Functional Analysis: Applications in Mechanics and Inverse Problems (Solid Mechanics and Its Applications)

Functional Analysis SOLID MECHANICS AND ITS APPLICATIONS Volume 100 Series Editor: G.M.L. GLADWELL Department of Civi...

Author: L.P. Lebedev | I.I. Vorovich | G.M. Gladwell

21 downloads 845 Views 13MB Size Report

This content was uploaded by our users and we assume good faith they have the permission to share this book. If you own the copyright to this book and it is wrongfully on our website, we offer a simple DMCA procedure to remove your content from our site. Start by pressing the button below!
Report copyright / DMCA form

DOWNLOAD PDF

Functional Analysis

SOLID MECHANICS AND ITS APPLICATIONS Volume 100 Series Editor:

G.M.L. GLADWELL Department of Civil Engineering University of Waterloo Waterloo, 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 researchers giving vision and insight in answering these questions on the subject of mechanics as it relates to solids. The scope of the series covers the entire spectrum of solid mechanics. Thus it includes the foundation of mechanics; variational formulations; computational mechanics; statics, kinematics and dynamics of rigid and elastic bodies: vibrations of solids and structures; dynamical systems and chaos; the theories of elasticity, plasticity and viscoelasticity; composite materials; rods, beams, shells and membranes; structural control and stability; soils, rocks and geomechanics; fracture; tribology; experimental mechanics; biomechanics and machine design. The median level of presentation is the first year graduate student. Some texts are monographs defining the current state of the field; others are accessible to final year undergraduates; but essentially the emphasis is on readability and clarity.

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

Functional Analysis Applications 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, Russia Fellow of the Russian Academy of Sciences and

G.M.L. GLADWELL Distinguished Professor Emeritus, Department of Civil Engineering, University of Waterloo, Canada Fellow of the Royal Society of Canada, and Fellow of the American Academy of Mechanics

KLUWER ACADEMIC PUBLISHERS NEW YORK, BOSTON, DORDRECHT, LONDON, MOSCOW

eBook ISBN: Print ISBN:

0-306-48397-1 1-4020-0667-5

©2003 Kluwer Academic Publishers New York, Boston, Dordrecht, London, Moscow Print ©2002 Kluwer Academic Publishers Dordrecht 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: and Kluwer's eBookstore at:

http://kluweronline.com http://ebooks.kluweronline.com

In memory of I.I. Vorovich

Table of Contents

1 Introduction 1.1 Real and complex numbers 1.2 Theory of functions 1.3 Weierstrass’ polynomial approximation theorem

1 1 9 14

2 Introduction to Metric Spaces 2.1 Preliminaries 2.2 Sets in a metric space 2.3 Some metric spaces of functions 2.4 Convergence in a metric space 2.5 Complete metric spaces 2.6 The completion theorem 2.7 An introduction to operators 2.8 Normed linear spaces 2.9 An introduction to linear operators 2.10 Some inequalities 2.11 Lebesgue spaces 2.12 Inner product spaces

19 19 25 27 29 30 32 35 40 45 48 51 58

3 Energy Spaces and Generalized Solutions 3.1 The rod 3.2 The Euler–Bernoulli beam 3.3 The membrane 3.4 The plate in bending 3.5 Linear elasticity 3.6 Sobolev spaces 3.7 Some imbedding theorems

65 65 74 78 83 85 88 90

4 Approximation in a Normed Linear Space 4.1 Separable spaces 4.2 Theory of approximation in a normed linear space 4.3 Riesz’s representation theorem 4.4 Existence of energy solutions of some mechanics problems 4.5 Bases and complete systems

99 99 103 106 110 113

viii

Table of Contents

4.6 4.7 4.8 4.9

Weak convergence in a Hilbert space Introduction to the concept of a compact set Ritz approximation in a Hilbert space Generalized solutions of evolution problems

120 126 128 132

5 Elements of the Theory of Linear Operators 5.1 Spaces of linear operators 5.2 The Banach–Steinhaus theorem 5.3 The inverse operator 5.4 Closed operators 5.5 The adjoint operator 5.6 Examples of adjoint operators

141 141 144 147 152 157 162

6 Compactness and Its Consequences compact 6.1 Sequentially compact 6.2 Criteria for compactness 6.3 The Arzelà–Ascoli theorem 6.4 Applications of the Arzelà–Ascoli theorem 6.5 Compact linear operators in normed linear spaces 6.6 Compact linear operators between Hilbert spaces

167 167 171 174 178 183 189

7 Spectral Theory of Linear Operators 7.1 The spectrum of a linear operator 7.2 The resolvent set of a closed linear operator 7.3 The spectrum of a compact linear operator in a Hilbert space 7.4 The analytic nature of the resolvent of a compact linear operator 7.5 Self-adjoint operators in a Hilbert space

195 195 199 201 208 211

8 Applications to Inverse Problems 8.1 Well-posed and ill-posed problems 8.2 The operator equation 8.3 Singular value decomposition 8.4 Regularization 8.5 Morozov’s discrepancy principle

219 219 220 226 229 234

Index

241

Preface This book started its life as a series of lectures given by the second author from the 1970’s onwards to students in their third and fourth years in the Department of Mechanics and Mathematics at Rostov State University. For these lectures there was also an audience of engineers and applied mechanicists who wished to understand the functional analysis used in contemporary research in their fields. These people were not so much interested in functional analysis itself as in its applications; they did not want to be told about functional analysis in its most abstract form, but wanted a guided tour through those parts of the analysis needed for their applications. The lecture notes evolved over the years as the first author started to make more formal typewritten versions incorporating new material. About 1990 the first author prepared an English version and submitted it to Kluwer Academic Publishers for inclusion in the series Solid Mechanics and its Applications. At that state the notes were divided into three long chapters covering linear and nonlinear analysis. As Series Editor, the third author started to edit them. The requirements of lecture notes and books are vastly different. A book has to be complete (in some sense), self contained, and able to be read without the help of an instructor. In the end these new requirements led to the book being entirely rewritten: an introductory chapter on real analysis was added, the order of presentation was changed and material was added and deleted. The last chapter of the original notes, on nonlinear analysis, was omitted altogether, the original two chapters were reorganized into six chapters, and a new Chapter 8 on 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 analysis needed for an understanding of the theory behind regularization methods for Inverse Problems had been assembled in the preceding chapters. In preparing that chapter the third author acknowledges his debt to Charles W. Groetsch and his beautiful little book Inverse Problems in the Mathematical Sciences. Chapter 8 attempts 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 lecture notes, it has this in common with them: it aims to cover only a part of functional analysis, not all of it in its most abstract form; it presents a ribbon running through the field. Thus Chapter 2 introduces metric spaces, normed linear space, inner product spaces and the concepts of open and closed sets and completeness. The concept of a compact set, which was introduced in Chapter 1 for real numbers, is not introduced until Chapter 4, and not discussed fully until Chapter 6. Chapter 3 stands somewhat apart from the others; it illustrates how the idea of imbedding, appearing in Sobolev’s theory, arises in continuum analysis. From Chapter 2 the reader may pass directly to Chapter 4 which considers the important problem of approximation, and introduces Riesz’s representation theorem 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, the presentation of the concepts of weak convergence, and of the adjoint operator in Chapter 5, are limited to inner product spaces. The theory of linear operators in discussed, but not covered (!), in Chapters 5 and 7. The emphasis here is on the parts of the theory related to compact linear operators and self-adjoint linear operators, It is the authors’ fervent wish that readers will find the book enjoyable and instructive, and allow them to use functional analysis methods in their own research, or to use the book as a jumping board to more advanced and/or abstract texts. The authors acknowledge the skill and patience of Xiaoan Lu in the preparation of the text. L.P. Lebedev I.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 in 2001. Professor Vorovich, who provided the initial stimulus for the book, had a scientific career spanning more than five decades; his books on Elasticity Theory, Contact Problems, and Nonlinear Theory of Shallow Shells laid the foundation for the work of generations of researchers. We, and his many colleagues and collaborators worldwide, miss his wise counsel and his store of knowledge. This new edition incorporates many small corrections passed on to us over the years. We are particularly grateful for the contribution of Professor Michael Cloud, of Lawrence Technological University; he has not only spotted many needed corrections, but also has played an invaluable role in resurrecting the original TeX files of the book, which were in danger of being lost in the intercontinental collaboration. L.P. Lebedev G.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 mathematics, and it, like others, must start from some body of assumed knowledge, otherwise, like Russell and Whitehead’s Principia Mathematica, it will have to start with the definitions of the numbers 1, 2 and 3. This first chapter is intended to provide an informal review of some fundamentals, before we begin in earnest in Chapter 2. We will start with the positive integers 1,2,3, ···; zero, 0, and the negative integers –1, –2, –3,···. From these we go to the rational numbers of the form where are integers (we can take but However we soon find that having just rational numbers is unsatisfying; there is no rational number such that For suppose there were such a number. If had a common factor (other than ±1) we could divide that out, and arrange that had no common factor – we say they are mutually prime. Our supposition is that But if is even, so is Thus for some integer and so that Therefore is even, and hence is even. Thus have a common factor, 2, contrary to hypothesis. This contradiction forces us to conclude that there is no rational number such that However, we can find a sequence of rational numbers whose squares get closer and closer to 2 as increases. Note: Sequence always means an infinite sequence. For let be obtained from using the formula

If we start from

We note that

we find the sequence

2.

1. Introduction

so that the

are all too big, i.e.

When

for

and

Thus

so that, given any small quantity we can find an integer N such that for all Thus the terms in the sequence get closer and closer to 2; we write this and say the sequence of a convergent sequence:

converges to 2. We can make a formal definition

Definition 1.1.1 The sequence to a if, given we can find we have

which we write as converges (depending on such that, for all

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 (rational) number to which the sequence converges. However, we note that the members of the sequence get closer to each other. For

so that as We 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 by the weaker statement

we can replace (1.1.3)

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 leads us 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, given there exists (depending on such that if then

This means that the sequence (1.1.2) is a Cauchy sequence. Clearly a convergent sequence is a Cauchy sequence. For if converges to then, given we can find such that if then Thus

The converse is false, because the sequence (1.1.2) is a Cauchy sequence which does not converge, to a rational number, and at this stage this is the only kind of number we have. What we would like to do now is to extend the definition of a number so that every Cauchy sequence is a convergent sequence. Think about the example. We would like to define a ‘number’ We could associate this ‘number’ with the sequence (1.1.2). But there are other sequences, for which the sequences of squares converge to 2, for example the sequences given by (1.1.1) which start from other rational values of say or or the truncated decimal sequence 1,1.4,1.41,1.414, etc. We must associate with all these sequences. This brings us to the concept of an equivalence class of Cauchy sequences. Definition 1.1.3 Two Cauchy sequences

Problem 1.1.2 Show that is equivalent to to then is equivalent to are equivalent, then so are

are said to be equivalent,

if

is equivalent are equivalent and

This justifies the use of the term equivalent; equivalent means essentially equal. The symbol has the properties if then if and then

4

1. Introduction

Now we can introduce Definition 1.1.4 With any Cauchy sequence we can associate all the Cauchy sequences equivalent to it. We call this class an equivalence class, A particular Cauchy sequence in is called a representative of the class. Equivalence classes divide all the Cauchy sequences into separate groups; a sequence cannot belong to two different equivalent classes. In fact we have Problem 1.1.3 If spectively, then there is an

belongs to different equivalence classes reand an N such that whenever

This means that two different equivalence classes are separated from each other 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 or This leads us to Definition 1.1.5 A real number is an equivalence class of Cauchy sequences of rational numbers. With this definition, any rational number is associated with the equivalence class containing the trivial (or so-called stationary) Cauchy sequence In a sense therefore the set of real numbers, denoted by includes all the rational numbers. With suitable definitions we can treat real numbers just like rational numbers, we can add, subtract, multiply and divide with them, thus Problem 1.1.4 Show that if the Cauchy sequences are representatives of and then and are Cauchy sequences, provided that for the last named We call the classes in which these 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 convergent sequence, converging to a real number. The definitions of a convergent sequence or 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 that the set of real numbers, is complete. We often think of real numbers as points


5

on a straight line, with negative real numbers on the left of zero, and positive numbers on the right. Saying that is complete means that the line has no holes; 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 a closed interval, the set of numbers satisfying written We have used the terms open and closed, but we need to define what we mean by an open set or a closed set. The essential feature of an open interval is not that the ends are excluded, but rather that any is itself the center of an open interval entirely contained in Thus in (0,1) the point is the center of the interval (0.985,0.995) contained in (0,1); clearly any point in (0,1) may be so viewed. On the other hand [0,1] is not open, because the points 0 and 1 in [0,1] are not the centers of open intervals entirely contained in [0,1]. This leads to Definition 1.1.6 A set S of real numbers (we say a set is open if every 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 and is 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 sequence converges to a point in S. Under this definition a closed interval is closed. For suppose that and If then and for all which contradicts the statement that Thus and similarly so that On the other hand the open interval (0,1) is not closed because the sequence 1/2, 1/3, 1/4, • • • in (0,1) converges to 0, which is not in (0,1). Problem 1.1.5 Show that if S is a closed set in sequence converges to a point Definition 1.1.8 A set M such that all satisfy interval [ – M , M].

then every Cauchy

is said to be bounded if there is a number Such a set is contained in the closed

Problem 1.1.6 Show that a Cauchy sequence Problem 1.1.5 states that for sequences and convergent sequence are synonymous).

is bounded (Note the terms Cauchy sequence

6

1. Introduction

Problem 1.1.7 Suppose converges to Show that any subsequence of also converges to Conversely, show that if is a convergent sequence, and a subsequence converges to then must converge to So far we have defined three terms relating to a set and bounded. Now we introduce

closed, open

Definition 1.1.9 A set is said to be compact if every sequence S contains a subsequence, converging to a point The fundamental theorem on compactness is the Bolzano–Weierstrass theorem, named after Bernard Bolzano (1781–1848) and Karl Theodor Wilhelm Weierstrass (1825–1897): Theorem 1.1.1 A set bounded.

is compact iff (if and only if) it is closed and

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 converges to some But therefore (Problem 1.1.7) the whole sequence must converge to S is closed. If S were not bounded, we could find a sequence such that and this sequence would have no Cauchy subsequence, and therefore no convergent subsequence, contrary to the supposition that S is compact; therefore S is bounded. Now we will show that if S is closed and bounded then it is compact. Since S is bounded, it may be contained in an interval I = [ – M , M]. Let be a sequence in S. We use the method of bisection. Bisect I into two closed intervals; one half, must contain an infinity of members of the sequence; choose one of them and call it bisect one half, must contain an infinity of numbers of the sequence; choose one of them, with and so 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 and bounded. What can be said about a set which is just bounded? If S is not closed we may close it by adding to it all the limit points of convergent sequences We have Definition 1.1.10 The closure of a set S is the set obtained by adding to S the limit points of all convergent sequences


7

If S is bounded, then every sequence in S will then contain a convergent sequence with a limit This limit will be in Note that any set S consisting of a finite set of real numbers is 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 at least one of say must appear an infinity of times; this will provide 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 max nor a min. (An example is provided by the set 0,1/2, –1/2, 2/3, –2/3,…) We must therefore proceed carefully. A set is said to be bounded above by if for all it is bounded below by if for all We may adapt the method of bisection used in the Bolzano–Weierstrass theorem to show that a set which is bounded above has a least upper bound or supremum, written with the properties

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

Choose Note that it may happen that and and take

Bisect

into

Choose and so on. We may verify that the sequence obtained in this way is a sequence which converges to M. Note that if we apply this process to a finite set of numbers then the terms in the sequence will eventually all be equal, to the maximum of the numbers in the set. This may happen with an infinite set which has a greatest member. We may show similarly that if S is bounded below then it has a greatest lower bound or infimum, written

8

1. Introduction

with the properties

As with the sup, we may construct a sequence of members of S which converges to inf S. Definition 1.1.11 A sequence is said to be monotonically increasing (decreasing) if for It is said to be strictly monotonic if the inequality is strict. Problem 1.1.8 Adapt the argument used in the Bolzano–Weierstrass theorem to prove that a monotonically increasing sequence that is bounded above by converges to a limit Similarly, if it is monotonically decreasing and bounded below by a it converges to a limit So far we have discussed only real numbers, which we intuitively place on the real line, However, for many purposes real numbers are inadequate, we need complex numbers of the form where are real and i.e. The set of all such numbers we call We can consider complex sequences in which each member of the sequence is a complex number. For these we write For complex sequences the terms appearing in Definitions 1.1.1, 1.1.2 must be interpreted as moduli of complex numbers. Thus if and

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

We can go further and consider points in a plane, i.e. in two dimensions; or in space, three dimensions; or even in N dimensions. The set of all N-tuples of real numbers is called the set of all N-tuples of complex numbers is called Note that for most purposes we can treat as a point in is specified by N complex numbers, i.e. 2N real numbers. We can then generalize all that we have said about closed, open, bounded and compact sets provided that we interpret as the Euclidean distance between the points with coordinates and in Definition 1.1.12 The Euclidean distance between in is

and

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 set

Definition 1.1.14 A set an open ball lying entirely in S.

and radius a in

is the

is open if every point of S is the center of

We must also generalize Definition 1.1.1 to give Definition 1.1.15 The sequence we can find such that for all

converges to a if, given we have

Most importantly, with these generalizations, the Bolzano–Weierstrass theorem 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 use functions of one or more variables. Displacements, velocities, loads and temperatures can be functions of points of a body, and of time. We need some definitions 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 a non-empty open set (Definition 1.1.14) in Definition 1.2.2 A rule which assigns a unique real (complex) number to every is 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 function is 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 L et be a function on Let The function is said to be continuous at if, given we can find depending on such that if then The function is said to be continuous on if it is continuous for every Problem 1.2.1 Show that is continuous on (0,1), but not bounded on (0,1). Thus, there is no number M such that for all One of the basic results of Calculus concerns functions that are continuous on 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 and bounded, i. e. a compact region is bounded, and achieves its supremum and infimum in Proof. Suppose were not bounded. Then there is a sequence such that Since and is compact, contains a subsequence converging to a point The function is continuous at Thus we can find such that if and then Choose N such that implies then i.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 sequence contains a subsequence converging to Choose Since is continuous at we can find such that if and then Choose such a and then choose N so that if then Then for all such we have

But is arbitrary, and may be taken arbitrarily large, so that On the other hand so that and achieves its supremum on We can prove in a similar fashion that it achieves its infimum. The theorem states that there exist

