A.N. Kolmogorov, S.v. Fomin-Elements of the Theory of Functions and Functional Analysis, Volume 1, Metric and Normed Spaces-GRAYLOCK PRESS (1963)

ELEMENTS OF THETHEORY OF

FUNCTIONS AND

FuNCTIONAL

ANALYSIS

Volume 1Metric and Normed Spaces

A.N. KoLmogorov and S. V. Fomin

ELEMENTS OF THE THEORY OF FUNCTIONS

AND FUNCTIONAL ANALYSIS

VOLUME 1

METRIC AND NORMED SPACES

OTHER GRAYLOCK PUBLICATIONS

ALEKSANDROV: Combinatorial TopologyVol. 1: Introduction. Complexes.Coverings. DimensionVol. 2: The Betti GroupsVol. 3: Homological Manifolds. TheDuality Theorems. Cohomology Groupsof Compacta. Continuous Mappings ofPolyhedra

KHINCHIN: Three Pearls of Number TheoryMathematical Foundations of QuantumStatistics

KOLMOGOROV and FOMIN: Elements of the Theory of Functionsand Functional AnalysisVol. 1: Metric and Normed SpacesVol. 2: Measure. The Lebesgue Integral.Hilbert Space

NOVOZHILOV: Foundations of the Nonlinear Theoryof Elasticity

PETROVSKII: Lectures on the Theory of IntegralEquations

PONTRYAGIN: Foundations of Combinatorial Topology

Elements of the Theoryof Functions

and Functional Analysis

VOLUME 1

METRIC AND NORMED SPACES

BY

A. N. KOLMOGOROV AND S. V. FOMIN

TRANSLATED FROM THE FIRST (1954) RUSSIAN EDITION

by

LEO F. BORONDepartment of Mathematics

University of Wisconsin

GRA YLOCK PRESSROCHESTER, N. Y.

1957

Copyright, 1957, byGRAYLOCK PRESS

Rochester, N. Y.

All rights reserved. This book, or partsthereof, may not be reproduced in anyform, or translated, without permis-sion in writing from the publishers.

Library of Congress Catalog Card Number 57-14021

Third Printing, July 1963

Manufactured in the United State8 of America

CONTENTS

Preface viiTranslator's Note ix

CHAPTER IFTJNDAMENTALS OF SET THEORY

1. The Concept of Set. Operations on Sets 1

2. Finite and Infinite Sets. Denumerability 33. Equivalence of Sets 64. The Nondenumerability of the Set of Real Numbers 85. The Concept of Cardinal Number 96. Partition into Classes 11

7. Mappings of Sets. General Concept of Function 13

CHAPTER IIMETRIC SPACES

8. Definition and Examples of Metrie Spaces 169. Convergence of Sequences. Limit Points 23

10. Open and Closed Sets 2611. Open and Closed Sets on the Real Line 3112. Continuous Mappings. Homeomorphism. Isometry 3313. Complete Metric Spaces 3614. The Principle of Contraction Mappings and its Applications 4315. Applications of the Principle of Contraction Mappings in Analysis 4616. Compact Sets in Metric Spaces 5117. Arzelà's Theorem and its Applications 5318. Compacta 5719. Real Functions in Metric Spaces 6220. Continuous Curves in Metric Spaces 66

CHAPTER IIINORMED LINEAR SPACES

21. Definition and Examples of Normed Linear Spaces 7122. Convex Sets in Normed Linear Spaces 7423. Linear Functionals 7724. The Conjugate Space 8125. Extension of Linear Functionals 8626. The Second Conjugate Space 8827. Weak Convergence 9028. Weak Convergence of Linear Functionals 9229. Linear Operators 95

V

Vi CONTENTS

ADDENDUM TO CHAPTER IIIGENERALIZED FUNCTIONS

CHAPTER IVLINEAR OPERATOR EQUATIONS

30. Spectrum of an Operator. Resolvents 11031. Completely Continuous Operators 11232. Linear Operator Equations. Fredholm's Theorems 116LIST OF SYMBOLS 122LIST OF DEFINITIONS 123LIST OF THEOREMS 123BASIC LITERATURE 125INDEX 127

PREFACE

The present book is a revised version of material given by the authors intwo courses at the Moscow State University. A course in functional analy-sis which included, on the one hand, basic information about the theory ofsets, measure and the Lebesgue integral and, on the other hand, examplesof the applications of the general methods of set theory, theory of func-tions of a real variable, and functional analysis to concrete problems inclassical analysis (e.g. the proof of existence theorems, the discussion ofintegral equations as an example of a special case of the theory of operatorequations in linear spaces, etc.) was given by the first author, A. N. Kol-mogorov, in the Department of Mathematics and Mechanics. A somewhatless comprehensive course was given by the second author, S. V. Fomin,for students specializing in mathematics and theoretical physics in theDepartment of Physics. A. A. Petrov's notes of A. N. Kolmogorov's lec-tures were used in a number of sections.

The material included in the first volume is clear from the table of con-tents. The theory of measure and the Lebesgue integral, Hilbert space,theory of integral equations with symmetric kernel and orthogonal sys-tems of functions, the elements of nonlinear functional analysis, and someapplications of the methods of functional analysis to problems arising inthe mathematics of numerical methods will he considered in later volumes.

A. KOLMOGOROVS. F0MIN

February 1954

vii

TRANSLATOR'S NOTEThis volume is a translation of A. N. Kolmogorov and S. V. Fomin's

Elementy Teorii i Funkcional'nogo Analiza., I. Metri3eskie i Nor-mirovannye Prostranstva. Chapter I is a brief introduction to set theoryand mappings. There is a clear presentation of the elements of the theoryof metric and complete metric spaces in Chapter II. The latter chapter alsohas a discussion of the principle of contraction mappings and its applica-tions to the proof of existence theorems in the theory of differential andintegral equations. The material on continuous curves in metric spaces isnot usually found in textbook form. The elements of the theory of normedlinear spaces are taken up in Chapter III where the Hahn-Banach theoremis proved for real separable normed linear spaces. The reader interested inthe extension of this theorem to complex linear spaces is referred to thepaper Extension of Functionals on Complex Linear Spaces, Bulletin AMS,44 (1938), 91—93, by H. F. Bohnenblust and A. Sobczyk. Chapter III alsodeals with weak sequential convergence of elements and linear functionalsand gives a discussion of adjoint operators. The addendum to Chapter IIIdiscusses Sobolev's work on generalized functions which was later gen-eralized further by L. Schwartz. The main results here are that everygeneralized function has derivatives of all orders and that every convergentseries of generalized functions can be differentiated term by term anynumber of times. Chapter IV, on linear operator equations, discussesspectra and resolvents for continuous linear operators in a complex Banachspace. Minor changes have been made, notably in the arrangement of theproof of the Hahn-Banach theorem. A bibliography, listing basic bookscovering the material in this volume, was added by the translator. Lists ofsymbols, definitions and theorems have also been added at the end of thevolume for the convenience of the reader.

Milwaukee 1957 Leo F. Boron

ix

Chapter I

FUNDAMENTAL CONCEPTS OF SET THEORY

§1. The concept of set. Operations on sets

In mathematics as in everyday life we encounter the concept of set. Wecan speak of the set of faces of a polyhedron, students in an auditorium,points on a straight line, the set of natural numbers, and so on. The con-cept of a set is so general that it would be difficult to give it a definitionwhich would not reduce to simply replacing the word "set" by one of theequivalent expressions: aggregate, collection, etc.

The concept of set plays an extraordinarily important role in modernmathematics not only because the theory of sets itself has become at thepresent time a very extensive and comprehensive discipline but mainlybecause of the influence which the theory of sets, arising at the end of thelast century, exerted and still exerts on mathematics as a whole. Here weshall briefly discuss only those very basic set-theoretic concepts which willbe used in the following chapters. The reader will find a significantly moredetailed exposition of the theory of sets in, for example, the books by P. S.Aleksandrov: Introduction to the General Theory qf Sets and Functions,where he will also find a bibliography for further reading, E. Kamke:Theory of Sets, F. Hausdorif: Mengenlehre, and A. Fraenkel: Abstract SetTheory.

We shall denote sets by upper case letters A, B, and their elementsby lower case letters a, b, . The statement "the element a belongs to theset A" will be written symbolically as a E A; the expression a A meansthat the element a does not belong to the set A. If all the elements of whichthe set A consists are also contained in the set B (where the case A = B isnot excluded), then A will be called a subset of B and we shall write A B.(The notation A B denotes that A is a subset of the set B and A B,i. e. there exists at least one element in B which does not belong to A. A isthen said to be a proper subset of the set B.) For example, the integers forma subset of the set of real numbers.

Sometimes, in speaking about an arbitrary set (for example, about theset of roots of a given equation) we do not know in advance whether or notthis set contains even one element. For this reason it is convenient to intro-duce the concept of the so-called void set, that is, the set which does notcontain any elements. We shall denote this set by the symbol 0. Every setcontains 0 as a subset.

If A and B are arbitrary sets, then their sum or union is the set C =A U B consisting of all elements which belong to at least one of the setsA and B (Fig. 1).

2 FUNDAMENTALS OF SET THEORY [CII. I

B A B

We define the sum of an arbitrary (finite or infinite) number of setsanalogously: if A a are arbitrary sets, then their sum A = U a A a is thetotality of elements each of which belongs to at least one of the sets A

The intersection of two sets A and B is the set C = A fl B which consistsof all the elements belonging to both A and B (Fig. 2). For example, theintersection of the set of all even integers and the set of all integers whichare divisible by three is the same as the set of all integers which are divisibleby six. The intersection of an arbitrary (finite or infinite) number of setsA a is the set A = fla A a of all elements which belong to all of the sets AIf A fl B = 0, we shall say that A and B are disjoint. The same term willapply to any collection of sets {A for which Afi fl = 0, /3 'y.

The operations of union and intersection are connected by the followingrelations:

(1) (AUB)flC=(AflC)U(BflC),(2) (AflB)UC= (AUC)fl(BUC).

We shall verify the first of these two relations. Let the element x belongto the set on the left side of equation (1). This means that x belongs to Cand moreover that it belongs to at least one of the sets A and B. But thenx belongs to at least one of the sets A fl C and B fl C, i.e. it belongs tothe right member of equation (1). To prove the converse, let x E (A fl C) U(B fl C). Then x E A fl C and/or x E B fl C. Consequently, x E C and,moreover, x belongs to at least one of the sets A and B, i. e. x E (A U B).Thus we have shown that x E (A U B) fl C. Hence, equation (1) has beenverified. Equation (2) is verified analogously.

We define further the operation of subtraction for sets. The differenceof the sets A and B is the set C = A\ B of those elements in A which arenot contained in B (Fig. 3). In general it is not assumed here that A B.

In some instances, for example in the theory of measure, it is convenientto consider the so-called symmetric difference of two sets A and B; the sym-metric difference is defined as the sum of the differences A \ B and B \ A(Fig. 4). We shall denote the symmetric difference of the sets A and B by

A

CaAUB CaAflBFIG. 1 FIG. 2

§2] FINITE AND INFINITE SETS. DENUMERABILITY 3

the symbol Its definition is written symbolically as follows:

= (A \ B) U (B \ A).

EXERCISE. Show that = (A U B) \ (A fl B).In the sequel we shall frequently have occasion to consider various sets

all of which are subsets of some fundamental set 5, such as for examplevarious point sets on the real line. In this case the difference S \ A is calledthe complement of the set A with respect to S.

In the theory of sets and in its applications a very important role isplayed by the so-called principle of duality which is based on the followingtwo relations.

1. The complement of a sum is equal to the intersection of the complements,

(3) S\UaAa = fla(5\Aa).2. The complement of an intersection is equal to the sum of the complements,

(4) S\flaAa = Ua (S\Aa).By virtue of these relations, we can start with an arbitrary theorem con-

cerning a system of subsets of a fixed set S and automatically obtain thedual theorem by replacing the sets under consideration by their comple-ments, sums by intersections, and intersections by sums. Theorem 1',Chapter II, §10, is an example of the application of this principle.

We shall now prove relation (3). Let x E S \ Ua A This means that xdoes not belong to the sum Ua Aa, i.e. x does not belong to any of the setsA Consequently, x belongs to each of the complements S \ A a andtherefore x E fla (S \ Aa). Conversely, let x E fla (S \ Aa), i.e. x belongsto every S \ A Consequently, x does not belong to any of the sets A a,i.e. it does not belong to their sum Ua A But then x E S \ Ua A a thisconcludes the proof of (3). We prove relation (4) analogously.

§2. Finite and infinite sets. Denumerability

In considering various sets we note that for some of them we can indi-cate the number of elements in the set, if not actually then at least intheory. Such, for example, is the set of chairs in a given room, the set ofpencils in a box, the set of all automobiles in a given city, the set of all

CeA\B C=MBFm. 3 FIG. 4

4 FUNDAMENTALS OF SET THEORY [CH. i

molecules of water on the earth, and so on. Each of these sets contains afinite number of elements, although the number may not be known to us.On the other hand, there exist sets consisting of an infinite number ofelements. Such, for example, are the sets of all natural numbers, all pointson the real line, all circles in the plane, all polynomials with rational co-efficients, and so forth. In this connection, when we say that a set is infinite,we have in mind that we can remove one element, two elements, and so on,where after each step elements still remain in the set.

When we consider two finite sets it may occur that the number of ele-ments in both of them is the same or it may occur that in one of thesesets the number of elements is greater than in the other, i.e. we can com-pare finite sets by means of the number of elements they contain. Thequestion can be asked whether or not it is possible in a similar way tocompare infinite sets. In other words, does it make sense, for example, toask which of the following sets is the larger: the set of circles in the planeor the set of rational points on the real line, the set of functions defined onthe segment [0, 1] or the set of straight lines in space, and so on?

Consider more carefully how we compare two finite sets. We can tacklethe problem in two ways. We can either count the number of elements ineach of the two sets and thus compare the two sets or we can try to es-tablish a correspondence between the elements of these sets by assigningto each element of one of the sets one and only one element of the otherset, and conversely; such a correspondence is said to be one-to-one. Clearly,a one-to-one correspondence between two finite sets can be established iiand only if the number of elements in both sets is the same. For instance,in order to verify that the number of students in a group and the numberof seats in an auditorium are the same, rather than counting each of thesets, one can seat each of the students in a definite seat. If there is a suffi-cient number of seats and no seat remains vacant, i.e. if a one-to-one cor-respondence is set up between these two sets, then this will mean that thenumber of elements is the same in both.

But it is obvious that the first method (counting the number of ele-ments) is suitable only for comparing finite sets, whereas the second method(setting up a one-to-one correspondence) is suitable to the same degree forinfinite as well as for finite sets.

Among all possible infinite sets the simplest is the set of natural numbers.We shall call every set whose elements can be put into one-to-one corre-spondence with all the natural numbers a den set. In other words,a denumerable set is a set whose elements can be indexed in the form of auinfinite sequence: a1, a2, , a,, , . The following are examples ofdenumerable sets.

1. The set of all integers. We can establish a one-to-one correspondence

§2] FiNITE AND iNFiNITE SETS. DENUMERABIL1TY

between the set of all integers and the set of all natural numbers in thefollowing way:

0 —1 1 —2 21 23 45,

where in general we set n *—÷ 2n + 1 if n � 0 and n *—÷ — 2n if n <0.2. The set of all positive even integers. The correspondence is obviously

n 2n.3. The set of numbers 2, 4, 8, ... , ... . Assign to each numberthe corresponding n; this correspondence is obviously one-to-one.

4. We now consider a slightly more complicated example: we shallshow that the set of all rationalnumber can be written in the form of an irreducible fraction a = p/q,q> 0. Call the sum n = p

I + q the height of the rational number a. It isclear that the number of fractions having height n is finite. For instance,the number 0/1 = 0 is the only number having height 1; the numbers 1/1and —1/1 are the only numbers having height 2; the numbers 2/1, 1/2,— 2/1 and —1/2 have height 3; and so forth. We enumerate all the rationalnumbers in the order of increasing heights, i.e. first the numbers with height1, then the numbers with height 2, etc. This process assigns some index toeach rational number, i.e. we shall have set up a one-to-one correspondencebetween the set of all natural numbers and the set of all rational numbers.

An infinite set which is not denumerable is said to be a nondenumerableset.

We establish some general properties of denumerable sets.1°. Every subset of a denumerable set is either finite or denumerable.Proof. Let A be a denumerable set and let B be a subset of A. If we

enumerate the elements of the set A: a1, a2, , and let n1,n2, ... be the natural numbers which correspond to the elements in B inthis enumeration, then if there is a largest one among these natural num-bers, B is finite; in the other case B is denumerable.

2°. The sum of an arbitrary finite or denurnerable set of denumerable setsis again a finite or denumerable set.

Proof. Let A1, A2, ... be denumerable sets. All their elements can bewritten in the form of the following infinite table:

au a12 a13 a14

a21 a22 a23 a24

aai a32

a41 a42 a43 a44

where the elements of the set A1 are listed in the first row, the elements

6 FUNDAMENTALS OF SET THEORY [CII. I

of A2 are listed in the second row, and so on. We now enumerate all theseelements by the "diagonal method", i.e. we take for the first element,a12 for the second, for the third, and so forth, taking the elements inthe order indicated by the arrows in the following table:

au a12 a13/ / /a21 a22 a23 a24'/a31 a32 a33/a41 a42 a43

It is clear that in this enumeration every element of each of the setsA receives a definite index, i.e. we shall have established a one-to-one cor-respondence between all the elements of all the A1, A2, .. . and the setof natural numbers. This completes the proof of our assertion.

EXERCISES. 1. Prove that the set of all polynomials with rational coeffi-cients is denumerable.

2. The number is said to be algebraic if it is a zero of some polynomialwith rational coefficients. Prove that the set of all algebraic numbers isdenumerable.

3. Prove that the set of all rational intervals (i.e. intervals with rationalendpoints) on the real line is denumerable.

4. Prove that the set of all points in the plane having rational coordi-nates is denumerable. Hint: Use Theorem 2°.

3° Every infinite set contains a denumerable subset.Proof. Let M be an infinite set and consider an arbitrary element in M.

Since M is infinite, we can find an element a2 in M which is distinct froma1 , then an element a3 distinct from both and a2 , and so forth. Continuingthis process (which cannot terminate in a finite number of steps since M isinfinite), we obtain a denumerable subset A = { a2, a3, of the setM. This completes the proof of the theorem.

This theorem shows that denumerable sets are, so to speak, the "smallest"of the infinite sets. The question whether or not there exist nondenumerableinfinite sets will be considered in §4.

§3. Equivalence of sets

We arrived at the concept of denumerable set by establishing a one-to-one correspondence between certain of infinite sets and the set of nat-ural numbers; at the same time we gave a number of examples of de-numerable sets and some of their general properties.

§3] EQUIVALENCE OF SETS 7

It is clear that by setting up a one-to-one correspondence it is possiblenot only to compare infinite sets with the set of natural numbers; it ispossible to compare any two sets by this method. We introduce the follow-ing definition.

DEFINITION. Two sets M and N are said to be equivalent (notation:M N) if a one-to-one correspondence can be set up between their ele-ments.

The concept of equivalence is applicable to arbitrary sets, infinite aswell as finite. It is clear that two finite sets are equivalent if, and only if,they consist of the same number of elements. The definition we introducedabove of a denurnerable set can now be formulated in the following way:a set is said to be denumerable if it is equivalent to the set of natural numbers.

EXAMPLES. 1. The sets of points on two arbitrary segments [a, b] and[c, d] are equivalent. A method for establishing a one-to-one correspondencebetween them is shown in Fig. 5. Namely, the points p and q correspond ifthey lie on the same ray emanating from the point 0 in which the straightlines ac and bd intersect.

2. The set of all points in the closed complex plane is equivalent to theset of all points on the sphere. A one-to-one correspondence a z can beestablished, for example, with the aid of stereographic projection (Fig. 6).

3. The set of all real numbers in the interval (0, 1) is equivalent to theset of all points on the real line. The correspondence can be established, forexample, with the aid of the function

y = (1/ir) arctan x +

It is clear directly from the definition that two sets each equivalent to athird set are equivalent.

Considering the examples introduced here and in §2 one can make thefollowing interesting deduction: in a number of cases an infinite set provesto be equivalent to a proper subset of itself. For example, there are "asmany" natural numbers as there are integers or even as there are of allrationals; there are "as many" points on the interval (0, 1) as there are onthe entire real line, and so on. It is not difficult to convince oneself of thefact that this situation is characteristic of all infinite sets.

C

FIG. 5 FIG. 6

0 N

8 FUNDAMENTALS OF SET THEORY [CH. 1

In fact., in §2 (Theorem 3°) we showed that every infinite set M has adeiiumerable subset; let this set be

A = {ai,a2,

We partitioii A into two denurnerable subsets

A1 = {ai , , a5 , } and A2 = {a2 a4 , a6 ,

Since A and A1 are denumerable, a one-to-one correspondence can be setup between them. This correspondence can be extended to a one-to-onecorrespondence between the sets

A1U(M\A)=M\A2 and AU(M\A)=Mwhich assigns to each element in M \ A this element itself. The set M \ A2is a proper subset of the set M. We thus obtain the following result:

Every infinite set is equivalent to some proper subset of itself.This property can be taken as the definition of an infinite set.EXERCISE. Prove that if M is an arbitrary infinite set and A is denumer-

able, then Al MU A.

§4. Nondenumerability of the set of real numbers

In §2 we introduced a number of examples of denumerable sets. Thenumber of these examples could be significantly increased. Moreover, weshowed that if we take sums of denumerable sets in finite or infinite num-ber we again obtain denumerable sets. The following question arises nat-urally: do there exist in general nondenumerable infinite sets? The follow-ing theorem gives an affirmative answer to this question.

THEOREM. The set of real numbers in the closed interval [0,1] is nondenumer-able.

Proof. We shall assume the contrary, i.e., that all the real numbers lyingon the segment [0, 1], each of which can be written in the form of an infinitedecimal, can be arranged in the form of a sequence

a12 ...a22 a23 ...

(1) a32 a33 ...

...

where each is one of the numbers 0, 1, ... , 9. We now construct thedecimal

(2) 0.b1b2b3

§51 CARDIXAL NUMBERS 9

in the following manner: for b1 we take an arbitrary digit which does notcoincide with , for b2 an arbitrary digit which does not coincide witha22, and so on; in general, for we take an arbitrary digit not coincidingwith a,Lfl. This decimal caiinot coincide with any of the decimals appearingin Table (1). In fact, it differs from the first decimal of Table (1) at least inthe first digit by construction, from the second decimal in the second digitand so forth; in general, since for all it, decimal (2) cannot coincidewith any of the decimals appearing in Table (1). Thus, the assumptionthat there is some way of enumerating all the real numbers lying on thesegment [0, 1] has led to a contradiction.

The above proof lacks precision. Namely, some real numbers can bewritten in the form of a decimal in two ways: in one of them there is aninfinite number of zeros and in the other an infinite number of nines; forexample,

= 0.5000 = 0.4999

Thus, the noncoincidence of two decimals still does not mean that thesedecimals represent distinct numbers.

However, if decimal (2) is constructed so that it contains neither zerosnor nines, then setting, for example, = 2, if a,Lfl = 1 and = I if 1,

the above-mentioned objection is avoided.So we have found an example of a nondenumerable infinite set. We shall

point out some examples of sets which are equivalent to the set of realnumbers in the closed interval [0, 1].

1. The set of all points on an arbitrary segment [a, bJ or the points of theopen interval (a, b).

2. The set of all points on a straight line.3. The set of all points in the plane, in space, on the surface of a sphere,

the points lying in the interior of a sphere, and so forth.4. The set of all straight lines in the plane.5. The set of all continuous functions of one or several variables.In Examples 1 and 2 the proof offers no difficulty (see Examples 1 and§3). In the other examples a direct proof is somewhat complicated.EXERCISE. Using the results of this section and Exercise 2, §2, prove the

existence of transcendental numbers, i.e., of real numbers which are notalgebraic.

§5. The concept of cardinal number

If two finite sets are equivalent, they consist of the same number ofelements. If M and N are two arbitrary equivalent sets we say that Al andN have the same cardinal number (or the same power or the same potency).rFhus cardinal number is what all equivalent sets have in common. For

10 FUNDAMENTALS OF SET THEORY [cH. I

finite sets the concept of cardinal number coincides simply with the conceptof number of elements in the set. The cardinal number of the set of naturalnumbers (i.e. of any denumerable set) is denoted by the symbol (read"aleph zero"). Sets which are equivalent to the set of real numbers between0 and 1 are said to be sets having the power of the continum. This poweris denoted by the symbol c.

As a rule, all infinite sets which are encountered in analysis are eitherdenumerable or have cardinal number c.

If the set A is equivalent to some subset of the set B but is not equivalentto the entire set B, then we say that the cardinal number of the set A isless than the cardinal number of the set B.

Logically, besides the two possibilities indicated, namely: 1) A equiva-lent to B and 2) A equivalent to a subset of B but not equivalent to all ofB, we shall allow two more: 3) A is equivalent to some subset of B and Bis equivalent to some subset of A; 4) A and B are not equivalent and inneither of these sets is there a subset equivalent to the other set. It is pos-sible to show that in Case 3 the sets A and B are themselves equivalent(this is the Cantor-Bernstein theorem) but that Case 4 is in fact impossible(Zermelo's theorem); however, we shall not give the proof of these tworather complicated theorems here (see, for example, P. 5. Aleksandrov:Introduction to the General Theory of Sets and Functions, Chapter I, §6 andChapter II, §6; E. Kamke: Theory of Sets; F. Hausdorif: Mengenlehre; andA. Fraenkel: Abstract Set Theory).

As was pointed out at the end of §2 denumerable sets are the "smallest"infinite sets. In §4 we showed that there exist infinite sets whose infinite-ness is of a higher order; these were sets having the cardinal number of thecontinuum. But do there exist infinite cardinal numbers exceeding thecardinal number of the continuum? In general, does there exist some"highest" cardinal number or not? It turns out that the following theoremis true.

THEOREM. Let M be a set of cardinal number m. Further, let 9)1 be the setwhose elements are all possible subsets of the set M. Then 9)1 has greater car-dinal number than the cardinal number m of the initial set M.

Proof. It is easy to see that the cardinal number of the set cannot beless than the cardinal number m of the initial set; in fact, those subsets of Meach of which consists of only one element form a subset of 9)1 which isequivalent to the set M. It remains to prove that these cardinal numberscannot coincide. Let us assume the contrary; then and M are equivalentand we can set up a one-to-one correspondence between them. Let a A,b B, be a one-to-one correspondence between the elements of the setM and all of its subsets, i.e., the elements of the set 9)1. Now let X be theset of elements in M which do not belong to those subsets to which they

§6] PARTITION INTO CLASSES 11

correspond (for example, if a E A then a X, if b B then b E X, and soforth). X is a subset of M, i.e. it is some element in By assumption Xmust correspond to some element x E M. Let us see whether or not theelement x belongs to the subset X. Let us assume x X. But by definitionX consists of all those elements which are not contained in the subset towhich they correspond and consequently the element x ought to be in-cluded in X. Conversely, if we assume that x belongs to X, we may concludethat x cannot belong to X since X contains only those elements which donot belong to the subset to which they correspond. Thus the element cor-responding to the subset X ought simultaneously to be contained in and notcontained in X. This implies that in general such an alement does notexist, i.e. that it is impossible to establish a one-to-one correspondencebetween the elements of the set M and all its subsets. This completes theproof of the theorem.

Thus for an arbitrary cardinal number we can in reality construct a setof greater cardinal number and then a still greater cardinal number, and soon, obtaining in this way a hierarchy of cardinal numbers which is notbounded in any way.

EXERCISE. Prove that the set of all numerical functions defined on a setM has a greater cardinal number than the cardinal number of the set M.Hint: Use the fact that the set of all characteristic functions (i.e. functionsassuming only the values 0 and 1) defined on 111 is equivalent to theset ofall subsets of M.

§6. Partition into classes

The reader can omit this section on a first reading and return to it inthe sequel for information as required.

In the most varied questions it occurs that we encounter partitions of aset into disjoint subsets. For example, the plane (considered as a set ofpoints) can be partitioned into straight lines parallel to the y-axis, three-dimensional space can be represented as the set of all concentric spheres ofdifferent radii, the inhabitants of a given city can be partitioned intogroups according to their year of birth, and so forth.

If a set M is represented in some manner as a sum of disjoint subsets,we speak of a partitioning of the set M into classes.

Ordinarily we encounter partitions which are obtained by means ofindicating some rule according to which the elements of the set M are com-bined into classes. For example, the set of all triangles in the plane can bepartitioned into classes of triangles which are congruent to one anotheror triangles which have the same area; all polynomials in x can be parti-tioned into classes by collecting all polynomials having th& same zeros intoone class, and so on.

12 FUNDAMENTALS OF SET THEORY [cii. I

Rules according to which the elements of a set are partitioned intoclasses can be of the most varied sort. But of course all these rules cannotbe entirely arbitrary. Let us assume, for example, that we should like topartition all real numbers into classes by including the number b in thesame class as the number a if, and only if, b> a. It is clear that no partitionof the real numbers into classes can be obtained in this way because ifb > a, i.e. if b must be included in the same class as a, then a < b, i.e. thenumber a must not belong to the same class as b. Moreover, since a is notlarger than a, then a ought not belong in the class which contains it! Weconsider another example. We shall see whether or not it is possible topartition all inhabitants of a given city into classes by putting two personsinto the same class if, and only if, they are acquaintances. It is clear thatsuch a partition cannot be realized because if A is an acquaintance of Band B is an acquaintance of C, then this does not at all mean that A is anacquaintance of C. Thus, if we put A into the same class as B and B intothe same class as C, it may follow that two persons A and C who are notacquaintances are in the same class. We obtain an analogous result if weattempt to partition the points of the plane into classes so that those andonly those points whose mutual distance does not exceed 1 are put into oneclass.

