Foundations of Mathematical Analysis (Pure & Applied Mathematics) 0824769198, 9780824769192

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Foundations of Mathematical

Analysis RICHARDJOHNSONBAUGH Chicago State University Chicago, Illinois

W. E. PFAFFENBERGER University of Victoria Victoria, British Columbia, Canada

MARCELDEKKER, INC.

New York and Basel

Library of Congress Cataloging in Publication Data

Johnsonbaugh, Richard, [date) Foundations of mathematical analysis. (Monographs and textbooks in pure and applied mathematics ; 62) Bibliography: p. Includes index. 1. Mathematical analysis-Foundations. I. Pfaffenberger, W. E.,joint author. II. Title. QA299.8.J63 515 80-25593 ISBN 0-824 7-6919-8

COPYRIGHT © 1981 by MARCEL DEKKER, INC. ALL RIGHTS RESERVED Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microfilming, and recording, or by any information storage and retrieval system, without permission in writing from the publisher. MARCEL DEKKER, INC. 270 Madison Avenue, New York, New York 10016 Current printing (last digit): 10 9 8 7 6 5 4 3 2 1 PRINTED IN THE UNITED STATES OF AMERICA

Preface

This book evolved from a one-year Advanced Calculus course that we have given during the last decade. Our audiences have included junior and senior majors and honors students, and, on occasion, gifted sophomores. The material is logically self-contained; that is, all of our results are proved and are ultimately based on the axioms for the real numbers. We do not use results from other sources, except for a fcw results from linear algebra which arc summarized in a brief appendix. Thus, theoretically, no prerequisites are necessary to understand this material. Realistically, the prerequisite is some mathematical maturity such as one might acquire by taking calculus and, perhaps, linear algebra. Our intent is to teach students the tools of modem analysis as it relates to furthcr study in mathematics, especially statistics, numerical analysis, differential equations, mathematical analysis, and functional analysis. It is our belief that the key to a sound foundation for the study of analysis lies in an understanding of the limit concept. Thus, after initial chapters on sets and the real number system, we introduce the limit concept using numerical sequences and series (Chapters IV and V). This is followed by Chapter VI on the limit of a function. We then move to the general setting of metric spaces (Chapter VII). Chapter VIII is a review of differential calculus. Chapter IX gives a detailed introduction to the theory of Riemann-Sticltjes integration. We then tum to the study of sequences and series of functions (Chapters X and XI), Fourier series (Chapter XII), the Riesz representation theorem (Chapter XIII), and the Lebesgue integral (Chapter XIV). The first seven chapters could be used for a one-term course on the Concept of Limit. Because we believe that an essential part of learning mathematics is doing mathematics, we have included over 750 exercises, some containing several Ill

iv

Preface

parts, of varying degrees of difficulty. Hints and solutions to selected exercises, indicated by an asterisk, are given at the back of the book. We would like to thank our colleagues, Dr. Rosalind Reichard, who taught this course from a preliminary version and gave us useful information, and Dr. Keith Rose, who read the manuscript and offered valuable criticism. Thanks also to our many students who studied this material and offered suggestions, and especially Mr. James Africh, who worked nearly every exercise and made many helpful comments. Our thanks also go to the secretarial staff at the University of Victoria, who over the years typed various versions of the manuscript. Of course, we assume joint responsibility for the book's strengths and weaknesses, and we welcome comment. Richard Johnsonbaugh W. E. Pfaffenberger

Contents

Preface

iii

I SetsudFacdoas

1

1. Sets 2. Functions

1 4

II 'TheReal Nam.herSystem

3. 4. 5. 6. 7.

m

9

9 12 14 17 20

The Algebraic Axioms of the Real Numbers The Order Axiom of the Real Numbers The Least-Upper-Bound Axiom The Set of Positive Integers Integers, Rationals, and Exponents

26

Set Equivalence

26 29

8. Definitions and Examples 9. Countable and Uncountable Sets IV Sequencesof Real Numbers

34

10. Limit of a Sequence 11. Subsequences 12. The Algebra of Limits

38

34

40

45

13. Bounded Sequences 14. Further Limit Theorems 15. Divergent Sequences 16. Monotone Sequences and the Number e

46

48 49 V

vi

Contents

17. 18. 19. 20. 21.

Real Exponents The Bolzano-Weierstrass Theorem The Cauchy Condition The lim sup and lim inf of Bounded Sequences The Jim sup and Jim inf of Unbounded Sequences

V InftniteSeries

22. 23. 24. 25. 26. 27. 28. 29.

The Sum of an Infinite Series Algebraic Operations on Series Series with Nonnegative Terms The Alternating Series Test Absolute Convergence Power Series Conditional Convergence Double Series and Applications

VI Umits of Real-ValuedFunctionsandContinaoasFunctions on the Real Line 30. 31. 32. 33. 34.

Definition of the Limit of a Function Limit Theorems for Functions One-Sided and Infinite Limits Continuity The Heine-Borel Theorem and a Consequence for Continuous Functions

VII Metric Spaces

35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47.

The Distance Function R",12 , and the Cauchy-Schwarz Inequality Sequences in Metric Spaces Closed Sets Open Sets Continuous Functions on Metric Spaces The Relative Metric Compact Metric Spaces The Bolzano-Weierstrass Characterization of a Compact Metric Space Continuous Functions on Compact Metric Spaces Connected Metric Spaces Complete Metric Spaces Daire Category Theorem

55 58 59 61 69

73 73 76 77 80 81 87 90 92 102

102 105 107 109 112 116

116 120 125 128 132 136 141 144 148 152 155 159 166

Contenta

VIII

vii

DifferentialCalcalusof the Real Line

171

48. Basic Definitions and Theorems 49. Mean-Value Theorems and L'Hospital's Rule 50. Taylor's Theorem

171 176 185

IX The Rie~tieltjes

Integral

189

51. Riemann-Stieltjes Integration with Respect to an 52. 53. 54. 55.

56. 57. 58. 59. X

Increasing Integrator Riemann-Stieltjes Sums Riemann-Stieltjes Integration with Respect to an Arbitrary Integrator Functions of Bounded Variation Riemann-Stieltjes Integration with Respect to Functions of Bounded Variation The Riemann Integral MeasureZero A Necessary and Sufficient Condition for the Existence of the Riemann Integral Improper Riemann-Stieltjes Integrals

190 204 210 213 219 225

230 234 238

Seqaeaces and Series of Fanctiom

245

60. Pointwise Convergence and Uniform Convergence 61. Integration and Differentiation of Uniformly Convergent Sequences 62. Series of Functions 63. Applications to Power Series 64. Abel's Limit Theorems 65. Summability Methods and Tauberian Theorems

245

XI Trtmlcendental Fanctiom

66. The Exponential Function 67. The Natural Logarithm Function 68. The Trigonometric Functions

249

253 259

262 265 268

268 271 274

XII Inner ProductSpaces andFourierSeries

280

69. Normed Linear Spaces 70. The Inner Product Space R3 71. Inner Product Spaces

280 285 288 293

72. Orthogonal Sets in Inner Product Spaces

viii

Contenta

73. Periodic Functions 74. Fourier Series: Definition and Examples 15. Orthonormal Expansions in Inner Product Spaces 76. Pointwise Convergence of Fourier Series in ~a,a + 2n] 77. Cesaro Summability of Fourier Series 78. Fourier Series in ~[a, a + 2n] 79. A Tauberian Theorem and an Application to Fourier Series

xm

XIV

NormedLinear Spaces andtbe .RieszR Theorem

295 298 302 308 315 322 331

tion 335

80. Normed Linear Spaces and Continuous Linear Transformations 81. The Normed Linear Space of Continuous Linear Transformations 82. The Dual Space of a Normed Linear Space 83. Introduction to the Riesz Representation Theorem 84. Proof of the Riesz Representation Theorem

339 343 346 349

The Lebespe IDtegral

355

85. The Extended Real Line O'-Algebras and Positive Measures Measurable Functions Integration on Positive Measure Spaces Lebesgue Measure on R Lebesgue Measure on [a, b] The Hilbert Spaces ~ 2(X, .,11, µ)

356 357 361 368 381 392 397

Appendix: Vector Spaces

405

References

409

Hints to Selected Exercises

411

Index

421

86. 87. 88. 89. 90. 91.

335

Foundations of Mathematical Analysis

I Sets and Functions

In this brief chapter we will summarize some of the fundamental notation and definitions which will be used throughout the text.

1.

Seta

As a starting point we describe what we mean by a set. We will not attempt to define the term set, but we demonstrate the use of the term by examples. The term set is used (roughly) to describe any collection of objects. For example, the set whose objects are the positive integers 1, 2, 3 may be denoted {l, 2, 3}

If a set consists of a finite number of objects, we may denote the set by listing its objects. If a set consists of infinitely many objects such a listing is impossible. In this case we may describe the set by naming a property common to all the objects in the set. For example, to describe the set of positive integers, we use the notation {x Ix is a positive integer} This notation is read "the set of all x such that xis a positive integer." (The bar "I" is read "such that.") In general, to describe the set of all objects having a particular property P, we write {xi x has property P} If x is an object in a set A, we write xeA 1

Seta and Functions

2

and say that xis an element (or a poinl) of A. If xis not an element of the set A, WC write

x¢A

For example, if Z denotes the set of integers, then I e Z, but ½¢ Z. We shall say two sets A and B arc equal and write A = B if A and B have the same clements. Thus A = B if whenever x e A, then x e B, and whenever x e B, then x e A.

De6litioa 1.1 If A and B arc sets, the union of A and B is the set

Au B

= {x Ix e A

or

x e B}

The intersection of A and B is the set

= {x Ix e A and x e B} = {I, 2, 3} and B = {2, 3, 4}, then A" B

For example, if A

{l,2,3,4}

AuB= An B

= {2, 3}

We may extend the definitions of union and intersection to more than two sets as follows. If d is a collection of sets, we define

ud

= {x Ix e A for

nd

= {x I x e d

some A e d}

for every A e d}

In case the family d is indexed by the positive integers, d A 3 , ••• }, we write Cl0

ud

nA

=

{Ai, A 2 ,

Cl0

= Ui A.

nd=

11=

11=

11

i

For example, if

d where then

Ai

= {I}, nd

A2

=

= {Ai, A 2 ,

= {I, 2}, ...

n A.=

,

• •• }

A 11 = {I, 2, ...

Cl0

a•l

, n}, ...

00

{1}

ud

= UA

11

11=

i

is the set of positive integers. Deftaitioa 1.2 The empty set is the set with no elements and is denoted

0.

Deftnidon 1.3 If every element of the set A is an element of the set B, we write A c: B and say that A is contained in B or that A is a subset of B.

1.

Seta

3

We write B => A if A c B and say that B contains A. We note that according to Definition 1.3, A c B allows the possibility that A = B. It follows immediately from our definitions that A = B if and only if A c Band B c A. We call A a proper subset of B if A c Band A#: B.

Definition1.4 If A and B are sets, the difference of A and B is the set A\B

= {x Ix e A and

x ¢ B}

For example, if A = {l, 2, 3} and B = {2, 3, 4}, then A\B = {1}. If we are working with sets all of which are subsets of some particular set U, we sometimes say that U is the universe in which we are working. For example, if we are working with sets of integers, we could agree that the universe U is the set Z of all integers.

Definition1.5 If we are working in a fixed universe U, and A A'=

c

U, we write

U\A

The set U\A is called the complement of A relative to U (or just simply the complement of A if the universe U is understood). For example, if U is the set of real numbers, we would state that the complement of the set of rational numbers is the set of irrational numbers. We now prove two theorems which are known as De Morgan's laws. These theorems can often be used to convert a statement about unions into a statement about intersections and vice versa. Theorem1.6

If A and B are subsets of a universe U, then

(Au B)'

= A'

ri

(A n B)'

B'

= A' u

B'

Proof. We prove only the first equation leaving the second as an exercise. Let x e (A u B)'. Then x ¢ A u B, and hence x ¢ A and x ¢ B. Thus x e A' and x e B' which implies x e A' n B'. Therefore, (A u B)'

C

A' ('\ B'

Next suppose x e A' n B'. Then x e A' and x e B', and hence x ¢ A and x ¢ B. Thus x ¢Au B, which implies x e (A u B)'. Therefore, A'('\ B'

which completes the proof.

■

C

(Au B)'

4

Sets and Functions

In a similar way one may prove the following theorem. Theorem 1.7 Let d be a collection of subsets of a universe U, and let d' = {A' I A e d}

Then Exercises

1.1 Let U = {1, 2, 3, 4, s, 6, 7, 8, 9, 10} A = {l, 2, 3, 4} B = {1, 3, S, 7, 9} Compute the following sets: (a) AuB (b) AnB (c) A\B (d) B\A (e) A' (0 B' 1.2 Prove the second equation of Theorem 1.6.

1.3 Prove Theorem 1.7. In exercises 4 to 13, A, B, and C are all subsets of a universe U. 1.4 Prove: Ac 0 if and only if A= 0 1.5 Prove: If A c Band B c C, then A c C 1.6 Prove: A u B = Bu A 1.7 Prove: Au (Bu C) = (Au B) u C 1.8 Prove: An(BuC) = (AnB)u(AnC) 1.9 Prove: Au (Bn C) = (A uB)n(A u C) 1.10 Prove: (A')' = A 1.11 Prove: If A c B, then B' c A' 1.12 Prove: A n A' = 0 1.13 Prove: A ¢ B if and only if An B' ¢. 0 ( ¢ means "is not a subset of") 1.14 Let Ube a universe. Let A be a subset of U and let d be a collection of subsets of U. Prove: (a) A n (ud) = u {A n XI Xe d} (b) Au(nd)=n{AuX]Xed}

2.

functions

The concept of a function is central to all of mathematics, and in this section we will give a precise definition of a function and prove several properties of functions. We begin with the concept of an ordered pair.

2.

Functions

6

Definidon2.1 The ordered pair of elements a and b, written (a, b), is the set (a, b)

= {{a}, {a, b}}

a is called the first element of (a, b) and b is called the second element of (a, b). The crucial property of an ordered pair is stated in the next theorem. The theorem states that two ordered pairs are identical if and only if they have the same first elements and the same second elements. Theorem2.2 Let (a, b) and (c, d) be ordered pairs. Then (a, b) and only if a = c and b = d. Proof. Suppose a= c and b = d. Then {a} and therefore (a, b) = (c, d). Conversely, suppose (a, b) = (c, d). Then {{a}, {a, b}}

= {c} and

= {{c}, {c, d}}

= (c, d)

{a, b}

if

= {c, d}, (2.1)

= b. Now (a, b) = {{a}, {a, b}} = {{a}, {a}} = {{a}} and so {{c}, {c, d}} = {{a}} Thus {a} = {c} = {c, d}, and therefore a = b = c = d. First, we consider the case that a

Now suppose a:/: b. From equation (2.1) we see that either {c} = {a} or {c} = {a, b}. Since a:/: b, we must have {c} = {a}, which implies that a= c. Again using equation (2.1), we have that either {a, b} = {c} or {a, b} = {c, d}. If {a, b} = {c}, then a = b = c, which is not the case, and thus {a, b} = {c, d}. It follows that b = c orb = d, but if b = c, we would have the contradiction b = c = a. Therefore b = d, and we have established the theorem. ■

Definidon2.3 If X and Y are sets, the Cartesian product of X and Y denoted X x Y is the set

X x Y

= {(x, y) Ix e X and ye

Y}

Definition2.4 Let X and Y be sets. A function from X into Y is a subset/ of X x Y satisfying (i) If (x, y) and (x, y') belong to/, then y = y'. (ii) If x e X, then (x, y) e/for some ye Y.

If/ is a function from X into· Y, we will write/: X

➔

Y. The crucial property of a function from X into Y is that with each [2.4(ii)] element x in X there is associated a unique [2.4(i)] element y in Y.

Definition2.5 Let/ be a function from X into Y. Let A c X and B c Y.

6

Seta and Functions

(i) X is called the domain off. Y is called the codomain off. (ii) If (x, y) E/, we write y = f(x) and call y the (direct) image of x under f. (iii) The range off is the set {/(x)

I XE

X}

(iv) The image of .A under f is the set/(.A) = {/(x) Ix (v) The inverse image of B under f is the set

1- 1(B) = {x lf(x)

E

.A}.

EB}

(vi) f is onto Y if f(X) = Y. (vii) f is one to one if f(x) = f(x') always implies that x

= x'

for x,x'

E X.

If f is a one-to-one function from X into Y, we may define the inverse function to f, denoted 1- 1 , from the range off onto X by the rule (y, x)

E/- 1 if and

only if (x, y)

E/

Notice that thefunctionf- 1 is defined only if /is one to one but that/- 1(B) is defined for an arbitrary function/ and for all sets B c Y. If g: X ➔ Y and /: Y ➔ Z, we define the composition f O g : X ➔ Z by the rule (f

If/: X ➔

O

g)(x)

= f(g(x))

for each

X

Y and .A c X, we define the function

fl .A = {(x, y) f

XE

E/1x E.A}

I .A is called the restriction off to .A, and/ is said to an extension off I .A to

X.

We illustrate the preceding definitions with an example.

Example. Let f

= {(I,

1), (2, 1), (3, 4)},

.A = {l, 2}, B

= {l}

The domain of/is {l, 2, 3} and the range of /is {l, 4}. The image of .A under / is the set/(.A) = {I}. The inverse image of B under f is the set/- 1(B) = {l, 2}. If we let Y = {l, 4},/is onto Y. The function/is not one to one. The restriction off to .A is the function fl .A = {(I, l), (2, l)}

Let

g

= {(l,

l), (2, 3)}

Then g is one to one and the inverse function to g is the function g-

1

= {(I, l), (3, 2)}

7

2. Functions

The composition f

O

g

is the function fog

=

{(I, t), (2, 4)}

The next theorem establishes several properties of inverse and direct images.

Theorem2.6 Let f be a function from X into Y. Let ..91be a collection of subsets of X, and let Cl be a collection of subsets of Y. Let C c Y. Then

= u{/(A) I A e.JJ/} (ii) 1- 1(uC/) = u{/- 1(C) IC e Cl} (iii) 1- 1(n'l) = n{/- 1 (C) IC e Cl} (iv)/- 1(C') = [/- 1(C)]' (i) /(u..91)

Proof. We prove part (ii) only. Let x e/- 1(u'l). Then/(x) e uC/, and thus /(x)e C 1 for some C 1 eC/. Thus xe/- 1(C 1), and therefore xe u{/- 1(C): Ce Cl}.We have proved that 1

/-

(uC/) c u{/-

1

(C)I CeC/}

Now let x e u{/- 1(C): Ce Cl},then x e/- 1(C 1) for some C 1 in Cl. Therefore /(x) e c., and hence /(x) e uC/. It follows that x e/- 1(u'l). We have proved that 1

u{/and part (ii) now follows.

(C) IC e Cl} c

1- 1(u'l)

■

In case JJ/ = {A, B} and Cl= {C, D} the conclusions of Theorem 2.6 may be written

= /(A) u /(B) 1- ccu D) = 1- 1(C) u 1- 1 1-•cc("\D) = 1- 1(C) n/- 1(D) in general one hasJtA. n B) = /tA) n/tB). /(A. u B) 1

It is not true that

Exercises

2.1

Let g

=

I=

{(1, 2), (2, 2), (3, 1), (4, 4)} {(1, 5), (2, 7), (3, 9), (4, 17)} {1,2}

A= Compute (a) The domain of g

(Verify.)

8

Sets and Functions

(b) The range of g (c) g(A) (d) g- 1(A) (e) / 0 g

(0 1- 1 2.2 Leth: X - Y, g: Y- Z, and/: Z - W. Prove that (f g) oh= f o (go h). 2.3 Let/: X- Yand Ac Xand B c Y. Prove (a) f(/- 1 (B)) c B (b) A c 1- 1 (/(A)) 2.4 Prove Theorem 2.6 (i), (iii), and (iv). 2.5 Let g: X- Yand/: Y-Z. (a) Prove that if g and/ are one-to-one functions, then/• g is a one-to-one function and (/ • g)- 1 = g- 1 • 1- 1 • (b) Prove that if g and/ are onto functions, then f O g is an onto function. (c) Prove that if A c Z, then (/ • g)- 1 (A) = g- 1 (f- 1 (A)). 2.6 Let/: X - Y. Prove that f is a one-to-one function if and only if O

f(A n B)

for all subsets A and B of X.

= /(A)

n/(B)

II The Real Number System

The central topic of study in this text is the real number system. We will define the real numbers by specifying which axioms or rules the real numbers are assumed to satisfy. In an appropriate theory of sets, one may construct a number system which satisfies these axioms [see Kelley (1955) and Landau (1960)). It can be shown that these axioms determine the real numbers (see Exercise 7.9).

3.

The Algebraic Axioms of the Real Numbers

We begin with a definition.

Definition3.1 A binary operation on a set X is a function from X x X into X.

Intuitively, a binary operation on a set Xis a rule which associates with each ordered pair of elements of X a unique element of X. Binary operations are often written +, ·, or •, and the value of the function at an ordered pair (x, y) is usually written x + y, x. y, or x•y. The particular binary operations with which we shall be immediately concerned are addition and multiplication of real numbers. Addition or multiplication of real numbers associates with each ordered pair of real numbers (a, b) another real number, namely the sum of a and b or the product of a and b.

Definition3.2 The real numbers R is a set of objects satisfying Axioms 1 to 13 as listed in the following: 9

The RNI Number Syatem

10

Axiom 1. There is a binary operation called addition and denoted that if x and y are real numbers, x + y is a real number.

+

such

Axiom 2. Addition is associative. (x

+ y) + z = x + (y + z)

for all x,y.z e R. Axiom 3. Addition is commutative.

x+y=y+x for all x,y e R. Axiom 4. An additive identity exists. There exists a real number denoted 0 which satisfies

for all xe R. Axiom 5. that

Additive inverses exist. For each x e R, there exists ye R such

The number y of Axiom 5 may be shown to be unique, and it is denoted -x. We define x - z as x + (-z) for all x and z in R. Any mathematical system which satisfies axioms 1 to 5 is called a commutative (or abelian) group. Using axioms I to 5 one may establish the usual rules for addition of real numbers. We give one example in the next theorem and several other additive properties are given as exercises. 1beorem 3.3 The additive identity of axiom 4 is unique, that is, if there exists O' e R such that x + O' = x for all x e R, then O = O'. Proof. Suppose there exists O' e R such that x + O' = x for all x e R. Then O + O' = 0. On the other hand, from axiom 4 we have O' + 0 = O'. Since addition is commutative (axiom 3),

• We continue by giving the axioms for multiplication of real numbers. Axiom 6. There is a binary operation called multiplication and denoted · such that if x and y are real numbers, then x·y (or xy) is a real number.

3. The Algebraic Axioms of the Real Numbers

11

Axiom 7. Multiplication is associative. (xy)z

= x(yz)

for all x,y,z e R.

Axiom 8. Multiplication is cummutative. xy =yx

for all x,y e R.

Axiom 9. A multiplicative identity exists. There exists a real number, different from 0, denoted I which satisfies x• 1

= x = l •x

for all x in R.

Axiom 10. Multiplicative inverses exist for nonzero real numbers. For any x e R with x #: 0, there exists y e R such that xy

= 1 = yx

The next axiom links the operations of addition and multiplication.

Axiom 11. Multiplication distributes over addition.

+ z) = xy + xz

x(y

(y

+ z)x = yx + zx

for all x,y ,z e R. As in the case of addition one may show that 1 and multiplicative inverses are unique. The multiplicative inverse of a nonzero real number x is denoted x- 1 or 1/x. We define x/y = x(y- 1) if y #: 0. Any mathematical system satisfying axioms 1 to 11 is called a field. Using axioms 1 to 11 one may establish all of the well-known algebraic properties of the real numbers. We give one example; others are given in the exercises. In subsequent sections we will assume that all of the well-known algebraic properties of the real numbers have been verified from the axioms. The interested reader may consult Landau (1960) to see how this might be accomplished. Theorem3.4 x · 0

= 0 for all x in R.

Proof. x·(0 + 0) = x·0 by axiom 4. On the other hand x·(0 0 + x·0 by axiom 11. Thus x·0 = x·0 + x·0. Now x•0

+ [-(x•0)] = (x•0 + x•0) + [-(x•0)]

+ 0)

= x·

12

The Real Number System

Again using axiom 5 and axiom 2, we have

= x•O + 0

0

and so by axiom 4 we have O = x·O. ■

ExerclsN

3.3 3.4 3.5 3.6 3.7 3,8

Prove that the additive inverse of axiom 5 is unique. Prove that (a) (x + y) + (z + w) = (x + (y + z)) + w for all x, y, z, w e R, (b) The sum x + y + z + w is independent of the manner in which the parentheses are inserted. = x for all x e R. Prove that -(-x) Prove that -(x + y) = -x - y for all x, ye R. Let x, y e R. Prove that xy = 0 if and only if x = 0 or y = 0. Let x, ye R. Prove that if xy = xz and x #- 0, then y = z. Prove that -(xy) = x(-y) = (-x)y for all x, ye R. Prove that ( - l)x = -x for all x e R.

4.

The Order Axiom of the Real Numbers

3.1 3.2

In this section we give the order axiom of the real numbers and derive several useful results.

Axiom 12. There is a subset P ofR called the positive realnumberssatisfying (i) If x and y are in P, then x + y and xy are in P. (ii) If x is in R, exactly one of the following statements is true: xeP

or

x=O

or

-xeP.

Using axiom 12 we can define the usual notation for order.

Definition4.1 Let x and y be real numbers. (i) x is negativeif - x is positive. (ii) x > y means x - y is positive. (iii) x ~ y means x > y or x (iv) x < y means y > x. (v) x s y meansy ~ x.

= y.

13

4. The Order Axiom of the Real Numbers

The inequality x > y (x < y) is read "xis greater (less) than y" and the inequality x ~ y (x s y) is read "x is greater (less) than or equal to y." Several order properties of the real numbers are given in the next theorem.

Theorem4.2 (i) (ii) (iii) (iv) (v)

1> If x If x If x If x

0.

> y and y > z, then x > z, x,y,z e R. > y, then x + z > y + z, x,y,z e R. > y and z > 0, then xz > yz, x,y,z e R. > y and z < 0, then xz < yz, x,y,z e R.

Proof We prove parts (i) and (ii) leaving the others as exercises. By axiom 12, exactly one of the following statements is true

I eP

or

1=0

or

-1 eP

By axiom 9, 1 ,;, 0. Suppose - I e P. Then by axiom 12, (- IX-1) = 1 e P (see Exercise 3.8) and hence I e P and -1 e P which contradicts axiom 12. Therefore I e P and (i) holds. If x > y and y > z, then x - y, y - z e P by Definition 4.1. By axiom 12, x - z = (x - y) + (y - z) e P and thus x > z. ■ We will employ the following notation throughout the remainder of the text.

Deftnitioa4.3 Let a,b e R with a < b. We define

= {x e RI a < x < b} [a, b] = {x e RI a S x s b} [a, b) = {x e RI a s x < b} (a, b)

= {xeRla < x Sb} ( - oo, a) = {x e RI x < a} (-oo,a] = {xeRlx Sa} (a,b]

[a, oo) = {x e RI a s x} {a, oo) = {xeRla

< x}

(-oo, oo) = R

We call (c, d) an open interval; [c, d] a closed interval; and either [c, d) or (c, d] a half-open interval, where cord are possibly ± oo. We conclude this section by defining the absolute value of a real number and deriving several results concerning absolute value.

14

The Reel Number System

Deftnition4.4 Let x be a real number. We define

lxl = {

x

-x

if

X ~

if

X

0

< 0

We ca.ll lxl the absolute value of x.

Theorem4.5 (i) Let e > 0. Then lxl < e if and only if -e < x < e and lxl s e if and only if - 8 $ X $ 8. (ii) x s lxl for all x e R. (iii) lxyl = lxl IYIfor all x,y e R. (iv) Ix + YI S lxl + IYIfor all x,y e R. Proof. We prove only (ii) and (iv) leaving the others as exercises. If x ~ 0, then x = lxl S lxl. If x < 0, then x = -lxl < lxl. In either casex s lxl. If x + y ~ 0, then Ix + YI = x + y =Slxl + IYI by (ii). If x + y < 0, then Ix+ YI = -(x + y) = -x - y s lxl + IYIby (ii). ■

Exercises

4.1 4.2 4.3 4.4 4.5

4.6 4.7 4.8 4.9 4.10

Prove parts (iii), (iv), and (v) of Th~rem 4.2. Prove parts (i) and (iii) of Theorem 4.S. Prove that if x ~ y and y ~ x, then x = y, x, ye R. Prove that if xy > 0, then either x > 0 and y > 0 or x < 0 and y < 0, x,yeR. Prove that if x > 0, then 1/x > 0, x e R. Prove that x 2 > 0 for all x e R with x #- 0. Prove that x 2 + y 2 ~ 2xy for all x, y e R. Prove that if 0 < xy and x < y, then 1/y < 1/x, x, ye R. Prove that if x ~ y + e for every e > 0, then x ~ y, x, y e R. Prove that if x + y > z and w > x, then w + y > z, x, y, z, w e R.

5. The Least-Upper-Bound Axiom

Before giving the final axiom for the real numbers we must make some definitions.

Definition5.1 A nonempty subset X of R is said to be bounded above (below)

&. The least-Upper-Bound Axiom

if there exists a real number a such that x ~ a (x ber a is called an upper (lower) bound for X.

~

16

a) for all x e X. The num-

Definition5.2 Let X be a nonempty subset of R. A number a in R is said to be a least upper bound for X if (i) a is an upper bound for X. (ii) If b is an upper bound for X, then a ~ b. A number a in R is said to be a greatest lower bound for X if (i) a is a lower bound for X. (ii) If b is a lower bound for X, then b ~ a. Part (ii) of Definition 5.2 concerning least upper bounds may be equivalently stated as follows: if b < a, then b is not an upper bound for X. By Definition 5.1, this last statement is equivalent to (ii') If b < a, there exists x e X such that b < x. In proving that a is a least upper bound for a set X, it is sometimes easier to use (ii') rather than (ii). Part (ii) of Definition 5.2 concerning greatest lower bounds may be rephrased in a similar way. The next theorem states that if a set has a least upper bound, it is unique.

Theorem 5.3 Let X be a subset of R. If a and b are least upper bounds for X, then a= b. Proof. By Definition 5.2, since a is a least upper bound for X and b is an upper bound for X, we have a ~ b. Similarly, b ~ a, and thus a= b. ■ Theorem 5.3 (and ·the corresponding result for greatest lower bounds) allows us to speak of the least upper bound (or the greatest lower bound) and justifies the following notation. If a is the least upper bound of a set X, we let a = lub X. Similarly, if bis the greatest lower bound of a set X, we let b = glb X. If X is finite set, we also denote lub X by max X, and we denote glb X by min X. For example, S u,.p lub (0, I) = 1 = lub [O, 1) and

li'1J glb (0, 1) = 0 = glb [O,1)

The least upper bo~nd- of a set X is also called the .rwJ,emumof X, and the greatest lower bound of a set X is also called the .infimumof X. The notation lub X = sup X and glb X = inf Xis also used. Axiom 13. A nonempty subset of real numbers which is bounded above has a least upper bound.

16

The Real Number System figure 6.1

Axiom 13 is called the completeness axiom for the real numbers. The real numbers are complete in the sense that there are no "holes" in the real line. Informally, if there were a hole in the real line (see Figure 5.1), the set of numbers to the left of the hole would have no least upper bound. The least-upper-bound axiom is the basis of many deep theorems in analysis concerning the real numbers. Many theorems about the real numbers (for example, Theorems 16.2 and 18.1) which involve the existence of a number with special properties ultimately rest on this axiom. Using axiom 13, we are able to prove the existence of greatest lower bounds. A nonempty subset of real numbers which is bounded below has a greatest lower bound. lbeorem 5.4

Proof. Let X be a nonempty subset of real numbers which is bounded below, and let Y be the set of lower bounds for X. Let c e X. Then y 5 c for y e Y. Thus Y is bounded above, and by the least-upper-bound axiom Y has a least upper bound a. We will show that a is the greatest lower bound of X. Let x e X. Then y 5 x for all y e Y, and thus x is an upper bound for Y. Since a is the least upper bound of Y, we have a 5 x. Therefore, a is a lower bound for X. Let b any lower bound for X. Then b e Y, and hence b 5 a. By Definition 5.2, a is the greatest lower bound of X. ■

Exercises

5.1 Let X be a set of real numbers with least upper bound a. Prove that if e > 0, there exists x e X such that a - e < x ~ a. 5.2 Prove that the greatest lower bound of a set of real numbers is unique. 5.3 Give another proof of Theorem 5.4 by considering the least upper bound of the set -X= {-xlxeX} 5.4 Give that if Xis a nonempty subset of real numbers which is bounded above then lub X = -glb (-X). 5.6 Let X and Y be nonempty subsets of real numbers such that X c Y and Y is bounded above. Prove that lub X ~ lub Y

8.

The Set of Positive Integers

17

5.7 Let X be a set of real numbers with least upper bound a. Lett~ 0. Prove that ta is the least upper bound of the set tX= {txlxe X} State and prove an analogous result if t < 0. 5.8 Let X and Y be sets of real numbers with least upper bounds a and b, respectively. Prove that a + b is the least upper bound of the set X + Y = {x + y I x e X, ye Y} 5.9• If/ is a real-valued function on (a, b) and c e (a, b), we say that/ is strictly increasing at c if there exists o > 0 such that if c - o < x < c, then /(x) < /(c) and if c < x < c + o,then/(c) < /(x). We say that/is strictly increasing on (a, b) if whenever x, ye (a, b) with x < y, we have/(x) < /(y). Prove that if f is strictly increasing at each point of (a, b), then/ is strictly increasing on (a, b).

8.

The Set of Positive Integers

We now isolate that subset of the real numbers known as the positive integers. We will define the positive integers as the smallest subset P of R having the properties that I e P and if n e P, then n + le P.

Definition6.1 A subset X of R is said to be a successor set (i) If I e X, (ii) If n e X, then n + l e X. Since R itself is a successor set, successor sets exist. Lemma 6.2 If .sil is any nonempty collection of successor sets, then n.si/ is a successor set.

Proof. Since I e A for every A e .sil, 1 e n.si/. Suppose n e n.si/. Then n e A for every A e .si/. Since every set A in .si/ is a successor set, n + 1 e A for every A e .si/, and thus n + l e n.si/. ■ Definition 6.3 The set P of positive integers is the intersection of the family of all successor sets. By Lemma 6.2, P is a successor set. P is the smallest successor set in the following sense: if X is a successor set, P c X. This is important but fairly -1be asterisk indicates that the entry is discussed in the Hints to Selected Exercises at the back of the book.

18

The Real Number System

easy to see. Let x e P. Then x belongs to every successor set, and in particular, x belongs to X. The theorem of mathematical induction follows almost immediately from Definition 6.3.

Theorem6.4 (MathematicalInduction) Suppose that integer n, we have a statement S(n). Suppose also that (i) S(1) is true. (ii) If S(n) is true, then S(n

+

for

each

positive

1) is true.

Then S(n) is true for every positive integer n. Let G

Proof.

=

{n e PI S(n) is true}. Then G c P. On the other hand, + 1 e G by (ii). Thus G is a successor set, and so

1 e G, and if n e G, then n P c G. Therefore G = P.

■

As an example of a proof by induction we will establish the following theorem.

Theorem6.5 If n is a positive integer, then n ~

1.

Proof. Let S(n) be the statement "n ~ l." Since 1 ~ 1, S(l) is true. If n ~ 1, then n + l > n ~ l; so if S(n) is true, then S(n + 1) is true. Therefore S(n) is true for every positive integer n by Theorem 6.4. ■

Theorem6.6 If m,n e P, then m

+ n e P.

Proof. Let S(m) be the statement "m + n e P for all n e P." The term 1 + n e P for all n e P since P is a successor set, and so S(l) is true. Next assume m + n e P for all n e P. Then (m + 1) + n = (m + n) + I e P for all n e P since P is a successor set (m + n e P implies (m + n) + I e P). Thus if S(m) is true, then S(m + 1) is true; so by Theorem 6.4, S(m) is true for all meP. ■ The next theorem we will prove is the well-ordering theorem for the positive integers (Theorem 6.10). The proof of this theorem depends on the fact that if n is a positive integer, there is no positive integer between n and n + 1 (Lemma 6.9), and in order to prove this result we must establish two lemmas. Lemma 6.7

Proof.

If n e P, then either n - 1

= 0 or n

- 1 e P.

Let

G

= {n e PI n -

1

= 0 or

n - 1 e P}.

