

elements

of M,

and

the

domain

of X is

(A (i)1 u A (i)0).

i= 1

Lemma (7.8) remains valid for measure spaces when interpreted appropriately Thus a simple function X can be written in the form

•

L ai XA(i)

where the complemented sets A (i) belong to M and are

i= I

disjoint, in the sense that XA(i)XA(j)=O on (A(i)1 uA(i)O)n(AU)1 uAU)°) whenever i j. (Note that if A and B are disjoint elements of M, then J.l(AAB)=O: this follows from (104) and (10.2).) Using an argument like that in the proof of Lemma (9.4), the reader can prove the following lemma.

*'

(106) Lemma. Let a l , ... , a. and b l , ... , bm be real numbers, and let A(l), ... ,A(n) and B(l), .. ,B(m) be elements of M. Suppose that n

m

i= I

j= I

L ai XA(i) ~ L b

j

F==

XB(j) on the set

(C\ (A (i) UA(i)0)) n CC\ (B(j)1 UBU)°)) I

n

Then

m

L aiJ.l(A(i))~ L bjJ.l(B(j)). j~

i= I

I

It follows from Lemma (10.6) that

J(tl

(10 7)

ai XA(i») ==

itl

ai J.l(A(i))

defines a mapping J from the set of simple functions into lR. For each simple function X we call J (X) the integral of X. In order to justify this use of the word 'integral', we need two estimates of J(X). (10.8) Lemma. Let f be a nonnegative simple function, and Ii a positive number Then there exists B in M such that BI u BO c dmnf, f < Ii on

BO, and

J.l(B)~2c

I

J(f).

10 Measure Spaces

285

n

Proof" We may assume that f

=

L a i XA(i)' where a p i~

""

an are non-

1

negative numbers and A (1), ... , A (n) are disjoint elements of M. Write {I, .. , n} as a union of two sets Sand T such that ai < E for all i in S, and ai > E/2 for all i in T. If G i < E for all i, then the desired conclusion is obtained by taking B=(0,dmn!). So we may assume that T is nonvoid. Let E= V A(i) and B=(E 1 ndmnf, EO ndmn!). Then iET

Bl uBo cdmnf Also, by (10.1.1) and (10.4), BEM and /-l(B)=/-l(E)= L/-l(A(i))

Consider any x in BO Since xEdmnfn(nA(i)O), either f(x)=O or ieT

there exists a unique i in S with XA(i)(X) = 1. In the latter case, f(x) =ai<E. Thus f<E on BO. D (10.9) Lemma. Let f be a simple function, let c be a positive number such that f ~ c on dmnf, and let A be an element of M such that f ~ 0

on A °n dmn! Then for each E > 0 there exists B in M such that Bl uBo cdmnf, f >E on Bl, and /-l(B)?;c-1(1(j)-2E/-l(A)). n

Proof We may assume that f=LaiXA(i)' where A(l), ... ,A(n) are i= 1

disjoint elements of M. Let E > 0 Write {I, ... , n} as a union of two disjoint sets Sand T, where ai < 2 E for each i in S, and ai > E for each i in T. To begin with, assume that T is nonvoid Let E= V ACi) and iET

B=(E1ndmnf, EOndmn!). Then B1uBocdmnf, f>E on Bl, BEM, and /-lCB) = /-l(E). Thus I(!)~I(XA!)

(by (10.6))

= Lai/-l(A 1\ A(i))+ Lai/-l(A 1\ A (i)) ieS

~2E L/-l(A

1\

A(i)) +C L /-l(A

ieS

1\

A (i))

ieT

~2E/-l(A)+C/-l(B).

Hence /-l(B)?;c-1(1(f)-2E/-l(A)). In the case where T is void we need only take B

=(0, dmn!).

D

(10.10) Theorem. Let (X, M, /-l) be a measure space, L the corresponding set of simple functions, and I the mapping defined on L by (10.7). Then (X, L, 1) is an integration space.

286

Chapter 6 Integration

Proof. To verify properties (1.1.1) and (114), we need only consider n

L CiXqi)

simple functions of the form

with C(1), .. :, C(n) disjoint ele-

i~1

ments of M, the appropriate arguments are then straightforward. To verify (1.1.3), choose A in M with Jl(A»O; then P==Jl(A)-1 XA belongs to L, and I(P) = 1. It remains to verify (1.1.2). Accordingly, let (fn) be a N

sequence of nonnegative elements of L, and let

f ==

L i~

element of L such that

aiXA(i)

be an

I

L I(fJ converges to a sum less than I(!). Write n

n

and N

A==

V A(i).

Then AEM, and f=O on AOndmnj Choose c>O so that.f~c on dmn f, and write and ex == c- l(r - 2 e Jl(A)).

Choose a strictly increasing sequence with n(l)=l, such that I(

n(k+ I ) - I

L /;

)

(n(k));;"~

~2-2k-lexe

I

of positive integers,

(k~2).

i~n(k)

Define n(k+

1)-1

L /;

gk==

(k~ 1).

By (10.8), for each k ~ 2 there exists Bk in M such that B~ u B~ cdmngk , gk<2- ke on B~, and Jl(Bk)~2k+le-1I(gk)~2-kex. Also, by (10.9), there exists B in M such that Bl u BO c dmnf n df!1n g l' f - g 1 >e on Bt, and Jl(B) ~ c- 1(l(f - g 1) - 2eJl(A)) ~ c- 1(r - 2eJl(A)) = ex.

Now, for j>i~2 we have

o~Jll62 (B - Bk~) -

~Jl (62 (B -Bk) -

Jll62 (B - Bk))

k62 (B - Bk))

(by (10.5))

10 Measure Spaces

~lllY2 Bk - kY2 Bk) ~Il{~~l+ V Bk)~ k~l+ Il(Bk)~ k~l+ t

t

1

Hence

t =- !~~

Ill0

2

1

287

2- k oc. 1

(B - BJ) exists, and 00

t~Il(B)- L 2-kOC~oc/2>0. k~2

It follows from (10.1.4) that

B1 n lC\ B~) contains an element x of X.

Since gk(x)<2- k e for each k~2, we see that Lf.(x) converges, and ili~ • 00

f(x)- Lf.(x)=f(x)-g1(X)- L gk(X) •

k= 2 00

>e- L 2- k e=e/2>0. k~2

This completes the verification of (1.1.2).

D

The complete extension (X, L 1 , I) of the integration space in Theorem (10.1 0) is called the integration space induced by the measure space (X, M, 11), and the elements of L1 are called integrable functions generated by (X, M, 11). We write 11 (A) =- I(XA) whenever the complemented set A is integrable with respect to I. Starting with an integration space, we obtain an induced measure space; in turn, this induces another integration space. The next theorem shows the relation between this and the original integration space. (10.11) Theorem. Let (X, L,I) be an integration space, with induced measure space (X, M, 11), and let (X, £1' 1') be the integration space induced by (X, M, 11). Then a function f is integrable with respect to I if and only if it is integrable with respect to 1', in which case /(f)=I'(f). Proof: Consider any function f integrable with respect to 1'. By definition of 1', there exists a representation (f.) of f by functions which are simple relative to the measure space (X, M, 11). Thus L 1(lf.1)

288

Chapter 6

Integration

= L 1'(lfnl) converges, f = L in on the set where L Ifni converges, and n

n

•

1'(f)= L1(f.). It follows from (2.15) that f is integrable with respect to n

I, and that I (f) = l' (f). Conversely, consider any function f integrable with respect to I. By (7.5), there is a sequence (Xn) of simple functions (relative to the measure space (X,M,J1)) such that 1(lf -Xnl)<2- n for each n. Hence 1(IXn+ 1 - x.l) converges; so that, by (2.15),

L n

X== X1 + L (Xn + 1 - Xn)

is integrable with respect to I, and lim 1(lx - Xnl) = O. It follows that n~oo

1(lf - xl)= 0; whence f = X on a full set F (relative to 1), and 1(f) = I(X) = lim I(Xn)· n~oo

On the other hand, since I and l' coincide on the set of simple functions relative to (X, M, J1), nlXn + 1 - Xnl) = 1(IXn + I - Xnl) con-

L

L n

verges, X is integrable with respect to 1', and 1'(X) = lim 1'(X.) = lim I(Xn) = 1(f) 11-00

n-oo

Now, by (3.3), A==(0,F) has measure 0 relative to I; so that AEM and 1'(XA) = J1(A) = 0 Thus, by (3.2), F is full relative to l' Hence f is integrable with respect to 1', and 1'(f) = 1'(X) = 1(f). D The following corollary is an immediate consequence of Theorem (10.11 ). (10.12) Corollary. Let (X, L, 1) be an integration space, with induced measure space (X, M, J1). Then (X, L, 1) is completely extended if and only if every integrable function generated by (X, M, J1) belongs to L.

If (X, M, J1) is an arbitrary measure space, with induced integration space (X,L1,I), then (X,L1,I) induces a measure space (X,M',J1'); in fact, AEM' if and only if XA is integrable with respect to I, in which case J1'(A) = I(XA) = J1(A). Moreover, for each A in M we have AEM' and J1(A) = J1'(A). Thus the new measure space (X,M',J1') can be regarded as an extension of the original one. In view of Corollary (10.12), it is natural to ask: under what conditions does this extension process give us back the measure space with which we started?

(1013) Definition. A measure space (X, M, J1) is complete if the following three conditions hold.

10 Measure Spaces

289

(10.13.1) If A is a complemented set, and B is an element of M such that XA=XB on BluB o, then AEM. (10.13.2) If (An) is a sequence of elements of M such that t==

;~~ J1. (Yl An)

exists, then

YAn belongs to M and has measure t.

(10.13 3) If A is a complemented set, and if B, C are elements of M such that B
= J1.(B), by Lemma (10.5). To avoid confusion in the next two results, recall that if (X, L I , I) is the integration space induced by a measure space (X, M, J1.), then we write J1.(A) == I(XA) for each complemented set A which is integrable with respect to I (10.14) Lemma. Let (X, M, J1.) be a measure space satisfying condition (10.13.2), let (X, L I ,!) be the integration space induced by (X, M, J1.), and let A be an integrable set relative to I. Then for each e > 0 there exist elements Band C of M such that BlcAl, COcA o, J1.(A)-e~J1.(B), and J1.(C)~J1.(A)+e.

Proof. Let (fn) be a representation of XA by simple functions. The proof hinges on the idea that, for a sufficiently large N and for a

suitable

r:J.

in (0, 1), the complemented set

(I. In>

r:J.)

will approximate

A as closely as we wish. ~I Given e > 0, let (n(k))~ I be a strictly increasing sequence of in-

tegers, with n(l)= 1, such that n(k+l)-l

L

I(I.t;I)<2- 2k - 2 e

(k~2).

i~n(k)

Define simple functions

gl,g2' ...

by

n(2)-1 gl==

L .t; i= 1

and n(k+I)-l

gk==

L 1.t;1

i~n(k)

(k~2).

290

Chapter 6 Integration m

We may assume that

g1

=

L ciXC(i)'

where C1 = 1 and C(1), ... , C(m)

i= 1

are disjoint elements of M. Choose IX in (1/2,3/4) so that IX =l= Ci (1 ~i~m). Then S= {i: Ci>lX} is nonvoid. Write K= V C(i) and iES

Then

so that A1 EM, by (1O.4). Using (10.8), for each k~2 construct Ak in M such that gk<2- k- 1 on A~, and 1L{Ak)~2k+2 I{gk)<2- ke. Then (1L

lY2 Ak) ) ~~

2

is a Cauchy sequence, so that t=

!~~ lLlY2Ak)

00

exists and is less than

L 2-

00

k

e. It follows from (lO.13.2) that

k~2

V Ak k~2

belongs to M and has measure less than e/2. Hence

is in M. 00 Consider any x in B1. Since gk{x)<2- k- 1 (k~2), the series L gk{X) converges to a sum less than 1/4. Also, k~ 2 n(2)-1

L Ifn{x)1 =

L

00

I.t;(x) I + L gk{X)

i~1

n

k~2

converges; so that xEdmnXA' 00

xix) = Lfn{x)~g1{X)- L gk{X»1X-1/4> 1/4, k~

n

and therefore xEA 1. Thus B1 c A 1. N~w suppose that lL{A) > lL{A 1) A-

2 00

+ 1L

lY2 A

k ).

V A k • Thus k~

1 00

1 =XA{y)~g1(Y)+ L gk{Y) k~2

~IX+ 1/4<3/4+ 1/4= 1.

Then there exists y in

10 Measure Spaces

291

Since this is absurd, we must have ,u(A) ~ ,u(A 1) + .u{Y/k)' It now follows that

,u(A)-e~J.l(A) -2J.llYzA k )

~J.l(A1) -

J.llYz Ak) ~J.l(B).

Next, consider the element

of M. Arguing as above, we can show that CO cAo. On the other hand, since B1 c A \ and the assumption J.l(A 1) >,u (A the existence of a point in B - A, we have J.l(A 1) ~ J.l (A

V V

kYz Ak) entails

kYZ A k). Hence

J.l(C)~,u(A1) + J.llYz Ak) ~J.l(A)+2J.llY2Ak)~J.l(A)+e.

D

(10.15) Theorem. Let (X,M,J.l) be a measure space, with induced integration space (X, L 1, I). Then (X, M, J.l) is complete if and only if every complemented set integrable with respect to I belongs to M. Proof: Suppose that (X, M, J.l) is complete, and consider any complemented set A integrable with respect to I. By (10.14), for each n in Z+ there exist elements B n, en of M such that B!cA 1, e~cAo, J.l(A) _n- 1~J.l(Bn)' and ,u(en)~J.l(A)+n-1. For each positive integer N we have XBN ~ XB1 V ... V XBN ~ XA on the full set

and therefore ,u(A) - N- 1 ~

J.llY1 Bn) ~,u(A). It follows from (10.13.2)

that S=V Bn belongs to M, and J.l(S)=,u(A); also, S1cUB!cA1. On n

n

the other hand, ,u(A)

~,ul~1 en)~J.l(A)+N-1

(N~l),

292

so that

Chapter 6

Integration

;~~ Jl C~

exists and equals Jl(A). Thus

1 C n)

;~~ Jl(Yl(C

1-

Cn)) =

;~~

(Jl(C 1 )

-

Jl(~l C n))

= Jl( C 1) - Jl(A).

V (C 1 -

By (1013.2),

Cn) belongs to M and has measure Jl(C 1 ) - Jl(A)

n

Hence T

=- 1\ C n=

C1 -

n

V (C 1 -

Cn) belongs to M and has measure

n

=- (Sl, AD U SO) and C =- (A 1 u Tl, TO); note that Band C are complemented sets with B < A < C, since Sl cAl and TOcAo. We have XB=XS on SIUSO; so that BEM and Jl(B)=Jl(S) =Jl(A), by (10.13.1) and (10.5). Likewise, CEM and Jl(C)=Jl(T)=Jl(A). It now follows from (10.13.3) that A EM. Conversely, suppose that every complemented set integrable with respect to I belongs to M. If A is a complemented set and B is an element of M such that XA = XB on Bl u BO, then since Bl u B O is a full set relative to I, A is integrable with respect to I; hence A EM. This proves (10.13.1). Condition (10132) holds in view of (3.8). To verify (10.13.3), consider a complemented set A and elements B, C of M, such that B
Problems 1 Without reference to any of the results of this chapter, prove that if 1 is a bounded interval in 1R and (in) is a sequence of bounded intervals such that 11nl converges to a sum less than Ill, then there

L

exists x in 1 such

th~t

x =j= y for all y in

UIn· n

2. Define the distance between integrable sets A and B to be p(A,B)=-Jl(A-B)+Jl(B-A), and call A and B equal if p(A,B)=O. Show that the integrable sets form a complete metric space.

Problems

293

3. Construct an integrable subset A of [0,1] relative to Lebesgue measure such that Jl.(l/\ A) > 0 and Jl.(l- A) > 0 for every proper subinterval J of [0, 1]. 4. Give an example of a compact subset of [0,1] that is not integrable with respect to Lebesgue measure 5. Let Jl. be a positive measure on JR, J == [a, b] a proper compact interval that is integrable relative to Jl., and 8 the set of all continuous functions from J into [0,1]. For each h in 8 let ii be an element of qJR) such that ii(x)=h(x) for all x in J. Then A(h)== SxJiidJl.

is independent of the choice of ii, and (8, A) is a profile for J. Show that if x is smooth for this profile, then the intervals [x, b] and (x, b] are integrable and have the same measure. 6. Let (X,p) be a compact space, and f: X -->JR a continuous mapping such that a == inf f < b == sup f Let 8 be the set of all continuous functions from [a, b] into [0,1], let x o , Xl be any two points of X, and for each h in It define A(h) == - inf {max {p(x, x o), P(Xl' x o)(l- h(f(x)))}: XEX}

Then (8, A) is a profile for [a,b]. Show that p(xo,X r) exists for each smooth point t for this profile, where Xr=={XEX: f(x)~t}. Show also that for each smooth point t and each 8> 0 there exists c5 > 0 such that Ip(x o, X r ) - p(x o, Xr")1 <8

whenever t' is smooth and It - t'l < c5 Use all this to give another proof of Theorem (49) of Chapter 4 7. Prove that if f is a nonnegative integrable function, then J(f) =sup {I(XAf): A is an integrable set}

8. Let Jl. be a positive measure on a compact space X. Let a be a positive constant, and n a positive integer, such that na > Jl.(X) Prove that there exists a subfinite set SeX such that for each x in - S there exists f~O in qX) with f(x) = 1 and SfdJl.
294

Chapter 6 Integration

10. Let [a,b] be a proper compact interval, and Fc[a,b] a full set with respect to Lebesgue measure on [a, b]. A bounded function f: F ~ IR is Riemann integrable if for each 8> 0 there exists 6> 0 such that whenever P == (a o , ... , am) and Q== (b o , ... , bn) are partitions of [a, b] with mesh less than 6, and Xl' ... ,Xm' YI' ... 'Y n are points of F with aO~xl ~al ~ ... ~am_l ~xm~am and bO~YI ~bl ~ ... ~bn_l ~Yn~bn' then IS(f, P) - S(f, Q)I ~ 8, where m

b

and S(f, Q) is defined similarly. Define the Riemann integral

Jf(x)dx a

of f, and show that f is Riemann integrable if for each 8> 0 there exist 6> 0 and a compact integrable set KeF with fl( - K) ~ 8, such that If(x)- f(Y)1 <8 whenever xEK, YEF, and Ix- yl <6. Show that the converse is not true. 11. A measure fl on a locally compact space X is a linear map of C(X) into IR such that for each f in C(X) there exists C f ~ 0 with Ifl(g)1 ~ C f whenever gE C(X) and Igl ~ If I· We write

Jf

dfl == fl(f)·

Show that a positive measure is a measure in this sense. 12. A real-valued function (J. defined on a dense subset A of an interval J == [a, b] is of bounded variation on J if there exists C ~ 0 such that

L 1(J.(XJ-(J.(Xi_I)I~c i= 1

whenever X O ' ••• , Xn are points of A with Xo <Xl < ... <x n ; the constant c is called a bound for the variation of f on J. Extend the definition of the Riemann-Stieltjes measure fla (given in Section 5) to cover the case where the function (J. is of bounded variation. 13. Show that an increasing function f. S ~ IR defined on a dense set S c IR is continuous at all but count ably many X in S, in the sense that for each 8> 0 there exists 6> 0 such that If(Y) - f(x)1 < 8 whenever YES and Iy-xi <6. What can be said when f is of bounded variation, rather than increasing 14. Give an example of a function of bounded variation on [0,1] that can not be written as the difference of two increasing functions.

