Elements of the Theory of Functions and Functional Analysis, Volume 2, Measure. The Lebesgue Integral. Hilbert Space

ELEMENTS OF THE THEORY OF FUNCTIONS AND FUNCTIONAL ANALYSIS Volume 2 Measure The Lebesgue Integral Hilbert Space

All. Kolmogorov and S. V. Fomin

ELEMENTS OF THE THEORY OF FUNCTIONS AND FUNCTIONAL ANALYSIS VOLUME

MEASURE. THE LEBESGUE INTEGRAL. HILBERT SPACE

OTHER GRAYLOCK PUBLICATIONS

KHINCHIN: Three Pearls of Number Theory Mathematical Foundations of Quantum Statistics PONTRYAGIN: Foundations of Combinatorial Topology

NOVOZHILOV: Foundations of the Nonlinear Theory of Elasticity KOLMOGOROV and FOMIN: Elements of the Theory of Functions and Functional Analysis. Vol. 1: Metric and Normed Spaces

PETROVSKIT: Lectures on the Theory of Integral Equations ALEKSANDROV: Combinatorial Topology Vol. 1: Introduction. Complexes. Coverings. Dimension The Betti Groups Vol. Vol. 3: Homological Manifolds. The Duality Theorems. Cohomology Groups of Compacta. Continuous Mappings of Polyhedra

Elements of the Theory of Functions and Functional Analysis VOLUME 2 MEASURE. THE LEBESGLTE INTEGRAL. HILBERT SPACE BY

A. N. KOLMOGOROV AND S. V. FOMIN TRANSLATED FROM THE FIRST (1960) RUSSIAN EDITION by HYMAN KAMEL AND HORACE KOMM

Department of Mathematics Rensselaer Polytechnic Institute

GI?A YLOCK

PRESS

ALBANY, N. Y. 1961

Copyright ©

1961

by GRAYLOCK PRESS

Albany, N. Y. Second Printing—January 1963

All rights reserved. This book, or parts thereof, may not be reproduced in any

form, or translated, without permission in writing from the publishers.

Library of Congress Catalog Card Number 57—4134

Manufactured in the United States of America

CONTENTS vii

Preface Translators' Note

ix

CHAPTER V MEASURE THEORY

33. The measure of plane sets 34. Collections of sets 35. Measures on semi-rings. Extension of a measure on a semi-ring to the minimal ring over the semi-ring 36. Extension of Jordan measure

1

15

20 23

37. Complete additivity. The general problem of the extension of 28 measures 38. The Lebesgue extension of a measure defined on a semi-ring with 31 unity 39. Extension of Lebesgue measures in the general case 36

CHAPTER VI MEASURABLE FUNCTIONS

40. Definition and fundamental properties of measurable functions.. 41. Sequences of measurable functions. Various types of convergence.

38 42

CHAPTER VII THE LEBESGUE INTEGRAL

42. The Lebesgue integral of simple functions 43. The general definition and fundamental properties of the Lebesgue integral 44. Passage to the limit under the Lebesgue integral 45. Comparison of the Lebesgue and Riemann integrals 46. Products of sets and measures 47. The representation of plane measure in terms of the linear measure of sections and the geometric definition of the Lebesgue integral 48. Fubini's theorem 49. The integral as a set function V

48 51 56

62 65

68 72 77

CONTENTS

CHAPTER VIII SQUARE INTEGRABLE FUNCTIONS

50. The spaceL2 51. Mean convergence. Dense subsets of L2 52. L2 spaces with countable bases 53. Orthogonal sets of functions. Orthogonalization 54. Fourier series over orthogonal sets. The Riesz-Fisher theorem. 55. Isomorphism of the spaces L2 and 12

79 84 88

...

91 96 101

CHAPTER IX ABSTRACT HILBERT SPACE. INTEGRAL EQUATIONS WITH SYMMETRIC KERNEL

56. Abstract Hilbert space 103 57. Subspaces. Orthogonal complements. Direct sums 106 58. Linear and bilinear functionals in Hilbert space 110 59. Completely continuous self adjoint-operators in H 115 60. Linear operator equations with completely continuous operators.. 119 61. Integral equations with symmetric kernel 120 SUPPLEMENT AND CORRECTIONS TO VOLUME 1 INDEX

123 127

PREFACE This book is the second volume of Elements of the Theory of Functions and Functional Analysis (the first volume was Metric and Normed Spaces,

Graylock Press, 1957). Most of the second volume is devoted to an exposition of measure theory and the Lebesgue integral. These concepts, particularly the concept of measure, are discussed with some degree of generality. However, in order to achieve greater intuitive insight, we begin with the definition of plane Lebesgue measure. The reader who wishes to do so may, after reading §33, go on at once to Ch. VI and then to the Lebesgue integral, if he understands the measure relative to which this integral is taken to be the usual linear or plane Lebesgue measure. The exposition of measure theory and the Lebesgue integral in this volume is based on the lectures given for many years by A. N. Kolmogorov

in the Department of Mathematics and Mechanics at the University of Moscow. The final draft of the text of this volume was prepared for publication by S. V. Fomin.

