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Undergraduate Texts in Mathematics Anglin: Mathematics: A Concise History
and Philosophy.
Croom: Basic Concepts of Algebraic
Topology.
Readings in Mathematics. Anglin/Lambek: The Heritage of
Thales.
Curtis: Linear Algebra: An Introductory
Approach. Fourth edition. Devlin: The Joy ofSets: Fundamentals
Readings in Mathematics. Apostol: Introduction to Analytic
Number Theory. Second edition.
of ContemporarySet Theory. Second edition. Dixmier: General Topology.
Armstrong: Basic Topology.
Driver: Why Math?
Armstrong: Groups andSymmetry.
Ebbinghaus/Flum/Thomas:
Axler: Linear Algebra Done Right.
Second edition.
Mathematical Logic. Second edition. Edgar: Measure, Topology, and Fractal
Reardon: Limits: A New Approach to
Real Analysis.
Geometry. Elaydi: An Introduction to Difference
Bak/Newman: Complex Analysis.
Equations. Second edition. Exner: An Accompaniment to Higher
Second edition. Banchoff!W' ermer: Linear Algebra
Through Geometry. Second edition. Berberian: A First Course in Real
Mathematics. Exner: Inside Calculus. Fine!Rosenberger: The Fundamental
Theory of Algebra.
Analysis. Bix: Conics and Cubics: A
Fischer: Intermediate Real Analysis.
Concrete Introduction to Algebraic
Flanigan/Kazdan: Calculus Two: Linear
and Nonlinear Functions. Second
Curves. Bremaud: An Introduction to
Probabilistic Modeling. Bressoud: Factorization and Primality
edition. Fleming: Functions ofSeveral Variables.
Second edition. Foulds: Combinatorial Optimization for
Testing. Bressoud: Second Year Calculus. Readings in Mathematics. Brickman: Mathematical Introduction
to Linear Prograrruning and Game Theory.
Undergraduates. Foulds: Optimization Techniques: An
Introduction. Franklin: Methods of Mathematical
Economics.
Browder: Mathematical Analysis:
An Introduction. Buskes/van Rooij: Topological Spaces:
From Distance to Neighborhood. Callahan: The Geometry ofSpacetime:
An Introduction to Special and General
Frazier: An Introduction to Wavelets
Through Linear Algebra. Gordon: Discrete Probability. Hairer/Wanner: Analysis by Its History. Readings in Mathematics. Halmos: Finite-Dimensional Vector
Spaces. Second edition.
Relavitity. Carter/van Brunt: The Lebesgue
Stieltjes Integral: A Practical Introduction
Cederberg: A Course in Modem
Halmos: NaiveSet Theory. Hammerlin!Hoffmann: Numerical
Mathematics. Readings in Mathematics. Harris/Hirst/Mossinghoff:
Geometries. Childs: A Concrete Introduction to
Higher Algebra. Second edition. Chung: Elementary Probability Theory
with Stochastic Processes. Third edition. Cox!Little/O'Shea: Ideals, Varieties, and A l g o rith ms Second edition. .
Combinatorics and Graph Theory. Hartshorne: Geometry: Euclid and
Beyond. Hijab: Introduction to Calculus and
Classical Analysis.
(continued after index)
M. Carter
B. van Brunt
The Lebesgue Stieltjes Integral
A Practioal Introduction
With 45 Illustrations
Springer
M. Carter B. van Brunt Institute of Fundamental Sciences Palmerston North Campus Private Bag 11222
Massey University
Palmerston North 5301
New Zealand
Editorial Board S. Axler
F.W. Gehring
K.A. Ribet
Mathematics Department
Mathematics Department
Mathematics Department
San Francisco State
East Hall
University of California
University
San Francisco, CA 94132 USA
University of Michigan
Ann Arbor, MI 48109 USA
at Berkeley
Berkeley, CA 94720-3840
USA
Mathematics Subject Classification (2000): 28-01
Library of Congress Cataloging-in-Publication Data Carter, M. (Michael), 1940-
The Lebesgue-Stieltjes integral : a practical introduction I M. Carter, B. van Brunt. p.
em.- (Undergraduate texts in mathematics)
Includes bibliographical references and index. ISBN 0-387-95012-5 (alk. paper) 1. Lebesgue integral. (tA312.C37
I. van Brunt, B. (Bruce)
II. Title. III. Series.
2000
515'.43-dc21
00-020065
Printed on acid-free paper.
© 2000 Springer-Verlag New York, Inc. All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer-Verlag New York, Inc., 175 Fifth Avenue, New York, NY 10010, USA), except for brief excerpts in connection with reviews or scholarly analysis. Usc in connection with any form of information storage and retrieval, electronic adaptation,
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Printed and bound by R.R. Donnelley and Sons, Harrisonburg, VA. Printed in the United States of America. 9 8 7 6 5 4 3 2 ] ISBN 0-387-95012-5 Springer-Verlag New York Berlin Heidelberg
SPIN 10756530
Preface
It is safe to say that for every student of calculus the first encounter with integration involves the idea of approximating an area by sum ming rectangular strips, then using some kind of limit process to obtain the exact area required. Later the details are made more precise, and the formal theory ofthe Riemann integral is introduced. The budding pure mathematician will in due course top this off with a course on measure and integration, discovering in the process that the Riemann integral, natural though it is, has been superseded by the Lebesgue integral and other more recent theories of integra tion. However, those whose interests lie more in the direction of applied mathematics will in all probability find themselves needing to use the Lebesgue or Lebesgue-Stieltjes integral without having the necessary theoretical background. Those who try to fill this gap by doing some reading are all too often put offby having to plough through many pages of preliminary measure theory. It is to such readers that this book is addressed. Our aim is to introduce the Lebesgue-Stieltjes integral on the real line in a nat ural way as an extension of the Riemann integral. We have tried to make the treatment as practical as possible. The evaluation of Lebesgue-Stieltjes integrals is discussed in detail, as are the key the orems of integral calculus such as integration by parts and change of v
vi
Preface
variable, as well as the standard convergence theorems. Multivariate integrals are discussed briefly, and practical results such as Fubini's theorem are highlighted. The final chapters of the book are devoted to the Lebesgue integral and its role in analysis. Specifically, func tion spaces based on the Lebesgue integral are discussed along with some elementary results. While we have developed the theory rigorously, we have not striven for completeness. Where a rigorous proof would require lengthy preparation, we have not hesitated to state important theo rems without proof in order to keep the book reasonably brief and accessible. There are many excellent treatises on integration that provide complete treatments for those who are interested. The book could also be used as a textbook for a course on in tegration for nonspecialists. Indeed, it began life as a set of notes for just such a course. We have included a number of exercises that extend and illustrate the theory and provide practice in the tech niques. Hints and answers to these problems are given at the end of the book. We have assumed that the reader has a reasonable knowledge of calculus techniques and some acquaintance with basic real analy sis. The early chapters deal with the additional specialized concepts from analysis that we need. The later chapters discuss results from functional analysis. It is intended that these chapters be essen tially self-contained; no attempt is made to be comprehensive, and numerous references are given for specific results. Michael Carter Bruce van Brunt Palmerston North, New Zealand
Contents
Preface I
2
3
Real 1.1 1. 2 1.3
v
Numbers Rational and Irrational Numbers . The Extended Real Number System Bounds . . . . . . . . . . . . . . . .
Some Analytic Preliminaries 2. 1 Monotone Sequences 2. 2 Double Series . . . . 2.3 One-Sided Limits . . 2.4 Monotone Functions 2.5 Step Functi0ns . . . . 2.6 Positive and Negative Parts of a Function 2. 7 Bounded Variation and Absolute Continuity The 3.1 3. 2 3.3
Riemann Integral Definition of the Integral Improper Integrals . . . A Nonintegrable Function
I
1 6 8 II
11 13 16 20 24 28 29
39
39 44 46
VII
VIII 4
Contents 49
The Lebesgue-Stieltjes Integral 4. 1 The Measure of an Interval . 4. 2 Probability Measures . . 4.3 Simple Sets . . . . . . . . 4.4 Step Functions Revisited 4.5 Definition of the Integral 4.6 The Lebesgue Integral
49 52 55 56 60 67
5 Properties of the Integral 5.1 Basic Properties . . . . 5.2 Null Functions and Null Sets 5.3 Convergence Theorems . 5.4 Extensions of the Theory 6
7
7I 71 75 79 81
Integral Calculus 6.1 Evaluation of Integrals . . . . . . . 6. 2 'TWo Theorems of Integral Calculus 6.3 Integration and Differentiation .
87 87 97 10 2
Double and Repeated Integrals 7.1 Measure of a Rectangle . . . . . . . . . . . . . . . . 7.2 Simple Sets and Simple Functions in Thro Dimensions . . . . . . . . . . . . 7.3 The Lebesgue-Stieltjes Double Integral . . 7.4 Repeated Integrals and Fubini's Theorem .
II3
8 The 8.1 8. 2 8.3 8.4 8.5 8.6 8.7 8.8 8.9
Lebesgue Spaces V Normed Spaces . . Banach Spaces . . . Completion of Spaces The Space L1 . . . The Lebesgue V . . Separable Spaces . . Complex V Spaces The Hardy Spaces HP Sobolev Spaces Wk,p .
113 114 115 115 I2 3
. . .
.
1 24 131 135 138 14 2 150 15 2 154 161
Contents
����--
--------------------------------------
9
Hilbert Spaces and L2 9 .1 Hilbert Spaces . . 9. 2 Orthogonal Sets . 9.3 Classical Fourier Series 9. 4 The Sturm-Liouville Problem 9.5 Other Bases for L2 •
•
.
.
.
•
1. X
1 65
165 17 2 180 188 199
.
10 Epilogue
203
10. 1 Generalizations of the Lebesgue Integral 10. 2 Riemann Strikes Back . 10.3 Further Reading . . . . .
.
Appendix: Hints and Answers
.
.
to
.
.
.
.
.
.
Selected Exercises
. 203 . 205 . 207 209
References
221
Index
225
Real Numbers CHAPTER
The field of mathematics known as analysis, of which integration is a part, is characterized by the frequent appeal to limiting processes . The properties of real numbers play a fundamental role in analysis . Indeed, it is through a limiting process that the real number system is formally constructed. It is beyond the scope of this book to recount this construction. We shall, however, discuss some of the properties of real numbers that are of immediate importance to the material that will follow in later chapters.
1.1
Rational and Irrational Numbers
The number systems of importance in real analysis include the nat ural numbers (N), the integers (Z), the rational numbers (Q), and the real numbers (JR). The reader is assumed to have some famil iarity with these number systems. In this section we highlight some of the properties of the rational and irrational numbers that will be used later. The set of real numbers can be partitioned into the subsets of rational and irrational numbers. Recall that rational numbers are
1
2
1. Real Numbers
numbers that can be expressed in the form min, where m and n are integers with n # 0 (for example �� �, -� (= -.,S ), 15 ( = \5), 0( = �)). Irrational numbers are characterized by the property that they cannot be expressed as the quotient of two integers. Numbers such as e, n, and v'2 are familiar examples of irrational numbers. It follows at once from the ordinary arithmetic of fractions that if r1 and r2 are rational numbers, then so are r1 + r2, r1 -r2, r1 r2, and r1/r2 (in the last case, provided that r2 =j:. 0). Using these facts we can prove the following theorem: Theorem 1 . 1 . 1 Ifr is a rational number and x. is an irrational number, then (i) r + x. is irrational; (ii) rx. is irrational, provided that r =j:. 0.
Proof See Exercises 1-1, No. 1.
0
A fundamental property of irrational and rational numbers is that they are both 11dense" on the real line. The precise meaning of this is given by the following theorem: Theorem 1 . 1 . 2 If a and b are real numbers with a < b, then there ex.ist both a rational number and an irrational number between a and b.
Proof Let a and b be real numbers such that a < b. Then b- a > 0, so v'2,!(b -a) > 0. Let k be an integer less than a, and let n be an integer such that n > v'il(b-a) . Then
,J2
- < b-a, n and so the succesive terms of each of the sequences 0
1 n
<- <
3 2 1 k + -, k + -, k + -, ... n n n
,J2
v'2
v'2
k + -, k + 2 - , k + 3 - , . . . n n n differ by less than the distance between a and b. Thus at least one term of each sequence must lie beween a and b. But the terms of the first sequence are all rational, while (by Theorem 1 . 1. 1) those of D the second are all irrational, so the theorem is proved.
3
1.1. Rational and Irrational Numbers ---- + I
I
-4
t
I t
-3
I
-2
-1
t
FIGURE 1.1
0
:
:
:
-+
1
I
2
I
Counting the integers
• 3
� 4
I
Corollary 1 .1 .3 If a and b are real numbers with a < b, then between a and b there exist infinitely many rational numbers and infinitely many irrational numbers. Proof This follows immediately by repeated application of Theo o rem 1.1. 2. An infinite set S is said to be countable if there is a one-to-one correspondence between the elements of S and the natural num bers. In other words/ S is countable if its elements can be listed as a sequence For example1 the set Z is countable because its elements can be listed as a sequence {a11 a21 a31 } by using the rule •
•
•
if n if n if n
= = =
1 2m1 m > 0 2m + 1 1 m > 0
so that a1 = 01 az = 1 1 a3 = 1 1 a4 = 2 1 and so on. The process of listing the elements of.Z as a sequence can be visualized by following the arrows in Figure 1.1 starting at 0. Much less obvious is the fact that the set Q is also countable. Figure 1. 2 depicts a scheme for counting the rationals. Th list the rationals as a sequence we can just follow the arrowed path in Figure 1. 2 starting at 0/1 = 01 and omitting any rational number that has already been listed. The set -
4
I.
Real Numbers I I ,I
I
-3/3
-2/3
-3/2
I
.f.
I
-1/3
+-
0/3 .-I/3
-2/2
-1/2
+-
0/2
+-
-311
-211
-Ill
011
-+
-3/-1
-2/-1
-1/-1
-3/-2
-2/-2-+ -1/-2
-3/-3
-2/-3
!
+-
!
!
!
!
!
-1/-3
FIGURE 1.2
2/3
3/3
1/2
2/2
3/2
Ill
211
311
3/-1
+--
t
t
t
t
0/-1
-+
11-1
-+
2/-1
-+
0/-2
-+
1/-2
-+
2/-2
1/-3
t
t
-+
0/-3
I I I
I
t
-+
2/-3
t
3/-2 3/-3
Counting the rationals
Q can thus be written as Q
=
{
0, 1,
�·- �· - 1, -2 , 2 , �' �'-�·-�·-�· -3 , 3, ... }.
The infinite sets N, Z, and Q are all countable, and one may won der whether in fact there are any infinite sets that are not countable. The next theorem settles that question:
Theorem 1 .1 .4 The set S ofall real numbers x such that 0 <
x <
1 is not countable.
Proof We use without proof here the well-known fact that any real number can be represented in decimal form. This representation is not unique, because N.n1 n2n3 . . .nk9999 . . and N.n1 n2n3.. .(nk + 1)0000 . . . are the same number (e.g. 2 .3 49999 . . . = 2.3 5); likewise N.999 . .. and N + 1 are the same number. We can make the representation unique by choosing the second of these representations in all such cases, so that none of our decimal expressions will end with recurring 9's. We will use a proof by contradiction to establish the theorem. Suppose S is countable, so that we can list all the elements of S as a sequence: .
1. 1. Rational
and
Irrational Numbers
5
Now, each element of this sequence can be represented in decimal form, say where for all n,j E N, Xnj is one of the digits 0, 1, 2 , ... , 9. The elements of S can thus be written in the form a1 = O.xux12XI3X14 .. . , , a2 = 0.x21X22X23X24 a3 = 0.X31X32X33X34
a4 = O.x41X42X43X44
We define a real number b
=
m·1 =
•
•
•
•
•
•
.
•
.
O.m1m2m3m4
{
.
, ,
•
•
, where for eachj
E N,
1 if X··']) -/.. r 1, 2 ifxff=l.
Suppose, for example, that our listing of elements of S begins a1 a2
a3 a4
Then:
=
0.8 3 712 4... ,
=
0.112 56 3 . .. ,
=
0.33. 3333 . . . , .. . , 0. 2 58614 .-
=
xu = 8 :f. 1 x22 = 1 X33 = 3 :f. 1 X44 = 6 :f. 1
so m1 = so m2 = so m3 = so m4 =
1, 2, 1, 1,
and so on. The decimal expansion of b therefore begins 0.12 11 . . . . It is clear that 0 < b < 1, so that b E S, and therefore we must have b = aN for some N E N. But by definition, the decimal expansion of b differs from that of a N at the Nth decimal place, so b :f. aN and we have a contradiction. We thus conclude that our original assumption D must be false, and S cannot be countable.
6
1. Real Numbers
It follows at once from this theorem that the set � is not count able. In fact, it is also not hard to deduce that the set of all real numbers belonging to any interval of nonzero length (however small) is not countable.
Exercises
1-1:
1. Use the method of proof by contradiction to prove Theorem 1.1.1.
2. Give examples to show that if XI and x2 are irrational numbers/ then XI + x2 and xix2 may be rational or irrational.
3. Since the set of all rational numbers is countable/ it follows easily that the set S* = {x : 0 < x < 1 and x rational} is countable. Thus1 if we apply the argument used in the proof of Theorem 1.1.4 to S* instead of S, something must go wrong with the argument. What goes wrong?
4. (a) Prove that the union of two countable sets is countable. (b) Use a proof by contradiction to prove that the set of all irrational numbers is not countable.
1.2
The Extended Real Number System
It is convenient to introduce at this point a notation that is useful in many parts of analysis; care/ however, should be taken not to read too much into it. The extended real number system is defined to be the set lRe consisting of all the real numbers together with the symbols oo and -oo, in which the operations of addition, subtraction, :rp.ultiplication, and division between real numbers are as in the real number system, and the symbols oo and -oo have the following properties for any XE
JR:
(i) -oo <
x
<
oo;
(ii) 00 + X = X + 00
(iii) oo + oo (iv) oo . x X >
=
0· I
= x
=
00 and
- 00
+ X = X + ( -00)
oo and -oo + ( -oo)
·
oo
=
oo
an d
( -oo)
= ·
x
-oo; =
x
· (-oo)
=
=
- 00;
-oo
for any
1.2.
(v) oo · x = x · oo X < 0· I
(vi) oo·oo 00.
=
=
The Extended Real Number System
-oo and ( -oo) · x
oo, oo ·( -oo)
=
(-oo)'oo
=
=
x ( -oo) ·
=
7
oo for any
-oo, and(-oo) ·(-oo)
=
The reader is warned that the new symbols oo and -oo are defined only in terms of the above properties and cannot be used except as prescnbed by these conventions. In particular, expressions such as 00 + ( -oo), c.a... oo ) + oo, 00 . 0, 0. oo , 0 . ( -00), and ( -oo) . 0 are meaningless. A number a E lRe is said to be finite if a E JR, i. e. if a is an ordinary real number. In all that follows , when we say that is an interval with endpoints a, b we mean that a and b are elements of lRe (unless specifically restricted to finite values) with a < b, and is one of the following subsets of JR:
I
I
(i) the
open interval {x E 1R: a
<
x
<
b}, denoted by (a, b);
(ii) the closed interval { x E 1R : a < x < b}, denoted by [a, b], where a and b must be finite; (iii) the closed-open interval {x E 1R [a, b), where a must be finite; (iv) the open-closed interval {x E 1R (a, b], where b must be finite .
:
a < x
:
a
<
<
b}, denoted by
x < b}, denoted by
Note that although the endpoints of an interval may not be finite, the actual elements of the interval are finite. Note also that for any a E JR, the set [a, a ] consists of the single point a, whereas the sets [a, a) and (a, a] are both empty. The interval (a, a) is empty for all a E lRe. The only change from standard interval notation is that intervals such as ( -oo, -3], ( -oo, oo), (-2 , oo), etc. are defined. (Intervals such as [ -oo, 3], [ -oo, oo], ( -2 , oo], etc. are not. )
8
1.3
I.
Real Numbers
Bounds
Let 8 be any nonempty subset of Re. A number c E lRe is called an upper bound of 8 if x < c for all x E 8. Similarly, a number d E lRe is called a lower bound of 8 if x > d for all x E 8. Evidently, oo is an upper bound and -oo is a lower bound for any nonempty subset oflRe. In general, most subsets will have many upper and lower bounds. For example, consider the set 81 = ( -3 , 2]. Any number c E lRe such that c > 2 is an upper bound of 81, and any number d E lRe such that d < -3 is a lower bound of 81. Note that there is a least upper bound for 81 (namely 2) and that in fact it is also an element of 81. Note also that there is a greatest lower bound (namely -3), which is not a member of 81. As another example, consider the set 82
=
{�:
n
} {
E N = 1,
�' �'. ·l
Here any c > 1 is an upper bound of 82, while any d < 0 is a lower bound. Note that no positive number can be a lower bound of 82, because for any d > 0 we can always find a positive integer n sufficiently large so that 1/n < d, and therefore d cannot be a lower bound of 82• Thus 82 has a least upper bound 1 an� a greatest lower bound 0. As a final example, let 83 ({]. Then oo is the only upper bound of 83 and -oo is the only lower bound. Thus 83 has a least upper bound oo and a greatest lower bound -oo. The following result ( often taken as an axiom), which we state without proof, expresses a fundamental property of the extended real number system: =
Theorem 1 .3 . 1
Any nonempty subset oflRe has both a least upper bound and a greatest lower bound in lRe.
c
The least upper bound of a nonempty set 8 lRe is often called the supremum of 8 and is denoted by sup 8; the greatest lower bound of 8 is often called the infimum of 8 and denoted by inf 8. The examples given above indicate that sup 8 and inf 8 may or may not be elements of 8; however, in the case where sup 8 or inf 8 is
1.3.
Bounds
g
finite, although supS and inf Sneed not be in S, they must at any rate be 11close" to S in a sense that is made precise by the following theorem:
Theorem 1 .3 .2
c
Let S :IRe be nonempty. (i) If M E IRe is finite, then M = supS if and only if M is an upper bound ofS and for each real number € > 0 (however smal� there exists a number xES (depending on €) such that M - € < x < M. (ii) If m is :fil).ite, then m infS if and only if m is a lower bound of S and for each real number € > 0 (however small) there exists a number xES (depending on €) such that m < x < m + €. =
Proof We shall prove part (i) and leave part (ii) as an exercise. Sup pose M = supS, where M is finite. Then M, being the least upper bound of S, is certainly an upper bound of S. Let € be any positive real number. Then M - € < M, and so M € cannot be an upper bound ofS, since M is the least upper bound. Thus there must exist a number x E S such that x > M - €, and since we know that M is an upper bound of S, we have M- € < x < M. Conversely, suppose that M is finite, M is an upper bound of S, and that for any real number € > 0 there exists a number xES such that M- € < x < M. Let K be any finite element of IRe with K < M. Then M- K > 0, so taking € = M - K we have that there exists an x E S such that M- (M- K) < x < M, i.e., K < x < M. Thus K ·-
cannot be an upper bound of S, and since -oo is obviously not an upper bound of S, it follows that M must be the least upper bound D ofS.
Exercises
1-3:
1. Give the least upper and greatest lower bounds of each of the following subsets of IRe, and: state in each case whether or not they are elements of the set in question:
(a) {x: 0 < x < 5} 2 (c) {x: x > 3} (e) {x : xis rational and x2 < 2} (g) {x: xis rational and positive}
{x: 0 < x < 5} {x: � > 2} (f) {x: x 3 + �, n E N}
(b) (d)
=
10
1. Real
2. IfS
Numbers
c :IRe has only finitely many elements , sayS= {x1,x2 ,
,Xn} , then clearlyS has both a greatest element and a least element, de noted by max{x1, x2 , . . . ,Xn} and min{x1, x2 , .. . ,Xn} , respectively. Prove: sup{xi, x2 , . . . , Xn}
=
max{x1,x2 , . . . , Xn},
inf{x1,x2 , . . . ,Xn}
=
min{x1,x2 , . . . ,xnl·
•
•
•
3 . Prove that if 81 and 82 are nonempty subsets of :IRe such that 82 , then sup sl < sup 82 and inf sl > inf 82 . sl
c
4. Let S be a nonempty subset of :IRe , and c a nonzero real number. Define S* by S* ={ex: xES}.
(a) Prove that if c is positive, then supS* inf(S*) = c(inf S).
=
c( sup S) and
(b) Prove that if c is negative, then supS* = c(infS) and inf(S*) = c( supS). 5. Prove part (ii) of Theorem 1 .3 .2.
CHAPTER
Some Analytic Preliminaries
Before we can develop the theory of integration, we need to re visit the concept of a sequence and deal with a number of topics in analysis involving sequences, series, and functions.
2.1
Monotone Sequences
Convergence of a sequence on :IRe can be defined in a manner anal ogous to the usual definition for sequences on JR. Specifically, a sequence {an} on :IRe is said to converge to a finite limit if there is a finite number a E :IRe having the property that given any posi tive real number E (however small) there is a number N E N such that lan- al < E whenever n > N. This relationship is expressed by an--+ a as n--+ oo, or simply an--+ a. The number a is called the limit of the sequence. If for any finite number M E :IRe there exists an N E N such that an > M whenever n > N, then we write an --+ oo as n--+ oo or simply an --+ oo, and the limit of the sequence is said to be oo; similarly, if for any finite number M E :IRe there exists anN E N such
11
12
2.
Some
Analytic
Preliminaries
that an < M whenever n> N, then we write an --+ -oo as n--+ oo or simply an --+ -oo, and the limit of the sequence is said to be -oo. Let {an} be a sequence of real numbers.The sequence {an} is said to be monotone increasing if an < an+l for all n eN, and mono tone decreasing if an> an+l for all n EN. For example: The sequence 1 , 2, 3, 4, . . . is monotone increasing. The sequence 1 , ! , , ... is monotone decreasing. The sequence 1 , 1 , 2, 2, 3, 3, .. . is monotone increasing. The sequence 1 , 1 , 1 , 1 , ...is monotone increasing and monotone decreasing. The sequence 1 , 0 , 1 , 0 , ... is neither monotone increasing nor monotone decreasing.
! �,
If a sequence {an} is monotone increasing with limit .e EIRe, we write an t .e (read "an increases to l"). If the sequence is monotone decreasing with limit .e eIRe, we write an _J.. .e (read "an decreases to l"). We shall frequently be studying sequences of functions. Let lfn} denote a sequence of functions fn : I --+ 1R defined on some interval I c JR. The sequence lfn} is said to converge on I to a function f if for each x E I the sequence lfn(x) } converges to f(x) , i.e., if the sequence is pointwise convergent. The notation used for sequences of functions is similar to that used for sequences of numbers: specif ically ,
fn--+ f on I means that for each x E I , fn(x) --+ f(x) . fnt f on I means that for each x E I , fn(x) t f(x) . fn _J.. f on I means that for each x E I , fn(x) -l- f(x) . •
The fundamental theorem concerning monotone sequences is the following:
Theorem 2 .1 .1
Let {an} be a sequence on llt (i) I fthe sequence {an} is monotone increasing, then ant sup {an} (ii) I f the sequence {an} is monotone decreasing, then an -J,. inf {an}
.
.
2 . 2 . Double Series
13
Proof We shall prove part (i) of the theorem, leaving the second part as an exercise. Let M = sup{ an}. The proof of part (i) can be partitioned into two cases depending on whether or not M is finite. Case 1 : If M = oo, then for any positive real number K, we know that K cannot be an upper bound of {an}, so there exists a positive integer N such that aN > K. Since the sequence is monotone increasing, it follows that an > aN > K for all n > N, and thus ant oo(= M) by definition. Case 2: Sup:eose M finite and let Ebe any positive real number. Then by Theorem 1.3.2 there exists a positive integer N such that
M-E < aN < M.
Since the sequence is monotone increasing and has M as an upper bound, it follows that
for all
n
> N. This implies that for all n > N, ! an -MI < E
and consequently an -+ M by definition. Since the sequence is D monotone increasing, this means that ant M as required.
Exercises 2-1 :
1.
Let S be a nonempty subset ofiR, with supS = M and inf S = m. Show that there exist sequences {an} and {bn} of elements of S such that an t M and bn .J, m.
2. Prove part ( ii) of Theorem
2.2
2. 1 .1 .
Double Series
Let {an} be a sequence on IRe. Recall that the infinite series L�=l am is said to converge if the sequence of partial sums {sn}, where Sn = L�=l am, converges to a finite number. If Sn -+ oo, then the series is said to diverge to oo; if Sn -+ -oo, then the series is said to diverge to -oo. Often, questions concerning the convergence of an infinite
14
2.
Some
Analytic
au a21
-!,
a31 a41
---+-
/
/ /
-!,
Preliminaries
a12 azz a32
/ /
a42
a13 a23
---+-
/
a14 a24
a33
a34
a43
a44
FIGURE 2 .1 au
---+-
a12
a13
a21
+--
azz
a23
a24
a31
---+-
a32
---+-
a33
a34
a41
+--
a42
+--
a43
-!,
-!,
-!,
t
---+-
a14
-!, -!,
t
+--
-!,
a44
FIGURE 2 . 2
series involve considering sequences {an} of nonnegative terms ( e.g., absolute convergence) . If the terms of the sequence {an} consist of nonnegative numbers, then the resulting sequence of partial sums is monotone increasing. Theorem 2.1. 1 thus implies that Sn t sup{sn} and therefore that either the series I:�=I � converges or it diverges to oo, according as sup{sn} is finite or oo. Consider the array of real numbers depicted in Figure 2. 1. This array can be written as a ( single) sequence in many ways. One way is to follow the arrowed path in the diagram. This gives the sequence
but this is obviously not the only way. Another scheme for constructing a sequence is given in Figure 2.2.
2.2. Double Series
15
For any way of writing this array as a single sequence A1, A2, A3, we can form the corresponding infinite series 2:}�1 Aj· We know from Riemann's theorem on the derangement of series [6] that in general, the convergence and limit of the series depends on the particular sequence {A n } used, but there are some situations in which every possible sequence leads to the same answer. When this is the case, it is sensible to introduce the notion of a "double series" 2:: n =l Clmn and consider questions such as convergence. This leads us to the following definition: If for all possible ways of writing the array {tlmn} a� a single sequence the corresponding series has the finite sum .e , then the double series L� n=l tlmn is said to converge to .e . If for all possible ways ofwriting the array as a single sequence the corresponding series either always diverges to oo or always diverges to -oo, then the double series is said to be properly divergent ( to oo or -oo as the case may be). In all other circumstances the double series is simply said to be divergent, and its sum does not exist as an element of IRe. As well as "summing" the array by writing it as a single sequence, we can "sum" it by first summing the rows and then adding the sums of the rows, giving the repeated series 2::=1 (L:1 amn)· Alterna tively, we can first sum the columns and then add the sums of the columns, giving the repeated series 2::1 (L�=l amn)· The relationship between convergence for a double series L�n=l amn and for the two related repeated series is, in general, complicated. For our purposes, however, we can focus on the par ticularly simple case where all of the entries in the array are nonnegative, i.e., tlmn > 0 for all n, m E N. In this case we have the following result, which is stated without proof: •
•
•
I
Suppose that for all n, m E N we have amn > 0, where tlmn E IRe. Then the double series L� n=l amn and the two repeated series Ln=l CLm=l amn) and Lm=l (Ln=; amn ) either all converge to the same finite sum or are all properly divergent to oo. Theorem 2 .2 . I
More details on double series can be found in [6].
f(x)
f+e t t-e
I
----------------
• �
f(x) lies between 1-e and f+E for all X E (t-8, t)
I
t
FIGURE 2.3
2.3
One-Sided Limits
Let f : IR --* IR be a function, and t and .e real numbers. Recall that limx-H fCx) = .e if and only if for any positive real number E, however small, there exists a positive real number 8 such that 0
< lx-t I <
8 ==:::}
If(x) - .e I < E .
We say that limx--+t f(x) = oo if for any number M there exists a 8 > 0 such that f(x) > M whenever 0 < lx -tl < 8. A similar definition can be made for limx--+tf(x) = - oo. In these definitions x can be either to the left or the right oft, i.e., x is free to approach t from the left or right ( or for that matter oscillate on either side of t). Often it is of use to restrict the manner in which. x approaches t, particularly if no information about f is available on one side of t, or t lies at the end of the interval under consideration. For these situations it is useful to introduce the notion of limits from the left and from the right. Such limits are referred to as one-sided limits. The limit from the left is defined as follows: limx--+r- f(x) = .e if and only if for any positive real number E there exists a positive real number 8 such that t-8 < X < t
=>
lf(x)-ll < E
2.3.
One-Sided Limits
17
f(x)
-
-
-
---
-
1I I
I
I+E for all x e (t, t+8)
- - - _I
I I I I I
t+8
t
FIGURE 2 .4
( cf. Figure 2.3). In this case we say that f(x) tends to .e as x tends to t from the left. Similarly, the limit from the right is defined as limx-+t+ f(x) = .e if and only if for any positive real number E there exists a positive real number 8 such that t
<X<
t
+ 8 ==} lf(x)- .fl <
E
( cf. Figure 2.4). In this case we say that f(x) tends to .e as x tends to t from the right. We can easily extend these definitions for cases where the limit is not finite, e.g., limx-+t- f(x) = oo if and only if for any positive real number M there exists a positive real number 8 such that t-
8
<x <
Example 2-3-1: Let f : JR. � JR. be defined as f(x) =
{
t
==}
-1 0 x/2
f(x)
>
M.
if x < 1, if x= 1, if x > 1.
Then limx--+1- f(x) =-1 and limx--+1+ f(x) = 1/2. This function is depicted in Figure 2.5.
18
2.
Some Analytic Preliminaries
f(x)
1/2 --Q
1
I
------------�--1--6
1
X
FIGURE 2.5
E�ple 2-3-2: Let fC x) = 1 /C x- 1) C cf. Figure 2.6). Then limx�r + fC x) = oo .
limx�r- fC x) = -oo and
The definition of a limit can be extended further to consider cases where � or � -oo. For example, let a E JR. Then = a if and only if for any positive real number E there limx�oo exists a number X such that
fC x)
x
oo
x
X> X ==>
lfC x) -a!<
E.
Definitions similar to the finite case can also be framed for = a, etc. = 00, and = The usual elementary rules for limits of sums, differences, products, and quotients of functions hold for one-sided limits just as for ordi = a and = nary limits. For example, if then = ab, etc. These = relations are proved the same way as for the ordinary limit case. It is also easy to prove that lim == l if and only = l and lim = l. + by and For succinctness, we shall ofte:n denote lim In some circumstances we will denote by + by and lim by
limx�oo fC x)
limx�-oo fC x)
limx�-oo fC x)
oo,
limx�t- fC x) limx�t- gC x) limx�t-(fC x)+gC x)) a+b, limx�r-(fC x)gC x)) x�t fC x)
x�t fC x) limx�oo fC x)
fCt+). fCoo-)
x�t fC x)
x� oo fC x)
iflimx�t- fC x) limx�t- fC x) f(t -) fCC- oo)+).
b,
2.3.
One-Sided Limits
19
f(x)
--------�0�-,,1--� I
I
X
FIGURE 2.6
f
t
One-sided continuity for a function at finite points is defined in terms of one-sided limits in the obvious way. We say that is continuous on the left at if ) is defined and finite, exists, = and continuous on the right at is defined and finite, = Evidently, is continuous exists, and at if and only if it is both continuous on the left and continuous on the right at i.e., if and only if = = There are several different ways in which a function can fail to all exist but are not all be continuous at a point. If equal, then is said to have a jump discontinuity at Thus, the function in Example 2-3-1 has a jump discontinuity at 1. A function may fail to be continuous at a point because the limit is not finite. The function of Example 2-3-2 is discontinuous at 1 not only because the limit is not finite but also because =I= +) and has not been defined. Yet another way in which a function can fail to be continuous at a point is when the right or left limits fail to exist. The next example illustrates this.
t f(t
andf(t-) f(t), f(t+) t t,
f(t+) f(t) .
f(t-) t iff(t) f
f(t-) f(t) f(t+).
f(t-), f(t), f(t+)
f
t.
f(1-) f(1
E�ple 2 -3-3:
f
Consider the function
f
:
f(x) =
1R
{
-+
f(t)
1R defined by
( x) �f x =I= 0, 1f x = 0.
sin 1/ , 0,
Figure 2.7 illustrates this function. Now, I sin 1 /
( x) l
<
1 and
20
2. Some
Analytic
Preliminaries
-1
FIGURE 2.7
sin(1 /x) = 0 if and only if 1 /x = mr, where n e Z- {0}, i.e., when x = 1 /(nn). Moreover, sin(1/x) = 1 if and only if 1/x = (4n + 1)n/2, where n E Z, i.e., x = 21((4n+ 1)n), and sin(1/x) = - 1 if and only if 1 /x = (4n + 3)n/2, where n E Z, i.e., x = 21((4n + 3)n). Near x = 0, x attains the values 1/(mr), 21((4n + 1)n), 2/((4n + 3)n) infinitely many times (for different n e Z), and thus it can be shown that nei therteo-) norteo+) exists, sot is discontinuous at 0. The function oscillates infinitely often in any interval ( -8, 8), 8 > 0. 2.4
Monotone Functions
Lett � � be a function. We say thatt is monotone increas ing ift(xi) < t(x2) whenever xi < x2• The functiont is said to be monotone decreasing ift(xi) 2::: t(x2) whenever xi < Xz. Ift is either monotone increasing or monotone decreasing, then it is said to be monotone. Some examples are: (i) The function in Example 2-3-1 is monotone increasing. (ii) The function lxl is neither monotone increasing nor monotone decreasing. :
--"*
2.4.
Monotone Functions
21
(iii) Constant functions are both monotone increasing and monotone decreasing. One can also speak of functions being monotone increasing or monotone decreasing on a particular interval rather than the en tire real line. For example, the function lxl is monotone decreasing on ( -oo, 0] and monotone increasing on the interval [0, oo) . In this section, however, we will restrict the discussion to functions that are monotone on the entire real line. The general case will be discussed in Section 2.7. The most important theorem on monotone functions is the following: Theorem 2 .4 . 1 Let f : 1R � 1R be a monotone function. Then, for all t E lR, f(t-) and f(t+) exist and are finite , and also f(oo-) and f(( -oo) +) exist, but are
not ne cessarily finite. Furthermore , for all t E IR, (i) iff is monotone increasing, then f(t-)
f(t) > f(t+) .
Suppose f is monotone increasing, and lett be any real num ber. Let m = inf{f(x) t < x } and M = sup{f(x) : x < t}. Now, f(t) is finite, and since f is monotone increasing, f(t) is a lower bound of {f(x) t < x} and an upper bound of{f(x) : x < t}. It follows that m and M are finite, and also M 0. By Theorem 1 .3.2, there exist x1 and xz, with t < x1 and t > x2, such that m :S f(xl ) < m + E and M - E < f(x2) <M. Since f is monotone increasing and m is a lower bound of{f(x) t < x}, it follows that t < x < x1 � m l f(x) - ml < E and similarly x2 < x < t ===> M - E < f(xz) < f(x) <M ===> lf(x) - Ml < E. Thus, by definition, f(t+) = m and f(t-) = M. Also, statement (i) follows from equation (2.1). Next, let A = inf{f(x) : x E IR}; here, A may be finite, or equal to -oo. If A is finite, an argument similar to that used previously Proof
:
:
:
22
2.
Some Analytic Preliminaries
shows that fCC -oo)+) = A. If A is -oo , let K be any negative real number. Then K is not a lower bound of { f(x) : x E JR.}, so there exists an x1 E JR. such that f(xi) < K. Sinc e f is monotone increasing, it follows that and so fCC -oo)+) = -oo = A in this case also. A similar argument shows that f(oo-) = sup{ f (x) : x E JR.}. The case where f is monotone decreasing can be proved in a similar way, or by considering the function -f (see Exercises 2-4, D No. 1). Corollary 2.4.2 (i )
(ii)
.
I ff is monotone increasing, anda,bare elements of lRe witha < b, then f(a+ ) < f( b - ). I ff is monotone decreasing, anda,bare elements oflRe witha < b, then f(a+) > f(b - ) .
We will prove part (i) of this theorem and leave the other part as an exercise. Let f be monotone increasing. From the proof of Theorem 2.4.1 we know that f (a+ ) = inf{ f(x) a < x} and f (b-) = sup{f (x) : x < b}. Since a < b, there exists a y E JR. such that a < y < b, and so f (a+) < f(y) and f(y) < f(b-), whence f(a+) < f(b-) D as required. If f is monotone, then for any real t we have by Theorem 2.4.1 that f(t-),f(t), and f(t+) all exist. It follows at once that the only discontinuities that a monotone function can·have are jump discontinuities. In general, a function f : JR. JR. may have any number of points of discontinuity. Indeed, the function f defined by atio?al, f(x) = 0,1 , 1�ff xx 1s�s :rrrat 10nal, is discontinuous at every real number. However, for monotone functions we have the following theorem:
Proof
:
--+
{
2.4.
Monotone Functions
23
Theorem 2.4.3 I f f : JR. � JR. is monotone, then the set of points at which f discontinuous is either empty, finite, or countably infinite.
is
Proo f Iff is monotone decreasing, then -f is monotone increasing (see Exercises 2-4, No. 1)) and has the same points of discontinuity as f, so it is sufficient to prove the theorem for the case where f is monotone increasing. Let Ebe the set of points at which f is discontinuous, and suppose Eis not empty:Then for each x a Ewe have f(x-) < f(x+), and so by Theorem 1.1 .2 there exists a rational number rx such that f(x-) < rx < f(x+) . Now by Corollary 2.4.2 we have xi < x2 ==> f(xi ) < f(xz), and it follows that if XI, xz E E are such that XI < x2 , then rx1 < rx2; thus, we have associated with each x E E a distinct rational number. Since the set of all rational numbers can be listed as a sequence, it follows that the set rx : x E E} can also be listed as a (finite or infinite) sequence. We can then list the elements of E in the same order as their associated rational numbers. Thus E (if not empty) is D either finite or countably infinite.
{
Although Theorem 2.4.3 places restrictions on the possible set of discontinuities of a monotone function, this set can nevertheless be quite complicated, and one must be careful not to make unjustified assumptions about it. For example, one might guess that the discon tinuities of a monotone function. must be some minimum distance apart, but the following example shows that this need not be so. Example 2-4-1 : Let f : JR. � JR. be defined as follows: f(x) =
{
0, 1/ (n + 1), 1,
ifx < 0, if 1/ (n + ifx > 1 .
1)
l
< x < i n, n
=
1 , 2, 3, .. . ,
Figure 2.8 illustrates this function. Clearly, f is monotone increas ing. It can be shown that f(O+) :t::: 0 (see Exercises 2-4, No. 3), so f has jump discontinuities at the countably infinite set of points { 1, �, �, t, . . } and is continuous a!t all other points. In fact, unlikely .
24
2 . Some Analytic Preliminaries
j{x) 1
o�-o-----1
I
FIGURE
2 .8
as it may seem , it is possible to construct a monotone increasing function that is discontinuous at every rational number! Exercises 2-4:
1.
2. 3.
Prove part (ii) of Theorem 2.4.1, by showing that iff is monotone decreasing, then -f is monotone increasing, and then applying part (i) . Prove part (ii) of Corollary Prove that f(O+)
2.5
=
2.4.2.
o in Example
2-4-1 .
Step Functions
Let I be any interval. A function e : I --* JR. is called a step func tion if there is a finite collection {h , lz, In} of pairwise disjoint intervals such that S = h U ]z U U In I and a set {c1, Cz, ... , Cn}
c ·
·
·
·
·
·
,
2.5 . Step Functions
8(x)
25
/=(-oo,oo) S=II U/z
C2 --
. I
• f
]
•
II
X -- - C1
!z
8(x)
S=II u !z u 13 u /4 u /5 = I A( B)= sum of hatched areas with appropriate signs 9
---- .
1 - 4 --- '1�1 �:��:/1-"+!/E�Y:�/4! C3 -- -
cz - -
C1
--
I= [a, b)
-
-
---
b
C
--------+---+-------�--����--��
C5 -
a
-
-
-
-
-
-
-
-
-
-
_,
��e(
:�0:
- - -o
X
)
](
FIGURE 2.9
of finite, nonzero real numbers such that B(x)
=
{ C0,j,
� f x E Ij , j
=
IfXEI - S.