M such that


11

Moreover assumes its supremum and infimum in such that

That is, there are

(we can thus say that has a maximum value, This means in particular that there is an that

and a minimum value, (either or ) such

Thus if is continuous and has compact support, it is bounded on the closed set if and is outside G, so that is bounded on A function that is continuous on a set which is not compact may not achieve its supremum and infimum. For example, in achieves its supremum, but not its infimum. Definition 1.2.5 A function on if, given we can find then

is said to be uniformly continuous such that if and

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

for Defini-

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

then it

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

This will give two sequences these sequences will contain subsequences respectively, and

so that

But

is continuous at

which contradicts (1.2.1).

so that

Since

is compact, each of converging to

12

1. Introduction

Theorem 1.2.2 states that if is continuous on a compact region then 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 tinuous, on (0,1).

is continuous, but not uniformly con-

A function that is uniformly continuous on a bounded domain is bounded, provided that the domain is sufficiently regular that, given one can go from one point to any other point in a finite number of steps of length less than For if it is uniformly continuous on we can find such that implies If we can get from to any other point in a (fixed) number of steps, of lengths then

This proof is valid if the domain is a connected finite union of star-shaped domains. A function which is uniformly continuous on an unbounded domain need not be bounded: is uniformly continuous but unbounded on Theorem 1.2.3 If is bounded and uniformly continuous on then 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 that be bounded. By continuous extension we mean that we can find a function defined on such that

and

is continuous on

Proof. Let and let be a sequence converging to then is a Cauchy sequence (in or converging to a number which we denote by Suppose and are sequences converging to respectively. Choose Since is uniformly continuous in we may choose such that implies Now choose large enough that:


and suppose that

so that

Corollary The extension Theorem 1.2.1, is bounded on

13

Then

and

is uniformly continuous on and thus, by and assumes its infimum and supremum, and

Proof. Clearly implies On the other hand, will (by Theorem 1.2.1) assume its supremum at some point There is a sequence converging to for which so that implies thus We conclude this section by proving Weierstrass’ theorem on uniformly convergent sequences of uniformly continuous functions. Let be a sequence of functions on For any particular we may consider the sequence For this value of the sequence will be a Cauchy sequence if, given we can find (depending on and such that implies Similarly, for a particular value of the sequence is said to converge to if given we can find (depending on and such that implies (Because and are complete and a Cauchy sequence is a convergent sequence.) In these statements may depend on as well as on If, for any it is possible to choose one depending on alone, which will work for all then the sequence is said to be a uniformly Cauchy sequence, or to converge uniformly to Weierstrass’ theorem is Theorem 1.2.4 A uniformly Cauchy sequence uniformly continuous on a compact region continuous function

of functions which are converges to a uniformly

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 in Choose and then choose such that for all and all we have

14

1. Introduction

Letting Thus if

Each function that

Thus for such

we find and

then

is uniformly continuous in implies

so that we may find

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 and bounded region may be uniformly approximated arbitrarily closely by a polynomial. This theorem states that, given that

we can find a polynomial

such

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] may be uniformly approximated arbitrarily closely by a polynomial. Proof. First we introduce the polynomials

where

We will need some identities relating to the binomial expansion

to obtain them we use the

1.3 Weierstrass’ polynomial approximation theorem

The first identity is obtained by putting

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

In each of these, put to obtain

15

then

to obtain

multiply the first by

and the second by

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) so that there is an such that for all Choose and then choose such that implies Now

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

1. Introduction

16

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

so that to make

when

Combining (1.3.3)–(1.3.5) we find

we need only take

We may use this theorem to prove Theorem 1.3.3 A uniformly continuous periodic function of period may 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 whole real line is the sup over

the sup over

1.3 Weierstrass’ polynomial approximation theorem

Proof.

17

The functions

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

which will be continuous in [–1,1]. Given find polynomials such that

for

In terms of the variable

for

we can, by Theorem 1.3.1,

these are

and we note that are trigonometric polynomials of the form (1.3.6) (with the These inequalities hold for but therefore for 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 the same procedure to the function and find a trigonometric polynomial such that

Both inequalities, (1.3.8) and (1.3.9), hold for all t; on replacing t by in (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 numbers Definition 1.1.5 Sets open: every point is an interior point closed: contains all its limit points

Definition 1.1.6 Definition 1.1.7

compact: every sequence contains a convergent subsequence Definition 1.1.9

Functions continuous:

Definition 1.2.4

uniformly continuous: Theorem 1.2.2.

Definition 1.2.5

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 create the world, then, as we all know, He created it according to the geometry of Euclid and the human mind with the conception of only 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 space by its coordinates in some Euclidean frame. We write and say is in which we write The Euclidean distance between two points is

Notice that we could use the notation x to indicate that is not a number, but a triplet of numbers. We will not do this because we want the reader to get used to the idea that the point or vector is the fundamental entity; its coordinates are secondary. Later we shall even use to denote different points if we have to specify their coordinates we will use notations such as for the coordinates of The context will (we hope) make the usage clear! We can generalize the idea of Euclidean space by defining an N -dimensional space consisting of vectors We define the Euclidean distance between by

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


20

then introduce an abstract concept which possesses these qualities. The essential qualities of Euclidean distance are that distance is real-valued; the distance between two different points is positive; the distance between and is the same 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 for with the properties:

Here and hereafter iff denotes if and only if. D1 is called the axiom of positiveness; D3 states that is reflexive; D4 is called the triangle axiom. We have Definition 2.1.1 A real valued function defined for a metric for if it satisfies D1–D4. Note that strictly we distinguish between a metric and its value

is called for

It is clear that given in (2.1.1) satisfies D1-D3. When N = 3, D4 is the familiar triangle inequality in 3-D geometry; it is quite difficult to prove it 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 equivalent metrics in if there exist two positive constants independent of such that for any

in

2.1 Preliminaries

Problem 2.1.1 Show that the metrics respectively, are equivalent.

and

21

given in (2.1.1) and (2.1.2)

Now we formally state the generalization of the concept of convergence discussed in Chapter 1 Definition 2.1.3 A sequence is said to converge to under the 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 complex numbers. We will use the double arrow for the type of convergence, just defined, of a sequence in a metric space. We will later call this convergence strong convergence to distinguish it from another kind of convergence, weak convergence, which we will introduce in Chapter 5, and will denote by a single arrow, For real and complex numbers there is no difference between the two types of convergence; thus we can write as Problem 2.1.2 Show that if in then in

and

are equivalent metrics in

and if

Problem 2.1.3 Show that

are possible metrics for

but no two of them are equivalent.

Problem 2.1.4 Show that in even though are not equivalent.

iff

in

(of Problem 2.1.3)

Problem 2.1.5 Consider system of particles in The configuration of the system is the set of triples of the Cartesian coordinates of the points named after René Descartes (1596–1650). Show that we can distinguish different configurations of the system by using a metric in where We can apply the notion of a metric not only to the sets of locations of a system of particles, but also to sets of velocities, accelerations, masses, or in fact to any finite set of parameters, forces, temperatures, etc. Now let us consider continuum problems. Take a taut string with fixed ends and length we can use a Fourier expansion

22


to describe a static displacement caused by some continuous load distribution. Any state of the string can be identified with the vector having an infinity of coordinates Now we have an infinitedimensional space of all For the moment we shall call this but we shall soon be more precise. Let us modify the metrics we introduced in so that we can determine the 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 right will converge. This means that the metric is appropriate to measure the distance between any two configurations of the string for which

The appropriate generalization of (2.1.2) makes use of the concept of the supremum 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 metrics for given by (2.1.1) and (2.1.2) are equivalent, but we can easily show that this 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. Thus and 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), and is finite, then will be finite; moreover, there is a constant that

such

We took the idea of Euclidean distance, and constructed the abstract notion of a metric. Now we use the examples of and our vaguely defined to construct the notion of a metric space. Definition 2.1.4 A metric space is a pair consisting of a set X (of points or elements) together with a metric a real valued function defined for any two points which satisfies D1–D4. We shall usually denote a metric space by X, with remaining implicit. We will generalize Definition 2.1.2 to X: Definition 2.1.5 Two metrics if there exist such that

of a space X are said to be etquivalent

We shall not distinguish between metric spaces consisting of the same elements, if their metrics are equivalent. Different problems in mechanics and physics require different types of metric spaces. Depending on the metric we choose, a solution of a problem may or may not exist, may be unique or non-unique, etc. The right choice of a metric space can be crucial for success. We now abolish the vaguely defined space and introduce some proper definitions. The spaces we are dealing with have elements which are infinite sequences, i.e. there are four examples of such spaces which 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


24

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

3. 4.

is the set of all sequences convergent to zero; the metric is the metric

of 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). We have the correspondence

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

We can also measure the distance between two configurations of the string by using the strain energy of the string. After reducing this to dimensionless form we 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 two configurations 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 called energy spaces. We have used the representation (2.1.4) to express these spaces as 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 concepts. 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 an of and denote it by is called the center of the neighborhood of is any subset M of X which contains an B of Conversely, we call an interior point of a set neighborhood of

A if M is a

Definition 2.2.2 A set S in a metric space X is said to be open if every point is 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 set if every neighborhood of contains at least one point of S, maybe just The set 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

Problem 2.2.3 Show that Definition 2.2.4 A point is called a limit point (or accumulation point) of S if every neighborhood of contains an infinity of points of S. We sometimes say the points cluster around The limit point may or may not belong to S.

26


For example, if S is the set of rational numbers on [0,1] with the metric then 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 neighborhood of 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 or an isolated point. Definition 2.2.6 Let X be a metric space. A set S is said to be closed in X if i. e. if it contains all its contact points, and in particular, all its limit points. The set

is called a closed ball of radius According to the last definition, it is a closed set. Note that we cannot say simply that a set is closed; we must state the metric space X for which it is closed. For example, suppose is the metric space of rational numbers under the usual metric Suppose S is the set of rational numbers such that S is closed in For if s is a limit point of S, then it must be rational, because it is in and we may show that it satisfy therefore it is in S. But suppose that is the set of real numbers under the same metric (i.e. ) and S is again the set of rational numbers such that Then we can find a set of points in S which cluster 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 S is 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 contains a point Definition 2.2.8 Let be a metric space. Let We may define a subspace of by the pair

is called the metric induced on Y by

of a point be a subset of X where, for

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 complement of S is the set of points in X which are not in S; it is denoted by X – S (or X\S). Problem 2.2.7 Let X be a metric space and S be a set in X. Show that S is open iff its complement is closed, and vice versa. Note that a finite set of points, is closed. The set has no limit points (see the crucial different from in Problem 2.2.4); its only contact points are 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 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 it need not be bounded. We therefore introduce Definition 2.3.2 are bounded on If we equip

in § 1.2.

is the subset of

then

consisting of functions which

with the metric

then we have a metric space. Clearly this verified as follows. Suppose and

satisfies D1–D3; D4 may be 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 pact support in

is the subset of

consisting of functions of com-

28


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 a subspace of We may also introduce Definition 2.3.4 is the subset of which are uniformly continuous on

consisting of those functions

Note that Theorem 1.2.3 states that if is bounded and uniformly continuous on i.e. in then it may be extended (uniformly) continuously to Problem 2.3.1 Take but

= (0,1). Show that if

then

When dealing with differentiable functions we often want to have some way of 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. tions which have continuous derivatives

is the set of funcfor

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

or but in order to ensure that these quantities are finite we must introduce subsets of 1

Uniformly continuously differentiable means that

is uniformly continuous.

2.4 Convergence in a metric space

29

We can introduce metric spaces and which are generalizations of and respectively. Thus for we take 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 which are uniformly continuous on We define to be the set of functions having continuous derivatives of all orders on to be the subset of

i.e.

and

of functions having compact support.

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

is a suitable metric for to the uniform metric (2.3.1).

i.e. it satisfies D1-D4, but is not equivalent

2.4 Convergence in a metric space Generally, from now on, means a point in a metric space, infinite sequence of such points.

refers to an

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

and say that the sequence is convergent, or converges to write

We will also

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 such that Take By definition, there exists N such that implies and But

30


This is a contradiction. The reader will note that this proof follows exactly the same lines as that used in § 1.1 for sequences of real numbers. Problem 2.4.1 Show that a sequence which is convergent in a metric space X is 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 limit by the definition (Definition 2.2.3 and Problem 2.2.5). In particular, if S is closed (Definition 2.2.6), then The 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 Cauchy sequence if, given there exists (depending on such that, if N, 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 and 1.1.2 which we introduced for sequences and later generalized (Definition 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. Now we can no longer assume that this is true. Indeed if X is the metric space of rational numbers under the metric the sequence (1.1.2) is a Cauchy sequence, but not a convergent sequence. In a general metric space a Cauchy sequence may not have a limit. Problem 2.4.3 Show that a convergent sequence in a metric space X is a Cauchy sequence. Problem 2.4.4 Show that if is a Cauchy sequence, and subsequence which converges to then converges to

has a

2.5 Complete metric spaces Definition 2.5.1 A metric space X is said to be complete if any Cauchy sequence 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 being 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 is a complete metric space; the counterexample (1.1.2) shows that the space of rational numbers with this metric is incomplete. Problem 2.5.1 Show that the space of complex numbers the metric is complete.

with

Weierstrass’ theorem on uniform convergence of uniformly continuous functions on (Theorem 1.2.4) may be interpreted as stating that is complete under the uniform metric (2.3.1). Note that uniform convergence of a sequence of 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]. Show that is a Cauchy sequence in the metric

.but that metric, i.e.

converges to the function

in the

Thus the in this example converge in the metric to the function which 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 incomplete under the metric (2.5.1). This is general: a space may be complete or incomplete 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) to give an alternative definition of dense, namely Definition 2.5.2 A set S is said to be dense in a metric space X if any of contains a point Weierstrass’ polynomial approximation theorem (discussed in § 1.2) states that if is a closed, bounded set in then the set of all polynomials is dense in with the metric (2.3.1).

32


The completeness of a metric space is of great importance since numerous passages to the limit appear in the justification of numerical methods, existence theorems, etc. Many of the spaces we have introduced up to now are not complete; 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 way in 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 and is said to be one-to-one if there is a rule which assigns a unique element to each element and vice versa. The correspondence is said to be isometric if

Definition 2.6.2 Two sequences to be equivalent if

and

in a metric space X are said

Theorem 2.6.1 For a metric space X, there is an isometric one-to-one correspondence 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 associate all the Cauchy sequences equivalent to it; these form a class, an equivalence class, F. A particular Cauchy sequence in F is called a representative of F. With any element we may associate the Cauchy sequence any equivalence class which contains such a sequence (and it can contain at most one) is called a stationary equivalence class. Denote the set of stationary equivalence classes by and the set of all equivalence classes by Now introduce a metric in by

where are representatives of the equivalence classes F and G respectively, 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 choice of representatives The triangle inequality gives

2.6 The completion theorem

33

so that Interchanging

and

we obtain

and hence

as since Cauchy sequence (in

are Cauchy sequences. Thus and the limit (2.6.1) exists because

is a is complete.

Problem 2.6.1 Show that the limit is independent of the choice of representatives 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 satisfy i.e. according to Definition 2.6.2, the same equivalence class: F = G. D3: D4: For we have

from F and G are in

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 and We now show that is dense in Let F be the equivalence class containing the Cauchy sequence and let be the stationary equivalence

34


class containing there is an N such that if

Choose then

Since

is a Cauchy sequence, Thus

In words, we may find a stationary class as close as we like to F; is therefore dense in Finally, we must show that is complete. Let be a Cauchy sequence in From each choose a representative Cauchy sequence in this sequence choose an element such that for all this is possible because is a Cauchy sequence. Let us show that is a Cauchy sequence. Let be the stationary equivalence class containing Then

so that, by the isometry (2.6.2),

since, by hypothesis, containing the sequence

is a Cauchy sequence. Let F be the equivalence class Then

This can be made as small as we like by choosing

Thus the Cauchy sequence complete.

large enough, so that

converges to F in the metric

Problem 2.6.2 Use the fact that under each of the metrics and

so that

is

is complete to show that is complete 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 so that 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 of X ) . Thus we can identify X with its completion, Since Theorem 2.6.1 is of great importance, let us emphasize some of its aspects. is a metric space whose elements are equivalence classes of Cauchy sequences from X; X is isometric with the set of all equivalence classes containing a stationary sequence for

2.7 An introduction to operators

35

In § 2.1 we discussed, in an informal manner, the construction of the real number from the rationals. A formal treatment of this construction would use the completion theorem. Note however that logically this must be done before the real numbers are used as the field for constructing metric spaces. If we can establish a property of a limit of any representative Cauchy sequence in a class F, we will say that the class F possesses this property. We will 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 metric spaces leads us to Definition 2.7.1 Let X and Y be metric spaces (they may be identical). A correspondence is called an operator from X into Y, if to each there corresponds no more than one The set of all those for which there exists a corresponding is called the domain of A and denoted by D(A); the set of all y arising from is called the range 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). We also 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 nonempty open set in .) Definition 2.7.2 A functional is a particular case of an operator, in which or 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 is said to be continuous at if, given there is a depending on such that if then If A is continuous at 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 from X into Y. Show that A maps Cauchy sequences in X into Cauchy sequences in 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


36

where A acts in a metric space X. A solution of (2.7.1) is called a fixed point of A. Typically (2.7.1) arises when we have an approximation procedure which yields 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 contraction operator (or a contraction mapping) in X, provided there exists a real number with such that

Clearly a contraction operator is a continuous operator. The fundamental theorem concerning contraction mappings is Banach’s fixed point theorem, due to Stefan Banach (1892–1945); it is Theorem 2.7.1 Let A be a contraction operator in a complete metric space X. Then: 1. A has only one fixed point 2. for any initial approximation imations

converges to mated by

the sequence of successive approx-

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

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

then Since this implies and therefore Now take an element and consider the iterative procedure (2.7.3). For we get successively


37

But

so that

It follows that element

is a Cauchy sequence. Since X is complete, there is an 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 then However the condition required by the theorem, namely is stronger than this, and is vital to the proof of Theorem 2.7.1 as is shown by a counterexample. Problem 2.7.2 Let let be the usual metric X is complete in the metric d. Let A be the mapping from X into X given by Show that if then but that A has no fixed point. We denote 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 A has a single fixed point a sequence of successive approximations (2.7.2) converges to independently of the choice of initial approximation with the rate

38


Proof. The operator the equation

meets all the requirements of Theorem 2.7.1, so that

has the unique solution such that sides of this last equation, so that

This means that so that

We can apply A to both

is also a solution to (2.7.6), but the solution is unique

i.e. equation (2.7.1) has the solution Noting that any fixed point of A is a fixed 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 approximations can be composed successively of elements of these sequences, we obtain the stated rate of approximation. We can apply Banach’s fixed point theorem to systems of linear algebraic equations. 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 initial approximation from


39

Problem 2.7.3 Consider a Volterra operator

with kernel operator in there exists

continuous on [0, a] x [0, a]. Show that A is a contraction for sufficiently small a > 0. Also show that for any finite such that is a contraction operator.

