l.p. lebedev, i.i. vorovich, g.m. gladwell,-functional analysis_ applications in mechanics and...

Functional Analysis

SOLID MECHANICS AND ITS APPLICATIONSVolume 100

Series Editor: G.M.L. GLADWELLDepartment of Civil EngineeringUniversity of WaterlooWaterloo, Ontario, Canada N2L 3GI

Aims and Scope of the Series

The fundamental questions arising in mechanics are: Why?, How?, and How much?The aim of this series is to provide lucid accounts written by authoritative researchersgiving vision and insight in answering these questions on the subject of mechanics as itrelates to solids.

The scope of the series covers the entire spectrum of solid mechanics. Thus it includesthe foundation of mechanics; variational formulations; computational mechanics;statics, kinematics and dynamics of rigid and elastic bodies: vibrations of solids andstructures; dynamical systems and chaos; the theories of elasticity, plasticity andviscoelasticity; composite materials; rods, beams, shells and membranes; structuralcontrol and stability; soils, rocks and geomechanics; fracture; tribology; experimentalmechanics; biomechanics and machine design.

The median level of presentation is the first year graduate student. Some texts are mono-graphs defining the current state of the field; others are accessible to final year under-graduates; but essentially the emphasis is on readability and clarity.

For a list of related mechanics titles, see final pages.

Functional AnalysisApplications in Mechanics and Inverse Problems

Edition

by

L.P. LEBEDEV

Professor, Department of Mechanics and Mathematics,Rostov State University, Russia & Department of Mathematics and Statistics,National University of Bogota, Colombia

I.I. VOROVICH †

Professor, Department of Mechanics and Mathematics,Rostov State University, RussiaFellow of the Russian Academy of Sciences

and

G.M.L. GLADWELL

Distinguished Professor Emeritus,Department of Civil Engineering, University of Waterloo, CanadaFellow of the Royal Society of Canada,and Fellow of the American Academy of Mechanics

KLUWER ACADEMIC PUBLISHERSNEW YORK, BOSTON, DORDRECHT, LONDON, MOSCOW

eBook ISBN: 0-306-48397-1Print ISBN: 1-4020-0667-5

©2003 Kluwer Academic PublishersNew York, Boston, Dordrecht, London, Moscow

Print ©2002 Kluwer Academic Publishers

All rights reserved

No part of this eBook may be reproduced or transmitted in any form or by any means, electronic,mechanical, recording, or otherwise, without written consent from the Publisher

Created in the United States of America

Visit Kluwer Online at: http://kluweronline.comand Kluwer's eBookstore at: http://ebooks.kluweronline.com

Dordrecht

In memory of I.I. Vorovich

Table of Contents

1 Introduction1.11.21.3

Real and complex numbersTheory of functionsWeierstrass’ polynomial approximation theorem

2 Introduction to Metric Spaces2.12.22.32.42.52.62.72.82.92.102.112.12

PreliminariesSets in a metric spaceSome metric spaces of functionsConvergence in a metric spaceComplete metric spacesThe completion theoremAn introduction to operatorsNormed linear spacesAn introduction to linear operatorsSome inequalitiesLebesgue spacesInner product spaces

3 Energy Spaces and Generalized Solutions3.13.23.33.43.53.63.7

The rodThe Euler–Bernoulli beamThe membraneThe plate in bendingLinear elasticitySobolev spacesSome imbedding theorems

4 Approximation in a Normed Linear Space4.14.24.34.44.5

Separable spacesTheory of approximation in a normed linear spaceRiesz’s representation theoremExistence of energy solutions of some mechanics problemsBases and complete systems

103106110113

9999

6565747883858890

19192527293032354045485158

119

14

viii Table of Contents

4.64.74.84.9

Weak convergence in a Hilbert spaceIntroduction to the concept of a compact setRitz approximation in a Hilbert spaceGeneralized solutions of evolution problems

5 Elements of the Theory of Linear Operators5.15.25.35.45.55.6

Spaces of linear operatorsThe Banach–Steinhaus theoremThe inverse operatorClosed operatorsThe adjoint operatorExamples of adjoint operators

6 Compactness and Its Consequences6.16.26.36.46.56.6

Sequentially compact compactCriteria for compactnessThe Arzelà–Ascoli theoremApplications of the Arzelà–Ascoli theoremCompact linear operators in normed linear spacesCompact linear operators between Hilbert spaces

7 Spectral Theory of Linear Operators7.17.27.37.47.5

The spectrum of a linear operatorThe resolvent set of a closed linear operatorThe spectrum of a compact linear operator in a Hilbert spaceThe analytic nature of the resolvent of a compact linear operatorSelf-adjoint operators in a Hilbert space

8 Applications to Inverse Problems8.18.28.38.48.5

Well-posed and ill-posed problemsThe operator equationSingular value decompositionRegularizationMorozov’s discrepancy principle

Index

120126128132

141141144147152157162

167167171174178183189

195195199201208211

219219220226229234

241

Preface

This book started its life as a series of lectures given by the second authorfrom the 1970’s onwards to students in their third and fourth years in theDepartment of Mechanics and Mathematics at Rostov State University. Forthese lectures there was also an audience of engineers and applied mechanicistswho wished to understand the functional analysis used in contemporary researchin their fields. These people were not so much interested in functional analysisitself as in its applications; they did not want to be told about functional analysisin its most abstract form, but wanted a guided tour through those parts of theanalysis needed for their applications.

The lecture notes evolved over the years as the first author started to makemore formal typewritten versions incorporating new material. About 1990 thefirst author prepared an English version and submitted it to Kluwer AcademicPublishers for inclusion in the series Solid Mechanics and its Applications. Atthat state the notes were divided into three long chapters covering linear andnonlinear analysis. As Series Editor, the third author started to edit them. Therequirements of lecture notes and books are vastly different. A book has tobe complete (in some sense), self contained, and able to be read without thehelp of an instructor. In the end these new requirements led to the book beingentirely rewritten: an introductory chapter on real analysis was added, the or-der of presentation was changed and material was added and deleted. The lastchapter of the original notes, on nonlinear analysis, was omitted altogether, theoriginal two chapters were reorganized into six chapters, and a new Chapter 8on applications to Inverse Problems was added. This last step seemed natural:it covers one of the interests of the third author, and all the functional analysisneeded for an understanding of the theory behind regularization methods for In-verse Problems had been assembled in the preceding chapters. In preparing thatchapter the third author acknowledges his debt to Charles W. Groetsch and hisbeautiful little book Inverse Problems in the Mathematical Sciences. Chapter 8attempts to fill in (some of) the gaps in the analysis given by Groetsch.

Although the final book bears only a faint resemblance to the original lec-ture notes, it has this in common with them: it aims to cover only a part offunctional analysis, not all of it in its most abstract form; it presents a ribbonrunning through the field. Thus Chapter 2 introduces metric spaces, normedlinear space, inner product spaces and the concepts of open and closed sets andcompleteness. The concept of a compact set, which was introduced in Chapter 1for real numbers, is not introduced until Chapter 4, and not discussed fully untilChapter 6.

Chapter 3 stands somewhat apart from the others; it illustrates how theidea of imbedding, appearing in Sobolev’s theory, arises in continuum analysis.From Chapter 2 the reader may pass directly to Chapter 4 which considersthe important problem of approximation, and introduces Riesz’s representationtheorem for linear functionals, and the concept of weak convergence in a Hilbert

x

space. In keeping with the aim of following a ribbon through the field, thepresentation of the concepts of weak convergence, and of the adjoint operatorin Chapter 5, are limited to inner product spaces.

The theory of linear operators in discussed, but not covered (!), in Chap-ters 5 and 7. The emphasis here is on the parts of the theory related to compactlinear operators and self-adjoint linear operators,

It is the authors’ fervent wish that readers will find the book enjoyableand instructive, and allow them to use functional analysis methods in theirown research, or to use the book as a jumping board to more advanced and/orabstract texts.

The authors acknowledge the skill and patience of Xiaoan Lu in the prepa-ration of the text.

L.P. LebedevI.I. Vorovich

G.M.L. Gladwell

July, 1995

Preface to the second edition

It is with sadness that we must report the death of Professor I.I. Vorovich in2001. Professor Vorovich, who provided the initial stimulus for the book, had ascientific career spanning more than five decades; his books on Elasticity Theory,Contact Problems, and Nonlinear Theory of Shallow Shells laid the foundationfor the work of generations of researchers. We, and his many colleagues andcollaborators worldwide, miss his wise counsel and his store of knowledge.

This new edition incorporates many small corrections passed on to us overthe years. We are particularly grateful for the contribution of Professor MichaelCloud, of Lawrence Technological University; he has not only spotted manyneeded corrections, but also has played an invaluable role in resurrecting theoriginal TeX files of the book, which were in danger of being lost in the inter-continental collaboration.

L.P. LebedevG.M.L. Gladwell

January, 2002

1. Introduction

Everyone writes as he wants to, and as he can.

Anton Chekhov, The Seagull

1.1 Real and complex numbers

A book must start somewhere. This is a book about a branch of applied math-ematics, and it, like others, must start from some body of assumed knowledge,otherwise, like Russell and Whitehead’s Principia Mathematica, it will have tostart with the definitions of the numbers 1, 2 and 3. This first chapter is in-tended to provide an informal review of some fundamentals, before we begin inearnest in Chapter 2.

We will start with the positive integers 1,2,3, ···; zero, 0, and the negativeintegers –1, –2, –3,···. From these we go to the rational numbers of the form

where are integers (we can take but However wesoon find that having just rational numbers is unsatisfying; there is no rationalnumber such that For suppose there were such a number.If had a common factor (other than ±1) we could divide that out, andarrange that had no common factor – we say they are mutually prime.Our supposition is that But if is even, so isThus for some integer and so that Therefore

is even, and hence is even. Thus have a common factor, 2, contraryto hypothesis. This contradiction forces us to conclude that there is no rationalnumber such that

However, we can find a sequence of rational numbers whosesquares get closer and closer to 2 as increases. Note: Sequence always meansan infinite sequence. For let be obtained from using the formula

If we start from we find the sequence

We note that

2 . 1. Introduction

so that the are all too big, i.e. for and

When Thus

so that, given any small quantity we can find an integer N such thatfor all Thus the terms in the sequence get closer

and closer to 2; we write this

and say the sequence converges to 2. We can make a formal definitionof a convergent sequence:

Definition 1.1.1 The sequence which we write as convergesto a if, given we can find (depending on such that, for allwe have

Notice that at present all the numbers in this definition, i.e.must be interpreted as rational numbers.

Problem 1.1.1 Show that a sequence cannot converge to two different limits,i.e. that a convergent sequence (one that has one limit) has a unique limit.

In our example the sequence converges to 2, but there is no (ratio-nal) number to which the sequence converges. However, we note that themembers of the sequence get closer to each other. For

so that asWe can also show that we can make the difference of any two members of

the sequence as small as we please merely by taking the indices large enough.For suppose then

We use (1.1.3) and (1.1.4). Since we can replace (1.1.3)by the weaker statement

1.1 Real and complex numbers 3

Thus

and so on, so that

We can make this as small as we please by taking large enough. This leadsus to the definition of a Cauchy sequence, after Augustin-Louis Cauchy (1789–1857).

Definition 1.1.2 A sequence is said to be a Cauchy sequence if, giventhere exists (depending on such that if then

This means that the sequence (1.1.2) is a Cauchy sequence. Clearly a con-vergent sequence is a Cauchy sequence. For if converges to then, given

we can find such that if thenThus

The converse is false, because the sequence (1.1.2) is a Cauchy sequence whichdoes not converge, to a rational number, and at this stage this is the only kindof number we have.

What we would like to do now is to extend the definition of a number sothat every Cauchy sequence is a convergent sequence. Think about the example.We would like to define a ‘number’ We could associate this ‘number’ withthe sequence (1.1.2). But there are other sequences, for which the sequences ofsquares converge to 2, for example the sequences given by (1.1.1) which startfrom other rational values of say or or the truncateddecimal sequence

1,1.4,1.41,1.414, etc.

We must associate with all these sequences. This brings us to the conceptof an equivalence class of Cauchy sequences.

Definition 1.1.3 Two Cauchy sequences are said to be equivalent,

Problem 1.1.2 Show that is equivalent to if is equivalentto then is equivalent to are equivalent and

are equivalent, then so are

This justifies the use of the term equivalent; equivalent means essentiallyequal. The symbol has the properties if then ifand then

4 1. Introduction

Now we can introduce

Definition 1.1.4 With any Cauchy sequence we can associate all theCauchy sequences equivalent to it. We call this class an equivalence class,

A particular Cauchy sequence in is called a representative of theclass.

Equivalence classes divide all the Cauchy sequences into separate groups;a sequence cannot belong to two different equivalent classes. In fact wehave

Problem 1.1.3 If belongs to different equivalence classes re-spectively, then there is an and an N such that whenever

This means that two different equivalence classes are separated from eachother in the sense stated in this problem. Moreover, we can show that if

belong to different equivalence classes then there is an N such that,for all either or In the former case we will write

in the latter Thus equivalence classes, like (rational) numbers,can be ordered: if are two classes then either or orThis leads us to

Definition 1.1.5 A real number is an equivalence class of Cauchy sequencesof rational numbers.

With this definition, any rational number is associated with theequivalence class containing the trivial (or so-called stationary) Cauchy sequence

In a sense therefore the set of real numbers, denoted by includesall the rational numbers. With suitable definitions we can treat real numbersjust like rational numbers, we can add, subtract, multiply and divide with them,thus

Problem 1.1.4 Show that if the Cauchy sequences are representa-tives of and then and are Cauchysequences, provided that for the last named We call the classes in whichthese sequences lie, and respectively.

Having defined real numbers we can think about sequences of real numbers,and in particular convergent sequences and Cauchy sequences of real numbers,and we can show that every Cauchy sequence of real numbers is a convergentsequence, converging to a real number. The definitions of a convergent sequenceor a Cauchy sequence of real numbers are precisely Definitions 1.1.1, 1.1.2, with

being interpreted as real numbers. We describe this by saying thatthe set of real numbers, is complete. We often think of real numbers as points


on a straight line, with negative real numbers on the left of zero, and positivenumbers on the right. Saying that is complete means that the line has noholes; every number, like or appears in

We now need to describe sets of points such as a finite set 1/2,0, –3/4;an open interval, the set of numbers satisfying written aclosed interval, the set of numbers satisfying written Wehave used the terms open and closed, but we need to define what we mean byan open set or a closed set. The essential feature of an open interval is not thatthe ends are excluded, but rather that any is itself the center ofan open interval entirely contained in Thus in (0,1) the pointis the center of the interval (0.985,0.995) contained in (0,1); clearly any pointin (0,1) may be so viewed. On the other hand [0,1] is not open, because thepoints 0 and 1 in [0,1] are not the centers of open intervals entirely containedin [0,1]. This leads to

Definition 1.1.6 A set S of real numbers (we say a set is open ifevery point in S is the center of an open interval lying entirely in S.

The concept of closed is linked to convergence. If S is a set of points andis a convergent sequence of points in S then the limit of the sequence may

or may not be in S. Thus we have

Definition 1.1.7 A set is said to be closed if every convergent sequenceconverges to a point in S.

Under this definition a closed interval is closed. For suppose thatand If then and

for all which contradicts the statement that Thus andsimilarly so that On the other hand the open interval (0,1) isnot closed because the sequence 1/2, 1/3, 1/4, • • • in (0,1) converges to 0, whichis not in (0,1).

Problem 1.1.5 Show that if S is a closed set in then every Cauchysequence converges to a point

Definition 1.1.8 A set is said to be bounded if there is a numberM such that all satisfy Such a set is contained in the closedinterval [ – M , M].

Problem 1.1.6 Show that a Cauchy sequence is bounded (NoteProblem 1.1.5 states that for sequences the terms Cauchy sequenceand convergent sequence are synonymous).

6 1. Introduction

Problem 1.1.7 Suppose converges to Show that any subsequenceof also converges to Conversely, show that if is a

convergent sequence, and a subsequence converges to then mustconverge to

So far we have defined three terms relating to a set closed, openand bounded. Now we introduce

Definition 1.1.9 A set is said to be compact if every sequenceS contains a subsequence, converging to a point

The fundamental theorem on compactness is the Bolzano–Weierstrass the-orem, named after Bernard Bolzano (1781–1848) and Karl Theodor WilhelmWeierstrass (1825–1897):

Theorem 1.1.1 A set is compact iff (if and only if) it is closed andbounded.

Proof. We first prove that if it is compact, then it is closed and bounded.Suppose is a convergent sequence. Since S is compact, the sequence

contains a subsequence which we write which convergesto some But therefore (Problem 1.1.7) the whole sequence mustconverge to S is closed. If S were not bounded, we could find a sequence

such that and this sequence would haveno Cauchy subsequence, and therefore no convergent subsequence, contrary tothe supposition that S is compact; therefore S is bounded.

Now we will show that if S is closed and bounded then it is compact. SinceS is bounded, it may be contained in an interval I = [ – M , M]. Let bea sequence in S. We use the method of bisection. Bisect I into two closedintervals; one half, must contain an infinity of members of the sequence;choose one of them and call it bisect one half, must contain aninfinity of numbers of the sequence; choose one of them, with andso on. The sequence obtained in this way is a Cauchy sequence. Since

and is complete, this Cauchy sequence is a convergent sequence,convergent to this limit will be in S because S is closed.

The Bolzano–Weierstrass theorem relates to a set which is closed andbounded. What can be said about a set which is just bounded? If S is notclosed we may close it by adding to it all the limit points of convergent se-quences We have

Definition 1.1.10 The closure of a set S is the set obtained by adding toS the limit points of all convergent sequences


If S is bounded, then every sequence in S will then contain a convergentsequence with a limit This limit will be in

Note that any set S consisting of a finite set of real numbersis closed, because its only limit points are and these are all in S.Since it is bounded it is compact. We note that in any sequence atleast one of say must appear an infinity of times; this willprovide the convergent subsequence converging to

A finite set of real numbers has a greatest and a least, written

An infinite set S of real numbers, even if it is bounded, may have neither a maxnor a min. (An example is provided by the set 0,1/2, –1/2, 2/3, –2/3,…) Wemust therefore proceed carefully. A set is said to be bounded above by

if for all it is bounded below by if for all Wemay adapt the method of bisection used in the Bolzano–Weierstrass theoremto show that a set which is bounded above has a least upper bound orsupremum, written

with the properties

Moreover there is a sequence which converges to M. For letand write There is at least one point, in ChooseBisect and denote left and right hand halves by and respectively.Choose as follows:

Choose Note that it may happen that Bisect intoand and take

Choose and so on. We may verify that the sequenceobtained in this way is a sequence which converges to M. Note that if we applythis process to a finite set of numbers then the terms in the sequence willeventually all be equal, to the maximum of the numbers in the set. This mayhappen with an infinite set which has a greatest member.

We may show similarly that if S is bounded below then it has a greatestlower bound or infimum, written

8 1. Introduction

with the properties

As with the sup, we may construct a sequence of members of S which convergesto inf S.

Definition 1.1.11 A sequence is said to be monotonically in-creasing (decreasing) if for It issaid to be strictly monotonic if the inequality is strict.

Problem 1.1.8 Adapt the argument used in the Bolzano–Weierstrass theoremto prove that a monotonically increasing sequence that is boundedabove by converges to a limit Similarly, if it is monotonically decreasingand bounded below by a it converges to a limit

So far we have discussed only real numbers, which we intuitively place on thereal line, However, for many purposes real numbers are inadequate, we needcomplex numbers of the form where are real andi.e. The set of all such numbers we call We can consider complexsequences in which each member of the sequence is a complex number. Forthese we write For complex sequences the termsappearing in Definitions 1.1.1, 1.1.2 must be interpreted as moduli of complexnumbers. Thus if and

With this change, the definition of a closed set remains as in Defini-tion 1.1.7 and the Bolzano–Weierstrass theorem still holds. In Definition 1.1.6the open interval must be replaced by an open disk. The open disk of radiusabout is the set of point satisfying

We can go further and consider points in a plane, i.e. in two dimensions; orin space, three dimensions; or even in N dimensions. The set of all N-tuples ofreal numbers is called the set of all N-tuples of complexnumbers is called Note that for most purposes we can treat as apoint in is specified by N complex numbers, i.e. 2N real numbers. We canthen generalize all that we have said about closed, open, bounded and compactsets provided that we interpret as the Euclidean distance be-tween the points with coordinates and in

Definition 1.1.12 The Euclidean distance between andin is

1.2 Theory of functions 9

In the definition of an open set, interval must now be replaced by open ball,according to

Definition 1.1.13 The open ball with center and radius a in is theset

Definition 1.1.14 A set is open if every point of S is the center ofan open ball lying entirely in S.

We must also generalize Definition 1.1.1 to give

Definition 1.1.15 The sequence converges to a if, givenwe can find such that for all we have

Most importantly, with these generalizations, the Bolzano–Weierstrass the-orem holds for sets (and also for thus we may state

Theorem 1.1.2 A set is compact iff it is closed and bounded.

Note: iff is an abbreviation for if and only if.

1.2 Theory of functions

To describe the behavior or a change in the state of a body in space, we usefunctions of one or more variables. Displacements, velocities, loads and tem-peratures can be functions of points of a body, and of time. We need somedefinitions and results from Calculus, based on the concepts we introduced in§1.1.

Definition 1.2.1 We use the symbol and the term domain to denote anon-empty open set (Definition 1.1.14) in

Definition 1.2.2 A rule which assigns a unique real (complex) number to everyis said to define a real (complex) function on Strictly we

distinguish between a function, and its value, at a point

Definition 1.2.3 The support of in written supp is defined as

10 1. Introduction

where the overbar means closure in as in Definition 1.1.10 . The functionis said to have compact support if supp is bounded, i. e. contained

in some ball in it is said to have compact support in if supp

Note that the support of a function is always closed, by definition.The reader is aware of the idea of a continuous function of a single variable,

we need to extend this to functions for

Definition 1.2.4 Let be a function on Let The functionis said to be continuous at if, given we can find dependingon such that if then The functionis said to be continuous on if it is continuous for every

Problem 1.2.1 Show that is continuous on (0,1), but not boundedon (0,1). Thus, there is no number M such that for all

One of the basic results of Calculus concerns functions that are continuouson the closure of a bounded domain i.e. a compact region.

Theorem 1.2.1 A real valued function that is continuous on a closed andbounded, i. e. a compact region is bounded, and achieves its supre-mum and infimum in

Proof. Suppose were not bounded. Then there is a sequencesuch that Since and is compact, contains asubsequence converging to a point The function is contin-uous at Thus we can find such that if and then

Choose N such that implies theni.e. This contradicts so

that must be bounded.Thus has a supremum M and infimum As shown in § 1.1, there

is a sequence such that The sequencecontains a subsequence converging to Choose Sinceis continuous at we can find such that if and then

Choose such a and then choose N so that if thenThen for all such we have

But is arbitrary, and may be taken arbitrarily large, so that Onthe other hand so that and achieves its supremumon We can prove in a similar fashion that it achieves its infimum.

The theorem states that there exist M such that


Moreover assumes its supremum and infimum in That is, there aresuch that

(we can thus say that has a maximum value, and a minimum value,This means in particular that there is an (either or ) such

that

Thus if is continuous and has compact support, it is bounded on theclosed set if and is outside G, so that isbounded on A function that is continuous on a set which is not compact maynot achieve its supremum and infimum. For example, inachieves its supremum, but not its infimum.

Definition 1.2.5 A function is said to be uniformly continuouson if, given we can find such that if andthen

When is uniformly continuous we can find a number for Defini-tion 1.2.4 which will work for every

Theorem 1.2.2 If is continuous on a compact region then itis uniformly continuous on

Proof. Suppose were not uniformly continuous on According to Defini-tion 1.2.5 this means that there is an such that for every we canfind such that while For such anwe can take and find such that while

This will give two sequences Since is compact, each ofthese sequences will contain subsequences converging torespectively, and

so that But is continuous at so that

which contradicts (1.2.1).

12 1. Introduction

Theorem 1.2.2 states that if is continuous on a compact regionthen it is uniformly continuous on but if the region is not compact then

there may be functions which are continuous, but not uniformly continuous.

Problem 1.2.2 Show that is continuous, but not uniformly con-tinuous, on (0,1).

A function that is uniformly continuous on a bounded domain is bounded,provided that the domain is sufficiently regular that, given one can gofrom one point to any other point in a finite number of steps of length lessthan For if it is uniformly continuous on we can find such that

implies If we can get from to any otherpoint in a (fixed) number of steps, of lengths

then

This proof is valid if the domain is a connected finite union of star-shapeddomains.

A function which is uniformly continuous on an unbounded domain neednot be bounded: is uniformly continuous but unbounded on

Theorem 1.2.3 If is bounded and uniformly continuous onthen it has a unique, bounded, continuous extension (or continuation)

to the closure of

Note that we do not demand that be bounded, but we do demand thatbe bounded. By continuous extension we mean that we can find a functiondefined on such that

and is continuous on

Proof. Let and let be a sequence converging to thenis a Cauchy sequence (in or converging to a number which we

denote bySuppose and are sequences converging to

respectively. Choose Since is uniformly continuous in we maychoose such that implies Now choose

large enough that:


and suppose that Then

so that and

Corollary The extension is uniformly continuous on and thus, byTheorem 1.2.1, is bounded on and assumes its infimum and supremum, and

Proof. Clearly implies On the other hand, will (byTheorem 1.2.1) assume its supremum at some point There is a sequence

converging to for which so that impliesthus

We conclude this section by proving Weierstrass’ theorem on uniformlyconvergent sequences of uniformly continuous functions.

Let be a sequence of functions on For any partic-ular we may consider the sequence For this valueof the sequence will be a Cauchy sequence if, given we can find(depending on and such that impliesSimilarly, for a particular value of the sequence is said to convergeto if given we can find (depending on and such that

implies (Because and are complete anda Cauchy sequence is a convergent sequence.) In these statements maydepend on as well as on If, for any it is possible to choose onedepending on alone, which will work for all then the sequence is saidto be a uniformly Cauchy sequence, or to converge uniformly to

Weierstrass’ theorem is

Theorem 1.2.4 A uniformly Cauchy sequence of functions which areuniformly continuous on a compact region converges to a uniformlycontinuous function

Proof. For any is a Cauchy sequence of real (or complex)numbers and, since (or is complete, defines a real (or complex) number

We must show that is uniformly continuous inChoose and then choose such that for all and all

we have

14 1. Introduction

Letting we find

Thus if and then

Each function is uniformly continuous in so that we may find suchthat implies

Thus for such we have

1.3 Weierstrass’ polynomial approximation theorem

This is the fundamental theorem

Theorem 1.3.1 Any function which is uniformly continuous on a closed andbounded region may be uniformly approximated arbitrarily closely bya polynomial.

This theorem states that, given we can find a polynomial suchthat

We will prove this result only for N = 1, i.e. for functions of one real variable,and suppose for simplicity that Thus we have

Theorem 1.3.2 A function which is uniformly continuous on [0,1] maybe uniformly approximated arbitrarily closely by a polynomial.

Proof. First we introduce the polynomials

where

We will need some identities relating to the to obtain them we use thebinomial expansion

1.3 Weierstrass’ polynomial approximation theorem 15

The first identity is obtained by putting then

Now differentiate (1.3.1) twice w.r.t. to obtain

In each of these, put multiply the first by and the second byto obtain

We now combine these to give

which is the identity that we will use in the following analysis.To prove that we can approximate uniformly by a polynomial, we

shall actually construct one, the Bernstein polynomial, due to Serge Bernstein(1880–1968),

Since is uniformly continuous on [0,1] it is bounded (Theorem 1.2.1) sothat there is an such that for all Chooseand then choose such that implies Now

For any given and there will be some values of for whichand the remaining values for which we distinguish these groupsby and and write

16 1. Introduction

Using the uniform continuity we can write

and using we can write

Now we use the inequality to give

and hence

which, with the identity (1.3.2) yields

since when Combining (1.3.3)–(1.3.5) we find

so that to make we need only take

We may use this theorem to prove

Theorem 1.3.3 A uniformly continuous periodic function of periodmay be uniformly approximated arbitrarily closely by a trigonometric polynomial

The theorem states that, given we may find such that

We note that since are both periodic with period the sup overthe whole real line is the sup over

1.3 Weierstrass’ polynomial approximation theorem 17

Proof. The functions

are both even functions of period With the substitution forwe can construct two functions given by

which will be continuous in [–1,1]. Given we can, by Theorem 1.3.1,find polynomials such that

for In terms of the variable these are

and we note that are trigonometric polynomials of the form(1.3.6) (with the These inequalities hold for but thereforefor all since all the functions involved are even functions of period

Equations (1.3.7) give

so that, on writing we find

and we note that is a trigonometric polynomial. We now apply exactly thesame procedure to the function and find a trigonometricpolynomial such that

Both inequalities, (1.3.8) and (1.3.9), hold for all t; on replacing t byin (1.3.9) we find

By combining (1.3.8), (1.3.10), and writing

we find

is a trigonometric polynomial, and the inequality holds for all

18 1. Introduction

Synopsis of Chapter 1: Numbers and Functions

Sequences of numbers

convergent to Definition 1.1.1

Cauchy sequence: Definition 1.1.2

Real numbers: equivalence classes of Cauchy sequences of rational numbersDefinition 1.1.5

Sets

open: every point is an interior point Definition 1.1.6

closed: contains all its limit points Definition 1.1.7

compact: every sequence contains a convergent subsequenceDefinition 1.1.9

Functions

continuous: Definition 1.2.4

uniformly continuous: Definition 1.2.5

Theorem 1.2.2. continuous on compact uniformlycontinuous

Weierstrass’ Theorem on: uniform convergence Theorem 1.2.4

uniform approximation by a polynomial Theorem 1.3.1

uniform approximation by a trigonometric Theorem 1.3.3

2. Introduction to Metric Spaces

And you must note this: if God exists and if He really did createthe world, then, as we all know, He created it according to thegeometry of Euclid and the human mind with the conception ofonly three dimensions in space.

Fyodr Dostoevsky, The Brothers Karamazov

2.1 Preliminaries

We have defined the symbols and as the sets of real and complex numbers,respectively. We can specify the position of a point in three-dimensional spaceby its coordinates in some Euclidean frame. We write

and say is in which we write The Euclideandistance between two points is

Notice that we could use the notation x to indicate that is not a number, buta triplet of numbers. We will not do this because we want the reader to get usedto the idea that the point or vector is the fundamental entity; its coordinates

are secondary. Later we shall even use to denote differentpoints if we have to specify their coordinates we will use notations such as

for the coordinates of The context will (we hope) make theusage clear!

We can generalize the idea of Euclidean space by defining an N -dimensionalspace consisting of vectors We define the Euclideandistance between by

The basic procedure in functional analysis is that we take a particular con-cept, for instance Euclidean distance, list (some of) its essential qualities, and

20 2. Introduction to Metric Spaces

then introduce an abstract concept which possesses these qualities. The essen-tial qualities of Euclidean distance are that distance is real-valued; the distancebetween two different points is positive; the distance between and is thesame as the distance between and distance satisfies the triangle inequality:the sum of the lengths of two sides of a triangle is always greater than the third.(There can be equality only when the triangle degenerates into a straight line.)

The process of abstraction leads us to the concept of a metric forwith the properties:

Here and hereafter iff denotes if and only if. D1 is called the axiom of posi-tiveness; D3 states that is reflexive; D4 is called the triangle axiom. Wehave

Definition 2.1.1 A real valued function defined for iscalleda metric for if it satisfies D1–D4.Note that strictly we distinguish between a metric and its value for

It is clear that given in (2.1.1) satisfies D1-D3. When N = 3, D4is the familiar triangle inequality in 3-D geometry; it is quite difficult to proveit for general N (See § 2.10.)

There is another metric in namely

This clearly satisfies D1–D3. To see that it satisfies D4 we note that

Thus

We need

Definition 2.1.2 Two metrics and are said to be equivalentmetrics in if there exist two positive constants independent of

such that

for any in

2.1 Preliminaries 21

Problem 2.1.1 Show that the metrics and given in (2.1.1) and (2.1.2)respectively, are equivalent.

Now we formally state the generalization of the concept of convergencediscussed in Chapter 1

Definition 2.1.3 A sequence is said to converge to underthe metric if, given there exists M such that for all we have

We say in and have

Note that we will use the single arrow for convergence of real and complexnumbers. We will use the double arrow for the type of convergence, justdefined, of a sequence in a metric space. We will later call this convergencestrong convergence to distinguish it from another kind of convergence, weakconvergence, which we will introduce in Chapter 5, and will denote by a singlearrow, For real and complex numbers there is no difference between the twotypes of convergence; thus we can write as

Problem 2.1.2 Show that if and are equivalent metrics in and ifin then in

Problem 2.1.3 Show that

are possible metrics for but no two of them are equivalent.

Problem 2.1.4 Show that in iff in (of Problem 2.1.3)even though are not equivalent.

Problem 2.1.5 Consider system of particles in Theconfiguration of the system is the set of triples of the Cartesiancoordinates of the points named after René Descartes (1596–1650). Showthat we can distinguish different configurations of the system by using a metricin where

We can apply the notion of a metric not only to the sets of locations of asystem of particles, but also to sets of velocities, accelerations, masses, or infact to any finite set of parameters, forces, temperatures, etc.

Now let us consider continuum problems. Take a taut string with fixed endsand length we can use a Fourier expansion


to describe a static displacement caused by some continuous load distribution.Any state of the string can be identified with the vector havingan infinity of coordinates Now we have an infinite-dimensional space of all For the moment we shall call this but we shallsoon be more precise.

Let us modify the metrics we introduced in so that we can determinethe distance between two vectors and We could take

provided that the sum converged. We notice that if

are two possible configurations of the string, then

This means that if the integral on the left is finite, then the sum on the rightwill converge. This means that the metric is appropriate to measure thedistance between any two configurations of the string for which

The appropriate generalization of (2.1.2) makes use of the concept of thesupremum of a set of real numbers, introduced in § 1.1:

But now there are differences. In Problem 2.1.3 we found that the two metricsfor given by (2.1.1) and (2.1.2) are equivalent, but we can easily show thatthis is not true for the generalized metrics for Call

Take then

2.1 Preliminaries 23

But the series in is divergent, so that is meaningless. Thusand are not equivalent metrics for a generalized infinite dimensional space,and there can be no constant (as in (2.1.3)) such that

Problem 2.1.6 Show that if are given by (2.1.8), andis finite, then will be finite; moreover, there is a constant suchthat

We took the idea of Euclidean distance, and constructed the abstract notionof a metric. Now we use the examples of and our vaguely defined toconstruct the notion of a metric space.

Definition 2.1.4 A metric space is a pair consisting of a set X (ofpoints or elements) together with a metric a real valued functiondefined for any two points which satisfies D1–D4. We shall usuallydenote a metric space by X, with remaining implicit.

We will generalize Definition 2.1.2 to X:

Definition 2.1.5 Two metrics of a space X are said to be etquivalentif there exist such that

We shall not distinguish between metric spaces consisting of the same ele-ments, if their metrics are equivalent.

Different problems in mechanics and physics require different types of metricspaces. Depending on the metric we choose, a solution of a problem may or maynot exist, may be unique or non-unique, etc. The right choice of a metric spacecan be crucial for success.

We now abolish the vaguely defined space and introduce some properdefinitions. The spaces we are dealing with have elements which are infinitesequences, i.e. there are four examples of such spaceswhich we shall name:

1. is the metric space of all bounded sequences; the metric is

2. is the set of all sequences such that the metric is


3. is the set of all convergent sequences; the metric is the metric of

4. is the set of all sequences convergent to zero; the metric is the metricof

Problem 2.1.7 Let M be the set of all directed straight lines in the plane.The straight line making angle with the -axis is given by the equation

Take

and show that is a metric in M.

We associated the metric (2.1.5) for the string with the integral (2.1.7). Wehave the correspondence

The integral on the left arises from the kinetic energy of the string; ifthen the (dimensionless) kinetic energy of the string is

We can also measure the distance between two configurations of the string byusing the strain energy of the string. After reducing this to dimensionless formwe have

If is given by (2.1.4), then

This suggests that we can use the metric

This will be a metric appropriate for measuring the distance between any twoconfigurations of the string having finite strain energy, i.e.