8. The Set of Positive Integers

19

We show that G is a successor set. Clearly I e G. Suppose n e G. Then (n + I) - I = n e P; son + l e G. Therefore G = P. (Why?) ■

Lemma6.8 If m,n e P and m < n, then n - me P. Proof We proceed by induction on m letting S(m) be the statement of the lemma. Suppose I < n, n e P. By Lemma 6.7, either n - l = 0 or n - le P. Since n ::;:1, n - l e P. Assume that if m,n e P and m < n, then n - m e P. Suppose m + 1 < n, m,n e P. By Lemma 6.7 and Theorem 6.5, n - le P. Since m < n - l, by induction we see that (n - I) - me P. Thus n - (m + l) e P, and we have completed the inductive step. ■ Lemma6.9 Let n be a positive integer. No positive integer m satisfies the inequality n < m < n + l. Proof Suppose there exists m e P such that n < m < n + l. By Lemma 6.8, m - n e P. On the other hand, since m < n + l, we have m - n < l, which contradicts Theorem 6.5. ■ We are now ready to prove the well-ordering theorem.

Theorem6.10 (Well-OrderingTheorem) If X is a nonvoid subset of the positive integers, then X contains a least element; that is, there exists a e X such that a :s; x for all x e X. Proof We use induction on n, letting S(n) be the statement: "If n e X, then X contains a least element." If 1 e X, then I is the least element of X by Theorem 6.5. Now assume S(n) is true and suppose n + le X. Since S(n) is true, Xu {n} contains a least element m. If m e X, then m is the least element of X. If m ¢ X, then m = n and n :s; x for all x e X. Since n ¢ X, we have n + l :s; x for all x e X by Lemma 6.9. In this case n + l is the least element of X. ■

Theorem6.11 The set of positive integers is not bounded above. Proof If P is bounded above, then by the least-upper-bound axiom, P has a least upper bound a. Since a - l is not an upper bound for P, there exists n e P such that a - l < n. But then a < n + l, which contradicts the assumption that a is an upper bound for P. ■

Corollary6.12 The set of real numbers is Archimedean ordered; that is,

20

The Reel Number System

if a and b are positive real numbers, there exists a positive integer n such that a< nb.

Proof. Since Pis not bounded above, there exists n e P such that a/b < n. Thus a< nb. ■

Corollary6.13 If t is a positive real number, there exists a positive integer N such that 1/N < a. Proof.

Take a= I, b

= tin

Corollary 6.12.

■

Exercises

1+ + +•••+

+

6.1 Prove the formula 2 3 n = n(n 1)/2. 6.2 Prove that if m and n are positive integers, then m·n is a positive integer. 6.3 Prove the binomial theorem: If a and b are real numbers and n is a positive

integer, then

+ (:)bb)·= (~)a-+ (;)a--lb + (;)a--2b2 +·..+ (n~ 1)ab•-l where (Z) = k!(/~ k)I'

(a+

6.4

6.5

6.6•

6.7 6.8

7.

Prove that if X is a nonempty subset of positive integers which is bounded above, then X contains a greatest element. Where is the flaw in the following "proof" by induction that any two positive integers are equal ? Theorem Let k and m be positive integers. If n is the maximum of k and m, then n = k = m. Proof. Let S(n) be the statement of the theorem. If I = max {k, m }, then k = m = I by Theorem 6.5. Suppose n + l = max {k, m}. Then n = max {k - I, m - I } so by induction n = k - l = m - 1, and thus n + I= k = m. ■ Prove the second form of mathematical induction: Suppose that for each positive integer n, we have a statement S(n). Suppose also that (a) S(l) is true. (b) If S(k) is true for all k < n, then S(n) is true. Then S(n) is true for all positive integers n. Deduce Theorem 6.4 from Theorem 6.10. Deduce Lemma 6.9 from Theorem 6.10.

Integers, Rationals, and Exponents

In this section we will define the integers, rational numbers, and rational exponents and collect some miscellaneous results.

7.

21

Integers, Rational•, and Exponent•

Definition 7.1 The set of integers,denoted Z, is the set

{0} u Pu -P where -P = {-n In e P}. It can be shown that Z is a commutative group (that is, Z satisfies axioms I to 5). We may define the rational numbers in terms of the integers.

Definition7.2 The set of rationalnumbers,denoted Q, is the set

{~lp,q e

z

and q #:

o}

It can be shown that Q is a field (that is, Q satisfies axioms I to 11). However we will shortly prove that Q ¢ R. It is convenient at this point to define x1, where x e Rand n e Z.

Definition7.3 Let x e R. We define X

1

=X

and

:,c1+1

= X·Jt1,

neP

If x #: 0, we define

xo

=1

an

d

x -11

I = x1'

nep

From this definition one can deduce the laws of integer exponents (see Exercise 7.5). We next show that there is no rational number x such that x 2 = 2 (Theorem 7.4), but there is a real number x such that x 2 = 2 (Theorem 7.5). These results show that Q is a proper subset of R. The fact that ./2, is not rational is one reason for extending Q to the larger set R. A real number which is not rational is said to be irrational. Tbeorem7.4 There is no rational number r satisfying r2

= 2.

Proof. Suppose there exists re Q satisfying r 2 = 2. Then r = p/q where p,q e Zand q #: 0. We may assume that not both p and q are even. Then p 2 = 2q 2 • This implies that p 2 , and hence also p is even. Thus p = 2n for some n e Z. Now 4n2 = 2q 2 ; so 2n2 = q 2 , and hence q is also even. This contradiction establishes the theorem. ■

We now prove the existence of nth roots in the set of real numbers. The idea of the proof is that the nth root of a should be the largest real number b such that b" s a. The least-upper-bound axiom will guarantee the existence of such a number. Theorem 7.5 If a is a nonnegative real number and n is a positive integer, there exists a real number b ~ 0 such that b" = a.

..

22

The Reel Number System

Proof. Let X

= {x e R I x ~

0 and x1 5 a}

X is nonempty since Oe X.

We argue by contradiction to show that Xis bounded above by a+ 1. Suppose there exists x e X such that x > a + 1. Then

a

~

x1 > (a+ 1)11 =

t (;)tt

1 0

2:: na

which is impossible; so Xis bounded above. By the least-upper-bound axiom, Xhas a least upper bound b. We will prove that b" = a. Either b" < a, b" > a, orb" = a. We show that the first two possibilities cannot occur. Suppose b" < a, and let ~ = a - b". Choose positive integers m0 , ••• , m,,_1 such that 6 (kn)b mi-Ii< n' 1

1:

Let m = max {m0 ,

•••

,

k

= 0, 1, ... , n-1

m,,_1 }. Then

-:t:(;)~ ;_,

+ b"

11-1

6

0. Choose a positive integer n such that b - a > l/n. Let m be the least positive integer such that b ~ m/n. If m = 1, clearly (m - 1)/n < b. If m > 1, m - 1 is a positive integer less than m, and thus (m - 1)/n < b. Since a - b < - l/n and b ~ m/n, we have a

= b + (a

- b)

1

m

m-1

< -n - n- ==-- n

Thus the rational number r = (m - 1)/n satisfies a < r < b. We have established the theorem in case b > 0. Suppose now that a and b are arbitrary real numbers with a < b. Choose a positive integer n such that b + n > 0. By the special case just established, there exists a rational number r' such that a + n < r' < b + n. Thus r = r' - n is a rational number satisfying a < r < b. ■ m-1

0

*

m

7iin

•

b

figure 7.1

24

The Real Number System

Lemma 7.9 The sum of a rational number and an irrational number is an irrational number.

Proof Suppose the lemma is false. Then there exists a rational number r and an irrational number s such that t = r + s is a rational number. Then s = t - r is the difference of two rational numbers and is therefore rational. This is a contradiction. ■ 1beorem 7.10 If a and b are real numbers with a < b, then there is an irrational number s such that a < s < b.

Proof By Theorem 7.8, there exists a rational number r 1 such that a ✓ "1. < r 1 < b - ✓ "1. By Theorem 7.4 and Lemma 7.9, r = r 1 + ✓ "1. is an irrational number. Since a < r < b we have completed the proof. ■ In Chapter III we will give a very different proof of Theorem 7.10.

Exercises

7.1 7.2 7.3 7.4 7.5

Prove Prove Prove Prove Prove (a)

that Z is a commutative group. that Q is a field. that there is no rational number r satisfying r 2 = 3. that there is no rational number r satisfying r 3 = 2. the laws of integer exponents:

,xi-+•

=

X-x", x e R, x '# 0, m, n e Z

1

(b) x" = x-•• x e R, x '# 0, n e Z (c) (d)

(xy)• = x"Y-, x, ye R, x '# 0 '# Y, n e Z (X-)" = X--, x e R, x '# 0, m, n e Z

(e)

(~r= ;.

x,yeR,

x '# 0 '# y, neZ

(f) If O < x < y, then x" < ,a, n e P (g) If n < m and x > l, then x" < X-, n, meZ. 7.6 (a) Do Exercise 1.S with x and y positive and m and n rational numbers. (b) Discuss Exercise 7.S, where x and y are arbitrary real numbers and m and n are rational numbers. 7.7 Complete the proof of Theorem 1.S by showing that b" > a is impossible. 7.8 Prove that if a > 0 and n is a positive integer there exists a unique b > 0 such that b" = a. 7.9• Prove that the set of rea:I numbers as defined by axioms I to 13 is unique in the following sense: If R' is a set which satisfies axioms I to 13, there exists a on~to-one function/ from R onto R' which satisfies

7.

Integers, Rationals, end Exponents

J(x

+ y) = /(x) + /(y)

26

for all x, ye R /(xy) = /(x)/(y) for all x, ye R /(x) < /()') if and only if x < y for x, ye R 7.10 Prove that no equilateral triangle in the plane can have all vertices with rational coordinates.

Ill Set Equivalence

When do two (possibly infinite) sets have the same number of elements? In this chapter we will give a definition (Definition 8.1) which answers this question. We will show that according to our definition there are as many positive integers as there are rational numbers, but that there are more real numbers than positive integers.

8.

Definitions and Examples

Suppose we are given two finite sets X and Y and are asked to compare their sizes. One way to do this is to count the number of elements in each set and compare the results. An alternative method is to pair off the elements in X with those in Y.

X

= {a, b, c, d}

Y

t t t t = {1, 2, 3, 4}

This second method generalizes to infinite sets, and thus we make the following definition.

Definition8.1 Let X and Y be nonempty sets. We say that Xis equivalent to Y and write X ~ Y if there exists a one-to-one function from X onto Y. Obviously X ~ X. (Why?) If X ~ Y, then Y ~ X. (Why?) The relation ~ is also transitive. If X ~ Y and Y ~ Z, then X ~ Z. (Why?) Sets are divided into two categories as specified in the following definition. 28

8.

Definitions and Exampl•

27

Deflnition8.2 A set Xis finite if either X = 0 or X ~ {l, 2, ... , n} for some positive integer n. A set which is not finite is said to be infinite. A set Xis countable if either Xis finite or X ~ P. A set Xis uncountableif Xis not countable. If a set X is countable and infinite there is a one-to-one function J from P onto X. Thus the elements of X may be listed

/(1), /(2), /(3), or as we sometimes write

J., /2, /3, • • • The set of even positive integers is countable, since the function f defined by /(n) = 2n is a one-to-one function from P onto {2, 4, 6, ... }. The correspondence or listing is 2 4 6 8

t t t t 1 2 3 4 The set of all integers is countable since the function f defined by

_(n; = {2

1)

J(n)

if n is odd

n

if n is even

is a one-to-one function from P onto Z. The correspondence or listing is

0 1 -1

2

-2

t t

t t

t

1 2

3 4

5

To prove that a set Xis countable from the definition, one must exhibit a listing like those given above (a one-to-one function from P onto X). In view of the remarks preceding Definition 8.2, one may also show that a set X is countable by showing that X is equivalent to a set which is known to be countable. To prove that a set Xis uncountable, one would have to show that there is no one-to-one function from P onto X. The usual method is to assume the existence of such a function and deduce a contradiction. A remarkable fact is that uncountable sets exist. We will see later that the real numbers are uncountable. At this point we will give a different example of an uncountable set.

Example8.3 A real sequence is a function from P into R. In this example

28

Set Equivalence

we will denote a real sequence as

(a 1, a 2 ,

a,. e R, n

••• )

= l, 2, . . .

Let

X

= {(ai,a

2 , ••

•)la,.e{O,

l},n

=

1,2, ... }

Thus X is the set of all real sequences which assume only the values O or 1. We show that Xis uncountable. If Xis countable, the elements of X may be listed:

1 1 0> 1 - (a< l• > a< 2• > a3••••

)

2 - (a 1, a 2, a 3•···

)

We will now define a sequence which is not in the list, thus deducing a contradiction. Let

b II

Then (b 1 , b2 , uncountable.

••• )

= {O 1

11 if a< >= 1 II

1·f

a11 -- o, n -- 1, 2, ...

is a sequence in X which is not in the list, and hence Xis

If Xis a set, we let P(X) denote the collection of all subsets of X. [P(X) is called the power set of X.]

Theorem8.4 Let X be a set. Then X ~

P(X).

Proof. If X = 0, then P(X) = {0}; so X ~ P(X). Therefore, suppose X #: 0 and X ~ P(X). Then there exists a one-to-one function f from X onto P(X). We now define a subset Yof X which is not in the range of/thus deducing a contradiction. Let Y= {xeXlx¢f(x)}

If Y ef(X), there exists ye X such that f(y) and we examine the two possibilities. 1. If ye Y, then y ¢/(y). Thus y ¢/(y) 2. If y ¢ Y, then y ef(y). Thus y ef(y)

= Y. The

=

=

Y. Either ye Y or y ¢ Y,

first case is impossible.

Y. The second case is impossible.

Thus Y ¢/(X), and we have the desired contradiction.

Corollary8.5 P(P) is uncountable.

■

9.

29

Countable and Uncountable Sets

Exercises

8.1 Prove that the set of even integers is countable. 8.2 Suppose that X and Y are equivalent sets. Prove that P(X) :::::P( Y). 8.3 Prove that every infinite set has a countably infinite subset. 8.4• Prove that a set Xis infinite if and only if Xis equivalent to a proper subset of itself. 8.5 Let X be a set. Let S be the set of all functions from X into {O,I}. (Such functions are ·called characteristic /unctions.) Prove that S :::::P(X). How does this exercise connect Example 8.3 and Corollary 8.S?

9. Countable and Uncountable Sets The following result should not be too surprising.

1beorem9.1 Any subset of the positive integers is countable. Proof. Let X be a subset of P. If Xis finite, there is nothing to prove; so assume Xis infinite. We define a function f from P to X as follows: Let /(1) be the least element in X. Let/(2) be the least element of X\{/(1)}. Continuing in this way, having defined /(1), /(2), ... , /(n), we let f(n + I) be the least element of X\{/(1),/(2), ... ,/(n)}. It is easy to verify that f is a one-to-one function from P onto X, and therefore Xis countable. ■

Corollary9.2 Any subset of a countable set is countable. The next theorem shows that the one to one requirement need not be checked in proving that a set is countable. 1beorem 9.3 Let X be a nonempty set, then X is countable if and only if there exists a function f from P onto X. Proof. If X is countable and infinite, there is nothing to prove. If X is finite, then X ~ {l, 2, ... , n} for some n e P; so X = {x1, x 2 , ••• , x,.}. Define /(i) = x 1 for i = I, 2, ... , n and /(j) = x,, for all j > n. Then f is a function from P onto X. Suppose f is a function from P onto a set X. If X is finite, there is nothing to prove; so assume Xis infinite. We define a function g from X onto a subset of P by the rule

for each xe X

30

Set Equivalence

Then g is a one-to-one (check this) function from X onto g(X). g(X) is an infinite subset of P; hence by Theorem 9.1, g(X) is countable. Thus X ~ g(X) ~ P and X is countable. ■ Suppose A 1 , A 2 , of A,. as

•••

are countably infinite sets. We may list the elements 11 11 11 a< 1•> a< 2, > a< 3•>

We now find that we may list the elements of

a\1 > .... a~•>

•..

U:=1 A,.:

a~1>.... ...

,/ ✓ a\2> a~2> a~2>

! /'

,/

a\3> a~3> a~3> ✓

The arrows indicate that 1 --+a\0

2--+ a~l) 3--+ a\2 > We have described a function from Ponto U:: 1 A,.; so by Theorem 9.3, U:..1 A,. is countable. This result is often summarized by saying that a countable union of countable sets is countable. We will now give another (tricky) proof of this theorem.

Lemma9.4 The set P x P is countable. Proof.

The function/: P x P --+P defined by

f(n, m)

= 2"3"'

is one to one, since factorization into primes is unique. Thus P x P is equivalent to a subset of P. By Theorem 9.1, P x P is•countable. ■

Theorem9.5 Let A 1 , A 2 ,

is a countable set.

•••

be a countable family of countable sets. Then

9.

Countable and Uncountable Seta

Proof

31

Since A,. is countable, we may list the elements of A.,, A,,:a~•>, a~•>, a~•>, ...

11 ~ 1s . firu"te, .n,, ~ = {a }, we Iet a< > -'•>"f · m. If .n,. ~ = (I n case .n,. 1 •... , a,. 1 = a_. 1 , ~ 0, redefine A.,,= A,. for some A,. :/:-0- If all A.111 = 0 then U!',.1 A.111 = 0 is countable.) The function/ defined by

f(n, m)

= a~•>

P x P, and it follows that maps P x P onto U.:°= 1 A.,.. By Lemma 9.4, P ~ we may map P onto U.:C ..1 A.11• By Theorem 9.3 U:..1 .A,,is countable. ■

Corollary9.6 If .A and B are countable sets, then A u B is a countable set. Corollary9.7 The set Q of rational numbers is countable.

Proof

For each positive integer n, let

A,. =

{· .. ' -n2' .=_!' n n~' !n ' n~'.. ·}

Then A.,,is countable for each positive integer n. (Why?) By Theorem 9.5, Q = U.:C ..1 A,, is countable. ■ Theorem9.8 Any interval of R is uncountable; in particular, R is un-

countable.

Proof Let / be an interval of R, and suppose / is countable. Then the elements of/ may be listed Choose a 1 ,b1 e R such that r 1 ¢ (a 1, b 1) c /. Choose a2 ,b2 e R such that a 1 < a 2 < b 2 < b 1 and r 2 ¢ (a 2 , b 2 ). Continue in this way so that having chosen a,.,b,.e R, we choose a,,+i,h,.+1 e R such that a,.< a,,+1 < b,,+1 < b,. and r,.+ 1 ¢ (a,.+i, b,.+1). The set {ai, a2 , ••• } is bounded above by b; so the least upper bound r of this set exists. It follows that a,. < r for all n e P. On the other hand, r < b,, for all n e P for if b,,, ;s; r for some me P, we would have a,. < b,,,+1 < b,,, ~

r

for all n e P. Thus b,,,+1 would be an upper bound for {ai, a2 , •• •}, but since b,,,+1 < ,, this is impossible. Therefore re (a,,,b,.) for all n e P. It follows that r :/:-r,. for every n e P and re /. This contradiction establishes the theorem. ■

32

Set Equivalence

Corollary9.9 Irrational numbers exist. Proof. If every real number were rational, R would be countable by Corollary 9. 7. This statement contradicts Theorem 9.8. ■

The proof of Corollary 9.9 caused considerable consternation when first give~ by Cantor, for it proves the existence of irrational numbers without exhibiting a single irrational number!

Corollary9.10 The set of irrational numbers is uncountable. Proof. Suppose Q' is countable. By Corollary 9.7,- Q is countable; so by Corollary 9.6, R = Q u Q' is countable. This contradicts Theorem 9.8. ■

We are now in a position to give another proof of Theorem 7.10. Theorem9.11 (Revisited) If a and b are real numbers with a < b, then there is an irrational number s such that a < s < b. Proof. If (a, b) c Q, then by Corollaries 9.7 and 9.2, (a, b) is countable, and this contradicts Theorem 9.8. ■ Exercises

9.1 Prove Corollary 9.2 (give all the details). 9.2 When is the function described before Lemma 9.4 one to one? 9.3 Let A and B be sets. Prove that if A is countable and B is uncountable, then A u B is uncountable. 9.4 Let A and B be sets such that A c B. Prove that if A is uncountable, then B is uncountable. 9.5 Deduce the fact that P x P is countable from Theorem 9.S. 9.6 Prove that if A and B are countable sets, then A x B is countable. 9.1• Prove that the set of all polynomials of degree n with rational coefficients is countable. 9.8• Prove that the set of all polynomials with rational coefficients is countable. 9.9 A real number r is said to be algebraic if r is a root of a polynomial with rational coefficients. If r is not algebraic, r is said to be transcendental.(It is known that e and n are transcendental.) (a) Prove that every rational number and every root of a positive rational number are algebraic numbers. (b) Prove that r is an algebraic number if and only if r is a root of a polynomial with integer coefficients.

9. Countable end Uncountable Sets

33

(c)• Prove that the set of algebraic numbers is countable. Deduce that transcendental numbers exist. 9.to• Prove that the plane is rwt the union of a countable family of straight lines. 9.11 Prove that the collection of finite subsets of P is countable. Deduce that the collection of infinite subsets of P is uncountable.

IV Sequences of Real Numbers

Analysis is concerned in one form or another with limits. We begin our study of real analysis with the study of limits of real sequences. 10.

Limit of a Sequence

Let X be a set. A sequence of elements of X is a "list" of elements from the set X. In this chapter we will be concerned with sequences of real numbers. We first make the notion of sequence precise.

Deflnition10.1 Let X be a set. A sequence of elements of X is a function from the set of positive integers into X. In particular, a real sequence (or sequence of reul numbers) is a function from P into R. The usual notation for a real sequence is {a,.}:°=1 , where a denotes the function from P into R and a,. is the value of the function at the positive integer n. The notations ai, a2 , ••• and {a,.} are also used to denote a real sequence. The number a,.is called the nth term of the sequence {a,.}:° .. 1 • A sequence may be defined by giving an explicit formula for the nth term. For example, the formula I a =" n defines the sequence whose value at the positive integer n is 1/n. The first three terms of this sequence are 01

34

= 1,

02

=

½,

03

=

¼

1O. Limit of • Sequence

36

A sequence may also be defined inductively. Thus equations 0,.+2

=

a,. + a,.+ t 2

•

n

= 1, 2, ...

define the sequence whose first six terms are

0,

½, i, i, t¼

I,

Consider the sequence {a,.}:.. 1 whose nth term is defined by the formula

n a,.=n+1 The first four terms of this sequence are The terms corresponding

½, i, i, t to n = I00, IO1, 102 are

m. m. m m

which are close to 1. For example, differs from I by only Th. It is clear that n/(n + I) is "close to" I "for all large positive integers n." For this reason we say that the sequence {n/(n + l)}:. 1 has limit I. In general, we say that a sequence {a,.}:.,1 has limit L if a,. is "close to" L "for all large positive integers n." To define the limit of a sequence, we need to make the concepts "close to" and "for all large positive integers n" precise. Since la,.- LI is the distance between a,. and L, we would agree that a,. is "close to" L if la,,- LI is small. How small? Given any positive numbers (no matter bow small) we wish to make la,.- LI < s for all large positive integers n. If we can do this, we can make a,.as close to L as we wish simply by making e small enough. "For all large positive integers n" means "for all n greater than (or equal to) some fixed positive integer N." Thus the condition we are seeking is that if e > 0, there exists a positive integer N such that la,.- LI < e, for all n ~ N. The positive integer N may depend on the e which we are given. We would expect that as e is taken smaller, we would have to choose N larger.

Deflnidon10.2 Let {a,.}:=1 be a sequence of real numbers. We say that {a,.}:=1 has limit Le R if for every e > 0, there exists a positive integer N, such that if n :2::N, then

la,.- LI< s Let us return to the sequence {n/(n + l)}:., i, and prove that {n/(n + I)}: .. 1

has limit I. If e > 0, we are required to show that for some positive integer N

_n

ln

+1

-1, 0. Then there exists a positive integer N such that if n ~ N, then la,.- LI < a. The inequality la,.- LI ~ a can hold for at most finitely many positive integers n (namely n = l, 2, ... , N - l). Thus a sequence {a,.},:0 .. 1 bas limit L if and only if for every s > 0, la,.- LI ~ s for finitely many positive integers n. Therefore, a sequence {a,.},:0. 1 does not have limit L if for some a > 0, la,.LI :2::s for infinitely many positive integers n. As an example, we prove that the sequence {n}:,'.1 does not have limit L for any real number L. Lets = I. Since In - LI :2::I for infinitely many positive integers n, {n}:. 1 does not converge to L for any real number L.

Exerci••• In Exercises IO.I to 10.8 prove the limit using Definition 10.2. 10.1 Jim!=

0

... oo n

10.2 Um

... oo ~

10.3 Jim-

1

... oo n

=0 1

+-=2

0

10.4 Jim (2 - !) = 2 ... oo n 10.5 Jim _n_= ... oo n + 2 10.,

Jim~=2 ... oon + 2

1

38

10.7

Sequences of Real Numbers

Jim (-1)" = 0 n

ft-+CIO

10.8

Jim~= n

0

ft-+CIO

Prove that the sequence {(-1)" }:'. 1 has no limit. 10.10 Prove that the sequence {(n + (1/n))}:'. 1 has no limit. 10.11 Let {a,.}:'-1 be a sequence with limit L. Suppose that {b,.}:'a 1 is a sequence such that for some positive integer N we have b,. = a. for every n ~ N. Prove that ,.Jim b,. = L. .. ao 10.9

10.12

Let {a,.}:'.1 be a sequence such that Jim.,.. ao a,. = L. Prove that Jim la.I= ILi ,. .. ao

10.13

What is wrong with the following "proof" of Theorem 10.3? Suppose the sequence {a11}:'. 1 has limits L and L'. Then L = Jim a,. and L' = Jim a,. 11-ao

so L

11.

11-00

= Jim a,. = L' 11-+CIO

■

Subsequences

Consider the sequence {a,.}!'.. 1 defined by a,. = 1/n which begins

1,

½, ¼,. ¼, ¼, ...

If we were to cross out every other term our sequence would become

1,

½, ¼, t, ...

The resulting sequence is called a subsequenceof the original sequence. The first term of this subsequence is the first term of the original sequence; the second term of this subsequence is the third term of the original sequence; the third term of this subsequence is the fifth term of the original sequence; etc. Thus the subsequence is defined by

n➔

} ➔

} ➔

01

2➔

3➔

03

=1 =¼

3➔

5➔

a5

=¼

2n - l ➔

I

a2,.-1

= 2n - 1

11. Subsequences

39

This subsequence is defined by composing the functions n-+ 2n - 1 and n-+ a,,. We now define subsequences of arbitrary sequences.

Definition11.1 Let {a11}:',. 1 be a sequence. Let f be a strictly increasing function from P into P. The sequence {a!}='= 1 is called a subsequenceof the sequence {a11}:. 1 • [The function/ is strictly increasingif f(m) < f(n) whenever m

< n.]

In our example, the function/ of Definition 11.1 is defined by /(n) = 2n - 1. The function f of Definition 11.1 specifies which terms of the original sequence we are to keep to form the subsequence. We now prove the "obvious" fact that if a sequence {a11}:= 1 has limit L, then any subsequence of {a,,}:'. 1 also has limit L.

Theorem11.2 Let {a11}:'=1 be a sequence with limit L. Then any subsequence of {a11} :. 1 has limit L.

Proof. Let {a,}:'=ibe a subsequence of {a11}:'=i· Let e > 0. There exists a positive integer N such that if n c:::N, then

la,.- LI< e If n c:::N, then/(n) c:::f(N) c:::N (verify), and therefore

la, - LI
0. There exists a positive integer N such that if n ~ N, then

la,.- LI
0. There exists a positive integer N 1 such that if n ~

Ni,

then

la,.I< 1 There exists a positive integer N 2 such that if n ~

N 2 , then

lb,.I< s

• 1beorem 12.6 Let {a11} and {b,.} be sequences such that lim11.. 00 b11 = M. Then lim 1 ➔ 0 a,.b11 = LM. Proof.

lim,.

➔

We write Di.b,.= (a11 - LXb11 - M)

+ a M + Lb,. - LM 11

By Theorem 12.1 and Corollary 12.4,

lim (a11 - L)

=0 =

lim (b11 - M)

0 a,. = Land

12. The Algebra of Llmlu

43

and using Lemma 12.5, we have lim (a. - L)(b. - M) ....

=0

00

By Theorem 12.3, lim a.M

= LM =

lim I.lJ.

R-+oo

R-+00

By Theorem 12.2, lim a_b. = lim [(a. - L)(b. - M)

= lim (a. ...

+ a,.M + I.lJ. -

LM]

+ lim a.M + lim I.lJ. + lim ( - LM)

L)(b. - M)

00

....

R-+OO

00

....

00

=0+LM+LM-LM=LM ■

Corollary12.7 Let {a.} be a sequence such that lim.... '° a. be a positive integer. Then lim.... 00 ~ = L".

= L,

and let k

Proof. The proof follows from Theorem 12.6 by induction on k.

■

As a preliminary to the quotient theorem for limits, we prove the following lemma.

=L

#: 0. Then

= ILl/2;there exists a positive integer N 1 such that

if n ;;::;Ni,

Lemma 12.8 Let {a.} be a sequence such that lim....00 a. a. #: 0 for all but finitely many positive integers n, and

lim R-+00

Proof. If t then

la. -

.!..- .!.

a. L

LI
0. There exists a positive integer N 2 such that if n ;;::;N 2 , then

la. -

LI
0. There exists a positive integer N such that if n ~ N, then

If n ~

N, then 6

la,.b,.I < MM = e

•

Exerclaea 13.1 Prove that if {a.} and {b.} are bounded sequences and c is a real number, then {ca.}, {a. + b.}, and {a,.b.} are bounded sequences. 13.2 Let {a.} be a sequence with limit 0. Prove that lim (-1)" a.= 0 ...

CIO

13.3 Give an example of sequences {a.} and {b.} such that {a.} is bounded and {b.} is convergent, but {a. + b.} and {a,.b.} are divergent. 13.4 Let a.= 1·3·5 .. ·(2n-1)/[2·4·6···(2n)]. Prove that {a.} is a bounded

sequence.

14.

Further Limit Theorems

In this brief section we collect some miscellaneous theorems which are often useful in proving limits. Lemma 14.1 Let {a,.} be a sequence such that O s a,. for every positive integer n and L = lim,....OD a•• Then O s L.

14.

Further Limit Theorem•

47

Proof. We argue by contradiction. Suppose L < 0. There exists a positive integer N such that if n 2: N, then

-L la,.-Ll< 2 -L

Thus

aN-L 0. There exists a positive integer N 1 such that if n 2:::N.,

L-e

0

(n

+ l)a"

for n

=

1, 2, •..

+ 1/Cn + 1) and b = 1 + 1/n in part ( 1 +l)■ + > ( 1 + n l 1 +n l + ¾]

(b)

Take a= 1

! r[ !

1

Ca) and prove that for n

= 1, 2, ...

Cc) Prove that (1

+n

!

1) • [ 1

!

!

2

for n = 1, 2, ... + n 1 + ¾]> (1 + n 1) • + ■ + 1 } is a decreasing sequence with limit e and {(1 + 1/n)

Cd) Prove that that e ~ 3. 16.7 Prove that every convergent sequence has a monotone subsequence. 16.8 Prove the nested interval theorem: If {[a., bJ} is a sequence of closed intervals such that. [a,,, bJ :::,Ca.+1, h ■ + 11 for n = 1, 2, ... then n[a,,, bJ is nonvoid. Find a necessary and sufficient condition that l [a., b,.] contains exactly one point. 16.9 Verify that 1½~ 2• ~ 3i

n:.

16.10

(a)

Let x and y be positive numbers. Let 0o for n

=

=

y, and let

1, 2, ...

Prove that {a.} is a decreasing sequence with limit ,./i. Cb) Generaliz.e (a) to nth roots. 16.11* Prove directly from Definition 10.2 that li.m,,.. 00 n 11" = 1. 16.12 Let a.= fii and b. = a ■ + 1 /a •. Prove that {b.};". 5 is increasing and find the limit. 16.13 Let {a,,} be any sequence of real numbers such that li.m,,.. "°na. = 0. Prove that •◄

16.14

lim (1 ao

+ !n + a.)"=

e

Ca) Prove that the positive sequence {a.} converges if the sequence {a ■ + a.} is bounded above by 1. Cb) Prove that if lim. .. oo a.+1/a. exists and is less than 1, then lim. .. 00 a.

ii =

0. Let a 1 > 1. Let a.+1 = 2 - 1/a. for n = 1, 2, .... Prove that {a.} is a bounded monotone sequence and find the limit. 16.16 Let {a.} be a positive sequence which satisfies a ■ + 2 = a.+1 + a,, for n = 1, 2, •••. Ca)• Assuming that lim. .. 00 Ca.+1/a,.) exists, prove that the value of this limit is (1 + ./3)/2. (b) Prove that li.m,,.. 00 Ca.+,/a,.) exists. 16.15

17.

AHi Exponenta

66

17. Real Exponents In this section we make use of the convergence criterion for monotone sequences (Theorem 16.2) to define aX,where a > 0 and xis a real (possibly irrational) number. This will complete the definition of exponents begun in Section 7. It is fairly clear that 3v'l should be defined as the limit of a sequence such as 3

1

31 •4 '

31 •41 ,

31 •414 t

'

•••

➔ 0 tr", where {r,.} is an increasing We will define aX,a > 0 and x real, as lim,. sequence of rational numbers with limit x. We will then verify that the laws of rational exponents carry over to real exponents. We first prove that a sequence such as {r,.} exists.

1beorem17.1 If x is ta real number, there exists an increasing rational sequence {r,.} with limit x. Proof.

Choose (Theorem 7.8) a rational number r 1 such that X -

1
0. Let x be a real number. Prove that lim. .. ao 0: = L".

18. The Bolzano-Weierstrass Theorem In Section 13 we proved that a convergent sequence is bounded (Theorem 13.2), and we noted that the converse of this theorem is false. The sequence {(- Jt} is bounded and divergent. Although {(- It} is divergent, it has a convergent subsequence, {(- 1)2"}. The Bolzano-Weierstrass theorem states that every bounded sequence has a convergent subsequence. This theorem states a fundamental property of the real numbers, and, in fact, it can be shown to be equivalent to the least-upper-bound axiom. 'Theorem) Every bounded real sequence

Theorem 18.1 ~Weientrass has a convergent subsequence.

Proof. Let {a,.} be a bounded sequence. Then there is a closed interval [c, d] such that a. e [c, d] for every positive integer n. Consider the two closed intervals

c+ [ c,-2-

d]'

One of these intervals must contain a,.for infinitely many positive integers n. We denote this interval by [c1 , d 1]. We repeat this process with the interval [c1 , di]. One of the intervals

[Ci, C1+2 di] '

[Ci+2 di ' d ] i

must contain a. for infinitely many positive integers n. We denote this interval by [c2 , d2 ]. Continuing this process, we obtain a sequence [c1 , di], [c2 , d2 ],

•••

of closed intervals such that

[ci, d 1]

::::>[c2 ,

d2 ] => [c3 , d3 ] => • • • k

= l, 2, ...

(18.1)

and each interval [ci, dJ contains a,. for infinitely many positive integers n. Choose a• positive integer n 1 such that a,.I e [ci, di]. Since [c2 , d2 ] contains a.for infirutely many positive integers n, there exists a positive integer n2 > n 1 such that a,.1 e [c2 , d 2 ]. Continuing this process we obtain elements a.1 ,

19. The Cauchy Condition

a,.2 ,

69

satisfying

•••

k

=

1, 2, ...

(18.2)

and Therefore, {a,.,.}f-1 is a subsequence of {a,.}. We will show that {a,..} converges. The sequence {c1J is monotone and bounded so by Theorem 16.2, {c.,} is convergent. Let L = lim.,_,00 c.,.Similarly, the sequence {d.,} converges to some number M. Using equation (18.1), we have M - L

= lim d., t-oao

Therefore L

lim c., = lim (d., - c.,) = lim d ~ c 2 t-oao t-oao

t-oao

=0

= M. Because of (I 8.2), we have c., s a,.,.s d.,

By the squeeze theorem (Theorem 14.3), {a,.,.}converges to L

= M. ■

We will use the Bolz.ano-Weierstrass theorem to derive a general condition for convergence of a sequence (Section 19) and to define a kind of limit applicable to any sequence (Sections 20 and 21). Exerci ... 18.1 Show (by example) that "bounded" cannot be omitted from the hypotheses of the Balzano-Weierstrass theorem. 18.2• Prove that every sequence has a monotone subsequence. 18.J• Let {a.} be a bounded sequence. Prove that if every convergent subsequence of {a.} has a limit L, then lim,... 00 a,,=L. 18.4 Let {a.} be a sequence. Prove that if every monotone subsequence of- {a,.} has limit L, then lim,... 00 a.=L. 18.5 Let {a.} be a sequence such that for some e > 0,

*

la,.- a.I ~ e for all n m Prove that {a.} has no convergent subsequence. 19. The Cauchy Condition We begin by deriving a property of convergent sequences. Theorem 19.1 Let {a,.} be a convergent sequence. Then for every s > 0, there exists a positive integer N such that if m,n ~ N, then

la..- a,.I< s

eo

Sequences of R•I

Proof. Suppose lim.... ao a. N such that if n ~ N, then

Therefore, if m,n ~

=

Numbers

L. Lett > 0. There exists a positive integer

N, we have e

t

la. - a.I s la.. - LI + la. - LI < 2 + 2 = e

•

The condition stated in the conclusion of Theorem 19.1 is known as the

Cauchy condition. Deftllitioll19.2 If {a.} is a sequence such that for every t > 0, there exists a positive integer N such that if m,n ~ N, we have

la. - a.I< then we call

e

{a.} a Cauchysequence.