1, 2, ... , n,
In other words, 8 is constant and nonzero on each interval Ij, and zero elsewhere in I. The set S on which e is nonzero is called the support of e. Note that S may be empty, so that the zero function on I is also a step function. Figure 2.9 illustrates some possible step-function configurations.
26
2.
Some Analytic Preliminaries
If the support of a step function e has finite total length, then we associate with e the area A(B) between the graph of e and the x-axis, with the usual convention that areas below the x-axis have negative sign (we often refer to A(B) as the 11area under the graph" of 8). Thus 4(8) exists for the step function e in Figure 2.9-2, but not for that in Figure 2.9-1 . If e1 , Bz , em are step functions on the same interval I all with supports of finite total length, and if a1 , a2 , , am are finite real numbers, then the function e defined by . . . I
I
•
•
•
B(x) = L OJBj (x) j=:;l m
for x E I is also a step function on I. The support of e has finite length, and
A(B) = l: OJA(Bj)· j::�l m
The fact that e is also a step function is a rather tedious and messy thing to prove in detail, but an example should be sufficient to indicate why it is true.
B
Example 2 -5-1 : Let 81 , z : [0, 3) � JR. be defined by f)
1
(X)
=
{ 21 , I
if 0 < X < 2, if 2 < x < 3 -
I
fJz (X)
=
(cf. Figure 2. 1 0) . Let e = 281 - Bz . Then B(x) =
{
{ -1 , 1,
if O < x < 1 1 : if 1 < x < Z, 3, if 2 < X < 3.
if O < X < 1 , if 1 < X < 3
3
(cf. Figure 2. 11 ) . Clearly, e is a step function. Note also that A(81) = 2(1 ) + 1(2) = 4, A(82) = - 1(1 ) + 2(1) = 1,
2.5.
Step Functions
27
2lh (x) - fh(x) 3
I I I I I I
2
C? l
9
0
1
2
A(8)
If f, g
1R
I I I I I I
1
FIGURE
as expected.
•
9
3
X
2 . 11
=
1( 3) + 1( 1) + 1( 3)
=
�A(81) - A (8z),
=
7
are such that f(x.) < g (x.) for all x. E I, we write simply 11[ < g on I:' The following properties of areas under graphs of step functions are geometrically obvious and straightforward to prove: :
I
--*
28
2 . Some Analytic Preliminaties
0 on I, and the support of () has finite total length , then A (B) > 0 . Also , A (O) 0 .
(i) If () >
=
(ii) If 81 and 82 both have supports of finite total length, and 81 < ()2 on I, then A(B1) < A(B2). Exercise s 2-5:
{
1 . L et 81 , 82 : ()1 (X)
=
[0, 3] � JR. be defined by if 0 < X < 1 1 I
-1
4,
I
I
if 1 < X < 2 if 2 < X < 3,
()2 (x)
I
=
{ 3,
2
I
if O < X < 1 if 1 < X < 3. -
-
I
Sketch the graphs of 81 , 82 , and 81 - 2 ()2 , and verify by direct calculation that A(B1 - 2 82) = A(81 ) - 2A(B2)·
2. Let 81 , 82 : JR. � JR. be defined by 81 (x)
=
{
0
1:
0,
if x < - 1 if -l < < 2, if x > 2,
;
82 (x)
=
{
0,
if x < 0, - 1 , if O < X < 0, if X > 3.
3,
Sketch the graphs of 81 , 82 , and 81 + 82 , and verify by direct calculation that A (B1 + 82) = A(B1) + A(B2) .
2.6
Positive and Negative Parts of a Function
Let I be any interval. For any function f : I � JR. we define the functions f+ : I � JR. and f- : I � JR., called the posjtive part and the negative part of f, respectively, as follows: f + (x)
=
max{f(x), 0} for all x
f - (x)
=
min{f(x) , 0}
We also define the function If I
:
lf l (x)
I� :t
E I, for all x E I.
JR. by
lf(x) l ,
for all x E I. These definitions are depicted graphically in Figure 2 . 1 2 . It is clear that for any function [ : I � JR., we have f = [+ + f -
2. 7. Bounded Variation and Absolute Continuity
f(x)
29
!fl(x)
Graph of !fl
J+"(x)
j-(x)
Graph off+"
Graph ofjFIGuRE
and lf l on ! .
=
2 .12
f+ - f - . It is also clear that 0 < f+ < lf l and - lf l < f - <
Exercises 2-6: If f, g I :
--+
0
JR., prove the following inequalities:
1 . If+ - g+ I < If - g l on I. 2 . If - - g - 1 < If - g l on ! .
3.
l l f l - l g l l < If - g l on I.
2.7
Bounded Variation and Absolute Continuity
For any (nonempty) interval I, a partial subdivision of I is a collection S = {h , h , . . . , In } of cltysed intervals such that:
30 (i) h
2.
U
Some Analytic Prel iminatties
Iz
U
·
·
·
U
In
C
I;
(ii) for any j, k = 1 , 2, . .. , n with j '# k, either h n Ij is empty or h n Ij consists of a single paint that is an endpoint of both Ij and Ik .
For example, if I = [ 0 , 3), then S = {[0, 1 ], [1 , � ], [2, � ]} is a partial subdivision of I. Let f : I --+ JR. b e a function, and let S = {h , Iz , . . . , In } be a partial subdivision of I. For each j = 1 , 2 , . . . , n, let Ij have endpoints � ' bi . We can associate with f , I, and S the quantity Vs (f, I) defined by
n
Vs (f, I) =
L lf (bi) - f (�) l . j=l
Consider now the set A (f, I) = { Vs(f, I) : S is a partial subdivision of I } . Obviously Vs (f, I) cannot be negative, so 0 is a lower bound of A(f, I) . The least upper bound of A(f, I) is called the total variation off over I , and denoted by V Cf, I); and we have 0 < V (f, I) < oo for any f and I . Example 2-7-1 : Let f : I --+ JR. be any step function. If f is constant on I, then evidently V(f, I) = 0. If not , then as x. increases through I, f(x.) has a finite number of changes in value. Let the absolute magnitudes of these changes be k1 , k2 , . . . , km . Now take any closed interval Ij = [ � , bi] I. If none of the changes in the value off(x.) occur within Ij , then f(x.) is constant on Ij and lf(bj) - f(aj) l = 0. If the changes numbered r1 , r2 , . . . , rp occur within Ij , then lf(bj) - f(� ) l < kr� · I f S is a partial subdivision of 1 , then since a given change in the value off(x.) can occur within at most one of the intervals h , Iz , . . . , In kr . Furthermore, if we that make up S, it follows that V8 (f, I) < choose S such that each change tin the value of f(x.) occurs within one of the intervals comprising S , and no interval has more than one change occurring within it , then V8 (f, I) = I kr . It follows that f has finite total variation given by
c
I:f=l
L�=l
I: �=
2 . 7. Bounded Vari�tion and Absolute Continuity
31
8(x) 3 2 1
0 -1
' I I I I I I
kz kt
..... 1 9
i:
k3
1
• I I
3 1
2
k4 X
I I I I
I I
0
FIGURE
0
2 .13
I: ;.n=l
where kr is the sum of absolute values of all changes in the value of f (x) . For instance, let 8 : [0, 4) -+ JR. be defined by
1, 3 81 (x) =
21
3, 1, -1
I
if O if 1 if x if 2 if 3
<x < <X < = 2, < X< < X <
1, 2, 3, 4.
(cf. Figure 2 . 1 3) . Numbering the changes in value of 8 (x) from left to right, their absolute magnitudes are k1 = � , k2 = � , k3 = 2, k4 = 2, respectively, and the sum of the absolute magnitudes of all kr t:: 6 . the changes is therefore Let s = { [ � , � ] , [ � , 2 ], [ 2 , � ] , [ 3, !n . Then
I: ;=l
3 1 3 V8 (8 , [ 0, 4)) = 1 8( ) - 8 ( ) 1 + 1 8 ( 2) - 8 ( ) 1 2 2 2 . 7 5 + 1 8 ( ) - 8C2) 1 + 1 8 ( ) - 8 ( 3) 1 2 2
32
2 . Some Analytic Preliminaries
3
+ + + +2+2,
= 12 - 11 1
=2
3 13 - 2 1
11 - 31
3 2
+
1 - 1 - 11
2:;=1
Vs (B,
kr . Note that exactly one of the four and so [0, 4)) = changes in the value of B(x) occurs within each of the four intervals making up S.
Example 2 - 7-2 : Let f : (0, 1 )
1R
--+
2. 7.)
be defined by f(x) = sin(1 /x) for all x E (0, 1 ). (The graph of this function is depicted in Figure For each j = 1, let
2, . . . , n,
Then S
Lj+\rr ' ! J . . ,In } = + ( l2 )rr ) Vs (f. = t, (� ) ( 2 ) ((j +2 ) = = ( 2 ) ((j +2 ) = = Ij
{h , ]z ,
.
=
j
·
is a partial subdivision of (O, 1), and we have
(0, 1))
- sin (i
sin
Now, ifj is even, then
. jn . s1n - - s1n
1)n
1 0 - (±1) 1
1,
1 (± 1) - 0 1
1.
while if j is odd,
. jn . s1n - - s1n
1)n
Therefore,
n Vs (f, (0, 1 )) ::::: L 1 = n, j= l and so for S of this form we have V8 (f, (0, 1 )) --+ oo as n --+ oo. It follows that V(f, (0, 1 )) = oo . If V(f, I) is finite for a particular function f : I --+ JR, we say that f has bounded variation (or is a function of bounded variation) on I. Example 2-7-1 shows that all step functions on I have bounded variation on I, while Example 2-7-2 is an example of a function that does not have bounded variation .
2 . 7. Bounded Variation and Absolute Continuity
33
There is a very important connection between functions of bounded variation and monotone functions, which we must now discuss. Recall first that f : -+ 1R is on if f(xi) < f(xz) whenever xi < x2 (x1 1 x2 E and on if f(xi) > f(xz) whenever XI < x2 (xt , x2 E in either case we say that is on A very slight modification of the appropriate part of the proof of Theorem 2 . 4 . 1 shows that iff is monotone on ! , where ! is an interval with endpoints then f(t-) exist and are finite for all such that < < and and also and exist (but are n:ot necessarily finite). Furthermore, are both finite ,if and only if sup{f(x) : x E I} and and inf{f(x) : x E I} are both finite.
I
ing I
f monotone· I.
f (t+ ) f(a+) f(b-;) f(a+) f(b-) L emma
t
monotone increasing I I), monotone decreas I) ; a, b, a t b,
2 . 7. 1
Let I be an interval, and f : I -+ lR a function of bounded variation on I. For any x E I, denote by Ix the �nterval {t : t E I, t < x} c I. Then (i) 0 < V(f, Ix) < V(f, I) for all E I; (ii) the function g : I -+ defined by g(x) = V(f, Ix) for all x E I is monotone increasing on I. 1R
k
Proof
Part (i) follows at once from the result proved in Exercises 1-3, No . 3, and the fact that any partial subdivision of is also a partial subdivision of To prove part (ii), let XI , x2 E be such that X I < x2 . Then c so any partial subdivision of is also a partial subdivision of Thus < which proves that D g is monotone increasing on
I. Ix1 Ix2, Ix2•
V(f, IxJ V(f, Ix2),
Ix I Ix1
I. Theorem Let I be any interval. Then a function f : I -+ R bounded variation on I if and only iff can be expressed a difference f = hi - hz , where the functions hi, h2 : I -+ R are both monotone increasing on I, and sup{hi (x) : x E I}, inf{hi (x) : x E I}, sup{hz (x) : x E I}, inf{h2 (x) : x E I} are all finite. Proof We prove first that iff has bounded variation on I, then it can be represented by the diffe rence hi - h z , where the hk are as claimed 2 . 7.2
as
has
34
2.
Some Analytic Preliminaries
f I. x I, h1, h2 : I -+ hi(x) V(f, Ix) Ix hz x) V(f, Ix) -f x) x I. f = hi - h2 xi, Xz I x2 , hz(xz) - hz(XI ) = V(f, Ix2) - V(f, IxJ - [f(xz) -f(xi)]. (2.2) Now, if S is any partial subdivision of Ix1, then S* = S {[xi, x2 ]} is a partial subdivision of Ix2 , and Vs (f, IxJ + !f (xz) -f(xi) 'I = Vs• (f, Ixz) < V(f, Ix2). Thus, Vs (f, IxJ < V(f, Ix2) - lf (xz ) -f(xi) I for all partial subdivisions S of Ix1, and so V(f, IxJ = sup {Vs (f, IxJ S a partial subdivision of Ix1} < V(f, Ix2) - !f (xz) -- f(xi) I . Henc e V(f, Ix2) - V(f, IxJ > lf (xz) - f(xi ) I > f(xz ) - f(xi), and it follows from equation (2 .2) that h2 (x2 )- h2 (xi) > 0, so h2 is monotone increasing on I. Finally, it can be shown (see Exercises 2-7, No. 1) that M = sup{f(x) : x E I} and = inf{f(x.) : x E I} are finite, and so by virtue of Lemma 2.7. 1 (i) we have that for any x E I, < f(x) < M and 0 < V(f, Ix) < V(f, I), i.e. , - M < -f (x) < and 0 < V(f, Ix) < V(f, I), and therefore < V(f, Ix) - f(x) < V(f, I) w]lere -M and V(f, I) - are both i.e. , - M < h2 (x) < V(f, I) finite. It follows that sup{h2 (x) : x E I} and inf{h2 (x) : x E I} are both finite as required. It remains to show that if f = H 1 - h2 , where hi and hz satisfy the conditions prescribed in the theorem, then f has bounded variation on I. We leave the proof of this as an exercise. Corollary If I is an interval with endpoints b and f : I -+ has bounded variation on I, then f(t-) and f(t+) exist and are finite for all t such that a t b, and also f(a+) and f(b-) exist and are finite. Proof This follows at once from Theorem 2.7.2, the corresponding in the theorem. Suppose has bounded variation on For each E 1R by = define as in Lemma 2 . 7 . 1 . Define , and ( = ( for each E Then certainly and Lemma 2.7. 1 shows that hi h.ts all the required properties. Also, if E are such that XI < then U
:
m
m,
-M
m
-m
m
•
2 . 7. 3
a,
<
<
m,
D
1R
properties of monotone functions, and the basic rules for limits.
D
2. 7. Bounded Variation and Absolute Continuity
35
h 1 (x) 1 0
1
h2(x)
2
X
�
f(x) 1 0
1
1
2
X
X --0:-+----;-1�--+2-� FIGURE 2 . 1 4
Example 2 -7-3: Let f( ) h1 , : [0, 2] JR by
,.,f�Zx---x�2 on the interval [0, 2]. Define the functions hz --+ { v'1 2x - x2 , ifif 01 < xX << 12, h1 (x) , , if 0 < X < 1 , hz (x) { 0, 2 1 - v'2x - x if 1 x < 2. Then f h1 - hz o n I, and the conditions prescnbed in Theo rem 2 .7.2 are certainly satisfied by h1 and h2 , so f has bounded x
=
=
=
<
<
=
variation on
[0, 2 ] (cf. Figure 2 . 1 4) .
Note that the expression of a particular function of bounded variation as a difference of monotone increasing functions is by no means unique. For instance, just replacing and by + k and + k , where k is a constant, gives an infinite number of different expressions of this kind; there are other possibilities as well. There is no necessary connection between continuity of a func tion and the property of having bounded variation. Since all step
h2
h1
hz h1
36
2.
Some
Analytic
Preliminaries
)
functions have bounded variation (see Example 2-7-1 , functions of b ounded variation need not be continuous. Conversely, a continuous function need not have bounded variation. For instance, the function tan(x) is monotone increasing and continuous on ( - rr/ 2 , rr/2) , but since its set of values does not have finite upper and lower bounds, it does not have bounded variation on this interval (See Exercises 2-7 , No . 3 ) . Indeed, Example 2-7-2 shows that even if the set of val ues of a continuous function has finite upper and lower bounds, the function need not have bounded variation (see also Exercises 2-7 , No. 5) . We say that a function f : ---+ 1R is absolutely continuous o n if given any E > 0, there exists a > 0 (depending on E) such that V8 (f, I) < E for all partial subdivisions S of for which the sum of the lengths of all the constituent intervals is less than It follows easily, by considering only partial subdivisions consisting of a single interval, that if a function is absolutely continuous on it is also continuous on We have also the following theorem:
I
8
I
8. I,
I.
Theorem 2 . 7.4 If is
I an interval with endpoints a, b and f I 1R is absolutely continuous on I, then f has bounded variation on I. Proof Let S = {h , Iz , . . . , In} be any partial subdivision of I such that Ij [� , bj] for j = 1 , 2 , . . . , n. Let E = 1 in the definition of absolute 0 such that Vs• (f, I) < 1 for all continuity. Then there exists a 81 partial subdivisions S* of I for which the sum of the lengths of all the constituent intervals is less than 81. Let N be the smallest positive integer greater than (b - a)/81. Then 1 /N 81/(b - a). Now take anyj = 1 , 2, . . . , n. D i e j into N subintervals o f equal length, Ij1 = [aj (= Xjo), Xjl] , Ijz = [xj1, Xjz ] , . . . , IjN = [Xj N 1 , bj (= XjN)] , a nd denote the length of Ijr by ljr (j = 1 , 2 , . . . , n; = 1 , 2 , . . . , N) . Then ljr = bj N- aj 8 1(bbj--a�) For each = 1 , 2 , . . . , N , let Sr = {I1 r, Iz r, . . . , Inrl· Then Sr is a partial subdivision of I, and the sum of the lengths of all its constituent finite
:
---+
=
>
vid i
<
( - ) r
<
r
.
2 . 7. Bounded Variation and Absolute Continuity
37
intervals is
� ljr n
LJ= I (bj - aj)
<
b a � (bj - aj) < 8 1, 81
n
_
since is the sum of the lengths of all the intervals that make up the partial subdivision S of I, and is therefore no greater than the length of I. Therefore, I) < 1 for each r 1 , 2, . , N. Now n
b-a
.
Vsr (f,
=
.
Vs (f, I) L= l lf (bj) -f(aj) l jn = L lf (xjN ) - [ (Xj(N-I ) ) + f(xj(N - I ) ) - · · · + f(xj2 ) - f (xjl ) j= l =
+ n
<
![ (Xjr) - [ (Xj(r-I) ) ! L L j= l r=l
N
=
(by the triangle inequality)
n
![ (Xjr) - [ (Xj(r-l) ) ! L L r=l j=l
N
=
N
[ (Xji ) - [ (Xjo) l
N
(f, I) < L 1 = N. V sr L r= l r=l
Since N is finite and independent of S, it follows that has bounded variation on I.
f
V(f, I) < N and D
Note that the converse of this theorem does not hold; as we have seen earlier, a function ofbounded variation need not even be continuous, let alone absolutely continuous. Note also that we have seen examples of continuous functions on finite intervals that do not have bounded variation, and therefore (by Theorem 2. 7 . 4) are not absolutely continuous. Thus absolute conti nuity, as the name suggests, is a stronger condition than continuity, in the sense that the set of absolutely continuous functions on an interval is a proper subset of the set of continuous functions on that
38
2.
Some Analytic Preliminaries
interval. Some examples of functions that are absolutely continuous are given in Exercises 2-7, No . 6. Exercises
1.
2- 7:
Prove that iff : I --+ JR has bounded variation on i, then sup{f(x) : x E I} and inf{f(x) : x E I} are both finite.
2 . Prove that iff, g : I --+ 1R have bounded variation on I, then so do kf (where k is a real number), f + g, and fg.
3 . Prove that a function f : I --+ 1R that is monotone on I has bounded variation on I if and only if sup{f(x) : x E I} and inf{f(x) : x E I} are both finite. Use this, together with the results proyed in the preceding exercise, to prove part (b) of Theorem 2 . 7. 2 .
4 . Express the step function f) defined in Example 2-7-1 as a dif
ference of two monotone increasing functions, and sketch the graphs of the two functions.
5 . Let f : [ -rr/2, rr/2 ] --+
f(x)
1R be
=
{
defined by
�
x sin( ),
0,
�f x #- 0, lf X
= 0.
(a) Prove that f is continuous o n [ -rr/2, rr/2]; you may assume that x sin(l /x) is continuous for all x #- 0, so all that has to be proved is that f is continuous at x = 0. (b) Use a method similar to that used i n Example 2-7-2 to show that f does not have bounded variation on [ -rr/2, rr/2].
on I if there function [ : I --+ 1R is said to be a exists a real number L such that lf(x i ) - f(xz ) l < L lx 1 - Xz l for all Xt , Xz E I . (a) Prove that any Lipschitz function o n I is absolutely continuous on I.
6. A
Lipschitz function
(b) Use this result to show that any linear function is abso lutely continuous on any interval and that the function x 2 is absolutely continuous on any interval with finite endpoints. (c) Use a proo f by contradiction to show that x2 is not absolutely continuous on ( - oo , oo ) .
C H A P T E R
The Riemann Integral
The development of a rigorous theory of the definite integral in the nineteenth century is associated particularly with the work of Augustin-Louis Cauchy (1 789-1 857) in France, and Bernhard Rie mann (1 826- 1 866) in Germany. In order to give some background to the modern theory, we will descnbe briefly a definition of an inte gral equivalent to that introduced by Riemann in 1 854 , and discuss some of the weaknesses in this definition that suggest the need for a more general theory.
3.1
Definition of the Integral
1R
[a, b] be any closed interval. A function [ : [a, b] � is said to be over [a, b] if and only if for any number E 0, there exist step functions g€, G€ : [a, b] � such that (i) g€ < f < G€ ; Let
Riemann integrable
1R
>
A(G€ ) - A(g€ ) < E (cf. Figure 3.1). A function [ : I � 1R for which the set {f(x) (ii)
: x E I} has finite upper and lower bounds will be called bounded on I . 39
40
3.
The Riemann Integral
f{x) hatched area ::; e b a
FIGURE 3 .1
From condition (i) it is clear that any Riemann integrable function over must be bounded on and that
[a, b]
[a, b]
1R
[a, b],
f on [a, b]} is finite. We call A (f) the off over [a, b], and it is usually denoted by J: f(x) dx. Note that if a function f : [a, b] is Riemann integrable, then condition (ii) implies that A(f) inf{A (G) : G : [a, b] is a step function and G f on [a, b]}.
A(f)
=
sup {A (g) : g :
�
is a step function and g <
Riemann integral
�
=
�
1R
1R
>
Riemann's definition extends the concept of the 11area under the graph" off to a wider class of functions than step functions by using step functions to approximate This is essentially the definition of the integral that is used in elementary calculus, and its properties are familiar to all students of the calculus. The main focus of our study will be the Lebesgue .: Stieltjes inte gral, which is a generalization of the Riemann integral. The Riemann integral, however, is still important, because many calculations involving the Lebesgue-Stieltjes integral involve the Riemann in tegral. Moreover, one must understand some basic facts about the Riemann integral in order to understand the relationship between the two integrals and appreciate the need for a more general theory. We will not embark on a detailed account of the Riemann integral. Instead, we will limit ourselves to a brief discussion of the integral,
f.
Definition of the Integral
3.1.
41
highlighting any properties that are of use later in establishing the relationship between the Riemann and Lebesgue-Stieltjes integrals. Given a bounded function f : I -+ JR, candidates for the step func tions g€, G€ in the above definition can be constucted by partitioning the interval and defining step functions based on the maximum and minimum values the functions assume in the subintervals. By a par tition P of an interval I = [a, b] we mean a finite set of numbers XQ , Xt , · · · , Xn , where •
a = Xo < XI < ·
·
·
< Xn =
b.
Let It = [x0, xl ], and for 1 < k < n let h = (xk - l , xk] denote the kth subinterval of I associated with the partition P. Let Ilk = Xk - Xk - I denote the length of the subinterval. I f f : I -+ R is bounded on I, then given any partition P of I, step functions gp , Gp : I -+ JR such that gp < f < Gp on I can readily be constructed. Let
Mk = sup f (x), mk X
Elk
= inf f(x), X
Elk
and define gp , Gp as follows:
gp (X) =
mt , mz ,
if X if x
E
E
]z ,
mn ,
if x
E
In,
It , Gp (x) =
E
It ,
if X if x
E
]z ,
Mn , if x
E
In
Mt , Mz ,
(3.1)
L� 1 Mktlk and Sp (f) (cf. Figure 3.2). Let Sp (f) = A ( Gp ) A (gp ) = L�=I mktlk; evidently, Sp(f) > §.p (f) for any partition P of I. A partition P' of I is called a refinement of the partition P of I if every xk in P corresponds to some xj in P'. Thus a refinement P' of P can be constructed from P by distnbuting additional partition points between those already occurring in P. -
Lemma
3. 1 . 1 [a, b] -+ JR is bounded on I =
Iff : for any refinement P' of P, and
[a, b] and ifP is a partition of I, then
42
3.
The Riemann Inte gral f(x )
A
A (gp) >- x j(x )
A
A (Gp) >- x FIGURE 3 . 2
Lemma 3 . 1 .2 f :
� is
Suppose that [a, b] --+ bounded on I = [a, b] and that P and P' are any two partitions ofI. Th en Sp (f) > Sp, (f) . •
[a, b]. a,
Let IT denote the set of all partitions of I = From Lemmas 3 . 1 . 1 and 3 . 1 .2 the set S = {Sp(f) : P E IT } is bounded below by x1 = The Sr if) , where I is the partition corresponding to Xo = set S must therefore have a finite lower bound if f is bounded on�I . S imilarly, if f is bounded on I, then the set S = {Sp(f) : P E IT } must have a finite upper bound, because it is bounded above by Sr(f) . The quantitie s infPen Sp(f) and supPen Sp(f) are finite and are called the upper and lower Riemann-Darboux integrals of f over I, respec tively. If, in addition, it is assumed that f is Riemann integrable over
b.
3 . 1 . Definition of the Integral
43
I, then it can be shown that A (f) = inf Sp(f) = sup Sp (f). Sp (f) = sup Sp(f) is commonly used Indeed, the condition inf in the definition of a Riemann integrable function. Iff : [a, b] � is Riemann integrable on I = [a, b], then there is a sequence of partitions {Pj}, Pj E IT , such that limr�oo S� (f) = A(f) and a sequence {:5c}, P k E IT , such that limk�oo S� (f) = A(f). Now, Pen
Pe n
Pe n
1R
Pen
for any two pa}iitions P and P', the set PUP' yields a partition Q that is the common refinement of P and P'. Lemma 3 . 1 . 1 and the definitions of infimum and supremum thus indicate that we can always find a limk�oo SF\ (f) and sequence {Pk} such that limk�oo SF\ moreover, we can assume that Pk+ l is a refinement of Pk , k 1 , 2 , . . . .
= A(f), =
(f) =
Theorem 3 . 1 . 3 1R
Iff : [a, b] � is Riemann integrable over I = [a, b], then there exists a sequence ofpartitions {Pk}, Pk IT, such that Pk+ l is a refinement of Pk, k = 1 , 2, . . . , and E
=
Let P {x0, x1 , . . . , Xn} be a partition of the interval norm of P, denoted by IIP II , is defined as IIP II
I = [a , b]. The
= k=l,2, max � k · ... , n
The norm of P is thus the maximum of all the lengths of all the subintervals formed by the partition P. It can be shown that 1R is Riemann integrable over then any sequence of partitions P { j } such that IIPj ll 0 as j oo will produce the Riemann integral of over i. e. ,
I� f
I,
�
Jim SP. (f) 1�oo 1
�
iff :
I,
S (f) = 1 b f = 1Jim � 00 i>
1
a
dx .
In general, it is not particularly convenient to prove that a given function is Riemann integrable directly from the definition. The following theorems are thus useful in this regard:
44
3.
The Riemann lnte gra.J.
Theorem 3 . 1 .4 JR
Iff : [a, b] � is monotone on I = [a, b], then it is Riemann integrable over I. 3.1.5 Iff : [a, b] � is continuous on I = [a, b], then it Riemann integrable over I. 3-1 : 1. Let f : [a, b] � be a bounded function on I = [a, b] and let be a refinement of the partition P of I. Prove that gp < gp' and Gp > GP' , where the step functions g and G are as defined in equation (3.1 ) . 2. Use Lemma 3.1.1 and the fact that Q = P U P' is a com mon refinement for any two partitions P and P' of I to prove Lemma 3.1.2. Theorem
JR
Exercises
3 .2
is
p'
JR
Improper Integrals
3-1
closed
is over inter The Riemann integral as defined in Section vals. The defintion of the integral can be extended to other intervals by using a limiting process leading to the theory of what are usu ally called 11improper integrals." We eschew a detailed account of improper integrals; instead, we give a brief description of the basic idea with examples. Suppose that is a continuous function on the interval By Theorem the function is Riemann integrable over any interval of the form where < < and we can enquire dx. If this limit "is finite, then about the existence of limc� a+ feb we say that it defines the improper integral off from to The im proper integral is denoted in the same way as the Riemann integral, i.e., by J: dx . If an improper integral exists, we also say that it converges. Improper integrals over other intervals such as etc. are defined in a similar way.
f 3.1.5
[c, b],
f
(a, b].
a c b, f(x)
a b.
f(x) [a, oo), (-oo, b],
Example 3-2 - 1 :
[a, b),
f(x) = 1/
x
1].
The function[ defined by Jx is continuous for all E (0, By Theorem is Riemann integrable in any closed subset of
3.1.5 f
3.2 .
(0,
Improper Integrals
45
1]. In fact, for any 0 < c < 1 ,
[ f(x) = [z.JX]� = 2(1 - .JC). Now, limc� o+ .fc1 f(x) dx = 2(1 - limc� o+ ,JC) = 2, and therefore the improper integral J01 f(x) dx exists. dx
E�ple 3-2 -2 :
The function f defined by f(x) = 1 /x2 does not have an improper integral from 0 to 1 . The function f is Riemann integrable over any interval [c, 1 ], 0 < c < 1 , because it is continuous there, but
f1
1c 1 �dx = [-�] c1 � - 1 , X
X
=
and thus limc� o+ c f(x) dx is not finite. Example 3-2 -3:
C
Let f be the function defined in Example 3-2-2 and consider the in terval [1 , ). The function f is Riemann integrable over any interval of the form [1 , c] , where 1 < c < and since c c lim 2_ dx = c�oo lim = c�oo lim 1 = 1, c�oo x2 c
oo
l1
oo,
the improper integral E�ple 3-2 -4:
f100
[ �J 1 ( �) 1/x2dx converges X
.
Let f : [1 , � JR be defined by f(x) = 1 /x. In any closed interval [1 , c] , c > 1 , we have that c1 - dx = [ log x ]� = log c.
oo)
oo
l1
X
Since log c � as c � not exist, i.e., it diverges.
oo,
the improper integral
f100 1 /x
dx
does
Although the definition of the Riemann integral can be extended to open or semiopen intervals, many of the results concerning the Riemann integral over a closed interval do not carry over in the extension. Example 3-2-2 indicates that continuity on a semiopen interval does not guarantee the existence of the improper integral.
46
3. The Riemann Inte gral
Examples 3-2-2 and 3-2-4 show that monotonicity does not imply Riemann integrability when the interval is not closed. If f [a, b] � JR. is Riemann integrable over [a, b], then it can be s:I:own that lf l is also Riemann integrable over [a, b]. For im dx. proper integrals, this is no longer true, i.e. , J: f(x) may converge but J: lf(x) l dx. may diverge. If J: f(x) dx. and J: lf(x)l dx. both con ver�e, then the improper integral is dx. called absolutely convergent. If f(x) dx. converges but J: lf(x)l diverges, then the improper integral is called conditionally convergent. The integral in Exam ple 3-2-3 is absolutely convergent. The next example requires more familiarity with improper integrals than assumed heretofore, but it provides a specific example of a conditionally convergent integral. :
fa
E �ple 3-2-5:
Letf : [rr, oo) � JR. be defined by f(x) = (sin x)/x. Over any closed in terval of the form [rr, c] , c > rr, the function [ is Riemann integrable, and (anticipating integration by parts) we have c sin x dx. c co� x cos x c dx..
l
7r
X
=
[-
X
]
7r
+l
7r
X
Now, l cos x/x2 1 < 1 /x2 for all x E [rr, c] , and it can be shown that limc�oo J: cos xlx2 dx. exists, since limc�oo J: 1/x2 dx. exists (the comparison test). On the other hand, it can be shown that limc�oo J: l sin xlx l dx. does not exist. The definition of the Riemann integral can thus be extended to intervals of integration other than closed intervals by using improper integrals. The modern approach, to be described in the next chapter, works with arbitrary intervals from the start, leading to a tidier the ory, but this is a relatively minor improvement. A more fundamental weakness of Riemann's approach is revealed in the next section.
3.3
A
Nonintegrable Function
Theorems 3. 1 .4 and 3.1 .5 indicate that the class of Riemann in tegrable functions is a large one. In fact, it can be proved that if f [a, b] � 1R is bounded on [a, b ] and the set of all points of dis:
3.3. A Nonintegrable Function
47
continuity of f in [a, b] is either empty, fi nite, or countably infinite, then f is Riemann integrable over [a, b]. However, if the set of points of discontinuity off is infinite but not countable, then f may not be Riemann integrable, as the following example illustrates. Example 3-3- 1 :
Let f : [ 0 , 1 ] --+
f Cx )
JR be
{ ._
defined by
1, 0,
if x is rational, x # 0, 1 , if x is irrational or x = 0 or x
=
1.
Suppose that f were Riemann integrable over [0, 1]. Then taking E = � in the definition of Riemann integrability, there must exist step functions g, G : [0, 1 ] --+ JR such that g < f < G on [0, 1] and A(G) - A(g) < ! · Now, we have seen that any interval of nonzero length contains infinitely many rational numbers and infinitely many irrational numbers. Thus we must have g(x) < 0 and G(x) > 1 for all but a finite number (possibly zero) of points x E [0, 1 ]. The values of g(x) and G(x) at a finite number of values of x do not affect the values of the areas A(g) and A( G), so we must have A (g) < 0 and A( G) > 1 . Thus A( G) - A(g) > 1 , which contradicts A ( G) - A(g) < ! , and so f cannot be Riemann integrable over [0, 1 ]. Note thatf is discontinuous point in the interval [0, 1 ] . at
every
The reader might rightly ask why we should b e concerned that this rather peculiar function does not have an integral in the Rie mann sense. The reason is connected with the following concern: Suppose the functions fn : [a, b] --+ lR are Riemann integrable over [a, b] for all n = 1 , 2, . . . , and fn --+ f on [a, b]. It is natural to hope that the property of integrability 11Carries over to the limit," so that we can be sure that f is also Riemann integrable over [a, b] (and that the integral of fn tends to the integral off as n tends to oo), but this may not be the case. This concern is important, because the solutions to many problems in the calculus such as differential equations are of ten obtained as the limit of a sequence of successive approximations. Unfortunately, there are sequences of functions that are Riemann integrable but that converge to a function that is not. It is necessary to find only one counterexample to destroy our hopes.
48
3. The Riemann Inte gral
We will now show that the nonintegrable function defined in the previous example is the limit of a sequence of Riemann inte grable functions. We know that the rational numbers in (0, 1 ) form a countably infinite set, so we can write the set of rationals in (0, 1 ) as { r 1 , r2 , r3 , . . . }. For each j = 1 , 2, . . . we subdivide [0, 1 ] into three subintervals: Ij1 = [0, rj), Ij2 = [rj, rj], and Ij3 = (rj, 1 ]. We then define ej : [0, 1 J � 1R by 8j(X) Now,
ej is
a
=
{
�
0 1, 0,
if X fX 1f x
I
=
step function for each j
E Ijl , E Ij2 , E Ij3 ·
1 , 2, . . . .
Define fn : [0, 1 ] �
by
fn
=
R
n
ei . L j l =
Now, each fn is also a step function, and is therefore Riemann integrable over [0, 1 ]. Furthermore, it is evident that fn (x)
=
{
0, 1,
� x is irrational or x E {0, 1 , rn+I , rn+2 . . . } , If X
E { ri , r2 , . . . rn} . I
Thus if x E [0, 1 ] is irrational or x = 0 or x = 1 , then fn(x) = 0 = f(x) for all n = 1 , 2 , . . . , and so fn (x) � f(x) as n � oo. If x E [0, 1 ] i s rational, say x = rN (N = 1 , 2, . . . ), then fn (x) = 1 = f(x) for all n > N, so again fn(x) � f(x) as n � oo. We have therefore established that the Riemann integrable sequence of functions fn converges to the nonintegrable function [ on [0, 1 ]. This example shows that integrability in the Riemann sense does not always carry over in the limit. It can be shown that it does under certain conditions, but these conditions are rather complicated. Be cause the modern theory (as we shall see) allows for a wider class of integrable functions, the conditions under which integrability car ries over to the limit are much simpler and easier to use. This is one of the most important ways in which the modern theory is an improvement over the older one.
C H A P T E R
The Lebesgue Stieltjes Integral
We now proceed to formulate the definition of the integral that we are going to study. It results from combining the ideas of two people. The French mathematician Henri Lebesgue (1 875- 1 941 ) , building on earlier work by Emile Borel (1 871 -1 956) on the measure of a set, succeeded in defining an integral (the Lebesgue integral) that ap plied to a wider class of functions than did the Riemann integral, and for which the convergence theorems were much simpler. The Dutch mathematician Thomas Stieltjes (1 856- 1 894) was responsible for the notion of integrating one function with respect to another function. His ideas were originally developed as an extension of the Riemann integral, known as the Riemann-Stieltjes integral. The subsequent combination of his ideas with the measure-theoretic approach of Lebesgue has resulted in a very powerful and flexible concept of integration.
4.1
The Measure of an Interval
Let a : 1R --+ 1R be a monotone increasing function, and let I be an interval with endpoints a, b. We define the a-measure of I , denoted 49
SO
4.
The Lebesgue-Stieltje s Integral
by JJ-a (I), as follows: JLa([a , b])
=
a (b+) - a (a -) ,
JLa((a, b])
=
a (b+) - a(a+) ,
JLa( [a, b))
=
a(b-) - a (a -) ,
and if a < b,
The 110pen interval" (a, a) is of course the empty set, and we define JLa ((a , a)) to be zero for any a E 1Re . The intervals (a, a] and [a, a) are also empty, but in those cases the fact that their a-measure is zero follows from the general definition, and need not be specified separately. It follows easily from Theorem 2 .4 . 1 (i) and Corollary 2.4.2(i) that JLa(I) > 0 for any interval I, and that if I and J are intervals with I c J, then JLa (I) < JLa U) . If a and b are finite, and a is continuous at both a and b, then a (b+ ) = a(b), and so we have a (a -) a (a+) a (a) and a(b - ) JJ-a(I) = a(b) - a (a) in all four cases. In particular, if a(x) = x for all x E JR, then JLa(I) = b - a is the ordinary length of the interval I . In general, the a-measure of an interval is just the change in the value of a over the interval in question; it can be thought of as a generalization of the notion of length. =
=
=
Example 4-1-1 :
Let a : 1R � 1R be defined by a(x)
=
0, x2 - 2x + 2,
3,
x + 2,
if x < 1 , if 1 < X < 2, if x 2 , if x > 2 =
(cf. Figure 4 . 1 ) . Then: JLa( [ 1 , 2 ] )
=
a (2+) - a ( 1 - )
=4-0
=
4,
JLa(( 1 , 2 ])
=
a (2+) - a ( 1 +)
=
=
3,
JLa ([ 1 , 2))
=
a( 2-) - a (1 - )
=2
=
2,
4-1 -
0
4.1.
4
The Measure of an Interval
a(x)
/
3
�IO j
FIGURE
51
• I
rl ?
4.1
>x
2
JLa(( 1 , 2)) = a(2-) - a(1 +) = 2 - 1
=
1,
JLa([2 , 3]) = a(3+) - a(2-) = 5 - 2 3, JLa((2, 3)) = a(3-) - a(2+) = 5 4 = 1 , JLa([2, 2]) = a(2+) - a(2-) = 4 - 2 = 2, JLa(( - 1 , 3)) = a(3-) - a( - 1 + ) = 5 - 0 = 5 , JLa ([ - 8 , � ]) = a( � + ) - a( - 8-) = 0 - 0 = 0 . =
-
It can b e seen from these examples that the a-measure of an interval takes account of a jump in the value of a at an endpoint if and only if that endpoint is included in the interval. Note also that it is the left- and right-hand limits of a at the endpoints that determine the measure, not the value of a at the enc;lpoints. Note finally that, as the following examples illustrate, an interval that has one or both endpoints infinite may have, but does not necessarily have, infinite measure :
JLa([2, oo)) = a(oo - ) - a(2- ) = oo - 2 = oo, JLa (( -oo, oo)) = a (oo-) - a (( -oo)+) = oo - 0 = oo , JLa (( -oo, 2]) = a(2+) - a (( -oo)+) = 4 - 0 = 4.
52
4. The Lebesgue-Stieltjes Integral
Exercises 4-1 :
{
1 . Let a : 1R -+ 1R be defined by =
a(x)
x, 1,
3 - e-x ,
if x < 0 , if x = 0 , if X > 0.
(a) Sketch the graph of a.
(b)
Find JLa((O, 1 )), JLa([O, 1 ]), JLa(( - 1 , 1 )), JLa([ O , 0]), JLa (( - oo , 1 )), JLa((O, oo) , JLa ([O, oo)) . 2. Let a : 1R -+ 1R be defined by
a(x)
0, 1, 4, 6,
=
if x if O if 1 if x
< < < >
0, X < 1, X < 2, 2.
(a) Sketch the graph of a.
(b) Find JLa( [- 1 , 2)) , JLa(( 1 , 00)) , JLa(( - oo , 4)) , JLa(( 0 , 2 ]) , JLa (( 1 /2 , 3/2 )) , JLa ([ 1 , 3]) , JLa (( 1 , 3 )) .
4.2
Probability Measures
A particularly important type of measure arises when the function a is a In this case, the variable x is referred to as a and for each real number X, the value a(X) is the probability that the random variable x has a value no greater than X:
probability distribution function. random variable,
•
a(X)
=
P(x < X) .
The corresponding a-measure is then called a probability measure, and has the property that for any interval I, JLa(I)
=
P(x
E
1) .
Any probability distribution function must necessarily satisfy the conditions a (( - oo) + ) = 0 and a( oo ) = 1 , and it follows from this
-
4.2.
Probability Measure s
53
a(x) 1
------� 0+----A�---------B �----� x FIGURE 4.2
that if JLa is a probability measure, then JLa(I) < 1 for any interval I. Example 4-2 -1 :
The uniform distribution on the interval [A , B] (A and A < B) is the probability distribution a defined by a(x)
=
{
0,
�=� ,
1,
if x < A , if A < x < if x > B
B
finite,
B,
(cf. Figure 4.2). Since a is continuous, we can say that if I is an interval with finite endpoints a, b, then JLa(I) = a(b) - a( a), so that if A < a < b < B, then b -A JLa(I) = B - A
-
a-A
B-A =
b-a B-A
=
length of I length of [A , Bf
Since the only changes in the value of a occur within the interval [A , B] , it follows that for any interval I, of I n [A , B] JLa(I) = length length of [A, B] · In this case JLa(I) can be interpreted as the probability that a random number generator, programmed to select a random number in the interval [A , B] , will in fact select a number in I. Example 4-2 -2 : A discrete distribution is a probability distribution that is con
stant except for jump discontinuities at a finite or countably infinite
54
4.
The Lebesgue-Stieltjes Integral
!
-1
lfx(x) oll/2
• I
r
> x
1
FIGURE 4 . 3
{
number of points. An example is the function a defined by a(x) =
0 , if x < - 1 , l2 1
1,
{
<x < 1 if - 1 if x > 1
I
(cf. Figure 4.3) . In this case we have J-L a (x) =
0,
}, 1,
,
if I contains neither 1 nor 1 if I contains 1 or - 1 but not both, if I contains both 1 and - 1 . -
This corresponds to a random variable x such that P(x = - 1 ) = P(x = 1) =
1 . 2
-
For example, x might be the outcome of tossing a coin if 11heads" is scored as 1 and "tails" as - 1 . Exercises 4-2 :
1 . If x is a random variable that can take only one value A (with probability 1 ) , what is the corresponding probability distribution function a? 2. If x is a random variable that can take exactly n values A1 , Az , . . . , An (where A1 < A2 < · · · < An) , each with a probability 1 In , what is the corresponding probability distribution function (f ?