The content of Volumes 1 and 2 is approximately that of the course Analysis III given by A. N. Komogorov for students in the Department of Mathematics. For convenience in cross-reference, the numbering of chapters and sections in the second volume is a continuation of that in the first. Corrections to Volume 1 have been listed in a supplement at the end of Volume 2. A. N. KOLMOGOROY

S. V. F0MIN

January 1958

vi'

TRANSLATORS' NOTE In order to enhance the usefulness of this book as a text, a complete set of exercises (listed at the end of each section) has been prepared by H. Kamel. It is hoped that the exercises will not only test the reader's understanding of the text, but will also introduce or extend certain topics which were either not mentioned or briefly alluded to in the original. The material which appeared in the original in small print has been enclosed by stars (*) in this translation.

ix

Chapter V

MEASURE THEORY The measure /1(A) of a set A is a natural generalization of the following concepts: of a segment 1) The length 2) The area 8(F) of a plane figure F.

3) The volume V(G) of a three-dimensional figure G. 4) The increment çc(b) — çc(a) of a nondecreasing function cc(t) on a half-open interval [a, b). 5) The integral of a nonnegative function over a one-, two-, or threedimensional region, etc. The concept of the measure of a set, which originated in the theory of functions of a real variable, has subsequently found numerous applications in the theory of probability, the theory of dynamical systems, functional analysis and other branches of mathematics. Tn §33 we discuss the concept of measure for plane sets, based on the area of a rectangle. The general theory of measure is taken up in The reader will easily notice, however, that all the arguments and results of §33 are general in character and are repeated with no essential changes in the abstract theory. §33. The measure of plane sets

We consider the collection of sets in the plane (x, y), each of which is defined by an inequality of the form a

x

a <x b, x 0 there are sets A', A" E A'

A

such that

m(A" \ A') ji(A),

m(A')

m(A"\A') = for arbitrary A', A" E

m(A")

m(A')

—

such that A'

A

h >0

A". Hence A

Conversely, if M(A)

then for arbitrary

> 0 there exist A', A" E A' A A", M(A) — m(A')

such that


0 is ar-

26

MEASURE THEORY

as the common value of the inner and outer measures:

= M(A) = Theorems

3 and 4 and the obvious fact that

= m(A)

=

(A E

imply THEOREM 5. The function /2(A) is a measure and is an measure m.

of the

The construction we have discussed above is applicable to an arbitrary measure m defined on a ring. The collection Sm2 = of elementary sets in the p'ane is essentially connected with the coordinate system: the sets of the collection consist

of the rectangles whose sides are parallel to the coordinate axes. In the transition to the Jordan measure

J

(2)

this dependence on the choice of the coordinate system vanishes: if { is a system of coordinates related to the original coordinate system { x1, by the orthogonal transformation

x21

xi cos a + x2 sin a + a1, —x1 sin a + x2 cos a

+

a2,

we obtain the same Jordan measure j(2) = j(m2)

=

denotes the measure constructed by means of rectangles with sides parallel to the axes This fact is justified by the following general theorem: THEOREM 6. In order that two Jordan extensions /11 = j(m1) and /22 = j(m2) of measures m1 and m2 defined on rings and coincide, it is neceswhere

sary and sufficient that

m1(A) =

/22(A)

on

m2(A)

/21(A)

on

The necessity is obvious. We shall prove the sufficiency. Suppose that A E . Then there exist A', A" E such that

A'

and mi(A') = mi(A").

A

A",

m1(A") — m1(A')


tx:f(x) n tx:fm(x)

exists a k such that f(x)

0 there exist a

42

[CII. VI

MEASURABLE FUNCTIONS

function co(x) continuous on [a, b] such that

co(x)} €. In other words, a measurable function can be made into a continuous function by changing its values on a set of arbitrarily small measure. This ,u{x:f(x)

property, called by Luzin the C-property, may be taken as the definition of a measurable function.* EXERCISES

1. For A XA(X) =

lifx

X let XA be the characteristic function of A defined by E A,XA(x) = Oifx E X\A.

a) XAnB(X) = XA(X)XB(X), XAUB(X)

= XA(X) + xn(x) — XA(X)XB(X),

= x0(x)

—

0,

I

—

XB(X)

Xx(X) =

,

1,

XA(X) XB(X) (x E X) if, and only if, A b) XA(X) is 2.

B.

if, and only if, A E

Suppose f(x) is a real-valued function of a real variable. If f(x) is

nondecreasing, then f(x) is Borel measurable. 3. Let X = [a, bJ be a closed interval on the real line. If f(x) is defined ii on X and X = where each is a subinterval of X, = 0 and f a step function.

Suppose that f is nondecreasing (or nonincreasing) on X. Show that all the functions of the approximating sequence of simple functions of Theorem 4 of this section are step functions. 4. Assume that X = [a, b] contains a non-Lebesgue measurable set A. Define a function f(x) on X such that I f(x) is Lebesgue measurable, but

f(x) is not. 5. Two real functions f(x) and g(x) defined on a set X are both ji-measurable. Show that {x:f(x) = g(x)} is

6. Let X be a set containing two or more points. Suppose that = {O, X}. Describe all measurable functions. 7. Let f(x) be a ji-measurable function defined on X. For t real define co(t) = ,u({x:f(x) < t}). Show that ço is monotone nondecreasing, continuço is called the co(t) = 0, and co(t) = ous on the left, distribution function of f(x). §41. Sequences of measurable functions. Various types of convergence

Theorems 5 and 7 of the preceding section show that the arithmetical operations applied to measurable functions again yield measurable func-