The examples introduced above point out those conditions which mustbe satisfied by any rule if it is to realize a partition of the elements of a setinto classes.

Let M be a set and let some pair (a, b) of elements of this set be "marked."[Here the elements a and b are taken in a definite order, i.e. (a, b) and(b, a) are two distinct pairs.] If (a, b) is a "marked" pair, we shall say thatthe element a is related to b by the relation p and we shall denote thisfact by means of the symbol a p b. For example, if we wish to partitiontriangles into classes of triangles having the same area, then a p b is to mean:"triangle a has the same area as triangle b." We shall say that the givenrelation is an equivalence relation if it possesses the following properties:

1. Refiexivity:aç,afor any elementa E M;2. Symmetry: if a b, then necessarily b a;3. Transitivity: if a b and b c, then a C.

Obviously every partition of a given set into classes defines some equiva-lence relation among the elements of this set.

In fact, if a b means "a belongs to the same class as b", then this relationwill be reflexive, symmetric and transitive, as is easy to verify.

Conversely, if a b is an equivalence relation between the elements ofthe set M, then putting into one class those and only those elements whichare equivalent we obtain a partition of the set M into classes.

In fact, let Ka be the class of elements in M which are equivalent to a

§71 MAPPINGS OF SETS. FUNCTIONS 13

fixed element a. By virtue of the property of reflexivity the elemeiit aitself belongs to the class Ka. We shall show that two classes Ka and Kbeither coincide or are disjoint. Let an element c belong simultaneously toKa and to Kb , i.e. C (p a and c (p b.

Then by virtue of symmetry, a (p c, and by virtue of transitivity,

(1) apb.Now ii x is an arbitrary element in Ka, i.e. x (p a, then by virtue of (1)

and the transitivity property, x (p b, i.e. x E Kb.Conversely, if y is an arbitrary element in Kb, i.e. if y (p b, then by virtue

of relation (1) which can be written in the form b (p a (symmetry!) and thetransitivity property, y (pa, i.e. y E Ka. Thus, two classes Ka and Kb havingat least one element in common coincide.

We have in fact obtained a partition of the set M into classes accordingto the given equivalence relation.

§7. Mappings of sets. General concept of function

In analysis the concept of function is introduced in the following way.Let X be a set on the real line. We say that a function f is defined on thisset if to each number x E X there is made to correspond a definite numbery = f(x). In this connection X is said to be the domain of the given func-tion and Y, the set of all values assumed by this function, is called itsrange.

Now if instead of sets of numbers we consider sets of a completely arbi-trary nature, we arrive at the most general concept of function, namely:Let M and N be two arbitrary sets; then we say that a function f is definedon M and assumes its values in N if to each element x E ill there is madeto correspond one and only one element in N. In the case of sets of auarbitrary nature instead of the term "function" we frequently use theterm "mapping" and speak of a mapping of one set into another.

If a is any element in M, the element b = f(a) in N which correspondsto it is called the image of the element a' (under the mapping f). The set ofall those elements in M whose image is a given element b E N is called theinverse image (or more precisely the complete inverse image) of the elementb and is denoted by r'(b).

If A is any set in M, the set of all elements of the form {f(a) :a E A 'iscalled the image of A and is denoted by f(A). In its turn, for every setB in N there is defined its inverse image f'(B), namely f'(B) is the setof all those elements in M whose images belong to B.

In this section we shall limit ourselves to the consideration of the mostgeneral properties of mappings.

We shall use the following terminology. We say that f is a mapping of

14 FUNDAMENTALS OF SET THEORY [cH. I

the set M onto the set N if f(M) = N; in the general case, i.e.when f(M) N we say that f is a mapping of M into N.

We shall establish the following properties of mappings.THEOREM 1. The inverse image of the sum of two sets is equal to the sum of

their inverse images:

f_i(A U B) = f'(A) Uf_i(B).

Proof. Let the element x belong to the set f'(A U B). This means thatf(x) E A U B, i.e. f(x) E A or f(x) E B. But then x belongs to at least oneof the sets f_i(A) and f (B), i.e. x E (A) U (B). Conversely, ifx E f_i(A) U f_i(B), then x belongs to at least one of the sets and

i.e. f(x) belongs to at least one of the sets A and B and conse-quently f(x) E A U B, whence it follows that x E U B).

THEOREM 2. The inverse image of the intersection of two sets is equal to theintersection of their inverse images:

ri(A fl B) =

fl E A fl f(x) E A f(x) Eand x E f_i(B), i.e. x E ri(A) fl ri(B).

Conversely, if x E f_i(A) fl i.e. x E ri(A) and x E r'(B), thenf(x) E A and f(x) E B, or in other words f(x) E A fl B.

Consequently x E fl B).Theorems 1 and 2 remain valid also for the sum and intersection of an

arbitrary finite or infinite number of sets.Thus, if in N some system of sets closed with respect to the operations of

addition and taking intersections is selected, then their inverse images inM form a system which is likewise closed with respect to these operations.

THEOREM 3. The image of the sum of two sets is equal to the sum of theirimages:

f(A U B) = f(A) U f(B).

Proof. If y E f(A U B), then y = f(x), where x belongs to at least one ofthe sets A and B. Consequently, y = f(x) E f(A) U f(B). Conversely, ify E f(A) U f(B) then y = f(x), where x belongs to at least one of the setsA and B, i.e. x E A U B and consequently y = f(x) E f(A U B).

We note that in general the image of the intersection of two sets does notcoincide with the intersection of their images. For example, let the mappingconsidered be the projection of the plane onto the x-axis. Then the seg-ments

O�x�1; y=O,O�x�1; y=l

do not intersect, but at the same time their images coincide.

§7] MAPPINGS OF SETS. FUNCTIONS 15

The concept of mapping of sets is closely related to the concept of par-titioning considered in the preceding section.

Let f be a mapping of the set A into the set B. If we collect into one classall those elements in A whose images in B coincide, we obviously obtain apartition of the set A. Conversely, let us consider an arbitrary set A andan arbitrary partitioning of A into classes. Let B be the totality of thoseclasses into which the set A is partitioned. If we correspond to each elementa E A that class (i.e. that element in B) to which a belongs, we obtain amapping of the set A into the set B.

EXAMPLES. 1. Consider the projection of the xy-plane onto the x-axis.rfhe inverse images of the points of the x-axis are vertical lines. Conse-quently to this mapping there corresponds a partitioning of the plane intoparallel straight lines.

2. Subdivide all the points of three-dimensional space into classes bycombining into one class the points which are equidistant from the originof coordinates. Thus, every class can be represented by a sphere of someradius. A realization of the totality of all these classes is the set of allpoints lying on the ray (0, oo). So, to the partitioning of three-dimensionalspace into concentric spheres there corresponds the mapping of this spaceonto a half-line.

3. Combine all real numbers having the same fractional part into oneclass. The mapping corresponding to this partition is represented by themapping of the real line onto a circle.

Chapter II

METRIC SPACES

§8. Definition and examples of metric spaces

Passage to the limit is one of the most important operations in analysis.The basis of this operation is the fact that the distance between any twopoints on the real line is defined. A number of fundamental facts fromanalysis are not connected with the algebraic nature of the set of real num-bers (i.e. with the fact that the operations of addition and multiplication,which are subject to known laws, are defined for real numbers), but de-pend only oii those properties of real numbers which are related to theconcept of distance. This situation leads naturally to the concept of "metricspace" which plays a fundamental role in modern mathematics. Furtheron we shall discuss the basic facts of the theory of metric spaces. The re-sults of this chapter will play an essential role in all the following discussion.

DEFINITION. A metric space is the pair of two things: a set X, whoseelements are called points, and a distance, i.e. a single-valued, nonnegative,real function p(x, y), defined for arbitrary a' aiid y in X and satisfying thefollowing conditions:

1) p(x, y) = 0 if and only if x =2) (axiom of symmetry) p(x, y) = p(y, x),3) (triangle axiom) p(x, y) + p(y, z) � p(x, z).

The metric space itself, i.e. the pair X and p, will usually be denoted byI? = (X, p).

In cases where no misunderstanding can arise we shall sometimes denotethe metric space by the same symbol X which is used for the set of pointsitself.

We list a number of examples of metric spaces. Some of the spaces listedbelow play a very important role in analysis.

1. If we setlo, if =

p(x, y) =if x y,

for elements of an arbitrary set, we obviously obtain a metric space.2. The set D1 of real numbers with the distance function

p(x, y) = I x — y

forms the metric space R1.3. The set of ordered n-tuples of real numbers x = (Xi, X2, ,

with distance function

p(x, 71) = (Yk — XkH

16

§8] DEFINITION AND EXAMPLES OF METRIC SPACES 17

is called Euclidean n-space The validity of Axioms 1 and 2 for isobvious. To prove that the triangle axiom is also verified in we makeuse of the Schwarz inequality

n 2 n 2 n 2(1) ( k=1 akbk) � k=1 ak k=1 bk

(The Schwarz inequality follows from the identityL\2 , n n L2\ 1 fl fl I L Lk=1 akuk) = k=1 ak I k=1 Uk ) — — u1a1)

which can be verified directly.) If

x = (x1., X2, . , y = (yr, Y2, , and z = (z1, z2, ,

then setting

yk—xk=ak, Zk—yk=bk,

we obtain

Zk — = ak + bk

by the Schwarz inequality

+ bk) = ak2 + 2 akbk + bk

n 2 n 22 k=1 ak k=1 k=1 bk

+i.e.

p2(x, z) Ip(x, y) + p(y, z)}2

or

p(x, z) � p(x, y) + p(y, z).

4. Consider space in which the points are again ordered n-tuplesof numbers (xi , X2, , and for which the distance function is definedby the formula

po(x, y) = max fl Yk — Xk I; 1 � k � n}.The validity of Axioms 1—3 is obvious. In many questions of analysis

this space is no less suitable than Euclidean spaceExamples 3 and 4 show that sometimes it is actually important to have

different notations for the set of points of a metric space and for the metricspace itself because the same point set can be metrized in various ways.

5. The set C[a, b] of all continuous real-valued functions defined on thesegment [a, 14 with distance function

(2) p(f, g) = sup g(t) — f(t) I; a � t � b}

18 METRIC SPACES [CH. II

likewise forms a metric space. Axioms 1—3 can be verified directly. Thisspace plays a very important role in analysis. We shall denote it by thesame symbol C[a, b} as the set of points of this space. The space of con-tinuous functions defined on the segment [0, 1] with the metric given abovewill be denoted simply by C.

6. We denote by 12 the metric space in which the points are all possiblesequences x = (x1 , X2, , . . .) of real numbers which satisfy thecondition Xk < and for which the distance is defined by meansof the formula

(3) p(x, y) = —

We shall first prove that the function p(x, y) defined in this way alwayshas meaning, i.e. that the series — Xk)2 converges. We have

n 2 k 2 V' n 2I (Yk — Xk) } � (L1k=1 Xk ) + Yk )

for arbitrary natural number n (see Example 3).Now let n By hypothesis, the right member of this inequality has

a limit. Thus, the expression on the left is bounded and does not decreaseas n consequently, it tends to a limit, i.e. formula (3) has meaning.Replacing x by — x in and passing to the limit as n cc, we obtain

2 2 4 2(4) { + } � +but this is essentially the triangle axiom. In fact, let

a= (al,a2,b = (b1, b2, b •..)c = (c1,c2, •..

be three points in 12. If we set

bk—ak=xk-, Ck—bkyk,then

Ck — = 71k + Xkand, by virtue of (4),

{ (Ck — ak)} � { — ak)} + { —

i.e.

p(a, c) � p(a, b) + p(b, c).

7. Consider, as in Example 5, the totality of all continuous functions onthe segment [a, b}, but now let the distance be defined by setting

Il 2(5) p(x, y)

= [Ja{x(t) — y(t) dt


This metric space is denoted by C2[a, b} and is called the space of con-tinuous functions with quadratic metric. Here again Axioms 1 and 2 in thedefinition of a metric space are obvious and the triangle axiom followsimmediately from the Schwarz inequality

{J' x(t)y(t) dt}1b

x2(t) dt f y2(t)

which can be obtained, for instance, from the following easily-verifiedidentity:

{J' x(t)y(t)dt} = 1b

x2(t) dt f y2(t)

— L f [x(s)y(t) — y(s)x(t)}2 ds dt.

8. Consider the set of all bounded sequences x = (xi, X2, , •)of real numbers. We obtain the metric space if we set

(6) p(x, y) = sup ilk — Xk I.

The validity of Axioms 1—3 is obvious.9. The following principle enables us to write down an infinite number

of further examples: if R = (X, p) is a metric space and M is an arbitrarysubset of X, then M with the same function p(x, y), but now assumed tobe defined only for x and y in M, likewise forms a metric space; it is calleda subspace of the space R.

(1) In the definition of a metric space we could have limited ourselves totwo axioms for p(x, y), namely:

1) p(x,y) = 0

if, and only if, x = y;

2) p(x, y) p(z, x) + p(z, y)

for arbitrary x, y, z.It follows that

3) p(x, y) � 0,4) p(x, y) = p(y, x)

and consequently Axiom 2 can be written in the form

2') p(x, y) p(x, z) + p(z, y).

(2) The set of ordered n-tuples of real numbers with distance

pp(x, y) I— Xk

13)1/P (p � 1)

20 METRIC SPACES [Cm II

also forms a metric space which we shall denote by Here the validityof Axioms 1 and 2 is again obvious. We shall check Axiom 3. Let

x = (x1 , , y = , , , and z = (z1 , . • ,

be points in If, as in Example 3, we set

yL—xL.=ak, zk—yk=bk,then the inequality

z) � pp(X, y) + pp(Y, z)

assumes the form

(7) + bkI

P)h/P

This is the so-called Minkowski inequality. Minkowski's inequality isobvious for p = 1 (since the absolute value of a sum is less than or equalto the sum of the absolute values) and therefore we can restrict ourselvesto considering the case p > 1.

In order to prove inequality (7) for p> 1 we shall first establish Holder'sinequality:

(8)I

Xk?Jk � Xj 1/k

where the number q is defined by the condition

(9) i/p + 1/q = 1.

We note that inequality (8) is homogeneous in the sense that if it issatisfied for any two vectors

x = (xi , , • , and y = , , • . ,

then it is also satisfied for the vectors Xx and where X and are arbitrarynumbers. Therefore it is sufficient to prove inequality (8) for the case when

n p n q(10) k=1 XkI = k=1 /1k I

= 1

Thus, we must prove that if Condition (10) is satisfied, then

(11) XkYk I � 1.

Consider in the the curve defined by the equation =

or equivalently by the equation = (see Fig. 7). It is clear from thefigure that for an arbitrary choice of positive values for a and b we haveSi + & � ab. If we calculate the areas and 82, we obtain

a b

Si = f = a"/p; 82 = f q—1 =


Thusab � e/p +

b = I I and summing with respect to k from 1 to n,we obtain

I� 1,

if we take (9) and (10) into consideration.Inequality (11) and consequently the more general inequality (8) are

thus proved. For p = 2 Holder's inequality (8) becomes the Schwarzinequality (1).

We now proceed to the proof of the Minkowski inequality. Consider theidentity

(IaI+IbIY= (IaI+IbIY'IaI+(IaI+IbIY'IbI.Setting a = Xk, b = Yk in the above identity and summing with respect tok from 1 to n, we obtain

p p—ik=1(I Ski + IYkI) = k=1(IXkI + IYkI) IxkI

p—i+ L1k=i(I Xk I +

I I) I

If we now apply Holder's inequality to each of the two sums on the rightof the above equality and take into consideration the fact that (p — 1)q =we obtain

fl1 )

� (I XkI + I I

Dividing both sides of this inequality by

Lt=l (I XkI + I Y'c

we obtain

(I Xk I + I )P} i/PI I We

FIG. 7


whence inequality (7) follows immediately. This also establishes the triangleaxiom for the space

(3) It is possible to show that the metric

po(x, y) = max {Iyk Xk I; 1 � k � n}introduced in Example 4 can be defined in the following way:

/ \ 1. p\1/p= Li'c=i Yk Xk )

(4) From the inequality

ab � ar/p + (i/p + 1/q = 1)

established in Example (2) it is easy to deduce also the integral form ofHolder's inequality

fbx(t)y(t) dt �

(fbdt)

(fbIdt),

which is valid for arbitrary functions x(t) and y(t) for which the integralson the right have meaning. From this in turn we obtain the integral formof Minkowski's inequality:

(fbIx(t) + y(t) P dt) � (fb

x(t) dt) + (f y(t) dt).(5) We shall point out still another interesting example of a metric

space. Its elements are all possible sequences of real numbers

x = (Xl,X2, ... ••)

such thatI

< oc, where p � 1 is any fixed number and the dis-tance is defined by means of the formula

00 p1/p(12) p(x, y) = ( IXk

I )

We shall denote this metric space byBy virtue of Minkowski's inequality (7) we have

Yk — XkI

for arbitrary n. Since the series00 p oo pk=1 I and k=1 I Yk I

converge by assumption, passing to the limit as n oo we obtain

(13)I Yk Xk � I Xk I Yk

P)1IP,

and so the series on the left side also converges. This proves that formula(12), which defines distance in actually has meaning for arbitrary

§9] CONVERGENCE OF SEQUENCES. LIMIT POINTS 23

x, y E At the same time inequality (13) shows that the triangle axiomis satisfied in The remaining axioms are obvious.

§9. Convergence of sequences. Limit points

In we shall establish some fundamental concepts which we shallfrequently use in the sequel.

An open sphere S(xo, r) in the metric space R is the set of all pointsx E R which satisfy the condition p(x, x0) < r. The fixed point x0 is calledthe center and the number r is called the radius of this sphere.

A closed sphere S[xo , r] is the set of all points x E R which satisfy thecondition p(x, x0) � r.

An €-neighborhood of the point x, denoted by the symbol O(x, €), is anopen sphere of radius and center x0.

A point x is called a contact point of the set M if every neighborhood of xcontains at least one point of M. The set of all contact points of the set Mis denoted by [M] and is called the closure of M. Since every point belongingto M is obviously a contact point of M (each point is contained in everyone of its neighborhoods), every set is contained in its closure: M [M].

THEOREM 1. The closure of the closure of M is equal to the closure of M:

[M].

Proof. Let x E Then an arbitrary €-neighborhood O(x, €) of xcontains a point x1 E [M]. Setting p(x, x1) = , we consider the sphereO(xi, €'). This sphere lies entirely in the interior of O(x, €). In fact, ifz E O(xi, €'), then p(z, x1) < €i; and since p(x, x1) = €1, then, bythe triangle axiom p(z, x) � €i + (€ — €i) = €, i.e. z E O(x, €). Since x1 E[M], O(x, €') contains a poiht X2 E M. But then X2 E O(x, €). Since O(x, €)is an arbitrary neighborhood of the point x, we have x E [M]. This com-pletes the proof of the theorem.

The validity of the following assertion is obvious.THEOREM 2. If M1 M, then [M1] [M].THEOREM 3. The closure of a sum is equal to the sum of the closures:

[M1 U M2] = [M1] U [M2].

Proof. Let x E [M1 U M2], i.e. let an arbitrary neighborhood O(x, €)contain the point y E M1 U M2. If it were true that x [M1] and x [M2],

we could find a neighborhood O(x, €') which does not contain points of M1and a neighborhood O(x, €2) which does not contain points of M2. Butthen the neighborhood O(x, €), where mm (€', €2), would not containpoints of M1 U M2. From the contradiction thus obtained it follows thatx is contained in at least one of the sets [M1] and [M2], i.e.

[M1 U M2] [M1] U [M2].


Since M1 U M2 and ]l'12 U M2, the converse inclusion followsfrom Theorem 2.

The point x is called a limit point of the set M if an arbitrary neighbor-hood of x contains an infinite number of points of M.

A limit point of the set M can either belong to M or not. For example, ifM is the set of rational numbers in the closed interval [0, 1], then everypoint of this interval is a limit point of 111.

A point x belonging to the set M is said to he an isolated point of thisset if x has a neighborhood O(x, €) which does not contain any points of Mdifferent from x.

THEOREM 4. Every contact point of the set ill is either a limit point of theset M or an isolated point of M.

Proof. Let x be a contact point of the set ill. This means that everyneighborhood O(x, €) of x contains at least one point belonging to M.Two cases are possible:

1) Every neighborhood of the point x contains an infiiiite number ofpoints of the set M. In this case, x is a limit point of M.

2) We can find a neighborhood O(x, €) of x which contains only a finitenumber of points of M. In this case, x will be an isolated point of the set M.In fact, let x1 , X2, • , be the points of M which are distinct from xand which are contained in the neighborhood O(x, €). Further, let be theleast of the positive numbers p(x, x1), i = 1, 2, ... , k. Then the neighbor-hood O(x, €o) obviously does not contain any point of M distinct from x.The point x itself in this case must necessarily belong to iTi since otherwiseO(x, €°) in general would not contain a single point of M, i.e. x would notbe a contact point of the set M. This completes the proof of the theorem.

Thus, the set [M} consists in general of points of three types:1) Isolated points of the set M;2) Limit points of the set M which belong to M;3) Limit points of the set M which do not belong to M.

[M] is obtained by adding to 111 all its limit points.Let xi, X2, . be a sequence of points in the metric space R. We say

that this sequence converges to the point x if every neighborhood O(x, €)contains all points starting with some one of them (i.e. if for every€> 0 we can find a natural number such that O(x, €) contains all points

with n > N€). The point x is said to he the limit of the sequenceThis definition can obviously he formulated in the following form: the

sequence converges to x if p(x, = 0.The following assertions follow directly from the definition of limit: I)

no sequence can have two distinct limits; 2) if the sequence convergesto the point x then every subsequence of } converges to the same point a.

The following theorem establishes the close connection between the

§9] CONVERGENCE OF SEQUENCES. LIMIT POINTS 25

Concepts of contact point and limit point on the one hand and the conceptof limit on the other.

THEOREM 5. A necessary and sufficient condition that the point x be a con-tact point of the set M is that there exist a sequence of points of theset M which converges to x; a necessary and sufficient condition that the point xbe a limit point of M is that there exist a sequence of distinct points of the setM which converges to x.

Proof. Necessity. If x is a contact point of the set M, then every neighbor-hood O(; 1/n) contains at least one point of M. These points form asequence which converges to x. If the point x is a limit point of M, everyneighborhood O(x, 1/n) contains a point E M which is distinct from allthe < n) (since the number of such points is finite). The points aredistinct and form a sequence which converges to x.

Sufficiency is obvious.Let A and B be two sets in the metric space R. The set A is said to be

dense in B if [A] B. In particular, the set A is said to be everywhere densein R if its closure [A] coincides with the entire space R. For example, theset of rational numbers is everywhere dense on the real line.

EXAMPLES OF SPACES CONTAINING AN EVERYWHERE DENSE DENTJMER-ABLE SET. (They are sometimes called "separable." For another definitionof such spaces in terms of the concept of basis see §10, Theorem 4.) Weshall consider the very same examples which were pointed out in §8.

1. The space described in Example 1, §8, is separable if, and only if, itconsists of a denumerable number of points. This follows directly fromthe fact that in this space [MI = M for an arbitrary set M.

All spaces enumerated in Examples 2—7, §8, are separable. We shallindicate a denumerable everywhere dense set in each of them and leave thedetails of the proof to the reader.

2. Rational points.3. The of all vectors with rational coordinates.4. The set of all vectors with rational coordinates.5. The set of all polynomials with rational coefficients.6. The set of all sequences in each of which all terms are rational and

only a finite (but arbitrary) number of terms is distinct from zero.7. set of all polynomials with rational coefficients.rfhe space of bounded sequeiices (Example 8, §8) is not separable. In

fact, let us consider all possible sequences consisting of zeros and ones. Theyform a set with cardinal number that of the continuum (since each of themcan be put into correspondence with the dyadic development of some realnumber which is contained in the interval [0, 1]). The distance between twosuch distinct points defined by formula (6), §8, is 1. We surround each ofthese points with a sphere of radius These spheres do not intersect. If

26 METRIC SPACES [CR. II

some set is everywhere dense in the space under consideration, then eachof the indicated spheres should contain at least one point of this set andconsequently it cannot be denumerable.

(1) Let A be an arbitrary set in the metric space R and let x be a pointin R. The distance from the point x to the set A is defined by the number

p(A, x) = inf {p(a, x); a E A}.

If x E A, then p(A, x) = 0; but the fact that p(A, x) = 0 does not implythat x E A. From the definition of contact point it follows immediatelythat p(A, x) = 0 if, and only if, x is a contact point of the set A.

Thus, the closure [A] of the set A can be defined as the totality of allthose points whose distance from the set A is zero.

(2) We can define the distance between two sets analogously. If A and Bare two sets in R, then

p(A,B) = inf {p(a,b);a E A,b E B}.

If A fl B 0, then p(A, B) = 0; the converse is not true in general.(3) If A is a set in the metric space R then the totality A' of its limit

points is called its derived set.Although the application to [M] once more of the operation of closure

always results again in [M], the equality (M')' = M' does not hold ingeneral. In fact, if we take, for example, the set A of points of the form1/n on the real line, then its derived set A' consists of the single point 0,but the set A" = (A')' will already be the void set. If we consider on thereal line the set B of all points of the form 1/n + 1/(nm) (n, m = 1, 2, . .

then B' = A U A', B" is the point 0, and B" is the void set.

§10. Open and closed sets

In this section we shall consider the more important types of sets in ametric space; these are the open and closed sets.

A set M in a metric space R is said to be closed if it coincides with itsclosure: [MI M. In other words, a set is said to be closed if it containsall its limit points.

By Theorem 1, §9, the closure of an arbitrary set M is a closed set.Theorem 2, §9, implies that [MI is the smallest closed set which contains M.

EXAMPLES. 1. An arbitrary closed interval [a, b] on the real line is aclosed set.

2. The closed sphere is a closed set. In particular, in the space C[a, b]the set of functions satisfying the condition I f I � K is closed.

3. The set of functions satisfying the condition I f I < K (open sphere)is not closed; its closure is the set of functions satisfying the conditionIfI�K.

§10] OPEN AND CLOSED SETS 27

4. Whatever the metric space R, the void set and the whole space Rare closed sets.

5. Every set consisting of a finite number of points is closed.The fundamental properties of closed sets can be formulated in the iorm

of the following theorem.THEOREM 1. The intersection of an arbitrary number and the sum of an

arbitrary finite of closed sets are closed sets.Proof. Let F = flaFa, where the Fa are closed sets. Further, let x be a

limit point of the set F. This means that an arbitrary neighborhood O(x, €)of x contains an infinite number of points of F. But then O(x, €) containsan infinite number of points of each Fa and consequently, since all the Faare closed, the point x belongs to each Fa ; thus, x E F = flaFa, i.e. F isclosed.

Now let F be the sum of a finite number of closed sets: F =and let x be a point not belonging to F. We shall show that x cannot be alimit point of F. In fact, x does not belong to any of the closed sets andconsequently it is not a limit point of any of them. Therefore for every iwe can find a neighborhood O(x, of the point x which does not containmore than a finite number of points of If we take the smallest of theneighborhoods O(x, ... , O(x, we obtain a neighborhood O(x, €) ofthe point x which does not contain more than a finite number of pointsof F.

Thus, if the point x does not belong to F, it cannot be a limit point ofF, i.e. F is closed. This completes the proof of the theorem.

The point x is said to be an interior point of the set M if there exists aiieighborhood O(x, €) of the point x which is contained entirely in M.

A set all of whose points are interior points is said to be an open set.EXAMPLES. 6. The interval (a, b) of the real line D' is an open set; in fact,

if a < a < b, then O(a, €), where mm (a — a, b — a), is containedentirely in the interval (a, b).

7. The open sphere S(a, r) in an arbitrary metric space R is an open set.In fact, if x E S(a, r), then p(a, x) < r. We set r — p(a, x). ThenS(x, €) S(a, r).

8. The set of continuous functions satisfying the condition I f I < K,where K is an arbitrary number, is an open subset of the space C[a, b}.

THEOREM 2. A necessary and sufficient condition that the set M be open isthat its complement R M with respect to the whole space R be closed.

Proo.f. If M is open, then each point x E M has a neighborhood whichbelongs entirely to M, i.e. which does not have a single point in commonwith R \ M. Thus, no point which does not belong to R \ M can be acontact point of R \ M, i.e. I? \ M is closed. Conversely, if R \ M isclosed, an arbitrary point of M has a neighborhood which lies entirely inM, i.e. M is open.

28 METB1C [cH. II

Since the void set and the whole space I? aie closed and are at the sametime complements of each other, the theorem proved above implies thefollowing corollary.

The void set and whole space 1? are open sets.The following important theorem which is the dual of Theorem 1 follows

from Theorem I and the principle of duality established iii §1 (the intei-section of (omplelnelits the complement of the sums, the sum of thecomplements equals the complement of the intersections).

THEOREM I'. The sum of an itrary nwnber and the intersection of anarbitrary finite number of open sets are o'pen sets.

A family {Ga} of open sets in I? is (ailed a basis in R if every open setin R can be represeiited as the sum of a (finite or infinite) number of setsbelonging to this family.

To check whether or not a given family of open sets is a basis we findthe followi i ig (riteli on useful.