4. 3.
4.3
A
Simple Sets
55
Simple Sets
simple set is a subset of lR that can be expressed as the union of a finite collection of disjoint intervals. If S is a simple set, say S = U� /i where !1 , h , . . . Jm are disjoint intervals, and if a : lR -+ lR is a monotone increasing function, then the a-measure ofS is defined by m J-La(S) = J-La(Ij) ·
L j= l
A
given simple set can, of course, be subdivided into disjoint in tervals in many different ways, but the value of its a-measure is independent of the way in which it is subdivided. Note also that (i)
J-La(S) > 0 for any simple set S;
(ii) if S and T are simple sets such that S
c
T, then J-La( S) < 1-La( T) .
Some other elementary properties of simple sets are explored in the exercises. Note finally that a simple set is said to be a-finite if it has finite a-measure. Exercises 4-3:
1 . It is true (though rather tedious and unenlightening to prove in general) that if S and T are simple sets, so are S U T, S n T, and S - T = {x : X E S and X e T} . Verify this for each of the following cases:
[1 , 3) U (4 , 8) , T = (2, 5 ] U ( 6, 7]; (b) S = (2, 3) U [5, 7] , T = [ 1 , 4] U [6, 8); (a) S =
(c) S = (1 , 2]
U [5, 6),
T = [2, 4] U (5, 7) .
2 . Prove that if S and T are disjoint simple sets, then J-La( S U T) = J-La(S) + J-La(T) for any monotone increasing function a : 1R -+ JR. Give examples to show that if S and T are not disjoint, then J-La(S U T) may or may not equal J-La(S) + J-La(T).
3 . Use what was proved in the preceding exercise to show that if S and T are simple sets such that T c S and T is a-finite, then J-La(S - T) = J-La(S) - J-La(T) for any monotone increasing function JR. (Note that T is required to be a-finite in order to avoid a : lR
-+
56
4.
The Lebesgue-Stieltjes Integral
having the meaningless expression oo - oo on the right-hand side.) 4 . Give examples to show that if T is not a subset of S, then JLa (S - T) may or may not equal JLa (S) - JLa (T).
4.4
Step Functions Revisited
Let a : lR � lR be a monotone increasing function. Let I be any interval, and let () : I � lR be a step function. It is clear from the relevant definitions that the support of () is a simple set. We say that () is a-summable if the support of () is a-finite. In that case we associate with () a real number A a(()) defined by
L Cj JLa (Ij) n
A a(()) =
j=1
using the notation introduced in Section 2-5 . If a(x) = x for all x E JR, so that JLa (Ij) is just the ordinary length of the interval Ii , then Aa (()) is just the area A(()) under the graph of e, as defined in Section 2.5. In general, A a (()) can be thought of as a generalized ��area" for which "lengths" along the x-axis are measured by a-measure rather than by ordinary length. Note that if the endpoints of I are both finite, then any step func tion () : I � lR is a-summable for all monotone increasing functions a : lR � JR. Example 4-4- 1 : Let a b e defined a s in Example 4-1 -1 , and let e1 defined by ()1 (X) = (cf. Figure 4 .4) . Then
{
-1, 2,
:•
if
0 < X < 1, if 1 < X < 3
JLa ([ O , 1 )) = a(1 - ) - a(O - ) = 0 - 0 = 0 , JLa ( [ 1 , 3 ]) = a (3 + ) - a(1 - ) = 5 - 0 = 5,
[ 0 , 3] � 1R be
4.4.
2
lh(x)
'
1
I I I I I I I
1
0
-1
2
Step Functions Revisited
' I
2
X
3
FIGURE 4.4
(h(x)
1
1
0
-1
I
I
•
2
X
3
FIGURE 4 . 5
and so
Aa(Bl)
=
( - 1 ) 0 + 2 (5)
=
10 .
Suppose we modify the definition of 81 very slightly, to give
Bz (x)
=
{
- 1 , if O < x < 1 , if 1 < x < 3 2,
(cf. Figure 4.5 ) . In this case JLa ([ 0, 1 ]) = a ( 1 + ) - a(O-) = 1 - 0 JLa (( 1 , 3]) = a (3 + ) - a ( 1 +) = 5 - 1 and so
Aa (Bz) = ( - 1) 1 + 2 (4) = 7.
= =
1, 4,
57
58
4.
The Lebesgue-Stieltjes Integral
O(x) 2 o------
------e l 0
-----+--��
x
FIGURE 4.6
Note that while the area A (8) under the graph is the same for these two functions, the values of Aa (81 ) and Aa (�) are different. This is because at the single point where 81 and 82 have different values, a has a discontinuity, and so the interval consisting of that single point has positive a-measure. Clearly, discontinuities in a complicate matters! E�ple 4-4-2 : Let a be the discrete distribution function defined in Example 4-2-2, and let 8 : lR � lR be defined by 8(x) = (cf. Figure 4.6 ) . Then
{ 1'
2,
if x < 0, if x > 0
+ + 1-La (( - 00 , 0]) = a(O ) - a(( -oo) ) = + 1-L a ((O, oo)) = a(oo-) - a ( O ) =
1
-
1 1 2 - 0 = 2' 1 1 2
=
2
,
and so
We conclude this section by listing a number of basic properties that are straightforward to prove and intuitively reasonable, so we will omit the proofs. (i) If 8 is a nonnegative a-summable step function, then Aa(8) > 0; also , Aa( O) = 0.
4.4.
Step Functions Revisited
59
a(x)
FIGURE 4 . 7
(ii) If 81 and 82 are a-summable step functions on the same interval I such that 81 < 82 on I, then Aa(81) < Aa(82)· (iii) If 8 is an a-summable step function, then so are 181, 8+, and 8-, and we have Aa(8) = Aa(8+) + Aa(8-) and Aa(l81) = Aa(8+) (iv)
Aa(8-). I f 81 , fh , . . , 8m are a-summable step functions on the same in terval I , a1 , a2, , am are finite real numbers, and 8 I --+ JR. is .
•
•
:
•
defined by
m 8(x) = L O-j8j(x) j= l for all X E I (i.e. , 8 = L:;, 1 aj8j), then 8 is also an a-summable step function on I, and
m Aa(8) = L ajAa(8j)· j= l Exercises 4-4: Let a be defined by
: JR. --+ JR.
a(x)
=
{�
x,
1,
if x < 0, if x > 0
(cf. Figure 4.7) . For each of the following step functions
8:
(a) sketch the graph of 8; (b) determine whether or not
Aa(8).
8 is a-summable,
and if it is, find
60 1.
4.
The Lebesgue-Stieltjes Integral
8 : ( -2, 1) -+ lR defined by O(x)
2.
=
=
-2 1,
8
:
( - ()() I 0)
8
:
'
< X < 0I if - 1 if O < x < l .
{ ::
<3 if - 1 < X if x > 3 .
=
{
if x < - 1 , if - 1 < X < 0.
8(x)
=
=
-
I
-+ lR defined by 8(x)
5.
{
if -2 < X < - 1 I 1 , if - 1 < x < 1 .
8 : [- 1 , oo) -+ lR defined by O(x)
4.
3,
8 : [- 1 , 1 ] -+ lR defined by 8(x)
3.
{
lR -+ lR defined by
O, 1,
{
-1 , 1,
if x < 0, if x > 0 .
Definition of the Inte gral
4.5
We are now in a position to set up the necessary machinery for defining the Lebesgue-Stieltjes integral. Throughout this section we to take I to be a given interval with endpoints b, and a : be a given monotone increasing function. on I. A se be a function that is : I Let quence is said to be admissible for if it satisfies all the following conditions:
f
-+ lR
81 , 82, 83,
(a)
(b) (c)
lR -+ lR a, nonnegative f
•
•
•
8j is an a-summable step function on I, for each j 8j (x) > 0 for each x I and each j 1 , 2, 3 , . . . ; 0 < f (x) < L:� 1 8j (x) for each x I.
1 , 2, 3, . . . ;
=
e
e
Theorem 4.5 . 1 admissible
An Proof
=
sequence exists for any nonnegative function f : I -+ JR.
There are two cases to consider.
61
4.5. Definition of the Inte gral
Case 1 . The endpoints
a, b of I are finite. In this case we define the
function ej : I � by 8j(X) 1 for all X E I (j 1 , 2, 3 . . . ) . Then ej is nonnegative, and since the endpoints of I are finite, ej is certainly a-summable for each j 1 , 2, 3 . . . . Since :L�1 8j(x) :L�1 1 oo , condition (c) above is also satisfied for any nonnegative function f:I� and so the sequence 81 , 82 , 83 , . . . is admissible for f.
JR
=
=
=
=
=
JR ,
Case 2 . Either a
=
=
- oo
or b oo or both. In this case we define the subinterval Ij of I as follows : If I is ( - oo, b), b finite, then Ij is (b - j, b) . If I is ( - oo b], b finite, then Ij is (b - j, b]. If I is (a, oo , a finite, then Ij is (a, a + J). If I is [a, oo , a finite, then Ij is [a, a + J). If I is ( - oo , oo), then Ij is ( -j,J). For each j 1 , 2, 3, . . . we then define ej : I � by
: ) )
=
ej(x)
=
{ 1,
0,
�f x E Ij ,
JR
lf x E I - Ij .
Then ej is nonnegative o n I and is a-summable, since its support Ij is an interval with finite endpoints. Further, for each x E I we have that ej (x) 1 for all sufficiently large values ofj, and so again 00 for each X E I. Thus the sequence el , e2 , e3 , . . . is l ej (X) D admissible for any nonnegative function f : I � =
L�
=
JR.
We associate with any nonnegative function f : I � extended real number defined by
La ([) La (f)
=
inf
JR an
{ f>a(llj) } , ]= l
where the greatest lower bound is taken over all sequences 81 , 82 , e3 , that are admissible for f. Since the set of all such sums :L�1 Aa (8j) is non-en1pty (by Theorem 4.5.1) and has 0 as a lower bound, it follows by Theorem 1 .3. 1 that (f) exists and (f) > 0 Note that (f) oo is a for any nonnegative function f : I � possibility. Note also that it follows easily from the definition that 0. •
.
•
La JR .
La(O)
La La
=
=
E�ple 4-5-1 :
JR
Let f : [0, 1 ] � be the function discussed in Section 3.3, and let the functions ej be as defined in that section. For each j 1 , 2 , 3, . . . , ej is =
62
4. The Lebesgue-Stieltjes Integral
�
a nonnegative a-summable step function on [0, 1 ], and 2:: 1 ej (x) = f (x) for each X E I , so the sequence el , Bz , e3 , . . is admissible for f. In particular, let a* be defined by a* (x) = x for all x E Then for each j = 1 , 2 , 3, . . . we have
.
JR..
A a* (Bj ) = JLa* ( [ rj , rj ] ) = rj - rj = 0,
and so Lj� l Aa* (Bj) = 0. It follows that La* (f) = 0. Note that although each ei is a step function, f = 2:: 1 ei is not a step function, since it is not possible to describe it by taking constant values on any finite set of subintervals of [0, 1 ].
�
Theorem 4.5 .2 we have: For any function f : I (i) La (f+ ) < La ( lf l) and La( -f-) < La ( lfl); (ii) La ( l af l) = l a i La( lf l) for any finite nonzero real number a (and for a = 0, provided that La( lf l) is finite).
--+ JR.
Proof (i) Clearly, any sequence fh , Bz , 83 , . . that is admissible for If I is also admissible for f+ and -f- . Thus the set of admissible sequences for If I is a subset of the set of admissible sequences for f+ and of the set of admissible sequences for -f- . Part (i) follows at D once from Exercises 1-3, No . 3 .
.
(ii) It is obvious that both sides are zero if a = 0 and La( lf l) is finite, so suppose a # 0 . Let 81 , Bz , 83 , . . be an admissible sequence for lf l , so that for all x E I,
.
0 < lf (x) l �
00
L Bj (x) . j=l
Then 0 < l af(x) l < l a l
for all x
E
L ej (X) = L l a i Bj (x) 00
00
j=l
j =l
I, and also
L Aa ( l a i Bj) = L l a iAa(Bj) = l a l L Aa(B;) , 00
00
00
j=l
j= l
j= l
and so l a l 81 , l a iBz , l a l 83 ,
. . . is an admissible sequence for l af l .
63
4 . 5 . Definition of the Integral
Conversely , if ()1 1 lh , is an admissible sequence for then is an a dmissible sequence for Thus, if S is the set of admissible sequences for Then from above and Exercises 1 -3, No . 4(a), we have
() 2 , . . , a f l l 1/ a b () ( i a 111 B2, 1l i a B3, ) l ( l ( ) ) l lf l . lf l . La (l af l ) � {�Aa(l al 9j) } � {t l a iAa(Oj)} l a l � { � Aa(Oj) } l a !La (lf l) . fLa(f) La (g) . JR. ar f(x) g(x) for all x ()2 , . f. g .
•
.
=i f
,
.
=i f
i f
=
=
Theorem 4 . 5 . 3 I f, g : I � <
D
e such that 0 <
<
e
Proof This result follows at once, since any sequence ()1 , that is admissible for is also admissible for
Theorem 4.5.4 Iffi , fz , f3 , . . . is a sequence of functions such that : I � each e I, then converges j = 1 , 2, 3, . . and
fj L:j�\ fj (x) for x La ( �fj ) � La( lfj l ) . L:� La ( lfj l ) fj La( ! Lf=l fj l ) LJ=l La(lfj l) .] L:� La ( lfj l ) L:� La(lfj l ) La(lfj l ) j LaBj2(l, fj l), lfj L Aa (Bjm ) La(lfj l ) m= l
I, then ()3 , . .
D
JR. for each
.
<
[Note that is it possible to have = oo under the condi 1 tions of the theorem. Note also that when the sequence 1 fz , 3 . . . is such that = 0 for all j > n, we obtain the important special case <
f, f,
= oo, so assume Proof The result is evidently true if 1 that is finite for each is finite. Then certainly 1 j = 1 , 2, 3, . . . . Thke any E > 0. By part (ii) of Theorem 1 .3 .2 and the definition of we have that for each = 1 , 2, 3, . . . there is a admissible for I and such that sequence ()j1 , ()j3 , •
•
•
00
<
+ 2 -j € .
64
4. The Lebesgue-Stieltjes Integral
It follows from the properties of double series of positive terms that (Section
2-2)
Lj=l (La(ljj l) 2-jE) = � La(l}j i) + E � ( 21 )j = � La(l}j l) ( _1 /1122 ) = Lj=l La (ljj l) E. jj x)l L� l x l = 1, 2, tjj x) t ljj (x)l t (� Ojm(x)) = J;, Ojm(x). = 1, 2, ) <
00
+
00
But since 0 < ( 3 , . . . , we have
0<
(
<
00
00
+E
00
+
<
l
E
Bjm (x) for all
I and each j
<
I f we write the double sequence ()jm (j , m 3 , . . . as a sin gle sequence (in any way we choose) , the result is a sequence 1jf3 , . . that is admissible for I jj I , and so
.
1/f1 , 1/fz ,
:L�1
La ( t(fj) ) � Aa('l/r;) = j�t Aa O ) t La (l}jl) E; La ( :L�1 (fj)l) - :L�1 La (ljj l) E E <
thus, that
I
(
<
jm
for any
<
>
.0 ,
+
and it follows
i.e. ,
as required.
0
65
4. 5 . Definition of the Integral
Theorem 4 . 5 . 5
If f1 , fz , f3 , . . . is a sequence of functions such that fn : I and D is finite for each n = 1 , 3, . . ., and if f : I is such La(l f n that La (lf -fnl ) 0 (as n then: (i) La(lf l ) , La(f+) and La( -f-) are all finite; (ii) La (If+ -f: I) 0, La( If- -f;- 1) 0 and La C IIf l - lfn II ) 0; (iii) La(lfnl) La(lf i ) ,La(fn+) La(f+) and.La(-fn- ) La(-f- ). there certainly exists a positive Proof (i) Since La(lf - fnl ) integer such that La (If -fN I ) is finite. Then La (If I ) La(lfN + f -fN I) < La(lfN I ) + La(lf -fN 1) , by Theorem 4.5 .4, and so La(lf l ) is finite, since La(lfN I ) and La(lf finite. The fact that La (f + ) and La(-f- ) are finite follows fatN Ionce ) are both from Theorem 4.5 2 (i) . (ii) From Exercises 2-6 we have that on I , It+ -f:l < If -fn L If- -f;l < If -fn L and llf l - lfnll < If -fnl · 0 by hypothesis, part (ii) follows at once using Since La(lf -fn D Theorem 4.5.3. � oo)
�
N
2,
�
�
� lR � lR
�
� 0,
�
�
�
•
=
.
�
(iii) We have, using Theorem 4.5.4:
LaCifnl ) La(lf + fn - fl) < La(lf l ) + La(lfn -fl ) (4 . 1 ) La(lf l ) + La(lf -fnl ) . Also, La(lf l ) = La( If -fn + fnl) < La( If -fnl ) + La( Ifni ) , and so (4 . 2 ) La(lfnl) > La(lf l ) - La(lf -fnl ) . From equations ( 4 . 1 ) and ( 4 . 2 ) we have La(lf l ) - La(lf -fnl ) < La(lfnl ) < La(lf l ) + La( If -fn l ) , I.e. , -La( If -fnl ) < La( Ifni ) -La(lf l ) < La( If -fnl ) , =
=
66
4 . The Lebesgue- Stieltjes Integral
i.e. ,
ILa(lfn l ) - La(lf l ) l < La( If -fnl ) . Since La(lf - fnl ) --+ 0 by hypothesis, it follows that La(lfnl ) --+ La(lf l ) . We have also fn+ f+ (f: -f+) and f+ (f+ -f:) fn+ , -fn- -f- ( -fn- - ( -f-)) and -f- (-f- - ( -fn-)) (-fn-) , and since La(lf+ - fn+ l ) --+ 0 and La( I - f- - ( -fn- ) 1 ) --+ 0 by part (ii), the rest of part (iii) can be proved by an argument similar to the =
+
=
preceding one.
+
+
=
=
+
o
--+
Now let a : JR. JR. be a monotone increasing function, and let f :I JR. be a function with the property that there is a sequence fh , fJ2 , fJ3 , . of a-summable step functions defined on I such that 0. It follows from part (i) of Theorem 4.5.5 are all finite. Under these conditions we write and
--+ La( If - fJn l ) --+ La (f+) La( -f-) .
that La(lf l) ,
.
Jt La(f"") - LaC-n da
=
f f f
and call this quantity the (Lebesgue-Stieltjes) integral of over I with respect to a. If a(x) = x for all x E JR., we write h dx instead of this special case is called the Lebesgue integral of over I. h It is important to be sure that h = A a (fJ) for any a-summable step function () : I JR., because only then can the integral justifi ably be regarded as an extension of the concept of the 11area under a graph" as defined for step functions. This is, in fact, true, but the proof is surprisingly hard, so we shall just state the result without proof:
f da;
--+
fJda
Theorem 4 . 5 . 6
For any a-summable step function () : I
--+ JR., we have h ()
da
a( ).
= A fJ
Exercises 4-5:
1 . Complete the proof of part (iii) of Theorem 4.5.5 by showing that and 2.
La (fn+) --+ La (f+) La( -fn-) --+ La( -f-). Let f be the function discussed in Section 3 . 3 and Example 4-5-1 , and let a* be defined by a* (x) for all JR.. Since f+ f and f- 0 , and we have shown in Example 4-5-1 that La*(f) 0, =X
=
X
E
=
=
4.6. The Lebesgue Integral
•f(f+ ) - La• ( La �0•1
67
it follows that -f-) = 0. Complete the proof of dx = 0, by showing that there is a sequence the fact that . . . of a* -summable step functions on [0 1 ] such that I) --+ 0. (This example shows that J; f dx may exist in the (If Lebesgue-Stieltjes sense in some cases where f is not Riemann integrable over I) .
el , Bz , e3 , La* -en
1
I
e I JR. is a nonnegative a-summable step function, then Aa(B) La(B).
3. Use Theorem 4.5.6 to show that if
:
--+
=
4.6
The Lebesgue Integral
The mathematical machinery required to define the Lebesgue Stieltjes integral is notably more complex than that needed for the Riemann integral. Here, we pause to discuss informally the defini tion of the Lebesgue integral and contrast it with that of the Riemann integral. Let I be an interval and f : I --+ JR. be some function, which for simplicity we assume to be nonnegative. Iff is Lebesgue integrable < oo and over I, then
Lx (f)
Recall that
1
Lx (f)
f dx
=
Lx (f). 00
=
inf
Bj LAj ( ), O = j
{ei} {Bj}
where the greatest lower bound is taken over all the sequences that are admissible for f. If { is an admissible sequence, then in < For a given admissible sequence particular, 0 < f is thus an upper bound for the 11area under the quantity plays a role in the Lebesgu e the graph of f:' and the quantity theory analogous to that of the upper Riemann-Darboux integral in that the Riemann theory. Suppose now we consider sequences satisfy conditions (a) and (b) for an admissible sequence (see p. 60), but satisfy the inequality
ei} L:�1 Bj (x) . (x) L�o A (Bj ) Lx (f)
{l/>j }
68 (c *)
4.
The Lebesgue-Stieltjes Integral
0 < :E� 1 A(¢j) < f(x)
for all x
E
I instead of condition (c). Let lx(f) = sup
L A (¢j) , 00
j=l
where the least upper bound is taken over all sequences satisfying conditions (a) , (b), and (c*). If lx(f) < oo , we could define another integral by
f
f dx = l,(f) . I
-
The quantity lx(f) is analogous to the lower Riemann-Darboux integral. Note that in the Riemann theory the upper and lower Riemann-Darboux integrals are always finite even if f is not Riemann integrable. This is because the definition of Riemann in tegrability is framed in terms of closed intervals on which f must be bounded. In the Lebesgue theory, no restrictions are placed on I, and f need not be bounded; consequently, neither lx (f) nor Lx (f) need be finite. If f is Riemann integrable, then condition (ii) of the definition requires that the upper and lower Riemann-Darboux integrals be equal. There is no analogue to this condition in the Lebesgue theory: It is not required that lx (f) = Lx(f) for the Lebesgue integral to exist, only that Lx (f) < oo. Prima facie, this may seem a weakness in the theory, as there is no particular reason to choose Lx(f) over lx (f) to define an integral, but it can be shown that the relationship lx(f) = Lx(f) is in fact a consequence of the condition Lx(f) < oo together with measurability of f (see p. 82) . In other words, if Lx(f) < oo and f is measurable, then lx(f) < oo and lx (f) = Lx(f) . A similar statement can be made if lx(f) < oo, and in this sense lx and Lx are always on the same mathematical 11footing" in the Lebesgue theory. The definitions of the Riemann and Lebesgue integrals diffe r fundamentally in the functions used to approximate the 11area under the graph," and it is this difference that affords greater generality for the Lebesgue integral. Recall that a function f is Riemann integrable on a closed interval I if for any E > 0 there are step functions gE , GE E. The definition of the GE and A (GE) - A(gE) such that gE f •
< <
<
4.6. The Lebesgue Integral
69
Lebesgue integral also uses step functions but in a different way. For example, if {Oj } is an admissible sequence, then the inequality in the Lebesgue definition analogous to the inequality < G€ in the Rie mann definition is < Although each f1 is a step function, the sum need not be a step function. A more general class of functions is thus allowed into the 11approximation." Coupled with this increased generality is the notion of measure, which is in itself a generalization of length. The generality is enough to make fu n ctions such as that descnbed in Section integrable under the Lebesgue definition ( cf.. Exercises No. The real value of the general ization, however, lies in results such as the dominated convergence theorem (Theorem which resolve some of the problems with the Riemann integral discussed at the end of Section The Lebesgue integral is a generalization of the Riemann integral. We state the following result without proof:
f
f Lj� 1 ej .
2:� 1 ej
3.3 2).
4-5, 5.3.3),
3.3.
If a function f : [a, b] JR. is Riemann integrable over the interval I = [a, b], then it is also Lebesgue integrable over I, and the two integrals are equal. Theorem 4.6 . 1
--+
The above result is used frequently to calculate Lebesgue in tegrals. Note that the generalization breaks down if the integral is improper. For example, the conditionally convergent integral in Ex ample is not Lebesgue integrable. (In fact, Theorem 5. 1 . 1 in the next chapter indicates that if a Lebesgue integral exists, it is al ways absolutely convergent.) If an improper integral is absolutely convergent, however, then it can be shown that it is also Lebesgue integrable and the integrals are equal.
3-2-5
Exercise 4-6 :
Let
f
:
[0, oo)
--+
f(x) = (Figure
4.8).
JR. be defined by
{ ��1
n'
ifn - 11 < x < n - � , n = 1, 2, . . . , if n - -2 x < n, n = 1, 2, . . . <
70
4.
The Lebesgue-Stieltj es Integral
f{x) C?
• I
0
d I I I I
-1
•
�O,c] f dx
I I I
•
• I I
12 I
0
C? I
•
.-13
0
X
0 FIGURE 4.8
f
for c a positive real number. (Since is a step (a) Evaluate function on [0 , c], this integral exists in both the Riemann and Lebesgue senses.)
f dx exists in the Riemann sense, as limc�oo J; fdx. limc�oo J; lf l dx does not exist; this shows that
(b) Show that J000
(c) Show that does not exist in the Lebesgue sense (cf. Theorem 5 . 1 . 1) .
�O, oo) f dx
C H A P T E R
Properties of the Integral
In this chapter we will examine some of the essential properties of the Lebesgue-Stieltjes integral, culminating in the convergence the orems that (as remarked in Chapter 3) are among the most important features of the Lebesgue-Stieltjes theory.
5.1
Basic Properties
Theorem 5 . 1 . 1
is integrable over I with respect to a, then so are f+, f-, and and we have
Iff : I � lf l ,
JR.
[r [r [r dct
=
dct
+
dct
aru1
[
lf r t1ct =
[r [r dct -
dct.
Proof
Since f is integrable over I with respect to a, there is a se quence 81 , Bz , 83 , . . of a-summable functions on I such that La ( lf Bn l ) � Since en is a-summable for each j = 3, so is IBn L and it follows from Exercises No. 3 , that La ( I Bn I ) is finite for each j = 1 , 3 , . . . . It then follows from part (ii) of Theorem that
0. 2,
.
4-5,
1, 2,
. . .
I
4.5.5
71
72
5 . Properties of the Integral
8: 1 ) � 0, La(lf - - 8; 1 ) � 0 and La(llf l - l 8nll ) � 0, so by l f + La( de finition f+ I f- I and If I are all integrable over I with respect to a . By definition, h f + da La (f + ) and h f - da -La( -f - ) I and so [t aa L (t+) L C n [r da + [r aa . =
= a
=
- a -
=
Also, we have by property (iii) (at the end of Section 5-4) that
l nl Thus La( l 8nl ) La (8: ) + La ( - 8;) by Exercises 4-5 , No . 3 . Letting n tend to we have by Theorem 4.5.5 part (iii) that La(lf l) LaCf+) + La(-r) if+ da - ir da. we have h lf l da B ut since If I + which proves the I , I , l f ) La(l f last part of the theorem. Aa ( 8 ) = Aa(8: ) - Aa(8; ) = Aa(8: ) + Aa( - 8; ) . =
oo,
=
=
=
=
o
Theorem
5 . 1 .2 = 1 , 2, 3, . . .
Iffor each n , fn : I � IR is integrable over I with respect to a, and if f : I � IR is such that La(lf - fn I ) � 0, then f is integrable over I with respect to a, and we have
[fn da ->- [t da, [tn+ aa _.. [r aa. [tn- aa [ aa, [ Ifn i da ->- [ lf l da. .....
r
n. La(lfn - 1/lj l ) � 0 j �
Proof
fn
Take any positive integer We know that is integrable, so by definition there is a sequence 1/11 , 1/12 , . . . of a-summable step as oo . Thus, for some functions such that 1/Jk ) < 1 / . Chpose such a k k = 1 , 2, 3, . . . we must have In this way we obtain a sequence 81 , 82 , . . . of and denote 1/Jk by a-summable step functions such that
8n.
La(lfn - l
La(lfn - 8n l ) < -n1 ,
n
Since for each = 1 , 2, . . . . Now take any E > there exists a positive integer N (E) such that E > N (E) =} < 2·
0.
n
n
La(lf -fnl )
(5 . 1 )
La(lf -fn l ) � 0, (5 .2)
5
.
I
.
Basic Properties
73
Also, from equation (5 . 1 ), we have that
n
>
2 E
::=}
1
n
<
E 2
::=} La Cifn - On ! )
By Theorem 4.5.4 we have that for each
Thus, if
n
>
<
E . 2
(5 .3)
n = 1 , 2, . . . ,
max{N (E), 2 / E} then equations (5 .2) and (5 . 3) imply that
La ( lf - Bnl )
<
E 2
E
+2
= E,
and so La ( If - On ! ) ---* 0 , and f is integrable over I with respect to a by definition. The rest of the theorem follows from Theorem 5 . 1 . 1 , Theo rem 4 . 5. 5 (iii) , and the definition of the integral (see Exercises 5-1 , o No . 1) . Theorem 5 . 1 . 3 Iff : I ---* lR
is integrable over I with respect to a, then
Proof
By definition,
thus,
1f da
by Theorem 5 . 1 . 1 . Theorem
5 . 1 .4
<
Laif" ) + La (-r)
= Jraa-Jraa = J lf l da ,
(Linearity of the Integral) 1 , 2, . . , m, I ---* lR
0
Iffor each} = fj : is integrable over I with respect to a, and � is a finite real number, then 2:}:1 �fj is integrable over I with .
74
5
.tTopernes
•
01 me
.m.u�grd.l
respect to a, and
,m
Proof For each j = 1 , 2, . . . we know that there exists a sequence Bj1, Bj2 , of a-summable step functions such that La ( lfj - Bin D � 0 as n � oo . By Theorem 4.5.6 we have that for each n = 1 , 2 , . . . , 1 (tJ=l tljOjn) da = Aa (tJ=l tljOjn) •
.
•
I
(5.4)
n = 1 , 2, . . . 0 < La ( t tljjj - t tljOjn ) = La ( t ai (jj - Ojn) ) J=l J=l J=l
Now for each
<
L ltiji La ( lfJ - Bjn l ) , m
j= l
by Theorems 4 . 5 .4 and 4.5.2(ii) . Since j = 1 , 2, . . , m , it follows that .
La
:L}= 1
La ( lfj - Bin D
( � aifi � ) -
tljOjn
-+
� 0 for each
0,
and so aifj is integrable over I with respect to a. by definition. as oo , It follows also, by Theorem 5 . 1 . 2 , that for each j = 1 , 2 , . . . , m , and that D Letting tend to oo in equation (5.4) gives the result.
h Bjn da � hfi da n � h CLi=1 ajBjn) da � h C:L}= 1 ajfj) da .
n
Theorem 5 . 1 . 5
Let f, g : I � lR be functions integrable over I with respect to a. (i) Iff > 0 on I, then h f da > 0. I, h g da < Ji f da . (ii) If g f <
on
then
5 . 2 . Null Functions and Null Sets
Proof Part (i) follows at once, since if f > 0 on I , then La (f) > 0 . For part (ii) we need only observe that
[
r aa =
[
rg + (f - g)J da =
by Theorem 5 . 1 .4. Part (i) implies that f da > g da .
h
h
Exercises
[
g aa +
[
75
h f da
=
rf - g) aa,
h (f -g) da > 0, and therefore 0
5-l :
1 . Complete the proof of Theorem 5 . 1 .2 . 1R
: I --+ we define the functions max if, g} : I --+ : I --+ 1R by (max{f, g})(x) = max {f(x), g(x)} for each x E I, (min{f, g}) (x) = min{f(x), g(x)} for each x E I.
2 . For any functions f, g 1R and min{{, g}
Prove that max{f, g} = f + (g - n+ and min{f, g} = f + (g - n- , and deduce that if f and g are both integrable over with respect to a, then so are max{f, g} and min{f, g } .
I
1R
3. Prove the 11first mean value theorem for integrals": If f : I --+ is integrable over I with respect to a, and if J-La (I) is finite, and if c 1 and c2 are finite real numbers such that c1 < f < c2 on I, then C! J! a(I) <
i
f da < C2 J!a(I) .
4 . Prove that if for each j = 1 , 2, . . . , m the function fj integrable over with respect to a, then
I
) fi ( r t r (tjj ) ;= 1 ]= 1
]I 5 .2
aa <
]I
, , aa
: I --+ 1R is
.
Null Functions and Null Sets
If f : I --+ 1R is such that La(lf l)
respect to a ) .
=
0,
we call f a null function (with
76
5.
Properties
of the Integral
Theorem 5 .2 . 1 I f
f : I --+ IR is a null function with respect to a, then f is integrable over I with respect to a and 1fda = l lf l da = 0 .
The sequence (h , fh , . . . defined by Bn 0 for all = 1 , 2, . . . is a sequence of a-summable functions such that La( lf - Bn = = 1 , 2, . . . , and so f is integrable. By Theo La( lf l) = 0 for all 0, and therefore h f = rem 4 .5 . 3(i) we have La (f+ ) = La( -f- ) o = by definition of these integrals. f
Proof
=
n
hI
n
I) da
=
I da 0 Corollary A function f I --+ IR is null with respect to a if and only iff is integrable over I with respect to a, and h If I da = 0. Now let S be any subset of We define the characteristic function of S to be the function xs (x) : IR --+ IR defined by 5 .2 . 2 :
Xs (x) _
{
JR.
1, 0,
if X E S, if x e s.
(5. 5)
We say that S is a null set (with respect to a) if xs is a null function (with respect to a) . Since xs is nonnegative, it follows at = 0 . It once from Corollary 5.2.2 that xs is null if and only if xs also follows from Theorem 4.5.3 and the definition of a null function that any subset of a null set is a null set. If f : is a function, and P is some property of f that holds everywhere in except possibly on some null subset of we say that P holds almost everywhere on (abbreviated to a.e.) . For example, f a.e:' if fn ( = 1 , 2 , . . . ) and f are functions defined on then 11fn f (x) for all x E except possibly for values of x means that fn (x) belonging to some null subset of
JR
I --+ IR I
n
da
I,
I
--+
I,
I
--+
I. Theorem (i) Iff : I --+ IR is null, then f = a. e. (ii) Iff : I --+ IR is such that f = 0 a. e. , then f null. (iii) Iff, g : I --+ IR are such that f = g a. e. , and iff is integrable over I, then g is also integrable over I, and h f da = h g da. 5 .2 .3
0
is
5 .2 . Null Functions and Null Sets
lR is null, so that La (lf l) as follows: define the se quence of sets A 1 , A 2 ,
Proof (i) Assume that f : I
�
.
A 1 = {X : X E I, If (X) I
{
A n = x : x E I,
�< n
•
=
77 0 . We
.
>
1},
lf(x) l <
1 n-1
},
for n = 2, 3 , . . . . Clearly, x E A n => 1 < n lf(x) l (n = 1 , 2, . . . ), and so for each n = 1 , 2, . . . we have 0 < XA n < n lf l on I, and therefore 0 < La (XA n ) < La (n lf l) = nLa (lf l) •
by Theorem 4 .5.3 by Theorem 4 . 5 .2(ii) .
Since f is null, it follows that La ( XA n ) = 0 for all n = 1 , 2, . . . . Now let S = {x : x E I, f(x) # 0} . For any x E S there exists a unique positive integer N such that x E AN, and therefore
() {
XA n X
=
1, 0,
if n = if n #
N,
N.
Thus Xs (x) = 1 = L �= l XAn (x) for each x E S. On the other hand , if X E S, then X E A n for each n = 1 , 2, . . . , and so Xs (x) = 0 = L �=l XA n (x) . Therefore, Xs = L �=l XA n , and so by Theorem 4.5.4 we have 0 < La (Xs ) <
Ll La (XAn ) Ll o 00
00
=
n=
n=
=
0,
and so xs is null, which proves part (i) . (ii) Assume that f = 0 a. e. Let S = {x : x E I, f(x) # 0} . Then S is a null set. For each n = 1 , 2, . . . define Bn = {x : x E I, n - 1 < lf(x) l < n} . Since Bn c S, it follows that Bn is null for each n = 1 , 2 , . . . . Now define fn = n xBn for each n = 1 , 2, . . . . For any x E S there is a uniqu e positive integer N such that x E BN , and therefore J!'n (x)
=
{
N, 0,
if n
=
if n #
N,
N.
Hence L�=I fn (x) = N > lf(x) l (since x E BN ), and so we can say that lf(x) l < L �=l fn (x) for all x E S. On the other hand, if X E I - s, then X e B n for each n = 1 , 2, . . . and so L �= l fn (x) = 0 = f(x) . Thus L �= 1 fn converges on I, and lf l < L�= 1 fn on I. B y I
78
5.
Properties of the Integral
Theorems 4 . 5.2(ii) , 4 . 5. 3 , and 4.5.4 we then have
0 < La(lf l) < La (nLl fn) 00
=
n= l
=L La(nXBn ) = L nLa (XBn ) n=l n=l = nL=l o = o, since B n is null for each n = 1 , 2 , . . . . Thus, La( lfl) = 0, and f is null as required. (iii) Since f = g a. e. , we have g - f = 0 a.e. , so by part (ii) , g - f is null. By Theorem 5.2 . 1 it follows that g - f is integrable and h (g - f) = 0. But g = f + (g - D on I, so by Theorem 5 . 1 .4, g is 00
00
00
da
integrable and
0
as required.
Part (iii) of the preceding theorem is particularly important. It shows that changing the values of a function on a null set does not affect the integral of the function. Two functions that are equal almost everywhere can be regarded as identical in the context of integration theory. Exercises 5-2 :
R be defined by a* (x) x for all x E R; Prove that 1 . Let a * : R any finite or countably infinite subset of R is null with respect to a * .
�
=
2 . Give an example of a finite subset S c R and a monotone in creasing function a : R R such that S is not null with respect to a.
�
5.3.
3.
Convergence Theorems
79
Let g : [0 , 1 ] --+ lR be defined by g (x)
Find
{ _
0, 1,
if x is rational, x =j:. 0, 1 , if x is irrational or x = 0 or x = 1 .
�O, I] g dx , explaining your reasoning in full.
I --+ lR are
4 . Generalize Theorem S . l .S(ii) by proving that if f, g : integrable over with respect to a, and g < a. e., then h da .
I
f
f
h g da <
5. Prove that the union of two null sets is a null set . •
5.3
Convergence Theorems
We now state the main convergence theorems for the Lebesgue Stieltjes integral. The proofs are, regrettably, too long and technically difficult to include here. The reader is referred to [31 ], [32], or [38] for the details. Theorem 5 . 3 . 1 fi , [2 ,
(Monotone Convergence Theorem)
Let . . . be a monotone sequence offunctions that are all integrable over I with respect to a, and are such that limn� Ch fn da) is finite. Let f : I --+ lR be such that fn --+ f a. e. Then f is also integrable over I with respect to a, and h fn da --+ h f da. oo
Suppose that {an } is a sequence of real numbers bounded below. Let kn = inf m ::: n llm · Then by Exercises 1 -3, No. 3, the sequence {kn } is monotone increasing, and so by Theorem 2 . 1 . 1 ,
� oo kn = nlim � oo (minf nlim ==:n elm)
exists in lRe. A consequence of the monotone convergence theorem is the following technical result, which we shall use in Chapter 8 : L emma 5 . 3 .2 [2 , . . .
(Fatou 's Lemma)
Let f1, be a sequence ofnonnegative functions that are all integrable over the interval I with respect to a and suppose that fn (x) --+ f(x) for all x E I except perhaps in a null set ofI. Then f integrable over I with respect to a if and only if limn� oo (infm:::n h fm(x) da) is finite, and in is
80
5.
Properties of the Integral
that case
l I
( j
f da < lim inf n� oo m==:n
I
)
fm (x) da .
Theorem 5 . 3 . 3 (Dominated Convergence Theorem) Let [I 1 [2 1 . . . be a monotone sequence of functions that are all integrable over I with respect to a and are such that for each n = 1 1 21 , Ifn i < A on I, where A : I --+ lR is integrable over I with respect to a. Let f : I --+ lR be such that fn --+ f a. e. Then f is also integrable over I with respect to a, and we have •
J
rnaa -+
J
r aa and
j
•
•
lfn - f l da -+ o .
As immediate consequences of Theorems 5. 3 . 1 and 5.3.31 obtained by applying these theorems to the sequence of partial sums of a series (see Exercises 5-3 1 No . 1)1 we obtain the following important results on the integration of series term by term. Theorem 5 .3 . 4 Let a1 1 a2 1 . . . be a sequence offunctions that are all integrable over I with respect to a, all have the same sign (i. e. , either � > 0 for all j = 1 1 21 aj da) converges. or aj < 0 for all j = 1 1 21 ), and are such that � = s a. e. Then s is also integrable over Let s : I --+ lR be such that I with respect to a, and •
•
'2::� 1 Cfr
'2::� 1
•
•
•
•
Theorem 5 . 3 . 5 Let a 1 1 a2 1 . . . be a sequence of functions that are all integrable over I 1 with respect to a, and are such that for each n = 1 1 21 •
n
L aj on I, where A : I be such that
2::� 1
•
< 1..
integrable over I with respect to a . Let s : I --+ lR aj = s a. e. Then s is also integrable over I with respect
--+
lR is
j=l
•
5 .4. Extensions of the Theory
fn(x)
1
--------- '
to a,
and
9
I
0
81
X
FIGURE
1sda
=
I
5.1
f (J�aa) . J=l
I
Exercises 5-3:
1.
(a) Deduce Theorem 5.3.4 from Theorem 5.3. 1 .
(b) Deduce Theorem 5.3 .5 from Theorem 5.3.3.
2 . For each n = 1 , 2, . . . let fn :
x
fn ( ) =
{ 0,
1,
JR
�
JR
be defined by
x.
if n < < n + 1 , otherwise
(Figure 5 . 1 ) . (a) Find the function f = limn __.oo fn .
(b) Show that limn__. oo (JJR fn dx) #- fJR f dx .
(c) State which hypothesis of the monotone convergence theo rem is not satisfied in this case, and explain in detail why it is not satisfied. Do the same for the dominated convergence theorem.
5 .4
Extensions of the Theory
In this section we descnbe briefly two important ways in which the definition of fr fda can be extended: (1) by allowing integration over sets other than intervals, and (2 ) by allowing a to be a function of bounded variation.
82
5.
Properties
of th e Integral
(I) We say that a function f : I --+ is a-measurable if there is a sequence fh , 82 , . . . of a-summable step functions on I such that limn� oo Bn = f a. e. Any a-summable step function is obviously a measurable, and so is the unit function 1, which takes the value 1 for each x E (see Exercises S-4 , No. 1 ) . From the properties of step functions and the elementary rules for limits, it follows easily that:
lR
lR
(i) the modulus and the positive and negative parts of an a-measurable function are a-measurable;
are a-measurable, then so are f ± g, fg, max {f, g}, (ii) iff, g : I --+ and min{f, g}. It is also true that f/g is a-measurable, provided that g =j:. 0 a. e., but this is a little harder to prove.
lR
It can be proved that a function f : I --+ is integrable over I with respect to a if and only if it is a-measurable and La ( If I ) is finite. A set X c lR is said to be a-measurable if its characteristic func tion x.x is a-measurable. If X is a-measurable and La ( Xx ) is finite, with respect to a, and we define the then xx is integrable over a-measure JLa (X) of X by
lR
lR
11-a(X) =
L
Xx da .
If X is a-measurable and La (Xx) = oo, we say that X has infinite a-measure and write JLa (X) = oo. This definition of measure is easily seen to be consistent with our previous definition of measure for simple sets. Note also that a set is null with respect to a if and only if it has a-measure zero . The concept of the measure of a set is of considerable importance in itself, quite apart from its connection with integration. wit'h respect to a, is integrable over --+ If a function f : is a-measurable, then fxx is a-measurable, and since and X c we can say that La ( lfx.x l) is finite and so fx.x is lfx.x l < lf l on inte grable over with respe ct to a. We can then define the integral of f over X with respect to a by
lR
lR lR
JR , lR
lR
5 .4. Extensions of the Theory
a(x)
83
a(x)
t
a
a
a� !
ae l FIGURE 5.2
a(x)
a(x) b
b
b� I
be l FIGURE 5 . 3
(2) Suppose a : I -+ JR. is a function ofbounded variation. If I is a proper subset of JR. with endpoints a, b, then we have by Corollary 2.7. 3 that a (a+ ) and a(b-) exist and are finite. We extend a to a function of bounded variation on JR. in the following way: (i) If a is finite, then define a (x) to be equal to a(a+ ) for all x < a in the case where a e I, and equal to a(a) for all x < a in the case where a E I (Figure 5 .2) . (ii) If b is finite, then define a (x) to be equal to a(b-) for all x > b in the case where b e I, and equal to a (b) for all x > b in the case b E I (Figure 5.3) . By Theorem 2.7.4, we can then express a as a difference
where a1 , a 2 : JR. -+ JR. are both monotone increasing. Now let J be any subinterval of I . If a function f : J -+ JR. is integrable over J with respect to both a1 and a2 , we say that f is
84
5.