It is worth noting that the problem of viscoelasticity can be formulated as an equation of Volterra type, but for functions that take values in a certain energy 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 X into 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 into Y. 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 some

is called the image of S. c) If

then the set of all image of

such that

d) If T is a set in Y, then the set of all

is called the inverse such that

is called

the 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 in D(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 into Y. 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 let be 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 radius around which is in T. Since A is continuous we can find such that implies All such are in T, so that the corresponding are in S. Therefore any is the center of an open ball in S, and S is open.


40

b) Suppose T is open implies that S is open. Suppose and T. Since T is open, is the center of an open ball of radius consisting of points 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 contains an open ball of radius therefore, if then is continuous at Theorem 2.7.3 Let X, Y be metric spaces, and A an operator from X into Y. 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 let be 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 at Choose There is a such that implies But since we can find N such that implies and thus Thus But T is closed so that which means 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 Suppose The set defined by image S is therefore closed. But hence Thus continuous.

but is closed. Its inverse and so that and which is impossible. Thus A is

2.8 Normed linear spaces Almost all the spaces we shall consider are linear spaces. This means that for every pair of elements in a linear space X and or a sum and a scalar product can be defined such that: 1. 2. 3. there is a zero element, 4. 5. 6. 7. 8.

such that

2.8 Normed linear spaces

If

41

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 valued function defined for every which satisfies the following norm axioms:

Definition 2.8.2 A linear space X is called a normed linear space if, for every a norm satisfying N1-N3 is defined. As with metrics, we can distinguish between a norm

and its value

for

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

and

We shall show below that any two norms in are equivalent. X is said to be real or complex, depending on whether the scalars taken from or In a normed linear space we may define a metric

such

are

Problem 2.8.1 Show that if the norm satisfies the axioms N1-N3, then given 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 means of (2.8.2), then Show that of Problem 2.1.3 does not satisfy 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 is called a subspace of X if it is itself a linear space, i. e. one which satisfies conditions 1–8 listed above, and has the norm on Y obtained by restricting the norm on X to the subset Y . The norm on Y is said to be induced by the norm on X.


42

Definition 2.8.5 Let X be a normed linear space, and Y is called a 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 is also a subspace. One of the most important kinds of subspace is a finite dimensional subspace. Before defining this we must introduce some concepts which are direct generalizations of concepts in linear algebra. Definition 2.8.6 Let X be a linear space. The elements said to be linearly independent (over the appropriate field equation implies otherwise dependent; in this case there is at least one of the as a linear combination of the others. Let

or

are if the

are said to be linearly which may be expressed

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 finite dimensional subspace Y of X, with its dimension, dim(Y) being defined by Definition 2.8.7 If are linearly independent then Otherwise 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 a non-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. If X is not finite dimensional we shall say that it is infinite dimensional. If dim(X) = N, any set of N linearly independent elements is called a basis for X. If is a basis for X, then any element of X has a unique 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 are equivalent. To prove this we first establish Lemma 2.8.1 Let X be a normed linear space, and linearly independent set of elements in X. There is a number for every set of scalars we have

Proof. Write (2.8.3) holds with any equivalent to

where

and

If

then all Suppose

be a such that

are zero, so that the inequality then the inequality (2.8.3) is

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 bounded set Since this set is compact, the sequence contains a subsequence converging to a number such that Denote the of this subsequence by 1,1; 1,2; ··· . Now consider the for this subsequence:

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


44

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 and we 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 ity (2.8.3) gives

On the other hand, the inequal-

so that This is half of the inequality (2.8.1); the other half may be obtained by reversing the 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 norm

be a closed and bounded region in

The space

with the

2.9 An introduction to linear operators

45

is a normed linear space; it is complete, and so it is a Banach space. This norm is called the uniform or infinity norm. We can apply the Completion Theorem to normed linear spaces; now we say 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. Problem 2.8.4 Show that the spaces spaces.

and

Problem 2.8.5 Show that the space norm

defined in § 2.3 is complete in the

corresponding to

defined in § 2.1 are Banach

given in (2.3.4).

Problem 2.8.6 Show that the space norm (2.8.4).

defined in § 2.3 is complete in the

Note that is larger than uniformly continuous on

in that its elements need not be

Problem 2.8.7 Take but

Construct

such that

2.9 An introduction to linear operators Suppose that X and Y are linear spaces over the same coefficient field either or Definition 2.9.1 A space S is said to be a linear subspace of a linear space X 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 its domain D(A) is a linear subspace of X and, for every and every For a linear operator, the image

is usually written

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

46


of X; a range R(A) which is a subspace of Y; and a null space N(A) consisting of 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 a subspace 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 which we have the simplifying result Problem 2.9.3 Let A be a linear operator from the normed linear space X into the normed linear space Y. Show that A is continuous at any point iff 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 X into the normed linear space Y. The operator A is continuous on D(A) iff there is a constant c, such that, for all we have

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

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 with states that there is some such that, if then For every the norm of is

so that

This states that if

But A is a linear operator, so that

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 operator from X into Y. If X is finite dimensional, then A is continuous. Proof. Let dim X = N and suppose set of elements in X. If we can write

is a linearly independent

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 and a 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 by

is a linear operator from into norm corresponding to (2.3.1) for

defined

Show that it is bounded. (Use the and (2.8.5) for


48

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 used to construct some new normed linear spaces. Lemma 2.10.1 Let with equality iff

Then

Proof. Consider for

so that, since Hence

where and we have for with equality only if

Then and Thus

we have If result.

then

If

we put

to obtain the required

Now we prove Holder’s inequality, due to Ludwig Otto Holder (1859-1937): Lemma 2.10.2 Let Then

and

Proof. Put

If AB = 0, then either A = 0 or B = 0. But A ( B ) can be zero only if all the (or are zero. In either case both sides of (2.10.1) are zero. Now if AB > 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 a constant M such that

Finally we prove Minkowski’s inequality, due to Hermann Minkowski (1864– 1909) Lemma 2.10.3 Let

Proof. The case

and

is trivial. Suppose

then

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

now use

then

defined by

to obtain (2.10.2).

We may use Minkowski’s inequality to generalize the metric in 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

intro-

50


provides a suitable norm in the space Problem 2.10.4 Verify that

and

of all sequences

with

are complete with the respective norms.

We can consider a contraction mapping in system (2.7.7), namely

Thus for the

we have

Applying the limiting form of the Holder inequality as and taking we obtain

to the inner sum,

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 inequality after 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 written out. So a limit passage demonstrates that corresponding inequalities hold for integrals. This is the way in which Minkowski’s and Hölder’s inequalities for integrals (see § 2.11) were derived.

2.11 Lebesgue spaces Up to now we have introduced only a few simple normed linear spaces. In this section we shall introduce some of the most important ones. Definition 2.11.1 Let Lebesgue space satisfying

be a domain in and let is the completion of the subspace,

The of

in the norm

There is much to note about this definition. First, we have used the abbreviation

Secondly, the integration is taken to be the ordinary Riemann integration named after Georg Friedrich Bernhard Riemann (1826–1866), over a domain in Thirdly, 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 Completion Theorem (Theorem 2.6.1), this means that elements in are equivalence classes of Cauchy sequences of continuous functions. Remember that is a Cauchy sequence in if

and that two sequences

and

are equivalent if

52


Let us examine elements in First, we say loosely that if then but this is not strictly accurate. The elements of are not functions, but equivalence classes of Cauchy sequences of functions. What we should say strictly is that if then there is an equivalence class which includes, as one of its Cauchy sequences, the sequence We need to label this equivalence class; we could label it or use the same label Let us explore this deeper. The function is in we call the equivalence class which includes the sequence 0,0,0, ···, the null class, and give it the label 0. But there are other Cauchy sequences in this class. Suppose for simplicity that N = 1 and Take 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 null class. In other words, the sequence is one of the (infinitely many) equivalent 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 the pointwise limit of the sequence of functions is

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


53

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

because with each of these functions we can associate a Cauchy sequence in the null class. We can generalize this with Problem 2.11.1 Take N = 1, cannot distinguish between

Suppose

Show that

Hint. In choosing a sequence take tent functions centered on each and choose the supports of the tents appropriately. Now we can return to the general case. Definition 2.11.2 The sequence quence if as

is said to be a null se-

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

For such functions In other words, if then, considered as an element of it is in the class 0. This generalizes Problem 2.11.1. A function which is zero except at a countable (see Definition 4.1.2 for the definition of countable) set of points is equal to zero a.e., but the converse is not true; it is possible for to be zero except in a set which is not countable, and still to be the limit of a null sequence according to (2.11.6). Such a set is called a set of measure zero. Countable sets are sets of measure zero, but there are sets of measure zero which are not countable. If a.e. on we say that and are equal almost everywhere. For such functions

54


The reader who is acquainted with the theory of Lebesgue integration due to Henri Leon Lebesgue (1875–1941), will realize that we have defined without using measure theory. That the two ways of approaching through completion of and by using measure theory, do lead to equivalent definitions, 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 cannot be accommodated in Riemann integration, but can be in Lebesgue integration. With some effort we could extend our argument to include them, but anyway such regions do not often occur in practice. Let us recapitulate. An element is an equivalence class of Cauchy 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 use Minkowski’s inequality in the form

Thus we have

since is, by definition, a Cauchy sequence in the norm norm Thus is a Cauchy sequence in the complete space has the limit The number

in the and so

is called the Lebesgue integral of

Problem 2.11.2 Show that K is independent of the choice of the representative sequence in the equivalence class Problem 2.11.3 Construct a sequence of tent functions such that but is a null sequence in We have defined as the completion of the space the subspace of satisfying (2.11.1), but there are other, more convenient ways to


55

define it; we can define it as the completion of any dense (Definition 2.2.7) subspace of this space, such as or Finally, Weierstrass’s polynomial approximation theorem (Theorem 1.3.1) states that if is bounded, then every continuous function in is the limit of a sequence of polynomials; thus we may apply the completion to the set of polynomials on to obtain We have defined for satisfying It may be shown that if 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 If then and

satisfy

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

(This integral inequality follows because a Riemann integral is the limit of a sum.) In this inequality, replace and by and 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


56

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

and say that if

then

and

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

imbedded in

is

and we write this

The imbedding defines an imbedding operator I from to this operator maps into the corresponding The operator I 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 the normed space Y, and write this if X is a subspace of Y, and the operator I from X to Y defined by for all is continuous. If for any

and

is bounded, the integral by

is uniquely determined


If

57

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 and Consider

where and respectively. Then

since and and, for large

Thus

are representative sequences of

where

and

are Cauchy sequences in the corresponding metrics

is a Cauchy sequence; we define

as its limit.


58

Problem 2.11.5 Show that is independent of the choice of representatives for 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 of distance and magnitude in We now consider the generalization of the inner product. Definition 2.12.1 Let X be a linear space over function uniquely defined for every pair product on X if it satisfies the following axioms:

as defined in § 2.8. The is called an inner

where and the overbar in P2 denotes complex conjugate. A linear space X with an inner product is called an inner product space. We can consider X over then the inner product must be real valued, P2 is 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, named after Hermand Amandus Schwarz (1843–1921). (This is also called the Cauchy– Schwarz or the Cauchy–Buniakowski inequality, after Victor Yakovlevich Buniakowski (1804–1899).) Note that for or for spaces like the Schwarz inequality 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 scalar. By P1 where equality occurs iff

Put

or

is zero. Let

and let

be a

We have

then

which is equivalent to (2.12.2). Problem 2.12.1 Use the Schwarz inequality to show that if the inner product satisfies P1–P3, then the norm defined by (2.12.1) will satisfy N1–N3. (Then by 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 characterizes norms that are derived from an inner product, as shown by the next problem. Problem 2.12.3 Show that if X is a normed linear space, with a norm which satisfies (2.12.3), then we can construct an inner product on X by taking

60


Show that and that the inner product satisfies P1-P3. (It is quite difficult to prove P3 for arbitrary (For a real normed linear space we omit the second pair of terms in the definition of Definition 2.12.2 By analogy with Euclidean space, we say that orthogonal if

and

are

Definition 2.12.3 Let X be an inner product space. A subspace S of X is a subspace of X which is itself a linear space (as in Definition 2.8.4) with the inner product on S obtained by restricting the inner product on X to The 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 of X 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; a closed 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 analysis itself. 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

61

and the real


is an inner product for on [0,1].

the set of continuously differentiable functions

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

Thus we have Problem 2.12.8 Show that if

then K is a continuous operator on

and

and


62

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 sequences in 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 sequence. 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 (2.11.2).

completion of

etc in norm

Sequences Cauchy:

as

convergent to x:

Definition 2.4.2. 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 it. Definition 2.2.7.

has a point

arbitrarily close to


63

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 Publications Inc., New York, 1975. This has extensive discussion on set theory, on measure theory and integration. A. Friedman, Foundation of Modern Analysis. Dover Publications Inc., New York, 1970. This covers much of the material in our book at greater depth and level of abstraction. In particular it has an extensive study of Lebesgue integration, and of the concept of the adjoint for spaces other than Hilbert spaces. L.V. Kantorovich and G.P. Akilov, Functional Analysis, Pergamon Press, 1982. This is an extensive work with copious references to the original literature and to other treatises. A comprehensive treatment of functional analysis at an abstract level may be found in K. Yosida, Functional Analysis, Springer-Verlag, New York, 1971. A brief, easily readable account of some aspects of functional analysis may be found 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 book at 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 is pleasant too. Anton Chekhov, The Seagull

Perhaps mathematical proofs too.

3.1 The rod Consider a perfectly elastic rod of length cross-sectional area Young’s modulus E (named after Thomas Young (1773–1829)), undergoing longitudinal displacement 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, and that is bounded, i.e.

We suppose that the rod is a cantilever, so that

We can use functions in

Thus

is fixed, i.e.

to define a metric, norm and inner product in the subset of satisfying (3.1.2), for which

66


Problem 3.1.1 Show that these d, and (·, ·) do satisfy the requirements for a metric, norm and inner product; in particular, that We call the set S of all space of functions with finite energy.

satisfying (3.1.2)–(3.1.3) the sub-

Problem 3.1.2 Show that S is an incomplete inner product space, by constructing a Cauchy sequence of functions in S which has a limit which is not in S. To create a complete space, we must apply the Completion Theorem; the energy space is the completion of S in the metric (3.1.4). We recall that an element in is a class, U, of Cauchy sequences in S which are equivalent in the metric (3.1.4). First we show that if and is thus in For if

then it is uniformly continuous in

then

Thus to make we need only take is uniformly continuous in it is in is a representative Cauchy sequence in a class

where Now suppose that we have

3.1 The rod

67

as This means that converges in the uniform norm, i.e. it converges uniformly. But by Weierstrass’ theorem 1.2.4, a uniformly convergent sequence of continuous functions on has a uniformly continuous limit. (This is equivalent to saying that is complete under the uniform metric.) This means that converges in the uniform metric to a limit function which is uniformly continuous on and Problem 3.1.3 Show that tive Cauchy sequence for U.

is independent of the choice of the representa-

We will use capital U to denote an element of the energy space, here and to denote an ordinary (continuous) function. What we have shown is that U is such a function, so that we could call it Problem 3.1.4 Show that

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

This is another example of an imbedding: every element can be identified with an element in such a way that the inequality (3.1.9) holds. The correspondence defines an operator, as defined in § 2.9, from into We call it the imbedding operator from into Clearly (3.1.9) shows that 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 that we could use lower case to denote it; it is not necessarily differentiable, but we can define a generalized derivative for it. If is a representative Cauchy sequence for U, then is a Cauchy sequence in and thus corresponds to an element in can call this the generalized derivative of U, and can use the notation with caution.