Theorem 19.1 states that a convergent sequence is a Cauchy sequence. The converse of Theorem 19.1 also holds; that is, if {a.} is a Cauchy sequence, then {a,.} is convergent. The Cauchy condition does not involve the limit, and thus we have a convergence test for arbitrary sequences which does not assume that the value of the limit is known in advance. Tbeorem 19.3 Let {a.} be a real sequence. Then {a,.} is convergent if and only if {a.} is a Cauchy sequence.

Proof. Theorem 19.1 states that if {a,.} is convergent, then {a,.} is a Cauchy sequence and thus we must prove the converse. Let {a,.} be a Cauchy sequence. We first show that {a,.}is bounded, so that we may apply the Bolzano-Weierstrass theorem. If e = 1, there exists a positive integer N such that if m,n ~ N, then

la,.- a,.!< 1 Thus if n ~

N, we have

la,.IS la,.- aNI+ laNI
0. There exists a positive integer N such that if m,n ~

N, then

e

la,..- a.I< 2 There exists a positive integer N' such that if k ~

N', then

e

la•• - LI< 2 Choose a positive integer K such that K then

~

N' and nx

e

~

N. Now if n

~

N,

e

•

la,.- LI s la,.- a,."I+ la,."- LI < 2 + 2 = e Exerciaes

19.1 Let {a.} and {b.} be Cauchy sequences, and let c be a real number. Prove that {a. + b,.}, {a. - b.}, {ca.}, and {a,.b.} are Cauchy sequences. 19.2 Let {a,.} be a sequence. Prove that {a,.} is a Cauchy sequence if and only if for every e > 0, there exists a positive integer N such that if n ~ N, then la,.- aNI < e. 19.3 Let O ~ « < 1, and let/ be a function from R into R which satisfies 1/(x) - /(y)I ~ «Ix - YI for all x, ye R Let a 1 e R, and let a.+1 = /(aJ for n = 1, 2, ... Prove that {a.} is a Cauchy

sequence. 19.4 Let {a.} be the sequence defined by the relations a 1 = 1 and a,.+ 1 (1/3•) for n = 1, 2, ... Prove that {a.} is a Cauchy sequence.

=

a.

+

20. The lim sup and lim inf of Bounded Sequences The "generalized limits" lim sup,. ➔ 0 a,. and lim inf,. ➔ 0 a,. are defined for arbitrary (not necessarily convergent) sequences {a,.}. In this section we will ➔ 0 a,. and lim inf,. ➔ 0 a,. for bounded sequences {a,.} and in define lim sup. Section 21 we will extend our definition to unbounded sequences. If {a,.} is a bounded sequence, the Bolzano-Weierstrass theorem assures ➔ 0 a,. is the us that {a.} has a convergent subsequence. The number lim sup,. maximum value obtainable as the limit of a convergent subsequence of {a,.} and lim inf,....00 a,.is the minimum value obtainable as the limit of a convergent subsequence of {a,.}.

Deftnition20.1 Let {a,.} be a bounded real sequence and let !l',, denote the

82

Sequences of R.. I Numbers

set of all L such that L=lima.., 1:-+ao

where {a..,} is a convergent subsequence of {a.}. We define lim sup a. = lub !i'. lim inf a,.= glb !i',,

and

11-+ao

The notations liiii ....ao a. and lim,....00 a,.are also used for lim sup .... ao a,.and lim inf.... 00 a,.,respectively. Let {a,.} be a bounded sequence. Then there exists a number M such that la.I s M for every positive integer n. By the Bolzano-Weierstrass theorem, the set !i' • is not empty. The set !i'. is bounded above by M; so by the least-upper-bound axiom, lub !i',, exists. Similarly, the set fi'. is bounded below by - M; so glb !i'" exists. Our first theorem follows immediately from Definition 20.1.

Theorem20.2 Let {a,.} be a bounded sequence. Then lim inf a,. s lim sup Jl ➔

Eumple.

ac,

- ➔

a,..

Cl()

Consider the sequence {a,.} defined by

a,. = (-1)",

n

= I, 2, ...

The subsequence {a2,.} has limit I; so I s lim sup a. 11-+oo

On the other hand, if {a,._}is any convergent subsequence of {a,.},

a,.,.s I,

k

= I, 2, ...

and thus Thus I is an upper bound for !i' •• and so lim sup a,. s I 1:-+oo

Therefore

lim sup a,.= 1

Similar methods show that Jim inf,....00 a,. = - I. The following theorem can be used to characterize lim sup,....00 a,. and lim inf,....oo a,.. That is if L and M are numbers satisfying the conclusionsof

20. The llm 1up and Um Inf of Bounded Sequences

Theorem 20.3, then L 20.3).

83

= lim sup,....ao a,.and M = lim inf,....ao a,.(see Exercise

'Theorem20.3 Let {a.} be a bounded sequence, and let L = lim sup,....ao a,. and M = lim inf,....ao a_. (i) If 8 > 0, there exist infinitely many positive integers n such that L 8 < a. and there exists a positive integer N 1 such that if n 2:: N 1 , then a.< L + 8. (ii) If a > 0, there exist infinitely many positive integers n such that a. < M 8

+ 8 and there exists a positive integer N,. such that if n

2:: N,., M -

< a_.

Proof. In this proof we use the notations a.,. and a,.,1 interchangeably. We establish part (i) leaving part (ii) as an exercise. Suppose that it is false that L - 8 < a. for infinitely many positive integers n. Then there exists a positive integer N 1 such that if n 2:: Ni, then a,. s L - s. Let {a,.,1 } be a convergent subsequence of {a,.}. If k 2:: Ni, we have a,.,1 s L - 8. By Theorem 14.2, lim1 ... ao a,.,1 s L - s. It follows that L s L - 8 which is a contradiction. Therefore L - e < a,.for infinitely many positive integers n. We next show that L + e s a,. for at most finitely many positive integers n, from which it follows that there exists a positive integer N such that if n 2:: N, then a,. < L + e. Suppose that L + a s a,. for infinitely many positive integers n. Then there exists a subsequence {a,.,1 } of {a,.} such that L + e s a,.,1 fork = 1, 2, .... By the Bolzano-Weierstrass theorem, {a,.,1 } has a convergent subsequence {a,.,1 ,1}. Thus L + e s a..1 ,1 for j = 1, 2, .... By Theorem 14.2, L + e s lim1...ao a,.,1 ,1 s L which is a contradiction. Therefore L + e s a,. for at most finitely many positive integers n. ■

If {a,.} is a convergent sequence, then {a,.} is bounded; so lim sup,....00 a,. and lim inf,....00 a,.are defined. It is easy to show that in this case, lim.,...00 a,. = lim sup,....00 0,. = lim inf,....ao a•.

Tbeorem20.4 (i) Let {a,.} be a sequence such that lim Then

a,.=L

lim sup a,. = L

= lim inf a,.

(ii) Let {a,.}be a bounded sequence such that

lim sup a,. = L = lim inf a,. 11-+00

Then

lim 11-+00

a,.= L

Sequences of R•I

84

Numben

Proof. (i) By Theorem 11.2, every subsequence of {a,.} has limit L. Thus !i',, = {L} and the conclusion follows. (ii) Let e > 0. By Theorem 20.3(i), there exists a positive integer N 1 such that if n ~ Ni, then a,. < L + e. By Theorem 20.3(ii), there exists a positive integer N 2 such that if n ~ N 2 , then L - e < a,..Therefore, if n :.?:max {Ni, N 2 }, then L-e

L = lim supb 11

and so

....

CIC>

L = lim inf b,.

Similarly,

l

➔

CIO

By Theorem 20.4(ii),

•

L =limb,. l

➔

CIO

We will later have occasion to use the following rather special result. 1beorem20.8 Let {a11} and {b11} be sequences such that

a.;?:0

lim

..... CIC>

and {b11} is a bounded sequence. Then Jim sup a,.b,.= lim sup a11 lim sup b11 l

➔

n CIO

➔

ao l

➔

= Jim a,. Jim sup b

11

CIO

Proof. The last equality follows from Theorem 20.4(i). Let L = lim 11... ao a•. If L = 0, then lim 11... ao a11b11 = 0 by Theorem 13.3, and the conclusion follows. Suppose L > 0. Let {b""}be a convergent subsequence of {b11}. Then

lim sup a,.b,.;?: lim a11,.b11,. = lim a,.,.lim b11,.

1 ➔

ao

Thus

l ➔

ao

l ➔

lim sup a,.b• L

..... ao

ao

l ➔

ao

lim b11,.

;?: n

➔

ao

It follows that lim sup a,.b,. L ;?: Jim sup bn

l ➔ CI)

l

➔

CIO

=L

lim b,.,. l ➔

ao

20. The llm sup •nd llm Inf of Bounded Sequences

Jim sup a,,b. ~

and hence

87

L Jim sup b.

Let {a_,.b. be a convergent subsequence of {a,,b,.}. Since {b.,.} is the product of the convergent sequences {a..,.b.,.} and { 1/a..,.}, it follows that {b-.} converges. .} 11

Therefore,

1 LI lim a.,.b.,. = lim - -(a.,.b.,.) A: ➔

0

t ..

Thus

00

lim a.,.b.,. ...

=

a.,. $

00

~

lim b.,.

t ..

00

lim sup b,. ,. .. 00

L lim sup b,. ,. .. 00

It follows that lim sup a,,b. ~ ...

and thus,

L lim sup b,. ,. .. 00

00

lim sup a,.b,.= L lim sup b,. = lim a. lim sup b,. ....

00 1

◄

co

1 ◄

= lim 1

➔

«>

sup a,. lim sup b,.

ao

1

➔

■

0

Exercises

20.1 Prove that lim inf,... ao (-1)" = -1. 20.2 Prove Theorem 20.3(ii). 20.3 Let L and M be numbers satisfying the conclusion of Theorem 20.3. Prove that L = Jim sup,,.. ao a. and M = Jim inf,... ao 20.4 Establish the inequality involving Jim inf in Theorem 20.S. 20.5 Establish the inequality involving Jim sup in Theorem 20.6. 20.6 Compute Jim sup,... ao a,. and Jim inf,... ao a,., where a.=

a,..

(a)

1 n

(b)

(1

(c)

( -1)" ( 1 - ~)

+

1r

Compute Jim sup,... ao a,. and Jim inf,... ao a,. and ft'•• where ai, a2, ... is an enumeration of the rational numbers in the closed interval [O, 1]. 20.8 Let {a.} be a bounded sequence. Prove that there exist subsequences {a,.,.} and {a.,,.} of {a.} such that Jim = Jim sup a. lim = lim inf a.

20.7

l ◄

ao

a,.,.

1

◄

c

l

◄

C10

a..,.

,. ....

ao

20.9 Let {a.} be a bounded sequence such that every convergent subsequence of {a.} has limit L. Prove that Iim,... ao a,. = L.

ea 20.10

Sequences of Real Numbers

Give an example of a sequence {a.}such that the sequence

{a'+ a2n+· · · +a.} but {a.} diverecs.

20.11

converecs Prove that if {a..,} is a subsequence of a bounded sequence {a.}, then lim sup a.., ~ Jim sup a. ◄

l

20.12

CI)

....

Cl)

Let {a.} be a bounded sequence. Prove that

Jim sup a. = -lim inf ( -a.) ....

20.13

(1)

Let {a.} and {b.} be sequences such that {a.} is convergent and {b.} is bounded. Prove that Jim sup (a. + b.) = Jim sup a. + Jim sup b. ■◄

and

20.14

Cl>

■◄

CIO

■◄

GO

+ bJ = Jim inf a. + Jim inf b.

Jim inf (a. ■◄

Cl>

■ -0

■◄

ao

Let {a.} be a bounded sequence. Suppose that for every bounded sequence {b.} we have lim sup (a. + b.) = Jim sup a,, + Jim sup b. ■◄

20.~

co

....

Cl)

■◄

CID

■◄

CID

Prove that {a.} is conver1CDt. Let {a.} and {b.} be bounded nonnegative sequences. Prove that

lim inf a. Jim inf b. ~ ■◄

GO

■ ,.GO

~

■◄

Jim inf a.b. ~ CO

■◄

Jim sup a.b. Cl>

■◄

lim sup a. Jim sup b. Cl)

...

Cl)

Show (by giving examples) that any of the inequalities of Theorem 20.6 and Exercise 20.15 may be strict. 20.17 Show (by giving examples) that the first and third inequalities of Exercise 20. t S may fail if the nonnegative hypothesis is omitted. 20.18 State and prove a theorem analogous to Theorem 20.8 where it is assumed that lim,... ao a,, ~ 0. 20.19 Let {a.} and {b.} be sequences such that lim. ...ao a. < 0 and {b.} is a bounded sequence. Suppose that lim sup a.b. = Jim sup a. Jim sup b.

20.16

■◄

20.20

GO

■◄

CIO

■◄

CIO

Prove that {b.} is convergent. Let {a.} be a sequence of positive numbers such that lim,,...ao a,. = L. Prove that lim (a,a2 • • • a,,)11• = L ,. .. ao

20.21 (Squeeze Theorem) Let {a.}, {b,.},and {c.} be bounded sequences such that

a.~ b. ~

c:

for every positive integern and Jim sup c. ~ Jim inf a. ■◄

CI

■◄

CID

21. The llm sup end lim Inf of Unbounded Sequences

89

Prove that -➔

20.22

Jim a,.= Jim b,. = Jim c., 01

l ➔

a0

l ◄

GO

Let {a.,} and {b.,}be sequences such that b11+1

=a,.+ a.,.,

for every positive integer n, and suppose that {b.,} is convergent. Prove that Jim,... .., (a,Jn) = 0. Give an example to show that {a.,} need not converge.

21. The lim sup and lim inf of Unbounded Sequences In this section we first give another characterimtion of lim sup,... 00 a,. and ➔ co a,. for bounded sequences {a,,} and then show how this charlim inf,. ➔ 0 a,, and acterimtion may be used to extend the definitions of lim sup, lim inf,. ➔ co a,, to unbounded sequences {a,.}. If {a,.} is a bounded sequence, then

A,. = tub {a,,,a,,+1,

•••

}

exists for every positive integer n. Since {a,.+i, a,,+2 ,

•••

}

c {a,,, a,,+i,

•••

}

it follows that A,,+1 s A,.

for n

= I, 2, ...

Because {a,,}is bounded, the monotone sequence {A,,} is also bounded and by Theorem 16.2, {A,.} is convergent. The next theorem states that lim,,.. 00 A,. = lim sup,. ➔ co a,, and similarly, Jim glb {a,,,a,,+ 1, 1

➔

••

0, we have lim,. ➔ 0 A,. S L. Therefore lim,. ➔ 0 A,. = L. ■ Let {a,.}be a sequence. If {a,.}is bounded above, we let A,. = lub {a,.,a,.+., ... }

for n

= 1, 2, ...

Then {A,.} is a decreasing sequence, and thus either {A,.} converges or lim,. ➔ 0 A,.= -oo. Guided by Theorem 21.I, we would define lim sup,. ➔ 00 a,. = lim,. ➔ 00 A,.. If { a,.} is not bounded above, we could let A,. = oo for n = 1, 2, . . . . If we want lim sup,. ➔ 00 a,. = lim,. ➔ 00 A,., we would have to define lim sup,. ➔ 00 a,. = oo in this case. This discussion motivates the next definition.

Definition21.2 Let {a,.}be a real sequence. (i) If {a,.}is not bounded above, we define lim sup,. (ii) If {a,.} is bounded above, we define lim sup,.

➔ ➔

0 a,. = oo.

0 a,. = lim,.

➔

0 lub {a,.,

a,.+i, .. . }.

(iii) If {a,.}is not bounded below, we define Jim inf,. (iv) If {a,.} is bounded below, we define lim inf,.

0 a,. = - oo. ➔ 0 glb {a,., 0 a,. = lim,.

➔ ➔

a,.+1t ••• }. By Theorem 21.1, Definition 21.2 is consistent with Definition 20.1.

Examples. (i) Let {a,.}be the sequence defined by a,. = n for n = 1, 2, .... ➔ 0 a,. = oo. Now {a,.}is bounded Since {a,.}is not bounded above, lim sup,. below and since glb {a,.,a,.+1 ,

•••

=n

}

for n

= 1, 2, ...

it follows that lim inf,. ➔ 0 a,. = oo. (ii) Let {a,.}be the sequence defined by

a,.={

-n

if n is even

0

Since {a,.}is not bounded below, lim inf,. Since lub {a,.,a,.+1 , it follows that lim sup,.

➔

•••

}

0 a,. = 0.

=0

if n is odd ➔

0 a,. = for n

-

=

oo. 1, 2, ...

21. The ltm sup end ltm Inf of Unbounded Sequences

71

Many of the theorems of Section 20 remain valid for unbounded sequences if we accept the following conventions. For any x in R, - oo < x, x < oo, and - oo < oo. The set R u {- oo} u {oo} with this ordering is called the extended real number system. The symbols "- oo" and "oo" are not real numbers, but are introduced for convenience. We conclude by giving an example of an extension of an earlier result (Theorem 20.2) to arbitrary sequences. 1beorem 21.3 If {a,.}is a sequence, then lim inf a. :s; lim sup a,. 1

➔

ao

1

➔

co

Proof. If {a,.} is not bounded above, then lim inf a,. :s; oo

= lim sup a,.

n-+GO

If {a,.} is not bounded below, then lim inf a.

=-

oo :s; lim sup a,.

If {a,.} is bounded above and below, the conclusion follows from Theorem 20.2. ■

Exercises 21.1 21.2

Prove the second equation of Theorem 21.1. Let A.= Jub {a., 0.+1, ••• } andB. = glb {a., a 0 i, ••• } forn = 1, 2, ..•• Compute A., B., Jim,... GO A., and Jim,... GO B., where a,, = (a) (-1)" 1 (b) n

(c) (d)

(e)

21.3

21.4

(1

+1r

(-1)"

n (-1)" ( 1

-1)

Give another proof of the Bolmno-Weierstrass theorem by showing that if {a.} is a bounded sequence, there exists a subsequence {a,..} of {a.} which converges to Ji.m,... GO Jub {a,,,0.+1, ••• }. (a) Let {a.} be a sequence which is bounded above. Prove that either {a,,} has a convergent subsequence in which case Jim sup ... GO a,,= Jub ft'. or {a,,}diverges to - 00 in which case Jim sup ... GO a. = - 00. (b) State and prove the result corresponding to (a) for Jim inf... GO a,..

72

Sequences of Real Number•

21.S Compute lim sup and lim inf of the following sequences: (a)

0, 1, 0, 2, 0, 3, ...

(b) · 1, -1, 2, -2, 3, -3, ... (c) -1, -2, -3, ... 21.6 Let {a,,} be a sequence. Prove that there exist subsequences {a111c} and {a.,.k} of {a,,} such that lim a,.k = lim sup a. and lim a,,.k = lim inf a. l

◄

GO

a

◄

CIO

1 ◄

0

a ◄

O

21.7 Prove Theorem 20.4 where L may assume the values - 00 or 00. 21.8 Prove Theorem 20.5 for arbitrary sequences {a.} and {b.}. 21.9 State and prove a version of Theorem 20.3 valid for arbitrary sequences.

V Infinite Series

The present chapter can be viewed as an extension of the ideas developed in Chapter IV on sequences. Infinite series play an important theoretical and practical role in analysis.

22. The Sum of an Infinite Series

In order to sum an infinite sequence of real numbers we must employ the notion of limit. We begin with the definition of an infinite series. Definition22.1 Let {a,.}be a sequence. For each positive integer n, let

s,. = a 1

II

+ a2 + •••+ a,. = L at A:•1

An infinite series is the ordered pair of sequences ({a,.}, {s,.}). The number a,. is called the nth term of the infinite series, and the number s,. is called the nth partial sum of the infinite series. Instead of using the cumbersome ordered pair notation for an infinite series, we will use the notation ao

La,.

or

11•1

to represent an infinite series. It should be understood that the symbolism I::..1 a,. denotes two sequences, namely the sequence {a,.} of terms of the infinite series and the sequence {s,.}of partial sums of the infinite series. 73

Infinite SeriN

74

It is important to distinguish between the sequence {a11} of terms and the sequence {s11} of partial sums of an infinite series E.~1 a,..We illustrate this with an example. For the infinite series I::. 1 ( -1)", we have a11 = ( -1)" and

s

II

={

-1 0

if n is odd if n is even

for n = I, 2, ...

The sequence {(-1) 211} is a subsequence of {( -1)"}. The nth term of the series I::. 1 ( - I)211 is 1, and the nth partial sum of this series is t,. = n for n = l, 2, .... Notice that although {(-1) 211} is a subsequence of {( - I)"}, {t,.} bears little resemblance to {s,.}. Sometimes it is convenient to begin the index of an infinite series with an integer other than 1. For example, we may consider the series E."°=o a,. or, in general, I::=, a,.,where p is an integer. The sequence of partial sums as defined in Definition 22.1 will be altered in the obvious way. To sum an infinite series, we simply add on more and more "terms." This corresponds to taking the limit of the sequence of partial sums. We make this notion precise in the following definition.

Deftnition22.2 Let I:: ..1 a,. be an infinite series. If the sequence of partial sums {s,.}(s,. = a 1 + · · · + aJ converges to L, we say that the infinite series L"°-1 a11 converges to L or that the infinite series I::=1 a11 has sum L. If the sequence {s,.}diverges, we say that the infinite series L"°= 1 a,. diverges. If the infinite series I::=1 a,, converges we also use the symbolism I::=1 a,, to denote its sum. We are using the notation I::'- 1 a11 in two very different ways. For an arbitrary sequence {a11} we will speak of the infinite series I::'. 1 a,,. Only for a convergent infinite series will we write I::=1 a,, = L, where L is the sum of the series I::'-1 a,..The context will always make clear whether we are speaking of the infinite series or the sum of a convergent infinite series. Our first theorem shows that if an infinite series is convergent, the terms of the series get small.

1beorem22.3 If the infinite series Li"°= 1 a,,converges, then a1

lim,

➔ "°

a,, = 0.

Proof. Suppose the infinite series L"°= 1 a,, converges to L. Let s,. = + · · · + a11 be the nth partial sum. Then S11+1 -

S11

= a,,+1

and so 0

=L

- L

= Jim s,,+1 -

lim s,. = lim (s,,+1 lim a11 = 0

Therefore 1

➔

l, then

L"°.1 a L 1a

converges absolutely. 11 diverges. 11

00 •

Proof. First suppose that L < l. Choose a number M such that L < M < l. Then there exists a positive integer N such that if n 2:: N, then

l'::11 1. There exists a positive integer N such that if n then ~

N,

la::. I>

1

laNI< laN+ 11< laN+2I < · ··

Thus

so that the sequence {an} does not converge to 0 (why?). By Theorem 22.3. L"°:. ■ 1 andiverges. The ratio test gives no information if limn-+colan+ 1 /anl= 1. The series 2 L = 1 1/n diverges and the series I::. 1 l/n converges, but 00

lim 1/(n + 1) = 1 = lim 1/(n \ 1)2 1/n n-+co 1/n

n-+co

Examples. The series L00•

1

1/n! converges since the ratio 1/(n + l)! 1/n!

1

=n + l

converges to 0. The series I::'-1 n !/~ converges since the ratio (n+l)!/(n+l)n+i n!/~ converges to 1/e

~

n+l

= (n +

1t+ 1 ~

=

1/2. Similar arguments show that

1 (1 + 1/n)"

r,.

00 •

1

~/n! diverges.

1beorem 26.7 (Root Test) Let {a,.} be a sequence and let L 11". (L = oo is allowed.) la,.1 (i) If L < l, then I::=1 anconverges absolutely. (ii) If L > l, then L00• 1 a,. diverges.

= lim SUPn-00

Proof. Suppose L < l. Choose a number M such that L < M < l. By Theorem 20.3, there exists a positive integer N such that if n ~ N, then 11 " < M la,.1

Thus if n

~

N, we have

la,.I
1 for infinitely many positive integers Next suppose L > l. Then la,.1 n. Thus lanl> 1 for infinitely many positive integers n, and hence the sequence {a,.} does not converge to 0. By Theorem 22.3, L00a 1 a,.diverges. ■

Infinite Serln

88 11 " As in the ratio test, iflim sup,....a)la,.I tion. We have

= 1, the root test givesno informa-

1)1/= I = lim :I1)1/ lim (n n 11

11-+00

,. ... a:,

but

11

(

La).1 1/n diverges and I::°-1 l/n

2

converges.

11 Examples. I:,."°. 1 2" diverges since lim,....00 (2") " = 2. La). 1 (½)" converges 11 11 since lim,....a)[(½)"]" = ½- La).1 n(½)" converges since lim,....a)[n(½)"] " = lim_...a:, nl/11,½ = ½,

Exercise• 26.1

Determine whether each of the series converges absolutely, converges conditionally, or diverges. CD (-1)•+1 CD (-1)11+1 (a) .~1 n 3 + n2 + n (b) .~1 n + ,Jn (c)

!: (-1)

■ +1(

Jn+l

(e)

- {ii)

!: n:i(-1)" 2"

(i)•

!: 2"(n n(-1)" + 1)

••1

f

(f)

••1

(g)

(d)

n

•-1

a•l

(-1)"

(2n - 1)!

.f:(-1)" E ( - l)"[n 11" - 1) .. 1iif+l7ii (h)• ••1 f (-l)"[e -(1 +!)"] (j)• .E(-1)" •• , n •• , Cl)

1.3.5 .. ,(2n -1) 2•4•6· • •(2n) 11 " < 26.2 Show (by example) that if {a,.}is a sequence such that lim inf,....00 la.1 1, then I:::'-1 a,, may either converge or diverge. Prove that if lim inf,,... la..111" > 1, then I:;:'. 1 a,, diverges. 26.3 Use Theorem 19.3 to prove Theorem 26.2. Cl)

a:

26.4 Prove that if I:::'-1 a,. converges absolutely, then I:::'. 1 converges. 26.S Let E:'-1 a. and !;:' .. 1b,. be absolutely convergent series. Prove that the converges. series I:::'-1 vla.b.l 26.6• Prove that the series !;:'. 1 a,,converges absolutely if and only if for every e > 0, there exists a positive integer N such that la ■ 1 + O ■ z +· · · + a,.,.I 1 1

then L ;'. I 0,, diverges_ 26.8 Let {a,,} be a sequence of positive numbers. Prove that if

Jim a.u = L ..... a. Jim a!1• = L

.....

then

Deduce •◄

. n I1Dl (nt)ll• ao

•

=e

Let {a.} be a sequence where each term is one of the integers 0, 1, 2, 3, 4, S, 6, 7, 8, 9. Prove that the series I::'. 1a,./10- converges and the sum is in the closed interval (0,1]. The term .a1a2•.• is called the decimal expansion of the number I::'-1 a./10-. 26.to• Prove that if x e (0, 11,there exists a sequence {a.}, whereeach term is one of 0, 1, 2, ••• , 9, such that !;;'. 1 a,./10- = x. 26.9

26.11* Let R,, = E:'-a ♦ 1 1/kl be the error in approximating I::-1 1/k!. Prove that (n

1

1/k! with

1 n +2 1 + 1) I < R,, < n + 1 (n + 1) I

Find the least n necessary to have R,, < decimal accuracy.)

27.

!;;'.

l x

10-

10

•

(This gives 10-place

Power Series

A power series is the infinite series generaliz.ation of a polynomial.

Deflnition27.1 Let t be a fixed real number. A power series (expanded about t) is an infinite series of the form

r a.(«> ra,,.. -o '

,,..o

«> r• -r--=-rr•=-r1t•O

I -

S

I I -

«> S

••0

s). Thus the sum by 1 I -

S

I I - r

Similarly, the sum by columns is I::°-.o (I:,:>.0a..•) = {1/(1 - r))(l/(1 - s)). In this example, the sum by rows and the sum by columns both exist and they are equal. In the proof of the next lemma, the foilowing fact will be used. If

94

Infinite Serlea

:I::'. 1 a,,,,,.converges

for every positive integer n, then converges for every positive integer k and

:I::'=1 CT:!= 1 a,,,,,.)

This result follows immediately from Theorem 23.1 since

00

IICI

L lfl•l

(a,,,,1

+ a,,,,2 + •••+ a,,,,J

Lemma 29.3 Let {a,,,,,.}be a double sequence of nonnegative terms. If the sum by rows of the double series :I:,,,,,. a,,,,,.exists, then the sum by columns also exists and they are equal.

:r:.

Proof. We are given that the series 1 a,,,,,.converges for every positive integer m and that the series :I::'=1 0. Suppose that if lxl < 1 and Ix - ti< R•, we have I:::'-ox 2• = I:::'-ob.(x - tY,. Prove that the radius of convergence of I:::'. ob.(x - t Y'is 1 - ItI, This exercise shows that, in general, the value R - ltl of Theorem 29.11 is the best possible. Let {o•.• } be a double sequence. We say that the double series I::•.• o•.• convergesto L if for every e > 0, there exists a positive integer N such that if k, I~ N, then (x -

29.11

a

I

I••1••1 L I:; a... - LI < 8 (a) Prove that under the hypotheses of Theorem 29.4, the double series converges. (b) Prove that the double series I::•.• a..• converges if and only if for every e > 0, there exists a positive integer N such that if k, I, p, q ~ N, then

t o•.• l 0, there exists a number x e X such that 0 < Ix - al < 6. In other words, a is an accumulation point of X if there exist points in X different from a which are arbitrarily near a. The number a may or may not belong to X. In Section 32, we will need the concept of left and right accumulation points. The definition is obtained by modifying Definition 30.1 so that the inequality 0 < Ix - al < 6 is restricted in the obvious way. For example, a is a left accumulation point of X if for every 6 > 0, there exists a number x e X such that 0 < a - x < 6. Definition30.2 Let/ be a function from a subset X of R into R, and let a be an accumulation point of X. We say that the limit off(x) as x approaches 102

30. Definition of the Umlt of • Function

103

a is L and write lim/(x)

=L

if for every 8 > 0, there exists 6 > 0 such that if 0 < Ix - al < 6 and x e X, then 1/(x) - LI < 8. The assumption that a is an accumulation point of X assures us that there are points x e X satisfying the inequality O < Ix - al < 6. As in the case of sequences (see Theorem 10.3) we can prove (Exercise 30.7) that if/ has a limit as x approaches a, the limit is unique. This justifies writinglimx... f(x) = L. The inequality Ix - al < 6 states that the distance between x and a is less than 6 and the inequality 1/(x) - LI < 8 states that the distance between f(x) and L is less than 8. The inequality 0 < Ix - al is equivalent to the statement x ::pa. Thus Definition 30.1 says (roughly) that limx... /(x) = L provided that the distance between/(x) and Lis small (less than 8) whenever the distance between x and a is small (less than 6) and x ::pa. The value /(a), if it is defined, is irrelevant as far as the limit at a is concerned. The inequalities 1/(x) - LI < 8 and Ix - al < 6 are equivalent, respectively, to L - 8 < /(x) < L + e and a - 6 < x < a + 6. Thus limx... f(x) = L provided given any open interval (L - e, L + e) there exists an open interval (a - 6, a + 6) such that if x e (a - 6, a + 6) and x ::pa, then/(x) e (L - e, L + e). See Figure 30.1.

Examples. Let/(x) = 3x - l. We will prove that limx_.1 /(x) = 2 directly from Definition 30.2. We are required to show that if e > 0, there exists l, > 0 such that if 0 < Ix - II < 6, then 1(3x - I) - 21 < e. Now 1(3x l} - 21 < e is equivalent to Ix - II < e/3, so if we take o = e/3 the limit will be established. The formal proof would be given as follows. Let e > 0. Set o = e/3 and suppose 0 < Ix - 11 < 6. Then 1(3x - 1) - 21 = 13(x - 1)1 < 36 = e Let/(x) = x 2 + x. We will prove that limx.. 2 /(x) nition 30.2. This time we are required to show Ix

+ 3llx -

21 =

= 6 directly from Defi-

l(x2+ x) - 61
0. Set Ix+ 31< 6; so ~

= min {t/6,

1} and suppose 0 < Ix - 11< ~- Then t

l(x2 + x) - 61= Ix+ 3llx - 21< 6 6 = t ExerclHS

In Exercises 30.1 to 30.5 prove the limit using Definition 30.2. 30.1 lim,.... /(x) = c, wherc/(x) = c for every real x. 30.2 lim (6x - 2) .... 2

30.3

limx2 =

....,

2

=

I

.... 2

10

4

30.4 limx 3 +x

30.5 Jim 2/x

=

+x+l=4

30.6 Let if x is rational if X is irrational Prove that lim,.... /(x) does not exist for any a. /(x)

= {~

31. Limit Theorems for Functions

105

30.7 Prove that if/ has limit L and limit M as x approaches a, then L = M. 30.8 Suppose lim.,.... /(x) = L > O. Prove that there exists o > 0 such that if 0 < Ix - al < then/(x) > 0.

o,

31. Limit Theorems for Functions In this section we will prove some limit theorems for functions which are analogous to those we derived earlier for sequences (see Chapter IV, Section 12). We begin by defining algebraic methods of combining real-valued functions.

Definition 31.1 Let/ and g be functions from a set X into R, and let c be a real number. We define

= 1/(x)I, x e X = cf(x), xeX + gXx) = f(x) + g(x).

(i) 1/l(x) (ii) (c/)(x) (iii) (/

= f(x)

- g(x), (v) (/• g)(x) = f(x)g(x), xeX

(iv) (/ - g)(x)

(vi) (l)cx) g

= /(x) g(x)'

x eX xeX

x e X and g(x) #: 0

The operations imposed on the functions in Definition 31.1 are said to be pointwise operations. For example, in Definition 31.1, (iii), to compute the value of the function/ + g at the point x, we simply add the values of/ and g computed at the point x. ➔ ./(x) = Land We may prove directly from Definition 30.2 that if lim., lim., ➔• g(x) = M, then lim., ➔ ./(x) + g(x) = L + M. Let e > 0. There exists cS1 > 0 such that if 0 < Ix - al < cS1 , then 1/(x) - LI < e/2. There exists cS2 > 0 such that if 0 < Ix - al < cS2 , then lg(x) - Ml < e/2. Let cS= min {cS 2 }. Now if 0 < Ix - al < cS,then 1 , cS 1/(x)

+ g(x)

- (L

+ M)I s

1/(x) - LI

+ lg(x)

- Ml
0. There exists 6 > 0 such that if O < Ix - al < 6, then 1/(x) - LI < e. There exists a positive integer N such that if n ~ N, then la,.- al < 6. Since a,. ¢ a for every positive integer n, we have O < la,.- al . Tberefore,if n ~ N, then

o < la..- al
a, x e X, such that s i < x :S c. By the definition of X, there exists Ii, ... , 111 e .f such that II

U /1

[a, x] c

1• 1

Therefore

[a, C]

C

[

l) /,] U ]

, .. 1

We conclude that c e X and the proof will be complete when we show that C = b. Suppose c < b. Since c e X, there exists Ii, ... , / 11 e .f such that II

[a, c] c

U /1

I• 1

In particular, c e 11 = (s2 , t 2 ) for some j, l c < d < b and c < d < t 2 • Then II

[a, d] c

U 11

I= i

:S j ~

n. Choose d such that

Real-Valued Function• and Contlnuoua Function•

114

and thus de X. Since d > c, we have a contradiction, an:i therefore c b. ■

=

The property of the closed interval [a, b] given in Theorem 34.2 is called compactness, and this property will be thoroughly investigated in Chapter

VII, Sections 42 to 44.

1beorem34.3 If /is continuous on [a, b], then/is

bounded on [a, b].