SEQUENCES OF FUNCTIONS. TYPES OF CONVERGENCE

43

tions. According to Theorem 2 of §40, the class of measurable functions, unlike the class of continuous functions, is also closed under passage to a limit. In addition to the usual pointwise convergence, it is expedient to define certain other types of convergence for measurable functions. In this section we shall consider these definitions of convergence, their basic properties and the relations between them. DEFINITIoN 1. A sequence of functions defined on a measure space X (that is, a space with a measure defined in it) is said to converge to a function F(x) a.e. if = F(x)

(1)

for almost all x E X [that is, the set of x for which (1) does not hold is of measure zero]. EXAMPLE. The sequence of functions

(x) = (— x) converges to the

function F(x) = 0 a.e. on the closed interval [0, 1] (indeed, everywhere except at the point x = 1). Theorem 2 of §40 admits of the following generalization. THEOREM 1. If a sequence

of /2-measurable functions converges to a

function F(x) a.e., then F(x) is measurable. Proof. Let A be the set on which

By hypothesis, ji(E \ A) =

0.

= F(x). The function F(x) is measurable on A by

Theorem 2 of §40. Since every function is obviously measurable on a set of

measure zero, F(x) is measurable on (E \ A); consequently, it is measurable on E. EXERCISE. Suppose that a sequence of measurable functions converges a.e. to a limit function f(x). Prove that the sequence converges a.e. to g(x) if, and only if, g(x) is equivalent to f(x). The following theorem, known as Egorov's theorem, relates the notions of convergence a.e. and uniform convergence. THEOREM 2. Suppose that a sequence of measurable functions f,1(x) converges to f(x) a.e. on E. Then for every > 0 there exists a measurable set E such that 1) > 2) the sequence

—

converges to f(x) uniformly on E6. Proof. According to Theorem 1, f(x) is measurable. Set

= Hence,

—f(x) I

whence ,u(Rj) —> 0

0.

It remains to verify that fnk(x) —*f(x) for

all x E E \ Q. Suppose that x0 E E \ Q. Then there is an io such that Then {x:Ifflk(x) —f(x)

§41]

SEQUENCES OF FUNCTIONS. TYPES OF CONVERGENCE

47

for all k io, i.e., fnk(xo) — f(xo) Since 6k

0

I
0, —

Show that if

fm(X)

I

>

=

0.

is fundamental in measure, then there exists a measur-

able function f(x) such that

converges in measure to f(x). Hint:

Use Theorem 4. be a sequence of measurable sets and let Xn be the characteris5. Let tic function of Show that the sequence is fundamental in measure

if, and only if,

6. If then

Am) = 0.

{gn(x)} converge in measure to f(x) and g(x), respectively. + gn(x)} converges in measure to f(x) + g(x).

Chapter VII

THE LEBESGUE INTEGRAL In the preceding chapter we considered the fundamental properties of measurable functions, which are a very broad generalization of continuous functions. The classical definition of the integral, the Riemann integral, is, in general, not applicable to the class of measurable functions. For instance, the well known Dirichlet function (equal to zero at the irrational points and one at the rational points) is obviously measurable, but not Riemann integrable. Therefore, the Riemann integral is not suitable for measurable functions. The reason for this is perfectly clear. For simplicity, let us consider functions on a closed interval. To define the Riemann integral we divide the interval on which a function f(x) is defined into small subintervals and, choosing a point in each of these subintervals, form the sum What we do, essentially, is to replace the value of f(x) at each point of the closed interval = [Xk, Xk+1J by its value at an arbitrarily chosen point of this interval. But this, of course, can be done only if the values of f(x) at points which are close together are also close together, i.e., if

f(x) is continuous or if its set of discontinuities is "not too large." (A bounded function is Riemann integrable if, and only if, its set of discontinuities has measure zero.) The basic idea of the Lebesgue integral, in contrast to the Riemann integral, is to group the points x not according to their nearness to each other on the x-axis, but according to the nearness of the values of the function at these points. This at once makes it possible to extend the notion of integral to a very general class of functions. In addition, a single definition of the Lebesgue integral serves for functions defined on arbitrary measure spaces, while the Riemann integral is introduced first for functions of one variable, and is then generalized, with appropriate changes, to the case of several variables. In the sequel, without explicit mention, we consider a o--additive measure

X of the al/2(A) defined on a Borel algebra with unit X. The sets A gebra are ji-measurable, and the functions f(x)—defined for all x E X— are also ji-measurable. §42. The Lebesgue integral of simple functions

We introduce the Lebesgue integral first for the simple functions, that is, for measurable functions whose set of values is countable. 48

LEBESGIJE INTEGRAL OF SIMPLE FUNCTIONS

§42]

Let f(x)

be

49

a simple function with values

•••, , It is natural to define the integral of f(x)

y5 for i

•

over

j).

(on) a set A as

E A,f(x) =

(1)

We therefore arrive at the following definition. DEFINITION. A simple function f(x) is over A if the series (1) is absolutely convergent. If f(x) is integrable, the sum of the series (1) is called the integral of f(x) over A. In this definition it is assumed that all the are distinct. However, it is possible to represent the value of the integral of a simple function as a sum of products Ck,u(Bk) without assuming that all the Ck are distinct. This can be done by means of the ii B, = 0 (i j) and that f(x) LEMMA. Suppose that A = Uk Bk, assumes only one value on each set Bk. Then

=

(2)

where the function f(x) is integrable over A if, and only if, the series (2) is absolutely convergent.