3. A necessary and sufficient condition that a system of open setsGa } be a basis in 1? is that for every open set G and for erery point x E G

a set Ga can be found in this sijstem such that x E Ga C G.Proof. If {Ga} is a basis, then every open set G is a sum of Ga's: G =Gaj , and consequently every point x in G belongs to some Ga contained

in G. Conversely, if the condition of the theorem is fulfilled, then {Ga} iSa basis. In fact, let G be an arbitrary open set. For each point x E G wecan find some Ga(X) such that x E Ga C G. The sum of these Ga(X) over allx E G equals G.

With the aid of this criterion it is easy to establish that in every metricspace the family of all open spheres forms a basis. The family of all sphereswith rational radii also forms a basis. On the real line a basis is formed,for example, by the family of all rational intervals (i.e. intervals withrational endpoints).

We shall say that a set is countable if it is either finite or denurnerable.1? is said to be a space with countable basis or to satisfy the second axiom

of countability if there is at least pne basis in R consisting of a countablenumber of elements.

THEOREM 4. A necessary and sufficient condition that R be a space withcountable basis is that there exist in R an everywhere dense countable set. (Afinite everywhere dense set occurs only in spaèes consisting of a finite setof points.)

Proof. Necessity. Let R have a countable basis{ }. Choose from each

G?L.an arbitrary point The set {x, } obtained in this manner is every-where dense in I?. In fact, let x be an arbitrary point in R and let O(x, €)be a neighborhood of x. According to Theorem 3, a set can be foundsuch that x E G,1 C O(x, €). Since contains at least one of the points of

OPEN AND CLOSED SETS 29

the set any neighborhood O(x, €) of an arbitrary point x E I? containsat. least one point from and this means that is everywhere densein II.

Sufficiency. If is a CoUntal)le everywhere dense set in R, then thefamily of spheres S(x7, , I/k) forms a countable basis iii 1?. In fact, the set.of all these spheres is countable (being the sum of a countable family ofcountable sets). Further, let G l)e an arbitrary open set and let x l)e artypoint in G. By the definition of au open set an m > 0 can be found suchthat the sphere S(x, 1/rn.) lies entirely in G. We now select a point .r,,0from the set {.r,1 such that p(.r, < 1/3m. Then the sphere ,1/2rn)contains the point .r and is contained in S(x, 1/rn) and consequently iiiG also. By virtue of Theorem 3 it follows from this that. the spheres

, 1/k) form a basis in I?.By virtue of this theorem, the examples introduced above 9) of

separable spaces are at. the same time examples of spaces with countablebasis.

W'e say that a system of sets Ala is a covering of the space I? if UMa 1?.

A covering consisting of open (closed) sets will be called an open (closed)covering.

rfHEOitEM 5. If I? is a metric space with countable basis, then we can select

a countable covering from each of its open coverings.Proof. Let be an arbitrary open covering of R. Thus, every point

e E I? is contained in someLet {G, } be a countable l)asis in R. Then for every x E 1? there exists a

E aFld an a such that x E (x) c Oa . The family of setsselected in this way is countable and covers R. If we choose for each

of the one of the sets Oa containing it., we obtain a countable sub-covering of the covering {Oa}.

It was already indicated above that the void set and the entire space 1?are simultaneously open and closed. A space in which there are no othersets which are simultaneously open and closed is said to be connected. Thereal line l?1 is one of the simplest examples of a connected metric space.But if we remove a fiuiite set of points (for example, one point) from R1,the remaining space is no longer connected. The simplest example of aspace which is riot connected is the space consisting of two points whichare at an arbitrary distance from one another.

(1) Let -1'Ik be the set. of all functionsf in C[a, b] which satisfy a so-calledLipschitz condition

— f(t2) � K — t2 ,

where K is a constant. The set Mk is closed. It coincides with the closureof the set of all differentiable functions which are such that I f'(t) < K.

30 METRIC SPACES [CII. ii

(2) The set M = UkMk of all functions each of which satisfies a Lipschitzcondition for some K is not closed. Since M contains the set of all poly-nomials, its closure is the entire space C[a, b].

(3) Let distance be defined in the space X in two different ways, i.e. letthere be given two distinct metrics pi(x, y) and p2(x, y). The metrics P1 andP2 are said to be equivalent if there exist two positive constants a and b suchthat a < [pi(x, y)/p2(x, y)] <b for all x y in R. If an arbitrary set M Xis closed (open) in the sense of the metric p1 , then it is closed (open) in thesense of an arbitrary metric p2 which is equivalent to p'

(4) A number of important definitions and assertions concerning metricspaces (for example, the definition of connectedness) do not make use ofthe concept of metric itself but only of the concept of open (closed) set, or,what is essentially the same, the concept of neighborhood. In particular, inmany questions the metric introduced in a metric space can be replaced byany other metric which is equivalent to the initial metric. This point ofview leads naturally to the concept of topological space, which is a gen-eralization of metric space.

A topological space is a set T of elements of an arbitrary nature (calledpoints of this space) some subsets of which are labeled open sets. In thisconnection we assume that the following axioms are fulfilled:

1. T and the void set are open;2. The sum of an arbitrary (finite or infinite) number and the inter-

section of an arbitrary finite number of open sets are open.The sets T \ G, the complements of the open sets G with respect to T,

are said to be closed. Axioms 1 and 2 imply the following two assertions.1'. The void set and T are closed;2'. The intersection of an arbitrary (finite or infinite) number and the

sum of an arbitrary finite number of closed sets are closed.A neighborhood of the point x E T is any open set containing x.In a natural manner we introduce the concepts of contact point, limit

point, and closure: x E T is said to be a contact point of the set M if everyneighborhood of the point x contains at least one point of M; x is said tobe a limit point of the set ill if every neighborhood of the point x containsan infinite number of points of M. The totality of all contact points of theset M is called the closure [MI of the set M.

It can easily be shown that closed sets (defined as the complements ofopen sets), and only closed sets, satisfy the condition [MI = M. As alsoin the case of a metric space [MI is the smallest closed set containing M.

Similarly, as a metric space is the pair: set of points and a metric, so atopological space is the pair: set of points and a topology defined in thisspace. To introduce a topology into T means to indicate in T those subsetswhich are to be considered open in T.

§11] OPEN AND CLOSED SETS ON THE REAL LINE 31

EXAMPLES. (4-a) By virtue of Theorem 1' open sets in a metric spacesatisfy Conditions 1 and 2 in the definition of a topological space. Thus,every metric space can be considered as a topological space.

(4-b) Let T consist of two points a and b and let the open sets in T be T,the void set, and the set consisting of the single point b. Axioms 1 and 2 arefulfilled. The closed sets are T, the void set, and the set consisting of thesingle point a. The closure of the set consisting of the point b is all of T.

(5) A topological space T is said to be metrizable if a metric can be intro-duced into the set T so that the sets which are open in the sense of thismetric coincide with the open sets of the initial topological space. Thespace (4-b) is an example of a topological space which cannot be metrized.

(6) Although many fundamental concepts carry over from metricspaces to topological spaces defined in (4), this concept turns out to be toogeneral in a number of cases. An important class of topological spacesconsists of those spaces which satisfy, in addition to Axioms 1 and 2, theHausdorif separation axiom:

3. Any two distinct points x and y of the space T have disjoint neigh-borhoods.

A topological space satisfying this axiom is called a Hausdorff space.Clearly, every metric space is a Hausdorif space. The space pointed out inExample (4-b) does not satisfy the Hausdorif axiom.

§11. Open and closed sets on the real line

The structure of open and closed sets in an arbitrary metric space canbe very complicated. We shall now consider the simplest special case,namely that of open and closed sets on the real line. In this case their com-plete description does not present much of a problem and is given by thefollowing theorem.

THEOREM 1. Every open set on the real line is the sum of a countable numberof disjoint intervals.

Proof. [We shall also include sets of the form (— oc), (a, oc), (— oc,as intervals.] Let G be an open set and let x E G. Then by the definitionof an open set we can find some interval I which contains the point x andbelongs entirely to the set G. This interval can always be chosen so thatits endpoints are rational. Having taken for every point x E G a correspond-ing interval I, we obtain a covering of the set G by means of a denumeráblesystem of intervals (this system is denurnerable because the set of allintervals with rational endpoints is denumerable). Furthermore, we shallsay that the intervals I' and I" (from the same covering) belong to oneclass if there exists a finite chain of intervals:

T TI ,1nl

32 METRIC SPACES [cH. II

(belonging to our covering) such that Ik intersects (1 � k � n — 1).It is clear that there will be a countable number of such classes. Further,the union of all the intervals which belong to the same class obviouslyagain forms an interval U of the same type, and intervals correspondingto distinct classes do not intersect. This completes the proof of the theorem.

Since closed sets are the complements of open sets, it follows that everyclosed set on the real line is obtained by removing a finite or denumerablenumber of open intervals on the real line.

The simplest examples of closed sets are segments, individual points,and the sum of a finite number of such sets. We shall now consider a morecomplicated example of a closed set on the real line, the so-called Cantorset.

Let F0 be the closed interval [0, 1]. We remove the open intervalfrom F0 and denote the remaining closed set by F1. Then we remove theopen intervals (-i, and from F1 and denote the remaining closed set(consisting of four closed intervals) by F2. From each of these four inter-vals we remove the middle interval of length and so forth. If we con-tinue this process, we obtain a decreasing sequence of closed sets

F set (since it is the intersection of the closedsets It is obtained from the closed interval [0, 1] by removing a de-numerable number of open intervals. Let us consider the structure of theset F. The points

(1) 0, 1,

which are the endpoints of the deleted intervals obviously belong to F.However, the set F is not exhausted by these points. In fact, those pointsof the closed interval [0, 1] which belong to the set F can be characterizedin the following manner. We shall write each of the numbers x, 0 � x � 1,in the triadic system:

x = ai/3 + a2/3 + •.. + +where the numbers can assume the values 0, 1, and 2. As in the case ofthe ordinary decimal expansion, some numbers allow two different develop-ments. For example,

= + (*)2 + ... + (*Y + = * + + + ... + + ...It is easily verified that the set F contains those, and only those, numbers

x, 0 � x � 1, which can be written in at least one way in the form of atriadic fraction such that the number 1 does not appear in the sequence

a2, , . Thus, to each point x E F we can assign the sequence

(2) , a2 , • •

§12] CONTINUOUS MAPPINGS. HOMEOMORPHISM. ISOMETRY

where is 0 or 2. The set of all such sequences forms a set having thepower of the continuum. We can convince ourselves of this by assigningto each sequence (2) the sequence

L LU1,U2,

where = 0 if = 0 and 1 if = 2. The sequence (2') can beconsidered as the development of a real number y, 0 � y � 1, in the formof a dyadic fraction. We thus obtain a mapping of the set F onto the entireclosed interval [0, 1]. This implies that F has the cardinal number of thecontinuum. [The correspondence established between F and the closedinterval [0, 1] is single-valued but it is not one-to-one (because of the factthat the same number can sometimes be formed from distinct fractions).This implies that F has cardinal number not less than the cardinal numberof the continuum. But F is a subset of the closed interval [0, 1] and conse-quently its cardinal number cannot be greater than that of the continuum.(See §5.)] Since set of points (1) is denurnerable, these points cannotexhaust all of F.

EXERCISE. Prove directly that the point belongs to t.he set F althoughit is not an endpoint of a single one of the intervals deleted. Hint: Thepoint divides the closed interval [0, 1] in the ratio 1:3. The closed interval[0, which remains after the first deletion is also divided in the ratio 1:3by the point and so on.

The points (1) are said to be points of the first type of the set F and theremaining points are said to be points of the second type.

EXERCISE. Prove that the points of the first type form an everywheredense set in F.

We have shown that the set F has the cardinal number of the con-tinuum, i.e. that it contains as many points as the entire closed interval[0, 1].

It is interesting to compare this fact with the following result: the sumof t.he lengths of all the deleted intervals is + + + i.e. exactly 1!

§12. Continuous mappings. Homeomorphism. Isometry

Let 1? = (X, p) and R' = (Y, p') be two metric spaces. The mapping fof the space R into R' is said to be continuous at the point x0 E R if forarbitrary 0 a > 0 can be found such that

p'[f(x), f(xo)] <

for all x such that

p(x, x0) < &

In other words, t.he mappingf is continuous at the point x0 if an arbitrary


neighborhood O(f(xo), €) of the point f(xo) contains a neighborhood O(xo, 5)

of the point x0 whose image is contained in the interior of O(f(xo), €).A mapping f is said to be continuous if it is continuous at each point of

the space R.If R' is the real line, then a continuous mapping of R into R' is called a

continuous function on R.As in the case of the mapping of arbitrary sets we shall say that f is a

mapping of R onto R' if every element y E R' has at least one inverseimage.

In analysis, together with the definition of the continuity of a function"in terms of neighborhoods", the definition of continuity "in terms ofsequences", which is equivalent to it, is widely used. The situation isanalogous also in the case of continuous mappings of arbitrary metricspaces.

THEOREM 1. A necessary and sufficient condition that the mapping f becontinuous at the point x is that for every sequence { x

converge to y = f(x).Proof. The necessity is obvious. We shall prove the sufficiency of this

condition. If the mapping f is not continuous at the point x, there exists aneighborhood O(y, €) of the point y = f(x) such that an arbitrary O(x, 5)contains points whose images do not belong to O(y, €). Setting = 1/n(n = 1, 2, . .), we select in each sphere O(x, 1/n) a point such that

O(y, Then x but the sequence } does not converge tof(x), i.e. the condition of the theorem is not satisfied, which was to beproved.

THEOREM 2. A necessary and sufficient condition that the mapping f of thespace R onto R' be continuous is that the inverse image of each closed set in R'be closed.

Proof. Necessity. Let M R be the complete inverse image of the closedset M' R'. We shall prove that M is closed. If x E [M], there exists asequence of points in M which oonverges to x. But then, by Theorem 1,the sequence } converges to f(x). Since E M' and M' is closed,we have f(x) E M'; consequently x E M, which was to be proved.

Sufficiency. Let x be an arbitrary point in R, y = f(x), and let O(y, €)be an arbitrary neighborhood of y. The set R' \ O(y, €) is closed (since itis the complement of an open set). By assumption, F = f'(R' \ O(y, €))is closed, and moreover, x F. Thus, R \ F is open and x E R \ F;consequently, there is a neighborhood O(x, 5) of the point x which is con-tained in R \ F. If z E O(x, then f(z) E O(y, €), i.e. f is continuous,which was to be proved.

REMARK. The image of a closed set under a continuous mapping is notnecessarily closed as is shown by the following example: map the half-open

§12] CONTINUOUS MAPPINGS. HOMEOMORPHISM. ISOMETRY 35

F—0 I

FIG. 8

interval [0, 1) onto a circle of the same length. The set 1) which is closedin [0, 1), goes over under this mapping into a set which is not closed (seeFig. 8).

Since in the case of a mapping "onto" the inverse image of the comple-ment equals the complement of the inverse image, the following theoremwhich is the dual of Theorem 2 is valid.

THEOREM 2'. A necessary and sufficient condition that the mapping f of thespace R onto R' be continuous is that the inverse image of each open set in R'be open.

The following theorem which is the analogue of the well-known theoremfrom analysis on the continuity of a composite function is valid for con-tinuous mappings.

THEOREM 3. If R, R', R" are metric spaces and f and are continuowsmappings of R into 1?' and R' into R", respectively, then the mapping z =

of the space R into R" is continuous.The proof is carried out exactly as for real-valued functions.The mapping f is said to be a homeomorphism if it is one-to-one and

bicontinuous (i.e. both f and the inverse mapping f' are continuous).The spaces R and R' are said to be homeomorphic if a homeomorphic

correspondence can be established between them.It is easy to see that two arbitrary intervals are homeomorphic, that an

arbitrary open interval is homeomorphic to R', and so forth.It follows from Theorems 2 and 2' of this section that a necessary and

sufficient condition that a one-to-one mapping be a homeomorphism is thatthe closed (open) sets correspond to closed (open) sets.

This implies that a necessary and sufficient condition that a one-to-onemapping be a homeomorphism is that the equality

=

hold for arbitrary M. (This follows from the fact that [M} is the intersec-tion of all closed sets which contain M, i.e. it is the minimal closed setwhich contains M.)

EXAMPLE. Consider the spaces = p) and = P0) (see §8,

1(4)


Examples 3 and 4). The following inequalities hold for the mapping whichassigns to an element in with coordinates Xi, x2 the element in

with the same coordinates:

pe(x, y) � p(x, y) � y).

Consequently an arbitrary €-neighborhood of the point x of the spacecontains a s-neighborhood Of the same point x considered as an element ofthe space and conversely. It follows from this that our mapping ofonto is a homeomorphism.

An important special case of a homeornorphism is an isometric mapping.We say that a one-to-one mapping y = f(x) of a metric space R onto a

metric space I?' is isometric if

p(x1, X2) = p'[f(x1), f(x2)}

for arbitrary x1, X2 E R. The spaces R and I?' themselves, between whichan isometric correspondence can be established, are said to be isometric.

The isometry of two spaces R and R' means that the metric relationsbetween their elements are the same and that they can differ only in thenature of their elements, which is unessential. In the sequel we shall con-sider two isometric spaces simply as identical.

(1) The concept of continuity of a mapping can be defined not only formetric but also for arbitrary topological spaces. The mapping f of thetopological space T into the topological space T' is said to be continuousat the point xo if for arbitrary neighborhood O(ye) of the point Ye = f(xo)there exists a neighborhood 0(xo) of the point x0 such that f(O(x0)) c O(ye).

Theorems 2 and 3 carry over automatically to continuous mappings oftopological spaces.

§13. Complete metric spaces

From the very beginning of our study of mathematical analysis we areconvinced of the important role in analysis that is played by the propertyof completeness of the real line, i.e. the fact that every fundamental se-quence of real numbers converges to some limit. The real line representsthe simplest example. of the so-called complete metric spaces whose basicproperties we shall consider in this section.

We shall call a sequence {x?L of points of a metric space R a fundamentalsequence if it satisfies the Cauchy criterion, i.e. if for arbitrary 0 thereexists an such. that , < for all n' � N€ , n" �

It follows directly from the triangle axiom that every convergent sequenceis fundamental. In fact, if converges to x, then for given 0 it ispossible to find a natural number N€ such that , x) < €/2 for allii � . Then , x) x) < for arbitraryn' � and n" � N€.

§ 13] COMPLETE METRIC SPACES 37

DEFINITION 1. If every fundamental sequence in the space I? convergesto an element in R, I? is said to be complete.

EXAMPLES. All the spaces considered in §8, with the exception of theone given in Example 7, are complete. In fact:

1. In the space consisting of isolated points (Example 1, §8) only thosesequences in which there is a repetition of some point, beginning with someindex, are fundamental. Clearly, every such sequence converges, i.e. thisspace is complete.

2. The completeness of the space R' of real numbers is known fromanalysis.

3. The completeness of the Euclidean space follows directly from thecompleteness of R'. In fact, let be a fundamental sequence; this meansthat for every 0 an N = can be found such that

En (Ic) (k) 2 2Xq ) <€

for all p, q greater than N. Then for each k = 1, 2, . .. ,

(k) (k)Xp Xq <€

f or all p, q > N, i.e. } is a fundamental sequence of real numbers.We set

=

and(1) (2) (n)x=(x ,x ).

Then it is obvious that

= x.

4. The completeness of the space is proved in an exactly analogousmanner.

5. We shall prove the completeness of the space C[a, b]. Let } be afundamental sequence in C[a, b]. This means that for each 0 thereexists an N such that I — Xm(t)

Ifor n, m> N and all t, a � t � b.

This implies that the sequence Xn(t) } converges uniformly and that itslimit is a continuous function x(t), where

IXn(t) — x(t)

I

f or all t and for all n larger than some N; this means thatx(t) in the sense of the metric of the space C[a, b].6. The space 12. Let }, where

(n) / (n) (n) (n)X = , X2 , •

be a fundamental sequence in 12.

38 METRIC SPACES [CIT. II

For arbitrary > 0 an N can be found such that2 (n) (m) oo (n) (m) 2(1) p (x , x ) = k=i (Xk — Xk ) < for n, m > N.

It follows from this that for arbitrary /c(n) (m) 2

(Xk Xk ) <€,i.e. for each k the sequence of real numbers } converges. Set

= Xk. Denote the sequence (x1, X2, , by x. Wemust show that

a) Xk2 < b) x) = 0.

To this end we shall write inequality (1) in the form/ (n) (m)\2

Xk )

EM (n) (m) 2 oo (n) (m) 2= k=i (Xk — Xk ) + k=M+i (Xk — Xk ) <

(M arbitrary). Since each of these two sums is nonnegative, each of themis less than €. Consequently

EM (n) (m) 2k=i(Xk —Xk ) < €.

If we fix m in this inequality and pass to the limit as n we obtain

EM (m) 2k=i(Xk —Xk ) � €.

Since this inequality is valid for arbitrary M, we can pass to the limit asM We then obtain

E0°(m) 2

k=i(Xk Xk ) � €.The inequality thus obtained and the convergence of the series

E Xk(m)2 imply that the series Xk2 converges; consequently x is anelement in 12. Further, since is arbitrarily small, this inequality meansthat

(m) oop(x , x) = { k=i (xk — xk ) } = 0,

i.e. x.7. It is easy to convince ourselves of the fact that the space C2[a, b] is

not complete. For example, the sequence of continuous functions

= arctan nt (— 1 � t � 1)is fundamental, but it does not converge to any continuous function (itconverges in the sense of mean square deviation to the discontinuous func-tion which is equal to —ir/2 for t < 0, ir/2 for t > 0, and 0 for t = 0).

EXERCISE. Prove that the space of all bounded sequences (Example 8,§8) is complete.

§ 13] COMPLETE METRIC SPACES 39

In analysis the so-called lemma on nested segments is widely used. In thetheory of metric spaces an analogous role is played by the following theoremwhich is called the principle of nested spheres.

THEOREM 1. A necessary and sufficient condition that the metric space 1? becomplete is that every sequence of closed nested spheres in R with radii tendingto zero have nonvoid intersection.

Proof. Necessity. Assume the space I? is complete and let S1, 82,be a sequeiice of closed nested spheres. Let dn be the diameter of thesphere . By hypothesis dn = 0. Denote the center of the sphere

by The sequence {Xn} is fundamental. In fact, if m > n, then ob-viously , < . Since 1? is complete, exists. If we set

x = xn,

then x E fln8n . In fact, the sphere Sn contains all the points of the givensequence with the exception perhaps of the points X1, x2, •,x n—i . Thus,x is a limit point of each sphere But since Sn is a closed set, we havethat x E for all n.

Sufficiency. To prove the sufficiency we shall show that if the space I? isnot complete, i.e. if there exists a fundamental sequence in R which doesnot have a limit, then it is possible to construct a sequeiice of closed nestedspheres in 1? whose diameters tend to zero and whose intersection is void.Let { } be a fundamental sequence of points in R which does not have alimit. We shall construct a sequence of closed spheres in the followingway. Let ni be such that p(Xni , Xm) < for all rn > ni. Denote by thesphere of radius 1 and center at . Further, let n2 > ni be such thatp(Xn2, Xm) < for all m > n2. Denote by 82 the sphere of radius withcenter Xn2. Since by assumption p(Xn1 , Xn2) < we have 82 C Si. Nowlet n3 > fl2 be such that p(Xn3, Xm) for all m > n3 and let 83 be asphere of radius with center Xn3, and so forth. If we continue this con-struction we obtain a sequence of closed nested spheres { }, wherehas radius n—i This sequence of spheres has void intersection; in fact, ifX E flkSk, then X = Xn. As a matter of fact the sphere Sk containsall points Xn beginning with Xnk and consequently p(x, Xn) < for alln> But by assumption the sequence {Xnl does not have a limit. There-fore = 0.

If the space R is not complete, it is always possible to embed it in anentirely definite manner in a complete space.

DEFINITION 2. Let R be an arbitrary metric space. A complete metricspace R* is said to be the completion of the space R if: 1) R is a subspaceof the space R*; and 2) R. is everywhere dense in R*, i.e. [RI = R*. (Here[R] naturally denotes the closure of the space R in R*.)

For example, the space of all real numbers is the completion of the spaceof rationals.

40 METRIC SPACES [cii. ii

2. Ereiy metric space has a completion and all of its completionsarc isometric.

Proof. \Ve lwgin by proving uniqueness. It is necessary to prove that if11* and R** are two completions of the space I?, then they are isometric, i.e.there is a one-to-one mapping of t.he space. R* onto R** such that 1)

= x for all a' I?; and 2) if and y* y**, then p(x*, y*) =p(x**, y**).

Such a mapping is defined in the following way. Let be an arbitrarypoint, of R*. Themi, by the definition of comp'etion, there exists a sequence

of points in I? which converges to But the sequence }can be

assumed to belong also to R**. Since is complete, } converges into some point. We set = It is clear that this correspondenceis one-to-one and does not depend on the choice of the sequence whichconverges to the point This is then the isometric mapping sought. Infact, by construction we have = x for all x E H. Furthermore, if welet.

—f in 11* and —f in R**,

{yn} y* J?* and {yn} y** in

then

p(x*, y*) = , ijn)

and at the same time

p(x**, y**) =,

Consequently

p(x*, y*) = p(x**, y**).

We shall now prove the existence of the completion. The idea involved inthe proof is the same as that in the so-called Cantor theory of real numbers.The situation here is essentially even simp'er than in the theory of realnumbers since there it is required further that one define all the arithmeticoperations for the newly introduced objects—the irrational numbers.

Let 1? be an arbitrary metric space. We shall say that two fundamentalsequences } and }

in R are equivalent (denoting this by } })

if = 0. This equivalence relation is reflexive, symmetricand transitive. It follows from this that all fundamental sequences whichcan be constructed from points of the space I? are partitioned into equiva-lence classes of sequences. We shall now define the space R* in the followingmanner. The points of R* will be all possible equivalence classes of funda-mental sequences and the distance between points in R* will be defined inthe following way. Let and be two such classes. We choose one repre-

§13] COMPLETE METRIC SPACES 41

sentative from each of these two classes, i.e. we select some fundamentalsequence {Xn } and } from each, respectively. We set

(2) p(x*, y*) = p(Xn, Yn).

We shall prove the correctness of this definition of distance, i.e. we shallshow that the limit (2) exists and does not depend on the choice of therepresentatives {Xn } E and I Yn } E

Since the sequelices } and{ } are fundamental, with the aid of the

triangle axiom we have for all sufficiently large n', n":

I , — , yn") I

= p(Xn' , yn') — , Yn") + p(Xn' , — p(Xn" , Yn")

— un") I + , — p(Xn" , Yn")

� p(yn' , yn") + p(xn' , < €/2 + €/2 =

Thus, the sequence of real numbers Sn = p(Xn, Yn) satisfies the Cauchycriterion and consequently it has a limit. It remains to prove that thislimit does not depend on the choice of {Xn} E and {Yn} E Let

{Xn}, {Xn'} E and {yn}, {yn'} E

Now

IXn} {Xn'} and {yn}

imply that

p(Xn , Yn) p(Xn', Yn')(

Ip(Xn, Yn) — p(Xn', Yn) + p(Xn', Yn) — p(Xn',

I , — p(Xn', Yn)( + p(Xn', — p(Xn', yn')!

� Xn') + P(Yn , Yn') 0,

i.e.

p(Xn , = p(Xn' , yn').

We shall now show that the metric space axioms are fulfilled in R*.Axiom 1 follows directly from the definition of equivalence of funda-

mental sequences.Axiom 2 is obvious.We shall now verify the triangle axiom. Since the triangle axiom is satis-

fied in the initial space R, we have

p(Xn, Zn) � p(Xn , Yn) + p(yn, Zn).


Passing to the limit as n —k we obtain

, , Yn) + ,

i.e.

p(x, z) � p(x, y) + p(y, z).

We shall now prove that R* is the completion of the space R. (We use thekeeping in mind that all completions of the space R are isometric.)

To each point x E R there corresponds some equivalence class of funda-mental sequences, namely the totality of all sequences which converge tothe point x.

We have:if

x = and y =

then

p(x, y) =

Consequently, letting the corresponding class of fundamental sequencesconverging to x correspond to each point x we embed R isometrically in thespace R*.

In the sequel we shall not have to distinguish between the space R itselfand its image in R* (i.e. the totality of all equivalence classes of convergentsequences) and we can consider I? to be a subset of R*.

We shall now show that I? is everywhere dense iii R*. In fact, letbe a point in R* and let 0 be arbitrary. We select a representativein i.e. we choose a fundamental sequence Let N be such that

Xm) < for all n, m > N. Then we have

, x*) = , Xm) �i.e. an arbitrary neighborhood of the point contains a point of R. Thuswe have [R] = R*.

It remains to be proved that the space R* is complete. We note, first ofall, that by the construction of R* an arbitrary fundamental sequence

(3) Xl,X2, n,

consisting of points belonging to R, converges in R* to some point, namelyto the point E R*, defined by the sequence (3). Further, since R is densein R*, then for an arbitrary fundamental sequence , ... ofpoints in R* we can construct an equivalent sequence x1 , X2, ,

consisting of points belonging to R. To do this it is sufficient to take forany point in I? such that < 1/n.

§ 14] CONTRACTION MAPPINGS AND APPLICATIONS 43

The sequence } thus constructed will be fundamental and by what wasproved above it will converge to some point E R*. But then the sequence

} also converges to This proves the theorem completely.

§14. Principle of contraction mappings and its applications

As examples of the applications of the concept of completeness we shallconsider the so-called contraction mappings which form a useful techniquefor the proof of various existence and uniqueness theorems (for example,in the theory of differential equations).