Problems

295

15. Let Jl be a positive measure on a locally compact space X, and let (Kn) be a sequence of integrable compact subsets of X such that (i) K 1 :::>K z :::> ... , (ii) diamKn--+O as n--+oo, and (iii) K==nKn is an integran

ble set containing exactly one element. Prove that Jl(K n) -+ Jl(K) as n --+ 00. 16. Give an example of a sequence (Kn) of compact sets that are integrable with respect to Lebesgue measure on [0, 1], such that (i) K 1 :::> K z :::> ... , (ii) K == Kn is an integrable compact set containing ex-

n n

actly one element, and (iii) the sequence (Jl(K n)) does not converge to Jl(K).

17. Let A be an integrable subset of IR. with respect to Lebesgue measure, and B a positive constant. Show that there exists a finite union 1 of bounded intervals with Jl(A - 1)+ Jl(1 - A) < B. 18. Prove that if 11' ... ,fn are measurable functions and ¢: IR. n--+ IR. is continuous, then ¢ 0 (f1' ... ,fn) is measurable. 19. Show how to define a neighborhood structure on the set M of measurable functions, corresponding to convergence in measure in such a way that a sequence (fn) converges in measure to a limit 1 if and only if all but finitely many terms of (fn) belong to any given neighborhood of f 20. Give an example of a sequence (fn) of integrable functions relative in to Lebesgue measure on [0,1], such that (fn) converges to measure, but (fn(x)) does not converge to at any point x of [0,1].

°

°

21. Let 1 be a function defined on a full set relative to an integral 1, such that - c v (xAI 1\ c) is integrable for all c > and all integrable sets A. Prove that 1 is measurable.

°

22. Two measurable functions 1 and g are equal if - n v (xAI 1\ n) and - n v (XAg 1\ n) are equal integrable functions for all n in Z+ and all integrable sets A. Prove that if a sequence of measurable functions converges in measure to two functions 1 and g, then 1 and g are equal as measurable functions. 23. Let B be an integrable set with positive measure relative to an integral 1. Prove that if a function is measurable with respect to l. then it is measurable with respect to 1B'

296

Chapter 6

Integration

24. Show that the complete extension of a positive measure on a locally compact space is u-finite. 25. Let J1. be a positive measure on a locally compact space X, and f a bounded integrable function which vanishes on the metric complement of a compact set. Show that there exist c > 0, and a sequence Un) of test functions with a common compact support, such that IIfnll ~c for all n, and Un) converges almost everywhere to f 26. An integral 1 is weakly u-finite if there is a sequence (K(n));;"~ 1 of integrable sets such that K(l)l c K(2)1 c.. and U K(n)l is a full set. n

Show that 1 is weakly u-finite if and only if there exists an integrable function h such that h > 0 on a full set 27. Let 1 be a weakly u-finite integral, and let Un) be a sequence of measurable functions converging in measure to a measurable function f Suppose that there is an integrable function g such that Ifni ~ g for each n. Show that f is integrable (whence, by the dominated convergence theorem, 1(lf - fnl) --> 0 as n --> (0). (Hint You may assume that g>O on a full set. Look at the proof of Theorem (7.11).) 28. Let 1 be a weakly u-finite integral, and f a nonnegative measurable function. Prove that f is integrable if and only if sup {I(n

1\

XAf): nEZ+, A is an integrable set}

exists. 29. Let 1 be a finite integral, f an integrable function, and S a locally compact subset of IR with nonvoid metric complement, such that J1.(A)-I/(XAf)ES for each integrable set A with positive measure. Prove that f(X)ES for all x in a full set. (Hint. Consider any r > 0 such that Sr={tEIR: p(t,S);:;:r} is locally compact, and show that p U(x), Sr);:;: r for all x in a full set) 30. Let (fn) and (gn) be sequences of nonnegative measurable functions. We say that (fn) dominates (gn) in measure if to each integrable set A and each G > 0 there corresponds N in 7l+ such that for each integer n;:;: N, there exists an integrable set B with Bl c A I, J1.(A - B) < G, and fn;:;: gn on Bl Prove that if (i) fn and gn are integrable for each n, (ii) Un) dominates (gn) in measure, and (iii) (gn) converges in measure to a nonnegative integrable function g, then for each G> 0 there exists N in 7l+ such that IUn) > I(g) - G for all n;:;: N.

Notes

297

31. Let X and Y be locally compact spaces, Jl a positive measure on X, and v a positive measure on Y. For each test function I on the locally compact product space X x y, yl-+ SI(x, y) dJl is a test function on Y, and so u(f) == S(S I(x, y) dJl) dv

is a well defined real number Show that u is a positive measure on X x Y, that u(f)= S(S f(x,y)dv)dJl

(IEC(X x Y)),

and that the complete extension of u is the product integral Jl x v as defined in Section 9. 32. Let I be a nonnegative measurable function with respect to a product integral I x J, and let the iterated integral c == J(l(f)) exist. Show that (l x J)(f) exists and equals c.

Notes The work of this chapter is based on the treatment of the integral given by Bishop and Cheng [l3]. The proof of Theorem(1.10) is due to Y-K Chan. Related to Theorem (1.10) is Problem 1, which originated with Newcomb Greenleaf. Strictly speaking, an integrable function is a pair (1, (fn)) consisting of a function I and a representation (In) of I by elements of L. It is simpler to concentrate on the function 1, and emphasize the representation (fn) when we need it. The existence of

;~~

Jll01 An)

in the hypotheses of Proposition

(3.8) is superfluous from a classical viewpoint. As Problem 15 shows,

it is necessary in the constructive setting. The obvious example of a profile is obtained by taking tC to be the set of all continuous functions from a proper compact interval [a, b] b

to [0, 1], and defining A(f) == Sf(x) dx (the Riemann integral) for each I in tC. The proof of Lemma (4.7) is a modification of one given by Y-K Chan.

298

Chapter 6

Integration

It can be shown that if J1. is a general measure (not necessarily positive) on JR, then there exist a set S consisting of all but countably many real numbers, and a function IX: S --> JR of bounded variation, such that J1. = J1.~ (where J1.~ is constructed as in Problem 12). Further information on general measures can be found in [23]. Note that whereas a function of bounded variation is classically defined at every point of the interval under consideration, constructively this is usually not possible. It is easy to give an example of a compact set K c JR and a positive measure J1. on JR with respect to which K is not integrable: take K == {OJ and let J1. be the point mass at a, where a = 0 is impossible but we do not know that a =l= O. A more interesting example is required in Problem 4. Note how the constructive definition (Definition (8.1)) of convergence almost everywhere differs from the classical definition (pointwise convergence on a full set). The classical definition would be of no use constructively. Lemma (8.12) extends Proposition (8.3) in the case where I is (Jfinite. Definition (10.13) differs slightly from the definition of a complete measure space given by Bishop and Cheng. Certain parts of measure theory in JR 2 have been developed by Brouwer, with a different approach from the one used here.

Chapter 7. Normed Linear Spaces

In Section 1 we introduce normed linear spaces and bounded linear mappings. Section 2 is concerned with finite-dimensional spaces and with the problem of best approximation by elements of a finitedimensional subspace. In Section 3 we discuss Lp spaces. we prove the completeness of L p , and determine the form of the normable linear functionals on Lp in case p> 1. (In contrast to the classical theory, a bounded linear functional need not have a norm.) We then apply these results to the proof of the Radon-Nikodym theorem. In Section 4 we prove the separation theorem and its corollary, the Hahn-Banach theorem, under the assumption that a certain convex set is located. Section 5 introduces the notion of a quasinorm, and discusses the quasinormed space Loo and !ts linear functionals. Next we prove the compactness of the unit ball of the dual, and a result (6.8) which makes it plausible that every constructively defined linear functional on the dual X* is determined by an element of X -a curious form of reflexivity! Section 7 deals with extreme points, and contains a proof of the Krein-Milman theorem; in contrast to the classical development, detailed estimates are needed. In the final section of the chapter we define Hilbert space and prove the spectral theorem for hermitian operators; we also prove versions of the Gelfand representation theorem for algebras of normable operators. Most classes of functions that arise in analysis are endowed with both a linear structure and a norm. This makes it natural to introduce the concept of a normed linear space.

1. Definitions and Examples To be able to discuss real and complex linear spaces simultaneously, we let IF denote either the real or the complex number field. The elements of IF are called scalars.

300

Chapter 7

Nonned Linear Spaces

(1.1) Definition. A linear space, or vector spac-e, over IF is an abelian group X with a multiplication operation (a,x)t---+ax from IF x X to X such that for all a, aI' a 2 in IF and x, XI' X2 in X,

(a l +a 2 )x=a l x+a 2 x, a(xI +x 2 )=ax I +ax 2 ,

a l (a 2 x) = (a l a 2 )x, and 1x=x.

The elements of X are called vectors. We assume the reader to be familiar with the elementary theory of linear spaces (1.2) Definition. A seminorm I lion a linear space X over IF is a function xt---+llxll from X to JR o + such that for all a in IF and X,XI' X2 in X, Ilaxll = lalllxli and IlxI +x 2 11 ~ IlxI11 + Ilx 211· The pair (X, II II) is called a seminormed linear space When there is no confusion over the seminorm, we often refer to X itself as a seminormed linear space. If x=O whenever Ilxll =0, then the seminorm II II is called a norm, and we refer to X as a normed linear space, or simply a normed space There is no loss of generality in confining our attention to norms, since a seminorm II lion a linear space X becomes a norm if the equality of XI and X2 in X is changed to mean that IlxI -x211 =0. A seminorm II lion a linear space X determines a pseudo metric p on X, given by

Clearly, p is a metric when II II is a norm A normed space X is thus a metric space. The standard inequality =1= on X is the inequality defined relative to the metric p. A vector X is nonzero if X=1= o. It is easily seen that when X is a normed space, addition and subtraction are continuous functions from X x X to X, and that multiplication of a vector by a scalar is a continuous function from IF x X to X. The notion of an infinite series with terms in a normed space X is defined in the natural way Tests for the convergence of such series, similar to tests for the convergence of series in JR, can be derived. We omit the details.

1. Definitions and Examples

301

The norm on a normed linear space X can be described geometrically in terms of the unit sphere Sc(O, 1) of X (Problem 1). (1.3) Definition. A mapping u of a vector space X into a vector space Y is said to be linear if

u(ax)=au(x) and

U(XI +XZ)=u(xI)+u(X Z) whenever aEIF and x, Xl' XzEX. (If Y=IF, then u is called a linear functional on X.) In case X and Y have norms and therefore metrics, we say that a linear mapping u: X ---+ Y is continuous if it is uniformly continuous on each bounded subset of X (1.4) Lemma. For each element

X

of a normed linear space X,

Ilxll = inf{lal- l : aEIF, a =1= 0, Ilaxll ~ 1}.

Proof. For each nonzero scalar a, Ilaxll ~ 1 ¢> lallixil ~ 1 ¢>

The result now follows.

lal- l ~ Ilxll·

D

(1.5) Proposition. The following are equivalent conditions on a linear map u of a normed linear space X into a normed linear space Y.

(i) (ii) (iii) (iv) (v)

u is continuous. u is uniformly continuous. u is bounded on the unit sphere of x. u is bounded on each bounded subset of x. There exists a positive number c, called a bound for u, such that Ilu(x)11 ~c Ilxll (XEX).

Proof. Suppose that u is continuous. Then u is uniformly continuous on the unit sphere S of X, and so there exists r > 0 such that Ilu(x)11 = Ilu(x)-u(O)11 ~ 1

(11xll ~r)

Consider an arbitrary x in X. For each nonzero scalar a with Ilaxll ~r we have Ilu(x)11 =lal-Illu(ax)11 ~Ial-l. Hence Ilu(x)1I ~r-Illxll, by (1.4). Thus (i) implies (v). It is clear that (v) implies (iv), and that (iv) implies (iii). Now assume (iii), and choose c>O so that Ilu(x)11 ~c for each x in S.

302

Chapter 7 Normed Linear Spaces

Consider arbitrary Xl' x 2 in X. For each nonzero scalar a with a(xi -X 2)ES, we have Ilu(x l )-u(x 2 )11 = Ilu(x l -x 2)11 =Ial-Illu(a(xi -x 2))11 ~c lal- l

It follows from (1.4) that Ilu(x l )-u(x 2)11 ~c Ilxl -x211.

Hence u is uniformly continuous, and (iii) implies (ii). Finally, it is obvious that (ii) entails (i). 0 In view of property (v) of Proposition (1.5), we often refer to a continuous linear map u: X --> Y as a bounded linear mapping on the normed linear space X. (1.6) Definition. The kernel, or null space, of a linear mapping u: X --+ Y between vector spaces X and Y is the set ker u == {XEX: u(x)=O}. In case Y has an inequality relation we say that u is nonzero.

=1=

and u(x) =1= 0 for some x in X,

Of particular interest are the bounded linear functionals on a normed linear space. These are attached to special linear subsets known as hyperplanes. (1.7) Definition. A hyperplane in a normed linear space X is a linear subset H of X with the property: there exist Xo in X and c > 0 such that Ilx-xoll ~c

(xEH)

and such that each element x of X can be written in the form (1.7.1)

x=axo+Y

with aEIF and YEH. The vector Xo is then said to be associated with H. If H is a hyperplane in X, then for each x in X the expression (1.7.1), with aEIF and YEH, is unique. (1 8) Proposition. If u is a nonzero bounded linear functional on a normed linear space X, then ker u is a hyperplane. Conversely, if H is a hyperplane with associated vector x o, theY,! there exists a unique bounded linear functional u on X with ker u = Hand u(x o) = 1.

1 Definitions and Examples

303

Proof" Let u be a nonzero bounded linear functional, and Xo any vector with u(x o)=1. By (1.5), there exists c>O such that clu(x)I;;;;llxll for all x in X. If x is any vector in ker u, this gives

Ilx -xoll

~c lu(x

-xo)1 =c lu(x)-u(xo)1 =c.

On the other hand, for each vector x in X we have x=ax o + y, where a=u(x) and y=x-u(x)x o ; clearly, YEkeru. Thus keru is a hyperplane with associated vector Xo' Conversely, consider a hyperplane H with associated vector xo' If the required linear functional u exists, then it must be given by u(x)=a, where x=ax o + y, a ElF, and YEH. Define u in this way. Clearly, u(x o)=l and H=keru. Choose c>O so that Ilx-xoll~c for all x in H. With x, a, and y as above, and 8> 0 arbitrary, we prove that (1.8.1 )

lu(x)1 ;;;;c- 1 Ilxll +8

Either (1.8.1) holds or lu(x)I>O. In the latter case, a = u(x)=t= 0 and so Ilxll = lallix o -( _a- 1 y)11 ~ lal c= lu(x)1 c. Hence lu(x)l;;;;c- 1 1Ixll, and (1.8.1) holds in this case also. Since 8 is arbitrary, it follows that lu(x)l;;;;c- 1 1Ixll. Thus u is bounded. D (1.9) Definition. Let u be a bounded linear map of a normed linear space X into a normed linear space Y. We say that u is normable if

Ilull =sup{llu(x)ll: XEX, Ilxll;;;; I} exists as a real number. In that case, Ilull is called the norm of u, and is the smallest bound for u. If X is nontrivial, in the sense that it contains a nonzero vector, and if u is normable, then Ilull =sup{llu(x)ll: XEX, Ilxll = I}. Although no general criterion of normability has been found, there is a very important criterion for normability of bounded linear functionals. (1.10) Proposition. A nonzero bounded linear functional u on a normed linear space X is normable if and only if ker u is located.

Proof: Assume that u is normable. Since u is nonzero, II u I > N -1 for some N in Z +. For each integer k ~ N let X k be a vector with I X k I = 1 and u(x k ) > Ilull-k- l . For each x in X and y in keru we have Ilx-YII~llull-llu(x)l. On the other hand, for each k~N the vector Zk=X-U(x)U(Xk)-l Xk

304

Chapter 7 Normed Linear Spaces

belongs to ker u, and Ilx - zkll = lu(x)1 U(X k)-1 < lu(x)l(11 ull - k- 1) -1. Thusp(x,keru) exists and equals Ilull- 1 Iu(x)l. Conversely, assume that ker u is located. Since u is nonzero, there exists a vector Xo with u(x o) = 1. As u is uniformly continuous (1 5), p(xo,keru»O. Thus

inf{llyll : u(y) = 1} = inf{ Ilxo -zll : zeker u} = p(xo, ker u) exists Also lu(z)1 >0 and Ilzll = 1, where z= Ilx o ll- 1xo. Hence sup{lu(x)l: Ilxll = 1} = sup {llyll-1: y = U(X)-1 x for some x with Ilxll = 1 and u(x)=l= O} =sup{ Ilyll-1: u(y)= 1} exists and equals p(xo, keru)-1. It now follows that u is normable, with Ilull =p(x o, keru)-1. 0 The next definition introduces the most useful type of normed linear space. (1.11) Definition. A normed linear space which is both separable and complete is called a Banach space.