Properties of the Integral
integrable over J with respect to a, and make the natural definition
jtaa = jtaa� - £taa2. J,
It can be proved that the value of f da does not depend on the particular way in which a is expressed as a difference of monotone increasing functions. In the same spirit , we define the a-measure of an interval J c I to be 1-La U)
= I-La1 U) - I-La2 U) .
From the elementary rules for limits , it is easy to see that 1-La U) can be descnbed in terms of the one-sided limits of a at the endpoints of J in precisely the same way as was done for monotone increasing a (Section 4-1 ). Of course , when a is a function of bounded variation , we must allow for the possibility that intervals may have negative a-measure. Note also (and particularly) that the theory of null sets is no longer valid in and null functions , as developed in Section this more general setting. For the most part we will continue to restrict ourselves to integra tion with respect to monotone increasing functions , but we will need the extension to functions of bounded variation when we discuss integration by parts in the next chapter.
5-2,
Exercises 5-4:
1 . Prove that the unit function 1 : JR. � JR. is a-measurable for any monotone increasing function a : JR. � JR.
2 . Define the functions f, a : [0, 2) � JR. as follows:
{-, x ={ -
f (x)
a( ) (Figure 5 .4) .
=
1,
1
X, 1
X,
�f 0 < x <
1, 1f 1 < X < 2,
if 0 < X < 1 , if 1 < X < 2
5-4.
Show (a) Extend a to a function on JR. as specified in Section that a, so extended , is a function ofbounded variation on JR.,
5 .4. Extensions of the Theory
f{x)
a(x)
1 ---o 0
1 1
--4-
-1
I I
•
85
--�--���
I
x
0
FIGURE
5.4
by expressing a as a difference of two monotone increasing functions a1 , a2 : IR --+ that satisfy the hypotheses of Theorem 2. 7.2 .
IR
(b) Find tLa ([O , 2)), tLa ([O , 1 ]), tLa ([ 1 , 1 ]) and tLa (( 1 , 2)). (c) Evaluate
�o, z) f da .
C H A P T E R
Integral Calculus
Having worked through the basic theory of the integral, we now turn to the actual techniques of integral calculus, that is, the practi calities of evaluating and manipulating Lebesgue-Stieltjes integrals. The emphasis here win be on understanding and applying the re sults. For this reason, the proofs of most the results stated in this chapter will be suppressed. The results are standard to the theory, and their proofs can be found in most texts on integration (e.g. , [31 ], [3 2 ] , [3 8]) . It should be noted, however, that many of these proofs pose a technical demand on the reader greater than what has been expected so far in this text.
6.1
Evaluation of Integrals
The actual evaluation of Lebesgue-Stieltjes integrals takes as its starting point the fact that for a closed interval [a, b] the ordinary Riemann integral f (x) dx. (if it exists) has the same value as the Lebesgue integral �a, b] f dx. (cf. Theorem 4.6 . 1). The same holds for improper Riemann integrals, provided that they are absolutely con-
J:
87
88
6.
Integral Calculus
vergent (as discussed at the end of Chapter 4). Lebesgue integrals that correspond to Riemann integrals can therefore be evaluated using all the elementary techniques with which you are presumed familiar. More general Lebesgue-Stieltjes integrals can usually be dealt with by reducing them to combinations of integrals that either are equivalent to Riemann integrals or are easy to evaluate. The theorems that follow provide the necessary tools. In all of them, we assume that a : JR. -+ JR. is a monotone increasing function.
Theorem 6 . 1 . 1
Ifthe interval I is a union ofa finite number ofpainvise disjoint intervals I = I1 U h U U In, then ·
·
·
in the sense that if one side exists, then so does the other, and the two are equal. Let a = L:;1 ciai, where for eachj = 1 , 2, , ai : JR. -+ JR. is a monotone increasing function and is a nonnegative finite real number. If a func tion : I -+ JR. is integrable over I with respect to each of a1 , a2 , . . . , am, then it is integrable over I with respect to a, and
Theorem 6 . 1 .2
Cj
f
.
.
.
1fda = tCjlfda j=l j . I
I
Theorem 6 . 1 . 3
(i) If a is continuous at a, then
1[a,b] fda = 1(a,b] fda
and
1[a,b) fda = 1(a,b) fda
in the sense that if one side of the equation exists, then so does the other, and the two are equal. (ii) If a is continuous at b, then
1[a,b] f = 1[a,b) f da and 1(a,b] da
f da =
r �(a,b)
fda
6. 1 .
Evaluation of Integrals
89
in the sense that if one side of the equation exists, then so does the other, and the two are equal. Theorem 6 . 1 .4
For any interval I, fr 1 da = JLa (I). Theorem 6 . 1 . 5
If a is constant on an open interval I, then fr f da =
0.
Theorem 6. 1 . 6
For any function f defined at a, fra,a] f da = f (a)[a (a+) - a (a - )]. Theorem 6 . 1 . 7
If a is differentiable at all points in an open interval I, then
in the sense that if one side exists, then so does the other, and the two are equal. Theorem 6 . 1 .8
Let I be an open interval, and let fJ : I � � be a monotone increasing function on I such that a (x) = {J (x) for all x E I. Then
(note that since I is open, the integral on the right is defined even if the domain of fJ does not extend beyond I). A few examples should be sufficient to show how these theorems are applied. The most important thing to note is that points of dis continuity of a must be dealt with separately, by treating them as single-point closed intervals and using Theorem 6.1 .6. Example 6-1 -1 : Let a : � � � be defined by
a (x) = (cf. Figure
6 . 1).
{ O1 ,,
if x < if x >
2, 2
90
6.
Integral Calculus
1
a(x)
•
FIGURE 6 . 1
fr
fc
fr
2 2 Then 2 • 31 x da = 2 • 21 x da + 2 •31 x2 da by Theorem 6 . 1 . 1 (note that a is discontinuous at 2) . Now, Theorem 6 . 1 . 6 implies that r x2 da = 2 2 (a(2 +) - a(2-)) = 4(1 -
}[2 , 2]
0) = 4,
and Theorem 6 . 1 . 3(ii) implies that
l(2,3] x2 da l(2,3) x2 da. =
The above equation and Theorem 6 . 1 .5 indicate that
l(2,3] � da
and thus
=
r x2 da = 4 +
}[2 ,3]
0
I
0 = 4.
Note that Theorem 6 . 1 . 5 requires the interval to be open, and we used Theorem 6. 1 . 3 (ii) to change the interval of integration to the open interval (2 , 3). The main feature to note from this example is that although a is constant on the closed interval [2 , 3], we cannot conclude that 3 x2 da = because of the discontinuity in a.
fr2 • 1
0,
Example 6-1 -2 : Let a : JR. � JR. b e defined by a(x) =
(Figure 6 . 2 )
.
{ 0, - e 3
-2x
'
if X < if x >
0,
0
6. 1 .
Evaluation of Integrals
91
a(x)
3
-----
2 1
0
X
FIGURE 6 .2
Then,
{
J[O,oo)
ex da
= =
{ ex da + {
J[O,O]
lco,oo)
e0(a(O+) - a(O - )) + {
lco,oo)
(by Theorems Applying Theorem
{
J[O,oo)
c d (3 - e-2x)
6. 1 .6 and 6.1 .8) .
6. 1 . 7 to the second integral gives
ex da = 1 ( 2 - 0) + {
lco,oo)
=
2+ f
=
2+
J[O,oo)
1
00
C (2e-2x) dx
2e-x dx (by Theorem 6.1 . 3 (i))
2e-x dx (an improper Riemann integral)
2 + lim00 [ - 2e-x]cQ C--+ = 2+ ( -2)) = 4 . =
C da (by Theorem 6. 1 . 1)
(0 -
Note again that although a(x) = 3 - e-2x for all x E [ 0, oo) , we cannot conclude that �O, oo) � da = �O,oo) � d (3 - e- 2x) . Again, the disc onti nuity in a at the endpoint of the interval is crucial; the condition in Theorem 6. 1 .8 that I must be open cannot be disregarded.
92
6.
Integral Calculus
FIGURE 6 . 3
2
[2 (X)
f1 (x) I I
1
• 1
2 FIGURE 6.4
Example 6-1-3: Let a : JR. -+ JR. be defined by
a (x) = (Figure 6 . 3) . Let fi , f2 : JR.
f1 (x)
-+
if X < 0, if x > 0
JR. be defined by X
{ 1,
=
{ x,x + 1 '
,
�Iff xX =j:.= 2,2 ,
{2 (x) =
{
X l, ,
if x =1- 0, if x = 0
(Figure 6.4 ) . Now, the set {2} has zero a-measure, and Theorem 5.2. 3 implies that � 1 3 fi da = � 1,3 x da. Thu s,
,1
f fi da = f
1[1,3]
1( 1, 3)
1
x d (x + 1) (by Theorems 6. 1 . 3 and 6. 1 .8)
6. 1 .
=
=
=
=
93
(by Theorem 6. 1 . 7)
{ x(l) dx 1(1,3)
13 x dx [�J:
Evaluation of Integrals
(a Riemann integral) by Theorem 6 . 1 . 3
�-�
=
4
.
However, we cannot say that � - 1 1 [2 da = � -1,1 x da, because the ,1 1 set {0} does not have zero a-measure. Instead, we proceed thus:
1r-1 , 1 1 r; 1r-1,o) h c1a
=
c1a
+ f h c1a + f h c1a
1ro,oJ
1co,1J
(by Theorem 6 . 1 . 1 )
=
j
1 o { x dx + {2 (0) (a(O + ) - a(O- )) + x dx -1 1o (by Theorem 6 . 1 . 6)
(using reasoning similar to that used in the first part to deal with the first and third integrals) . Therefore,
[J
[]
2 o 2 1 x x 1. [2 da + 1 (1 - 0) + 2 2 -1 [-1,1 ] 0 Note that in contrast, � -1 1 x da 0 , the calculation being the same ,1
1
=
=
=
as the preceding one except that {2 ( 0 ) is replaced by 0 , the value of the integrand at 0. Example
6-1-4: Let a : 1R -+ 1R be defined by
if x < 1 , 2 x - 2x + 2 , if 1 < X < if x 2, 3, if x > 2 X + 2, 0,
a(x) (Figure 6 .5) . Then ,
f
1[0 , 3)
x2 da
=
=
2,
=
x2 da + f x2 da + f x2 da 1(1, 2) 1[1,1] 1[0, 1) + f x2 da + f x2 da (by Theorem 6 . 1 . 1 ) 1(2 ,3) 1[2 , 2] f
94
6.
Integral Calculus
a(x)
4
/ I I
3
• I I I
2
~
1
0
2
1
FIGURE 6.5
=
f
l
co , 1 )
x2 da + 1 2 (a( 1 + )
- a(l - )) +
+ 2 2 (a(2 + ) - a(2 -)) +
6.1 .3
(by Theorems =
+
(2 ,3)
=
=
+
10 9
-
6
x2
[1 , 2 ] 2
=
( 1 , 2)
[
i
c1 , 2)
dx.
(2 ,3) (i), 6. 1 .6, and
- 2x + 2)
6.1 .8)
x2 (2x - 2) dx. + 4 (4
(by Theorems
(2.0 -
( 2.x'l -
x2 d (x2
x2 d (x + 2)
6 . 1 .5 1 1 zil ) dx + 8 + f x2 dx 1 1 1 9+1 ) dx + i dx 9 [� - z; J : + [ � J :
+
=
i
0 + 1 ( 1 - 0) +
X
- 2)
and
[2 ,3]
6. 1 .7) •
(by Theorem
2x
2
3
x2
6 . 1 .3)
(Ri emann integral s)
6. 1 .
Evaluation of Integrals
95
a(x)
I
3
2
1 1
0
FIGURE 6.6
X
Exercises 6-1 :
{
1 . Let a : 1R -+ 1R be defined by a(x) =
e3x. ' 2, 2x + 1 ,
if x < 0, if 0 < x < 1 , if x > 1
(cf. Figure 6.6 ) . Let f : 1R -+ 1R be defined by f(x) =
{
e -2-x. , x,
if x < 1 , if x > 1
(cf. Figure 6. 7) . Evaluate h f da for each of the following inte rvals
J:
(i) ( - 1 , 0) (iv) ( - 1 , 1 ]
(ii) [ - 1 , 0] (v) ( 1 , 3]
(iii) ( - 1 , 1) (vi) ( -oo, 0)
2 . Let [x] denote the integer part of x, i. e., the largest integer n such that n < x; for example, [2 .71 ] = 2, [3] = 3, [- 1 . 82] = - 2 . (a) Sketch the graph of the function [x] : 1R -+ JR. (b) Evaluate the following integrals:
96
6.
Integral Calculus
f{x)
FIGURE 6 . 7
+ 1 ) d [x]
(ii) �O , S ] ef d (x + [x] )
(iii) �114,5141 [x] d [2x]
(iv) �11 4,5141 [ 2x] d[x]
(i) �o, s](x2
3 . Let a : 1R -+ 1R be a probability distribution function correspond ing to a random variable x. We define the mean of x (also called the expectation or the expected value) to be
E(x)
=
j
(-oo,oo)
x da .
(a) Calculate the mean of the uniformly distnbuted random variable defined in Example 4-2-1 . ·
(b) Calculate the means of the random variable s defined in Example 4-2-2 and Exercises 4-2 , No. 2. (c) If x is a random variable that can take exactly n values AI , A 2 , . . . , An (where AI < A2 < < An ) with probabil itie s P I , P2 , . . . , Pn , respectively (where LJ= I Pi = 1 ) , find the corresponding probability distnbution function a, and the mean of this random variable. ·
·
·
6.2.
6 .2
Two Theorems of Integral Calculus
97
Two Theorems of Integral Calculus
In this section we will look at the form taken in the Lebesgue Stieltjes theory by two theorems that are familiar to you in the context of elementary integral calculus. An important aid in evaluating elementary integrals is the 11Change of variable" theorem
1b a
f[u(t)]u' (t) dt
1u(b) u(a)
=
f(x) dx,
where we have made the substitution x = u(t) in order to simplify the integral on the left. The same thing can be done within the Lebesgue-Stieltjes theory. First we need a definition. We say that a function u : 1R � 1R is strictly increasing on an interval I if u(xi ) < u(x2) for all XI , x2 E I such that xi < x2 . Suppose now that u : 1R --* 1R is continuous and strictly increasing on I, and write u(I)
=
{u(x) : x E I} .
Then it is easy to see that u(I) is an interval. For example, if u(x) = x3 + 1 for all x E JR , then u(( -2, 1 )) = ( - 7 , 2), u([3, oo)) = [ 2 8 , oo) , u(( -oo, - 1 )) = ( -oo, 0) , and so on. We can now state the "change of variable" theorem for the Lebesgue-Stieltjes integral. Theorem 6 .2 . 1 (Change of Variable) Let I be any interval, and u : 1R --* 1R be a function that is continuous and strictly increasing on the interval I. Then
l(f
o
I
u) du
=
1 u(I)
f dx ,
(f
where f o u denotes the composition off and u, defined by o u)(x) = f[ u(x) ] for all x E I. If, in addition, u is differentiable on I, then this result can be written in the form
l(f I
o
u)u' dx
=
1 u(I)
f dx .
98
6.
Integral Calculus
Finally, if a : lR
--+ lR is monotone increasing, then f
J1
(f o u) d(a o u) =
{ f da . fu(I)
All three results hold in the sense that if one side exists, then so does the other, and the two are equal.
I
The condition that u should be strictly increasing on is not really a restriction in practice. If u is not strictly increasing, the interval of integration can usually be split up into subintervals on which u is either strictly increasing or strictly decreasing, or constant, and each of these can be dealt with separately (note that if u is strictly decreasing then -u is strictly increasing, so the theorem can still be used with the obvious modifications). When evaluating integrals in practice, therefore, one is usually safe, provided that u is continuous (and for the most part u is also differentiable) . The second matter we will consider here is integration by parts. Recall that for the Riemann integral, the technique of integration by parts centers on the formula
t
fg' dx +
t f'g dx = fig]: .
For the Lebesgue-Stieltjes integral, this result takes the basic form
1r ag + [g ar = JLfg (I),
where f and g can be allowed to be functions of bounded variation, using the approach outlined in Chapter 5. However, a correction term is needed in order to take account of cases where f and g have points of discontinuity in common. In order to understand the need for a correction term, it is suf ficient to consider the simplest case, where is the closed interval [a , a] = {a} consisting of a single point. We then have •
I
1 f = f(a)[g(a+) - g(a-)] = f(a)p,g({a}), 1 g df g(a)[f(a+) - f(a-)] g(a)ttr({a}) , {a}
{a }
dg
=
P,fg ({a}) = (fg)(a+) - (fg)(a-)
=
=
f (a+)g(a+) - f (a-)g(a- ) .
6.2.
Thus,
(1
{a}
f dg +
Two Theorems of Integral Calculus
99
1 ) {a}
g df - JLtg({a})
= f (a)JLg({a}) + g (a)JLr({a}) - [f(a+)g (a+) - f(a-)g(a-) ] = f (a)JLg({a}) + g (a)JLr({a}) - [f(a+)g(a+) - f(a+)g(a-) + f (a+)g(a- ) - f (a-)g (a-)] = f(a)JLg({a}) + g (a)JLr({a}) - [f ca+) JLg({a}) - g (a-)JLr({a})] = (f (a) _:_ f (a+))JLg({a}) + (g(a) - g (a-)) JLr({ a}) = A (a) , say. Then also
A (a) = [f(a) - [ (a-) + [ (a-) - f(a+ )]tLg({a}) + [g (a) - g (a+) + g(a+) - g (a-)]tLr({a}) = [f(a) - f (a-)]tLg({a}) - JLr({a})JLg({a}) + [g (a) - g (a+)]tLr({a }) + JLg({a})JLr({a } ) = [f(a) - f (a-)]tLg({a}) + [g (a) - g (a+)]tLr({a}) . Adding gives
2A (a) = [2f (a)-f(a+)-f (a-)]tLg({a }) +[2g (a)-g (a-)-g (a+)]tLr({a }) , and so finally,
1 A (a) = [f (a) - - (f(a+) + f (a-)) ]tLg({a}) 2 1 + [g (a) - - (g (a+) + g (a-))]tLr({a}) . 2 We then have
1 f + 1 g df = JLtg({a}) + A (a) . {a}
dg
{a}
Note first that iff is continuous at a , then f(a) = � (f (a+ ) +f (a-)) and JLr({a}) = 0, and so A (a) 0; similarly if g is continuous at a . Thus it is possible for A to be nonzero only if both f and g are discontinuous at a . Three important special cases are : =
= � (f (a+ ) + [(a- )) and g(a) = �(g(a+ ) + g (a - )) , then A (a) = 0.
(i) If [ (a)
100
6.
Integral Calculus
(ii) Iff and g are both continuous on the right at a , so that f (a+) f (a) and g (a+) = g (a) , then
=
1 1 A (a) = - (f(a+) - f (a-))JLg({a}) + - (g (a+) - g (a-))JLr({a}) 2 2 JLr({a})JLg({a}) . =
(iii) Iff and g are both continuous on the left at a , so that f(a - ) f (a) and g (a - ) = g (a) , then
=
1 1 A (a) = 2. (f(a-) - f(a+))JLg({a}) + 2. (g (a-) - g (a+))JLr({a}) -JLr({a})JLg({a}) . =
It is now easy to see why the general theorem for integration by parts takes the form given below. Note that if f and g are functions of bounded variation, then by Theorems 2 . 7. 2 and 2 .4.3 the set of points of discontinuity off and g is either empty or countably infinite. Theorem 6.2 .2 (Integration by Parts) Let f, g : I � 1R be functions of bounded variation, and let S denote the set ofpoints at which f and g are both discontinuous. Then
i I
where
f dg +
i df I
g
=
JL[g(I) + L A (a) , ae s
In particular, (i) If S is empty, or iff(a) = � (f (a+ ) + f (a-)) and g(a) g(a- )) for all a E S, then
ir + ig ag
df
=
=
�(g (a+ ) +
JL[g(IJ .
(ii) Iff and g are continuous on the right at all points of S, then
i I
f dg +
ig I
df
=
JL[g(I) + L JLr({a})JLg({a} ) . ae s
6.2 .
Two Theorems of Integral Calculus
101
(iii) Iff and g are continuous on the left at all points of S, then
+{
1r fdg 1r gdf {
Example 6-2-1 : Consider the integral
�0,3)
= /.Lfg(I)
x2 da
-L a es
f.Lr ( { a })f.Lg ( { a }) .
that was discussed in Example 6-1-4. Using integration by parts, bearing in mind that has no points of discontinuity, we have
� da { a d x2 = x2 + �0,3) �o.� ( )
{
x2
f.Lx2 a ( [O , 3))
x2 ) ) - x2 ) ) )
= ( a (3- ( a (O = 9a (3-) - 0 = 9 (5 = 45.
Thus,
da 45 - { a d x2) x2 1[0,3) 1[0,3) 6 1 .3 = 45 - 1 a d � ) (0,3) 45 6 . 1 .7 1(0,33) a(2x) dx by 6.1 .3 45 - la 2xa dx 3 2 [ 45 - /, 2x (� - 2x+ 2) dx + L 2x(x +2) dxJ 3 3 2 2 4 4 x3 45 - [ � - - + 2� ] - [ __:_ + 2x2 ] 3 2 3 2 3 45 [8 : + 8 - G - : + 2) J - [1 8 + 1 8 - c : + 8) J f
(
=
(
by Theorem . Theorem
=
by Theorem
=
=
1
=
=
_
=6
10 9
as before.
I
_
(a Riemann integral)
1 02
6.
Integral Calculus
I
�------�/\_________
----�----+---4---�� x
t
a
b
L___ __}\____ ___., ---y-v
It
Jt ·
FIGURE 6.8
Exercises 6-2 :
1 . Use integration by parts to evaluate the integrals in Example 6-1 -1 and Exercises 6-1 , No. 2(b) (i) . 2.
(a) Investigate what happens when integration by parts is tried as a method for evaluating the integral in Example 6-1 -2. Which hypothesis of Theorem 6.2.2 does not hold in this case?
�o , oo)
e-x da, where (b) Use integration by parts to evaluate a is as defined in Example 6-1 -2. Check your answer by evaluating the integral directly.
6.3
Integration and Differentiation
We now examine some connections between differentiation and in tegration. Here we restrict ourselves to the Lebesgue rather than the Lebesgue-Stieltjes integral; that is to say, integration is with respect to x throughout. For the first theorem, we need some notation. Let I be any in terval with endpoints a, b and let t be any real number such that a < t < b. We denote by It , lt the intervals defined by It = {x : x E I, x < t} , lt = {x : x E I, t < x} (cf. Figure 6.8). Theorem 6 . 3 . 1 (Fundamental Theorem of Calculus) f dx. and (i) If f : I -+ 1R is integrable over I, then both F(t) = G(t) = Jlt f dx. are absolutely continuous on I and differentiable a. e. on I, and F' (t) = f(t), G' (t) = -f(t) a. e. on I. If, in addition, f is continuous on I, then "a. e.n can be replaced by "everywheren in the preceding statement.
ht
6.3.
103
Integration and Differentiation
(ii) IfF : I --+ IR is absolutely continuous on I, then it is differentiable a. e. on I, F' is integrable over I, and F(t) = frt F' dx + C for all t E I, where C is constant on I. E�ple 6-3- 1 :
The error function erf(t) is defined as an ordinary Riemann integral erf(t) = 2 {o t e-x dx ,Ji J
2
for all t > 0. It is important in statistics (in connection with the nor mal distribution), and it also arises in the context of certain partial differential equations connected with heat flow. The complementary error function e rfc(t) is defined by erfc(t) = 2 e-x dx for all t t
>
0,
>
dt
d
100
0.
2
,Ji t
It follows at once from Theorem 6.3.1(a) that for all
[erf(t)
]
=
2 e-t and dtd [erfc(t)] = 2 t2 e- . ,Ji ,Ji 2
-
E�ple 6-3-2 : Let a : lR --+ lR be a probability distnbution function. If there exists a function f : lR --+ lR such that f > 0 on lR and a (t) = oo t] f dx for all t E then f is call a density of a. From Theorem 6.3.1, we know that this happens if and o:p.ly if a is absolutely continuous, and that in this case a' = f a.e. on Clearly, discrete distnbutions, as
fc
JR,
JR .
-
,
defined in Section 4-2, do not have densities, since their distnbution functions are discontinuous. If the functions f and g are both densities of a, then f = g a. e., so in this sense we can say that the density of a, if it exists, is unique. As an example, take the case of the uniform distnbution defined in Example 4-2-1 : 0, if X < A, ;:� , if A < x < B, a (x) = 1 , if x > B (Figure 6. 9) . Here we can take the density f of a to be the derivative of a where it exists (which is everwhere except at A and B), and
{
104
6.
Integral Calculus
a(x) 1 0
A
--------+---�--�--�•
x
FIGURE 6.9
f{x)
1
/(B--A)
r--- r-------�r FIGURE 6 . 10
define
f
(Figure
(arbitrarily) to be zero at A and B, giving
f { 0\ ( ) X
6 .10) .
=
B=A ,
� X < A or X > B,
lf A < X < B
The previous theorem dealt with differentiation of an integral that has a variable interval of integration. It is also important to be able to differentiate functions of the form
t
g(t) [f(t, x) dx, =
g'(t)
where the variable appears in the integrand, not in the interval of integration. It is natural to ask whether we can find by inter changing the order of the differentiation and integration operations. Thus g' (t)
=
! [[f(t,x) dx] [ �f(t,x) dx, =
6.3.
Integration and Differentiation
105
where in the right-hand integral, x. is held constant while differenti ation is carried out with respect to t. Simple examples suggest that this is correct. Consider, for example, g (t)
=
1 0
1
sin ( 2 t + 3x.) dx
=
By direct differentiation, g' (t) hand, �
1 0
=
1 a
=
- sin ( 2 t + 3x.) dx at
[-�3 -3 1
cos ( 2 t + 3x.)
cos ( 2 t + 3) +
-
J x=O x=1
1
3
cos 2 t .
� sin (2 t + 3) � sin 2t. On the other
=
=
= =
1 2[ 3
1
0
-
2
-
3
2 cos (2 t + 3x.)dx
sin ( 2 t + 3x.)
sin (2 t +
' g (t)
3)
J x=O x=1
-3
2 - sin 2 t ,
as expected. The general theorem, which tells us that this process, called //differentiation under the integral:' is legitimate, is as follows: Theorem 6 . 3 .2
(Differentiation Under the Integral)
Let J and J be any intervals. Let the real-valued function f(t, x.) be such that (i) f(t, x.) is defined for all t E J, x. E I; (ii) f(t, x.) is integrable with respect to x. over I; (iii) For each t E J, ;tf (t, x.) exists a. e. on I; (iv) For each closed subinterval J* c J, there exists a function A : I --+ 1R such that A is integrable over I, and I � f(t, x.) l < A(x.) for all x. E I and t E J*. Then for each t E J, al at[ (t , x.) is integrable with respect to x. over I, and
� [[ t(t, x) dx] [ :/(t, x) dx. =
Example 6-3-3: Consider the function g (t) J; ln(1 + t cos x.) dx, where -1 < t < 1 . Note that since I cos x. l < 1 for all x., we have that I t cos x. l < l t l < 1 =
106
6.
Integral Calculus
for all t E ( -1 , 1). Thus 1 + t cos x > 0 for all x and all t E ( -1 , 1), and so both ln(1 + t cos x) and a ln(1 + t cos x) = 1 +cosx t COS X are continuous for all x and all t E ( -1 , 1). Thke any closed subinterval [a, b] c ( -1 , 1 ), and let k = max{laL lbl}. We have that for all x and all t E [a, b], It cos xl < k < 1 , and so 1 1 + t cos xl 1 - lt cos x l 1 - k > 0. Thus cosx < 1 1 + t COS X - 1 - k for all x and all t E [a, b]. Hence all the conditions of Theorem 6.3.2 are satisfied in this case, where f(t, x) = ln(1 + t cos x), I = (O, n], J = ( -1, 1), and A. (x) = 1 /(1 - k) for each J* = [a, b] c J . By Theorem 6.3.2 we have therefore 1( cos x 1( a ' g (t) = -at ln (1 + t cosx) dx = 1 + t cos x dx for all t E ( -1 , 1 ). Note first that g' (O) = cosx dx = [ sin x] � = 0. � UL
>
>
--
1
1
o
o
( 1( ( [
f
) )
If t E ( -1, 1) is not zero, then 1 1 + t COS X - 1 g (t) - -t 1 + t cos x dx 0 1( 1 - 1 + t1cos x dx - -1t 0 x=7r tan(x/2) (1 t) 2 . arctan = � nt .J1 - t2 x=O .J1 - t2 (The final equality is not obvious, but it can be verified using any reasonably comprehensive table of integrals.) Now, as x --+ Jr- , tan(x/2) --+ oo , and so (1 - t) tan(x/2) _;__-;::::=: ::= :::;; -� J 1 - t2 --+ 00 1
11r
(
)] )
6.3.
as x --+ rr- , and thus
Integration and Differentiation
)
(
10 7
arctan (1 - t) tan(2x/2) --+ rr 2 ,J1 - t as x --+ rr-. Also, arctan(O) = 0, and so g (t) = t1 - --; 1 == ,J= t2 ==r for t '# 0, t e ( -1 , 1). Thus we have finally • 1, 1) and t -:f:. 0, g (X) - 0,7 ( ../11-tJ' ifif tt =E (-1 0. Since g(O) = J;On 1 ) dx = J; O dx = 0, we have that for all t e (-1 , 1)
(
1
1
_
j(
{
g(t) = g(t) - g (O) =
j(
-
)
{ g' (x) dx
by the fundamental theorem of calculus; thus, t g(t) = { x 1 - 1 2 dx. ,J1 - x Jo Now (using a table of integrals) we have that 2 1 1 1 x .J + _ _ _ l l n I I dx = x + n +C X x J 1 - x2 X 2) ,J1 1 x(1 x + = n +C X j(
f (.!_ -
(
)
)
__
(note that 1 + ,J1 - x2 > 0 for lx l < 1) . Hence, g (t) = rr ( In(1 + .j1 - x2 )] � = rr (ln(1 + .j1 - t2 ) - In 2) , and so finally 2 ln( l + t cosx) dx = rr ln + l - t for -1
<
t<
1" 1.
C
�
)
108
6.
Integral Calculus
In the most general case, a function defined by an integral may have the variable appearing both in the limits of integration and in the integrand, for example g(t) =
1
t2
t
e� dx.
Such a function can be differentiated by using a combination of the fundamental theorem of calculus and differentiation under the integral. Consider an integral of the form g(t) =
],
I(t)
f(w(t), x) dx,
where 1(t) is the closed interval [u(t), v (t) ]. Here we assume that for some interval 10, u, v, w : 10 -+ lR are differentiable functions of t such that for some interval h we have w(t) E 11 for all t E 1o. We assume also that for each t E 10, f(w, x) satisfies the conditions of Theorem 6 . 3 .2 for w E 11 and x E 1 (t). Let h(u, v, w) = �u,v) f(w, x) dx. Then g(t) = h(u(t), v(t), w(t)), so by the chain rule g' (t)
= =
ah du au dt
ah dv
a h dw
+ av dt + aw dt
du dv f(w, v) - + + -f(w, u) dt dt
)
(1
dw a , dx x f(w, ) dt [u,v) aW
where the fundamental theorem of calculus has been used to find ahi au and ah i av , and differentiation under the integral to find ah/ aw. In particular, if we take w(t) = t for t E 10, then dwldt = 1 , and we obtain the following result, often referred to as Leibniz's rule: d dt
[ 1[u(t),v(t)]
f(t, x) dx
]
=
�
�
+ f(t, v(t)) -f(t, u(t)) dt dt +
-f(t, x) dx .
a [u(t) ,v(t)] at
1
(6 . 1)
Iff happens to be independent of t, then we obtain the important special case d dt
[1
[u(t),v(t)]
f(x) dx
J
=
dv du t)) ( f( (u(t)) + 1 v f d . at t
(6 .2)
6.3.
Integration and Differentiation
Consider the function g(t) = Jt' etx dx for t 1 . Here !0 and since w(t) = t, we have also 11 = [1 , oo) . Take any t [a, b] be any subinterval of [1 , oo) . Then >
>
for all w obtain
E [a, b]
and x
E [t, t2 ],
=
-e" + 2 te" +
{ [�"' J:' -
- et2
+ 2 teil +
{
{
(
and let
f t2 xetx dx t [" e; dx } integrating by parts
etx [teil - e� ] - [ t2 t� t2 efl p e f3 t2 = - e t2 + 2 te + te - e t2 - t2 f3 t2 1 f3 � . = 3 te - 2 e - 2 e - e t
=
1,
[1 , oo) ,
so we can apply Le:ibniz's rule to
d d g' (t) = - e t(t) - (t) + et(t2) - (t 2 ) + dt dt •
=
109
)
[
}
]}
In this particular case we can check by evaluating g(t) directly: g(t) =
thus,
[ t ] t �t (e" - e" ) ; etx t'
=
as before. E�ple 6-3-4:
Consider the differential equation
11 0
6.
Integral Calculus
with initial conditions y(O) = c1 , y' (O) = c2 . Assume that g is continuous on the interval [0, oo) . For all t 0, define >
y(t) = c1 + c2 t +
{(t - x)g(x) dx.
Clearly, y(O) = c1 . The function f(w, x) = (w - x)g(x) certainly satisfies conditions (i) and (ii) of Theorem 6.3.2, for w E [0, oo) and x E [0, t ] (t 0) . Further, for each t E [0, oo) we have that a f(w, ) = aw x l g(x) l (independent of w) is continuous and therefore integrable over [0, t]. It follows by Leibniz's rule that >
y' (t) = c2 + (t - t)g(t) +
Thus y' (0) = calculus that
=
c2 ;
c2 +
1 ' g(x)dx.
{ g(x) dx
also, we have by the fundamental theorem of y" (t) = g (t),
and so the function y(t) defined above is the solution of the given initial value problem. As a final comment on the relationship between the integral and the derivative, we point out a serious gap in the Lebesgue theory. Recall that an antiderivative (sometimes called an indefinite inte gral) of a function f is a function F such that F' = f. It turns out that there exist functions that are 11integrable" in the sense of having an antiderivative at all points of a certain interval but are not Lebesgue integrable on that interval. An example of such a function is given in Section 10. 1 , where we revisit this matter. Exercises 1.
6-3:
Use L'Hospital's rule to find limo+ erf(t) and (b) t-+ limoo t erfc(t). (a) t-+ t
6.3.
2.
By
Integration and Differentiation
Ill
writing erf(x) as 1 x erf (x) and integrati ng by parts, show that {t erf(x) Jo
dx
1 t erf(t) - ,ft (1 - e-t2 ) . 3. Use a table of integrals to show that =
dx - -Jt2 - 1
('r 1 j0 t - COS X
rr
for all t > 1 . By differentiating both sides of this equation with respect to t, evaluate �
4.
5.
{1r 1 Jo (t - cos x)2
dx.
Given thatg(t) = J� sin(x-t) dx, find g' (t)by using Leibniz's rule. Check by evaluating g (t) directly and then differentiating. Find g' (t) if g(t) =
1 t2 1 sin(tx) dx, t
X
where t 1 . 6. Assuming that g is continuous on the interval [0, oo), show that the function 1 {t 1 y(t) = k c2 sin(kt) + c1 cos(kt) + k o g(x) sin { k(t - x) } dx, J for t 0, satisfies the differential equation d2 y � dt2 + y = g(t), where k > 0, together with the initial conditions y(O) = c1 , y ' (O) = C2 . 7. Assuming that g is continuou s on the interval [0, oo), show that the function 1 t (t - x) "g (x) y(t) = n! o >
>
-
lo
dx
satisfies the differential equation dn+l y dtn+l
=
g (t) together with the initial conditions 1 y(O) = 0, yC \0) = 0,
. . . I
y C") (O) = 0.
Double and Repeated Integrals
C H A P T E R
Lebesgue-Stieltjes integrals of functions of more than one variable can be defined using the same approach as was used in Section 4.5 for functions of one variable. For the sake of simplicity we will discuss only functions of two variables. The process for functions of more than two variables is completely analogous. 7. 1
Measure of a Rectangle
We define a rectangle to be a set of the form 11 12 c JR2 , where 11 and h are intervals. For monotone increasing functions a1 , a2 1R --+ 1R we define the a1 a2 -measure of 11 12 , denoted by J.ta1 x a2 (11 12 ), by x
:
x
x
x
l-ta1 x a2 (11 X h ) = J.ta1 (fi ) X J.ta2 (h) . For example, if a1 and a2 are the functions defined in Exercises 4-1 ,
problems 1 and 2 , respectively, then l-t a1 ((0, 1 )) l-tct2 ((0, 1 ))
= =
1 1 - e -1 , J.ta1 ([0, 1 )) = 3 - e - , 0, l-tct2 ([0, 1 )) = 1 , 11 3
11 4
7. Double and Repeated Integrals
and therefore JL a1 x a2 (( 0 ,
1) X (0, 1)) = 0, JLa1 x a2 ((0, 1) e- 1 . JLa 1 x a2 ( [0, 1) X ( 0, 1)) =
7.2
3-
X
( 0 , 1 ))
=
1 - e- 1 ,
Simple Sets and Simple Functions in Two Dimensions
A simple set in JR 2 is a subset ofJR2 that can be expressed as the union
of a finite collection of disjoint rectangles. Just as for simple sets in JR, we can define the measure of a simple set in JR2 . tf a1 , a2 : JR JR are monotone increasing functions and S is a simple set of the form --+
s
m
=
Uul j h ,j), j= l x
where h,1 !2 ,1 , h, 2 h , 2 , . . . , l1, m I2 ,m are pairwise disjoint rectangles, then the a1 a2 -measure of S is defined by m JLa1 x a2 ( S) L JL a1 x a2 (Ilj X h,j ) x
x
x
x
=
j= l
·
The properties of simple sets in JR2 and their measures are the same as those described in Section 5-3 for simple sets in JR. We can now define simple functions of two variables by analogy with step functions of one variable (see Sections 2-5 and 4-4). We could continue to use the term 1 Step functions:' but customary usage restricts this term to functions of one variable. JR is a simple function if there is a simple A function e : JR2 set •
--+
{
and a list (c1 , c2 , . . . 8 (x, y)
=
s , en
Cj ,
0,
)
UCil ,j h,j) j= l n
=
X
of finite, nonzero real numbers such that if (x, y) E l1 ,j X h ,j , (j 1 , 2, . . . , n), =
if (x, y) E JR2 - S.
7 .4.
Repeated Integrals and Fubini's Theorem
11 5
The set S is called the support of e. The properties of step functions given in Se ction 2 -5 carry over without difficulty to simple functions. If a1 , a2 : lR --+ lR are monotone increasing functions, we define the generalized ��volume" A a 1 x a2 (8) in a way exactly analogous to the definition of Aa(8) in Section 4-4.
7.3
The Lebesgue- Stieltj es Double Integral
Let S be a subset of JR2 and let f : S --+ JR be a function. We extend the definition of f to JR2 by defining f(x , y) to be zero if (x , y) E JR2 S. Let a1 , a2 : JR --+ JR be monotone increasing functions. The Lebesgue-Stieltjes double integral of f, denoted by -
L, f f
d (Ott
X
012 ),
is defined by a process that is almost word-for-word the same as that used for the single-variable integral in Section 4-5. The only change is that a-summable step functions on are replaced by a1 x ar summable simple functions on R2 • All the elementary properties analogous to those proved in Sections 5-l and 5-2 carry over and are proved in the same way, and the same goes for the convergence theorems of Section 5-3 and the definitions of measurable functions and measurable sets given in Section S-4. In practice, the evaluation of double integrals is invariably done, as in elementary calculus, by converting them to repeated integrals .
I
7.4
Repeated Integrals and Fubini's Theorem
Let f : JR2 --+ JR be a function. For any y E JR we define the single variable function f(·, y) : x --+ f(x, y), and for any x E JR we Hkewise define the function f(x , ·) : y --+ f(x , y) . Let a1 , a2 be monotone
11 6
7.
Double and Repeated Integrals
increasing functions. If for each y e R, JIRf ( · , y) da 1 exists, then this defines a function [2 : y --+ firt fC· , y) da 1 . If fut h da2 exists, we call this a repeated integral of f and write it as f·�.f da 1 da2 . If for each x e JR, fJR. f(x , · ) da2 exists, we define f1 : x --+ JJR. f (x , ·) da2 , and if [I da 1 exists, it gives us the repeated integral of f with the f da2 da 1 . In most cases opposite order of integration, written calculation shows that JJR f da 1 da2 and f da2 da 1 have the same value, but this is not always the case. Consider, for example, the improper Riemann repeated integrals
fiR.
fiR.
fiR.
11 1
o
We have
fiR. fiR. f f iR. iR.
o
1
y --3 - dx dy and (x + Y) X
_
1 1 (x + 1
1
o
o
X -
y dy dx. Y) 3
However,
11 1
o
1 x - y - dx = -3 dy ) ( x + o Y =
1 1 (xy+- x 1 1 (xx +- y 1
1
o
-
1 2'
o
1
1
o
o
Y) 3
dx dy (interchanging x and y)
y) 3
•
dx dy
and so the two repeated integrals have different values. It turns out, though we shall not prove it, that this cannot hap pen if either repeated integral is absolutely convergent. We can easily verify that this condition does not hold in our example. We
7 4 .
.
Repeated Integrals and Fubini's Theorem
11 7
y
y <x -�--�-. x
0
1
FIGURE 7.1
investigate
lx - y l - dx dy 1 1 1 1 -(x + y)3 o
o
by splitting the region of integration into two parts, one where y < x and one where y > x (cf. Figure 7. 1) . Then,
10 1 10 1 lx +-YY)l3 dx dy 10 1 1 1 + Yy)3 dx dy + 10 1 10y y+ Y)3 dx dy [ {[ c� y)2 - (x �y)3 ) dx - (x �y) 3 ) ax } dy 2 x y) � c ( - f1 y x= {[ 1 1 - x + y + (x + y)2 J x=y1 x= y } 1 y - [ X + Y + (x + Y) 2 J x=O dy 110 { 1 -+1y + (1 +y y)2 + -21y - 2_ - (-2.. + 2_ + � - � ) } dy 4y 2y 4y y y ( [ - (1 � y)2 + ;J - � + � 1 1 � dy , 2 2 0 y (X
___ _;_
_
y
=
X(X
(X
=
=
o
=
=
=
dy
-X
11 8
7.
Double and Repeated Integrals
y
y =x
2
-
-
-
-
\
-
-
FIGURE 7.2
and so this integral (and likewise J; J01 C�;�� dy dx) does not converge. The fundamental theorem that relates double and repeated in tegrals using the absolute convergence condition is called Fubini's theorem: Theorem 7 .4. 1 (Fubini's Theorem)
If a 1 , a2 : JR --+ JR are monotone increasing functions and f : JR2 --+ JR is a 1 x a2 -measurable, then the existence of any one of the integrals
LJ lf l d(<>t
X
L L lf l
<>2 ) ,
da , da2 ,
L L lf l da2
da , ,
implies the existence and equality of the integrals
LJ f d(a,
x
LL r aa aa2 , LL r aa2 aa .
<>2 ) ,
,
,
In practice, the functions that arise are almost always measurable, so Fubini's theorem justifies the use of repeated integrals to evaluate double integrals, provided that one of the repeated integrals is abso lutely convergent. The details can be very messy, so we will confine ourselves to one example. E�ple 7-4- 1 :
Let S
{ (x , y) : 0 < x < f : JR2 --+ JR be defined by =
f(x , y) =
{
2,
0 < y < x}
1 + xy, 0,
(see Figure
if (x , y) E S , if (x , y) E JR2 - S .