Let R be an arbitrary metric space. A mapping A of the space R intoitself is said to be a contraction if there exists a number a < 1 such that

(1) p(Ax, Ay) � ap(x, y)

two points x, y E R. Every contraction mapping is continuous. Infact, if —f x, then, by virtue of (1), we also have —f Ax.

THEOREM (PRINCIPLE OF CONTRACTION MAPPINGS). Every contractionmapping defined in a complete metric space R has one and only one fixedpoint (i.e. the equation Ax = x has one and only one solution).

Proof. Let x0 be an arbitrary point in R. Set x1 = Ax0 , = Ax1 = A2x0,and in general let = = A We shall show that the sequence

}is fundamental. In fact,

, = p(A A �� x1) + p(xi , + . + p(Xm_,_i ,

� , + a + a2 + + &p(xo, xi)11/(1 — a)}.

Since a < 1, this quantity is arbitrarily small for sufficiently large n. SinceR is complete, exists. We set x = . Then by virtue of thecontinuity of the mapping A, Ax = A = Axl =xn+1 = x.

Thus, the existence of a fixed point is proved. We shall now prove itsuniqueness. If Ax = x, Ay = y, then p(x, jj) � ap(x, y), where a < 1;this implies that p(x, y) = 0, i.e. x = y.

The principle of contraction mappings can be applied to the proof of theexistence and uniqueness of solutions obtained by the method of successiveapproximations. We shall consider the following simple examples.

1. y = f(x), where f(x) is a function defined on the closed interval [a, b]satisfying the Lipschitz condition

f(x2) — f(xi) < K X2 —

with K < 1, and mapping the closed interval [a, b] into itself. Then f is acontraction mapping and according to the theorem proved above the


sequence x0, Xi = f(xo), X2 = f(xi), converges to the single root of theequation x = f(x).

In particular, the condition of contraction is fulfilled if f'(x)( � K < 1on the closed interval [a, b].

As an illustration, Figs. 9 and 10 indicate the course of the successiveapproximations in the case 0 <f'(x) < 1 and in the case —1 <f'(x) <0.

In the case where we are dealing with an equation of the form F(x) = 0,where F(a) < 0, F(b) > 0 and 0 <K1 � F'(x) � K2 on [a, b], a widelyused method for finding its root amounts to setting 1(x) = x — XF(x)and seeking a solution of the equation x = f(x), which is equivalentto F(x) = 0. In fact, since f'(x) = 1 — XF'(x), 1 — XK2 � f'(x) � 1 — XK1and it is not difficult to choose X so that we can apply the method of suc-cessive approximations.

FIG. 9

FIG. 10

§14] CONTRACTION MAPPINGS AND APPLICATIONS 45

2. Let us consider the mapping y = Ax of the space into itself givenby the system of linear equations

= + (i = 1, 2, , n).

If Ax is a contraction mapping, we can apply the method of successiveapproximations to the solution of the equation x Ax.

Under what conditions then is the mapping A a contraction? The answerto this question depends on the choice of the metric in (It is easy to seethat with the metric (b) is a metric space.)

a) p(x,y) = max — yil;.l � i �p(y', y") = — = max1

J— xf)I

�J

x,' — x7" <J

max, x5' — x/'I

=Ip(x', x").

This yields

(2)J J

a < 1

as the condition of contraction.

b) p(x, y) = — (;

p(y', it) = I — y/' I = I— x5")f

� E IIxa' — x," � max5 I p(x', x").

This yields the following condition of contraction:

(3)

c) p(x, y) = —

Here

p2(y', y") = { — � x")

on the basis of the Schwarz inequality.Then

(4)

is contraction condition.Thus, in the case where one of the Conditions (2)—(4) is fulfilled there

exists one and only one point x = (x1, x2, , such that

= + be.,

46 METRIC SPACES {CH. II

where the successive approximations to this solution have the form:(0) 0 0 0x = (x1,x2,

= (x11, x21, •

(k) k k kx = (Xl,X2,

wherek n k—i= —I-

(Consequently any one of the Conditions (2)—(4) implies that

a12

a21 a22 — 1 • • • 0.)

Each of the Conditions (2)—(4) is sufficient in order that the mappingy = Ax be a contraction. As concerns Condition (2) it could have beenproved that it is also necessary in order that the mapping y = Ax be acontraction (in the sense of the metric a)).

None of the Conditions (2)—(4) is necessary for the application of themethod of successive approximations. Examples can be constructed inwhich any one of these conditions is fulfilled but the other two are not.

IfI

< 1/n (in this case all three conditions are fulfilled), then themethod of successive approximations is applicable.

If 1/n (in this case all three sums equal 1), it is easy to seethat the method of successive approximations is not applicable.

§15. Applications of the principle of contraction mappings in analysis

In the preceding section there were given some of the simplest examplesof the application of the principle of contraction mappings in one- andn-dimensional spaces. However, the most essential applications for analysisof the principle of contraction mappings are in infinite-dimensional func-tion spaces. Further on we shall show' how with the aid of this principlewe can obtain theorems on the existence and uniqueness of solutions forsome types of differential and integral equations.

I. Let

(1) dy/dx = f(x, y)

be a given differential equation with the initial condition

(2) y(x0) = Yo,

§15] CONTRACTION MAI'PINGS iN ANALYSIS 47

where f(x, y) is defined and continuous in some plane region G which con-tains the point (x0 , yo) and satisfies a Lipschitz condition with respect to

f(x, yi) — f(x, � M Yi —

We shall prove that then on some closed interval x — < d thereexists a unique solution y = of the equation (1) satisfying the initial

(2) (Picard's theorem).Equation (1) together with the initial condition (2) is equivalent to the

integral equatioii

(3) = Yo + f f(t, dt.

Since the function f(x, y) is continuous, we have f(x, y) � k in someregion G' G which contains the point (x0, yo). Now we select a d > 0such that the following conditions are fulfilled:

1) (x,y) E �d, y — kd;

2) Md < 1.

Denote by C* the space of continuous functions which are definedon the closed interval x — < d and are such that — � frd

with the metric , = —

It is easy to see that C* is a complete space. (This follows, for instance,from the fact that a closed subset of a complete space is a complete space.)Let us consider the mapping L' = Açc defined by the formula

= Yo + Jf(t, dt,

where x — � d. This is a contraction mapping of the complete spareinto itself. In fact, let ço E x — < d. Then

= kd

and consequently A ((7*) C*. Moreover, we have

— p2(X)I � f(t, p1(t)) — f(t, p2(t)) dt

<MdI

— c02(x)

Since Md < 1, the mapping A is a contraction.From this it follows that the operator equation = Ap (and consequently

equation (3) also) has one and only one solution.II. Let


(4) p/ (x) , çoi(x), • • , i = 1, 2, • • • , ii;

be a given system of differential equations with initial conditions

(5) = /101 1 = 1, 2, • • • , ii;

where the functions f1(x, • . . , are defined and continuous in someregion G of the space such that G contains the point (x0 , • , yon)and satisfy a Lipschitz condition

. .

,— f1(x, • .

� M max{I yj(l) — yj(2); 1 � i n}.

We shall prove that then on some closed interval x — < d thereexists one and only one system of solutions yi = ço1(x) satisfying system (4)and the initial conditions (5).

System (4) together with the initial conditions (5) is equivalent to thesystem of integral equations

(6) = + f f1(t, p1(t), ••• , dt; i = 1, , n.

Since the function is continuous in some region G' G containing thepoint (x0 , Yoi , • • , you), the inequalities f1(x, • • K, whereK is a constant, are fulfilled.

We now choose d > 0 to satisfy the following conditions:

1) (x, , . . , E G if I x — � d, — Yoi Kd,

2) Md < 1.

We now consider the space whose elements are ordered systems= (p1(x), . • , consisting of n functions which are defined and

continuous for all x for which x — < d and such that — <Kd, with the metric

= —

The mapping = A given by the system of integral equations

= Yoi + ff(t, p1(t), ... , dt,

is a contraction mapping of the complete space (L* into itself. In fact,

(x) — (x)=

ft, ••• ,— f(t, ••• dt

and consequently

(x) —(x) I � Il/Id (x) — (x)

§15] CONTRACTION MAPPINGS IN ANALYSIS 49

Since Md < 1, A is a contraction mapping.It follows that the operator equation = has one and only one solu-

tion.III. We shall now apply the method of contraction mappings to the proof

of the existence and uniqueness of the solution of the Fredholm nonhomo-geneous linear integral equation of the second kind:

(7) f(x) = A f K(x, y)f(y) dy +

where K(x, y) (the so-called kernel) and are given functions, f(x) isthe function sought, and A is an arbitrary parameter.

We shall see that our method is applicable only in the case of sufficientlysmall values of the parameter A.

We shall assume that K(x, y) and co(x) are continuous for a � x � b;a � y � b and consequently that K(x, y)J � M. Consider the mappingg = Af, i.e. g(x) = A fab K(x, y)f(y) dy + ço(x), of the complete spaceC[a, b] into itself. We obtain

p(gi, = max g1(x) — g2(x) A M(b — a) max 1' — 12

Consequently, the mapping A is a contraction for A < 1/M(b — a).From this, on the basis of the principle of contraction mappings, we can

conclude that the Fredholm equation has a unique continuous solution forevery A I < 1/M(b — a). The successive approximations to this solution:fo(x), f1(x), •.. , ... have the form

= A j K(x, dy +

IV. This method is applicable also in the case of nonlinear equations ofthe form

(8) f(x) = A f K(x, y, 1(y)) dy +

where K and are continuous. Furthermore K satisfies the condition

for A < 1/M(b — a) since here again for the mapping g = Af of thecomplete space C[a, b] into itself given by the formula

g(x) A f K(x, y, 1(y)) dy +

the inequality


max gi(x) — g2(x)I I I M (b — a) max fi — f2

holds.V. Consider the Volterra type integral equation

(9) f(x) = A f K(x, y)f(y) dy +

which differs from an equation of Fredhoim type in that the upper limitin the integral is the variable quantity x. This equation can be consideredas a particular case of the Fredholm equation if we complete the definitionof the function K(x, y) for y > x by means of the equation K(x, = 0(fory > x).

In contrast to the Fredholm integral equation for which we were requiredto limit ourselves to small values of the parameter A the principle of con-traction mappings (and the method of successive approximations basedon it) is applicable to Volterra equations for all values of the parameterA. We note first of all that the principle of contraction mappings can begeneralized in the following manner: if A is a continuous mapping of acomplete metric space R into itself such that the mapping is a contractionfor some n, then the equation

Ax = x

has one and only one solution.In fact, if we take an arbitrary point x E R and consider the sequence

(k = 0, 1, 2, a repetition of the argument introduced in §14yields the convergence of this sequence. Let x0 = ThenAx0 = x0. In fact, Ax0 = Since the mapping is a contrac-tion, we have

� � ... < aGp(Ax, x).

Consequently,kn knp(A Ax, A x) = 0,

i.e. Ax0 = x0.Now consider the mapping

g(x) = A f K(x, y)f(y) dy + Af.

If and f2 are two continuous functions defined on the closed interval[a, b], then

IAfi(x) — Af2(x) = A f K(x, y)[fi(y) — dy � A Mm(x — a),

§16] COMPACT SETS IN METRIC SPACES 51

(M = max IK(x, m = max (Ifi — f21)),

A2fi(x) — A2f2(x)I

� X2 Mm(x — a)2/2,

I — � — a)7n1 �X the number n can be chosen so large that

— < 1,

i.e. the mapping will be a contraction. Consequently, the Volterraequation (9) has a solution for arbitrary X and this solution is unique.

§16. Compact sets in metric spaces

A set M in the metric space R is said to be compact if every sequence ofelements in M contains a subsequence which converges to some x E R.

Thus, for example, by virtue of the Bolzano-Weierstrass theorem everybounded set on the real line is compact. Other examples of compact setswill be given below. It is clear that an arbitrary subset of a compact set iscompact.

The concept of total boundedness which we shall now introduce isclosely related to the concept of compactness.

Let M be any set in the metric space R and let be a positive number.The set A in R is said to be an &net with respect to M if for an arbitrarypoint x E M at least one point a E A can be found such that

p(a, x) <

For example, the lattice points form 2k-net in the plane. A subset M ofR is said to be totally bounded if R contains a finite &-net with respect to Mfor every > 0. It is clear that a totally bounded set is bounded since if an&net A can be found for M consisting of a finite number of points, then A isbounded and since the diameter of M does not exceed diameter A +

2 below will show, the converse is not true ingeneral.

The following obvious remark is often useful: if the set M is totallybounded, then its closure [M} is totally bounded.

It follows at once from the definition of total boundedness that everytotally bounded metric space R with an infinite number of points is sepa-rable. In fact, construct a finite (1/n)-net in R for every n. Their sum overall n is a denumerable set which is everywhere dense in R.

EXAMPLES. 1. For subsets of Euclidean n-space total boundedness coin-cides with ordinary boundedness, i.e. with the possibility of enclosing agiven set in the interior of some sufficiently large cube. In fact, if such acube is subdivided into cubicles with diagonal of length then the ver-

52 METRIC SPACES [CII. II

tices of these cubicles will form a finite c-net in the initial cube and there-fore also in an arbitrary set which lies in the interior of this cube.

2. The unit sphere S in the space 12 is an example of a hounded set whichis not totally bounded. In fact, consider in S points of the form

Ci = (1,0,0, •••),= (0, 1,0,

The distanee 1)et\veeIl any two such points and Cm (in n) is Fromthis it is clear that in S there can he no finite &-net for any �

Consider in 12 the set H of points x = (x1, x2, , whichsatisfy the following eonditions:

lxii � 1, X21 � , X7j �This set is called the fundamental parallelopiped in the space 12. It is anexample of an infinite-dimensional totally bounded set. To prove thatthis set is totally bounded we proceed as follows.

Let > 0 he given. Choose n so that < Associate with eachpoint

(1) x = (Xi , . .

in II the point

(2) = (a'1 , , , 0, 0, .

from the same set. Theti we have

p(x, x*) = � < <

The subset 11* of points of the form (2) of II is totally hounded (since it isa bounded set. in n-dimensional space). We choose a finite inIt is clear that it be at the same time an &net for II.

The following theorem establishes the interrelation among the conceptsof compacti iess, completeness, and total boundedness.

Tn 1. A necessary and sufficient condition that a subset M qf a corn.-plete metric space R 1w compact is that M be totally bounded.

Proof. Necessity. We shall assume that ill is not totally bounded, i.e.we shall assume that for some > 0 a finite &net cannot. be found in M.We take an arl)itrary point x1 in ill. since, by assumption, there is no finitec-net in ill, a point x2 C ill can be found such that p(Xi , x2) � Further,them'eexistsa poimmtx3 C ill such that p(xi, x3) � and p(x2, � (otherwisethe points and a2 would form an &net. in ilfl, and SO oil; continuing thisprocess we coimstnmct a sequence Xi , , , . of points in R whichsatisfy time (olmdltion p(x,,, , � for in n. It is (lear that it is impos—sil)le to select ammy convergent subsequemmee from such a sequence.

§17] ARZELA'S THEOREM AND APPLICATIONS 53

Sufficiency. Let R be complete and let M be totally bounded. We shallprove that in M we can select a convergent subsequence from every se-quence. Let be a sequence of points in M. Set

€1 = 1, = • • • , = 1/k,

and construct for every €k a coiiespondiiig finite iii it:(k) (k) (k)

a1 , a2 • • • ,

Describe about each of the poiimts which form a 1-imet iii a sphere ofradius 1. Since these spheres cover all of M and are finite in number, atleast one of them, let us call it 81, contains an infinite subsequence x1U),

(1) (1)X2 , , , ••• of the sequence }. Further, about each of the pointswhich form a 4-net in R we describe a sphere of radius Since the numberof these spheres is again finite, at least one of them, let us call it S2 , con-tains an infinite subsequence •••,x • • of the sequemmce

}. Further, we find a sphere S3 of radius containing au infinitesubsequence • • • of the sequence }, and so forth.

Now we choose from the sequences(1) (1) (1)Xi ,X2 , ,

(2) (2) (2)x1 ,x2 , •••

the "diagonal" subsequence(1) (2) (n)

.t 1 , , • • • , ii , .

this subsequence is fundamental because all its terms beginning withlie in the interior of the sphere of radius 1/n. Since the space R is

complete, this subsequence has a limit. This completes the proof of thetheorem.

The following generalization of Theorem I is often useful.THEOREM 2. A necessary and sufficient condition that a subset Ill of a

complete metric space R be compact is that for every � 0 there exist in R acompact €-net for M.

Proof. The necessity is trivial; we shall prove the sufficiency. Let > 0be given and let. t be a compact (€,/2)-net. for M. We choose a finite (€/2)-net in A. It is clear that. it will be a fiuuite €-net for ii; consequently, byvirtue of the preceding theorem, time set if is comflI)act.

§17. Arzelà's theorem and its applications

Tue applicatioum of Theorems 1 and 2 of the puecediumg sectioui, whichyield necessary and sufficient conditions for conmpactness to individual spe-cial cases, is not always simple. For sets, situated in some given special


space, special criteria for compactness can be given which are more suitablefor practical application.

In analysis one of the more important metric spaces is the space C[a, b].For subsets of this space the most important and frequently employed

criterion of compactness is given by Arzelà's theorem.In order to formulate this theorem it is necessary to first introduce the

following concepts.A family {ço(x) } of functions defined on a closed interval is said to be

uniformly bounded if there exists a number M such that < M forall x and for all belonging to the given family.

A family of functions is said to be equicontinuous if for every > 0there is a > 0 such that

— <

for all x1, X2 such that — X2I

< and for all cc in the given family.ARZELA'S THEOREM. A necessary and sufficient condition that a family of

continuous functions defined on the closed interval [a, b} be compact in C[a, b}is that this family be uniformly bounded and equicontinuous.

Proof. Necessity. Let the set f be compact in C[a, b}. Then, by Theorem 1,§16, for each positive there exists a finite , c02, , inEach of the functions 'pt, being a continuous function on a closed intervalis bounded: I cci � M1.

Set M max + By the definition of an for everyE c1 we have for at least one

pep, = maxI

— <Consequently

I I < I I + <M1 + <M.Thus, cf is uniformly bounded.

Further, since each of the functions is continuous and consequentlyuniformly continuous on [a, b], for a given there exists a such that

— I <if

I— X2

I<

Set mm Then forIxi — X2

I< and for any cc E taking cci

so that çcj) < we have:

Icc(xi) — cc(x2)

I

= çc(xi) — çc2(xi) + çc(xi) — cc(x2)

I— + —

I+ cc(x2)

< + + =

The equicontinuity of is thus proved.

§17] ARZELA'S THEOREM AND APPLICATIONS 55

Sufficiency. Let be a uniformly bounded and equicontinuous family offunctions. In accordance with Theorem 1, §16, in order to prove its compact-ness in C[a, b] it is sufficient to show that for arbitrary 0 there existsfor it in C[a, b] a finite €-net. Let

I� M for all E

and let ô> 0 be chosen so that

I— ce'(x2)

I< €/5 for xi — < Xi, E [a, b];

and for all E Subdivide the segment [a, b] on the x-axis by means ofthe points x0 = a, xi, X2, , = b into intervals of length less than ôand construct vertical lines at these points of subdivision. We subdividethe segment [—M, M] on the y-axis by means of the points y0 = —M,Yi, y2, , = M into intervals of length €/5 and construct horizontallines at these points of subdivision. Thus we subdivide the rectanglea x � b, — M � y � M into cells with horizontal sides of length lessthan ô and vertical sides of length €/5. We now assign to every function

E a polygonal arc i/'(x) with vertices at the points (Xk, yz), i.e. at ver-tices of the constructed net and deviating at the points Xk from the func-tion by less than €/5 (the existence of such a polygonal arc is obvious).

Since, by construction,

— < €15,

— < €/5,

and

I—

I< €15,

we have

I— < 3€/5.

Since the function is linear between the points Xk and Xk+i, we have

— i,t'(x) < 3€/S for all x, Xk � x � Xk+i.

Now let x be an arbitrary point of the closed interval [a, b] and let Xkbe the subdivision point chosen which is closest to x from the left. Then

—I I

— + —I + I

—I <€.

Consequently the polygonal arcs form an €-net with respect toTheir number is finite (since only a finite number of polygonal arcs can bedrawn through a finite number of points); thus, is totally bounded. Thisproves the theorem completely.

Arzelà's theorem has many applications. We shall demonstrate its

56 METRIC SPACES [Cu. II

application in the example of the following existence theorem for ordinarydifferential equations with continuous right member.

THEOREM 2 (PEANO). Let

dy/dx = f(x, y)

be a given differential equation. If the function f(x, y) is continuous in aclosed region G, then at least one integral curve of the given equation passesthrough each interior point (x0, Yo) of this region.

Proof. Since f(x, y) is continuous in the closed region, it is bounded:

f(x,y)J <M.We draw straight lines with slopes M and —M through the point (x0 , yo).

Further, we draw vertical lines x = a and x = b so that the two trianglescut off by them with the common vertex (x0, Yo) lie entirely in the interiorof the region G.

We now construct a so-called Euler polygonal arc for the given equationin the following way: we draw from the point (x0, Yo) a straight line withthe slope f(xo, yo). On this straight line we take a point (x1, y') and drawthrough it a straight line with the slope f(x1, y'). On this straight linewe take a point (x2, Y2) and so forth. We now consider a sequence of Eulerpolygonal arcs L1, L2, ... , •.. , passing through the point (x0, Yo)such that the length of the greatest of the links of the polygonal arc Lktends to zero as k —* oo. Let cok(x) be the function whose graph is the po-lygonal arc Lk. The functions c02, , ... possess the followingproperties: they are 1) defined on the same segment [a, b], 2) all bounded,and 3) equicontinuous.

On the basis of Arzelà's theorem, we can choose a uniformly convergentsubsequence from the sequence {cok(x) Let this sequence be

. . , .

We set ço(x) = It is clear that co(xo) = Yo. It remains toverify that ço(x) satisfies the given differential equation on the closed inter-val [a, b]. To do this it is necessary to show that for arbitrary E > 0

ço(x") — co(x')— f(x',ço(x')) <E

provided the quantity x" — x' is sufficiently small. To prove this itis sufficient in turn to establish that for sufficiently large k

(k) FF (k) F

(x) ço (x) — f(x', <E

if only the difference x" — x' is sufficiently small.

§18] COMPACTA 57

Since f(x, y) is continuous in the region G, for arbitrary E > 0 an > 0can be found such that

f(x', y') — E <f(x, y) <f(x', y') + E, (y' =

if

Ix — <2, and

I ii — ii' I<4Mg.

The set of points (x, y) E G which satisfy both these inequalities formsome rectangle Q. Now let K be so large that for all k > K

Iço(x) —

I<

and all links of the polygonal arc Lk have length less than Then ifI

— x < all the Euler polygons y = for which k > K lieentirely in the interior of Q.

Further, let (ao, b0), (ai, b1), , be vertices of the po-lygonal arc y = where

(we assume x" > x' for definiteness; we can consider the case x" < x'analogously). Then

(k) (a1) — (x') = f(ao , bo) (a1 —

(k) (k)cc — cc = , — ad), i = 1, 2, , fl — 1

(k) (k)cc (x ) — cc (an) = — an).

From this it follows that for — I < n,

[f(x', y') — €](ai — x') < — < [f(x', y') + €](ai —

[J(x', y') — — < —

< [f(x', y') + — i = 1, 2, ••• , n — 1;

[f(x', y') — €](x" — < — < [f(x', y') + E](x" — an).

Summing these inequalities, we obtain

[f(x', y') — E](x" — x') < — < [f(x', y') —I-- E](x" —

which was to be proved.The solution cc(X) thus obtained will in general not be the unique solution

of the equation y' = f(x, y) which passes through the point (x0, yo).

§18. Compacta

In §16 we said that a subset M of a metric space R is compact if everysequence of elements in M has a subsequence which converges to some x E R.

58 METRIC SPACES [CII. II

In this connection the limit point x might belong but it also might notbelong to the set M. If from every sequence of elements in M it is possibleto select a subsequence which converges to some x belonging to M, then theset M is said to be compact in itself. For the space R itself (as moreoveralso for all its closed subsets) the concepts of compactness and compactnessin itself coincide. A compact metric space is called a compactum for brevity.

Inasmuch as compactness in itself of a set is an intrinsic property ofthis set which does not depend on the metric space in which it is embedded,it is natural to limit oneself to the study of compacta, i.e. to consider eachsuch set simply as a separate metric space.

THEOREM 1. A necessary and sufficient condition that a set M compact in ametric space R be a compactum is that M be closed in R.

Proof. Necessity. Suppose M is not closed; then we can find in M a se-quence which converges to a point x M. But then this sequencecannot contain a subsequence which converges to some point y E M, i.e.M cannot be a compactum.

Sufficiency. Since M is compact, every sequence { c M contains asubsequence which converges to some x E R. If M is closed, x E M, i.e. Mis compact in itself.

The next corollary follows from this and Theorem 1, §16.COROLLARY. Every closed bounded subset of Euclidean space is a compact urn.THEOREM 2. A necessary and sufficient condition that a metric space be a

compactum is that it be: 1) complete and totally bounded.The proof of this theorem is made by a verbatim repetition of the proof

of Theorem 1, §16.THEOREM 3. Every compacturn K contains a countable everywhere dense set.Proof. Since a compactum is a totally bounded space, K contains a finite

(1/n)-net: a1, a2, , am, for every n. The union of all these €-nets is afinite or denumerable set. This and Theorem 4, §10, imply the followingcorollary.

COROLLARY. Every compactum has a countable basis.THEOREM 4. A necessary and sufficient condition that the metric space R be

a compactum is that either of the following two conditions holds:1) An arbitrary open covering { of the space R contains a finite sub-

covering;2) An arbitrary system { of closed sets in R with the finite intersection

property has a nonvoid intersection. (A system of sets is said to have the finiteintersection property if an arbitrary finite number of these sets has nonvoidintersection.)

Proof. We note first of all that the equivalence of the two conditionsformulated follows directly from the principle of duality Infact, if {Oal is an open covering of the space R, then { R\ Oal is a system

§18] COMPACTA 59

of closed sets satisfying the condition

fl(R\Oa) = 0.

The condition that we can select a finite subcovering from {} is equiva-

lent to the fact that the system of closed sets {R \ Oa} cannot have thefinite intersection property if it has a void intersection.

We shall now prove that Condition 1 is necessary and sufficient that Rbe a compactum.

Necessity. Let R be a compactum and let { } be an open covering of R.We choose in R for each n = 1, 2, a finite (1/n)-net consisting of thepoints and we enclose each of these points with the sphere

1/n)

of radius 1/n. It is clear that for arbitrary n

R = 1/n).

We shall assume that it is impossible to choose a finite system of setscovering K from { }. Then for each n we can find at least one sphere

1/n) which cannot be covered by a finite number of the sets Oa.We choose such a sphere for each n and consider the sequence of theircenters Since R is a compactum, there exists a point E R whichis the limit of a subsequence of this sequence. Let be a set of {Oa} whichcontains Since is open, we can find an 0 such that €} cWe now choose an index n and a point so that < €/2,1/n < €12. Then, obviously, 1/n) c , i.e. the sphere1/n) is covered by a single set The contradiction thus obtained provesour assertion.

Sufficiency. We assume that the space R is such that from each of itsopen coverings it is possible to select a finite subcovering. We shall provethat R is a compactum. To do this it is sufficient to prove that R is completeand totally bounded. Let 0. Take a neighborhood O(x, €) about eachof the points x E R; we then obtain an open covering of R. We choose fromthis covering a finite subcovering O(xi, €), ... , €). It is clear that thecenters x1, ••• , of these neighborhoods form a finite €-net in R. Since

0 is arbitrary, it follows that R is totally bounded. Now let { be asequence of nested closed spheres whose radii tend to zero. If their inter-section is void, then the sets R \ form an open covering of R from whichit is impossible to select a finite subcovering. Thus, from Condition 1 itfollows that R is complete and totally bounded, i.e. that R is compact.

THEOREM 5. The continuous image of a compactum is a compactum.Proof. Let V be a compactum and let V = ço(X) be its continuous image.

Further, let {Oa} be an open covering of the space V. Set Ua =


Since the inverse image of an open set under a continuous mapping is open,it follows that { } is an open covering of the space X. Since X is a com-pactum, we can select a finite subcovering U1, U2, ••' , from thecovering { }. Then the corresponding sets 01, 02, •• , form a finitesubcovering of the covering { Oa }.

TIIEOREM 6. A one-to-one mapping of a compactum which is continuous inone direction is a homeomorphism.

Proof. Let cc be a one-to-one continuous mapping of the compactum Xonto the compactum Y. Since, according to the preceding theorem, thecontinuous image of a compactum is a compactum, the set cc(M) is a com-pactum for an arbitrary closed M C X and consequently cc(M) is closed inY. It follows that the inverse image under the mapping c5' of an arbitraryclosed set M C X is closed; i.e. the mapping çc1 is continuous.

REMARK. The following result follows from Theorem 6: if the equation

(3) dy/dx = f(x, y),

where the function f(x, y) is continuous in a closed bounded region Gwhich contains the point (xo, Yo) for every Yo belonging to some closedinterval [a, b], has a unique solution satisfying the initial condition y(xo) =Yo , then this solution is a continuous function of the initial value Yo.