68


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

is properly defined if and Cauchy sequences for F, U respectively, and

For if

are

then

But

so that limit W. Moreover, if

as

Thus

is a Cauchy sequence with

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

A virtual displacement is one that is sufficiently smooth and which satisfies the geometric constraints imposed on the body. Generally, if the particles of a body are displaced, the internal forces, and maybe also the external forces, will be affected. In computing and we ignore 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 are called the virtual work done by the internal and external forces, respectively. For the rod there is only one kind of displacement: longitudinal. A virtual displacement is one that is sufficiently smooth, e.g. or at least and which satisfies the geometric constraint

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

the virtual work

The internal forces are longitudinal forces acting on the cross-section of the rod. At section the strain is and the corresponding stress is A virtual displacement of particles of the rod induces a virtual longitudinal strain The work done by the stress in the volume 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 from which we may deduce the classical differential equation governing the rod, and also 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) to obtain

This must hold for all sufficiently smooth satisfying therefore hold for the subset of these satisfying equation 3.1.16) reduces to

it must also. For such

70


Under the condition (3.1.15), the bracketed term in the integrand is in We now prove Lemma 3.1.1

If

for all functions

and

satisfying

Proof. Suppose there is a generality we may assume closed interval with support

This is a contradiction. Thus so that and thus

then

such that without loss of Since is continuous at there is a in which Choose a function such that in then

in

but

is continuous in

in

Note that the conclusion would follow if equation (3.1.17) held just 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 equation (3.1.16), which holds for all satisfying and not just for those 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 the conditions (3.1.15). In fact we can set up the equation

for and We may suppose that has derivatives of all orders, i.e. is in and has some compact support where Thus and in fact but and its derivatives need not be zero at We could call this set Under these conditions the integral on the right may be bounded using the Schwartz inequality:

3.1 The rod

so that it is defined when limit

71

Note that the first integral is defined as the

for a representative Cauchy sequence of functions in the equivalence class U; note that convergence of the is in the norm (3.1.4), so that equation (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 since the elastic rod is a conservative system we can also derive the governing equations 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 and

If

satisfying

For if we take then

for all

then

where


72

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 satisfying Now 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 Remember the definition of a continuous linear functional given in equation (2.9.1). For if is a Cauchy sequence for and is one for then, 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 imization of J(U) as before; thus if then

Since

is arbitrary, we must have

We can consider the min-

3.1 The rod

for all

73

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 problem as the minimization of a functional

where is an appropriate norm, and is a continuous linear functional in 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 the minimizer satisfies for all We 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 other hand, if (3.1.20) holds for then (3.1.25) will hold for every since is dense in Before leaving this problem we note that the natural boundary condition (3.1.19) which occurs in the classical problem, has no meaning in the generalized problem because need not exist. (We can assign a meaning to it by introducing distributions.) We have considered the rod with the cantilever end condition We 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 not a norm, for implies only or There are various ways to circumvent this difficulty. We can construct a new space of equivalence classes; thus will belong to the same equivalence class if for (Note that will actually be constant in since we showed that (3.1.3) implies that is uniformly continuous on Note that for the free rod, equation (3.1.5) does not hold, since but the analysis (3.1.6) does still hold if Thus we can show that is bounded and uniformly continuous on and therefore in All the which are constant in belong to the same equivalence class, which we will call the zero class. Now implies which means 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

74


Now

and equation (3.1.29) imply In either case the energy functional variant under the transformation satisfies the condition

in equation (3.1.26) must be inand this will be so iff

This we recognize as the condition of static equilibrium for the external loads applied to the rod. Provided (3.1.30) holds, we may make the necessary changes in the analysis given above to define the energy space for the free rod and the generalized solutions. Problem 3.1.6 Show that if the rod is free at both ends, the Principle of Virtual Work (3.I.14) gives the condition (3.1.30). Problem 3.1.7 Write down the Principle of Virtual Work for a cantilever rod with a distributed load and end load at Derive the classical end condition, 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 after Leonhard Euler (1707–1783) and Daniel Bernoulli (1700–1782), of a beam in flexure. In this simplified model, which is adequate for the analysis of thin straight beams in flexure in a principal plane, it is assumed that plane sections of the beam normal to its axis remain plane and normal to the axis of the deformed beam. If the beam is deformed in the plane, then the analysis is based on the assumptions that the elastic displacements of a particle at are given by

where is measured from the neutral axis of the section. If the beam has length and 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 neutral axis Again we choose units so that E = 1, suppose all quantities are dimensionless, 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 in satisfying (3.2.4), for which

Thus

We call the set S of all in satisfying (3.2.4), (3.2.5), the subspace of functions with finite energy for the bar. Again S is an incomplete inner product space; the completion of S in the metric (3.2.6) is called the energy space We now examine the elements in We can argue as in § 3.1 that if is a representative Cauchy sequence for then converges in the uniform norm to a limit function v(x) which is continuous in and We may now go further with Problem 3.2.1 Show that if is a representative Cauchy sequence for then is a Cauchy sequence in the uniform norm, and thus has a continuous limit Show that Thus we can say and 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 identified with a function in such a way that (3.2.9) holds. We say

76


and note that (3.2.9) states that the imbedding operator from into is continuous. We have shown that any element may be identified with a continuously differentiable function We can define a generalized second derivative for as the element in corresponding to the Cauchy sequence Problem 3.2.3 Show that the work functional

is continuous if define terms

and

Show also that if

we can

corresponding to work done by concentrated forces and moments. To analyze the beam we use the Principle of Virtual Work. The virtual work done by the stress on the virtual strain induced by the virtual displacement is, with E = 1,

Thus the Principle gives

for all sufficiently smooth

e.g. for

satisfying

In the classical analysis we assume

then two integrations by parts in (3.2.11) give

Now we argue as before. If satisfies terms in (3.2.14) vanish, and Lemma 3.1.1 states that ential equation

then the integrated satisfies the differ-

3.2 The Euler–Bernoulli beam

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

Now by choosing a

with

and by choosing a

with

77

satis-

we deduce that

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 beam problem is the satisfying

for

Alternatively, 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 a norm, for implies only i.e. We divide the into equivalence classes, putting into the same class if We call the class of all those for some the zero class. Alternatively, in any equivalence class we may choose the unique member satisfying

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

78


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 defined if 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 Principle of Virtual Work. Problem 3.2.7 Carry out the classical and the generalized analysis for a cantilever beam loaded by a concentrated force and moment at the end in addition to the distributed loading

3.3 The membrane Consider a taut membrane stretched with uniform tension T across a domain The strain energy of the membrane is

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

where is the boundary of We 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 for the clamped membrane We now examine elements To do so we establish Friedrichs’ inequality, due to Kurt Otto Friedrichs (1901–1982). Suppose is 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 functions by zero to give functions The inequality (3.3.6) applied to and is equivalent to the same inequality applied to and

Since for

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

and

i.e.

is imbedded in

Moreover if

i.e.

is a representative Cauchy sequence for then and are representative Cauchy sequences in for elements which we call and respectively. Thus if then U, and are all elements of Note that for the rod we showed that if then it was continuous in i.e. in For the membrane we have proved only that if then

80


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

For the classical analysis we assume that putting where and

and on

Then we deduce that

This is the equation we would get from the Principle of Virtual Work. In the classical 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).) Now the extension of Lemma 3.1.1) to shows that must satisfy the differential equation 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 Friedrich’s inequalities give

Thus we can write J(U) as

because Hölder’s and

3.3 The membrane

and proceed as in § 3.1 to find that the minimizer

for all

81

satisfies

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 necessarily zero on the boundary, so that on using the divergence theorem (3.3.11) we obtain Arguing as in § 3.1, i.e. first taking (3.3.12). Therefore,

for all

on

we find that

must satisfy

which implies

This is the natural, or so-called Neumann boundary condition for the free membrane. To derive the generalized solution for the free membrane we first note, as we expect, that (3.3.3) is not a norm on because implies only When we considered the rod we placed and in the same equivalence class if We could do that because we had shown that if then they were continuous on But now, if it is not necessarily continuous on it is an equivalence class of Cauchy sequences of functions in the norm (3.3.3). We must therefore proceed differently. We use Poincaré’s inequality, due to Jules Henri Poincaré (1854–1912). If then

for some constant m independent of Now suppose that norm (3.3.3), and that

is a Cauchy sequence of functions in

Then Poincaré’s inequality shows that and that

is a Cauchy sequence in

in the

82


Thus if then (see (2.11.6)); it is zero in moreover

so that so that

is zero almost everywhere does constitute a norm, and

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

Problem 3.3.1 Use the divergence theorem to show that (3.3.10) is a necessary condition for the existence of a solution of (3.3.12) satisfying the boundary condition (3.3.14). So far we have considered only static problems. Now consider the free vibration problem for the clamped membrane. If the membrane has uniform mass/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 the eigenvalues of the equations

of the membrane are related to

by the equation

Problem 3.3.2 Show that if makes

then the classical solution of (3.3.22)

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

where

3.4 The plate in bending

83

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

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

is the tensor

and we have used Einstein’s double suffix summation convention, due to Albert Einstein (1879–1955). We assume that the tensor is positive definite in the 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

Thus

i.e.

and the clamped boundary condition (3.4.3) yields and

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 Let us consider properties of elements Since on Friedrich’s inequality gives

84


But

on and then to

so that on applying Friedrich's inequality first to

we find

This means that if

is a Cauchy sequence in

then are all Cauchy sequences then it and its first and second generalized deriva-

in Thus if 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 by zero outside the plate, and suppose that lies in some square then

so that on using Hölder’s inequality we find

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

(We encourage the reader to study the analysis for the rod, beam, membrane and plate, 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 form of 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 a norm. Now we must use Poincaré’s inequality twice to obtain

3.5 Linear elasticity

For any function

we can find constants

85

such that

satisfies

We can thus take Cauchy sequences that

so that W, also is

composed of such functions and show

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

so

The argument which we used for the clamped plate to show that if then 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 so that, in particular, if then on

3.5 Linear elasticity Consider an elasticity body occupying a bounded region energy of the body in Cartesian coordinates is

The strain

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

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

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

such that, for

86


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

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

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

then so that (3.5.4) is an inner product. The completion of S in the norm given by (3.5.4) is the energy space for the body. To describe properties of elements of we establish Korn’s inequality. To prove this we need to apply the divergence theorem to derivatives of u; this is why we require Korn’s inequality. For a vector function

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

With the notation

so that

But

we have

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 obtain

to these integrals to

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 the first derivatives all belong to Note that the construction of the energy is the same if the boundary condition (3.5.5) is given only on some part of of the boundary, i.e.


88

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 circumvent the difficulty associated with the zero in the energy space we introduce the restrictions These make the ‘zero’ of the energy space actually zero, and ensure that Korn’s inequality remains valid for the vector functions.

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

Definition 3.6.1 A semi-norm, on a linear space X is a real valued function 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 and clamped 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 Clearly if is a representative Cauchy sequence for then is a Cauchy sequence s in for any such that We can therefore take Cauchy sequence to define elements which we label We recall that is the set of functions having continuous derivatives of all orders in and having compact support in i.e. their supports, which are closed, lie inside We define to be the completion of in the norm is a subspace of The spaces form the generalization of the energy spaces for the clamped membrane and plate. We will now show that the semi-norm | · | m,p is a norm on this will generalize what we found earlier: was a norm for was a norm for To do this we need an inequality which is a generalization of Friedrich’s inequality (3.3.5); it, like (3.3.17), is called Poincaré’s inequality; it is proved by generalizing the proof of Friedrich’s inequality. Poincaré’s inequality. Let be a bounded domain in constant C, depending on and such that

Moreover

defines a norm on

Proof. Suppose is the ‘box’ use the abbreviation

which is equivalent to N . Suppose

Then

since

There is a positive

Now use Hölder’s inequality to give

90


with

so that

Now integrate over

to give

and thus

This proves (3.6.6) for But is dense in so that the inequality holds for all If is not a box, then we can enclose it in a box and extend functions by zero to the box to give functions The inequality (3.6.6) applied to and is equivalent to the same inequality applied to and To show that defines a norm on we need only verify N1. Thus we need to show that if and then This follows immediately from (3.6.6). We also see that if and then so that F is zero almost everywhere. To see that and are equivalent norms on we note 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 space X has been imbedded in another space Y, according to Definition 2.11.4. In this section 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

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

91

and

while In § 2.5 we showed that under the norm (3.7.1) is a complete normed linear space, i.e. a Banach space. We can show similarly that is a Banach space. The first imbedding we note is one between and Clearly is 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 to the definition of it is the set of functions that are bounded and uniformly continuous on This means that, given we can find such that if and then The important condition which distinguishes uniform continuity from ordinary continuity is that we can find one which will fit any two We now introduce Definition 3.7.1 Let is the subspace of consisting of those functions which satisfy a Hölder condition, that is there is a constant K (depending on such that

(Note that in the notation of § 2.3.) Such functions are said to be Hölder, or Lipschitz continuous, after Rudolf Otto Sigismund Lipschitz (1832–1903), if Problem 3.7.1 Show that if satisfies (3.7.4), then it is uniformly continuous on Construct a function which is uniformly continuous on (0,1), but does not satisfy a Hölder condition, for any on (0,1). Note that there is no point in considering Problem 3.7.2 Show that if r necessarily constant. Theorem

3.7.1

in Definition 3.7.1, because

satisfies (3.7.4) with

is a Banach space with the norm

then

is


92

Proof. is defined in (3.7.1). We need to prove first that (3.7.5) does satisfy the norm axioms in Definition 2.8.1. This is straightforward. Secondly, we must show that a Cauchy sequence of functions has a limit in Suppose is such a Cauchy sequence, in the norm Equation (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 that implies

Choose Choose we can find M > N such that

Since

Thus

as

so that

But 1 is arbitrary, so that

Thus

sequence

converges to in the norm of and if then satisfies the Hölder condition with Therefore also satisfies the Hölder condition, so that the Cauchy converges to i.e. is a Banach space.

The inequality (3.7.6) gives the imbedding

We can generalize

according to


93

Definition 3.7.2 Let is the subspace of consisting of those functions with derivatives satisfying a Hölder condition, 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 inequality (3.7.8) gives the imbedding

is complete; it is a Banach space; the

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 show that behaves, in a way, like a fractional as increases from 0 to 1, an becomes nearly differentiable one more time, i.e. nearly in One would therefore expect that would be imbedded in i.e.

94


This is true for many domains e.g. all convex or star-shaped domains, and in fact for all domains for which there exists a constant M > 0 such that any two points may be joined by a piecewise straight line with total length 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 imbeddings for Sobolev spaces. Our first result was (3.1.10). For this we have a domain in with 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


95

The imbedding theorems all have the form

where X is another Banach space. The theorems show that if a function then the amount by which it is ‘constrained’ depends essentially on the value of in relation to the dimension N of the space Thus we will find that if we can say that must be bounded, or even uniformly continuous. As one would expect, some of the results hold for an arbitrary domain in others hold only if has special properties. We will not prove the theorems, nor will we state them in their full generality; for this the reader may see Adams (1975). The proofs are not so much difficult, as intricate; they require carefully chosen integrations by parts and applications 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 of domains. Theorem 3.7.2 Let and

be a domain in

if

then

and

then

if

if

if

be a nonnegative integer,

then

if

if

let

then

then

then


96

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 is a finite cone C such that each point is the vertex of a finite cone contained in and congruent to C. (Note that need not be obtained from C 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 have the Lipschitz property, according to Definition 3.7.4 The domain has the Lipschitz property if for each point there is a neighborhood in which the boundary is the graph of a Lipschitz continuous function, according to Definition 3.7.1. We now state Theorem 3.7.3 Let be a domain in a non-negative integer, and hold with replaced by Corollary 1. If

having the cone property, let be then the imbeddings (3.7.17)–(3.7.20)

is a non-negative integer, and

then

Proof. If thus

then for The assumed imbedding gives

Theorem 2.8.1 states that any two norms in

so that

are equivalent; thus


Corollary 2. If

97

is a non-negative integer, and

then

Proof. If thus

then for The assumed imbedding gives

so that

so that in

Theorem 3.7.4 Let be a domain in let be a non-negative integer, and (3.1.22) hold with replaced by Corollary

If

having the Lipschitz property, then the imbeddings (3.7.21)–

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).

98


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 a function; see (3.6.1), (3.6.5). Imbedding:

means

and

see Theorem 3.7.2.

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.

4. Approximation in a Normed Linear Space

They say there was a fish who said two words in such a strange language that for three years scientists have been trying to understand 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 people standing outside we can do one of two things: Count the number of chairs, whether

and the number of people,

and see

Start seating the people, and continue until all the chairs are filled, or all the people are seated, whichever comes first. The second procedure has the advantage that it avoids counting; it relies on establishing a one-to-one correspondence between chairs and people. This leads to Definition 4.1.1 Two sets are said to have equal power if there is a one-toone correspondence between their elements. The set of positive integers 1,2,3, ··· is the set containing an infinity of elements which has the least power. Definition 4.1.2 A set which has the same power as the set of positive integers is said to be countable (enumerable). Theorem 4.1.1 The union of a finite number, or a countable set, of countable sets is countable. Proof. It suffices to show how to enumerate the elements of the union. This is clear from the diagram in Fig. 4.1.1. The countable sets are etc. We take them in the order

100


etc. In this way we will cover all the elements in the union, and we can 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 coefficients is 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 a real 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 points form a continuum. We can extend this to say that the set of points in forms an 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 process only a finite set of numbers. An operator using a computer expects a solution to a problem to be approximated with a certain accuracy by the sequence of numbers used by the computer. If is an arbitrary element of an infinite set X and we want to use a computer to find an approximation to it, then we must be certain that every element of X can be approximated by elements of another set 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 countable subset which is dense in X. We call such a subset a countable dense subset. In other words, X is separable if there is a countable set such that for every there exists a sequence such that as Equivalently, for any there is, for each an element (depending on in an of The set of real numbers in [0, 1] is a separable metric space, since the set of rational numbers in [0, 1] is a countable dense subset. There is a more important example: Weierstrass’ theorem on polynomial approximation (Theorem 1.3.1) states that if is a bounded domain in then the set of polynomials (with real or complex coefficients) is dense in in the uniform norm. If is the set of polynomials with rational coefficients, then Problem 4.1.2 Show that

is dense in

in the uniform norm.

This means that is separable, because is countable. Putting this together with Weierstrass’ theorem, we can deduce that is a countable dense subset of so that is separable, again in the uniform norm. However, not all spaces are separable, because Lemma 4.1.1 The space of all bounded functions on [0, 1] equipped with the norm

is not separable. Proof. It is sufficient to construct a subset M of the space whose elements cannot be approximated by functions from a countable set. Let be an arbitrary point 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 intersection is empty. This means that every element of M is an isolated point; there is just one element in each ball of radius 1/3 about If a set S is to be dense in M, then each of these balls must contain at least one element of S. But the set of balls with real values of has the same power as


102

the continuum, i.e. it is not countable (Theorem 4.1.2). Therefore there is no countable set which is dense in M . Therefore M , and a fortiori (meaning ‘all the more so’) the set of bounded functions, is not separable. We will now proceed to show that the Lebesgue spaces Sobolev spaces are separable. We prove a general result:

and the

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 , the original incomplete space composed of elements the space of stationary sequences 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, it has a countable dense subset D. Let be the space of stationary sequences for If and then we can find such that Let and then

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

which,like

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 is sufficient to prove that is separable in But is dense in in 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 that is 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 arguments we can show that is dense in in the metric (2.3.4), and hence also 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 is separable. The result is of great importance, for the following reason. We have only a few convenient countable sets of functions which we may use to show that various spaces are separable: the space of polynomials with rational coefficients; the space of trigonometric polynomials with rational coefficients; etc. In general, the elements of these sets will not satisfy the boundary conditions imposed on functions in energy spaces, for example, so that we cannot use them to show that the energy spaces are separable. To circumvent this difficulty we can take a wider space, containing the space under consideration, and show that it is separable; Problem 4.1.3 shows that the subspace is separable. In § 3.6 we showed 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 problem of approximation in a normed linear space: Given and with find numbers to minimize

The problems of best approximation of a continuous function by an order polynomial, by a trigonometric polynomial, or by some other specified functions, all have this form. Our analysis will depend on a well known result from the theory of continuous functions, which we stated as Theorem 1.2.1. Now return to (4.2.1). We suppose that are linearly independent. This means that the equation

implies We show that the problem of minimizing (4.2.1) has a solution: we prove the existence theorem Theorem 4.2.1 For any and

there is an

depending on

such that

104

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 and are continuous in (or in if X is a complex space). The function real homogeneous function of degree 1, i.e.