Recall that a metric space is separable if it has a countable dense set There are no important constructively defined normed linear spaces that are not separable As far as existing mathematics is concerned, there is no loss of generality in postulating separability, and there is a great gain in power and convenience. The fact that every metric space can be completed leads us to hope that the same is true of every normed linear space X. This turns out to be the case, and the completion of X as a normed linear space can be identified with its completion as a metric space. Here is an outline of the construction. We take X to consist of all Cauchy sequences in X, with termwise addition and termwise scalar multiplication. Two elements (x n) and (Yn) of X are equal if lim Ilx n- Ynll =0 Clearly X is a linear space. Under the norm n- 00

it is a complete normed linear space, called the completion of X. The inclusion map t·: X ~X defined by i(x)=(x n), where Xn=X for each n, preserves norms and realizes X as a dense linear subset of X. We thus have the following result.

1 Definitions and Examples

305

(1.12) Proposition. If X is a normed linear space, then there exist a complete normed linear space g and a norm-preserving linear inclusion map t from X onto a dense subset of g

In case X is separable, g is also separable and therefore a Banach space. The bounded linear functionals on X are in one-one correspondence with the bounded linear functionals on g. if u is any bounded linear functional on X, and (xn) is any point of g, then (u(xn)):;O~ 1 is a Cauchy sequence in IF whose limit u((x.)) defines an extension of u to X. An important example of a Banach space is the space C,u(X, IF) of continuous functions vanishing at infinity on a locally compact space (X, p). In this case, the norm is given by Ilfll =sup{lf(x)l: XEX}

(fE Coo(X, IF)).

Recall from (611) of Chapter 4 that if (Y, d) is a one-point compactification of X with point at infinity wand inclusion map i. X ---+ Y, then there is a norm-preserving algebra isomorphism ff-+f" which identifies Coo(X,IF) with the metric space G of all elements f in c(Y,IF) with f(w)=O. As with every normed linear space, we are interested in the bounded linear functionals on CDC (X, IF). It will suffice to consider the case IF=lR, since a bounded linear functional u on Coo(X,
(fl,f2 E Coo (X)),

where U 1 and U 2 are bounded linear functionals on Coo (X). Consider therefore a bounded linear functional u on Coo (X). For each g in C(Y) write Tg=g-g(w). Then Tg belongs to G, and so can be identified with an element of Coo(X) The map u T is thus a bounded linear functional on C(Y) that vanishes on the constant functions. Conversely, every bounded linear functional v on C(Y) that vanishes on the constant functions defines a bounded linear functional u on Coo(X) by u(f)=v(f"), and v=uoT. An interesting special case arises when X is taken to be the set 7l + of positive integers. In this case we can describe the bounded linear functional u explicitly. For each n in 7l + let (jn be the element of C oo (71+) defined by (jn(n) = 1 and (j.(m) =0 if m=l=n, and write 0

306

Chapter 7 N onned Linear Spaces

L f (n) c5 n converges to 00

F or each f in Coo (7l +) the series

f: indeed, if

<;

n= 1

is an arbitrary positive number, and N. a positive integer such that If(n)1
Ik-.t!(n)c5nll<1: (k~N.). Since u is uniformly continuous, if follows that 00

(1.13)

L anf(n)

u(f)=

(fEC oo (7l+))·

n= 1

Now, for each pair m, N of positive integers there exists h: 7l+ ~ { -1,0, 1} such that hen) = 0 for all n > N, and such that for 1 ~ n ~ N,

h(n) = -1

=

h(n)=1

= an>m- 1 ,

h(n)=O

= lanl <2m-

and Clearly, hE Coo (7l+), Ilhll

~1,

an < -m-l,

1.

and

IJl an h(n)1 = lu(h)1 ~ c, where c>O is any bound for u. Thus

ntl lanl =

Intl anh(n)1 + ntl lanl~c+2Nm-l. h(n)= 0

Holding N fixed and letting

m~ 00, we

obtain

N

(1.14)

L rani ~ c

(N E7l+).

n= 1

Conversely, if (an) is any sequence of positive numbers satisfying (1.14), then (1.13) defines a bounded linear functional u on C()()(7l+) with bound c. The details are left as an exercise.

2. Finite-Dimensional Spaces The simplest instances of Banach spaces are the euclidean spaces IFn, with norm

These spaces are finite-dimensional, in the following sense.

2 Finite-Dimensional Spaces

307

(2.1) Definition. A normed linear space X is finite-dimensional if there exist nonzero vectors e l' ... , en and bounded linear functionals U l ' ... , un such that n

(2.1.1)

X

=

L Ui(X) ei

(XEX)

i= 1

and (2.1.2)

ui(e)=O

(1 ~i,j~n, i=t=j).

The coordinate functionals U l ' ... ' un are uniquely determined by conditions (2.1.1) and (2.1.2). The numbers ul(x), ... ,u.(x) are the coordinates of X; the vectors e l' ... ,e. are the basis vectors of X; and the set {e l , ... , e.} is a basis of X. The JDteger n is called the dimension of X, and X is said to be ndimensional. The reader may verify that the dimension does not depend on the choice of the basis of X. We consider the trivial vector space {OJ to be finite-dimensional, with dimension O. Under the conditions of Definition (2.1), the mapping (J defined by (J(x)=(ul(x), ... ,u.(x»

(XEX)

is a linear isomorphism of X with the linear space IFn. It is uniformly continuous, because each u i is uniformly continuous. The inverse mapping takes the element (a l , ... , a.) of IFn to the vector a l e l + ... + a. e. in X. It also is uniformly· continuous, by virtue of the inequality n

Ileal e l + ... +anen)-(a'l e l + ... +a~e.)II~

L lai-a;llleJ. i= 1

Hence (J is a metric equivalence. Since (J maps bounded subsets of X to bounded subsets of IFn, and since IFn is locally compact, so is X. Hence X is separable and complete. Thus every finite-dimensional normed linear space is a locally compact Banach space. It follows from this and (6.2) of Chapter 4 that if X is a linear subset of a normed linear space V, and if X is finite-dimensional relative to the restriction of the norm on V, then X is a subspace of V in the sense of the next definition. (2.2) Definition. By a subspace of a normed linear space V we mean a closed located linear subset of V. (2.3) Proposition. The unit sphere of a finite-dimensional Banach space X is compact.

308

Chapter 7

N armed Linear Spaces

Proof. This is obvious if X = {O}; so we may assume that X has positive dimension n. First note that (2.3) of Chapter 5 holds with
normed linear space into a normed linear space is normable. Proof. This follows from (2.3), with reference to (4.3) of Chapter 4.

0

Before proving a converse of Proposition (2.3) we need a simple lemma. (2.5) Lemma. Let x l'

... , Xn be elements of a normed linear space V, S the set of all linear combinations of x l' ... , x., and e a positive number. Then there is a finite-dimensional subspace X of V with Xc S, such that P (Xi' X) < e for each i.

Proof We construct finite-dimensional spaces Xo=={O}, X 1, ... ,X. such that XjcS and P(xi,X}<e for 1;;£i;;£j;;£n. Indeed, having found Xj for some j (O;;£j O. In the former case, we take X j+ 1 == Xj' in case p(Xj+ l ' X}>O, we take X j+1 to be the set of all linear combinations of elements of Xju {xj+ d. To complete the proof, we need only take X==X •. 0 (2.6) Theorem. A

locally compact normed linear space is finite-

dimensional. Proof. Let V be a locally compact normed linear space, and let K c V be a compact set containing the unit sphere of V. Let {Xl> .. ,x.} be a -!- approximation to K, and construct a finite-dimensional subspace X of V such that p(Xi,X)
~

in

2 Finite-Dimensional Spaces

309

belongs to K, so that IIY-Xill <± for some i. Choosing ~i in X so that Ilxi-~;l1 <±, we have Ilx-~lllly-x;l1 = II(x-~-llx-~llx;)11

~ Ilx-(~+ Ilx-~11 OII-llx-~llllxi-O

~p(x,X)-Hx-~11

=15

-± Ilx -

~II.

Thus Ilx-~11 ~c5(lly-xJ +±)-l >c5(±+±)-1 =215,

contrary to our choice of~. It follows that 15=0. Hence XEX, as X is closed. Since x is an arbitrary element of X, we have V=X, so that V is finite-dimensional. D We have already observed that a finite-dimensional linear subset F of a normed linear space X is located. According to a classical theorem of approximation theory, we can say more: to each x in X there corresponds an element b of F that is closest to x, in that p(x, b) = p(x, F). This theorem is not true constructively, but there is an interesting constructive substitute, for which we need three lemmas. For convenience, we write IFe=:= {te: tEIF} for each e in X. If e is nonzero, then IF£ is a I-dimensional subspace of X. (2.7) Lemma. Let x, e be elements of a normed linear space X over JR, with e nonzero, and for each 15 > p(x, JR e) let S(c5)=:={tEJR: Ilx-tell ~c5}. Then

if

S (c5) is compact, it is a proper compact interval of JR.

Proof. Suppose that S(c5) is compact, and write m =:= inf S(c5), M=:=supS(c5). Choose c5' so that p(x,JRc) Ilx - tell IS Ulllformly continuous on JR, we have Ilx-mell =c5= Ilx-M ell and Ilx-m' ell =15'= Ilx-M' ell. Since S(c5')cS(c5), it follows that m<m'~M'<M. Thus m<M. Now consider any t in [m,M]. We have t=(I-1:)m+1:M, where O~1:=:=(t-m)(M

_m)-l

~

1.

310

Chapter 7 Normed Linear Spaces

Hence Ilx-tell

~(1-,)

Ilx-mell +r Ilx-Mell =(1-,)fJ+,fJ=fJ

and so tES(fJ). Thus S(fJ) = [m,Ml

D

(2.8) Lemma. Let x, e be elements of a normed linear space X over IR, with e nonzero, and let d be a nonnegative number. Suppose that max{ Ilx -tell, Ilx -t' ell} >d whenever t, t' are distinct real numbers. Then there exists that IIx-~ell >d entails p(x, IRe»d.

~

in IR such

Proof. We may assume that p(x, IRe) < d + 1. For each c5 > p(x, IRe) let S(fJ) = {tEIR: IIx-tell

~fJ}.

Construct recursively an increasing sequence A.: Z+ --+{O, I} and sequences (an), (b n), (c5 n) of real numbers, as follows. Set A(1)=0. Choose fJ I so that p(x,IRe)1 be an integer, and suppose that A(n-1), an_I' bn_ l , and fJ n_ 1 have been constructed so that p(x,1Re)
=>

p(x,IRe)
A(n) = 1

=>

p(x, IRe»d.

and Moreover, if A(n)=O, then by (2.8.2), we have O
-all

2 Finite-Dimensional Spaces

311

and therefore (2.8.3) Clearly, (2.8.3) also holds if A(n) = 1. It follows that (an) is a Cauchy sequence and therefore converges to a limit ~ in IR. Suppose that I x - ~ e I > d. Then there exists a positive integer N such that Ilx-aNell >d+N- 1 . Were A(N)=O we would have the contradiction

IIx-aN"'11 =bNd. D In the proof of our next theorem we use the notion of a quotient space, which will appear again later in the chapter.

(2.9) Definition. Let Y be a located linear subset of a normed linear space X. Then Ilxllx/y=p(x, Y) (XEX) defines a seminorm on X. Taken with this seminorm and the corresponding equality relation = XIY, X is a normed linear space - called the quotient space of X by Y, and written XIY. (2.10) Lemma. Let X be a normed linear space, n a positive integer, and

F an (n + 1)-dimensional subspace of X. Let {e 1 , ... , en + d be a basis of F, and Y= IFe n + 1 . Then F is an n-dimensional subspace of the quotient space XIY, with basis {e1, ... ,en }.

Proof" Let the coordinate functionals for the basis {e1, ... ,en+tl of F be U l ' ... , Un + l ' and let c > 0 be a common bound for these functionals n

For each x in F we have x=x/Y i

*' j). If also x'

L ui(x)e i, where ui(e)=O i~

EF

(1 ~i, j~n,

1

and x = X/Y x', then as Y is closed, n

L (ui(x) -

ui(x')) ei E Y,

i= 1

and so for 1 ~k~n we have uk(X)-Uk(x')

=

Uk

(tl (ui(x) -Ui(X'))ei) =0.

Hence Uk is a function on F relative to the equality relation on XIY. Since Uk is linear relative to = X/Y, it remains to prove that Uk is bounded on F relative to I Ilx/y' To this end, consider an arbitrary

312

Chapter 7 N armed Linear Spaces

element x of F with II x II X/y ~ 1. Since p (x, Y) < 2, there exists a in IF such that Ilx-aen + 1 11 <2. Thus luk(x)1 = luk(x -aen + 1)1 ~

Hence

Uk

cllx-aen + 1 11 <2c.

has bound 2c relative to II Ilx/y.

0

We now introduce two basic notions of approximation theory. (211) Definition. Let F be a nonvoid subset of a metric space (X, p), and a an element of X. We say that a has at most one closest point in F if for all x and x' in F with x =1= x' there exists y in F such that

max {p(a, x), p(a, x')} > p(a, y). On the other hand, any element b of F such that p(a,b)~p(a,x)

(xEF)

is called a closest point to a in F. If every point of X has a closest point in F, then F is located in X.

We can now prove a constructive version of the fundamental theorem of approximation theory. (2.12) Theorem. If F is a finite-dimensional subspace of a normed linear space X over JR, and a is an element of X with at most one closest point in F, then a has a unique closest point in F. Proof We may assume that F has positive dimension. If F has dimension 1, then F=lRe for some nonzero e in X. We can therefore apply (2.8) with x=a and d=p(a,F), to find ~ in lR with lIa-~ell = p(a, F).

Now let n be a positive integer, suppose we have proved the desired result for all n-dimensional subspaces of a real normed linear space, and consider the case where F is (n + l)-dimensional. Let {e), .. ,en+d be a basis of F, and let Y=JRen+I' By (2.10), F is an ndimensional subspace of the quotient space XIY, with basis {e),. , en}. Also, for each x in X we have (2.12.1 )

p(x, F) = inf{ Ilx - z - te n + III: zEF, tElR} = inf{p(x -Z, JRen + I): zEF}

=infUlx-zllx/y: zEF}.

3 The Lp Spaces and the Radon-Nikodym Theorem

Consider arbitrary elements ZI' Z2 of F with Ilzl -z21Ix/y>O. If t, distinct real numbers, then for k = 1, 2 we have

313 t'

are

and therefore

Applying (2.8) with x=a-zb .e=en+b and d=p(a,F), we construct 'k in IR such that (2.12.2)

Ila-zk-'ken+lll >p(a,F) => Ila-zkllx/y=p(a-z k, IRe n+ 1 »p(a,F).

Since

II(zl +'l e n+l)-(z2+'2 e n+l)11 = II(ZI -z2)+('I-'2)en+ 1 11 ~ Ilzl -z21Ix/y>O,

we have

max {ila -Zl - , I e n+ III, Iia -Z2 - '2en+ III} > pea, F) Hence, by (2.12.1) and (2.12.2), max{lla -zlllx/y, Ila-z 21Ix/y} > pea, F)=inf{lla-zllx/y: zEF}. Since ZI' Z2 are arbitrary, it follows that a has at most one closest point in the n-dimensional subspace F of XIY. By our induction hypothesis, there exists Zo in F with p(a-z o, Y)= Ila-zollx/y=inf{lla-zllx/y: zEF} =p(a, F).

Applying (2.8) with x=a-z o, .e=en+ l , and d=p(a,F), we now construct a real number to such that Ila-zo-toen+111 =p(a,F).

=

Then b Zo + to e n + I is a closest point to a in F relative to the original norm on X. Finally, if b' is any closest point to a in F, then assuming that b =l= b' , we have max{lla -bll, Ila-b'll} > pea, F), a contradiction. Hence b = b'.

D

3. The Lp Spaces and the Radon-Nikodym Theorem Throughout this section, unless we state otherwise, I will be a completely extended, a-finite integral on x.

314

Chapter 7 Normed Linear Spaces

Our next goal is to introduce certain basic normed spaces of measurable functions. This requires some preliminary inequalities. (3.1) Lemma. Let a, b be nonnegative numbers, and cx, p positive numbers with cx + p = 1. Then (3.1.1 )

Proof: For fixed values of b, cx, and p, consider f(a)= cxa + pb -a~bfJ.

We have f'(a) = cx - cxa~-l bfJ = cx(l- (ba-ll)

if a =1= O. Thus f'(a)~O if a>b, and f'(a)~O if O O. By continuity, (3.1.1) is valid for a~O. D Successive applications of Lemma (3.1) lead to (3.2) Corollary. If ai' ... , a. are nonnegative numbers, and cx l ' positive numbers with ex 1 + ... + cx. = 1, then

... '

cx. are

The proof of our next lemma is another simple application of differentiation, and is left to the reader. (3.3) Lemma. Let x, y be nonnegative numbers, and let 0 < t < 1. Then (x + y)' ~x' + y'.

.t.

(3.4) Theorem. Let fl' ... be nonnegative integrable functions, and cx l ' ... ' cx. positive numbers whose sum is 1. Then ff' . .. f.~n is integrable, and

(3.4.1) Proof: The integrability of ff' .. f.~n follows from (3.2), and (7.11) of Chapter 6. Let Cl> ... , Cn be positive numbers with Ci ~ I(f;) (1;£ i;£ n). By (3.2), I((cllflt' ... (c;;lf.)~n);£1(CXl Cllfl + ... +cx.c;;lfn)

= cx l c l l 1(fl) + ... + CXnc;; 1 I(fn) ~ CX l + ... + CX n = 1.

3 The Lp Spaces and the Radon-Nikodym Theorem

315

Thus

l(ff' ... g.n)~c~' ... c~n. Letting ci -+l(J;) (1 ~ i ~ n), we obtain (3.4.1).

D

Inequality (3.4.1) is called HOlder's inequality. It is a basic tool in the theory of certain spaces - the Lp spaces - which we now define. (3.5) Definition. For each p ~ 1 the set Lp consists of all measurable functions f such that Ifl P is integrable. The norm of a function f in Lp is (3.5.1 ) Elements f and g of Lp are equal if f(x) = g(x) on a full set. We now derive a remarkable inequality relating certain Lp spaces, which is actually a reformulation of Holder's inequality. (3.6) Proposition. Let p> 1, q> 1, and

r

1 +q-1

= 1. Let fELp and

gELq. Then fgEL 1 and l(fg)~ Ilfllp Ilgll q.