7.2) and let
7. 4 .
Repeated Integrals and Fubini's Theorem
11 9
Let a 1 and a2 be the functions defined in Exercises 4- 1 , problems 1 and 2, respectively. We will evaluate fiRz f f d (a 1 a2 ) by evalu ating the repeated integral fiR fJR f da1 da2 , and check by evaluating fiR fiR f da2 da1 . Since the integrand is nonnegative, our procedure will be justified (by Fubini's theorem) if one of the repeated integrals exists. x
We have that
•
f(· , y)(x)
hence, h (JJ) =
=
Ja rc· , y) dal
{
{=
Jry, 21 ( 1 + xy) da1 1
1 + xy,
if 0 < y < X < 2,
0,
otherwise;
if 0 < y < 2,
0, otherwise 1 (a 1 (0 +) - a 1 (0-)) + f02 1 tJ3 - e -x ) dx ,
f:(l + xy) !C3 - e -x) dx, 0
I
3 -
0,
if O
otherwise e- 2 ,
< y <
otherwise
if y = 0,
0,
if y = 0,
if y
= 0,
otherwise.
2,
1 20
7.
Double and Repeated Integrals
Therefore,
LL r aa1 da2 L tz daz =
=
(3 - e- 2 )(o:2 (0 +) - o:2 (0- ))
+
2 { ( -e-2 - 3ye-2 + e-Y + y2 e-Y + ye-Y ) � (4) dy dy
Jl
+ 0 (0: 2 (2+ )
- 0: 2 (2 - )) = (3 - e- 2 )( 1 ) + ( - 4 e- 2 + 3e- 1 )(3 ) =
3 + 9e- 1 - 1 3e- 2.
We have that f(x, ·) (y) =
{
1 + xy ,
if 0 < y < x < 2 ,
0,
otherwise;
hence, ft(x)
=
=
L t (x, ·) daz { fr0,o,x] (1 xy) da2 , +
if O < x < 2 , otherwise
7.4.
Repeated Integrals and Fubini's Theorem if 0
121
< X < 1,
1 (a2 (0+) - a2(0-)) + j0\1 + xy)� (1) dy
+ (1 + x)(a2 (1 +) - a2 (1 -)) + JtCI + xy) � (4) dy , if 1 < x < 2 ,
1 1 (a2(0+) - a2(0-)) + f0 (1 + xy) � (1 ) dy 2 + ci + x)(a2(1 +) - a2(1 -)) + f1 (1 + xy) � (4) dy + (1 + 2x) (a2(2 +) - a2(2 -)) , 0, 1,
otherwise if O < x < 1 ,
4 + 3x , if 1 < X < 2, -
6 + 7x, if x = 2, 0,
Therefore,
otherwise.
if x = 2,
122
7.
Double and Repeated Integrals
Exercise 7-4:
Let S = { (�, y) defined by
: 0 <
f (� , y)
�< =
2,
0 <
y<
{ � sin y, 0,
2 - � } and let f
if (�, y) e S , if (�, y) E JR2 - S .
:
JR2
--+
lR be
Let a 1 and a2 be the functions defined in Exercises 4-1 , problems 1 and 2, respectively. Verify that aa , aa2 aa2 aa , . r r ll ll =
C H A P T E R
The Lebesgue Spaces V
There are many mathematical problems for which the solution is a function of some kind, and it is often both possible and convenient to specify in advance the set of functions within which the solution is to be sought. For example, the solution to a first-order differential equation might be specified as being differentiable on the whole real line. The set of functions differentiable on the whole real line has the useful property that sums and constant multiples of functions in the set are also in the set. In fact, this set of functions has the structure of a vector space, where the 11Vectors" are functions. Beyond the algebraic properties associated with vector spaces, many problems are solved by use of series or sequences of functions, and it is desirable that any 11limits" also be in the set. Some of the most useful sets of functions have this property. We have seen in Section 3-3 that the limit of a sequence of Riemann integrable functions does not necessarily yield a Riemann integrable function, and this signals that the sets defined using the Riemann integral may not be suitable for many applications. In contrast, sets defined using the Lebesgue integral have the desirable 11limit properties!' There are a number of sets of functions that are vector spaces that are of importance in subjects such as differential and integral equations, real and complex function theory, and probability theory, 123
1 24
8.
The Lebesgue Spaces
V
along with the fields of applied mathematics where these subjects play a significant role. Some of the most important of these function sets make use of the Lebesgue integral in their definitions, and so it is appropriate to discuss them here. In this chapter we aim to give the reader an overview of some of these function sets. We neither go into all the technical details nor attempt a comprehensive survey. References are given where more detail can be found if desired. 8.1
Normed Spaces
The reader has probably encountered the concept of a finite dimensional vector space. These spaces are modeled after the set of vectors in JR.n . Vector spaces, however, can be defined more gen erally and need not be finite-dimensional. Indeed, most the vector spaces of interest in analysis are not finite-dimensional. Recall that a vector space is a nonempty set X equipped with the operations of addition 1 + ' and scalar multiplication. For any elements f, g, h in X and any scalars a , f3 these operations have the following properties: (i) f + g E X ; (ii) f + g = g + f ; (iii) f + (g + h) = Cf + g) + h ; (iv) there is a unique element 0 (called zero) in X such that f + O = f for all f E X; (v) for each element f E X there is a unique element (-f) E X such that f + (-f) = 0 ; (vi) af E X; (vii) a(f + g) = af + ag ; (viii) (a + fJ)f = af + {Jf ; (ix) (a {J)f = a ({Jf) ; (x)
1
·
f = f.
For our purposes, the scalars will be either real or complex numbers. We shall use the term complex vector space if the scalars are complex numbers when there is some danger of confusion.
8. 1 .
Normed Spaces
1 25
Example 8-1 - 1 : The set of vectors { (x1 Xz , . . . , Xn) : xk E R, k = 1 , 2, . . . , n} is denoted by Rn . Let x = (xl l Xz , . . . , Xn ) and y = (Y b Y2 , . . . , Yn ) be vectors in Rn . If addition is defined by 1
X + Y = (X I + YI , X2 + Y2 ,
· · · ,
Xn + Yn)
and scalar multiplication by
ax = (ax1 , ax2 , . . . , axn ) for any a E R,. then Rn is a vector space. Similarly, the set en = { (zi , Z2 , . . . , Zn ) : Zk E C, k = 1 , 2, . . . , n} of complex vectors is a complex vector space when addition is defined by z
+w
=
(zl + WI , Z2 + w2 ,
for any vectors z = (z1 , z2 , . . . , Zn ) , multiplication by
w =
. . . I
Zn + Wn )
(w1 , w2 , . . . , Wn ) ,
and scalar
ax = (ax 1 , ax2 , . . . , axn ), where a E C. The vector spaces Rn and e n are essentially the prototypes for more abstract vector spaces. E�ple 8-1 -2 : Let C[a, b] denote the set of all functions f : [a, b] --+ R that are continuous on the interval [a, b]. If for any f, g E C[a, b], addition is defined by
(f + g) (x)
= f(x)
+ g (x),
and scalar multiplication by for a E
R,
(af)(x) = af(x)
then it is not difficult to see that
C[a, b] is a vector space.
E�ple 8-1 -3 : Let .e 1 denote the set of sequences {an } in R such that the series I:: � 1 I an I is convergent, and define addition so that for any two elements A = {an }, B = { bn },
A + B = { an + bn },
126
8.
The Lebesgue Spaces
V
and scalar multiplication so that Then l 1 is also a vector space. The above examples show that the elements in different vector spaces can be very different in nature . More importantly, however, there is a significant difference between a vector space such as JR.n and one such as C[a, b] having to do with 1 dimension." The space �n has a basis: Any set of n linearly independent vectors in JR.n such as e 1 = ( l , O, . . . , O), e2 = (O, l , . . . , O) . . . , en = (O, O, . . . , l ) forms a basis. The concept of dimension is tied to the number of elements in a basis for spaces such as JR.n , but what would be a basis for a space like C[a, b]? In order to make some progress on this question we need first to define what is meant by a linearly independent set when the set itself might contain an infinite number of elements. We say that a set is linearly independent if every finite subset is linearly independent; otherwise, it is called linearly dependent. If there exists a positive integer n such that a vector space X has n linearly independent vectors but any set of n + 1 vectors is linearly dependent, then X is called finite-dimensional. If no such integer exists, then X is called infinite-dimensional. We will return to the question of bases for certain infinite-dimensional vector spaces in Chapter 9 . A subspace of a vector space X is a subset of X that is itself a vector space under the same operations of addition and scalar multiplication. For example, the set of functions f [a, b] JR. such that f is differentiable on [a, b] is a subspace of C[a, b]. Given any vectors x 1 , x2 , , Xn in a vector space X , a subspace can always be formed by generating all the linear combinations involving the Xk , i.e ., all the vectors of the form a 1 x 1 + a2 x2 + + anXn , where the a k 's are scalars . Given any finite set S c X the subspace of X formed in this manner is called the span of S and denoted by [S]. If S C X has an infinite number of elements, then the span of S is defined to be the set of all finite linear combinations of elements of S. Vector spaces of functions such as C[a, b] are often referred to simply as function spaces. After the next section we shall be concerned almost exclusively with function spaces, and to avoid rep--+
:
.
•
•
•
·
·
·
8. 1 .
1 27
Normed Spaces
etition we shall agree here that for any function space the operations of addition and scalar multiplication will be defined pointwise as was done for the space C[a, b] in Example 8-1-2 . Vector spaces are purely algebraic objects, and in order to do any analysis we need to further specialize. In particular, basic con cepts such as convergence require some means of measuring the � distance" between objects in the vector space. This leads us to the concept of a norm. A norm on a vector space X is a real-valued func tion on X whose value at f E X is denoted by ll f II and that has the following properties: (i) ll f ll > 0 ; (ii) llf II 0 if and only iff 0; (iii) ll af ll lal ll f ll ; (iv) ll f + g i l < llf ll + ll g ll (the triangle inequality). Here, f and g are arbitrary elements in X , and a is any scalar. A vector space X equipped with a norm II II is called a normed vector =
=
=
·
space.
Example 8 - 1 -4:
For any x E JR.n let II
·
II e be
defined by
ll x ll e = { (x� + (x2 i + + (xn i l 1 1 2 . I II e is a norm on JR.n . This function is called the Euclidean on JR.n . Another norm on JR.n is given by ·
Then norm
·
·
·
llx ll r Example 8-1 -5: The function II II oo ·
=
lx1 l + lx2 l + · · · + lxn l ·
given by ll f ll oo
=
sup
xe[a,b]
lf (x) l
is well-defined for any f E C[a, b], and it can be shown that II ll oo is a norm for C[a, b]. Alternatively, since any function f in this vector space is continuous, the function If I is Riemann integrable, and thus ·
128
8.
The Lebesgue Spaces
IJ'
y
g I I f I - - - - � - - IJf- glloo I I
---r--� a--------------------�b----�� X FIGURE 8 . 1
the function II
·
II R
given by ll f i i R =
t
lf (x) l dJI
is well-defined on C[a, b]. It is left as an exercise to show that II IIR is a norm on C[a, b]. The above examples indicate that a given vector space may have several norms leading to different normed vector spaces. For this reason, the notation (X, 11 · 11) is often used to denote the vector space X equipped with the norm II · 11 . Once a vector space is equipped with a norm II II , a generalized distance function (called the metric induced by the norm II II ) can be readily defined. The distance d (f, g) of an element f E X from another element g E X is defined to be d (f, g) = ll f - g il . The distance function for the normed vector space (JR.n , II li e) cor responds to the ordinary notion of Euclidean distance. The distance function for the normed vector space (C[a, b], II ll oo ) measures the maximum vertical separation of the graph off from the graph of g (Figure 8. 1 ) Convergence can be defined for sequences in a normed vector space in a manner that mimics the familiar definition in real anal·
·
·
·
·
.
8.1 .
Normed Spaces
1 2g
ysis. Let (X, II II) be a normed vector space and let lfn } denote an infinite sequence in X . The sequence lfn } is said to converge in the norm if there exists an f E X such that for every E > 0 an integer N can be found with the property that ll fn - f I < E whenever n > N . The element f is called the limit of the sequence lfn } , and the re lationship is denoted by lim n oo fn f or simply fn � f . Note that convergence depends on the choice of norm: A sequence may con verge in one norm and diverge in another. Note also that the limit f must also be an element in X. In a similal spirit, we can define Cauchy sequences for a normed vector space. A sequence lfn } in X is a Cauchy sequence (in the norm II · II ) if for any E > 0 there is an integer N such that ·
-+
=
ll fm - fn ll < E
whenever m > N and n > N . Cauchy sequences play a vital role in the theory of normed vector spaces . As with convergence, a se quence lfn } in X may be a Cauchy sequence for one choice of norm but not a Cauchy sequence for another choice. It may be possible to de fi ne any number of norms on a given vector space X. Two different norms, however, may yield exactly the same results concerning convergence and Cauchy sequences. Two norms II · ll a and II · l i b on a vector space X are said to be equivalent if there exists positive numbers a and f3 such that for all f E X,
a llf ll a < ll f ll b < ,B II f ll a · If the norms I · ll a and ll · ll b are equivalent, then it is straightforward to show that convergence in one norm implies convergence in the other, and that the set of Cauchy sequences in (X, II II a) is the same as the set of Cauchy sequences in (X, II li b) · Equivalent norms lead to the same analytical results. Identifying norms as equivalent can be an involve d process. In finite-dimensional vector spaces, however, the situation is simple: All norms defined on a finite-dimensional vector space are equiva lent. Thus the two norms de fined in Example 8-1 -4 are equivalent. The situation is different for infinite-dimensional spaces. For exam ple, the norms II · II R and ll · ll oo defined on the space C[a, b] in Example 8-1-5 are not equivalent. We elucidate further this comment in the next section, when we discuss completeness. ·
·
1 30
8.
The Lebesgue Spaces
V'
Exercises 8-1 : 1 . Let Q denote the set of rational numbers. Show that Q is a vector space, provided that the scalar field is the rational numbers. 2.
(a) Prove that the function II · II R define d on 8-1 -5 satisfies the conditions of a norm.
C[a, b] in Example
(b) Suppose that the set C[a, b ] is extended to R[a, b], the set of all functions f : [a, b ] � JR. such that If I is Riemann integrable. Show that II · IIR is not a norm on R[a, b ]. 3 . Let e n [a, b ] denote the set of functions f : [a, b ] � JR. with at least n continuous derivatives on [a, b]. Show that the functions 11 · 11 1 ,00 and I · II I , I defined by
llf l h, oo = sup lf(x) l + sup lf'(x) l , xe[a, b] xe[a,b] ll f ii i , I = are norms on the space
j[a,b] { lf(x) l + lf'(x) l } dx,
C1 [a, b].
4 . The number � can be approximated by a sequence {an } of ra tional numbers. Let S0 = { 1 , 2, . . . , 9} and choose ao as the largest 2 2 element of So such that a5 < 2 . Since 1 = 1 < 2 = 4, we have ao = 1 . Let S1 = { 1 . 1 , 1 .2 , . . . , 1 . 9} and choose a1 as the largest element of S1 such that ar < 2 . Thus, a 1 = 1 .4 . Let S2 = { 1 .41 , 1 .42 , . . . , 1 .49} and choose a2 as the largest element in S2 such that a� < 2 . This gives a2 = 1 .41 . Following this pro cedure for the general n show that the resulting sequence {an} must be a Cauchy sequence.
5. Suppose that I · lla and II · l i b are equivalent norm& for the vector space X . Prove that the condition
a llfll a < ll f ll b < ,B ll f ll a , where a and ,8 are positive numbers, implies that there exist positive numbers y and o such that
Yllf ll b < llf ll a < o llf ll b ·
8.2 .
8.2
Banach Spaces
1 31
Banach Spaces
The de finitions for convergence and Cauchy sequences for the normed vector space (IRn , II li e) correspond to the familiar defini tions given in real analysis. Various results such as the uniqueness of the limit can be proved in general normed vector spaces by essen tially the same techniques used to prove analogous results in real analysis. The space (IRn , II li e), however, has a special property not inherent in the de finition of a norme d vector space. It is well known that a sequenc;; e in (IR, II li e) converges if and only if it is a Cauchy sequence. This result does not extend to the general normed vector space. It is left as an exercise to show that every convergent sequence in a normed vector space must be a Cauchy sequence. The converse is not true. The following examples illustrate the problem for finite and infinite-dimensional spaces. ·
·
·
E�ple 8-2-1: The set Q of rational numbers, equipped with the Euclidean norm II . II e restricted to the rational numbers, is a normed vector space, provided that the scalar field is the rational numbers (Exercises 8-1 , No . 1 ) . It is well known that the number ,J2 is not a rational number. The sequence { an } de fined in Exercises 8-1 , No . 4, is a Cauchy se quence, which in the normed vector space (IR, ll · ll e) can be shown to converge to the limit ,JZ. This sequence is also a Cauchy sequence in Q, but it cannot converge to an element in Q and is therefore not convergent in Q. E�ple 8-2 -2 : Consider the normed vector space of functions lfn } defined by
fn (X) =
{
1, 1 - z nx 0,
( C[ -1 , 1 ],
if - 1 < X < 0 if O < x < 1 / z n , if 1 ; z n < x < 1 . -
I
II · II R ) and the sequence
-
I
The function fn is depicted in Figure 8.2, and it is clear that C[ - 1 , 1] for all positive integers n .
fn E
1 32
8.
The Lebesgue Spaces
IJ'
fn(x)
FIGURE 8.2
Now,
Ifn i = fn , and ll fn ii R =
therefore, for any
m
L fn(x) dx = 1 + zn�J ;
> N, n > N, we have that
1 1 1 ll fm - fn ii R = zm+l - zn+l < zN .
Given any E > 0 , any positive integer N such that N > - log2 E suffices to ensure that ll fm - fn ii R < E whenever n > N and m > N. The sequence is thus a Cauchy sequence. It is clear geometrically that fn approaches the function f defined by
f(x) =
{ 0,1 ,
if -1 < x < 0, if O < X < 1 .
Indeed, for any fixe d x0 e [ - 1 , 1 ] the sequence of real numbers lfn(xo ) } converges to f(xo) (in the I · l i e norm), i.e., lfn } is point wise convergent to f . The function f , however, cannot be a limit in C[- 1 , 1] for lfn } because f E C[-1 , 1]. Although the pointwise limit function f cannot b e a limit in C[ -1 , 1] for lfn } , this does not extinguish the possibility that there is some other function g e C[ -1 , 1] that is the limit. We will show now that no such g exists. Suppose, for a contradiction, that fn -+ g E C[ - 1 , 1] in the II · IIR norm. Then
I=
L lf(x) - g(x) l dx = L I Cf(x) - fn(x)) + Cfn(X) - g(x)) l dx
8.2 .
<
L
Banach Spaces
lfn (X) - f (x) l dx +
= zn�l + = It + Ig.
L
L
1 33
lfn (x) - g(x) l
dx
lfn (X) - g(x) l dx
The quantity It can be made arbitrarily small by choosing n suffi ciently large. By hypothesis fn � g in the II · IIR norm, so that Ig can be made arbitrarily small for n large. Now, I < It + Ig , and I does not depend op n. This implies that I 0. Since g E C[ - 1 , 1] and f is continuous on the intervals [ - 1 , 0), ( 0 , 1 ], the condition I 0 implies that f g for all x E [ - 1 , 0) and x E ( 0, 1 ]. Therefore, limx-+ o- g(x) 1 and limx-+o+ g(x) 0, so that limx-+ o g(x) does not exist, contradicting the assumption that g is continuous on the inter val [ -1 , 1 ]. The Cauchy sequence lfn } therefore does not converge in the II · IIR norm. Suppose that the space C[ - 1 , 1] is equipped with the II · II 00 norm defined in Example 8-1-5 instead of the II · II R norm. If {hn } is a Cauchy sequence in the ll · ll oo norm, then {hn } converges pointwise to some limit function h. The difference here is that because {hn } is a Cauchy sequence in the II · II 00 norm, it can be shown that {h n } in fact con verges uniformly to h, and a standard result in real analysis implies that a uniformly convergent sequence of continuous functions CC?n verges to a continuous function. In other words, the limit function h must be in C[ - 1 , 1 ]. It is left as an exercise to verify that the se quence lfn } defined in this example is not a Cauchy sequence in the II · ll oo norm .
=
A
=
=
=
=
normed vector space is called complete if every Cauchy sequence in the vector space converges. Complete normed vec tor spaces are called Banach spaces. In finite-dimer1sional vector spaces, completeness in one norm implies complete n ess in any norm, since all norms are equivalent. Thus, spaces such as (:!Rn , II · II e ) and (:!Rn , II · II T ) are Banach S, paces. On the other hand, Example 82-1 shows that no norm on the vector space Q can be defined so that the resulting space is a Banach space. For finite-dime nsional vector spaces, completeness depends entirely 011: the vector space ; for infinite-dime nsional vector spaces Example a-2-2 shows that
1 34
8.
The Lebesgue Spaces
Y
completeness depends also on the choice of norm. The space (C[ - 1 , 1 ], ll · ll oo) is a Banach space, whereas , the space (C[ - 1 , 1 ], II · IIR) is not. If the norms ll · lla and ll · llb are equivalent, then the correspond ing normed vector spaces are either both Banach or both incomplete, since the set of Cauchy sequences is the same for each space and convergence in one norm implies convergence in the other. The two norms II · IIR and II · II oo on C[ - 1 , 1] are evidently not equivalent. Given a set S c X , if (X, I · II) is a normed vector space, a new subset S called the closure can be formed by requiring that f E S if and only if there is a sequence lfn } of vectors in S (not necessarily distinct) such that fn � f (in the norm I · II). If S S, then S is called a closed set. The subset S is called complete if every Cauchy sequence in S converges to a vector in S. For Banach spaces the concepts of completeness and closure are linked by the following fundamental result: =
Theorem 8.2 . 1
Let (X, I I II) be a Banach space and S c X. The set S is closed if and only if S is complete. In particular, if S is a subspace, then it forms a normed vector space (S, 11 · 11 ) in its own right, and if it is closed, the above theorem indicates that ( S, 11 · 11) is a Banach space. This observation leads to the following ·
corollary: Corollary 8.2 .2 Let (X, II · II) be a
Banach space and S
c X.
Then ([S], II · II) is also a
Banach space. E xercises 8-2 :
1 . Let (X,
II II) be a normed vector space and suppose that {an } is a sequence in X that converges in the norm to some element a E X. Prove that {an } must be a Cauchy sequence.
2. Let
·
lfn } be the sequence defined in Example
8-2-2 . Prove that lfn } is not a Cauchy sequence in the normed vector space (C[- 1 , 1 ] , II ll oo ) ·
·
8.3.
8.3
Completion of Spaces
1 35
Completion of Spaces
If a normed vector space is not complete, it is possible to 11expand" the vector space and suitably redefine the norm so that the resulting space is complete. In this section we discuss this process, and in the next section we apply the result to get a Banach space with a norm defined by the Lebesgue integral. Before we discuss the main result, however, we need to introduce a few terms. In functional analysis, a function T : X -+ Y that maps a normed vector space X to a normed vector space Y is called an operator . •
Example 8-3-1:
Let the set C 1 [a, b] and the norm II lh , oo be as defined in Exercises 8-1 , No . 3 . Every function in C 1 [a, b] has a continuous derivative. If T is the operator corresponding to differentiation d/ dx , then T maps every element in C 1 [a, b] to a unique function continuous on the interval [a, b]. Thus T maps C 1 [a, b] into C[a, b]. The definition of an operator is not norm dependent, but for illustration, we can regard T as mapping the space (C 1 [a, b], ll · lh , oo) into the space (C[a, b], ll · ll oo) · ·
Much of functional analysis is concerned with the study of oper ators. For a general discussion the reader can consult [ 2 5]. Here, we limit ourselves to a special type of operator that preserves norm. An operator T from the normed vector space (X, I llx) into the normed vector space (Y, II II y ) is called an isometry if for all x1 , x2 E X ·
·
In essence, an isometry preserves the distance between points in X when they are mapped to Y. An isometry must be one-to-one. If there exists an isometry T : X -+ Y such that T is onto (so the inverse T 1 : Y -+ X exists), then the normed spaces (X, II · llx) and (Y, 11 . I y ) are called isometric. If two spaces are isometric, then completeness in one space implies completeness in the other. -
Example 8-3-2 :
Th e differentiation operator of Example 8-3-1 is clearly not an isom etry from (C 1 [a, b], I lh,oo) into the space (C[a , b], I ll oo), sin ce in ·
·
1 36
8.
The Lebesgue Spaces I!
ge neral, llf - g lh ,oo
=
sup lf (x) - g (x) l + sup lf ' (x) - g ' (x) l xe [a,b] xe [a,b] > sup lf ' (x) - g' (x) l = llf ' - g ' ll oo · xe [a,b]
Example 8-3-3:
Let H(D ( c; r)) denote the set of all functions holomorphic (analytic) in the closed disk D ( c; r) = {z E C : l z - c l < r} , r > 0. This set forms a complex vector space, and the function II · li e defined by llf ll c
=
zED(c;r)
sup lf (z) l
provides a norm for the space. In fact, it can be shown that (H(D (c; r)) , ll · ll c) is a Banach space. Let Trp ,b be the operator mapping H (D (a; r)) to H (D (a - b ; r)) defined by Trp, b[ = eitl>f (z - b), where ¢ E � is a constant. The operator Trp, b is a one-to-one and onto mapping from (H(D (a ; r)), II · ll a ) to (H (D (a - b ; r)), II · ll a-b ), and since II T¢,bf - Trp ,bg ll a-b
=
zeD(a-b;r)
sup
zeD(a-b;r)
sup
=
zeD(a;r)
leitl>f (z - b) - eitl>g (z - b) l lf(z - b) - g(z - b) I
sup lf (z) - g (z) l
=
llf - g il a ,
the operator is also an isometry. The two normed spaces are thus isometric. Given an incomplete normed vector space (X, 11 · 11 ), it is natural to enquire whether the space can be made complete by enlarging the vector space and extending the definition of the norm to cope with the new elements. The paradigm for this process is the completion of the rational number system Q to form the real number system R. This example has two features, which, loosely speaking, are as follows:
8.3.
Completion of Spaces
1 37
(i) the completion does not change the value of the norm where it was originally defined, i.e., ll r ll e = ll r ll e, for any rational number r; (ii) the set Q is dense in the set
lR (cf. Section 1-1 ) .
The first feature is obviously desirable: We wish to preserve as much as possible the original normed vector space, and any extended defi nition of the norm should not change the value of the norm at points in the original space. The second feature expresses the fact that the extension of tne set Q to the set is a 11minimal'' one: Every el ement added to Q is required for the completion. We could have 11completed" Q by including all the complex numbers to form the complex plane C, which is complete, but this is overkill. The completion of the rational numbers serves as a model for the general completion process. Feature (i) can be framed for general normed vector spaces in terms of isometries. In order to discuss feature (ii) in a general context, however, we need to introduce a general definition of density. Let (X, II II) be a normed vector space and W c X. The set W is dense in X every element of X is the limit of some sequence in W . Density is an important property from a practical viewpoint. If W is dense in X, then any element in X can be approximated by a sequence in W to any degree of accuracy. For example, the sequence {an } of Example 8-2-1 consists purely of rational numbers and can be used to approximate ,J2 to within any given (nonzero) error. A fundamental result in the theory of normed vector spaces is that any normed vector space can be completed. Specifically, we have the following result:
lR
·
if
Theorem 8 . 3 . 1
Given a normed vector space (X, I I llx), there exists a Banach space (Y, II II y ) containing a subspace ( W, II II y ) with the following properties: (i) ( W, II II y) is isometric with (X, II llx ); (ii) W is dense in Y. The space ( Y, II II y) is unique except for isometries. In other words if (Y, II II y-) is also a Banach space with a subspace (W, II II y-) having properties (i) and (ii), then (Y, II II y-) is isometric with (Y, II II y ) . ·
·
·
·
·
·
·
·
·
·
1 38
8.
The Lebesgue Spaces
LP
The space ( Y, II · II y ) is called the completion of the space (X, ll · llx). The proof of this result would lead us too far astray from our main subject, integration. We refer the reader to [ 25] for the details. Exercises 8-3 :
1.
(a) Suppose that Z is dense in that Z is dense in Y.
W, and W is dense in
Y. Prove
(b)
Suppose that the completion of (X, ll · llx) is (Y, ll · ll y ) and that P is dense in X . Prove that (Y, II · II y) is also the completion of (P, II · llx) . 2. Let P[ a, b ] denote the set of polynomials on the interval [a, b ], and let PQ[a, b ] denote the set of polynomials on [a, b] with rational coefficients. Prove that PQ[a, b] is dense in P[a, b]. 3.
Weierstrass' s theorem asserts that any function in C[a, b] can be approximated uniformly by a sequence of polynomials, i.e. , P[a, b] is dense in C[a, b] with repect to the II ll oo norm. Use Ex ercises 8-3 , No. 2, to deduce that any function in C[a, b] can be approximated uniformly by a sequence in PQ[a, b]. ·
8 .4
The Space
L1
Having made our brief foray into functional analysis, we are now ready to return to matters directly involved with integration. Ex ample 8-2-2 shows that the normed vector space (C[a, b], II IIR) is not complete. We know, however, that this space can g e completed, but it is not clear exactly what kinds of functions are required to complete it. In this regard, the norm itself can be used as a rough guide. Clearly, a function f need not be in C[a, b ] for the Riemann integral of If I to be defined. This observation suggests that perhaps the appropriate vector space would be R[a, b], the set of all functions f : [a, b] � lR such that If I is Riemann integrable. This 11expansion" of C[a, b] to R[a, b] solves the immediate problem, since the sequence lfn } in Example 8-2-2 would converge to a function f E R[a, b], but it opens the floodgates to sequences such as that defined in Section ·
8.4. The
Space
1 39
L1
3-3-1 that do not converge to functions in R[a, b] . Although 11 II R is not a norm on R[a, b] (Exercises 8-2, No. 2(b)) , this problem can be overcome. Any hopes of using R[a, b] to complete the space, how ever, are dashed by Example 3-3-1 , because this example indicates that (R[a, b ], II IIR) is not complete. Recall that Example 4-3-1 motivated us initially to seek a more •
·
general integral to accommodate functions such as
f(x) •
=
{ 01 ,,
�f x �s �atio?al, x -j. 0, 1 , if x
IS Irrational or x
=
0, 1 .
Eventually, we arrived at the Lebesgue integral. The function f plays a role in the completion of (R[a, b ], I IIR) analogous to that played by the number -J2 in the completion of (Q, II · l ie) · The Lebesgue integral essentially leads us to the appropriate space and isometry for the completion of (R[a, b ], II IIR) (and (C[a, b ], II · IIR)). Let A I [a, b ] denote the space of all functions f : [a, b] -+ lR that are (Lebesgue) integrable on the interval [a, b] and let II II I be the function defined by ·
·
·
llf l h
=
r lf (x) l dx .
J[a,b]
The set A I [a, b] forms a vector space, but I II I is not a norm on it because there are nonzero functions g in A I [a, b ] such that ll g lh = 0 , i.e . , if g = 0 a. e. then ll g l h = 0. Functions that fail to be norms solely because they cannot satisfy this condition are called se:minorms, and the resulting spaces are called se:minormed vector spaces. Notions such as convergence and Cauchy sequences are defined for seminormed vector spaces in the same way they are defined for normed vector spaces. The problem with the seminorm on A I [a, b] is not insurmount able. The essence of the problem is that ll g ll = llf ll whenever f = g a.e. (Theorem 5.2.3 (iii)) . The set A I [a, b], however, can be parti tioned into equivalence classes based on equality a.e. Let L I [a, b] denote the set of equivalence classes of A I [a, b]. An element F of L I [a, b ] is thus a set of functions such that if[I , fz E F, fq�n f1 = fz a . e. Since any element f of F can be used to represent the equivalence ·
1 40
8.
The Lebesgue Spaces I!
class, we use the notation F = [f]. 1 Addition is defined as
[f] + [g] = [f + g],
and scalar multiplication as
a[f ] = [af ]. The set L1 [a, b] forms a vector space, and if II 11 1 is defined by ·
ll [f] ll
=
{ lf(x) l dx , lra , b]
then (L1 [a, b], I li t) is a normed vector space. The candidate for the completion of the space (C[a, b], II · IIR) (and the space (R[a, b], II · IIR)) is the space (L1 [a, b], II l h) . In the nota tion of the previous section, we have X = C[a, b], I llx = I IIR , Y = L1 [a, b], and I I II Y = I l h . Let W = {[f ] E L1 [a, b] : [f] contains a function in C[a, b]} , and let T be the operator that maps a function f E C[a, b] to the element [f] E W . Now, every element of C[a, b] has a corresponding element in W , and no equiv alence class in W contains two distinct functions in C[ a, b ]; therefore, T is a one-to-one, onto operator from C[a, b] to W . Moreover, Theorem 4.6. 1 implies that ·
·
·
·
II Tf l h
·
·
=
ll [f] l h
=
. t if(x) l
{ lf (x) l dx lra ,b] dx = llf ii R .
so that T is an isometry. The space W is thus isometric with C[a, b]. To establish that (L1 [a, b], II · 11 1) is the completion of (C[a, b], I · IIR) it remains to show that W is dense in L1 [a, b] and that GL} [a, b], I · lh ) is a Banach space. We will not prove that W is dense in L1[a, b]. The reader is referred to [ 37 ] for this result. We will, however, sketch a proof that (L1 [a, b], I · lh ) is complete. Theorem
8.4. 1
The normed vector space (L1 [a, b], I · li t ) is a Banach space. 1 Although this is standard notation, there is some danger of confusion with the
notation used for the span that takes sets as arguments
8.4. The Space L1
1 41
Proof We prove that the seminormed space (A 1 [a, b] , II · lh ) is complete. The completeness of (L 1 [a, b] , I lh ) then follows upon ·
identification of the functions with their equivalence classes in L 1 [a, b]. Let lfn } be a Cauchy sequence in (A 1 [a, b] , I II I ) . Given any € > 0 there is thus an integer N such that llfn - fm II < € whenever m > N and n > N. In particular, there is a subsequence lfnk l of lfn l with the property that ·
Let gm
=
m
L lfnk+l - fnk l , k=l
and let g = limm �oo gm denote the pointwise limit function. Note that g(x) need not be finite for all x e [a, b] . Let [a, b] = h U h , where 11 denotes the set of all points such that g (x) < oo . We will show that Iz must be a null set. Now, gm e A 1 [a, b] and ll gm ll 1
= =
1
[a,b]
lgm(x) l dx <
t1 k=l
m
m
[a,b]
L utnk+l - tnk ll l L z k <
k=l
k=l
1
lfnk+1 (x) - fnk (x) l dx <
1.
The sequence {gm } is a monotone sequence of functions in A 1 [a, b], and the above inequalities indicate that limm�oo ll gm l1 1 < 1 . The monotone convergence theorem (Theorem 5 .3. 1) implies that g e A 1 [a, b] and ll gm ll 1 --* ll g l1 1 ; hence, ll g lh < 1 . Since ll g ll 1 is finite, g(x) < oo a.e. , and so the set Iz must be null. The series fn1 (x) +
L Cfnk+1 (x) - fnk (x)) 00
k=l
must therefore be absolutely convergent for almost all x . This series thus defines a function f, the pointwise limit, almost everywhere. Eventually, f will be identified with an equivalence class in L 1 [a, b], so the fact that f is defined only a.e. is not a real concern.
1 42
8. The Lebesgue Spaces lJ'
We have shown that fnk -+ f; we need to show that fn -+ f in the 11 · 11 seminorm and that f E A 1 [a1 b]. Since lfn } is a Cauchy sequence/ 1 for any € > 0 there is an integer N such that ll fn - fm lh
=
1
[a, b]
lfn (X) - fm (X) I dx < €
for any m > Nl n > N. Let k be sufficiently large so that nk > N and let m = n k . Then for n > N 1
( 1
0 < lim inf k� oo m ?:. k
[a,b]
)
lfn (X) - fnm (X) I dx
< €1
and so Fatou l s lemma (Lemma 5 . 3 .2) implies that for n > N 1 lfn -f l is integrable over [a1 b] and a , b] lfn (x) - f (x) l dx < €. Therefore/ fn - f E A 1 [a1 b]1 and so f E A 1 [a1 b]. Moreover/ llfn - f ll 1 -+ 0 as n -+ 00 1 so that the sequence lfn } converges to f in the II · I I I seminorm. The space ( A 1 [a1 b ] 1 II · 11 1 ) is thus complete. The completeness of this space implies the completeness of the space (L1 [a1 h ] 1 II · II I ) � since each Cauchy sequence in L1 [a1 b] can be represented by a Cauchy D sequence in A 1 [a1 b].
fr
8.5
The Lebesgue Spaces IY
// The norm defined for the space L1 [a1 b] is a "natural choice in appli cations where the average magnitude of a function is of conspicuous importance. The function llf l l 1
=
1
[a,b]
lf (x) l dx
is the continuous analogue of the norm II · II r defined in Example 81 -4 for Rn . If we seek a continuous analogue for the Euclidean norm in Rn we are led to the function II · 1! 2 defined by llf ll 2
=
{1
[a,b]
lf (x) l 2 dx
}
1 12
I
8. 5 . The Lebesgue lJ'
1 43
and more generally/ if we seek a continuous analogue to a general mean norm for Rn 1 lip ll x ll = { l xi iP + !x2 !P + · · · + lxn !P } for p
>
1 1 we are led to a function II · l iP defined by 11P llf ii P = lf(x) IP dx [a,b]
}
{1
In this manner/ vector spaces for which these functions define norms or seminormS' come into prominence. The space (L 1 (a 1 b]1 ll · l l 1 ) serves as a prototype for all the Lebesgue spaces. Let AP [a1 b] 1 1 < p < 00 1 denote the set of measurable functions f such that
1
[a,b]
lf (x) IP dx
<
oo .
Now/ l lf llp = ll g ll p for any J; g E AP [a 1 b] such that f = g a.e. 1 so we know that I · li is at best a seminorm for AP [a 1 b] . This problem P can be easily remedied by using equivalence classes. A more serious concern is that AP [a 1 b] may not even be a vector space. In partic ular/ if J; g E AP [a 1 h] 1 it is not clear that f + g E AP [a1 b] . Moreover/ it is not obvious that I · l i will satisfy the triangle inequality. As it P turns out/ the sets AP [a 1 b] are vector spaces and ll · ll p is a seminorm 1 on them for 1 < p < oo . This follows from Minkowski S inequal ity/ which is derived from another inequality of importance called 1 H6lder S inequality (versions of these results are given below for the corresponding IJ' spaces) . Let Il'[a1 b] denote the set of the equivalence classes of AP [a 1 b] modulo equivalence a. e. 1 and for [f] E Il' [a 1 b ] define the function II · li p by ll [f] ll p =
1
{1
[a,b]
lf(x) !P dx
}
Theorem 8 . 5 . 1 (H6lder S Inequality) Let E LP [a 1 b] and G E Lq[a1 b], where 1 < p Then FG E L 1 [a1 b] and
F
11P
<
oo and l ip + 1/ q =
1.
1 44
8.
Theorem
Let 1 and
<
p
The Lebesgue Spaces U'
(Minkowski ' s Inequality) < oo and suppose that F, G E If[a, b] . 8.5.2
Then F + G E If[a, b]
The proofs of these inequalities can be found in most texts on functional analysis, e.g. , [38] . In the Holder inequality, the product is the pointwise product of functions, i.e. , if F = rf] , G = [g] , then FG = rfg], where (fg)(x) = f (x)g(x) . That If[ a, b] is a vector space and I I · l ip defines a norm on it follows from Minkowski ' s inequality. As with the space (L 1 [a, b] , II · l h ) , the normed vector spaces (If[a, b] , II · l ip ) are complete. This result is a generalized version of the classical Riesz-Fischer theorem. Theorem
8.5.3
The normed vector spaces p < 00 .
(If[a, b] , I I
·
l i p) are Banach spaces for 1 <
A detailed proof of this result can be found in [1 7] and [1 8]. The proof for the case 1 < p < oo is similar to that for the case p = 1 . Essentially, the civilized behavior of the Lebesgue integral (as manifested in the monotone convergence theorem) is responsible for completeness. The Lebesgue integral thus yields an entire family of Banach spaces. To simplify notation, we shall refer to the Banach space (If [a, b] , II · l i p ) simply as Ll[a, b] unless there is some ambiguity re garding the norm. These Banach spaces are collectively referred to as the Lebesgue or If spaces. We also follow the common (and con venient) practice of blurring the distinction between AP [a, b] and If[ a, b] by treating elements of V[a, b] as functions. We trust the reader to make the correct technical interpretation ana to remember in this context that "f = g " means f = g a. e. Suppose that f E L2 [a , b] . The constant function g = is also in L2 [a, b], and therefore Holder' s inequality implies that the function f · = f is in L1 [a, b] . In addition, we have that
1
1
llf lh < llf ll 2 11 1 llz = (b - al12 11f llz . This observation shows that L 2 [a , b] c L 1 [a, b] . The calculation 11Works" because p = q = 2 in Holder's inequality and g = 1 is
8.5 . The Lebesgue
U'
1 45
finite p 1 p 1.
integrable on any interval of length. We can repeat this argu ment for the general > because g E Lq[a, b] for any q. Thus, I .LP[a, b] c L [a, b] for all > The next result indicates that if 2 < P I < P z , then LP [a, b] c LP1 [a, b] .
1 Let 1 < PI < P2 and suppose that f Theorem 8 . 5 . 4
E
.LP2 [a, b].
Then
I Ip - 1P llf ii P1 < ll f iiP2 (b - a) 1 1 2 ,
nd consequently f .LP1 [a, b]. Proof If pi = pz , the result is trivial. Suppose that 1 < PI p2 , and let k = p2 /p1, m = kl(k - 1). Now, a
E
•
{
lra,b]
<
! lf (x) !Pl l k dx =
{
J[a,b]
lf (x) !P2 dx < oo ,
since f E LP2 [a, b], and therefore lf (x) IP1 E Lk[a, b] . Holder's in equality with E L m [a, b] yields the = k, q = m, and g = inequality II If (X) lp 1 I II I < II If (X) l p1 II k II l ii m
p
1
·
·
and
Consequently, ( llf 11Pl y1 < ( 1! f iiP2 Y 1 (b - a) I -p 1 /p2 ,
and the inequality follows.
0
Two positive numbers and q are called conjugate exponents q = if Holder's inequality suggests that the Banach spaces .LP[a, b] and Lq[a, b] are related if p and q are conjugate exponents. In fact, the spaces are intimately related. A linear functional is an
1/p + 1/ 1.
p
1 46
8. The Lebesgue Spaces
V'
operator 1 from a normed vector space (X, 11 · 1[ ) to the scalar field (1R or C) of the vector space such that for any [I, fz E X 1 ([1 + fz ) = l(fi ) + l(fz), and for any scalar a
The functional 1 is bounded if there is a number c > 0 such that 11 (f) I < c ! I f II for all f E X. The norm of a bounded functional 1 is defined as the smallest number c such that 11 (f) I < c ! If II for all f E X and denoted by I ll 11 . Since = 0 is a legitimate choice for a scalar, we see that 1(0 · f1 ) = 1(0) = Ol(f1 ) = 0 , i.e., 1(0) = 0. The norm of 1 is thus given by a
111 11
= sup ll(f) l f=/0 fEX
ll f II
(any choice of c > 0 satisfies 101 < c l i O !! ) . Exam.ple
8-5- 1 :
Let
l (f)
=
1
[a,b]
k(x)f(x) dx,
where k E L2 [a, b] , and let X = L2 [a, b] . Now, 1 is evidently a linear functional, and Holder's inequality implies that for any f E L2 [a, b] , ll(f) l <
thus,
1
[a,b]
•
l k(x)f(x) l dx = llkf ll 1 < l ! k ll z llf ll z ; I ll II
< II k H z .
I n fact, IIl ii = H k l! z . To see this, note first that if Il k l i z = 0 , then k = 0 a.e., and therefore kf = 0 a.e. for any f E L2 [a, b] ; hence, l(f) = 0 for all f E L2 [a, b] and IIl ii = 0 . If Il k liz :f:. 0 , choose f = kl l l k ll z E L2 [a, b] .