Iii fact, since the function f(x, y) is continuous in a closed boundedregion, it is bounded and consequently the set P of solutions of equation(3) corresponding to initial values belonging to the closed interval [a, b]is uniformly bounded and equicontinuous. Moreover, the set P is closed.In fact, if {ccn(x) } is a sequence of solutions of equation (3) which con-verges uniformly to a function cc(x), then cc(x) is also a solution ofequation (3) since, if

= f(x, con(x)),

then passing to the limit as n —* we obtain

cc' = f(x, co(x)).

We have

cc(xo) = lim E [a, b].

In virtue of Arzelà's theorem and Theorem 1 of this section it followsfrom this that P is a compactum.

We set the point cc(xo) of the closed interval [a, b] into correspondencewith each solution cc(x) of equation (3). By assumption this correspondenceis one-to-one. Moreover, since

max (x) —I I

(x0) — ,

§18] COMPACTA 61

tile mapping co(x) —* co(xo) is continuous. By virtue of Theorem 6 the inversemapping is also continuous and this then signifies the continuous de-.pendence of the solutions on the initial conditions.

Now let be the set of all continuous mappings y = f(x) of a com-pactum X into a compactum Y. We introduce distance into Cxy by meansof the formula

p(f, q) = sup (p[f(x), g(x)]; x E XI.It is easy to verify that in this way Cx y is transformed into a metric

space.THEOREM 7 THEOREM OF ARzELA). A necessary and suqi-

cient condition that a set D C Cxy be compact in Cxy is that in D the func-tions y = f(x) be equicontinuons, i.e. that for arbitrary e > 0 there exist aô> 0 such that

(1) p(x', x") < ô

implies

(2) p[f(x'), f(x")] < e

for arbitrary f in D and x', x" in X.Proof. We embcd in the space of all mappings of the conipactum

X into the compacturn V with the same metric

p(f, g) = sup g(x)]; x E XI

which was introduced in C1 and prove the compactness of the set D inSince Cxy is closed in the compactiiess of the set D in

will imply its compactness in Cx y. (That Cx y is closed in M1 y follows fromthe fact that the limit of a uniformly convergent sequence of continuousmappings is also a continuous mapping. The indicated assertion is a directgencralization of the known theorem in analysis and is proved in exactlythe same way.)

Let e > 0 be arbitrary and choose ô such that (1) implies (2) for all f inD and all x', x" in X. Let the points x1, x2, , form a (ô/2)-net in X.It is easy to see that X can be represented as the sum of nonintersectingsets such that x, y E implies that p(x, y) < ô. In fact, it is sufficientto set,,for example,

= ô/2) \ S(x5, ô/2).

We now consider in the compactum V a finite e-net y2, , ; wedenote the totality of functions g(x) which assume the values yj on thesets by L. The number of such functions is clearly finite. We shall showthat they form a 2e-net in with respect to D. In fact, let f E D. For


every point in xi, , we can find a point in , Ym such that

p[f(xj, y5} < e.

Let the function g(x) E L be chosen so that g(xj) = Then

g(x)} � p[f(x), + g(xj} + p[g(x), g(xj)} < 2e

if i is chosen so that x E e2..From this it follows that p(f, g) < 2e and thus the compactness of D in

and consequently in Cxy also is proved.

§19. Real functions in metric spaces

A real function on a space R is a mapping of R into the space R' (thereal line).

Thus, for example, a mapping of into R' is an ordinary real-valuedfunction of n variables.

In the case when the space R itself consists of functions, the functions ofthe elements of R are usually called functionals. We introduce severalexamples of functionals of functions f(x) defined on the closed interval[0, 1]:

F1(f) = supf(x);

F2(f) = inff(x);

F3(f) = f(xo) where x0 E [0, 1];

F4(f) = co[f(xo), f(xi), , where E [0, 1]

and the function co(yi, ... , is defined for all real yi;

F5(f) = f f(x)] dx,

where co(x, y) is defined and continuous for all 0 � x � 1 and all real y;

F6(f) = f'(xo);

F7(f) = f [1 +

f If'(x)

Idx.

Functionals can be defined on all of R or on a subset of R. For example,in the space C the functionals F1 , F2, F3, F4, F5 are defined on the entirespace, F6(f) is defined only for functions which are differentiable at thepoint x0, F7(J) for functions for which [1 + is integrable, and F8(f)for functions for which I f'(x) is integrable.

§19] REAL FUNCTIONS IN METRIC SPACES 63

The definition of continuity for real functions and functionals remainsthe same as for mappings in general (see §12).

For example, F1(f) is a continuous functional in C since

p(f, g) = sup I f — gI

and I sup f — sup g I � sup I f — g I;

F2, F3, F5 are also continuous functionals in C; F4 is continuous in thespace C if the function cc is continuous for all arguments; F6 is discontinuousat every point in the space C for which it is defined. In fact, let g(x) besuch that g'(xo) = 1, I g(x) <e and f = fo + g. Then f'(xo) = fo'(xo) + 1and p(J, fo) < e. This same functional is continuous in the space Cu) offunctions having a continuous derivative with the metric

p(f, g) = sup [ff — gI+ If' — g' II;

F7 is also a discontinuous functional in the space C. In fact, let fo(x) 0and = (1/n) sin 2irnx. Then fo) = 1/n —* 0. However,is a constant (it does not depend on n) which is greater than andF7(f0) = 1.

Consequently, F7(J) is discontinuous at the point fo.By virtue of this same example F8(f) is also discontinuous in the space

C. Both functionals F7 and F8 are continuous in the spaceThe following theorems which are the generalizations of well-known

theorems of elementary analysis are valid for real functions defined oncompacta.

THEOREM 1. A continuous real function defined on a compactum is uni-formly continuous.

Proof. Assume f is continuous but not uniformly continuous, i.e. assumethere exist and such that

I—

I< 1/n and

I— J �

From the sequence } we can choose a subsequence {xflk } which con-verges to x. Then als& {xnk'} —> x and either

If(x) — I � e/2 or

I f(x) — I � e/2, which contradicts the continuity of f(x).THEOREM 2. If the function f(x) is continuous on the compactum K, then

f is bounded on K.Proof. If f were not bounded on K, then there would exist a sequence

} such that We choose from } a subsequence whichconverges to x: {xflk } —* x. Then in an arbitrarily small neighborhood of xthe function f(x) will assume arbitrarily large values which contradicts thecontinuity of f.

THEOREM 3. A function f which is continuous on a compactum K attainsits least upper and greatest lower bounds on K.

Proof. Let A = sup f(x). Then there exists a sequence } such that

A > > A — 1/n.

64 METRiC SPACES [CII. II

We choose a convergent subsequence from : x. By continuity,f(x) = A. The proof for ml f(x) is entirely analogous.

Theorems 2 and 3 allow generalizations to an even more extensive classof functions (the so-called sernicontinuous functions).

A function f(x) is said to be lower (upper) semicontinuous at the pointx0 if for arbitrary e > 0 there exists a ô-neighborhood of a'0 in which f(x) >f(xo) — e, <f(xo) + e).

For example, the function "integral part of x", f(x) = E(4, is uppersemicontinuous. If we increase (decrease) the value of f(.ro) of a continuousfunction at a single point x0 , we obtain a function which is upper (lower)seniicontinuous. If f(x) is upper sernicontinuous, then —f(x) is lower semi-continuous. These two remarks at once permit us to construct a largenumber of examples of sernicontinuous functions.

We shall also consider functions which assume the values ± If f(xo) =— oc, then will be assumed to be lower semicontinuous at x0 and uppersemicontinuous at x0 if for arbitrary h > 0 there is a neighborhood of thepoint x0 in which f(x) < — h.

If f(x0) = + then f(x) will be assumed to be upper semicontinuous atx0 and lower semicontinuous at x0 if for arbitrary h> 0 there is a neighbor-hood of the point x0 in which f(x) > h.

The upper limit J(xo) of the function f(x) at the point x0 is the limE.0sup [f(x); x E S(xo , e)} The lower limit is the inf [f(x);

x E S(xo, e)} The difference cof(xo) = J(xo) — f(xo) is the oscillation ofthe function f(x) at the point xo . It is easy to see that a necessary andsufficient condition that the function f(x) be continuous at the point x0 isthat cof(xo) = 0, i.e. that J(xo) = f(xo).

For arbitrary f (x) the function J(x) is upper semicontinuous and thefunction .[(x) is lower semicontinuous. This follows easily from the defini-tion of the upper and lower limits.

We now consider several important examples of semicontinuous func-tionals.

Let f(x) be a real function of a real variable. For arbitrary real a and bsuch that f(x) is defined on the closed interval [a, b} we define the totalvariation of the function f(x) on [a, b} to be the functional

Vab(f) = supI

—I

where a = x0 <x1 <X2 < = b and the least upper bound istaken over all possible subdivisions of the closed interval [a, b}.

For a monotone function Vab(f) = I f(b) — f(a) I. For a piecewise mono-tone function Vab(f) is the sum of the absolute values of the increments onthe segments of monotonicity. For such functions

§19] REAL FUNCTIONS IN METRIC SPACES 65

supI

—I

is attained for some subdivision.We shall prove that the functional Vab(J) is lower semicontinuous in the

space M of all bounded functions of a real variable with metric p(f, g) = sup

If(x) — g(x)

I(it is clear that C is a subspace of the space M), i.e. that

for arbitrary f and e > 0 there exists a & such that > Vab(f) — e forp(f, g) < &.

We choose a subdivision of the closed interval [a, b] such that

I— I > Vab(f) — e/2.

Let & = e/4n. Then if p(g, f) < &, we have

I—

I — I— g(xi_i) I < e/2

and consequently

� I— I > Vab(f) —

In the case Vab(f) = the theorem remains valid since then for arbi-trary H there exists a subdivision of the closed interval [a, b] such that

I— > H

and & can be chosen such that

I— I — I

g(xj) — g(xi_i) <

Then > H — e, i.e. � H, so that =

The functional is not continuous as is easily seen from the follow-ing example. Let f(x) 0, gn(x) = (1/n) sin nx. Then p(gn, f) = 1/n, but

= 2 and = 0.Functions for which Vab(f) < are said to be functions of bounded (or

finite) variation. The reader can find more information about the propertiesof such functions in the books by Aleksandrov and Kolmogorov: Introduc-tion to the Theory of Functions of a Real Variable, Chapter 7, §7; Natanson:Theory of Functions of a Real Variable, Chapter 8; and Jeffery: The Theoryof Functions of a Real Variable, Chapter 5.

We shall define the length of the curve y = f(x) (a � x � b) as thefunctional

Lab&) = sup {— + —

where the least upper bound is taken over all possible subdivisions of theclosed interval [a, b]. This functional is defined on the entire space M. Forcontinuous functions it coincides with the value of the limit

lim {— + — as — I —p 0.

66 METRIC SPACES [C1. II

Finally, for functions with continuous derivative it can be written in theform

fb

The functional Lab(f) is lower semicontinuous in M. This is provedexactly as in the case of the functional Vab(J).

Theorems 2 and 3 established above generalize to semicontinuous func-tions.

THEOREM 2a. A finite function which is lower (upper) semicontinuous on acompacturn K is bounded below (above) on K.

In fact, let f be finite and lower semicontinuous and let inf f(x) = — oc•

Then there exists a sequence } such that < —n. We choose asubsequence {xnk} —* xo. Then, by virtue of the lower semicontinuity of f,f(xo) = — which contradicts the assumption that f(x) is finite.

In the case of an upper semicontinuous function the theorem is provedanalogously.

THEOREM 3a. A finite lower (upper) semicontinuous function defined on acompactum K attains its greatest lower (least upper) bound on K.

Assume the function f is lower semicontinuous. Then by Theorem 2 ithas a finite greatest lower bound and there exists a sequence } suchthat f(x) + 1/n. We choose a subsequence {xnk} —* Thenf(xo) = inf f(x) since the supposition that f(xo) > inf f(x) contradicts thelower semicontinuity of f.

The theorem is proved analogously for the case of an upper semicon-tinuous function.

Let K be a compact metric space and let CK be the space of continuousreal functions defined on K with distance function p(f, g) = sup f — g J.

Then the following theorem is valid.THEOREM 4. A necessary and sufficient condition that the set D ç CK be

compact is that the functions belonging to D be uniformly bounded and equi-continuous (Arzelà's theorem for continuous functions defined on anarbitrary compactum).

The sufficiency follows from the general Theorem 7, §18. The necessityis proved exactly as in the proof of Arzelà's theorem (see §17).

§20. Continuous curves in metric spaces

Let P = f(t) be a given continuous mapping of the closed intervala � t � b into a metric space R. When t runs through the segment froma to b, the corresponding image point P runs through some "continuouscurve" in the space R. We propose to give rigorous definitions connectedwith the above ideas which were stated rather crudely just now. We shall

§20] CONTINUOUS CURVES IN METRIC SPACES 67

QQ2FIG. 11 FIG. 12 FIG. 13

consider the order in which the point traverses the curve an essentialproperty of the curve itself. The set shown in Fig. 11, traversed in thedirections indicated in Figs. 12 and 13, will be considered as distinct curves.As another example let us consider the real function defined on the closedinterval [0, 1] which is shown in Fig. 14. It defines a "curve" situated onthe segment [0, 1] of the y-axis, distinct from this segment, traversed oncefrom the point 0 to the point 1, since the segment [A, B] is traversed threetimes (twice upward and once downward).

However, for the same order of traversing the points of the space weshall consider the choice of the "parameter" 1 unessential. For example,the functions given in Figs. 14 and 15 define the same "curve" over they-axis although the values of the parameter I corresponding to an arbitrarypoint of the curve can be distinct in Figs. 14 and 15. For example, in Fig. 14to the point A there correspond two isolated points on the 1-axis, whereasin Fig. 15 to the point A there correspond on the 1-axis one isolated pointand the segment lying to the right (when I traverses this segment the pointA on the curve remains fixed). [Allowing such intervals of constancy of thepoint P = f(I) is further convenient in the proof of the compactness ofsystems of curves.]

We pass over to formal definitions. Two continuous functions

P = and P = f2(I")

FIG. 14 FIG. 15


defined, respectively, on the closed intervals

a' � t' � b' and a" � t" � b"are said to be equivalent if there exist two continuous nondecreasing func-tions

= coi(t) and t" = ço2(t)

defined on a closed interval a � t � b and possessing the properties

coi(a) = a', coi(b) b',

c02(a) = a", ço2(b) =

fi[coi(t)} = f2[co2(t)}

for all t E [a, b].It is easy to see that the equivalence property is reflexive (f is equivalent

to f), symmetric (if fi is equivalent to f2, then f2 is equivalent to fr), andtransitive (the equivalence of fi and f2 together with the equivalence off2 and f3 implies the equivalence of fi and fe). Therefore all continuous func-tions of the type considered are partitioned into classes of equivalent func-tions. Every such class also defines a continuous curve in the space R.

It is easy to see that for an arbitrary function P = f1(t') defined on aclosed interval [a', b'] we can find a function which is equivalent to it andwhich is defined on the closed interval [a", b"] = [0, 1]. To this end, it issufficient to set

= coi(t) = (b' — a')t + a', t" = co2(t) = t.

(We always assume that a < b. However we do not exclude "curves"consisting of a single solitary point which is obtained when the functionf(t) is constant on [a, b]. This assumption is also convenient in the sequel.)Thus, we can assume that all curves are given parametrically by means offunctions defined on the closed interval [0, 1].

Therefore it is expedient to consider the space CIR of continuous mappingsof the closed interval I = [0, 1] into the space R with the metric p(f, g) =supt p[f(t), g(t)].

We shall assume that the sequence of curves L1, L2, . .,L the curves can be represented parametrically in

the form

and the curve L in the form

P=f(t),so that —*Oasn—-*

§201 CONTINUOUS CURVES IN METRIC SPACES 69

We obtain Theorem 1 if we apply the generalized Arzelà theorem (Theo-rem 7, §18) to the space CIR.

THEOREM 1. If the sequence of curves L1, L2, , lying in thecompactum K can be represented parametrically by means of equicontinuousfunctions defined on the closed interval [0, 1], then this contains aconvergent subsequence.

We shall now define the length of a curve given parametrically by meansof the function P = f(t), a � t � b, as the least upper bound of sums ofthe form

where the points are subject only to the following conditions:

It is easy to see that the length of a curve does not depend on the choiceof its parametric representation. If we limit ourselves to parametric repre-sentations by functions defined on the closed interval [0, 1], then it iseasy to prove by considerations similar to those of the preceding sectionthat the length of a curve is a lower semicontinuous functional of f (in thespace CuR). In geometric language this result can be expressed in the formof such a theorem on semicontinuity.

THEOREM 2. If the sequence of curves converges to the curve L, then thelength of L is not greater than the greatest lower bound of the lengths of thecurves

We shall now consider specially curves of finite length or rectifiable curves.Let the curve be defined parametrically by means of the function P = f(t),a � t � b. The function f, considered only on the closed interval [a, TI,where a � T � b, defines an "initial segment" of the curve from the pointPa = f(a) to the point PT = f(T). Let 5 = co(T) be its length. It is easilyestablished that

P = g(s) = f[ço1(s)]

is a new parametric representation of the same curve. In this connection sruns through the closed interval 0 � s � S, where S is the length of theentire curve under consideration. This representation satisfies the require-ment

p[g(si), g(s2)] � 82 — SiI

(the length of the curve is not less than the length of the chord).Going over to the closed interval [0, 1] we obtain the parametric repre-

sentation

P=F('r)=g(s), r=s/S


which satisfies the following Lipschitz condition:

p[F(ri), F(r2)] � S —

We thus see that for all curves of length S such that S � M, where Mis a constant, a parametric representation on the closed interval [0, 1] bymeans of equicontinuous functions is possible. Consequently, Theorem 1 isapplicable to such curves.

We shall show the power of the general results obtained above by apply-ing them to the proof of the following important proposition.

THEOREM 3. If two points A and B in the compacturn K can be connectedby a continuous curve of finite length, then among all such curves there existsone of minimal length.

In fact, let V be the greatest lower bound of the lengths of curves whichconnect A and B in the compactum K. Let the lengths of the curves L1,L2, ... , ... connecting A with B tend to V. From the sequenceit is possible, by Theorem 1, to select a convergent subsequence. ByTheorem 2 the limit curve of this subsequence cannot have length greaterthan V.

We note that even when K is a closed smooth (i.e. differentiable a suffi-cient number of times) surface in three-dimensional Euclidean space, thistheorem does not follow directly from the results established in usualdifferential geometry courses where we restrict ourselves ordinarily to thecase of sufficiently proximate points A and B.

All the arguments above would take on great clarity if we formed of theset of all curves of a given metric space R a metric space. This can be doneby introducing the distance between two curves L1 and L2 by means of theformula

p(Li, L2) = ml , f2),

where the greatest lower bound is taken over all possible pairs of parametricrepresentations

P = fi(t), P = f2(t) (0 � t � 1)

of the curves L1 and L2, respectively.The proof of the fact that this distance satisfies the axioms of a metric

space is very straightforward with the exception of one point: there is somedifficulty in proving that p(Li, L2) = 0 implies that the curves L1 and L2are identical. This fact is an immediate consequence of the fact that thegreatest lower bound in the formula which we used in the definition of thedistance p(Li, L2) is attained for a suitable choice of the parametric repre-sentations fi and f2. But the proof of this last assertion is also not verystraightforward.

Chapter III

NORMED LINEAR SPACES

§21. Definition and examples of normed linear spaces

DEFINITION 1. A set R of elements x, y, z, ••• is said to be a linearspace if the following conditions are satisfied:

I. For any two elements x, y E R there is uniquely defined a third ele-ment z = x + y, called their sum, such that

1) x + y = y + x,2) x + (y + z) = (x + y) + z,3) there exists an element 0 having the property that x + 0 = x for all

x E R, and4) for every x E R there exists an element — x such that x + (— x) = 0.II. For an arbitrary number a and element x E R there is defined an ele-

ment ax (the product of the element x and the number a) such that1) a(8x) = (a8)x, and2) 1•x = x.III. The operations of addition and multiplication are related in the

following way:1) (a + f3)x = ax + f3x, and2) a(x + y) ax + ay.Depending on the numbers admitted (all complex numbers or only the

reals), we distinguish between complex and real linear spaces. Unless other-wise stated we shall consider real linear spaces. In a linear space, besidesthe operations of addition and multiplication by scalars, usually there isintroduced in one way or another the operation of passage to the limit.It is most convenient to do this by introducing a norm into the linear space.

A linear space R is said to be normed if to each element x E R there ismade to correspond a nonnegative number x which is called the normof x and such that:

1) x = 0 if, and only if, x = 0,2) ax = al3) tx +yJl � +

It is easy to see that every normed space is also a metric space; it is sufficientto set p(x, y) = x — y JJ. The validity of the metric space axioms followsdirectly from Properties 1—3 of the norm.

A complete normed space is said to be a Banach space, a space of Banachtype, or, more briefly, a B-space.

EXAMPLES OF NORMED SPACES. 1. The real line with the usual arith-metic definitions is the simplest example of a normed space. In this casethe norm is simply the absolute value of the real number.

71

72 NORMED LINEAR SPACES [OH. III

2. Euclidean n-space, i. e. the space consisting of all n-tuples of realnumbers: x = (x1, x2, ••• , in which the norm (i. e. the length) of thevector is defined to be the square root of its scalar square,

1 =

is also a normed linear space.In an n-dimensional linear space the norm of the vector x = (Xi, X2,

can also be defined by means of the formula

1X = I

IP)1fP, (p � 1).

We also obtain a normed space if we set the norm of the vector x =(Xi , x2, , equal to the max { Ixk ; 1 � k � n}.

3. The space C[a, b] of continuous functions with the operations of addi-tion and multiplication by a scalar which are usual for functions, in which

ttf(t)tI

is a normed linear space.4. Let C2[a, b] consist of all functions continuous on [a, b] and let the

norm be given by the formula

It= dt).

All the norm axioms are satisfied.5. The space 12 is a normed linear space if we define the sum of two

elements x = (Er, E2, ...) and y = (ni., fl2, ...) in 12

to be

and let

ax = (aEi, aE2, • ,

and

Itxlt =

6. The space c consisting of all sequences x = (xj, x2, , ...) ofreal numbers which satisfy the condition = 0.

Addition and multiplication are defined as in Example 5 and the normis set equal to

001.

7. The space m of bounded sequences with the same definitions of sum,product, and norm as in the preceding example.

§21] DEFINITION AND EXAMPLES OF NORMED LINEAR SPACES 73

In each of these examples the linear space axioms are verified withoutdifficulty. The fact that the norm Axioms 1—3 are fulfilled in Examples1—5 is proved exactly as was the validity of the metric space axioms in thecorresponding examples in §8, Chapter II.

All the spaces enumerated in the examples, except the space C!2[a, bj,are Banach spaces.

DEFINITION 2. A linear manifold L in a normed linear space R is any setof elements in R satisfying the following condition: if x, y E L, then ax +

E L, where a and B are arbitrary numbers. A subspace of the space Ris a closed linear manifold in R.

REMARK 1. In Euclidean n-space R'4 the concepts of linear manifoldand subspace coincide because every linear manifold in is automaticallyclosed. (Prove this!) On the other hand, linear manifolds which are notclosed exist in an infinite-dimensional space. For example, in 12 the set Lof points of the form

(1) x = (x1 , k , 0, 0, .

i.e. of points which have only a finite (but arbitrary) number of nonzerocoordinates, forms a linear inanif old which is not closed. In fact, a linearcombination of points of form (1) is a point of the same form, i.e. L is alinear manifold. But L is not closed since, for instance, the sequence ofpoints

(1,0,0,0,..),It

1 11,

belonging to L, converges to the point (1, , •••), which doesnot belong to L.

REMARK 2. Let x1, x2, ... be elements of a Banach space Rand let M be the totality of elements in R which are of the formfor arbitrary finite n. It is obvious that M is a linear manifold in R. We shallshow that [M] is a linear subspace. In view of the fact that [M] is closed itis sufficient to prove that it is a linear manifold.

Let x E [M], y E [MI. Then in an arbitrary €-neighborhood of x we canfind an E M and in an arbitrary €-neighborhood of y we can find a E M.We form the element ax + f3y and estimate ax + j3y — axE — II:

ax + j3y — —

� �from which it is clear that ax + !3y E [M].


The subspace L = [M] is said to be the subspace generated by the elementsx1,x2,

§22. Convex sets in normed linear spaces

Let x and y be two points in the linear space R. Then the segment con-necting the points x and y is the totality of all points of the form ax + 3y,where a � 0, � 0, and a + = 1.

DEFINITION. A set M in the linear space R is said to be convex if, giventwo arbitrary points x and y belonging to M, the segment connecting themalso belongs to M. A convex set is called a convex body if it contains at leastone interior point, i.e. if it contains some sphere completely.

EXAMPLES. 1. in three-dimensional Euclidean space, the cube, sphere,tetrahedron, and halfspace are convex bodies; but a triangle, plane, andsegment are convex sets although they are not convex bodies.

2. A sphere in a normed linear space is always a convex set (and also aconvex body). In fact, consider the unit sphere S: x � 1.

If xo, y0 are two arbitrary points belonging to this sphere: x0 1,

II yo � 1, then

� Ilaxoll + = atixoll � a+ 1,

i.e.

axo+f3y0ES3. Let R be the totality of vectors x = E2) in the plane. Introduce

the following distinct norms in R:

x 12 = (Ei + E2Y; x = max I, I

IIxIti= Ed 1E21) (p>1).Let us see what the unit sphere will be for each of these norms (see Fig. 16).

FIG. 16

§22] CONVEX SETS IN NORMED LINEAR SPACES 75

In the case1

x 112 it is a circle of radius 1, in the case x it is a squarewith vertices (± 1, ± 1), in the case II x Iii it is a square with Vertices(0, 1), (1, 0), (—1, 0), (0, —1). If we consider the unit sphere correspondingto the norm x and let p increase from 1 to then this "sphere"deforms in a continuous manner from the square corresponding to x tothe square corresponding to x Had we set

(1) =

for p < 1, then the set II x � 1 would not have been convex (for example,for p = 2/3 it would be the interior of an astroid). This is another expres-sion of the fact that for p < 1 the "norm" (1) does not satisfy Condition 3in the definition of a norm.

4. Let us consider a somewhat more complicated example. Let 4S be theset of points x = (Ei , , . . , • in 12 which satisfy the condition

2 2 � 1.This is a convex set in 12 which is not a convex body. In fact, if x, y Eand z = ax + where a, � 0 and a + = 1, then by virtue of theSchwarz inequality (Chapter II),

+ = a2 + 2af3 + 22

� a2 + 2af3 + ,32

= + � (a + ç3)2 = 1.

We shall show that contains no sphere. is symmetric with respectto the origin of coordinates; hence, if contained some sphere 5', it wouldalso contain the sphere 8" which is symmetric to 5' with respect to theorigin. Then being convex, would contain all segments connectingpoints of the spheres 5' and 5", and consequently it would also containa sphere S of the same radius as that of 5', with the center of S at the ori-gin. But if contained some sphere of radius r with center at the origin,then on every ray emanating from zero there would lie a segment belongingentirely to 4S. However, on the ray defined by the vector (1, 1/2, 1/3,1/n, ...) there obviously is no point except zero which belongs to

4S Prove thatset convex body.

2. Prove that 4S is not contained in any subspace distinct from all of 12.3. Prove that the fundamental parallelopiped in 12 (see Example 3, §16)

is a convex set but not a convex body.We shall now establish the following simple properties of convex sets.THEOREM 1. The closure of a convex set is a convex set.Proof. Let M be a convex set, [M] its closure and let x, y be two arbitrary

76 NORMED LiNEAR SPACES [cii. iii

points in [M]. Further, let be an arbitrary positive number. Points a, bcan be found in 111 such that p(a, x) < and p(b, y) < Then p(ax +aa + < for arbitrary nonnegative a and j3 such that a + j3 = 1,and the point aa + belongs to M since M is convex. Since 0 is ar-bitrary, it follows that ax + !3y E [M], 1. e. [M] is convex also.

THEOREM 2. The intersection of an arbitrary number of convex sets is aconvex set.

Proof. Let M = flaMa , where all Ma are convex sets. Further, let x andy be two arbitrary points in ill. These points x and y belong to all Ma.Then the segment connecting the points x and y belongs to each Ma andconsequently it also belongs to M. Thus, M is in fact convex.

Since the intersection of c'osed sets is always closed, it follows thatthe intersection of an arbitrary number of closed convex sets is a closed convexset.

Let A be an arbitrary subset of a normed linear space. We define theconvex closure of the set A to be the smallest closed convex set containing A.

The convex closure of any set can obviously be obtained as the intersec-tion of all closed convex sets which contain the given set.

Consider the following important example of convex closure. Letx1, X2, , be points in a normed linear space. We shall say thatthese n + 1 points are in general position if no three of them lie on onestraight line, no four of them lie in one plane, and so forth; in general, nok + 1 of these points lie in a subspace of dimension less than k. The convexclosure of the points x1 , , • , which are in general position is calledan n-dimensional simplex and the points x1, , themselves arecalled the vertices of the simplex. A zero-dimensional simplex consists of asingle point. One-, two-, and three-dimensional simplexes are, respectively,a segment, triangle, tetrahedron.

If the points x1, x2, , are in general position, then any k + 1of them (k <n) also are in general position and consequently they generatea k-dimensional simplex, called a k-dimensional face of the given n-dimen-sional simplex. For example, the tetrahedron with the vertices ej, e2, e3,e4 has four two-dimensional faces defined respectively by the triples ofvertices (e2, e3, e4), (e1 , e3, e4), (e1 , e2, e4), (e1, e2, e3); six one-dimensionalfaces; and four zero-dimensional faces.