Proof: Applying (3.4) with f1=lfI P, f2=lglq, IX1=rl, and we see that If gl, and therefore f g, is integrable, and that

IX 2 =q-1,

l(fg)~I(1fgl)=I(lfIP«1Iglq«2)

~I(lfIP)«1 1(lglqt 2

= Ilfllp Ilgll q.

D

HOlder's inequality leads to another remarkable inequality (3.7.1), known as Minkowski's inequality. (3.7) Theorem. Let p ~ 1, and let f, gELp. Then f (3.7.1)

+ g is

in L p' and

Ilf +gllp~ Ilfllp+ Ilgll p·

Proof Since f + g is measurable and If +gIP~(2max{lfl,

Igl})P~2P(lfIP +

IgjP),

f

+ g belongs to L p ' by (7.11) of Chapter 6. Let 0 < IX < r 1. Then by (3.3), If+gIP«~lfIP«+lgIP«. By (3.4), the functions IfIP«lf+gIP(l-«) and IgIP«lf+gI P(1-«) are integrable, and

1(lf + gIP)=l(lf +gIP« If + gIP(1-«) ~ 1(lfl P« If + gIP(1-«) +1(lgIP« If + gIP(1-«) ~ (l(lfI P)« + l(1gIP}«) l(1f + gIP}1-«.

316

Chapter 7

Nonned Linear Spaces

Thus (3.7.2) Now for each e>O, either (3.7.3)

Ilf +gllp~e+ Ilfllp+ Ilgll p

Ilf+gllp>O In the latter case, dividing both sides of (3.7.2) by Ilf + gll~(I-a) and then letting (X---+p-l gives Ilf + gllp ~ I flip + Ilgllp. There-

or

fore (3.7.3) holds in all cases. Since e>O is arbitrary, inequality (3.7.1) now follows. D It is now clear that Lp is a normed linear space relative to the Lpnorm I lip given by (3.5.1). Note that the equality relation corresponding to I lip is the same as the equality on Lp defined in (3.5). So far, we have not made use of the assumption that I is a-finite.

This property of I comes into play when we prove that Lp is complete. (3.8) Lemma. If (f.) is a Cauchy sequence in the normed linear space Lp (where p ~ 1), then (f.) is Cauchy, and hence convergent, in measure. Proof: For each e>O there exists N in Z+ such that I(lfm-f.n~el+P

(m,n~N).

Consider integers m, n with m ~ n ~ N. By (8.20) of Chapter 6, there exist (X in (e,2e) and a measurable set A such that Ifm-f.I<(X on Al and Ifm-f.I~(X on AO. Then LA~(X-llfm-f.1 on the full set A 1 uAo; so that -A is an integrable set, by (7.11) of Chapter 6. Moreover, e 1 +P~I(lfm-f.IP)~eP J1.( -A);

whence J1.( - A) ~ e. It follows that the sequence (f.) is Cauchy in measure. Reference to (8.16) of Chapter 6 completes the proof. D (3.9) Theorem. For each p ~ 1 the space Lp is complete with respect to the metric induced by the Lp-norm. Proof Let U.) be a Cauchy sequence in Lp. By (3.8), and (8.16) of Chapter 6, there is a subsequence (f.J~ 1 converging almost uniformly, and pointwise on a full set, to a measurable function f. By passing to another subsequence, we may assume that Ilf.k+ 1-f.kllp~2-k for each k. Write gl=lfll and gk= If.11 + If.2 -f. 1I + ... + If'k -f.k- 11

(k~2).

3 The Lp Spaces and the Radon-Nikodym Theorem

317

For j>k we have (3.9.1)

Therefore (gk);:'~ 1 is a Cauchy sequence in Lp; so that, by (3.8), and (8.16) of Chapter 6, some subsequence converges almost uniformly, and pointwise on a full set, to a measurable function g. Since the sequence (gf);:'~ 1 is increasing, it converges almost uniformly, and pointwise on a full set, to gPo Hence gP is measurable. Moreover, we see from (3.9.1) that Illgjllp-llgkllpl ~2-k+l

0>k),

so that (1Igkllp)r'~ 1 is a Cauchy sequence in JR. It follows that the sequence (I(gmr'= 1 converges in JR. Therefore, by the monotone convergence theorem, gP is integrable. Since IfnY ~ g~ ~ gP for each k, we have IfIP~gP. Therefore Ifl P is integrable, by (7.11) of Chapter 6. Also, If-fnY~(2max{lfl, Ifn.l})P~(2g)P,

and so Ilf-fnJp-+O as k-+oo, by the dominated convergence theorem. Since (fn) is a Cauchy sequence in Lp, it follows that Ilf-inllp-+O as n-+ 00. Hence Lp is complete. 0 (3.10) Lemma. Let f be a nonnegative element of Lp (where p ~ 1), and let e > o. Then there exists a simple function g such that 0 ~ g ~ f and

Ilf-gllp<e. Proof By (8.10) of Chapter 6, there is an integrable set K such that Ilf -x Kfll p<e/4. By (4.15) and (7.2) of that chapter, there exist c>O and an integrable set A such that Ale Kl, IlxK-Af lip < e/4, and o~ XA f ~ c. Then II f - XA f I p < e/2. By (7.4) of Chapter 6, there exists a simple function g such that O~g~XAf ~g+H1

+ J.L(A))-l/PeXA

on a full set. Thus 0 ~ g ~ f on a full set, and Ilf - gllp~ IlxAf - gllp +e/2 ~t(1 + J.L(A))-l/p eJ.L(A)l/p +e/2<e.

0

The first corollary follows immediately. (3.11) Corollary. The set of simple functions is dense in Lp. (3.12) Corollary. Let f be a nonnegative element of Ll n L 2 , and let e > o. Then there exists a simple function g such that 0 ~ g ~ f, Ilf-glll~e, and Ilf-gI12~e.

318

Chapter 7

Normed Linear Spaces

Proof: By (3.10), there exist simple functions gl' g2 with O~gl ~f, Ilf-gllll~E, and Ilf-g2112~E. It suffices to take g=gl v g2· 0 0~g2~f,

Of particular interest is the case where I arises from a positive measure on a locally compact space X. (Recall that I is then a-finite.) (3.13) Lemma. If I is the complete extension of a positive measure on a locally compact space X, then the set C(X) of all test functions on X is dense in Lp (where p ~ 1).

Proof. Since the set of simple functions is dense in L p ' it will suffice to show that if A is any integrable set and E any positive number, then there exists f in C(X) with II XA - f II p < E. To do this, let Un) be a

"Ifk converges to XA in 00

representation of XA in C(X) Then the series measure. Define test functions gn by gn=2A (-2V

kt/k)

k= 1

(nEZ+).

Then the sequence (gn) converges to XA in measure, and for each n,

on a full set. Hence, by the dominated convergence theorem, I(lxA - gnl)---+O as n---+ 00. Now choose a positive integer N so that

I(lxA - gNI) < 31 - p EP. Then IlxA -gNII:=I(lxA -gNIP- 1 IxA -gNI) ~ 3P - 1 I(lxA - gNI) <EP

and therefore IlxA -gNllp<E.

0

(3.14) Proposition. If I is the complete extension of a positive measure on a locally compact space X, then Lp is a Banach space.

Proof. In view of (3.9), it will suffice to prove Lp separable. To this end, let (Kn) be an I-basis of compact subsets of X such that each bounded set Be X is contained in some Kn. Let (gJ be a dense sequence in C(X). Consider an arbitrary f in Lp and an arbitrary E> O. By (3.13), there exists g in C (X) with II f - g II p ~ E/2. Choose III

3 The Lp Spaces and the Radon-Nikodym Theorem

319

and n in Z+ so that Km is a support for g, and

Ilg - gnll ~te(1 + fJ,(K m))-l/P• Then

Ilf-gllp+ Ilg-XK",gnllp ~ e/2 + I (XK", Ig - gnI P)l/ P

Ilf-XK",gnllp~

~ e/2 +te(1

+ fJ,(K m))-l/ p fJ,(Km)l/p

~e

Hence the countable set {XK,gj: i,jeZ+} is dense in Lp.

0

We now return to the case where I is an arbitrary a-finite integral. To smooth the way to the incisive geometrical inequalities of Clarkson for Lp spaces, we prove some lemmas. The first provides an inequality which replaces Holder's inequality when 0 < p < 1. (3.15) Lemma. Let p, q be real numbers with 0 0 on a full set, such that f g and gq are integrable. Then fP is integrable, and

(3.15.1)

I(f g) ~ I(fP)l/ PI (gq)l/q.

Proof' Define u=.(fg)P, v=.g-P, r=.p-1, and s=. _qp-1. Then r>1, s> 1, and r- 1 +S-l = 1. By (3.6), uv is integrable, and I(uv)~I(ur)l/r I(vs)l/s.

In other words, fP is integrable, and I(fP)~I(fg)P I(gq)-p/q.

By (4.13) of Chapter 6, I(gq»O. Hence

I (f P) I (gq)p/q ~ I (f g)p, from which (3.15.1) follows immediately.

D

(3.16) Lemma. Let p > 0, and let f, g be nonnegative measurable functions such that fP and gP are integrable. Then for each e > 0 there exists an integrable set 8 such that I (X _ sfP) < e, I (X _s gP) < e, and f + g > 0 on 8 1•

Proof: By (4.11) and (4.14) of Chapter 6, we can find r>O such that the complemented sets A=. (fP ~ r) and B =. (gP ~ r) are integrable, I(X_AfP)<e, and I(X __ BgP)<e. Clearly, f+g~r1/p on A 1uB1. It therefore suffices to take S =. A v B. 0

320

Chapter 7

Nonned Linear Spaces

Our next lemma gives a replacement for Minkowski's inequality in the case where 0 < p < 1. (3.17) Lemma. If 0 < p < 1, and if f and g are nonnegative measurable functions such that fP and gP are integrable, then (f + g)p is integrable, and (317.1) Proof. The inequality (f + g)p ~ fP + gP implies that (f + g)P is integrable. Since f(f + g)p-1 ~ (f + g)P, it follows that f(f + g)p-1 is integrable; similarly, g(f + g)p-1 is integrable. Using (3.16) and an approximation argument, we see that it is enough to consider the case in which f +g>O on a full set. Write q=pj(p-1). By (3.15), I((f + g)p)=I(f(f +g)P-1)+I(g(f +g)p-1) ~ I(fP)l/ p I((f + g)p)l/q + I(gP)l/ p I((f + g)p)l/q.

Since I((f + g)p) > 0 by (4.13) of Chapter 6, we can divide through by I((f + g)p)l/q to obtain (3.17.1). 0 (3.18) Lemma. If O
(t>O)

defines an increasing function of t. Proof· Write IX=lnc. Then for each t>O, 00

f(t)=t- 1(exp( -IXt)-l)=

L (_1)"IX n t n - 1jn! n= 1

so that 00

f'(t)=

L (-1)"(n-1)IX n t n - 2 jn! n= 2

Since IX < 0, all the terms of the last series are positive, and so f'(t) > O. The result now follows. 0 (3.19) Lemma. Let x, y, p, and q be real numbers, with p -1 Then

(3.19.1) for

p~2,

and

(3.19.2) {or 1
+ q -1 = 1.

3 The Lp Spaces and the Radon-Nikodym Theorem

321

Proof. Assume first that p ~ 2, so that 1 < q ~ 2 By continuity and symmetry, we may take Ixl > Iyl >0. Dividing by Ixl q and setting c=:lyx- 1 1 (so that 0
(3.l9.3)

Expanding the left side of (3.19.3) in powers of c by means of Taylor's theorem, we get 00

2

L: ((2k) !)-1 q(q -1) '"

(q -2k + 1)c 2k

00

L: ((2k)!)-I(q-1) ... (q-2k)C 2kp

-2

k= I 00

- 2

L: ((2k _l)!)-I(q -1) ... (q -

2k + 1)C(2k-l)p;

k= I

that is, 00

(3.19.4)

2L:((2k)!)-I(q-1) .. (q-2k+1) k= 1

. (qc2k_(q_2k)C2kp_2kc(2k-l)p).

Each of the products (q -1) ... (q - 2k + 1) is nonnegative, because 1 < q ~ 2. Consider k in 7l +, and write IX =: 2 k(p -1). Then by (3.l8), qc 2k _(q _ 2k)C 2kp _ 2kc(2k-l)p =2kpC 2kp (rx- 1 c-~ -(1X- 1 _ p-l) _ p-l c-P) = 2kpC 2kP (1X -I(C-~ -1) - p -I(C - P-1)) ~ 0,

since 1X=:2k(p-1)~2(p-1)=p+p-2~p. Thus each term in the expansion (3.19.4) is nonnegative. Hence (3.l9.l) is valid Assume next that 1 < p ~ 2. Then q ~ 2, and thus by the case already considered, we have Ix + yiP + Ix - yiP ~ 2(lxl q + lyIQ)P-I.

Substituting (x + y)/2 for x and (x - y)/2 for y gives Ixl P + lylP ~ 2(~)Q(P-1)(lx + ylQ + Ix _ yIQ)P-l =2 1 - P(lx + ylQ + Ix - yIQ)P-I,

which is equivalent to (3.192).

D

We now prove Clarkson's basic inequalities for the Lp norms. (3.20) Theorem. Let f and g be elements of L p ' where p> 1 and p-l + q - I = 1 Then for p ~ 2 we have (3.20.1)

2(llfll~+ Ilgll~)q-l ~ Ilf +gll~+ Ilf -gll~

322

Chapter 7 N ormed Linear Spaces

and Ilf +gll:+ Ilf -gll:~2(llfll~+ Ilgll~)P-1,

(3.20.2)

while for 1 < p < 2 the reverse inequalities hold. Proof. Inequality (3.20.2) arises from (3.20.1) when we replace f with (f + g)/2 and g with (f - g)/2. It is therefore enough to prove (3.20.1). Consider first the case p?;2. Write s==p/q, so that s?;1. From Minkowski's inequality and (3.19) we get Ilf + gll~+ Ilf -gll~= II If +glqll.+ Illf -glqll. ?; II If + glq +

If -

glqll.

?; 211(lf1 P+ IgIP)q-111. =2(llfll:+ Ilgll:)q(P =2(llfll:+ Ilgll:)q-1, as desired. In case 1
~ 211(lf1 P + IgIP)q-111.

=2(llfll:+ Ilgll:)q-t, as desired.

0

(3.21) Definition. A normed linear space V is uniformly convex if there exists an operation {) from 1R + to (0, 1), called a modulus of uniform convexity for V, such that IIt<x+ y)11 ~{)(6) whenever 6>0, x and yare elements of V with Ilxll = Ilyll = 1, and Ilx - yll ?; 6. Theorem (3.20) has the following important consequence. (3.22) Corollary. For each p> 1 the normed linear space Lp is uniformly convex.

Proof: Let f, g belong to Lp, and let 6>0. Without loss of generality, take 6<2. Assume that Ilfllp= Ilgll p= 1 and that Ilf -gllp?;6. Let q == p/{p -1). By (3.20), if p?; 2, we have II-HI + g)llp =H f + gllp ~1-{2P _6P)1(p ={1-{1-6)P)IIP,

3 The Lp Spaces and the Radon-Nikodym Theorem

323

while if 1 < p < 2, we have 11~(f +g)llp=Hf +gllp;;i;!(2q-eq)1/q=(1-(~e)q)1/q.

Thus if 1

(3.22.1) where

c5(e) == max {(l- (~e)p)l/p, (1- (!e)q)l/ q}.

In case p>1 is arbitrary, we again have (3.22.1): for if 11~(f+g)II>c5(e), then both cases 1 < p < 2 and 2;;i; p are ruled out, which is absurd. D We can use Theorem (3.20) and its Corollary (3.22) to characterize the normable linear functionals on Lp for p> 1. To do this we need two preliminary results, the first of which is a general result about normable linear functionals on a uniformly convex, complete normed linear space. (3.23) Proposition. Let u be a nonzero normable linear functional on a uniformly convex, complete normed linear space B. Then there exists x in B with Ilxll = 1 and u(x)= Iluli. Proof. Let (xn) be a sequence of vectors with norm 1 in B, such that (u(xn)) converges to Iluli. Let c5 be a modulus of uniform convexity for B, and consider an arbitrary e>O. Choose N in 7l+ so that

HI + c5(e)) Ilull < u(xn) Then for m,

n~N

(n~ N).

we have

Ilullllt(xm+xn)11 ~ilu(xm+xn)1 -xn)1 ~u(xm) -ilu(xm) -llulll-ilu(xn) -llulll >i(1 + c5(e)) Ilull -*(1- c5(e)) I ull -*(1- c5(e)) I ull =c5(e) Iluli. Since u is nonzero, it follows that 11~(xm + xn)ll > c5(e), and therefore that I Xm - Xn I ;;i; e. Thus (x k) is a Cauchy sequence and so converges to a vector x in B. Clearly Ilxll = 1 and u(x)=lIuli. D ~u(xm) -ilu(xm

(3.24) Lemma. Let r> 1 and r - 1 + s - 1 = 1. Then there exists c > 0, depending only on r, such that (3.24.1) whenever O;;i;e;;i; 1.

(2(1 +er )'-1_1)1/';;i; 1 +cer

324

Chapter 7

Normed Linear Spaces

Proof For a fixed positive number c, and for 0::;; e::;; 1, consider fee) == (1

+ ce')' -

2(1 + e'),-I + 1.