Proof. By Lemma 34.1, for each c e [a, b], there exists an open interval le containing c such that/ is bounded on le. Since

VeIc e [a, b]}

[a, b] c u by Theorem 34.2, there exist

Ci, •••

,

[a, b] c

c. in [a, b] such that

• U le, 1

I•

Now /is bounded on

1beorem34.4 If/ [a, b] such that

Ui=ile,, and

therefore/is

bounded on [a, b]. ■

is continuous on [a, b], there exist points c and d in /(c)

s /(x) s /(d)

for all x in [a, b]. That is, if/is continuous on [a, b], then/attains and a minimum on [a, b].

a maximum

Proof. By Theorem 34.3,/is bounded on [a, b]. Let M

= lub {/(x) I x e [a, b]}

We must show that/(d) = M for some din [a, b]. Suppose this fails. Then /(x) < M for every x in [a, b]. Let 1 . g(x) = M _ f(x) for x e [a, b] By Theorem 33.2, g is continuous on [a, b]. By Theorem 34.3, g is bounded on [a, b], and thus there exists a positive number N such that g(x) < N for every x in [a, b]. We have 1 M-f(x)

< N

for every x in [a, b] which reduces to /(x) < M-

I

N

(34.1)

34. Th• Heine-Borel Theorem

116

for every x in [a, b]. Since M - l/N < M, inequality (34.1) contradicts the fact that M is the least upper bound of the set {f(x) I x e [a, b]}. Therefore /(ti) = M for some din [a, b]. The existence of a minimum is proved in a similar way. ■ In Section 42, we will generalize Theorems 34.3 and 34.4 (see Theorem 42.6 and Corollary 42.7).

Exercises

34.1

34.2

34.3

34.4

34.S 34.6

Prove that if /is continuous on [a, b], there exists a point c in [a, b] such that f(c) = glb {/(x)lx e [a, b]} thus completing the proof of Theorem 34.4. (a) Use Exercise 30.8 and the Heine-Borel theorem to prove that if f is continuous on [a, b] and/(x) > 0 for every x in [a, b], then there exists e > 0 such that /(x) ~ e for every x in [a, b]. (b) Prove the statement in part(a) by considering the function g(x) = 1//(x) Give another proof of Theorem 34.3 by completing the following argument. Hf is not bounded on [a, b], there exists a sequence {a,,} in [a, b] such that 1/(a.)I > n. By the Bolzano-Weierstrass theorem there exists a convergent subsequence {a-.} of {a,,}. Let c = liJDa ◄ ao a,... Then /(c) = liD& ◄ aol(.J, which is a contradiction. Give an example of a collection J of open intervals such that uJ :::>(a, b) and such that for no finite subcollection ,I of J do we have u,I :::>(a, b). Deduce the Bolzano-Weierstrass theorem (Theorem 18.1) from the HeineBorel theorem (Theorem 34.2). Prove that if f is continuous on [a, b] and e > 0, there exists b > 0 such that if Ix - YI < b and x, ye [a, bl, then 1/(x) - /(y)I < a.

........

VII Metric Spaces

In this chapter, we present an introduction to the theory of metric spaces. Our results generalize much of the material in Chapters IV, V, and VI, which were concerned with real numbers. The theory of metric spaces generalizes those parts of our earlier work which relied on the notion of distance.

36. The Distance Function

If x and y are real numbers, we may interpret Ix - YI geometrically as the distance between x and y. (See Figure 35.1.) Using the notion of distance, we may rephrase the definition of the limit of a sequence as follows: lim,....00 a,.= L if and only if for every s > 0, there exists a positive integer N such that the distance between a,. and L is less than e if n ~ N. The definition of the limit of a function also involves the notion of distance. Letting d(x, y) = Ix - yl denote the distance between x and y, we may say that limx .....f(x) = L if and only if for every e > 0, there exists 6 > 0 such that if d(x, y) < 6, then d(f(x), L) < e. Thus both of our fundamental definitions of limit involve the notion of distance, and we would expect to be able to define limits in arbitrary sets provided an adequate notion of distance were defined. The distance function d(x, y) = Ix - YI is a function from R x R into [O,oo) which satisfies (i) d(x, y) = 0 if and only if x = y (ii) d(x, y) = d(y, x) for all x,y e R (iii) d(x, z) ~ d(x, y) + d(y, z) for all x,y,z e R It turns out that properties (i), (ii), and (iii) above are enough to determine

118

36. The Distance Function

117

'/

X

J

1

l•-'11

Figure 36.1

a useful notion of distance in an arbitrary set and we are led to the following definition.

Deftnition35.1 Let M be a set. A metric on M is a function d from M x M into [O,oo) which satisfies (i) d(x, y) = 0 if and only if x = y (ii) d(x, y) = d(y, x) for all x,y e M (iii) d(x, z) s d(x, y) + d(y, z) for all x,y,z e M Deftnition35.2 A metric space is an ordered pair (M, d), where Mis a set and d is a metric for M. A metric space is (roughly) a set M together with a distance function (metric) d on M. Property (iii) of Definition 35.1 is called the triangle inequality because of its geometric interpretation in the plane. (See Example 35.4.)

Example35.3 The pair (R, d) is a metric space, where d is defined by d(x, y) = Ix - yl. The metric d is called the usual (or absolute value or Euclidean) metric for R. The space R is understood to have the usual metric unless otherwise specified. Let R 2 = R x R. Geometrically, R 2 is the plane. Example35.4 We will show that d(x, y) where x Now

= (xi, x2 ),

y

= J(xi

- Yi)2

+ (x2

- Y2)2

= (Yi, y 2) e R 2 defines a metric on R 2 .

d(x, y)

= J(xi

- Yi)2

+ (x 2 -

Y2)2

=0

if and only if x 1 = Yi and x 2 = y 2 • Therefore, d(x, y) = 0 if and only if = y, and hence part (i) of Definition 35.l holds. Part (ii) of Definition 35.1 holds since

x

d(x, y)

= J(xi = J(yi

- Y1)2

- Y2)2

-

-

+ (x2 2 X1) + (Yz

X2)

2

= d(y, x)

Finally, we must establish the triangle inequality [part (iii) of Definition

118

Metric Spaces

35.1]. As is often the case, the triangle inequality is the most difficult part of Definition 35.1 to verify. Let x = (x., x 2 ), y = (y., y 2), z = (z., z 2) e R 2 • We must prove that

J(x

z 1) 2

1 -

+ (x 2

-

z2 ) 2

= d(x, z) S d(x, y) + d(y, z) = J(x1 - Y1)2 + (x2 - Y2)2 + J(y1 -

Z1)2

+ (Y2 -

Z2)

2

(35.1)

Letting a 1 = (x 1 - yJ, b 1 = (y 1 - z 1), and writing x 1 - z1 = (x, - y,) (y 1 - zJ, i = l, 2, inequality (35.1) becomes

J(a

1

+

+ b1) 2 + (a2 + b2 ) 2 s Ja~ + a~ + Jb~ + b~

Squaring both sides and simplifying, we obtain the equivalent inequality

a1b1

+ a2 b2 s J(a~ + a~Xb~+ b~)

We show that (35.2) for any real numbers ai, a 2 , b., b 2 thus establishing the triangle inequality. Squaring both sides of inequality (35.2) and simplifying, we obtain the equivaµent inequality 0

s

aib~ - 2a 1b1a 2b2 + a~bi

which in tum may be written 0 S (a 1b2 - a 2b 1) 2

(35.3)

Since (35.3) holds for all real numbers a., a 2 , bi, b 2 , (35.2) and hence also (35.1) are established. Thus the triangle inequality holds ford, and therefore dis a metric for R 2 • The metric d of Example 35.4 is called the usual (or Euclidean)metric for R 2 • This metric gives the usual notion of distance in the plane. The triangle inequality is illustrated in Figure 35.2. We will generalize Example 35.4 to n-space in the next section.

y X

z

Figure 36.2

3&. Th• Distance Function

118

Example35.5 We let / 1 denote the set of all real sequences {tZi.}such that L~ 1 a,.converges absolutely. We will show that

d({a,.},{b,.})=

CIO

L la,.- b,.I

(35.4)

••l

defines a metric on / 1 • The series in equation (35.4) converges by the comparison test since la,.- b,.IS la,.I+ lb,.Ifor every positive integer n. It is easy to verify that parts (i) and (ii) of Definition 35.1 hold ford. We will verify the triangle inequality. Let {a,.}, {b,.},and {c,.}be elements of / 1 • Then

la..- c,.Is la,.- b,.I+ lb,.- c,.I for every positive integer n, and hence t

L la,.-

n•l

t

L la..-

c,.IS

11•1

b,.I+

t

L lb,.-

11""1

c,.I

CIO

Ll la,.- c,.I

d({a,.}, {c,.})=

11•

GO

S

CIO

L la..- b,.I+ 11•1 L lb,.- c,.I ••1

= d({a.},{b,.})+ d({b,.}, {c,.}) Therefore d is a metric on /1 •

Example35.6 Let X be a set. It is easily verified that d(x, y)

= {~

if X

=y

if X

~

y

defines a metric on X. The metric d is called the discrete metric for X, and we will refer to the metric space (X, d) as X with the discrete metric.

ExerciHa 35.1 Verify that the function d of Example 35.5 satisfies Definition 35.1 (i) and (ii). 35.2 Prove that the function d of Example 35.6 is a metric. 35.3 Let d be a metric on a set M. Prove that ld(x, z) - d(y, z)I s; d(x,y) for all x, y, z e M.

120

Metric Spaces

Let M be a set and let d bea function from M x Minto R which satisfie.1the three properties of Definition 35.1. Prove that d: M x M - (0, oo). 35.5 Let (M,d) be a metric space and let Xbe a subset of M. Prove that (X,dlX x X) is a metric space. 35.6 Let/"" denote the set of all bounded real sequences, and let c0 denote the set of all real sequences which converge to 0. (a) Prove that / 1 c c0 c I"". (b) Prove that the containments in (a) are proper. (c) Prove that 35.4

35.7

d'({a,,}, {b.}) = tub {la. - b.l Ine P} define.. a metric on /"" [and thus by (a) on c0 and / 1 also]. (d) Let d be the metric on / 1 defined in Example 35.5. Prove that d'(x, y) ~ d(x, y) for x, ye / 1. Let (M, d) be a metric space. Prove that

d'(x, y)

=

1

~(~(:~ y)

d"(x, y)

=

min {d(x, y), 1}

define metrics on M. Prove that d' and d" are bounded by 1. 35.8 Let (Mi, d 1) and (M,., d,.) be metric spaces. Prove that (M1 x M,., d) is a metric space, where d is defined by the formula d[(xi, x,.), (yi, y,.)J = d1(X1,Y1) + d,.(x,.,y,.) The space (M1 x M,., d) is called the product metric space. 35.9 Let H"" denote the set of all real sequences {011} such that la.I~ 1 for every positive integer n. H"" is called the Hilbert cube. (a) Let {a,,}, {b.} e H"". Prove that the series

f; la.2-b.!

••l

(b)

converges. Prove that d({a,,}, {b.})

= ..~1 ja.

r b.,j

define.. a metric on H"".

38. R•, / 2 , and the Cauchy-Schwarz Inequality We first generalize Example 35.4 ton-space. We let

R•

= {(xi, x 2 ,

•••

,

x,.) I x, e R for i

= 1, ...

, n}

The space R• is called real Euclideann-space.In case n = 1, R 1 is identified with the real line; in case n = 2, R 2 is identified with the plane; and in case n = 3, R3 is identified with ordinary 3-space.

38. R•, / 2 , and the Cauchy-Schwarz Inequality

We will prove that d(x,y)

=

Jf

121

(36.1)

(xa;- Ya:>2

t•l

where x = (x., ... , x,.),y = (y., ... , y,.) e R11 defines a metric on R11• For n = I, d is the absolute value metric for the line; for n = 2, d is the usual metric of Example 35.4 for R 2 ; and for n = 3, d gives the usual notion of distance in 3-space. The function d of equation (36.1) is called the usual (or Euclidean) metric for R11• To prove that the triangle inequality holds in R", we must show that if (x., ... , x,.), (y., ... , y,.), and (z., ... , z,.) e R",then

Jt

zJ2

(Xa; -

t•l

S

f

(Xa; -

t•l

Ya:) +

f (Ya:-

2

t•l

zJ2

Letting a11 = x 11 - Ya:and b11 = y 11 - za:for k = I, ••• , n and arguing as in Example 35.4, we can show that the triangle inequality will be established if we can prove that

This famous inequality is known as the Cauchy-Schwarzinequality.

1beorem 36.1 (Cauchy-SchwarzInequalityfor R") Let b., ... , b,.be real numbers. Then

J(f

f

aa:ha:I S

It•l

lc•l

a? (

f

t•l

a., ...

, a,.

and

b?

Proof. If b11 = 0 for I s k s n, the inequality follows immediately (both sides are zero). Thus assume that b11 :/: 0 for some k, l s k s n. Then

If x is any real number, we have 0 S

L"

i•l II

so that

0 S

L i•l

for any x. If we let II

A=

La:, t=l

(a11 - xba;)2 II

a? - 2x

L t•l

II

aa;b,,+ x2 L b: izl

Metric Spaces

122

and take x

= B/C,

then

0s A - 2(~)B + (~Y C

which reduces to

0 s AC - B2 ■

and the inequality now follows.

1beorem36.2 The equation d(x, y) where x

= (x.,

.. . , x..), y

=

= (y.,

Jt 1; ..

1

(x 1

-

yJ 2

... , y,.) e R"defines a metric on R".

Proof. Parts (i) and (ii) of Definition 35.1 are easily seen to hold ford, and part (iii) follows from Theorem 36.1. ■ We next generalize R"to "infinite-tuples" which are sequences and generalize the metric of Theorem 36.2 to the function 00

L (a1 1:-=1

d({a,.}, {b,.}) =

bJ

2

(36.2)

I::'..1 (a1 -

b1) 2 must converge, and thus we must restrict our attention to those sequences {a1 } for which the series I::°= 1 af converges. We let / 2 denote the set of all real sequences {a1 } for which the series 2 00 I:a: = 1 af converges. We next show that if {a1 },{b1 } e / , then the series I:t°-1 {a1 - bJ 2 converges, and thus equation (36.2) defines a function from 2 2 / x / into [O, oo). Actually we will deduce this result from another theorem which is important in its own right.

If equation (36.2) is to be meaningful, the series

'lbeorem 36.3 lutely.

Proof.

If {a 1}, {b1 } e / 2 , then the series

I:a: C.O,. converges 1 aa:h1:

By the Cauchy-Schwarz inequality (Theorem 36.1),

f

l:=1

J(f

la1:llh1:I s

1:-=1

f

2 2 ( la1:l lh1:1 1:•1

for every positive integer n. Therefore

f

(-f-b ....... :)

laa:h1:I s J...,...( L-.-a: .......

l:=1

l:=1

s

l:•1

Ct. a:)Ct. b:)

abso-

36. R•, /2, end the Cauchy-Schwarz Inequality

123

for every positive integer n. By Theorem 24.1, solutely. ■

L:'°.. 1 aJ,,,,converges ab-

Corollary36.4 If {a,,,},{b,,,} e / 2 , then the series L:'°.. 1 (a1

Proof. The series I:.00•

1

-

bJ 2 converges.

(a1:- bJ 2 is the sum of the three convergent series

r a:,

r aJJ1:, 00

00

-2

1:•l

l:•l

■

and therefore converges by Theorem 23.1. 1beorem36.S The equation

defines a metric on / 2 • 2 Proof. The series Lt= 1 (a1:- b1;) converges by Corollary 36.4. Parts (i) and (ii) of Definition 35.1 are easily seen to hold for d, and thus we verify only the triangle inequality. Let {a1;},{b1;},{c1;} e / 2 • From the triangle inequality for R• we have

for every positive integer n. Taking limits we have

Jf

d({a1;},{c1;})==

(a1:- cJ 2

1:•l

~

f

l:•l

(a1:- bJ

2

+

Jf l:•l

(b1:- cJ

= d({a1:},{b1;})+ d({b1:},{c1;})

2

•

By taking limits in Theorem 36.1, we can derive the Cauchy-Schwarz inequality for / 2 • 1beorem36.6 (Cauchy-Schwarz Inequalityfor 12 ) series Lt• 1 aJ,1;converges absolutely and

If {a1;}, {b1:}e / 2 , then the

Proof. The series I:.00• 1 aJ,1;converges absolutely by Theorem 36.3. The conclusion now follows by taking limits in Theorem 36.1. ■

124

Metric Spaces

Exercises 36.1 36.2

36.3 36.4 36.5

Give the details of the proof of Theorem 36.2. Prove that parts (i) and (ii) of Definition 35.1 hold for the metric of Theorem 36.S. Prove that / 1 c /2 c c0 and that the containments arc proper. Prove that if .E:"-1 ai converges, then .E;".1 a,./n converges absolutely. Let a1, ... , a. be real numbers. Prove that

ja1+·~·+a

.., ~

✓ af

+·~.+a]

This exercise shows that the absolute value of the arithmetic average does not exceed the root mean square average. 36.6* Show that

36.7• 36.8 36.9 36.10

36.11

36.12 36.13

if and only if there exist numbers s and t such that sa1 = tb1 for 1 ~ k ~ n. Interpret this result geometrically for n = 2. Show that the function/(x) = A - 2Bx + Cx 2 , C > 0, has a minimum at X = B/C. Let {a.} e / 1 and {b,.} e /"". Prove that {a,.b..} e / 1 • Let {a ..} e c0 and {b,.} e I"". Prove that {a,.b,.} e c0 • Give an example of sequences {a,.} and {b,.} such that {a.} e c0 and {b,.}e /"", but {a,.b,.}1112. Let {a,.}, {b,.} e /"". Prove that {a,.b,.} e /"". Give an example of sequences {a,.} and {b,.} such that {a,.}, {b,.} e I"",but {a,.b,.} 11c0 • Let {a,.} be a sequence such that {a,.b,.} e / 1 for every sequence {b,.} e / 1 • Prove that {a.} e I"". Show (by example) that the above statement is false if /"" is replaced by co. State and prove other results obtained by replacing either or both /1's in Exercise 36.11 by any of /1, / 2 , I"",or co. This alternative proof of the Cauchy-Schwarz inequality for R" was published by Tolstcd (1964). (a) Prove that for any real numbers a and b a2 + b2 2 ab~ (b)

Let a1, ... , a,. and b1, ... Set A = (.E~al) 112 and B in (a) to derive

, b,. be real numbers.

= (.E1bi) 112 • Take a = a1/A and b = b1/B

atbt~ ! (al + M) AB

2

R

B

(c) Sum the inequalities in (b) to derive the Cauchy-Schwarz inequality (Theorem 36.1).

37. Sequences In Metric Spaces

126

37. Sequences in Metric Spaces If (M, d) is a metric space, we will often abbreviate (M, d) to M. When we refer to a metric space M, it should be understood that there is a metric defined on M, but that we are not mentioning it explicitly. A sequence in a metric space M is simply a function from P into M, and we denote such a sequence as {a,.}:• ., or more simply {a,.}.(We will some.. 1 or {a}.)The definition times find it convenient to use the notation {a};i of the limit of a sequence in a metric space is a direct generalization of the definition of the limit of a real sequence (Definition 10.2).

Deftnition37.1 Let {a,.}be a sequence in a metric space (M, d). We say that {a,.}convergesto (or has limit) L, where Le M, and write Jim

,... ao

a,.=L

if for every e > 0, there exists a positive integer N such that if n ~ N, then d(a,.,L) < e. If {a,.} has a limit, we call {a,.} a convergentsequence. If {a,.} has no limit, we call {a,.}a divergentsequence.

As in the case for real sequences, we can show that the limit of a sequence in a metric space, if it exists, is unique, and so we are justified in writing lim,._00 a,. = L. One can also prove that if {a,.}is a constant sequence (a,. = L for every n), then Jim,._00 a,. ==L. Another important fact is that if a sequence {a,.}in a metric space converges to L, then every subsequence of {a,.}converges to L. In each case the proof is an imitation of the proof of the corresponding result for real sequences. In an arbitrary metric space we will not be able to prove generalizations of the algebraic theorems of Chapter IV, Section 12, for in an arbitrary metric space, there may be no algebraic operations defined. We now examine the meaning of sequential convergence in some specific metric spaces. In the metric space (R, d), where d(x, y) = Ix - yl is the usual metric, Definition 37.1 is identical to Definition 10.2, and thus a real sequence {a,.} converges to L in the metric space (R, d) if and only if {a,.} converges to L according to Definition 10.2. Let X be a set with the discrete metric d. Suppose that {a,.} converges to L in (X, d). Taking e = ½,there exists a positive integer N such that if n ~ N, then d(a,.,L) < t. But this implies that d(a,.,L) = 0 if n ~ N. Therefore, a,. = L if n ~ N. We will say that a sequence {a,.} is eventually constant if there exists a positive integer N such that a,. = aN if n ~ N. We have just shown that if {a,.}converges in (X, d), then {a,.}is eventually constant. It is easy to verify that the converse of this statement also holds.

Metric Spaces

128

0 Let {a}: = 1 be a sequence in R". Then each element a-- (a lt lt•••t

a}f.1 for 1 s j s n and convergence of the original sequence {a}f.1 in R"is given in the next theorem. Theorem 37.2 Let {a= a1 for j = 1, ... , n. {a,a) < 8

I"' . ... a(le) -- a 1or an d th us I1m1; J• -- 1, ••. , n. 00 1 1 Now suppose that lifflt...00 a}t>= a1 for j = 1, ... , n. Let e > 0. For each j, 1 s j s n, there exists a positive integer N1 such that if k ~ N1, then

8

la, - a11 < -:T,, (t)

If k

~

max {N1 ,

•••

,

N,.}, then

2


0 and let x e M. We let B.(x)

= {ye MI d(x, y)
0), and thus M is open. If 0 is not open, there exists x e 0 such that B,(x) ¢ 0 for any e > 0. However x ,J0, and thus we have a contradiction. ■ We next show that open balls are open sets. The proof of this theorem (Theorem 39.4) for R2 is illustrated in Figure 39.2. Let M be a metric space. Let x e M and let s > 0. Then the open ball B,(x) is an open subset of M.

Theorem39.4

Proof. Let (M, d) be a metric space. We must show that if ye B,(x), there exists b > 0 such that B,(y) c B,(x). Let ye B,(x). Let b = e - d(x, y). Then b > 0. (Why?) We will show that B,(y) c B,(x). Letze B,(y). Then d(y, z) < b. Now

/

/

- ,,•-;--,

··f,LV

8= e:-d(x,y)

I ' y , ,- - - -e:_- • • ' • - , \

I

X

/

Figure 39.2

134

Metric Spaces

d(x, z) s d(x, y)

< d(x,y)

+ d(y, +~

z)

=e and hence z e B,{x).

■

The next theorem gives the relationship between closed and open sets.

neorem 39.5 Let M be a metric space and let X and only if X' is closed.

c

M. Then X is open if

Let (M, d) be a metric space. Suppose that Xis an open subset of M. Let x be a limit point of X'. We must show that x e X'. If x ¢ X', then x e X. Since X is open there exists an open ball B,(x) such that Bc(x) c X. Since x is a limit point of X', there exists a sequence {x,.} of points in X' such that lim,... 00 x,. = x. Therefore, there exists a positive integer N such that d(xN, x) < e. Now xN e B,(x), and so xN e X n X', which is impossible. Thus x e X', and therefore X' is closed. Suppose X' is closed. If X is not open, there exists x e X such that for any e > 0, B,(x) ¢ X. Equivalently, we have B,(x) n X' ::/: 0 for every e > 0. For each positive integer n, there exists a point x,. in B 11n(x) n X'. Now {x,.} is a sequence of points in X', and since d(x,., x) < 1/n, we have lim,... 00 x,. = x. Thus x is a limit point of X' and since X' is closed, x e X'. This contradiction yields the theorem. ■ Proof.

Let X be a set with the discrete metric. In Section 38 we proved that every subset of Xis closed. It follows from Theorem 39.5 that every subset of X is open. The next theorem is analogous to Corollary 38.6 and Theorem 38.8. Theorem 39.6 Let M be a metric space. (i) If Ui, ... , U,. are open subsets of M, then U1 n U2 n · · · n U,. is an open subset of M. (ii) If If/ is a collection of open subsets of M, then u If/ is an open subset of M. Let Ui, U2 , ••• , U,. be open subsets of M. By Theorem 39.5 it is enough to show that (U 1 n · · · n U,.)' is closed. By Theorem 39.5, each of Ui, U 2, ..• , U~ is closed and by Corollary 38.6, Ui u U 2u · · · u U~ is closed. By Theorem 1.7, (U 1 n· · ·n U,.)' = Ui u· · ·u U~. Thus (U 1 n· · · n UJ' is closed, and hence Vi n · · · n U,. is open. Part (ii) is proved similarly using Theorems 38.8 and 1.7. ■ Proof.

An arbitrary intersection of open sets need not be open. For example,

39. Open Sets

136

{O}=

1) n (-1 -,n n 00

•• 1

is not open in R. If a subset of a metric space is not open, this does not mean in general that it is closed. Indeed, there are subsets of R which are neither open nor closed. For example, the half-open interval (0, 1) is neither open nor closed inR. If X is a set and ff is a collection of subsets of X satisfying (i) X, 0 eff (ii) The union of a subcollection of ff is a member of ff (iii) The intersection of a finite subcollection of ff is a member of ff then ff is called a topology for X. By Theorems 39.3 and 39.6, the collection of open subsets of a metric space M is a topology for M.

Exercises

39.1 Prove that a half-open interval not of the form ( - 00, b] or [a, 00) is neither closed nor open in R. 39.2 Prove Theorem 39.6 (ii). 39.3 Deduce Theorem 39.6 from Definition 39.2. 39.4 Let M be a metric space such that Mis a finite set. Prove that every subset of Mis open. 39.S Prove that the interior of a rectangle in R2 {(x, y)la < x < b, c < )' < d} is an open subset of R2 • 39.6 Prove that if X and Y are open subsets of R, then X x Y is an open subset of R2 • State and prove a generalization to R•. 39.7 Let/be a continuous function from R into R. Prove that {xl/(x) > 0} is an open subset of R. 39.8 Let X be an open, nonempty subset of R. Prove that there exists a unique countable set of open intervals {(a., bJ }:".,such that Cl)

(a)

U (a,,, b.) = X ••

1

*

(b) (a,,,, b.) n (a., b.) = 0 if n m (a. = - 00 and b. = 00 may occur) 39.9 Let X be a subset of a metric space M. Prove that Xis an open subset of M if and only if Xis the union of open balls. 39.10 Let X be a subset of a metric space M. Prove that Xis a closed subset of M if and only if whenever x is a point in M such that BJ..x)n X 0 for every e > 0, then X E X.

*

138

Metric Spaces

39.11 If Xis a subset of a metric space M, we say that a point x in Xis an interior point of X if B,(x) c X for some e > 0, and we let x 0 denote the set of interior points of X. Let M be a metric space. Prove the following: (a) X 0 c X for X c M. (b) Xis open if and only if X 0 = X, X c M. (c) (X 0 ) 0 = X 0 for X c M. (d) X 0 is open for all X, X c M. (e) If X C y C M, then x° C yo, (f) (Xn Y) 0 = X 0 n Y 0 for X, Y c M. (g) If Y is an open subset of M such that Y c X c M, then Y c X 0 • (h) If X c M, then x 0 = u { YI Y c X and Y is open}. (i) ((X')-)' = X 0 , for all X c M. 39.12 If Xis a subset of a metric space M, we define the boundary of X to be the set

ax= ..t" ex,-. Let M be a metric space. Prove the following: (a) ax is closed for all X, X c M.

(b) Xu oX = ..f for all X, X c M. (c) X\oX = x 0 for all X, X C M. (d) If X is a proper nonempty subset of R, then aX

40.

* 0.

Continuous Functions on Metric Spaces

We will define continuity of a function in an arbitrary metric space analogously to the situation for the real line. Guided by Theorem 33.3 we make the following definition.

Definition40.1 Let (M 1, d 1) and (M 2 , d2 ) be metric spaces, let a e Mi, and let/be a function from M 1 into M 2 • We say that/ is continuousat a if for every e > 0, there exists o > 0 such that if d 1{x, a) < o,then di(f(x),f(a)) < e. We say that/is continuouson M 1 if /is continuous at every point of M 1 • In case M 1 = M 2 = R and d 1 = d 2 is the usual metric for R, Definition 40.1 gives exactly the definition of continuity for real-valued functions defined on a subset of R (see Theorem 33.3).

Examples. Let/: / 1 -+ R be defined by /({a,.}:,. 1) = a 1 • Let {a,.}: .. 1 e / 1 • We will prove that f is continuous at every point of / 1 • Let e > 0 and take o = e. If O < d({x,.}, {a,.}) < o, then I::=1 Ix,. - a,.I< o, so that lx 1 a 1 1 s :r:,.1 Ix,. - Cli.l< b = e, and thus 1/({x,.}) - /({a,.})I < e, and we have established continuity. Let /: / 1 -+ / 2 be defined by /(x) = x, for x e / 1 • Let {a,.}:, 1 e / 1 • We will

40. Continuous Functions on Metric Spaces

137

prove that I is continuous at every point of / 1 • Let d 1 and d 2 denote the metrics on / 1 and / 2 , respectively. It is easy to verify that d 2 (x, y) :s; d 1 (x, y) for x,y e / 1• Let e > 0. Set 6 = e. If O < d 1(x, {a,.}) < 6, then d 2 (l(x), {a,.}) = d 2 (x, {a,.}) :s; d 1 (x, {a,.} < 6 = 8, and we have established continuity.

Theorem40.2 Let f be a function from a metric space M 1 into a metric space M 2 • Let a e M 1 • Then/is continuous at a if and only if whenever {x,.}is a sequence in M 1 such that lim,....cox,. = a, then lim,....cof(x,.) = /(a). Proof. Let (Mi, d 1) and (M 2 , d2 ) be metric spaces, let/be a function from M 1 into M 2 , and let a e M 1 • First, suppose that/ is continuous at a. Let {x,.}be a sequence in M 1 such that lim,....cox,. = a. Let e > 0. There exists 6 > 0 such that if d 1(x, a) < 6, then d2(/(x),f(a)) < 8. Since lim,....cox,. = a, there exists a positive integer N such that if n ~ N, then d 1(x,., a) < 6. If n ~ N, then d 1(x,., a) < 6, and hence d2 (/(x,.),/(a)) < s. Now suppose that whenever {x,.}is a sequence in M 1 such that lim,....cox,. = a, then lim,....co/(x,.) = /(a). If/ is not continuous at a, there exists 8 > 0 such that for any 6 > 0, d 1(x, a)< 6 for some x e Mi, but di(f(x),f(a)) ~ 8. Thus for every positive integer n, there exists x,. e M 1 such that d 1(x,., a)

I n

0 for all :xe M. Prove that there exists T > 0 such that/(:x) > T for all :xe M. 42.7 Let X be a compact subset of Rand let ye R. Prove that the set {x + YI x e X} is compact. 42.8 Let/be a continuous, real-valued function on a metric space M which is never zero. Provethatthecollectionofopensets Uforwhicheither/(x) > 0 for x e U or f(:x) < 0 for x e U is an open cover of M. 42.9 Call an open cover tfl of a metric space M an additive cover if whenever U, Ve tfl, we have U u Ve tfl. Prove that M is compact if and only if every additive open cover of M contains M (Johnsonbaugh, 1977). 42.10 Let {X.} be a sequence of compact subsets of a metric space M with X1 :::,X2 ::J X3 ::J • • •. Prove that if U is an open set containing nX., then there exists x. c U. 42.11 Let /be a function on [a, b]. Let K be a compact subset of[a, b] on which / is continuous. Suppose there exists c > 0 such that for each x e K, there exists h,. > 0 with

1/(:x +

hh~ -/(x)I


0. We first show that the special open cover If/

=

{B,(x) Ix e M}

has a finite subcover.

Lemma43.2 Let M be a metric space in which every sequence has a convergent subsequence. Lets > 0. Then there exist x 1 , ••• , x,. e M such that M

= B,(x 1)

u• • •U B,(xJ

43. The Bolzano-Welerstraas Characterization

149

Proof. Let d denote the metric on M. Let x 1 e M. If B.(x1) = M, we stop. Otherwise, there exists x 2 e M\Bix 1), and thus d(x 2 , x 1) ~ e. If M = B,(x1) u B.(x 2), we stop. Otherwise, there exists x 3 e M\(B.(x 1) u B,(x 2 )), and thus d(x 3 , x 1) ~ e and d(x 3 , x 2 ) ~ e. This process must stop after a finite number of steps, for otherwise we obtain a sequence {x,.}in M such that d(x., x,.)

~

if n #=m

e

Such a sequence cannot have a convergent subsequence (verify).

■

Let M be a metric space in which every sequence has a convergent subsequence, and let If/ be an open cover of M. If for some e > 0, each open ball B.(x) were contained in some U e If/, then M would be compact. For

M

= B.(x1) u · · · u

B.(xJ

by Lemma 43.2 and since B,(x 1) is contained in U 1 for some U 1 e If/, i I, ... , n, we would have

M

= B.(x 1)

u• • •u B.(x.)

c

=

U1 u• • •U U,.

in which case M would be compact. Our next lemma shows that such an e exists. The proof of Lemma 43.3 is illustrated in Figure 43.1.

Lemma43.3 Let M be a metric space in which every sequence has a con-

u

,, ,' I ',

..n, /

.,,,,.,---

',a,,n ,. I

'

I

nf

....

' , a.,.,

)

\

I I I

I

'

flgure43.1

- - --

,.

I ✓

Metric Spaces

160

vergent subsequence. If 'fl is an open cover of M, there exists e > 0 such that if x e M, B,(x) c U for some U e 'fl. Proof. Suppose the conclusion is false. Then for every e > 0, there exists x e M such that Bix) ¢ U for any U e 'fl. Taking e = 1/n, there exists x" e M such that B 11,.(x") ¢ U for any U e 'fl. By our hypothesis, {x,.} has a convergent subsequence {xn.,}, and we suppose lim1 ... x"" = x. Now x e U for some U e 'fl, and since U is open, there exists e > 0 such that B,(x) c U. Choose a positive integer k such that CX)

d(x,..,,x)

< e/2

and

c U which where d denotes the metric for M. We will show that B 11n,.(xn.,) will be a contradiction. Suppose ye B 11n..(xn.,). Then d(y, xn,.) < 1/n,. Now

1 d(y, x) ::5 d(y, x,.,.) + d(x""' x) < n,

+ d(x""'

x)
}~1 • Now {a\"J,>}~1 is a subsequence of the convergent sequence {a\"J>}j,. 1 and thus converges.

162

Metric Spaces

1 We continue in this way producing a subsequence {bJ;o. 1 of {aJ:°= 1 such that {b}">}:°=tconverges for j = l, ... , n. By Theorem 37.2, {b 0. The refore 1- 1(x) = x 11" is continuous on [0, oo). If n is an odd positive integer, the argument can be modified to show that/- 1{x) = x 11" is continuous on R. Let/be a continuous function from a metric space (M 1, d1) into a metric space (M2 , d2 ). Let 8 > 0. For each ye M 1, there exists 6 > 0 such that if d 1(x, y) < 6, then d2 (/(x),f(y)) < 8. In general, 6 will depend upon y. However if there exists 6 > 0 which is independent of y, we say that f is uniformly continuous on M. Definition44.4 Let (M 1, d 1) and (M 2 , d2) be metric spaces, and let/ be a function from M 1 into M 2 • We say that/is uniformly continuous on M 1 if for every 8 > 0, there exists 6 > 0 such that if d 1(x, y) < 6, then d2 (f(x), f(y))
0 we may take 6 = 8. If Ix - YI < 6, then 1/(x) - /(y)I = Ix - YI < 6 = s. The function f(x) = x 2 is not uniformly continuous on R. We will show that fore= 1, there is no 6 such that if Ix - yl < 6, then 1/(x) - /(y)I < I.

164

Metric Spaces

For suppose such a

oexists. Choose a positive integer n such that n

Let y)(x

X

= n

+ o/2,y

210- 0/2 ~

2

= n. Then Ix -

+ y) = (o/2)(2n + o/2) ~ I.

YI= o/2< o. But

lx 2

-

y 2 1 = (x -

Therefore / is not uniformly continuous

on R. We will show that a continuous function on a compact metric space is uniformly continuous.

Theorem44.5 If/ is a continuous function from a compact metric space M 1 into a metric space M 2 , then/ is uniformly continuous on M 1 • Let / be a continuous function from a compact metric space (Mi, d 1) into a metric space (M 2 , d2). Let s > 0. For each z e M 1 , there exists Oz > 0 such that if d 1(x, z) < o&, then di(/(x),f(z)) < s/2. The collection Proof.

{B1 .(z) I z e Mi}

is an open cover of M 1 • By Lemma 43.3, there exists o > 0 such that for any x e M, B,(x) c B 1.(z) for some z e M 1• Suppose d 1 (x, y) < o. Now B,(x) c B1.(z) for some z e M 1 • Since x,y e B,(x), we have x,y e B 1.(z). Thus d 1(x, z) < Oz and d 1(y, z) < Oz, and therefore d2 (f(x),f(z)) < r./2 and d 2 (f(y),f(z)) < s/2. Therefore, if d 1(x, y)


IS

L

~

N, then

la1"'>- a1">1 = d({a},{a}:=1 is a

48. Complete Metric Spec..

181

Cauchy sequence in R. By Theorem 19.3, {af">}:,.1 is convergent. We let

a, -- 1· 1m 11... 00 a,(11) .