2.2 Sets in a metric space 25

Certain spaces based on energy integrals (2.1.10), (2.1.11) will be calledenergy spaces. We have used the representation (2.1.4) to express these spacesas spaces with elements which are infinite sequences with appropriate metrics.We will return to these spaces in Chapter 3 and treat them as spaces of functions

after we have developed the necessary terminology and techniques.

2.2 Sets in a metric space

By analogy with Euclidean space we can introduce some definitions and con-cepts.

Definition 2.2.1 In a metric space X the set of points

is called the open ball of radius about We also call it anof and denote it by is called the center of the Aneighborhood of is any subset M of X which contains anB of Conversely, we call an interior point of a set if M is aneighborhood of

Definition 2.2.2 A set S in a metric space X is said to be open if every pointis the center of an of radius contained in S. Thus

every point of an open set S is an interior point.

Definition 2.2.3 A point is called a contact point of a setif every neighborhood of contains at least one point of S, maybe just Theset of all contact points of S is called the closure of S and is denoted by

Clearly since every point of S is a contact point.

Problem 2.2.1 Show that the empty set is both open and closed.

Problem 2.2.2 Show that if then


Definition 2.2.4 A point is called a limit point (or accumulationpoint) of S if every neighborhood of contains an infinity of points of S. Wesometimes say the points cluster around The limit point may or may notbelong to S.


For example, if S is the set of rational numbers on [0,1] with the metricthen every point of [0,1], whether rational or not, is a limit

point of S.

Problem 2.2.4 Show that is a limit point of S iff every neighborhoodof contains at least one point of S different from

Definition 2.2.5 A point is called an isolated point of S if there is a(sufficiently small) neighborhood of containing no other point of S.

Problem 2.2.5 Show that every contact point of S is either a limit point oran isolated point.

Definition 2.2.6 Let X be a metric space. A set S is said to be closed in Xif i. e. if it contains all its contact points, and in particular, all its limitpoints.

The set

is called a closed ball of radius According to the last definition, it isa closed set.

Note that we cannot say simply that a set is closed; we must state the metricspace X for which it is closed. For example, suppose is the metric space ofrational numbers under the usual metric Suppose S is the setof rational numbers such that S is closed in For if s is a limitpoint of S, then it must be rational, because it is in and we may showthat it satisfy therefore it is in S. But suppose that is the set ofreal numbers under the same metric (i.e. ) and S is again the set of rationalnumbers such that Then we can find a set of points in S whichcluster around the irrational number This number is in but is not in S;therefore S is not closed in

Definition 2.2.7 Suppose S, T are two sets such that The set Sis said to be dense in T if By definition, therefore, S is dense in

Problem 2.2.6 Show that S is dense in T iff any of a pointcontains a point

Definition 2.2.8 Let be a metric space. Let be a subset of XWe may define a subspace of by the pair where, for

is called the metric induced on Y by

2.3 Some metric spaces of functions 27

Definition 2.2.9 Let X be a metric space, and S be a set in X. The com-plement of S is the set of points in X which are not in S; it is denoted byX – S (or X\S).

Problem 2.2.7 Let X be a metric space and S be a set in X. Show that S isopen iff its complement is closed, and vice versa.

Note that a finite set of points, is closed. The set has no limitpoints (see the crucial different from in Problem 2.2.4); its only contact pointsare points of S; thus and S is closed.

2.3 Some metric spaces of functions

We introduced the concept of a continuous function on a domain in § 1.2.Now we introduce

Definition 2.3.1 is the set of continuous functions on

Problem 1.2.1 provides a counterexample to show that if thenit need not be bounded. We therefore introduce

Definition 2.3.2 is the subset of consisting of functions whichare bounded on

If we equip with the metric

then we have a metric space. Clearly this satisfies D1–D3; D4 may beverified as follows. Suppose and then

so that

Thus given by (2.3.1), is a metric, called the maximum or uniform metric.Another way to circumvent the fact that functions in are not bounded

is to introduce

Definition 2.3.3 is the subset of consisting of functions of com-pact support in


By Definition 1.2.3 the support of is a closed and bounded set G =and Theorem 1.2.1 states that a continuous function defined on such a

set is bounded. Thus with the metric (2.3.1) is a metric space, and is asubspace of

We may also introduce

Definition 2.3.4 is the subset of consisting of those functionswhich are uniformly continuous on

Note that Theorem 1.2.3 states that if is bounded and uniformly con-tinuous on i.e. in then it may be extended (uniformly) continuouslyto

Problem 2.3.1 Take = (0,1). Show that if thenbut

When dealing with differentiable functions we often want to have some wayof measuring the distance between the derivatives of two functions.

We cannot just replace by in, say, because

Problem 2.3.2 Show that is not a metric on the

set of uniformly continuously differentiable functions on [0,1], because D2 fails.Show that it is a metric on the subset of those functions satisfying

To obtain suitable metrics we proceed as follows. Introduce the abbreviation

and

Definition 2.3.5 Let be a non-negative integer. is the set of func-tions which have continuous derivatives for

There are (at least) two possible metrics we can use:

or

but in order to ensure that these quantities are finite we must introduce subsetsof

1 Uniformly continuously differentiable means that is uniformly continuous.

2.4 Convergence in a metric space 29

We can introduce metric spaces and which aregeneralizations of and respectively. Thus for wetake the subset of for which are bounded on for

we take the subset of consisting of functions of compact support;for we take the subset of of functions for whichare uniformly continuous on We define to be the set of functions

having continuous derivatives of all orders on i.e. and

to be the subset of of functions having compact support.

Problem 2.3.3 Show that the metrics in (2.3.3) and (2.3.4) are equivalentmetrics for all these metric spaces.


is a suitable metric for i.e. it satisfies D1-D4, but is not equivalentto the uniform metric (2.3.1).

2.4 Convergence in a metric space

Generally, from now on, means a point in a metric space, refers to aninfinite sequence of such points.

Definition 2.4.1 In a metric space X, an infinite sequence is said tohave a limit if, given there exists an integer (i. e. dependingon such that if then In other words, if then allmembers of the sequence belong to an of We write

and say that the sequence is convergent, or converges to We will alsowrite

This notion is a direct analogue concept of convergence introduced in § 1.2,and possesses similar properties.

Theorem 2.4.1 A convergent sequence has a unique limit.

Proof. Let be two different limits of the convergent sequence suchthat Take By definition, there exists N such that

implies and But


This is a contradiction.

The reader will note that this proof follows exactly the same lines as thatused in § 1.1 for sequences of real numbers.

Problem 2.4.1 Show that a sequence which is convergent in a metric space Xis bounded, i.e. all the elements in the sequence lie in a ball of finite radius.

Clearly, if is a convergent sequence in a set then its limitby the definition (Definition 2.2.3 and Problem 2.2.5). In particular, if

S is closed (Definition 2.2.6), thenThe definition of a convergent sequence states that there is a limit point

We need a wider concept, and this is provided by

Definition 2.4.2 A sequence in a metric space X is said to be a Cauchysequence if, given there exists (depending on such that, ifN, then

Problem 2.4.2 Show that a Cauchy sequence in a metric space X is bounded.

Definitions 2.4.1 and 2.4.2 are the analogies of the Definitions 1.1.1 and1.1.2 which we introduced for sequences and later generalized (Def-inition 1.1.5) to sequences in In Chapter 1, in dealing with sequences in

or we found that every Cauchy sequence is a convergent sequence. Nowwe can no longer assume that this is true. Indeed if X is the metric space ofrational numbers under the metric the sequence (1.1.2) is aCauchy sequence, but not a convergent sequence. In a general metric space aCauchy sequence may not have a limit.

Problem 2.4.3 Show that a convergent sequence in a metric space X is aCauchy sequence.

Problem 2.4.4 Show that if is a Cauchy sequence, and has asubsequence which converges to then converges to

2.5 Complete metric spaces

Definition 2.5.1 A metric space X is said to be complete if any Cauchysequence in X has a limit in X; otherwise it is said to be incomplete.

In other words, complete metric spaces are precisely those in which be-ing a Cauchy sequence is a necessary and sufficient condition for convergence;completeness guarantees the existence of a limit.

2.5 Complete metric spaces 31

The space of all real numbers with the metric isa complete metric space; the counterexample (1.1.2) shows that the space ofrational numbers with this metric is incomplete.

Problem 2.5.1 Show that the space of complex numbers withthe metric is complete.

Weierstrass’ theorem on uniform convergence of uniformly continuous func-tions on (Theorem 1.2.4) may be interpreted as stating that is completeunder the uniform metric (2.3.1). Note that uniform convergence of a sequenceof functions is precisely convergence in the metric (2.3.1).

Problem 2.5.2 The functions

are bounded and uniformly continuous on (0,1), i.e. they are in C[0,1]. Showthat is a Cauchy sequence in the metric

.but that converges to the function in themetric, i.e.

Thus the in this example converge in the metric to the functionwhich is not uniformly continuous in (0,1), i.e. is not in C[0, 1].

We conclude that C[0,1] is complete under the metric (2.3.1), but is in-complete under the metric (2.5.1). This is general: a space may be complete orincomplete depending on the chosen metric.

Problem 2.5.3 Show that a subspace in a complete metric space (X,is a complete metric space iff Y is closed in X. See Definition 2.2.8.

We can use the definition of dense (Definition 2.2.7 and Problem 2.2.6) togive an alternative definition of dense, namely

Definition 2.5.2 A set S is said to be dense in a metric space X if anyof contains a point

Weierstrass’ polynomial approximation theorem (discussed in § 1.2) statesthat if is a closed, bounded set in then the set of all polynomials isdense in with the metric (2.3.1).


The completeness of a metric space is of great importance since numerouspassages to the limit appear in the justification of numerical methods, exis-tence theorems, etc. Many of the spaces we have introduced up to now are notcomplete; how they can be completed is the subject of the next section.

2.6 The completion theorem

The way in which we complete a metric space is a direct extension of the wayin which we introduced real numbers to complete the set of rational numbers.Before introducing the theorem we need two definitions:

Definition 2.6.1 A correspondence between two metric spaces andis said to be one-to-one if there is a rule which assigns a unique

element to each element and vice versa. The correspondenceis said to be isometric if

Definition 2.6.2 Two sequences and in a metric space X are saidto be equivalent if

Theorem 2.6.1 For a metric space X, there is an isometric one-to-one cor-respondence between X and a set which is dense in a complete metric space

called the completion of X.

Proof. With any given Cauchy sequence of elements of X we can associateall the Cauchy sequences equivalent to it; these form a class, an equivalenceclass, F. A particular Cauchy sequence in F is called a representative of F.With any element we may associate the Cauchy sequenceany equivalence class which contains such a sequence (and it can contain atmost one) is called a stationary equivalence class. Denote the set of stationaryequivalence classes by and the set of all equivalence classes by Nowintroduce a metric in by

where are representatives of the equivalence classes F and G re-spectively, to obtain the needed correspondence.

To complete the proof we must show that (2.6.1) does define a proper metric,and that is complete.

First we show that the limit (2.6.1) exists, and is independent of the choiceof representatives The triangle inequality gives

2.6 The completion theorem 33

so that

Interchanging and we obtain

and hence

as since are Cauchy sequences. Thus is aCauchy sequence (in and the limit (2.6.1) exists because is complete.

Problem 2.6.1 Show that the limit is independent of the choice of representa-tives

Now we verify that in (2.6.1) obeys the axioms D1-D4.Dl:D2: If F = G, then they contain the same representative Cauchy sequences.

Taking the same sequence from both we have

Conversely, if then any two sequences from F and Gsatisfy i.e. according to Definition 2.6.2, are inthe same equivalence class: F = G.

D3:D4: For we have

The passage to the limit gives

for the equivalence classes F, G, H containing respectively.Let be elements of X. Let F, G be the stationary equivalence classes

containing and respectively, then

This establishes an isometry between X andWe now show that is dense in Let F be the equivalence class con-

taining the Cauchy sequence and let be the stationary equivalence


class containing Choose Since is a Cauchy sequence,there is an N such that if then Thus

In words, we may find a stationary class as close as we like to F; istherefore dense in

Finally, we must show that is complete. Let be a Cauchy sequencein From each choose a representative Cauchy sequence in thissequence choose an element such that for allthis is possible because is a Cauchy sequence. Let us show thatis a Cauchy sequence. Let be the stationary equivalence class containing

Then

so that, by the isometry (2.6.2),

since, by hypothesis, is a Cauchy sequence. Let F be the equivalence classcontaining the sequence Then

This can be made as small as we like by choosing large enough, so that

Thus the Cauchy sequence converges to F in the metric so that iscomplete.

Problem 2.6.2 Use the fact that is complete to show that is completeunder each of the metrics and in (2.1.1), (2.1.2) respectively.

It is interesting to consider what happens if the original space X is complete.In that case every Cauchy sequence in X will have a (unique) limit in X sothat we can set up a one-to-one correspondence between any Cauchy sequence(i.e. a representative of an element of and its limit point (i.e. element ofX ) . Thus we can identify X with its completion,

Since Theorem 2.6.1 is of great importance, let us emphasize some of itsaspects. is a metric space whose elements are equivalence classes of Cauchy

sequences from X; X is isometric with the set of all equivalence classescontaining a stationary sequence for

2.7 An introduction to operators 35

In § 2.1 we discussed, in an informal manner, the construction of the realnumber from the rationals. A formal treatment of this construction would usethe completion theorem. Note however that logically this must be done beforethe real numbers are used as the field for constructing metric spaces.

If we can establish a property of a limit of any representative Cauchy se-quence in a class F, we will say that the class F possesses this property. Wewill present many examples of this procedure in the following sections.

2.7 An introduction to operators

The reader is familiar with the idea of a function; its generalization to metricspaces leads us to

Definition 2.7.1 Let X and Y be metric spaces (they may be identical). Acorrespondence is called an operator from X intoY, if to each there corresponds no more than one The set of allthose for which there exists a corresponding is called the domainof A and denoted by D(A); the set of all y arising from is called therange of A and denoted by R(A). Thus

We say that A is an operator on D(A) into Y, or on D(A) onto R(A). Wealso say that R(A) is the image of D(A) under A.

(Note that the term domain of A must be distinguished from domain, a non-empty open set in .)

Definition 2.7.2 A functional is a particular case of an operator, in whichor we call these real or complex functionals respectively.

In accordance with the classical definition of continuity, we have

Definition 2.7.3 Let A be an operator from X into Y. The operator A issaid to be continuous at if, given there is a dependingon such that if then If A is continuousat every point of an open set then it is said to be continuous on M.

Problem 2.7.1 Let X,Y be metric spaces, and A a continuous operator fromX into Y. Show that A maps Cauchy sequences in X into Cauchy sequencesin Y. Show also that if then

If the operator acts on X into X, we say that A acts in X.Many problems in mechanics can be formulated in the form


where A acts in a metric space X. A solution of (2.7.1) is called a fixed pointof A. Typically (2.7.1) arises when we have an approximation procedure whichyields a new approximation from an old one by means of the equation

Definition 2.7.4 An operator A acting in a metric space X is called a con-traction operator (or a contraction mapping) in X, provided there existsa real number with such that

Clearly a contraction operator is a continuous operator. The fundamentaltheorem concerning contraction mappings is Banach’s fixed point theorem, dueto Stefan Banach (1892–1945); it is

Theorem 2.7.1 Let A be a contraction operator in a complete metric spaceX. Then:

1.

2.

A has only one fixed point

for any initial approximation the sequence of successive approx-imations

converges to the solution to (2.7.1); the rate of convergence is esti-mated by

Proof. We first show that A has no more than one fixed point. If there weretwo points such that

then

Since this implies and thereforeNow take an element and consider the iterative procedure (2.7.3).

For we get successively


But

so that

It follows that is a Cauchy sequence. Since X is complete, there is anelement such that

Let us estimate

Thus

and is the fixed point of A.Passing to the limit in (2.7.5) as we obtain the estimate (2.7.4).

We note that Definition 2.7.2 implies that if thenHowever the condition required by the theorem, namely

is stronger than this, and is vital to the proof of Theorem 2.7.1 as isshown by a counterexample.

Problem 2.7.2 Let let be the usual metricX is complete in the metric d. Let A be the mapping from X

into X given by Show that if thenbut that A has no fixed point.

Corollary Suppose that A is an operator in a complete metric space X and,for some natural number N, is a contraction operator. Then the operator Ahas a single fixed point a sequence of successive approximations (2.7.2)converges to independently of the choice of initial approximationwith the rate

We denote


Proof. The operator meets all the requirements of Theorem 2.7.1, so thatthe equation

has the unique solution such that We can apply A to bothsides of this last equation, so that

This means that is also a solution to (2.7.6), but the solution is uniqueso that

i.e. equation (2.7.1) has the solution Noting that any fixed point of A is afixed point of we see that the solution of (2.7.1) must be unique. Finally,we note that for each of the N sequences

the estimate (2.7.4) holds. Since the whole sequence of successive approxima-tions can be composed successively of elements of these sequences, we obtainthe stated rate of approximation.

We can apply Banach’s fixed point theorem to systems of linear algebraicequations. Suppose we want to solve a system

The corresponding operator A is defined by

How we treat this system depends on the space in which we seek the solution.If we take the space of bounded sequences with metric

then we see that A is a contraction operator if

and If this result holds, then we can find a solution to(2.7.7) by the method of successive approximations beginning with any initialapproximation from


Problem 2.7.3 Consider a Volterra operator

with kernel continuous on [0, a] x [0, a]. Show that A is a contractionoperator in for sufficiently small a > 0. Also show that for any finite

there exists such that is a contraction operator.

It is worth noting that the problem of viscoelasticity can be formulated asan equation of Volterra type, but for functions that take values in a certainenergy space. The same iteration method can be used to solve this problem,and the result is quite similar.

We conclude this section by showing that a continuous operator A from Xinto Y has two complementary properties relating to open sets and closed sets.First we need

Definition 2.7.5 Let X, Y be metric spaces, and A an operator from X intoY.

a) If and then is called the image of

b) If S is a set in X, then the set of all such that for someis called the image of S.

c) If then the set of all such that is called the inverseimage of

d) If T is a set in Y, then the set of all such that is calledthe inverse image of T.

Clearly the image of D(A) is R(A) and the inverse image of R(A) is D(A).Note that if its inverse image may be a single element or a set inD(A). If then its inverse image is the empty set.

Now we prove

Theorem 2.7.2 Let X, Y be metric spaces, and A an operator from X intoY. A is continuous iff the inverse images of open sets are open sets.

Proof, a) Suppose A is continuous. Let T be an open set in Y, and letbe its inverse image. If S is empty, it is open. Suppose S is non-empty and

then T is open so that there is an open ball of radiusaround which is in T. Since A is continuous we can find such

that implies All such are in T, so that thecorresponding are in S. Therefore any is the center of an open ballin S, and S is open.


b) Suppose T is open implies that S is open. Suppose andT. Since T is open, is the center of an open ball of radius consisting ofpoints of T. Choose The open ball is an open set around

Its inverse image is an open set around Since this set is open, it containsan open ball of radius therefore, if thenis continuous at

Theorem 2.7.3 Let X, Y be metric spaces, and A an operator from X intoY. A is continuous iff the inverse images of closed sets are closed sets.

Proof. a) Suppose A is continuous. Let T be a closed set in Y, and letbe its inverse image. If S is empty it is closed. Suppose S is non-empty and

is a convergent sequence in S, converging to A is continuous atChoose There is a such that impliesBut since we can find N such that implies andthus Thus But T is closed so that whichmeans i.e. S is closed.

b) Suppose the inverse image of every closed set in Y is a closed set in X.Suppose A is not continuous at This means that there is an for

which there is a sequence such that but

Suppose The set defined by is closed. Its inverseimage S is therefore closed. But and so that andhence Thus which is impossible. Thus A iscontinuous.

2.8 Normed linear spaces

Almost all the spaces we shall consider are linear spaces. This means that forevery pair of elements in a linear space X and or a sumand a scalar product can be defined such that:

1.

2.

3.

4.

5.

6.

7.

8.

there is a zero element, such that

2.8 Normed linear spaces 41

If then X is said to be a linear space over

Definition 2.8.1 is called a norm in a linear space X if it is a real valuedfunction defined for every which satisfies the following norm axioms:

Definition 2.8.2 A linear space X is called a normed linear space if, forevery a norm satisfying N1-N3 is defined.

As with metrics, we can distinguish between a norm and its valuefor

Definition 2.8.3 Let X be a normed linear space. Two norms andon X are said to be equivalent if there exist positive numbers suchthat

We shall show below that any two norms in are equivalent.X is said to be real or complex, depending on whether the scalars are

taken from orIn a normed linear space we may define a metric

Problem 2.8.1 Show that if the norm satisfies the axioms N1-N3, thengiven by (2.8.2) satisfies the axioms D1-D4.

This shows that a normed linear space is a metric space.

Problem 2.8.2 Show that if a normed linear space is given a metric by meansof (2.8.2), then Show that of Problem 2.1.3 does notsatisfy this equation.

We defined a subspace of a metric space in Definition 2.2.8. Now we introduce

Definition 2.8.4 Let X be a normed linear space, and suppose Y iscalled a subspace of X if it is itself a linear space, i. e. one which satisfiesconditions 1–8 listed above, and has the norm on Y obtained by restricting thenorm on X to the subset Y . The norm on Y is said to be induced by thenorm on X.


Definition 2.8.5 Let X be a normed linear space, and Y is calleda closed subspace of X if it is closed as a set with the norm induced by X,and is a subspace of X.

Note that a closed subspace is a particular kind of closed set: one that isalso a subspace.

One of the most important kinds of subspace is a finite dimensional sub-space. Before defining this we must introduce some concepts which are directgeneralizations of concepts in linear algebra.

Definition 2.8.6 Let X be a linear space. The elements aresaid to be linearly independent (over the appropriate field or if theequation

implies otherwise are said to be linearlydependent; in this case there is at least one of the which may be expressedas a linear combination of the others.

Let be elements of a linear space X. The set of all elements

with the norm on X, satisfies Definition 2.8.4 for a subspace of X; it is a finitedimensional subspace Y of X, with its dimension, dim(Y) being defined by

Definition 2.8.7 If are linearly independent thenOtherwise there will exist N, satisfying such that N of

are linearly independent while any N +1 are linearly dependent,in which case dim(Y) = N.

Clearly we can extend this to define a finite dimensional space X:

Definition 2.8.8 The linear space X is said to be finite dimensional if there is anon-negative integer N such that X contains N linearly independent elements,but any set of N + 1 elements is linearly dependent. We write dim(Y) = N. IfX is not finite dimensional we shall say that it is infinite dimensional.

If dim(X) = N, any set of N linearly independent elementsis called a basis for X. If is a basis for X, then any element of X has aunique representation

We now use the Bolzano–Weierstrass theorem to prove

2.8 Normed linear spaces 43

Theorem 2.8.1 In a finite dimensional normed space X any two norms areequivalent.

To prove this we first establish

Lemma 2.8.1 Let X be a normed linear space, and be alinearly independent set of elements in X. There is a number such thatfor every set of scalars we have

Proof. Write If then all are zero, so that the inequality(2.8.3) holds with any Suppose then the inequality (2.8.3) isequivalent to

where and Suppose this is false then, given any we

can find a combination

such that

This means that we can find a sequence of the form

such that

Since we have for We consider the se-

quence this is an infinite sequence in the closed and boundedset Since this set is compact, the sequence contains a subse-quence converging to a number such that Denote the ofthis subsequence by 1,1; 1,2; ··· . Now consider the for this subsequence:

The sequence contains a subsequence converging to anumber Proceeding in this way we eventually find a subsequence


of such that

Thus

Now since not all the can be zero. Since is an independent

set, On the other hand, since is a subsequence of andwe must have and hence This is a contradiction.

Hence inequality (2.8.4) holds.

We may now prove Theorem 2.8.1

Proof. Let be a basis for X. Then any may be written

so that

where On the other hand, the inequal-ity (2.8.3) gives

so that

This is half of the inequality (2.8.1); the other half may be obtained by reversingthe roles of and

Definition 2.8.9 A complete normed linear space is called a Banach space.

Problem 2.8.3 Show that a closed subspace of a Banach space is complete, i.e.is a Banach space.

Let be a closed and bounded region in The space with thenorm

is a normed linear space; it is complete, and so it is a Banach space. This normis called the uniform or infinity norm.

We can apply the Completion Theorem to normed linear spaces; now wesay that the completion is in the appropriate norm.

Note that when we apply the Completion Theorem to a normed linear space,the resulting complete space is a linear space.

2.9 An introduction to linear operators

For a linear operator, the image is usually written

2.9 An introduction to linear operators 45

Problem 2.8.4 Show that the spaces and defined in § 2.1 are Banachspaces.

Problem 2.8.5 Show that the space defined in § 2.3 is complete in thenorm

corresponding to given in (2.3.4).

Problem 2.8.6 Show that the space defined in § 2.3 is complete in thenorm (2.8.4).

Note that is larger than in that its elements need not beuniformly continuous on

Problem 2.8.7 Take Construct such thatbut

Suppose that X and Y are linear spaces over the same coefficient fieldeither or

Definition 2.9.1 A space S is said to be a linear subspace of a linear spaceX if S is linear space and S is a subset of X.

Definition 2.9.2 The operator A is a linear operator from X into Y if itsdomain D(A) is a linear subspace of X and, for every andevery

Now suppose that X and Y are normed linear spaces and A is a linearoperator from X into Y. The operator has a domain D(A) which is a subspace


of X; a range R(A) which is a subspace of Y; and a null space N(A) consistingof such that which also is a subspace of X.

Problem 2.9.1 Verify that D(A),N(A) are subspaces of X, and R(A) is asubspace of Y.

Problem 2.9.2 Show that if A is a linear operator from X into Y, and D(A)is finite dimensional, then R(A) is finite dimensional, and

We shall be concerned largely with continuous linear operators, for whichwe have the simplifying result

Problem 2.9.3 Let A be a linear operator from the normed linear space X intothe normed linear space Y. Show that A is continuous at any pointiff it is continuous at one point say

This leads to

Theorem 2.9.1 Let A be a linear operator from the normed linear space Xinto the normed linear space Y. The operator A is continuous on D(A) iffthere is a constant c, such that, for all we have

The infimum of such constants c is called the norm of A and denoted byThus

Proof. We need to consider the continuity of A only at If (2.9.1) holds,then Definition 2.7.3 shows that A is continuous at

Conversely, if A is continuous at then Definition 2.7.3 withstates that there is some such that, if then For every

the norm of is

so that But A is a linear operator, so that

This states that if then

2.9 An introduction to linear operators 47

which is (2.9.1) with

A linear operator satisfying (2.9.1) is said to be bounded. Thus we have

Corollary 2.9.1 A linear operator is continuous iff it is bounded.

We also have the important result

Theorem 2.9.2 Let X,Y be normed linear spaces and A be a linear operatorfrom X into Y. If X is finite dimensional, then A is continuous.

Proof. Let dim X = N and suppose is a linearly independentset of elements in X. If we can write

uniquely. Then

Now apply Lemma 2.8.1 to give

where Thus

so that A is continuous. •

Problem 2.9.4 Let X be a finite dimensional normed linear space anda linear functional on X. Show that is continuous.

There are classes of differential and integral operators that are continuous,as given by the following problems.

Problem 2.9.5 The differential operator, with constant coefficients definedby

is a linear operator from into Show that it is bounded. (Use thenorm corresponding to (2.3.1) for and (2.8.5) for


Problem 2.9.6 Show that the integration operator defined by

is continuous from C(0,1) into C(0,1). Is it continuous from C(0,1) into

2.10 Some inequalities

In this section we shall derive some important inequalities which will be usedto construct some new normed linear spaces.

Lemma 2.10.1 Let Thenwith equality iff

Proof. Consider where and Thenso that, since we have for and

for Hence with equality only if Thuswe have

If then If we put to obtain the requiredresult.

Now we prove Holder’s inequality, due to Ludwig Otto Holder (1859-1937):

Lemma 2.10.2 Let andThen

Proof. Put

If AB = 0, then either A = 0 or B = 0. But A (B ) can be zero only if allthe (or are zero. In either case both sides of (2.10.1) are zero. Now ifAB > 0, then by the inequality in Lemma 2.10.1,

so that

2.10 Some inequalities 49

Problem 2.10.1 Show that equality holds in Holder’s inequality iff there is aconstant M such that

Finally we prove Minkowski’s inequality, due to Hermann Minkowski (1864–1909)

Lemma 2.10.3 Let and then

Proof. The case is trivial. Suppose then

Now apply Holder’s inequality to each sum, using defined byso that then

now use to obtain (2.10.2).

We may use Minkowski’s inequality to generalize the metric in intro-duced in § 2.1, in fact

Problem 2.10.2 Show that if

satisfies the conditions for a norm in

Problem 2.10.3 Show that if then


provides a suitable norm in the space of all sequences with

Problem 2.10.4 Verify that and are complete with the respective norms.

We can consider a contraction mapping in Thus for thesystem (2.7.7), namely

we have

Applying the limiting form of the Holder inequality as to the inner sum,and taking we obtain

so that

This means that A is a contraction mapping in if and

If this is the case, then we can solve (2.10.3) by iteration.

Problem 2.10.5 Show that if and then

with equality only if of the are zero. This is called Jensen's inequalityafter Johan Ludvig William Valdemar Jensen (1859-1925) (See Hardy,Littlewood and Polya (1934), p. 28).

2.11 Lebesgue spaces 51

Riemann (and Lebesgue) integrals are approximated by finite Riemann(Lebesgue) sums, for which many of the above finite inequalities can be writtenout. So a limit passage demonstrates that corresponding inequalities hold forintegrals. This is the way in which Minkowski’s and Hölder’s inequalities forintegrals (see § 2.11) were derived.

2.11 Lebesgue spaces

Up to now we have introduced only a few simple normed linear spaces. In thissection we shall introduce some of the most important ones.

Definition 2.11.1 Let be a domain in and let TheLebesgue space is the completion of the subspace, ofsatisfying

in the norm

There is much to note about this definition. First, we have used the abbre-viation

Secondly, the integration is taken to be the ordinary Riemann integration namedafter Georg Friedrich Bernhard Riemann (1826–1866), over a domain inThirdly, the fact that (2.11.2) does constitute a norm, i.e. it satisfies N1-N3,

follows from the generalization of Minkowski’s inequality (2.10.2) to integrals:

The space is the completion of According to the CompletionTheorem (Theorem 2.6.1), this means that elements in are equivalenceclasses of Cauchy sequences of continuous functions. Remember that

is a Cauchy sequence in if

and that two sequences and are equivalent if


Let us examine elements in First, we say loosely that ifthen but this is not strictly accurate. The elements

of are not functions, but equivalence classes of Cauchy sequences offunctions. What we should say strictly is that if then there is anequivalence class which includes, as one of its Cauchy sequences, the sequence

We need to label this equivalence class; we could label itor use the same label Let us explore this deeper. The function

is in we call the equivalence class which includes the sequence0,0,0, ···, the null class, and give it the label 0. But there are other Cauchysequences in this class. Suppose for simplicity that N = 1 andTake the sequence of continuous tent functions

as shown in Fig. 2.11.1. The function has compact support

and

Thus the sequence is in the same equivalence class as 0,0,0, ..., the nullclass. In other words, the sequence is one of the (infinitely many) equiv-alent Cauchy sequences for the element labeled 0 in Equation (2.11.5)shows that the limit of the norm of tends to zero; on the other hand thepointwise limit of the sequence of functions is

This limiting function is not continuous, i.e. it is not in Nor is it inbecause as we pointed out, elements of are equivalence classes,


not functions. But if we said, loosely, that if thenwe can say, in the same loose way, that But, since is inthe null class, treats as 0. In other words cannot distinguishbetween the (ordinary) functions

because with each of these functions we can associate a Cauchy sequence in thenull class. We can generalize this with

Problem 2.11.1 Take N = 1, Suppose Show thatcannot distinguish between

Hint. In choosing a sequence take tent functions centered on eachand choose the supports of the tents appropriately.

Now we can return to the general case.

Definition 2.11.2 The sequence is said to be a null se-quence if as

Definition 2.11.3 The function is said to be zero almost everywhere(we write this if there is a null sequence suchthat

For such functions

In other words, if then, considered as an element of it isin the class 0.

This generalizes Problem 2.11.1. A function which is zero except at a count-able (see Definition 4.1.2 for the definition of countable) set of points isequal to zero a.e., but the converse is not true; it is possible for to bezero except in a set which is not countable, and still to be the limitof a null sequence according to (2.11.6). Such a set is called a set of measurezero. Countable sets are sets of measure zero, but there are sets of measure zerowhich are not countable.

If a.e. on we say that and are equal almosteverywhere. For such functions


The reader who is acquainted with the theory of Lebesgue integration due toHenri Leon Lebesgue (1875–1941), will realize that we have defined with-out using measure theory. That the two ways of approaching throughcompletion of and by using measure theory, do lead to equivalent def-initions, is a matter which is discussed in more specialist books, e.g. Adams(1975). We have used Riemann integration over as the basis for constructing

but we must admit that there are some exotic regions in which cannotbe accommodated in Riemann integration, but can be in Lebesgue integration.With some effort we could extend our argument to include them, but anywaysuch regions do not often occur in practice.

Let us recapitulate. An element is an equivalence class ofCauchy sequences To define we take a Cauchy

sequence in the equivalence class and consider the sequence

To show that is a Cauchy sequence (of nonnegative numbers) we useMinkowski’s inequality in the form

Thus we have

since is, by definition, a Cauchy sequence in the norm in thenorm Thus is a Cauchy sequence in the complete space and sohas the limit

The number is called the Lebesgue integral of

Problem 2.11.2 Show that K is independent of the choice of the representativesequence in the equivalence class

Problem 2.11.3 Construct a sequence of tent functionssuch that but is a null sequence in

We have defined as the completion of the space the sub-space of satisfying (2.11.1), but there are other, more convenient ways to


define it; we can define it as the completion of any dense (Definition 2.2.7) sub-space of this space, such asor Finally, Weierstrass’s polynomial approximation theorem (Theo-rem 1.3.1) states that if is bounded, then every continuous function inis the limit of a sequence of polynomials; thus we may apply the completion tothe set of polynomials on to obtain

We have defined for satisfying It may be shown thatif then (2.11.2) does not constitute a norm, in fact:

Problem 2.11.4 Show that if and if then

We now obtain the first example of what is called an imbedding theorem.

Theorem 2.11.1 Suppose is a bounded domain, and satisfyIf then and

Proof. Let Holder’s inequality (2.10.1) gives Hölder’s inequalityfor Riemann integrable functions , namely

(This integral inequality follows because a Riemann integral is the limit of asum.) In this inequality, replace and byand respectively, where We obtain

where

But

so that

and on taking roots of both sides, we obtain (2.11.7) .The inequality (2.11.9) shows that if is a Cauchy sequence in

then it will also be a Cauchy sequence in and


Moreover, if is an equivalence class of Cauchy sequences in thenthis will also be an equivalence class in Thus we can proceed to thelimit in the inequality

and say that if then and

We can show this schematically as in Fig. 2.11.2. We say that is

imbedded in and we write this

The imbedding defines an imbedding operator I from to this op-erator maps into the corresponding The operatorI is linear, and the inequality (2.11.7) states that it is continuous (or bounded).We can make this formal.

Definition 2.11.4 We say that the normed space X is imbedded in thenormed space Y, and write this if X is a subspace of Y, andthe operator I from X to Y defined by for all is continuous.

If and is bounded, the integral is uniquely determined

for any by


If then Hölder’s inequality gives

so that on the passage to the limit we have

In what follows, we shall frequently deal with integrals of the form

For example, work done by external forces is of this form.Let us determine the integral when where

and Consider

where and are representative sequences of andrespectively. Then

since and are Cauchy sequences in the corresponding metricsand, for large

Thus is a Cauchy sequence; we define as its limit.


Problem 2.11.5 Show that is independent of the choice of representativesfor and and so is properly determined.

The passage to the limit in the Hölder inequality

shows that the Hölder inequality holds for elements

when

2.12 Inner product spaces

The concepts of metric and norm in a linear space generalize the notions ofdistance and magnitude in We now consider the generalization of the innerproduct.

Definition 2.12.1 Let X be a linear space over as defined in § 2.8. Thefunction uniquely defined for every pair is called an innerproduct on X if it satisfies the following axioms:

where and the overbar in P2 denotes complex conjugate. A linearspace X with an inner product is called an inner product space.