We have F(e) = re' -I [c s(1 + ce')'- 1_ 2(s -1)(1 + ery- 2]. If 1 <s<2, then

F(e)?;, re'- I [c s - 2(s -1)]; while if s ?;, 2, then F(e)?;, re' -I [c s - 2(s -1 )2' - 2]. It follows that if s> 1 and c>max {2s- l (s-I), 2S - l s- l (s-I)}, then F(e)?;,O (O::;;e::;; 1). In that case, for O::;;e::;; 1 we have f(e)?;,f(O) =0, from which (3.24.1) follows immediately D (3.25) Theorem. Let p> 1 and p -

1+ q -I

= 1. For each

g in L q, define

u g : Lp-+lR by

(3.25.1 )

Then u g is a normable linear functional on L p' and Ilugll = Ilgll q Conversely, if u is any normable linear functional on L p' then u = U g for some g in L q. Proof Consider ug , where gEL q. By (36), ug is well defined on L p' and (3.25.2)

Since ug is clearly linear, it is therefore a bounded linear functional on Lp. Now the function f==(g+)q,p_(g-)q,p belongs to L p' and fg?;,O and Ifl P = Iglq on a full set Thus ug(f) = J(fg) = JOgl I +qp - ') = J(lglq) = IlgllqJ(lglq)l-q-' = IlgllqJ(lfIP)I'P= Ilgllqllfll p Together with (3 25 2), this implies that u g is normable, with norm Ilgll q. Conversely, consider a normable linear functional U on Lp. Choose ( >0 so that (2(1 + e,),-I _1)I/s::;; 1 +ce' whenever 0<e<1, rE{p,q}, and r- l +s- I =1. Assume first that llull >0, and consider an arbitrary e in (0,1). By (3.9), (3.22), and (3.23),

3 The Lp Spaces and the Radon-Nikodym Theorem

325

there exists I in Lp with IIIllp=1 and u(f)=lIuli. Choose g in Lq so that Ig~O and IIIP=lglq on a full set. Then IIgllq=1 Let h be an arbitrary vector in Lp with IIhllp= 1 and u/h) =0. Then

ug(f +eh)=ug(f -eh)=ug(f)=I(fg)=I(IIIP)= 1. Therefore, as lIugll = IIgll q= 1, we have III +ehllp~ 1 and III -ehllp~ 1 In case p ~ 2, (3.20) gives

III +ehll~;2;2(1 +eq)p-1-1 and therefore III + ehllp;2; 1 + ceq. Similarly, if 1

111+ ehll:;2; 2(1 + eP)q-l -1 and therefore

I I + eh I p~ 1 + c cp. It follows that for all

p> 1,

III + chllp;2; 1 +d, where

t

== min {p, q}. Therefore Iu(f + ch)1 ~ (1 + d) lIuli

and so

u(h);2; c l(lu(f + eh)l- u(f) ;2;c 1 ((1 +d) lIull-liull)

=d- 1 I1ull. It follows that

(3.25.3) Now for any h in Lp,

ug(h-u g(h)f)=u g(h)(1-u g(f)=O. Hence, by (3.25.3),

lu(h) - u II ull g(h)1 = lu(h - ug(h)f)1 ~ d-1llull (lIhllp + IUg(h)llIfllp) ~d

- 1 I u I/( I h I p + I g I q I h I p)

and so (3.25.4) Now consider the case of general u. For each positive integer n we have either lIuli >0 or lIuli
(3.25.5)

lu(h)-ugJh)1 ;2;n- 1 Ilhll p

(hELp).

In case lIuli
326

Chapter 7 Normed Linear Spaces

positive integers j and k we then have

Ilgj-gkllq= Ilugj-ugJ

~rl +k- 1.

Hence (gn) is a Cauchy sequence in Lq By (3.9), there exists g in Lq such that Ilug-ugJ = Ilg-gnllq----.O as n----'oo. It follows from this and (3.25.5) that u = ug, as we wanted.

D

As we shall see later, certain bounded linear functionals on Lp (p> 1) are not normable, and therefore do not come from elements of L q . Nevertheless, we shall prove that the normable linear functionals on any Banach space are dense in a natural metric (to be described later) on the space of bounded linear functionals The case p = q = 2 of Theorem (3.25) is of particular interest, and has a very important application in the proof of the Radon-Nikodym theorem. This will be the main topic in the rest of this section, in which we no longer require that I be O'-finite. The Radon-Nikodym theorem relates two integrals I and J defined relative to the same underlying set X. To avoid confusion, we use notation like Lp(l), and expressions like '[-measurable' and 'Jfull', the meaning of which should require no explanation. If I and J are the complete extensions of two integrals defined on the same subset L of ff(X), then we call L a common initial set for I and J. In that case, a function f defined on a subset of X is [integrable (respectively, J-integrable) if and only if there exists an 1representation (respectively, J-representation) of f by elements of L. (3.26) Definition. Let [ and J be completely extended integrals with a common initial set L. We say that J is absolutely continuous relative to [ if there is an operation b from 1R + to 1R + such that J(XA) ~ 8 whenever 8> 0, A is a complemented set that is integrable relative to both [ and J, and [(XA) ~ b(8) The operation b is then called a modulus of absolute continuity for J relative to I. Clearly, the identity map continuity for [ relative to [.

81--+8

of 1R + is a modulus of absolute

(3.27) Lemma. Let I and J be completely extended integrals with a common initial set L, such that J is absolutely continuous relative to I Let f be an I -integrable function, and (In) an [-representation of f in L.

Then the sequence

Ctl f=

measure relative to J.

fk

1

of J -integrable functions is Cauchy in

3 The Lp Spaces and the Radon-Nikodym Theorem

327

Proof" Let () be a modulus of absolute continuity for 1 relative to 1. Given 1'>0, let lX=min {E/2, {)(E)}, and choose N. in '1.+ so that 00

L

I(lfk!) < 1X2.

k=N.+I Let m and n be integers with the complemented set

m>n~N.,

and choose r in (IX, 2 IX) so that

is both 1- and I-integrable Then m

I(xd~r-I

L

I(lfkl)
and so I(xd~E. With A any I-integrable set and B=A-C, we now see that BlcAI, B is I-integrable, I(XA_B)~E, and

Ik=n+1 fkl
f

n m

on the intersection of BI and the I-full set dmn fk' Since A and k=n+1 are arbitrary, the result follows. 0

I'

(3.28) Proposition. Let I, 1 be completely extended integrals with a common initial set L, such that 1 is u-finite, and absolutely continuous relative to I. Then every I-full set is I-full, and every I-integrable function is I-measurable.

Proof: Let fELI(l), and let Un) be an I-representation of f in L. Since 00

(Ifni) is an I-representation of the function that the sequence ctl lfkl Since Ifkl

~0

f=

I is Cauchy

L Ifkl, we see from i~ Imeasure relative

L Ifkl

to 1.

for all k, it follows from (8.16) of Chapter 6 that the

00

series

(3.27)

00

converges pointwise on a I-full set F. Hence

L fk

k= I k= I converges to f on F, so that Fcdmnf and therefore dmnf is I-full. Since f is arbitrary, we see that every I-full set is I-full. On the other hand, by (3.27), and (8.16) of Chapter 6, there exist a strictly increasing sequence (n)i= I of positive integers, aI-measurable function g, and a I-full set G, such that wise to g on G. It follows that is I-measurable. D

ttl

fk )~= I converges point-

f = g on the I-full set

F n G; whence

r

128

Chapter 7

Nonned Linear Spaces

(329) Corollary. If 1 is finite, and absolutely continuous relative to I, then every bounded function in L 1 (I) is 1 -integrable.

Proof. This follows from (3.28), and (7.11) of Chapter 6.

D

(3.30) Corollary. If 1 is a-finite, and absolutely continuous relative to I, and if the complemented set A has measure 0 with respect to I, then A has measure 0 with respect to 1.

Proof: By (3.2) of Chapter 6, AD is I-full, whence, by (3.28), it is 1-full. The result now follows from (3.3) of Chapter 6. D (3.31) Lemma. Let I and 1 be completely extended integrals with a common initial set, and let K be the complete extension of the integral fI-+I(f)+l(f) on L 1(I)nL 1(1). Then L 1(K)=L 1(I)nL 1(1); every K-full set is both I-full and 1-full, and I, 1 are both absolutely continuous relative to K.

Proof: Let fEL1(K), and let (f.) be a K-representation of f in L1(I) 00

L I(lf.l)

00

L l(lf.1) converge; so that by (2.15) of Chapter 6, the K-full set F on which L If.1 converges is both I-full and 1-full, and L f. belongs to L 1(I)nL 1(1). Since f = L i. on n Ll (1). Then the series

and

co

n= 1

00

.~

OC'

1

.~

1

F, it follows that fEL1(I)nLl(1). Hence Ll(K)=Ll(I)nLl(l). It is now straightforward to establish the remaining conclusions of the lemma. D The integral K in Lemma (3.31) is called the integral sum of I and 1. (3.32) Lemma. If 1 is absolutely continuous relative to I, then so is the integral sum K of I and 1.

Proof" Let b be a modulus of absolute continuity for 1 relative to I, and let e>O. Consider a complemented set A that is both 1- and Kintegrable, such that I(XA) ~min {e/2, b(e/2)}. By (3.31), A is 1-integrable, and K(XA) = I(XA) + l(XA) ~ e/2 + e/2 = e. Hence K is absolutely continuous relative to I.

0

(3.33) Corollary. Let 1 be finite, and absolutely continuous relative to I, and let K be the integral sum of I and 1. Then a set is K-full if and only if it is I-full.

3 The L. Spaces and the Radon-Nikodym Theorem

329

Proof. Let F be an I -full set, and choose a bounded function f in L 1(I) such that dmnf c F. Then fEL 1(K), by (3.29) and the definition of K. Hence F is K-full. On the other hand, every K-full set is I-full,

by (3.31).

D

Under the conditions of Lemma (331), if also the two integrals I and 1 are finite, then lEL 1(K); so that, by (3.6), LzCK)cL 1(K) and 11(f)1 ~ l(lfl) ~ K(lf!) ~ K(I)t K(f2)t

(f EL2(K».

Hence 1 is a bounded linear functional on L2(K). If this bounded linear functional is normable - that is, if sup {ll(f)I: fELzCI)n L 2(1), I(f2) +1(f2) ~ I} exists - then we say that the integral 1 is normable relative to I. The first conclusion of our next theorem is a constructive substitute for the classical Lebesgue decomposition of measures. (334) Theorem. Let I and 1 be completely extended, finite integrals with a common initial set L, such that 1 is normable relative to I. Then there exist a nonnegative I-measurable function fo and a sequence (S.) of complemented sets with S~ c S~ c ... , such that

is both 1- and l-integrable for each n, lim I(Ls) =0, and US! ~et, and .~ <X(ii) for ead1 f ;/1 L 1U)nL 1(1) and each n, Xs.ffoEL1(i) WId l(Xsjfo) =l(XsJ)·

(i)

S.

;S Qn [-(rtfl

Proof. Let K be the integral sum of I and 1. Since 1 is normable, we obtain from (3.25) a function g in L 2(K) such that 1(f) = K(fg) for each f in L2(K). In particular, o~l(g-)=K(g-

g)=K( _(g-)2) ~o.

Hence K«g-)2)=0, and so g~o on a K-full set F. Note that F is 1full, by (3.31). On the other hand, if f is any function in LzCK), then (3.34.1)

1(f) = K(fg)=I(fg) +l(fg)

= I(fg) + K(fg2) = I(fg) + I(fg 2) +1(fg2)

for each positive integer N.

BO

Chapter 7 Nonned Linear Spaces

Since K is finite, we can find a strictly increasing sequence (sn) of positive numbers converging to 1 such that

Sn==(g<sn) is K-integrable, and therefore both 1- and J-integrable, for each n. Clearly, S~cS~c .... By (3.34.1), for all n, N in Z+ we have

J(1)~J(LsJ ~I(X_s"

r.

gk)

k= 1

~I(~_s"

r.

s:),

k= 1

so that Hence (3.34.2)

I(LsJ~J(1)s;1(1-sn)-+O

It follows from (3.9) of Chapter 6 that measure O. Thus

G==US;

1\ -

as n-+oo. Sn is I -integrable, with 1-

n

is an I-full set; so that the set FnG (on

n

which O~g< 1) is I-full. Let

00

lo(x)==

L gn(x)

(xEFnG)

n= 1

For each I ~ 0 in L 2 (K) and all positive integers nand N, we see from (3.34.1) that Since O~J(XsJgN)~S:J(f)-+O

as N-+oo,

it follows from the monotone convergence theorem that Xs"lloEL1(I) and (3.34.3)

Clearly, this holds for any I in LiK). Taking 1==1, we see in particular that XsJo is I-integrable. Since (3.34.2) implies that (Xs" 10)='= 1 converges I-almost everywhere to 1 0, it follows that 10 is 1measurable. Now consider any positive integer n and any I in Ll(I)nLl(J) Using (3.11), construct a sequence (Ik) of K-simple functions converging to I in the L1(K)-norm. Then each Ik is both 1- and J-simple, and (ft.) converges to I in both the Ll(I)- and the Ll(J)-norms. Since Xs 110 is I-measurable, and

"

IXsJlol ~

C~l s:) III

3 The Lp Spaces and the Radon-Nikodym Theorem

331

on the I-full set FnGndmnf, we see that XsJfoEL1(1). Also, I(IXsJfo - XsJJol)

~ (~1 s~ ) I(lf - fkl)

and J(lxsJ - Xsjkl) ~ J(lf - fkl)

for each k. Hence, by (3.34.3), I(XsJfo) = lim I(XsJJo) k~oo

=

lim J(XSjk) k~oo

=J(xsj)·

This completes the proof.

0

We can now prove the weak Radon-Nikodym theorem. (3.35) Corollary. Under the conditions of Theorem (3.34), absolutely continuous relative to 1, then for each is I-integrable, and I(ffo)=J(f).

if

also J is

f in L 1(I)nL 1(J), ffo

Proof. It will suffice to consider a nonnegative element f of Ll(I)nLl(J). Since J is absolutely continuous relative to 1, we have lim J(x-s) =0. Hence (Xsnf):' 1 converges to f almost everywhere

relative to J; so that, by (3.34) and the monotone convergence theorem, I(xsJfo)=J(Xsj)-+J(f) as n-+CX). Again

by

the monotone convergence theorem, the sequence converges almost everywhere relative to 1, and pointwise on an I-full set, to an I -integrable function g with I(g) = J(f). Clearly, g= ffo on the I-full set

(Xsnffo):'

1

(U S!)ndmnf ndmnfondmng. n

The result now follows.

0

if also J is absolutely continuous relative to I, then fo is I-integrable, and I(fo) =J(l).

(3.36) Corollary. Under the conditions of Theorem (3.34),

Proof: This follows from (3.35) by taking

f == 1.

0

In order to extend Corollary (3.35), we prove some lemmas.

332

Chapter 7

Normed Linear Spaces

(337) Lemma. Under the conditions of Theorem (334), suppose also that i is absolutely continuous relative to I. Let F be a ijull set, and let r be a positive number such that the complemented set S == (fo ~ r) is I -integrable. Then (Sl n F) u SO is an I -full set. Proof By (3.3) of Chapter 6, the complemented set E == (0, F) is integrable relative to i. Let (cPn) be a i-representation of XI:. hy elements of L. Since Xs~r-lfo on the I-full set SlUSO, we have (by (3.35)) I(IXscP.l) ~ r- 1 I(lcPnl fo) = r- 1 i(lcPnD

(nEZ+)

00

Hence

L I(IXscPnD

00

converges, and so the set G on which

n= 1

L IXscPnl n= 1

converges is I-full. Since Gc:(Sl nF)uSO, the desired result now follows. 0 (3.38) Lemma. Let fo and g be nonnegative elements of Ll (I) such that gfoELl(I). Then (gAn)foELl(I) for each n in Z+, and I((gAn)fo)--> I(gfo) as n -+ 00. Proof" For each n in Z+, (g A n)fo is measurable with respect to I, and Hence (gAn)foELl(I). In view of the monotone convergence theorem, it will suffice to prove that ((g A n)fo) converges in measure to gfo . Consider therefore any e > 0 and any integrable set A. By (7.2) of Chapter 6, there exist c > 0 and an integrable set K such that K1c:Al, I(XA_K)<e/2, and O~fo~c on Kl. Since I(gAn)-->I(g) O~(gAn)fo~gfo.

as n --> 00, the sequence (g A n) converges to g in measure Thus for all sufficiently large n there exists an integrable set B such that Bl c: Kl, I(XK_B) <e/2, and Ig-gAnl
if xEdmn f =0 if fo(x)
Then g is I(gfo)=i(g).

and fo(x) ~ r,

is I-integrable, in which case

Proof. By (3.37), dmng is I-full; hence dmng is i-full, by (3.28). Suppose first that g is i-integrable, and let (gn) be a i-representation

3 The L. Spaces and the Radon-Nikodym Theorem

333

of g by elements of L. Then ao

L Xsg,,=Xsg=g 11=

1 00

00

on the l-full set F on which Llg,,1 converges. Thus LXsgJo=gfo on the set (Sl nF)uS o, which is I-full by (3.37). Now by (3.29), S is a l-integrable set. Hence for each n in '1.+ the function Xsg" belongs to L l (1)nL l (l); so that, by (335), I(IXsgJol) = l(IXsg"l) ~l(lg"l). ao

It follows that the series

11=

tX)

<Xl

L I(IXsgJol) and 1

00

L l(IXsg"i) converge; 11=

1

whence L XsgJo is I-integrable, L Xsg" is l-integrable, and PI=

1

11=

1

,,~

1

Thus gfo is I-integrable, and I(gfo)=l(g). Conversely, suppose that gfoEL1(I). Since g=(fovr)-lg!o on the I-full set dmng, we see from (7.7) and (7.10) of Chapter 6 that g is 1measurable. Since also O~g~r-lgfo on dmng, g is I-integrable. Hence g 1\ n is l-integrable for each positive integer n, by (329). Replacing f by f 1\ n in the first part of the proof, we see that (g 1\ n)fo ELl (1) and I((g 1\ n)fo) = l(g 1\ n). Hence, by (3.38) and the monotone convergence theorem, (g 1\ n):."'~ 1 converges almost everywhere relative to 1, and pointwise on a l-full set, to a l-integrable function h with l(h) = I(gfo). Since (g 1\ n) converges to g on the 1 -full set dmn g, we see that gEL1(1) and l(g) = I(gfo). 0 We now prove the constructive Radon-Nikodym theorem. (3.40) Theorem. Let I and 1 be completely extended, finite integrals with a common initial set, such that 1 is absolutely continuous and normable relative to I. Then there exists a nonnegative I-integrable function fo such that for each f in Ll(I)nLl(l), ffoEL1(1) and I(ffo) =l(f). Let (rll)~ 1 be a decreasing sequence of positive numbers converging to 0, such that the complemented set All (fo ~ rll) is I -integrable

=

334

Chapter 7 Normed Linear Spaces

for each n. Let f be a real-valued function defined on a J jull set, and for each n in 7l+ define fn(x) = f(x) =0

if if

xEdmnf and fo(x)~rn' fo(x)
Then fEL1(J) if and only if (i) fJoEL1(J) for each n in 7l+, and (ii) converges in measure, relative to I, to an I -integrable function g. In that case, (fnfo) converges to g almost everywhere relative to I, I(g)=J(f), g= ffo on a J-full set, and there exists an Ijull set G such that g(x)=O whenever XEG and fo(x) =0. (fnfo)~ 1

Proof: The first part of the theorem follows from (3.34)-(3.36). By (3.29), -An is J-integrable for each n. Hence J(X-AJ=I(X-AJo)~rnI(l)--+O

as n--+oo;

so that /\ - An has measure 0 relative to J, and therefore U A~ is Jfull. Sinc~ f = fn on A~ n dmn f for all n ~ N, and since J(Xcan be made arbitrarily small, we see that (fn) converges to f almost everywhere relative to J, and pointwise on the J -full set (U A~) n dmn f

:N)

n

We may assume that f~O. Suppose first that fEL1(J). For each n in 7l+, dmnfn is both I- and J-full, by (3.37) and (3.38). Also, fn=XAJ on the J-full set dmnfndmnfn, so that fnELdJ). By (3.39), fJo EL 1(I) and I (fn!o) = J(fn)' Since (fn) converges to f almost everywhere relative to J, the monotone convergence theorem implies that I(fnfo)--+J(f) as n--+oo. Hence, again by the monotone convergence theorem, (fnfo) converges almost everywhere relative to I, and pointwise on an I-full set G, to a function g in L 1(J) with I(g)=J(f). Clearly, g=ffo on the set Gn(UA~)ndmnf, which is J-full in view of n

(3.28). Also, g(x)=O whenever XEG and fo(x) =0.