Proof. It is easy to see that each set An = tx:x is

E

A,f(x) =

the union of all the sets Bk for which Ck =

= Since

Therefore,

(Bk) =

Yn

Ck/2(Bk).

the measure is nonnegative, I

Yn

ii(An) = En

=

Yn I

I

Cic

I

that is, the series Yn,u(An) and Ek ck,u(Bk) are either both absolutely convergent or both divergent. We shall now derive some properties of the Lebesgue integral of simple functions. A)

+ fg(x)

=

L

+

where the existence of the integrals on the left side implies the existence of the integral on the right side.

To prove A) we assume that f(x) assumes the values f1 on the sets

50

[CII. VII

THE LEBESGIJE INTEGRAL A, and

that g(x) assumes the values gj on the sets G

(3)

=

(4)

J2 =

Lfx

A;

hence

= =

Then, by the lemma,

(5)

=

L {f(x) +

g(x)}

+

=

n Gd).

But = =

that the absolute convergence of the series (3) and (4) implies the absolute convergence of the series (5). Hence so

J=

J1

+ J2.

B) For every constant k, k ff(x)

= L {kf(x)}

the existence of the integral on the left implies the existence of the integral on the right. (The proof is immediate.) C) A simple function f(x) bounded on a set A is integrable over A, and where

Lfx dH M on A. (The proof is immediate.) EXERCISES

1. If A, B are measurable subsets of X, then

fIxA(X) A, then 2. If the simple function f(x) is integrable over A and B f(x) is integrable over B. 3. Let F0 = [0, 1]. Define the simple function f(x) on F0 as follows: On open intervals deleted in the nth stage of the construction of the the Cantor set F let f(x) = n. On F let f(x) = j1

is linear Lebesgue measure.

0.

Compute f f(x)

where

§43]

DEFINITION AND PROPERTIES OF THE LEBESGUE INTEGRAL

51

§43. The general definition and fundamental properties of the Lebesgue integral DEFINITION. We shall say that a function f(x) is integrable over a set A if there exists a sequence of simple functions integrable over A and uniformly convergent to f(x). The limit

J

(1)

IA

is denoted by IA

and is called the integral of f(x) over A. This definition is correct if the following conditions are satisfied:

1. The limit (1) for an arbitrary uniformly convergent sequence of simple functions integrable over A exists. 2. This limit, for fixed f(x), is independent of the choice of the sequence 3. For simple functions this definition of integrability and of the integral is equivalent to that of §42. All these conditions are indeed satisfied.

To prove the first it is enough to note that because of Properties A), B) and C) of integrals of simple functions,

ffn(x)

—

f fm(X)dH

sup

—fm(x) I;x

E

A).

To prove the second condition it is necessary to consider two sequences

and {f *(x)) and to use the fact that — IA

ff*(x) dH — f(z)

1;

x E A] + sup [Ifn*(x) — f(x) x E I;

Finally, to prove the third condition it is sufficient to consider the se= f(x). We shall derive the fundamental properties of the Lebesgue integral. THEOREM 1.

IA

=

Proof. This is an immediate consequence of the definition. THEOREM 2. For every constant k,

52

THE LEBESGUE INTEGRAL

kff(x)

=

IA

[cH. VII

{kf(x)J

where the existence of the integral on the left implies the existence of the integral on the right.

Proof. To prove this take the limit in Property B) for simple functions. THEOREM 3.

IA

where the existence tegral on the right.

IA {f(x) + g(x)}

+ f g(x) of the

integrals on the left implies the existence of the in-

The proof is obtained by passing to the limit in Property A) of integrals of simple functions. THEOREM 4. A function f(x) bounded on a set A is integrable over A. The proof is carried out by passing to the limit in Property C). THEOREM 5. If f(x) 0, then

0,

IA

on the assumption that the integral exists.

Proof. For simple functions the theorem follows immediately from the definition of the integral. In the general case, the proof is based on the possibility of approximating a nonnegative function by simple functions (in the way indicated in the proof of Theorem 4, §40). COROLLARY 1. If f(x) g(x), then IA

M on A, then

COROLLARY 2. If m f(x)

IA THEOREM 6. If

dii,

IA

fAfl

where the existence of the integral on the left implies the existence of the integrals and the absolute convergence of the series on the right.

§43]

DEFINITION AND PROPERTIES OF THE LEBESGUE INTEGRAL

We first verify the theorem for a simple function 1(x) which

Proof.

assumes

53

the values

,Yk,

Yi,

Let

Bk = {x:x E A,f(x) Bflk = {x: x E

YkL

,

f(x) =

=

Yk

Yk}.

Then

f f(x)

=

A

(1)

=

fAfl

dii.

is absolutely convergent if f(x) is integrable, the series the measures are nonnegative, all the other series in (1) also converge absolutely. If f(x) is an arbitrary function, its integrability over A implies that for 0 there exists a simple function g(x) integrable over A such that every Since and

11(x) — g(x)

(2)

I

1/€. : n no) be a sequence of integrable simple functions converging Let

+ €, and let uniformly to the function no} be a sequence of simple functions converging uniformly to f(x). These sequences are chosen so that they satisfy the inequalities — 0, fTh(x) — f(x) < 1/n. + €] 0, /3 > 0, f(x) is integrable on [0, b] with respect to Lebesgue measure. By Ex. 3, is defined a.e. on [0, b]. Show

that Hint: Use the result: for p > 0, q > 0, —

f[0,1]