We can consider X over then the inner product must be real valued, P2is replaced by

and X is called a real inner product space.

If it is clear from the context, the terms real or complex will be omitted.Let us consider some properties of X. Let us introduce by

2.12 Inner product spaces 59

To show that we really have a norm, we prove the Schwarz inequality, namedafter Hermand Amandus Schwarz (1843–1921). (This is also called the Cauchy–Schwarz or the Cauchy–Buniakowski inequality, after Victor Yakovlevich Buni-akowski (1804–1899).) Note that for or for spaces like the Schwarzinequality reduces to a special case of the Holder inequality (2.10.1) with

Theorem 2.12.1 For any

where, for equality occurs iff

Proof. The inequality holds if either or is zero. Let and let be ascalar. By P1

where equality occurs iff We have

Put then

which is equivalent to (2.12.2).

Problem 2.12.1 Use the Schwarz inequality to show that if the inner productsatisfies P1–P3, then the norm defined by (2.12.1) will satisfy N1–N3. (Thenby Problem 2.8.1, the metric (2.8.2) will satisfy D1-D4).

We conclude that an inner product space is a normed linear space.

Problem 2.12.2 Show that if are in an inner product space, then

This equation, called the parallelogram law, is important because it char-acterizes norms that are derived from an inner product, as shown by the nextproblem.

Problem 2.12.3 Show that if X is a normed linear space, with a norm whichsatisfies (2.12.3), then we can construct an inner product on X by taking


Show that and that the inner product satisfies P1-P3. (It is quitedifficult to prove P3 for arbitrary (For a real normed linear space we omitthe second pair of terms in the definition of

Definition 2.12.2 By analogy with Euclidean space, we say that and areorthogonal if

Definition 2.12.3 Let X be an inner product space. A subspace S of X is asubspace of X which is itself a linear space (as in Definition 2.8.4) with theinner product on S obtained by restricting the inner product on X toThe inner product on S is said to be induced by the inner product on X.

Definition 2.12.4 Let X be an inner product space. A closed subspace ofX is a set S which is a subspace of X and which, as a set, is closed under the(metric corresponding to the) inner product induced on S.

Note that closed set in X and closed subspace in X are not synonymous; aclosed subspace is a closed set, but not necessarily vice versa.

Definition 2.12.5 A complete inner product space is called a Hilbert space,after David Hilbert (1862–1943), and is denoted by H.

Problem 2.12.4 Show that a closed subspace of a Hilbert space is complete,i.e. it is a Hilbert space.

Let us consider some examples of Hilbert spaces.

The space For we define the inner product by

The space is the ancestor of all Hilbert spaces, and of functional analysisitself. It was introduced by Hilbert in his justification of Dirichlet’s Principle,named after Gustave Peter Lejeune Dirichlet (1805–1859).

Sometimes we use the real here the inner product is

The space The inner product is

Problem 2.12.5 Verify the axioms P1–P3 for and


Problem 2.12.6 What are the Schwarz inequalities for and the real


is an inner product for the set of continuously differentiable functionson [0,1].

We may bring together the concepts of a linear operator (§ 2.9), a Lebesguespace (2.11) and the Schwarz inequality (2.12.2) in the consideration of theFredholm integral operator, named after Ivar Fredholm (1866–1927):

Thus we have

Problem 2.12.8 Show that if and

then K is a continuous operator on and


Synopsis of Chapter 2: Metric Spaces

Spaces

metric: has metric satisfying D1–D4 in § 2.1, includes

normed linear: has norm satisfying N1–N3 in § 2.8, includes

inner product: has inner product satisfying P1–P3 in § 2.12.

Completion of a space X: space of equivalence classes of Cauchy sequencesin X. § 2.6.

Complete space: every Cauchy sequence has a limit belonging to the space.Definition 2.5.1. In a complete space, Cauchy sequence convergent se-quence.

Banach space: complete normed linear space. Definition 2.8.9.etc; see § 2.8.

Hilbert space: complete inner product space. Definition 2.12.3.

Lebesgue space completion of etc in norm(2.11.2).

Sequences

Cauchy: as Definition 2.4.2.

convergent to x: as Definition 2.4.1.

Sets

open: every point is an interior point. Definition 2.2.2.

closed: contains all its limit points. Definition 2.2.6.

dense: X is dense in Y if every has a point arbitrarily close toit. Definition 2.2.7.


References

The classical text on functional analysis is

F. Riesz and B. Sz.-Nagy, Functional Analysis, Frederick Ungar Publishing Co.,New York, 1955.

Among the many other excellent treatises we mention

A.N. Kolmogorov and S.V. Fomin, Introductory Real Analysis, Dover Publi-cations Inc., New York, 1975. This has extensive discussion on set theory, onmeasure theory and integration.

A. Friedman, Foundation of Modern Analysis. Dover Publications Inc., NewYork, 1970. This covers much of the material in our book at greater depthand level of abstraction. In particular it has an extensive study of Lebesgueintegration, and of the concept of the adjoint for spaces other than Hilbertspaces.

L.V. Kantorovich and G.P. Akilov, Functional Analysis, Pergamon Press, 1982.This is an extensive work with copious references to the original literature andto other treatises.

A comprehensive treatment of functional analysis at an abstract level may befound in

K. Yosida, Functional Analysis, Springer-Verlag, New York, 1971.

A brief, easily readable account of some aspects of functional analysis may befound in

C.W. Groetsch, Elements of Applicable Functional Analysis, Marcel Dekker,New York, 1980.

An exemplary text book which covers much of the material in the present bookat much greater depth, and which has many examples and references is

E. Kreyszig, Introductory Functional Analysis with Applications. Robert E.Krieger Publishing Company, Malabar, Florida, 1989.

3. Energy Spaces and Generalized Solutions

Yes. It’s a pleasant feeling, writing ... and looking over proofs ispleasant too.

Anton Chekhov, The Seagull

Perhaps mathematical proofs too.

3.1 The rod

Consider a perfectly elastic rod of length cross-sectional area Young’smodulus E (named after Thomas Young (1773–1829)), undergoing longitudinaldisplacement There is only a single strain and a single stress

so that its strain energy

We choose units so that E = 1, suppose all quantities are dimensionless, andthat is bounded, i.e.

We suppose that the rod is a cantilever, so that is fixed, i.e.

We can use to define a metric, norm and inner product in the subset offunctions in satisfying (3.1.2), for which

Thus

66 3. Energy Spaces and Generalized Solutions

Problem 3.1.1 Show that these d, and (·, ·) do satisfy the requirementsfor a metric, norm and inner product; in particular, that

We call the set S of all satisfying (3.1.2)–(3.1.3) the sub-space of functions with finite energy.

Problem 3.1.2 Show that S is an incomplete inner product space, by con-structing a Cauchy sequence of functions in S which has a limit which is notin S.

To create a complete space, we must apply the Completion Theorem; theenergy space is the completion of S in the metric (3.1.4). We recall that anelement in is a class, U, of Cauchy sequences in S which are equivalent inthe metric (3.1.4).

First we show that if then it is uniformly continuous inand is thus in

Thus to make we need only take whereis uniformly continuous in it is in Now suppose that

is a representative Cauchy sequence in a class we have

For if then

3.1 The rod 67

as This means that converges in the uniform norm, i.e. itconverges uniformly. But by Weierstrass’ theorem 1.2.4, a uniformly convergentsequence of continuous functions on has a uniformly continuous limit. (Thisis equivalent to saying that is complete under the uniform metric.) Thismeans that converges in the uniform metric to a limit functionwhich is uniformly continuous on and

Problem 3.1.3 Show that is independent of the choice of the representa-tive Cauchy sequence for U.

We will use capital U to denote an element of the energy space, hereand to denote an ordinary (continuous) function. What we have shown isthat U is such a function, so that we could call it


The expression on the left of (3.1.8) is the uniform, or infinity, norm ofas in (2.8.4), so that we can write

This is another example of an imbedding: every element can be identifiedwith an element in such a way that the inequality (3.1.9) holds.The correspondence defines an operator, as defined in § 2.9, from intoWe call it the imbedding operator from into Clearly (3.1.9) showsthat the operator is bounded, i.e. continuous. See Theorem 2.9.1. As in § 2.11,we write the imbedding

We have shown that if then it is continuous in so thatwe could use lower case to denote it; it is not necessarily differentiable,but we can define a generalized derivative for it. If is a representativeCauchy sequence for U, then is a Cauchy sequence in andthus corresponds to an element in can call this the generalizedderivative of U, and can use the notation with caution.


The functional expressing the total energy of the rod contains the termcorresponding to the work done by the external forces. The functional

is properly defined if and For if areCauchy sequences for F, U respectively, and

then

But

so that as Thus is a Cauchy sequence withlimit W.

Moreover, if we can accommodate terms of the form

corresponding to work done by concentrated forces, for

To find the displacement of the rod due to a distributed load we usethe Principle of Virtual Work, as we now describe.

The Principle of Virtual Work began historically as a principle in statics ofrigid bodies; it was extended to dynamics by using d’Alembert’s principle, dueto Jean le Rond d’Alembert (1717–1783), and then generalized to statics anddynamics of continua. In brief the Principle states that when a solid body isloaded by external forces, the work done by the internal forces in any virtualdisplacement is equal to the work done by the external forces, i.e.

A virtual displacement is one that is sufficiently smooth and which satisfies thegeometric constraints imposed on the body.

Generally, if the particles of a body are displaced, the internal forces, andmaybe also the external forces, will be affected. In computing and weignore such effects; in making the displacements, we assume that the internal

3.1 The rod 69

and external forces are kept constant; for this reason the terms arecalled the virtual work done by the internal and external forces, respectively.

For the rod there is only one kind of displacement: longitudinal. A virtualdisplacement is one that is sufficiently smooth, e.g. or atleast and which satisfies the geometric constraint

For a rod loaded by a distributed (longitudinal) force the virtual workdone by the external force is

The internal forces are longitudinal forces acting on the cross-section of therod. At section the strain is and the corresponding stress is

A virtual displacement of particles of the rod induces a virtual longi-tudinal strain The work done by the stress in thevolume element under the virtual strain is

so that, with E = 1,

and the Principle (3.1.11) gives

We now show that we can treat this equation as the fundamental equation fromwhich we may deduce the classical differential equation governing the rod, andalso a generalized solution of problem.

For the classical analysis we assume that

Under these conditions we can perform an integration by parts in (3.1.14) toobtain

This must hold for all sufficiently smooth satisfying it musttherefore hold for the subset of these satisfying also. For such

equation 3.1.16) reduces to


Under the condition (3.1.15), the bracketed term in the integrand is inWe now prove

Lemma 3.1.1 If and

for all functions satisfying then

Proof. Suppose there is a such that without loss ofgenerality we may assume Since is continuous at there is aclosed interval in which Choose a functionwith support such that in then

This is a contradiction. Thus in but is continuous inso that and thus in

Note that the conclusion would follow if equation (3.1.17) heldjust for a subset, say of

This lemma, applied to equation (3.1.17), states that

This is the classical differential equation for the rod. Now return to equa-tion (3.1.16), which holds for all satisfying and not just forthose satisfying the extra condition Choose so that

we have

This is the classical free (or natural) end condition.To set up the generalized problem we note that equation (3.1.14) is well

defined for a much larger class of than those satisfying theconditions (3.1.15). In fact we can set up the equation

for and We may suppose that has derivatives of allorders, i.e. is in and has some compact support where

Thus and in fact butand its derivatives need not be zero at We could call this set Underthese conditions the integral on the right may be bounded using the Schwartzinequality:

3.1 The rod 71

so that it is defined when Note that the first integral is defined as thelimit

for a representative Cauchy sequence of functions in the equivalenceclass U; note that convergence of the is in the norm (3.1.4), so thatequation (3.1.21) states that

The integral on the right of (3.1.20) is defined when for

and, since is continuous on and has compact support, it is bounded.We call the solution which satisfies (3.1.20) for all

the generalized solution of the rod problem.

Problem 3.1.5 Show that U is unique.

We derived equation (3.1.16) from the Principle of Virtual Work, but sincethe elastic rod is a conservative system we can also derive the governing equa-tions using the Principle of Minimum Energy. If the classical conditions (3.1.15)hold, then the solution of the rod problem is the minimizer of the total energy

on the subset of satisfying For if we take whereand then

If for all then


which is precisely equation (3.1.14).We have shown that under the classical condition (3.1.15), we can derive the

virtual work equation (3.1.14) by minimizing (3.1.22) for satisfyingNow we show that we can derive the generalized problem (3.1.20) by

minimizing

for and If then the first term is while if

the second term is a continuous linear functional in Rememberthe definition of a continuous linear functional given in equation (2.9.1). For if

is a Cauchy sequence for and is one forthen, by definition

The limit on the right does exist, because

Putting in (3.1.7) we find that

Thus

But converges to F in and converges to U in so that

This means that we can write J(U) as

where is a continuous linear functional in We can consider the min-imization of J(U) as before; thus if then

Since is arbitrary, we must have

3.1 The rod 73

for all When written in full, this is

This has the same form as equation (3.1.20), except that in (3.1.20),while in (3.1.26),

This is an example of a general result. If we can express the classical problemas the minimization of a functional

where is an appropriate norm, and is a continuous linear functionalin that norm, then we take the generalized solution to be the minimizer of J(U)in the appropriate energy space, E (in the present case and find that theminimizer satisfies

for allWe note that equations (3.1.20), (3.1.25) are equivalent. For if (3.1.25) holds

for every then (3.1.20) will hold for any On the otherhand, if (3.1.20) holds for then (3.1.25) will hold for everysince is dense in

Before leaving this problem we note that the natural boundary condi-tion (3.1.19) which occurs in the classical problem, has no meaning in the gen-eralized problem because need not exist. (We can assign a meaning to itby introducing distributions.)

We have considered the rod with the cantilever end conditionWe now consider what changes must be made if the rod is free at both ends;there is no constraint on First we note that defined by (3.1.4) is nota norm, for implies only or There arevarious ways to circumvent this difficulty. We can construct a new space ofequivalence classes; thus will belong to the same equivalence class if

for (Note that will actually beconstant in since we showed that (3.1.3) implies that is uniformlycontinuous on

Note that for the free rod, equation (3.1.5) does not hold, sincebut the analysis (3.1.6) does still hold if Thus we can show that

is bounded and uniformly continuous on and therefore inAll the which are constant in belong to the same equivalence class,which we will call the zero class. Now implies whichmeans that is in the zero class.

Instead of choosing equivalence classes as the elements in the energy space,we may choose the unique element in an equivalence class such that


Now and equation (3.1.29) implyIn either case the energy functional in equation (3.1.26) must be in-

variant under the transformation and this will be so iffsatisfies the condition

This we recognize as the condition of static equilibrium for the external loadsapplied to the rod. Provided (3.1.30) holds, we may make the necessary changesin the analysis given above to define the energy space for the free rod and thegeneralized solutions.

Problem 3.1.6 Show that if the rod is free at both ends, the Principle of VirtualWork (3.I.14) gives the condition (3.1.30).

Problem 3.1.7 Write down the Principle of Virtual Work for a cantilever rodwith a distributed load and end load at Derive the classical endcondition, to replace (3.1.19), at and derive the generalized problem.

3.2 The Euler–Bernoulli beam

We can extend the analysis of a rod to the Euler–Bernoulli model, named afterLeonhard Euler (1707–1783) and Daniel Bernoulli (1700–1782), of a beam inflexure. In this simplified model, which is adequate for the analysis of thinstraight beams in flexure in a principal plane, it is assumed that plane sectionsof the beam normal to its axis remain plane and normal to the axis of thedeformed beam. If the beam is deformed in the plane, then the analysis isbased on the assumptions that the elastic displacements of a particleat are given by

where is measured from the neutral axis of the section. If the beam has lengthand Young’s modulus E, then the only strain is

and so that the strain energy of the beam is

3.2 The Euler–Bernoulli beam 75

where is the second moment of area of the cross-section about the neutralaxis Again we choose units so that E = 1, suppose all quantities aredimensionless, and that is bounded, i.e.

For definiteness we first assume that the beam is a cantilever, so that

We use to define an inner product in the subset of functions insatisfying (3.2.4), for which

Thus

We call the set S of all in satisfying (3.2.4), (3.2.5), the subspaceof functions with finite energy for the bar. Again S is an incomplete innerproduct space; the completion of S in the metric (3.2.6) is called the energyspace

We now examine the elements in We can argue as in § 3.1 that ifis a representative Cauchy sequence for then con-

verges in the uniform norm to a limit function v(x) which is continuous inand

We may now go further with

Problem 3.2.1 Show that if is a representative Cauchy sequence forthen is a Cauchy sequence in the uniform norm, and thus

has a continuous limit Show that Thus we can sayand

Problem 3.2.2 Use the result of the last problem to show that

The expression on the left is the norm corresponding to the metric (2.3.4)with which we label say, so that

This is another example of an imbedding. Each element in can be identifiedwith a function in such a way that (3.2.9) holds. We say


and note that (3.2.9) states that the imbedding operator from intois continuous.

We have shown that any element may be identified with a con-tinuously differentiable function We can define a generalizedsecond derivative for as the element in corresponding tothe Cauchy sequence

Problem 3.2.3 Show that the work functional

is continuous if and Show also that if we candefine terms

corresponding to work done by concentrated forces and moments.

To analyze the beam we use the Principle of Virtual Work. The virtualwork done by the stress on the virtual straininduced by the virtual displacement is, with E = 1,

Thus the Principle gives

for all sufficiently smooth satisfying

e.g. for In the classical analysis we assume

then two integrations by parts in (3.2.11) give

Now we argue as before. If satisfies then the integratedterms in (3.2.14) vanish, and Lemma 3.1.1 states that satisfies the differ-ential equation

3.2 The Euler–Bernoulli beam 77

Thus the integrand in (3.2.14) is identically zero so that for general satis-fying just (3.2.12), we have

Now by choosing a with we deduce that

and by choosing a with we deduce that

These are the natural end conditions for the classical problem.To set up the generalized problem we argue as in § 3.1, that (3.2.11) makes

sense for and The generalized solution for the beamproblem is the satisfying

forAlternatively, we can set up a variational formulation for the generalized

solution, starting from

Proceeding as in § 3.1 we obtain the generalized problem

which we can interpret if and Again (3.2.18), (3.2.20)are equivalent.

If the beam is free at both ends, then again (3.2.6) does not provide anorm, for implies only i.e. We dividethe into equivalence classes, putting into the same class if

We call the class of all those forsome the zero class. Alternatively, in any equivalence class we may choosethe unique member satisfying

Again, if the work functional is to be invariant under the transformationthen must satisfy


which we recognize as the condition for force and moment equilibrium.

Problem 3.2.4 Show that the integrals in equation (3.2.18) are properly definedif and

Problem 3.2.5 Show that equations (3.2.18) and (3.2.20) are equivalent.

Problem 3.2.6 Deduce the equilibrium conditions (3.2.22) from the Principleof Virtual Work.

Problem 3.2.7 Carry out the classical and the generalized analysis for a can-tilever beam loaded by a concentrated force and moment at the endin addition to the distributed loading

3.3 The membrane

Consider a taut membrane stretched with uniform tension T across a domainThe strain energy of the membrane is

(Here denotes the gradient vector.) We choose units so that T = 1, andconsider first the clamped membrane with boundary condition

where is the boundary ofWe can use to define a metric, norm and inner product on the subset S

of satisfying (3.3.1), for which

Thus

The completion of the space S in the norm (3.3.3) we call the energy space forthe clamped membrane We now examine elements To do so weestablish

Friedrichs’ inequality, due to Kurt Otto Friedrichs (1901–1982). Supposeis a bounded domain in and then there is a constant

such that

3.3 The membrane 79

for all

Proof. First suppose is the square Then

and

so that

Now integrate over to give

and hence

If is not a square, we can enclose it in a square and extend functionsby zero to give functions The inequality (3.3.6)

applied to and is equivalent to the same inequality applied to and

Since is dense in S in the norm (3.3.3), Friedrich’s inequality holdsfor This means that if is a representative Cauchy sequence for

then is a representative Cauchy sequence forand

i.e. is imbedded in i.e.

Moreover if is a representative Cauchy sequence for thenand are representative Cauchy sequences in

for elements which we call and respectively. Thus ifthen U, and are all elements of

Note that for the rod we showed that if then it was continuous ini.e. in For the membrane we have proved only that if

then


We can base our analysis of the membrane on the Principle of MinimumEnergy, or the Principle of Virtual Work. For the former we suppose that themembrane is loaded with a distributed load so that the total energy is

For the classical analysis we assume that and Thenputting where and on we deduce that

This is the equation we would get from the Principle of Virtual Work. In theclassical case we use the identity

and the divergence theorem to give

The integral over the boundary is zero, so that equations (3.3.10), (3.3.11) give

(Note that we need to use the divergence theorem in (3.3.11).) Nowthe extension of Lemma 3.1.1) to shows that must satisfy the differentialequation

For the generalized problem we consider J(U) on the energy space, i.e.and for then

is a linear functional in U. It is also continuous in because Hölder’s andFriedrich’s inequalities give

Thus we can write J(U) as

3.3 The membrane 81

and proceed as in § 3.1 to find that the minimizer satisfies

for all Thus

This has the same form as (3.3.10).Now consider the free membrane. Starting from the minimization of the

total energy we deduce (3.3.10) as before, but now is not necessarilyzero on the boundary, so that on using the divergence theorem (3.3.11) weobtain

Arguing as in § 3.1, i.e. first taking on we find that must satisfy(3.3.12). Therefore,

for all which implies

This is the natural, or so-called Neumann boundary condition for the free mem-brane.

To derive the generalized solution for the free membrane we first note, aswe expect, that (3.3.3) is not a norm on because implies only

When we considered the rod we placed and in thesame equivalence class if We could do that because wehad shown that if then they were continuous on Butnow, if it is not necessarily continuous on it is an equivalenceclass of Cauchy sequences of functions in the norm (3.3.3). Wemust therefore proceed differently. We use

Poincaré’s inequality, due to Jules Henri Poincaré (1854–1912). Ifthen

for some constant m independent of

Now suppose that is a Cauchy sequence of functions in in thenorm (3.3.3), and that

Then Poincaré’s inequality shows that is a Cauchy sequence inand that


Thus if then so that is zero almost everywhere(see (2.11.6)); it is zero in so that does constitute a norm, andmoreover

Again we can set up a generalized problem and derive (3.3.14), provided ofcourse that

Problem 3.3.1 Use the divergence theorem to show that (3.3.10) is a necessarycondition for the existence of a solution of (3.3.12) satisfying the boundarycondition (3.3.14).

So far we have considered only static problems. Now consider the freevibration problem for the clamped membrane. If the membrane has uniformmass/unit area and is executing free vibration

with frequency then the strain and kinetic energy of the membrane are

It is well known that natural frequencies of the membrane are related tothe eigenvalues of the equations

by the equation

Problem 3.3.2 Show that if then the classical solution of (3.3.22)makes

stationary. Hence show that a generalized statement of the eigenvalue problemis the integro-differential equation

where

3.4 The plate in bending

3.4 The plate in bending 83

For a general anisotropic and nonhomogeneous plate occupying a domainand undergoing small out-of-plane bending, the strain energy has the form

in Cartesian coordinates. Here take the values 1,2; is the tensorof elastic constants of the plate, satisfying

and we have used Einstein’s double suffix summation convention, due to AlbertEinstein (1879–1955). We assume that the tensor is positive definite inthe sense that there is a constant such that for any we have

We assume first that the plate is clamped, so that

We show that on the set S of all satisfying (3.4.3), and

does constitute a norm. First consider N1. We have

so that implies i.e.

Thus and the clamped boundary condition (3.4.3) yieldsand

Problem 3.4.1 Show that the norm (3.4.4) satisfies the axiom N3 of § 2.8.

The completion of S in the norm (3.4.4) is called the energy space Letus consider properties of elements Since on Friedrich’sinequality gives


But on so that on applying Friedrich's inequality first to

and then to we find

This means that if is a Cauchy sequence in thenare all Cauchy sequences

in Thus if then it and its first and second generalized deriva-tives are all in

But we can say more; we can argue for the plate (a fourth order system)in as we did for the rod (a second order system) in Extend byzero outside the plate, and suppose that lies in some squarethen

so that on using Hölder’s inequality we find

This means that a Cauchy sequence in is a Cauchy sequence in the uniformnorm on so that it converges to a function Thus

(We encourage the reader to study the analysis for the rod, beam, membrane andplate, and try to find the patterns that are appearing with regard to imbedding.)

As with the membrane we can set up the generalized problem in the formof the minimization of a functional J(W) and obtain (3.1.25). We note

Problem 3.4.2 Show that the external work functional

is continuous in if and

For the free plate we find, as we expect, that does not constitute anorm. Now we must use Poincaré’s inequality twice to obtain

3.5 Linear elasticity 85

For any function we can find constants such that

satisfies

We can thus take Cauchy sequences composed of such functions and showthat

so that W, and are all in Moreover, if soalso is so that W is zero almost everywhere, and

The argument which we used for the clamped plate to show that ifthen does not hold for the free plate; equation (3.4.8) does not hold,however, the result still holds. This means that is continuous on sothat, in particular, if then on

3.5 Linear elasticity

Consider an elasticity body occupying a bounded region The strainenergy of the body in Cartesian coordinates is

where is the tensor of elastic constants, take the values 1,2,3 andwe use Einstein’s summation convention. is the strain component

where is the displacement vector. The tensor of elastic constantsis symmetric, in the sense

and positive definite, in the sense that there is a constant such that, forall we have


On the set S of twice continuously differentiable vector-functions u(x),(i.e. whose components are in displacements of points of thebody, introduce a metric with inner product

This fulfills P2 and P3. For P1 we note that impliesfor all and, as is known from the theory of elasticity, this means

where a, b are constant vectors, and × denotes the vector product. (This isthe so-called general form of the vector of infinitesimal motion of the body as arigid body.) If we restrict u by the boundary condition

then so that (3.5.4) is an inner product. The completion of S inthe norm given by (3.5.4) is the energy space for the body.

To describe properties of elements of we establish Korn’s inequality. Toprove this we need to apply the divergence theorem to derivatives of u; this iswhy we require

Korn’s inequality. For a vector function

Proof. The integral on the R.H.S. of the inequality is

With the notation we have

so that

But

3.5 Linear elasticity 87

where we have used the double suffix summation on the right. Now

and since the divergence theorem gives

and thus

Now apply the elementary inequality to these integrals toobtain

Thus

Corollary For a vector function

Proof. Friedrich’s inequality applied to shows that

On the other hand, the positive-defmiteness condition (3.5.3) shows that

The corollary to Korn’s inequality shows that if then and thefirst derivatives all belong to

Note that the construction of the energy is the same if the boundary con-dition (3.5.5) is given only on some part of of the boundary, i.e.


The inequality (3.5.6) remains valid, but its proof is more complicated.If we consider an elastic body with a free boundary, we meet difficulties

similar to those we encountered with the free membrane or plate. To circumventthe difficulty associated with the zero in the energy space we introduce therestrictions

These make the ‘zero’ of the energy space actually zero, and ensure that Korn’sinequality remains valid for the vector functions.

3.6 Sobolev spaces

The Sobolev spaces, due to Sergei L’vovich Sobolev (1908–1989), which we in-troduce in this section, can be considered as mathematical generalizations of theenergy spaces that we have introduced in the previous five sections. They canalso be regarded as generalizations of the Lebesgue spaces; in spacesthe metric measures not only the distance between two functions, but also thedistance between their derivatives.

Let be a domain, a non-empty open set, in We recall the definition

Let be a non-negative integer, and let denote the set of functionswhich have bounded continuous derivatives

Definition 3.6.1 A semi-norm, on a linear space X is a real valuedfunction satisfying N2, N3 of Definition 2.8.1, and with N1 replaced by

and if (not iff

Introduce the semi-norm

where denotes the Lebesgue norm (2.11.2). Thus if

if N = l,

if N = 2,

3.6 Sobolev spaces 89

We note that the energy norms for the cantilever rod, clamped membrane andclamped plate were similar to the semi-norms

Now we introduce the norm

so that if, for instance N = 2, then

We define to be the completion of in the norm Clearlyif is a representative Cauchy sequence for then isa Cauchy sequence s in for any such that We can thereforetake Cauchy sequence to define elements which we label

We recall that is the set of functions having continuous derivativesof all orders in and having compact support in i.e. their supports, whichare closed, lie inside We define to be the completion of inthe norm is a subspace of The spacesform the generalization of the energy spaces for the clamped membrane andplate. We will now show that the semi-norm | · | m,p is a norm on thiswill generalize what we found earlier: was a norm for was anorm for To do this we need an inequality which is a generalization ofFriedrich’s inequality (3.3.5); it, like (3.3.17), is called Poincaré’s inequality; itis proved by generalizing the proof of Friedrich’s inequality.

Poincaré’s inequality. Let be a bounded domain in There is a positiveconstant C, depending on and such that

Moreover defines a norm on which is equivalent to

Proof. Suppose is the ‘box’ N . Supposeuse the abbreviation

Then

since Now use Hölder’s inequality to give


with so that

Now integrate over to give

and thus

This proves (3.6.6) for But is dense in so thatthe inequality holds for all If is not a box, then we can encloseit in a box and extend functions by zero to the box to givefunctions The inequality (3.6.6) applied to and is equivalentto the same inequality applied to and

To show that defines a norm on we need only verify N1. Thuswe need to show that if and then This followsimmediately from (3.6.6). We also see that if and then

so that F is zero almost everywhere. To see that andare equivalent norms on we note that

Problem 3.6.1 Show that is a norm which is equivalent to on

3.7 Some imbedding theorems

In these first three chapters we have introduced a number of function spaces,and at various times, e.g. in § 2.11, § 3.1, § 3.2, we have shown that one spaceX has been imbedded in another space Y, according to Definition 2.11.4. In thissection we will draw these results together, and make some generalizations.

In § 2.3 we introduced two families of function spaces:and based on the uniform metric (2.3.1), so that

and based on the metric (2.3.3), so that

3.7 Some imbedding theorems 91

(We could also use (2.3.4) instead of (2.3.3).) The spaces andare subsets of in fact

while

In § 2.5 we showed that under the norm (3.7.1) is a complete normedlinear space, i.e. a Banach space. We can show similarly that is a Banachspace.

The first imbedding we note is one between and Clearlyis a subspace of i.e. and

so that the operator from to is bounded. Thus

We now define another family of function spaces. To do so, we return tothe definition of it is the set of functions that are bounded and uniformlycontinuous on This means that, given we can find such thatif and then The important conditionwhich distinguishes uniform continuity from ordinary continuity is that we canfind one which will fit any two

We now introduce

Definition 3.7.1 Let is the subspace of consistingof those functions which satisfy a Hölder condition, that is there is a constantK (depending on such that

(Note that in the notation of § 2.3.) Such functions aresaid to be Hölder, or Lipschitz continuous, after Rudolf Otto SigismundLipschitz (1832–1903), if

Problem 3.7.1 Show that if satisfies (3.7.4), then it is uniformly contin-uous on Construct a function which is uniformly continuous on (0,1), butdoes not satisfy a Hölder condition, for any on (0,1).

Note that there is no point in considering in Definition 3.7.1, because

Problem 3.7.2 Show that if r satisfies (3.7.4) with then isnecessarily constant.

Theorem 3.7.1 is a Banach space with the norm


Proof. is defined in (3.7.1). We need to prove first that (3.7.5) doessatisfy the norm axioms in Definition 2.8.1. This is straightforward. Secondly,we must show that a Cauchy sequence of functions has a limitin

Suppose is such a Cauchy sequence, in the norm Equa-tion (3.7.5) shows that

This means that is a Cauchy sequence in since is complete,converges to a function The statement that is a Cauchy

sequence in means that given we can find N such thatimplies

Choose Choose Since aswe can find M > N such that

Thus so that

But 1 is arbitrary, so that

Thus converges to in the norm of and if thensatisfies the Hölder condition with Therefore

also satisfies the Hölder condition, so that the Cauchysequence converges to i.e. is a Banach space.

The inequality (3.7.6) gives the imbedding

We can generalize according to


Definition 3.7.2 Let is the subspace of consist-ing of those functions with derivatives satisfying a Höldercondition, that is, there is a constant K (depending on such that

To provide a norm for we introduce the notation

and then define

Again we see that

We can prove, as before, that is complete; it is a Banach space; theinequality (3.7.8) gives the imbedding

The next imbedding is given in

Problem 3.7.3 Suppose that Show that if then

and

Hence show that

This means that if then

We now have three imbeddings: (3.7.3), (3.7.9) and (3.7.11). These showthat behaves, in a way, like a fractional as increases from 0 to 1,

an becomes nearly differentiable one more time, i.e. nearly inOne would therefore expect that would be imbedded in

i.e.


This is true for many domains e.g. all convex or star-shaped domains, andin fact for all domains for which there exists a constant M > 0 such thatany two points may be joined by a piecewise straight line with totallength not exceeding M times the length Thus, in Fig. 3.7.1,

The number may depend on but one M applies for all

Problem 3.7.4 Show that if (3.7.12) holds, then

for

We now consider another set of imbeddings, those relating to Sobolev spaces.We start by interpreting the imbeddings that we found in § 3.1–3.4 as imbed-dings for Sobolev spaces.

Our first result was (3.1.10). For this we have a domain inwith N = 1. According to (3.6.5) we may write (3.1.10) as

or

The next result is that for the beam, (3.2.8), which gives

The results for the clamped membrane and plate (3.3.8), (3.4.10) give


The imbedding theorems all have the form

where X is another Banach space. The theorems show that if a functionthen the amount by which it is ‘constrained’ depends essentially on

the value of in relation to the dimension N of the space Thus we willfind that if we can say that must be bounded, or even uniformlycontinuous. As one would expect, some of the results hold for an arbitrarydomain in others hold only if has special properties.

We will not prove the theorems, nor will we state them in their full gener-ality; for this the reader may see Adams (1975). The proofs are not so muchdifficult, as intricate; they require carefully chosen integrations by parts andapplications of Hölder’s integral inequality.

We will state the theorem, generally called Sobolev’s imbedding theorem,in three parts, first for an arbitrary domain, then for two restricted types ofdomains.

Theorem 3.7.2 Let be a domain in let be a nonnegative integer,and

if then

if then

if and then

if then

if then

if then


Problem 3.7.5 Identify the imbeddings (3.7.13), (3.7.16) as special cases of(3.7.21), and (3.7.15) as a special case of (3.7.18).

This theorem holds for an arbitrary domain. Now we introduce

Definition 3.7.3 The domain has the cone property if there isa finite cone C such that each point is the vertex of a finite conecontained in and congruent to C. (Note that need not be obtained fromC by parallel translation, just by rigid motion.)

Most ‘ordinary’ domains bounded or unbounded, have the cone property;certainly balls, cubes or parallelopipeds do. If is bounded, then a sufficient,but by no means necessary, condition for to have the property, is that it havethe Lipschitz property, according to

Definition 3.7.4 The domain has the Lipschitz property if foreach point there is a neighborhood in which the boundary is the graphof a Lipschitz continuous function, according to Definition 3.7.1.

We now state

Theorem 3.7.3 Let be a domain in having the cone property, let bea non-negative integer, and then the imbeddings (3.7.17)–(3.7.20)hold with replaced by

Corollary 1. If is a non-negative integer, and

then

Proof. If then for so thatthus The assumed imbedding gives

Theorem 2.8.1 states that any two norms in are equivalent; thus


Corollary 2. If is a non-negative integer, and

then

Proof. If then for so thatthus The assumed imbedding gives

so that in

Theorem 3.7.4 Let be a domain in having the Lipschitz property,let be a non-negative integer, and then the imbeddings (3.7.21)–(3.1.22) hold with replaced by

Corollary If is a non-negative integer and

then

The proof follows the same lines as Corollary 2 above.

Problem 3.7.6 Identify (3.7.14) as a special case of (3.7.26).

References

A full treatment of Sobolev spaces is to be found in

R.A. Adams, Sobolev Spaces, Academic Press, 1975.

See also

J.-P. Aubin, Applied Functional Analysis, John Wiley, New York, 1979.


Synopsis of Chapter 3: Energy spaces

Energy space: completion of space of functions with bounded strain energy;see (3.1.4); (3.2.6); (3.3.3); (3.4.4).