Conversely, suppose that conditions (i) and (ii) obtain. Then by (3.39), fnEL1(J) and J(fn)=I(fJO) for each n in 7l+. Hence, by the monotone convergence theorem, (J(fn» converges to I (g), and therefore (fn) converges almost everywhere relative to J, and pointwise on a J-full set H, to a J-integrable function h with J(h)=J(g). Clearly, f=h on the J-full set Hn(UA~)ndmnf; so that fEL1(J) and J(f) =I(g). D n

4. The Extension of Linear Functionals In many situations it is essential to extend a linear functional on a subspace of a normed linear space X to one on X itself. In order to

4 The Extension of Linear Functionals

335

make this extension possible, we need to look more closely at the geometry of X (4.1) Lemma. Let K be a bounded located subset of a normed linear space X, a an element of X, and oc a positive number. Then the set

S=={(1-t)a+tx:O
°

Proof. Choose M > so that I x I arbitrary vector y in X. Define

~M

for all x in K. Consider an

¢(t)==p(y, {(1-t)a+tx: xEK}) =tp(t- l y-t- l (1-t)a,K)

(tElR.+).

F or all t, t' > 0, since

Ily -(1-t)a -txll ~ Ily -(1-t')a -t' xii +(IIall + Ilxll) It -t'l

(xEK),

we have ¢(t) ~ ¢(t') +(Ilall + M) It-t'I. Hence ¢ is uniformly continuous on 1R. +. Thus p(y, S) exists, as the infimum of ¢ on the totally bounded interval (0, oc). D A subset S of X is called a cone if for all x, y in Sand t > 0, the vectors x + y and t \' belong to S. An important type of cone is associated with convex subsets of X. Recall that K c X is convex if (1 - t)x + t Y belongs to K whenever x, y E K and ~ t ~ 1. In that case, the set c(K)== {tx: t>O,xEK}

°

is a cone, the cone generated by K. To see this, it is enough to show that if Xl' X2EC(K), then Xl +X2EC(K). Writing Xl == tl Yl and x 2 == t 2Y2 with t l , t2 >0 and Yl' Y2EK, we have Xl +X2 =(t l + t 2)z, where Z==tl(t l +t 2)-1 Yl +t 2(t l +t 2)-1 YlEK. Hence

Xl

+x 2 belongs to c(K).

(4.2) Lemma. Let K be a bounded, located, convex subset of a normed

linear space X, whose distance from located.

°is positive. Then the cone c(K) is

Proof: For each t>O write tK=={tX:XEK}.

336

Chapter 7

Normed Linear Spaces

Then for each Y in X we have p(y, tK) = tp(t- 1 y, K) ~ t p(O, K) -IIYII,

Since p(O,K»O,we can choose or: > 1 so that p(y,tK»p(y,K) whenever t > or: - 1. It now follows that p(y, c(K»

which exists by (4 1).

= inf{p(y, t K): 0 < t < or:},

D

The fundamental geometric fact about normed linear spaces is that under certain conditions two convex sets can be separated by a normable linear functional. This result, which we now prove, is called the separation theorem. (43) Theorem. Let F and G be bounded convex subsets of a separable normed linear space X, whose algebraic difference {y-x: xEF, YEG} is located, and whose mutual distance

d==:inf{ Ily -xii: xEF,YEG} is positive. Then for each E > 0 there exists a normable linear functional u on X of norm 1 such that (4.3.1)

Reu(y»

Re u(x)+d -E

(xEF,YEG)

Proof We may assume that E < d. Consider first the case IF = JR.. By a generalization of (2.3) of Chapter 5, the bounded, open, convex set K==:{y-X-Z:XEF,YEG, Ilzll
is located, and p (0, K) = E/2. The core of our proof will be the construction of successive enlargements of K, leading to a half-space Koo. The hyperplane bordering Koo will then determine a unique norm able linear functional u of norm 1, such that u is positive on K", and therefore on K. As we shall see, the fact that u is positive on K will ensure that (4.3.1) holds. Fix any vector Xo with -xo in K. Then p(xo, c(K»

inf{ lit x -xoll : t >0, xEK} = inf{(t + 1) I t(t + 1)-1 x +(t + 1)-1( -xo)11 : t >0, xEK}

=

~

p(O, K) = E/2.

Let (X n)::"= 1 be a sequence of vectors which is dense in X. We construct inductively a sequence K ==: K °c K 1 C ••• of bounded, located, open, convex sets such that the following properties hold for all n ~ 1 : (4.3.2)

4 The Extension of Linear Functionals

(43 3)

p(O,Kn»O

(4.3.4)

max {p(xn' c(Kn)), p( -xn, c(Kn))} < n-l.

337

Assume that K o, ... , K n _ 1 have been constructed, and note that c(K n_ 1) is located, by (4.2). If either p(xn,c(Kn_1))
and K;; =.{ -tx n+(1-t)y:0
Each of these sets is located (by (41)), bounded, open, and convex. If xEK n_ 1, then as K n_ 1 is open, we can find ).,>0 so that Since we have K n_ 1c K,;; similarly, K n_ 1c K;; We show that at least one of K,;, K;; satisfies our requirements for Kn. Clearly, both K,; and K;; satisfy (4.3.4). To prove that they satisfy (4.3.3), first note that for each x=.tx n+(l-t)y in K,; we have

Ilxll ~(1-t)p(0,Kn_1)-t Ilxnll· Hence there exists O
'r

in (0, 1) such that I x I ~ t p (0, Kn _ 1) whenever

Ilxll =t Ilxn+t-1(1-t)yll ~

t p( -x n , c(K n _

1 ))

~(2n)-lt~(2n)-1'r

whenever t > 'r, it follows that p(O, K,;) ~min {tp(O, K n_ 1), (2n)-1 'r} > 0.

Hence K,;, and similarly K;;, satisfies (4.3 3). Now suppose that max {p(xo, c(K,;)), p(x o, c(K;;))} < IX =. (1 - 2 -n- 1) p(x o, c(Kn_ 1)). Choose Zl in c(K,;) and Z2 in c(K;;) so that

338

Chapter 7

Normed Linear Spaces

Then ZI=tIXn+YI and Z2=-t 2 Xn+Yz, YI Ec(K n _ I ), and Y2EC(Kn_I)· We compute

where

tl>O,

t 2 >0,

(1 - 2 -n-I)(t 1 + t 2) p(Xo, c(Kn_ 1))

> tl Ilxo -( -t 2 Xn + Yz)[f + t211 xo-(tl Xn + YI)II ~ II(t l +t 2)X O -(t 2YI +t l Y2)[f =(tl +t 2) Ilx o-(t 2(t1 +t 2)-1 YI +tl(t l +t 2)-1 Yz)11 ~(tl

+t 2 )p(x O,c(Kn _

I

»·

This contradiction ensures that max {p(xo, c(K;;», p(xo, c(K;;»} > (1 - 2 -n) p(xo, c(K n_ I»' Therefore either K;; or K;; satisfies (4.3.2). In the former case define Kn == K;;, and in the latter define Kn == K;; This completes the construction of the sequence (Kn)~ o. The set 00 K", == U c(Kn) n= 1

is an open convex cone By (4.3.2), (p(xo, c(Kn»)~ 1 is a Cauchy sequence. As c(Kn_l)cc(K n) for each n, the limit of this sequence is p (xo, K ",). Moreover, again by (4 3.2), p(xo, c(Kn)) >

and so Let

COy

-2- k ») p(xo, c(Ko)) >8/8

(nEZ+)

p(xo,K",)~8/8.

K_oo=={XEX: -xEKoo}'

Since -XoEKoo, it follows that xoEK_ oo ' By (4.3.4), KoouK_ oo is dense in X. Also, Koo and K_", are disjoint, in the sense that if xEKoo and YEK_ oo , then Ilx-YII >0: for, choosing n so that xEc(K n) and -YEc(K n), we see that x-YEc(Kn); so that x-y=tz for some t>O and z in K n, and therefore Ilx-YII ~tp(O,Kn»O, by (4.3.3). Let N be the intersection of the closure of K 00 and the closure of K_ oo • Then N is closed. Since both K", and K_ oo are cones, N is a cone; since also XEN if and only if -xEN, N is a linear subset of X. Consider an arbitrary vector x in Koo' We show that there is a unique vector ¢(x) in NnL, where L== {tx+(1-t)xo:

O~t~

1}

is the line segment joining Xo to x. As K 00 and K _ '" are open, there exists b in (0,1) such that x+zEK oo and xo+zEK_oo whenever Ilzll
4 The Extension of Linear Functionals

339

Assume first that YoEKoo. Then x+<5- 1 (y-yo) belongs to Koo, by the choice of <5 Since K 00 is convex, x 00 = (1 + <5) - 1 Yo + <5 (1 + <5) - 1 (x + <5 - 1 (y - Yo)) belongs to Koo. Also, x 00 = <5 (1 + <5) - 1X + (1 + <5) - 1 Y, so that xooEL and Ily-x oo ll=<5(1+<5)-11Ix-yll. Similarly, if YEK_ oo ' then X_ ro =(1 + <5)-1 Yo +<5(1 +<5)-l(XO+ <5- 1 (y - Yo)) belongs to K_ooIlL, and Ily-x_ ro ll=b(l+<5)-lllx o -YII. Since <5 can be chosen arbitrarily small, (KrouK_aJIlL is dense in L. Thus if (J. and f3 are arbitrary real numbers with 0 ~ IX < f3 ~ 1, then there exists A in (1X,f3) such that x'=Ax+(1-A)X oEK ro uK_ ro . Either A belongs to the set or x'EKooIlL; in the latter case, since xEKrollL and KrollL is convex, f3 is an upper bound of A It now follows from (4.3) of Chapter 2 that (= sup A exists. Since K 00 and K _ ro are open, we have 0 < C< 1. Moreover, since Kroll Land K _ 00 Il L are convex, and (K 00 u K _ 00) Il L is dense in L, {tx+(1-t)x o: O~t
It follows that (an) is a Cauchy sequence and therefore converges to a limit a in JR. Writing z=x-ax o, we have

whence Zn~Z as n~oo, and so ZEN =N.

340

Chapter 7 Normed Linear Spaces

As N c Ken and Ilxo -¢(x)11 ~ Ilxo -xii for each x in K oo ' we see that p(xo, N) exists and equals the positive number p(xo' KaJ Thus N is a hyperplane with associated vector Xo Moreover, if x is any element of X and we choose a in JR and z in N with x =ax o + z, then p(x, N) exists and equals lal p(xo' K 00)' Hence N is located. Let v: X --JR be the bounded linear functional with ker v = Nand v(xo) = 1, and define u= -lIvll-lv. (Note that v is normable, by (1.10).) Then u is a real linear functional on X with norm 1, and u(x o) < 0. Also, if xEKoo and ¢(x)=(x+(1-0xo with 0«<1, then u(x)=

-(-l(1-0u(x o»0.

In particular, u is positive on K Consider arbitrary vectors x and y with x in F and y in G. Since Ilull=l, there exists a vector z with Ilzll u(x + z) > u(x) +d -B.

This completes the proof of the theorem for the case IF = JR Now consider the case IF =
Since v is real linear, to show that u is complex linear we need only note that u(ix) = v(i x) - i v(i 2x) = v(i x) + i v(x) = i u(x)

(XE X).

Let XEX, and suppose that lu(x)l> Ilxll. Then with 1X=lu(x)lu(x)-l we have V(IXX) - i v(iocx} =u(ocx} = IXU(X} = lu(x}l;

whence V(iIXX}=O, as v maps X into JR. Thus v(ocx}=lu(x}l>

Ilxll = IIIXXII,

which is absurd as Ilvll = 1. Hence lu(x)1 ~ Ilxll, and so u has bound l. On the other hand, if B>O, Ilxll = 1, and Iv(x}1 > I-B, then lu(x)12=v(x}2+v(ix}2>(I-B}2,

as v (x), V(iX)EJR. Hence u is normable, and Ilull = 1. Finally, (4.3.1) holds since Re u = v. 0

4 The Extension of Linear Functionals

341

(4.4) Corollary. Let Xo be a vector in a separable normed linear space X. Let F be a bounded convex subset of X such that aXEF whenever xEF, aEIF, and lal ~ 1. Suppose that {xo -x: xEF} is located, and that

O 0 there exists a normable linear functional u on X of norm 1, such that u(xo»lu(x)l+d-E

(xEF).

Proof We may assume that O<E
Re v(x o) > Re v(x) + d -E

(xEF).

In particular, since OEF we have v(xo)=l=O. Thus u=v(xo)-llv(xo)lv is a normable linear functional on X of norm 1, u(xo)=lv(xo)I>Rev(x)+d-E

(xEF),

and u(xo»d-E. For each x in F either u(x o) > lu(x)1 +d-E

or u(x) =1= O. In the latter case, z=v(x)-llu(x)lxEF

and so u(xo) > Re v(z)+d-E =lu(x)l+d-E.

0

(45) Corollary. Let Xo be a vector in a nontrivial separable normed linear space X, and E a positive number. Then there exists a normable linear functional u on X of norm 1 with u(x o) > II Xo II - E

Proof" In case xo=l=O, apply (4.4) with F={O} and G={x o }. In general, either Xo =1=0 or, as we· may assume, Ilxoll <E/3. Choose y in X with y =1= 0 and II Xo - y I < E/3, and then construct a normable linear functional u with Ilull=1 and u(y»llyll-E/3. Then u(x o) ~ u(y) -llxo - yll

> Ilyll-E/3-E/3

> Ilxoll-E.

0

We can now prove the Hahn-Banach theorem on the extension of linear functionals. In this connection, recall the definition (2.9) of a quotient space.

342

Chapter 7

Normed Linear Spaces

(4.6) Theorem. Let Y be a linear subset of a separable normed linear space X, and v a nonzero linear functional on Y whose kernel is located in X. Then for each 6>0 there exists a normable linear functional u on X such that u(y)=v(y) for all y in Y, and Ilull ~ Ilvll +6.

Proof First observe that, by (1 10), v is normable as a linear functional on the normed linear space Y. Let N == ker v, and consider the quotient space X/N, with norm given by

IlxllxlN== p(x, N)

(XEX).

Let Yo be any vector in Y with v(Yo) = 1 For each z in N we have Ilyo -zll ~ Ilvll- 1 v(Yo -z)= Ilvll- 1 Hence IIYollxIN=P(Yo,N)~ Ilvll- 1 .

For each normable linear functional u on X/N, write IlullxIN==sup{lu(x)l: XEX, IlxllxlN~ 1} If 6> 0, then by (4.5), there exists a normable linear functional U o on X/N such that Iluoll xlN = 1 and uo(Yo»(llvll +6)-1. Thus u==U O(YO)-l UO is a normable linear functional on X/N with IlullxlN< Ilvll +6 and u(Yo)=1. Since II IlxIN~ II II, we see that when considered as a linear functional on X, u has bound Ilull XIN. In fact, u is norm able on X. To see this, consider an arbitrary positive number t>, and choose x in X with IlxllxlN<1 and u(x» IlullxIN-{). There exists y in N with Ilx - yll < 1 Then x and x - yare equal elements of X/N, so that u(x- y)=u(x» IlullxIN-{).

Since t> is arbitrary, it follows that u is norm able on X, with Ilull = IlullxlN< Ilvll +6. Finally, for each y in Y we have y-v(Y)Yo in N, and therefore o=u(y -v(Y)Yo) =u(y) - v(y) - that is, u(y) = v(y).

0

5. Quasinormed Linear Spaces; the Space Leo Since it is not true that every bounded linear functional on a normed linear space X is normable, the dual space of X cannot be considered as a normed linear space as it is in the classical theory. In order to discuss dual spaces without distorting the theory beyond recognition, we weaken the notion of a norm to that of a quasinorm.

5 Quasinormed Linear Spaces, the Space L",

343

(5.1) Definition. Let X be a linear space over IF. A quasinorm on X is a family (II IIJiEI of seminorms on X such that for each x in X, {llxll i : iEI} is a bounded subset of IR. Taken with this quasinorm, X becomes a quasinormed linear space. The standard equality and inequality relations on the quasinormed linear space X are defined by

x=y

if and only if Ilx-ylli=O for all i in I

X9=Y

if and only if Ilx-ylli>O forsomeiinI.

and We may assume without loss of generality that the equality given by the quasinorm coincides with the original equality on X An element x of X is said to be normable if (5.1.1)

- the norm of x - exists. Trivially, a normed linear space X is a quasinormed linear space in which every element is normable. Conversely, if every element of a quasinormed linear space (X,(II IIJiEl) is normable, then X is a normed linear space with norm given by (5.1.1). A less trivial example is given by the set Hom (X, Y) of bounded linear maps of a normed linear space X into a normed linear space Y; in this case, we have a natural quasinorm (II Ilx)xEs' where S is the unit sphere of X and (5.2)

(uEHom(X, Y),XES).