From equation (46.1), we have

if n

~

N. Thus for any positive integer p, if n ~

N,

p

L l:=1 where T

= e + Lt=1 la1N>j. Taking

lai">I< T the limit on n, we have

for every positive integer p. By Theorem 24.1, {a1J e / 1 • Again using equation (46.1), we have for any positive integer p,

if m,n

~

N. Taking the limit on m, we have

for

n ~

N

Taking the limit on p, we have

d({a1J,{a1">}) = if n ~

00

L l:•1

la1:- a1">1 Se

N, and hence {a},:o. 1 converges to {a1 }:'.,

1

in / 1 • ■

Let (M, d) be a metric space. We say that a metric space is a completion of (M, d) if

Definition46.6 (M., d 1)

1. (M 1 , d 1) is a complete metric space 2. Mc M •. 3. d(x, y) = d1 (x, y) for all x and yin M.

1beorem 46.7 Every metric space (M, d) has a completion. Moreover there exists a completion (Mi, d 1) of (M, d) such that M = Mi, i.e., Mis dense in M 1 (see Definition 47.1). Proof.

Let (M, d) be any metric space. Define

!i'

= {{x,.}:.,,I {x,.}:., is a Cauchy

sequence in (M, d)}

Metric Spacea

182

We say that two Cauchy sequences {x,.} and {y,.} in (M, d) are equivalent and write {x,.} {y,.}if lim,,...ao d(x,.,y,.) = 0. is an equivalence relation on !i', that is, satisfies 1. {x,.} {x,.}for all {x,.}e !i'. 2. If {x,.} ~ {y,.}, then {y,.} ~ {x,.}for {x,.}and {y,.}in !i'. 3. If {x,.} ~ {y,.} and {y,.} ~ {z,.}, then {x,.} ~ {z,.}for {x,.}, {y,.}, and {z,.} in !i'. (1) to (3) are easily verified. The relation ~ partitions !i' into disjoint equivalence classes X, i.e., {x,.} and {y11} are in X if and only if {x,.} ~ {y11}. We define

~

~

~

~

M1

= {XI

Xis an equivalence class of the set !i' under the equivalence relation ~}

We now make M 1 into a metric space by defining d 1 : Xx d 1(X, Y)

=

[O,oo) by

X ➔

Jim d(x,.,y,.) 11-+co

where {x,.};'.1 e X and {y,.}:'- 1 e Y. I. (Mi, d 1) is a metric space. 4. d 1(X, Y) e [O,oo) and is independent of the elements {x,.}and {y,.}chosen from X and Y, since first if {x,.} e X and {y,.}e Y, we have for each e > 0 there exists a positive integer N such that d(x,., x111) < e/2 and d(y,., y,,.) < e/2 for all n,m ~ N. Therefore using the triangle inequality we get CLAIM

+ d(x y + d(y,,., y,.) d(x,,., x,.) + d(x,., y,.) + d(y,., y,,.)

d(x,., y,.) S d(x,., x111)

and

d(x 111, y 111) S

111,

111)

which implies d(x,., y,.) - d(x,,.,y,,.) S d(x,., x,,.) + d(y,,.,y,.) d(x,,.,y,,.) - d(x,., y,.) S d(x,., x,,.) + d(y,,., y,.)

and this implies ld(x,., y,.) - d(x,,.,y,,.)I s d(x,., x,,.) + d(y,,., y..)
0, there exists a Cauchy sequence {x,.}e X such that the diameter of {x.} in (M, d) is less thane, i.e., p({x,.}) < e. Choose any {y11}:. 1 e X. Since {y,.} is a Cauchy sequence, there exists a positive integer N such that for m,n ~ N, d(y,., y,,.) < e. Now the subsequence {yN+il~ 1 has diameter less thane in (M, d). Define {x,.}:_1 by x, = YN+I for i = l, 2, .... Now {x,.} e X since lim,... 00 d(x,., y,.) = lim.... 00 d(YN+ ■ • y,.) = 0. For each X,. in our given sequence choose {x}">}j;,1 e X. such that p({x}">J.:_ Recall di(X,., X111) = 1) < 1/n. lim, .. 00 d(x}">, xi•>). Consider the sequence {x\•>}:_1 • Let e > 0. Since {X,.}:°= 1 is a Cauchy sequence in M 1 , there exists a positive integer N such that if m,n ~ N, d 1(X,., X111) < e/3. Choose a positive integer L, such that L > N and 1/L < e/3. Now for each m,n ~ L we have that d 1(X,., X.) < · 1·1m,.. 00 "'..,(x , x,) < e/3. Th us t h ere exists . a pos1t1ve . . integer . k, B/3, 1.e., such that d(x1">, x1"'>)< e/3. Now d(x\"' 1,

x\"1)

1) + d(x1"1, x\" 1) + d(x1"'1, x1" 1) s d(x\"'>, xi"' I e l e e e

n

y,,.E B,..(y.)

(47.1)

Since 611 s 1/n, we have d(y,,., y 11) < 1/n if m > n, and therefore {y 11} is a Cauchy sequence. Since M is a complete metric space, lim,... 00 y 11 = y for some y in M. It follows from (47.1) that ye B1.,(y.)- for n = l, 2, .... Since 6. s sJ2, we have ye B,,.(y,.)- c: Ba,J1(Y 11)-

for n = 1, 2, .... thus d(x, y) < e.

Therefore yen:=

1

c: B.,.(y.) c: U11

U11• Finally, ye B. 1(y 1) c: B1 (x), and

■

Deflnidon 47.3 Let M be a metric space and let X be a subset of M. We say that X is nowheredense (in M) if x-' - = M.

........

Metric Spaces

168

For example, a point of R is nowhere dense in R. However, a point of P is not nowhere dense in P. These examples show that the concept of a set X being nowhere dense is defined only relative to a metric space of which X is a subset. This situation is different from such concepts as compactness or connectedness. A set X is compact or connected in and of itself. It is immaterial in which metric space it resides. Before giving the next definition we note that a closed set X is nowhere dense in M if its complement is dense in M.

Definition47.4 Let M be a metric space and let X be a subset of M. We say that X is of first category (in M) if X is a countable union of nowhere dense sets. We say that Xis of second category (in M) if Xis not of first category (in M). Just as the definition of X being nowhere dense is relative to the metric space M of which X is a subset, so too are the concepts of first and second category relative to the containing metric space M. It is easy to see that a countable union of sets of first category (in M) is of first category (in M). We pointed out before that a point is nowhere dense in R, and thus any countable subset of R is of first category in R. In particular, the set of rational numbers is of first category in R. If the set of irrational numbers were of first category in R, then R = Q u Q' would be of first category. This is not, however, the case as the Daire category theorem shows. Theorem47.5 (DaireCategoryTheorem) If M is a nonempty complete metric space, then M is of second category (in M). Proof. Suppose that M is of first category. Then M = U:°1 A,., where each A,. is nowhere dense in M. Now M = U:'-1 A; and thus 0 = I (A;'). Now each set A; I is an open dense subset of M; so by Theorem 47.2, 0 = 1 (A;') is dense in M. This contradiction establishes the theorem. ■

n:.

n::

The Daire category theorem has many applications in analysis. The usual application is to show that a point x of a complete metric space exists with a particular property P. A typical argument runs as follows. Let X = {x e MI x does not have property P}. By some argument, we show that Xis of first category. Since Mis of second category, there exists x e Mn X'. Thus x has property P. We give one interesting application of the Baire category theorem.

Definition47.6 Let M be a metric space and let x e M. We say that xis an isolated point of M if for some e > 0, B,(x) n M = {x}.

47. Bair• Category Theorem

189

1beorem47.7 If Mis a complete metric space with no isolated points, then M is uncountable.

U:.

Proof Suppose Mis countable. Then M = 1 {a,.}, and each set is nowhere dense, since M has no isolated points. Thus Mis of first category, and this contradicts Theorem 47.5. ■

Corollary47.8 If Mis a compact metric space with no isolated points, then M is uncountable. Proof

A compact metric space is complete. Now apply Theorem 47.7.

■

Exerclae•

47.1 Use the Baire category theorem to prove that irrational numbers exist. 47.2 Do Exercise 9.10 using the Daire category theorem. 47.3 Prove that {O}is of first category in R, but that {0} is of second category in itself. 47.4 Assuming Theorem 47.S (Baire category theorem), deduce Theorem 47.2. 47.5 If Mis a metric space and Xis a subset of M, we say that Xis a G, set if X is a countable intersection of open sets. Prove that the set of rational numbers is not a G, set in R. 47.6 If/ is a real-valued function which is bounded on an open interval / containing a, we let w,(./) = lub {/(x) I x e /} - glb {/(x) I x e /} and w,(.a) = glb {w,(.J) IJ is an open interval containing a} (a) Prove that/is continuous at a if and only if w,(.a) = 0. (b) Prove that ifs > 0, then {x I w,(.x) < e} is open. (c)• Prove that there is no function / on R such that / is continuous at every rational point and discontinuous at every irrational point. In Section 33 (Exercise 33.4), we defined a function which is continuous at every irrational point and discontinuous at every rational point. 47.7 (a) Let M 1 be a compact metric space and let (M2, d) be a complete metric space. Let 'l(M., M 2) denote the set of all continuous functions from M, into M 2, Prove that d 1(/, g) = lub {d(/(x), g(x)) I x e M1} definesa metric on 'l(M1, M2) and that 'l(M1, M2) is a complete metric space. (b)* Let M 2 and (M1, d) be metric spaces and let/ be a continuous function from M 1 into M 2. We say that/ is an s-mapping if whenever x, x' e M 1 and /(x) = /(x'), we have d(x, x') < B. (''f is within e of being one to one.") Prove the following result due to Hurewicz (see Hurewicz and Wallman, 1941). Let M 1 be a compact metric space and

..

170

Metric Spaces

let M2 be a complete metric space. If for every e > 0, the set G, of a-mappings of M1 into M2 is dense in 'l(M1, M2), then M 1 is homeomorphic to a subset of M2. 47.8* Prove that the closed interval [a, b] cannot be written as a disjoint union of closed intervals of length lea than b - a. 47.9• Lct/(x,y) be a function from R 2 into R such that for each xeR,/.(y) = /(x,y) is a polynomial of finite degree in y (the degree dependent only on x). Also for each ye R, /,(x) = f(x, y) is a polynomial of finite degree in x (the degree dependent only on y). Prove that/(x, y) is a finite polynomial in x and y, that is, for all x and y

VIII Differential Calculus of the Real Line

In this chapter we review many of the topics which are covered in introductory calculus texts.

48.

Basic Definitions and Theorems

Deftnitioa48.1 Let / be a real-valued function defined at all points of an open interval (a, b) in R. For each x in (a, b), we define

f'(x)

= lim/(y) 7-0"

- /(x) y - X

if this limit exists. lf/'(x) is defined, we say /is differentiable at x, and we call the numberf'(x) the derivative off at x. lfj'(x) is not defined, we say /is not differentiable at x. If/ is differentiable at each point of (a, b), we say that/ is differentiable on (a, b). In dealing with functions which are defined only on closed intervals [a, b] it is sometimes useful to use the concept of right and left derivatives at the endpoints band a; so we have the following definition. Deftnitioa48.l Let / be a real-valued function defined at all points of a closed interval [a, b] in R. For each x in [a, b], we define J:(x)

=

lim f(y) - f(x) y - X

7-0"+

and

J;(x)

=

lim /(y) - /(x) , .. ". y - X 171

172

Differential Calculus of the Real Line

if these limits exist. J;(x) and f,(x) are called the right and left derivatives of

J at x (when they are defined). By Theorem 32.2 we see that f'(x) exists if and only if J;(x) and J,(x) exist and/;(x) = f,(x) [ =f'(x)]. We say a real-valued function/is differentiable on the closed interval [a, b] with a < b if /is differentiable on (a, b) andJ;(a) and/,(b) both exist.

Examples48.3 (a) Let.f(x) = lxl on [ -1, 1).Then/;(O) = 1 and/,(0) = -1 both exist, but/'(O) does not exist, since/;(O) :f,:/',(O). Note that f'(x)

=-

f'(x)

=1 = -1

J:(-1)

1

for all -1 < x < 0 for all O < x < 1 and

f:(l)

=1

(Verify.) (See Figure 48.1.) (b) Let 0 for - 2 ~ X S - 1 J(x) = { +Jl - t' for -1 < x S I (See Figure 48.2.)

(1,1)

(-1,1)

-1

Figure 48.1

1

0

I I

(0,1)

I

------+---1

-2

I

I

1

--Figure 48.2

48. Basic Definition• and Theorems

173

Verify that J;(-1) = 0 and J;(-1) does not exist and that therefore f'(-1) does not exist. Verify that /'(x) = 0 for all -2 < x < - l and f'(x) = -x/ ✓ 1 - x 2 for all -1 < x < l. Also show that/;(l) does not exist.

If/ is differentiable at a point c, then/ is continuous at c.

1beorem48.4

Proof.

We are given that lim/(y) 7 ➔ c

- f(c) = f'(c) y - C

exists. But

lim/(y)

- f(c) = lim/(y)

7 ➔ c

7 ➔ c

- f(c) •(y - c) Y- C

= lim/(y)-/(c).lim(y7 ➔ c

So lim,

➔ cf(y)

= f(c),

Y-

C

c) =f'(c)•0

= 0

7 ➔ c

which says/ is continuous at c. ■

The converse of Theorem 48.4 is not true. For example, Example 48.J(a) is continuous at 0, but not differentiable at 0. Theorem48.5 Suppose f and g are defined on an open interval (a, b) and are differentiable at a point c in {a, b). Then/+ g, f· g, and fig [g(c) :p 0)

are differentiable at c and

+ g)'(c) = f'(c) + g'(c) (/•g)'(c) = f'(c)•g(c) + f(c)•g'(c)

(a) (/

(b) (c)

(!.)'(c) =

g(c)•f'(c)

g

Proof.

- g'(c)•f(c) [g(c)]l

[g(c) "F OJ

(a) is left as an exercise.

'( ) (b) (/ •g ) c

. (/•g)(y) - (/·gXc) Y- C

= 1, 1m ➔ c

= lim (t(y) , ➔ c

=f(c)•g'(c)

[g(y) - g(c)] Y- C

. f(y)•g(y) = ,11m ➔ c Y-

+ g(c)

f(c)•g(c) C

[/(y) - /(c)]) Y- C

+ g(c)f'(c)

The last equality holds by Theorem 48.4 and the algebraic properties of limits.

174

Differential Calculus of the Real Line

(c) We have

(£)'

(c)

= Jim (f!gXy)

g

- (f/g)(c)

Y-

7-+c

= Jim [ 1 .. c

-

= lim/(y)/g(y)

C

7-+c

- f(c)/g(c) Y- C

1 . ( (c) /(Y) - f(c) _ f(c). g(y) - g(c))] g(y)•g(c) g Y - C y - c - f(c)•g'(c)

g(c)•f'(c)

•

[g(c)]2

Examples. (a) If /(x) = c for all x e R (where c is a fixed constant), then f'(x)

= 0 for all x e R, since f'(x)

= Jim /(y)

- f(x) y - X

7-+1&

(b) If /(x)

==x for all x e R. Then/'(x) f'(x)

= Jim ~ 7-+1&

y -

==0 X

==1 for all x e R, since

==lim/(y) - f(x) ==Jim Y - x ==1 y -

7-+1&

X

7-+1&

y -

X

(c) If /(x) ==x 2 ==x· x for all x e R. Then by Theorem 48.5(b) f'(x) == 1 · x + x · 1 ==2x for all x e R. Using induction on n and Theorem 48.5(b ), we may prove the following. (d) If /(x) ==x' for n a positive integer and for all x e R, then f'(x)

= nx'-1

for all xe R

Theorem48.6 (Ibe Chain Role) Suppose f is a function on (a, b) andf'(c) exists at some c e (a, b). Suppose g is defined on an interval/, which contains the range off, and suppose g is differentiable at the point f(c) in I. If we define K(x) ==(g O fXx) ==g(f(x)) for all x in (a, b), then K is differentiable at c and

==g'(f(c))·f'(c)

K'(c)

Proof.

K'(c)

= Jim K(y) 7-+c

- K(c) ... Jim g(f(y))

Y-

C

7-+c

- g(f(c))

Y-

C

CASE 1. If there exists an e > 0 such that for all y with O < IY - cl < e, we have /(y) #: f(c), then since f is differentiable at c, f is continuous at c (Theorem 48.4); so

lim/(y) 7-+c

= f(c)

48.

Beale Definition• end Theorem•

176

Therefore K'(c)

= lim g(J(y))

- g(J(c)) == Y- C

r•c

==

lim

g(J(y)) - g(J(c))

Y-

r•c

O< 17-cl

X

/(xi:) - f(x)

/i-(x) == lim g(f(x,:)) - x X

==

Ii

s-o/(%)

m

r-o/(%)

f(x.) - f(x)

1 1 .,. -[/(x.) - /(x)]/(x. - x) f'(x)

The last equality follows by (49.5) and the fact that/'(x)

exists.

■

If we had known that g was differentiable at/(x), since

g(f(x)) = x

for all x in (a, b)

(49.6)

we could have applied Theorem 48.6 to the left side of (49.6) and usual differentiation rule on the right to get

g'(f(x))-f'(x) or

= 1

g'(f(x)) ..,.f'~x)

(49.7)

...

182

Dlfferentlal Calculus of the Real Line

= x2

Example. Let f(x) g(y) =

= g'(f(x)) = f'~x) =

g'(y)

= x.

since Jy = g(y) g'(f

= 2x > 0 for all x in (0, 2). = (0, 4). By Theorem 49.10,

on (0, 2). Then f'(x)

Jy is the inverse off on (f(0),/(2))

=

✓ "J)

t/2

! = 27Y

For example, if y = 3, then /( ✓ "J)

= 3; so g'(3) =

✓ l

The following theorem, which arises from Theorem 49.5, gives a simple procedure for the evaluation of limiting values of quotient functions.

1beorem49.11 (L'Hospltal'sRule) Let f and g be differentiable on the interval [a, b), with g'(x) ::/:0 on [a, b). (i)

If

lim f(x) ➔

(ii)

If

➔

lim g(x)

= oo

➔

. f'(x) bm --,--() g

X

lim f(x) g(x)

z ➔ b-

or

= oo

lim g(x) z

z ➔ b-

=0

b-

and

b-

and if then

and z

lim f(x) z

=0

b-

z

➔

b-

=L

=L

The end "point" b may be finite or "oo," and L may be finite or "oo."

Proof. We will prove the theorem under hypothesis (i), and we shall assume that b is finite; therefore for convenience we take b = l. Since limz ...1 - /(x) = limz ...1 - g(x) = 0, we define /(I) = g(l) = 0 and therefore have f and g continuous on [a, I]. Applying Theorem 49.5, we have that for each a < x < l, there is a point x• with x < x• < I such that f'(x*) g'(x*)

/(I) - f(x) = g(l) - g(x)

=L

lim f'(x*)

Since

f(x) = g(x)

z• ➔ 1 - g'(x*)

we may choose {),.> 0 so that

/(x) 1g(x)

LI=1/'(x*)- LI) (c) /(0)

«> )

=0

(d) f'(x) is increasing on (0, «>). Define g(x) = f(x)/x for all x in (0, Q) ).

49.5

Evaluate . 1 - cos x 2 (a) hm

.,... o

- g(b)f'(c)] g(aXf(b)h'(c) - h(b)/'(c)J

•

X 3 SID X

«>)

and prove that g is increasing on (0,

186

60. Taylor'• Theorem (b)

Jim

.....o+

r

(1be functions sin x, cos x, log x and

r

are defined in Chapter XI.) 49.6 Suppose/ is defined on [a, b] and c e (a, b) and suppose/' exists in an open interval containing c and that (/')'(c) exists. Show that lim f(c

+ 2h) -

2ji(c

+ h) + f(c) = (/')'(c)

11--0

Give an example to show that the limit above may exist even though /'(c) does not exist. 49.7 Let/be a function on (a, b) such that/' is bounded on (a, b). Prove that there exists c '4=0 such that g(x) = x + cf(x) is one to one. 49.8 Suppose that/is nonnegative andr exists on (0, 1). Suppose also that there exist points c '4=d in (0, 1) such that /(c) = 0 = /(d). Prove that r(x) = 0 ~or some x e (0, 1).

50. Taylor's Theorem In this section we look at the problem of approximating a real-valued function /by a polynomial function P. That is, given/ on (a, b), find a polynomial P such that 1/(x) - P(x)I is small for all x in (a, b). Of course, there are functions / where it is impossible to find P as above, but we shall develop a method applying to certain types of functions/, which produces a polynomial P, where we have some information about the accuracy of the approximation.

Deftnitioa50.1 Let/be

a real-valued function on (a, b).j< on a subset of (a, b) defined by / is the function j< 2 >(x) = (/,•••

are defined similarly as the successive derivatives of/ on (a, b). For convenience we definej< 0 >(x) = /(x) for all x in (a, b).

We always have (a, b)=> domain J=,....

/=, domain j< 2>=,· · · => domain

Deftnidoa50.2 A real-valued function/ is said to be of class c• on (a, b)

if j(x) exists and is continuous for all x in (a, b). Clearly, if/ is of class c•, then / must be of classes c•- 1 , c•- 2 , ••• , If / is of class c• for all positive integers n, then/ is said to be of class C..,. Let/ e c• on (a, b) and let c e (a, b). Among all the polynomials of degree

c•.

...

188

Differential Calculus of the Real Line

n, there is exactly one polynomial P,, such that P!,">(e)= J(e) , A,,= - 1-

n.

Therefore the nth degree Taylor polynomial for/ at e is P,,(x)

J(e)

= f(e) + /(t) =

g(x) - (n + l}!M

for all x in (a, b)

(50.7)

Therefore, our proof will be complete if we can find a number t between c and d such that g(c) = J(c)= 0 that g(d) = 0. But since g(d) = 0

and g(c) = 0, Our choice of M shows the mean-value theorem states there is a number t 1 between c and d such that g 0. Since/is continuous on [a, b],/is uniformly continuous on [a, b] (Corollary 44.6); so there exists 6 > 0 such that if Is - ti < 6,

197

61. Integration with an lncreaalng Integrator

then 8

l/(s) - /(t)I < 2[a(b) - a(a)]

Let P = {x0 , Xi, l, ... ,n.Then

•••

,

x,.} be a partition such that x 1 - x 1_

1

< ~

for i =

8

M, -

for i

=

1, ...

m, S

2[a(b) - a(a)]

, n (verify). Thus II

U(f, P) - L(f, P)

= L (M 1 I• 1

8

s---2[a(b)-

m 1)&a 1

r &a, a(a)] "

1• 1

8

=-- < 2

8

By Theorem 51.8,/e ~Ja, b]. ■ The remainder of this section is devoted to deriving properties of the set ~Ja, b] where a is an increasing function on [a, b].

Theorem51.11 Let a be an increasing function on [a, b]. If/ e £1.[a,b], then/ e ~Jc, d] for every closed subinterval [c, d] of [a, b] and

for all c in (a, b).

Proof. Let c e (a, b). Let P be a partition of[a, b] and let P* = Pu {c}. Then P 1 = P* n [a, c] and P2 = P* n [c, b] are partitions of [a, c] and [c, b], respectively. Now

U(f, P)

~

U(f, P*)

= U(f,

P 1)

+ U(f,

P 2 ) 2!: rf

It follows that

J:/da

2!: rfda

Similarly,

+ rfda

f.1t1a s J:1t1a + £1t1a

da + rfda

198

The Rlemann-Stleltj•

Therefore,

Integral

rfdar/dar/dar/da +

2!

2!

+ i:1c1a 2! J:1c1a + £1c1a 2! f.1c1a

r/da = l:fda

Thus

r/da = if®

and

and hence/e lf.[a, c] n lf.[c, b]. Furthermore,

Let [c. d] be a closed subinterval of [a, b]. By the above result f e lf .[a. d] and another application of the above result yields/ e lf.[c. d]. ■

Theorem51.12 Let f be a bounded function on [a. b] and let exbe an increasing function on [a, b]. Suppose that m s /(x) s M for all x in [a, b]. If f e lf.[a, b] and g is continuous on [m. M]. then g O f e lf.[a, b]. Proof. Let h = g f and let k satisfy k > lg(t)I for all t in [m. M]. Let e > 0. Since g is uniformly continuous on [m, M]. there exists 6, 0 < 6 < e/ (2k + cx(b)- cx(a))such that if Is - ti s b, then O

lg(s) - g(t)I < 2k

e

+ cx(b)-

cx(a)

Since f e lf.[a, b]. by Theorem 51.8, there exists a partition P

(51.1)

= {x0 ,

x., ... , x 11} of [a, b] such that U(f, P) - L(f, P) < 62

M 1 = lub {/(x) I x e [x1- u xJ}

Let

m1 = glb {/(x) I x e [x,_ 1 , xJ}

M 1*

= lub {h(x) I x e [x,_

mr= glb {h(x) I for i

=

l, ...

XE

i,

xJ}

[x,_ 1• xJ}

, n. Let A

= {i I I s i s

B

= {i I I

n and M 1 - m 1

:s; i :s; n and M 1 - m 1 2! 6}

Ifie A, we have by inequality (51.1) that M 1* If i E B. we have M,* :s; 2k. Now

mr

< b}

m1:s; e/(2k + cx(b)-

cx(a)).

61. Integration with en lncrNalng Integrator

6

> Acx,S ~ > (M, ,;. =

- m,)Acx,S

199

• L (M, ,-1

- m 1)Aa1

U(f, P) - L(.J.P) < 62

and it follows that

Therefore, U(h, P) - L(h, P) ..

L•

,_ 1

(M,* - mr)acx,

mr)aa, +~

=~

(M,* -

s 2k

+ cx(b)-

8

cx(a)~ Aa,

8

:s;;2k

(M,* - mr)acx,

+ 2k ~

Aa,

•

+ cx(b)- a(a) 1~ 1 Acx,+ 2kb 8

< 2k + cx(b)- cx(a)(cx(b)- a(a)) 8 + 2k-----2k + a(b) -

By Theorem 51.8, h e tf Ja, b].

a(a)

=8

■

Theorem51.13 Let exbe an increasing function on [a, b]. If f and g are in tlJa, b] and c e R, then

f. = f. f.(/ =

J:f da. +

g da..

(iv) If f(x) s g(x) for all x in [a, b], then

f da. s

g da..

(i) cf e R.[a, b] and (ii) f (iii)

cf da.

+ g e R.[a, b] and

c

f da..

+ g) da.

f· g e R.[a, b].

(v) lfleR.[a,b]andlf.fda.ls

f. f. f.

f.1flda.,

Proof. (i) Let g(x) = ex. By Theorem 51.12,g O f(x) Suppose c ~ 0. Let P be a partition of [a, b]. Then U(cf, P)

= cU(f,

P)

~

c J:f dx

= cf(x) is in tlJa,

b].

200

The Rlemann-StleltJes Integral

and it follows that

r.

~

cf da

J: f.

Similarly,

C

f.tda

cf da Sc f.tda

Therefore,

=C

cf da

J:/da

The case c < 0 is treated in a similar way. (ii) For any partition P of [a, b], we have

I.JJ, P)

+ L(g, P)

5

I.1J+ g, P) s U(f + g, P)

5

U(f, P)

+

U(g, P)

(verify). Let e > 0. By Theorem 51.8, there exist partitions Sand T of [a, b] such that

U(f, S) - I.JJ, S) < Let P

=S u

e

2

and

u(g,

n - L(g, n < 2 8

T. Then we have

+ g, P)

U(f By Theorem 51.8,/

-

e

e

I.JJ+ g, P) < 2 + 2 = e

+ g e lf.[a, b]. Also

U(f, P)

s

U(f, S) < I.JJ, S)

+;s

f.tda

+;

Similarly, Therefore,

J:(/+

g) da

s

U(f

+ g, P) s

< f.tda

U(f, P)

+

+ f.uda + e

Since e > 0 was arbitrary, we have

f.(/+ Similarly,

g) dcxs

f.(f+g)da~

J:/da + J:

g da

f.tda+

f.udcx

U(g, P)

61. Integration with an Increasing Integrator

and therefore

201

J:(/+ g) = J:tda + f. da

g da

(iii) It follows from Theorem 51.12 that if/ e lf.[a, b], then / 2 e lfJa, b] for if/ e lf.[a, b], since g(x) = x2 is continuous, we have g O f= / 2 e lf.[a, b]. Now let/, g e lf.[a, b]. By parts (i) and (ii),/+ g,f - g e lfJa, b]. By the g) 2, (/ - g)2 e lf.[a, b]. By parts (i) and (ii), result above(/+

f•g

= ¼((/ + g)2

is in lf .[a, b]. (iv) Suppose/ e -'-'.[a, b] and/(x) [a, b]. Then ~

_ (/-

g)2]

0 for x in [a, b]. Let P be a partition of

~ UJ, J:/da

P)

~

0

If/, g e lf.[a, b] and g(x) ~ f(x) for x in [a, b], then g(x) - f(x) in [a, b] and by our result above ~

0 for x

J:Cg-/)da~O By parts (i) and (ii)

J:gt1a-J:1t1a

f.cg-f)da=

J:

and therefore (v) Since g(x) equation

g da

= lxl is continuous,

~

f.

f da

g O f=

1/1 e -'-'.[a, b]. Define c by the

(The number c is either I or -1.) Now cf(x) s 1/(x)I for all x in [a, b]; so by part (iv) we have

Using part (i) we have ■

Theorem51.14 (Mean-ValueTheoremof tbe Integral)

If/ is continuous on

The Rlemann-Stleltjea Integral

202

[a, b] and a. is increasing on [a, b], there exists c e [a, b] such that

J:/

= f(c)[a.(b) -

dt,.

a.(a)]

Proof. If a.(b) = a.(a), then any value of c in [a, b] gives the desired conclusion. Suppose a.(b) > a.(a).By Theorem 34.4,/attains its maximum Mand minimum m on [a, b]. For the partition P = {a, b} we have

m[a.(b) - a.(a)] = L(.f, P) S

J:/

dt,. s U(f, P) = M[a.(b) - a.(a)]

~

Therefore,

m S a.(b)•_ a.(a) S M

By the intermediate-value theorem (Theorem 45.6), there exists such that

c

in [a, b]

•

c-~

/( ) - a.(b) - a.(a)

Exercises

In Exercises 51.l to 51.5/is a bounded function on [a, b] and a.is an increasing function on [a, b]. 51.1 Prove that if Pis a partition of [a, b], then L(.I, P) ~ U(f, P). 51.2 Prove that if P and Sare partitions of [a, b] and Sis a refinement of P, then L(f, S) ~ L(f, P), thus completing the proof of Lemma 51.5. 51.3 Prove that/ e dl.[a, b] if and only if for every e > 0, there exist partitions P and S of [a, b] such that U(f, P) - L(f, S) < e 51.4 Let c e (a, b). Prove that

J.fd« =

51.5 Prove that

r/

dac+

LI

dac

and

J"/ ti« = JcI ti« + J"/ dac

,

,

n f.1t1«= -I:dcx

thus completing the proof of Theorem 51.13(i). Let ct be an increasingfunction on [a, b]. Prove that if/, g e aJa, b], then

J: (I+ u> dcx~ J:/dcx + J:u dcx thus completing the proof of Theorem 51.13(ii). 51.10 (a) Prove that if/ is a continuous function on [a, b], there exists a point c in (a, b) such that J:/(x) dx = /(cXb - a) (b) Give an example to show that for an arbitrary increasingfunction cton [a, b], part (a) may fail. 51,11 Prove that if/ is a continuous nonnegative function on [a, b] and J:t(x) dx =0, then / (x) = 0 for all x in [a, b]. 51,12

Let

ct(X)= /(x) = {~ g(x)

= {~

(a) Is/eaJ0,2]?Ifso,compute (b) Is g e aJO,2]? If so, compute 51.13

J:/dcx.

J:

g dct.

(c) Do (a) and (b) above with ct(x)= x. Let ct1 and ct2 be increasingfunctions on [a, b]. (a) Prove that ct1 + ct2 is increasingon [a, b]. (b) Let/e a. 1[a, b]" a. 2 [a, b]. Prove that/e a.

J:/d(ct1 51.14

if0S:x P), then

n - /I
0. By Theorem 51.8, there exists a partition P such that U(f, P) - L(J, P) < s Let P* be a refinement of P. Then U(f, P*) - L(J, P*) ;s; U(f, P) - L(f, P) < e

nS

L(J, P*) ;s; S(f, P*,

Now

L(J, P*)

and and hence

~

U(f, P*)

J:/

da. ;s; U (f, P*)

Is(/.n - J:1 dal< P*,

e

for any points

r

It follows that / = I!f da.. Now suppose the condition holds. We may assume a(b) > a(a). Let e > 0. There exists a partition P = {x0 , x 1 , ••• , x 11} of [a, b] such that

IS(f. P,

n - II < e4

for all points T. For each i, I :S i ;s; n, there exists If e [x,_ 1 , xJ such that

e

M, - 4[a(b) - a(a)]

Thus

U(f, P) -

Let T*

= {tr,

...

8

J:.M, -

4 ==

II

(

•

< f(t, )

f:

)

4[a(b) - a(a)] aa, :S

, 1:}. Then U(f, P) -

e

4 ;s; S(f,

P, T*)

Similarly, there exist points T** such that S(f, P, T**)

~ L(f,

P)

+

i

II

l~l

f(t,*)aa,

The Rlemann-Stleltjes Integral

208

Therefore, U(f, P) - L(f, P)

$

[s(f, P, T*) +

i]+

[-s(f,

P, T**) +

s IS(f, P, T*) - S(f, P, T**)I +

i]

8

2

s IS(f, P, T*) - II + IS(/, P, T**) - II + ; 8

8

0, there exists 6 > 0 such that for any partition P of [a, b] with norm P < 6 and for any points T, we have IS(f, P, T) - II < e

We cannot prove a theorem analogous to Theorem 52.2 for the kind of convergence described in Definition 52.3. That is, convergence of RiemannStieltjes sums as described in Definition 52.3 is not equivalent to the existence of the Riemann-Stieltjes integral. For example, let if-lsx a.(a). Lets > 0. Since/is continuous on [a, b],/is uniformly continuous on [a, b]. Thus, there exists b > 0 such that if Is- ti S b, then 8

1/(s) - /(t)I < 2(a.(b)- ex(a)]

208

The Rlemann-Stleltjea Integral

Let P = {x0 ,

Xi, •..

,

x,.} be a partition with norm P < s(f, P,

o.Let

n = E• 1 0. There exists a partition Q = {y0 , Yi, ••• , y11} of [a, b] such that U(f, Q)
0. Choose a number k

62. Rlemenn-Stleltju Suma

209

such that/(x) + k > 0 on [a, b]. By the above argument, there exists c51 > 0 such that if norm P < c5.,then

U(f Since U(f

+ k, P) =

+ k, P)
0 such that if norm P < c52 , then U(-f, P) < Since U(-f, P)

=

r(-/)da.+;

-L(f, P), we conclude that L(f, P) >

J:/da -;

if norm P < c52 • Let c5= min {c5.,c52 }. If norm P
dot+ f:g dat

If/e RS[a, b), then c/e RS[a, b] and

f>f=f:f C

(c) If/, g e RS[a, b] and/(x} ~

g(x) for all x e [a, b], then

f:1dot~ f: dot g

(d)

Prove or disprove: if/ e RS[a, b] and at is increasing on [a, b), then/ e RS[c, d] for every subinterval [c, d] of [a, b] and

f:fdot=f:idot + f:idot 53.

Riemann-Stieltjes Integration with Respect to an Arbitrary Integrator

In this section we extend the definition of the Riemann-Stieltjes integral I!f da. to arbitrary (not necessariiy increasing) functions ex.Motivated by Theorem 52.2, we make the following definition.

Definition53.1 Let f and exbe bounded functions on [a, b]. We say that f is Riemann-Stieltjes integrable with respect to exon [a, b], and write/ e lf«[a, b], if there exists a number / having the property that for every e > 0, there exists a partition P of [a, b] such that if P* is a refinement of P (P* => P), then IS(f. P*,

n - II




Proof. then

211

0. There exists a partitionP

f.

cfdr,. =

1

C

f.f

of[a, b]suchthatif

dr,.

P* => Pi,

js(f, P*, n - J:1t1aj P 2 , then

!scg, P*, n - J: g dal P, then

+ g e lf.[a, b] and

J: + (f

g) dcx=

J: + J: f dcx

g da ■

The second equation is proved in a similar way.