Generalized solution: solution in the energy space; see (3.1.28), (3.2.18),(3.3.13) etc.

Sobolev space: complete space with norm which measures derivatives of afunction; see (3.6.1), (3.6.5).

Imbedding: means and see Theorem 3.7.2.

4. Approximation in a Normed Linear Space

They say there was a fish who said two words in such a strangelanguage that for three years scientists have been trying to under-stand it.

N.V. Gogol, Notes of a Madman

4.1 Separable spaces

If we want to know whether a room holds enough chairs to seat some peoplestanding outside we can do one of two things:

Count the number of chairs, and the number of people, and seewhether

Start seating the people, and continue until all the chairs are filled, or allthe people are seated, whichever comes first.

The second procedure has the advantage that it avoids counting; it relies onestablishing a one-to-one correspondence between chairs and people. This leadsto

Definition 4.1.1 Two sets are said to have equal power if there is a one-to-one correspondence between their elements.

The set of positive integers 1,2,3, ··· is the set containing an infinity ofelements which has the least power.

Definition 4.1.2 A set which has the same power as the set of positive integersis said to be countable (enumerable).

Theorem 4.1.1 The union of a finite number, or a countable set, of countablesets is countable.

Proof. It suffices to show how to enumerate the elements of the union. Thisis clear from the diagram in Fig. 4.1.1. The countable sets are

etc. We take them in the order

100 4. Approximation in a Normed Linear Space

etc. In this way we will cover all the elements in the union, and wecan place the elements so ordered in a one-to-one correspondence with 1; 2,3;4,5,6; etc.

Corollary 4.1.1 The set of rational numbers is countable.

Problem 4.1.1 Show that the set of all polynomials with rational coefficientsis countable.

Georg Cantor (1845–1918) proved

Theorem 4.1.2 The set of real numbers in the interval [0,1] is not countable.

The proof can be found in any textbook on set theory or of functions of areal variable. Cantor’s Theorem implies that the set of points (real numbers) in[0,1] does not have the same power as the set of positive integers; these pointsform a continuum. We can extend this to say that the set of points in formsan N-dimensional continuum.

We shall not discuss Cantor’s theory of sets, which is a special subject.Our interests lie in applications of the notion of countability to metric spaces.Modern mechanics depends heavily on computer ability. A computer can processonly a finite set of numbers. An operator using a computer expects a solutionto a problem to be approximated with a certain accuracy by the sequence ofnumbers used by the computer. If is an arbitrary element of an infinite set Xand we want to use a computer to find an approximation to it, then we must becertain that every element of X can be approximated by elements of anotherset which is finite or, at least, countable. This leads to:

4.1 Separable spaces 101

Definition 4.1.3 X is called a separable space if it contains a countablesubset which is dense in X.

We call such a subset a countable dense subset. In other words, X is sepa-rable if there is a countable set such that for every there existsa sequence such that as Equivalently,for any there is, for each an element (depending onin an of

The set of real numbers in [0, 1] is a separable metric space, since the setof rational numbers in [0, 1] is a countable dense subset.

There is a more important example: Weierstrass’ theorem on polynomialapproximation (Theorem 1.3.1) states that if is a bounded domain inthen the set of polynomials (with real or complex coefficients) is densein in the uniform norm. If is the set of polynomials with rationalcoefficients, then

Problem 4.1.2 Show that is dense in in the uniform norm.

This means that is separable, because is countable. Puttingthis together with Weierstrass’ theorem, we can deduce that is a count-able dense subset of so that is separable, again in the uniformnorm.

However, not all spaces are separable, because

Lemma 4.1.1 The space of all bounded functions on [0, 1] equipped with thenorm

is not separable.

Proof. It is sufficient to construct a subset M of the space whose elements can-not be approximated by functions from a countable set. Let be an arbitrarypoint in [0,1]. Construct a set M of functions defined as follows:

The distance from to is

Take a ball of radius 1/3 about If the intersectionis empty. This means that every element of M is an isolated point; there isjust one element in each ball of radius 1/3 about If a set S isto be dense in M, then each of these balls must contain at least one elementof S. But the set of balls with real values of has the same power as


the continuum, i.e. it is not countable (Theorem 4.1.2). Therefore there is nocountable set which is dense in M . Therefore M , and a fortiori (meaning ‘allthe more so’) the set of bounded functions, is not separable.

We will now proceed to show that the Lebesgue spaces and theSobolev spaces are separable. We prove a general result:

Theorem 4.1.3 The completion of a separable metric space is separable.

Proof. In the notation of Theorem 2.6.1, there are three metric spaces: M , theoriginal incomplete space composed of elements the space of stationarysequences the space of equivalence classes X of Cauchy sequences

where with the metric (2.6.1), namely

In Theorem 2.6.1 we showed that is dense in . Since M is separable, ithas a countable dense subset D. Let be the space of stationary sequences

for If and then we can find suchthat Let and then

so that is dense in which in turn is dense in ; therefore which,likeD, is countable, is dense in , and is separable.

Using this general result we may now prove

Theorem 4.1.4 If is bounded, then is separable.

Proof. is the completion of in so, by Theorem 4.1.3, it issufficient to prove that is separable in But is dense inin so it is sufficient to show that is separable in Weierstrass’theorem states that is dense in in the uniform norm. Thus, if

and we can find such that

This implies

so that is dense in in the norm, and is separable in

Weierstrass’ theorem states that is dense in so thatis separable, in the uniform norm. From that we showed that and its

4.2 Theory of approximation in a normed linear space 103

completion in the norm, are separable. Using similar argumentswe can show that is dense in in the metric (2.3.4), and hencealso in the norm, so that is separable.

We conclude this section with the almost trivial result

Problem 4.1.3 Show that any subspace E of a separable metric space X isseparable.

The result is of great importance, for the following reason. We have onlya few convenient countable sets of functions which we may use to show thatvarious spaces are separable: the space of polynomials with rational coef-ficients; the space of trigonometric polynomials with rational coefficients; etc.In general, the elements of these sets will not satisfy the boundary conditionsimposed on functions in energy spaces, for example, so that we cannot use themto show that the energy spaces are separable. To circumvent this difficulty wecan take a wider space, containing the space under consideration, and show thatit is separable; Problem 4.1.3 shows that the subspace is separable. In § 3.6 weshowed that the energy spaces we introduced were subspaces of Sobolev spaces;since the Sobolev spaces are separable, so too are the energy spaces.

4.2 Theory of approximation in a normed linear space

The first problem we will consider is relatively simple, the so-called general prob-lem of approximation in a normed linear space: Given andwith find numbers to minimize

The problems of best approximation of a continuous function by an orderpolynomial, by a trigonometric polynomial, or by some other specified functions,all have this form. Our analysis will depend on a well known result from thetheory of continuous functions, which we stated as Theorem 1.2.1.

Now return to (4.2.1). We suppose that are linearly indepen-dent. This means that the equation

implies

We show that the problem of minimizing (4.2.1) has a solution: we prove theexistence theorem

Theorem 4.2.1 For any there is an depending on such that

and


Proof. Define and let The triangle

inequality gives

We use this to give the following chain of inequalities:

This shows that both of the functions andare continuous in (or in if X is a complex space). The function is areal homogeneous function of degree 1, i.e.

First consider on the unit ball This is a closed and bounded,

i.e. compact, set in so, by Theorem 1.2.1, the real continuous functionwill assume its minimum value at some point on the unit ball.This minimum value must be nonzero, since the are linearly independent.

Thus if then

We now show that the minimizing of the theorem must lie in a ball ofradius For the inequality (4.2.2) gives

On the unit ball, therefore, outside the ball of radiusand But so the minimum valueof must be inside the ball of radius R; since this is a closed and boundedset we see that the minimum will actually be attained; we can say min in thestatement of the theorem, not just inf.

4.2 Theory of approximation in a normed linear space 105

We called the general problem in the theory of approximation relativelysimple; we mean that the problem of solvability is simple, not that concreteapplications of this theory are simple.

We have shown that the problem of minimizing (4.2.1) has a solution; in ageneral normed linear space the solution is not unique; it is unique when X isa strictly normed space.

Definition 4.2.1 A normed linear space is called strictly normed if the equal-ity

implies and

Most of the normed linear spaces which appear in applications are in factstrictly normed. First we have

Lemma 4.2.1 An inner product space is strictly normed.

Proof. Suppose then For acomplex space this may be written

so that But the Schwarz inequality (2.12.2) states that

so that and thus But then Theorem 2.12.1 givesand hence so that and

Now we pose

Problem 4.2.1 Use the properties of the Minkowski inequality to show that thespaces and are strictly normed when

We also need the general notion of a convex set; we have

Definition 4.2.2 A set M in a linear space is said to be convex if, for anytwo elements each element with is also inM. (Thus when are in a convex set M , the segment joining and liescompletely in M .)

Problem 4.2.2 Show that the closed unit ball innormed linear space X is convex.

When the set M of elements


is a closed convex set, so that the problem of minimizing (4.2.1) can be viewedas finding which is closest to

We now combine the notions of strictly normed space and convex set to givea uniqueness theorem. First, however, we combine Definitions 2.7.2 and 2.9.2into

Definition 4.2.3 Let X be a linear space. An operator (linear operator) fromX into or is called a real or complex functional (linear functional).

Theorem 4.2.2 Let X be a strictly normed linear space, let and letbe a closed convex set. There is no more than one which

minimizes the functional on M.

Proof. If there is clearly only one minimizer, Supposeand that, if possible, there are two minimizers, and Thus

Since M isconvex, so that

On the other hand

Thus

But X is strictly normed so that this equality impliesand (4.2.3) implies so that This

is a contradiction, so that there can be at most one minimizer.

In this section we have shown that the problem of minimizing (4.2.1) alwayshas a solution, and that this problem, as well as the more general problem ofTheorem 4.2.2 has at most one solution if the space is strictly normed. Thus theminimizing problem (4.2.1) has a unique solution in a strictly normed space.

We now proceed to study the problem stated in Theorem 4.2.2 when X isa Hilbert space.

4.3 Riesz’s representation theorem

Riesz’s theorem, due to Frédéric Riesz (1880–1956), is the most important ofa number of results we shall obtain in this chapter about approximation in aninner product, and in particular a complete inner product, i.e. a Hilbert, space.

4.3 Riesz’s representation theorem 107

First we show the existence of a solution to the problem of Theorem 4.2.2.We have

Theorem 4.3.1 Let H be a Hilbert space, let and let be aclosed convex set. There is a unique which minimizes the functional

on M.

Proof. The uniqueness is proved in Theorem 4.2.2; in Theorem 4.2.1 we showedthe existence in the special case in which M is a finite dimensional subspaceof a normed linear space; now we will establish the existence for an arbitraryclosed convex set in a Hilbert space.

Let be a minimizing sequence for i.e.

Such a sequence exists by the definition of infimum. If we can show thatis a Cauchy sequence, then, because M is closed and a closed set in a completespace is itself complete, (Problem 2.5.3) it will have a limit in M , and this willbe the minimizer.

Since H is an inner product space, the parallelogram law (2.12.3) holds.Thus

and therefore

Since we can write where asSince M is convex, and

Thus

so that is a Cauchy sequence, having a limit

Now suppose that M is not just a closed convex set, but a closed subspace(Definition 2.8.4). If and M is convex, then for

When M is a subspace, then for any Thusthere is a unique minimizer i.e.

Take an arbitrary we consider the real valued function


of the real variable Because M is a subspace, for all andhas its minimum at Thus But

Replacing by we get so that

This means that is orthogonal to every and leads us to

Definition 4.3.1 Let H be a Hilbert space, a linear subspace, andThe element is said to be orthogonal to M if is orthogonal to

every i.e. for all Two subspaces aresaid to be mutually orthogonal if for all and Wewrite

Definition 4.3.2 Let H be a Hilbert space, and be mutually or-thogonal subspaces. We say that H has an orthogonal decomposition intoM and N if any can be uniquely represented in the form

We can state the result already obtained in equation (4.3.1) as the so-calleddecomposition theorem for a Hilbert space:

Theorem 4.3.2 Let H be a Hilbert space, and a closed subspace.Then there is a closed subspace orthogonal to M , such that H has anorthogonal decomposition into M and N , as in (4.3.2).

Proof. Suppose Let N be the set of all such that anyis orthogonal to every We show that N is not empty, is

a linear subspace of H , and is closed. Equation (4.3.1) shows that N is notempty. N is a linear subspace because for allimplies for all Thus if then

Suppose is a Cauchy sequence. Since H is complete,will have a limit and

so that and N is closed.The analysis leading to (4.3.1) shows how to construct the projection

M of an arbitrary is orthogonal to M and Thedecomposition is unique because

4.3 Riesz’s representation theorem 109

implies

But and M , N are mutually orthogonal, i.e.

so that and

The element is the projection of on M . We may consider theprojection as defining a projection operator P on H onto M , according to

Definition 4.3.3 Let H be a Hilbert space and M a closed subspace of H . Theprojection operator P on H onto M is defined by where

Clearly when

Theorem 4.3.2 has widespread applications. One of them is Riesz’s repre-sentation theorem:

Theorem 4.3.3 Let H be a Hilbert space, and be a continuous linearfunctional on H. There is a unique such that

and

Proof. Let M be the kernel of the set of satisfying Weshow that M is a closed linear subspace. If then

but F is linear so that andthus is a subspace. If is a Cauchy sequence,then it has a limit But since F is continuous,

so that and M is closed.Since M is a closed linear subspace of H we can apply Theorem 4.3.2.

There is a subspace orthogonal to M , and any can be uniquelyrepresented as

We now show that N is a one-dimensional space, i.e. any may bewritten as where is a fixed element in N . Let then

Butwhich means But M, N are mutually orthogonal so that being inboth, must be zero. Thus and N is one-dimensional.

Take an element and define by

Any element can be represented uniquely as


and Therefore

whereIf there were two representing element then

so that and taking we would findi.e.

Finally,

but so that

The proof above was given for a complex Hilbert space, but the result holdsfor real spaces also.

The meaning of the theorem is that any continuous linear functionalon H can be identified with a unique element The set of continuouslinear functionals on H is called the dual of H , and is denoted by H*.Riesz’s theorem gives a one-to-one correspondence between elementsand

4.4 Existence of energy solutions of some mechanicsproblems

In this section we discuss some applications of Riesz’s theorem. We recall thatin Chapter 3 we introduced generalized solution for several mechanics problemsand reduced these problems to that of finding a solution to the abstract equation

(see for example (3.1.28)) in an energy space. (We will use lower case letters,rather than script capitals to denote elements of the energy space.) (There weresome restrictions on the forces to ensure the continuity of the linear functional

in the energy space.) The following theorem concerns the solution of thesegeneralized problems.

Theorem 4.4.1 Let be a continuous linear functional on a Hilbert spaceH . There is a unique element which satisfies (4.4.1) for every

Proof. By Riesz’s representation theorem, there is a unique elementsuch that the continuous linear functional can be written in the form

and so equation (4.4.1) becomes

4.4 Existence of energy solutions of some mechanics problems 111

Writing this as

we see that this is zero for all iff i.e. This is theunique solution of (4.4.1).

Now let us consider another application of Riesz’s theorem. In Problem 3.3.2we set up the integro-differential equation

as the generalized statement of the eigenvalue problem (3.3.22). We need to findand corresponding so that satisfies (4.4.2) for every

Problem 4.4.1 Show that if and satisfy (4.4.2) for everythen is real.

First we consider the term

the inner product of in the space for fixed as a linearfunctional in for The Schwarz inequality (2.12.2) gives

while Friedrich’s inequality (3.3.7) gives

This inequality states that is a continuous functional in the Hilbert spaceBy Riesz’s theorem, has a unique representation

where, from now on, we implicitly take all inner products and norms inWhat have we found? For any there is a unique element suchthat (4.4.4) holds. The correspondence defines an operatoracting from to

Let us study some properties of this operator. First we show that it is linear.Let

Then on the one hand we have


and on the other hand

Combining these, we have

But is an arbitrary element of so that

and K is linear operator.Now let us rewrite the inequality (4.4.3) using K ; it is

Taking we have

so that

and K is a continuous operator.Now return to equation (4.4.2) which we can write as

But is an arbitrary element of so that equation (4.4.6) is equivalent to

with a continuous linear operator K .The inequality (4.4.5) shows that

so that, if satisfies the conditions of Definition 2.7.4 for acontracting mapping. The contracting mapping theorem (Theorem 2.7.1) statesthat has a single fixed point, which, since K is linear, is Thismeans that if the only solution of equation (4.4.7) is i.e.equation (4.4.7), and therefore (4.4.2), has no eigenvalues satisfying

4.5 Bases and complete systems 113

Problem 4.4.2 Set up a generalized problem similar to (4.4.2), for the freevibration of a membrane (with fixed boundary) with variable, but bounded, massdensity. Show that the problem can still be reduced to the form (4.4.7), whereK is a continuous linear operator in

Problem 4.4.3 Carry out the analysis of the free vibration of a plate, withclamped boundary, and with variable, but bounded, mass density. Show that theproblem can be reduced to the form (4.4.7), where K is a continuous linearoperator in

4.5 Bases and complete systems

If a linear space X has finite dimension there are linearly independentelements called a basis for X , such that every element hasa unique representation

where the are scalars. We now generalize this definition to an infinite di-mensional normed space X .

Definition 4.5.1 Let X be a normed linear space. A system of elementsis said to be a basis for X if any element has a unique

representation

with scalars Note that the meaning of (4.5.1) is: if then

It is clear that a basis is a linearly independent system since theequation

has the unique solution

Problem 4.5.1 Show that if a normed linear space X has a basis, then it isseparable. (Show that there is a countable set of linear combinations of the form

with and rational coefficients which is dense in X .)

Consider the normed linear space C[0, 1] of continuous functions on [0, 1]under the metric (2.3.1), and remember that convergence in this metric means


uniform convergence. We ask whether the powers form a basis forC[0, 1]. If they did, then any continuous function in C[0, 1] could beexpanded as a uniformly convergent power series

in [0, 1]. But this means that is analytic, and there are clearly continuousfunctions which are not analytic. Therefore the powers do not constitute a basisfor C[0, l].

Problem 4.5.2 Construct a function which cannot be expressedas a uniformly convergent power series.

Even though the powers do not form a basis for C[0, 1], Weierstrass’ theoremstates that they do have properties similar to those of a basis: we can find apolynomial arbitrarily near, in the uniform norm, to any function in C[0, 1].This leads us to the next definition

Definition 4.5.2 Let X be a normed linear space. A countable systemis said to be complete in X if for any and any there is a

finite linear combination of the such that

We can also refer to a system of elements that is complete in a subset S ofa normed linear space X . This simply means that for any and element ofS, we can find a finite linear combination of elements of the system such thatthe distance between the element and the sum is less than

Let us be clear about the distinction between a basis and complete system.For the former, the depend only on if we are given and want tomake we simply take large enough, i.e. take more of the Forthe latter, the values of the as well as the value of will depend on if oneset of coefficients makes (4.5.2) true for one value of and wedecrease to we will not only have to take more i.e.but also, maybe, have to change to

Weierstrass’ theorem states that the powers are complete inC[0, 1], and generally, that the composite powers are completein where

Problem 4.5.3 Generalize the last result to find a system which is complete inand when

The problem of the existence of a basis for a particular normed linear spacecan be very difficult, but there is a special case when this problem is fully solved:


when the space is a separable Hilbert space. Those who are familiar with thetheory of Fourier Series, after Jean Baptiste Fourier (1768–1830), will see thatit will largely be repeated here in abstract terms.

We begin with

Definition 4.5.3 Let H be a Hilbert space. A system of elements issaid to be orthonormal if, for all integers

There are many advantages in using an orthonormal system of elements asa basis.

If we have an arbitrary linearly independent system of elementsin a Hilbert space H , they will span a subspace We may form an orthonor-mal basis for by using the familiar Gram–Schmidt process, named afterJórgen Pedersen Gram (1850–1916) and Erhard Schmidt (1876–1959):

so that

Applying the Gram–Schmidt procedure to subsets of monomials in thespaces L(a, b) , we get systems of polynomials that are called orthogonal poly-nomials. Orthogonal polynomials are widely used in mathematical physics.

Problem 4.5.4 Show that constructed in the Gram–Schmidt process willbe orthonormal iff are linearly independent.

If H is a, separable Hilbert space, then, by definition, it has a countabledense subset From this we may, by the Gram–Schmidt process, constructan orthonormal set which is dense in H ; this will be a complete orthonormalsystem in H .

Although there are Hilbert spaces which are not separable, the importantones, and are separable. The following theorem is based onthe premise ‘If H has a complete orthonormal system’.

Theorem 4.5.1 Let H be a Hilbert space. If H has a complete orthonormalsystem then it is a basis for H ; any element has a uniquerepresentation

so that

1.

2.

3.


called the Fourier series for The numbers are called theFourier coefficients of

Proof. First we consider the problem of best approximation of an elementby elements of the subspace spanned by In § 4.2 we

showed that this problem has a unique solution; now we show that it is

Indeed, take an arbitrary element i.e.

Then

But

and so that

which shows that takes its minimum value when i.e. whenThus

which gives

This inequality states that the sequence of partial sums of the series

is bounded above; it therefore converges, and we have Bessel’s inequality , afterFriedrich Wilhelm Bessel (1784–1846),


This means that the sequence of partial sums is a Cauchy sequence, for

We have not yet used the completeness of the system so that, inparticular, the results (4.5.6), (4.5.7) hold for any orthonormal systemH . The completeness of the system means that if and wecan find a number and coefficients such that

But then the inequality (4.5.4) for that shows that

which means that the sequence converges to in the norm of H . This isthe meaning of equation (4.5.3).

When the system is complete, we can sharpen (4.5.6). Indeed equa-tion (4.5.4) means that

so that

This is called Parseval’s equality, after Marc Antoine Parseval (1755–1846).

Now we introduce

Definition 4.5.4 Let H be a Hilbert space. A system is said to beclosed in H if the system of equations

implies

Theorem 4.5.2 An orthonormal system in a Hilbert space H is closed iff it iscomplete.

Proof. Suppose the orthonormal system is complete in H , and thatwith then


so that the system is closed. Now suppose that is closed. The system

with and is a Cauchy sequence in H ; because

H is a complete space, this sequence has a limit

and

But is closed, so that equation (4.5.10) implies i.e. is given by(4.5.3), so that is actually a basis for H .

Problem 4.5.5 Show that an orthonormal system in a Hilbert space H is closediff it is a basis for H .

Problem 4.5.6 Show that if a system is complete in a set S that is dense ina Hilbert space H , then it is complete in H . (Hint: For any elementthere is an element that is closer than to For there is a finitelinear sum of the system elements whose distance from s is less thanThe distance between and is less than which means that the system iscomplete in X).

An important application of this application concerns By defini-tion this is complete; it can be obtained as the completion of in thenorm. The functions

are orthogonal in Thus Bessel’s inequality states that ifand

then

One of the consequences of the convergence of the infinite series on the left isthat

But

so that if is real then


i.e.

The results in (4.5.12) are usually given the name, the Riemann–Lebesguelemma; they hold for We will now prove

Theorem 4.5.3 The system given by (4.5.11) is complete in

Proof. By Problem 4.5.6 it is sufficient to show that there is a dense set inin which the system is complete. The set of functions with

compact support in is dense in . These functions are continuouson and so, since supp is closed and bounded, are uniformlycontinuous in Since these functions satisfy they maybe continued to the whole real line as functions of period We may there-fore apply Weierstrass’ trigonometric polynomial approximation theorem (The-orem 1.3.3) to them. This means that, given we may find a trigonometricpolynomial of the form (1.3.6) which we may write in the form

such that

This completes the proof.

If then we may extend by taking

and then, where is given by (4.5.11),

Similarly, if we take


then we find

Theorem 4.5.4 A Hilbert space has a countable orthonormal basis iff it isseparable.

Proof. Problem 4.5.1 shows that if it has a basis it is separable. On the otherhand if H is separable it has a countable set which is dense in H ; we now applythe Gram–Schmidt process to this set to produce an orthonormal basis.

4.6 Weak convergence in a Hilbert space

Suppose is a sequence in and has components1,2, · · · , N. The sequence converges iff each of the sequencesconverges. In a Hilbert space with orthonormal basis the Fourier co-efficients play the part of components, but now there is a differencebetween the two kinds of convergence, as shown by the following example. Let

be an orthonormal basis for H , then for every

Thus the sequence of the components of tends to zero, but itselfdoes not converge, since

We need to introduce a new kind of convergence. For a general normed linearspace we have

Definition 4.6.1 The sequence in a normed linear space X is said to bea weak Cauchy sequence if, for every continuous linear functional onX, the sequence is a Cauchy sequence, in The sequence issaid to converge weakly to if, for every continuous linear functional

on X ,

We use for strong convergence, i.e. for weakconvergence.

Problem 4.6.1 Let X be a normed linear space. Show that if is a(strong) Cauchy sequence, then it is a weak Cauchy sequence. Show also that if

then

4.6 Weak convergence in a Hilbert space 121

Although it is possible to consider weak convergence in a general normedlinear space, as in Definition 4.6.1, we shall usually consider it only in an innerproduct space. In such a space, if then

is a linear functional on X . Then we can easily show

Problem 4.6.2 Let X be a normed linear space. Show that a sequenceX cannot have two distinct weak limits.

If X is a complete inner product space, i.e. a Hilbert space H , then Riesz’srepresentation theorem (Theorem 4.3.3) states that every linear functional onH has the form (4.6.1) for some This yields

Theorem 4.6.1 Let H be a Hilbert space. A sequence is a weakCauchy sequence if, for every is a Cauchy sequence. Thesequence converges weakly to if, for every

Theorem 4.6.2 If and then

Proof. Consider

But and so that

and

As we will see later, it is often easier, when discussing numerical methods, toestablish weak convergence rather than strong convergence. This is why the lastresult is important, and why weak convergence will be a major preoccupationin this presentation.

Problem 4.6.3 Show that in a finite dimensional Hilbert space,implies This implies that in a finite dimensional space weak conver-gence and strong convergence are synonymous.

Theorem 4.6.3 A weak Cauchy sequence in a Hilbert space is bounded.


Proof. Suppose, on the contrary, that is a weak Cauchy sequence whichis unbounded. Let denote the closed ball with center radiusWe show that if then there is a sequence of pointssuch that We take

then so that and

since the numerical sequence is bounded, because is weaklyconvergent.

We now obtain a contradiction. Take By the above argu-ment, we can find and such that

By the continuity of the inner product, we can find a ballsuch that this inequality holds for all

Now apply the same argument to and find with and aball such that

Repeating this procedure ad infinitum , we find a nested sequence of closed ballssuch that

Since H is a Hilbert space, there is at least one element which belongs toeach and

Thus we have a continuous linear functional for whichis not a Cauchy sequence, i.e. a subset of the weak Cauchy sequence

is not itself a weak Cauchy sequence. This is impossible.

A corollary of the proof of this theorem is the following statement.

Corollary If is an unbounded sequence in a Hilbert space H , then thereis a and a subsequence such that as

Proof. Let us introduce the sequence For any with unit normthe numerical sequence is bounded and thus we can select a convergentsubsequence. If there exist such a unit element and subsequence for


which then the statement of the Corollary is valid for thesubsequence and if or if

If, on the other hand, we cannot find such and then we haveas for any which means that tends weakly to zero.

In this case we demonstrate the Corollary using the second part of the proof ofthe above theorem. In it, the existence of an element and subsequencesuch that as is a consequence of two facts:

is unbounded, which is the case;1.

the numerical set when runs over any is unbounded.2.

The proof of the latter we give under the additional condition thatas and this will complete the proof of the Corollary.

First, the element belongs to Next,

Since is finite and as we have asneeded.

We will use the corollary to prove the Principle of Uniform Bounded-ness, contained in

Theorem 4.6.4 Let be a family of continuous lin-ear functionals defined on a Hilbert space H . If then

Proof. Riesz’s representation theorem states that each has the form

The condition of the theorem is therefore

If then the Corollary to Theorem 4.6.3 would state that there

is an and a subsequence such that which wouldcontradict (4.6.2).

Problem 4.6.4 Use Theorem 4.6.4 to prove that if is a sequenceof continuous linear functionals on H , and if for every the sequence

is a Cauchy sequence, then there is a continuous linear functionalon H such that


and

The following theorem gives a convenient check for weak convergence.

Theorem 4.6.5 A sequence is a weak Cauchy sequence in a Hilbert spaceH iff

is bounded in H , i.e. there is a M such that1.

for any from a system which is complete in H, the numericalsequence is a Cauchy sequence.

2.

Proof. The necessity of the conditions follows from the definition of weak con-vergence and Theorem 4.6.4.

Now we prove the sufficiency. Suppose conditions 1 and 2 hold. Take anarbitrary continuous linear functional defined, because of the Riesz represen-tation theorem, by an element and consider the numerical sequence

The system is complete; given we can find

a linear combination such that

Then

Since, by 2, each of the sequences is a Cauchysequence, we can find a number R such that

so

This means the sequence is a Cauchy sequence.

Problem 4.6.5 Show that a sequence is weakly convergent to in H iff

is bounded in H ;1.


for any from a system which is complete in H ,2.

Since weak convergence differs from strong convergence we need to definethe terms weakly closed and weakly complete

Definition 4.6.2 Let X be a normed linear space. A set is said to beweakly closed in X if all its weak limit points are in S . Thus ifthen implies

Definition 4.6.3 Let X be a normed linear space. X is said to be weaklycomplete if every weak Cauchy sequence (Definition 4.6.1) converges weaklyto an element

We first prove the important

Theorem 4.6.6 A Hilbert space (a complete inner product space) is weaklycomplete.

Proof. Suppose is a weak Cauchy sequence. For any we maydefine the linear functional Theorem 4.6.5 states thatis bounded, i.e. for all so that

Thus F is continuous and, by Riesz’s representation theorem,

where This means that is a weak limit of

Corollary A (strongly) closed ball about zero in a Hilbert space H is weaklyclosed.

Let S be the (strongly) closed ball Suppose andas in the theorem. Then and i.e. and S

is weakly closed.We now prove

Theorem 4.6.7 Let X be an inner product space. A weakly closed setis closed. A closed set need not be weakly closed.

Proof. Let be a (strongly) convergent sequence in S converging toWe need to prove that The sequence converges weakly tobecause, if is any continuous linear functional on X , then


But S is weakly closed, so that is closed.For a counterexample we take X to be the set S to be

This is (strongly) closed, for and implies However,the Riemann–Lebesgue Lemma (equation (4.5.12)) shows that ifthen

If therefore we take

then since

But converges weakly to 0, (i.e. because if we have

since 0 is not in S, S is not weakly closed.

Even though a strongly closed set in an inner product space need not beweakly closed, we can use the orthogonal decomposition in Theorem 4.3.2 toobtain:

Problem 4.6.6 Show that a (strongly) closed subspace M of a Hilbert spaceH is weakly closed.

There is also the more difficult

Problem 4.6.7 Show that a (strongly) closed convex subset S of a Hilbertspace H is weakly closed.

The corollary to Theorem 4.6.6 is an example of this; a closed ball is a closedconvex set.

4.7 Introduction to the concept of a compact set

We introduced the concepts of weakly closed and weakly complete in § 4.6. Nowwe introduce

Definition 4.7.1 Let X be an inner product space. The set is saidto be weakly compact if every sequence in S contains a subsequence whichconverges weakly to an element

4.7 Introduction to the concept of a compact set 127

We now prove

Theorem 4.7.1 Let H be a Hilbert space. A set is weakly compact iffit is bounded and weakly closed.

Proof. We will show that if it is bounded and weakly closed, then it is weaklycompact, i.e. that any sequence contains a weakly convergent sub-sequence. Since S is weakly closed we know that the weak limit of such asubsequence will be in S.

Let be a sequence in S , and M be the closed linear subspace spannedby Since M is a closed linear subspace of H we may (by Theo-rem 4.3.2) decompose H into M and N which are mutually orthogonal. If

then we can write where and Ifthen if then Thus it issufficient to consider for The subspace M , being a closedsubspace of a Hilbert space, is a Hilbert space (Problem 2.12.4). It is clearlyseparable, and so has an orthonormal basis By Theorem 4.6.5 it is suf-ficient to show that there is a subsequence of such that, for each

the numerical sequence is convergent. We proceed as follows. Thesequence in is bounded and therefore contains a convergent se-quence The sequence is bounded and therefore containsa convergent sequence. Continuing in this way we obtain, at the ith step, aconvergent sequence The subsequence is suchthat, for each fixed the sequence is convergent. Therefore isa weakly convergent sequence converging to some

We leave the (easier) converse to

Problem 4.7.1 Show that a weakly compact set in a Hilbert space is boundedand weakly closed.

Look back at Theorems 4.2.2 and 4.3.1. They show that if H is a Hilbertspace and is a closed convex set, then if there is (existence) aunique which minimizes on M . Problem 4.6.7 statesthat a (strongly) closed convex set is weakly closed, so that we can replace‘closed convex set’ in Theorem 4.3.1 by ‘weakly closed convex set.’ However, wecan use the weak compactness of a weakly closed and bounded set in a Hilbertspace to provide a separate proof. Thus we have

Problem 4.7.2 Let H be a Hilbert space, let and let M be a boundedand weakly closed set in H , then there is a such that

If, in addition M is convex, then is unique.


We will use the concept of weak compactness in our discussion of the Ritzprocedure in the following section.

4.8 Ritz approximation in a Hilbert space

We return to the problem of Theorem 4.2.2, but now suppose X is a Hilbertspace H.

Thus let H be a Hilbert space, M be a closed subspace of H , andFind the unique minimizer of

forWe consider the problem in four steps due to Walter Ritz (1878–1909).

Step 1. Set up the approximation problem and study its solutions

We solve the problem approximately using the so-called Ritz method.Assume that M has a complete system This will certainly be the case

if H is separable. Suppose that any finite subsystem is linearly in-dependent. Let be the subspace spanned by Theorem 4.2.1states that there is an which minimizes on call one suchminimizer For convenience we now suppose that H is a real Hilbert space.We can argue as in § 4.3. Thus the real function

of the real variable takes a minimum value at and since is differ-entiable,

Thus is orthogonal to eachWriting

we obtain a set of simultaneous linear equation for the namely

Since are linearly independent the solution to this equation isunique. For if it were two solutions

4.8 Ritz approximation in a Hilbert space 129

their difference

would satisfy

Thus so that But since the are

linearly independent, this means and thusHence the solution is unique.

Step 2. An a priori estimate of the approximation

An a priori estimate is one which can be obtained without actually knowingthe approximation, or even whether it exists.

We begin with the definition of

As we have

from which we obtain

which is the required estimate.

Step 3. Weak passage to the limit

By (4.8.2), the sequence is bounded. By Theorem 4.7.1, containsa weakly convergent subsequence whose weak limit since M ,being a closed subspace, is weakly closed (Problem 4.6.6).

For any fixed we can pass to the limit in the equation (4.8.1),namely

and obtain

This passage is possible because is a continuous (linear) functional.


Now consider where is an arbitrary but fixed element of M .The system is complete in M , therefore, given we can find a finite

linear combination such that

Then

where, in the last step, we used the inequalities (4.8.2) and (4.8.3). Therefore,for any we have

Finally, by considering values of for and we obtain

This implies that is the solution to the problem.

Step 4. Study the convergence of the sequence of approximations

We have shown that there is a subsequence which converges weakly toWe will show that the whole sequence converges weakly to and then

that it converges strongly toSuppose, if possible that does not converge weakly to This means

that there is an such that does not converge to Removefrom any subsequence such that converges to Re-name the remaining sequence Since the set is bounded, it has,by the Bolzano–Weierstrass theorem, a convergent subsequence andthe limit of this sequence will not be Thus

But for the sequence we can repeat Step 3, and find a subsequencewhich is weakly convergent to a solution of the problem. Theorem 4.3.1 statesthat this minimizer is unique, Thus

This contradicts (4.8.5). Thus converges weakly to i.e.