First consider

on the unit ball

is a

This is a closed and bounded,

i.e. compact, set in so, by Theorem 1.2.1, the real continuous function will 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 of radius For the inequality (4.2.2) gives

On the unit ball, therefore, outside the ball of radius and But so the minimum value of must be inside the ball of radius R; since this is a closed and bounded set we see that the minimum will actually be attained; we can say min in the statement 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 relatively simple; we mean that the problem of solvability is simple, not that concrete applications of this theory are simple. We have shown that the problem of minimizing (4.2.1) has a solution; in a general normed linear space the solution is not unique; it is unique when X is a strictly normed space. Definition 4.2.1 A normed linear space is called strictly normed if the equality

implies

and

Most of the normed linear spaces which appear in applications are in fact strictly normed. First we have Lemma 4.2.1 An inner product space is strictly normed. Proof. Suppose complex space this may be written

so that

then

For a

But the Schwarz inequality (2.12.2) states that

so that

and thus and hence

But then Theorem 2.12.1 gives so that and

Now we pose Problem 4.2.1 Use the properties of the Minkowski inequality to show that the spaces 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 any two elements each element with is also in M. (Thus when are in a convex set M , the segment joining and lies completely in M.) Problem 4.2.2 Show that the closed unit ball normed linear space X is convex. When

the set M of elements

in

106


is a closed convex set, so that the problem of minimizing (4.2.1) can be viewed as finding which is closest to We now combine the notions of strictly normed space and convex set to give a uniqueness theorem. First, however, we combine Definitions 2.7.2 and 2.9.2 into Definition 4.2.3 Let X be a linear space. An operator (linear operator) from X into or is called a real or complex functional (linear functional).

Theorem 4.2.2 Let X be a strictly normed linear space, let be a closed convex set. There is no more than one minimizes the functional on M. Proof. If there is clearly only one minimizer, and that, if possible, there are two minimizers, and

Since M isconvex,

and let which

Suppose Thus

so that

On the other hand

Thus But X is strictly normed so that this equality implies and (4.2.3) implies so that is a contradiction, so that there can be at most one minimizer.

This

In this section we have shown that the problem of minimizing (4.2.1) always has a solution, and that this problem, as well as the more general problem of Theorem 4.2.2 has at most one solution if the space is strictly normed. Thus the minimizing 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 is a Hilbert space.

4.3 Riesz’s representation theorem Riesz’s theorem, due to Frédéric Riesz (1880–1956), is the most important of a number of results we shall obtain in this chapter about approximation in an inner 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 a closed 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 showed the existence in the special case in which M is a finite dimensional subspace of a normed linear space; now we will establish the existence for an arbitrary closed 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 that is a Cauchy sequence, then, because M is closed and a closed set in a complete space is itself complete, (Problem 2.5.3) it will have a limit in M , and this will be the minimizer. Since H is an inner product space, the parallelogram law (2.12.3) holds. Thus and therefore

Since we can write Since M is convex,

where

as

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 Thus there is a unique minimizer i.e.

Take an arbitrary

we consider the real valued function

108


of the real variable Because M is a subspace, has its minimum at Thus But

Replacing

by

This means that

we get

is orthogonal to every

for all

and

so that

and leads us to

Definition 4.3.1 Let H be a Hilbert space, a linear subspace, and The element is said to be orthogonal to M if is orthogonal to every i.e. for all Two subspaces are said to be mutually orthogonal if for all and We write Definition 4.3.2 Let H be a Hilbert space, and be mutually orthogonal subspaces. We say that H has an orthogonal decomposition into M 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-called decomposition 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 an orthogonal decomposition into M and N , as in (4.3.2).

Proof.

Suppose Let N be the set of all such that any is 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 not empty. N is a linear subspace because for all implies 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 The decomposition is unique because

4.3 Riesz’s representation theorem

109

implies and M , N are mutually orthogonal, i.e.

But

so that

and

The element is the projection of on M . We may consider the projection 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 . The projection 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 representation theorem: Theorem 4.3.3 Let H be a Hilbert space, and functional on H. There is a unique such that

be a continuous linear

and Proof. Let M be the kernel of the set of satisfying We show that M is a closed linear subspace. If then but F is linear so that and thus 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 uniquely represented as We now show that N is a one-dimensional space, i.e. any may be written as where is a fixed element in N . Let then But which means But M, N are mutually orthogonal so that being in both, must be zero. Thus and N is one-dimensional. Take an element and define by

Any element

can be represented uniquely as

110


and

Therefore

where If there were two representing element so that and taking i.e. Finally,

but

then we would find

so that

The proof above was given for a complex Hilbert space, but the result holds for real spaces also. The meaning of the theorem is that any continuous linear functional on H can be identified with a unique element The set of continuous linear functionals on H is called the dual of H , and is denoted by H*. Riesz’s theorem gives a one-to-one correspondence between elements and

4.4 Existence of energy solutions of some mechanics problems In this section we discuss some applications of Riesz’s theorem. We recall that in Chapter 3 we introduced generalized solution for several mechanics problems and 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 were some restrictions on the forces to ensure the continuity of the linear functional in the energy space.) The following theorem concerns the solution of these generalized problems. Theorem 4.4.1 Let be a continuous linear functional on a Hilbert space H . There is a unique element which satisfies (4.4.1) for every

Proof. By Riesz’s representation theorem, there is a unique element such 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 unique solution of (4.4.1).

iff

i.e.

This is the

Now let us consider another application of Riesz’s theorem. In Problem 3.3.2 we set up the integro-differential equation

as the generalized statement of the eigenvalue problem (3.3.22). We need to find and corresponding so that satisfies (4.4.2) for every Problem 4.4.1 Show that if then is real.

and

satisfy (4.4.2) for every

First we consider the term

the inner product of functional in for

in the space for fixed as a linear The Schwarz inequality (2.12.2) gives

while Friedrich’s inequality (3.3.7) gives

This inequality states that By Riesz’s theorem,

is a continuous functional in the Hilbert space has a unique representation

where, from now on, we implicitly take all inner products and norms in What have we found? For any there is a unique element such that (4.4.4) holds. The correspondence defines an operator acting 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


112

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 a contracting mapping. The contracting mapping theorem (Theorem 2.7.1) states that has a single fixed point, which, since K is linear, is This means 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 free vibration of a membrane (with fixed boundary) with variable, but bounded, mass density. Show that the problem can still be reduced to the form (4.4.7), where K is a continuous linear operator in Problem 4.4.3 Carry out the analysis of the free vibration of a plate, with clamped boundary, and with variable, but bounded, mass density. Show that the problem can be reduced to the form (4.4.7), where K is a continuous linear operator in

4.5 Bases and complete systems If a linear space X has finite dimension there are linearly independent elements called a basis for X , such that every element has a unique representation

where the are scalars. We now generalize this definition to an infinite dimensional normed space X . Definition 4.5.1 Let X be a normed linear space. A system of elements is 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

It is clear that a basis equation

then

is a linearly independent system since the

has the unique solution Problem 4.5.1 Show that if a normed linear space X has a basis, then it is separable. (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

114


uniform convergence. We ask whether the powers C[0, 1]. If they did, then any continuous function expanded as a uniformly convergent power series

form a basis for in C[0, 1] could be

in [0, 1]. But this means that is analytic, and there are clearly continuous functions which are not analytic. Therefore the powers do not constitute a basis for C[0, l]. Problem 4.5.2 Construct a function as a uniformly convergent power series.

which cannot be expressed

Even though the powers do not form a basis for C[0, 1], Weierstrass’ theorem states that they do have properties similar to those of a basis: we can find a polynomial 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 system is 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 of a normed linear space X . This simply means that for any and element of S, we can find a finite linear combination of elements of the system such that the 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 to make we simply take large enough, i.e. take more of the For the latter, the values of the as well as the value of will depend on if one set of coefficients makes (4.5.2) true for one value of and we decrease 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 in C[0, 1], and generally, that the composite powers are complete in where Problem 4.5.3 Generalize the last result to find a system which is complete in and when

The problem of the existence of a basis for a particular normed linear space can be very difficult, but there is a special case when this problem is fully solved:


115

when the space is a separable Hilbert space. Those who are familiar with the theory of Fourier Series, after Jean Baptiste Fourier (1768–1830), will see that it 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 said to be orthonormal if, for all integers

is

There are many advantages in using an orthonormal system of elements as a basis. If we have an arbitrary linearly independent system of elements in a Hilbert space H , they will span a subspace We may form an orthonormal basis for by using the familiar Gram–Schmidt process, named after Jórgen Pedersen Gram (1850–1916) and Erhard Schmidt (1876–1959):

1. 2.

so that so that

3. Applying the Gram–Schmidt procedure to subsets of monomials in the spaces L(a, b) , we get systems of polynomials that are called orthogonal polynomials. Orthogonal polynomials are widely used in mathematical physics. Problem 4.5.4 Show that constructed in the Gram–Schmidt process will be orthonormal iff are linearly independent.

If H is a, separable Hilbert space, then, by definition, it has a countable dense subset From this we may, by the Gram–Schmidt process, construct an orthonormal set which is dense in H ; this will be a complete orthonormal system in H . Although there are Hilbert spaces which are not separable, the important ones, and are separable. The following theorem is based on the premise ‘If H has a complete orthonormal system’. Theorem 4.5.1 Let H be a Hilbert space. If H has a complete orthonormal system then it is a basis for H ; any element has a unique representation

116


called the Fourier series for

The numbers

are called the

Fourier coefficients of Proof.

First we consider the problem of best approximation of an element by 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

which shows that Thus

so that

takes its minimum value when

i.e. when

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 , after Friedrich Wilhelm Bessel (1784–1846),


This means that the sequence of partial sums

117

is a Cauchy sequence, for

We have not yet used the completeness of the system so that, in particular, the results (4.5.6), (4.5.7) hold for any orthonormal system H . The completeness of the system means that if and we can 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 is the meaning of equation (4.5.3). When the system is complete, we can sharpen (4.5.6). Indeed equation (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 closed in H if the system of equations

is said to be

implies Theorem 4.5.2 An orthonormal system in a Hilbert space H is closed iff it is complete. Proof. Suppose the orthonormal system with then

is complete in H , and that

118


so that the system is closed. Now suppose that with

and

is closed. The system

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 (4.5.3), so that is actually a basis for H .

i.e.

is given by

Problem 4.5.5 Show that an orthonormal system in a Hilbert space H is closed iff it is a basis for H .

Problem 4.5.6 Show that if a system is complete in a set S that is dense in a Hilbert space H , then it is complete in H . (Hint: For any element there is an element that is closer than to For there is a finite linear sum of the system elements whose distance from s is less than The distance between and is less than which means that the system is complete in X). An important application of this application concerns tion this is complete; it can be obtained as the completion of norm. The functions

are orthogonal in and

By definiin the

Thus Bessel’s inequality states that if

then

One of the consequences of the convergence of the infinite series on the left is that

But

so that if

is real then


119

i.e.

The results in (4.5.12) are usually given the name, the Riemann–Lebesgue lemma; 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 in in which the system is complete. The set of functions with compact support in is dense in . These functions are continuous on and so, since supp is closed and bounded, are uniformly continuous in Since these functions satisfy they may be continued to the whole real line as functions of period We may therefore apply Weierstrass’ trigonometric polynomial approximation theorem (Theorem 1.3.3) to them. This means that, given we may find a trigonometric polynomial of the form (1.3.6) which we may write in the form

such that This completes the proof. If

and then, where

then we may extend

is given by (4.5.11),

Similarly, if we take

by taking

120


then we find

Theorem 4.5.4 A Hilbert space has a countable orthonormal basis iff it is separable.

Proof. Problem 4.5.1 shows that if it has a basis it is separable. On the other hand if H is separable it has a countable set which is dense in H ; we now apply the 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 components 1,2, · · ·, N. The sequence converges iff each of the sequences converges. In a Hilbert space with orthonormal basis the Fourier coefficients play the part of components, but now there is a difference between 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 does not converge, since

components of

tends to zero, but

itself

We need to introduce a new kind of convergence. For a general normed linear space we have Definition 4.6.1 The sequence in a normed linear space X is said to be a weak Cauchy sequence if, for every continuous linear functional on X, the sequence is a Cauchy sequence, in The sequence is said to converge weakly to if, for every continuous linear functional on X ,

We use convergence.

for strong convergence, i.e.

for weak

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 normed linear space, as in Definition 4.6.1, we shall usually consider it only in an inner product 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 sequence X cannot have two distinct weak limits.

If X is a complete inner product space, i.e. a Hilbert space H , then Riesz’s representation theorem (Theorem 4.3.3) states that every linear functional on H has the form (4.6.1) for some This yields Theorem 4.6.1 Let H be a Hilbert space. A sequence is a weak Cauchy sequence if, for every is a Cauchy sequence. The sequence 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, to establish weak convergence rather than strong convergence. This is why the last result is important, and why weak convergence will be a major preoccupation in this presentation. Problem 4.6.3 Show that in a finite dimensional Hilbert space, implies This implies that in a finite dimensional space weak convergence and strong convergence are synonymous. Theorem 4.6.3 A weak Cauchy sequence

in a Hilbert space is bounded.

122


Proof. Suppose, on the contrary, that is a weak Cauchy sequence which is unbounded. Let denote the closed ball with center radius We show that if then there is a sequence of points such that We take

then

so that

and

since the numerical sequence is bounded, because is weakly convergent. We now obtain a contradiction. Take By the above argument, we can find and such that

By the continuity of the inner product, we can find a ball such that this inequality holds for all

Now apply the same argument to ball such that

and find

with

and a

Repeating this procedure ad infinitum , we find a nested sequence of closed balls such that

Since H is a Hilbert space, there is at least one element each and

which belongs to

Thus we have a continuous linear functional for which is 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 there is a and a subsequence such that as

Proof. Let us introduce the sequence For any with unit norm the numerical sequence is bounded and thus we can select a convergent subsequence. If there exist such a unit element and subsequence for


123

which then the statement of the Corollary is valid for the subsequence and if or if If, on the other hand, we cannot find such and then we have as for any which means that tends weakly to zero. In this case we demonstrate the Corollary using the second part of the proof of the above theorem. In it, the existence of an element and subsequence such that as is a consequence of two facts: 1.

is unbounded, which is the case;

2. the numerical set

when

runs over any

is unbounded.

The proof of the latter we give under the additional condition that as and this will complete the proof of the Corollary. First, the element belongs to

Since needed.

is finite and

as

Next,

we have

as

We will use the corollary to prove the Principle of Uniform Boundedness, contained in Theorem 4.6.4 Let be a family of continuous linear 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 contradict (4.6.2).

such that

which would

Problem 4.6.4 Use Theorem 4.6.4 to prove that if is a sequence of continuous linear functionals on H , and if for every the sequence is a Cauchy sequence, then there is a continuous linear functional on H such that


124

and

The following theorem gives a convenient check for weak convergence. Theorem 4.6.5 A sequence H iff 1.

is a weak Cauchy sequence in a Hilbert space

is bounded in H , i.e. there is a M such that

2. for any sequence

from a system which is complete in H, the numerical is a Cauchy sequence.

Proof. The necessity of the conditions follows from the definition of weak convergence and Theorem 4.6.4. Now we prove the sufficiency. Suppose conditions 1 and 2 hold. Take an arbitrary continuous linear functional defined, because of the Riesz representation 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 sequence, we can find a number R such that

is a Cauchy

so This means the sequence

is a Cauchy sequence.

Problem 4.6.5 Show that a sequence 1.

is bounded in H ;

is weakly convergent to

in H iff


2. for any

from a system

125

which is complete in H ,

Since weak convergence differs from strong convergence we need to define the terms weakly closed and weakly complete Definition 4.6.2 Let X be a normed linear space. A set is said to be weakly closed in X if all its weak limit points are in S . Thus if then implies

Definition 4.6.3 Let X be a normed linear space. X is said to be weakly complete if every weak Cauchy sequence (Definition 4.6.1) converges weakly to an element

We first prove the important Theorem 4.6.6 A Hilbert space (a complete inner product space) is weakly complete.

Proof. Suppose is a weak Cauchy sequence. For any we may define the linear functional Theorem 4.6.5 states that is 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 weakly closed.

Let S be the (strongly) closed ball as in the theorem. Then is weakly closed. We now prove

and

Suppose i.e.

and and S

Theorem 4.6.7 Let X be an inner product space. A weakly closed set is closed. A closed set need not be weakly closed.

Proof. Let be a (strongly) convergent sequence in S converging to We need to prove that The sequence converges weakly to because, if is any continuous linear functional on X , then


126

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 if then

If therefore we take then

But

since

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 be weakly closed, we can use the orthogonal decomposition in Theorem 4.3.2 to obtain: Problem 4.6.6 Show that a (strongly) closed subspace M of a Hilbert space H is weakly closed.

There is also the more difficult Problem 4.6.7 Show that a (strongly) closed convex subset S of a Hilbert space H is weakly closed.

The corollary to Theorem 4.6.6 is an example of this; a closed ball is a closed convex set.

4.7 Introduction to the concept of a compact set We introduced the concepts of weakly closed and weakly complete in § 4.6. Now we introduce Definition 4.7.1 Let X be an inner product space. The set is said to be weakly compact if every sequence in S contains a subsequence which converges 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 it is bounded and weakly closed.

is weakly compact iff

Proof. We will show that if it is bounded and weakly closed, then it is weakly compact, i.e. that any sequence contains a weakly convergent subsequence. Since S is weakly closed we know that the weak limit of such a subsequence will be in S. Let be a sequence in S , and M be the closed linear subspace spanned by Since M is a closed linear subspace of H we may (by Theorem 4.3.2) decompose H into M and N which are mutually orthogonal. If then we can write where and If then if then Thus it is sufficient to consider for The subspace M , being a closed subspace of a Hilbert space, is a Hilbert space (Problem 2.12.4). It is clearly separable, and so has an orthonormal basis By Theorem 4.6.5 it is sufficient to show that there is a subsequence of such that, for each the numerical sequence is convergent. We proceed as follows. The sequence in is bounded and therefore contains a convergent sequence The sequence is bounded and therefore contains a convergent sequence. Continuing in this way we obtain, at the ith step, a convergent sequence The subsequence is such that, for each fixed the sequence is convergent. Therefore is a 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 bounded and weakly closed. Look back at Theorems 4.2.2 and 4.3.1. They show that if H is a Hilbert space and is a closed convex set, then if there is (existence) a unique which minimizes on M . Problem 4.6.7 states that 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, we can use the weak compactness of a weakly closed and bounded set in a Hilbert space to provide a separate proof. Thus we have Problem 4.7.2 Let H be a Hilbert space, let and weakly closed set in H , then there is a

If, in addition M is convex, then

is unique.

and let M be a bounded such that

128


We will use the concept of weak compactness in our discussion of the Ritz procedure 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 Hilbert space H. Thus let H be a Hilbert space, M be a closed subspace of H , and Find the unique minimizer of

for

We 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 independent. Let be the subspace spanned by Theorem 4.2.1 states that there is an which minimizes on call one such minimizer 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 entiable,

Thus Writing

takes a minimum value at

and since

is differ-

is orthogonal to each

we obtain a set of simultaneous linear equation for the

namely

Since are linearly independent the solution to this equation is unique. For if it were two solutions

4.8 Ritz approximation in a Hilbert space

129

their difference

would satisfy Thus

so that

linearly independent, this means Hence the solution is unique.