Our next theorem demonstrates that the roles off and exmay be interchanged. Theorem 53.3 (Integration by Parts) Let f and exbe bounded functions on [a, b]. If f e lf.[a, b], then exelf Jla,b] and

J:/

da

Proof. then

= f(b)cx(b) -

J:

f(a)cx(a) -

exdf

Let e > 0. There exists a partition P of [a, b] such that if P* => P,

II

Let

S(rl, P*, T**)

= L

cx(t,)[f(x1)

f(x,_

-

1)]

I• 1

be a Riemann-Stieltjes sum for ex,where P* => P. Now

f(b)cx(b) - f(a)cx(a) =

II

II

1•1

1=1

L f(x1)cx(x1)- L f(x1-

1>cx(x,_ 1)

212

so

The Alemann-Stleltjes Integral

/(b')a.(b) - /(a')a.(a) - S(oc,P*, T**)

- L" /(x 1•1

)[oc(xJ- oc(t1)] + " f(x,_ 1)[oc(t1)

L

1

-

1•1

which is a Riemann-Stieltjes sum S(/, p••, T 0 t 2 , ••• , t,.}. Now P 0 :::>P and we have

js(oc,p•, T**) - [t(b')a.(b) - f(a')a.(a) -

=

•),

oc(x,_1)]

= p• u {ti,

where po

J:/ I da]

IJ:1

doc- S(f, p••, T 0

•)1
1, ot is constant. Prove that if p = 1, then ote BV[a, b].

&&. Integration with Functions of Bounded Variation

219

(b) Give an example of a function otwhich satisfies a uniform Lipschitz condition of order p < 1 on [a, bl, but ot'1BV[a, b]. (c) Give an example of a function « e BV[a, b] which satisfies no uniform Lipschitz condition on [a, b]. (d) Prove that if ote BV[a, b] and if g satisfies a uniform Lipschitz condition of order 1 on [c, d] :J ot{[a,bD,then g o ote BV[a, b]. (e) Use (d) above and an argument like that given in the proof of Theorem 51.13 to prove that if«, Pe BV[a, b], then «· P,lotl,and 1/P(if P is bounded away from zero) are also in BV[a, b].

55.

Riemann-Stieltjes Integration with Respect to Functions of Bounded Variation

In this section we extend the results of Section 51 concerning integration with respect to increasing functions to integration with respect to functions of bounded variation. Our main tools are Theorem 54.8 which states that if a e BV(a, b], then a is the difference P - y of the increasing functions P(x) = V!a and y = P - a and Theorem 55.2 which states that if r./1

■

dP

The last theorem of this section shows, that in certain cases, a RiemannStieltjes integral may be reduced to a Riemann integral. Theorem 55.7 Let exbe continuous and differentiable on [a, b]. If/, a' e lf[a, b], then/ e lf.[a, b] and

J: / da = J: /(x)cx'(x)

dx

(Note since a' is bounded on [a, b], then exe BV[a, b].) Proof. LetP(x) ==x. By Theorem 55.5,/·ex' e lf[a, b]. Suppose 1/(x)I < M for all x in [a, b]. In this proof we let S7 denote the Riemann-Stieltjes sum for the integrator y. Let a > 0. There exists a partition P of [a, b] such that if p• =>P, then

js,(fa', p•, r•) - J: /(x)cx'(x) dxl < i 1s,(a', P*, T**) - J: ex'(x)dxl < 4~

and Let P*

=

{x0 ,

Xi,

•••

,

x,.} =>P and let t 1 e [x1_i, xJ for i

(55.5) (55.6)

=

1, ... , n.

&&. Integration with Function• of Bounded Variation

= {t1 ,

Let T

•••

,

223

We show that

111}.

js.(J, p•.n - f. f(x)a.'(x) dxl < e

thus establishing the theorem. By the mean-value theorem, there exist s; e [x 1_ i, xJ such that .da1 = «(xJ - «(x,_ 1) = «'(s,)(x, 1) for i = I, ... , n. Let

x,_

v,

for i

=

={

s,

if «'(s 1)

-

«'(tJ

II

if «'{s1)

-

a.'(ti) < 0

if «'{s1)

-

«'(t 1)

if «'(s 1)

-

a.'(t1) < 0

= {vi, ...

1, ... , n. Let T*

II

L

l=l

, v11} and

~

0

:2:::0

r•• = {ui, ...

II

la.'(s1)

x,_ = L [«'(v

«'(t 1)l(x, -

-

1)

l•l

1) -

= s,(«', P*,

«'(u 1)](x 1 -

, u,.}. Then

x,_ 1)

T*) - s,(a.', P*, T**) (55.7)

by inequality (55.6). Now

js.(J, P*, n - J:f(x)a.'(x)

If

l,t

dxl ==

f(t,) A«, - J:1(x)«'(x) dxl

f(t,)a.'(t,Xx, - x,_

==

I• 1

+

f /(t )[«'(s 1

l•l

~

+

L•

I• 1

£

IJ(t 1)lla.'(s1)

= J: + /(a)

d«

!5.6

Let

ci

Jim

r--o

I (b)

Jcd«

S(/, P, T)

= J• Id« •

Let ci e BV[a, b] and let/ be a bounded function on [a, b]. Prove that if Jim,,.,.. ho S(f,P, T) exists, then/e 91.[a, b] and Jim

aora

55.8

d«

e B V[a, b] and let / be continuous on [a, b]. Prove that ..,.

!5.7

+ A(a)/(a)

dx

1

S(f, P, T)

l'-+0

Let ci e BV[a, b] and suppose that /e 91.[a,b], then Jim

.. ,. , ...o

ci

= ctId«

J•

is continuous on [a, b]. Prove that if

S(f, P, T)

=

J•Id« •

Let ci e BV[a, b]. Prove that if/, g e 91.[a,b], then max{/, g} and min{/, g} are in 91.[a,b]. 55.10 Let ci e BV[a, b]. Suppose that/e 91.[a,b] and there exists a positive number M such that 1/(x)I~ M for x in [a, b]. Prove that 1/fe 91.[a,b]. !5.9

58. The Riemann Integral

In this section we prove the fundamental theorem of integral calculus and a change of variable formula for the Riemann integral.

Theorem56.1 (FundamentalTheoremof lntepal Calculus) (i) Let/be continuous and differentiable on [a, b] and suppose/' e 9'[a, b]. Then

J:

f'(x) dx

(ii) Let/ e 9'[a, b]. Let F(x)

for x in (a, b] and let F(a)

=

= /(b)

- /(a)

J:

/(t) dt

= 0. Then Fis

continuous on [a, b] and Fis

.....

The Rlemann-StleltjN Integral

228

differentiable at every point at which/is continuous. If xis such a point, we have F'(x) = f(x). Proof. Part (i) restates Corollary 55.8. Suppose 1/(x)I < M for all x in [a, b]. Lett > 0. Let 6 6 with x < y, we have IF(x) - F(y)I

=

IJ:

/(t) dtl

~

11/(t)I

J:

~

dt

M dt

= e/M. If Ix =

YI
0. There exists 6 > 0 such that if IY - xi < 6, then 1/(y) - f(x)I < s. Suppose 0 < IY - xi < 6. Without loss of generality, we assume y > x. (The case for y < xis similar.) By Theorem 51.l3(iv), since /(x) - e < f(z) < J(x) + s for all z between x and y, we have [/(x) - s](y - x)

~

F(y) - F(x)

) Ji(x-e~----~

Therefore,

=

J:

~

/(t) dt

[f(x)

+ e](y -

x)

Ji(x+e )

F(y) - F(x) y-x

which implies that F(y) - F(x) - f(x)I I y-x

and thus F'(x)

= f(x).

~

e

■

Corollary56.2 Let/be a continuous function on [a, b]. (i) There exists a function Fsuch that Fis continuous on [a, b], differentiable on [a, b], and F'(x) = f(x) for all x in [a, b]. (ii) If G is continuous on [a, b], differentiable on [a, b], and G'(x) = f(x) for all x in [a, b], then

J:

f(x) dx

= G(b)

- G(a)

Proof. Part (i) follows from Theorem 56.1(i). Let G be as in part (ii). By Theorem 56. l(i),

J:

/(x) dx

=

J:

G'(x) dx

= G(b)

- G(a)

■

In our subsequent work we will need the following change of variable formula. Another change of variable formula where g is not assumed to be increasing is given in Exercise 56.3.

68. The Riemann Integral

227

Theorem56.3 Let/ e bt[a,b] and let g be a strictly increasing function from [c, cl) onto [a, b] such that g is differentiable on [c, d] and g' e bt[c, d). Then (/o g)·g' e bt[c,d] and

J:f(x) dx= J: /(g(t))g'(t)

dt

Suppose 1/(x)I < M for x in [a, b]. Let e > 0. There exists a partition P 1 of [a, b] such that if Pf => P., we have Proof

lsu;

PT,T1}

J:/(x)< j dxl

-

There exists a partition P 2 of [c, d] such that if P! => P 2 , we have

IS(g', Pt T

2) -

g'(x)

[

dxl< 3~

(56.1)

Since g is a strictly increasing function from [c, d] onto [a, b], it follows that P 3 = g- 1(P 1) is a partition of [c, cl). Let P = P 2 u P 3 • Let P* = {y0 , y., . .. , y11} be a partition of [c, d] containing P. Let s 1 e [y,_., yJ for i = l, ... , n and let T* = {s., s 2 , ••• , s11}. We show that

IS{(/ g)•g', P*, T*} 0

J:/(x)< dxl

which will prove the theorem. Let x 1 = g(y,) for i = 0, ... , n and t 1 = g(s,) for i P!

= g(P*) = {x 0 , x.,

...

e

= l,

... , n. Then

, x 11}

is a partition of [a, b] which contains P 1 and t I e [x,- 1 , x J for i Therefore,

= 1, ... , n. (56.2)

where T1 = {ti, ... , t11}. By the mean-value theorem x 1 1 = g(y,) - g(y,_ y 1_ 1) where u, e (Y,-i, y,) for i = 1, ... , n. Now

x,_

S((/o g)•g', P*, T*) -

=

L f(g(s,))g'(s,) II

L f(g(s 11

1"'1

L\y 1 -

J"f(x)

(56.3)

dx

•

1))g'(s 1)

= g'(u1XY1 -

J:/(x) dx

I• 1

=

1)

L\y 1 -

L f(t,) II

l•l

L\x,

+

L f(t,) II

l=l

L\x, -

J"f(x)

dx

a

A

.

228

The Rlemann-StleltJ•• Integral

•

L f(g(s,))g'(s,)

=

•

L f(g(s 1))g'(u 1) Ay,

Ay 1 -

I• 1

+ S(f, PT, T1)

'"" 1

- J:/(x) dx = L• f(g(s,))[g'(s 1)

-

I• 1

g'(u,)] Ay 1 + S(f, PT, T1 )

-

Jbf(x)

dx

11

We have

If

f(g(s,))[g'(s,) - g'(u,)]

I• 1

Ay,I :s; f

1/(g(s,))llg'(s,) - g'(uJI Ay,

I• 1

=s;M

s

•

L lu'(s,) I• 1

g'(u,)I Ay,

M(~)-2.e 3M 3

(56.4)

where we have used inequality (56.l) and an argument like that given in the proof of Theorem 55.7. Inequalities (56.2), (56.3), and (56.4) combine to give IS((/

0

g)•g', P*, T*) -

J:

f(x) dxl < e

For example, if f e at[a, b] and we let g(t)

J:

f(x) dx

=

= t + k,

r.:: f(t

■

we have

+ k) dt

Exercises

56.1 Let I be an interval of R and Jet / be a function on I. We say that / is absolutely continuouson / if for every e > 0, there exists o > 0 such that if {(01:,b1)};. 1 isa finite set ofopen subintervals of /with (a,. b,) n (aJ, bJ) = (2J for i j and 1 (b1: - 01:)< o,then I::., I/(b1:) - /(01:)I < e. (a) Prove that if exis absoJuteJycontinuous on [a, b], then exis continuous and of bounded variation on [a, b]. (b) Prove that if ex is absolutely continuous on [a, b], then ex= /J- y, where fJ and y arc increasing and absoJuteJy continuous on [a, b]. 56.2 Let exe BV{a, b] and suppose that/e SIJa, b]. Let

*

.E:.

F(x)

=

J:td«

for x in (a, b] and Jet F(a) = 0. (a) Prove that Fe BV{a, b]. (b) Prove that if exis continuous at x, then Fis continuous at x. (c) Prove that if exis absolutely continuous on [a, b], then Fis absolutely

229

68. The Riemann Integral

continuous on [a, b]. (See Exercise 56.1 for the definition of absolutely continuous.) (d) Prove that if« is increasing on [a, b], then Fis differentiable at each point at which« is differentiable and/is continuous; and that if xis such a point, we have F'(x) = f(x)a.'(x). 56.3 Let g' be continuous on [c, d] and let/ be continuous on g([c, d]). Let a = g(c), b = g(d). Prove that f.J(x) dx

= J:J(g(t))g'(t)

dt

56.4 Let g be continuous on [a, b] and suppose that / is increasing on [a, b]. (a)

J:

Prove that

/(x)g(x)

dx

= /(a)·

J:

g(x) dx

+ f(b)·

J:

g(x) dx for

some c in [a, b]. (b) (Bonnet's Theorem) Prove that if in addition to the hypotheses in part (a), / is nonnegative, then J:!(x)g(x) dx 56.S

= /(b)·

J:

g(x) dx

for some c in [a, b]. (a)• Prove the trapezoid rule: If/" exists on [c - h, c 3 c:H/= h[f(c + h) + f(c - h)] - 21'f>h

+ h], then

J.,_,.

(b)

56.6

(a)

where c - h < : < c + h. Subdivide the interval [0, 1) into 10 equal parts and use the trapezoid rule on each subdivision to estimate JAdx/(1 + x 2 ). Estimate the error. (We shall see later that JAdx/(1 + x 2 ) = n/4.) Prove Simpson's rule: If /'

J.,_,./= h3(/(c c:H

56.7

4

'

exists on [c - h, c

+ h) + 4/(c) + /(c

- h)] -

+ h], then

J' 4 '("h' 90

(b)

Repeat Exercise 56.S(b) replacing the trapezoid rule with Simpson's rule.

(a)

Derive the rectangle rule, which approximates J~!U(x) dx by 2h/(c), together with a remainder estimate similar to the trapezoid rule and Simpson's rule [see Exercises 56.S(a) and 56.6(a)]. Repeat Exercise 56.S(b) replacing the trapezoid rule with the rectangle rule.

(b)

56.8 Prove that the following form of the integration by parts formula is valid for Riemann integrals: If/', g' e bY[a,b], then J:f'g

= /(b)g(b)

-/(a)g(a)

- J:Jg'

56.9 Suppose that/ is (n + 1)-times differentiable on [c, d] and J X .....

67.

Measure Zero

and

231

r 11..1::;e L 31.. = -2 < e t

00

00

11•1

11•1

Therefore, X is of measure zero. The method of th~ last example may be used to prove that a countable union of sets of measure zero is of measure zero.

1beorem57.2 If {X11} is a countable family of sets of measure zero, then U:°-1 X11 is of measure zero. Proof. Let t > 0. For each positive integer n, there exists a countable 1 n•>::> x..and L~ 1 111•>1 < family {Ii11>J~1 of open intervals such that t/3 11• The family vi•>I i,n E P}

u~

is countable (Theorem 9.5) and covers

U:''"' 1 X

11

and ■

Having given several examples of sets of measure zero, we now give an example of a set which is not of measure uro. Theorem 57.J

The closed interval [a, b] is not of measure zero.

Proof. It is easy to show by induction that if {/1 }:. intervals which covers [a, b], then

1

is a.finite set of open

(57.1) Suppose [a, b] is of measure zero. Taking t = (b - a)/2, there exists a countable family {/1 } of open intervals which covers [a, b] such that b - a 2

00

L 1:-1

1111 < -

By the Heine-Borel theorem (Theorem 34.2), some finite subcollection {/ 1J}j .. 1 covers [a, b]. Now • co b-a

L

J• 1

111:,I < L l/1:I< -2l•

1

which contradicts (57.1). Therefore [a, b] is not of measure zero.

■

Theorem 57.3 furnishes yet another proof that irrational numbers exist. For if [O, 1) contains only rational numbers, by Theorem 57.2, [O, 1) is countable, hence of measure zero. However, this contradicts Theorem 57.3.

232

The Rlemenn-Stleltjn Integral

J

i

i

E-4:J--;i;r---13~.

J

•••

Figure &7.1

. A remarkable fact is that uncountable sets of measure zero exist. The standard example is the Cantor set (see Figure 57.1).

Deflnidon57.4 From the closed interval [O,I] we remove the open middle third / 1 = (½,i) yielding the set [O,¼]u [i, 1) From each of the closed intervals [O,¼]and [i, I] we remove the open middle thirds (¼,j-) and (t, f) yielding the set

[O,¼]u [t, We let 12

¼]u [i, t] u [f,

= (f, ½)u (i, f ). At the nth

l]

stage we have closed intervals

Ji, 12, . .. 'Jl'.

From each of the closed intervals J., we remove the open middle third /Ct> for k = l, . . . , 2", yielding the set (JI \J(l >)U (J2\J(l)) U • • • U (J2,.\J(l"))

We let I,. =

Uf: 1 1.The Cantor set [O,

C is the set

11\lJI,. ... 1

Some properties of the Cantor set are immediately obvious. Since C is the intersection [O, l] n ( () /~) 11•1

of closed sets, C is closed (Theorem 38.8). Since C c [O, l], C is bounded. By Theorem 43.9, C is compact.

Theorem57.S The Cantor set is uncountable and of measure zero.

67. Meaaure Zero

233

Proof. We first show that C is of measure zero. At the nth stage (see Definition 57.4), C is contained in the union of closed intervals

12',

11 U 12 U • • • U

where lltl = 1/3• fork = l, ... , 2•. Lets > 0. Choose a positive integer n such that (t)" < s/2. Cover each closed interval lt by an oper. interval It such that lltl < 1/3• + s/2•+1, for k = 1, ... , 2•. Then Vt}l:1 is a finite set of open intervals which covers C and

tf1 lltl
0 2

and we have a contradiction. Therefore, /(x) = 0 almost everywhere in [a,

bJ. ■

238

The Rlemann-StleltjN Integral

Corollary58.7 Let/ e 9'[a, b] and let F(x)

=

J:1

for x in (a, b] and let F(a) = 0. Then F is differentiable almost everywhere in [a, b] and F'(x) = f(x) almost everywhere in [a, b]. Proof By Theorem 56.1(ii), F'(x) = f(x) for each x in [a, b] at which/ is continuous. By Theorem 58.5, f is continuous almost everywhere in [a, b] and the conclusion now follows. ■

Exercises

58.1 Let/e lf(a,b] and suppose that J!/ = 0 for all x in (a,b]. Prove that/(x) = 0 almost everywhere in [a, b]. 58.2* Let/, g e lf(a, b] and suppose that f(x) = g(x) almost everywhere in [a, b]. Prove that J:I= I: g. S8.3 Using Theorem S8.S prove that if /e BV[a, b], thcn/e lf(a, b]. 58.4 State and prove an analogue of Theorem S8.S valid for Riemann-Stieltjcs integrals. 58.5 Using Theorem S8.S prove that if/, g e lf[a, b], his continuous on the range of/, and c e R, then/ + g, cf, f· g, and h o / e lf(a, b]. Prove that if in addition 1//is bounded on [a, bl, then 1//e lf[a, b].

69.

Improper Riemann-Stieltjes Integrals

So far in this chapter, all of our functions have been bounded and all of our integrals have been computed over closed, bounded intervals. In this section we relax these restrictions by defining improper Riemann-Stieltjes integrals.

Deftnition59.1 Let f and a be functions defined on the interval [a, oo). Suppose that f e £1.[a,b] for every b > a. The improper Riemann-Stieltjes integral (of the first kind) I: f da is the ordered pair (/, F), where F(b)

= J:/dx

for b > a. In a similar way, we define the improper integral

I"'- f a:,

dx.

This definition is analogous to the definition (Definition 22.1) of an

69. Improper Rlemenn-Stleltju

Integrals

239

infinite series. The infinite series

was defined to be the ordered pair ({a,.}, {s,.}), where s,. = a 1 + · · · + a,., n = l, 2, . . .. The function f of Definition 59.I corresponds to the function a of Definition 22.1 and the function F of Definition 59.l corresponds to the function s of Definition 22.1. The function F is sometimes called the partial integral in analogy with the partial sum of an infinite series. Convergence of an improper integral is defined in terms of convergence of the partial integral F(b) = I!f da.

De8nition59.2 Let I: f da be an improper integral. If lim6 _, 00 f.J da exists we say that the improper integral I: f da converges,and we write

f

t/a

J.

00 /

= lim 6 ➔ ao

J t/a 6

/

•

If li~_, 00 I!J da does not exist, we say that the improper integral I: f da diverges. Convergence and divergence of the improper integral I._00 f da is defined in a similar way. If the improper integrals I._00 f da and I: f da are both convergent, we define

Examples. The improper integral lim

a-oao

The improper integral lim

Ii (1/x 2) dx converges to l since

J•..;dx = lim ( I - !) =I a a-ooo

l X

Ii If.Ji dx diverges since

a-oao

+

JI vX l

dx

r-

= lim 2(Jaa-oao

1)

= oo

Let f(x) = x. Then I:' f and 00 f diverge (for every a), and hence the improper integral I~00 f diverges. However, lim__,00 I._.J= 0. We call lim__,00 I._.Jda (if this limit exists) the Cauchyprincipal valueof the improper integral I~00 f da. We have just seen that the Cauchy principal value may exist for a divergent integral. It is easy to show that if the improper integral I~00 f da converges, then the Cauchy principal value is equal to the number I~oofda = I._oof da + I: f da.

240

The Rlemann-StleltjH Integral

Many of the theorems for infinite series have analogues for improper integrals. The methods of proof are also often similar. For example, if the improper integrals I:' f da. and I:' g da. are convergent, then the improper integral I:' (f + g) da. is convergent and (59.1)

This follows from Theorem 53.2 (for ordinary Riemann-Stieltjes integrals) by taking limits. Corresponding to an absolutely convergent series, we have the concept of an absolutely convergent improper integral.

Deftnidon59.3 We say that the improper integral I:' f da. converges absoI:' Ill da. converges and/ e at.[a, b] for all b > a. We say that the improper integral I:' f da. converges conditionally if I:' If I da. diverges, but I:'f da.converges. lutely if

The next theorem parallels Theorem 24. l for infinite series. Theorem59.4 Let ex be an increasing function on [a, oo) and let / be a nonnegative function on [a, oo). Then the improper integral I:' f dexconverges if and only if the function F(b) = I!f da.is bounded on [a, oo). Proof.

Suppose Fis bounded on [a, oo). Let L

= lub {F(b) I be

[a, oo)}.

Lets > 0. There exists a number b such that L - e < F(b). If x L - s < F(b)

$

F(x)

$

L < L

~

b, then

+e

and thus I:' f da. = L. Now suppose that I:' f da.converges. If b e [a, oo), we have

J:/da.J~

f da.

$

and therefore Fis bounded on [a, oo).

■

Theorem59.5 (ComparisonTest) Let ex be an increasing function on [a, oo). Suppose that/ and g are functions such that O $ f(x) $ g(x) for x in [a, oo). If I:' g da. converges and / e at.[a, b] for all b > a, then I:' f da. converges and

68. Improper Riemann-Stleltjes Integrals

Proof

241

The proof follows from Theorem 59.4 and the inequality

J:/ f. J~ da

s

g dcxs

■

g da

We can now show that an absolutely convergent improper integral is convergent (for increasing integrators). Theorem59.6 Let exbe an increasing function on [a, co). If the improper integral I: f da converges absolutely, then J: / da converges.

I:

Proof If x ~ a, then O s 1/(x)I - f(x) s 21/(x)I. By Theorem 59.5, (1/1 - /) da converges. By equation (59.1), I: f da converges. ■

We now give an example of a conditionally convergent improper integral. Let

~

if n is an odd positive integer

= 2m is an even positive integer and m is odd -1 if n = 2m is an even positive integer and m is even. Extend exlinearly to [l, co). (See Figure 59.1). Let P(x) = Jf cx(t)dt for x ~ l. cx(n)=

{

Then O s P(x)

if n

s I for x

~

l. Using integration by parts, we have

rb cx(x)dx = P(b) + rb ~ dx b X-

J1 X

J.

Since P is bounded on [I, co), limb.. ao (P(b)/b) = 0. Since O s (P(x)/x 2) s l/x 2 for x in [1, co) and J'f (l/x 2) dx is convergent, by Theorem 59.5, J'f (P(x)/x 2) dx is convergent. Therefore, J'f (cx(x)/x)dx is convergent. Since (1/x) ~ l/(2k - 1) if x s 2k - 1 and J~::~lcx(x)Idx = l, we have r211-•1oc(x}1 X dx

J.

II

rli-11%(X)I dx

= l:~lJ21:-3 X

~

II

l:~l

1 2Jc- l

for every positive integer n. The series :Ef..2 1/(2k - 1) diverges, and therefore the sequence of partial sums {I::. 2 l/(2k - l)} is unbounded (Theorem

-1 Figure 68.1

The Relmann-StleltjN Integral

242

24.1). Therefore, the function F(b) = Jtl«(x)/xl dx is unbounded. By Theorem 59.4, Ji lcx(x)/xl dx diverges. Thus the improper integral Ji («(x)/ x) dx is conditionally convergent. Our next theorem demonstrates a link between infinite series and improper integrals. See Exercise 59.10 for another result.

1beorem 59.7 (Integral Test) Let f be a nonnegative decreasing function on [I, oo). Then the infinite series L,,00• 1 /(n) converges if and only if the improper integral Ji f(x) dx converges. Proof. Since/(k

+

I) 5. f(x) 5. f(k) fork 5. x s k

f(k

+ 1) s.

1+ 1

/(x) dx

+

I, we have

s f(k)

Therefore, The proof now follows from Theorems 24.1 and 59.4.

■

We close this section by defining the improper integral of the second kind and giving two examples. Analogues of the theorems about improper integrals of the first kind are valid for improper integrals of the second kind (see Exercise 59.4).

Deflnidoa59.8 Let f and exbe functions defined on the half-open interval [a, b). Suppose that f e lf.[a, c] for a < c < b. The improper RiemannStieltjes integral (of the second kind)

is the ordered pair (J, F), where

F(c)

=

J:1 dcx

for a < c < b. In a similar way we define the improper Riemann-Stieltjes integral

J:./

dcx

If

lim F(c) c-eb·

=

lim c-eb-

exists, we say that the improper integral

Jcf dcx •

r.-f da converges, and we write

69. Improper Rlemann-Stleltjea Integrals

243

b·

J f da. = lim F(c) c .. b-

11

If limc.. t· F(c) does not exist, we say that the improper integral ,r:/ da. diverges. In a similar way, we define convergence and divergence of the improper integral I!+Ida..

Examples. The improper integral

/J+(1/.Jx)

J+dx = lim 2 -

dx converges to 2, since

1

lim

c .. o+

The improper integral

C

VX

c-+o+

2Jc

=2

I~+(l/x2) dx diverges, since

J 1

lim

c .. o+

..;. c X-

dx

=

!-

lim

c .. o+ C

I

=

oo

Exercises

Prove that the improper integral If (1/x'')tix diverges if p ~ 1 and converges if p > 1. 59.2 Determine the values of p and q for which the following improper integrals converge OD xP-1 dx o 1 + xq 59.3 Prove equation (59.1). 59.4 State and prove analogues of Theorems 59.4, 59.5, and 59.6 for improper integrals of the second kind. 59.5 Prove that if/ e ht[a, b), we have

59.1

J

• f=f·1=f"-1 f •+ • •

[For this reason, if at(x) = x, we usually write

I J,,f= f"I= J"•+

59.6

59.7 59.8 59.9 59.10

•

•

if either improper integral converges.) Give an example of a function/ such that the improper integral If f(x) dx conver.ges, but lim~.. 00 /(x) does not exist. State and prove an integration-by-parts formula valid for improper integrals. Prove an analogue of Theorem 26.4 valid for improper integrals. State and prove analogues of Corollaries 28.4 and 28.7 valid for improper integrals. Let/ be a decreasing nonnegative function on [l, ao).

The Rlemann-Stleltju

244

Integral

(a) Prove that if the series E:'- d(n) converges to L, then

1.,t f(n) - LI < J: /(x) dx

(b) Prove that if the improper integral

If"/(t) I

dt -

LI
a. Let P(x) = v:a. for x >a.Prove that if the integrals I:'Ida. and I:'I dP converge absolutely, then I:'Ida. converges. 59.13 (a) Prove that the integral It 111- 1 r'dt converges for positive x. The function r(x) = It 111- 1e-'dt, 0 < x < ao, is called the gamma function. (b) Prove that r(x + 1) = xr(x) for positive x. Deduce that r(n + 1) = n! for n = 1, 2, 3, ....

X Sequences and Series of Functions

In this chapter we shall investigate convergence of sequences and series of functions. The results of this chapter will extend the results of Chapters IV and V. In particular, we shall be concerned with the following problem. If each element of a sequence {!..} of functions has a particular property, when does the limit function/ have the same property? For example, we might ask whether the limit function/ is continuous if each function/.. is continuous. We begin by defining convergence of a sequence of functions.

80.

Pointwise Convergence and Uniform Convergence

We shall be concerned with sequences of real-valued functions defined on some set X. We shall denote a sequence as {f..}:. 1 (or sometimes just {f..}). Thus for each positive integer n we have a real-valued function/.. defined on a set X. The set X will not, in general, be a subset of the real numbers.

Definition60.1 Let {!..} be a sequence of functions on a set X. Let / be a ➔ 0 f.(x) = function on X. We say that{!..} convergespointwiseto/ on Xiflm,. /(x) for every x in X. Example60.2 Let f..(x)

= x'

Then {!..} converges pointwise to/ on [O, 1), where /(x)

= {~

Osx 0, there exists a positive integer N such that if n ~ N, then IJ.(x) - /(x)I < a

for every x in X. The crucial part of Definition 60.3 is that 1/.(x) - /(x)I < e for every x in X (if n ~ N). That is, N depends on a, but not on x. It is clear that if a sequence of functions converges uniformly it also converges pointwise (verify). We now prove that continuity is preserved under uniform convergence. 1beorem 60.4 Let {/.} be a sequence of functions which converges uniformly to a function / on a metric space M. If each /. is continuous at a point a in M, then/ is continuous at a.

Proof. Let d be the metric for M. Let a > 0. There exists a positive integer N such that 1/N(x) - /(x)I
0 such that if d(x, a) < b, then

Now if d(x, a) < b, then 1/(x) - /(a)I S 1/(x) - fN(x)I 8

8

8

0. There exists a positive integer N such that if n 2:: N, then d(f., /) < 8. This means d(/.,/) = lub {1/.(x) - /(x)I Ix e [a, b]} < 8, or equivalently, 1/.(x) /(x)I < 8 for every x in [a, b]. By Definition 60.3, this says that/. converges uniformly to/ on [a, b]. Conversely, suppose {/.} is a sequence in C[a, b] which converges uniformly to a function/ on [a, b]. By Corollary 60.5 we have that/ e C[a, b]. ➔ 0 /. =/in the metric of C[a, b]. We must show that lim,. Let e > 0. Since/. converges uniformly to/ on [a, b], there exists a positive integer N such that if n 2:: N, then lf.(x) - /(x)I < 8/2 for every x e [a, b]

Thus if n 2:: N, e/2 is an upper bound for the set {11,.(x)- /(x)I Ix e [a, b]}

and so d(f.,/)

:s; e/2 < e. Therefore, lim,.

➔

0 /.

=f

■

248

Sequences end Serles of Functions

Corollary60.8 C[a, b] is a complete metric space in the metric of Example 60.6. The proof is left as an exercise (Exercise 60.10).

Proof

■

Exercises

60.1 Prove directly from Definition 60.3 that the sequence in Example 60.2 does not converge uniformly. 60.2 Let f.(x) = 1/(1 + n 2 x 2 ) and g.(x) = nx(l - x)-, x e (0, 1). Prove that {/.} and {g.} converge pointwise but not uniformly on (0, 1). 60.3 Prove that if the sequence {/.} converges uniformly to the function f on a set X, then {/.} converges pointwise to/. 60.4 Give an example of a sequence of continuous functions (on a metric space) which converges pointwise, but not uniformly to a continuous function. 60.S Let {/.} be a sequence of bounded functions on a set X. Prove that if {/.} converges uniformly to/ on X, then/ is bounded. Give an example to show that this statement is false if uniform convergence is replaced by pointwise convergence. 60.6 We say that a sequence{/,,} is uniformly boundedon X if there exists M such that IJ,,(x)I < M for all x e X and for every positive integer n. Suppose that {/.} and {g.} converge uniformly to/and g, respectively, on

X. (a)

60.7

60.8•

60.9 60.10•

Prove that if each function/. is bounded, then/ is bounded and {f,,} is uniformly bounded. (b) Prove that {f,, + g.} converges uniformly to/+ g on X. (c) Prove that if c e R, then {cf,,} converges uniformly to cf on X. (d) Prove that if {/,,} and {g.} are uniformly bounded sequences on X, then {f,,g.} converges uniformly to/· g. (e) Give an example to show that statement (d) is false if the boundedness hypothesis is removed. (Cauchy conditionfor uniform convergence)Let {/.} be a sequence of functions on a set X. Prove that there exists a function/ such that {/.} converges uniformly to/ on X if and only if for every e > 0, there exists a positive integer N such that if m, n ~ N, then 1/.(x) - /.(x)I < e for every x in X Prove Dini's theorem. If a sequence {/.} of conti.nuous functions converges pointwise to a continuous function/ on a compact metric space M and if f.(x) ~ /,,+ 1 (x) for every x in Mand for every positive integer n, then {/.} converges uniformly to f on M. Prove that C[a, b] is a metric space with d defined in Example 60.6. Prove that C[a, b] is a complete metric space with d defined in Example 60.6.

248

81. Integration and Differentiation

81. Integration and Differentiation of Uniformly Sequences

Convergent

We first prove that the analogue of Corollary 60.5 holds for RiemannStieltjes integrable functions.

Theorem61.1 Let

ix be a function of bounded variation on [a, b] and let {/.} be a sequence of functions in £1Ja, b] which converges uniformly to a function/. Then/ e £1.[a,b] and

lim

•◄ ao

r /.da. = J" J. •f da.

The conclusion of Theorem 61.1 may be written lim

• ◄ ao

J"f,, da. = J.r (• lim / J da. ◄ ao •

where lim_.. ao/. denotes the uniform limit of the sequence {/.}. Proof. First assume that ix is increasing. If ix(a) = ix(b), there is nothing to prove; so assume ix(a) < ix(b). Let e > 0. There exists a positive integer N such that 8

1/(x) - IN(x)I

< 3[ix(b) - ix(a)]

for all x in [a, b]. It follows that for any partition P of [a, b] 8

IUJ - IN, P)I ~ 3

(61.1)

Since IN e £1.[a,b], by Theorem 51.8 there exists a partition P of [a, b] such that

(61.2) Let M,

= lub/

M,•= lub/N M,••= lub (/ If

on

xJ

[x 1_i,

on [x,_ 1 ,xJ - IN)

on [x1_ 1, x 1]

x e [x1_ i, xJ, then f(x)

and hence

= [/(x)

- IN(x)]

+ IN(x) ~

Mr·+ Mr

Sequences and Serl•

260

of Functions

It follows that U(f, P) s U(f - IN, P)

UJ. P)

Similarly

~

L(f - IN, P)

+ U(JN,P) + L(/N, P)

(61.3) (61.4)

Using inequalities (61.l) through (61.4), we have

UJ. P) s U(f- IN, P)

U(f, P) -

e

e

IN, P)

- L(f-

+

U(IN, P) - L(JN, P)

e

S3+3+3=t

By Theorem 51.8,le lfJa, b]. We next show that

f.1,, da. = f.1 dtX.

lim •◄

•

ao

•

(where exis increasing). Let a > 0. There exists a positive integer N such that if n ~

N, then

8

1/,,(x) - l(x)I < 2[ex(b)- ex(a)] By Theorem 5l.13(iv), if n

~

J:I/,,- II Thus if n

~

N, then

da. s

f.

~

2[ex(b) ex(a)]da. = ;

N, then

If exis of bounded variation on [a, b], we let ex= ex1 - ex2 be the decomposition of exas the difference of two increasing functions as in Theorem 54.8. By Theorem 55.2, I,, e lf. 1 [a, b] n lf. 2 [a, b] for every positive integer n. By our preceding result, I e lf. 1[a, b] n lf. 2 [a, b], and hence I e lf.[a, b] by Corollary 53.4. Furthermore,

lim •➔

er)

f./,,da. = lim (f./,,da.1- f./,,da.2) •

• ➔

co

•

•

81. Integration and Differentiation

261

If {f,.} is a sequence of functions which converges pointwise to a function/ on [a, b] and eachf,. is in El[a,b], it may happen that/ fails to be in El[a,b]. For example, let {ri, , 2 , ••• } be an enumeration of the rational numbers in [a, b]. Let if x e {ri, ... , ,,.} if x e {ri, ...