THEOREM 3. A simplex with the vertices x1, x2, , is the totality ofall points which can be represented in the formI \ n+1 n+1

X = k=1 akxk ; ak , k=1 ak =

Proof. In fact, it is easy to verify that the totality of points of the form(2) represents a closed convex set which contains the points x1, x2,

. On the other hand, every convex set which contains the points xj , x2,

§23] LINEAR FUNCTIONALS 77

must also contain points of the form (2), and consequently thesepoints form the smallest closed convex set containing the points x1, x2,

,xn+1.

§23. Linear functionals

DEFINITION 1. A numerical function f(x) defined on a normed linearspace R will be called a functional. A functional f(x) is said to be linear if

f(ax + = af(x) +where x, y E R and a, are arbitrary numbers.

A functional f(x) is said to be continuous if for arbitrary 0 a ô > 0can be found such that the inequality

I— f(x2) <

holds whenever

II — II < ô.

In the sequel we shall consider only continuous functionals (in particularcontinuous linear functionals) and for brevity we shall omit the word"continuous".

We shall establish some properties of linear functionals which followalmost directly from the definition.

ThEOREM 1. If the linear functionalf(x) is continuous at some point xo E R,then it is continuous everywhere in R.

Proof. In fact, let the linear functional f(x) be continuous at the pointx = x0. This is equivalent to the fact that —* f(xo) when —* Xo.

Further, let

y y f(y) — f(xo).

But y + x0 —* x0. Consequently, by assumption, — y + x0)

f(xo). Thus,

f(y) — f(xo) = f(y).

A functional f(x) is said to be bounded if there exists a constant N suchthat

(1)I

f(x) I < N x

for allx ER.ThEOREM 2. For linear functionals the conditions of continuity and bounded-

ness are equivalent.Proof. We assume that the linear functional f(x) is not bounded. Then for


arbitrary natural number n we can find an element E R such thatI

> n We shall set = It Then1

= 1/n,i.e. 0. But at the same time

I I = I H 1) I= (i/n

1 II) I

f(x) the point x = 0.Now let N be a number which satisfies Condition (1). Then for an arbi-

trary sequence —* 0 we have:

I I � N 1 II —*0,

i.e. f(x) is continuous at the point x = 0 and consequently at all the re-maining points also. This completes the proof of the theorem.

DEFINITION 2. The quantity

ff11 = sup {If(x) x x O}

is called the norm of the linear functional f(x).EXAMPLES OF LINEAR FUNCTIONALS ON VARIOUS SPACES. 1. Let be

Euclidean n-space and let a be a fixed nonzero vector in For arbitraryx E we set f(x) = (x, a), where (x, a) is the scalar product of the vectorsx and a. It is clear that f(x) is a linear functional. In fact,

f(ax + = (ax + a) a(x, a) + a) = af(x) + 13f(y).

Further, by virtue of Sehwarz's inequality

(2) If(x) = I � lxii laM.Consequently, the functional f(x) is bounded and is therefore continuous.From (2) we find

If(x)I/IIxIt � It alLSince the right member of this inequality does not depend on x, we have

sup f(x) I/Il x a

i.e. f a But, settiiig x = a we obtain:

If(a) = (a, a) = a 112, i.e. (If(a) a II) = a

rrhelefore II fit = II aIf a is zero, then f is the zero linear functional. Hence f II = a in

this case also.2. The integral

I = j x(t) dt

§23] LINEAR FUNCTIONALS 79

(x(t) is a continuous function on [a, b}) represents a linear functional on thespace C[a, b]. Its norm equals h — a. In fact,

li J � maxjx(t) I(h — a),

where equality holds when x = constant.3. Now let us consider a more general example. Let yo(t) be a fixed con-

tinuous function on [a, b]. We set, for arbitrary function x(t) E C[a, bi,

f(x) = f x(t)yo(t) dt.

This expression represents a linear functional on C[a, b} because

f(ax + = j (ax(t) + dt

b b

= a f x(t)yo(t) dt + j y(t)yo(t) dt = af(x) +

This functional is bounded. In fact,b b

I f(x) I = f x(t)yo(t) dt x f yo(t)I

dt.

Thus, the functional 1(x) is linear and bounded and consequently it iscontinuous also. It is possible to show that its norm is exactly equal to

f4. We now consider on the same space C[a, b] a linear functional of

another type, namely, we set=

i.e. the value of the functional for the function x(t) is equal to the valueof this function at the fixed point . This functional is frequently en-countered, for example, in quantum mechanics where it is usually writtenin the form

= f — t0) di,

where is the "function" equal to zero everywhere except at the point= 0 and such that its integral equals unity (the Dirac s-function). The

s-function can be thought of as the limit, in some sense, of a sequence offunctions çôn(t) each of which assumes the value zero outside someborhood —* 0 as n oc) of the point t = 0 and such that the integralof the limiting function equals 1.

80 NORMED LINEAR SPACES [CH. III

5. In the space 12 we can define a linear functional as in by choosingin 12 some fixed element a = (a1, a2, , an., and setting

(3) f(x) =

Series (3) converges for arbitrary x E 12 and

(4) x a

Inequality (4) transforms into the identitya a

a a linear func-tional on the space R. We shall assume f(x) is not identically zero. The setL1 of those elements x in R which satisfy the condition f(x) = 0 form asubspace. In fact, if x, y E L1, then

f(ax + = af(x) + = 0,

i.e. ax + E Lf. Further, if x and E L1, then by virtue of thecontinuity of the functional f,

f(x) = = 0.

DEFINITION 3. We say that the subspace L of the Banach space R hasindex (or deficiency) s if: 1) R contains s linearly independent elementsx1, x2, , x8 which do not belong to L with the property that everyelement x E R can be represented in the form

x = + a2x2 + + + y, y E L;

and 2) it is impossible to find a smaller number of elements which possessthe indicated properties.

In the case of a finite-dimensional space R the index plus the dimensionof the subspace L is equal to the dimension of the whole space.

THEOREM 3. Let f(x) 0 be a given functional. The subspace L1 has indexequal to unity, i.e. an arbitrary element y E R can be represented in the form

(5) y=Xxo+x,where x E L1, x0 L1.

Proof. Since x0 Lf, we have f(xo) 0. If we set X = f(y)/f(xo) andx = y — {f(y)/f(xo)lxo, then y = Xx0 + x, where

f(s) = f(y) — (f(y)/f(xo))f(xo) 0.

If the element x0 is fixed, then the element y can be represented in theform (5) uniquely. This is easily proved by assuming the contrary. In fact,let

y = Xx0 + x,y = X'x0 + x';

§24] THE CONJUGATE SPACE 81

then

(X — X')xo = (x' — x).

If X — = 0, then obviously, x — x' = 0. But if X — X' 0, then= (x' — x)/(X — X') E Lf , which contradicts the condition that Xo Lf.Conversely, given a subspace L of R of index 1, L defines a contiuuous

linear functional f which vanishes precisely on L. Indeed, let x0 L. Thenfor any x E x = y + Xx0, with y E L, Xo L. Letf(x) = X. It is easilyseen that f satisfies the above requirements. If 1, g are two such linearfunctionals defined by L, then 1(x) = ag(x) for all x E R, a a scalar. Thisfollows because the index of L in R is 1.

We shall now consider the totality Mf of elements in R which satisfythe condition 1(x) = 1. M1 can be represented in the form M1 = Lf + x0,where x0 is a fixed element such that f(xo) = 1 and Lf is the totality ofelements which satisfy the condition 1(x) = 0. In analogy with the finite-dimensional case it is natural to call Mf a hyperplane in the space R. It iseasy to verify that the hyperplanes 1(x) = 1 and = 1 coincide if, andonly if, the functionals I and coincide. Thus, it is possible to establish aone-to-one correspondence between all functionals defined on R and allhyperplanes in R which do not pass through the origin of coordinates.

We shall now find the distance from the hyperplane 1(x) = 1 to theorigin. It is equal to

d = inf{!fxII;f(x) = 11.

For all x such that 1(x) = 1 we have

1 � 1111111 x i.e. lxi � 1/!!f I!;therefore d � f Further, since for arbitrary e > 0 an elemeut xsatisfying the condition 1(x) = 1 can be found such that

1> (11ff! - e) txit follows that

d = inf {II xli <1/(Iif Ii - e);f(x) = 11.

Consequently,

d =

i.e. the norm of the linear functional f(x) equals the reciprocal of the magnitudeof the distance of the hyperplane f(x) = 1 from the origin of coordinates.

§24. The conjugate space

It is possible to define the operations of addition and multiplication by ascalar for linear functionals. Let and f2 be two linear functionals on a

82 NORMED LINEAR SPACES [CII. III

normed linear space R. Their sum is a linear I = Ii + 12 suchthat 1(x) = fi(x) + f2(x) for arbitrary x E R.

The product of a linear functional 1' by a number a is a functional f = af'such that

f(x) = afi(x)

for arbitrary x E R.It is easy to verify that the operations of addition and multiplication by

a scalar of functionals so defined satisfy all the axioms of a linear space.Moreover, the definition we gave above of the norm of a linear functionalsatisfies all the requirements found in the definition of a normed linearspace. In fact,

1) f > 0 for arbitrary I 0,

2)11 af II = J a I I! I3) +12 II = sup {Jfi(x) +12(X) 1/lIx II}

� sup { (I fi(x) I + I 12(x) 1)/lI x

� sup {Jfi(x) x II} + sup {jf2(x) x II}

= Ill'! + 112 I!.

Thus, the totality of all linear functionals on a normed space R itselfrepresents a normed linear space; it is called the conjugate space of R andis denoted by R.

THEOREM 1. The conjugate space is always complete.Proof. Let } be a fundamental sequence of linear functionals. By the

definition of a fundamental sequence, for every e > 0 there exists an Nsuch that II — fm < e for all n, in > N. Then for arbitrary x E R,

— fm(x)I

— fm Il Jx U <e lxi.e. for arbitrary x E R the numerical sequence fn(x) converges.

If we set

1(x) = fn(X),

then 1(x) represents a linear functional. In fact,

1) f(ax + 13y) = + /3y)

= [afn(x) + = af(x) +2) Choose N so that II In — fn+p < 1 for all n > N. Then

II 1 <II In II + 1

for all p. Consequently, fn÷p(x) � (Ii In II ± 1)11 x I!.


Passing to the limit as p —* oc, we obtain

I � (lIft II + 1)11 x

i.e. the functional f(x) is bounded. We shall now prove that the functionalf is the limit of the sequence 1' , 12, , ... . By the definition of thenorm, for every 0 there exists an element such that

— I II � — 1)/Il + €/2

= I II) — I!) + €/2;

since

I!) =

it is possible to find an no(€) such that for n > n0

I II) — Xe II) I < €/2,

so that for n > n0 the inequality

IIfn -111<is fulfilled.

This completes the proof of the theorem.Let us emphasize once more that this theorem is valid independently of

whether the initial space R is complete or not.EXAMPLES. 1. Let the space B be finite-dimensional with basis e1 , e2,

Then the functional 1(x) is expressible in the form

(1) 1(x) =

where x = and =Thus, the functional is defined by the n numbers 1', ,

which arethe values of I on the basis vectors. The space which is the conjugate of thefinite-dimensional space is also finite—dimensional and has the same dimen-sion.

The explicit form assumed by the norm in the conjugate space dependson the choice of norm in B.

a) Let1

x = We have already shown that then

Ill! =

i.e. the conjugate of an Euclidean space is itself Euclidean.

b) Let I! xII

= supj.Then

1(x) = � lix I!.

From this it follows that

lift! �

84 NORMED LINEAR SPACES {CH. Ill

The norm I! f cannot be less than since if we set

1+1 0,

= —1 <0,

then the following equality is

1(x) I = E lid = Ifd) lix I!.c) If x

1 = (E I I Ii = where

i/p + i/q = 1. This follows from the Holder inequality

I I� q)1/q

and from the fact that the equality sign is attained [for = (sgn2. Let us consider the space c consisting of sequences x = (xi, x2,

•..) which are such that —f 0 as n —f where 1 x = supnIf a functional in the space c is expressible by means of the formula

(2) 1(x) = Er=' i I

<

then it has norm

11111 =

The inequality If II � fi Iis obvious. On the other hand, if

= a, then for every e > 0 it is possible to find an N such that> a—e.

We now set

+i iff,, > 0—i <0 n � N.

xn =0

o if n> NThen

1(x)I

= > awhence it follows that I = a.

We shall prove that all functionals in the space c have the form (2).We shall set

= (0, 0, . . , i, 0, . .

i.e. denotes the sequence in which the n-th entry is unity and the re-maining are zeros.


Let the functional 1(x) be given; we denote by If

(3) x = (x1 , , ,

x = x1e1 + x2e2 + + and f(x) =

The sum is <oo for every bounded linear functional. Ifwere = then for every H it would be possible to find an N

such that

> H.

We construct the element x in the following way:

1

if 0

Oifn> NThe norm of such an element is equal to unity, and

11(x) I = = > H H xwhich contradicts the assumption concerning the boundedness of the func-tional.

The set of elements of the form (3) is everywhere dense in the space c.Therefore the continuous linear functional is uniquely defined by its valueson this set. Thus, for every x

f(x) =

The space which is conjugate to the space c consists of sequences

(1', , ...) satisfying the conditionI I

<

3. Let the space consist of sequences

x = (xi,x2, ... <

with norm x =It can be proved that the space conjugate to this space is the space of

bounded sequences

1= (11,12, •.. ,fn,with norm = supn

I In I.

In all the examples of finite-dimensional spaces introduced above, thespace which is conjugate to the conjugate space coincides with the initial

86 NORMED LINEAR SPACES [CII. III

space. This is always so in the finite-dimensional case. However, as Exam-ples 2 and 3 show, in the infinite-dimensional case the space conjugate tothe conjugate space may not coincide with the initial space.

We consider eases when this coincidence holds also in infinite-dimensionalspace.

The space 12 consists of sequences

x = (x1 , . xn )

with < and norm x = All functionals in the12 have the form

f(x) = Lx.We shall prove this assertion.To each functional there is set into correspondence the sequence Ii , 12,

of its values on the elements e1, e2, •• •.. definedexactly as in Example 2, above.

Tf the functional is bounded, then < We shall assume thecontrary, i.e. we shall assume that for every H there exists an N such that

= U � H.If we apply the functional under consideratioii to the element

= (11 , f2 , , , 0, ), 1

X = U,we obtain

f(x) = = U � x

contrary to the assumption that the functional is bounded.Since the functional I is linear, its values on the elements of the form

x = (x1, X2, 0, •••) are easily found; on all other elements ofthe space the values of f are found from continuity considerations and wealways have

f(x) =

The norm of the functional f equals This is established withthe aid of the Schwarz inequality.

5. The space is the space of all sequences of the form00 pit)) 00 p1/p= (x1 , , ( x7 ) < , x = ( j=i x2 )

rfhe conjugate space of is the space lq , where i/p + i/q = 1. The proofis analogous to the preceding proof. Hint: Use Holder's inequality.

§25. Extension of linear functionals

THEOREM (HAHN-BANACH). Every linear functional f(x) defined on alinear subspace G of a normed linear space E can be extended to the entire

§251 EXTENSION OF LINEAR FUNCTIONALS 87

space with preservation of norm, i.e. it is possible to construct a linear func-tional F(x) such that

1) F(x) = f(x), x E G,

2) IF lIE =

Proof. The theorem will be proved for a separable space B, although inactuality it is valid also in spaces which are not separable.

First, we shall extend the functional to the linear subspace G1 obtainedby adding to G some element G. An arbitrary element of this subspaceis uniquely representable in the form

y=txo+x, xEG.If the functional sought exists, then

F(y) = tF(xo) + f(s)

or, if we set — F(so) = c, then F(y) = f(x) — ct.

In order that the norm of the functional be not increased when it iscontinued it is necessary to find a c such that the inequality

(1) f(x) — ctj < Hf!! lx + tx0 II

be fulfilled for all x E G.If we denote the element x/t by z (z E G), the inequality (1) can be re-

writtenif(z) - ci � if H Hz + xo ii.

This inequality is equivalent to the following inequalities:

-hf II liz + <f(z) - c � If II liz +or, what amounts to the same thing,

f(z) + Hf!! liz + I! � c � f(z) — If ii I! z +

for all z E G. We shall prove that such a number c always exists. To dothis we shall show that for arbitrary elements z', z" E G we always have

(2) f(?) + if 1liz" + � f(z') — if! II z' +

But this follows directly from the obvious inequality

f(z') — f(z") < ii f liii z' - z" = if it 1

z' + xo — (z" + Xø)1

ii f liii z' + + I! f 1 z" + 1.

We introduce the notation:

c' = inf {f(z) + Ii f ii!! z + Ii; z E GI,c" = sup tf(z) — Ii f If z + 1; z E GI.

88 NORMED LINEAR SPACES [Cu. III

It follows from inequality (2) that c" � c'.We take an arbitrary c such that c" � c � c'. We set

Fi(x) = f(x) — ct

for the elements of the subspace G1 = {G U xo}. We obtain the linearfunctional F1, where 1 F1 1= 1 1 f.

The separable space E contains a denumerable everywhere dense setXi X2, , ••• . We shall construct the linear subspaces

= {GUx0},

G2 = {G1Ux1},

= U

and define the functional F on them as follows: we construct functionalswhich coincide with on and which have norm equal to f J.

Thus, we obtain the functional F defined on a set which is everywheredense in E. At the remaining points of E the functional is defined by con-tinuity: if x = then F(x) =

F(x) < f x is valid since

I I = I� If If 1 II = x

This completes the proof of the theorem on the extension of a functional.COROLLARY. Let x0 be an arbitrary nonzero element in R and let M be an

arbitrary positive number. Then there exists a linear functional 1(x) in Rsuch that

If II = M and f(xo) = 1111 II

In fact, if we set f(txo) = tM xo we obtain a linear functional withnorm equal to M which is defined on the one—dimensional subspace of ele-ments of the form tx0 and then, by the Hahn-Banach theorem, we can extendit to all of R without increasing the norm. The geometric interpretation ofthis fact is the following: in a Banach space through every point x0 therecan be drawn a hyperplane which is tangent to the sphere ff x = II I!.

§26. The second conjugate space

Inasmuch as the totality of linear functionals on a normed linearspace I? itself represents a normed linear space it is possible to speak of thespace of linear functionals on E, i.e. of the second conjugate space withrespect to R, and so forth. We note first of all that every element in Rdefines a linear functional in R. In fact, let

= f(xo),

§26] THE SECOND CONJUGATE SPACE 89

where x0 is a fixed element in R and I runs through all of I?. Thus, to eachf E R there is set into correspondence some number In this connec-tion we have

(au + 1312) = afi(xo) + 13f2(xo) = (1') + (12)

and

I u/'xo(f) ii I liii II (boundedness),

i.e. is a bounded linear functional onBesides the notation 1(x) we shall also use the more symmetric notation:

(1) (1, x)

which is analogous to the symbol used for the scalar product. For fixedI E R we can consider this expression as a functional on R and for fixedx E R as a functional on R.

From this it follows that the norm of every x E R is defined in two ways:firstly, its norm is defined as an element in R, and secondly, as the normof a linear functional on R, i.e. as an element in Let x denote thenorm of x taken as an element in R and let x 112 be the norm of x takenas an element in R. We shall show that in fact x I! = 1

x 12. Let I be anarbitrary nonzero element in R. Then

(1, x) I � x lix 11 1(1, x) I/Ill I!;

since the left member of the last inequality does not depend on 1, we have

lxii � sup {I(f,x)J/IIfl!;fE = !IxI!2.

But, according to the corollary to the Hahn—Banach theorem, for everyx E R a linear functional 10 can be found such that

(b, x) = hi! I!x

Consequently,

sup {I (f,x) i/Ill ll;f E R} = lix!!,

i.e. x !2 = 1x

This proves the following theorem.THEOREM. R is isometric to some linear manifold in R.Inasmuch as we agreed not to distinguish between isometric spaces this

theorem can be formulated as follows: R C R.The space R is said to be reflexive in case R = R. If R, then R is

said to be irreflexive.Finite-dimensional space and the space 12 are examples of reflexive

spaces (we even have R = R for these spaces).The space c of all sequences which converge to zero is an example of a


complete irreflexive space. In fact, above Examples 2 and 3) weproved that the conjugate space of the space c is the space 1 of numericalsequences (x1 , X2, •• , ••.) which satisfy the condition

I I <to which in turn the space m of all bounded sequences is conjugate.

The spaces c and m are not isometric. This follows from the fact that cis separable and m is not. Thus, c is irreflexive.

The space C[a, b] of continuous functions on a closed interval [a, b] isalso irreflexive. However, we shall not stop to prove this assertion. (Thefollowing stronger assertion can also be proved: No normed linear spaceexists for which C[a, b] is the conjugate space.)

A. I. Plessner established that for an arbitrary normed space R thereexist only two possibilities: either the space R is reflexive, i.e. R = =

= ... ; = = ; or the spaces R, are all distinct.The space lp(p > 1) is an example of a reflexive space (since =

where 1/p + 1/q = 1, we have lp = lq = lv).

§27. Weak convergence

The concept of so-called weak convergence of elements in a normed linearspace plays an important role in many questions of analysis.

DEFINITION. A sequence { } of elements in a normed linear space Rconverges weakly to the element x if

1) The norms of the elements are uniformly bounded: ft II � M,and -

2) f(x) f E R.(It can be shown that Condition 1 follows from 2; we shall not carry outthis proof.)

Condition 2 can be weakened slightly; namely, the following theorem istrue.

THEOREM 1. The sequence { converges weakly to the element x if1) � M, and2) f(x) for every I E where is a set whose lineai' hull is

everywhere dense in R.Proof. It follows from the conditions of the theorem and the definition

of a linear functional that if is a linear combination of elements inthen —*

Now let be an arbitrary linear functional on R and let } be a sequenceof functionals which converges to each of which is a linear combinationof elements in We shall show that —* Let 111 be such that

� M(n = 1,2, Ixl! � M.Let us evaluate the difference

I— I. Since given an

arbitrary e > 0 a K can be found such that for all k > K,

— II < e;

§27] WEAK CONVERGENCE 91

it follows from this that

co(xn) — 4p(x)I I

— I + I cok(xn) —I

+ I — 4p(x) � + + (Pk(Xn) — cok(x)

But by assumptionI

— cok(x) —* 0 as n —* Consequently,

Ico(xn) — 4p(x) I —*0 as n —* oO.

If the sequence converges in norm to x, that is, if — xfJ

—* 0

as n —* oo, then such convergence is frequently called strong convergenceto distinguish it from weak convergence.

If a sequence } converges strongly to x, it also converges weakly tothe same limit. In fact, if — x —* 0, then

If(x) f II — x —* 0

for an arbitrary linear functional f. The converse is not true in general:strong convergence does not follow from weak convergence. For example,in 12 the sequence of vectors

e1 = (1,0,0, ...),e2= (0,1,0,...),

e3= (0,0,1, ...),

converges weakly to zero. In fact, every linear functional f in 12 can berepresented as the scalar product with some fixed vector

a = (ai a2 , , , f(x) = (x, a);

hence,

= (es, a) =

Since —* 0 as n —* oo for every a E 12, we have lim = 0 for everylinear functional in 12.

But at the same time the sequence { } does not converge in the strongsense to any limit.

We shall investigate what weak convergence amounts to in severalconcrete spaces.

EXAMPLES. 1. Infinite-dimensional space weak and strong convergencecoincide. In fact, consider functionals corresponding to multiplication bythe elements

e1 = (1,0,0, •..e2 = (0, 1,0, ...


If converges weakly to x, then

(Xk , = xk —* (i = 1, 2, • • • , n),

i.e. the first coordinates of the vectors Xk tend to the first coordinate of thevector x, their second coordinates tend to the second coordinate of thevector x, and so forth. But then

n (i) (i) 2p(xk , x) = i=1 (Xk — X ) —* 0,

i.e. {xkl converges strongly to x. Since strong convergence always impliesweak convergence, our assertion is proved.

2. Weak convergence in 12. Here we can take for the set linear com-binations of whose elements are everywhere dense in 12, the totality ofvectors

e1 = (1,0,0, •..),e2= (0,1,0,...),e3= (0,0,1, ...),

If x = (Ei, E2, , is an arbitrary vector in 12, then the valuesassumed at x by the corresponding linear functionals are equal to (x, =

i.e. to the coordinates of the vector x. Consequently, weak convergenceof the sequence in 12 means that the numerical sequence of the k-thcoordinates of these vectors (k = 1, 2, •'') converges. We saw above thatthis convergence does not coincide with strong convergence in 12.

3. Weak convergence in the space of continuous functions. Let C[a, b] bethe space of continuous functions defined on the closed interval [a, b]. Itcan be shown that the totality of all linear functionals , each of whichis defined as the value of the function at some fixed point t0 (see Example 4,§23) satisfies the conditions of Theorem 1, i.e. linear combinations ofthese functionals are everywhere dense in C[a, b]. For each such functional

the condition —* is equivalent to the condition —*

x(to).Thus, weak convergence of a sequence of continuous functions means

that this sequence is a) uniformly bounded and b) convergent at everypoint.

It is clear that this convergence does not coincide with convergence innorm in C[a, b], i.e. it does not coincide with uniform convergence of con-tinuous functions. (Give a suitable example!)

§28. Weak convergence of linear functionals

We can introduce the concept of weak convergence of linear functionalsas analogous to the concept of weak convergence of elements of a normedlinear space R.

§28] WEAK CONVERGENCE OF LINEAR FUNCTIONALS 93

DEFINITION. A sequence of linear functionals converges weakly tothe linear functional f if

1) fJare uniformly bounded; i.e. � M, and

2) f(x) for every element x E R.(This is usually called weak* convergence.—Trans.)

Weak convergence of linear functionals possesses properties which areanalogous to properties stated above of weak convergence of elements,namely strong convergence (i.e. convergence in norm) of linear functionalsimplies their weak convergence, and it is sufficient to require the fulfillmentof the condition f(x) not of all x E R, but only for a set of elementslinear combinations of which are everywhere dense in R.

We shall consider one important example of weak convergence of linearfunctionals. Above Example 4), we spoke of the fact that the "ô-func-tion", i.e. the functional on C[a, b], which assigns to every continuous func-tion its value at the point zero, can "in some sense" be considered as thelimit of "ordinary" functions, each of which assumes the value zero outsidesome small neighborhood of zero and has an integral equal to 1. [We assumethat the point t = 0 belongs to the interval (a, b). Of course, one can takeany other point instead of t = 0.] Now we can state this assertion pre-cisely. Let {con(t) be a sequence of continuous functions satisfying thefollowing conditions:

(1) 1) çon(t) = 0 for t > 1/n, � 0,

2) f dt = 1.

Then for an arbitrary continuous function x(t) defined on the closedinterval [a, b], we have

ri/n

Jipn(t)x(t) dt = J

dt x(O) as na —i/n

In fact, by the mean-value theorem,ri/fl ri/n

Jdt =

Jçon(t) dt = — 1/n � � 1/n;

—i/n —i/n

when n —* 0 and —* x(0).The expression

f con(t)x(t)

represents a linear functional on the space of continuous functions. Thus,the result we obtained can be formulated as follows: the ô-function is thelimit of the sequence (1) in the sense of weak convergence of linear func-tionals.

94 XORMED LINEAR SPACES III

The following theorem plays an important role in various applications ofthe concept of weak convergence of linear functionals.

THEOREM 1. If the normed linear space R is separable, then an arbitrarybounded sequence o.f linear on I? a weaklysubsequence.

Proof. Choose in R a denurnerable everywhere dense set

{x1,x2, •..If is a bounded (in norm) sequence of linear functionals on R, then

f1(x1), f2(x1), • • ,

is a bounded numerical sequence. Therefore we can select from asubsequence

.1',J1,J2,

such that the numerical sequence\ -(Fl \Ji J2 kXl1,

converges. Furthermore, from the subsequence we can select a sub-sequence

4FF 4FF (F,Ji J2 • • J

such that.eFFI \ (FF1 \

J1 J2

converges, and so forth. Thus, we obtain a system of sequences4' 4'FJi J2 /(F, .t' FFJi J2 Jn

each of which is a subsequence of the one preceding. Then taking the "di-agonal" subsequence f1F, 13FFF , we obtain a sequence of linear func-tionals such that 12FF

• . converges for all n. But then fiF(x)also converges for arbitrary x E R. This completes the proof of

the theorem.The last theorem suggests the following question. Is it possible in the

space E, conjugate to a separable space, to introduce a metric so that thebounded subsets of the space become compact with respect to_this newmetric? In other words, is it possible to introduce a metric in I? so thatconvergence in the sense of this metric in R coincides with weak con-vergence of elements in R considered as linear functionals. Such a metriccan in fact be introduced in R.

Let { be a denumerable everywhere dense set in R. Set

(2) ,=

— II

§29] LINEAR OPERATORS 95

for any two elements fi , f2 E This series converges since its n-th termdoes not exceed (J} f' + The quantity (2) possesses all theproperties of a distance. In fact, the first two axioms are obviously satisfied.We shall verify the triangle axiom.

Since

I— f3(xn)

I

= — f2(xn) + f2(xfl) —

� — f2(xfl) + 1f2(xn) —,

we have

� , + P(f2 , fe).

Direct verification shows that convergence in the sense of this metric isin fact equivalent to weak convergence in

Now Theorem 1 can be formulated in the following manner.THEOREM 1'. In a space R, which is the conjugate of a separable space,

with metric (2), every bounded subset is compact.