But since the

are

and thus

Step 2. An a priori estimate of the approximation An a priori estimate is one which can be obtained without actually knowing the 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, contains a 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.


130

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

This implies that

for

and

we obtain

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 to We will show that the whole sequence converges weakly to and then that it converges strongly to Suppose, if possible that does not converge weakly to This means that there is an such that does not converge to Remove from any subsequence such that converges to Rename the remaining sequence Since the set is bounded, it has, by the Bolzano–Weierstrass theorem, a convergent subsequence and the limit of this sequence will not be Thus

But for the sequence we can repeat Step 3, and find a subsequence which is weakly convergent to a solution of the problem. Theorem 4.3.1 states that this minimizer is unique, Thus

This contradicts (4.8.5). Thus

converges weakly to

i.e.

4.8 Ritz approximation in a Hilbert space

Now we prove that tion (4.8.1) states that

converges strongly to

i.e.

131

Equa-

Thus

so that But so that tion (4.8.4) states that

But Therefore,

so that equa-

Now we may use Theorem 4.6.2 to state that To 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 by Riesz’s representation theorem, we can write

so that Since is fixed, the problem of minimizing imizing

is equivalent to that of min-

for

This problem has the unique, obvious, solution To apply the Ritz method we suppose, as before, that is a complete system in H such that any finite set is linearly independent. We take the Ritz approximation as

and find the equations

for tional

Note that we express in terms of the given funcThe result of the earlier analysis gives us

132


Theorem 4.8.1 For each equations (4.8.7) have the unique solution When is a continuous linear functional, the sequence of Ritz approximations defined by (4.8.6) converges strongly to the unique minimizer of the 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 the analysis given here provides justification for the application of the Ritz method to these problems.

4.9 Generalized solutions of evolution problems Consider the heat transfer equation

Here

is the temperature, the time, and the position in 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 that as a function of and as a function of i.e. it has 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) by assuming that satisfied the restrictive conditions we stated, but we now consider it in its own right. In what space(s) should we treat it? The functions are defined for and i.e. the domain and they 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 therefore introduce 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 problem (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 establish an 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 then Friedrich’s inequalities, we find

Now use the elementary inequality

to obtain


134

Now consider the first term in (4.9.9). If and, for every we have

then

and

Thus

Now use the triangle inequality

This inequality holds for and it shows that

to give

but it therefore holds for

is a uniformly continuous function of on (0, T) and may therefore (by Theorem 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


135

which we may rewrite as

This is the needed a priori estimate. It shows that the generalized solution is unique. For if there were two generalized solutions and then their difference, would satisfy equations (4.9.9), (4.9.3) with F = 0 and 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 we can always add a function which is zero a.e., to the solution.) We could use the Galerkin procedure on the domain Q , but instead we will separate the variables, using a complete system of functions in space to reduce the partial differential equation (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 Hilbert space, and therefore, by Theorem 4.5.3 it has a countable orthonormal basis Apply the Gram–Schmidt procedure to this sequence to construct a system 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. Faedo and Boris Grigor’evich Galerkin (1871–1945). To do that we return to equation (4.9.4), take given by (4.9.13) and to obtain the equation

where We define the which minimizes

Faedo–Galerkin approximation as the solution of (4.9.14)

136


over

for If then the general theory of first order differential equation with constant coefficients shows that, for given equation (4.9.14) for has a unique solution which is continuous in [0, T]. The equation (4.9.14) then shows that We now show that we can establish properties of and more importantly of when Suppose that Multiply equation (4.9.14) by and sum over 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 inequality 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 also. To show this, we multiply (4.9.14) by

is bounded and sum over

Thus


137

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 a closed linear subspace of the complete space Problem 4.6.6 and Theorem 4.7.1 show that is weakly compact, i.e. it contains a subsequence which converges weakly to Return once more to equation (4.9.14), and suppose

Multiplying (4.9.14) by we find

and summing over

and integrating over time,

138


Since, as we have shown, all the integrals are continuous functions with respect to in W , we can pass to the limit in the subsequence and find

for any Hence

given by (4.9.17). But, by Problem 4.9.1, such

are dense in W.

for all But this is equation (4.9.5), so that satisfies the first of the two conditions for the generalized solution, stated in Definition 4.9.1, and therefore, as we showed earlier, satisfies (4.9.8). Now we may repeat step 4 of § 4.8 to show that the whole sequence converges weakly to in W . Actually, the convergence of the approximation is stronger than we have established. 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 the subspace of product

which is the completion of satisfying (4.9.2) in the norm corresponding to the inner

Show that this is a proper inner product, and that a sequence which converges weakly 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 illustrates how the spaces in which a set of approximations to a given problem converges weakly, or strongly, must be chosen to fit the problem under consideration.


139

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 written Theorem 4.3.3. Orthogonal decomposition of H: Basis:

Complete: Closed : :

Definition 4.3.2.

Definition 4.5.1.

is complete if

Definition 4.5.2.

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 : rem 4.6.1.

as

Theo-

: 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 converging weakly to Definition 4.7.1 : in H , S is weakly compact iff S is bounded and weakly closed. Theorem 4.7.1

5. Elements of the Theory of Linear Operators

Introduce public education with moderation, avoiding bloodshed if possible. (From Service Regulations relating to the Kindness of Mayors 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 only those parts which we shall use in later applications. We recall the basic definitions from § 2.9 of an operator, and in particular a 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 constant c 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 a normed space, so that implies This holds for all so that A = 0. N2: This is evident. N3: The chain of inequalities


142

implies

As in any normed linear space, we can introduce the notion of convergence

in Definition 5.1.1 The sequence of continuous linear operators is said to converge to A if as in such a case we shall say that converges uniformly to A. (Note that c in (5.1.1) is a uniform bound for Theorem 5.1.1 If X is a normed linear space and Y is a Banach space, then is a Banach space. Proof. We recall (Definition 2.8.9) that a Banach space is a complete normed linear space. Let be a Cauchy sequence in , i.e. We need to show that there is a continuous operator For any

the sequence

But Y is complete, so there is

such that

is a Cauchy sequence in Y , since such that

Thus to every there is a such that (5.1.3) holds. This correspondence defines a linear operator A on X into Y , such that It now remains to show that i.e. A is continuous. To do this we note that since is a Cauchy sequence, the sequence of norms is bounded, i.e. and so

which means that A is continuous, i.e. plete. In the Banach space its sum by convergent if the numerical series

and

is com-

we can introduce a series

and define

A series

is said to be absolutely

is convergent.

5.1 Spaces of linear operators

143

Problem 5.1.1 Let X be a normed linear space, Y be a Banach space, and the sequence

In

Show that if

is absolutely convergent, then

is uniformly convergent.

Now consider operators A on X into X , and denote we can introduce the product of operators by

by

The product possesses the usual properties of a (numerical) product, except commutativity:

where I is the identity operator. If A and B are continuous, then so is AB , since

and so Problem 5.1.2 Suppose respectively. Show that If

then we write

converge uniformly to converges uniformly to AB . 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 considered has been uniform convergence, in the uniform norm (5.1.2). Now we will consider a weaker convergence, which is called pointwise or strong convergence. (It is called strong because it is stronger than yet a third, weak, convergence, but it is actually weaker than uniform convergence, as we will soon show.) Definition 5.1.2 A sequence if, whenever

If

is said to converge strongly to

converges uniformly to A , then it converges strongly to A , for


144

To construct an example to show that strong convergence does not necessarily imply uniform convergence we suppose that X = Y = H , a separable Hilbert space. By Theorem 4.5.1, and the remark preceding it, H has an orthonormal basis Define the orthonormal projection operator

from H onto the subspace of H spanned by basis for H , we have

Since

is a

This means Thus the sequence

converges strongly to I. But if

then

so that This means that the sequence a uniformly convergent sequence.

is not a uniform Cauchy sequence; it is not

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 . We shall 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 a normed space X , whose range lies in a Banach space Y , and which is bounded on D(A), i.e. for all Then there is a continuation or extension of A to X , denoted by

1.

for all

2. 3. where

is defined as

such that

5.2 The Banach–Steinhaus theorem

Proof. If construct such that

145

then Suppose but We as follows. Since D(A) is dense in X , there is a sequence The sequence is a Cauchy sequence, since

Y is a Banach space, it is therefore complete, so that the sequence has a limit, which we call (Show that the limit is independent of the choice of sequence Now it suffices to prove 3. The norm of A, is the infimum of M such that for Thus

and passing to the limit we find

which means that is continuous and so that

But on D(A) we have

This theorem is important. It states that if Y is a Banach space, then A may 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, is complete also (Problem 2.8.3). From now on we shall suppose, unless we state otherwise, that D(A) = X , i.e. that Now we prove the Banach–Steinhaus theorem, after Banach, and Hugo Dyonis Steinhaus (1887–1972) Theorem 5.2.2 Let X be a normed linear space, Y a Banach space and a sequence of continuous linear operators in such that

1.

for all

2.

exists for all

then the sequence for all

a subspace which is dense in X ,

converges strongly to an operator as

Proof. The linear operator A, defined on

is bounded on

i.e.

by the relation

indeed 1 implies

Using the construction of Theorem 5.2.1, we can extend A to X , keeping the norm unchanged. We call this extension and will now show that if then

146


Let be such that On the other hand, since

then by definition we have

Choose we can find N such that because of 2 we can find such that This means that when

Take when

When X in Theorem 5.2.2 is a Banach space, we can replace condition 1 by the statement that, for all the set according to the principle of uniform boundedness: Theorem 5.2.3 Let X be a Banach space, Y a normed linear space and If for every the set is bounded, then the set is bounded. Proof. If there is a closed ball on which the bounded, i.e. there is a constant K such that

then

is in

is uniformly bounded on X . For if

so that

and

are uniformly

is in X , then

Thus

so that Now we must show that we can find a ball on which the are uniformly bounded. 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 continuity there is a ball with such that for all The are not uniformly bounded in this ball. Therefore, there is an

5.3 The inverse operator

and such that with and this way we find sequences points such that

But

when and

147

and thus a ball Proceeding in such that and

is a Cauchy sequence in the Banach space X so that it has a limit this limit point is in each so that

which contradicts the statement that

is bounded on X .

5.3 The inverse operator 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 X into Y . If for any there is no more than one such that then A is said to be a one-to-one operator. In this case the correspondence from Y to X defines an operator; this operator is called the inverse of A, and is denoted by Problem 5.3.1 Show that

and

Problem 5.3.2 Show that the operator exists iff the equation has the 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 from X into Y . Show that if exists and dim then dim We are interested not only in the solvability of equation (5.3.1), but also in knowing whether the inverse operator is continuous. There is a simple result given by Theorem 5.3.1 The operator is bounded on R(A) iff there is a constant c > 0, such that, if then

148


Proof. Necessity. Suppose exists and is bounded on R(A). Then there is a constant such that for all Putting and we get (5.3.2). Sufficiency. If then (5.3.2) implies i.e. Thus has 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 will be 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 equation 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 inverse of the operator A? Without loss of generality we can assume that are linearly independent. If equation (5.3.3) is soluble, the solution has the form

Substituting this into (5.3.3) and equating the coefficients of find the equations

where

to zero, we


Provided that the determinant of coefficients, zero, we may solve equations (5.3.4) to give

149

of equation (5.3.4) is not

so that

In this equation each

Since hence, when

is a linear combination of

we have we can find

and

for some c > 0, and 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 order there 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 operator does not exist. The compose the spectrum of the integral operator (see Chapter 6). Now consider a simple boundary value problem

where

Its solution is

150


We can phrase this problem in terms of linear operators. The direct problem is: given differentiate it twice to obtain f. This operator which takes to form f we can call A; its domain is and its range is a pair consisting of a function and a vector such that The inverse operator is that which obtains from 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 bounded.

in this example exists and is

Theorem 5.3.2 Let X and Y be Banach spaces. Suppose A is continuously invertible and Then A + B has the inverse and

Proof. The equation can be reduced to

By the condition of the theorem The contraction mapping theorem (Theorem 2.7.1) shows that equation (5.3.8) has a unique solution for any This means that the inverse exists, and its domain is Y . We now obtain the inequality (5.3.6). From it follows that and 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 iteratively: so that


151

Theorem 5.3.1 gives a necessary and sufficient condition for an operator to 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 linear operator 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 continuous linear operator on X onto Y, then is a continuous linear operator on Y onto X. Proof.

is a linear operator, for if there is an such that and 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 continuous at 0. Thus we must show that if there is a sequence such that and then Suppose This means that for some given any N we can find an for which The open mapping theorem states that the open set around zero in X is mapped onto an open set around zero in Y. Since this set is open, there is an open ball in it. Since we can find such that if then Thus if is the map of an in On the other hand we showed that if then for some is the map of an in Thus is the map of two distinct so that A is not one-to-one, contrary to our hypothesis. Thus and is continuous . Note that the condition that A be an operator on X onto Y may be replaced by the statement that R(A) is closed in a Banach space Y, for then R(A) will itself be a Banach space, and A will be an operator on X onto R(A). If A is to have an inverse, then N(A) must be empty. If X, Y are Hilbert spaces and then we may circumvent this constraint by decomposing into N(A) and and considering the restriction of A to 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 Banach space with respect to two norms and and if there is a constant such that for all then there is a constant

such that

for all

152


i.e. the norms

and

are equivalent.

5.4 Closed operators Closed operators have somewhat limited interest, and the reader is advised to skip 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 A from 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, by using Theorem 5.2.1, we may continue A to the closure of D(A); then if we define Thus a continuous linear operator is closed. To show that a closed linear operator need not be continuous we take the following counter-example. Let X = Y = C[0, 1] under the uniform norm, and let A be the differentiation operator: We show that A is closed. The domain of A is Suppose uniformly and uniformly. We need to show that is differentiable, and that Suppose that and consider

Clearly

To bound the magnitude of the first term we use the mean value theorem, which states that for some and find

between a and b. We apply the theorem to the function

for some

between

and

and hence

5.4 Closed operators

Now choose such that

For such

Since implies

converges uniformly to

we can find

the inequality (5.4.1) shows that

But find

as

The function

is differentiable at c so that there exists implies

The sequence implies

153

so that taking the limit

converges to

Choose

then if

in (5.4.2), we

so that we can find

such that

such that

we have

Note that this is a proof of the familiar result: A series (i.e. a sequence of partial sums) of differentiable functions can be differentiated term by term if the differentiated series converges uniformly. Thus is closed, but it is not continuous because if then, on [0,1],

so that is unbounded as Now 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 product space X x Y is a normed linear space with elements where in which

A possible norm for X x Y is

154


All the norms (5.4.3) are equivalent. Definition 5.4.3 Let A be a linear operator from X into Y. Thus and 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 if its 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 a closed operator is not necessarily continuous. However, now we prove the dosed graph theorem: Theorem 5.4.1 Let X and Y be Banach spaces and let A be a linear operator on X into Y (i.e. D(A) = X, but If A is closed, then it is continuous. Proof. The graph G(A) is a closed linear subspace in the product space X × Y with the norm (i.e. with in (5.4.3)). But a closed linear subspace of a normed linear space 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 means that 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 satisfy the conditions of this theorem. The domain of A is and this is not a Banach space under the uniform norm for C[0,1]; it is not a closed linear subspace of X = C[0,1]. However, note

5.4 Closed operators

Problem 5.4.2 Let X be the subspace of functions and let so that

155

satisfying

Show that A is a closed linear operator from X onto C[0, 1], but is not continuous. 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) onto C[0,1], in agreement with Theorem 5.4.1 and that is continuous, in agreement with Theorem 5.3.4. Problem 5.4.4 Let X and Y be normed linear spaces, and A be a closed linear operator 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 linear operator from X into Y. Show that if D(A) and R(A) are closed in X and Y respectively, 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. (Note that 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 operator could be extended to the closure of its domain D(A). We now consider whether and how a general (i.e. not necessarily bounded) linear operator may be extended so that it becomes a closed linear operator. First, we point out once again the difference between a continuous and closed operator. Suppose is a Cauchy sequence in D(A). If A is a continuous linear operator, then is 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 a useful deduction from the definition of a closed operator. Suppose and are two Cauchy sequences in X both converging to then and cannot converge to different limits. For according to the definition,

This last condition by itself does not imply that A is closed, but it does ensure that A has a closed extension according to


156

Lemma 5.4.1 Let X and Y be Banach spaces and A be a linear operator from X 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. Then imply 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 condition of the lemma, this is uniquely determined by we call it B is clearly a linear operator; we show that it is closed. Let be a sequence such that and We must show that How do we construct For every we choose a sequence such that and define From each

as the unique element in Y such that sequence

we choose a member

which we call

such that

Then But then, by the way we constructed B, we have that B is closed.

and

so

As an application of this lemma we consider the extension of the operator

with coefficients where is a domain in We consider A from into The domain of A is the range of A is in We examine the conditions of the lemma and show that 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 of the lemma, i.e. both in the norm. For any integration by parts to obtain

we can perform successive

5.5 The adjoint operator

because all the integrated terms involving with boundary. Now we take the passage to the limit in

157

vanish on the 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 the theorem are fulfilled and A has a closed extension. The approach we have used here can bring us to generalized derivatives; it is equivalent to that used by Sobolev.

5.5 The adjoint operator We shall introduce the idea of adjoint operator for an operator from a Hilbert space into a Hilbert space although it can be considered for operators acting from a general normed space X into a normed space Y . Let A be a continuous linear operator on into , i.e. in The inner product is, for fixed a functional on Because 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 defines an operator A* on into i.e.

The correspondence

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

158


Proof. Using the Schwarz inequality, we get

But using (5.5.1), we may write

Putting

we have

Thus is continuous and

so that A* is continuous. Since A* is continuous, (A*)*

Thus so that

for all

Therefore,

for all

and

Applying the first part of the proof to A*, we find

Definition 5.5.1 Let be Hilbert spaces, and A an operator in The null space of A, denoted by N(A) is the set of such that N(A) is a closed subspace of and is thus a Hilbert space. orthogonal complement of N(A) in Theorem 5.5.1 Proof. Suppose such that

so that Therefore

and

i.e.