, ,,.}' '"' [a, b]

Thenf,. e El[a,b] for every positive integer n, but the pointwise limit function

if x is rational if x is irrational is not in El[a,b]. Some difficulties ofthis sort can be eliminated by considering a more general integral such as the Lebesgueintegral (see Chapter XIV). See Exercises 61.1 to 61.4 for further examples. No analogue of Corollary 60.5 and Theorem 61.1 holds for differentiation. If we let f,.(x)

x"

=n

then {f,.} converges uniformly to O on [O, 1] (verify). However,/~(1) = 1 ++ 0. Other problems are also possible. We will show later (see Section 77) that there exists a sequence {f,.} of infinitely differentiable functions on [O,1]which converges uniformly to a function/that is nowhere differentiable in [O, 1]. The foilowing theorem on differentiable functions is often quite useful.

Theorem61.2 Let {f,.} be a sequence of differentiable functions on (a, b). Suppose that (i) /~ is continuous on (a, b). (ii) {f,.} converges pointwise to f on (a, b). (iii) {/~} converges uniformly on (a, b). Then/is differentiable on (a, b) and/~ converges uniformly to/'

on (a, b).

Proof. Let c e (a, b) and let x e (a, b). Let g denote the limit of {/~}. Then/~ e El[c,x] (or El[x, cDsince/~ is continuous. By Theorem 61.1, (61.5) By the fundamental theorem of calculus

J:J: =

f,.(x) - f,.(c)

262

Sequences and Serles of Functions

Using hypothesis (ii), we have Jim

a ◄ c:o

ric1: = Jim [f,.(x) Jc

- f,,(c)]

= /(x)

- /(c)

(61.6)

11~00

Combining equations (61.5) and (61.6), we have

f: = g

f(x) - f(c)

Differentiating each side of this equation, we have g(x)

= f'(x) ■

Exercises 61.5 and 61.6 give other examples of phenomena that can occur with sequences of differentiable functions. Exercise 61.7 gives a less restrictive theorem on limits of differentiable functions. Exercises

61.1

Give an example of a sequence{/..} of functions in .st(a,b] which converges pointwise but not uniformly to a function/ on [a, b] in .st(a, b] such that

Jim

s·I,, = f.I

.... ao •

61.2

•

Give an example of a sequence{/,,} of functions in ~[a, b] which converges pointwise to a function/ in ~[a, b] but

s· * s·

Jim I,. I ao • • 61.3 Give an example of a sequence {/,,} of functions and a function / such that (a) I,, e ~[a, b] for every positive integer n a ➔

(b) /e ~[a, b] (c) lim ... ""J=/,, =

I=t

(d) lim,... ""f,.(x) does not exist for any x e [a, b]. 61.4* Let {I,.} be a uniformly bounded sequence of functions in ~[a, b] which converges pointwise to a function/ in ~[a, b]. Prove that l

Jim ➔

GO

s·I,. = s·I a

a

61.S Give an example of a sequence {/..} of differentiable functions which con-

verges uniformly on R to a differentiable function/ such that {/..'}converges pointwise but not uniformly to/' on R. 61.6 Let f,,(x) = x 1 + 1 1 0. Since I::.S1 M,. converges, there exists a positive

268

Sequences and Serles of Functions

integer N such that if n > m ~

N, then

M,,.+1

If n > m

~

+···+

M,. < e

N and x e X, then

lu,,.+1Cx)+ · · · + u,.(x)I S lu,,.+1(x)I

< M,,.+1 and so by Theorem 62.5,

+ · · · + lu,.(x)I + · · · + M,. < e

1::,.1 u,.converges uniformly

■

on X.

As an example of the use of the results of this section, we will give an example of a function which is continuous on R but differentiable at no point of R. Let/ 0 (x) be the distance from x to the integer nearest x. (See Figure 62.1.) We observe that/ 0 is continuous on R,/ 0 (x + N) = / 0 (x) for every integer N, and l/ 0 (x)I S ½ for every x e R. Let fi(x) = / 0 (4"x)/4" for x e R and k = l, 2, .... (See Figure 62.2.) Thenfi is continuous on Rand lfi(x)I s 1/(2·4"). By the Weierstrass Mtest (Theorem 62.6), Lt=oh converges uniformly on R. Since each ft is continuous on R, the limit function, which we denote by F, is continuous on R (Theorem 62.2). We will show that Fis nowhere differentiable. Let us define a "roof" of/,,. to be the graph of/,,. over [n/(2 · 4'"), (n + 1)/ (2·4'")] for some integer n. (See Figure 62.3.) We note that if x and y are under a roof of/,,., then (x,f,,.(x)) and (y,f,,.(y))

,,

''

I

2

I

a

I

Figure 82.1

Figure 82.2

4

82. Serln of Functions

267

.1

a

0

1

1

I

Figure 82.3

lie on a line segment which has slope either + 1 or - 1 and hence [/ 111(x) f,,.(y)]/(x - y) = ± 1. (See Figure 62.2.) Furthermore, in this case, x and y are under a roof of ft for k < m, so that ft(x) - ft(y) = x-y

±1

for k ~

m

Let a e R. Let m be any positive integer. Then a e [n/(2·4•- 1), (n + 1)/ (2·4•- 1)] for some integer n. Since the length of this interval is 1/(2·4•- 1), by taking h111 = +1/4 111 or -1/4•, we can assure that a and h,,.e[n/(2·4•- 1), (n + 1)/(2· 4• - 1)]. We showed above that, in this case, ft(a

+ h...) -

ft(a)

h,,.

= ±1

for k < m

If k 2:: m, "( '" a

+

h,,J

"( ) _ / 0 (4"a + 4"h,,.)- / 0 (4"a) 411 a -

- '"

(4"a + N) ------r--4 / 0

- / 0 (4"a)

for some integer N. It follows that ft(a + h,,J - ft(a) Therefore, F(a + h..) - F(a) h"'

=0

for k 2:: m.

f ft(a + h,,.) - ft(a) h"' = 1•ft(a + h,,.) - ft(a) _ "'f ± 1 =

l•O

1

ll•O

hlll

llaO

Thus [F(a + h111) - F(a)]/h111 is an odd integer if mis odd and [F(a + h..) F(a)]/h,,.is an even integer if mis even. It follows that lim,,.....00 [F(a + h,,.) F(a)]/h.. does not exist. If F were differentiable at a, then this limit would exist. Consequently, Fis differentiable at no point of the real line.

Sequences end Serlea of Function•

268

Exerclsea 62.1 62.2

Prove that the series E:°-1 1/(n 2 + x 2 ) converges uniformly on (0, oo). Prove that the series I:;,:0.1 x" (1 - x) converges pointwise but not uniformly on (0, 1). Prove that E:°. 1 (-l)"x"(l - x) converges uniformly on [0, 1).

62.3 Let E:'-1 u,.and 1:;:.1 11,. be series of functions on a set Xsuch that lu,.(x)I~ 111,.(x)I for every positive integer n and for every x e X. Prove that if E:°-1 111.Iconverges uniformly on X, then E:'-1 "• converges uniformly on X. 62.4• Prove Dini's theorem for uniform convergence of series: Let {u.} be a sequence of nonnegative continuous functions on a compact metric space M. If E:'-1 u,, converges pointwise to a continuous function / on M, then E:°-1 "• converges uniformly to/ on M. 62.5 Fill in the details of the following proof of the Tietze extension theorem for metric spaces. Theorem Let/ be a continuous function from a closed subset C of a metric space Minto (-1, 1). Then there exists a continuous extension of /from M into (-1, 1), (that is, there exists a continuous function g from M into (-1, 1] such that g IC=/). Proof (a) Let A= {xe C 1/(x) ~ -½} and B = {x e Cl/(x) ~ -l}. Modify Exercise 40.14(c) to conclude that there exists a continuous function /1 from M into [ -¼.·Usuch that /1 = -1 on A and ½on B. (b) Show that 1/(x) - /1(x)I ~ j for all x e C. (c) Repeat (a) and (b) with/replaced by/ - /1 to find a continuous function /2 from M into [ -j, i] such that /2 is -j on {x e CI /(x) /1(x) ~ -i} and /2 is j on {x e CI /(x) - /1(x) ~ i} and for all x e C lf(x) - C/1(x) + /i(x)JI ~ j (d) Show that there exists a sequence {/.} of continuous functions on M such that (i) the range off. is contained in [ - 2•- 1/3", 2•- 1/3"]; (ii)/.+ 1 is -2"/3 ■ + 1 on {xeCl/(x)-[/1(x) +···+/.(x)] ~ -2"/3 ■ + 1 } and 1 f.+ 1 is 2"/3 ■ + on {x e C 1/(x) - C/1(x)+ · · · + /.(x)] ~ 2•13•+1}; and (iii) 1/(x) - C/1(x) + · · · + f,,(x)JI ~ (,J)"for all x e C. (e) Prove that E:'. 1 f. converges uniformly to a continuous function on M. (f) Prove that g = 1:;:'. 11,, is a continuous extension of /to Minto (-1,

1]. 62.6 Let M be a metric space. Prove that M is compact if and only if every continuous real-valued function on M is bounded. 62.7 Suppose that 1:;:. 1 u,. converges pointwise on a set X. Suppose also that {u.} converges uniformly to zero on X. Must E:'. 1 u,.converge uniformly?

269

13. Appllcatlona to Power Serles

83. Applications to Power Series We recall that a power series is a series of the form I::°-oa.(x - t)" (see Sections 27 and 29). If R is the radius of convergence of a power series a.(x - t)" converges absolutely if Ix - ti < R L:°-o a,.(x - t)", then L"°-o and diverges if Ix - t I > R (see Section 27). We will use the results of Section 62 to prove that a power series may be integrated or differentiated term by term for Ix - ti < R.

I::°-oa.(x - t)" be a power series and let R be its radius of convergence. Lct/(x) = I::°-oa.(x - t)", Ix - ti < R. Then (i) If O < S < R, then I::°-oa.(x - t)" converges uniformly to/ on [t - S, t + S]. (ii) / is continuous on (t - R, t + R). (iii) If a, b e (t - R, t + R), a < b, and « is of bounded variation on [a, b], then/ e lf.[a, b], I::°-oa. J!(x - t)" d«(x) converges and Theorem63.1 Let

J,,/ da. = f a.J.r (x •

t)" da.(x)

11•0

(iv) / is differentiable on (t - R, t absolutely if Ix - t I < R, and

+

R),

I::'=1 na,.(x -

t)"-

1

converges

CX)

f'(x)

=

L 11•

na,.(x - t)"-

1

,

Ix - ti
b,. ~

lim c.,.,,.= 1 + 1 +

,. .. co

lim,,....00 a,,. = exp I.

1,+ .. •+ m. ..!,= a,..

2•

■

For each positive integer n we have by equation (66.1),

e" = e•e• • •e = exp I-exp l • exp 1 = exp (1 + I + • • • + 1) = exp n (66.6) 00

270

Transcendental Functions

and by equation (66.3), 1

1

e-• = - - -= exp (-n) e" exp n

(66.7)

Therefore for any rational number , = m/n (where n > 0), we have by equations (66.1) and (66.6) or (66.7) (exp ,

r = exp ,

... exp , = exp nr = ti"

therefore, exp, = {e"')11" = e'. We conclude that exp,=

e'

for any rational number ,

(66.8)

Let IX be any real number. Choose {,,.},:.1 any increasing sequence of rational numbers such that lim r,. =

IX

By Section 17 and equation (66.8), we have e1X =

lime'" = lim exp,,. = exp (lim ,,.) = exp ix

Therefore exp IX =

e1X

for any real number

IX

(66.9)

By equations (66.8) and (66.9) we have the following theorem.

Theorem66.3 exp x = r for every x e R. Since e > 2 and r is strictly increasing, we have limx...00 r = oo. By equation (66.3) limx...-ao r = 0. By the intermediate value theorem (Corollary 45.6) it follows that the range of r is (0, oo). In Figure 66.1 we have sketched the graph of r. We now summariz.ethe above results. Theorem66.4 (i) r = "'I:.':-o x"/nl for all Xe R. (ii) r is continuous and differentiable for all x in R. (iii) (r)' = r for all x e R. (iv) r is strictly increasing on Rand r > 0 for all x e R. (v) r+' = r·e' for all x and yin R. (vi) limx...00 r = oo and limx...-ao r = 0. Exercise•

66.1 Suppose/: R - R and f'(x) = f(x) for all x e R and /(0) = 1. Prove that /(x) = r for all x e R.

87. The Natural Logarithm Function

271

------------~..----------• 1

Figure 88.1

66.2 Prove lim. ...'° (1 + x/nY' = e" for all x e R. 66.3 Suppose/(x)·/(y) = f(x + y) for all x and yin R. Assuming/is differentiable on Rand not :zero, prove that/(x) =~".where c is some constant 66.4 Do Exercise 66.3 with differentiablereplaced by continuous. 66.5 Evaluate Jgo e- 0 and a* 1. Define lo&, to be the inverse function to OZ.We call log.x the logarithm of x to the base a. Compute the derivative of log.x. 67.6 Let a> 0 and let/(x) = x9 for x in (0, oo). Prove that/'(x) = ax--1• 67.7 (a) Let /(x) = log (I + x). Apply the mean-value theorem to/ on [0, 1/n] to conclude that

n log ( 1 (b)

+ 1)=

1

tc

where 0 < c < l/n. Use (a) to prove that

- +1-1 < log (1+ !)n < !n

n (c) Let

c. = 1 + ½+ •••+ ~ D.

=

log n

c. -n1

c.

By considering Cu1 and D0 1 - D. and (b), prove that {C.} is decreasing and {D.} is increasing. (d) Prove that {C.} is bounded below and that {D.} is bounded above. Deduce that {C.} and {D.} converge to a common limit. This limit is known as Euler's constant and is denoted y. 67.8 (a) Derive the following expansions valid for lxl < 1: log (1

+ x) = Eao (-1\&+1~ .... ___,_,,,__ •·•

k

274

Tranacendental Function•

f~ lo (1+ x) = 2f x21-1 g 1-x 1-12k-i log (1 - x)

(b)

= -

lt•l

1'

■ +i x 21 - 1 /(2k - 1) in approximating log Show that the error 2 :E'((1 + x)/(1 - x)) with n terms of the last expansion in part (a) is at most 2x2a-1

(2n

(c)

+ 1)(1 -

x 2)

•

Derive the expansion ao 1 log 2 = 2 1~ 1 J:li-1(2k _ l)

Show that if we take six terms of this expansion, the error is 1~ than I- x 10- 6 • Compare this result with Exercise 25.S(e).

68. The Trigonometric Functions

We begin by defining two power series functions, which tum out to be the functions sin x and cos x.

Deflnltion68.1 For each x e R we define

co

and

C(x)

= 11•0 L (-1)"

x2" x2 x4 x6 = 1 - - + - - - + • •• (2n)! 2! 4! 6! -

(68.2)

These series converge absolutely for all values of x by the ratio test. Since the reader is probably familiar with the properties of the functions sin x and cos x, we will proceed in justifying Definition 68.5 by showing that S(x) and C(x) have all those familiar properties and that they are the unique functions with those properties. From (68.1) and (68.2) we see that C(O) = 1 C( -x)

= C(x)

S( - x)

S(O) = 0

= - S(x)

(68.3) for all x e R

(68.4)

and by Theorem 63.l(iv) C'(x) = -S(x)

S'(x) = C(x)

for all x e R

+ [C(x)] 2 = 1 for all x e R. = [S(x)] 2 + [C(x)] 2 for all x e R.

(68.5)

Lemma68.2 [S(x)]2 Proof.

Let g(x)

By (68.5) and the

88. The Trigonometric Functions

276

chain rule g'(x)

= 2S(x)C(x) + 2C(xX -

=0

S(x))

for all x e R

By Theorem 49.9, g(x) is a constant function on R; however, g(O) = 1; so g(x)

=1

for all x e R ■

Lemma68.3 If/ is a real-valued function on R satisfying/•(x) all x e R and/(0) = 0 and/'(O) = b, then

= bS(x)

f(x) Proof.

If we let U(x)

and

=·- f(x)

for

for all x e R

= f(x)S(x) + f'(x)C(x)

V(x)

= f(x)C(x)

- f'(x)S(x)

it is easily verified that

=0 =

U'(x)

for all x e R

V'(x)

Therefore, we have b

and

f(x)C(x)

= f(x)S(x) + f'(x)C(x) = f'(x)S(x) for all x e R

Using Lemma 68.2, we conclude that bS(x)

= S(x)[f(x)S(x) + f'(x)C(x)]

= /(x){l

- [C(x)]2 }

+ f'(x)S(x)C(x)

= f(x) - C(x)lf(x)C(x)

- f'(x)S(x)]

= f(x)

for all x e R

Theorem68.4 If/ is a real-valued function on R satisfying J•(x) for all x e R and/(0) = a andf'(O) = b, then f(x)

= aC(x) + bS(x)

= bS(x)

for all x e R and

for all x e R

Theorem 68.4 justifies the following definition.

Deftnitioa68.5 We define for all x e R the functions sin x

= S(x)

= - f(x)

for all x e R

Proof. Let g(x) = f(x) - aC(x). Then g•(x) = -g(x) g(O) = 0 and g'(O) = b; so by Lemma 68.3, we have g(x)

■

cos x

= C(x)

and we call these functions the sine and cosine, respectively.

■

278

Tranacendental Function•

We proceed by proving many of the formulas involving sin x and cos x.

+ y) = sin x cosy + sin y cos x for all x and y in R. Rand let/(x) = sin (x + y) for all x in R. Then/•(x) =

1beorem68.6 sin (x

Proof. Fix yin - f(x) for all x e R and /(0) we have

f(x)

= sin y

= sin y cos x + cosy

1beorem 68.7 cos (x

= cosy;

and f'(0)

for all x and y in R

sin x

+ y) = cos x cosy

so by Theorem 68.4 ■

- sin x sin y for all x and y in R.

Proof. Fix yin Rand let g(x) = sin (x + y) for all x e R. By Theorem 68.6, g(x) = sin (x + y) = sin x cosy + sin y cos x; so g'(x) = cos (x + y) = cos x cosy - sin y sin x for all x and y in R. ■

We now define the real number ,r. It is easily verified (see Exercise 25.3) that since cos2 we have lcos2-

=

2 2 24 1 - -+2! 4!

-···

(1-;;)Is ;;=j

that is, cos 2 is within t of - 1. This implies that cos 2 < 0. Now since cos 0 = I and cos x is continuous on R, we conclude by the intermediate-value theorem that cos x = 0 for at least one x in (0, 2). Let xi denote the smallest number in (0, 2) such that cos Xi = 0. The existence of Xi is guaranteed by the completeness of Rand the continuity of cos x. (Verify!)

Definition68.8 We define 1C

= 2x,

where Xi is the smallest positive real number such that cos Xi = 0. By this definition, cos (,r/2) [cos (n/2)] 2

= 0, and

since

+ [sin (n/2)] 2 = 1, we conclude

that sin (n/2) is 1 or -1.

However, cos x > 0 in (0, ,r/2); so equation (68.5) implies that sin x is strictly increasing in (0, ,r/2); therefore, sin (,r/2) = I, since sin 0 = 0. Using Theorems 68.6 and 68.7, we find that cos ,r = cos {1t/2 + n/2) = -1, and similarly sin ,r = 0, cos 2,r = I and sin 2,r = 0. Using these facts and Theorems 68.6 and 68.7, we have cos(x

+ 21t) = cosx

sin (x

+ 2,r) =

sin x

for all x e R

(68.6)

277

88. The Trigonometric Functions

ooe(•)

Figure 88.1

and cos(x+i)=-sinx

sin(x+i)=cosx

forallxeR

(68.7)

The graphs of sin x and cos x are sketched in Figure 68.1. We conclude this section with the definition of the other trigonometric functions. The formulas and interrelationships of these functions are left as exercises. (See Exercises 68.1 to 68.6.)

Definition68.9 We define the following functions for each x e R, where the expression makes sense sin x tanx = -cosx

(tangent)

cosx cotx = -.-

(cotangent)

SID X

secx

I

= -cosx

1 cscx = -.-

SID X

(secant) (cosecant)

Transcendental Functio

278

■

Exercises

68.1

Determine the domain and range of each of the functions in Definition

68.9. Prove that (tan x) 1 + 1 = (sec x) 1 and (cot x) 1 + 1 = (csc x) 1 . Find the derivative of each of the functions in Definition 68.9. 68.4 Sketch the graph of each of the functions in Definition 68.9. 68.S Define appropriately restricted inverse functions for each of the trigonometric functions. 68.6 Find the derivative of the functions defined in Exercise 68.5. 68. 7 Evaluate the following limit: 68.2 68.3

r

sinx+cosx-r

log (1 + x 2) 68.8 (Wallis' Product) Let a,, = f3 11 sin•x dx, n "~

= 2, 3, .... Prove that {a.} is decreasing. Prove that 1·3·5 · · · (2n - 1) 1t 0111 = 2·4·6 · · · (2n) • 2 2 · 4 · · • (2n - 2) and Oi11-l = 3.5 • • • (2n - 1) (a) (b)

(c)

Prove that

2·4 · · · (2n) < 1 ·3 · • · (2n - 1). ~ < 2·4 · · · (2n - 2) 3.5 · · · (2n + 1) 2·4 · · · (2n) 2 3.5 · • · (2n - 1) and, hence, conclude that 2n 1t ( 2· 4 · · · (2n) ) i 1 1t 2n + 1 . 2 < 1· 3 · · · (2n - 1) 2n + 1 < 2 (d) Derive Wallis' product:

)i

( 2·4 · · · (2n) 1 1t l ·3 · · · (2n - 1) 2n + 1 = 2 68.9 (Stirling's Formula) (a) Let aa:= I and ha:= log k in the summation by parts formula (Theorem 28.1) to derive I; log k = - l:•1 k log (1 + -k1) + n log n l:•1 (b) Take three terms in the expansion for log (1 + x) (sec Exercise 67.8) to derive

~

.

1/

log ( 1 +

¾)= ¼- ~

+

W

where O < c1 < 1. Combine the equations above to obtain • 1 •- 1 1 1 ,._ 1 C .!: log k = n log n - (n - 1) + - !: - - - .!: ,:\ l:•1 21:-1k 31:-1K-

88. The Trigonometric Function•

279

(c) Using the result of Exercise 67.7, which established the existence of the limit lim: .. ., U:::. 1 1/k - log n), prove that lim •➔ CO

±

1 2

I log k \a:•1

- n log n - - log n

+ n)

exists. (d) Prove that

. .....,_..,._ nle" un 1 ... ., n ■ ./n exists. (e)• Let a.=

n!e"/(

■

.[ii). Prove that = 2·4 ... (2n) I a2. l • 3 · · · (2n - 1) ./2n and deduce that

a:

l

(f)

Jim 5. ... ., a2. Prove Stirling's formula: lim n ! e"

,... «>n

68.10

■ Jn

=

+I

[2n(2n

+ 1)] 112

n2

./2TC

= ./2n

Stirling's formula states that n! is approximately n ■ ./2nTC/e"in the sense that the ratio of these two terms approaches 1. [The argument of (a) through (d) is due to Shohat (1933).] (a) Derive the expansion «>

arctan x

= .~o(-It

xli+

2k

1

+ 1 , lxl
0, there exists a positive integer N such that if n ~ N, we have llx,. - xii

= d(x,,, x)
0, there exists L > 0 such that in every interval of

\

298

Inner Product Spacee and Fourier Serl•

•

.....

•

•

••

•+Ip

Figure 73.2

length L there exists a point a for which 1/(a + t) - /(a)I < e for every t e R. (a) Prove that any continuosu periodic function is almost periodic. (b) Prove that an almost periodic function is uniformly continuous on R. (c) Prove that if/and g are almost periodic functions, then/+ g is almost periodic. (Compare with Exercise 73.1).

74.

Fourier Series: Definitions and Examples

We begin by extending Theorem 70.3 (which is concerned with R 3) to an arbitrary inner product space. The proof of Theorem 74.1 is identical to the proof of Theorem 70.3.

1beorem 74.1 Let {x1 , ••• , x,,} be an orthonormal set in an inner product space Vand let x e V. Suppose there exist scalars ci, ... , c,, such that II

X

Then

= 11:=1 L CtX11: fork

=

l, ...

,n

Proof. ■

74. Fourier Serl•:

Definition• and Examplea

299

Theorem 74.1 states that if a vector x in an inner product space Vis a linear combination of the members of an orthonormal set {xi, ... , x.}, then x is the unique linear combination x

•

=L

1:•1

(x, x 1)x1:

Let {xi, x 1 , ••• } be a countable orthonormal set in an inner product space V and let x e V. Motivated by Theorem 74.1, we can form the infinite series 01)

L

(x, xJx,.

·- 1

and ask when 01)

L (x, x.)x.

x =

••1

This is one of the central topics studied in the theory of Fourier series.

Deftnition74.2 Let X = {xi, x 1 , ••• } be a countable orthonormal set in an inner product space V and let x e V. The infinite series 01)

L

••1

(x, xJx,.

is called the Fourier series of x (relative to X). The coefficient (x, x,.) is called the nth Fourier coefficient of x (relative to X). The Fourier series of a function/ in £f[a, a + 2n] relative to the trigonometric set is

00 L (f, ■ =0

I [+le /(t) ,.) 0. Then there exists h e CP[a, b]

hll
J2= L

all

11•0

Our example shows that it is possible to find such a function/for

L:.1 1/n

2

•

the series

328

Inner Product Specn end Fourier Serl•

Unfortunately, there are series for which no such function in ~[a, a + 2n] exists which will sum the series. This difficulty can be overcome if we extend the class ~[a, b] to the Lebesgue integrable functions on [a, b] to the extent converges, there exists a Lebesgue integrable function f such that if l:,.00.. 0 that (J, tf,,.)= a,. (see Theorem 91.9). We conclude this section by proving a theorem about Fourier series analogous to Theorem 63.l(iii) for power series. Our theorem states that whether or not the Fourier series off converges, the Fourier series may be integrated term by term with the correct result.

a:

Theorem78.8 Let/e ~[a, a+ 2n]. Let a °+ 2

.... oo

L (a. cos nx + b,.sin nx)

be the Fourier series off. If a

s

s

c

a

+ 2n, then

J: + ..t.J:

(a,. cos nt

~o dt

converges uniformly to

I:f

on [a, a

the series

+ b,. sin nt)

dt

+ 2n].

Proof. Let {s,.}be the sequence of partial sums of the Fourier series of/. We must show that

converges uniformly to I~f on [a, a + 2n]. This is equivalent to showing that U: (s,. - /)} converges uniformly to zero on [a, a + 2n]. Let a > 0. By Theorem 78.5, there exists a positive integer N such that if n ~ N, 8

Us,. -/11 < J2n If x e [a, a

+ 2n] and n ~

N, then

= lls,.-/IIJ1.i < e where we have used the Cauchy-Schwarz inequality.

■

78.

+ 2,r)

Fourier Sarias in .61[a,a

329

= x for

Example. The Fourier series for f(x)

11+ 1

2(-1)

~

~

n is

7t ~

X ~

x

.

L ---smnx 11=1 n (X)

- n

By Theorem 78.8, JC

J•

tdt

2(- 1)11+ 1

JJC

n

•

= L --(X)

•·

1

for -

sinntdt

1t

(78.1)

and convergence is uniform on [-n, n]. It is easily verified that 11 1

00

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

+

•

Using the fact that

sinntdt

= L

7t2

(X)

+L

••1

4( -

n

11•1

n

l)II

2

00

2(-1)"

cosnx - 2

2

I::. 1 l/n 2 = n2/6, equation

x2=- 3 Taking x

00

JJC

(78.1) becomes

for -

cosnx

1 L 2 •• 1n

,r ~

X ~

1t

(78.2)

= n, we again derive I::°-1 l/n 2 = n 2 /6. Taking x = 0, we have (X)

L ••1

(-1)"+1

n

7t2

=-

2

12

By Theorem 75.3, equation (78.2) gives the Fourier series for g(x) [ - ,r, ,r]. Thus by Corollary 78.6(ii)

f

4

-1 x 4 dx=-+fn n -• 2 This equation reduces to

= x 2 on

00

16 L =-.i ,.. 1 n

Exercises

78.1 Let /e lf[a, b] and let e > 0. Prove that there exists a polynomial p such that Ill- PII< e 78.2 Let/ e C[a, b] and let e > 0. Prove that there exists a step function g on [a, b] such that lu(x) - /(x)I < e for all x e [a, b]

Inner Product Spaces and Fourier Serles

330

78.3

Let/e

!l[a, b]. Suppose that f.J(x)x" dx

for n = 0, 1, 2, .... 78.4

Prove that/=

=

0

0 almost everywhere in [a, b].

Apply Corollary 78.6(ii) to the functions of Exercise 74.3.

78.5

Prove the following expansions: 00 x2 n2 cos nx (a) - = nx - - + 2 I:: -:r for all O ~ x ~ 2n 2 3 ■ -1 n (b) x n 4 cos (2n - 1).x for all O ~ x ~ n = 2 - ff ·- 1 (2n - 1)2 00 1 (-l)•cosnx for all -n (c) .x sin x = 1 - - cos x - 2 .E 2 2 ■ -2 n - 1 When is convergence uniform? ao 1 7r6 78.6 Prove that .E ::i = 945 ■ -1 n

r;

~

x

~

n

78.7 (a)• Prove that for no positive integer m is it true that

1

ao .~1

(b)

7r3

ii!=;;,

State and prove a generaliz.ation for the series .E:-'-1 (1/nl), where k is an odd positive integer.

78.8 Let f e !l[a, a

+ 2n]. Let ' + .~1 (a. cos nx + b. sin nx)

be the Fourier series of/. Let c e [a, a + 2n]. (a)• Prove that the series .E:'.1 [(b,Jn) cos nc - (a.In)sin nc] converges absolutely. (b) Let

K= E (b• cos nc - a.sin nc) n n ■ -1

and

g(x)=

I"I-C

aox - K 2

for all x e [a, b]

Prove that

-cao 2

+ ■ f: (a. sin nx -1 n

- b. cos nx) n

is the Fourier series of g and that the Fourier series of g converges uniformly to g on [a, a + 2n]. (c) Prove that 1 c2aa ao a~ + b~

[+2.g 2-- -2

:: ,. •

+ ••1 .E

n2

78.9 Compare the properties of Fourier series and power series. 78.10 Is it possible to sum the series ,E:',.1 (1/n 3 ) by the methods of this section?

79. A Tauberian Theorem and an Application

331

79. A Tauberian Theorem and an Application to Fourier Series Fejer's theorem (Theorem 77.5) states that the Fourier series of a function/ in CP[a, a + 2n] is uniformly Cesaro summable to f. Thus a Tauberian theorem of the form, "If 1 f,. is uniformly Cesaro summable to f and a Tauberian condition is satisfied, then L:-'= 1 f,. converges uniformly to f," will have an application to Fourier series. The Tauberian condition is that {nf,.} is uniformly bounded. The theorem is due to G. H. Hardy.

:r:..

'Theorem79.1 (Hardy) If L:-'-1 f,. is uniformly Cesaro summable to f on a set X and if the sequence {nf,.} is uniformly bounded on X, then 1 1,.

:r:.

converges uniformly to f on X. Proof. There exists a number M such that lnl,.(x)I < M for every positive integer n and for every x in X. Let II

= L ft(x)

s11(x)

t•l

and let for x in X. Then if m > n, we have

(m - n)s,..(x) - (m - n)f(x)

= m[u,,.(x) -

f(x)] - n[u,.(x) - f(x)]

Ill

+ L

(k - 1 - n)ft(x)

(verify)

(79.1)

t"'11+l

Let e > 0. Choose e* > 0 such that e* < 1

and

e*

+ 2.Je* + 2MJe•
n .!:: N*, we have, by (79.1),

m

ls,..(x) - f(x)I S m _ n lu,..(x) - f(x)I

+ --

1

m -

+m

n _ nlu.(x) - f(x)I

Ill

L n1:•11+2

(k - 1 - n)lft(x)I




(79.'l\ therefore >

6

-

NJ? 6

> 1 (79.4)

Inequalities (79.3) and (79.4) imply that there exists a positive integer n satisfying

m

1+ Now

• N

< 1+

2Ja•

N

m

< n < 1+

.J?

m

2Ja• :s;;1 + 2J?'< n
0 such that if llxll < 6, then IIT(x)II < 1. Let x E V, x ::;:8. Then

ll2tx11 xii=- ~


0. Let 6 = e/(K + 1). Suppose llx - yll < 6. If x -:/:y, ll(x - y)/llx - YII II = 1, and thus

llr(n; =~11)11 s K

Thus

IIT(x) - T(y)II

=

IIT(x -

= y, then

On the other hand, if x implies (i). ■

y)II s Kllx - YII s K6 < e

IIT(x) - T(y)II

=0
N we have 0 ::5/(x) - s,.(x) < 1/2". It follows that lim s,.(x)

•--ao

= f(x)

By cases I, 2, and 3, we have lim,....00 s,.(x)

= f(x)

for all x e X.

■

Exercises

In these exercises, unless specified otherwise, X is a set and JI is a CT-algebrain X. 87.1 Deduce Corollary 87.6 from Theorem 87.5. 87.2 Prove that if/: X - [O, 00] is bounded and measurable, then the sequence {.r,.}:". 1 in Theorem 87. IO converges uniformly to/. 87.3 Let/be a real-valued function on X. Prove that/is a measurable function if and only if for each closed set Cc RJ- 1(C) e JI. Show that this statement remains valid if closed is replaced by compact. 87.4 Let/be a real-valued measurable function and c be a real number. Prove 0) are measurable directly from that cf, Ifl, and 1// (provided /(x) Theorem 87.2. 87.S Let X be any set and .A= {;, X}. Prove that the class of measurable functions is exactly those functions which are constant on X. 87.6 Let X be a set and .A be a CT-algebrain X. (a) Prove that a simple function s = !:1-1 c,XA 1 on X, where the {c1}

*

87. MNaurable Function•

87.7

87.8

87.'J 87.10

87.11 87.12

387

are distinct and not zero. is measurable with respect to .A if and only if Ai. A2, .••• A. arc all elements of .A. (b) Prove that the sum of any two simple functions is a simple function. Let X be a set. let .A be a a-algebra in X. and let/ be a function from X into [ - ao. ao]. Prove that the following are equivalent. (a) / is measurable. (b) {xl/(x) > r} e .A for every re Q. (c) {xl/(x) < r} e .A for every re Q. (d) {xi /(x) ~ r} e .A for every re Q. (e) {xl/(x) ~ r} e .A for every re Q. Let X be a set and .A be a a-algebra in X. Let {.f.}:'. 1 be a sequence of measurable real-valued functions on X. Prove that the set A = {x e XI {.f.(x)}:'. 1 converges in R} is in .A. , Let/ and g be measurable finite-valued functions on X with g(x) 0 for all x e X. Prove that fl g is measurable. Let/ and g be measurable functions on X. Prove that the sets {xl/(x) ~ g(x)} and {xl/(x) = g(x)} are measurable. Prove that if/ is a real-valued measurable function. so is I/ I".c > 0. Let .A be a a-algebra in a set X and let Ee .A. Let .A• be the relative a-algebra (see Exercise 86.9). (a) Let/: X -- [-ao. ao] be .A-measurable. Prove that /IE is .6 rmeasurable. (b) Let/: E -- [ - ao• ao] be .A rmeasurable.

*

Define

g(x)

= {/(Ox)

xeE xeX\E

Prove that g is .A-measurable. 87.13• Prove the following theorem due to F. Riesz. Let (X, .A.µ) be a measure space and let/ and {/.} be measurable functions on X such that for every ~

> o.

lim µ( {x 11/(x) - .f.(x)I ~ ... ao

~ })

=0

Then there exists S e .A with µ(S) = 0 and a subsequence {/.,.} such that lim.t.. ao/.,.(x) = /(x) for every x e X\S. 87.14• Let (X • .A,µ) be a measure space with µ(X) < ao. Let{/.} be a sequence of measurable functions on X which converges to/ on X. Prove that for any e > 0, there exists a set Ee .A with µ(E) < e such that {/.} converges to / uniformly on X\E.

388

The Lebesgue Integral

Integration on Positive Measure Spaces

88.

We now develop a theory of integration on any positive measure space (X, ..II, µ). The simplicity of the development and the power of the results are the real beauty of this part of mathematics. In the following X is a set, ..II is a a-algebra in X, and µ is a positive measure on ..II. We begin by defining the integral of a nonnegative simple measurable function on X.

Definition88.1 Let s be a nonnegative simple measurable function in X, s = :I:j. 1 c,xA,• where A, e ..II and c1 ~ 0 for i = 1, ... , n. We define the integral of s over X with respect to µ to be

J

s dµ

=

X

r,

l• l

c,µ(.A 1)

Notice that Ix s dµ ~ 0. Consider (P, 9'(P), µ), where µ is counting measure. Supposes = {s11} is a nonnegative simple function on P. If s = :I:i'..1 c,xA,• where c1 > 0 for I= 1, ... , m,

J P

~

M

sdµ =

L l•l

c,µ(A

1)

=

L s,. 11•1

This sum is finite if and only if for some N, s,. = 0 for n ~ N. The reasoning for our choice of the number Ix s dµ as our integral of s over Xis clear, since we are averaging s over the set X and weighing the various disjoint subsets of X, where s takes on different values, by means of our measureµ. We used a similar method when integrating step functions over an interval [a, b] with the Riemann integral. We use the following definition to extend Definition 88.2 to the class of all nonnegative measurable functions on X.