4.8 Ritz approximation in a Hilbert space 131

Now we prove that converges strongly to i.e. Equa-tion (4.8.1) states that

Thus

so that

But so that But so that equa-tion (4.8.4) states that Therefore,

Now we may use Theorem 4.6.2 to state thatTo conclude this section we note that we can apply the argument above to

the problem of minimizing

in a Hilbert space H , where is a continuous linear functional. For byRiesz’s representation theorem, we can write

so that

Since is fixed, the problem of minimizing is equivalent to that of min-imizing

for This problem has the unique, obvious, solutionTo apply the Ritz method we suppose, as before, that is a complete

system in H such that any finite set is linearly independent. Wetake the Ritz approximation as

and find the equations

for Note that we express in terms of the given func-tional The result of the earlier analysis gives us


Theorem 4.8.1 For each equations (4.8.7) have the unique solutionWhen is a continuous linear functional, the sequence of Ritz

approximations defined by (4.8.6) converges strongly to the unique minimizer ofthe quadratic functional

The problems considered in Chapter 3 and set in the various energy spaces— which were all separable Hilbert spaces — fall into this category, and theanalysis given here provides justification for the application of the Ritz methodto these problems.

4.9 Generalized solutions of evolution problems

Consider the heat transfer equation

Here is the temperature, the time, and the positionin a domain with boundary containing heat sources

To pose the problem, we need boundary conditions, say

and an initial condition

To obtain a generalized statement of the problem we first suppose thatas a function of and as a function of i.e. ithas continuous second derivatives in space and a continuous derivative in time;and that and Now suppose that and

and satisfies (4.9.2). Multiply (4.9.1) by and use the identity

and Gauss’ divergence theorem, to obtain

The integral over the boundary is zero, so that

Now integrate in time over (0, T) to obtain

4.9 Generalized solutions of evolution problems 133

where we use the abbreviation

This is the basis for the generalized solution; we derived it from (4.9.1) byassuming that satisfied the restrictive conditions we stated, but we nowconsider it in its own right. In what space(s) should we treat it? The functions

are defined for and i.e. the domain andthey satisfy (4.9.2). We recall the definition of it is the completion of

in the norm, i.e.

where

Straightforward application of the Schwarz inequality shows that equation(4.9.5) may be interpreted for and We thereforeintroduce

Definition 4.9.1 Let W be the subspace of satisfying (4.9.2),the subspace of satisfying (4.9.2); and The

element is called the generalized solution of the heat transfer prob-lem (4.9.1) with the Dirichlet boundary condition (4.9.2) if it satisfies (4.9.5)for every and

First we show that the generalized solution is unique. For this, we establishan a priori estimate for a solution. Let be a generalized solution of (4.9.5).Put in (4.9.5) to obtain

Consider the terms in this equation separately. Using first Schwarz’s and thenFriedrich’s inequalities, we find

Now use the elementary inequality to obtain


Now consider the first term in (4.9.9). If then andand, for every we have

Thus

Now use the triangle inequality to give

This inequality holds for but it therefore holds forand it shows that

is a uniformly continuous function of on (0, T) and may therefore (by Theo-rem 1.2.3) be extended continuously to [0,T], so that

Now return to equation (4.9.9); we have

so that putting (4.9.10), (4.9.11) together in (4.9.9) we find


which we may rewrite as

This is the needed a priori estimate. It shows that the generalized solutionis unique. For if there were two generalized solutions and then theirdifference, would satisfy equations (4.9.9), (4.9.3) with F = 0and respectively. This would mean that, for this the left hand side of(4.9.12) would be zero, so that would be zero a.e. in Q. (See Definition 2.11.3)

Note that because is a uniformly continuous function of on

(0, T), so is by the Schwarz inequality, so that equation (4.9.8)

will hold.Having shown that there cannot be more than one generalized solution, we

show that there is one, which will thus be the generalized solution. (Of course wecan always add a function which is zero a.e., to the solution.) We could use theGalerkin procedure on the domain Q , but instead we will separate the variables,using a complete system of functions in space to reduce the partial differentialequation (4.9.1) to a system of ordinary differential equations in time.

Consider the Sobolev space the subspace of satisfying(4.9.2). As a corollary to Theorem 4.1.4 we showed that is separable,and therefore, by Problem 4.1.2, is separable. is a Hilbertspace, and therefore, by Theorem 4.5.3 it has a countable orthonormal basis

Apply the Gram–Schmidt procedure to this sequence to construct asystem orthonormal in i.e.

Problem 4.9.1 Show that the set of all finite combinations

with is dense in W .

We now define the Faedo–Galerkin approximation , after S. Faedoand Boris Grigor’evich Galerkin (1871–1945). To do that we return to equa-tion (4.9.4), take given by (4.9.13) and to obtain the equation

where

We define the Faedo–Galerkin approximation as the solution of (4.9.14)which minimizes


over forIf then the general theory of first order differential equa-

tion with constant coefficients shows that, for given equa-tion (4.9.14) for has a unique solutionwhich is continuous in [0, T]. The equation (4.9.14) then shows that

We now show that we can establish properties of and moreimportantly of when

Suppose that Multiply equation (4.9.14) by and sumover to obtain

which may be written

Integrating over we obtain

This means that

The right hand side is bounded, independently of because

But so that the second term is bounded. Thus by using this in-

equality in conjunction with (4.9.15) we see that each of and

is bounded. We wish to show that is bounded in the

norm (4.9.6). To do this we must show that is bounded

also. To show this, we multiply (4.9.14) by and sum over Thus


which we can write as

Integrating this in time we find

Applying the Schwarz inequality, we find

Substituting this into (4.9.16), we obtain

which implies that is bounded.

We have now shown that the sequence is bounded in W ; W is aclosed linear subspace of the complete space Problem 4.6.6 and Theo-rem 4.7.1 show that is weakly compact, i.e. it contains a subsequence

which converges weakly toReturn once more to equation (4.9.14), and suppose

Multiplying (4.9.14) by and summing over and integrating over time,we find


Since, as we have shown, all the integrals are continuous functions with respectto in W , we can pass to the limit in the subsequence and find

for any given by (4.9.17). But, by Problem 4.9.1, such are dense in W.Hence

for all But this is equation (4.9.5), so that satisfies the first ofthe two conditions for the generalized solution, stated in Definition 4.9.1, andtherefore, as we showed earlier, satisfies (4.9.8).

Now we may repeat step 4 of § 4.8 to show that the whole sequenceconverges weakly to in W .

Actually, the convergence of the approximation is stronger than we haveestablished. We formulate a set of problems.

Problem 4.9.2 Show that the n th approximation to the solution of (4.9.5)satisfies

and that it is possible to pass to the limit to obtain

Problem 4.9.3 Introduce a new Hilbert space which is the completion ofthe subspace of satisfying (4.9.2) in the norm corresponding to the innerproduct

Show that this is a proper inner product, and that a sequence which convergesweakly to an element of W converges weakly to the same element in

Problem 4.9.4 Use Problems 4.9.2, 4.9.3 to show that

This means that (i.e. strongly) in the norm of and illustrateshow the spaces in which a set of approximations to a given problem convergesweakly, or strongly, must be chosen to fit the problem under consideration.


Synopsis of Chapter 4: Approximation

Separable : has countable dense subset. Definition 4.1.3.

: if bounded.

Linear functional: linear operator with values in or Definition 4.2.3.

Riesz’s representation: continuous linear functional on H can be writtenTheorem 4.3.3.

Orthogonal decomposition of H: Definition 4.3.2.

Basis: Definition 4.5.1.

Complete: is complete if Definition 4.5.2.

Closed : is closed if Definition 4.5.4.

: is closed in H iff it is a basis. Problem 4.5.5.

: H has basis iff separable. Theorem 4.5.3.

Weak Cauchy sequence in H : as Theo-rem 4.6.1.

: Strong Cauchy implies weak Cauchy. Problem 4.6.1.

: with implies Theorem 4.6.2.

: Bounded. Theorem 4.6.3.

: Strongly complete implies weakly complete. Theorem 4.6.6.

: Weakly closed implies strongly closed. Theorem 4.6.7.

: A strongly closed subspace of H is weakly closed.

Weak Compactness : every sequence in S contains a subsequence converg-ing weakly to Definition 4.7.1

: in H , S is weakly compact iff S is bounded and weakly closed. Theo-rem 4.7.1

5. Elements of the Theory of Linear Operators

Introduce public education with moderation, avoiding bloodshedif possible. (From Service Regulations relating to the Kindness ofMayors issued by Lieutenant-Colonel Prysch.)

M.E. Saltykov-Shchedrin, History of a Town

5.1 Spaces of linear operators

This chapter aims to present some results from the theory of linear operators.We cannot pretend to give a full treatment of this vast field; we shall select onlythose parts which we shall use in later applications.

We recall the basic definitions from § 2.9 of an operator, and in particulara linear operator on a normed linear space X into a normed linear space Y .Remember that such an operator is continuous, bounded, if there is a constantc such that

The infimum of these constants is the norm of A; thus

We now consider the set of continuous linear operators A on X into Y (i.e.and show that they form a new normed linear space

which we denote by

Lemma 5.1.1 is a normed linear space, with the norm (5.1.2).

Proof. is clearly a linear space. We must verify that the norm in (5.1.2)satisfies the axioms N1-N3 of § 2.8.N1: Clearly If then for all But Y is anormed space, so that implies This holds for allso that A = 0.N2: This is evident.N3: The chain of inequalities

142 5. Elements of the Theory of Linear Operators

As in any normed linear space, we can introduce the notion of convergencein

Definition 5.1.1 The sequence of continuous linear operatorsis said to converge to A if as in such a case we shallsay that converges uniformly to A. (Note that c in (5.1.1) is a uniformbound for

Theorem 5.1.1 If X is a normed linear space and Y is a Banach space, thenis a Banach space.

Proof. We recall (Definition 2.8.9) that a Banach space is a complete normedlinear space. Let be a Cauchy sequence in , i.e.

We need to show that there is a continuous operator such that

For any the sequence is a Cauchy sequence in Y , since

But Y is complete, so there is such that

Thus to every there is a such that (5.1.3) holds. This corre-spondence defines a linear operator A on X into Y , such that It nowremains to show that i.e. A is continuous. To do this we note thatsince is a Cauchy sequence, the sequence of norms is bounded,i.e. and so

which means that A is continuous, i.e. and is com-plete.

In the Banach space we can introduce a series and define

its sum by A series is said to be absolutely

convergent if the numerical series is convergent.

implies

5.1 Spaces of linear operators 143

Problem 5.1.1 Let X be a normed linear space, Y be a Banach space,

and Show that if is absolutely convergent, then

the sequence is uniformly convergent.

Now consider operators A on X into X , and denote byIn we can introduce the product of operators by

The product possesses the usual properties of a (numerical) product, exceptcommutativity:

where I is the identity operator. If A and B are continuous, then so is AB ,since

and so

Problem 5.1.2 Suppose converge uniformly torespectively. Show that converges uniformly to AB .

If then we write etc.

Problem 5.1.3 Let X be a Banach space, and Show that the series

converges uniformly in

Up to now, the convergence of sequences we have consideredhas been uniform convergence, in the uniform norm (5.1.2). Now we will considera weaker convergence, which is called pointwise or strong convergence. (It iscalled strong because it is stronger than yet a third, weak, convergence, but itis actually weaker than uniform convergence, as we will soon show.)

Definition 5.1.2 A sequence is said to converge strongly toif, whenever

If converges uniformly to A , then it converges strongly to A , for


To construct an example to show that strong convergence does not necessar-ily imply uniform convergence we suppose that X = Y = H , a separable Hilbertspace. By Theorem 4.5.1, and the remark preceding it, H has an orthonormalbasis Define the orthonormal projection operator

from H onto the subspace of H spanned by Since is abasis for H , we have

This means

Thus the sequence converges strongly to I. But if then

so that

This means that the sequence is not a uniform Cauchy sequence; it is nota uniformly convergent sequence.

Problem 5.1.4 Use Bessel’s inequality to prove that

5.2 The Banach–Steinhaus theorem

Suppose A is a linear operator whose domain is dense in a normed space X . Weshall show how to continue A into the whole of X , with the following theorem.

Theorem 5.2.1 Let A be a linear operator whose domain D(A) is dense in anormed space X , whose range lies in a Banach space Y , and which is boundedon D(A), i.e.

Then there is a continuation or extension of A to X , denoted by such that

for all

where is defined as

1.

2.

3.

for all

1455.2 The Banach–Steinhaus theorem

Proof. If then Suppose but Weconstruct as follows. Since D(A) is dense in X , there is a sequencesuch that The sequence is a Cauchy sequence, since

Y is a Banach space, it is therefore complete, so that the sequence hasa limit, which we call (Show that the limit is independent of the choice ofsequence Now it suffices to prove 3. The norm of A, is the infimumof M such that for Thus

and passing to the limit we find

which means that is continuous and But on D(A) we haveso that

This theorem is important. It states that if Y is a Banach space, then Amay be continued to the closure of its domain. If X is complete, i.e.a Banach space, then being a closed subspace of a Banach space, iscomplete also (Problem 2.8.3). From now on we shall suppose, unless we stateotherwise, that D(A) = X , i.e. that

Now we prove the Banach–Steinhaus theorem, after Banach, and Hugo Dy-onis Steinhaus (1887–1972)

Theorem 5.2.2 Let X be a normed linear space, Y a Banach space anda sequence of continuous linear operators in such that

1. for all

2. exists for all a subspace which is dense in X ,

then the sequence converges strongly to an operator i.e.for all as

Proof. The linear operator A, defined on by the relation

is bounded on indeed 1 implies

Using the construction of Theorem 5.2.1, we can extend A to X , keeping thenorm unchanged. We call this extension and will now show that ifthen


Let be such that then by definitionOn the other hand, since we have

Choose we can find N such that Takebecause of 2 we can find such that whenThis means that when

When X in Theorem 5.2.2 is a Banach space, we can replace condition 1by the statement that, for all the set according to theprinciple of uniform boundedness:

Theorem 5.2.3 Let X be a Banach space, Y a normed linear space andIf for every the set is bounded, then the

set is bounded.

Proof. If there is a closed ball on which the are uniformlybounded, i.e. there is a constant K such that

then is uniformly bounded on X . For if is in X , then

is in so that and Thus

so that

Now we must show that we can find a ball on which the are uniformlybounded. We suppose we cannot, and derive a contradiction.

Choose The are not uniformly bounded in B(0,1).There is therefore and such that By continuitythere is a ball with such that for allThe are not uniformly bounded in this ball. Therefore, there is an

1475.3 The inverse operator

and such that and thus a ballwith and when Proceeding in

this way we find sequences and such that andpoints such that

But is a Cauchy sequence in the Banach space X so that it has a limitthis limit point is in each so that

which contradicts the statement that is bounded on X .


Now we are interested in solving the equation

where A is a linear operator, is given, and is unknown.

Definition 5.3.1 Let X , Y be normed linear spaces, and A an operator from Xinto Y . If for any there is no more than one such thatthen A is said to be a one-to-one operator. In this case the correspondencefrom Y to X defines an operator; this operator is called the inverse of A, andis denoted by

Problem 5.3.1 Show that and

Problem 5.3.2 Show that the operator exists iff the equation hasthe unique solution show that, if it exists, is a linear operator.

When exists we can apply Problem 2.9.3 to prove

Problem 5.3.3 Let X , Y be normed linear spaces and A a linear operator fromX into Y . Show that if exists and dim then dim

We are interested not only in the solvability of equation (5.3.1), but alsoin knowing whether the inverse operator is continuous. There is a simple resultgiven by

Theorem 5.3.1 The operator is bounded on R(A ) iff there is a constantc > 0, such that, if then


Proof. Necessity. Suppose exists and is bounded on R(A). Then there isa constant such that for all Puttingand we get (5.3.2).Sufficiency. If then (5.3.2) implies i.e. Thushas the unique solution and so, by Problem 5.3.2, exists. Putting

in (5.3.2) we find

i.e. Therefore is bounded.

If we say that A is continuously invertible. Clearly A willbe continuously invertible iff R(A) = Y , and there is a c > 0 such that (5.3.2)holds.

Let us consider some examples. We begin with the Fredholm integral equa-tion of the second kind:

We can write this as

where A is the integral operator given by

Suppose the kernel is degenerate, i.e.

and that are in C[a, b]. What can we say about the inverseof the operator A?

Without loss of generality we can assume that are linearly inde-pendent. If equation (5.3.3) is soluble, the solution has the form

Substituting this into (5.3.3) and equating the coefficients of to zero, wefind the equations

where

5.3 The inverse operator 149

Provided that the determinant of coefficients, of equation (5.3.4) is notzero, we may solve equations (5.3.4) to give

so that

In this equation each is a linear combination of and

Since we have for some c > 0, andhence, when we can find such that

or, in terms of the uniform norm on C[a, b],

This means that if then (Remember that C[a, b],like any is a complete space under the uniform norm (See § 2.5)) and

Now consider the case Since is a polynomial of orderthere are no more than distinct (possibly complex) roots for which

For the equation

has a non-trivial solution Thus for such the equation

has a non-trivial solution, so that, according to Problem 5.3.2, the inverse oper-ator does not exist. The compose the spectrum of the integral operator(see Chapter 6).

Now consider a simple boundary value problem

where Itssolution is


We can phrase this problem in terms of linear operators. The direct problemis: given differentiate it twice to obtain f. This operator whichtakes to form f we can call A; its domain is and its range is apair consisting of a function and a vector such that

The inverse operator is that which obtainsfrom by using (5.3.5). If we use the norms

and

we see that exists and is bounded.

Problem 5.3.4 Show that the operator in this example exists and isbounded.

Theorem 5.3.2 Let X and Y be Banach spaces. Suppose Ais continuously invertible and Then A + B has the inverse

and

Proof. The equation

can be reduced to

By the condition of the theorem The contractionmapping theorem (Theorem 2.7.1) shows that equation (5.3.8) has a uniquesolution for any This means that the inverse exists, andits domain is Y .

We now obtain the inequality (5.3.6). From it follows thatand so Equation (5.3.7) shows that

for any we have

Problem 5.3.5 Show that the solution of equation (5.3.8) may be found itera-tively: so that


Theorem 5.3.1 gives a necessary and sufficient condition for an operatorto be bounded. However, there is a deeper result which we will not prove,

namely Banach’s open mapping theorem. (For a proof, see Friedman (1970), p.141, listed in the references to Chapter 2)

Theorem 5.3.3 Let X, Y be Banach spaces, and let A be a continuous linearoperator from X onto Y. Then A maps open sets of X onto open sets of Y.

One of the corollaries of this result is

Theorem 5.3.4 Let X, Y be Banach spaces and let A be a one-to-one contin-uous linear operator on X onto Y, then is a continuous linear operatoron Y onto X.

Proof. is a linear operator, for if there is an such thatand the definition of a one-to-one operator implies that is unique.

To show that is continuous, it is sufficient to show that it is continuousat 0. Thus we must show that if there is a sequence such thatand then Suppose This means that for somegiven any N we can find an for which The open mappingtheorem states that the open set around zero in X is mapped onto anopen set around zero in Y. Since this set is open, there is an open ballin it. Since we can find such that if then Thusif is the map of an in On the other hand we showedthat if then for some is the map of an inThus is the map of two distinct so that A is not one-to-one, contraryto our hypothesis. Thus and is continuous .

Note that the condition that A be an operator on X onto Y may bereplaced by the statement that R(A) is closed in a Banach space Y, for thenR(A) will itself be a Banach space, and A will be an operator on X ontoR(A). If A is to have an inverse, then N(A) must be empty. If X, Y are Hilbertspaces and then we may circumvent this constraint bydecomposing into N(A) and and considering the restriction of Ato This will be a one-to-one operator on onto R(A) so that,if R(A) is closed, this operator will have a continuous inverse on R(A) onto

Problem 5.3.6 Use Theorem 5.3.4 to show that if a linear space X is a Banachspace with respect to two norms and and if there is a constantsuch that

then there is a constant such that

for all

for all


i.e. the norms and are equivalent.

5.4 Closed operators

Closed operators have somewhat limited interest, and the reader is advised toskip this section on a first reading. Closed operators do appear in Chapter 8.

Definition 5.4.1 Let X and Y be normed linear spaces. A linear operator Afrom X into Y is called closed iff the three statements

together imply

If A is a continuous linear operator and Y is a Banach space, then, byusing Theorem 5.2.1, we may continue A to the closure of D(A); then if

we define Thus a continuous linear operator isclosed. To show that a closed linear operator need not be continuous we take thefollowing counter-example. Let X = Y = C[0, 1] under the uniform norm, andlet A be the differentiation operator: We show that A is closed. Thedomain of A is Suppose uniformlyand uniformly. We need to show that is differentiable, andthat

Suppose that and consider

Clearly

To bound the magnitude of the first term we use the mean value theorem, whichstates that

for some between a and b. We apply the theorem to the functionand find

for some between and and hence

5.4 Closed operators 153

Now choose Since converges uniformly to we can findsuch that implies

For such the inequality (5.4.1) shows that

But as so that taking the limit in (5.4.2), wefind

The function is differentiable at c so that there exists such thatimplies

The sequence converges to so that we can find such thatimplies

Choose then if we have

Note that this is a proof of the familiar result: A series (i.e. a sequence ofpartial sums) of differentiable functions can be differentiated term by term ifthe differentiated series converges uniformly.

Thus is closed, but it is not continuous because ifthen, on [0,1],

so that is unbounded asNow we consider closed operators from another point of view, by using the

concept of the graph of an operator. First we introduce

Definition 5.4.2 Let X and Y be normed linear spaces. Then the productspace X x Y is a normed linear space with elements where

in which

A possible norm for X x Y is


All the norms (5.4.3) are equivalent.

Definition 5.4.3 Let A be a linear operator from X into Y. Thusand The graph of A is the subset G(A) of X × Y defined by

Thus G(A) = D(A) x R(A).

Definition 5.4.4 A linear operator A from X into Y is said to be closed ifits graph is a closed linear subspace of X x Y.

Problem 5.4.1 Show that Definitions 5.4.1 and 5.4.4 are equivalent.

When we discussed Definitions 5.4.1, we showed by counterexample that aclosed operator is not necessarily continuous. However, now we prove the dosedgraph theorem:

Theorem 5.4.1 Let X and Y be Banach spaces and let A be a linear operatoron X into Y (i.e. D(A) = X, but If A is closed, then it iscontinuous.

Proof. The graph G(A) is a closed linear subspace in the product space X × Ywith the norm

(i.e. with in (5.4.3)). But a closed linear subspace of a normed linearspace is a Banach space. Consider the continuous linear operator B on G(A)onto X, defined by

Since B is a one-to-one operator, we may apply Theorem 5.3.4 to it. This meansthat is a continuous linear operator on X onto G(A), i.e.

But with the norm in G(A) given by (5.4.4) we have

which implies for some c > 0. Thus A is continuous.

Note that the counterexample of on X = C[0,1] does not satisfythe conditions of this theorem. The domain of A is and this is nota Banach space under the uniform norm for C[0,1]; it is not a closed linearsubspace of X = C[0,1]. However, note

5.4 Closed operators 155

Problem 5.4.2 Let X be the subspace of functions satisfyingand let so that

Show that A is a closed linear operator from X onto C[0, 1], but is not contin-uous. Show that exists and is continuous.

Problem 5.4.3 Suppose that we norm D(A) in the last example with the norm

Show that A is a continuous linear operator on the Banach space D(A) ontoC[0,1], in agreement with Theorem 5.4.1 and that is continuous, in agree-ment with Theorem 5.3.4.

Problem 5.4.4 Let X and Y be normed linear spaces, and A be a closed linearoperator from X into Y. Show that if exists, then it is closed.

Problem 5.4.5 Let X and Y be normed linear spaces, and A be a linearoperator from X into Y. Show that if D(A) and R(A) are closed in X and Yrespectively, then G(A) is closed, but that the converse is not true.

Problem 5.4.6 Show that the closed graph theorem may be viewed as follows:If D(A) and G(A) are closed, then A is a continuous linear operator. (Notethat there is no loss in generality in taking D(A) = X, a Banach space.)

In Theorem 5.2.1 we showed that a continuous (i.e. bounded) linear operatorcould be extended to the closure of its domain D(A). We now considerwhether and how a general (i.e. not necessarily bounded) linear operator maybe extended so that it becomes a closed linear operator. First, we point outonce again the difference between a continuous and closed operator. Suppose

is a Cauchy sequence in D(A). If A is a continuous linear operator, thenis a Cauchy sequence in Y. If therefore then we define

as If A is merely closed and is a Cauchy sequence in D(A),then is not necessarily a Cauchy sequence in Y, but we can make auseful deduction from the definition of a closed operator. Suppose and

are two Cauchy sequences in X both converging to then andcannot converge to different limits. For according to the definition,

This last condition by itself does not imply that A is closed, but it does ensurethat A has a closed extension according to


Lemma 5.4.1 Let X and Y be Banach spaces and A be a linear operator fromX into Y. Let be an arbitrary sequence such that

Then A has a closed extension iff 1 implies

Proof. Suppose A has a closed extension B, and satisfies 1. Thenimply so that

Now suppose that 1 implies We will construct a closed extension,B, of A. The domain of B will not simply be the closure of the domain of A.Rather, iff there is a sequence such that

and there is a such that By the conditionof the lemma, this is uniquely determined by we call it B is clearlya linear operator; we show that it is closed.

Let be a sequence such that andWe must show that How do we construct For every we choosea sequence such that and define

as the unique element in Y such that From each

sequence we choose a member which we call such that

Then

But then, by the way we constructed B, we have and sothat B is closed.

As an application of this lemma we consider the extension of the operator

with coefficients where is a domain in We considerA from into The domain of A is therange of A is in We examine the conditions of the lemma and showthat A has a closed extension. To do so we consider the set of trial functions

the set of functions in with compact support;like is dense in Let satisfy the conditions ofthe lemma, i.e.

both in the norm. For any we can perform successiveintegration by parts to obtain

5.5 The adjoint operator 157

because all the integrated terms involving with vanish on theboundary. Now we take the passage to the limit in to obtain

Since the are dense in the separable Hilbert space the analysis of§ 4.6 shows that as an element of Thus the conditions of thetheorem are fulfilled and A has a closed extension.

The approach we have used here can bring us to generalized derivatives; itis equivalent to that used by Sobolev.

5.5 The adjoint operator

We shall introduce the idea of adjoint operator for an operator from a Hilbertspace into a Hilbert space although it can be considered for operatorsacting from a general normed space X into a normed space Y .

Let A be a continuous linear operator on into , i.e. inThe inner product is, for fixed a functional onBecause A is linear, F is a linear functional; it is bounded because

so that

By Riesz’s representation theorem can be written

where is uniquely defined by F, i.e. by A and The correspondencedefines an operator A* on into i.e.

which we call the adjoint of A. We note the fundamental equation

Problem 5.5.1 Show that A* is a linear operator.

Problem 5.5.2 Show that and

Lemma 5.5.1 The adjoint A* of a continuous linear operator A is continuous,i.e. moreover


Proof. Using the Schwarz inequality, we get

But using (5.5.1), we may write

Putting we have

Thus so that A* is continuous. Since A* is continuous, (A*)*is continuous and

Thus

so that for all Therefore, for alland

Applying the first part of the proof to A*, we find

Definition 5.5.1 Let be Hilbert spaces, and A an operator inThe null space of A, denoted by N(A) is the set of such thatN(A) is a closed subspace of and is thus a Hilbert space. is theorthogonal complement of N(A) in

Proof. Suppose This means that there is a sequencesuch that and If then and

Thus

Theorem 5.5.1

so that i.e. is orthogonal to every it is inTherefore


Now suppose We will show that is a closedsubspace of the Hilbert space By equation (4.3.1) there is a unique element

in such that

and for all Put then for allso that, in particular, and thus

But since for all it is zero for all that is

Therefore so that Thus equation (5.5.3) states that isnot orthogonal to a non-zero element y is not in Therefore

and so By changing A to A* we obtain thesecond part of the theorem.

Problem 5.5.3 Let be Hilbert spaces, and A an operator inShow that

The first half of this proof, implies Inother words, if the equation

has a solution for a certain then must be orthogonal to all solutionsof For

Thus is a necessary condition for (5.3.1) to have a solution. It willbe a sufficient condition only if R(A) is closed, so thatIn this case the so-called Fredholm alternative holds:

either the equation has a solution for all i.e.so that i.e. the equation has no nontrivial

solution.

or the equation has solutions, in which case theequation has a solution iff is orthogonal to all the solutionsi.e.

Problem 5.5.4 Show that if are Hilbert spaces and thenR(A) is closed iff R(A*) is closed.


This means that when R(A) is closed there is a similar Fredholm alternativefor A*.

Definition 5.5.2 The operator A in a Hilbert space H is said to be self-adjoint if A* = A.

Theorem 5.5.2 Let H be a Hilbert space, and A a continuous self-adjointlinear operator in H, then

Proof. Write Using the Schwarz inequality, we get

If then so that by the definition of

Suppose and then

But A is self-adjoint, so that and

On the other hand

Combining (5.5.6) and (5.5.7), we deduce that

for all and all real Putting we obtain

Take to obtain

Thus which with (5.5.5) yields

or

for all


We conclude this section with two useful results related to the adjoint. In§ 4.6 we introduced the concept of weak convergence in a Hilbert space; thesequence is said to converge weakly to (We write if

for every A continuous linear operator in a Hilbertspace H maps a strongly convergence sequence into a strongly convergentsequence Thus, if then equivalently,if then We could say that A is strongly continuous, butwe usually say just continuous.

Definition 5.5.3 Let H be a Hilbert space, and A a linear operator in H. Ais said to be weakly continuous if it maps weakly convergent sequences intoweakly convergent sequences, i.e. implies

Lemma 5.5.2 Let H be a Hilbert space, and A a linear operator in H. If Ais continuous, then it is weakly continuous.

Proof. Suppose (i.e. weakly) in H, so that for allWe must show that this is so because

since A* is a continuous operator. Therefore and A is weaklycontinuous.

Lemma 5.5.3 Let H be a Hilbert space and A be a continuous linear operatorin H. If and then

Proof.

Choose A weakly convergent sequence is bounded (Theorem 4.6.3); thusChoose so that if then Then

if we have

Choose so that if then Then ifwe have


5.6 Examples of adjoint operators

A matrix operator in

We recall that is the metric space of sequences with norm

We consider the matrix operator where

To find a bound for its norm we use the Holder inequality. Thus

so that

The adjoint operator is defined by (5.5.1), so that

and where

We note that a bounded A is self-adjoint iff i.e. iff the matrix A isHermitian, after Charles Hermite (1821–1901).

An integral operator

in Note that B acts on the function f; it is a linear operator actingin the space of functions. If then B isbounded in Indeed, using the Holder inequality, as we have just done,we find

We consider an integral operator, or so-called Fredholm operator

5.6 Examples of adjoint operators 163

Just as with the matrix operator we may derive the adjoint as

so that B is self-adjoint when

In particular, if is real and symmetric, then B is self-adjoint.

Stability of a thin plate

Saint Venant’s equation, named after Barré de Saint Venant (1797–1886), gov-erning the deflection of an isotropic thin plate due to in plane forces is (S.P.Timoshenko and J.M. Gere, 1961, Theory of Elastic Stability, McGraw-Hill)

where are in-plane forces per unit length satisfying the appropriateequilibrium conditions, and D is the flexural rigidity of the plate. Usually thisequation is considered for and We derive ageneralized form of the equation. Suppose that the boundary of the plate isclamped, so that

We take a test function satisfying

multiply equation throughout by and integrate over use the divergencetheorem and the boundary conditions satisfied by and to obtain

where

and from now on, in this section an undistinguished norm or inner product istaken to be that in thus


We now show that we may consider equation (5.6.1) for andThe problem (5.6.1) thus reduces to finding a non trivial

element satisfying the equation for everyWe first show that for fixed is a bounded linear functional

in Holder’s integral inequality gives

But

and on using Holder’s inequality we find

and we recognize the quantities on the last line as the Sobolev semi-normsand given in equation (3.6.1). Thus

We showed that the norm is and Corollary 1 used with equation(3.7.18) and N = 2, shows that

so that

and hence

so that is a bounded linear functional in in By Riesz’s represen-tation theorem there is an element such that

5.6 Examples of adjoint operators 165

The correspondence defines a linear operator on such that

The operator G is clearly linear, and bounded since the inequality (5.6.3) gives

Putting we find

so that

Since is symmetrical in and we have

which means that G is self-adjoint. The equation (5.6.1) becomes

or in other words


Synopsis of Chapter 5: Linear Operators

: space of continuous linear operators on X into Y.

: is a normed linear space Lemma 5.1.1.

: Y is Banach implies is Banach’s Theorem 5.1.1.

: uniform Cauchy

: strong Cauchy

: uniform implies strong; strong does not imply uniform Definition 5.1.2.

Banach–Steinhaus Theorem: strong limit of Theorem 5.2.2.

Inverse operator : exists iff implies Problem 5.3.2.

: bounded if Theorem 5.3.1.

Banach’s open mapping theorem: Theorem 5.3.3.

Closed operator : imply and Defini-tion 5.4.1.

: continuous implies closed.

: graph is a closed linear subspace Definition 5.4.4.

Closed graph theorem: if D(A) = X, X,Y are Banach; closed implies con-tinuous Theorem 5.4.1.

Adjoint operator : (5.5.1)

: Lemma 5.5.1.

Weakly continuous : implies

: continuous implies weakly continuous Lemma 5.5.2.

References

A fuller account of linear operator theory is given in

A.W. Naylor and G.R. Sell, Linear Operator Theory in Engineering and Science,

and also in the books by Yosida, and Kantorovich and Akilov, cited at the endof Chapter 2.

6. Compactness and Its Consequences

I don’t understand new ideas, I don’t even understand why it isnecessary to understand them.

M.E. Saltykov-Shchedrin, History of a Town

6.1 Sequentially compact compact

We introduced the term compact for a set in Definition 1.1.9; we gen-eralized it for a set and proved the Bolzano-Weierstrass theorem(Theorems 1.1.1, 1.1.2) which states that a set is compact iff it isclosed and bounded.

Unfortunately, the Bolzano-Weierstrass theorem cannot be generalized sothat it applies to all metric spaces, as the following counterexample shows. Wedefined the metric space in § 2.1. It is the set of all sequences such

that with the metric

This space is complete (Problem 2.8.4) and the closed ball B

is closed and bounded. Consider the sequence

This is in B, but since

the sequence cannot contain a Cauchy subsequence, and therefore cannotcontain a convergent subsequence. Thus, since not all closed and bounded sets

168 6. Compactness and Its Consequences

in a general metric space X possess the property that any sequence contains aconvergent subsequence, we must introduce a new term to describe those whichdo. For many years the term that was used, was compact, (this was the term weused in § 1.1) but that is now used for a property related to the Heine–Boreltheorem, due to Heinrich Eduard Heine (1821-1881) and Emile Borel (1871-1956), that we discuss below. Now therefore the term that is used is sequentiallycompact. However the outcome of the analysis in this section is that the twoterms are identical in meaning:

sequentially compact compact.

We start with

Definition 6.1.1 A set S in a metric space X is said to be sequentiallycompact if every sequence in S contains a subsequence which convergesto a point

Theorem 6.1.1 Let X be a metric space, and let S be a sequentially compactset in X, then S is closed, bounded and complete.

Proof. Using Definition 2.2.8 we may treat S as a metric space with the metricinduced by X. The proofs of all three properties follow similar lines:

closed. Let be a convergent sequence in S. Since S is sequentially com-pact, contains a subsequence which converges to But there-fore the whole sequence must converge to Thus S contains all its limitpoints, and so is closed.

complete. Let be a Cauchy sequence in S. Since S is sequentially com-pact, must contain a subsequence which converges to Buttherefore, by Problem 2.4.4, converges to Therefore any Cauchy se-quence in S has a limit in S, so that S is complete.

bounded. Suppose S were not bounded. Choose Now choose suchthat this is possible because S is unbounded. Now choose sothat and and so on. The sequence can containno Cauchy sequence, which contradicts the statement that S is sequentiallycompact.

Theorem 6.1.1 states that

sequentially compact closed and bounded.

The counterexample in equation (6.1.1) shows that

closed and bounded sequentially compact

1696.1 Sequentially compact compact

for all metric spaces. In fact we will show later (Theorem 6.2.2) that in a Banachspace X (a complete normed linear space)

closed and bounded sequentially compact

only if the dimension of X is finite.In § 2.2 we defined a domain as a non-empty open set in its closure,is thus a closed set. The reader will notice that closed and bounded sets (or

regions) in figured largely in earlier chapters; we can now call these setssequentially compact; by the end of this section we shall have justified using theterm compact for them, as in fact we did in § 1.2.