This means that there is a sequence If then and Thus


it is in

is the


159

Now suppose We will show that is a closed subspace of the Hilbert space By equation (4.3.1) there is a unique element in such that

and

for all so that, in particular,

But since

Put

for all

it is zero for all

Therefore so that not orthogonal to a non-zero element and so second part of the theorem. Problem 5.5.3 Let Show that

then

for all

and thus

that is

Thus equation (5.5.3) states that is y is not in Therefore By changing A to A* we obtain the

be Hilbert spaces, and A an operator in

The first half of this proof, other words, if the equation has a solution for a certain of For

implies

then

In

must be orthogonal to all solutions

Thus is a necessary condition for (5.3.1) to have a solution. It will be a sufficient condition only if R(A) is closed, so that In this case the so-called Fredholm alternative holds: either the equation so that solution. or the equation equation i.e.

has a solution for all i.e. the equation

i.e. has no nontrivial

has solutions, in which case the has a solution iff is orthogonal to all the solutions

Problem 5.5.4 Show that if are Hilbert spaces and R(A) is closed iff R(A*) is closed.

then

160


This means that when R(A) is closed there is a similar Fredholm alternative for A*. Definition 5.5.2 The operator A in a Hilbert space H is said to be selfadjoint if A* = A. Theorem 5.5.2 Let H be a Hilbert space, and A a continuous self-adjoint linear operator in H, then

Proof. Write

If

Using the Schwarz inequality, we get

then

so that by the definition of for all

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

Take

and all real

Putting

to obtain

or

Thus

which with (5.5.5) yields

we obtain


161

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; the sequence is said to converge weakly to (We write if for every A continuous linear operator in a Hilbert space H maps a strongly convergence sequence into a strongly convergent sequence Thus, if then equivalently, if then We could say that A is strongly continuous, but we usually say just continuous. Definition 5.5.3 Let H be a Hilbert space, and A a linear operator in H. A is said to be weakly continuous if it maps weakly convergent sequences into weakly convergent sequences, i.e. implies

Lemma 5.5.2 Let H be a Hilbert space, and A a linear operator in H. If A is continuous, then it is weakly continuous. Proof. Suppose (i.e. weakly) in H, so that We must show that this is so because

since A* is a continuous operator. Therefore continuous.

for all

and A is weakly

Lemma 5.5.3 Let H be a Hilbert space and A be a continuous linear operator in H. If and then

Proof.

Choose if

Choose

A weakly convergent sequence is bounded (Theorem 4.6.3); thus Choose so that if then Then we have

so that if we have

then

Then if

162


5.6 Examples of adjoint operators A matrix operator in We recall that

is the metric space of sequences

We consider the matrix operator

with norm

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 Hermitian, after Charles Hermite (1821–1901).

i.e. iff the matrix A is

An integral operator We consider an integral operator, or so-called Fredholm operator

in Note that B acts on the function f; it is a linear operator acting in the space of functions. If then B is bounded in Indeed, using the Holder inequality, as we have just done, we find

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), governing 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 appropriate equilibrium conditions, and D is the flexural rigidity of the plate. Usually this equation is considered for and We derive a generalized form of the equation. Suppose that the boundary of the plate is clamped, so that

We take a test function

satisfying

multiply equation throughout by and integrate over theorem and the boundary conditions satisfied by and

use the divergence to obtain

where

and from now on, in this section an undistinguished norm or inner product is taken to be that in thus

164


We now show that we may consider equation (5.6.1) for and The problem (5.6.1) thus reduces to finding a non trivial element satisfying the equation for every We 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-norms and given in equation (3.6.1). Thus

We showed that the (3.7.18) and N = 2,

norm is

and Corollary 1 used with equation shows that

so that

and hence so that is a bounded linear functional in in tation theorem there is an element such that

By Riesz’s represen-

5.6 Examples of adjoint operators

The correspondence

defines a linear operator on

165

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


166

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 Inverse operator : exists iff : bounded if

implies

Theorem 5.2.2. Problem 5.3.2.

Theorem 5.3.1.

Banach’s open mapping theorem: Theorem 5.3.3. Closed operator : tion 5.4.1.

imply

and

Defini-

: 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 continuous Theorem 5.4.1. Adjoint operator : : Weakly continuous :

(5.5.1) Lemma 5.5.1. 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 end of Chapter 2.

6. Compactness and Its Consequences

I don’t understand new ideas, I don’t even understand why it is necessary 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 generalized 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 is closed and bounded. Unfortunately, the Bolzano-Weierstrass theorem cannot be generalized so that it applies to all metric spaces, as the following counterexample shows. We defined 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 cannot contain a convergent subsequence. Thus, since not all closed and bounded sets

168


in a general metric space X possess the property that any sequence contains a convergent subsequence, we must introduce a new term to describe those which do. For many years the term that was used, was compact, (this was the term we used in § 1.1) but that is now used for a property related to the Heine–Borel theorem, due to Heinrich Eduard Heine (1821-1881) and Emile Borel (18711956), that we discuss below. Now therefore the term that is used is sequentially compact. However the outcome of the analysis in this section is that the two terms 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 sequentially compact if every sequence in S contains a subsequence which converges to a point Theorem 6.1.1 Let X be a metric space, and let S be a sequentially compact set 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 metric induced by X. The proofs of all three properties follow similar lines: closed. Let be a convergent sequence in S. Since S is sequentially compact, contains a subsequence which converges to But therefore the whole sequence must converge to Thus S contains all its limit points, and so is closed. complete. pact, therefore, quence in

Let be a Cauchy sequence in S. Since S is sequentially commust contain a subsequence which converges to But by Problem 2.4.4, converges to Therefore any Cauchy seS has a limit in S, so that S is complete.

bounded. Suppose S were not bounded. Choose Now choose such that this is possible because S is unbounded. Now choose so that and and so on. The sequence can contain no Cauchy sequence, which contradicts the statement that S is sequentially compact. 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

6.1 Sequentially compact

compact

169

for all metric spaces. In fact we will show later (Theorem 6.2.2) that in a Banach space 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 sets sequentially compact; by the end of this section we shall have justified using the term compact for them, as in fact we did in § 1.2. The newer definition of compact is based on the concept of a covering, and the 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 point is an interior point; the simplest example is an open interval Definition 6.1.2 Consider a collection Their union

is the set of all

for some

of open sets in a metric space X.

The collection

is said to cover a set

if

The Heine–Borel theorem in its classical form is Theorem 6.1.2 Any cover of a closed interval by a collection of open sets has a finite sub-cover. In other words, there is a finite subsequence 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; one half, say must have no finite cover; let be a point in (say, its right hand end point); bisect one half, must have no finite cover; let be a point in and so on. Each of the intervals obtained in this way has no finite cover. On the other hand, the sequence is a Cauchy sequence in I. Since is complete, has a limit Since I is closed, converges to a point since is contained in one of the sets since is open, is an interior point of so that contains a neighborhood 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

170


Problem 6.1.1 Show that any cover of a closed and bounded set a collection of open balls (or open sets) has a finite sub-cover.

by

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 every cover 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 tion 6.1.2) iff it is closed and bounded.

is compact (according to Defini-

By comparing Theorem 6.1.2 and Problem 6.1.2 we see that the terms compact and sequentially compact are equivalent in we shall now show that they are equivalent in any metric space. Theorem 6.1.3 A set S in a metric space X is compact iff it is sequentially compact. Proof. Suppose that S is compact (according to Definition 6.1.3), but not sequentially compact. Thus there is an infinite sequence with no subsequence converging to a point in S. This means that the points of do not cluster about any point of S. Thus each point can be covered by an open ball which contains at most one point of This provides an open cover for S, which has a finite sub-cover Since can have 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 is an 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 of open sets, it cannot be covered by the ball of radius about Therefore we can choose such that For the same reason we can choose outside balls of radius around and i.e. so that Continue in this way to define such that The sequence must possess a Cauchy sequence for 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 compact if its closure is compact.

is said to be pre-

6.2 Criteria for compactness

171

Thus if S is precompact any sequence in 5 contains a subsequence which converges 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 set of N balls with and is said to be a finite of 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

Definition 6.2.2 Let X be a metric space. A set bounded if it has a finite for every

for S.

is said to be totally

Clearly every finite set is totally bounded. Also, if any infinite set S has a finite then one of the balls must contain an infinity of elements. We are now ready for Hausdorff ’s compactness criterion, due to Felix Hausdorff (1868–1942), which we may state as Theorem 6.2.1 Let X be a complete metric space. A set iff it is closed and totally bounded.

is compact

Proof. We argue very much as in Theorem 6.1.3. Suppose S is compact, then it is closed (Theorem 6.1.1). Suppose it is not totally bounded. This means that there is an for which S has no finite Choose Choose such that choose so that and so on, as before. The sequence must possess a convergent subsequence so that for sufficiently large This is a contradiction. Now suppose that S is closed and totally bounded, and let be a sequence 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 Choose one of them and call it Take and construct a finite of One of these balls, denoted by must contain an infinity of elements of which are in Choose one of these and call it Continuing this procedure we obtain a subsequence of Since and are, by construction, in the ball which has radius we have

Then the triangle inequality gives

172


which means that is a Cauchy sequence. X is complete, so that this Cauchy sequence has a limit it is a convergent sequence. S is closed so that This theorem balances the counterexample used at the beginning of this section (equation (6.1.1)) to show that a closed and bounded set is not necessarily compact. In a general metric space, boundedness is not sufficient to ensure compactness; the set must be (closed and) totally bounded. Problem 6.2.1 Let X be a metric space. Show that if a set pact (Definition 6.1.4), then it is totally bounded.

is precom-

Note that we do not need completeness of X to show that S is totally bounded. On the other hand, we do need the completeness of X to show that a closed and totally bounded set is compact, or equivalently, that a totally bounded set is precompact. For consider Problem 6.2.2 Let X be the set of rational numbers with the metric Let S be the set of rational numbers such that Show that S is closed in X and totally bounded, but not compact; X is not complete. For normed linear spaces, the question ‘when is a closed and bounded set compact?’ is answered by Theorem 6.2.2 Let X be a normed linear space. Every closed and bounded set is compact iff the dimension of X is finite. Note that the theorem states that if the dimension of X is finite, then every closed and bounded set is compact. On the other hand, it also states that if the dimension of X is infinite, then not every closed and bounded set is compact, i.e. there is at least one which is not compact. It is easy to prove the first part, but to prove the second, converse part, we need Lemma 6.2.1 Let X be a normed linear space, and Then for any such -that that and

be a closed subspace, there is an element such

6.2 Criteria for compactness

173

(Note X – S, sometimes written X\S , is the set of elements in X which are not in S.) Proof. Since

there is an element

Let

First we show that d > 0. For if d = 0, then there is a sequence such that as this means that since S is closed. This contradicts the supposition that Thus d > 0. According to the definition of infimum (§ 1.1), for any is a such that The

and so there

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 there is a finite set of elements such that any can be represented in the form

We now show that the result holds only if the dimension of X is finite. We will 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 element such that and denote by the space spanned by i.e. the set of all elements where If then by Lemma 6.2.1, there is such that and Denote by the linear space 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 obtain a sequence such that if This sequence on the unit ball cannot contain a Cauchy subsequence, and therefore cannot contain a convergent sequence. Thus if X is infinite dimensional, the unit ball cannot be compact.

174


Corollary 6.2.1 In an infinite dimensional space the closed unit ball around zero is not compact. Corollary 6.2.2 A bounded set in a finite dimensional normed linear space is precompact – its closure is compact. Note how the second half of this proof mimics the counterexample used at the beginning of the section. We conclude this section by showing how the terms compact and separable are linked, by Theorem 6.2.3 A compact set S in a metric space X, and in particular a compact metric space, is separable. Proof. We recall Definition 4.1.3, that 5 is separable if it contains a countable subset M which is dense (Definition 2.2.7) in 5. Suppose S is compact. Take the sequence where By Theorem 6.2.1, for each S has a finite Thus there is a finite sequence of open balls where and of course N depends on which covers S. The collection of all is the required countable dense set M,‘ since for any and any there is an such that Thus S is separable.

6.3 The Arzelà–Ascoli theorem We start with a note on terminology. There are three terms which we have used 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 into another Y; a functional (Definition 2.7.2) is a mapping from X into or a function (Definition 1.2.2) is a mapping from into or Thus in our usage, a function is a particular functional, which in turn is a particular operator. We warn the reader, however, that some authors use function and operator interchangeably, while others equate function and functional; we will retain the distinctions we have made. In § 1.2, we proved some of the classical theorems in the theory of functions of a real variable, in particular, Theorems 1.2.1 and 1.2.2; the last named states that if f ( x ) is continuous on a compact region then it is uniformly continuous on We now show that we can generalize these results to continuous functionals defined on a compact set Y of a metric space X. First we need some definitions and preliminary results. Definition 6.3.1 Let X be a metric space, and Y a subset of X. A functional f (real or complex valued) defined on Y is said to be continuous at if, 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 is continuous at every Definition 6.3.2 Let X be a metric space, and f be a functional defined on Y. The functional f is said to be uniformly continuous on Y if, given we can find such that and imply

When is uniformly continuous on Y, there is one which makes for any two satisfying when f is only continuous, depends on We now prove Theorem 6.3.1 Let X be a metric space, Y a compact subset of X, and f be a continuous functional defined on Y, then f is uniformly continuous on Y. Proof. Suppose the assumption were false. This means that for some we cannot find a such that and implies Thus for this and each we can find such that but But Y is compact so that the sequences converging to and

as

, so that

contain subsequences

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 f a real-valued continuous functional defined on Y. Show that f is bounded and attains its maximum and minimum values. Note that Theorem 6.3.1 and Problem 6.3.1 are the generalizations of Theorem 1.2.2 and Theorem 1.2.1 respectively. Theorem 6.3.1 concerns a simple functional f ; now we consider a family of functionals. We use the notation or for the family; we use rather than 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 family of functionals defined on Y. The family is said to be uniformly bounded if there is a constant c > 0 such that for all and all

176


Definition 6.3.4 Let X be a metric space, and on Y. The family is said to be equicontinuous if, given such that if then

for all

be defined there is a

and all

Problem 6.3.2 Let X be a metric space, Y a compact subset of X, and a 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 a closed and bounded region of (and thus uniformly continuous on On the basis of Theorem 6.3.1 and Problem 6.3.1 we can define C(Y) for a compact set 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. Show that C(Y) is a complete metric space. The Arzelà–Ascoli theorem links the precompactness of a family of functionals, i.e. a set in C(Y), to the conditions of uniformly bounded and equicontinuous. 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 and a family of continuous functionals on Y, i. e. The family precompact in C(Y) iff is uniformly bounded and equicontinuous.

is

Proof. Suppose is precompact in C(Y). By Problem 6.2.1 it is totally bounded. By Definition 6.2.2, it has an for every Thus it has an for Thus there is a finite family of functionals in 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. Thus there is a such that for all and all so that 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 any there is a for which

But is a finite family, and so, by Problem 6.3.2, is equicontinuous. Thus there 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 in contains 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 sufficient to 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 Since the sequence of numbers is bounded, it contains a Cauchy subsequence Now consider the sequence at i.e. it contains a Cauchy subsequence Continuing in this way we obtain, at the th step, a subsequence which is a Cauchy subsequence at each of We now show that the sequence with is a uniformly Cauchy sequence. Choose The sequence being a subset of is equicontinuous. Thus we can find such that if then

for all Since Y is compact, it is precompact and therefore totally bounded (Theorem 6.2.1). Let the finite set of balls be a finite of Y with radius Since the set is dense in Y, there is a member of this set, say in each ball This means that for any we have

178


Thus if M = then the set forms a finite for Y with radius But is a Cauchy sequence at each of the finite set of points Therefore we can find N such that if then

Now suppose and choose if we have

such that

Then

(Note that we use (6.3.3) to bound the first and third term, (6.3.4) to bound the second.) Thus is a Cauchy sequence in C(Y). Since C(Y) is a complete metric space (Problem 6.3.3), will converge to a continuous functional in C(Y); this functional will lie in the closure of Thus is compact, i.e. is precompact.

6.4 Applications of the Arzelà–Ascoli theorem We first prove Theorem 6.4.1 Let functions defined on precompact in

be a compact region in Let If is uniformly bounded in is compact if it is closed.

be a family of then is

Proof. Note that the theorem concerns functions defined on It is sufficient to prove that is uniformly bounded and equicontinuous. The metric in is (2.3.4), namely

where

Thus if it is uniformly bounded in

is uniformly bounded in

under the metric

We suppose is sufficiently regular so that if ciently close together, then the segment joining and we have

and is in

are suffiin that case

6.4 Applications of the Arzelà–Ascoli theorem

179

The chain rule gives

But

for all

so that

is uniformly bounded in

all

so that

all

and all

Thus

is equicontinuous in

It is worthwhile to restate this result for the simplest particular case. Let and 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, due to 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 satisfies the condition of Theorem 6.4.1. To do so we divide the real line around into segments of length and define on one segment in terms of its values on the previous segment (to the left). In order to start the process we must define on we define it as

180


In

we define it as

and generally, in

we define

Putting the formulae (6.4.4) together for we have

Equation (6.4.2) shows that if

we see that for

then

This means that the arguments for f in equation (6.4.3) lie in Q, so that if we have

We now show, by induction, that this result holds for It holds for Suppose it holds for (6.4.4) gives

so that it holds for

This establishes that

which means that the family is uniformly bounded on differentiating equation (6.4.5) we see that

Since

and

for then

satisfy (6.4.6), the arguments of

On

are in Q, so that

Thus the family satisfies the condition of Theorem 6.4.1. Therefore is precompact, and has a Cauchy subsequence, Since we are using the uniform norm in the compact space this subsequence will converge, uniformly, to a continuous function Thus we may pass to the limit in equation (6.4.5) with replaced by and find that satisfies

6.4 Applications of the Arzelà–Ascoli theorem

This means that

181

is a solution of (6.4.1) on

Problem 6.4.1 Generalize Theorem 6.4.2 to the Cauchy problem for the system of N ordinary differential equations

[Hint: Put and write the equations as

In practice we do not solve the Cauchy problem (6.4.1) by using the process described in Theorem 6.4.2; instead we use a numerical method such as Euler’s method . In the simplest version of this method we divide the interval into equal segments of length and construct a piecewise linear function which is given by

when relation

We determine the

from the recurrence

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 and equicontinuous when f satisfies the conditions of Theorem 6.4.2. Now we argue as before. For in (6.4.10), we have

Thus that

But

so that

we may now prove as before and thus

is a piecewise linear function, so that


182

and the family is uniformly bounded. Now we must show that it is equicontinuous. If are in the same interval then

Now suppose that

are in different intervals, thus where We write

If we use the equation (6.4.9) for sion we find

and for

and simplify the resulting expres-

Now

and

Thus

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 the equation for and let we find that the limiting function satisfies 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 solution of 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 questions may be answered by assuming some differentiability of There are even more difficult questions connected with the finite difference solution of boundary value problems for partial differential equations. Few of these procedures have been completely justified, those which have are so-called variational-difference methods, related to finite element methods; they are justified by modifying the energy 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 functions in the norm. Thus if is a bounded domain in and then a family offunctions in is precompact iff it is uniformly bounded and equicontinuous 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. There our presentation dealt largely with continuous (bounded) linear operators. Compared to the theory of linear operators in a finite dimensional space, the theory of continuous operators in an infinite dimensional space is complicated. Many of the linear operators encountered in practice possess an additional property which make them, in some sense, very like linear operators in a finite dimensional 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 from X into Y is said to be compact (or completely continuous,) if it maps bounded 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 sequence such that By the definition of a compact operator, the infinite set lies in a compact set, (Definition 6.1.1) therefore it contains a convergent subsequence, and a convergent sequence is bounded (Problem 2.4.1), i.e. This contradicts Thus a compact operator is continuous, but the converse is false. For example, 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 if the space is finite dimensional (see Corollary 6.2.1 of Theorem 6.2.2).