§29. Linear operators

1. Definition of a linear operator. Boundedness and continuity.Let R and R' be two Banach spaces whose elements are denoted respec-

tively by x and y. Let a rule be given according to which to each x in someset X R there is assigned some element y in the space R'. Then we saythat an operator y = Ax with range of values in R' has been defined on theset X.

DEFINITION 1. An operator A is said to be linear if the equality

A(aixi + a2x2) = a1Ax1 + a2Ax2

is satisfied for any two elements x1, x2 E X and arbitrary real numbersa1 ,

DEFINITION 2. An operator A is said to be bounded if there exists a con-stant M such that

�MJJxJJ

for all x E X.DEFINITON 3. An operator A is said to be continuous if for arbitrary> 0 there exists a number > 0 such that the inequality

— x" < (x', x" E X)implies that

Ax' — Ax" JR' <

Only linear operators will be considered in the sequel. If the space R'is the real line, the operator y A (x) is a functional, and the formulated


definitions of linearity, continuity and boundedness go over into thecorresponding definitions introduced in §23 for functionals.

The following theorem is a generalization of Theorem 1, §23.THEOREM 1. Continuity is equivalent to boundedness for a linear operator.Proof. 1. Assume the operator A is bounded. The inequality x' — x" <implies that

(1) Ax' — Ax"II

A(x' — x") � M x' — x" �M is the constant occurring in the definition of boundedness. If we

take < inequality (1) yields fJ Ax' — Ax"fJ < i.e. the operator A

is continuous.2. Assume now that the operator A is continuous. We shall prove that A

is bounded by c'ontradiction. We assume that A is not bounded. Then thereexists a sequence

(2) xl,x2, ,xn,such that

> nIIxnJJ.

Set = xn/n II II; it is obvious thatIf

ff = 1/n, i.e. —* 0 asn —* oo. Consider the sequence

AZn = Axn/n II

which is the map of the sequence {Zn} under A. The norm of each elementAZn is not less than 1:

ft AZn = II Axn I/n II � n xn 1/ nIl xn II = 1.

Since for every linear operator, A (0) = 0, and Zn = 0, we obtain acontradiction of our hypothesis that the operator is continuous. Conse-quently, the operator A must be bounded.

EXAMPLE. The general form of a linear operator mapping a finite-dimen-sional space into a finite-dimensional space. Given an n-dimensional space

with basis e1, e2, , en, every point of this space can be representedin the form x = x1e1.

A linear operator A maps into the finite-dimensional space Rm withthe basis e1', e2', , em'.

Let us consider the representation with respect to this basis of the imagesof the basis vectors of the space Rn:

= a11e1'.

Now let y = Ax,n n V'm ,' ,

y = Ax = = =

§29} LINEAR OPERATORS 97

where

(3) d, = a13x1.

It is clear from formula (3) that to determine the operator A it is suffi-cient to give the coefficient matrix with entries as,.

A linear operator cannot map a finite-dimensional space into a space ofgreater dimension since all linear relations among the elements are preservedfor their images.

2. Norm of an operator. Sum and product of operators. Product of anoperator by a scalar.

DEFINITION 4. Let A be a bounded linear operator. This means thatthere exist numbers M such that

(4) Ax <M xfor all x E X. The norm A of the operator A is the greatest lowerbound of the numbers M which satisfy condition (4). It follows from thedefinition of the norm of an operator that Ax A liii x But ifM < A then there exists an element x such that Ax > M If x

THEOREM 2. If A is an arbitrary linear operator, then

IA = sup {IIAxII; lxii = 11 = sup {IIAxII/IIxLl; lxii 01.

Proof. Introduce the notation

a = sup {IlAxll; lxii = 11 = sup {IIAxlI/llxll; lxii 0}.

We shall first prove that A � a. Since a = sup {Ax x ii; x 01,

for arbitrary 0 there exists an element x1 not zero such thatAx1 xi ii > a — or Ax1 > (a — €) which implies that

a — A and hence that A � a because is arbitrary.The inequality cannot hold. In fact, if we let A If — a = €, then a <A — €12. But this implies that the following inequalities hold for an

arbitrary point x:

Ax x a < II A II — €12,

or

Ax � (Ii A — €12) x

i.e. A is not the greatest lower bound of those M for which ff AxIf

M x It is clear from this contradiction that A = a.In the sequel we shall make use of the above expression for the norm of

an operator as equivalent to the original definition of the norm.DEFINITION 5. Let A1 and A2 be two given continuous linear operators

which transform the Banach space E into the Banach space E1. The sum


of these two operators is the operator A which puts the element y E E1defined by the formula y = A1x + A 2x into correspondence with the ele-ment x E E. It is easy to verify that A = A1 + A2 is also a linear operator.

THEOREM 3. The following relation holds for the norms of the operatorsA1, A2 and A = A1 + A2:(5) A 1 Ai II —F— H A2

Proof. It is clear that

Ax = Ii Aix + A2x A1x + A2x � A1 + 1 A2 II) It x

whence inequality (5) follows.DEFINITION 6. Let A1 and A2 be continuous linear operators where A1

transforms the Banach space E into the Banach space E1 and A2 transformsthe Banach space E1 into the Banach space E2 . The product of the operatorsA1 and A2 (denoted by A = A2A') is the operator which sets the elementz E E2 into correspondence with the element x E E, where

z = A2(A1x).

THEOREM 4. If A = A2A1, then

(6) 1A

HA2 A1

Proof. Ax = 1A2(A1x) II < A2 A1x A2 JJJ A1 x JJ,

whence the assertion of the theorem follows.The sum and product of three or more operators are defined by iteration.

Both operations are associative.The product of the operator A and the real number k (denoted by kA)

is defined in the following manner: the operator kA puts the element k(Ax)of the space E1 into correspondence with the element x E E.

It is easy to verify that with respect to the operations of addition andmultiplication by a scalar introduced above the bounded linear operatorsform a linear space. If we introduce the norm of the operator in the wayindicated above we can form a normed linear space.

EXERCISE. Prove that the space of bounded linear operators which trans-form the space E1 into a complete space E2 is complete.

3. The inverse operator.Let us consider the operator T which transforms the Banach space E

into the Banach space E1

Tx=y, xEE, yEE1.DEFINITION 7. The operator T is said to have an inverse if for every

y E E1 the equation

(7) Tx—yhas a unique solution.

§29] LiNEAR OPERATORS 99

To each y E E1 we can put into correspondence the solution of equation(7). The operator which realizes this correspondence is said to be the inverseof T and is denoted by T'.

THEOREM 5. The operator T' which is the inverse of the linear operator Tis also linear.

Proof. To prove the linearity of T' it is sufficient to verify that theequality

rjrl (aiyi + a2y2) = aiT'y1 + a2T'y2

is valid. Denote Tx1 by yi and Tx2 by Y2. Since T is linear, we have

(8) T(aixi + d2x2) aiyi + a2y2.

By the definition of the inverse operator,

= xi, T'y2 = x2;

whence, multiplying these equations by and a2 respectively, and adding,we obtain:

aiT'yi 4- a2T'y2 = 4- a2x2

On the other hand, from (8) and from the definition of the inverse operatorit follows that

aiXi + a2X2 = T'(ai Yi -f- a2y2)

which together with the preceding equalities yields

T'(aiyi + a2Y2) = a1T'y1 —I— a2T'y2

THEOREM 6. If T is a bounded linear operator whose inverse T' exists,then T' is bounded.

We shall need the following two lemmas in the proof of this theorem.LEMMA 1. Let M be an everywhere dense set in the Banach space E.

Then an arbitrary element y E E, y 0, can be developed in the series

where Yk E M and II II � 3 II y

Proof. We construct the sequence of elements Yk in the following way:we choose so that

(9) II — b/i II ii 11,/2.

This is possible because inequality (9) defines a sphere of radius y 1/2with center at the point y, whose interior must contain an element of M(since M is everywhere dense in E). We choose Y2 E M such that

II — —1 II II/4,y3 such that y — — Y2 — Y3 II y 1/8,

and in general, such that — — — II II y

100 NORMED LINEAR SPACES [cii. III

Such a choice is always possible because M is everywhere dense in E.By construction of the elements Yk,

1 — 0 as n —*

i.e. the series converges to y.To evaluate the norms of the elements Yk we proceed as follows:

1 = - y + y1 ffy' - y H + II y 1 = 3 1 y 11/2,

� -Yi -y211 + �Finally, we obtain

H Yn1

II Yn + Yn-i + + Yi — Y + Y — Yi — — Un—i H

This proves Lemma 1.LEMMA 2. If the Banach space E is the sum of a denumerable number of

sets: E = Mn, then at least one of these sets is dense in some sphere.Proof. Without loss of generality, we can assume that

ç M3 çWe shall assume that all the sets are nowhere dense, i.e. that in the

interior of every sphere there exists another sphere which does not con-tain a single point of Mk , k = 1, 2,

Take an arbitrary sphere So ; in it there exists a sphere Si which does notcontain a single point of M1 ; in there exists a sphere 52 which does notcontain a single point of M2 ; and so forth. We obtain a sequence of nestedspheres which can be chosen so that the radius of the sphere Sn convergesto zero as n In a Banach space such spheres have a common point.This point is an element of E but it does not belong to any of the setsThis contradicts the hypothesis of the lemma and proves Lemma 2.

Proof of Theorem 6. In the space E1 let us consider the sets Mk, whereMk is the set of all y for which the inequality II If � k y holds.

Every element of E1 is contained in some Mk, i.e. E1 = ByLemma 2, at least one of the Mn is dense in some sphere So. In the interiorof the sphere So let us consider the spherical shell P consisting of the pointsz for which

13 < LI z — Yo < a,

where

O<f3<a,


If we translate the spherical shell P so that its center coincides with theorigin of coordinates, we obtain the spherical shell P0.

We shall show that some set is dense in P0. Consider z E P; thenZ Yo E Po. Furthermore, let z E By virtue of the choice of z and Yowe obtain:

II — Yo) II II T'z II + T'y0 II

� z + 1) < n(1I + 2 Yo II)

=

The quantity n(1 + 2 Yo does not depend on z. Denote n(1 +2 Yo by N. Then by definition z — yo E MN and MN is dense in Pobecause MN is obtained from as was P0 from P, by means of a trans-lation by Yo and is dense in P. Consider an arbitrary element y in E1.It is always possible to choose X so that j3 < Xy <a. For Xy we can con-struct a sequence Yk E MN which converges to Xy. Then the sequence(1/X)yk converges to y. (It is obvious that if Yk E MN, then (1/X)yk Ef or arbitrary real 1/X.)

We have proved that for arbitrary y E E1 a sequence of elements of MNcan be found which converges to y, i.e. that MN is everywhere dense in E1.

Consider y E E1 ; by Lemma 1, y can be developed in a series of elementsin MN:

where II < 3 IIConsider in the space E the series formed from the inverse images of the

Yk, i.e. from Xk = T'yk:

= X1 + X2 + + +This series converges to some element x since the following inequality

holds:JJ = 1

� N JJ < 3N II and consequently,lxii � � = 3lIyIIN.

By virtue of the convergence of the series and the continuity ofT we can apply T to the series. We obtain:

Y1+Y2+ Y,

Whence X = T'y. We have

1x = Il T'y ii � 3N II ii

and since this estimate is valid for arbitrary y, it follows that T' is bounded.THEOREM 7. An operator which closely approximates an operator whose

inverse exists has an inverse, i.e. if T0 is a linear operator which has an in-

102 NORMED LINEAR SPACES [ci-r. iii

verse and which maps the space E into the space E1, and is an operatorwhich also maps E into E1, where < 1 / T0' then the operatorT T0 + has an inverse.

Proof. Let y E E1 . We wish to find a unique x0 E E such that

y = Tx0 = T0x0 +If we apply the operator T0' to this equation, we obtain

(10) T0'y = xo +if we denote T01y by z E E and by A, then equation (10) can bewritten in the form

z = xo + Ax0,

where A is an operator which maps the Banach space E into itself andIA <1.

The mapping x' = z — Ax is a contraction mapping of the space E intoitself; consequently, it has a unique fixed point which is the unique solutionof equation (10) and this means that the operator T has an inverse.

THEOREM 8. The operator which is the inverse of the operator T = I — A,where I is the identity operator and the operator A (of E into E) has norm lessthan 1 ( A

1< 1), can be written in the form

(11) (1— =

Proof. Consider the following transformation of the space E into itself:

y=Tx, xEE, yEE.The mapping x' = y + Ax is a contraction mapping of the space E into

itself by virtue of the condition that A iI < 1.

We solve the equation x = y + Ax by means of the iterations: x +1 =y + . If we setx0 = 0, we obtainx1 = y;x2 = y + Ay;x3 = y + Ày +A2y; ... ; = y + Ay + A2y + +

As n cc, tends to the unique solution of the equation x = y + Ax,i.e. x = Aky, whence

—1 oo k(I—A) k=OAy,

which yields equation (11).4. Adjoint operators.Consider the linear operator y = Ax which maps the Banach space E

into the Banach space E1 . . Let g(y) be a linear functional defined on E1i.e. g(y) E E1 . Apply the functional g to the element y = Ax; g(Ax), as iseasily verified, is a linear functional defined on E; denote it by f(x). Thefunctional f(x) is thus an element of the space E. We have assigned to each


functional g E E1 a functional f E E, i.e. we have obtained an operatorwhich maps E1 into E. This operator is called the ac/joint operator of theoperator A and is denoted by A* or by f = A*g.

If we use the notation (f, x) for the functional f(x), we obtain (g, Ax) =(f, x), or (g, Ax) = (A*g, x). This relation can be taken for the definitionof the adjoint operator.

EXAMPLE. The expression for the adjoint operator in finite-dimensionalspace. Euclidean n-space is mapped by the operator A into Euclideanin-space En'. The operator A is given by the matrix (a13).

rfhe mapping y = Ax can be written in the form of the system

yj = , i = 1, 2, . . , in.

The operator f(x) can be written in the form

f(x) — f3x3.

The equalities

f(x) g(Ax) = =1 gjajjxj = a'1 gjai)

imply that f = A*g, it follows that the 'operator A*is given by the transpose of the matrix for the operator A.

We shall now list the basic properties of adjoint operators.1. The adjoint operator of the sum of two linear operators is equal to the

sum of the adjoint operators:

(A + B)* 11* + B*.

Let fi = A*g, f2 = B*g, orf1(x) = g(Ax), f2(x) = g(Bx); then (fi + f2)(x)g(Ax + Bx) = g[(A + B)x], whence (A + B)* = A* + B*.

2. The adjoint operator of the operator kA, where k is a scalar multi-plier, is equal to the adjoint operator of A, multiplied by k:

(kA)* = kA*.

The verification of this property is elementary and is left to the reader.3. 1* = I, i.e. the adjoint of the identity operator on E is the identity

operator on E.THEOREM 9. The operator A*, the ac/joint of a linear operator A which

maps the Banach space E into the Banach space E1 , is also linear and14*11 f.

Proof. The linearity of the operator A* is obvious. We shall prove theequality of the norms. By virtue of the properties of the norm' of a,n operatorwe have:

104 NORMED LINEAR SPACES [CH. Ill

whence II f A g or A*g1

A g and consequently,

(12) A IL

Let x E E and form yo = Ax E E1 it is clear that Yo = 1.

From the corollary to the Hahn-Banach theorem, there exists a func-tional g such that g = 1 and g(yo) = 1, i.e. çj(Ax) = Ax

From the inequalities Ax = (g, Ax) I = I (A*g, x) A*g II x

IIA*II MCII = IIA*II lxii we obtain IA �

bined with inequality (12), yields A = II A*

ADDENDUM TO CHAPTER III

Generalized Functions

In a number of cases in analysis and in its various applications, for ex-ample in theoretical physics, the need arises to introduce various "general-ized" functions in addition to the "ordinary" functions. A typical exampleof this is the well-known ô-function which we have already mentionedabove Example 4).

We wish to emphasize, however, that these concepts, which arediscussed briefly in this addendum, did not in any sense originate iii an at-tempt to generalize the concepts of analysis merely for the sake of general-izing. Rather, they were suggested by perfectly concrete problems. More-over, essentially the same concepts were used by physicists for quitesome time before they attracted the attention of mathematicians.

The method of introducing generalized functions which we shall usebelow originated in the work of S. L. Sobolev, published in 1935—36. Later,these ideas were developed in a somewhat extended form by L. Schwartz.

Consider on the real line the set D of functions ço(x) each of whichvanishes outside some interval (where for each cc there is a correspondinginterval) and has derivatives of all orders. The elements of D can be addedand multiplied by a scalar in the usual way. Thus, D is a linear space.We shall not introduce a norm into this space; however, in D one can de-fine in a natural way the convergence of a sequence of elements. We shallsay that cc if: 1) there exists an interval in the exterior of which all

and cc are equal to zero and 2) the sequence of derivatives of orderk (k = 0, 1, 2, ...) (where the derivative of order zero is understood to beas usual the function itself) converges uniformly to on this interval.The fact that this concept of convergence is not connected with any normdoes not give rise to any inconveniences.

We now introduce the concept of generalized function.DEFINITION 1. A generalized function (with values on the real line

— < t < is any linear functional T(çc) defined on the space D. Thus,T(çc) satisfies the following conditions:

1. T(açoi + 13cc2) = aT(çci) + f3T(ço2);2. If cc,, cc (in the sense indicated above), then T(co).

We now consider several examples.1. Let f(t) be an arbitrary continuous function of t. Then, since every

function çc(t) E D vanishes outside some finite interval, the integral

(1) = dt

exists for all E D. The expression (1) represents a linear functional on105

106 ADDENDUM TO CHAPTER III [cii, ni

D (it is left to the reader to verify that Conditions 1 and 2 of Definition 1 areactually fulfilled), i.e. (1) represents a generalized function. Thus, every"ordinary" continuous function is at the same time a generalized functionalso.

In this connection, two distinct continuous functions will define distinctfunctionals on D, i.e. they will represent distinct generalized functions.Obviously, it is sufficient to show that every continuous function f(t) whichis not identically zero defines a linear functional which is not equal to zero,i.e. to show that there is a function ço(t) E D for which

f(t) 0, an interval (a, j3) can be found on which f(t) is differentfrom zero and consequently keeps the same sign (for instance, positive).Let us consider the function

—1/( )2

This function is positive on the interval (a, j3); it assumes the value zeroat t = a, j3 and this is true of its derivatives of all orders. Conse-quently, the function which is equal to on theinterval (a, f3) and is zero outside this interval, has continuous derivativesof all orders, i.e. it belongs to D. Furthermore,

00

f dt> 0,

since the function under the integral sign is positive.2. We now set

(2) T(ço) =

Formula (2) defines a linear functional on D, i.e. a generalized function.This is none other than the ô-function which we have already mentionedabove. (In §23 we considered the a-function as a functional defined on thespace of all continuous functions. The advantages of the point of viewadopted here will be manifest in the sequel.)

3. Set=

This generalized function is called the derivative of the ô-function and isdenoted by ô'. (A more general definition of the derivative of a generalizedfunction will be given below.)

It is sometimes convenient to denote the generalized function, as in thecase of an ordinary function, by the symbol f(t) and say that the linearfunctional T is "defined" by generalized function f(t). The value T(ço)of a linear functional T for each function E D in this context is more

ADDENDUM GENERALIZED FUNCTIONS 107

conveniently written in the form of a "scalar product" (f, cc). If f(t) is a"proper" function,

= dt.

Frequently, especially in physics books, such an integral notation isused also for generalized functions, for example, we find:

ô(t)cc(t) dt =

however, it is necessary to remember here that such an integral is only asymbolic representation of the value of the corresponding functional andthat it has no other meaning.

We shall now introduc.e the concept of the limit of a sequence of general-lized functions.

DEFINITION 2. The sequence { } is said to converge to the generalizedfunction T if T(cc) uniformly on every "bounded" set in D. In thisconnection, the set {cc} c D is said to be bounded if: 1) there exists a finiteinterval outside of which all the cc of this set equal zero and 2) it is possibleto choose constants M0 , , M2, such that

IccI<Mo; lcc'I<Mi;forallccE {cc}.

It is easily verified that in the sense of this definition the ô-function is thelimit, for example, of such a sequence of "proper" functions:

n/2 for — 1/n < t < 1/n,=

0 for all other values of t.

In exactly the same way it is easy to construct a sequence of "proper"functions converging in the indicated sense to the generalized function t5'.In general, it can be proved that every generalized function is the limit ofsome sequence of linear functionals defined by "proper" functions.

We shall now formulate the definition of the derivative of a generalizedfunction.

We consider first of all the linear functional T defined by means of a dif-ferentiable function:

T(cc) = dt.

It is natural to call the functional defined by means of the formula00

= (cc) = f f'(t)cc(t) dt

the derivative T' of T.

108 ADDENDuM TO CHAPTER III [CH. III

Making use of the formula for integration by parts and taking into con-sideration the fact that every function ço(t) is differentiable and equal to zeroin the exterior of some finite interval, we can rewrite the last expressionin the form

=— dt

in which the derivative f'(t) is no longer present in any way.This discussion brings us to the following definition.DEFINITION 3. Let T be a generalized function. Its derivative T' is a

functional defined by the formula,I \ I

It is clear that T' has meaning for all cc E D and represents a generalizedfunction.

Derivatives of the second, third and higher orders are defined analo-gously.

The validity of the following assertions follows directly from the defi-nition.

THEOREM 1. Every generalized function has derivatives of all orders.THEOREM 2. If the sequence { T

the sense of Definition 2), then the sequence of derivatives{ } converges

to T'. In other words: every convergent series of generalized functions can bedifferentiated term by term any number of times.

It is now clear that the functional (see Example 3) which we calledthe derivative of the s-function is in reality the derivative of in the senseof Definition 3.

The concept of the derivative of a generalized function can beintroduced in a somewhat different way. Namely, we define the translationTh of the functional T (where h is an arbitrary number) by setting

Th(cc(t)) = T(çc(t — h))

(for example, = çc(—h), W(cc) = —çc'(—h), and so forth). Then it isnot difficult to verify that the limit

[(Th — T)/h}

exists and is equal to the derivative T' of the functional T.4. If the functional T(çc) is defined by means of a differentiable function

f(t), then the functional T' is defined by the function f'(t), i.e. the genera-lized derivative of a function coincides with its ordinary derivative if thelatter exists.

5. Let T be defined by the function

ADDENDUM GENERALIZED FUNCTIONS 109

1 for t > 0,

Ofort<0.Then

= f f(t)ç,(t) dt = f dt

and

=— f dt =

i.e. T' is the s-function.6. If f(t) is a piecewise continuous function having a derivative at all

points of continuity, its generalized derivative at points of continuitycoincides with its ordinary derivative and at each point of discontinuity

= t0 it is equal to the s-function — to) multiplied by the magnitude ofthe jump of the function at this point.

7. If

(3) f(t) = (sin nt)/n,

then the generalized derivative of f(t) equalsV"oo 1cos nt = — + ir — 2kir).

We emphasize that the series on the left in the last equation converges in thesense of convergence for generalized functions (Definition 2). Series (3)can be differentiated term by term any number of times.

The last example shows that the concept of generalized function allows usto assign a perfectly definite meaning to the sum of a series which divergesin the ordinary sense. The same remark applies also to many divergentintegrals. This situation is frequently encountered in theoretical physicswhere in each individual case special methods are used to give a definitemeaning to a divergent series or to a divergent integral.

In an analogous way, we can introduce generalized functions of severalindependent variables. To do this, it is merely necessary to take as theinitial space D the space of all functions of n variables which are differen-tiable an infinite number of times and each of which vanishes outside somesphere. All the concepts introduced above carry over automatically to thiscase. It is easy to verify that every generalized function of several inde-pendent variables has partial derivatives of all orders and that the resultof differentiating with respect to several variables does not depend on theorder of differentiation.

Chapter ITT

LINEAR OPERATOR EQUATIONS

§30. Spectrum of an operator. Resolvents

Throughout this entire chapter we shall consider bounded linear operatorswhich map a (generally speaking complex) Banach space E into itself. Anoperator adjoint to an operator of the type indicated will, obviously, mapthe conjugate space E into itself.

In the study of linear operators in finite-dimensional space an importantrole is played by the concepts of characteristic vector and characteristicvalue. In the case of n-dimensional space the concept of the characteristicvalue of an operator can be introduced by means of the following equivalentmethods:

1) The number X is said to be a characteristic value of the operator Aif there exists a nonzero vector x (characteristic vector) such that Ax = Xx.

2) The number X is said to be a characteristic value of the operator Aif it is a root of the characteristic equation

DetIA—XIJ =0.The first of these definitions carries over without any changes to the

case of infinite-dimensional space, i.e. it can be considered as the definitionof characteristic values of an operator in infinite-dimensional space.

The second of the definitions formulated above does not carry over di-rectly to the infinite-dimensional case since the concept of determinantdoes not have meaning in infinite-dimensional space. However, this seconddefinition can be changed in form in the following manner. The fact thatthe determinant of the matrix A — )J is zero is equivalent to the fact thatthis matrix has no inverse, i.e. we can say that the characteristic values Xare those numbers for which the inverse of the operator A — )J does notexist. Taking this as our point of departure we introduce the followingdefinition.

The totality of those values X for which the inverse of the operator A —does not exist is called the spectrum of the operator A. The values of X forwhich the operator A — XI has an inverse are said to be regular; thus, thespectrum consists of all nonregular points. In this connection, the operator

= (A — XI)1 itself is called the resolvent of the operator A. In n-dimen-sional space the concepts of characteristic value and nonregular, or singular,point coincide. In the general case, as the example to be introduced belowwill show, this is no longer the situation. The spectrum necessarily containsall the characteristic values but it may contain, besides these, other num-

110

§30] SPECTRUM OF AN OPERATOR. RESOLVENTS 111

bers also. The set of characteristic values is called the point spectrum of theoperator A. The remaining part of the spectrum is called the continuousspectrum. (This is not the usual terminology. The remaining part of thespectrum consists of two parts: 1) Those values of X for which A — XIhas an unbounded inverse with domain dense in E form the continuousspectrum. 2) The remainder of the spectrum, consisting of those values of Xfor which A — XI has an inverse whose domain is not dense in E, is calledthe residual spectrum.—Trans.)

If the point X is regular, i.e. if the inverse of the operator A — XI exists,then for sufficiently small ô the operator A — (X + ô)í also has an inverse(Theorem 7, §29), i.e. the point X + ô is regular also. Thus, the regularpoints form an open set. Consequently the spectrum, which is its comple-ment, is a closed set.

THEOREM 1. The inverse of the operator A — V exists for arbitrary X forwhich X I> A fJ.

Proof. Since we obviously have

A — XI = — x (i — A),

then

= (A — XI)' = —(i/X) (I — A/2)' = —(1/X)

This series converges for A/X H < 1 (see Theorem 8, §29), i.e. theoperator A — XI has an inverse. In other words, the spectrum of theoperator A is contained in a circle of radius A with center at zero.

EXAMPLE. Let us coiisider in the space C the operator A defined by theformula

A(x(t)) =

where /1(t) is a fixed continuous function. We have

(A — XI)x(t) = (/1(t) — X)x(t),

whence

(A — XI)'x(t) = { — XI }x(t).

rfhe spectrum of the operator A under consideration consists of all Xfor which /2(t) — X is zero for some t lying between 0 and 1, i.e. the spectrumis the totality of all values of the function /1(t). For example, if /1(t) = t,then the spectrum is the closed interval [0, 1]; in this case characteristicvalues are absent, i.e. the operator defined as multiplication by t is anexample of an operator with a purely continuous spectrum.

(1) Every linear operator, defined in a Banach space which has at least

112 LINEAR OPERATOR EQUATIONS [cii. IV

one nonzero element, has a nonvoid spectrum. Operators exist for whichthe spectrum consists of a single point (for example, the operator definedas multiplication by a scalar).

(2) Theorem 1 can be sharpened in the following way. Let r =inf then the spectrum of the operator A lies entirely in a circlewith radius r and center at zero.

(3) The resolvent operators and corresponding to the pointsand X are commutative and satisfy the relation

— = (/1 —

which is easily verified by multiplying both sides of this equality by(A — XI) (A — /21). From this it follows that the derivative of withrespect to X, i.e.

—

exists and is equal to

§31. Completely continuous operators

DEFINITION. An operator A which maps a Banach space E into itself issaid to be completely continuous if it maps an arbitrary bounded set into acompact set.

In finite-dimensional space every bounded linear operator is completelycontinuous since it maps an arbitrary bounded set into a bounded set andin finite-dimensional space every bounded set is compact.

But in the case of infinite-dimensional space there always exist operatorswhich are bounded (i.e. continuous) but which are not completely con-tinuous; such, for example, is the identity operator in infinite-dimensionalspace.

An extensive class of completely continuous operators y = Ax in thespace C[a, b] can be written in the form

(1) y(s) = f K(s, t)x(t) dt.

THEOREM 1. Formula (1) defines a completely continuous operator in thespace C[a, b] if the function K(s, t) is bounded on the square a � s

a � t � b and all points of discontinuity of the function K(s, t) lie on a finitenumber of curves

t = 10k(5), k = 1, 2. ... , n,

where the are continuous functions.REMAm. The distribution of points of discontinuity on the straight

lines s = constant is essential here. For example, for a 0, b = 1, the

§31] COMPLETELY CONTINUOUS OPERATORS 113

kernel

Ii for s <K(s, t) =

for s �has for points of discontinuity all the points of the square with s = Inthis case transformation (1) maps the function x(t) = 1 into a discontinuousfunction.