The newer definition of compact is based on the concept of a covering, andthe Heine–Borel theorem as we will now describe.

We defined an open set in Definition 2.2.2: it is a set in which every pointis an interior point; the simplest example is an open interval

Definition 6.1.2 Consider a collection of open sets in a metric space X.Their union

is the set of all for some The collection is said to cover a setif

The Heine–Borel theorem in its classical form is

Theorem 6.1.2 Any cover of a closed interval by a collec-tion of open sets has a finite sub-cover. In other words, there is a finitesubsequence of these open sets which we can number l , 2 , . . . , N such that

Proof. We use the method of bisection, as in the Bolzano–Weierstrass theorem.Suppose the theorem were false, so that I had no finite cover. Bisect I; onehalf, say must have no finite cover; let be a point in (say, its righthand end point); bisect one half, must have no finite cover; let be apoint in and so on. Each of the intervals obtained in this way hasno finite cover. On the other hand, the sequence is a Cauchy sequencein I. Since is complete, has a limit Since I is closed,

converges to a point since is contained in one of thesets since is open, is an interior point of so that contains aneighborhood of this neighborhood will contain all for sufficiently large

this means that has a finite cover, namely This is a contradiction.

The Heine–Borel theorem may be extended to give


Problem 6.1.1 Show that any cover of a closed and bounded set bya collection of open balls (or open sets) has a finite sub-cover.

This leads to the new definition of a compact set, namely

Definition 6.1.3 A set S in a metric space X is said to be compact if everycover of S by a collection of open sets has a finite sub-cover.

In this terminology, the Heine–Borel theorem, as extended in Problem 6.1.1,states that a closed and bounded set in is compact.

Problem 6.1.2 Show that a set is compact (according to Defini-tion 6.1.2) iff it is closed and bounded.

By comparing Theorem 6.1.2 and Problem 6.1.2 we see that the terms com-pact and sequentially compact are equivalent in we shall now show thatthey are equivalent in any metric space.

Theorem 6.1.3 A set S in a metric space X is compact iff it is sequentiallycompact.

Proof. Suppose that S is compact (according to Definition 6.1.3), but notsequentially compact. Thus there is an infinite sequence with nosubsequence converging to a point in S. This means that the points of donot cluster about any point of S. Thus each point can be covered by anopen ball which contains at most one point of This provides anopen cover for S, which has a finite sub-cover Since canhave at most one point in each such ball, is finite, which is impossible.

Now suppose that S is sequentially compact, but not compact, so there isan infinite cover of the set which does not contain a finite sub-cover. Choose

and a point Since S cannot be covered by a finite collection ofopen sets, it cannot be covered by the ball of radius about Therefore wecan choose such that For the same reason we can choose

outside balls of radius around and i.e. so thatContinue in this way to define such that Thesequence must possess a Cauchy sequencefor sufficiently large This is a contradiction.

With this theorem we have achieved our goal:

sequentially compact compact.

We will sometimes need the concept of a precompact set, given in

Definition 6.1.4 Let X be a metric space. A set is said to be pre-compact if its closure is compact.

6.2 Criteria for compactness 171

Thus if S is precompact any sequence in 5 contains a subsequence whichconverges to

6.2 Criteria for compactness

We start by recalling the Definition 2.2.1 of an open ball. Then we introduce

Definition 6.2.1 Let X be a metric space, and suppose A finite setof N balls with and is said to be a finiteof S, if every element of S lies inside one of the balls i.e.

The set of centers of a finite is called a finite for S.

Definition 6.2.2 Let X be a metric space. A set is said to be totallybounded if it has a finite for every

Clearly every finite set is totally bounded. Also, if any infinite set S has afinite then one of the balls must contain an infinity of elements.

We are now ready for Hausdorff ’s compactness criterion, due to Felix Haus-dorff (1868–1942), which we may state as

Theorem 6.2.1 Let X be a complete metric space. A set is compactiff it is closed and totally bounded.

Proof. We argue very much as in Theorem 6.1.3. Suppose S is compact, thenit is closed (Theorem 6.1.1). Suppose it is not totally bounded. This means thatthere is an for which S has no finite Choose Choose

such that choose so that and soon, as before. The sequence must possess a convergent subsequenceso that for sufficiently large This is a contradiction.

Now suppose that S is closed and totally bounded, and let be asequence in S. We will show that we can select a convergent subsequence from

which converges to a point in 5, so that S must be sequentially compact,and therefore compact. Take and construct a finite for S.One of the balls, say must contain an infinity of elements of Chooseone of them and call it Take and construct a finiteof One of these balls, denoted by must contain an infinity of elementsof which are in Choose one of these and call it Continuing thisprocedure we obtain a subsequence of Since andare, by construction, in the ball which has radius we have

Then the triangle inequality gives


which means that is a Cauchy sequence. X is complete, so that thisCauchy sequence has a limit it is a convergent sequence. S is closed sothat

This theorem balances the counterexample used at the beginning of thissection (equation (6.1.1)) to show that a closed and bounded set is not nec-essarily compact. In a general metric space, boundedness is not sufficient toensure compactness; the set must be (closed and) totally bounded.

Problem 6.2.1 Let X be a metric space. Show that if a set is precom-pact (Definition 6.1.4), then it is totally bounded.

Note that we do not need completeness of X to show that S is totallybounded. On the other hand, we do need the completeness of X to show thata closed and totally bounded set is compact, or equivalently, that a totallybounded set is precompact. For consider

Problem 6.2.2 Let X be the set of rational numbers with the metricLet S be the set of rational numbers such that

Show that S is closed in X and totally bounded, but not compact; Xis not complete.

For normed linear spaces, the question ‘when is a closed and bounded setcompact?’ is answered by

Theorem 6.2.2 Let X be a normed linear space. Every closed and bounded setis compact iff the dimension of X is finite.

Note that the theorem states that if the dimension of X is finite, then everyclosed and bounded set is compact. On the other hand, it also states that if thedimension of X is infinite, then not every closed and bounded set iscompact, i.e. there is at least one which is not compact. It is easy to prove thefirst part, but to prove the second, converse part, we need

Lemma 6.2.1 Let X be a normed linear space, be a closed subspace,and Then for any such -that there is an element suchthat and

6.2 Criteria for compactness 173

(Note X – S, sometimes written X\S , is the set of elements in X whichare not in S.)

Proof. Since there is an element Let

First we show that d > 0. For if d = 0, then there is a sequencesuch that as this means that and so

since S is closed. This contradicts the supposition thatThus d > 0. According to the definition of infimum (§ 1.1), for any thereis a such that

The required by the lemma is

Clearly and for any we have

We are now ready to prove Theorem 6.2.2

Proof. To say that X has finite dimension means (Definition 2.8.7) that thereis a finite set of elements such that any can be representedin the form

We now show that the result holds only if the dimension of X is finite. Wewill show that if X is infinite dimensional, then it has a closed bounded set,the closed unit ball around zero, which is not compact. Take an elementsuch that and denote by the space spanned by i.e. the setof all elements where If then by Lemma 6.2.1, there is

such that and Denote by the linearspace spanned by and If then, by the same lemma, we can find

such that and

If X is infinite dimensional then we can continue this procedure to obtaina sequence such that if This sequence on theunit ball cannot contain a Cauchy subsequence, and therefore cannot contain aconvergent sequence. Thus if X is infinite dimensional, the unit ball cannot becompact.


Corollary 6.2.1 In an infinite dimensional space the closed unit ball aroundzero is not compact.

Corollary 6.2.2 A bounded set in a finite dimensional normed linear space isprecompact – its closure is compact.

Note how the second half of this proof mimics the counterexample used atthe beginning of the section.

We conclude this section by showing how the terms compact and separableare linked, by

Theorem 6.2.3 A compact set S in a metric space X, and in particular acompact metric space, is separable.

Proof. We recall Definition 4.1.3, that 5 is separable if it contains a countablesubset M which is dense (Definition 2.2.7) in 5. Suppose S is compact. Takethe sequence where By Theorem 6.2.1, for each

S has a finite Thus there is a finite sequence of open ballswhere and of course N depends on which covers

S. The collection of all is the required countable dense set M,‘ since forany and any there is an such that Thus S isseparable.

6.3 The Arzelà–Ascoli theorem

We start with a note on terminology. There are three terms which we haveused widely in this book: operator, functional and function; all are mappings.An operator (Definition 2.7.1) is a mapping from one metric space X intoanother Y; a functional (Definition 2.7.2) is a mapping from X into ora function (Definition 1.2.2) is a mapping from into or Thus inour usage, a function is a particular functional, which in turn is a particularoperator. We warn the reader, however, that some authors use function andoperator interchangeably, while others equate function and functional; we willretain the distinctions we have made.

In § 1.2, we proved some of the classical theorems in the theory of functionsof a real variable, in particular, Theorems 1.2.1 and 1.2.2; the last named statesthat if f ( x ) is continuous on a compact region then it is uniformlycontinuous on We now show that we can generalize these results to continu-ous functionals defined on a compact set Y of a metric space X. First we needsome definitions and preliminary results.

Definition 6.3.1 Let X be a metric space, and Y a subset of X. A functionalf (real or complex valued) defined on Y is said to be continuous atif, given we can find such that and implies

6.3 The Arzelà–Ascoli theorem 175

The functional f is said to be continuous on Y if it iscontinuous at every

Definition 6.3.2 Let X be a metric space, and f be a functionaldefined on Y. The functional f is said to be uniformly continuous on Yif, given we can find such that and imply

When is uniformly continuous on Y, there is one which makesfor any two satisfying when f is only continuous,

depends onWe now prove

Proof. Suppose the assumption were false. This means that for some wecannot find a such that and impliesThus for this and each we can find such that

but

But Y is compact so that the sequences contain subsequencesconverging to and

as , so that Hence

because f is continuous at This contradicts (6.3.1).

Problem 6.3.1 Let X be a metric space, Y a compact subset of X and fa real-valued continuous functional defined on Y. Show that f is bounded andattains its maximum and minimum values.

Note that Theorem 6.3.1 and Problem 6.3.1 are the generalizations of The-orem 1.2.2 and Theorem 1.2.1 respectively.

Theorem 6.3.1 concerns a simple functional f ; now we consider a family offunctionals. We use the notation or for the family; we use ratherthan because the family need not be countable. We define two new terms:uniformly bounded, and equicontinuous

Definition 6.3.3 Let X be a metric space, and be a familyof functionals defined on Y. The family is said to be uniformly bounded ifthere is a constant c > 0 such that for all and all

Theorem 6.3.1 Let X be a metric space, Y a compact subset of X, and f bea continuous functional defined on Y, then f is uniformly continuous on Y.


Definition 6.3.4 Let X be a metric space, and be definedon Y. The family is said to be equicontinuous if, given there is asuch that if then

for all and all

Problem 6.3.2 Let X be a metric space, Y a compact subset of X, anda finite family of continuous functions on Y. Show that

is uniformly bounded and equicontinuous.

In § 2.3 we defined to be the set of functions continuous on aclosed and bounded region of (and thus uniformly continuous on Onthe basis of Theorem 6.3.1 and Problem 6.3.1 we can define C(Y) for a compactset as the set of continuous (and therefore uniformly continuous)functionals on Y. This C(Y), like is itself a metric space, with

and we can talk about compact and precompact sets in this metric space.

Problem 6.3.3 Let X be a metric space, and Y be a compact set in X. Showthat C(Y) is a complete metric space.

The Arzelà–Ascoli theorem links the precompactness of a family offunctionals, i.e. a set in C(Y), to the conditions of uniformly bounded andequicontinuous. We state the Arzelà–Ascoli theorem, due to Cesare Arzelà(1847–1912) and Guilio Ascoli (1843–1896), as

Theorem 6.3.2 Let X be a metric space, Y a compact set in X anda family of continuous functionals on Y, i. e. The family isprecompact in C(Y) iff is uniformly bounded and equicontinuous.

Proof. Suppose is precompact in C(Y). By Problem 6.2.1 it is totallybounded. By Definition 6.2.2, it has an for every Thus it hasan for Thus there is a finite family of functionalsin C(Y) such that for any there is a for which

But is a finite family, and so, by Problem 6.3.2, is uniformly bounded. Thusthere is a such that for all and all sothat

for all and all i.e. is uniformly bounded.

6.3 The Arzelà–Ascoli theorem 177

We now show that if is precompact, then it is equicontinuous.Choose By Problem 6.2.1, is totally bounded, and thus has a

finite Thus there is a finite family such that, for anythere is a for which

But is a finite family, and so, by Problem 6.3.2, is equicontinuous. Thusthere is a such that if then

for all and all Thus for any we have

Thus is equicontinuous.We must now show that if is uniformly bounded and equicontinuous,

then it is precompact. Thus we must show that any sequence of functionals incontains a subsequence converging to a functional in C(Y) (actually

to a functional in Since C(Y) is complete (Problem 6.3.3) it is sufficientto find a Cauchy subsequence. Since the norm we are using is the uniform norm(6.3.2), a Cauchy sequence is one that is a uniform Cauchy sequence on Y, i.e.given we can find N such that if then

To find such a sequence we use the fact that Y is compact. Since it is compact,it is separable (Theorem 6.2.3). Let be a countable set which is dense in Y.Take a sequence of functionals and consider the sequence at Sincethe sequence of numbers is bounded, it contains a Cauchy subsequence

Now consider the sequence at i.e. itcontains a Cauchy subsequence Continuing in this way we obtain, atthe th step, a subsequence which is a Cauchy subsequence at each of

We now show that the sequence withis a uniformly Cauchy sequence.

Choose The sequence being a subset of is equicontin-uous. Thus we can find such that if then

for all Since Y is compact, it is precompact and therefore totallybounded (Theorem 6.2.1). Let the finite set of balls bea finite of Y with radius Since the set is dense in Y, thereis a member of this set, say in each ball This means that for anywe have


Thus if M = then the set forms a finite forY with radius But is a Cauchy sequence at each of the finite set ofpoints Therefore we can find N such that if then

Now suppose and choose such that Thenif we have

(Note that we use (6.3.3) to bound the first and third term, (6.3.4) to boundthe second.) Thus is a Cauchy sequence in C(Y). Since C(Y) is acomplete metric space (Problem 6.3.3), will converge to a continuousfunctional in C(Y); this functional will lie in the closure of Thusis compact, i.e. is precompact.

6.4 Applications of the Arzelà–Ascoli theorem

We first prove

Theorem 6.4.1 Let be a compact region in Let be a family offunctions defined on If is uniformly bounded in then isprecompact in is compact if it is closed.

where Thus if is uniformly bounded in

it is uniformly bounded in under the metric

We suppose is sufficiently regular so that if and are suffi-ciently close together, then the segment joining and is in in that casewe have

Proof. Note that the theorem concerns functions defined on Itis sufficient to prove that is uniformly bounded and equicontinuous. Themetric in is (2.3.4), namely

6.4 Applications of the Arzelà–Ascoli theorem 179

The chain rule gives

But is uniformly bounded in so that

for all all all and all Thus

so that is equicontinuous in

It is worthwhile to restate this result for the simplest particular case. Letand be the set of functions defined on [0,1] such that

and are uniformly continuous on (0,1), i.e. and

then being closed, is compact in the metric of C[0, 1], i.e. in

We now use the result to prove the following local existence theorem, dueto Giuseppe Peano (1858–1932), for the Cauchy problem

Theorem 6.4.2 Let denote the rectangle

Let be continuous for and bounded there by M, i. e.Let then there is a solution to the

Cauchy problem (6.4.!) on the segment

Proof. We will construct a family of functions which satisfiesthe condition of Theorem 6.4.1. To do so we divide the real line around intosegments of length and define on one segment in terms of itsvalues on the previous segment (to the left). In order to start the process wemust define on we define it as


In we define it as

and generally, in we define

Putting the formulae (6.4.4) together for we see that forwe have

Equation (6.4.2) shows that if then

This means that the arguments for f in equation (6.4.3) lie in Q, so that ifwe have

We now show, by induction, that this result holds for forIt holds for Suppose it holds for then

(6.4.4) gives

so that it holds for This establishes that

which means that the family is uniformly bounded on Ondifferentiating equation (6.4.5) we see that

Since and satisfy (6.4.6), the arguments of are in Q, so that

Thus the family satisfies the condition of Theorem 6.4.1. Therefore isprecompact, and has a Cauchy subsequence, Since we are using theuniform norm in the compact space this subsequence will converge,uniformly, to a continuous function Thus we may pass tothe limit in equation (6.4.5) with replaced by and find thatsatisfies

1816.4 Applications of the Arzelà–Ascoli theorem

This means that is a solution of (6.4.1) on

Problem 6.4.1 Generalize Theorem 6.4.2 to the Cauchy problem for the systemof N ordinary differential equations

[Hint: Putand write the equations as

In practice we do not solve the Cauchy problem (6.4.1) by using the processdescribed in Theorem 6.4.2; instead we use a numerical method such as Euler’smethod . In the simplest version of this method we divide the intervalinto equal segments of length and construct a piecewise linearfunction which is given by

when We determine the from the recurrencerelation

Since z(t) is not differentiable at the knots we cannot use Theorem 6.4.1,but must show, directly, that the family is uniformly bounded andequicontinuous when f satisfies the conditions of Theorem 6.4.2. Now we argueas before. For in (6.4.10), we have

Thus so that we may now prove as beforethat and thus

But is a piecewise linear function, so that


and the family is uniformly bounded. Now we must show that it is equicon-tinuous. If are in the same interval then

Now suppose that are in different intervals, thuswhere We write

If we use the equation (6.4.9) for and for and simplify the resulting expres-sion we find

Now andThus

Thus the family is equicontinuous on so that, by the Arzelà–Ascoli theorem, there is a subsequence which is a Cauchy sequence.As before, this subsequence will converge uniformly to a continuous function

Now if then

The expression on the right is a finite Riemann sum. Thus if we write down theequation for and let we find that the limiting functionsatisfies

and so is the solution to the Cauchy problem (6.4.1).

Problem 6.4.2 Justify Euler’s method for the Cauchy problem (6.4.8).

Euler’s method is of course not used in the actual computational solutionof differential equations. There are various finite difference methods which are

6.5 Compact linear operators in normed linear spaces 183

used for which the question of convergence is open. Some of these questionsmay be answered by assuming some differentiability of There are evenmore difficult questions connected with the finite difference solution of boundaryvalue problems for partial differential equations. Few of these procedures havebeen completely justified, those which have are so-called variational-differencemethods, related to finite element methods; they are justified by modifying theenergy space techniques which we considered in Chapters 3 and 4.

In the Arzelà–Ascoli theorem we consider functional, and in particular,functions in the uniform norm. However there is a similar result for functionsin the norm. Thus if is a bounded domain in and then afamily offunctions in is precompact iff it is uniformly bounded andequicontinuous in the norm of

6.5 Compact linear operators in normed linear spaces

We laid the foundations of the theory of linear operators in Chapter 5. Thereour presentation dealt largely with continuous (bounded) linear operators. Com-pared to the theory of linear operators in a finite dimensional space, the theoryof continuous operators in an infinite dimensional space is complicated. Manyof the linear operators encountered in practice possess an additional propertywhich make them, in some sense, very like linear operators in a finite dimen-sional space. This is the property that we will explain in this section.

Definition 6.5.1 Let X, Y be normed linear spaces. A linear operator A fromX into Y is said to be compact (or completely continuous,) if it mapsbounded sets of X into compact sets of Y.

We first prove

Theorem 6.5.1 A compact linear operator is continuous.

Proof. Suppose A is not continuous. This means that there is a bounded se-quence such that By the definition of a compactoperator, the infinite set lies in a compact set, (Definition 6.1.1) there-fore it contains a convergent subsequence, and a convergent sequence isbounded (Problem 2.4.1), i.e. This contradicts

Thus a compact operator is continuous, but the converse is false. For exam-ple, the identity operator f defined by is continuous, but not compact;it maps a ball, a bounded set, into the same ball, and a ball is compact only ifthe space is finite dimensional (see Corollary 6.2.1 of Theorem 6.2.2).


Problem 6.5.1 Let X,Y be normed linear spaces. Show that a linear operatorA from X into Y is compact iff it maps bounded sets of X onto precompactsets of Y.

Problem 6.5.2 Let X, Y be normed linear spaces. Show that if A, B are com-pact linear operators from X into Y, then so are

Problem 6.5.3 Let X be a normed linear space and i.e. A,Bare continuous linear operators in X. Show that if A is compact, then AB andBA are compact. (Note that B need not be compact.)

The following theorem gives some simple sufficient conditions for a linearoperator to be compact.

Theorem 6.5.2 Let X, Y be normed linear spaces and A be a linear operatorfrom X into Y.

a)

b)

If A is continuous and dim R(A) is finite, then A is compact.

If dim D (A) is finite, then A is compact.

Proof, a) Let be a bounded sequence in X. The map of is bounded,since dim R(A) is finite; the corollary to Theorem 6.2.2states that a bounded set in a finite dimensional space is precompact. ThereforeA is compact by Problem 6.5.1.

b) If dim D(A) is finite, then Problem 2.9.5 states that dim R(A) is finite, andTheorem 2.9.2 states that A is continuous, thus b) reduces to a).

This theorem leads us to

Definition 6.5.2 Let X, Y be normed linear spaces and A be a linear operatorfrom X into Y. If R(A) is finite dimensional, then A is said to be a finitedimensional operator.

We may thus rephrase Theorem 6.5.1 to state that a finite dimensionallinear operator is compact.

If are linear functionals on X then the operator A given by

where is a finite dimensional linear operator from X into Y. If theare continuous, (in particular if dimX is finite – Problem 2.9.4) then A

is compact.


Problem 6.5.4 Let X, Y be normed linear spaces. Show that a linear operatorA from X into Y is compact iff it maps the unit ball in X into a compact setin Y.

As an example of a compact linear operator, we consider the operator Adefined by

acting in C[0,1]; it is compact when i.e. when K isuniformly continuous with respect to s and A is continuous because K isbounded, i.e. and

By Problem 6.5.4, it is sufficient to show that the map of the unit ball in C(0,1)is precompact. By the Arzelà–Ascoli theorem, we need only show that this setis uniformly bounded and equicontinuous. The inequality (6.5.2) shows that itis uniformly bounded, for implies

Suppose then

Choose K is uniformly continuous in and Thus we can findsuch that for all and all satisfying

Thus for such

Thus for is uniformly bounded and equicontinuous, andtherefore precompact. Therefore A is compact.

Problem 6.5.5 Show that if then the operator A in(6.5.1) is compact in

If A is a compact linear operator on X into Y, then since it is continuous,it is in We show that the set of compact linear operators inis closed, in the norm (5.1.2) of Thus we prove

Theorem 6.5.3 Let X be a normed linear space and Y be a Banach space.If a sequence of compact linear operators converges uniformly(i.e. in the norm (5.1.2)) to A, then A is compact.


Proof. Let S be a bounded set in X. Choose and then choose sothat for every The operator is compact; thereforethe map of S under is precompact. Therefore, by Theorem 6.2.1,it is totally bounded. Therefore there is a finite setsuch that every point in lies in a ball of radius around one of

Choose then choose so that

then

This means that the set is totally bounded, and therefore, again by The-orem 6.2.1, precompact. (Notice that we need Y to be complete.) Thus A iscompact.

We may apply Theorem 6.5.3 to show that the infinite dimensional matrixoperator A defined by

is compact if

We can easily show that

Now define a finite dimensional operator by

The operator is finite dimensional and therefore compact. But

so that


Thus is bounded and so that A, like is compact.Using this theorem we can weaken the conditions of Problem 6.5.5; all we

need is that and then the operator A will becompact in The space is the completion of C([0,1] x[0,1]). Thus there is a sequence of functions suchthat

as To apply Theorem 6.5.3 we need to show that the operator Adefined by

is continuous, i.e. in and that if

then converges uniformly to A. The boundedness of A follows from

Thus

Similarly

Thus

and Theorem 6.5.3 shows that A is compact in

Problem 6.5.6 Let be a bounded domain in and letShow that the Fredholm integral operator A defined by

is compact in

We have shown that if then the operator Agiven by


is compact in i.e. if then and A maps abounded set into a compact set.

In § 5.5 we showed that the functions form acomplete orthonormal system for Therefore the functions

form a complete orthonormal system forThe operator A maps to

But and the Riemann–Lebesgue lemma (4.5.12) states that

Thus the sequence converges weakly to zero. But A is compact so thatthe map, of the bounded set must be precompact. Therefore

must contain a subsubsequence converging toBut then so that, since the are complete,Thus, for the sequence there is no constant as required by (6.3.2),such that

so that the integral operator does not have a bounded inverse. We can generalizethis result to give

Theorem 6.5.4 Let X, Y be normed linear spaces, and A be a compact linearoperator on X, i.e. in and onto Y, i.e. R(A) = Y. If A has abounded linear inverse on Y onto X, then X is finite dimensional.

Proof. Let S be a closed and bounded set in X and let be the image ofS under A. Let be a sequence in S. By Problem 6.5.1 the image,of under A is precompact. Therefore contains a subsequenceconverging to By hypothesis is bounded and has domain Y. Thusthere is an such that and

Thus but and S is closed, so that Therefore Sis compact. Therefore any closed and bounded set in X is compact so that, byTheorem 6.2.2, X is finite dimensional.

Corollary 6.5.1 If X, Y are normed linear spaces and A is a compact linearoperator on X onto Y then A has a continuous inverse iff dim X is finite.

Then, of course, Theorem 6.5.1 shows that dim Y is finite and is com-pact also.

6.6 Compact linear operators between Hilbert spaces 189

It is possible for a compact linear operator from X into Y to have a con-tinuous inverse even when X is infinite dimensional by having R(A) strictlycontained in Y, i.e. there are which are not in the range of A.

6.6 Compact linear operators between Hilbert spaces

So far we have been considering compact linear operators on a normed linearspace X into a normed linear space Y. Now we consider a compact linearoperator from a Hilbert space into a Hilbert space For Hilbert spaceswe have two extra concepts which we have introduced: weak convergence ofa sequence (Definition 4.6.1); and the adjoint operator A*, defined in (5.5.1).(Actually both these concepts can be given broader definitions which apply ingeneral normed linear spaces.)

Regarding the adjoint we have

Lemma 6.6.1 Let A be a continuous linear operator on a Hilbert spaceinto a Hilbert space If A* A is compact, then A is compact.

Proof. Let S be a bounded set in The operator A* A is a compact operatorin It therefore maps S into a precompact set. Thus there is a sequence

such that is a convergent sequence, and

But and is bounded so thatThus is a Cauchy sequence in but is complete so

that is a convergent sequence. Thus A maps S into a precompact set,and A is compact.

Corollary 6.6.1 If A is compact, so is A*.

For if A is compact, then, by Problem 6.5.3, AA* = (A*)*A* is compact.Therefore, by Lemma 6.6.1, A* is compact.

Regarding weak convergence we prove

Theorem 6.6.1 Let be Hilbert spaces, and A a continuous linear op-erator from into A is compact iff it takes every weakly convergentsequences in intoa strongly convergent sequence in

Proof. Suppose is a weakly convergent sequence in A weakly con-vergent sequence is bounded (Theorem 4.6.3). Therefore, since A is compact,

contains a subsequence such that converges strongly to


an element in the complete space On the other hand, since A is contin-uous, it takes the weakly convergent sequence into a weakly convergentsequence (Lemma 5.5.2). Since this sequence contains a subsequence

which converges weakly to the whole sequence must converge weaklyto

We show that converges strongly to Suppose if possible that thereis a subsequence which does not converge to This meansthat there is an and a sequence such that

The sequence is bounded so that, since A is compact, we can find asubsequence such that converges to someBut then must converge weakly to However is a subse-quence of which converges weakly to and the weak limit is unique(Problem 4.6.2), so that Thus we have the contradictory statements

This contradiction forces us to the conclusion thatWe have shown that a compact linear operator takes a weakly convergent

sequence into a strongly convergent sequence. Now we show that if A takesevery weakly convergent sequence into a strongly convergent sequence, then itis compact.

Let S be a bounded set in and A(S) its image under A. According tothe definition of a compact operator (Definition 6.5.1 or Problem 6.5.1) we needto show that A(S) is precompact (i.e. its closure is compact). Take a sequence

and consider a sequence such that Thesequence being in S, is bounded. A bounded set in is weakly precom-pact (i.e. its weak closure is weakly compact) (Theorem 4.7.1). Thereforecontains a weakly convergent subsequence (converging to some in theweak closure of S). By assumption, A takes this subsequence into a stronglyconvergent sequence Therefore contains a strongly con-vergent subsequence A(S) is precompact and A is compact.

We now show that in a separable Hilbert space any compact operator maybe approximated uniformly by a sequence of finite dimensional operators.

Theorem 6.6.2 Let H be a separable Hilbert space and A be a compact operatorin H. Then there is a sequence of finite dimensional operators such that

Proof. Since H is separable it has (Theorem 4.5.3) an orthonormal basisAny can be written


and then

Let be the finite dimensional operator defined by

Let is compact (Problem 6.5.2); we show that as

Consider

By the definition of the supremum, there is a maximizing sequence suchthat and as The set is boundedand weakly closed (Corollary to Theorem 4.6.6) and therefore weakly compact(Theorem 4.7.1). Therefore contains a subsequence which convergesweakly to some such that Since Theorem 6.6.1 showsthat so that But so that

But, on returning to equation (6.6.1), we see that

so that We wish to show that as i.e. thatsince A is compact, it is sufficient to show that (weakly). To

do this we take an arbitrary and find

In Theorem 6.5.4 we showed that if a compact linear operator A on anormed linear space X onto a normed linear space Y has a bounded inverse

by Parseval’s equality.since


then X is finite dimensional. This result holds only if A is one-to-one. Wenow prove a companion result which does not require that A be one-to-one.

Theorem 6.6.3 Let A be a compact linear operator on a Hilbert space intoa Hilbert space If R(A) is closed, then it is finite dimensional.

Proof. If R(A) were infinite dimensional then we could form an orthonormalsequence in R(A). For this sequence, Let N(A] be thenull space of A. Decompose into N(A) and The restriction,of A to is a continuous one-to-one linear operator on the Hilbert space

onto the Hilbert space R(A). Therefore, by Theorem 5.3.4, it has abounded inverse Therefore maps the sequence ontoa bounded set in But like A, is compact so that it maps this boundedset in into a precompact set in R(A). Thus must contain a convergentsubsequence. But this is impossible since Therefore R(A) cancontain no infinite orthonormal sequence. It must be finite dimensional.


Synopsis of Chapter 6: Compactness

Sequentially compact : every sequence in S contains a subsequence whichconverges to Definition 6.1.1.

: a sequentially compact set is closed, bounded and complete. Theo-rem 6.1.1.

Compact : every cover of S by a collection of open sets has a finite sub-cover.Definition 6.1.3. Theorem 6.1.3.

Criteria for compactness : in a complete space S is compact iff it is closedand totally bounded. Theorem 6.2.1.

: in a finite dimensional space S is compact iff it is closed and bounded.Theorem 6.2.2.

A compact set is separable: Theorem 6.2.3.

Continuous functionals : on a compact Y

: continuous. Theorem 6.3.1.

: uniformly bounded , Definition 6.3.3; equicontinuous, Definition 6.3.4.

: precompact family iff uniformly bounded and equicontinuous. Theo-rem 6.3.2.

Compact linear operator : maps bounded sets into compact sets. Defini-tion 6.5.1.

: maps bounded sets onto precompact sets. Problem 6.5.1.

: if inverse is bounded, then X is finite dimensional. Corollary 6.5.1.

: implies Theorem 6.6.1.

7. Spectral Theory of Linear Operators

Half hero and half ignoramusWhat’s more, half scoundrel, don’t forgetBut on this score the man gives promiseHe’s apt to make a whole one yet.

Alexander Pushkin, On Count M.S. Vorontzov(The count was Pushkin’s superior in Odessa; in common par-

lance, a scoundrel is an operator.)

7.1 The spectrum of a linear operator

In continuum mechanics we often encounter operator equations of the form

in a Banach space X, where is a linear operator depending on a real orcomplex parameter The most important example is the equation governingthe steady vibration of an elastic body with frequency namely

In particular, the natural vibration of a string are governed by the boundaryvalue problem

We now introduce

Definition 7.1.1 Let A be a linear operator in a normed linear space X, i.e.from X into X. The resolvent set is the set of complex numbers forwhich is a bounded operator with domain which is dense in X.Such points of are called regular points. The spectrum, of A isthe complement of

If i.e. if is not a regular point, there are three possibilities:

7. Spectral Theory of Linear Operators196

the range of is dense in exists, but is unbounded.We say that belongs to the continuous spectrum of A;

exists, but its domain is not dense in X. We say that belongsto the residual spectrum of A;

does not have an inverse. In this case, according to Problem 5.3.2,there is an satisfying We say that is an eigenvalue ,and any such we call an eigenvector of A.

1.

2.

3.

The theory can be developed for any of the three forms of the basic equation:We shall start with the first,

then go to the second and the third.

Problem 7.1.1 Extend Definition 7.1.1 to the third form of the basic equation.

Problem 7.1.2 Let be an eigenvalue of a continuous linear operator A in anormed linear space X. Show that the set of all eigenvectors corresponding to

is a closed linear subspace of X.

We consider some examples:

A matrix operator acting in This operator has only a point spectrumconsisting of no more than eigenvalues. All other points of the complexplane are regular points.

1.

The differentiation operator acting in Any point in the com-plex plane belongs to the point spectrum, since for any the equation

2.

3.

has a solution Thus the operator has no regular points.

The boundary value problem

where

where is the square

We consider the third problem in If then, givenwe can find N such that

7.1 The spectrum of a linear operator 197

Thus the set S of all such is dense in and

Consider equation (7.1.3) for Suppose but is not on thenegative real axis. The unique solution of (7.1.3) is

To show that we need to show that is bounded awayfrom zero, i.e. there is a such that

If then

Thus if then we may take If and thenThus (7.1.5) holds with

Thus

so that

This means that if equations (7.1.3), (7.1.4) is written as the operator equation

then the inverse operator is a bounded linear operator on i.e.according to Definition 7.1.1, belongs to the resolvent set.

What can we say about the remaining i.e. those on the negative realaxis? If where are integers, then is a solutionof

so that is an eigenvalue.


Problem 7.1.3 Show that if where are integers, then there isno solution of (7.1.3), (7.1.4) for

Consider the remainder, M, of the negative real axis, i.e. the set ofwhich cannot be represented in the form where are integers.Again consider for The set of points of the formis dense in M. In other words, if then there is a sequence where

such that Take

then the corresponding is

Thus the norms of and are related by

so that is unbounded as This means that although theinverse operator exists on S which is a dense subset of it is unbounded.Thus according to 1 following Definition 7.1.1, belongs to the continuousspectrum. Note that the problem (7.1.3), (7.1.4) has no residual spectrum.

Now we consider the so-called coordinate operator in defined by4.

Clearly has no eigenvalues. If then the equation

i.e.

has the unique solution

in Thus belongs to the resolvent set. If then the inverse(7.1.7) is defined for, i.e. has the domain of, functions of the form

where This domain is not dense in whichmeans that points belong to the residual spectrum.

Problem 7.1.4 Consider the coordinate operator in and show thatis the continuous spectrum.

7.2 The resolvent set of a closed linear operator 199

7.2 The resolvent set of a closed linear operator

Now we place a limitation on A; it is not just a linear operator, but a closedlinear operator, as discussed in § 5.4.

Theorem 7.2.1 Let A be a closed linear operator acting in a Banach space X.For any the resolvent operator

is a continuous linear operator defined on X.

Proof. Let D, S denote the domain and range of Thus

By Definition 7.1.1 of the resolvent set, is bounded on S. Thus thereis C > 0 such that

If there is an such that so that (7.2.1) gives

Now suppose is an arbitrary element in X. Since S is dense in X, wecan find such that

Since we can find such that

and thus

Applying the inequality (7.2.2) to we find

But since is a Cauchy sequence. Therefore, is a Cauchysequence; since X is a Banach space, there is an such that

Now we apply Definition 5.4.1 to the closed operator and deduce thatand

200 7. Spectral Theory of Linear Operators

But was an arbitrary element in X; thus the range of i.e. S , thedomain of is X. Thus the inequality (7.2.1) holds on X, so that

is continuous on X.