8.5 . The Lebesgue
If
1 47
Now, ll Cf) l
=
1
[a,b]
1
=
ll k ll 2
k(x)
1
[a , b]
k(x) dx ll k ll 2 k2 (x) dx = ll k ll 2 .
Since ll f ll 2 = 1 for this choice, we see that II l ii = ll k ll 2 . The above example is typical of bounded linear functionals on the Lf[a, b] spaces for 1 < p < oo . The following representation theorem expresses the situation: Theorem
8.5.5
Let p and q be conjugate exponents with 1 < p < oo, and suppose that 1 : Lf[a, b] � 1R is a bounded linear functional. Then there exists a unique element g E L q [a, b] such that lCf) = for all f E If[ a, b] . Moreover;
1
[a,b]
II l i i
=
f(x)g (x) dx ,
ll g ll q ·
The proof of this fundamental result can be found in [37]. In the more general context of functional analysis, the dual space of a normed space (X, I · II) is defined to be the set of all bounded linear functionals on X. The above theorem indicates that Lq [a, b] is the dual space of If[ a, b] when p and q are conjugate exponents with 1 < p < 00 . Conjugate exponents and dual spaces bring to the fore two ex ceptional cases. Each Banach space If[a, b] is paired with its dual space Lq [a, b], where l ip + 1/q = 1 , 1 < p < oo . The conjugate exponent of p = 2 is q = 2, and so L2 [a, b] is its own dual space. The space L2 [a, b] is clearly the only member of the Lf[a, b] spaces with this property, and it turns out that L2 [a, b] has several other special properties not shared by the other Lf(a, b] spaces. The space L2 [a, b] is special because it forms what is known as a Hilbert space. We postpone further discussion about the special properties of L2 [a, b] until the next chapter.
1 48
8.
The Lebesgue Spaces lJ'
The other exceptional case is the space L 1 [a, b]. There is no conjugate exponent for p = 1 , yet there are certainly bounded lin ear functionals defined on L 1 [a, b]. What then is the dual space of L 1 [a, b]? If l k (x) l < M < oo for all x E [a, b], then the functional J defined by J (f) = {
lra,b]
k (x)f(x) dx
is a bounded linear functional, since IJ (f) l < M {
lra,b] lf
(x) l dx = M ll f lh .
In fact, the condition that l k (x) l < M < oo for all x E [a, b] can be relaxed to l k (x) l < M < oo a. e. in view of the Lebesgue integral. This example serves as a guide to what might be expected as a dual space for L 1 [a, b]. Let f [a, b] � be a measurable function. A number J-L is an essential upper bound for If I if lf (x) l < J-L a. e. If If I has an essential upper bound, then it can be shown that there exists a least such bound. This least bound is denoted by ess sup If 1 . Let A 00 [a, b] denote the set of all measurable functions on [a, b] such that ess sup If I < oo and define the function I · I 00 by ll f ll oo = ess sup lf l . As with the AP [a, b] spaces, A 00 [a, b] is a vector space and II ll oo is a seminorm on it. Let L00[a, b] denote the set of equivalence classes of A 00[a, b] modulo equality a .e. Then (L00 [a, b], 11 · 11 00 ) is a normed vec tor space. The following theorem summarizes some of the properties of L00 [a, b]: :
--*
·
Theorem
8.5.6
(i) (L00[a, b], II
ll oo )
is a Banach space. (ii) The dual space of L 1 [a, b] is L00[a, b]. (iii) If f E L 1 [a, b] and g e L00 [a, b], then fg E L 1 [a, b] and ·
llfg I l l < ll f II I ll g I I 00
(an extension of the Holder inequality). (iv) Iff E L00 [a, b], then f E If[a, b] for all 1 < p
< oo.
8.5 . The Lebesgue If
149
The proofs of parts (i) and (ii) can be found in [33]. The proofs of parts (iii) and (iv) are left as exercises. Ifp and q are conjugate exponents with p > then the dual space of LP[a, b] is L q [a, b] and vice versa. In the language of functional analysis, the £P spaces are reflexive for p > In contrast, the dual space of L00[a, b] is not L 1 [a, b] . The dual space of L00(a, b] turns out to be a space of measures. More details concerning the dual space of L00 [a, b] can be found in [34]. The space L 1 [a, b] is thus unusual among the IF[ a, b] spaces in that it is not reflexive. We have t4us far been concerned with £P spaces where the inter val of integration is bounded. The definitions for £P spaces can be extended to include unbounded intervals of integration. If I c IR is any interval, the space LP(I) consists of the set of functions f : I � lR (i.e. , equivalence classes) such that
1.
1,
All the results stated for the LP[a, b] spaces carry over to the general LP(I) spaces with the notable exception of the inclusion results. The proof of Theorem 8.5.4 relies crucially on the fact that the interval is bounded. If the interval is not bounded, then results such as L 2 (I) c L1 (I) are not valid. For example, consider the functions [I and h defined by
fi (x) =
.;.
12 (x) =
{ 0,1 / y'x, { 0,1 x, /
�f O < X <
If X >
if 0
1,
1, 1.
1,
< X<
if x >
Now, f1 E L1 (0, oo), but [2 E L1 (0, oo); alternatively, but [2 E L 2 ( 0, oo) . Hence L2 (0, oo) does not contain L1 (0, oo) does not contain L2 (0, oo). Exercises 8-5 : 1.
Suppose that the function r positive on the interval [a, b].
[0,
1]
f1 E L2 (0, oo) L 1 (0, oo), and
� IR is continuous and
1 50
8. The Lebesgue Spaces
(a) Prove that for
Y
1 < p the function r II r ll f llp =
(
{
J[a,b]
)
· l i P defined by
r (x) lf (x) IP dx
1�
is a norm on ll[a, b].
(b) Prove that the norm r ll liP is equivalent to the norm II · lip · 2 . Let 1 < PI < Pz and let I be a bounded interval. Prove that llf llp 1 < 1 + ll[ ll p2 and from this deduce that Il2 (I) c Il1 (I) . ·
3.
The
Fredholm integral operator K is defined by (Kf)(x) =
{
,
J[O l ]
k (x, ;)[(;) � '
where k : [0, 1 ] x [0, 1 ] --+ � is a given function. Suppose that k is bounded in the square [0, 1 ] x [0, 1 ]. Prove that the operator K maps functions in L 2 [0, 1 ] to functions in L2 [0, 1 ].
4 . Prove parts (iii) and (iv) of Theorem
5.
If f E
L00 [0, 1 ],
8.5.6.
prove that
ll f ll 1 < llf ll z < · · < llf ll n < ll f ll n+l < · · · < ll f ll oo · ·
8.6
Separable Spaces
A Banach space (X, I I II) is separable if X contains a countable set that is dense in X in the II · II norm. The Banach space (�, II li e) , for example, is a separable space. To establish this clairp. we note that the rational numbers form a countable set that is dense in the real numbers . Separability is a desirable property, because the existence of a countable dense set simplifies problems concerning the represen tation of functions (e.g. , bases for spaces) and the approximation of functions. In the Hilbert space theory discussed in Chapter 9 , separability is a key ingredient in establishing the existence o f an orthonormal basis (cf. Theorem 9.2.4) . Fortunately, the If spaces are separable. ·
·
8 . 6 . Separable Spaces
Theorem
1 51
8.6 . 1
The V spaces are separable for 1 < p <
oo.
The proof of this result can be found in (1 7]. We limit ourselves here to a brief informal discussion of the ideas underlying the proof and focus on sets that are dense in the V spaces. These sets are of interest in their own right for the purpose of approximation . The construction of the Lebesgue-Stieltjes integral in Chapter 4 brings to the fore one class of functions that must be dense in any V space, viz. the set of functions consisting of x-summable step functions (Section 4.5) . This density relationship is used to prove the Riemann-Lebesgue theorem (Theorem 9.3. 4) in the next chapter. Unfortunately, this set is not countable. There are, of course, other sets dense in the I.l spaces. The next result (which we state without proof) provides a sample of such a set. Theorem
8. 6.2
Let ego (I�) denote the set offunctions f : R --+ R with derivatives of all orders and with supports offinite length. The set ego (JR.) is dense in each V (R) space for 1 < p < oo. Similar results are available for the V(I) spaces where I is an interval. The set ego (R) is not countable, but if the above theorem is cou pled with Weierstrass's theorem concerning the approximation o f continuous functions by polynomials (Exercises 8-3 , No . 3) , a count able dense set can be derived. Suppose that f E cgo(R) and choose some E > 0. Now, f has a support of finite length, and hence there is some number f3 > 0 such that f = 0 outside the interval -{3, {3] . Certainly f (restricted to the interval -{3, f3]) is in the space -{3, {3], and thus there is a polynomial PE such that
[ C[
[
sup IPE (x) - f(x) l <
xe[ -,8,,8]
E
(8. 1 )
( Z JJR) l iP
PQ[
Moreover, we know from Exercises 8-3-2 that the set -{3, f3] of polynomials with rational coefficients is dense in P [ -{3, {3] , and hence we can find a polynomial with rational coefficients satisfying inequality (8. 1 ). In fact, we can find a polynomial q€ with rational co efficients such that inequality (8. 1 ) is satisfied and q€ ( -{3) = q€ ({3) = 0. Now the polynomial q€ can be extended to a function gE = q€ · X[-,8,,8]
1 52
8. The Lebesgue Spaces V
defined on �- Here
X [ - tJ , tJ]
X [ _ tJ , tJ]
is the characteristic function defined by
(x) =
{ 0,1 ,
Thus,
llgE - f l i
p = { �{ lgE (x) - f(x) !P dx } <
sup
xe[ - tJ, tJ]
if x E [ - ,8, ,B], otherwise .
l lp
=
lgE (x) - f(x) I (2,8) 1 1P
{1
�A�
lgE (x) - f(x) IP
dx
}
l lp
< E.
Now, the set of rational numbers is countable, and this can be used to establish that the set PQt - ,B, ,8] is also countable. If we define the set r = {g : g = q · X [ -t! ,tJJ for some q E PQ[ - ,8, ,8] and some ,B = 1 , 2 , . . . } , then it can be 1?hown that r is a countable set in ego�). and hence The above inequality indicates that r is dense in it is dense in V' (�) . Thus, V' (�) contains a countable dense set.
Cjf(IR),
8.7
Complex
£P
Spaces
In this section we pause to make a modest generalization of V' spaces to include complex-valued functions of a real variable. A fuller treat � be some interval ment of these spaces can be found in [1 7]. Let and let f : -+
Ic
I
[t(t)dt [ =
Re f(t) dt + i
[ Im f(t)dt.
We wish to construct spaces analogous to AP (I) and V' (I) . The natu ral generalization of the norm function is the function II · li defined P by
lif l i
p
=
{ [ lf(t)[P dt}
l ip ,
8. 7.
Complex V Spaces
1 53
where for 1 = Re f - iim f (the complex conjugate) , lf l 2 = ff . The next theorem shows that lf !P is integrable if and only if I Re f iP and I I m f iP are both integrable. The proof rests on the inequalities I Re f l < lf L I Im f l < lf l , and lf l < I Re f l + I Im f l , and is left as an exercise. Theorem 8 . 7 . 1 Let f : I --+
c
•
if and only if
1
l
lf(t) IP dt < oo
1 Re f(t) 1P dt < oo and
l
l lm f(t) IP dt < oo .
Let A�(I) denote the set of functions f : I --+ 1 , then the dual space of L�(I) is the space Lt (I). If J is a bounded linear functional from L�(I) to
J (f)
=
1 (t
f )g (t ) dt
for all f E L�(I). The space Lc (I) can be defined in a manner anal ogous to that used to define L00 (I), and this space is the dual of Lt (I) .
1 54
8.
The Lebesgue Spaces Y
The Lt (I) spaces often arise in applications involving line in tegrals in the complex plane. Recall that if y is some curve in C represented parametrically by some piecewise smooth function for t E to , tl ] , then the line integral of a function = + defined on y is given by
z (t) x(t) iy (t)
[
F
1 F(z) rlz 1 t1 F(z(t))z' (t) dt. =
y
to
The definition of a line integral can be extended using the Lebesgue integral in an obvious way, and thus line integrals can be defined for a more general class of functions. Note that a curve y may be parametrized any number of ways. For example, the variable may be replaced by any smooth function g : so , s! ] � to , l ] to give a new parametrization z(s) = provided that g s ¥= 0 for all s E [s o , s l ] . If g s > 0 in the interval [s0 , sl ] then the parametriza tion is orientation preserving, and it is well known that the value of the line integral in the Riemann setting is invariant under smooth orientation preserving reparametrizations. In the Lebesgue setting, Theorem 6 . 2 . 1 ensures that this is also the case. Thus, provided that the orientation of y is specified, we can use the notation fr without ambiguity. In this context we may also use the notation Lt (Y) unless a specific parametrization of y is required. A particularly interesting case is that in which y is a simple is holomorphic (ana closed curve (no self-intersections) and lytic) in the region enclosed by y but not necessarily on y itself. In the next se ction , we investigate a class of spaces known as Hardy spaces and show that the Lt (I) spaces are closely related to spaces of holomorphic functions.
'( )
z(g (s)) ,
[
[ t '( )
t
F (z)rlz
F(z)
8.8
The Hardy Spaces
HP
Hardy spaces are complex normed vector spaces of functions holo morphic on a region of the complex plane. These spaces are closely related to the Lebesgue spaces and share the same properties. The functions in a Hardy space are quite different in nature from ele ments in an Lt(I) space. Aside from the fact that elements in the
8.8. The Hardy Spaces HP
1 55
latter set are equivalence classes , the underlying functions in an L�(I) space need not be continuous. In contrast, the functions in a Hardy space are infinitely differentiable. That these spaces should be so closely linked is remarkable. Hardy spaces play an impor tant role in complex function theory and harmonic analysis. These spaces also arise in applications such as control theory. The discus sion in this section presumes some knowledge of complex analysis, but it is primarily a descriptive account and no proofs are given. For a basic reference on complex analysis the reader is directed to [2 ] , [10], or [ 3 9 ] . A much fuller discu$sion of Hardy spaces along with the proofs of various results in this section can be found in [20], [21 ], and [37]. Let n C denote a region (Le. , a nonempty, connected open set) and let A( n) denote the set of functions holomorphic on n. It is clear from the elementary properties of holomorphic functions that A( n ) is a vector space. A candidate for a norm on A( n) is the function II II defined by
c ·
00
llfll oo
= s up zen
lf (z) l .
Although this function satisfies the conditions for a norm such as the triangle inequality, it is not finite for all f E A( n ) . There are functions in A( n) with singularities on the boundary an of n. If the set A( n) is restricted to the functions f for which llfll oo < oo, then the resulting space is a normed vector space. Let
H00 ( n) = {f E A(Q) llf ll oo :
<
oo } .
The set H00 ( n) is still a vector space, and now II II 00 is a norm on it. In fact, the space (H00 ( n), II · ll oo ) is a Banach spac e. The value of ll f l l oo for a given f E H 00 ( n) is less tractable than the analogous function defined fo:r C[a, b] (Example 8-1 -5) owing to the two-dimensional nature of the region n and the fact that n is an open set. A fundamental result in complex analysis known as the maximum modulus principle, however, mitigates these problems. C be a bounded region with boundary an and suppose that Let n f E A( n) and also that f is continuous in the set n = n u an. Under these conditions, the maximum modulus theorem implies that the function lfl assumes its maximum value o n the boundary an and ·
c
1 56
8.
The Lebesgue Spaces 1l
that a nonconstant function in A(O) cannot have local maxima for lf l in 0 . The maximum modulus principle cannot b e applied directly to all the functions in H00 (0), sinoe functions in this set need not be continuous on Q u an. Nonetheless, the norm llf ll oo of a function in H00 (0) reflects the values of lf(.Z) I as z approaches the boundary of the region, and this suggests that the norm can be calculated using a limiting argument. The Riemann mapping theorem implies that arbitrary regions such as S can be conformally mapped to a unit disk in the complex plane. We can thus limit our investigati�ns mostly to the study of functions holomorphic on the unit disk centred at the origin. Let D(O; r) = {z E c : l z l < r} , D(O; r) = {z E c : l z l < r} , aDr = {z E c : l z l = r} , and for simplicity, denote D(O; 1), D(O; 1 ) , aD1 by D, D, and aD, respectively. If 0 < r < 1 , then, taking Q = D(O; r) , the maximum modulus principle implies that sup lf (z) l = sup lf (z) l . ze BDr
zeD(O;r)
The function Moo is defined by
Moo (r, f)
=
sup lf( reilP ) I ,
¢e [0,2rr]
and it is evident from the maximum modulus principle that for a fixed f E H00 (D) the function M00 is monotonic increasing in r and that sup lf(z) l = Moo (r, f) .
zeD(O;r)
We thus expect that for any f E !H00 (D),
lim Moo Cr, f) = llf ll oo ,
r--+ 1 -
and in fact, this relationship can be proved formally. The function Moo suggests a generalization for norms analogous to the IY norms. Let Mp be the function defined by
Mp (r, f)
=
{
2._ 2rr
1·
[0, 2 rr]
lf (rei
}
1/p
,
I S7
8.8. The Hardy Spaces HP
and let HP (D) be defined as
HP(D)
= {f E A(D) : sup Mp (r, f) r<
1
<
oo}.
It can be shown that for 1 < p < oo the Mp functions satisfy Holder and Minkowski-type inequalities. Specifically, if p and q are con jugate exponents with p > 1 , then for any functions f E HP (D), g E Hq (D), M1 (r, fg) < Mp (r, f)Mq (r, g),
and if h If p
>
E
HP(D) , then
Mp (r, f + h) < Mp (r, f) + Mp (r, h) .
1 , the function
I
·
l ip defin¢d by
llf l lp = s�p Mp (r, f) r<: 1
is therefore a norm on HP (D) . (Note that we need not form equiva lence classes because the functions involved are holomorphic, and thus if f = g a.e. , then f = g .) As with M00 (r, f), it can be shown that for all f E HP (D), The normed vector spaces (HP(JJ), II · l i p) are called Hardy or HP spaces. As with the Lebesgue spaces, we refer to the normed vector space (HP(D), I · lip) as HP(D) . 'Ule Hardy spaces share the same properties as the Lebesgue spaceS'. The next theorem gives a sample of some of these properties. Theorem 8.8. 1 (i) Holder's Inequality: If 1 < p < oo and q is the conjugate exponent ofp, then for any f E HP(D), g E Hq (D) we have that fg E H1 (D) and llfg l h < llf ll p llg ll q ·
(ii) Mink.owski's Inequality: For any f, g
E
HP (D) with p
>
1,
ll f + g ll p < llf ll p + ll g ll p ·
(iii) Completeness: The normed vector spaces (HP (D) , Banach spaces. ·
II
· lip ) are
1 58
8.
The Lebesgue Spaces V
(iv) Inclusion: For any 1 <
P 1 :5 p2
< oo,
The two inequalities in the above theorem are direct consequences of the corresponding inequalities involving the Mp functions. A proof of completeness can be found in [21 ]. The inclusion result is estab lished using the same approach as used to prove a similar result in Theorem 8.5. 4 . It is not a coincidence that the HP spaces mimic the IY spaces. These spaces are closely related through the 1'boundary values" of the functions in HP(D). Cauchy's integral formula is a remarkable mani festation that a holomorphic function is determined by its boundary values. Recall that for any f e A(D(O; r)), r > 0, the Cauchy integral formula states that
f (z)
=
� 2m
1.
8Dr
f (w) dw WZ
for any z e D(O; r) . Here, the circle aDr is oriented anticlockwise. The value off on the boundary aDr thus determines the function f uniquely in the interior D(O; r). We cannot apply Cauchy's integral formula directly on the boundary of D because f need not be holo morphic on aD, but we can still apply the formula on any boundary aDr when 0 < r < 1 . This suggests that any function f e HP(D) can be 11identified" with a function f defined on aD a. e. by some limiting process as r -+ 1 - . Let
fr(z)
==
f (rz) .
. If 0 < r < 1 , then fr(z) is holomorphic on D and the Cauchy integral formula implies that
fr(z)
=
1
fr(�) �. � 2m . aD � - z
Intuitively, we should be able to define a function f : aD -+ C by considering the function limr--+ 1 - fr . The next result shows that this approach does lead to an isometry from the HP(D) spaces to the LIJ:: (aD) spaces. The notation for the norms for these spaces is the
8.8.
The Hardy Spaces HP
1 59
same by convention. Here, we shall use the notation II IIHP , and II ll v to distinguish the norm f
·
Theorem 8.8.2 L_et 1 < p < oo and suppose that f E [f] E L�(8D) such that: (i) ll f ll v = ll f ii HP · (ii) limr --* 1 - llf - fr llv = 0. (iii) f (z) =
� 2m
HP(D). Then there is an element
1
[ (;)
aD � - z
�.
The above result indicates that there is an isometry from HP(D) to L�(8D). These spaces are not isometric, since the set L�(8D) contains elements with no corresponding members in HP(D). Let '!!P (8D) L�(8D) denote the set of elements lf] E L�(8D) such that f corresponds to some function f E HP(D). The complex normed vector spaces ('J-lP(8D), II ll v) and (HP(D), II II HP ) are isometric. Now, it can be shown that the space (1-lP(8D), II ll v) is a closed subspace of the Banach space (L�(8D), II ll v ), and this means that (1-lP(8D), II ll v) is also a Banach space. Since ('J-lP(8D), II ll v) is iso metric to (HP(D), II IIHP ), the ]atter space must also be a Banach space. Properties of the HP(D) spaces such as completeness are gen erally proved by establishing the analogous results for the 'J-lP(8D) spaces, and in this sense the Hardy spaces "inherit" the qualities of the IY spaces. The HP(D) spaces can be generalized to sets of functions holo morphic in arbitrary regions or the complex plane via conformal transformations. Perhaps the most frequently encountered Hardy spaces aside from the HP(D) spaces are the spaces of functions holo morphic on a half-plane. Specifically, the Hardy spaces of functions analytic on the right half-plane n 0 = {z E C : Re z > 0 } are of interest owing to (among other things) their connection with the Laplace transform. The Hardy $paces in the half-plane share most of the properties of the HP(D) spaces. Indeed these properties are sometimes established by using a Mobius transformation such as
c
·
·
·
·
·
·
·
z=
w-1 w+1
--
1 60
8.
The Lebesgue Spaces IJ'
w
E no; the circle This transformation maps points z E D to points aD is mapped to the imaginary axis. If g E A(D) , then the function f defined by f(
w) g (ww +- l1)
(8. 2)
=
is in A(no), and in this manner we can 11transport" some of the properties of the HP(D) spaces. We return to this comment at the end of the section. Let p Mp (x, n = lf(x iy) IP dy ' and define the function
I
{L +
·
r
l i p by
ll [ ll p = sup Mp (x, f) . x > (l The Hardy spaces HP(no) are defined as
HP (no) = {f E A(ll o)
:
l l f ll p < oo } .
In contrast with the set D, the set llo is not bounded in the complex plane, and this complicates matters because along any line Ix = {z E C : Re z = x, x > 0} the function fx defined by
fx (y) = f (x
+
iy)
mu st be in IY (Ix) · Moreover, we cannot appeal directly to the max imum modulus principle to evaluate llf ll p because the region is unbounded and the function may have a singularity at the point at infinity. The bound for Mp (x, f) for large x is as important as that for small x for the general f E A(IT 0 ). It turns out, however, that the requirement that ll [ ll p < oo forces f to tend uniformly to zero as z tends toward the point at infinity along any path inside any fixed half-plane of the form n8 = {z E C : Re z > > 0} . It can thus be shown that Mp (x, f) is a decreasing function of x and that
8
lim Mp (x, f) = llf ll p · x-+O+ As with the HP(D) spaces, the boundary values of functions In HP(no) can be mapped to elements in an IY space. The
8.9.
Sobolev Spaces
Wk,p
1 61
Lebesgue space associated with HP(no) is the space L�(IR), and the corresponding Cauchy (Poisson) integral for this case is
fx (Y) For each x
>
x ( f(it) dt. rr }R x2 + (Y - t) 2
=
0 the function fx is in L� (IR) , and llfx - f llv � 0
o+ . Moreover, we have that ll f II HP = llf llv . As remarked earlier, the sets D and n 0 are related by a Mobius transformation, and we can expect that the sets of functions in one space can be used to generate the functions in the other space. If g e HP(D), then the function f as defined in equation (8. 2) is certainly in A(n 0 ), but it is not clear whether f e HP(no) and whether all the functions in HP(no) can be generated by functions in HP(D). The final result of this section gives a concrete connection between the two spaces (cf. [ 2 1 ]). as
x�
Theorem 8 . 8 . 3 Let g e A(D) and define f as in equation (8.2). Ifp :::: 11 then f if and only if there is a function G e HP(D) such that
Equivalently�
g (z)
=
(1
f (w)
=
(1 + w) 21PF(w).
HP(no)
- zi1P G(z). the function g is in HP(D) if and only if there is a function
F e HP(no) such that
8.9
e
Sobolev Spaces
Wk,p
Sobolev spaces are another class of function spaces based on the Lebesgue integral. These function spaces have found widespread applications in differential equations and feature norms that not only ��measure" the modulus of a function but also the modulus of its derivatives in an Il' setting. In this section we give a brief glimpse
1 62
8.
The Lebesgue Spaces
LP
of these imp ortant spaces and ho p efully whet the reader' s appetite for a more serious study. A detailed account of the theory can be found in [ 1 ] . Let C 1 (I) denote the set of functions f mapping the interval ! IR to IR such that exists and is continuous for all E J. The set C 1 (I) forms a vector space, and the function defined by
c
x I · lh ,oo l f lh , oo lf(x) l lf'(x)l (C 1 (J), I l h , oo ) I · l h ,p 1 1 P l f lh .v { l lf(x)IP dx } + {llf'(x)IP dx } P l [ l p + l f'llp , (C 1 (I), l f l h ,p) · 1 < (C(I), I · l i p) (C 1 (I) , l f l h ,p) f'
= sup xei
+ sup xei
can be shown to be complete. is a norm. The space by More generally, we can define functions ·
=
1
1
=
and investigate the corresponding normed vector spaces For is not complete. The com p < oo we know that pletion of this space essentially led us to use the Lebesgue integral instead of the Riemann integral and "expand" the vector space to has IY(I) . Now we know that the normed vector space a completion, but it is not clear what new concepts will be involved in finding it. Evidently, we need to enlarge our vector space to include functions that have a first derivative in IY (I) , but is this adequate? That f must have a derivative in the classical sense requires f to be continuous, and this seems a strong restriction in the Lebesgue setting. The completion of spaces such as (C1 (I) , leads to the concept of distributions (generalized functions) a.nd generalized derivatives. A discussion about these functions would lead us far can be com astray. Suffice it to 'say that spaces Hke (C 1 (I) , pleted, but the 11Cost'' is the introduction of an entire new class of objects that are not functions in the classical sense. Once again, the completion of a space yields a new mathematical concept. The Sobolev spaces are the completions of spaces like (C 1 (I) , The completion of (C 1 (I), is denoted by or simply W 1 ·P (I). If Ck(J) denotes the set of func (W1 ·P (J), is defined tions with kth-order continuous derivatives, and
I · lh ,p)
I l h ,p) ·
l f 1l h.,p1)1·, p)
I · l h ,p) I · l k,p
8.9 .
by
Sobolev Spaces
Wk,p
1 63
k
11r 11 k,p
=
then the Sobolev space ( W k.P (J), of the space ( C k (I), II · ll k,p) ·
L 11rm lip, j =O
I ll k,p) is defined as the completion ·
Hilb ert Spa ces and L 2
C H A P T E R
9.1
Hilbert Spaces
Hilbert spaces are a special class of Banach spaces. Hilbert spaces are simpler than Banach spaces owing to an additional structure called an inner product. These spaces play a significant role in functional analysis and have found widespread use in applied mathematics. We shall see at the end of this section that the Lebesgue space L 2 (and its complex relative H 2 ) is a Hilbert space. In this and the next section, we introduce some basic definitions and facts concerning Hilbert spaces of immediate interest to our discussion of the space L 2 . Fur ther details and proofs of the results presented in these sections can be found in most books on functional analysis, e . g. , [25]. Let X be a real or complex vector space. An inner product is a scalar-valued function ( ·} on X x X such that for any f, g, h E X and any scalar a the following conditions hold: ·,
(i) {f, f}
0; (ii) {f, f} = 0 if and only if f = 0; (iii) if + g , h } = (f, h } + (g, h } ; >
(iv) {f, g} = (g, f} ; (v) (af, g} = a (f, g} .
1 65
1 66
9
•
Hilbert Spaces and L2
A vector space X equipped with an inner product ( · , · ) is called an inner product space and denoted by (X, ( · , · ) ) . Note that in general, (f, g) is a complex number; however, condition (i) indicates that (f, f) is always a real nonnegative number. Note also that conditions (iii) and (iv) imply that
(f, g + h )
=
(f, g) + (f, h ) .
In general, (f, f3g) = f3(f, g) # f3 (f, g) , and consequently the inner product is not in general a bilinear function. The special case arises frequently in applications that X is a real vector space and ( ·) is real valued. These spaces are referred to as real inner product spaces, and for these spaces the inner product is a bilinear function. ·
Example 9-1 -1 : e n and lRn (zl , Z2 , Let X = en and for any z e n define ( , ) by =
·
. . . I
Zn ) , w = (wl l W2 ,
,
. . . I
Wn )
E
·
n
( z , w ) = L zj Wj .
j= l
Then it is straightforward to verify that ( · , ) is an inner product Xn ) , y = on e n . Similarly, if X = ]Rn and for any X = (XI , X2 , n (Yl i Y2 , · . . , y n ) E lR the function ( ·, ·) is defined by n (x, y) = L xjyj , ·
. . . I
j =l
then ( · , · ) defines an inner product on R n . The definition of the inner product is modeled after the familiar inner product (dot product) defined for lR n .
Example 9 - 1 -2 : l2 Let .e2 denote the set of complex sequences {an } such that the series and scalar multiplication are :L�= l l an 1 2 is convergent . If addition defined the sam e way as for the space .e1 in Example 9-1 -3 , then l2 is a vector space. Suppose that a = {an } , b = { bn } E l2 , and let Cn = max(an , bn ) · Then the series L �= l l cn l 2 is convergent , and hence the series :L� an b n is absolutely convergent. An inner product on this
1
9.1.
Hilbert Spaces
.
167
vector space is defined by
L an bn . 00
(a, b} =
n =l Let (X, ( · , ·}) be an inner product space and let the function defined by
llf ll
=
I ·I
:
X � lR be
llfJ).
We will show that II · II as defined above is a norm hence justifying our notation. The.conditions for the inner product ensure that II · II meets the requirements for a norm except perhaps the triangle inequality. In order to establish the triangle inequality we need the following result, which is of interest in its own right: Theorem 9 . 1 . 1 (Schwarz ' s Inequality) Let (X, ( · , ·}) be an inner product space. Iff, g E X, then
I if, g } I < llf I l l ig 11 . The proof of this inequality is left as an exercise. Now,
llf + g ll 2 = if + g, f + g } = (f, f} + (f, g } + (f, g} + (g , g } = 11!11 2 + 2 Re (f, g } + ll g ll 2 < 11! 11 2 + 2 l (f, g } l + llg ll 2 , and Schwarz ' s inequality implies that
llf + g ll 2 < 11! 11 2 + 2 llf ll llg ll + ll g ll 2 = C llf ll + ll g lli . Thus,
llf + g i l < llf ll + ll g ll , and hence I · I defines a norm on X. Given any inner product space (X, ( · , ·}) we can construct a normed vector space (X, II · II) . The function II · II is called the norm induced by the inner product. The normed vector space may or may not be complete. If (X, I · II) is a Banach space, then the inner
1 68
9.
Hilbert Spaces and L2
product space (X, ( · , ·)) is called a Hilbert space. A Hilbert space is thus an inner product space that is complete in the norm induced by the inner product. The inner product spaces (l�n , ( · , ·)) and ((C:: n , ( · , ·)) are examples of finite-dimensional Hilbert spaces. Although we do not show it here, the inner product space (£ 2 , ( · , ·)) of Example 9-1 -2 is an infinite-dimensional Hilbert space. It is of interest to enquire whether a given normed space (X, II · II) can be identified with an inner product space (X, (· , ·)). In other words, given a norm 11 · 11 on X can an inner product ( · , · ) be defined on X such that II · II corresponds to the norm induced by ( · , · ) ? Suppose that the norm II · II can be obtained from some inner product ( · , ·) . Then for any f, g E X ,
llf + g ll 2 + llf - g ll 2
=
(f + g, f + g)
+ (f - g, f - g) = (f, f) + (f, g) + (g, f) + (g, g) + if, f) + if, -g) + ( -g, f) + (g, g) = 2 (11! 11 2 + ll g ll2 ) ;
thus, if a norm II · II can be obtained by some inner product, then it must satisfy the equation
llf + g ll 2 + llf - g ll 2 = 2 ( 11! 11 2 + l lg ll 2 ) .
This condition is also sufficient. This equation is called the paral lelogram equality. The name comes from an elementary relation in plane geometry. If x and y are two vectors in R2 that are not parallel, then they can be used to define a parallelogram. Here, the quantities ll x ll and IIY II correspond to the lengths of the vectors x and y respectively and hence the lengths of the sides of the parallel ogram. The quantities ll x + y ll and ll x - y ll correspond to the lengths of the diagonals. The parallelogram equality is useful for (among other things) showing that certain norms cannot be obtained from an inner product.
E xample 9 - 1 -3 : Consider the normed vector space (C[a, b], ll · ll oo) defined in Example 8-1 -5 . We shall use the parallelogram equality to show that the norm II · II 00 on C[ a, b] cannot be obtained by an inner product. Suppose, for contradiction, that the norm II · II 00 can be obtained by an inner product. Then the parallelogram equality must be satisfied for any
9. I .
Hilbert Spaces
1 6g
choice of f, g E C[ a, b] . Let f and g be the functions defined by
f (x) = 1 , g (x) = Now,
x-a
b-a
.
llf ll oo = 1 , ll g ll oo = 1 , llf + g ll oo =
and
llf - g II oo = consequently,
sup 1
x-a + b - a = 2,
sup 1
-
xe[a,b]
xe[a,b]
x-a b-a
= 1;
llf + g ll� + llf - g ll� = 4 + 1 = 5 , and
2 ( llf ll�
+ llg ll�) = 4 .
As the parallelogram equality is not satisfied for these functions, the normed vector space (C[a, b] , II lloo ) cannot be obtained from an inner product space. ·
The parallelogram equality can be used to show that the only IY space that might arise from an inner product space is L 2 • Theorem 9 . 1 .2 The only Lebesgue norm II · l ip that can be obtained from an inner product is the L 2 norm II · 1! .
2
Proof We first establish the result for IY[ - 1 , 1 ] spaces. A modest change of the proof leads to the result for general IY [a, b] spaces and IY(I) spaces where I is an unbounded interval. Suppose that the norm II · l ip on IY[ - 1 , 1 ] can be obtained from an inner product. Then the parallelogram equality indicates that
llf + g i l; + llf - g i l; = 2 ( l i t H; + llg ll;)
for all f, g E IY[ - 1 , 1 ]. Consider the functions f and g defined by
f (x) = 1 +
x,
g (x) = 1 - x.
9. 1 .
Hilbert Spaces
1 71
Now, I
S (p) = 1 -
p+1 - log (p + 1) = - log (p + 1 ) < 0, p+1
and thus S is a monotonic strictly decreasing function of p. Since S ( 1 ) = 1 - 2 log 2 < 0, it follows that S (p) < 0 for all p > 1 and consequently that Q' (p) < 0 for all p > 1 . This means that Q is a monotonic strictly decreasing function of p, and therefore the equation Q(p) = 0 can have at most one solution. Therefore, the parallelogram equality is satisfied only if p = 2 . A slight modification of the functions f, g for the general interval [a, b] leads to the proof of the re sult for the LP[a, b] spaces. If I is not a bounded interval, then a suitable restriction of the functions can be used to establish the result for the LP(I) spaces. For example, suppose that I = ( - oo , oo) . Then we can choose the functions •
f(x) =
g(x) =
{ {
1 + x, 0,
if x e [- 1 , 1 ] , otherwise,
1 - x, if X E ( - 1 , 1 ) , otherwise, 0, 0
and the proof then is exactly the same.
It is not difficult to see that for any interval I, L 2 (I) is a Hilbert space. We know from Chapter 8 that this space is complete with respect to the II 11 norm, and the inner product ( · , ) defined by ·
2
if, g) =
J
·
t(x)g(x) dx
for all f, g E L 2 (J) induces this norm. The complex space L�(I) is also a Hilbert space with an inner product ( ·) defined by (f. g) =
1
·
,
f(t)g(t) dt.
The conjugate exponent p = q = 2 is special throughout the function spaces based on the Lebesgue integral. One can show, for example, that the only Hilbert space among the Hardy spaces is H 2, and a sim ilar statement can be made about the Sobolev spaces. Th � se spaces have found widespread use and have many special properties not enjoyed by the other £P (HP , w k,p ) spaces, p # 2, because they
1 70
9.
Hilbert Spaces and £2
These functions are (in equivalence classes) in U [ - 1 , I ] for all p ::: I . Now,
llf ll� = ll g ll� = and
1[-1,1] 1[-1,1]
ll f + g i l� = -
_
llf - g i l�
= =
I I + x!l' dx = I I - x iP dx =
1[-1,1] I
2p+1
1
1-1 1 1-1
I + x + (I
-
I +x
- (I
iP dx =
-
,
:zP+1
,
p+I
( I - xf dx =
p+ I
1
dx
x) IP dx =
I
1[-1 ,1] I 1 2p+1 1 0
2p+1
( I + xf dx =
x) IP dx =
:zP+1
p+I
.
1-1 2P 1
1-1 l2x jl'
dx
The parallelogram equality implies that
(2P+ 1 ) 21p +
( )
zP+1 2/p
p+I
=
2
{( ) ( ) } 2p+1 2 /p
p+I
+
:zP+1 2/p
p+I
,
i.e. , (p +
I)21P - 3 = 0 .
Note that p = 2 is a solution to this equation. We will show that this is the only solution for p > I . Let Q(p) = (p + I ilp
- 3.
Then
where S (p)
=p-
(p + 1) log(p + 1).
•
1 72
9.
Hilbert Spaces and L2
are Hilbert spaces. In the next section we investigate some special properties of Hilbert spaces. For the remainder of this chapter we shall denote an inner prod uct space (X, ( · , · )) simply by X, unless there is some danger of confusion, and the norm II · II on X will always be the norm induced by the inner product unless otherwise stipulated.
Exercises 9-1 :
1 . Verify that the function ( · , ) defined in Example 9-1 -2 satisfies the conditions for an inner product. ·
2.
Let X be an inner product space and suppose f, g E X with g =1 0. Prove that l (f, g/ ll g ll ) l < llf ll and use this to prove the Schwarz inequality.
3 . Let X be an inner product space. Prove that for any elements f, g, h E X,
ll h - f ll 2 + ll h - g ll 2 = �2 llf - g ll 2 + 2 llh - �(f + g) ll 2 . 2 This relation is known as the Appolonius identity.
9 .2
Orthogonal Sets
The paradigm for a finite-dimensional inner product space is the space R n discussed in Example 9-1 -1 . In this space, the inner prod uct can be used to measure the angle between two vectors, i. e. , (x , y) = ll x ii ii Y II cos ¢, where
,
9 .2 .
Orthogonal Sets
1 73
This relationship is denoted by f ..L g . As we shall see, orthogonality plays an important part in Hilbert space theory . E xample 9-2-1 : Consider the inner product space L 2 [ -Jr, Jr] , and let fn denote the functions defined by fn (x) = sin(nx) for n = 1 , 2, . . . . Now, lfn , fm ) =
=
1-rr,rr] L: [
fn(x)fm(x) dx
sin(nx) sin(mx) dx.
J rr lrr -rr -rr
Integration by parts indicates that
[
1 ifn , fm ) = - - sin(nx) cos(mx) n
m
= -
lrr -rr m
n
m
+-
cos(nx) cos(mx) dx
cos(nx) cos(mx) dx,
and integration again by parts yields ifn , fm) = =
!!._ m
{ [�
(:)
m
2
sin(mx) cos(nx)
lfn , fm) .
J -rr 1-rr 1r
1r
+ .::. m
sin(nx) sin(mx)
ax}
If n ::j= m , then 1 - (n lm) 2 ::/= 0, and consequently ifn , fm ) = 0 . If n = m , then 2 ifn . fm ) = llfnll =
We therefore have that fn
..L fm
L:
s in2 (nx) dx = rr.
unless m =
n.
Suppose that f and g are elements in the inner product space X. Then ll f + g ll 2 = if + g , f + g) = 11! 11 2 + 2 Re (f, g) + llg ll 2 •
If f ..L g then (f, g) = 0, and we thus have an extension of Pythagoras's theorem:
1 74
9.
Hilbert Spaces and L2
We pause here to introduce two terms applicable to a general vector space. Let X be a vector space, and let Y and Z be subs paces of X. If for each f E X there exist elements g E Y and h E Z such that f = g + h, then we say that X is the vector sum of Y and Z , and denote this relationship by X = Y + Z . If, in addition, the elements g E Y and h E Z are determined uniquely for every f E X , then we say that X is the direct sum of Y and Z, and write X = Y ffi z . Given a set S c X , where (X, ( · , · )) is an inner product space, another set s l. can be formed by taking all the elements of X that are orthogonal to every element of S, i.e. , the set sl. = {f E X : (f, g) = 0 for all g E S} . The set sl. is called the orthogonal complement of S. E xample 9-2 -2 : Let e1 , ez , e3 be three linearly independent vectors in JR3 such that (ej , ek) = 0 forj, k = 1 , 2 , 3 unless} = k. If S = {v E 1R3 : v = ae 1 + be2 for some a, b E JR} (i. e. , the span of {e1 , e2 }), then sl. = {w E 1R3 : w = ce3 for some c E 1R} . Since the set { e 1 , e 2 , e3 } forms a basis for 1R3 We have that 1R3 = s ffi s J. , I
Example 9-2-3: Let £ 2 be the inner product space defined in Example 9-1 -2 , and let e1 E £ 2 be the sequence {1 , 0, 0, . . . } . Let S be the subspace defined by S = { a E £ 2 : a = ae1 for some a E C} , and define the set B by B = {b = { bn } E £ 2 : b1 = 0 } . Then (a, b) = 0 for all a E S and all b E B, and consequently B c sl. . In fact, B = sl. , for if there is a sequence c = {e n } with c1 # 0, then (a, c) = ac1 , and ac1 is not zero for all a E C. Since any sequence d = {tin} can be written as {d1 , 0, 0, . . . } + {0, d2 , d3 , } , and this representation is unique, we 2 l. have that £ = s E9 s . The inner product space JR.3 and the set S of Example 9-2-2 demon strate two geometrical properties both of which extend to any finite dimensional inner product space. First, given any vector x E 1R3 there is a unique vector v E S at a minimum distance from x , i.e. , ll x - v ii < ll x - v ii for all v E S \ {v} . Second, any vector x E 1R3 can be represented in the form x = v + w, where v E S and w E sl. are uniquely determined, i.e. , R3 = S EB Sl. . Any subspace of 1R3 has these properties, and one can enquire whether these properties persist in •
•
•
9.2.
Orthogonal Sets
1 75
the infinite-dimensional case. The space £2 in Example 9-2-3 demon strates the second property, and it is straightforward to show that the set S in this example also has the 11minimum distance" property. Unfortunately, these properties do not carry over to the general in ner product space. Under the crucial assumption of completeness, however, these properties do extend to infinite-dimensional inner product spaces.