An element of Hom(X, Y) is normable relative to this quasinorm if and only if it is normable in the sense of Definition (1.9). Thus if X is finite-dimensional, Hom(X, Y) is actually a normed linear space, by Corollary (2.4). With every quasinormed linear space X we associate various topological notions which are equivalent to familiar ones when X is a normed linear space. Thus, for example, the open sphere of radius r > 0 about a point x in X is the set Sex, r) consisting of all y in X such that there exists r' < r with II x - y II i < r' for all i. The neighborhood structure of X is defined by taking the neighborhoods to be the open spheres. When applied to subsets of X, the notions of open set, closed set, dense set, interior, and closure refer to this neighborhood structure. A subset Y of X is closed if and only if it contains all points that are limits of sequences of points of Y, in the sense of (5.3) Definition. A sequence (xn) of elements of a quasinormed linear space (X, (II II ;)iEI) converges to an element x of X if for each B > 0

344

Chapter 7

N armed Linear Spaces

there exists N. in '1.+ such that Ilx -x.lli~8

(n~N., iEI).

The point x is then called the limit of (x.) in X, and we write either lim x. =X .~oo

or x. -+x

as n-+ 00.

The closed sphere of radius r ~ 0 about a point x in a q uasinormed linear space (X, (II II i)iEf) is the closed set Sc(x,r)={YEX: Ilx-ylli~r for each i in I}.

Of special interest will be the unit sphere Sc(O, 1) (5.4) Definition. A Cauchy sequence of elements of a quasinormed linear space (X,(II IUiEl) is a sequence (x.) such that for each 8>0 there exists a positive integer N. with Ilxm-x.lli~8

(m,n~N., iEI).

A subset A of X is said to be complete if every Cauchy sequence of elements of A converges to a limit in A. Naturally, we are interested in bounded linear maps between quasinormed linear spaces. (5.5) Definition. Let u be a linear map of a quasinormed linear space (X, (II II;liEf) into a quasinormed linear space (Y,(II Ilj)jEJ). We say that u is bounded if there exists a positive number c, called a bound for u, such that if XEX, 8>0, and jEJ, then c Ilxll i > Ilu(x)ll j -8 for some i in I. We say that u is uniformly continuous if for each 8>0 there exists <»0 such that if x, X'EX and Ilu(x)-u(x')ll j >8 for some j in J, then Ilx-x/lli>b for some i in I. The following result generalizes Proposition (1.5). (56) Proposition. A linear map between quasinormed linear spaces is bounded if and only if it is uniformly continuous. Proof Let (X,(II IIJiEf) and (Y,(II II)jEl) be quasinormed linear spaces, and u: X -+ Y a linear mapping. Suppose that u is bounded, with bound c>O. Let x, X'EX, 8>0, jEJ, and Ilu(x)-u(x')ll j >8. Choose i in I so that c Ilx -x' Iii> Ilu(x -x')ll j -8/2 >8/2.

Then Ilx -x' Iii >(2C)-1 8. Hence u is uniformly continuous.

5 Quasinonned Linear Spaces, the Space La>

345

°

Conversely, assume that u is uniformly continuous. Choose r> so that for each j in J, if Ilu(x)ll j > 1, then there exists i in 1 with Ilxll;>r. Consider x in X, E>O, and j in J. Either Ilu(x)ll j >E/2 or Ilu(x)llj<E. In the former case, setting a =(llu(x)ll j -E/2)- \

we have Ilu(ax)llj=lalllu(x)llj> 1; whence IlaxII;>r for some i. Then Ilxll i = lal- 1 IlaxII; > r lal- 1 > r(llu(x)ll j -E) and therefore (56.1)

In case Ilu(x)llj<e, inequality (56.1) clearly holds for all i in 1. Hence r- 1 is a bound for u. 0 For all quasinormed linear spaces X and y, we shall use the notation Hom(X, Y) to denote the set of all bounded linear mappings of X into Y. We shall consider Hom(X, Y) as a quasinormed linear space with the quasinorm (II 11.}"es given by Ilullx= Ilu(x)11

(uEHom(X, Y), XES),

where S is the unit sphere of X. The neighborhood structure corresponding to this quasinorm on Hom(X, Y) will be called the uniform neighborhood structure on Hom(X, Y). Of special interest is the case Y=IF. (5.7) Definition. The dual space of a quasinormed linear space X is the quasinormed linear space Hom(X,IF) of all bounded linear functionals on X. The dual space is also written X*. From a classical point of view, two quasinorms on a vector space X are essentially the same if they give rise to the same norm. We capture this notion in a more general form by means of (5.8) Definition. Let (X, (II II ;);El) and (Y, (II UjEJ) be quasinormed linear spaces, and u: X -+ Y a linear map. We say that u is a linear isometry if (i) for all e > 0, x in X, and i in 1, there exists j in J such that Ilu(x)ll j > Ilxlli-E, and (ii) for all E>O, x in X, and j in J, there exists i in 1 such that Ilxll;> Ilu(x)llj-E If u: X -+ Y is a linear isometry, then it is injective, in the sense that u(x)=I=u(x') whenever x, X'EX and x=l=x'. Also, u is norm-preserving, in

346

Chapter 7 N ormed Linear Spaces

that if XEX is norm able, then u(x) is normable in Y and Ilu(x)11 = Ilxll Finally, 1 is a bound for u, and so U is uniformly continuous. For example, if (X, II II) is a nontrivial separable normed space, then the identity map is a linear isometry of (X, II II) onto (X, (II Ilu)uEs')' where S* is the unit sphere of X* and Ilxllu=lu(x)1

(UES*).

This follows from Corollary (4.5) and the definition of S*. For another example, consider a u-finite integral I on a set X, and positive numbers p, q with p> 1 and p- 1 + q-l = 1. By Theorem (3.25), the linear map gl-+U g defined by u g (f)=I(fg)

(fELp, gELq)

is a linear isometry of Lq onto the normed linear space of normable elements of L~. In case p = 1, the normable elements of L~ are similarly related to the normable elements of a certain quasinormed linear space. In order to discuss this space, for the rest of this section we take I as a u1inite integral on X. (5.9) Definition. A real-valued function f defined on a full subset of X is essentially bounded relative to I if there exists a nonnegative number c, called a bound for f, such that If I ~ c on a full set. Two essentially bounded functions f and g are equal if f = g on a full set. Taken with this notion of equality, the collection of all essentially bounded, measurable functions relative to I is a set, which we denote by Loo. Let d denote the set of all integrable sets with positive measure, where two elements of d are equal if their characteristic functions are equal elements of L l . If AEd, fELoo, and c~O is a bound for f, then since If XA I ~ C XA on a full set, IlfII A =Jl.(A)-ll(lfl XA) exists and is at most c. Thus (II IIA)A""" is a quasinorm on Loo. From now on we shall consider Loo as a quasinormed linear space relative to (II IIA)A""". Note that the equality induced on Loo by this quasinorm coincides with the notion of equality introduced in Definition (5.9). We shall write II f II 00 for the norm of a norm able element f of Loo. Thus Ilfll oo =sup{Jl.(A)-l 1(lfl XA): XAEL1' Jl.(A) > O} if the right side exists. The next two lemmas will enable us to prove the completeness of Loo·

5 Quasinormed Linear Spaces, the Space Loo

347

(5.10) Lemma. If fELoo, and IlfllA~c for each A in .91, then Ifl~c on a full set.

Proof: By (8.20) of Chapter 6, for each n in 7l+ there exist cn III (c,c+n- 1) and a measurable set An' such that Ifl0, then IlfIIA~,u(A)-1 I(cnXA)=cn>c,

which is absurd. Hence ,u(Ki-An)=O for each i; so that, by (8.10) of Chapter 6, - An is integrable, and ,u( - An) = O. Thus A~ is a full set for each n, and so F ==: A~ is a full set. Clearly, If I ~ C on F. D

n n

(5.11) Lemma. If (fn) is a Cauchy sequence in L oo ' then (fn) converges in measure to a measurable function.

Proof. Consider any integrable set A and any E > O. Either ,u(A) > 0 or ,u(A) < E. In the latter case, setting B ==: A - A, we have B integrable, BlcAl, ,u(A-B)<E, and Ifm-fnl<E on Bl for all m, n in 7l+. In case ,u(A) >0, write <5==:(2,u(A))-IE2 and choose N in 7l+ so that I (Ifm - fnl XA) ~ <5,u(A)

(m, n ~ N).

Consider integers m, n with m, n ~ N. By (8.20) of Chapter 6, there exist a measurable set K and a number a in (E/2, E) such that Ifm - fnl < a on Kl and Ifm - fnl ~ a on KO. Then B ==: A A K is an integrable set, Bl CAl, and Ifm - fnl < E on Bl. Also,

,u(A -B)~ a-I I(lfm -fnl XA) ~2E-l <5 ,u(A) = E. Thus we see that (fn) is Cauchy in measure. The result now follows from (8.16) of Chapter 6. D (5.12) Theorem. The quasinormed linear space Loo is complete.

Proof" Let (fn) be a Cauchy sequence in Loo. Choose NI in 7l+ so that for all n ~ Nl and A in .91, II fn - fN ,II A~ 1 and therefore II fn II A~ l+llfN,II A. Choose c>1 so that IlfnIIA~c-1

(1~n~NI'

Then for each n in 7l+ we have IlfnllA~c

so that Ifni ~c on a full set, by (5.10).

(AEd);

AEd).

348

Chapter 7

Normed Linear Spaces

By (5.11), and (8.16) of Chapter 6, there is a subsequence

(jnJ,{,~

1

of

(jn) that converges almost uniformly, and pointwise on a full set, to a

measurable function f Clearly, If I ~c on a full set, so that fEL"". It only remains to prove that (jn) converges to f relative to the quasinorm on Loo. To do this, consider an arbitrary positive number Il, and choose N in Z + so that Ilfm-fnIIA~1l/3

(m,n?;N, AEd).

Fix A in d. There exists an integrable set B such that Bl cAl, Jl(A-B)«6c)-1IlJl(A), and (jnJ'{'~l converges to f uniformly on Bl. Hence there existsj?;N such that It,.j-fIXB~1l13. Thus Ilfnj -f IIA = Jl(A)-l(I(lfnj -fl XB) + I(lfnj -fl XA-B))

~Jl(A)-l (~Jl(B)+2CJl(A-B)) ~21l/3

So for all m?;N we have Ilfm -fIIA~ Ilfm -fn)iA + IIfnj -filA ~ 11/3

+ 2 11/3 =

Il.

Since AEd is arbitrary, it follows that (jn) converges to f in L oo ' as we required. 0 The first part of the proof of Theorem (5.12) shows that a Cauchy sequence (jn) in L"" is bounded, in the sense that the terms fn have a common bound c. Classically, Loo can be identified with the dual space of L l . Although this is not true constructively, there is an important result along those lines. (5.13) Theorem. For each gin L oo ' define u g : Ll-+IR by

ug(f)=I(jg)

(fELl)·

Then l(g)=u g defines a linear isometry l of Loo into L"i. Moreover, is normable, then so is u g , and II u g II = II g 1100·

if

g

Proof: Let gEL oo ' and let c>O be a bound for g Then for each f in L l , fg is measurable and Ifgl ~c If I; so that l(jg) exists, and (5.13.1)

lug(f)I=II(jg)1 ~I(c Ifl)=c Ilflll·

Since u g is linear, it follows that ugEL"i.

5 Quasinormed Linear Spaces, the Space L""

349

The map 2 is clearly linear. To show that it is an isometry, consider first I' > 0 and A in d. We construct 1 in LI such that 11/111~1 and Ilf II )1> IlgllA-e Either IlgIIA<e, when we can take 1=.0; or, as we a~~lIJllc. IlgII A>eI2. By (820) of Chapter 6, there exist a measurable set B and a number b in (0,1'/2) such that Igl > b on Bl and Igl ~ bon BO. Thus

el2 < Ilgll A = tL(A) -I (I (Igl XA -B) + I (Igl XA AB)) ~ b tL(A)-1 tL(A - B) + c tL(A) - I tL(A 1\ B)

< 1'/2 + c tL(A)-1 tL(A 1\ B). Hence tL(A

1\

B) > O. Now define

1 on

the full set

F=.dmngndmnXAAB

by [(x) = tL(A

1\

B)-I

if xEAlnBlndmng and g(x»b,

= -tL(AI\B)-l

if xEAlnBlndmng and g(x)< -b,

=0

if xE(A I\B)Ondmng

Then

1 = tL(A 1\ B)-l (gXB + X-B)-l

Igl XAA B

on F; whence 1 is measurable, by (7.7) and (7.10) of Chapter 6. Since also 1/1=tL(AI\B)-I XAAB on F, we see that IELI and that 11/111=1. Also, Ig=tL(A 1\B)- l lgIXAAB on F, and so lug(f)I=I(fg) =tL(A

1\

B)-I I(lgIXAAB)

~ tL(A)- I (I (Igl XA) - I (Igl XA-B)) ~ Ilgii A-tL(A)-1 btL(A -B)

> Ilgii A-e. To complete the proof that 2 is a linear isometry, we must show that for each I' > 0 and each 1 in LI with III III ~ 1, there exists A in d with (5.13.2)

Consider such e and f There exist a measurable set B and a number a such that lug(f)I-ea on BI, and Igl~a on BO. We have IUg(f)I~I(l/gI)

=I(l/gl X-B) +I(l/glxB)

~all/ill +cI(l/lxB)

< IUg(f)1 +cI(lfl XB)'

l50

Chapter 7 Normed Linear Spaces

Hence 1(lfl XB) > O. It follows from (4.16) of Chapter 6 that If IXB > 0 on A 1 for some set A with positive measure. Since Ale B1, we have

as we required. Thus A. is a linear isometry. Now assume that g is norm able. Then taking c=llglloo in (5.13.1), we see that Ilg 1100 is a bound for u g • On the other hand, given E > 0, choose A in d so that IlglIA> Ilglloo -E/2. Since A. is an isometry, there exists fin L1 with IIfl11 ~ 1 and lug(f)I> Ilgll A -E/2 > Ilgll 00 -E. Hence ug is norm able, and Ilugll = Ilgll 00·

0

Note that there exist normable linear functionals on L1 which are not of the form ug with g in La;) (Problem 29).

6. Dual Spaces We have already introduced the quasinormed space Hom(X, Y) of bounded linear maps of a quasinormed space X into a quasinormed space y. When X and Yare normed spaces and we are considering the unit sphere of Hom(X, Y), the topological structure of greatest interest is not derived from the standard quasinorm on Hom(X, Y), but is induced by metrics corresponding to certain norms which we now define. (6.1) Definition. Let X and Y be normed spaces, with X separable. For each dense sequence (Xn):."'~ 1 in X, we define the corresponding double norm 111111 on Hom(X, Y) by 00 IIlulli == L 2-"(1 + IIx nll)-l Ilu(xn)11 (uEHom(X, n~

1

Y».

Although a double norm makes Hom(X, Y) into a normed linear space, different dense sequences on X give rise to different double norms, which will induce different metric topologies on Hom(X, Y). However, these norms induce the same metric topology on a subset B of Hom(X, Y) that is bounded, in the sense that there exists a common bound for all the linear maps in B. (6 2) Proposition. Let X and Y be normed spaces, with X separable Let (xn) and (x~) be dense sequences in X, with corresponding double norms

6 Dual Spaces

111111 and III III', respectively, on Hom(X, Y). Then III III and equivalent metrics on each bounded subset of Hom(X, Y).

III III'

351

induce

Proof Let B be a bounded subset of Hom(X, Y), and let c>O be a common bound for the elements of B. Given e > 0, choose N so that 00

L n~N+

2- n < e/4c. Choose also integers kl' ... , kN' and then

c5 > 0, so that

1

c5 «e/4)2-kn(1 + Ilxd)-1 (1 ~n~N). Let be elements of B with IIlu I -u 2111
Ilx~-xd <e/8c and U2

II(u l -u 2)(x k JII
UI

and

(nEZ+)

and hence N

IIIu 1-u 2111' ~

L 2-

00

-u2)(x~)11 +

L

+ Ilx~II)-1 2c Ilx~11

n

II(u l

n

(ll(u l -u2)(x~-xkJII + II(u l -u 2)(x kJII) +2c

2- n (1

I

n~

N

~

L 2-

00

n~1

L

2- n

n~N+l

N

<

L 2- n (2c Ilx~-xd +c52kn (1 + Il xd»+e/2 n~

I

N

<

L 2-

n

(e/4+e/4)+e/2

<e. Thus the mapping u 1---+1I tinuous on B. Likewise, tinuous on B relative to the metrics induced on X

of (X, 111111) into (X, 111111') is uniformly conthe inverse of this map is uniformly conthe metric induced by III III'. In other words, by III III and III III' are equivalent. D

In view of Proposition (6.2), there is some justification for our abuse of language in referring to 'the' double norm III ilion Hom(X, Y). (6.3) Proposition. Let X and Y be normed spaces, with X separable. Then for each x in X the map u I---+u(x) is uniformly continuous on bounded subsets of Hom(X, Y), relative to the double norm. Proof: Let XEX be arbitrary. Construct a dense sequence (x n) in X with x I == x, and let III III be the corresponding double norm. The result now follows from the inequality

Ilu I (x) -u 2(x)11 ~2(1 + Ilxll)lllu l -u 2 111 and (6.2).

D

(u l , u2 EHom(X, Y»

352

Chapter 7

Normed Linear Spaces

For a converse to Proposition (6.3) in the case where Y==:IF, see Corollary (6.9) below (6.4) Corollary. Let X and Y be normed spaces, with X separable; let X 0 be a finite-dimensional subspace of X, and for each u in Hom(X, Y) let u D denote the restriction of u to X 0 Then the map u I---+U D of Hom(X, Y) into the normed space Hom(X 0' Y) is uniformly continuous on bounded subsets of Hom(X, Y), relative to the double norm.

Proof Let B be a bounded subset of Hom(X, Y), and f. a positive number. Let c>O be a common bound for the elements of B. By (2.3), there exists an E/4c approximation {x~, ... , x~} to the unit sphere So of Xo' By (6.3), we can choose <'5>0 so that lu(x~)-v(xDI<E/2 whenever u,vEB, Illu-vlll
l(u D- vD)(x)1 ~ I(u -

Since

XES o

+ I(u - v)(x - x~)1 ~E/2+2c Ilx -x~ I <E/2+E/2=f. is arbitrary, it follows that IluD-vDII ~f.. 0 v)(x~)1

(6.5) Proposition. Let X and Y be normed spaces, with X separable and Y complete. Then the unit sphere of Hom(X, Y) is complete in the metric induced by the double norm.