Proof of Theorem 1. We note that under the conditions of the theorem,integral (1) exists for arbitrary s in the closed interval a � s � b. Let M =sup K(s, t)

Ion the square a � s � b, a � t � b. Denote by G the set

of points (s, t) for which t — cok(s) < f/l2Mn holds for arbitrary k =1, 2, ... , n; denote by F the complement of G with respect to the squarea � s � b, a � t � b. Since F is closed and K(s, t) is continuous on F,there exists a ô such that for points (s', t), (s", t) in F for which -— < ô

the inequality

IK(s', t) — K(s", t)

I< f/3(b — a)

holds.Now let s' and s" be such that — I < ô. Then

1y(s') — y(s")

I � IK(s', t) — K(s", t) x(t) dt

can be evaluated by integrating over the sum of the intervals

It —

I< f/l2Mn,

t — I <(denote this sum by A) and over the complement B of A with respect tothe closed interval [a, b]. The length of A does not exceed f/3M. Therefore

L I K(s', t) — K(s", t)Ix(t) dt � (2f/3) x

The integral over B, obviously, may be estimated by

L I K(s', t) — K(s", t)I

x(t)Idt � (f/3) II x

Theref ore,

y(s') — y(s")I � x

We have proved the continuity of y(s) and the equicontinuity of thefunctions y(s) corresponding to the functions x(t) with bounded norm. Theuniform boundedness of y(s) corresponding to the x(t) with bounded norm

114 LINEAR OPERATOR EQUATIONS [CH. IV

follows from the inequalities

y(s)(�fThe Volterra Operator. If we assume that K(s, t) = 0 for t > s, then the

operator (1) takes on the form

(2) y(s) K(s, t)x(t)

We shall assume the function K(s, t) to be continuous for t � s. The validityof the hypothesis of Theorem 1 is easily verified for such a kernel. Conse-quently this operator is completely continuous.

We shall now establish several properties of completely continuousoperators.

THEOREM 2. If is a sequence of completely continuous operators on aBanach space E which converges in norm to an operator then the operatorA is completely continuous.

Proof. Let us consider an arbitrary bounded sequence of elements in E:< c. It is necessary to prove that the sequence

contains a convergent subsequence.The operator A1 is completely continuous and therefore we can select a

convergent subsequence from the sequence { A Let

(3) xi, x2, • ,

be the inverse images of the members of this convergent subsequence. Weapply the operator A2 to each member of the subsequence (3). Since A2 iscompletely continuous, we can again select such a convergent subsequencefrom the sequence { }, the inverse images of whose terms are, say,

(2) (2) (2)x1 ,x2 , ,

We apply the operator A3 to this sequence } and then in an analo-gous manner select a subsequence x2, • • , , and so on.

We now form the diagonal sequence(1) (2) (n)

Xi ,X2 , ,Xn ,

Each of the operators A1, A2, . .. , ... transforms this sequenceinto a convergent sequence. If we show that the operator A also transformsthis sequence into a convergent sequence, then this will also establish thecomplete continuity of the operator A. Let us evaluate the norm of thedifference — A4xm(m):

— II 1—

+A — Ak

(m) + II — AkXm(m)II

§31] COMPLETELY CONTINUOUS OPERATORS 115

We choose k so that A — Ak < f/(2c) and N such that the relation

I

— AkXm(m) < f/2holds for arbitrary n > N and in > N. (This is possible because the se-quence {A kXn } converges.) Then U — AXm(m)

1

< i.e. the se-quence is fundamental.

Since the space E is complete, this sequence converges; this proves thetheorem.

THEOREM 3. The adjoint operator of a completely continuous operator iscompletely continuous.

Proof. Let the operator A map the Banach space E into itself. Then theoperator A *, adj oint to A, maps the space E into itself. Inasmuch as anarbitrary bounded set can be enclosed in a sphere, the theorem will beproved if we prove that the image of the sphere belonging to the space E(i.e. the set is compact. It is obviously sufficient to carry out the prooffor the case of the unit sphere.

We introduce into the space E the auxiliary metric obtained by settingfor any two functionals g', g" in E,

(4) p(g', g") = sup {I g'(y) — g"(y) ; y E AS}.

Since the operator A is completely continuous by hypothesis, the image A Sof the unit sphere S is compact in the space E. If y E AS, then y (I

A IL Consequently if g E then g(y)I H III y II

A i.e. thefunctionals g(y) E are uniformly bounded in the sense of metric (4).

Further, we have

Ig(y') — g(y") III y' — y"

1 II — y"

i.e. the functionals g(y) E are equicontinuous. In accordance with Arzelà'stheorem, §17, this means that the set is compact in the sense of metric(4). But since

II— � sup {I g'(Ax) — g"(Ax) ; x E S},

the compactness of the set in the sense of metric (4) implies the compact-ness of the set in the sense of the original metric of the space E.

THEOREM 4. If A is a completely continuous operator and B is a boundedoperator, then the operators AB and BA are also completely continuous.

Proof. If M is a bounded set in E, then the set BM is also bounded.Consequently, the set ABM is compact and this means that the operatorAB is completely continuous. Further, if M is bounded, then AM is com-pact; but then in virtue of the continuity of B, the set BAM is also com-pact. This completes the proof of the theorem.

COROLLARY. A completely continuous operator A cannot have a boundedinverse in an infinite-dimensional space E.

116 LINEAR OPERATOR EQUATIONS [CH. IV

In fact, in the contrary case the identity operator I = AA' would becompletely continuous in E, which is impossible.

We shall now study the spectrum of a completely continuous operator.THEOREM 5. Every completely continuous operator A in the Banach space E

has for arbitrary p > 0 only a finite number of linearly independent charac-teristic vectors which correspond to the characteristic values whose absolutevalues are greater than p.

Proof. Let us assume that this is not the situation, i.e. let us assume thatthere exists an infinite number of linearly independent characteristic vec-tors (n = 1, 2, . . .), satisfying the condition

(5) =I

> p > 0, (n = 1, 2, ..Consider in E the subspace E1 generated by the vectors In this

subspace the operator A has a bounded inverse. In fact, vectors of theform

(6) x = + cE2X2 + + cxkXk

are everywhere dense in E1. For these vectors,

Ax = cEiXiXi + a2X2X2 + + cxkXkXk

and consequently,

(7) A'x = (ai/Xi)xi + (a2/X2)x2 + ... + (cxk/Xk)xk

and

A'x � (l/p) x

Consequently the operator A', defined for vectors of the form (6) bymeans of formula (7), can be extended by continuity (and consequentlywith preservation of norm) to all of E1. The operator A, being completelycontinuous in E, is completely continuous in E1 also. But according to thecorollary to the preceding theorem a completely continuous operator cannothave a bounded inverse in infinite-dimensional space. The contradictionthus obtained proves the theorem. It follows from this theorem that everynonzero characteristic value of a completely continuous operator has onlyfinite multiplicity and these characteristic values form a bounded set whichcannot have a single limit point distinct from the origin of coordinates.We have thus obtained a characterization of the point spectrum of acompletely continuous operator. It will follow from the results of the nextsection that a completely continuous operator cannot have a continuousspectrum.

§32. Linear operator equations. The Fredhoim theorems

In this section we shall consider equations of the form

(1) y=x—Ax,

§32] LINEAR OPERATOR EQUATIONS. FREDHOLM'S THEOREMS 117

where A is a completely continuous operator which maps a Banach space Einto itself. We shall show that for such equations a number of theorems arevalid which are analogous to the known theorems for systems of linearalgebraic equations. First of all the corresponding theory was developedby Fredholm for integral equations of the form

(2) = f(t) + f K(s,

t) are given continuous functions.It was established In the preceding section that the Fredholm operator

y(t) = f K(s, t)x(s) ds

is completely continuous; thus, equation (1) actually is a generalization ofthe Fredholm integral equation (2).

THEOREM 1. Let A be a completely continuous operator which maps theBanach space B into itself. If the equation y = x — Ax is solvable for arbi-trary y, then the equation x — Ax = 0 has no solutions distinct from the zerosolution.

Proof. Let us assume the contrary: suppose there exists an element x1 0such that x1 — Ax1 Tx1 = 0. Those elements x for which Tx = 0 forma linear subspace E1 of the space E. Let us denote by the subspaceconsisting of the elements which satisfy the condition = 0.

It is clear that the subspaces form a nondecreasing subsequence

E1çE2ç...We shall show that the equality sign cannot hold at any point of this

chain of inclusion relations. In fact, since by assumption x1 0 and theequation y = Tx is solvable for arbitrary y, we can find a sequence of ele-ments distinct from zero x1, x2, , , such that

Tx2 =

Tx3

=

The element belongs to the subspace for each n but it does not belongto the subspace In fact,

n n—i =Txi=0,but

= = •.. = Tx2 = x1 0.

118 LINEAR OPERATOR EQUATIONS [cii. IV

All the subspaces are linear and closed; therefore for arbitrary nthere exists an element Yn+i E such that

H Yn+i = 1 and p(Yn+i, �where p(yn+i, denotes the distance from Yn+i to the space i.e.

p(Yn+i, = ml{ 1

— x x E

[In fact, if we let p(xn+i, = a, then we can find an element E

such that II — < 2a; at the same time we clearly have —

= , = a. We can then set Yn+i = — —

Consider the sequence {Ayk }. We have (assuming p > q):

— Ayq = 1— (yq + — Tyq)

II �since + — Tyq E It is clear from this that the sequence tAyk}cannot contain any convergent subsequence, which contradicts the com-plete continuity of the operator A. The contradiction thus obtained provesthe theorem.

COROLLARY 1. If the equation y = x — Ax is solvable for arbitrary y,then it has a unique solution for each y, i.e. the operator I — A has aninverse in this case.

In fact, if the equation y = x — Ax had two distinct solutions for some y,say x1 and x — Ax = 0 would have a nonzero solution

— which contradicts Theorem 1.In the sequel it is convenient to consider together with the equation

y = x — Ax the equation h I — A*f which is adjoint to it, where A*is the adjoint of A and h, I are elements of the Banach space E, the conju-gate of E.

For the adjoint equation we can formulate the following result.COROLLARY 2. If the equation h = f — A*f is solvable for all h, then the

equation I — 0 has only the zero solution.This assertion is obtained from Theorem 1 if we recall that the operator

adjoint to a completely continuous operator is completely continuous(Theorem 2, §31) and a space conjugate to a Banach space is itself a Banachspace.

THEOREM 2. A necessary and sufficient condition that the equationy x — Ax be solvable is that the following condition be fulfilled: f(y) = 0for allf for whichf — A*f = 0.

Proof. 1. If we assume the equation y = x — Ax is solvable, thenf(y) = f(x) — f(Ax) = f(x) — A *f(x), i.e. f(y) = 0 for all f which satisfythe condition f — A*f = 0.

2. Now letf(y) equal zero for allf which satisfy the equationf — A*f = 0.

For each of these functionals f we shall consider the set L1 of elements for

§32] LINEAR OPERATOR EQUATIONS. FREDIIOLM'S THEOREMS 119

which f takes on the value zero. Then our assertion is equivalent to thefact that the set flL1 consists only of the elements of the form x — Ax, i.e.it is necessary for us to prove that an element yi which cannot be repre-sented in the form x — Ax cannot be contained in flL1. rFO do this weshall show that for such an element y' we can construct a functionalwhich satisfies the conditions

f1_A*fi=O.These conditions are equivalent to the following:

0, f1(x — Ax) = 0 for all x.

In fact,

(f' — (x) = fi(x) — A*fi(x) — f1(Ax) = — Ax).

Let G0 be the subspace consisting of all elements of the form x — Ax.Consider the subspace {G0 , i.e. the set of elements of the form z + ay1z E G0, and define the linear functional f' on this subspace by setting

f1(z + ayi) = a.

Extending this functional to the whole space E (which is possible byvirtue of the Hahn-Banach theorem) we do obtain a linear functionalsatisfying the required conditions. This completes the proof of the theorem.

COROLLARY. If the equation f — = 0 does not have nonzero solu-tions, then the equation x — Ax = y is solvable for all y.

We shall now establish the analogue of Theorem 2 for the adjoint equa-tion.

THEOREM 3. A necessary and sufficient condition that the equation

(3) h=f_A*fbe solvable is that h(x) = 0 for all x for which x — Ax 0.

Proof. 1. If the equation Ii f — A*f is solvable, then

h(x) = f(x) — A*f(x) = f(x — A x),

i.e.h(x) = Oifx — Ax = 0.2. Now let h(x) = 0 for all x which satisfy the equatioii x — Ax = 0.

We shall show that equation (3) is solvable. We shall construct the func-tional f on the set F of all elements of the form y = x — Ax by setting

(4) f(Tx) = h(x).

This equation in fact defines a linear functional. First of all we note thatthe value of the functional f is defined uniquely for each y because ifTx1 = Tx2 , then h(x1) = h(x2). It is easy to verify the linearity of the func-

120 LINEAR OPERATOR EQUATIONS [cm IV

tional (4). This functional can be extended to the whole space E. We obtainf(Tx) = T*f(x) = h(x),

i.e. this functional is a solution of equation (3).COROLLARY. If the equation x — Ax = 0 has 110 nonzero solutions, then

the equation I — = h is solvable for all h.The following theorem is the converse of Theorem 1.THEOREM 4. If the equation x — Ax = 0 has x = 0 for its only solution,

then the equation x — Ax = y has a solution for all y.Proof. If the equation x — Ax = 0 has only one solution, then by virtue

of the corollary to the preceding theorem the equation f — = h issolvable for all Ii. Then, by Theorem 1, Corollary 2, the equation f — A*f =0 has only the zero solution. Hence the Corollary to Theorem 2 impliesthe solvability of the equation y = x — Ax for arbitrary y.

Theorems 1 and 4 show that for the equation

(1) y=x—Axonly the following two cases are possible:

1) Equation (1) has a unique solution for each y, i.e. the operator I — Ahas an inverse.

2) The equation x — Ax 0 has a noiizero solutioll, i.e. the numberX = 1 is a characteristic value for the operator A.

It is clear that all the results obtained above for the equation y = x — Axcarry over automatically also to the equation y = x — XAx, where X is anarbitrary number. (For such an equation the adjoint equation will beh f — XA *f.) It follows that either the operator I — XA has an inverse orthat the llumber 1/X is a characteristic value of the operator A. In otherwords, in the case of a completely continuous operator, an arbitrary numberis either a regular point or a characteristic value, i.e. a completely con-tinuous operator has only a pOillt spectrum. The point spectrum of a com-pletely contiiiuous operator was studied in the preceding section (Theo-rem 5).

THEOREM 5. The dimension n of the space N whose elements are the solu-tions of the equation x — Ax = 0 is equal to the dimension of the subspace N*whose elements are the solvtions of the equation f — = 0.

Proof. We note first that ill virtue of the complete continuity of theoperators A and A*, the subspaces N and N* are finite-dimensional. Letx1, x2, , } be a basis for the subspace N and let , ... , } be

a basis for the subspace N*. According to the Hahn-Baiiach theorem on theextension of a lillear functional, n functionals c02, , can be con-structed which satisfy the COllditiOns

1 for i ==

0 for j j.

§32] LINEAR OPERATOR EQUATIONS. FREDHOLM'S THEOREMS 121

We now consider the fuiictioiials , , and construct for eachof them the corresponding subspace defined by the coiiditioii = 0.Since the functionals are linearly independent, the corresponding setsare distinct; and since each of the subspaces has iiidex 1, a relation ofthe form L, c is likewise impossible.

Let us consider the sets

The set consists of those elements which belong to all exceptIt is clear that the sets are disjoint. Since all the functionals 11, 12,

• . ., except assume the value zero on and since is a linear mani-fold, it is possible to choose a point in each M1 to satisfy the followingconditions:

1 for i =(5)

0 for i I.

After these auxiliary constructions we pass directly to the proof of theassertion of the theorem.

We assume first that ii > n.We construct the operator

R(x) = Ax +This operator is completely continuous since it is the sum of a finite numberof completely continuous operators. Further, let us consider the operator

W(x) = x — R(x) = x — Ax —

We shall show that if W(xo) = 0, then x0 = 0.

Let us apply the functional to W(xo):

(6) = — = —

Since, by hypothesis, satisfies the equation = 0, = 0 for allx (i = 1, 2, . . , Besides, = 0 since, by assumption, Wx0 = 0.Therefore from (6) it follows that

(7) 0 (1 = 1, 2, . n)

But then the equation Wx0 0 implies that Tx0 = 0, i.e. xo E N; sincethe elements x1, x2, , form a basis in N, we have

Xo = + a2x2 + . +

122 LINEAR OPERATOR EQUATIONS [CE. Iv

If we apply the functional coi to this equation and make use of equations(7), we see that — This and (7) imply that x0 = 0. We have shownthat the equation Wx = 0 has only the zero solution. From this it follows(Theorem 4) that the equation Wx = y is solvable for all y and, in particu-lar, for y

Let x be a solution of the equation Wx = Then, on the one hand,it follows from the fact that = 0 and from (5) that

— — coj(x)zjJ = 0;

on the other hand, according to (5), that

= 1.

The contradiction thus obtained proves that v n.Let us now assume that n > v. We construct the operator W* by setting

(8) W*f = T*f —

We shall prove that if = 0, then f = 0. We have

_ 0 (i 1, 2, •.. , n)

for all f since E N. Therefore, if 0, applying both sides of (8) towe obtain:

(9) = 0 (i = 1, 2, ... ,

Thus, the equation = 0 reduces to T*f = 0, i.e. f E N*. Conse-quently, f can be written in the form

+I3vfv.

If we apply both sides of this equation to the element we see that= From this and (9) it follows that f = 0. This implies that the

equation = g is solvable for all g.Let f be such that W*f Then, on the one hand,

(W*f) = (T*f) — = 0,

and on the other,

cov+i(xv+i) = 1.

We have thus arrived at a contradiction, which proves that n v. Sofinally, n = v. This completes the proof of the theorem.

LIST OF SYMBOLSPage Page

E 1 C[a,b] 171 C2[a, b] 19

SYMBOLS. DEFINITIONS. TH EOREMS 123

Page

1

U 1

fl 2

A\B . 2

AiXB

m 10

p(x, y) 16

R= (X,p) 16

it 17

20C 18

8(xo ,r)8[xo, r]O(x, e)[M]J(xo)

ii x

ii,

C

Inii xRR

LIST OF DEFINITIONSPage

Equivalence of sets 7Metric space 16Complete space 37Completion of a space 39Linear space 71Linear manifold 73Subspace 73Convex set and convex body 74Functional and linear functional 77Norm of a linear functional 78Index of a subspace of a Banach space 80Weakly convergent sequence of elements 90Weakly convergent sequence of linear functionals 93Linear operator 95Bounded operator 95Continuous operator 95Norm of a bounded linear operator. 97Sum of continuous linear operators . 97Product of continuous linear operators 98Inverse of an operator 98Generalized function 105Convergence of a sequence of generalized functions 107Derivative of a generalized function 108Completely continuous operator 112

LIST OF THEOREMSCHAPTER 1 Page

§2 Theorem 10§2 Theorem 2°

3

10

10

Page2323232364

64

71

18

22

72727482

8899

103

124 THEOREMS

Page§2 Theorem 30 6§4 Theorem (Nondenuinerability of the set of real numbers) 8§5 Theorem (Cardiiialitv of the set of all subsets of a set) 10§7 Theorem 1 14§7 Theorem 2 14§7 Theorem 3 14

ChAPTER II§9 Theorem 1 23§9 Theorem 2 23

Theorem 3. 23§9 Theorem 4 24§9 Theorem 5.. 25§10 Theorem 1. 27§10 Theorem 2. 27§10 Theorem 1' 28§10 Theorem 3. 28§10 Theorem 4. 28§10 Theorem 5 29§11 Theorem 1. 31§12 Theorem 1. 34§12 Theorem 2. 34§12 Theorem 3. 35§13 Theorem 1. 39§13 Theorem 2. 40§14 Theorem (Principle of contraction mappimmgs) 43§16 Theorem 1 52§16 Theorem 2. 53§17 Theorem 1 (Arzelà) 54§17 Theorem 2 (Peano) 56§18 Theorem 1. 58§18 Theorem 2. 58§18 Theorem 3. 58§18 Theorem 4. 58§18 Theorem 5. 59§18 Theorem 6. 60§18 Theorem 7 61

§19 Theorem 1 63§19 Theorem 2 63§19 Theorem 3 63§19 Theorem 2a§19 Theorem 3a 66§19 Theorem 4 (Arzelà's theorem for contimmimous functions defined on an arbitrary

compactum) 66§20 Theorem 1 69§20 Theorem 2 69§20 Theorem 3 70

CI-LAPTER III

§22 Theorem 1 75§22 Theorem 2 76

THEOREMS. BASIC LITERATURE 125

§22 Theorem 3 76§23 Theorem 1 77§23 Theorem 2 77§23 Theorem 3 80§24 Theorem 1 82§25 Theorem (Hahn-Banach) 86§27 Theorem 1 90§28 Theorem 1 94§28 Theorem 1' 95§29 Theorem 1 96§29 Theorem 2 97§29 Theorem 3 98§29 Theorem 4 98§29 Theorem 5 99§29 Theorem 6 99§29 Theorem 7 101§29 Theorem 8 102§29 Theorem 9 103

ADDENDUM TO CHAPTER III

Theorem 1 108Theorem 2 108

CHAPTER IV

§30 Theorem 1 111§31 Theorem 1 112§31 Theorem 2 114§31 Theorem 3 115§31 Theorem 4 115§31 Theorem 5 116§32 Theorem 1 117§32 Theorem 2 118§32 Theorem 3 119§32 Theorem 4 120§32 Theorem 5 120

BASIC LITERATUREALEKSANDROV, P. S.

Introduction to the General Theory of Sets and Functions, Moscow-Leningrad,Gosudarstv. Izdat. Tehn.-Teor. Lit., 1948.

BANACH, S.des Operations Linêaires, Warsaw, 1932 (Monografje Matematyczne, Vol.

I).FRAENKEL, A. A.

Abstract Set Theory, Amsterdam, North-Holland, 1953.Einleitung in die Mengenlehre, New York, Dover, 1946.

GRAVES, L. M.The Theory of Functions of Real Variables, 2nd edition, New York, McGraw-Hill,

1956.HALL, D. W. AND SPENCER, G. L.

Elementary Topology, New York, Wiley, 1955.

126 BASIC LITERATURE

HAUSDORFF, F.Mengenlehre, 3rd edition, New York, Dover, 1944.

HILLE, E.Functional Analysis and Semi-Groups, New York, AMS Coil. Pub., Vol. XXXI,

1948.JEFFERY, R. L.

The Theory of Functions of a Real Variable, Toronto, University of Toronto Press,1951.

KAMKE, E.Theory of Sets, New York, Dover, 1950.

KANT0R0vIcH, L. V.Functional Analysis and Applied Mathematics, Los Angeles, National Bureau of

Standards, 1952.KELLEY, J. L.

General Topology, New York, Van Nostrand, 1955.KURATOWSKI, K.

Introduction to the Theory of Sets and Topology, Warsaw, 1955 (Biblioteka Mate-matyczna).

L0vITT, W. V.Linear Integral Equations, New York, Dover, 1950.

LYUSTERNIK, L. A. AND SoBoLEv, V. I.The Elements of Functional Analysis, Moscow-Leningrad, Gosudarstv. Izdat.

Tehn.-Teor. Lit., 1951.NATANSON, I. P.

Theory of Functions of a Real Variable, New York, TJngar, 1955.NEWMAN, M. H. A.

Elements of the Topology of Plane Sets of Points, 2nd edition, Cambridge, 1951.RIE5z, F. AND SZ.-NAGY, B.

Functional Analysis, New York, Ungar, 1955.W.

General Topology, Toronto, University of Toronto Press, 1952.SILOV, G. E.

Introduction to the Theory of Linear Spaces, Moscow-Leningrad, Gosudarstv. IzdatTehn.-Teor. Lit., 1952.

WILDER, R. L.Introduction to the Foundations of Mathematics, New York, Wiley, 1952.

ZAANEN, A. C.Linear Analysis, New York, Interscience, 1953.

Adjoint— operator 103— equation 118Arzelà's theorem 54

Banach space 71Basis 28Bounded— functional 77— linear operator 95

variation 65B-space, see Banach space

Cantor set 32Cardinal number 9— — of a denumerable set 10— — of the continuum 10Center of a sphere 23Characteristic value of an operator 10Closed

interval 26— set 26— sphere 23Closure of a set 23Compact 51— in itself 58— metric space, see Compacturn— sets in a metric space 51Compactum 58Complement of a set 3Complete— inverse image 13— metric space 37Completely continuous operator 112Completion of a space 39Conjugate space 82Connected space 29Contact point 23Continuous— curve in a metric space 67— functional 63, 77— linear operator 95— mapping 33, 34, 36— spectrum of an operator 111Contraction mapping 43Convergence— of a sequence of points 24— of a sequence of linear functionals 93—, weak 90

INDEXConvex-- body 74

closure 76— set in a normed linear space 74Countable set 28Covering 29—, closed 29—, open 29Curves 68

of finite length 69

Deficiency, see IndexDenumerability 6Denumerable set 4, 7Derivative of a generalized function

106, 108Diagonal method 6Difference of sets 2Dirac function 79, 106Distance— between sets 26

function 16— to a set 26Domain of definition of a function 13

Element of a set 1€-neighhorhood 23€-net 51Equicontinuity 54Equivalence—— of sets 7

relation 12Equivalent— functions 68-— metrics 30— sequences 40

sets 7

Euclidean n-space 17Euler polygonal arc 56Everywhere dense set 25Extension of a linear functional 86 if.

Face, see k-dimensional fareFinite— intersection property 5S— set 4

Fredholm— integral equation 49— theorems 117 if.

127

128 INDEX

Function 13—, general concept of 13Functional 62, 77—, bounded 77-—, continuous 63, 77—, geometric interpretation of a 80—, linear 77Functions 13—, continuous 34—, equivalent 68—, generalized 105 if.— of bounded variation 65—, real 62—, semicontinuous 64Fundamental sequence 36

General position of points 76Generalized function 105Greatest lower bound 66

Hahn-Banach theorem 86Hausdorif— separation axiom 31— space 31

Height of a rational number 5Holder's inequality 20, 22Homeomorphic spaces 35Homeomorphism 35Hyperplane 81

Image 13Index 80Infinite set 4, 8Interior point 27Intersection of sets 2Inverse 99— image 13

— operator 98Irreflexive space 89Isolated point 24Isometric spaces 36

k-dimensional face 76Kernel 49

Least upper bound 66Length of a curve 65, 69Limit—, lower 64— of a sequence 24— point 24—. upper 64

Linearfunctional 77

— manifold 73— operator 95— operator equation 116 if.— space 71

Lipschitz condition 29, 43Lower limit 64

Mapping 13—, continuous 33, 34, 36—, contraction 43—, homeomorphic 35— into 14—, isometric 36

of sets 13—onto 14Metric— space 16

Metrizable space 31Minkowski's inequality 20, 22

n-dimensional simplex 76Nondenumerable set 5Norm 71— of a functional 78— of an operator 97Normed linear space 71, 98

One-to-one correspondence 4Open— set 27— sphere 23Operations on sets 1 if.Operator 95—, adjoint 103—-, bounded 95—, continuous 95—, inverse 98—, linear 95Oscillation 64

Partition into classes 11Peano's theorem 56Picard's theorem 47Plessner 90Point spectrum of an operator 111Polygonal arc 56Potency, see Cardinal numberPower, see Cardinal number

INDEX 129

Product of operators 98Proper subset 1

Quadratic metric 19

Radius of a sphere 23Range of variation of a function 13Real functions in metric spaces 62 if.Rectifiable curves 69Reflexive space 89Reflexivity 12Regular point 110Residual spectrum 111Resolvent 110

Schwartz 105

Schwarz inequality 17, 19Second— conjugate space 88— countability axiom 28Segment 74Semicontinuous function 64Set 1—, Cantor 32

—, closed 26—, compact 51

—, compact in itself 58—, convex 74

—, dense 25—, denumerable 4, 7—, derived 26

—, everywheredense 25—, finite 4—, infinite 4, 8—, open 27

—, nondenumerable 5

—, totally hounded 51—, void 1

Simplex 76Space—, complete 37—, conjugate 82—, connected 29—, Hausdorif 31

—, linear 71

—, metric 16

——, n-dimensional (Euclidean) 17—, n-dimensional fi0fl

17

—, n-dimensional 20—, normed 71

— of continuous functions C 18— of continuous functions CIa, bl 17, 72— of continuous functions C2[a, b] 19, 72-— of continuous functions with quad-

ratic metric 19-— of real numbers R' 16, 72

— of sequences c 72— of sequeuces 12 18, 72—— of sequences 22— of sequences 72—, reflexive 89

—, separable 25—, topological 30— with countable basis 28Spectrum—, continuous 111

— of an operator 110—, point 111

-—, residual 111

Subset 1proper 1

Subspace 73Sum— of operators 97— of sets 1, 2Symmetric difference of sets 2Symmetry 12

System of sets with finite intersectionproperty 58

Theorem on nested spheres 39Topological space 30Total— boundedness 51

— variation 64Totally bounded set 51Transitivity 12

Uniform boundedness of a family offunctions 54

Union 1Upper limit 64

Variation, see Total variationVertices of a simplex 76\Tolterra-— integral equation 50

operator 114

Weak convergence 90, 93Weak* convergence 93

Documents

A.N. Kolmogorov, S.v. Fomin-Elements of the Theory of Functions and Functional Analysis, Volume 1, Metric and Normed Spaces-GRAYLOCK PRESS (1963)