= p(p)p(q)/J1(p

5. (INTEGRATIoN BY PARTS.) Let X = V is linear Lebesgue measure. where

+

b] and let /.L =

[0,

® /h11,

Suppose that f(x), g(x) are integrable over X. If

F(x) for x E

[0,

G(x) =

=f[Ox]

[

[Ox]

1], then

f F(x)g(x)

= F(b)G(b)

L f(x)G(x)

The result may be demonstrated as follows:

a) Let E =

{ (x,

y): (x, y) E X X Y, y

Show that E is

urable. Hence XE is /h-measurable and H(x, y) = XE(X, y)g(x)f(y) is also /2-measurable.

b) Show that H(x,

y)

is integrable over X X Y

(Apply Ex. 1.) c) Apply Fubini's theorem to obtain

with

respect to

77

THE INTEGRAL AS A SET FUNCTION

§49]

f IF(x)g(x)

=

f

H(x, y)

=

XXY

f f(y) (f

g(x)

[Ybl

Y

This will yield the stated result. §49. The integral as a set function

the assumption that

f

f(x) d/2 as a set function on = a Borel algebra with unit X and that I f(x)

We shall consider the integral F(A) is

exists.

Then, as we have already proved: 1. F(A) is defined on the Borel algebra ASH. 2. F(A) is real-valued.

3. F(A) is additive, that is, if

A=

E ASH)'

then

F(A) = 4.

F(A) is absolutely continuous, that is, /1(A) = 0 implies that

F(A)

0.

We state the following important theorem without proof: RADON'S THEOREM. If

a set function F(A) has properties 1,

2, 3 and

4, it

is representable in the form

F(A) =

fAfx

We shall show that the function f = fact, if

F(A) for all A

E

= IA fi(x)

is uniquely defined a.e. In

=

then

1/ni.

THE LEBESGUE INTEGRAL

78

=

[CH. VII

0

for

= {x:f2(x)

—

fi(x) > 1/ni.

Since

{x:fi(x)

UnAnUUmBm,

it. follows that 0.

This proves our assertion. EXERCISES

1. With the notation of this section, suppose that f(x) 0 and let v(A)

f f(x)

Then the conditions listed before Radon's theorem

can be paraphrased by saying that v(A) is a completely additive, absolutely continuous measure on the Borel algebra Show that if g(x) is integrable then over X with respect to ii, .

f

f(x)g(x) (A E Sn). = 2. If v(A) is a completely additive measure on the Borel algebra 0 there exists a > 0 then v may have the following property: For and /2(A) < imply v(A) < It is easy to see that if such that A E v has this property, then ii is absolutely continuous with respect to /1, i.e., 0 implies v(A) = 0. Show, conversely, that if v is absolutely conproperty. tinuous with respect to then v has the above fA

g(x) dv

Chapter VIII

SQUARE INTEGRABLE FUNCTIONS One of the most important linear normed spaces in Functional Analysis

is filbert space, named after the German mathematician David filbert, who introduced this space in his research on the theory of integral equations. It is the natural infinite-dimensional analogue of Euclidean n-space. We became acquainted with one of the important realizations of filbert space in Chapter Ill—the space 12, whose elements are the sequences x

(x1, •..

)

satisfying the condition

< can now use the Lebesgue integral to introduce a second, in certain n=lxn

We

respects more convenient, realization of filbert space—the space of square integrable functions. In this chapter we consider the definition and funda-

mental properties of the space of square integrable functions and show that it is isometric (if certain assumptions are made about the measure used in the integral) to the space 12. We shall give an axiomatic definition of filbert space in Chapter IX. §50.

The space L2

In the sequel we consider functions f(x) defined on a set R, on which a

measure /2(E) is prescribed, satisfying the condition /h(R) < oo• The functions f(x) are assumed to be measurable and defined a.e. on R. We shall not distinguish between functions equivalent on R. For brevity, instead of

we write simply f.

DEFINITIoN 1. We say that f(x) is a square integrable (or summable) function on R if the integral

f f2(x) exists (is finite). The collection of all square integrable functions is denoted by L2. The fundamental properties of such functions follow. THEOREM 1. The product of two square integrable functions is an integrable function. 79

80

SQUARE INTEGRABLE FUNCTIONS

[CH. VIII

The proof follows immediately from the inequality

12 (x) + g (x)] 2

f(x)g(x) I

and the properties of the Lebesgue integral. COROLLARY 1. A square integrable function f(x) is integrable. 1 in Theorem 1. THEOREM 2. The sum of two functions of L2 is an element of L2. Proof. Indeed,

For, it is sufficient to set g(x)

+ 2 f(x)g(x) + g2(x)

[f(x) + g(x)]2

and Theorem 1 implies that the three functions on the right are summable. THEOREM 3. If f(x) E L2 and a is an arbitrary number, then a f(x) E L2. Proof. If f E L2, then

=

f

2ff2()d


0 such that I f(x) < a a.e. on R. The number a is called an essential upper bound of f on R. For an essentially bounded function f, let m = inf { a : a an essential upper bound of The number m is called the essential supremum of f: m = ess. sup f. a) Show that ess. sup f is the smallest essential upper bound of f on R. be the collection of essentially bounded functions on b) Let L00(R, R. If we put f IJ = ess. sup f, show that Lc,3 becomes a normed linear space.