For functions of a complex variable we have

Definition 7.2.1 Let G be a domain in The function is said to beholomorphic in G iff at every point it has a power series expansion

with non-zero radius of convergence.

We now show that, when treated as a function of the resolvent operatorfor a closed operator A is a holomorphic function of on

according to

Theorem 7.2.2 Let A be a closed linear operator acting in a Banach spaceX. The resolvent set is a domain (an open set) of and isholomorphic with respect to in

Proof. Suppose Theorem 7.2.1 states that is a continuouslinear operator on X. Thus the series

is convergent in the circle of and thus is aholomorphic function of in this circle. Multiplying the series by

we obtain I, i.e. (7.2.3) is

Problem 7.2.1 Let A be a closed linear operator in a Banach space X. Forany show that satisfies the Hilbert identity

Let B be a bounded linear operator in X. The series

is convergent if Multiplying it by we obtain I, i.e.

7.3 The spectrum of a compact linear operator in a Hilbert space 201

Note the difference between this result and that provided by Theorem 7.2.2.Equation (7.2.4) holds outside the circle of radius because B is bounded;on the other hand, the series (7.2.3) converges because is boundedwhen

Problem 7.2.2 Let B be a bounded linear operator in a normed linear spaceX. Show that the spectral radius of B, defined by

exists and that the expansion

is valid in the domain

Problem 7.2.3 Let be a continuous operator in a normed linear spaceX, and suppose is holomorphic with respect to in Show that theresolvent set of is open and is a holomorphicoperator in

7.3 The spectrum of a compact linear operator in aHilbert space

For compact linear operators in a Hilbert space, we can describe the spectrumfully. The first results in this direction are due to Fredholm; he studied theintegral operator and established that its spectrum had properties similar tothose of a matrix operator. The theory was extended to compact operators in aBanach space by Riesz and Pavel Julius Schauder (1899–1940); we will describeit for operators in a Hilbert space. The theory is of great importance as itdescribes the vibrations of bounded elastic bodies.

It transpires that we must consider the free and forced vibration problemstogether; that is, in abstract terms, we suppose is a compact linear oper-ator in a Hilbert space H and consider the eigenvalue equation

and the non-homogeneous equation

We need to introduce the adjoint operator We know (Corollary 6.6.1)that if is compact, so is We introduce the definitions in

Definition 7.3.1 The null space and range of will be denoted by


respectively. In a similar manner we will use

to denote the null space and range of

Problem 7.1.2 shows that N, N* are closed subspaces of H.

Lemma 7.3.1 and are finite dimensional.

Proof. Let S be a closed and bounded set in and suppose Ais compact; A maps bounded sets on precompact sets (Problem 6.5.1); there-fore is precompact; but so that is precompact;therefore S is closed and precompact, and therefore compact. Therefore everyclosed and bounded set in is compact so that, by Theorem 6.2.2,is finite-dimensional. We may argue similarly for

Definition 7.3.2 Let denote the orthogonal complements ofon H, respectively.

We recall the orthogonal decomposition of a Hilbert space (Definition 4.3.1)and remember that an orthogonal complement is always closed. Thus

being closed subspaces of a Hilbert space H, are themselves Hilbertspaces.

Lemma 7.3.2 There are constants such that

for all

Proof. The right hand inequality holds because being compact, isbounded. Let us prove the left hand inequality. Suppose that there is no such

for all This means that there is a sequencesuch that and as Because is com-pact, the sequence contains a Cauchy subsequence. But this meansthat must also contain a Cauchy subsequence because

Let us rename this Cauchy subsequence Since is complete,will converge to since we have On the otherhand, and A is continuous, so that But


so that i.e. But so that isnot zero, but it belongs to both of the mutually orthogonal setsThis is impossible. Therefore the left hand inequality holds.

The inequalities (7.3.3) state that, in the Hilbert space the normis equivalent to

and that the inner product is equivalent to

We now prove

Theorem 7.3.1 and

Proof. To prove the first result we must show that the equation

has a solution iffSuppose that, for some equation (7.3.7) has a solution i.e.

Let then, by using equation (5.5.1) we find

Thus is orthogonal to every i.e. Therefore

Now suppose that The functional is linear and continuouson H, and therefore on The space is a Hilbert space with innerproduct given by (7.3.6). Therefore, by Riesz’s representation theoremthere is such that

This equality holds for but it also holds for also. For ifwe may write where and and use

to give

Put then


so that

In other words, if then equation (7.3.7) has a solution, i.e.thus and The second part follows

similarly from Lemma 7.3.2 applied to

Note the distinction between this theorem and Theorem 5.5.1: isclosed, whereas R(A) is not necessarily closed.

Lemma 7.3.3 Let be the null space of i.e.

Then

is a finite-dimensional subspace of H;

for

there is an integer such that3.

2.

1.

Proof. 1. can be written in the form

where is a compact linear operator. The result follows from Lemma 7.3.1.This is evident.2.

3. First we note that if for some then for allIndeed if then i.e.

Thus Thereforeand so that x Thus

Now suppose, on the contrary, that there is no such that Thenthere is a sequence such that and is orthogonalto i.e. is an orthonormal sequence. is bounded and iscompact so that contains a Cauchy subsequence, but this leads to acontradiction. Indeed we have

where

Now for

2057.3 The spectrum of a compact linear operator in a Hilbert space

Thus and are orthogonal and

which means that cannot contain a Cauchy sequence.

Theorem 7.3.2 iff

Note that, since we know that iff

Proof. Suppose that but Take Sincewe can solve successively the infinite system of equations

The sequence has the property

Thus but so that, contrary to Lemma 7.3.3, there isno such such that This shows that if then

Now suppose that then and by The-orem 7.3.1. But therefore, by the same proof we have just used, applied to

we deduce that and again by Theorem 7.3.1,

Corollary 7.3.1 If then is continuous.

Proof. If then and Thus the inequali-ties (7.3.3), and in particular the left hand inequality, holds on H. By Theo-rem 5.3.1, this means that is a continuous operator on H.

Corollary 7.3.2 A compact linear operator in a Hilbert space H has onlya point spectrum.

Proof. Suppose is not an eigenvalue, then and thusand The first states that the domain of is H. Usingthe second in (7.3.3), we see that is a bounded linear operator inH. Thus, according to Definition 7.1.1, is in the resolvent set of A.

Theorem 7.3.3 The spaces have the same dimension.

Proof. Let the dimensions of and be and respectively, andsuppose that Choose orthogonal bases andfor N and N* respectively. Introduce the auxiliary operator by


The operator C, being the sum of a compact operator and a finite-dimensional(and therefore compact (Theorem 6.5.3)) operator, is compact. Letbe the null space and range of we will show that For if

then

Since and all the terms in (7.3.8)are mutually orthogonal, and therefore zero. Thus

The first equation states that the second states that is orthogonalto all the basis elements of therefore Therefore andso, by Theorem 7.3.2 applied to C, Therefore the equation

has a solution, But then

This contradiction shows that On the other hand is the adjointof so that Thus

Remark. In this proof we used the operator which is continuouslyinvertible in H. The same holds for the operator defined by

with any small this operator has a continuous inverse, and

Thus if we replace the equation by the close equationwe can solve the latter for any An operator like is called a regularizer,such operators are widely used in inverse problems.

We pause to consider the meaning of the results of Theorem 7.3.1-7.3.3.Again we have a case of the Fredholm alternative:


either the equation

has a solution for all this means and thereforeso that the solution is unique; in which case so

that and the equation

has no solution.

or the equation

has a finite dimensional space of solutions spanned by inwhich case

has a solution iff (This states thatThe solution is not unique, because

Thus if span then

The results we have established in this section apply to the general equa-tions (7.3.1), (7.3.2). We conclude this section by deriving an extra result whichholds for the important special case

Problem 7.3.1 Suppose that is a continuous linear operator in aHilbert space H. Show that if are eigenvalues of A, then

This simply states that an element x cannot be an eigenvectorcorresponding to two different eigenvalues. It implies that if areeigenvalues of A, and from each of we take a linearlyindependent set of eigenvectors, then their union will be linearly independent.

Lemma 7.3.4 The set of eigenvalues of a compact linear operatorhas no finite limit point in

Proof. Suppose that there is a sequence of eigenvalues such thatFor each eigenvalue take an eigenvector Let be the subspace

spanned by By Problem 7.3.1, and Wecan apply the Gram–Schmidt process to and find an orthonormal sequence

, i.e. such thatThe sequence is bounded; A is compact, so that is

precompact, i.e. it contains a Cauchy subsequence. We now show that thisis impossible. Indeed


where

We now show that if then Sinceit is sufficient to consider only the first two terms in Since

we have

Since and and are orthogonal,thus

so that cannot contain a Cauchy sequence.

Combining the results we have obtained in this section we can state that ifis a compact linear operator in a Hilbert space H, then:

A has only a point spectrum;

each eigenvalue has only a finite dimensional space of eigenvectors;

if then in addition

two eigenvalues cannot have a common eigenvector;

the point spectrum (if there is one) is countable and has no finitelimit point in

(a)

(b)

1.

2.

3.

N.B. Nowhere have we shown that a compact linearoperator in a Hilbert space has an eigenvalue. All our statements have had theform ‘if is an eigenvalue . . . ’ or have been negative statements as in (a), (b)above. In 7.5 we will show that a self-adjoint compact linear operator has atleast one eigenvalue, and then that it has, in a precise sense to be stated, a fullset of eigenvectors.

7.4 The analytic nature of the resolvent of a compactlinear operator

We know (Theorem 7.2.2) that the resolvent of a closed operator isa holomorphic operator-function of in the resolvent set. What is its behaviornear the spectrum? We can answer this question for a compact linear operator.

7.4 The analytic nature of the resolvent of a compact linear operator 209

We begin the study with a finite-dimensional operator in a Hilbert space;such an operator is compact (Theorem 6.5.2) and has the general form

where we suppose are linearly independent. The equation

is

Its solution has the form

and, on substituting this into (7.4.1) we find

Since the are linearly independent we have

This system may be solved by Cramer’s rule to give

and thus

The solution is a ratio of two polynomials in of degree not more thanAll which are not eigenvalues of A are points where the resolvent is

holomorphic; thus they cannot be roots of If is an eigenvalue ofthen If this were not true, then (7.4.3) would be a solution of (7.4.2)for any and this would mean that was not an eigenvalue (Remember

in Theorem 7.3.2 implies Thus the set of all roots ofcoincides with the set of eigenvalues of and so each eigenvalue of

is a pole of finite multiplicity of the resolventWe now consider a general case:

Theorem 7.4.1 Every eigenvalue of a compact linear operator A in a sepa-rable Hilbert space is a pole of finite multiplicity of the resolvent


Proof. We showed in Theorem 6.6.2 that a compact linear operator A in aseparable Hilbert space may be approximated arbitrarily closely by a finitedimensional operator Thus if we choose we can find such that

where The equation

takes the form

In the circle we can write

We note that Problem 5.1.1 shows that the series is uniformly convergent if thenumerical series

is convergent; this is so if in this circle is a holomorphicoperator-function in by Theorem 7.2.2.

We apply the operator to equation (7.4.4) and find

Write in the form (7.4.1), i.e.

and put

then equation (7.4.5) becomes

This looks like equation (7.4.2) except that, instead of being independent ofas were, and are holomorphic functions of in the circle

For any satisfying the elements arelinearly independent, since are linearly independent andis continuously invertible. So, by analogy with (7.4.3), for the solutionto (7.4.7) is

7.5 Self-adjoint operators in a Hilbert space 211

We note that and are the same functions of andas and are of and respectively. Thus

depends on explicitly as a polynomial of degree no greater than but alsodepends on implicitly through the quantities see equation (7.4.6).

If is not an eigenvalue of A then, according to Theorem 7.2.2, the solution(7.4.8) is holomorphic in some neighborhood of and so Ifis an eigenvalue, then For we may choose so that sothat if were not zero, the equation would be soluble for all and sofor all which is impossible. This means that the set of eigenvalues of A lyinginside any circle coincides with the set of zeros of lying insidethis circle.

7.5 Self-adjoint operators in a Hilbert space

Many important problems in continuum mechanics may be phrased as prob-lems relating to a self-adjoint linear operator. The theory for such operators isparticularly straightforward. We will take the eigenvalue problem in the form

We recall that A is self-adjoint if We start with two simple results:

Problem 7.5.1 If A is self-adjoint, then is real for all

Problem 7.5.2 If A is self-adjoint, eigenvalues of A (if there are any) arereal, and eigenvectors corresponding to distinct eigenvalues are orthogonal.

This, combined with Corollary 7.3.1 shows that a self-adjoint compact linearoperator A in a Hilbert space has a spectrum which is a real point spectrum—ifit has one at all. We will now show that A has at least one eigenvalue. We startwith a definition and a lemma.

Definition 7.5.1 A functional on a Hilbert space H is called weaklycontinuous if it takes weakly convergent sequences into (strongly) convergent(numerical) sequences.

Problem 7.5.3 Show that a weakly continuous functional is continuous.

Thus if and then

Lemma 7.5.1 A real valued weakly continuous functional on a Hilbert space Hassumes its maximum and minimum values in any ball

Proof. Let sup There is a sequence such that and

The set is bounded and weakly closed by the


Corollary to Theorem 4.6.6. Therefore, by Theorem 4.7.1 it is weakly compact.Thus the sequence contains a subsequence converging weakly tosome such that By definition of a weakly continuous functional,

The proof for the minimum point is similar.

To use this lemma for operators we prove

Lemma 7.5.2 Let A be a self-adjoint compact linear operator in a Hilbertspace. is a real valued weakly continuous functional on H.

Proof. By Problem 7.5.1, is real valued. Let be weakly convergentto Then

A is compact and so that i.e. Aweakly convergent sequence is bounded (Problem 4.6.3) so that,and

On the other hand so that and thusas and is weakly continuous.

Problem 7.5.4 Show that if A is a self-adjoint operator, then

Theorem 7.5.1 A non-zero self-adjoint compact operator A in a Hilbert spaceH has at least one, non-zero, eigenvalue.

Proof. By Lemmas 7.5.1, 7.5.2, assumes its maximum and minimumvalues on Let these be

then

This must be non-zero because, by homogeneity and Problem 7.5.4,

Therefore there is an such that and


Now consider the functional

The range of values of for coincides with the range of forThus

We will show that is an eigenvector of A. Indeed consider whereis an arbitrary, but fixed, element of H, as a real valued function of the real

variable It is differentiable in some neighborhood of and takes itsminimum value at so that

But

so that (7.5.1) gives

or

Replacing by we get

so that

Since is an arbitrary element of H, we have

Having shown that A has at least one eigenvalue, we now prove

Theorem 7.5.2 A non-zero compact self-adjoint operator A in a Hilbert spaceH has a finite or infinite sequence of orthonormal eigenvectors cor-responding to non-zero eigenvalues which is com-plete in the range R(A) of the operator A, i.e. for every the Parsevalequality


holds.

Proof. By Theorem 7.5.1 there is an eigenvector withwhere

Rename the Hilbert space the operator as let be the space spannedby and decompose into and The space is a Hilbert space.If then for

This means that we may define a new operator in by

This operator is called the restriction of to it is clearly a self-adjointcompact linear operator in the Hilbert space If this operator is not identi-cally zero we may apply Theorem 7.5.1 to it, and find an eigenvector suchthat

Since wehave and

We now continue this process; we let be the space spanned by decomposeinto and call the restriction of to and find an eigenvectorand eigenvalue and so on.First consider the case in which the process stops. That means that there

is an integer for which the restriction of to is identically zero, i.e.

In this case we obtain a finite orthonormal sequence of vectorscorresponding to non-zero eigenvalues moreover

Suppose and consider

and


We have so that and hencesatisfies (7.5.2) so that with Problem 7.5.4, i.e. Thus

so that

Now consider the case in which the process does not stop. We have aninfinite sequence of vectors and a corresponding sequence of non-zeroeigenvalues where According to Lemma 7.3.4 wemust have Choose and then choose N so that if

then Take Suppose and consider given by (7.5.3);so that

Thus

so that, as before

or equivalently

which implies Parseval’s equality

We now obtain another result by making a further assumption concerningA; thus we introduce

Definition 7.5.2 A self-adjoint continuous linear operator A in a Hilbert spaceH is called strictly positive if for all and iff


For a strictly positive, compact, self-adjoint operator in a Hilbert space theprocess described in Theorem 7.5.2 can stop only if H itself is finite dimensional.This leads to

Theorem 7.5.3 Let A be a strictly positive, compact, self-adjoint operator inan infinite dimensional Hilbert space H. There is an orthonormal systemwhich is a basis for H, and A has the representation

Proof. Let and consider

where is the orthonormal sequence of eigenvectors, as in Theorem 7.5.2.We showed in Theorem 4.5.1 that is a Cauchy sequence. We wish to provethat its (strong) limit is zero. Assume that it is not, i.e. Since

we have

But as so that passage to the limit gives

which is a contradiction since A is strictly positive. Therefore and

so that forms a basis for H, and moreover

This theorem shows that one can have a strictly positive compact self-adjoint operator only in a separable Hilbert space. (See Theorem 4.5.3.)

Corollary Under the condition of the Theorem 7.5.3 we can introduce a norm

and a corresponding inner product


The completion of H with respect to this norm is called

Problem 7.5.5 Using the notation of Theorem 7.5.3, show that withis an orthonormal basis for

As an example, consider the eigenvalue problem

This has eigenvalues and eigenfunctions

Let W be the Hilbert space of functions with the innerproduct

Remember that Problem 3.6.1 shows that is a norm for The prob-lem (7.5.4) can thus be posed as

where the operator A is defined as

Thus in the space

which means that But is an orthonormal basis

for In the language of Problem 7.5.5,

so that

which thus forms a basis for


Synopsis of Chapter 7: Spectral Theory

The spectrum of a linear operator. Definition 7.1.1.

For a compact linear operator in a Hilbert space

Definition 7.3.1, 7.3.2 and Theorem 7.3.1.

For self-adjoint operators in a Hilbert space

Eigenvalues are real. Problem 7.5.2.

Eigenvectors corresponding to distinct eigenvalues are orthogonal. Prob-lem 7.5.2.

A non-zero operator has at least one, non-zero, eigenvalue. Theorem 7.5.1.

The eigenvectors of a non-zero compact self-adjoint operator are com-plete in

One can have a strictly positive (Definition 7.5.2) compact self-adjointoperator only in a separable Hilbert space. Theorem 7.5.3.

Applications to Inverse Problems8.

As an orthodox mathematician, he believes his formula morethan his eyes and common sense, and doesn’t see the incongruity init.

Academician A.N. Krylov, on a formula in an article byLevi-Civita.

Well-posed and ill-posed problems8.1

Most problems in mechanics and physics have the form ‘Find the effect of thiscause.’ There are numerous examples: Find how this structure is deformed whenthese forces are applied to it. Find how heat diffuses through a body when aheat source is applied to a boundary. Find how waves are bent, or absorbed, asthey pass through a nonhomogeneous medium.

At the turn of this century the French mathematician Jacques SalomonHadamard (1865–1963) identified three characteristics of what he called a well-posed problem, which we paraphrase as:

existence, i.e. the problem always has a solution;

uniqueness, i.e. the problem cannot have more than one solution;

stability, i.e. a small change in the cause will make only a small change in theeffect.

Much of the research in theoretical mechanics and physics during this centuryhas been devoted to showing that, under specified conditions, the traditionalproblems in these fields do in fact possess these properties.

Traditional cause and effect problems, with attendant studies of the accu-racy and stability of approximate solutions still dominate mechanics. However,during the last three or four decades (since about 1960) there has been a grow-ing recognition that there are important problems which fail to have some orall of the defining properties of a well-posed problem; they are called ill-posedproblems. Many, but not all, are concerned with questions of the form ‘whatis the cause of this effect?’ Since, in many cases, each problem in this subclassmay be associated with a direct problem ‘what is the effect of this cause?’

220 8. Applications to Inverse Problems

they are, somewhat loosely, called inverse problems. It should be noted thata problem is called an inverse one only because of its relation to another thatwe call direct; in some cases the choice of which to call direct and which tocall inverse is arbitrary. Also, not all inverse problems are ill-posed, nor are allill-posed problems, inverse problems.

It may be shown that many ill-posed and/or inverse problems may be re-duced, perhaps after some linearization, to the operator equation

where belong to normed linear spaces X, Y, and A is an operator from Xinto Y. The simplest, and most common form that this equation takes is theFredholm integral equation of the first kind, namely

or more particularly, when

Problem 8.1.1 Show that the integral equation

may be reduced, by changes of variables, to equation (8.1.2).

In § 8.2 we start by recapitulating the various results which we have foundso far regarding the operator equation (8.1.1).

8.2 The operator equation

In § 2.7 we defined a continuous operator from a metric space X into a metricspace Y. The Definition 2.7.3 is a straightforward generalization of the defi-nition of a continuous function of a real variable. In § 2.7 we concentrated oncontraction mappings (Definition 2.7.4), but at the end of the section we showedthat a distinguishing mark of a continuous operator is that the inverse images(Definition 2.7.5) of open (closed) sets in Y are open (closed) sets in X.

In § 2.9 we defined a linear operator from a normed linear space X into anormed linear space Y. Thereafter all the operators that we have consideredhave been linear. The important Problem 2.9.3 shows that a linear operatoris continuous iff it is continuous at 0, and consequently that an operator iscontinuous iff it has a bounded norm, in the sense of equation (2.9.2).

8.2 The operator equation 221

We returned to the theory of continuous (i.e. bounded) linear operators inChapter 5. In Theorem 5.2.1 we showed that a linear operator which is boundedon a domain which is dense in a normed space X, and whose range lies in aBanach space Y, can be extended, without increasing its norm, to the wholespace X. This means that if R(A) lies in a Banach space, there is no loss ofgenerality in assuming that D(A) is closed (If it is not, then Theorem 5.2.1shows that we can extend A to its closure Further, if X is a Banachspace then being a closed subspace of a Banach space, is itself a Banachspace. In this case there is no loss of generality in supposing that A is definedon X, i.e. D(A) = X, for D(A) being a closed subspace of a Banach space, isa Banach space; it is this Banach space that we call X.

In § 5.3, we considered whether a continuous linear operatori.e. on X into Y, had an inverse. A necessary and sufficient condition forto exist is (Problem 5.3.2) that there should not be two distinct and

such that the null space N(A) must be empty. InTheorem 5.3.1 we proved that the operator is a continuous linear operatoriff

The most important results concerning were derived from Banach’sopen mapping theorem (Theorem 5.3.3), that if X, Y are Banach spaces andA is a continuous linear operator on X onto Y, then A maps open sets of Xonto open sets of Y. (This is a much deeper result than the straightforwardTheorem 2.7.2) Note that, as we said in the previous paragraph, since Y isa Banach space, there is no loss of generality in assuming that A is definedon X, i.e. D(A) = X. By contrast, it is a restriction, and a prerequisite ofthe theorem, that A be an operator from X onto a complete space Y. As wepointed out earlier, after the proof of Theorem 5.3.4, when Y is complete, wecan replace ‘R(A) = Y ’ by ‘R(A) is closed’. (The old Y is replaced by thecomplete space R(A) From the open mapping theorem we derivedthe fundamental Theorem 5.3.4. This states that if X,Y are Banach spacesand A is a one-to-one continuous linear operator on X onto Y, then A has acontinuous inverse on Y onto X. This suggests that the equation (8.1.1)should present no difficulty: the solution is simply However, if weaccept this suggestion readily it is because we have underestimated the powerof the restriction ‘onto Y’ i.e. R(A) is closed. The difficulties start to becomeapparent when we consider compact operators.

The formal definition of a compact operator was given in Definition 6.5.1;equivalently we may use Problem 6.5.1: A compact operator maps boundedsets onto precompact sets. In § 6.5 we showed that we may consider the integraloperator A in equation (8.1.2) either in C[0,1] or in in both casesit is compact. The fundamental result concerning compact operators is Theo-rem 6.5.4: if a compact operator has a bounded inverse then X must be finitedimensional.

There are two important corollaries of this results:

Corollary 8.2.1 If X, Y are normed linear spaces and X is infinite dimen-sional, a compact linear operator A from X onto Y cannot have a boundedinverse.

This agrees with what we found for equation (6.5.4): is infinitedimensional and A does not have a bounded inverse.

In § 5.3 we showed, on the basis of the open mapping theorem, that a one-to-one continuous linear operator A from a Banach space X onto a Banachspace Y does have a bounded inverse. From this we can deduce that if indeedit is onto Y, then the only way it can fail to satisfy Theorem 5.3.4 is that it isnot one-to-one. This leads to

Corollary 8.2.2 If X, Y are Banach spaces and X is infinite dimensional,there is no one-to-one compact linear operator from X onto Y.

This means that if X, Y are Banach spaces, X is infinite dimensional, andA is a one-to-one compact linear operator from X into Y, then R(A) cannotbe Y, and indeed cannot be closed, for then R(A) being a closed subspace ofa Banach space, would itself be a Banach space.

Let us apply these results to the Fredholm operator (8.1.2), i.e. (6.5.1),which, under the conditions stated there, is a compact linear operator fromthe infinite dimensional Banach space into the Banach spaceCorollary 8.2.2 states that if A is one-to-one, i.e. there is no (non-zero) function

such that

then R(A) cannot be i.e. there is a which cannot beexpressed in the form

for any In other words the solution of (8.1.2) cannot be uniqueand exist for all one or other of uniqueness and existence mustfail, perhaps both, as in the simple example

where lie in the null space, and a solution exists only if

Moreover, Corollary 8.2.1 states that the solution to equation (8.1.2) is notstable; this is the meaning of the statement that even if it exists (i.e.N(A) = 0) is unbounded. This means that we can find a sequence of functions

such that and For the integral operator in

(8.1.2) the functions are such a sequence, in agreement



with the Riemann–Lebesgue lemma. Explicitly, this means that given wecan take sufficiently large that

is so small that

We describe this situation by saying that small amplitude high frequency noisein may cause large errors in the solution.

We conclude that the operator equation faces us with three diffi-culties: R(A) does not exhaust Y, i.e. there are which are not in therange of A; A may not be one-to-one, i.e. the operator may have a null space;even if the operator has an inverse, this inverse may not be continuous. Thereare various ways in which one or more of these difficulties may be overcome, aswe shall now discuss.

The first result we prove is due to Andrei Nikolaevich Tikhonov (1906–1994):

Theorem 8.2.1 Let X, Y be normed linear spaces and A be a continuous one-to-one operator from X into Y. Let S be a compact subspace of X and letbe the restriction of A to S, then is continuous.

Note that this does nothing to the difficulty that R(A) is strictly containedin Y, and it assumes that A is one-to-one; it simply ensures that the inverseoperator, which will have a range within R(A), has a continuous inverse.

Proof. According to Theorem 5.3.1 we must show that if then there isa constant such that

Suppose this were not so. Then we could find a sequence such thatand Since and S is compact, there is a subsequence

converging to and

Thus but so that A is not one-to-one as we assumed. Therefore(8.2.3) holds and is continuous.

To apply this theorem to the integral equation (8.1.2) we must restrict thefunction toa compact subspace of C[0,1]. To do so we use Theorem 6.4.1,which states that we must ensure that the are uniformly bounded in

[0,1]. Note how this restriction excludes the functions

for which the are not uniformly bounded.


Tikhonov’s theorem deals with the continuity of the inverse by restrictingthe domain, and hence also the range of A. We now consider how we can enlargethe range of A and also deal with the fact that A may not be one-to-one. Todo so we will assume that the operator A is a continuous linear operator on aHilbert space into a Hilbert space The closure of the range of A isa closed subspace of Theorem 4.3.2 states that may be decomposed into

and its orthogonal complement (because the orthogonalcomplement is always closed). Thus the closure of is orin other words, the subspace of is dense in We showhow we can extend the inverse operator from R(A) to Suppose

then its projection onto is actually in R(A). Thismeans that there is an such that

This being the projection of onto is the element of which isclosest to , i.e. according to Theorem 4.3.1

The decomposition Theorem 4.3.2 states that any may be written

Here By saying that we state that there is an suchthat

Any such is called a least squares solution of the equation, because it min-imizes the norm But Problem 5.5.3 states that(Remember that a null space and an orthogonal complement are both automat-ically closed.) This means that so that

There will thus be a unique least squares solution iff has no null solution.This occurs iff A has no null solution. (For if then while if

then so thatSuppose A does have a null space, so that the solution of (8.2.4) is not

unique. There will then be a subset M of solutions satisfying (8.2.3). Thissubset is closed and convex. (It is convex because

and imply We may apply Theorem 4.3.1 toM. This shows that there is a unique which minimizes on M. Wetake this to be the generalized solution of equation (8.2.1); it gives a uniquesolution for which is a dense subspace of This solution


is called the least squares solution of minimum norm. (Of course we can useTheorem 4.3.1 to find the least squares solution which is closest, in the normof to some other element )

The mapping from into D(A) which associatesto the unique least squares solution of minimum norm, is called the

Moore-Penrose generalized inverse of A, after Eliakim Hastings Moore (1862–1932) and Roger Penrose (1932- ).

We circumvented the difficulty that A may have a null space by choosingthe of minimum norm. Another way of proceeding is to restrict toThis however gives exactly the same solution, as shown by

Problem 8.2.1 Let A be a continuous linear operator from to SupposeShow that is the unique least squares solution in

Problem 8.2.2 If represents the restriction of A toshow that for any where is theprojection of on

What have we achieved so far? We started with a continuous linear op-erator which could have a null space (i.e. need not be one-to-one) and whichcould have a range R(A) which was not dense in We have constructed ageneralized inverse of A which is defined on a dense subspace of andwhich yields a unique for any The critical question is whetherthis generalized inverse is a continuous operator. In general it is not; it is merelya closed operator, as discussed in § 5.4. We prove

Theorem 8.2.2 Let be Hilbert spaces, and A be a continuous linearoperator from into The generalized inverse from into isa closed operator. It is continuous iff R(A) is closed.

Proof. We recall Definition 5.4.1, and reword it for our case. is closed iffthe three statements

together imply

Problem 8.2.1 states that is the unique solution of inThus and is closed, imply

Also and imply Butimplies so that

and Thus and is a solution ofAgain Problem 8.2.1 states that Thus is

a closed operator.Now suppose that is continuous. Let be a convergent sequence

in R(A), converging to Let then Since is


continuous and is closed On the other handso that and Thus and

Therefore R(A) is closed.Now suppose R(A) is closed, then so thatis a closed linear operator on into so that, by Theorem 5.4.1, it is

continuous.

Note that Theorem 8.2.2 leaves us with a most unsatisfying result if, asoften the case, A is a compact operator. For we showed in Theorem 6.6.2 thatif A is compact then its range will be closed iff it is finite dimensional. For theFredholm integral equation this means that the equation must be degenerate.

We still have not achieved the construction of a stable ‘inverse’ for thegeneral non-degenerate integral equation.

8.3 Singular value decomposition

In this section we suppose that A is a compact linear operator on a Hilbert spaceinto a Hilbert space As we showed in § 6.6, this means and are

compact self-adjoint linear operators in and respectively. We considerthe eigenvalues and eigenvectors of these operators. Both operators are non-negative in the sense that and for

and Thus theireigenvalues will be non-negative. Note that in these equations, as elsewhere inthis section and the remainder of the chapter, we use the same symbol todenote inner products in and If we used subscripts to distinguish them,then, for example, the last equation would read

The operators and have the same positive eigenvalues. For supposethat is an eigenvector of corresponding to thenso that Then so that is aneigenvector of corresponding to and similarly vice versa.

We may now use Theorem 7.5.2. This states that since it is self-adjoint, has a finite or infinite sequence of orthonormal eigenvectorscorresponding to positive eigenvalues and that the arecomplete in the closure of the range of We note that Theo-rem 5.5.2 shows that and that

for if then on the other hand ifthen so that )

Let and then

and

8.3 Singular value decomposition 227

so that

Thus the form an orthonormal set of eigenvectors for and Theo-rem 7.5.2 states that they are complete in the closure

The system is called a singular system for the operator A,and the numbers are called singular values of A.

The null space N(A) is a closed subspace of so that, according to § 4.3,any element of may be written

and is the projection of on N(A): Since the are complete inwe may write

where Hence

This is called the singular value decomposition (SVD) of the operator A.We now return to the equation

If then this equation has a solution. We recall that the arecomplete in (Theorem 5.5.1). Thus if

then

This imposes a restriction on for it implies

Conversely, if and


then any element

where isa solution of (8.3.6).We conclude that equation (8.3.6) has a solution iff and the

condition (8.3.8) is fulfilled. The condition (8.3.8) is called Picard’s existencecriterion, after Charles Emile Picard (1856–1941).

We note that will either have a finite number of eigenvalues, or aninfinite sequence of eigenvalues. Since the eigenvectors span

the former case will hold only when R(A) is finite dimensional, i.e. A isdegenerate. In the latter case we will have as This means that,in order for (8.3.6) to have a solution, i.e. Picard’s existence criterion to hold,

must tend to zero faster thanWe showed earlier that when gives the unique

element in which satisfies (8.2.5). Equation (8.3.9) shows that when Ais compact this solution is the one obtained by taking Thus

We note that if we denote this by then

in the notation of (8.2.4). We conclude that in taking the generalized inversewe do two things: replace by its projection onfind the unique such that Equation (8.3.10) shows thatif there are an infinity of singular values then is unbounded because, forexample, while

In order to obtain an approximation to we may truncate the expan-sion (8.3.10) and take the approximation as

then as However, the question arises as to how manyterms to take in the expression. For that we must consider the error in the data.Suppose that, instead of evaluating equation (8.3.11) for we actually evaluateit for some nearby such that We will obtain a bound for thedifference between the formed from which we will call and the ‘true’

formed from we will estimate We have

8.4 Regularization 229

This means that

This bound on the solution error illustrates the characteristic properties of asolution to an ill posed problem: for fixed the error decreases with but fora given the error tends to infinity as The inequality (8.3.12) impliesthat in choosing an say corresponding to a given data error we mustdo so in such a way that

Thus there are two conflicting requirements on it must be large enough tomake small, but not so large as to make large. A choice of

such that

is called a regular scheme for approximating

8.4 Regularization

As before, let A be a compact linear operator on into The generalizedinverse gives a ‘solution’ of (8.3.6) for all a densesubspace of which satisfy Picard’s criterion. However is not continuousunless R(A) is finite dimensional (and then closed). The unboundedness ofarises because the tend to zero, and this in turn can be attributed to thefact that does not have a bounded inverse. The operator arose inequation (8.2.7); this in turn followed from (8.2.6) and (8.2.5). To find wefirst found the closest to in the sense of (8.2.5), then, if there were morethan one corresponding to that we choose the having minimum norm.Now, instead of doing this, we will choose a positive parameter and find the

which minimizes

forTo do this we set up a new Hilbert space with elements

where We define the inner product in this space by


so that

Problem 8.4.1 Show that equation (8.4.2) does define a proper inner product,i.e. one that satisfies P1-P3 of § 1.2, and that H is a Hilbert space, i.e. acomplete inner product space.

We can now imitate for this new space H what we did with in § 8.2. Forthe continuous linear operator A which takes into inducesanother continuous linear operator, which we will call which takesinto in H. It is continuous, because

The range, of this new operator is the set of those suchthat and it is a closed subspace of H.

Problem 8.4.2 Show that if a sequence converges toin the norm (8.4.3), then i.e. that is closed.

The Hilbert space H may be decomposed into and its orthogonalcomplement According to Theorem 4.3.1 this means that for any

there is an such that is the element of R(A) which isclosest to i.e.