Theorem 9 .2 . I Let S be a closed subspace of the Hilbert space X and let f E X. Then there exists a unique g E S at a minimum distance from f. Note that if S is a closed subspace, then S is a Hilbert space in its own right. •
Theorem 9 .2 .2 (Projection Theorem) Let S be a closed subspace of the Hilbert space X. Then sJ.. is also a closed subspace in X, and X = S E9 sJ.. . Moreover, iff E X is decomposed into the form f = g + h, where g E S and h E SJ.. , then g is the unique element in S closest to f. The projection theorem provides a key to understanding bases in Hilbert spaces. The study of bases in general Banach spaces is of limited value, but there is a rich and useful theory for bases in Hilbert spaces. As our main focus will eventually be on the space L 2 , which is separable, we limit our general discussion to bases for separable Hilbert spaces. In Example 9-2-2 , consider the set K = {e 1 , e2 , e3 } . The set KJ.. consists of all the vectors x E R3 such that {x, ek) = 0 for k = 1 , 2 , 3, but the only vector with this property is x = 0 ; thus KJ.. = 0 . Now, the set K forms a basis for R3 , and this is characterized by the condition KJ.. = 0 . Motivated by this observation, we can extend the concept of an orthogonal basis to general (separable) Hilbert spaces. Let N be a set of elements in the Hilbert space X. If NJ.. = 0 , then N is called a total set1 in X. If M is a countable set in X and {a, b) = 0 for all a , b E M , a # b , then M is called an orthogonal set in X. If, in addition, ll a ll = 1 for all a E M , then M is called an orthonormal 1 These sets are also called "complete" in the literature. We avoid this term, since 11complete" in this context has no conne ction with 11Complete" as used in Banach space theory.
1 76
9.
Hilbert Spaces
and L2
set in X. For our purposes, we are interested primarily in the case where M has an infinite number of elements but is countable . {4>n } be an orthonormal set in the Hilbert space X , and Let M let f E X. The numbers 4>n ) are called the Fourier coefficients of f with respect to M , and the series L�= l 4>n ) 4>n is called the Fourier series of f (with respect to M ) . Now, for any m E N, =
if,
llf - Lif, 4>n }4>n l l 2 if Lif, lf>n }l/>n, f - Lif, lf>n }l/>n } n=l n=l n=l (f, f} - if, Lif, lf>n }lf>n } - {Lif, lf>n}l/>n , f } n=l n=l m
0 < =
=
=
m
m
-
m
+ =
if,
m
{ L if, lf>n}l/>n, L if, lf>n }lf>n} m
m
n=l n=l llfll 2 - 2 L (f, lf>n } (f, 4>n } + II Lif, 4>n }4>n ll 2 n=l n=l 11[ 1 2 - 2 L l (f, 4>n} l 2 + II Lif, 4>n}4>n ll 2 , n=l n=l m
m
m
m
and Pythagoras's equation implies that II L (f, lf>n }n }l/>n 1 2 n=l n=l L I ({,
=
=
=
m
m
m
and therefore we have
l lf - L if,
0 < =
m
9 .2 . Orthogonal
Sets
1 77
Hence,
m
L I if, c/>n) l 2 < llf ll 2 , n=
l
and since the sequence {am } = CL::= l l (f, c/>n } 1 2 } is a monotonic in creasing sequence bounded above, we have that {am } converges to some a E JR. and Bessel's inequality holds:
L l (f, c/>n } l 2 < llf ll 2 · 00
n=
l
Bessel's inequality indicates that the Fourier series of f with re spect to M is absolutely convergent (in the norm) . This means that the Fourier series is convergent (in the norm) and that its sum is independent of the order in which terms are added. Thus, the series :L �= 1 (f, cl>n} cl>n converges to some function PMf E X. Now, iff E [M]..L ( = M..L), all the Fourier coefficients off are zero, and thus PMf = 0 . Alternatively, iff E [M ] , then it can be shown that PMf = f. For the general f E X, the set [M] is a closed subspace in the Hilbert space X, and the projection theorem indicates that f = PMf + h, where PMf E [M] and h�M]..L. The function PMf is thus the 11closest" approximation in [M] to f. An orthonormal set M = { c/>n } in the separable Hilbert space X is called an orthonormal basis of X if for every f E X,
f
=
L if, c/>n } c/>n · 00
n=
l
If M is an orthonormal basis, then m
lim " if, c/>n } c/>n = f, � n= l
m� oo
and using the derivation ofBessel's inequality (and continuity of the norm function) we have that
1 78
9. Hilbert Spaces
and £2
We thus arrive at Parseval's formula 00
ll f ll 2 = L 1 Pn } l 2 • n=l Suppose now that f e M..L. Since M is an orthonormal basis of X, f = L � 1 (f, l/>n}l/>n and (f, n} = 0 for all l/Jn E M; consequently, by Parseval's formula we have that ll f ll = 0, and the definition of a norm implies that f = 0 . In this manner we see that if M is an orthonormal basis, then M..L = 0 , i.e. , M is a total orthonormal set. Yet another implication of M being an orthonormal basis is that [M ] = X. This follows from the projection theorem, since [M ]..L c M..L = 0 , and so ---------..l X = [M] EB [ M ] = [ M ] EB 0 = [ M ] . It is interesting that the above implications actually work the other way as well. For example, if M is an orthonormal set in X such that Parseval's formula is satisfied for all f e X, then it can be shown that M is an orthonormal basis. In summary we have the following result: Theorem 9.2 . 3 Let X be a separable Hilbert space and suppose that M = {n } is an orthonormal set in X. Then the following conditions are equivalent: (i) M is an orthonormal basis; (ii) ll f ll 2 = L � 1 I (f, n } 1 2 for all f E X; (iii) M is a total set; (iv) [M ] = X. It can be shown that every separable Hilbert space has an orthonormal basis. Formally, we have the following result: Theorem 9 .2 . 4 Let X be a separable Hilbert space. Then there exists an orthonormal basis for X. The proof of this result is based on a Gram-Schmidt process anal ogous to that used in linear algebra to derive orthonormal bases. Parseval's formula is remarkable in this context because it essentially identifies all separable Hilbert spaces with the space e 2 . Now, f.. 2 is a separable Hilbert space, and given an orthonormal basis M = {n }
9 . 2 . Orthogonal
Sets
1 79
on a general separable Hilbert space X there is a mapping T : X -+ .e2 defined by the Fourier coefficients (f, cl>n } . In other words, T maps f E X to the sequence { (f, cl>n } } e .e2 . Parseval's formula shows that T is an isometry from X to .e2 . On the other hand, given any sequence {an } E i2 , the function defined by 2:� 1 an cl>n is in X, since X is a Hilbert space and the series is convergent. We thus have that any s e parable Hilbert space is isometric to the Hilbert space .e2 . In fact, an even stronger result is available, viz. , all separable complex Hilbert spaces are isomorphic to .e2 . This means that at an 11algebraic level" a complex separable Hilbert space is indistinguishable from the space .e2 , i. e. , at this level there is only one distinct space. A similar state ment is true for real Hilbert spaces, where the space .e2 is replaced by its real analogue. In passing we note that any Hilbert space X possesses an 110r thogonal basis," though it may not be countable. The cardinality of the set is called the Hilbert dimension of the space . If X and X are two Hilbert spaces both real or both complex with the same Hilbert di mension, then it can be shown that X and X are isomorphic. This is a generalization of the situation in finite-dimensional spaces, where, for example, all n-dimensional real Hilbert spaces are isomorphic to
JRn .
Exercises
9-2:
1 . Show that for any integers
and
1
[ -Jr, Jr]
m,
n
cos(mx) cos(nx) dx
1
[ -Jr,Jr]
=
{
rr O,
I
if m =I= n, if m = n,
sin(mx) cos(nx) dx = 0 .
2 . The first three Legendre polynomials are defined by Po (x) P1 (x) = x, and Pz (x) = � (3 x2 - 1). (a) Show that the set P on £2 [- 1 , 1 ] .
=
=
1,
{P0, P1 1 P2 } forms an orthogonal set .
(b) Construct an orthonormal set M from the set P and find the Fourier coefficients for the function f(x) = ef .
1 80
9.
Hilbert Spaces and
L2
3 . The Rademacher functions are defined by n Tn (x) sgn(sin(2 rrx)) =
for n
=
1 , 2 , . . . , where sgn denotes the signum function sgn Cx)
=
(a) Show that the set M
=
{
-1, 1,
if x < 0, if x > 0.
{rn } is an orthogonal set on £ 2 [0 , 1 ] .
(b) Iff is de:finedbyf(x) cos(2rrx) , show that f .l rn for all n 1 , 2 , . . . and deduce that M cannot form a total orthogonal set. =
=
4. Let X be an inner product space and let M orthonormal basis of X. Show that for any f, g e X,
=
(f, g >
9.3
=
{ lf>n } be an
L ({, k> {g, lf>k> . 00
k=l
Classical Fourier Series
We saw in Section 9-1 that £ 2 is a Hilbert space, and we know from Theorem 8. 6 . 1 that all IY spaces are separable. Theorem 9.2.4 thus implies that the space £ 2 must have an orthonormal basis. It turns out that the classical Fourier series (i. e. , trigonometric series) lead to an orthonormal basis for L 2 [a, b] . In this section we study classical Fourier series and present some basic results with little detail. There are many specialized texts on the subject of Fourier series such as [1 5] and [45 ] , and we refer the reader to these works for most the details. A particularly lively account of the theory, history, and applications of Fourier series can be found in [24] . For convenience we focus primarily on the space £ 2 [ -rr, rr] and note here that the results can be extended mutatis mutandis to the general closed interval. A classical Fourier series is a series of the form 00 1 ao + (an cos(nx) + bn sin(nx)) , 2 n =l
L
(9 . 1 )
9 .3. Classical Fourier Series
1 81
where the an 's and bn 's are constants . We know from Example 9-2-1 that for any nonzero i ntegers m, n ,
1.
[ -1r,1r]
{
sin(mx) sin(nx) dx =
�f m # n ,
O,
n, 1f m = n .
We also know from Exercises 9-2, No . 1 , that for all integers m, n ,
1.
[-1r,1r]
and
{
cos(mx) cos(nx) r1x =
1.
[ -1l',1l']
if m # n , if m = n ,
0' n,
sin(mx) cos(nx) r1x = 0 .
In addition, it is evident that for any integer n ,
1.
[ -1l',1l']
sin(nx) r1x =
1.
[ -1l',1l']
cos(nx) r1x = 0 .
Let
lPn (x) =
cos(nx)
� , 1/ln (x) =
sin(nx)
� ,
for n = 1 , 2 , . . . . Then the above relationships indicate that the set S = {1 / $} U
( -) - 1. .J2i .J2i �Jr 1. �Jr 1. f,
1
1
=
I
(f, l/Jn } =
(f, 1/ln } =
y
y
[ -1r,1r]
[ -1l',1l']
[-1r,1l']
f(x) dx
I
f(x) cos (nx) dx, f(x) sin(nx) dx,
and the Fourier series is
(
)
00 1 + � L ( (f, lPn )lPn + y 2n y 2n n =l
1 Psf = f, �
(f, 1frn ) 1frn ) '
(9 .2)
1 82
9. Hilbert Spaces
and
£2
which is equivalent to expression (9 . 1 ) with the familiar coefficient relations
i-Jr,Jr] an = � 1· bn = � i ao = �Jr
[
f (x) dx ,
Jr
[ - Jr, Jr]
f(x) cos(nx) dx,
Jr
[ -Jr ,Jr]
f(x) sin(nx) dx.
Fourier series can be expressed in a tidier, more symmetric, form using the relation einx = cos(nx) + i sin(nx). Using this relation, the series (9 . 1 ) can be written in the form 00
"""' c �
n einx ,
- 00
where the
(9. 3)
Cn are complex numbers defined by Cn = 1 2n
-
The set of functions B
i
[ -1r, 1r1
= { .8n } , where .Bn (x)
.
f(x) e znx dx.
einx
= $'
forms an orthonormal set for L� [ -n, n], and for any f E L� [ -n, n] (and consequently for any f E L2[ -n, n]) the corresponding Fourier series is
L{f, .8n ).8n· 00
PBf =
(9 .4)
-oo
Now, for any f E L2[ -n, n], Bessel's inequality guarantees that the Fourier series (9 .2) (and (9 .3)) converges in the II 11 norm to 2 some function Psf E L2[ -n, n]. The central question here is whether the set S forms an orthonormal basis for L2[ - n, n], so that Psf = f a . e. In fact, it can be shown that S forms an orthonormal basis for -n, n]. If we combine this fact with Theorem 9 . 2 . 3 , we have the following result: ·
L2[
9 .3 . Classical Fourier Series
1 83
Theorem 9 . 3 . 1 Let f e L 2 [ -rr, rr] and let
k=n
=
L (f, f3k) f3k ·
k= -n
Then: (i) ll sn - f ib --+ 0 as n --+ oo; (ii) Parseval's relation is satisfied: ,
ll f ll � = l (f, 1 } 1 2
+ L (I if, l/Jn } l 2 + I if, 1/lk} l 2 ) 00
n=l
rr = � + rr L ea� + b�) 2 n=l oo
= L l (f, f3n) l 2 = 00
-oo
00
2rr L l cn l 2 . -oo
From the last section we know that all separable Hilbert spaces are isomorphic to the Hilbert space .f} , and Parseval's relation is a manifestation of this relationship. This observation leads to the following result:
Theorem 9 . 3 .2 (Riesz-Fischer) Let {en } e l 2 . Then there is a unique function f e L 2 [ -rr, rr] such that the Cn are the Fourier coefficients for f. Note that 11Unique" in the above theorem means that the sequence {en } determines an equivalence class of functions modulo equality a. e . An immediate consequence of Theorem 9 . 3 . 1 and the projection theorem (Theorem 9.2.2 ) is the following result:
Theorem 9 . 3 . 3 Let f e L 2 [ -rr, rr]. Then for any € > 0 there exists a positive integer n and a trigonometric polynomial of degree n, say On = L� -n q kf3k such that l i On - f ll 2 < €. Moreover; among the trigonometric polynomials of
1 84
9. Hilbert Spaces
and
L2
degree n, the closest approximation to f in the II 11 2 norm is that for which the q k correspond to the Fourier coefficients. In other words, trigonometric polynomials can be used to approxi mate any function in L2 [ -n, n], and the Fourier coefficients provide the best approximation among such polynomials in the II 11 2 norm . Theorem 9 . 3 . 1 guarantees that the Fourier series will converge in the II · ll 2 norm, but this is not the same as pointwise convergence, and an immediate question is whether or not a Fourier series for a function f E L 2 [ -n, n] converges pointwise to f . More explicitly, for a fixed x E [ -n, n] does the sequence of numbers {sn (x)} con verge, and if so does sn(x) -+ f(x) as n -+ oo? The answer to this question is complicated, and much of the research on Fourier series revolved around pointwise convergence. Some simple observations can be made. First, the orthonormal basis defining the Fourier series is manifestly periodic, so that if sn(x) converges for x = -n, it also converges for x = n, and s( -n) = s(n). The existence of a Fourier series does not require that f( -n) f(n) , so pointwise convergence will fail at an endpoint unless f satisfies this condition. More gen erally, we see that any two integrable functions f and g such that f = g a.e. produce the same Fourier coefficients, so that for a spe cific function f, we expect that the best generic situation would be that s n (x) -+ f(x) a. e. We shall not 11plumb the depths" of the vast re sults concerning pointwise convergence of Fourier series; however, we will discuss two results of interest in their own right that make crucial use of the Lebesgue integral. Prima facie, it is not obvious that the coefficients an and bn in the Fourier series have limit zero as n -+ oo . Given that arguments such as x = 0 lead to series such as L�=o an , this is clearly a concern. An elegant result called the Riemann-Lebesgue theorem resolves this concern and is of interest in its own right. We state the result here in a form more general than is required for the question at hand. ·
·
=
•
Theorem 9 . 3 . 4
(Riemann-Lebesgue) Let I c JR be some interval and f E L 1 (I). If { .A. n} is a sequence of real numbers such that An -+ oo as n -+ oo, then
as
1
f (x) cos(J.. n x) dx
n -+ oo .
--->
0 and
1
f(x) sin(Anx) dx ---+ 0
9 .3 . Classical Fourier
Series
1 85
Proof
The proof of this result is of particular interest because the convergence theorems of Chapter 5 are of little help, and we must return to the definition of the integral itself. We sketch here the proof for the cosine integral . Suppose first that I = [a, b] and that f : I -+ 1R is bounded on I. If M denotes an upper bound for lf l , then
1
f(x) cos(Anx) dx <
[a, b]
= =
1 M1 M
M
An
cos(AnX) dx
[a,b] b
cos(Anx) dx
l sin(Anb) - sin(Ana) l
2M
< . An -
�
Since An -+ oo, as n -+ oo, we have that a, b] f(x) cos(AnX) dx -+ 0. This calculation shows in particular that the theorem holds for any step function having a support of finite length. Now, the Lebesgue integral is defined in terms of a sequence of a-summable functions, with a = x, and by definition these functions have supports of finite length. If f E L1 (I), then we know that there is a sequence Bj of x-summable functions such that
as j -+ oo . Choose any
Now,
J
l
lf(x) - 9j (x) l dx --+ 0
€ > 0, [ lf(x) - Bj (x) l dx 11
and select a j sufficiently large so that <
€. 2
t(x) cos(An x) dx
<
1 1
(f(x) - 9j (x) + 9j (x)) cos(Anx) dx
(f(x) - 9j (x) ) cos(Anx) dx +
1
�(x) cos(Anx) dx
1 86
9.
Hilbert Spaces and
< <
1 €
2
£2
1f (x) - Bj (x) l dx + +
1 I
1
Bj (X) cos(Anx) dx
8j (x) cos(.A.nx) dx .
From the above discussion we know that
1
Oj (x) cos(Anx) dx
--*
0
as n � oo for any j, so that there is an integer N such that
whenever n
whenever n
> N.
> N,
€ < 2
Therefore,
1 1
€
f(x) cos(.A.nx) dx <
2
+
€
2
=
€
and by definition this means that
as n � oo .
1
f(x) cos(Anx) dx
--*
0 0
The Riemann-Lebesgue theorem with I = [ -n, n] and An = n shows that the Fourier coefficients tend to zero as n � oo. This theorem is also crucial in proving another notable result known as the Riemann localization theorem, which we shall not prove. Theorem 9 . 3 . 5 (Riemann Localization) Let f E L 1 [ - n, n] and x E [ -n, n] . Then for any fixed 8 with 0 < 8 < n, sn (x) � f(x) if and only if •
_2_ f
nlim � oo 2n
}[O,o]
f(x + t) + f(x -
t
t)
sin((n +
� )t) dt = 0 . 2
Here it is assumed that f has been extended periodically to a func tion on 1R so that f is defined for arguments x ± t that may not be in [ -n, n] . What is interesting in the above theorem is that the number can be arbitrarily small, and this means that the pointwise con vergence of the Fourier series depends only on the values that f
8
9.3.
Classical Fourier Series
1 87
assumes in a small neighb orhood of x . Given th at the Fourier coef ficients in the series defining sn depend on the values f assumes in the entire interval [ -n, n], this result is remarkab le. We can study Fourier series outside the comfortable space L2 [ -n, n]. Naturally, we lose the results that rely on L2 [ -n, n] being a Hilbert space such as Parseval's relation, but in the larger space L 1 [ -n, n] the Fourier coefficients are still well-defined, and results such as the Riemann-Lebesgue theorem are still valid. We even have the following uniqueness result: �
Theorem 9 . 3 . 6 Iff, g E L1 [ - n, n] have the same Fourier coefficients, then f = g a. e.
What is needed, however, is some result that shows that the partial sums of the Fourier series for a function f E L1 [ -n, n] converge in the ll · ll 1 norm to f , but this is where things go wrong . Iff E L 1 [ - n, n], then in general we do not have that ll sn - f lh -+ 0 as n -+ oo . It can be shown, however, that for
an =
l
n- 1
sk , nL k=1
-
-+ 0 as n -+ oo for any f E L1 [ - n, n] . The quantity an is called the Cesaro mean of the partial sum sequence {snl · This is a weaker convergence result, since an effectively measures an av eraged partial sum. The sequence {an} may converge even if { sn l diverges, and if {sn} converges to some s, then {an} also converges to some function s E L1 [ -n, n] . In the space L1 [ -n, n] this is the sharpest result we can get. In fact, it can be shown that there are functions in L1 [ -n, n] such thart {sn (x) } diverges a.e. Though most of the above results concerning pointwise convergence were estab lished by the early twentieth century, it was not until the 1 96 0 's that Carleson [7] proved that is f E L2[ -n, n] then {sn (x) } -+ f (x) a.e. From this perspective, the space L2 [ -n, n] is the natural space in which to study Fourier series.
llan - f lh
Exercises 9-3 : 1 . Let f (x) = x on the interval [ -n, n] . (a) Determine the Fourier series for f.
1 88
9.
Hilbert Spaces and
L2
(b) Show that the series obtained by differentiating term by term the Fourier series in part 1 (a) is a divergent series. 2 . The complex Fourier series for a function f E L2 [ -A, A], A > o, is given by
PMf (x) = L cn eimr:x!A , 00
- 00
where
c, =
A}
� L: r(x)e-;"""IA
dx,
for n = 0, ± 1 , ±2, . . . , and it can be shown that the set M { suitably normalized forms an orthonormal basis for 2 L [ -A , A].
eimr:xl
(a) Find the complex Fourier series for the function g [ - � , �] -+ R defined by g (X)
_
-
{
-1,
1,
:
if - �
< X < 0, lf 0 < X < 21 · •
(b) Use Parseval ' s relation and the Fourier series in 2(a) to prove that 00 1 7r 2
an , bn
= L 2n + 1 2 a· ( ) n=O
3 . Suppose that f has a continuous derivative on the interval [ -rr , rr] , and let denote the Fourier coefficients off. Use integration by parts to prove the Riemann-Lebesgue theorem for this special case .
9 .4
The Sturm-Liouville Problem
Theorem 9 . 2 . 4 indicates that for any interval I the Hilbert space L2 (I) must have an orthonormal basis. If I is bounded, then the trigonometric functions provide such a basis. Theorem 9 .2.4 does not, however, preclude the existence of other orthonormal bases, and there are, in fact, any number of orthonormal bases available. If
9 .4 . The Sturm-Liouville Problem
1 89
is not bounded, then the trigonometric functions are not even in the space L2 (I), and the classical Fourier expansions are no longer valid. Nonetheless, Theorem 9 .2.4 guarantees the existence of an orthonormal basis for any interval I . In practice, many of the orthonormal bases for L2 (I) arise as solutions to boundary value problems of the Sturm-Liouville type. A regular Sturm-Liouville problem consists in determining a solution y to a differential equation of the form I
� (r(x) :)
+
(q(x) + J...p (x))y = 0
(9.5)
on some interval [a, b], satisfying boundary conditions of the form (9.6) k1y(a) + k2y' (a) = 0, l1 y (b) + l2y' (b) = 0. Here, r, q, and p are given functions, ).. is a complex parameter, and the kj's and �'s are constants such that ki + k� -:j:. 0, li + l� # 0. In addition, it is assumed for the r¢gular Sturm-Liouville problem that the functions r and p are nonzero in the interval [a, b]. Sturm -Liouville problems are important in theory and appli cations of differential equations. As a consequence, this class of boundary value problems has been studied exhaustively for some 150 years. Relatively accessible accounts of the theory can be found in [5], (9], [11 ], [23], and [40]. In addition, many applications-oriented texts such as [3], [8], and [26] contain short summaries of the the ory and various applications. We do not attempt here to replicate the general theory in any detail or depth, but merely focus on the Sturm -Liouville problem as a 1 machine" for producing orthogonal sets in L2 . Equation (9.5) is often written in the abbreviated form (9.7) £y = -)..p (x)y, where the linear operator £ is defined by C.y =
� (r x) :) + q(x)y. (
(9.8)
If y E C2 [a, b], r E C' [a, b], and q E C[a, b] then Ly E C[a, b], and in particular Cy E L2 [a, b]. For linear operators such as £ there exists another operator £* called the (Hilbert) adjoint. The adjoint operator
1 9Q
9 . Hilbert Spaces and
L2
satisfies the relation (9 .9) for all Y b Y2 E L2 [a, b], which also satisfy the boundary conditions (9 .6). The peculiar form of the operator £ for the Sturm-Liouville problem ensures that the operator is self-adjoint, i. e., £ = £* . This can be verified directly as follows: ( £y1 , Y2 ) - (y 1 , £y2 )
= =
} f[a,b] ( la,bJ ; ( ( ; ) ( ; ) ) [ ( ! ! ) J: y2 (x)£y1 (x) - Y1 (x) £y2 (x) dx r(x)
= r(x) Y2 (x)
Y l (x) Y2 (x) - r (x)
Yl (x) - Y! (x)
Y2 (x) y1 (x)
dx
Y2 (x)
= 0.
The final equality follows from the condition that y1 and y 2 satisfy the same boundary conditions (9 .6). (The specific details can be found in [26]). As we shall see shortly, the self-adjointness of £ is the key to obtaining orthogonal sets. For any value ). E C, the Sturm-Liouville problem always has one obvious solution, viz . , the triVial solution y = 0 . Generically, this is the only solution available for a given value of ). , but there may be some values of ). for which nontrivial solutions exist. These values are called eigenvalues, arid the associated nontrivial solu tions are called eigenfunctions. The set of all eigenvalues for the problem is called the spectrum2 of the problem. The theory of the existence and distribution of eigenvalues to problems such as the Sturm-Liouville problem forms a significant component of linear functional analysis known as spectral analysis. Most introductory texts on functional analysis such as [ 25] devote a few chapters to the general theory. A comprehensive and detailed discussion of the subject can be found in [ 1 2]. 2More generally, the spectrum is the set of points where the inverse operator (the resolvent) (£ - AIF1 is not well-defined.
9 .4. The Sturm-Liouville Problem
1 91
The regular Sturm-Liouville problem has fairly tractable spectral properties, as the next theorem illustrates.
Theorem 9.4.1 Suppose that r E C1 [a, b], p, q E C[a, b], and that r (x) > O andp(x) > 0 for all x E [a, b]. Then: (i) the spectrum for the Sturm-Liouville problem is an infinite but countable set; (ii) the eigenvalues are all real; (iii) to each eigenvalue there corresponds precisely one eigenfunction (up to a� constant factor), i.e., the eigenvalues are simple; (iv) the spectrum contains no finite accumulation points.
Thus, under the co nditions of the above theorem, the Sturm Liouville problem always has an infinite (but countable) number of eigenfunctions. The spectrum is a countable set, so we can regard it as a sequence
and therefore Ym LYn - YnLYm = (Am - An)p(x)YmYn ·
The above equation indicates tl$t
1
[a, b)
(Ym(x) £Yn(x) - Yn(x) £Ym (x) ) dx
= (ym , LYn ) - Wn , LYm} = (A.m - A n)
1
[a l b]
p(x)Ym(X)Yn (x) dx .
Now, the eigenfunctions are real, so that (ym , £yn ) = ( £yn , Ym ) , and since £ is self-adjoint, (Am - An)
1
[a,b)
p(X)Ym (X)Yn (x}dx = ( £yn , Ym } - (yn , £ym } = 0 .
1 92
9 . Hilbert Spaces and
L2
By hypothesis, the eigenvalues are distinct (A.m =f A.n) , and the above c alculation shows that
{
f[a, b]
P(X)Ym (X)Yn (x) dx = 0 .
(9 . 1 0 )
If p = 1 , the above equation im;plies that Ym j_ Yn for m =f n , and thus the eigenfunctions are orthogonal. By hypothesis we have that p (x) > 0 , so that in any event the set of functions {.J.PYn} is orthogo nal. As the eigenfunctions are by definition nontrivial solutions, we know that IIPYn l i z =f 0 , so that this set of functions can always be nor malized to form an orthonormal set in L z [a , b] . The Sturm-Liouville problem thus produces eigenfundtions from which orthonormal sets can be derived. Given a continuous function p positive on the interval [a, b] , it is always possible to define another norm for Lz [a , b] , viz . , p ii Y II z =
{ f[a,b] {
p(x)yz (x) dx
}
l!Z
(9 . 11 )
Since p is positive and continuous on [a, b] , there exist numbers Pm and PM such that 0 < Pm < p (x) :5. PM for all x E [a, b] . This implies that Pm ii Y IIz :S p iiY II z < PM I I Y II z ,
so that the p ll · ll z norm is equivaletnt to the ll · ll z norm. Consequently, the vector space Lz [a , b] equipped with the p ll · l i z norm is a Banach space, and any convergence results valid in (L z [a , b] ,p II · li z ) are still valid in (L z [a , b], II · li z ) (and vice versa). If we define ( · , · )p by
(y 1 , Yz )
=
{ p b] f[a,
(x)yl (x)yz (x) dx ,
then it is readily verified that (·, ·)p is an inner product and that p ll · liz is the norm induced by this inn�r product. Thus the inner product space (L z [a , b], ( · , · )p ) is a Hilbert space. We use the (standard) nota tion L z ([a , b ] , p) to denote this Hilbert space, with the abbreviation L z [a, b] for the space L z ([a, b] , 1 ) . If equation (9 .5) satisfies the conditions of Theorem 9.4 . 1 , the above arguments indicate tliat the normalized eigenfunctions .
9 . 4 . The Sturm-Liouville Problem
193
{ynlp ii Yn ll2 } form an orthonormal set in the Hilbert space L2 ([a, b], p). In fact, this set forms an orthonormal basis for L 2 ([a, b], p) .
Theorem 9.4.2
Suppose that equation (9. 5) sati$fies the conditions of Theorem 9. 4. 1 . Then the collection of normalized eigenfunctions forms an orthonormal basis for the Hilbert space L 2 ([a, b] , p).
Example 9-4-1 : Fourier Sine Series
Consider the differential equation
y" (x) + AY = 0,
with boundary conditions
y(O) =
0,
(9 . 1 2)
y(1r) = 0 .
(9. 1 3)
If A < 0 , then the general soluti9n to the differential equation (9 . 1 2) is
y (x) = Ae,../-fx + Be - Hx
'
where A and B are constants. Th e boundary conditions (9 . 1 3) , how ever, indicate that A = B = 0 , so that only the trivial solution is available in this case. Thus, this problem does not have any negative eigenvalues. If A = 0, then the general solution is
Y(x) :;: A� + B, where A and B are constants. Again, the boundary conditions imply that A = B = 0 , so that A = 0 ' cannot be an eigenvalue. If A > 0 , then the general solution is
y (x) = A cos( vfix) + B sin(...fix), where A and B are constants. 'the condition y(O) = 0 implies that A = 0, and the condition y(1r) ::::::; 0 implies that B sin(Ji.1r) = 0 .
(9 . 1 4)
Equation (9 . 1 4) is satisfied for B1 f. 0 only if A = n 2 for some integer n, and in this case equation (9 . 1 2) has the nontrivial solution Yn (x) ;:: sin(nx) .
(9 . 1 5)
1 94
9 . Hilbert Spaces and
L2
The set {n2 } corresponds to the spectrum. Hence from Theorem 9.4.2 we know that the set {yni iiYn ll 2 } ::±: {yn .../27i} forms an orthonormal basis for L2 [ 0 , n']. Example
9-4-2 : Mathieu FunctionJs
C onsider the differential equation (9.16) y" + (A. - 20 c!os(2x))y = 0 , along with the boundary conditions (9. 13). Equation (9.16) is called Mathieu's equation, and e is some fixed number. Here, r(x) = p(x) = 1 and q(x) = - 2 e cos(2x). Note that when e = 0 equation (9.16) reduces to equation (9.12). Now, unlike the previous example, we cannot solve equation (9.16) in closed form, and it is clear that the eigenvalues will depend on the parameter e. Nonetheless, the above results indicate that for any e , there is a set {A.n (B)} of eigenvalues with corresponding eigenfunction;s that when normalized will yield an orthonormal basis for L2 [0 , n']. The solutions to equation (9.16) are well-known special functions called (appropriately enough) Mathieu functions. The intricate details concerning these functions can be found in [28] or [41]. Suffice it h�re to say that corresponding to the spectru m {A.n (B)}, Mathieu's equation has eigenfunctions sen (x, B) that are periodic with period 2Jr aJi.d reduce to sine functions3 when e = 0 . The Mathieu functions { s�n (x, B)} thus form an orthogonal basis for L2 [ 0 , Jr]. The Sturm-Liouville problem; can be posed under more gen eral conditions. These generalizations lead to bases for L2 that are widely used in applied mathematics and numerical analysis. The ge neralizations commonly made �orrespond to either: (i) relaxing the conditions on p and r at the endpoints so that these functions may vanish (or even be discontinuous) at x = a or x = b (or both); or (ii) posing the problem on an unbounded interval. The general solutions to the differential equations with these modifications are usually unbounded on the interval, and the homo3 The
notation
sen
comes
from
Whittaker and Watson
" sine-elliptic:' There are "cosine-elliptic" functions
ce"
(41 ]
and
denotes
with analogous properties.
9.4.
The Sturm-Liouville Problem
1 95
geneous boundary conditions (9 . 6) are often replaced by conditions that ensure that the solution is bounded, or that limit the rate of growth of the function as x apprpaches a boundary point of the inter val. These generalized versions of the Sturm-Liouville problem are called singular Sturm-Liouville problems. The theory underlying sin gular Sturm-Liouville problems and the corresponding results are more complicated than those fd>r the regular Sturm-Liouville prob lems. For example, the spectrui!;n may consist of isolated points or a continuum, and not every poirl.t in the spectrum need correspond to an eigenvalue. The singular Sturm-Liouville problem is studied in some depth in and More general references such as and give less detailed but clear, succinct accou.nts of the basic theory. We content ourselves here with a few exa�ples that lead to well-known bases for L2 . The special functions atising in these examples have been studied in great detail by numer�us authors, and we direct the reader to the aforementioned reference s as a starting point.
[9]
[40].
[5]
[11]
Legendre Polyn�mials Consider the Legendre differential equation
Example 9 -4-3 :
=
(9.17) � (c1 - �)�) + Ay o on the interval ( - 1, 1). Note that r (x) 1 - x2 is zero at x = ±1, so that this equation leads to a sin$Ular Sturm-Liouville problem. The general solution to equation (9 . ! 1 7) can be found by using a power =
series method, which seeks solutions of the form 00
n y(x) ::::: L,: anx , n =O
(9.18)
( [5]
where the an 's are constants cf, for details of the method) . Sub stituting power series into the Legendre differential equation and equating the coefficients of: xn to zero for n 2, . . . yields the recursive relation 71r(n + A an + Z = . + 2) (n +
(9.18)
=
1) 1)(n
1,
Once ao and a1 are specified, the :above relation determines the other an 's uniquely. Specifically, the r� cursive relation defines two linearly
1 96
9.
Hilbert Spaces and
L2
independent solutions Y e and Yo ; corresponding to the choices ao = 1 , a1 = 0 and ao = 0, a1 = 1 , respectively. Here ye is an even s olution and Yo is an odd solution. Suppose now that we require the s olution to be bounded as x -+ ± I . If A :;¢ n(n + 1 ), we can apply the ratio test to the series defining Ye (or : y0) to establish that the radius of c onvergence is 1 , and it can be sh�wn that ye (and y0) are unbo unded in the interval ( - 1 , 1). We thus need A = n(n + 1) for some positive integer n to get bounded solution$. The eigenvalues for the problem are thus A n = n(n + 1 ) . The contesponding eigenfunctions are the polynomials formed by the trunc.jlted series for Ye and Yo (modulo a scaling factor). Specifically, the eigenfunctions Pn correspon ding to An are defined by Pn (x)
where M
�
= �( - 1)
m
, (2 n - 2m) ! n - 2m x n ' 2 m ! Jn - m)!(n - 2m)!
= n/2 , or (n - 1)/2, whichever is an integer. For example, Po (x)
= 1,
P2 (x)
= � (3x2 - 1 ) ,
·
P1 (x)
=
P3 (x)
= � (Sx3 - 3x) .
x,
The peculiar form of the polynomial coefficients is standard: The polynomials have been scaled so ' that Pn ( 1 ) = 1 for all n. This cor responds to choosing the last coefficient Cn in the polynomial Pn to be
cn -
_
: (2n)! zn (n!)2
and using the recursive relation to get the lower-order coefficients. The functions Pn are called the L�gendre polynomials.. Note that the proof of orthogonality follows im:nh.ediately from the self-adjointness of the operator. The boundary values for the Pk do not matter, because r(1) = r( - 1) = 0. Th� Legendre polynomials, suitably normalized, form an orthonorma] basis for L2 [ - 1 , 1 ] . '
E xample 9-4-4: Hermite Polynomials
The Hermite differential equation is d2 y dy -2 - 2x-- + AY dx dx
= 0.
(9 . 1 9)
9.4.
The Sturm-Liouvill e Problem
This equation is not in the self-adjoint form of equation since
and
e-x =I= 0 for all x 2
197
(9 .5), but
(9.19) is equivalent to d ( x2 dy )·· e- + Ae-x2y = 0. (9 .20) E JR , equatton
dx
dx :
The singular Sturm-Liouville pnoblem consists in finding solutions on the interv�l -oo, oo) such that y to equation does not grow exponentially as ±oo. Substituting the power series into the equivalent equation yields the recursive relation
(9. 2 0)
ly(x)l ( x --+ (9.18) (9 .19) an+2 = (n +2n1)(n- A+ 2) an . As with the Legendre equation,, this recursive relation defines an even and an odd solution. If A tj; 2n, then the ratio test indicates that the series converges for all x· JR. The solutions in this case will have exponential growth. In or¢ler to meet the growth condition, we must therefore have that A 2n for some n = 0, 1, 2, . . . . If A = 2n, then the series (9.18) repuces to a polynomial of degree n. The eigenfunctions are commotXly given in the form Ho (x) = 1, dn ( 2 ) Hn (X) = (-1 ) � axn e-x and called Hermite polynomials. The first few Hermite polynomials are H2 (x) = 4x2 - 2, H3 (x) = 8x3 - 12x, H4 (x) = 16x4 - 48x2 + 12. Note that for the self-adjoint equation (9.20), we have that p (x) e-x , and therefore for m # n, ( x dx 0 . x) H (x)H e m( n oo oo lc- , ) E
·=
n
I
=
2
2
=
1 98
9.
Hilbert Spaces and
L2
It can be shown that the Hermite polynomials form an orthogonal basis for If is defin¢d by
L2 (� , e-x2 ). en 1 -x2 2H ( ) en (x) j2nn!:jii then it can be shown that the s�t {en } is an orthonormal basis for L2 (�, e-x2 ). Example 9-4-5 : Laguerre Polynomials 2 A basis for L ( 0 , oo) can be deriived from the Laguerre differential equation 2y d X dx2 + (1 - �) dydx + Ay 0 ' (9. 21) !
=
e
1
n x
,
=
which in self-adjoint form is
(9. 22) The singular Sturm-Liouville prqblem consists in solving equation < on the interval ( 0 , oo) subjebt to the conditions limx-+o+ x oo, limx-+oo = 0 . As with the Hermite equation, the growth conditions lead to a restriction o:rjl the values of A. For the Laguerre differential equation, the eigenv�ues are A = for = .... The standard representation for fu.ese eigenfunctions is
(9 .22)
e- y (x)
y(x)
I
n n n 1, 2,
Lo(x) 1, n Ln (X) n!l d n (xn -x). =
=
�·
---+
dx.
e
Ln are polynomials of degree n; they are called Laguerre polynomials. The first few Laguerre polynoJTI.ials are L1(x) 1 - x, L (x) 1 - 2x + �x2 , L3 (x) 1 - 3x + �x2 - ix3 , L14 (x) 1 - 4x + 3x2 - � x3 - 214 x4 • Laguerre polynomials fonm an orthogonal basis for the space 2L ((The 0 , oo), e-x)
The functions =
=
�
=
=
The above examples are a small sample of the many 11Special functions" in mathematics that c
9 . 5 . Other Bases for
L2
1 99
well-known special-function bases for L2 is extensive. Aside from the references given above, the reader is also directed to the monograph of [1 9] , which lists most the bases for L2 in common use and gives another perspective on total set$ for L2 . Exercises 9-4:
1 . Using the properties of the jnner product and self-adjoint op erators, prove that a Stur:rq.-Liouville problem satisfying the conditions of Theorem 9.4. 1 must have only real eigenvalues. 2. Solve equation (9 . 1 2) subject to the boundary conditions y(O) = 0, y'(1) = 0 and determine the eigenvalues and corresponding eigenfunctions.
3.
Verify by direct calculation that the Hermite polynomials Ho, H1 , and H2 of Example 9-4-4 fmtm . an orthogonal set in the space 2 2 L (R, e-x ).
9 .5
Other Bases fot
L2
The Sturm-Liouville problem cfin be used to generate numerous bases for L 2 spaces; however, atjly solution to equation (9 .5) must be at least twice differentiable, and consequently, any basis derived from a Sturm-Liouville problem must consist of 11Smooth" functions with at least two derivatives. Giyen the nature of the functions in L2 , one expects that bases consis�ng of nonsmooth functions should also be available. In this section we p resent a brief example of such a basis for the space L2 [0, 1 ]. The Haar functions hn : [0 , 1 ] � R are defined by h 1 (x) = 1 ,
l- 1 l- 1 /2 ) 2 k/2 , if X E [ zr , � ,
0 , . otherwise,
2 00
9 . Hilbert
Spaces
and
where .e = 1 , 2, . . . , 2k and k = 0 and .e = 1 , so that
h 2 (x) If n = 3 , then
k
=
L2 = 0, 1 ,
{
1,
. . . . For example, if n
if x
e
[0 ,
2, then
�),
- 1 ,, if x E [ ! , 1].
k = 1 and .e = 1, and thus
2 1 12 if x E [ 0 l4 ) I
I
-2L12 if X E [ l4 1 l ] 2 I
If n =
=
4, then k = 1 , .e = 2, and
I
I
0, if x E C ! , 1] . 0,
if X E [ 0 ,
!),
2 1 ;(2 if X E [l 1 �] 2 4 I
-2U2 , if X E
I
(�, 1 ].
The support of the Haar function h2 k +l is the interval [ (l 1)12k , l/2k]. Figure 9 . 1 depicts th d functions h5 through h8 • The set of Baar functions H :::j: {hn } forms an orthonormal basis for L 2 [0, 1 ] , The proof that H is a t?tal set can be found in [1 9]. Here, we show that H is an orthonorm*l set in L2 [0, 1 ]. The normality of the Haar functions is simple to �stablish. Evidently, ll h 1 11 2 = 1 ; if n = 2k + .e , then
Hence ll hn ll 2 = 1 for all n = 1 , 2, . � . . To establish orthogonality note first that h 1 l_ hn for all n = 2 , 31 , . . . . Suppose that k is fixed, and
9 . 5 . Other Bases for
2 0
-2
2 0
-2
hs (x)
2
I
I I I I I I I I I
1 14
1
I I I I
X
h7(x)
2
---- I I I I
1 /2
I I I I
1 I I I
X
3/4 I I I I
-
I I I I
I I I I I
1 14 1I I I I
1 12
FIGURE
1
I I I I
X
- - --
h g(x) - - - - - - -I I I I
0
----- -
201
h 6(X)
0
-2
L2
-2
I I I I
1
-------- -
9.1
consider the functions h2k+l for lj = 1 , 2 , . . . , 2 k . The support of h2k+e is the interval Ik,e = [ (.f - 1 )f2 k , .e/2 k ] , and if .e1 ¥= .e2 , the set Ik, e1 n ik,e2 contains at most one point. Therefore, h2k+e 1 (x) h2k+e2 (x) = 0 a.e., so that
r
1[0, 1 ]
h2k+l1 (x) h2k+lz (x)
.
dx = 0.
Consequently, we have that h2k� 1 ..L h2k+lz for all l1 # l 2 . Suppose now that k1 > k2 and consider: the functions h2k1 +e 1 , h2kz +lz with supports Ik1 , e 1 , h2,e2 , respectively.! Since k1 > k2 , the length of Ik1 ,e 1 is at most half that of Ik2 , e2 , and sinpe any endpoint in a support for a Haar function must be of the forth m/2 n for m, n = 0, 1 , 2 . . . , the set Ik1 , e 1 n Ik2,e2 will be one of the follbwing : I
(i) the empty set;
(ii) a set consisting of a single �ndpoint of Ik1 ,e1 ; (iii) the set h1 , e 1 •
202
9 . Hilbert
Spaces and L2
Cas es (i) and (ii) indicate that h2k1 te1 (x) h2k-z +e2 (x) = 0 a.e. and there fore h2ki +l 1 ..L h2k-z +lz · For case (c), note that in Ik1 , e 1 1 the function h2k-z+lz does not change sign; henc�,
i
[0, 1 )
hzq +l , (x) h z>z +t, (x) dx
1=
±2 !:,
:t::
0.