5. Let LP(R, /2), p 1, be the set of measurable functions f defined on R for which I f(x) is integrable over R. a) If a, b are real numbers, show that (The condition p > 1 is essential here.)

b) Show then that and that f E

f+gE

lip

[JR f(x)

f

.

We

implies that f = is a normed space with

is a linear space, i.e., f, g E

and a real imply af E

shall shortly see that

Define

as norm.

6. a) Suppose p > 1. Define q by the equation i/p + 1/q = 1. p and q are called conjugate exponents. Let v = f(u) = Then u = g(v) = Verify that the hypotheses of Young's inequality

and that F(u) =

Ex. 5) are satisfied and that therefore

un/p, G(v) =

uv u9p + with equality if, and only if,

=

b) (HoLDER INEQUALITY.) Suppose f E

g E Lq(R,

with

p, q conjugate exponents. Show that

f(x)g(x)

E

L1(R,

= L

and

JR f(x)g(x)

d/2

1/q

i/p

(JR Jf(x)

(JR

g(x)

= This result may be obtained as follows: It is trivial if Hg

=

0.

Otherwise, put

JJ

f

0 or

84

{CH. VIII

SQUARE INTEGRABLE FUNCTIONS

u = If(x)

/ If

v

g(x) I / IJ

=

Ilq

in the result of a), and integrate over R (see vol. 1, p. 20). c) (MINK0wSKI'S INEQUALITY.) If f, g E ii), then

If or, in terms of integrals,

(fR

f(x) + g(x)

If 1 1 + g

=

(L I f(x)

P

0,

P

+

(f

I g(x)

dy).

then the result is clear. If f + g 1T > 0, observe

that I

f(x) + g(x)

I f(xfl f(x) + g(x)

g(x) In',

I

If(x) + g(x)

E Lq.

Apply Holder's inequality to each term on the right to obtain 1/q

(fRIf+

fRI

It is now clear that

+

with norm

p > 1. Note also that if p =

2,

f

then q =

a normed linear space for and Holder's inequality re-

is 2,

duces to the Schwarz inequality. §51. Mean convergence. Dense subsets of L2

The introduction of a norm in L2 determines a new notion of convergence for square integrable functions:

(inL2)

f

—

f(x)]2

= 0.

This type of convergence of functions is called mean convergence, or, more precisely, mean square convergence. Let us consider the relation of mean convergence to uniform convergence and convergence a.e. (see Chapter VI). THEOREM 1. If a sequence (x)1 of functions of L2 converges uniformly to f(x), then f(x) E L2, and 'is mean convergent to f(x). Proof. Suppose that 0. If n is sufficiently large,

f(x) 1).

b) The corresponding result obtains for 0, then — f

3. a) If

112

2

J

f(x)

1'

I

2

b) The corresponding result obtains for (p > 1). converge to f(x) in the mean and suppose g(x) E L2. converges to f(x)g(x). Then b) More generally, if 0, then — g f 1k 0 and

4. a) Let

112

c) Similar results obtain for 1/q = 1, p> 1.

gn, g E J}, with i/p +

fE

d) Let R = [a, b], and let be linear Lebesgue measure. Then converges to f in the mean implies that

f

[ax0]

Hint: Choose

f

[a,xo]

f(x)

(a x0 b).

88

SQUARE INTEGRABLE FUNCTIONS

g(x) =

Ii

ECH. VIII

(ax<xo), (x > xo).

Show that g E L2. §52. L2 spaces with countable bases

The space L2 of square integrable functions depends, in general, on the choice of the space R and the measure To designate it fully it should be written as L2(R, /2). The space L2(R, is finite-dimensional only in exceptional cases. The spaces L2(R, which are most important for analysis are the spaces which have infinite dimension (this term will be explained below). To characterize these spaces, we need an additional concept from the theory of measure. We can introduce a metric in the collection of measurable subsets of the space R (whose measure we have assumed to be finite) by setting

p(A, B) =

B).

B) = 0 (that is, we consider sets which are the same except for a set of measure zero to be indistinguishable), then the set together with the metric p becomes a metric

If we identify sets A and B for which /2(A

space. DEFINITION. A measure /h is said to have a countable contains a countable dense set. space

if the metric

In other words, a measure /2 has a countable base if there is a countable set

(n= 1,2,...) of measurable subsets of R (a countable base for the measure /2) such that R and for every measurable M for which 0 there is an Ak E



[Ck = (g, cok)].

Ck

Then, by the Riesz-Fisher theorem, there exists a function f E (f,

The function f — inequality

g

=

Ck,

g

such that

(f, 1)

is orthogonal to all the functions coi.. In view of the

(f, f) = f—

L2

Ck2

cannot be equivalent to 4'(x) =

0.

< (g, g), This proves the theorem.

EXERCISES

be an orthonormal set in L2 and suppose f E L2. Verify is orthogonal to all linear combinations if, that f — (1 k n). and only if, ak (1, be an orthonormal set in L2 and let F 2. Let L2 be dense in L2. If Parseval's equality holds for each f E F, then it holds for all g E L2,

1. Let

i.e., {ccn(x)1 is closed.

be the nth

This may be proved as follows: Let partial sum of the Fourier series of f E L2. a) 1ff, g E L2, then —

= II

If —

—

g II.

b) Parseval's equality holds for g if, and only if, II

g — sn(g)

II = 0.

c) Now use the hypothesis of the exercise. be complete orthonormal sets in L2(R, /2). Let /2 /2 and consider L2(R X R, :n, m = 1, 2, •} is orthonormal a) The set { Xnm(X, y)

3. Let

in L2(R x

2)

R, b) The set { Xnm(X, y) } is complete.