The decomposition Theorem 4.3.2 applied to H states that anymay be written

Here m is the projection of onto Since there is ansuch that so that

In § 8.2 we used the result, proved in Problem 5.5.3, that Touse this we must first define the adjoint of the operator from to H. Wenote that for any i.e. for any the functional

is a continuous linear functional on the Hilbert space Therefore, by Riesz’srepresentation theorem (Theorem 4.3.3), there is an element of which wecall such that


Thus

so that

Since this holds for all we have

Now we return to equation (8.4.5) and use the result thatto give

which, with given by (8.4.6) is

or

which has the unique solution

We will now show that as the solution of this equation tends tofor those (satisfying Picard’s condition) for which exists. We note that

But so that But we showed that the spanso that we may write

Substituting this into (8.4.7) we find

so that

and hence

To show that we proceed in two steps, first we show that this operatorwhich gives in terms of is bounded. We note that since the are positiveand tend to zero we can find such that


Thus when

while if

so that if we may write

Now we show the convergence of to for those for which Picard’sexistence criterion holds. We note that

so that

Choose Since the series in (8.4.12) converges, the sum from tomust tend to zero as N tends to infinity. Therefore we can find N such thatthat sum is less than and

But now the sum on the right is a finite sum, and we can write

Finally, we choose so that then so that

We have proved that, for any is a continuous operator and that, forthose for which exists, converges to as

Now suppose that the data, is subject to error. This means that insteadof solving equation (8.4.7) for we are actually solving it for some nearbysuch that We wish to obtain a bound for the difference betweenthe formed from which we will call and the ‘true’ formed from

we wish to estimate We have


so that, by proceeding as in (8.4.10), (8.4.11), we have

where Since the series converges, we may, for any givenfind N such that the sum from N + 1 to is less than Now

so that

and hence, since this is true for all we must have

Again, this bound on the solution error illustrates the characteristic prop-erties of a solution to an ill-posed problem: for fixed the error decreases with

but for a given the error tends to infinity as The inequality (8.4.14)implies that in choosing an say corresponding to a given data errorwe must do so in such a way that

When we choose so that (8.4.15) holds, the difference between andsatisfies the inequality

and we have already shown that the second term tends to zero with A choiceof such that

is called a regular scheme for approximatingThe inequality (8.4.16) gives a bound for the error in The error has

two parts, the first is that due to the error in the data, while the second is thatdue to using rather than the limit as It is theoretically attractive toask whether we can choose the way in which depends on i.e. so thatboth error terms are of the same order.


To bound the second term we return to the inequality (8.4.13). This holdsfor arbitrary If we take we find

so that

This means that if we use the simple choice then the first term in(8.4.15) will be of order while the second will be of order On the otherhand if we take then and will both be of orderso that

8.5 Morozov’s discrepancy principle

We continue to assume that A is a compact linear operator from to Thechoice is theoretically attractive, but difficult to apply. Morozov(1984) put forward a discrepancy principle in which the choice of a is made sothat the error in the prediction of i.e. is equal to the error inthe data, i.e.

We will show that for any there is a unique value of satisfying (8.5.1).First we note how we choose for a given we choose it using (8.4.4).

The element is replaced by the closest to it in the norm of H;we could do this because unlike R(A), is always closed. This meansthat, in computing there is no loss of generality in assuming thati.e. that Therefore we assume that the error in the data isless than or equal to and that the signal to noise ratio is greater than unity:

Decompose into and Theorem 5.5.1 states thatand equation (8.3.3) states that the are complete in

Thus

where is the projection of on Equation (8.4.10) gives Tofind we may apply A term by term to get

because Thus

8.5 Morozov’s discrepancy principle 235

and

This equation shows that is a monotonically increasingfunction of for To show that there is a unique value of such that

we must show that

Since and thus

On the other hand, by Parseval’s equality

This proves the required result.We conclude by showing that choosing according to the discrepancy

principle does provide a regular scheme for approximating i.e.

Again, without loss of generality we may take so that there is aunique which we call such that Since we haveshown that is uniquely determined by we may write as

First we show that the are bounded. We find as the minimum of

for all Thus if then

so that in particular

But we choose so that so that

while

from which we conclude that


i.e. the are bounded.Now suppose that is a sequence converging to and thatEach such pair will determine an and a corresponding

which we will call We now show that there is a subsequence of forwhich the converge to

The sequence lies in the closed ball with center O and radius inThe corollaries to Theorem 4.6.6, and Theorem 4.7.1 state that a closed

ball in a Hilbert space is weakly compact. Therefore, there is a subsequenceof which converges weakly to some i.e. such thatEquation (8.4.9) shows that is a closed

subspace of and therefore also weakly closed, by Problem 4.6.6. ThusLet be the pair corresponding to then

Lemma 5.5.2 states that a continuous linear operator in a Hilbert space is weaklycontinuous, so that according to Definition 5.5.3, it maps a weakly convergentsequence into a weakly convergent sequence. Therefore A maps whichconverges weakly to into a sequence which converges weakly to But

converges strongly to and therefore weakly to The weak limit isunique (Problem 4.6.2) so that Thus and But,by Problem 8.2.1 the unique element with these properties is Thus

and converges weakly to i.e. We now show that thereis a subsequence of which converges strongly to According toTheorem 4.6.2, in order to show that it is sufficient to show that

We know that so that lies in thecompact set of Therefore there is a subsequence of suchthat and On the other hand, since we have

so that Therefore and hence and infact and We conclude that Morozov’s discrepancy principledoes provide a regular scheme for solving equation (8.3.6) when A is a compactlinear operator from into


Conclusion

A work of fiction usually has an ending: the murderer is unmasked, theprince and princess live happily ever after, or Romeo and Juliet lie dead. Some-times, however, the writer purposely leaves the reader in suspense: the hero liftsup the telephone and starts to dial, the door is flung open, or a shot rings out.

The end of this book is even less satisfying. There is no end; the story isleft for the reader to continue. The theory described in this book has alreadybeen applied to numerous problems, but there are countless more possible ap-plications and extensions which have been described elsewhere, and many moreextensions and applications remain to be discovered. May happiness attend yoursearch.


Synopsis of Chapter 8: Inverse Problems

Well-posed problems: existence, uniqueness, stability

The operator equation: If X is infinite dimensional a compact operator Afrom X onto Y cannot have a bounded inverse. Corollary 8.2.1

Tikhonov’s Theorem 8.2.1 If 5 is compact then is continuous.

Generalized inverse : is closed. Theorem 8.2.2.

: is continuous iff R(A) is closed.

Singular Value Decomposition

Regularization

The solution of (8.4.4) (8.4.13)

The effect of error (8.4.14)

Morozov’s discrepancy principle


References

The first book devoted to ill-posed problems was

A.N. Tikhonov and V.Y. Arsenin, Solution of Ill-Posed Problems, John Wiley,New York, 1977. This is invaluable as a guide to the early literature. It usesthe methods of functional analysis, and has many instructive examples from thetheory of Fredholm integral equations. The reader who has studied the presentbook will have more than sufficient background knowledge in functional analysisto understand it.

Classical treatment of the abstract theory of ill-posed problems is to be foundin the rather difficult

V.A. Morozov, Methods of Solving Incorrectly Posed Problems, Springer-Verlag,New York, 1984.

Perhaps the best introduction to the theory of the inverse problems we havestudied in this chapter is

C.W. Groetsch, Inverse Problems in the Mathematical Sciences, Vieweg, Braun-schweig, 1993. This motivates the study of inverse problems by many examplestaken from different areas of mathematics, physics and engineering. It providesa very brief summary of functional analysis and then applies it to the inverseproblem stated as a Fredholm integral equation of the first kind, or more gen-erally as the equation The principal aim of Chapter 8 has been toexpand on Groetsch’s treatment, trying to fill in some of the steps which he leftto the reader. The book has a valuable guide to the literature.

Index

241

a-priori estimate, 129absolute convergence

in a Hilbert space, 106Faedo–Galerkin, 135in a normed space, 103Ritz, 128

Arzelà, Cesare, 176Arzelà–Ascoli theorem, 176Ascoli, Guilio, 176axiom

triangle, 20axioms

inner product, 58metric, 20norm, 41

ballopen, 9, 25

Banach space, 44Banach’s fixed point theorem, 36Banach’s open mapping theorem, 151,

221Banach, Stefan, 36Banach–Steinhaus theorem, 144, 145basis for

normed linear space, 113Bernoulli, Daniel, 74Bernstein polynomial, 15Bernstein, Serge, 15Bessel’s inequality, 116Bessel, Friedrich Wilhelm, 116Bolzano, Bernard, 6Bolzano–Weierstrass theorem, 6Borel, Emile, 168bounded

set of real numbers, 5uniformly, 175

bounded above, 7bounded below, 7Buniakowski, Victor Yakovlevich, 59

Cantor’s theorem, 100Cantor, Georg, 100Cauchy problem, 179

for N equations, 181Cauchy sequence, 3, 30

weak, 120stationary, 4

Cauchy, Augustin-Louis, 3Cauchy–Schwarz inequality, 59class

equivalence, 32null, 52

closedinterval, 5set, 26set of real numbers, 5

closed system, 117closure, 25

of a set of real numbers, 6compact

sequentially, 168set, 170set of real numbers, 6

compact linear operator, 183limit of, 185

compact linear operatorsin a separable Hilbert space, 190product of, 184

compact operatorresolvent of, 208

compact support, 10complete system, 114

of series of operators, 142accumulation point, 25approximation

242 Index

completenessof IR, 5

completionof a metric space, 32

cone property, 96contact point, 25continuation

of operator, 144continuity

of a function, 10of an operator, 46of inverse operator, 147

continuousfunctional, 174

continuous function, 10contraction mapping, 36contraction operator, 36convergence

in a metric space, 29of linear operators, 142pointwise, 143strong, 143uniform, 143weak, 120

convergent sequence, 2, 29convex, 105correspondence

isometric, 32countable, 99countable dense subset, 101cover, 169

d’Alembert, Jean le Rond, 68degenerate, 148derivative

generalized, 67, 157direct problem, 219Dirichlet, Gustave Peter Lejeune, 133distance

Euclidean, 19domain, 9

of operator, 35dual space, 110

eigenvalue, 196Einstein’s double suffix summation con-

vention, 83Einstein’s summation convention, 85Einstein, Albert, 83

elastic bodywith free boundary, 88

energy space, 25for clamped membrane, 78for elastic body, 86separability of, 103

equal almost everywhere, 53equicontinuous, 176equivalence class, 32

of Cauchy sequences, 4representative of a, 4stationary, 32

equivalentnorms, 41

equivalent Cauchy sequences, 3equivalent metrics, 20equivalent sequence, 32Euclidean distance, 9, 19Euler’s method

justification of, 181Euler, Leonhard, 74Euler–Bernoulli beam, 74evolution problems, 132extension

of a function, 12of an operator, 155of operator, 144

external forceswork of, 68, 84

Faedo, 135Faedo–Galerkin approximation, 135family of functionals, 175

equicontinuous, 175uniformly bounded, 175

finiteset of real numbers, 5

fixed point, 36Fourier

expansion, 21Fourier coefficients, 116Fourier series, 116Fourier, Jean Baptiste, 115Fredholm alternative, 159, 206Fredholm integral equation, 220Fredholm integral operator, 162, 187Fredholm operator, 61Fredholm, Ivar, 148

Index 243

Friedrichs’ inequality, 78Friedrichs, Kurt Otto, 78function

of compact support, 10continuous, 10definition of, 9extension of, 12support of, 9tent, 52uniformly continuous, 11

functional, 35complex, 35continuous, 174real, 35uniformly continuous, 175weakly continuous, 211work, 68, 76, 84

Galerkin, Boris Grigor’evich, 135generalized solution, 224

for eigenvalue problem, 82for free vibration of a membrane, 113for Neumann problem, 82for plate, 84for the rod, 71

Generalized solutionsfor evolution problems, 132

Gram, Jórgen Pedersen, 115Gram–Schmidt process, 115graph

of an operator, 153

Hölder condition, 91Hölder continuous, 91Hölder’s inequality, 48

for integrals, 55Hölder, Ludwig Otto, 48Hadamard, Jacques Salomon, 219Hausdorff, Felix, 171heat transfer equation, 132Heine, Heinrich Eduard, 168Hermite, Charles, 162Hilbert identity, 200Hilbert space, 60

orthogonal decomposition of, 108separable, 115, 144, 190

Hilbert space, approximation in, 106Hilbert, David, 60

ill-posed problem, 219image, 35imbedding, 55, 67, 75induced

inner product, 60metric, 26normed, 41

inequalityFriedrichs’, 78Poincaré’s, 81Bessel’s, 116Cauchy–Schwarz, 59Friedrich’s, 87Hölder’s, 48Hölder’s integral, 55Jensen’s, 50Korn’s, 86, 87Minkowski’s, 49Minkowski’s integral, 51Poincaré’s, 89Schwarz, 59triangle, 20

infimum, 7inner product

induced, 60inner product space, 58integral

Lebesgue, 54Riemann, 51

integral equationFredholm, 220

integral operatorcompact, 185

interior point, 25inverse operator

continuity of, 147inverse problem, 220isometric, 32

Jensen’s inequality, 50Jensen, Valdemar, 50

kernel, 109degenerate, 148

kinetic energyof membrane, 82

Korn, 86Korn’s inequality, 86

244 Index

Lebesgue integral, 54Lebesgue space, 51

separability of, 102Lebesgue, Henri Léon, 54limit, 29limit point, 25linear elasticity, 85linear functional

kernel of, 109Linear operator

space of, 141linear operator, 45

bounded, 47continuous, 35domain of, 45norm of, 46, 141

linear operatorsproduct of, 143

linearly independent, 42Lipschitz continuous, 91Lipschitz property, 96Lipschitz, Rudolf Otto Sigismund, 91

mappingcontraction, 36

matrixHermitian, 162

matrix operatorinfinite dimensional, 186

maximum metric, 27maximum value, 11membrane, 78

clamped, 78metric, 20

axioms, 20equivalent, 20induced, 26maximum, 27uniform, 27

metric spacecomplete, 30completion of, 32incomplete, 30separable, 101

minimum value, 11Minkowski’s inequality, 49Minkowski, Hermann, 49Moore, Eliakim Hastings, 225

Moore–Penrose generalized inverse, 225Morozov, 234Morozov’s discrepancy principle, 234

natural boundary condition, 73, 81natural end condition, 70natural end conditions, 77natural frequencies

of clamped membrane, 82neighborhood, 25

25norm, 41

of an operator, 46axioms, 41equivalent, 43, 152induced, 41Sobolev, 89

normed linear space, 41basis for, 113strictly normed, 105

normsequivalent, 41

null class, 52null sequence, 53null space, 158

openball, 25interval, 5set, 25set of real numbers, 5

open ball, 9operator, 35

compact, 183continuation of, 144integral, 162adjoint, 157bounded, 47closed extension of, 155closed linear, 152, 154continuous, 35continuously invertible, 148contraction, 36coordinate, 198domain of, 35eigenvalue of, 196extension of, 144fixed point of, 36Fredholm, 61

Index 245

Fredholm integral, 162, 187graph of, 153imbedding, 56, 67, 76integral, 61, 148inverse, 147linear, 45matrix, 162norm of, 46null space of, 158projection, 144range of, 35residual spectrum of, 196resolvent set of, 195self-adjoint, 160spectrum of, 195strictly positive, 215weakly continuous, 161

orthogonal, 60mutually, 108

orthogonal decomposition, 108orthogonal system, 115orthonormal, 115

parallelogram law, 59Parseval’s equality, 117Parseval, Marc Antoine, 117Peano’s local existence theorem, 179Peano, Giuseppe, 179Penrose, Roger, 225Picard’s existence criterion, 228Picard, Charles Emile, 228plate, 83

stability of, 163Poincaré’s inequality, 81, 89Poincaré, Jules Henri, 81point

accumulation, 25contact, 25fixed, 36interior, 25isolated, 26limit, 25

principleof uniform boundedness, 146

Principle of Minimum Energy, 71Principle of Virtual Work, 68, 76problem

direct, 219

ill-posed, 219inverse, 220well posed, 219

productscalar, 40

product space, 153

range of operator, 35real number, 4regular points, 195representative sequence, 32resolvent operator, 199resolvent set, 195Riemann integral, 51Riemann, Georg Friedrich Bernhard, 51Riemann–Lebesgue lemma, 119Riesz’s representation theorem, 106, 110Riesz, Frédéric, 106Ritz method, 128Ritz, Walter, 128rod

cantilever, 65free, 73

Saint Venant, 163Schauder, Pavel Julius, 201Schmidt, Erhard, 115Schwarz inequality, 59Schwarz, Hermand Amandus, 59semi-norm, 88sequence

Cauchy, 30Cauchy, 3convergent, 2, 29null, 53representative, 32

setclosed, 26, 60closure of, 25compact, 170convex, 105countable, 99cover of, 169dense, 26, 31linearly dependent, 42of measure zero, 53open, 25separable, 174sequentially compact, 168

Index246

weakly closed, 125set of real numbers

bounded, 5closed, 5compact, 6finite, 5infimum of, 7open, 5supremum of, 7

singular system, 227singular value decomposition, 227singular values, 227Sobolev norm, 89Sobolev space, 88

separability of, 102Sobolev’s imbedding theorem, 95Sobolev, Sergei L’vovich, 88solution

generalize, 224least squares, 225minimum norm, 225

spaceinfinite dimensional, 42separable, 101, 174Sobolev, 88Banach, 44complete, 30dual, 110energy, 25finite dimensional, 42Hilbert, 60incomplete, 30inner product, 58Lebesgue, 51linear, 40product, 153real inner product, 58separable, 99strictly normed linear, 105weakly complete, 125

spaceinner product, 58

spectrumresidual, 196continuous, 196of compact linear operator, 201, 208of coordinate operator, 198of differential operators, 196

of linear operators, 195of membrane equation, 196

stationary Cauchy sequence, 4Steinhaus, Hugo Dyonis, 145strain energy

of linearly elastic body, 85of membrane, 78, 82of plate, 83of rod, 65of string, 24

subspaceclosed, 42, 60dimension of, 42finite dimensional, 42linear, 45of a metric space, 26

subspacesmutually orthogonal, 108

supremum, 7system

closed, 117complete, 114orthogonal, 115orthonormal, 117

tent function, 52Theorem

Weierstrass’ polynomial approxima-tion, 14

Weierstrass’ uniform convergence, 13Banach’s open mapping, 221

theoremBanach’s fixed point, 36Weierstrass’ polynomial approxima-tion, 31

Arzelà–Ascoli, 176Banach’s open mapping, 151Banach–Steinhaus, 144, 145Bolzano–Weierstrass, 6Cantor’s, 100closed graph, 154, 155contraction mapping, 36Heine–Borel, 169imbedding, 55Riesz’s representation, 106, 109, 110Sobolev’s imbedding, 95Tikhonov, 223

Tikhonov’s theorem, 223

Index 247

Tikhonov, Andrei Nikolaevich, 223total energy

of rod, 71triangle axiom, 20triangle inequality, 20

uniform boundednessprinciple of, 123

uniform convergenceof sequence of operators, 143

uniform metric, 27uniformly bounded, 175uniformly continuous

functional, 175

variational formulationfor rod, 71

virtual displacement, 68Virtual Work

Principle of, 68, 76

weakCauchy sequence, 120Convergence, 120

weakly closed, 125weakly complete, 125Weierstrass’

polynomial approximation theorem,14

uniform convergence theorem, 13Weierstrass’ polynomial approximation

theorem, 31Weierstrass, Karl Theodor Wilhelm, 6well posed problem, 219work

of external forces, 68, 84

Young, Thomas, 65

zero almost everywhere, 53

MechanicsSOLID MECHANICS AND ITS APPLICATIONS

Series Editor. G.M.L. Gladwell

Aims and Scope of the Series

The fundamental questions arising in mechanics are: Why?, How?, and How much? The aim ofthis series is to provide lucid accounts written by authoritative researchers giving vision and insightin answering these questions on the subject of mechanics as it relates to solids. The scope of theseries covers the entire spectrum of solid mechanics. Thus it includes the foundation of mechanics;variational formulations; computational mechanics; statics, kinematics and dynamics of rigid andelastic bodies; vibrations of solids and structures; dynamical systems and chaos; the theories ofelasticity, plasticity and viscoelasticity; composite materials; rods, beams, shells and membranes;structural control and stability; soils, rocks and geomechanics; fracture; tribology; experimentalmechanics; biomechanics and machine design.

1.

2.3.4.5.6.

7.

8.9.

10.

11.

12.13.

14.15.16.17.

18.

19.

20.

21.22.23.

R.T. Haftka, Z. Gürdal and M.P. Kamat: Elements of Structural Optimization. 2nd rev.ed., 1990ISBN 0-7923-0608-2

J.J. Kalker: Three-Dimensional Elastic Bodies in Rolling Contact. 1990 ISBN 0-7923-0712-7P. Karasudhi: Foundations of Solid Mechanics. 1991 ISBN 0-7923-0772-0Not publishedNot published.J.F. Doyle: Static and Dynamic Analysis of Structures. With an Emphasis on Mechanics andComputer Matrix Methods. 1991 ISBN 0-7923-1124-8; Pb 0-7923-1208-2O.O. Ochoa and J.N. Reddy: Finite Element Analysis of Composite Laminates.

ISBN 0-7923-1125-6M.H. Aliabadi and D.P. Rooke: Numerical Fracture Mechanics.J. Angeles and C.S. López-Cajún: Optimization of Cam Mechanisms. 1991

ISBN 0-7923-1175-2

ISBN 0-7923-1355-0D.E. Grierson, A. Franchi and P. Riva (eds.): Progress in Structural Engineering. 1991

ISBN 0-7923-1396-8R.T. Haftka and Z. Gürdal: Elements of Structural Optimization. 3rd rev. and exp. ed. 1992

ISBN 0-7923-1504-9; Pb 0-7923-1505-7J.R. Barber: Elasticity. 1992 ISBN 0-7923-1609-6; Pb 0-7923-1610-XH.S. Tzou and G.L. Anderson (eds.): Intelligent Structural Systems. 1992

ISBN 0-7923-1920-6E.E. Gdoutos: Fracture Mechanics. An Introduction. 1993 ISBN 0-7923-1932-XJ.P. Ward: Solid Mechanics. An Introduction. 1992M. Farshad: Design and Analysis of Shell Structures. 1992

ISBN 0-7923-1949-4

H.S. Tzou and T. Fukuda (eds.): Precision Sensors, Actuators and Systems. 1992

J.R. Vinson: The Behavior of Shells Composed of Isotropic and Composite Materials. 1993

ISBN 0-7923-1950-8

ISBN 0-7923-2015-8

ISBN 0-7923-2113-8H.S. Tzou: Piezoelectric Shells. Distributed Sensing and Control of Continua. 1993

ISBN 0-7923-2186-3W. Schiehlen (ed.): Advanced Multibody System Dynamics. Simulation and Software Tools.1993C.-W. Lee: Vibration Analysis of Rotors. 1993D.R. Smith: An Introduction to Continuum Mechanics. 1993G.M.L. Gladwell: Inverse Problems in Scattering. An Introduction. 1993

ISBN 0-7923-2192-8ISBN 0-7923-2300-9ISBN 0-7923-2454-4ISBN 0-7923-2478-1



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G. Prathap: The Finite Element Method in Structural Mechanics. 1993 ISBN 0-7923-2492-7J. Herskovits (ed.): Advances in Structural Optimization. 1995M.A. González-Palacios and J. Angeles: Cam Synthesis. 1993W.S. Hall: The Boundary Element Method. 1993J. Angeles, G. Hommel and P. Kovács (eds.): Computational Kinematics. 1993

A. Curnier: Computational Methods in Solid Mechanics. 1994D.A. Hills and D. Nowell: Mechanics of Fretting Fatigue. 1994B. Tabarrok and F.P.J. Rimrott: Variational Methods and Complementary Formulations inDynamics. 1994E.H. Dowell (ed.), E.F. Crawley, H.C. Curtiss Jr., D.A. Peters, R. H. Scanlan and F. Sisto: AModern Course in Aeroelasticity. Third Revised and Enlarged Edition. 1995

A. Preumont: Random Vibration and Spectral Analysis. 1994J.N. Reddy (ed.): Mechanics of Composite Materials. Selected works of Nicholas J. Pagano.1994A.P.S. Selvadurai (ed.): Mechanics of Poroelastic Media. 1996Z. Mróz, D. Weichert, S. Dorosz (eds.): Inelastic Behaviour of Structures under VariableLoads. 1995R. Pyrz (ed.): IUTAM Symposium on Microstructure-Property Interactions in CompositeMaterials. Proceedings of the IUTAM Symposium held in Aalborg, Denmark. 1995

ISBN 0-7923-3427-2M.I. Friswell and J.E. Mottershead: Finite Element Model Updating in Structural Dynamics.1995D.F. Parker and A.H. England (eds.): IUTAM Symposium on Anisotropy, Inhomogeneity andNonlinearity in Solid Mechanics. Proceedings of the IUTAM Symposium held in Nottingham,U.K. 1995J.-P. Merlet and B. Ravani (eds.): Computational Kinematics ’95. 1995L.P. Lebedev, I.I. Vorovich and G.M.L. Gladwell: Functional Analysis. Applications in Mech-anics and Inverse Problems. 1996

Mechanics of Components with Treated or Coated Surfaces. 1996

D. Bestle and W. Schiehlen (eds.): IUTAM Symposium on Optimization of Mechanical Systems.Proceedings of the IUTAM Symposium held in Stuttgart, Germany. 1996

D.A. Hills, P.A. Kelly, D.N. Dai and A.M. Korsunsky: Solution of Crack Problems. TheDistributed Dislocation Technique. 1996V.A. Squire, R.J. Hosking, A.D. Kerr and P.J. Langhorne: Moving Loads on Ice Plates. 1996

ISBN 0-7923-3953-3A. Pineau and A. Zaoui (eds.): IUTAM Symposium on Micromechanics of Plasticity andDamage of Multiphase Materials. Proceedings of the IUTAM Symposium held in Sèvres,Paris, France. 1996A. Naess and S. Krenk (eds.): IUTAM Symposium on Advances in Nonlinear StochasticMechanics. Proceedings of the IUTAM Symposium held in Trondheim, Norway. 1996

D. and A. Scalia: Thermoelastic Deformations. 1996ISBN 0-7923-4193-7ISBN 0-7923-4230-5

ISBN 0-7923-4188-0

ISBN 0-7923-3848-0

ISBN 0-7923-3830-8

ISBN 0-7923-3700-X

ISBN 0-7923-3849-9

ISBN 0-7923-3594-5ISBN 0-7923-3673-9

ISBN 0-7923-3431-0

ISBN 0-7923-3397-7

ISBN 0-7923-3329-2ISBN 0-7923-3041-2

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J.R. Willis (ed): IUTAM Symposium on Nonlinear Analysis of Fracture. Proceedings of theIUTAM Symposium held in Cambridge, U.K. 1997A. Preumont: Vibration Control of Active Structures. An Introduction. 1997

G.P. Cherepanov: Methods of Fracture Mechanics: Solid Matter Physics. 1997

D.H. van Campen (ed.): IUTAM Symposium on Interaction between Dynamics and Control inAdvanced Mechanical Systems. Proceedings of the IUTAM Symposium held in Eindhoven,The Netherlands. 1997N. A. Fleck and A.C.F. Cocks (eds.): IUTAM Symposium on Mechanics of Granular and PorousMaterials. Proceedings of the IUTAM Symposium held in Cambridge, U.K. 1997

J. Roorda and N.K. Srivastava (eds.): Trends in Structural Mechanics. Theory, Practice, Edu-cation. 1997Yu. A. Mitropolskii and N. Van Dao: Applied Asymptotic Methods in Nonlinear Oscillations.1997C. Guedes Soares (ed.): Probabilistic Methods for Structural Design. 1997

D. A. Pineau and A. Zaoui: Mechanical Behaviour of Materials. Volume I: Elasticityand Plasticity. 1998D. A. Pineau and A. Zaoui: Mechanical Behaviour of Materials. Volume II: Visco-plasticity, Damage, Fracture and Contact Mechanics. 1998L.T. Tenek and J. Argyris: Finite Element Analysis for Composite Structures. 1998

Y.A. Bahei-El-Din and G.J. Dvorak (eds.): IUTAM Symposium on Transformation Problemsin Composite and Active Materials. Proceedings of the IUTAM Symposium held in Cairo,Egypt. 1998I.G. Goryacheva: Contact Mechanics in Tribology. 1998O.T. Bruhns and E. Stein (eds.): IUTAM Symposium on Micro- and Macrostructural Aspectsof Thermoplasticity. Proceedings of the IUTAM Symposium held in Bochum, Germany. 1999

F.C. Moon: IUTAM Symposium on New Applications of Nonlinear and Chaotic Dynamics inMechanics. Proceedings of the IUTAM Symposium held in Ithaca, NY, USA. 1998

R. Wang: IUTAM Symposium on Rheology of Bodies with Defects. Proceedings of the IUTAMSymposium held in Beijing, China. 1999Yu.I. Dimitrienko: Thermomechanics of Composites under High Temperatures. 1999

P. Argoul, M. Frémond and Q.S. Nguyen (eds.): IUTAM Symposium on Variations of Domainsand Free-Boundary Problems in Solid Mechanics. Proceedings of the IUTAM Symposiumheld in Paris, France. 1999F. J. Fahy and W.G. Price (eds.): IUTAM Symposium on Statistical Energy Analysis. Proceedingsof the IUTAM Symposium held in Southampton, U.K. 1999H.A. Mang and F.G. Rammerstorfer (eds.): IUTAM Symposium on Discretization Methods inStructural Mechanics. Proceedings of the IUTAM Symposium held in Vienna, Austria. 1999

ISBN 0-7923-5591-1

ISBN 0-7923-5457-5

ISBN 0-7923-5450-8

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ISBN 0-7923-5276-9

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ISBN 0-7923-4392-1



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P. Pedersen and M.P. Bendsøe (eds.): IUTAM Symposium on Synthesis in Bio Solid Mechanics.Proceedings of the IUTAM Symposium held in Copenhagen, Denmark. 1999

S.K. Agrawal and B.C. Fabien: Optimization of Dynamic Systems. 1999

A. Carpinteri: Nonlinear Crack Models for Nonmetallic Materials. 1999

F. Pfeifer (ed.): IUTAM Symposium on Unilateral Multibody Contacts. Proceedings of theIUTAM Symposium held in Munich, Germany. 1999E. Lavendelis and M. Zakrzhevsky (eds.): IUTAM/IFToMM Symposium on Synthesis of Non-linear Dynamical Systems. Proceedings of the IUTAM/IFToMM Symposium held in Riga,Latvia. 2000J.-P. Merlet: Parallel Robots. 2000J.T. Pindera: Techniques of Tomographic Isodyne Stress Analysis. 2000G.A. Maugin, R. Drouot and F. Sidoroff (eds.): Continuum Thermomechanics. The Art andScience of Modelling Material Behaviour. 2000N. Van Dao and E.J. Kreuzer (eds.): IUTAM Symposium on Recent Developments in Non-linearOscillations of Mechanical Systems. 2000S.D. Akbarov and A.N. Guz: Mechanics of Curved Composites. 2000M.B. Rubin: Cosserat Theories: Shells, Rods and Points. 2000S.Pellegrino and S.D. Guest (eds.): IUTAM-IASS Symposium on Deployable Structures: Theoryand Applications. Proceedings of the IUTAM-IASS Symposium held in Cambridge, U.K., 6–9September 1998. 2000A.D. Rosato and D.L. Blackmore (eds.): IUTAM Symposium on Segregation in GranularFlows. Proceedings of the IUTAM Symposium held in Cape May, NJ, U.S.A., June 5–10,1999. 2000A. Lagarde (ed.): IUTAM Symposium on Advanced Optical Methods and Applications in SolidMechanics. Proceedings of the IUTAM Symposium held in Futuroscope, Poitiers, France,August 31–September 4, 1998. 2000D. Weichert and G. Maier (eds.): Inelastic Analysis of Structures under Variable Loads. Theoryand Engineering Applications. 2000T.-J. Chuang and J.W. Rudnicki (eds.): Multiscale Deformation and Fracture in Materials andStructures. The James R. Rice 60th Anniversary Volume. 2001S. Narayanan and R.N. lyengar (eds.): IUTAM Symposium on Nonlinearity and StochasticStructural Dynamics. Proceedings of the IUTAM Symposium held in Madras, Chennai, India,4–8 January 1999S. Murakami and N. Ohno (eds.): IUTAM Symposium on Creep in Structures. Proceedings ofthe IUTAM Symposium held in Nagoya, Japan, 3-7 April 2000. 2001W. Ehlers (ed.): IUTAM Symposium on Theoretical and Numerical Methods in ContinuumMechanics of Porous Materials. Proceedings of the IUTAM Symposium held at the Universityof Stuttgart, Germany, September 5-10, 1999. 2001D. Durban, D. Givoli and J.G. Simmonds (eds.): Advances in the Mechanis of Plates and ShellsThe Avinoam Libai Anniversary Volume. 2001U. Gabbert and H.-S. Tzou (eds.): IUTAM Symposium on Smart Structures and Structonic Sys-tems. Proceedings of the IUTAM Symposium held in Magdeburg, Germany, 26–29 September2000. 2001 ISBN 0-7923-6968-8

ISBN 0-7923-6785-5

ISBN 0-7923-6766-9

ISBN 0-7923-6737-5

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ISBN 0-7923-6604-2



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Y. Ivanov, V. Cheshkov and M. Natova: Polymer Composite Materials - Interface Phenomena& Processes. 2001R.C. McPhedran, L.C. Botten and N. A. Nicorovici (eds.): IUTAM Symposium on Mechanicaland Electromagnetic Waves in Structured Media. Proceedings of the IUTAM Symposium heldin Sydney, NSW, Australia, 18-22 Januari 1999. 2001D.A. Sotiropoulos (ed.): IUTAM Symposium on Mechanical Waves for Composite StructuresCharacterization. Proceedings of the IUTAM Symposium held in Chania, Crete, Greece, June14-17, 2000. 2001V.M. Alexandrov and D.A. Pozharskii: Three-Dimensional Contact Problems. 2001

J.P. Dempsey and H.H. Shen (eds.): IUTAM Symposium on Scaling Laws in Ice Mechanicsand Ice Dynamics. Proceedings of the IUTAM Symposium held in Fairbanks, Alaska, U.S.A.,13-16 June 2000. 2001

U. Kirsch: Design-Oriented Analysis of Structures. A Unified Approach. 2002

A. Preumont: Vibration Control of Active Structures. An Introduction ( Edition). 2002

B.L. Karihaloo (ed.): IUTAM Symposium on Analytical and Computational Fracture Mechan-ics of Non-Homogeneous Materials. Proceedings of the IUTAM Symposium held in Cardiff,U.K., 18-22 June 2001. 2002S.M. Han and H. Benaroya: Nonlinear and Stochastic Dynamics of Compliant Offshore Struc-tures. 2002A.M. Linkov: Boundary Integral Equations in Elasticity Theory. 2002

L.P. Lebedev, I.I. Vorovich and G.M.L. Gladwell: Functional Analysis. Applications in Me-chanics and Inverse Problems ( Edition). 2002 ISBN 1-4020-0667-5; Pb 1-4020-0756-6

ISBN 0-7923-7008-2

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Kluwer Academic Publishers – Dordrecht / Boston / London

ICASE/LaRC Interdisciplinary Series in Science and Engineering

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J. Buckmaster, T.L. Jackson and A. Kumar (eds.): Combustion in High-Speed Flows.1994M.Y. Hussaini, T.B. Gatski and T.L. Jackson (eds.): Transition, Turbulence andCombustion. Volume I: Transition. 1994

ISBN 0-7923-3084-6; set 0-7923-3086-2

ISBN 0-7923-2086-X

M.Y. Hussaini, T.B. Gatski and T.L. Jackson (eds.): Transition, Turbulence andCombustion. Volume II: Turbulence and Combustion. 1994

ISBN 0-7923-3085-4; set 0-7923-3086-2D.E. Keyes, A. Sameh and V. Venkatakrishnan (eds): Parallel Numerical Algorithms.1997 ISBN 0-7923-4282-8T.G. Campbell, R.A. Nicolaides and M.D. Salas (eds.): Computational Electromag-netics and Its Applications. 1997V. Venkatakrishnan, M.D. Salas and S.R. Chakravarthy (eds.): Barriers and Chal-lenges in Computational Fluid Dynamics. 1998M.D. Salas, J.N. Hefner and L. Sakell (eds.): Modeling Complex Turbulent Flows.1999 ISBN 0-7923-5590-3

ISBN 0-7923-4733-1

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KLUWER ACADEMIC PUBLISHERS – DORDRECHT / BOSTON / LONDON

l.p. lebedev, i.i. vorovich, g.m. gladwell,-functional analysis_ applications in mechanics and...

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