Definition88.2 Let/ be a nonnegative extended real-valued function on X. We define the integral off over X with respect to µ to be

JIdµ x

= sup

{t

s dµ I s

is a simple measurable function on X with

s st}

o s

Since O = Ix O dµ is in the set on the right, Ixf dµ is well defined. It is easily verified (Exercise 88.1) that Definitions 88.1 and 88.2 are consistent. Returning again to (P, 9'(P}, µ), where µ is counting measure, suppose

88. Integration on Positive Measure Spaces

389

{a.} is a nonnegative sequence. First, suppose I: a. number. Choose N such that M < I::. 1 a,,.Define

a,.

n> N

s a; so

Then s is a simple function, s

M
0. Choose N such that

L.:°-N+ 1 a. < a.

= {0

sII Then s is a simple function, 0

11•1

< Since

n s N n > N 00

La.=La,.+

11=1

e. Let

s s s a, and N

00

(88.1)

11•1

J.

s dµ

I::-'.. 1 a. < I• s dµ + e for

> .-71'+1 +es

a,.

J.

adµ+

e

every e > 0, it follows that

f a,. SJ• adµ

••1

This inequality combined with (88.1) shows that

f a.= J• adµ

••1

Definition88.3 Let / be a measurable function on X. Let /+ and

1- be defined as in Corollary 87.6. Since/+ and 1- are measurable, nonnegative functions, lxf+ dµ and fxf- dµ are defined. If at least one of the numbers l x f + dµ or Jx 1- dµ is finite, we define the integral off over X with respect

370

The Lebeague Integral

toµ to be

t/+

=

L/dµ

dµ - Jx1- dµ

H both Ix/+ dµ and Ix/- dµ are oo, we say that Ix/ dµ is not defined. H both lx / + dµ and Ix 1- dµ are finite, then Ix f dµ is finite, and in this case we say that/ is Lebesque integrable on X with respect to µ, and we let !i'(X, JI,µ) denote the set of all suchf It is easily verified (Exercise 88.2) that Definitions 88.2 and 88.3 are consistent. Again consider (P, a'(P), µ),whereµ is counting measure. Let {a.} be an = p. and a; = q., wherep. and q. are defined in arbitrary sequence, then Section 26. The proof of Theorem 26.2 shows that 1:.:°1 p. and 1:-"°. 1 q. converge if and only if 1:-"'. a. converges absolutely. Since I. a+ dµ = 1 1:.:°1 p. and I. a- dµ = L:'=1 q., it follows that {a.} e !i'(P, a'(P), µ) if and only if 1:-'°.. 1 a. converges absolutely, and in this case we have

a:

f a..

J

adµ=

•=1

p

Theorem88.4 (i) Hf, g e !i'(X, JI,µ) and/(x) s g(x) for x e X, then Ixfdµ

Ixg dµ

S

(ii) H/is measurable, µ(X) < oo, and for some a, be R we have a b for x e X, then aµ(X)

s

t/

dµ

s

s /(x) s

bµ(X)

(iii) Hf is measurable and bounded on X and if µ(X) < oo, then/ e !i'(X, JI,µ).

(iv) H/e !i'(X, JI,µ) and c e R, then c/e !i'(X, JI,µ) and

Ixcfdµ = c Jxfdµ (v) If µ(X)

= 0, then every measurable function/ on Xis in !i'(X, Ixfdµ

JI, µ) and

=0

Proof We prove (i) only. The other parts are proved similarly and are left as exercises(see Exercise 88.3). Suppose O s /(x) s g(x) for x e X. If s is a measurable simple functic~

88. Integration on Poaltlve Meaaure SpacN

with O s s

s /, then also O s s s

371

g; hence

s ]i.gdµ

fxsdµ

Thus Ix g dµ is an upper bound for the set of numbers

jx s dµ with

Os s

sf

j

It follows that xi dµ

Jx g dµ.

s

Let/, g e !i'(X, JI, µ) with f(x) s g(x) for x e X. It is easily verified that ~ g-. By the special case above

/+ s g+ and/-

fxt+ dµ S

jxg+ dµ

~

Lf-dµ

Therefore

jx f dµ = jx 1+dµ - Jx ,-

dµ s

jx g

+ dµ

-

L L =J g-dµ

g - dµ

x g dµ

■

If/ e !i'(X, JI, µ), we sometimes wish to integrate/ on a set E e JI. This integral, denoted I Bf dµ, is defined to be Ix fxs dµ. It is easy to show (see Exercise 88.4) that if Ix f dµ is defined, then also Ix fxs dµ is defined so that IBf dµ is defined for each set Ee .,II. Our immediate goal is to prove the Lebesgue monotone convergence theorem (Theorem 88.6). The proofs of other important theorems such as the linearity of the integral (Theorem 88.11) and other powerful convergence theorems (Theorems 88.13, 88.14, and 88.15) will depend on Theorem 88.6. First we must establish a lemma. Lemma88.5 Lets be a nonnegative measurable simple function. Let {E11} be a sequence of sets in .,II with E,. c E11+ 1 for n = l, 2 ... and LJ,:0. 1 E,. = X. Then lim

1 ◄

co

f

B,,

s dµ

=

j s dµ X

Proof We show that l(E) = Is s dµ defines a positive measure on JI. Since l(.0) = 0, we need only verify the countable additivity of l. Supposes ==Li- 1 c,xA.,•For Ee .,II we have

I

E

sdµ •

=

f sxsdµ Jx

•

J (r X

1•1

C1XA. 1)xsdµ

JrX (I•f 1 C1XA.1nB)dµ ==I•f 1 c,µ(A., () E)

372

The LebNgue Integral

Let {£ 1} be a pairwise disjoint sequence in .,It. Letting E we have

f

= LJ~1 £ 1,

Ec,µ(A. E) = ,t1 c{µ( A.," C91 E.))] - Ec,[f µ(A., EJ] =- f [f c,µ(A, E.)] = ■ f f s dµ = f ).(£.) ••1 J... ••1

).(E) -=

=

s dµ

E

1 ri

l•l

ri

■ •1

l•l

ri

•• 1

I• 1

Thus ). is a positive measure on .,If. Let {E.} be a sequence in .,If with E,. c E.+ 1 for n U:.1 E,, = X. By Theorem 86.S(v), lim

... ao

Js B,,

dµ

= ...Jim).(£.) = ).( LJE,,) = l(X) = ao •• 1

= I,

Js X

2, ... and

■

dµ

1beorem88.6 (Lebesgue MonotoneConvergenceTheorem) Let {/.} be a sequence of nonnegative measurable functions on Xsuch thatf.(x) s J.+1(x) for n = 1, 2, ... and xe X. Letf(x) = lim,,.. 00 /.(x). Then/is measurable on Xand Jim

■ -ao

Proof.

JJ. X

dµ

=

J/ X

dµ

By Corollary 87.6(iii), / is measurable on X. Since

fx.r. jx/.+ dµ S

1

dµ

Jim,,.. 00 Ix f. dµ exists in [O,oo]. Denote this limit by L. Since J. s / for n = I, 2, ... , we have IxJ. S Ix/for Letting n -+ oo, we obtain

n = I, 2, .... (88.2)

Let t e (0, I) and let s be a simple measurable function with O :S s :Sf. Set

E,. = {x e XI J.(x) ~

ts(x)}

373

88. Integration on Poaltive Meaaure Spaces

Then Enc En+l for n

= l,

2, ... and X

lim

n ◄

Jf.

Now

co

and letting n

ts dµ

=

J

ts dµ

dµ

~

J I,,

dµ

~

J

En

X

ts dµ

Bn

oo, we obtain ➔

J

L = lim • ◄

Letting t

J

Bn

X

= LJ!°. 1 E •. By Lemma 88.5,

f,, dµ

co

JC

~

J

lim

• ◄

co

ts dµ =

Bn

J

ts dµ = t

X

L

~

s dµ

JC

l, we obtain L c!!:Ix s dµ. By the definition of ➔

J

Ix/ dµ, we have

Ixfdµ

which, together with inequality (88.2), establishes the theorem.

■

As an application of Theorem 88.6, let {a., 1} be a nonnegative double sequence. Let f,,(m) = :I:r.1 a,., 1• Consider the measure space (P, a'(P), µ), where µ is counting measure. Then {/,,} is an increasing sequence of nonnegative functions on P. Also,/(m) = lim. .. cof,,(m) = :I:~1 a_,1• By Theorem 88.6, lim

•◄ co Now

Ir/

dµ

r I,, dµ = Jr

I

(88.3)

f dµ

p

= :I::. 1 (l:~ 1 a...J.Also,

JI,, P

dµ

= ••1f

(f a.,1) • f ( f a,.,1) 1•1

1•1

••1

Thus equation (88.3) may be rewritten ) - L co(coL a.., ) Lco(co••1L a,.,, ••1 1•1

1•1

We earlier established this result as Lemma 29.3. Before establishing our next major result, which is essentially a generalization of Lemma 88.5, we must establish a special case of the linearity of the integral. Lemma 88.7

If/ and

g are nonnegative measurable functions on X, then

Ix(f + g) dµ = IXf dµ + IXg dµ

374

The Lebesgue Integral

First suppose/ and g are measurable simple functions. We may

Proof. write

where {..41} and {B1} are pairwise disjoint and u..41 = X f + g = :E~1 :Ej.1 (a1 + b1)1,,.1na,• Therefore

J (/+

±f

g) dµ -

(a,

1•1/•l

X

II

-=

= uB 1•

Then

+ b1)µ(A.1f"'I B1)

Ill

Ill

L a, JmlL µ(A., f"'I B1) + J•lL 1•1

Ill

b1 L µ(A,

B1)

f"'I

l•l

Now suppose that/ and g are nonnegative measurable functions on X. By Theorem 87.10 there exist sequences of simple measurable functions, 0 S s 1 s s 2 s ···and 0 s 11 s 12 s ···such that lim,....GO s,.(x) = f(x) and lim,....GO l,.(x) = g(x) for all x e X. Clearly lim,,...GO (s,, + 1,.Xx) = f(x) + g(x) for all x e X; so by Theorem 88.6 and the special case above

J (/+ X

g) dµ =

J X

lim (s,. + t,.) dµ = lim 11... GO

(J

11-+GO

j t,.dµ) = j (lim s,.) dµ + j (lim = lim

,. ... GO

X

s,,dµ

+

X ,.... GO

X

(s,. + 111) dµ

J =J

= ,.lim ...

X

X ,. ... GO

J

GO

1,,)dµ

X

X

s,.dµ

+ ,.lim ...

GO

f dµ

+

j

X

J X

t,.dµ

g dµ

■

Theorem88.8 Let/be a nonnegative measurable function on X. Then l(E)

=

L/

dµ

defines a positive measure on .,II. If/e !i'(X, .,II,µ), A.(E)= fEf dµ defines a countably additive set function on .,II. Proof. Let/be a nonnegative measurable function on X. Since l(0) = 0, we need only verify the countable additivity of l. Let {£ 1} be a pairwise disjoint sequence in .,II and let E = 1 £ 1• Define

Uj;.

Then {.f,.} is a sequence of nonnegative measurable functions on X such that

88. Integration on Poaltlve Meaaure Spacea

J. ~ J.+1 for n = l, 2, ....

37&

By Theorem 88.6, since lim....00 /.

lim

JJ. = J dµ

11-+oo X

X

= fxs,

Jx. dµ

By Lemma 88.7

JJ. dµ = Jf ( f fxs,)dµ = t J fxs, dµ = t l(E X

l(E)

Thus

I• 1

X

=

J

I• 1

fxs dµ

X

= lim

I• 1

X

f l(E

•-+oo I• l

1)

=

1)

f l(EJ

I• 1

If /e !i'(X, JI,µ), one shows that l(E) = Isl dµ defines a countably additive set function on JI by applying the result just established to/+ and

1-. ■

With modest additional hypotheses, a converse of Theorem 88.8 can be proved. This converse is a special case of a more general result known as the Lebesgue-Radon-Nikodym Theorem (see Hewitt and Stromberg, 1965, p. 315). One can show that ifµ and A.are positive measures on the u-algebra JI satisfying µ(X) < oo and l(X) < oo and further µ(£) = 0 implies l(E) = 0, then there exists a nonnegative measurable function f on X such that A.(E) = j.Jdµ

for every Ee JI. In Section 57 we said that a property P holds almost everywhere if the set of points on the line where P does not hold is of measure zero. We now generalize this definition to an arbitrary measure space.

Definition88.9 If a property P holds for every x e E\A and if µ(A) = 0 (where E, A e JI, of course), then we say that P holds for almost all x e E with respect to µ or that P holds µ-almost everywhere on E or that P holds

µ-a.e. on E. An important example is given by the next theorem.

Theorem88.10 If/, g e !i'(X, JI, µ) and f L.fdµ Proof.

= g µ-a.e. on X, then

= Ixgdµ

Let E

= {x e X lf(x)

#: g(x)}

378

The LebesgueIntegral

By definition, µ(E) = 0. Now

J

X\B

fdµ

=

J

X\B

g dµ

Therefore, by Theorem 88.8,

J/ dµ = J /dµ + J/dµ = J X\E

JC

B

g dµ +

JC\B

J B

g dµ =

J

g dµ

JC

■

We next show that the integral is linear. Although it may appear that this fact should foilow immediately from the definition it does not. Our proof depends directly on Theorem 88.4 and indirectly on Theorems 87.10 and 88.6. Theorem88.11 If/, g e Z(X, .II,µ), then/+

Jx(/+

g)dµ

g

e Z(X, .II,µ) and

= Jxfdµ + Ixgdµ

This result has been established (Lemma 88.7) for / and g nonnegative. Suppose/~ 0 and g s 0. Let A = {x e XI (/ + g)(x) ~ O} and B = {x e XI(/+ g)(x) < O}. Then/+ g,f, and -g are nonnegative on .A. Therefore, by Theorem 88.4(iv) and Lemma 88.7, Proof.

Ltdµ=

L

0, a contradiction. Thus µ{A.11) = Ofor n = l, 2, .... Now ) A = U:°-1 A,.; so Os µ(A)= µ(LJ:'A s :E:°µ(A..) = 0, and there11 1 1 fore µ(A.) = 0.

■

Exercin•

In these exercises, unless specified otherwise, (X, JI, µ) is a positive measure space.

88.1

Give the details of the argument that Definitions 88.1 and 88.2 are consistent for nonnegative simple measurable functions.

380

The Lebesgue Integral

Show that Definitions 88.2 and 88.3 are consistent for nonnegative measurable functions. 88.3 Prove Theorem 88.4, parts (ii), (iii), (iv), and (v). 88.4 Prove that if Ixfdµ is defined, then also Ixfxsdµ isdefinedforeach Ee Jt. Prove also that if/ e !i'(X, Jt, µ), then/x. e !i'(X, Jt, µ)for each Ee Jt. 88.5 Let (X, Jt, µ) be a positive measure space. Let Ee Jt. Let (E, Jt., l),where l(A) = µ(An E), be the relative measure space (see Exercise 86.9). (a) Let/e !L'(X, Jt, µ). Prove that/lEe !i'(E, Jt., l) and 88.2

f.JIEd.l

= fxtx.dµ

(b) Let/e !i'(E, Jt•, l). Define g(x)

= {/~)

xeE xeX\E

Prove that g e !L'(X, Jt, µ) and

=

= fxux.dµ

fxudµ

f.1t1.t

88.6 Let (X, Jt, µ) be a positive measure space with µ(X) < 00. Suppose lim,,.. ..,f,,(x) = /(x) for every x e X and for some M, If,,(x)I < M for n = 1, 2, • • • and for every x e X. Prove that lim

,. .. ..,

f /.dµ = f Idµ X

X

88.7 Show that Theorems 88.6, 88.13, and 88.15 remain valid if for all x is replaced by for almost all x with respect µ. 88.8 Let (X, JI,µ) be any positive measure space. Suppose / e !i'(X, .K, µ). Prove that/(x) must be finite µ-almost everywhere on X. 88.9 Let X = {l, 2, 3, 4, 5 }, Jt = a'(X), andµ be counting measure. Let E = {1, 2} and define I,, = xs if n is odd, I,, = 1 - X• if n is even. What is the relevance of this example to Fatou's lemma? 88.10 Let (X, Jt, µ) be any positive measure space. Suppose µ(X) < 00 and {l,,}:'. 1 is a sequence of bounded real measurable functions on X and/. - / uniformly on X. Prove that lim

■ -+ex,

f f.dµ = f fdµ X

JC

88.11 Show that the hypothesis "µ(X) < 00" cannot be omitted from Exercise 88.10. 88.12 Let (X, Jt, µ) be any positive measure space. Suppose /e !L'(X, Jt, µ). Prove that for every e > 0 there exists a o > 0 such that

f.Ill dµ < e

whenever µ(E)
I. Then V c U:°- 1 I.,.; so by Definition 89.2, we have m(V) ~ l::.. 1 m*{/.,.)= l:~..1 m*(/.,.) = Le., m*(/.) = m•( V). Therefore m(V) s m*(V)

for all open Vin R

(89.2)

If m*(V) = oo, then it is clear that m*(V) = m(V). So suppose m*(V) < oo, then V = U-, I,. with I.,.= (a,.,b.,.)and a.,.,b,. e R (both finite) and the In pairwise disjoint. Now suppose {u,.}:°.1 is any collection of open intervals or 0 such that V c 1 un. If J is finite, it is clear that m*(V) = L.., (bn -

U:°-

384

The Lebelgue Integral

aJ = that

L~=1 (b. -

:r:=m*(e1J(verify). But if J = {l, 2•...

aJ :5

L (b. '

1

L m*(e1.) Cl)

a,,) :5

■ •l

Therefore

m*(V) =

for I=

} we have

I, 2, ...

■ •l

L (b. ••I

L (b. Cl)

a.) =

L (b. '

a.) = lim

■ •l

I ◄

«>

a.)

■ •l

Cl)

L m*(e1J ••l We conclude, since m(V) is the inf of the sums :5

L«>. 1 m*(e1.),that (89.3)

m*( V) :5 m( V) Equations (89.2) and (89.3) give us that m*(V)

R.

= m(V) for

all V open in

■

=

Theorem89.4 If A,, e a'(R) for n

m(

I, 2, ... , then

LJA.):5 ••1f

••1

m(AJ

By Definition 89.2, it is clear that for any sets A

Proof.

c

Bin R we have

m(.A.):5 m(B)

U:.

Since Aa:c = oo for some k, it follows that oo = 1 A., if m(.A.1:) m(.A.J :5 m(LJ:., 1 A.); therefore,

m(

UA.)= oo = f m(AJ

■ •1

■ •l

Therefore, we assume m(.A.J < oo for n = I, 2, .... Let e > 0. By Definition 89.2 for each positive integer n we may choose a covering u1•>}:°-1 of A. 11 such that l:t'°-1 m(11•>):5 m(.A..)+ e12•. Then 1 l 11 >is a countable covering of LJ:. 1 A,.; so by Definition 89.2, we have

u:.. u~

m(.Q

1

A.) .t 1:t m(/1•>) .t m(A,,) +; ..)=.tim(.A.J+ :5

1

:5

1

1

(

e

Since e > 0 was arbitrary, we have

m(

UA,.) f m(A,.)

••1

:5

••1

•

We have constructed a function m: a'(R)-. [O,oo] with the following properties:

89.

Lebesgue Measure on A

386

(i) m(Qf) = 0. (ii) m(A) s m(B) if A c B c R. (iii) m(LJ:°-1 AJ s l::°- 1 m(A.J for any sequence {A,.}:°-1 of subsets of R. We now return to the abstract setting, because the development of a a-algebra on which m is countably additive does not depend on R, but only on the properties stated above.

Definition89.5 Let X be a set. A function µ from 9'(X) into [0, oo] is called an outer measureif (i) µ(Qf) = 0. (ii) µ(A) s µ(B) whenever A c B c X. (iii) µ(LJ:. 1 A.J s l::°- 1 µ(AJ for all sequences {A.,.}:°1 of subsets of X. This last property is called the countable subadditivity ofµ. We note that countable additivity demands the pairwise disjointness of the sets involved, while countable subadditivity does not. The following definition is due to Caratheodory and is the key to defining a useful a-algebra on which µ will be a positive measure.

Definition89.6 Let X be a set and let µ be an outer measure on 9'(X). A subset S of X is said to be µ-measurableif µ(T)

= µ(Sn

T)

+ µ(S' n

T)

for all Tc

X

Let JI" denote the family of all µ-measurable subsets of X. Since T

= (S n

T) u (S' n T), it follows from the subadditivity ofµ that

µ(T) s µ(Sn T)

+ µ(S'

n T)

Thus the equation in Definition 89.6 holds precisely when µ(T) :.:::µ(S n T)

+ µ(S' n

T)

A subset S of X is not µ-measurable if there exists a set T µ(T) > µ(S n T)

+ µ(S' n

c X

with

T)

Notice that in this case µ is not even finitely additive on the disjoint pair {Sn T, S' n T}. It can be shown (see Hewitt and Stromberg, 1965) that there exists a non-m-measurable set S c R, where m is given in Definition 89.2. Thus mis not even finitely additive on 9'(R). Again, we see that we must restrict m to a a-algebra which is properly contained in 9'(R). We now proceed to show that JI,, is a a-algebra in X and µ is a positive measure on JI,,.

The Lebeegue Integral

388

Every subset S of X such that µ(_S)= 0 is in JI,..

'Theorem 89.7

Proof. Let S c X and µ(S) = 0. Let T be any subset of X. Then by Definition 89.5(ii), since S (') Tc S, we have O s µ(_S(') T) s µ(S) = 0, which implies µ(S (') T) = 0. By (iii) and (ii) of Definition 89.5, we have µ(T)

s

µ(S (') T)

+ µ(S'

(') T)

= µ(S'

(') T)

s

µ(T)

Therefore, it follows that µ(T)

so Se JI,..

= µ(S (') T) + µ(S'

(') T)

■

'Theorem89.8 If S e JI,,, then S' e .,II,,. Trivial.

Proof.

■

Theorem89.9 If A. and B are in

JI,,,then

A. (') B' is in JI,..

Proof. It suffices to prove that if E c A. (') B' and F c (A. (') B')', then µ(Eu F) = µ(E) + µ(F). Now F = (F (') B) u (F (') B') and Be .,II";so µ(E)

+ µ(F) = µ(E) + µ((F (') B) u (F (') B'))

= µ(E) + µ(F (') B) + µ(F

(') B')

(89.4)

Since E c A. and F (') B' c A.' and A. e .,II,,, we have µ(E)

+ µ(F (') B') + µ(F (') B) = µ(Eu

(F (') B'))

+ µ(F

(') B)

(89.5)

Again Eu (F (') B') c B' and F (') B c B; so µ(Eu (F (') B'))

+ µ(F (') B) = µ(Eu

= µ(Eu

(F (') B') u (F (') B))

(89.6)

F)

Combining equations (89.4), (89.5), and (89.6) µ(E)

Lemma89.10 Then

Let {A.11}:,.

1

+ µ(F) = µ(E u

■

F)

be a sequence of pairwise disjoint sets in .,II,,.

for all Tc

I::.

X

(89.7)

Proof. By Definition 89.S(iii), we have µ(T) s 1 µ(T (') A,.) + µ(T (') CU:°A.,.)'). If µ(T) = oo, then equation (89. 7) follows immediately. 1

88. Lel>Ngue Measure on R

387

Therefore suppose that µ(T) < oo. We first show that µ(T)

t

=

11• I

µ(Tr. A,.) + µ(T n (

lJA,.)')

for all p e P

(89.8)

11• I

We use induction on p. For p = 1, equation (89.8) becomes µ(T) =µ(Tr. A 1) + µ(Tr. At). But this follows since .A1 e JI,,.. Suppose that equation (89.8) is true for a positive integer p and all Tc X. Since A,+ 1 e JI,., we have µ(T)

= µ(Tr. +

t

11•1

+ µ(Tr.

A,+ 1)

µ(Tr. A~+1 n

A~+ 1)

AJ +

= µ(Tr.

A,+

1)

µ(Tr. A~+ 1 n (

lJA

(89.9)

11)')

11•1

The last equality follows from the inductive hypothesis applied to the set Tr. A~+I· Now since .A11 ~ A~+1 for n ,;, p + 1, equation (89.9) can be written µ(T)

= µ(Tr.

A,+ 1)

-:t:

+

t

µ(Tr. AJ

11•1

+ µ(Tr.

A~+1 n (

+µ(Tr.co: A,.)')

µ(Tr.AJ

lJA

11°1

11)')

which is equation (89.8) for p + l. The sequence of numbers {µ(T n 1 A,.)')};'=1 is a nonincreasing sequence bounded below by the number µ(Tr. CU:. 1 A,.)'). Thus it has a limit, which is greater than or equal to µ(Tr. (U:.1 .A11)'). Taking limits in the equality (89.8), we get

+ m_X[.,.,,_

s

1 ,.,.,.

1

= glb {/(x) Ix e [x 1_ i, x,)} for i = I, 2, ... , n. Sinceeachoftheintervals[x = 1,2, ... ,n - I and[x._ 1 ,xJ 1_i,x,)fori is measurable, s is a simple measurable function on [a, b]. Now O s s s /; so by Definition 88.2, we have wherem,

L(f, P)

t

=

m,(x, - x 1_ 1)

=f

JC•.bl

I• 1

s dm s

f / dm

JC•.bl

for every partition P of [a, b]. By Definition 51.3, we have

r·/(x)

J.

dx

r t dm

$

Jca,bl

Let/ e ~[a, b]. Since/ is bounded, there exists a number M such that 0 s /(x) + M for all x in [a, b]. By the above argument, we have

r•(/(x)

+

J•

M) dx

s

f

(/ + M)

J

dm

[a,b]

which reduces to

J• /(x) b

$

J

/dm

C•,bJ

Repeating the argument above for - /, we have

J•

-/(x) dx s

•

J (-/)

dm

[a,bl

which reduces to

J•

J/

/(x) dx 2!::

•

Therefore, J:/(x) dx

= lc.,1, 1/ dm for all/

dm

C•,bl

e ~[a, b]. ■

Exercise•

90.1 Define /.(x) = nxon (0, l]. Prove that {/.} converges pointwise to /(x) = 0 on (0, l] and that each function/. and/is in .5r[0, l], yet

~

fco,11/. * fco.,/

90. Lebesgue Meaaure on [a, b]

386

Explain why each convergence theorem (Jbcorems 88.6, 88.13, and 88.1.S) does not apply here. 90.2 Let n be a positive integer. Select the unique positive integer I such that 2 1- 1 S: n S: 21 - 1. Let r = n - 2 1- 1 • Let/.= xu1-1+(r/21-l),21-l+((r+1)/21-I)] on [0, 1). Let/(x) = 0 on [0, 1). Prove that (a) f.,f e 9'[0, 1) for n = 1, 2, .... (b) lim.....ao Jco.1)/.= /co.1i/, (c) Jim.,...aof.(x) does not exist for any x e [0, 1). Prove that there exists a subsequence {/,.11} of{/.} such that {/ 1111} converges to f m-almost everywhere. 90.3* Let/ e 9'[a, b]. Define

F(x)=f

xe[a,b]

f C••"l

Prove that Fis absolutely continuous, uniformly continuous, and of bounded variation on [a, b]. (The definition of absolute continuity is given in Exercise 56.1.)

90.4

(a) Prove that 9'[a, b] is a vector space if we define operations pointwise. (b) Prove that for f e 9'[a, b]

Ir•.• ]I/I

11/11 =

dm

defines a norm on the vector space 9'[a, b], except that 11/11 = 0 implies 0 is replaced by II/II = 0 implies/= 0 m-almost everywhere. The space 9'[a, b] with this norm is often denoted 9' 1[a, b]. (c) Let g e 9'[a, b], where g = h m-almost everywhere and h is bounded. (In this case, g is said to be essentially bounded.)Define

I=

T(/}

=

f

C•,tl

fe 9'[a, b]

fg

Prove that T is a continuous linear functional on 9'[a, b]. (Actually, every continuous linear function on 9'[a, b] is of this form, but the proof is rather involved. (See Hewitt and Stromberg, 1965.) 90.S (a)• Suppose/, I/I e £t[c, b] for each c satisfying a< c < b and that the I/I converges. improper Riemann integral Prove that/ e 9'[a, b] and that

n.

f J-f•I C•,•l

••

(b) Use (a) to compute Jco,1 1/. where

I (x) = { 1/ ,/i

O < x s; 1 x=O 90.6 State and prove a result similar to that of Exercise 90.S(a) for improper Riemann integrals of the first kind. 0

90.7 Let/be a bounded function on [a, b]. Fix a nonnegative integer n. Let li = (b - a)/2" and x 1 = a + ili for i = 0, 1, ..• , 2". Define

398

The Lebesgue Integral

= glb {/(x)lx e (x,_ 1, x 1)} M, = lub {/(x)lx e (x,_1, x,)} and for x e [a, b], define x e (x,_ 1, x,) L,,(x) = {: otherwise M, XE(X1-1,X1) U,,(x) = { Mo otherwise (a) Prove that L,,(x) ~ L,,+1(x) ~ u. ♦ 1(x) ~ U,.(x) for n = 1, 2, .••. (b) Prove that L,, and U,, are measurable functions for n = 1, 2, .... (c) Prove that {L,,} and {U,,}converge pointwise to measurable functions (hereafter, denoted Land U, respectively). (d) Prove that/e lf[a, b] if and only if L(x) = U(x) m-almost everywhere on [a, b]. (e) Prove that if x t,. x, for any ; corresponding to any n, then L(x) = U(x) if and only if/ is continuous at x. 90.8 Let/ be a bounded, measurable function on [a, b]. Let (c, d) =>fP, then L(f, P) ~ L(f, S) ~ U(f, S) ~ U(f, P) and that for any partitions P and S of [c, d], we have L(f, S) ~ U(f, P) Define

I£:/= lub {L(/, P)} JP:/= glb {U(/,P)}. (c) Prove that g:J= fc•.•1/=lf!f. 90.9• Let F(x, y) be a function on [a, b] x (c, d) and suppose that for each x e [a, b], the function g,,(y) = F(x, y) is differentiable on (c, d). Letting D,.F (x, y) = g;(y) suppose further than for some M, ID,.F(x, y)I < M for all x e [a, b] and ye (c, d). Suppose also that for each ye (c, d) the function /.,(x) = F(x, y) is m-measurable in [a, b]. Set G(y) = fc•.•J F(x, y) dm(x) D,.F(x, y) dm(x). and prove that G'(y) = fc•.•J

91. The Hilbert Spaces ~ 2(X, .A, µ)

397

91. The Hilbert Spaces 2 2 (X, JI, µ) We complete our text with a discussion of a large class of Hilbert spaces. Let (X, JI, µ) be a positive measure space and define 2 2 (X, JI,µ)= {/1/ is a real measurable function on X and Ix 1/12 dµ < oo, with the identification that/ = g if and only if/ = g almost everywhere on X}.

Deftnition91.1

We shall prove that 2 2 (X, JI, µ) is an inner product space and is complete, i.e., that 2 2(X, JI, µ) is a Hilbert space. Lemma 91.2 If/, g e 2 2 (X, JI,µ), then/g e 2(X, JI,µ). By Theorems 87.4 and 87.8 1/gl is measurable. Let

Proof.

A

= {x e X 11/(x)I

2:: lg(x)I}

Then A e JI and O

~

J X

1/gl dµ

=

2

S Jx 1/1 dµ +

Therefore,

Jge 2(X,

J A

1/gl dµ +

J

A'

1/gl dµ

~ Jf

,t

2

1/1 dµ +

J ,t•

2

191 dµ

Ix191 dµ < oo 2

JI, µ).

■

1beorem 91.3 2 2 (X, JI,µ) is an inner product space with inner product

(f, g)

=

Ixfg dµ

for all/, g e 2 2(X, JI,µ)

Proof. Using Lemma 91.2 and the properties of the integral everything follows as long as we can show that 2 2 (X, JI,µ) is a vector space. The only difficulty is in proving that if/, g e 2 2 (X, JI,µ), then/+ g e 2 2 (X, JI,µ). But using Lemma 91.2 again,

IxIf+

2

91 dµ S

Ix1/1 dµ + 2 Ix1/gl dµ + Ix191 dµ < oo 2

2

Note that with our identification, (/, /) = 0 if and only if / = 0 almost everywhere on X. ■ We also note that in 2 2 (X, JI,µ) we have 11/11= Ux1/12 dµ) 112 and that the Cauchy-Schwarz inequality can be written as Ix 1/gl dµ S 11/11 · llgll for all/, g e 2 2 (X, JI,µ).

Theorem91.4 2 2 (X, JI,µ) is a Hilbert space.

The LebNgue Integral

398

Proof. We must show that !l' 2 (X, .,If,µ) is complete. Let {/.} be a Cauchy sequence in !l' 2 (X, .,If, µ). There is a subsequence {/.,} j. 1 with n 1 < n2 < · · · such that

for j

=

I, 2, ...

(91.2)

and

Set

By equation (91.2) and using the fact that the triangle inequality holds for II·II, we see that 119 < I for k = l, 2, .... Now using Theorem 88.14 1 11 (Fatou's lemma), we have

JxIf lf.,.

2 11911 = f

J•l

2 1 -

f ilim inf/ f If..,•• - t.))dµ Jx .. V•l 2

.f..,11 dµ

==

CID

Jf (J•f 1 If.,., - J.,1) dµ = 2

s lim inf i ..

CID

JC

lim inf

i ..

Therefore 11911 s I. This implies, in particular, that so that the series

f. 1(x) +

2 119 1 11:5

I

CID

g(x)
.} in 9' 2 [a, a + 2n]. Therefore, we may develop Fourier series in 9' 2 [a, a + 2n] relative to {,.}. In fact, throughout the remainder of this section, we will assume that all Fourier series are relative to this orthonormal set. Our immediate goal will be to show that the trigonometric set is a complete orthonormal set in 9' 2[a, a + 2n]. Along the way we shall derive a result is (Theorem 91.6) which is interesting in its own right. The proof that {cf>.} complete in 9' 2 [a, a + 2n] (Theorem 91.7) parallels the proof of Theorem 78.5.

Theorem91.5 If f e 9' 2 [a, b] and e > 0, there exists a simple function he 9' 2 [a, b] such that II/ - hll < e. Proof. First suppose that/is nonnegative on [a, b] and e > 0. By Theorem 87.10 there exists an increasing sequence {s,.}of nonnegative simple functions such that lim_...00 s.(x) = f(x). Now 1/(x) - s,.(x)l2 ~ [f(x)] 2 and since 2 / e !l'[a, b], we may use the Lebesgue dominated convergence theorem (Theorem 88.15) to conclude that lim a-+oo

J

If - s.12 =

C•,bJ

J

lim

C•,bJ11-+00

2 1/ - s,.1 =0

In particular, there exists n such that II/- s,.11< e. For an arbitrary f e 9' 2 [a, b] and e > 0, choose simple functions s and t, as above, such that II/+ - sll < e/2 and 11/-- tll < e/2. Then h = s - tis a simple function and

It follows from the next result that the completion, in the sense of Theorem 46.7, of !M[a,b] relative to the norm 11/11 = (J/ 2)½,is 9' 2 [a, b]. Similarly (see Exercises 91.4 and 91.5), it can be shown that the completion of ~[a, b] relative to the norm 11/11 = I Ill is !l'[a, b]. neorem 91.6 If f e 9' 2 [a, b] and e > 0, there exists a step function EM[a, b] such that 11/- ull < e.

ge

Proof First assume that f = Xa, a characteristic function, where E is a measurable subset of [a, b]. Let e > 0. By Theorem 89.13(i) there exists an open set U such that E c U and m(U) < m(E) + e2/2. We may write

The Lebesgue Integral

400

U = U:. 1 U,,, where {U,,} is a collection of pairwise disjoint open intervals. Since m( U) = 1 m( U,,) and m( U) is finite, the series 1 m( U,,) con2 verges. Therefore, there exists N such that I::.N+im(U,,) < e /2. Let V = 1 u,,. Then Xv is a step function. We will show that llxv- XEII < t. First, note that since V\E c U\E, m( V\E) s m( U\E) < e2/2. Also, since

I::.

I::.

u~..

u,,,m(E\V)

E\V C u:.N+1 Ica,b) I

= Ic.,b)\(YvB) =

I

J +j Y\B

t

+

B\Y


0. By Theorem 91.6 there exists a step function g e a[a, a + 21t]such that t

11/- gll < 2 Let {t,,}be the sequence of partial sums of the Fourier series of g. By Theorem 78.5, lim,,.. 00 Ilg - t,,11 = 0. Thus, there exists a positive integer N such that if n 2!::N,

Ilg - t,,11