Hint: Use Fubini's theorem and the criterion of Theorem 3 for completeness.

ISOMORPHISM OF L2 AND 12

§551

101

§55. Isomorphism of the spaces L2 and 12

The Riesz-Fisher theorem immediately implies the following important THEOREM. The space L2 is isomorphic to the space 12.

[Two Euclidean spaces R and R' are said to be isomorphic if there is a one-to-one correspondence between their elements such that y 0. Choose an s such that 1/s < p(x, vms) p(X, Em) + P(Em,

vms)

< 1/2s + 1/28 =

1/8

(As,

=

Proof. — I

(Au)

+

—

(Au) I.

—

But

=

—

—

I

—

1111

II

I

and I

(As,

Since

— (As,

I

the numbers

H

= are

—

I

bounded and

A

0,

—

—>0.

—

I

—

This proves the lemma. LEMMA 2. If a functional I

where

point

I,

I

A is a bounded self-adjoint linear operator, assumes a maximum at a of the unit sphere, then '1)

=0

implies that

= (Eo,An) = =

Proof. Obviously, lEo

1.

0.

Set

+ = a is an arbitrary number. From EM

=

1

it follows that

= 1.

Since

=

(1

+ a2

1

n

+

+

§59]

117

COMPLETELY CONTINUOUS SELF-ADJOINT LINEAR OPERATORS

it follows that + O(a2) + = for small values of a. It is clear from the last relation that if (A 0, then a can be chosen so that This contradicts the > . hypothesis of the lemma. It follows immediately from Lemma 2 that if I I assumes a maximum then is an eigenvector of the operator A. at = Proof of the theorem. We shall construct the elements cok by induction, in the order of decreasing absolute values of the corresponding eigenvalues: I

To construct the element and

we consider the expression

show that it assumes a maximum on the unit sphere. Let

S=

= I (As,

sup

and suppose that

is a sequence such that II

1 1 and

Since the unit sphere in H is weakly compact, { contains a subsequence weakly convergent to an element In view of Theorem 1, §58, 1, and by Lemma 1, II

I

We take

as ce'i. Clearly, II

= S.

(An,

I! = 1. Also

= whence

Xii =

I

I

(Açci,çoi)

=

Now suppose that the eigenvectors ••• ,(Pn

corresponding to the eigenvalues

xl, •..

,Xfl

have already been constructed. We consider the functional on the elements of

= H

e

,

, pn)

S.

118

ABSTRACT HILBERT SPACE. INTEGRAL EQUATIONS

[CH. IX

1. (that is, the set orthogonal to and such that I! Ii , is an invariant subspace (a subspace which is mapped into itself) of A , is invariant and A is self-adjoint]. Applying the [since M(çoi,

above arguments to

we obtain an eigenvector Wn+1 of A in Ma'.

Two cases are possible: 1) after a finite number of steps we obtain a in which (As, for all n. subspace = 0; 2) (As, 0 on In the first case Lemma 2 implies that A maps into zero, that is, consists of the eigenvectors corresponding to X = 0. The set of vecis finite. tors In the second case we obtain a sequence {con} of eigenvectors for each of (like every ortho0. We show that which 0. The sequence

normal sequence) is weakly convergent to zero. Therefore, = converge to zero in the norm, whence 0. Let

=

1

M' = If

E M' and E

0.

0, then (AE,

for all n, that is,

= 0.

(As,

1} = 0) to M', we Hence, applying Lemma 2 (for sup { (As, I; A maps the subspace M' into zero. obtain From the construction of the set tcon} it is clear that every vector can be written in the form I

= Eic

1

Ckcok +

(As' = 0).

Hence

A

be a continuous linear operator of H into H. Suppose that

c H, f E H and

converges converges weakly to f. Show that weakly to Af. 2. In the second paragraph of this section it is stated that in H the norm bounded sets are precisely the weakly compact sets. Show that this is true as follows:

a) If A H is norm bounded, i.e., there exists an M > 0 such that If II < M for allf E A, then Theorem l'of §28 (see vol. 1) shows that A

is weakly compact (see the statement preceding Theorem 1 in §58).

LINEAR OPERATOR EQUATIONS

§60]

b) If A is weakly compact, show that A is §58 for this purpose.

119

norm bounded. Use Ex. 4 of

3. In the fourth paragraph of this section it is stated that the following two properties of an operator A on H are equivalent: a) A maps every weakly compact set into a norm compact set. b) A maps every weakly convergent sequence into a norm convergent sequence.

Prove that a) and b) are equivalent. 4. Let A be a continuous (bounded) linear operator of H into H with the additional property that A (H), the range of A, is contained in a finitedimensional subspace of H. Then A is completely continuous. Hint: The Bolzano-Weierstrass theorem holds in 5. Let A be a completely continuous operator, T = I — A and suppose H, M = {x: Tx = 0}. Show that M is a finite-dimensional subspace of H.

§60. Linear equations in completely continuous operators We consider the equation (1)

where A is a completely continuous self-adjoint operator, scribed and E H is the unknown.

E H is pre-

Let 'P1,

the eigenvectors of A corresponding to the eigenvalues different from zero. Then can be written as be

(2)

+ n',

=

where

=

0.

We shall seek a solution of (1) of the form

+

=

(3) where A E' = 0.