12
1
Iq ,l 1
h2k! +l 1 (x) dx
Thus, for case (iii), h2ki +e 1 ..L h2k-l+fz · The above arguments indicate that hm ..L hn for m =j:. n , and the set H is consequently orthonormal. Series of Haar functions may be used to represent measurable functions. If f : [0, 1 ] --+ Re is fi:riite a. e. on the interval [ 0, 1 ] and measurable, then there exists a senes of the form L�I CXn hn (x) that converges a.e. on the interval [0 , 1 � to f. In passing we note that the Rademacher functions rn of Exercises 2 9-2-3 do not form a basis for £ [ 0 , 1 ] ;i. however, another set of functions called the Walsh functions Wn cart be formed from the products of Rademacher functions. The set ofWalsh functions {wn } forms a basis 2 for £ [ 0, 1 ], and these functions ha�e applications in probability and communication theory. For a short discussion of these functions the reader is referred to [1 9 ] . I
'
Epilogue C H A P T E R
10 . 1
Generalizations of the Lebesgue Integral
The Lebesgue-Stieltjes integral may be generalized in any number of ways to accommodate, for example, higher dimensions. Rather than pursue extensions of this nature, we choose to describe infor mally and briefly a generalization of the Lebesgue integral whose origins go back to Newton. Newton regarded the integral of a function as being the an tiderivative of that function. Formally, a function f : [a, b] � 1R is said to be Newton integrable on the interval [a, b] if there exists a differentiable function F defined on [a, b] such that F' (x) = f(x) for all x E [a, b] . The definition of the Newton integral is evidently very limiting: not even step functions, strictly speaking, have Newton in tegrals. On the other hand, there are functions that have Newton integrals but are not Lebesgue integrable. Consider, for example, the function F defined by
F(x)
=
{
x2 sin(l /x2 ) 0
if X ;;j:= 0 , if X = 0 . 203
204
10. Epilogue
Now, F is evidently differentiable at every point in the interval (0, 1 ]. At x = 0 the definition of the derivative can be used to establish that F is diffe rentiable there as well. Consequently, the function f defined by f(x) = F' (x) is Newton integrable in the interval [0, 1 ] . It can be shown, however, that this function is not Lebesgue integrable on [0, 1 ]. It is interesting to note that the improper Riemann integral also exists for f on the interval (0, 1 ]. The essence of the problem with the Lebesgue integral is that the theory is restricted to absolutely convergent integrals. Recall that if a function f is Lebesgue integrable, then the function If I must also be Lebesgue integrable. This seems quite a harsh restriction, and it filters out conditionally convergent Riemann integrals as well as certain Newton integrals. Newton's definition of an integral is very natural to the student of elementary calculus and useful, particularly in fields such as dif ferential equations. The class of Newton integrable functions is not contained by the class of Lebesgue integrable functions, and this awkward situation was soon realized by mathematicians. Within fifteen years of L ebesgue's pioneering work two mathematicians ar rived independently at a generalization of the Lebesgue integral that would mend the awkward gaps in the definition where improper Riemann integrals and/ or Newton integrals exist, but the Lebesgue integral does not. In 1912, A. Denjoy made a generalization directly from the Lebesgue integral. The definition of the Denjoy integral proved a complicated affair, and as a result, some of its potential for applications and generalizations was lost. In 1914, 0. Perron devised another integral that would also remedy the problem. Rather than start with the Lebesgue integral, Perron devised a definition based on upper and lower integrals defined by function� whose deriva tives are respectively greater than and less than the given function. Roughly, a function A is said to be a major function off on the inter val [a, b] if its derivative A' satisfies A ' (x) > f(x) for every x E [a, b]. A function B is said to be a minor function for f if -B is a major func tion for -f. A function [ is said to be Perron integrable on the interval [a, b] if f has both major and minor functions, and
- oo < inf {A(b) - A (a)} = sup{B (b) - B(a) } < oo .
10.2 . Riemann Strikes Back
2 05
Here, the infimum is taken over all major functions of f on [a, b], and the supremum is taken over all minor functions of f on [a, b] . The common value of the infimum and supremum is defined to be the Perron integral of f on [a, b] . Note that Perron's extension immediately includes the class of Newton integrable functions. This is because the primitive F of a Newton integrable function f is at the same time a major and a minor function. Although the Perron integral patched up the gaps with the Lebesgue integral, it suffered from the same problems as the Denjoy integral. The Perron integral and the Denjoy integral are defined very differently, and for a time it was thought that they characterized different functions as 11integrable!' As it turns out, however, these integrals are equal, i.e. , f is Denjoy integrable if and only if it is Perron integrable. The integral is now commonly referred to as the Denjoy-Perron integral.
10 .2
Riemann Strikes Back �
The reader who has followed these chapters on the Lebesgue Stieltjes integral is doubtless convinced of the vast superiority of the Lebesgue integral over the humble Riemann integral. Indeed, the Riemann integral is denigrated by many authors as merely a mathematical object of 11historical" interest. At best, the integral is used as a pedagogical tool to introduce a 11rigorous definition" of the integral in an elementary course in analysis. For example, in the 1 930s Norbert Wiener [42] wrote in the introduction to his book on the Fourier integral,t1However, the Riemann integral is of rela tively little importance in the theory of Fourier series and integrals, save as the classical ci d1nition applying to continuous and 1Step-wise continuous' functions!' It is certainly true that the Lebesgue integral has won many resounding victories over the Riemann integral in fields such as Fourier analysis; it is also true that the Riemann inte gral suffers analytical deficiencies that are absent with the Lebesgue integral, but the victory is not complete, and good ideas (once ac cepted) are difficult to extinguish completely in mathematics. There is still an important realm where the Riemann integral reigns, viz. ,
206
10. Epilogue
conditionally convergent integrals. The (improper) Riemann inte gral does not suffer the same restrictions as the Lebesgue integral regarding absolute convergence, and this annoying fact perhaps mo tivated Lebesgue and some of his contemporaries (e.g. , Denjoy and Perron) to search for a more all-embracing definition of an integral. It is interesting to note that though many mathematicians have joined in the funeral chorus for the Riemann integral, the study of the Riemann integral has never really left the curriculum of mathematicians, owing to certain pedagogical advantages it has over the Lebesgue integral, and the need for conditionally convergent integrals in applications. Riemann's approach to integration was as revolutionary as Lebesgue's; Riemann effectively divorced the integral from differ entiation and brought the geometrical properties of the integral into focus. In the late 1950s, the Riemann approach was vindicated by R. Henstock and J. Kurzweil. Working separately, these mathemati cians extended the Riemann integral to include Lebesgue integrable functions. Even more impressive, their Riemann-based definition of an integral also captured the more general Denjoy-Perron in tegral. The resulting integral is now called the Henstock-Kurzweil integral, and it has since been shown to be equivalent to the Denjoy Perron integral, 1 i. e. , f is Denjoy-Perron integrable if and only if it is Henstock-Kurzweil integrable. The key to their success in the extension was replacing Riemann's uniformly fine partitions of the integration interval with locally fine partitions. The use of locally fine partitions can be readily motivated by numerical examples. Con sider, for example, the function f defined by f (x) = x- 1 sin(1 /x) in some interval [E, 100], where E is some small positive number. The graph of f shows that the function oscillates rapidly in the inter val [E, 1], but it then decreases steadily to 0 in the interval [1 , 100] . I f we wished to approximate efficiently and accurately the integral 1 It is tempting now to refer to the common integral as the Denjoy-Perron
if some proof requires the
Henstock- Kurzweil integral, but one ne eds some patience in reading and typing such an appellation, and thus we eschew it. Often,
integral, then it is referred to as the D enjoy-Perron integral or the Henstock
Kurzweil integral, depending on which definition is to be used for the purposes of the proof.
1 0 . 3 . Further Read ing
2 07
of f over the interval [E , 100], we would be inclined to use a much finer partition of the interval [E , 1 ] than in the interval [1 , 100] . In deed, given some freedom, we would probably devise even more refined partitions for subintervals near the endpoint at x = E. The Henstock-Kurzweil approach allows the partition refinements in the integration interval to be nonuniform in the limit and thus 11ex tra fine" where needed. Although the Henstock-Kurzweil integral is equivalent to the Denjoy-Perron integral, the Riemann-type defini tion of the former makes it a more tractable concept and thus more amenable to applications and generalizations. In passing we note that these general integrals do not lead imme diately to complete function spaces. The extension of the Lebesgue integral to include conditionally convergent integrals (the Denjoy Perron integral) or the generalization of the Riemann integral to include the Lebesgue integral (the Henstock- Kurzweil integral) opens floodgates that neither Riemann nor Lebesgue can close without help .
10 . 3
Further Reading
Our approach to the Lebesgue-Stieltjes integral in this book has been pragmatic and arguably ostrich-like. Major results such as the mono tone convergence theorem were stated without proof, and we seldom entertained generalizations or abstractions. The purist might right fully claim that this approach is demeaning to the subject; however, there is little harm in viewing some of the cornucopia before a sub stantial investment in further study is made (if desired) . Moreover, the Lebesgue-Stieltjes integral is no longer the exclusive preserve of mathematicians: It is an important tool in any subject that uses inte grals. At any rate, there are numerous excellent texts that cover the Lebesgue-Stieltjes integral in depth resplendent with proofs and ab stractions. Some elementary texts pitched approximately at the same level as this book include those by Pitt [30], [31 ], Priestley [32], and Weir [43], among others. These books develop the Lebesgue integral from the central concept of measure (which we did not emphasize) . Pitt's books also cover applications of the Lebesgue-Stieltjes integral
20 8
10. Epilogue
to geometry, harmonic analysis, and probability theory. Priestley's book is a particularly lively account of the integral2 with many practical comments and examples. There are several advanced accounts of integration theory avail able. Classic specialist references include Halmos [14] and Taylor [38], but many advanced analysis books such as Royden [36] and Rudin [37] also cover integration in depth. In addition, since the IY spaces loom large in functional analysis, most books on this subject devote some time to the Lebesgue integral. Riesz and Nagy [35], for example, devote nearly a quarter of their book to the Lebesgue and Stieltjes integrals; other authors such as Hutson and Pym [22] and Yosida [ 44] give concentrated but general accounts of the theory. The reader is encouraged to explore these references as curiosity or the n eed for more refined details dictates. Some of the reference cited above discuss the 11post Lebesgue" integrals of Denjoy et al. These integrals have formed a nucleus o f specialist literature somewhat apart from the normal texts on integration. Lee [27] and Pfeffer [29] provide a quite accessible reference on the Henstock-Kurzweil integral and its equivalence to the Denjoy-Perron integral. Gordon [13] also provides a basic self-contained account of these integrals along with the Lebesgue integral. The reader is also directed to the article by Bartle [4] for an introductory account. Finally, the history of integration is interesting in its own right. Most accounts (such as ours) contain fleeting glimpses of the de velopment of the integral. Hawkins [1 6] traces the history of the Riemann and Lebesgue integrals along with some early applications of the Lebesgue integral.
2Where else would you find yetis and nonmeasurable functions discussed in the s ame paragraph?
Appen dix: Hints and Answers to Selecte d Exercises.
Exercises 2.
3.
1 -1
For example, .J2+(1 -.J2) = 1 is rational, while .J2+.J2 = 2.J2 is irrational; ( .J2)(.J2) = 2 is rational, while ( .J2)(1 +.J2) = .J2+2 is irrational.
Hint: Can you be sure that b is an element of S* ?
4.
(a) Hint: Look at the proof that the set of all integers is countable.
(b)
Hint: If the set of all irrational numbers were countable, what would that imply about the union of the set of all irrational numbers and the set of all rational numbers?
Exercises I.
1 -3
Let lub and glb denote the least upper bound and the greatest lower bound respectively.
(a) lub = 5, glb = 0 , both are in the set.
(b) (c)
lub = 5, glb = 0 , 5 is in the set but 0 is not. lub = +oo, glb = -oo, neither is in the set.
(d) lub =
t, glb
= 0 , neither is in the set.
209
210
4.
Exercises
= viz, glb = - viz, neither is in the set. lub = 4, glb = 3, 4 is in the set but 3 is not. (g) lub = +oo, glb = 0 , neither is in the set. Hint: Show that sup S2 is an upper bound of 81 , and that inf S2 is
(e) (f)
3.
Appendix: Hints and Answers to Selected
lub
a lower bound of sl .
Hint: For (a) , show that c(sup S) satisfies the definition of sup S*, in other words, show that c(sup S) is an upper bound of S* and that c( sup S) < B for any upper bound B of S* . Follow a similar strategy to show that c(inf S) = inf S*, and also for part (b) .
Exercises 2-1
1 . Hint: 1b choose a sequence { an } such that an t M, consider sep arately (i) M finite, (ii) M = oo. For (i), choose a1 E S such that M - 1 < a1 < M (using Theorem 1 . 3 .2) . Then choose a2 E S such that max{M - � , ad < a2 < M , a3 E S such that max {M - � ' a2} < a3 < M , and so on. For (ii), choose a1 E S such that a1 > 1 . (Why is this always possible?) Then choose a2 E S such that a2 > max{2, ad, a3 E S such that a3 > max{ 3 , a2 }, and so on. Use a similar approach to choose a sequence bn such that bn ,!.. m.
Exercises 2-4 3.
Hint: Given E > O, choose n such that l /(n+l) < E. It follows from the definition of f that 0 < x < l in => 0 < f(x) < l l (n + 1) < E.
Exercises 2-6 1.
2.
3.
Hint: Prove that !f+(x) - g+ (x) l < lf (x) - g (x) l for all x E I by considering the four cases f (x) > 0 and g (x) > 0 , f (x) > 0 and g (x) < 0 , f (x) < 0 and g (x) > 0, f(x) < 0 and g (x) < 0 . Similar to 1 .
Hint: For each x E I, lf(x) l lf (x) - g (x) + g (x) l < lf(x) - g (x) l + l g (x) l , therefore lf (x) - g (x) l > lf (x) l - lg (x) l . Interchanging [ and g gives lf(x) - g (x) l > lg(x) l - lf(x) l . Sin c e l lf(x) l - lg(x) l l mu st equal either lf(x) l - lg(x) l or lg (x) l - lf(x) l , the result follows. =
Appendix:
Hints and Answers to
Selected Exercises
211
Exercises 2-7
I. Hint: Suppose f has bounded variation on I . Choose a point a e I, and let x be any point in I. Denote by Ix. the closed interval with endpoints a and x. Use the fact that Vx. } is a partial subdivision of I, together with the definition of bounded variation, to obtain the required result.
2 . Hint: For fg , use
n
n
L lt(bj)g (bj) - t (llj)g (llj) l = L lt (bj)g (bj) - t(bj)g (aj) j== l
3.
•
j== l +f (bj)g (�) - f (aj)g (aj ) l n < L lt(bj) l lg (bj) - g (llj ) l j== l n + L lg (aj) l lf (bj) - f (ai) l , j == l
and then use the result of Exercise 1 to obtain bounds for lf(bj ) l and lg ( llj) l that are independent ofj.
Hint: In view of Exercise 1 , for the first part you need prove only that if sup{f(x) : x E I} and inf {f (x) : x E I} are both finite (and f is monotone on I), then f has bounded variation on I. Using the notation for partial subdivisions introduced earlier, you can assume without loss of generality that
al < bl < a2 < b2 < . . . < an < bn . If f is monotone increasing on I, then
f (a! ) < f (b! ) < f (a2 ) < f (b2 ) <
·
·
·
< f (an ) < f (bn) ,
and therefore for any partial subdivision of I we have
n
n
j==
j== n
Ll lt(bj) - t (llj) l = Ll lfcbj) - t(llj)) <
n
L lf (bj) - f(aj)) + L2 lf (aj) - f(bj- ! ))
j== j==l = f (bn ) - f (a! ) < sup{f(x) : x e I} - inf{f(x)
: x E I}.
Appendix: Hints and Answers to Selected Exercises
212 A
similar argument shows that if f is monotone decreasing, then n
L lf (bj) - f (aj) l < f(al ) - f(bn ) j=l
< sup{f(x) : x
4.
e
I} - inf{f(x) : x
e
I}.
The first part then follows easily, and the second part is a straightforward application of the results already proved.
Hint: Use the construction descnbed in the proof of Theorem 2 .7.2 .
5. (a) Hint: Use the fact that l x sin(1 /x) l < l x l for all x # 0 .
6.
(b)
Hint: If an interval I has finite endpoints a , b , then for any X 1 , X2 e I we have l xf - x� l = l x 1 + x2 l l x 1 - x2 l < (2 m ax { l a l , l b l }) l x l - x2 1 . (c) Hint: Let f (x) x 2 , and assume that f is absolutely continu ous on I = ( - oo, oo). Then by definition (choosing E = 1 in the definition), there exists a 8 > 0 such that V8 (f, I) < 1 for all partial subdivisions S of I for which the sum of the lengths of all the constituent intervals is less than 8. Considering in particular partial subdivisions consisting of a single interval [n, n + 812] (where n is a positive integer) leads to the desired contradiction. =
Exercises 1.
3- 1
Suppose that x e [a , b]. Then x e h for some Ik e P and x e Ik' for some Ik' e P'. Now, P' is a refinement of P, so that Ik' c Ik. The result follows from the general inequalities inf{f{x) : x e Ik' } > inf{f(x) : x e Ik} and sup{f(x) : x e Ik' } < sup{f(x) : x e Ik} .
Q = P U P' is a refinement o f both P and P', Lemma 3 . 1 .2 implies that Sp(f) > SQ(f) and �Cf) > Q_p, Cf) . Since SQ (f) > SQ (f) for any partition Q, we have that
2 . Since the partition
Sp(f) > SQ(f) > � (f) 2: Sp� (f),
and the lemma thus follows.
Appendix: Hints and Answers to Selected Exercise s
21 3
Exercises 4-I
(b) t-La((0 , 1)) 1 - e-I , t-La([ 0 , 1]) = 3 - e-I , t-La((- 1 , 1)) 4 - e -1 , t-La([O, 0]) = 2 , t-La(( -oo, 1)) = oo , t-La((O, oo)) = 1 , t-La( [ O, oo)) = 3 . 2 . (b) t-La([- 1 , 2)) = 4, t-La(( 1 , oo)) = 2, J.La((-oo, 4)) = 6, tla((O, 2]) = 5 , tla(( � , �)) 3 , t-La([1, 3]) = 5, J.La(( 1 , 3)) = 2. =
I.
=
Exercises 4-2 I . a(x)
2 . a(x)
= =
{ 1,
{
Q,
0,
iln,
1,
if x
< A, if x > A.
if X < A J I if Ai < x < Ai+l (i if X > An .
= 1, 2 , . . . , n - 1 ),
Exercises 4·3
S U T = [1, 8). S n T = (2 , 3) U (4, 5] U (6, 7]. S - T = [1 , 2] U (5, 6] U (7, 8) . (b) S U T = [1, 4] U [5 , 8). S n T = (2, 3) U [6 , 7] . S - T = [5, 6) . (c) S U T = (1, 4] U [5 , 7) . S n T = [2 , 2] U ( 5 , 6 ) . S - T = (1 , 2) u [5, 5] .
I . (a)
Exercises 4·4 I. 2.
3.
4. 5.
A a(B) = 0. A a(B) = 0. A a(B) = 3 . A a(B) = � a
·
is not a-summable.
Exercises 4-5 2. Hint: The difficulty with this one is that it is too easy! Since La*(lf l ) = La* (f) = 0 , you can just take Bn to be the zero function on [0 , 1 ] , for each n = 1 , 2, . . . .
21 4
Appendix: Hints and Answers to Selected Exercises
Exercise 4-6
(a) Let n be the integer part of c, 1.e. , the largest integer not exceeding c. Then
(b)
1 tax - {
f.=!! n+l '
l +n-c n+l '
[O, c]
if n < c < n + � , if n + 1 / 2 < c < n + 1 .
Hint: Following on part (a), show that
l 0
c
f dx <
1 2 (n + 1 ) '
where n is the integer part of c.
(c)
Hint: Show that if n is the integer part of c and c > 1 , then
Exercises 5-l
lo
c
lf l dx
1 2
1 3
1 n
=1+-+-+...+-+
c-n . n+1
2 . Hint: Tb prove that max{f(x) , g(x)} = f (x) + (g - n+ (x) for all x E I, consider separately the cases f (x) > g(x) and f (x) < g(x); similarly, for min{f(x), g(x) } . 3 . Hint: Use Theorem 5. 1 . 5(ii) and Theorem 4 . 5. 6 . 4. Hint: Use Theorem 5. 1 .4 and Theorem 5 . 1 .3. Exercises 5-2 1.
Hint: Use the same approach as was used in Example 4-5-2 .
3. Hint: Use the fact that g = 1 -f , where f is the function defined in Section 3 . 2 , or use the fact that g = 1 a. e. to show that I ] g dx = 1.
fro,
4. Hint: Define g* on I by g * (x) so that g
5.
= g* a.e.
=
2 . (a) f
= 0.
g(x) , f(x),
if g(x)
< f(x), if g(x) > f(x), '
and g* < f on I.
Hint: Show that xsur <
Exercises 5-3
{
Xs
+ XT on JR.
Appendix: Hints and Answers to Selected Exercises
21 5
Exercises 5-4
1.
2.
i
H nt To obtain a sequence 81 , 02 ,
of a-summable step functions on lR such that limn-+oo en 1 on JR, define - n x n, On(X) = 0,1 , ifother wise.
(b)
(c)
:
{
.
•
.
=
< .
J.La([O, 2 )) = - 1 , J.La([O, 1 ]) = -1 .
<
0, J.La([ l , 1 ])
=
- 1 , J.La(( l , 2)) =
3.
Exercises 6-1
1. 2.
(i) 3 ( 1 - e- 1 ) , (ii) 4 - 3e- 1 , (iii) 4 - 3e- 1 , (iv) 4 - 3e- 1 + e-2, (v) 8 + e-2 , (vi) 3. (b) (i) 61 , (ii) e + e2 + e3 + e4 + 2e5 , (iii) 1 , (iv) 2. � (A + B). (b) 0, l in L�= l Ai · 0,
3 . (a)
(c) a (x) =
{
ifx < A 1 , 2:J= 1 pj , if Ai x < Ai+ l ifx > An. 1, <
(i
=
1 , 2,
.
. . , n - 1),
M ean = L�=l PiAi· Exercises 6-3
1.
5.
limt-+o+ erf(t)/t = 2 / ....(i. limt-+oo terfc(t) = 0. � sin(t3) - � sin(t2).
Exercise 7-4
Each repeated integral has the value 9 sin 1 + 4 sin 2.
Exercises 8-1
1 . (a)
Properties (i), (iii), and (iv) are straightforward to establish. The real problem is showing that property (ii) is satisfied. Ev ide ntly, if f = 0, then llf ii R = 0. Suppose now that llfiiR = 0 but that f =f: 0. Since f is not identically zero on the interval 0. [a, b], there is some number c e [a, b] such that lf (c) l Since f is continuous on [a, b] this means there is some in terval [a , ,8] c [a, b] containing c such that If (x) I 0 for all >
>
21 6
Appendix: Hints and Answers to Selected Exercises
Now llf ii R = J: lf (x)l dx > Jt lf (x)l dx > 0, which contradicts the hypothesis that llf iiR = 0. Therefore, II IIR satisfies property (ii). (b) Hint: Let c be any number in the interval [a , b] and let f be the function defined by f(x) = 0 if x =I= c and f(c) = 1 . What is the norm of this function? Hint: Properties (i) and (ii) follow from the inequalities l lf l h , oo > llf ll oo and llf l h , 1 > llf iiR · Property (iv) follows from the inequality x
E [a, fi] .
·
3.
SUPxe[a,b] lf(x) + g(x) l < SUPxe[a,b] lf(x) l + SUPxe[a,b] lg(x) l . n 4. Hint: ISn + 1 - Sn l < 1/10 , and if n > m , then ISn - Sm l = I (Sn Sn - 1 ) + (Sn - 1 - Sn - ) + · · · + (Sm+ 1 - Sm) l . 2 5 . Since llf llh < fi llf ll a we can choos e y = 1 / fi. Similarly, we can choose 8 = 1 /a . Exercises 8-2
1.
Suppose that an � a as n � oo . Then, for any E > 0 there is an N such that ll am - a ll < E/2 whenever m > N. Now, ll am - al l = ll (am - an ) + (an - a) ll > ll am - an II - ll an - a ll , and thus llam - an ll - ll an - a ll < ll am - a ll <
so that if n
> N,
then I am - an I -
2.
E 2
<
E , 2
E 2.
Thus, for any E > 0 there is an N such that II am - an II < E whenever n, m > N . Let n be any positive integer. Note that SUPxe[- 1 , 1J ifn (x) fn+ 1 (x) l is achieved at x = 1 1 2 n+ 1 , where fn+ 1 ex) = 0 and fn(x) = 1 - 2 m - (m+ 1 ) = � · Thus llfn(x) - fn+l (x) ll oo = � for all n, and lfn } cannot be a Cauchy sequence.
Exercises 8-3
1 . (a)
Let y be any element in Y and choose any E > 0. Since W is dense in Y, there is a w E W such that ll w - Y ll < E/2 . Similarly, since Z is dense in W , there is a z E Z such that
App endix:
l i z - w ll
2.
Hints and Answers to Selected Exercises
21 7
< e/2 . Consequently, for any y E Y and any E > 0 a z E Z such that l i z - y ll < E , i.e., Z is dense in Y .
there is (b) Use part (a) and the definition of completion. Hint: Use the fact that the set of rational numbers is dense in the set of real numbers.
Exercises 8-5
Hint: 1b show the triangle inequality r llf +g llp < r ll f llp + r llgllp apply the Minkowski inequality with F = r1 1Pf and G = r 1 1Pg . (b) See the discussion after equation (9.11 ) . Since f E £2 [0 , 1 ], we have that f E £ 1 [0 , 1 ]; since k is bounded, there is an M < oo such that l k (x, y) l < M for all (x, y) E [0 , 1] [0 , 1 ]. Hence,
1 . (a)
3.
x
I CKD (x) l < { lk (x , �) l l f(�) � < M llf lh , j[O,l]
and consequently
II Kf ll� < { M2 llfll � � = M2 llfll � .
4.
j[O,l]
Part(iii): Suppose that f E £ 1 [a, b] and that g E V"'[a, b]. Then { lf (x)g(x) l ax < llgll oo {a lf (x) l ax J[ ,b]
J[a, b]
=
ll g ll oo llf lh¥
< 00 ,
and therefore fg E £ 1 [a , b] . Part(iv): Hint: First establish the inequality
5.
{
{
J[a, b]
lf(x) IP ax < ll f ll� (b - a).
Hint: Note that f lf (x) IP ax
J [a , b ]
} { 1/p
<
f
J[a,b]
ll f ll� ax
}
1/p .
21 8
Appendix:
Hints and Answers to Selected Exercises
Exercises 9-1
1.
Let h = g/ ll g ll . Th en 0
3.
< llf - (f, h ) h ll 2 = (f - (f, h ) h, f - (f, h ) h ) = llfll 2 - (f, h ) (h , f) - (f, h ) (f, h ) + (f, h ) (f, h ) = llfll 2 - I if, h ) 1 2 ;
thus, l (f, h) l < ll f i L and hence l (f, g) ·l < ll f ll ll g ll for g f:: 0 . (If g = 0 , then the inequality follows immediately.) Hint: Use the parallelogram equality, or verify by direct calcula tion.
Exercises 9-2
2 . (a)
1b
normalize P0(x), note that II Po 11 2 = J� 1 1 2 dx = 2 ; thus, let
4.
For example, 1 1 1 1 P1 (x)P2 (x) dx = 1 x-(3x2 - 1) dx -1 -1 2
ll f + gll 2 = :E l ak + bk l 2 , k=1 00
Now,
ll f + ig ll 2 = :E l ak + ibk l 2 . k=1 00
ll f + g ll 2 = ll fll 2 + 2Re (f, g) + ll g ll 2 ,
Appendix :
Hints and Answers to Selected Exercises
so that
11[11 2 + 2Re if, g ) + llgll 2 = :E l ak + bkl 2 k=l
21 9
00
=
:E lak l 2 + 2 :E Re (akbk) + :E l bk l 2 00
k=l
00
00
k=l
k=l
11[11 2 + 2 :E Re (akbk) + llg ll 2 ; k=l consequently, Re (f, g) = L: � 1 Re (akbk) · A similar argument ap plied to llf + ig ll indicates that Im (f, g) = L:�1 Im (akbk), and hence the result follows. =
Exercises 1 . (a) 2 . (a)
9-3
PMx = 2(sin(x) - � sin(2x) + t sin(3x) - i sin(4x) + · · · . PMg (x) = L �oo Cne2nmx, where -2i if n 1s. o dd, n;r Cn = 0, , if n is even.
{
(b) llgll 2 = �� �� g2 (x) dx = 1
3.
00
L �oo l cn l2 = 2 L � 1 (2k! l )Jr 2 ·
( ) 2 Since f' is continuous on the interval [ -7r, 7r] , there exist numbers M an d M' such that lf (x) l < M and lf' (x) l < M' for all x e [ -7r, 7r] . Now, Jr Jr f(x) cos(An X) dx = _!_ [sin(Anx)]�Jr - _!_f' (x) sin(AnX) dx, An Jr -Jr An and therefore r(x) cos(Anx) dx
1
L: <
=
1
: ( l [f(x) sin(Anx)]�, \ + L: !f'(x) sin(A,x) ! dx)
l nl
- (2M + 2JCM') . -< -1 AI n l Since I A nl � oo as n --+ oo , J�Jr f(x) cos(A nx) dx � 0 as n � oo . The limit for the sine integral can be established using the same arguments.
220
Appendix:
Exercises 9-4
1.
2.
Hints and Answers to Selected Exercises
Suppose A is an eigenvalue for the operator £ . Then £y = - Apy , and {£y , y) = {- Apy , y) . If £ is self-adj oint, then {£y, y) = {y, £y); thus, {£y , y) = -A (py, y) = {y, £y) = {y, - Apy) = -A lJjy , y) . Since p is a real-valued function, we must have that A = A , i.e, A is real. l 2 The eigenvalues are An = u2n; )rr ) for n = 0 , 1 , 2 . . . ; l correspo nding eigenfunctions are l/Jn = sin ( (2n; )rr x) .
References
[1 ] Adams, R.A., Sobolev Spaces, Academic Press, 1975. [2] Ahlfors, L., Complex Analysis, 2nd edition, McGraw-Hill Book Co., 1 966. [3 ] Arfken, G., Mathematical Methods for Physicists, 2nd edition, Academic Press, 1970. [4] Bartle, R.G., 1 A. return to the Riemann integral", Amer. Math. Monthly, 103 (1996) pp. 625-632 [5] Birkhoff, G. and Rota, G., Ordinary Differential Equations, 4th edition, John Wiley and Sons, 1 989. [6] Bromwich, T.A., An Introduction to the Theory of Infinite Series, Macmillan and Co., 1926. [7] Carleson, L. 1 Convergence and growth of partial sums of Fourier Series" Acta Math. , 11 6, (1 966) pp. 1 35-157 [8] Churchill, R.V:, Fourier Series and Boundary Value Problems, 2nd edition, McGraw-Hill Book Co., 1963. [9] Coddington, E.A. and Levinson, N., Theory of Ordinary Differential Equations, McGraw-Hill Book Co., 1955. [10] Conway, J.B., Functions of One Complex Variable I, 2nd edition, Springer-Verlag, 1 978. 221
222
References
[11] Courant, R. and Hilbert, D., Methods of Mathematical Physics, volume 1, John Wiley and Sons, 1 953. [12] Dunford, N. and Schwartz, J.T., Linear Operators, parts I, II, III, John Wiley and Sons, 1 971 . [13] Gordon, R.A., The Integrals of Lebesgue, Denjoy, Perron, and Henstock, American Math Soc., 1994 [14] Halmos, P.R., Measure Theory, Springer-Verlag, 1974. [15] Hardy, G.H. and Rogosinski , WW., Fourier Series, 3rd edition, Cambridge University Press, 1956 [16] Hawkins, T., Lebesgue's Theory of Integration, Its Origins and Development, The University of Wisconsin Press, 1970. [17] Heider, L.J. and Simpson, J.E., Theoretical Analysis, W.B. Saunders Co., 1967. [18] Hewitt, E. and Stromberg, K. , Real and Abstract Analysis, Springer-Verlag, 1969. [19] Higgins, J.R., Completeness and Basis Properties of Sets of Special Functions, Cambridge University Press, 1977. [20] Hille, E., Analytic Function Theory, volume II, Ginn and Co., 1959. [21] Hoffman, K. , Banach Spaces of Analytic Functions, Prentice-Hall, 1962. [2 2] Hutson, V. and Pym, J.S., Applications of Functional Analysis and Operator Theory, Academic Press, 1 980. [23] Ince, E.L., Ordinary Differential Equations, Longmans, Green and Co., 1 927. [24] Korner, T.W., Fourier Analysis, Cambridge University Press, 1 988. [25] Kreyszig, E., Introductory Functional Analysis with Applications, John Wiley and Sons, 1 978. [26] Kreyszig, E., Advanced Engineering Mathematics, 4th edition, John Wiley and Sons, 1 979 . [27] Lee, P. , Lanzhou Lectures on Henstock Integration, World Scientific, 1 989. [28] McLachlan, N.W., Theory and Application of Mathieu Functions, Oxford University Press, 1947. �
References
223
(29] Pfeffer, W.F. , the Riemann Approach to Integration: Local Geometric Theory, Cambridge University Press, 1993. [30] Pitt, H. R., Integration, Measure and Probability, Oliver and Boyd, 1 963. [31] Pitt, H. R., Measure and Integration for Use, Oxford University Press, 1 985. [32] Priestley, H.A., Introduction to Integration, Oxford University Press, 1 997. [33] Pryce, J. D. Basic Methods of Linear Functional Analysis, Hutchinson and Co., 1 973. [34] Richtmyer, R.D. Principles of Advanced Mathematical Physics, volume I , Springer-Verlag, 1978. [35] Riesz, F. and Nagy, B., Functional Analysis, Frederick Ungar Publishing Co., 1955. [36] Royden, H.L., Real Analysis, Macmillan and Co., 1 963. [37] Rudin, W., Real and Complex Analysis, 2nd edition, McGraw-Hill Book Co., 1974. (38] Thylor, A.E., General Theory of Functions and Integration, Blaisdell Publishing Co., 1965. (39] Titchmarsh, E . C., The Theory of Functions, 2nd edition, Oxford University Press, 1 939. [40] Titchmarsh, E. C., Eigenfunction Expansions, Part I, 2 nd edition, Oxford University Press, 1 962. [41] Whittaker, E.T. and Watson, G.N., A Course of Modem Analysis, 4th edition, Cambridge University Press, 1 952. [4 2 ] Wiener, N., The Fourier Integral and Certain of Its Applications, Cambridge University Press, 1 933. [43] Weir, A.J. , Lebesgue Integration and Measure, Cambridge University Press, 1 973. (44] Yosida, K. Functional Analysis, 6th edition, Springer-Verlag, 1 980. [45] Zygmund, A., Trigonometric Series, volumes I and II, Cambridge University Press, 1 959.
Index
almost everywhere (a.e.), 76
anti-derivative, no
Appolonius identity, 1 72
continuity one-sided, 1 9 continuous absolutely, 36 convergence
Banach space, 1 33
in the norm, 1 29
Lebesgue, 1 44
of a double series, 1 5
separable, 1 50
of a sequence, ll
Sobolev, 1 62
of a series, 1 3
HP spaces, 1 57
of an improper integral, 44
Bessel's inequality, 1 77
pointwise, 1 2
bound
uniform, 1 3 3
essential upper, 1 4 8 greatest lovver, 8
Denjoy integral, 204
least upper, 8
dense, 2 density of distnbution, 103
Cauchy integral formula, 1 58 Cesaro mean, 1 8 7 characteristic function, 76 , 1 52 conjugate exponents, 1 4 5
divergence of a series, 1 3 proper, 1 5 dominated convergence theorem, 80
225
dual space, 1 4 7
Haar functions, 1 99 Hardy spaces, 1 57
eigenfunctions, 1 90
Henstock-Kurzweil integral, 206
eigenvalue s, 1 90
Hermite
error function, 102 complementary, 103 extended real number system, 6
differential equation, 1 97 polynomials, 1 97 Hilbert adjoint operator, 1 9 0
Fatou's Lemma, 8 0 Fourier
dimension, 1 79 Hilbert space, 1 68
coefficients, 1 76 complex series, 1 82
improper integral, 44
series, 1 76
absolutely convergent, 46
series, classical, 1 81
conditionally convergent, 46
sine series, 1 9 3
convergence of, 44
Fredholm integral operator, 1 50
infimum, 8
Fubini's theorem, 11 8
inner product, 1 65
function
inner product space, 1 66
a-measurable, 82 characteristic, 76 , 1 52
real, 1 66 integral
error, 1 0 2
Denjoy, 204
Lipschitz, 3 8
Henstock-Kurzweil, 206
major, 204
improper, 44
monotone, 20
indefinite, 110
negative part, 28
Newton, 203
null, 75
Perron, 205
positive part, 28
Riemann, 39
simple, 1 1 5
Riemann-Darboux, 42
step, 24 strictly increasing, 97 function spaces, 1 26 functional
interval closed, 7 open, 7 irrational numbers, 2
bounded, 1 46 linear, 1 45
jump discontinuity, 1 9
norm, 1 46 fundamental theorem of calculus, 102
Laguerre differential equation, 1 98 polynomials, 1 98
Holder's inequality, 1 43 HP spaces, 1 58
Lebesgue integral definition, 66
Index
-------
generalization of Riemann, 69 Lebesgue-Stieltjes integral change of variable, 97 definition, 6 6
22 7
Euclidean, 1 27 indu ced by inner product, 1 6 7 of a partition , 43 normed vector sp ace
differentiation under the integral, 105
closed sub set of, 1 34 compl ete, 1 33
double, 11 5
complete subset of, 1 34
first mean value theorem, 75
completion of, 1 37
integration by parts, 100
definition, 1 2 7
linearity of, 74 �
dual space, 1 4 7
repeated, 1 1 6
isometric, 1 35
Legendre
nul
differential equation, 1 95
function, 75
polynomials, 1 79, 1 9 6
set, 76
Leibniz's rule, 108 limit as
x -+
±oo, 1 8
operator adjoint, 1 90
for a function, 1 6
de finition, 1 3 5
of a sequence, 11
Fredholm integral, 1 50
one-sided, 1 6
isometry, 1 3 5
Lipschitz function , 38
resolvant, 1 9 0 self-adjoint, 1 90 orthogonal
Mathieu equation, 1 94
complement, 1 74
functions, 1 94
definition, 1 72
maximum modulus principle,
orthonormal
1 55
basis, 1 77
measure of a rectangle, 11 3 of an interval,
set, 1 75
SO
set, 1 76 ostrich, 207
probability, 52 Minkowski's inequality, 1 44 HP spaces, 1 58 monotone convergence theorem, 79
parallelogram equality, 1 68 Parseval's formula, 1 78 partial subdivision, 30 partition locally fine, 2 0 6
Newton integral, 203
norm of, 43
norm
of an interval, 41
definition, 1 2 7 equivalent, 1 2 9 , 1 30
refinement of, 41 Perron integral, 205
2 2 8--------�m=d=a= probability
orthogonal, 1 75
density, 103
orthonormal, 1 76
discrete distnbution, 54
simple, 55, 11 4
distribution function, 52
total, 1 75
measure, 52
signum function, 1 80
uniform distribution, 53
simple function, 11 5
Pythagoras's theorem, 1 73
Sobolev space, 1 62 span, 1 26
Rademacher functions, 1 80
spectrum, 1 90
random variable, 52
step function, 24
mean, 96 rational numbers, 1
a-summable, 56 Sturm-Liouville problem
rectangle, 11 3
regular, 1 89
Riemann
singular, 1 95
-Lebesgue theorem, 1 84
sum
localization theorem, 1 8 6
direct, 1 74
Mapping Theorem, 1 5 6
vector, 1 74
theorem o n derangement of series, 1 5 Riemann integral definition, 3 9
support of a Haar function, 201 of a step function, 25 of simple function, 11 5 supremum, 8
Schwarz's inequality, 1 67 seminorm, 1 39
variation
separable, 1 50
functions of bounded, 32
sequence
total, 30
admissible, 60
vector space
Cauchy, 1 29
complex, 1 24
convergence, 11
definition, 1 24
monotone, 1 2
finite-dimensional, 1 26
series term by term integration, 80 set a-finite, 55 a-measurable, 82
infinite-dimensional, 1 26 normed, 1 2 7 subspace, 1 26 vector spaces seminormed, 1 3 9
closure of, 1 34 complete, 1 75
Walsh functions, 202
countable, 3
Weierstrass, 1 38
dense, 1 3 7 nul, 76
yeti, 208
Undergra duate Texts in Mathematics
(continuedfrom page ii)
Hilton/Holton!Pedersen : Mathematical Reflections : In a Room with Many Mirrors.
Iooss/Joseph: Elementary Stability and Bifurcation Theory. Second edition.
Isaac : The Pleasures of Probability. Readings in Mathematics.
James: Topological and Uniform Spaces.
Janich : L inear Algebra. Janich: Topology. Kemeny/Snell: Finite Markov Chains .
Owen: A First Course in the Mathematical Foundations of Thermodynamics .
Palka : An Introduction to Complex Function Theory.
Pedrick: A First Course in Analysis . Peressini/Sullivan!Uhl : The Mathematics of Nonlinear Programming .
Prenowitz/Jantosciak: Join Geometries . Priestley: Calculus : A Liberal Art. S econd edition.
Protter/Morrey : A First Course in Real Analysis. Second edition.
Kinsey: Topology of Surfaces. Klambauer: Aspects of Calculus .
Protter/Morrey: Intermediate Calculus .
Lang : A First Course in Calculus . Fifth
Roman : An Introduction to Coding and
edition.
Lang: Calculus of Several Variables . Third edition.
Lang: Introduction to Linear Algebra. Second edition.
Lang: Linear Algebra. Third edition . Lang: Undergraduate Algebra. Second edition.
Lang: Undergraduate Analysis. Lax/Burstein/Lax: Calculus with Applications and Computing.
Volume 1 . LeCuyer: College Mathematics with APL.
Lidi/Pilz: Applied Abstract Algebra. Second edition.
Logan: Applied Partial Differential Equations .
Macki-Strauss : Introduction to Optimal Control Theory.
Malitz: Introduction to Mathematical Logic .
Marsd en/Weinstein: Calculus I, II, III. S econd edition.
Martin: The Foundations of Geometry Marti n : Geometric Constructions. Martin: Transformation Geometry: An and the Non-Euclidean Plane.
Introduction to Symmetry.
Millman/Parker: Geometry: A Metric Approach with Models. Second edition.
Moschovakis : Notes on Set Theory.
Second edition. Information Theory.
Ross: Elementary Analysis : The Theory of Calculus .
Samuel : Proj ective Geometry. Readings in Mathematics.
Scharlau/Opolka : From Fermat to Minkowski .
Schiff: The Laplace Transform: Theory and Applications .
Sethuraman : Rings, Fields, and Vector Spaces : An Approach to Geometric Constructability.
Sigler: Algebra. Silverman/Tate: Rational Points on Elliptic Curves.
Simmonds : A Brief on Tensor Analysis . S econd edition.
Singer: Geometry: Plane and Fancy. Singerrfhorpe: Lecture Notes on Elementary Topology and Geometry .
Smith : Linear Algebra . Third edition. Smith: Primer of Modern Analysis. Second edition .
Stanton/White: Constructive Combinatorics.
Stillwell : Elements of Algebra: Geometry, Numb ers, Equations .
Stillwell: Mathematics and Its History. Stillwell: Numbers and Geometry. Readings in Mathematics.
Strayer: Linear Programm i ng and Its Applications .
Undergraduate Texts in Mathematics Thorpe: Elementary Topics in Differential Geometry .
Toth : Glimpses of Algebra and Geometry. Readings in Mathematics.
Troutman: Variational Calculus and Optimal Control . Second edition .
Valenza:
Linear Algebra: An Introduction
to Abstract Mathematics .
Whyburn!Duda: Dynamic Topology .
Wilson: Much Ado About Calculus .