Proof" Let (xn) be a dense sequence in X, and III III the corresponding double norm. Let B be the unit sphere of Hom(X, Y), and (un) a Cauchy sequence in B relative to the metric induced by 111111. By (6.3), (un(x))~~ 1 is a Cauchy sequence in Y for each x in X. Defining u(X) ==: lim un(x)

(XE X),

n~oo

we obtain an element u of B. Let

E

> 0, and choose in turn positive

00

L

integers N, v so that

k~N+

2~k<E/4

Ilu(x k)-u n(x k)II<E/2

Then for

n~

and

1

(n~v,l~k~N).

v we have

Illu-unlll ~

N

00

k~l

k~N+l

L 2~k Ilu(xk)-un(xk)11 +2 L

2~k

N

<

I

k~

2~k(E/2)+E/2<E. 1

Thus Illu-unlll-tO as n-->oo, and so B is double-norm complete.

0

6 Dual Spaces

353

We now focus our attention on the unit sphere S*={UEX*: lu(x)l~ Ilxll for all x in X}

of the dual X* of a separable normed space X. The following lemma extracts some of the sting from the proofs of the succeeding theorems. (6.6) Lemma. Let (xn) be a dense sequence in a separable normed space X, and IIIIII the corresponding double norm on X* Then for each e>O there exists a finite-dimensional subspace X 0 of X such that Illu - vIII < e whenever u, v belong to the unit sphere of X* and lu(x) - v(x)1 < e/3 for all x in the unit sphere of X o. 00

Proof" Choose a positive integer N with

L

2 -n < e/6. By (2.5), there

n=N+ 1

exists a finite-dimensional subspace X 0 of X such that p(x n , X 0) < min {1, e/6}

For each n (1 ~n~N), choose Then

x~

(1 ~ n ~ N).

in Xo with

Ilxn-x~11

<min{1,e/6}.

Let u, v be elements of the unit sphere of X* such that lu(x)-v(x)l<e/3

(xEXo,llxll~l)

Then N

00

Illu-vlll~ L2- n(1+llx nll)-1Iu(x n)-v(x n)I+2 11=

L

2- n

1

N

~

L 2-n(l(u- v)«1 + Ilxnll)-l x~)1 H=

1

+(1 + Ilxnll)-11(u-v)(Xn-X~)I)+E/3 N

<

L 2-n(E/3+21Ixn-x~II)+E/3 11=

1

N

~

L 2- n(2e/3)+E/3<E.

D

n= 1

(67) Theorem. The unit sphere S* of the dual of a separable normed space X is compact in the double norm, and the normable elements are dense in S*. Proof" Bearing in mind (6.5), we see that it will suffice to prove that for each E> 0 there exists an E approximation to S* in the double norm with each element normable. To this end, construct a finitedimensional subspace X 0 of X as in (6.6). If X 0 = {OJ, then Illu - 0111 < E

354

Chapter 7

Normed Linear Spaces

for all u in S* (so that {OJ is an e approximation to S*) We may therefore assume that X 0 has positive dimension Then every element of X~ is normable, by (2.4), and the quasinormed space X~ is actually a finite-dimensional Banach space. Thus the unit sphere S~ of X~ is compact, by (2.3). Moreover, for each nonzero u in X~, the closed subset ker u is located in X 0' by (1.10). Therefore since X 0 is locally compact, so is keru. Hence keru is located in X. Let {u?, ... , u~} be an e/3 approximation to S~ with 0 < Ilu~ II < 1 for each k. By the Hahn-Banach theorem (4.6), there exist norm able linear functionals U l ' .. ,Un in S* such that Uk(X)=U~(x)

(XEX o, 1 ~k~n).

Let UES*. Then the restriction of u to X 0 belongs to lu(x)-u~(x)1 <e/3

S~,

so that

(XEX o, Ilxll ~ 1)

for some k with 1 ~ k ~ n. Hence, by our choice of X 0' we have IIlu-uklll<e. Thus {u 1 , ... ,un } is an e approximation to S* with each element normable. 0 As we pointed out in Chapter 3, Brouwer contends that every realvalued function on a compact space is continuous. Although Brouwer's proof is unsatisfactory, his contention has not been disproved. Therefore, if we wish to examine linear functionals on X*, it is reasonable to require that they be continuous on its unit sphere S*. The next two results describe the form of all such functionals. (6.8) Theorem. Let X be a separable normed space, S* the unit sphere of X*, and qJ a linear functional on X* that is continuous on S* relative to the double norm. Then for each e > 0 there exists Xo in X such that IqJ(u)-u(xo)l<e for all u in S*.

Proof: First observe that since S* is compact in the double norm, qJ is a normable element of Hom(X*,lF). Let e be a fixed positive number. Either II qJ II < e, in which case we take Xo == 0; or, as we now assume, IIqJlI >0. By (6.7), there exists a normable element U o of X* with Iluoll = 1 and qJ(uo)=I=O. We may assume that qJ(U o) = 1. Define closed sets N*=={UES*: qJ(u)=O}

and Nr*=={uES*:lqJ(u)l~t}

(t>O).

We now prove that N* is totally bounded. There exist arbitrarily small t > 0 such that Nr* is compact in the double norm. Given such t,

6 Dual Spaces

let {u\, ... ,u m } be a t approximation to u~ =:0 (1

Then

u~EN*.

+ t)-l (Uk -

~*.

For

l~k~m

355

write

cp(uk)U O).

If uEN*, and k is chosen so that IIlu-uklll
IIlu - u~1I1 ~ Iliu - uklll + IIlu k -u~1I1 < t + IIltu k + cP(uk)U O III ~ t+ t Illuklli + Icp(uk)llIluolll ~t+t+t=

3t.

Therefore {U'l' ... ' u~} is a 3 t approximation to N*. Since t can be arbitrarily small, N* is totally bounded. It now follows that Ilxll o =:0 sup {lu(x)1 : uEN*} exists for each x in X, and that II 110 is a seminorm on X with II II 0 ~ II II· Let oc be an arbitrary positive number. By (6.6) and the continuity of U on S*, there exist 15>0 and a finite-dimensional subspace X 0 of X such that Icp(u)1 < oc/2 whenever UES* and lu(x)1 < 15 for all x in the unit sphere So of (X 0' II II). Since II II 0 ~ II II, the map x f-+ IIxll 0 is uniformly continuous on X o. Since X 0 is finite-dimensional, it follows that f3=:oinf{llxllo: XEX o, Ilxll = I} exists. We show that 13 < oc. Since either 13 > 0 or 13 < oc, we may assume that 13>0. Then Ilxll~f3-11Ixllo for each x in Xo. It readily follows that X 0 is a finite-dimensional Banach space relative to the norm II 110; that (6.8.1)

Ilullo =:0 sup {lu(x)l: XEX 0' Ilxll 0 ~ I}

exists for each bounded linear functional U on X 0; and that X~ is a finite-dimensional Banach space relative to the norm II 110 given by (6.8.1). Let S~ be the unit sphere of X~ relative to the norm I II 0 For each U in N* the restriction u b of u to X 0 is in S~, by definition of II 110· By (6.4), the map Uf-+U b of N* into S~ is uniformly continuous. Since N* is compact, its image K under this map is a totally bounded subset of S~. Clearly, K is convex, and aVEK whenever vEK, a Ell', and lal ~ 1. We show that K is dense in S~. To this end, let u be any element of S~, and suppose that O<po=:oinf{llu-vllo: vEK}. By (4.4), there exists a bounded linear functional I/J on I/J(u»sup{II/J(v)l: vEK}.

X~

such that

156

Chapter 7

Normed Linear Spaces

Since (by elementary linear algebra) the mapping which takes an arbitrary x in Xo to the linear functional vf-+v(x) is an algebraic isomorphism of X 0 with the dual of X~, there exists y in X 0 such that ljJ(v) = v(y) for each v in X~ Thus u(y) > sup {lv(y)l: vEK}

= Ilyll o.

This contradicts the fact that UES~. Hence Po = 0, and therefore K is dense in S~. For each x in Xo with Ilxllo~l, we have II{hll ~1 and therefore

lf3u o(x)1 = lu o(f3x)1 ~ Iluoll = 1. Thus the restriction of f3u o to X 0 belongs to S~. Since K is dense in S~ and I I 0 ~ I II, there exists u in N* such that 1(f3u o -u)(x)I<2b

(xEXo,llxll~l).

Now, since f3u o and u both belong to S* (it is clear that f3 ~ 1), we have ·Hf3u o -U)ES* Hence Icp(~(f3uo -u))1 < rY./2

by our choice of X 0' and therefore f3 = f3 cp(u o) - cp(u) = 2 cp(~(f3uo - u)) < rY..

Thus f3 < rY. in all cases We may therefore choose ~ in X 0 so that lu(~)1 <0( whenever uEN*. For each u in S*, since

WI = 1 and

(1 + I cp 11)- 1 (u - cp (U)Uo)E N*,

we therefore have (6.8.2) By (4.5), there exists

U1

in S* with

~=U1W ~ Icp(u 1 ) uoWI + (1 + I cp II) 1(1 + I cp 11)- 1 (u 1 - cp(ul)uo)(~)1

+(1 + Ilcpll)rY.· Therefore luo(~)1 > til cp 11- 1 if rY. is sufficiently small. In that case, write ~ Ilcpllluo(~)1

xo=UO(~)-l~.

Then u o(x o)=1. Hence for each

U

in S* we have, by

(6.8 2),

Iu(x o) - cp(u)1 = IUoWI- 1 I(u - cp(u)uo)(~)1 < 3 I cp I (1 + I cp II) rY. < E if rY. is sufficiently small. This completes the proof. 0 When X is complete, Theorem (6.8) can be strengthened in a very satisfactory manner.

7 Extreme Points

357

(6.9) Corollary. Let X be a Banach space, S* the unit sphere of X*, and cP a linear functional on X* that is continuous on S* relative to the double norm. Then there exists a vector Xo in X such that cp(u)=u(xol [or each u in X*.

Proof. Repeatedly applying (6.8), with reference to (6.3), construct a sequence (Xn)~= 1 in X such that (6.9.1)

For each integer

n~2

and each u in S*, we then have

lu(xn)l;;;:! Icp(u) -U(XI + ... +xn)1 + Icp(u) -U(XI + ... + xn_1)1 <2- n + 2 •

By (4.5), we can choose u in S* so that lu(xn)1 is arbitrarily close to I Xn II. Hence It fo11o:s that

series

Lx

k

ct I= Xk

(n ~ 2).

I

is a Cauchy sequence in X; whence the

converges to a sum Xo in X. Letting n--+ 00 in (69.1), we

k= 1

see that cp(u)=u(xo) for each u in S*, and hence for each u in X*.

0

7. Extreme Points The vertices of a convex polygon are extreme, in the sense that each line segment which lies inside the polygon and contains a vertex has that vertex as an end point. Our geometric intuition leads us to hope that every point inside a convex polygon is an average of extreme points. As we shall see, this is almost the case. But first we define the notion of extreme point in a very general context, and show that such points exist. (7.1) Definition. Let X be a normed space over JR, K a compact

convex subset of X, and Xo a point of K. We say that Xo is an extreme point of K if for each e > 0 there exists (j > 0 such that I x - y I ;;;:! e whenever x, YEK and Ilt(x+ y)-xoll ;;;:!(j. If u is a bounded linear functional on X, we write M(u, K)=:sup{u(x): xEK}

and S(u, 0(, K) =: {xEK: u(x)~ O(}.

358

Chapter 7 Normed Linear Spaces

If S is a compact subset of X, then diam u(S)=:sup{lu(x)-u(y)l: X,YES} is called the variation of u on S. We now prove two lemmas which will enable us to construct extreme points. (7.2) Lemma. Let K be a compact convex subset of a normed space X over IR, and let u, v be elements of X*. Then there exist w in X* and c < M (w, K) such that u - w has arbitrarily small bound, Sew, c, K) is compact, and the variations of u and v on sew, c, K) are arbitrarily small.

Proof. Let (jE(O,l) be arbitrary. Using (4.9) of Chapter 4, choose real numbers a, [3 with M(u, K)_(j2
[3=:!(a+M(u,K)) so that the sets S.=:S(u,a,K) and Sp=:S(u,[3,K) are compact, and p(S., Sp) is arbitrarily small, where p is the usual metric on the set of compact subsets of X. Since v is uniformly continuous, we may assume that p(S., Sp) is small enough to make

M(v, S.) - M(v, Sp) < (j2. Let Xo be any point in Sp with v(xo»M(v,S.)_(j2. For each x in K with u(x)
a)(u(x o) - u(x))- 1 X + (a - u(x))(u(x o) - u(x))- 1 Xo

lies in K (by convexity), and u(x') = a. Therefore M(v,S.)~v(x')

> (u(xo) - a)(u(xo) - u(x))- 1 vex)

+ (a -

u(x))(u(x o) - u(x))- 1 (M (v, S.) _

Thus (7.2.1)

M(v, S.) > vex) - (a - u(x))(u(x o) - oc)- 1 (j2 ~ vex) - (a - u(x))([3 - a)- 1 (j2 (XE K, u(x) < a).

Now define an element w of X* by

w=:u+(j-l([3-a)v.

(j2).

7 Extreme Points

359

Since fJ - rx. < ~2, we can give u - w an arbitrarily small bound b} taking ~ sufficiently small. On the other hand, we have M(w, K) ~ w(x o» fJ + ~-1 (fJ - rx.)(M(v, S~)- ~2),

so that there exists c with fJ + ~- 1(fJ - rx.)(M(v, Sa) - ~2)< C< M(w, K) and S=S(w,c,K) compact. By (7.2.1), for each x in K with u(x)
(rx. -u(x)) ~ + ~-1 (fJ - rx.)M(v, Sa) < rx.+ ~-1(fJ - rx.)(M(v, Sa) - ~2)+ ~(fJ - rx.) < fJ + ~- 1(fJ - a.)(M(v, Sa) - ~2)
u(x)~oc

and therefore v(x)

u(x)~M(u,K)
and v(x) = ~(fJ - rx.)- 1(w(x) - u(x)) ~ ~(fJ -

rx.)-1 (c - M(u, K))

= ~(fJ - oc)- 1(c - fJ) + ~(fJ - rx.)- 1(fJ - M(u, K)) > M(V,Sa)-~2 -~, the last step following from our choice of fJ and c. Thus, by choosing ~ sufficiently small, we can make the variations of u and v on S arbitrarily small D (7.3) Lemma. Let X be a normed space over JR, K a compact convex subset of X, and Xo a point of K. Let v be a bounded linear functional on X such that (i) v(xo)=M(v,K), and (ii) for each e > 0 there exists rx. < M (v, K) such that S (v, rx., K) is compact and has diameter less than e. Then Xo is an extreme point of K. Proof. Let w be a modulus of continuity for v on K. Consider an arbitrary positive number e, and choose oc < v(x o) so that S(v, rx., K) is compact and has diameter less than e/2. Let x and y be points of K with 11!(x+ y)-xoll ~wrx..

l60

Chapter 7 Normed Linear Spaces

Since the assumption max {v(x), v(y)} < i(v(x o) + ex) entails the contradiction v(i(x + y» =~(v(x) + v (y» < i(v(x o)+ ex),

we may assume that v (x) > ex. Then both x and i(x + y) belong to S(v, ex, K), so that Ilx- yll =21Ix-i(x+ y)11 ~2diam S(v, ex, K)<e. Since e is arbitrary, it follows that

Xo

is extreme.

0

Under the conditions of Lemma (7.3), we say that the bounded linear functional v determines the extreme point Xo of K. (7.4) Proposition. Let K be a compact convex subset of a separable normed space X over JR., u an element of X*, and {) a positive number. Then there exists an extreme point Xo of K determined by an element v of X* such that v - u has bound {). Proof" Let (u.) be a sequence of elements of the unit sphere S* of X* which is dense in S* relative to the double norm. We construct inductively a sequence (v.) of elements of X*, and a sequence (ex.) of real numbers, such that for each n, (7.4.1)

M(v.,K)-n- 1 <ex.<M(v.,K)

(7.4.2)

S(v.,exn,K) is compact

and such that for each n ~ 2, (7.4.3)

S(v., ex., K)c S(vn_ l' ~ex._1 +1 M(v n_ l' K), K)

(7.4.4)

the variation of u. on S(v n , ex n , K) is at most n -

(7.4.5)

Iv.(x)-v n_ 1 (x)1 <2- n- 1 min {M(v j , K)-ex j : 1 ~j~n-1}

1

for all points x in K and (7.4.6)

v n -v._ 1 has bound 2-·{).

To start the construction, set V 1 == u and choose ex 1 so that M(u, K) -1<ex 1 <M(u,K) and S(u,ex 1 ,K) is compact. Now assume that v 1 , •.. , V n _ 1 and ex 1, ... , ex n _ 1 have been constructed to satisfy the appropriate parts of (7.4.1)-(7.4.6). By (7.2), we can find v. in X* and c<M(v.,K) so that V n-V._ 1 has arbitrarily small bound, S(v.,c,K) is

7. Extreme Points

361

compact, and the variation of Un on S(vn,c,K) is less than n~l. We may assume that Vn satisfies (7.4.5) and (7.4.6), and that there exists t<M(vn,K) such that vn~ 1 (x);~~ tIXn~ 1 +1 M(vn~ l' K)

whenever XE K and vn(x) ~ t. To complete our inductive construction, it only remains to choose IXn > max {c, t} so that (7.4.1) and (7.4.2) obtain. It follows from (7.4.6) that for x in X and m > n, m

Ivm(x)-vn(x)1 ~

I

j~n+

Ivix)-Vj~ 1 (x)1 1

Hence V(X)= lim vn(x)

(XEX)

defines an element v of X* such that v-v n has bound In particular, V-U=V-V 1 has bound b. We now show that

2~nb

for all n.

(7.4.7) By (7.4.1) and (7.4.3), we have S(v n, IXn' K)c S(Vn~ l' IXn~ l' K)

(n ~ 2).

It will therefore suffice to show that diam S (v n , IXn' K) is arbitrarily small for some n. To this end, let a be any positive number, and choose n in Z+ with n~l IIxi-Xjll-