Substitution of (2) and (3) into (1) yields

+

—

+

=

This equation is satisfied if, and only if,

= — that

is, if

=

77',

=

'1',

ABSTRACT HILBERT SPACE. INTEGRAL EQUATIONS

120

=

(4)

=

[CH. IX

1/c),

—

= 1/c).

0

The last equality gives a necessary and sufficient condition for a solution

of (1), and (4) determines the solution. The values of = 1/c remain arbitrary. those n for which

corresponding to

§61. Integral equations with symmetric kernel

The results presented in the preceding section can be applied to integral equations with symmetric kernel, that is, to equations of the form

+f K(t, s)f(s) ds,

f(t) =

(1)

where K(t, s) satisfies the conditions

1) K(t, s) = K(s, t), 1b

2) f

K2(t, s) dt ds < oc.

The application of the results of §60 to equations of the form (1) is based on the following theorem:

If a function K(t, s) defined

THEOREM. Let R be a space with measure on R2 = R X R satisfies the conditions

K(t, s) = K(s,

(2)

f K2(t, s)

(3)

t)

< oc

(M2 =

®

then the operator

g = Af defined on L2(R,

by

the formula

g(t)

=

f K(t, s)f(s) dM3

is completely continuous and seif-adjoint. be Proof. We shall denote the space L2(R, simply by L2. Let a complete orthonormal set in L2. The collection of all possible products /'n(t)/'m(5) is a complete orthonormal set of functions in R2 (see Ex. 3, §54), and

(4)

K(t, s) = Em

amn/'n(t)/'m(s)

121

INTEGRAL EQUATIONS WITH SYMMETRIC KERNEL

§611

converges to K in the norm of L2 (R2,

in the mean [i.e.,

where

amn = anm

(in view of (2)), and >2m


0 there is an m0

amn2

am2

converges, for every

such

that

2

m=m0+1 am < Cm

Now suppose that ing

2

2

II g(x) —

converge to Cm

2

HIlL

is weakly convergent to f. Then the correspondevery m. Hence the sum

for

converges in the mean to the sum Cm%llm(X)

for arbitrary fixed m0. In view of the inequality (7) and the boundedness g(k) = it follows that converges of the norm (x)1 (where in the mean to g(x). This proves that A is completely continuous. Multiplying (4) by (5), integrating with respect to and comparing the result

with (6), we see that (Af)(s) =

f K(s, t)f(t)

122

ABSTRACT HILBERT SPACE. INTEGRAL EQUATIONS

[CH. ix

This and Fubini's theorem imply that (Af, g)

=

f (f K(s, t)f(t)

=

f f(t) (f K(s, t)g(s)

= (f,Ag), that is, A is self-adjoint. This proves the theorem. Hence the solution of an integral equation with symmetric kernel satisfying conditions (2) and (3) reduces to finding the eigenfunctions and eigenvalues of the corresponding integral operator. The actual solution of the latter problem usually requires the use of some approximation method, but such methods are outside the scope of this book. EXERCISES

1. Let R = [a, b] be an interval on the real line, linear Lebesgue measure, H = L2(R, ia), K(t, s) as in the theorem of §61. Then by Theorem 1 of §59, the operator A determined by K has an orthonormal sequence of eigenof eigenvalues functions {conl corresponding to a sequence { 0. h(t) in the mean, with Further, for f(t) E H, f(t) + E=i (f,

Ah=0, i.e., f

= (Af)(s) =

in the mean.

Suppose now that there is a constant M such that f I K(t, s) 12

<M2

for all I E [a, b]. An example is furnished by K(t, s) = I — s a< Show that the series for g(s) will converge uniformly and absolutely (pointwise) to g(s). Hints: For uniform convergence, apply the condition and Schwarz's inequality. For absolute convergence, show that mean convergence of a series with orthogonal terms is equivalent to convergence of the series with 112) and that therefore the convergence is indepositive terms (Ek pendent of the order of the terms. I

SUPPLEMENT AND CORRECTIONS TO VOLUME 1 (1) p. 28, 1. 23. Substitute Ga(X) for Ga. (2) p. 46, 1. 13* (1. 13 from bottom). Replace "the method of successive approximations is not applicable" by "the method of successive approximations is, in general, not applicable". (3) p. 49, after 1. 10* insert: "An arbitrary continuous function may be chosen for fo(x) ". (4) p. 50, 1. 1*. Replace the X after the inequality sign by I X I. (5) p. 51, 1. 2. Replace M by M2. p. 51, 1. 3. Replace M by (two times). Replace by I

(two

times). p. 51, 1. 5. Replace by I (6) p. 56, 1. 6. Replace "closed region" by "closed bounded region". p. 59, 1. 2*. Replace the first occurrence of V by X. (7) p. 61, after 1. 9 insert: "A mapping y = f(x) is said to be uniformly continuous if for every > 0 there is a 5 > 0 such that p(f(xi),f(x2)) < for all x1, x2 for which p(xi, x2) < & The following theorem holds: Every continuous mapping of a compactum into a compactum is uniformly continuous. This theorem is proved in the same way as the uniform continuity of a function continuous on a closed interval".

(8) p. 61, 1. 18. After "Proof" insert: "We shall prove the necessity first. If D is compact, D contains a finite ,fN. Since each fi, mapping is continuous, it is uniformly continuous. Therefore, there is a

5 > 0 such that < p(x1

If f E D, there exists an

,

x2)