184

Problem 6.5.1 Let X,Y be normed linear spaces. Show that a linear operator A from X into Y is compact iff it maps bounded sets of X onto precompact sets of Y. Problem 6.5.2 Let X, Y be normed linear spaces. Show that if A, B are compact linear operators from X into Y, then so are Problem 6.5.3 Let X be a normed linear space and i.e. A,B are continuous linear operators in X. Show that if A is compact, then AB and BA are compact. (Note that B need not be compact.) The following theorem gives some simple sufficient conditions for a linear operator to be compact. Theorem 6.5.2 Let X, Y be normed linear spaces and A be a linear operator from X into Y.

a) If A is continuous and dim R(A) is finite, then A is compact. b) 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 6.2.2 states that a bounded set in a finite dimensional space is precompact. Therefore A 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, and Theorem 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 operator from X into Y. If R(A) is finite dimensional, then A is said to be a finite dimensional operator. We may thus rephrase Theorem 6.5.1 to state that a finite dimensional linear 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 the are continuous, (in particular if dimX is finite – Problem 2.9.4) then A is compact.


185

Problem 6.5.4 Let X, Y be normed linear spaces. Show that a linear operator A from X into Y is compact iff it maps the unit ball in X into a compact set in Y. As an example of a compact linear operator, we consider the operator A defined by

acting in C[0,1]; it is compact when uniformly continuous with respect to s and bounded, i.e. and

i.e. when K is A is continuous because K is

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 set is uniformly bounded and equicontinuous. The inequality (6.5.2) shows that it is uniformly bounded, for implies Suppose then

Choose such that

K is uniformly continuous in for all Thus for such

and Thus we can find and all satisfying

Thus for is uniformly bounded and equicontinuous, and therefore precompact. Therefore A is compact. Problem 6.5.5 Show that if (6.5.1) is compact in

then the operator A 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 in is 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.


186

Proof. Let S be a bounded set in X. Choose and then choose so that for every The operator is compact; therefore the map of S under is precompact. Therefore, by Theorem 6.2.1, it is totally bounded. Therefore there is a finite set such 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 Theorem 6.2.1, precompact. (Notice that we need Y to be complete.) Thus A is compact. We may apply Theorem 6.5.3 to show that the infinite dimensional matrix operator A defined by

is compact if

We can easily show that

Now define a finite dimensional operator

The operator

so that

by

is finite dimensional and therefore compact. But


187

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 be compact in The space is the completion of C([0,1] x [0,1]). Thus there is a sequence of functions such that as defined by

To apply Theorem 6.5.3 we need to show that the operator A

is continuous, i.e. in

then

and that if

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 let Show that the Fredholm integral operator A defined by

is compact in We have shown that if given by

then the operator A

188


is compact in i.e. if then and A maps a bounded set into a compact set. In § 5.5 we showed that the functions form a complete orthonormal system for Therefore the functions form a complete orthonormal system for The 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 that the map, of the bounded set must be precompact. Therefore must contain a subsubsequence converging to But 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 generalize this result to give Theorem 6.5.4 Let X, Y be normed linear spaces, and A be a compact linear operator on X, i.e. in and onto Y, i.e. R(A) = Y. If A has a bounded 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 of S under A. Let be a sequence in S. By Problem 6.5.1 the image, of under A is precompact. Therefore contains a subsequence converging to By hypothesis is bounded and has domain Y. Thus there is an such that and

Thus but and S is closed, so that Therefore S is compact. Therefore any closed and bounded set in X is compact so that, by Theorem 6.2.2, X is finite dimensional. Corollary 6.5.1 If X, Y are normed linear spaces and A is a compact linear operator 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 pact also.

is com-

6.6 Compact linear operators between Hilbert spaces

189

It is possible for a compact linear operator from X into Y to have a continuous inverse even when X is infinite dimensional by having R(A) strictly contained 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 linear space X into a normed linear space Y. Now we consider a compact linear operator from a Hilbert space into a Hilbert space For Hilbert spaces we have two extra concepts which we have introduced: weak convergence of a 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 in general normed linear spaces.) Regarding the adjoint we have Lemma 6.6.1 Let A be a continuous linear operator on a Hilbert space into 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 operator in 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 that Thus 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 operator from into A is compact iff it takes every weakly convergent sequences in intoa strongly convergent sequence in Proof. Suppose is a weakly convergent sequence in A weakly convergent sequence is bounded (Theorem 4.6.3). Therefore, since A is compact, contains a subsequence such that converges strongly to

190


an element in the complete space On the other hand, since A is continuous, it takes the weakly convergent sequence into a weakly convergent sequence (Lemma 5.5.2). Since this sequence contains a subsequence which converges weakly to the whole sequence must converge weakly to We show that converges strongly to Suppose if possible that there is a subsequence which does not converge to This means that there is an and a sequence such that

The sequence is bounded so that, since A is compact, we can find a subsequence such that converges to some But then must converge weakly to However is a subsequence 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 that We have shown that a compact linear operator takes a weakly convergent sequence into a strongly convergent sequence. Now we show that if A takes every weakly convergent sequence into a strongly convergent sequence, then it is compact. Let S be a bounded set in and A(S) its image under A. According to the definition of a compact operator (Definition 6.5.1 or Problem 6.5.1) we need to show that A(S) is precompact (i.e. its closure is compact). Take a sequence and consider a sequence such that The sequence being in S, is bounded. A bounded set in is weakly precompact (i.e. its weak closure is weakly compact) (Theorem 4.7.1). Therefore contains a weakly convergent subsequence (converging to some in the weak closure of S). By assumption, A takes this subsequence into a strongly convergent sequence Therefore contains a strongly convergent subsequence A(S) is precompact and A is compact. We now show that in a separable Hilbert space any compact operator may be 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 operator in 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 basis Any can be written


191

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 such that and as The set is bounded and weakly closed (Corollary to Theorem 4.6.6) and therefore weakly compact (Theorem 4.7.1). Therefore contains a subsequence which converges weakly to some such that Since Theorem 6.6.1 shows that so that But so that But, on returning to equation (6.6.1), we see that

so that

We wish to show that as since A is compact, it is sufficient to show that do this we take an arbitrary and find

since

i.e. that (weakly). To

by Parseval’s equality.

In Theorem 6.5.4 we showed that if a compact linear operator A on a normed linear space X onto a normed linear space Y has a bounded inverse

192


then X is finite dimensional. This result holds only if A is one-to-one. We now 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 a Hilbert space If R(A) is closed, then it is finite dimensional.

into

Proof. If R(A) were infinite dimensional then we could form an orthonormal sequence in R(A). For this sequence, Let N(A] be the null 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 a bounded inverse Therefore maps the sequence onto a bounded set in But like A, is compact so that it maps this bounded set in into a precompact set in R(A). Thus must contain a convergent subsequence. But this is impossible since Therefore R(A) can contain no infinite orthonormal sequence. It must be finite dimensional.


193

Synopsis of Chapter 6: Compactness Sequentially compact : every sequence in S contains a subsequence which converges to Definition 6.1.1. : a sequentially compact set is closed, bounded and complete. Theorem 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 closed and 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. Theorem 6.3.2. Compact linear operator : maps bounded sets into compact sets. Definition 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 ignoramus What’s more, half scoundrel, don’t forget But on this score the man gives promise He’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 parlance, 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 or complex parameter The most important example is the equation governing the steady vibration of an elastic body with frequency namely

In particular, the natural vibration of a string are governed by the boundary value 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 for which is a bounded operator with domain which is dense in X. Such points of are called regular points. The spectrum, of A is the complement of If

i.e. if

is not a regular point, there are three possibilities:


196

1. the range of We say that

is dense in exists, but is unbounded. belongs to the continuous spectrum of A;

2.

exists, but its domain is not dense in X. We say that to the residual spectrum of A;

3.

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.

belongs

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 a normed linear space X. Show that the set of all eigenvectors corresponding to is a closed linear subspace of X. We consider some examples: 1. A matrix operator acting in

consisting of no more than plane are regular points.

This operator has only a point spectrum eigenvalues. All other points of the complex

2. The differentiation operator

acting in Any point in the complex plane belongs to the point spectrum, since for any the equation

has a solution

Thus the operator has no regular points.

3. The boundary value problem

where

is the square

We consider the third problem in we can find N such that

where

If

then, given

7.1 The spectrum of a linear operator

Thus the set S of all such

is dense in

Consider equation (7.1.3) for Suppose negative real axis. The unique solution of (7.1.3) is

To show that from zero, i.e. there is a

If

Thus if

we need to show that such that

197

and

but

is not on the

is bounded away

then

then we may take If Thus (7.1.5) holds with

and

then

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 real axis? If where are integers, then is a solution of

so that

is an eigenvalue.

198


Problem 7.1.3 Show that if no solution of (7.1.3), (7.1.4) for

where

are integers, then there is

Consider the remainder, M, of the negative real axis, i.e. the set of which cannot be represented in the form where are integers. Again consider for The set of points of the form is dense in M. In other words, if then there is a sequence where such that Take

then the corresponding

Thus the norms of

and

is

are related by

so that is unbounded as This means that although the inverse 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 continuous spectrum. Note that the problem (7.1.3), (7.1.4) has no residual spectrum. 4. Now we consider the so-called coordinate operator in

Clearly

has no eigenvalues. If

defined by

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 which means that points belong to the residual spectrum. Problem 7.1.4 Consider the coordinate operator in is the continuous spectrum.

and show that

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 closed linear 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

By Definition 7.1.1 of the resolvent set, is C > 0 such that

If

there is an

Thus

is bounded on S. Thus there

such that

so that (7.2.1) gives

Now suppose is an arbitrary element in X. Since S is dense in X, we can 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 Cauchy sequence; since X is a Banach space, there is an such that

Now we apply Definition 5.4.1 to the closed operator and

and deduce that

200


But was an arbitrary element in X; thus the range of i.e. S, the domain 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 holomorphic in G iff at every point

The function is said to be it has a power series expansion

with non-zero radius of convergence. We now show that, when treated as a function of the resolvent operator for 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 space X. The resolvent set is a domain (an open set) of and is holomorphic with respect to in Proof. Suppose Theorem 7.2.1 states that linear operator on X. Thus the series

is a continuous

is convergent in the circle of and thus is a holomorphic 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. For any 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 bounded when Problem 7.2.2 Let B be a bounded linear operator in a normed linear space X. Show that the spectral radius of B, defined by

exists and that the expansion

is valid in the domain Problem 7.2.3 Let X, and suppose resolvent set operator in

be a continuous operator in a normed linear space is holomorphic with respect to in Show that the of is open and is a holomorphic

7.3 The spectrum of a compact linear operator in a Hilbert space For compact linear operators in a Hilbert space, we can describe the spectrum fully. The first results in this direction are due to Fredholm; he studied the integral operator and established that its spectrum had properties similar to those of a matrix operator. The theory was extended to compact operators in a Banach space by Riesz and Pavel Julius Schauder (1899–1940); we will describe it for operators in a Hilbert space. The theory is of great importance as it describes the vibrations of bounded elastic bodies. It transpires that we must consider the free and forced vibration problems together; that is, in abstract terms, we suppose is a compact linear operator 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


202

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 A is compact; A maps bounded sets on precompact sets (Problem 6.5.1); therefore is precompact; but so that is precompact; therefore S is closed and precompact, and therefore compact. Therefore every closed 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 on H, respectively.

denote the orthogonal complements of

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 Hilbert spaces. Lemma 7.3.2 There are constants

such that

for all Proof. The right hand inequality holds because being compact, is bounded. Let us prove the left hand inequality. Suppose that there is no such for all This means that there is a sequence such that and as Because is compact, the sequence contains a Cauchy subsequence. But this means that must also contain a Cauchy subsequence because

Let us rename this Cauchy subsequence Since will converge to since we have hand, and A is continuous, so that

is complete, On the other But


so that

i.e.

But

so that

203

is

not zero, but it belongs to both of the mutually orthogonal sets This is impossible. Therefore the left hand inequality holds. The inequalities (7.3.3) state that, in the Hilbert space is equivalent to and that the inner product

the norm

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 iff Suppose that, for some equation (7.3.7) has a solution Let then, by using equation (5.5.1) we find

Thus


i.e.

i.e.

Therefore

Now suppose that The functional is linear and continuous on H, and therefore on The space is a Hilbert space with inner product given by (7.3.6). Therefore, by Riesz’s representation theorem there is such that

This equality holds for we may write

to give

Put

then

but it also holds for where and

also. For if and use

204


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: closed, whereas R(A) is not necessarily closed. Lemma 7.3.3 Let

be the null space of

is

i.e.

Then

1.

is a finite-dimensional subspace of H;

for

2. 3. there is an integer Proof. 1.

such that

can be written in the form

where is a compact linear operator. The result follows from Lemma 7.3.1. 2. This is evident. 3. First we note that if for some then for all Indeed if then i.e. Thus Therefore and so that x Thus Now suppose, on the contrary, that there is no such that Then there is a sequence such that and is orthogonal to i.e. is an orthonormal sequence. is bounded and is compact so that contains a Cauchy subsequence, but this leads to a contradiction. Indeed we have

where Now

for


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 we can solve successively the infinite system of equations

The sequence

205

Since

has the property

Thus but so that, contrary to Lemma 7.3.3, there is no such such that This shows that if then Now suppose that then and by Theorem 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 inequalities (7.3.3), and in particular the left hand inequality, holds on H. By Theorem 5.3.1, this means that is a continuous operator on H. Corollary 7.3.2 A compact linear operator a point spectrum.

in a Hilbert space H has only

Proof. Suppose is not an eigenvalue, then and thus and The first states that the domain of is H. Using the second in (7.3.3), we see that is a bounded linear operator in H. 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, and suppose that Choose orthogonal bases and for N and N* respectively. Introduce the auxiliary operator by

206


The operator C, being the sum of a compact operator and a finite-dimensional (and therefore compact (Theorem 6.5.3)) operator, is compact. Let be the null space and range of we will show that For if then

Since and are mutually orthogonal, and therefore zero. Thus

all the terms in (7.3.8)

The first equation states that the second states that is orthogonal to all the basis elements of therefore Therefore and so, by Theorem 7.3.2 applied to C, Therefore the equation

has a solution,

But then

This contradiction shows that of so that Thus

On the other hand

Remark. In this proof we used the operator invertible in H. The same holds for the operator

with any small

is the adjoint

which is continuously defined by

this operator has a continuous inverse, and

Thus if we replace the equation by the close equation we 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:


207

either the equation has a solution for all this means so that the solution is unique; in which case that and the equation

and therefore so

has no solution. or the equation has a finite dimensional space of solutions spanned by which case

in

has a solution iff (This states that The solution is not unique, because Thus if span then

The results we have established in this section apply to the general equations (7.3.1), (7.3.2). We conclude this section by deriving an extra result which holds for the important special case Problem 7.3.1 Suppose that is a continuous linear operator in a Hilbert space H. Show that if are eigenvalues of A, then This simply states that an element x cannot be an eigenvector corresponding to two different eigenvalues. It implies that if are eigenvalues of A, and from each of we take a linearly independent set of eigenvectors, then their union will be linearly independent. Lemma 7.3.4 The set of eigenvalues of a compact linear operator has no finite limit point in Proof. Suppose that there is a sequence of eigenvalues such that For each eigenvalue take an eigenvector Let be the subspace spanned by By Problem 7.3.1, and We can apply the Gram–Schmidt process to and find an orthonormal sequence , i.e. such that The sequence is bounded; A is compact, so that is precompact, i.e. it contains a Cauchy subsequence. We now show that this is impossible. Indeed


208

where We now show that if then Since it is sufficient to consider only the first two terms in Since

we have

Since thus

and

so that

and

are orthogonal,

cannot contain a Cauchy sequence.

Combining the results we have obtained in this section we can state that if is a compact linear operator in a Hilbert space H, then: 1. A has only a point spectrum;

2. each eigenvalue has only a finite dimensional space of eigenvectors; 3. if

then in addition

(a) two eigenvalues cannot have a common eigenvector; (b) the point spectrum (if there is one) is countable and has no finite

limit point in N.B. Nowhere have we shown that a compact linear operator in a Hilbert space has an eigenvalue. All our statements have had the form ‘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 at least one eigenvalue, and then that it has, in a precise sense to be stated, a full set of eigenvectors.

7.4 The analytic nature of the resolvent of a compact linear operator We know (Theorem 7.2.2) that the resolvent of a closed operator is a holomorphic operator-function of in the resolvent set. What is its behavior near 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 than All which are not eigenvalues of A are points where the resolvent is holomorphic; thus they cannot be roots of If is an eigenvalue of then 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 of coincides with the set of eigenvalues of and so each eigenvalue of is a pole of finite multiplicity of the resolvent We now consider a general case: Theorem 7.4.1 Every eigenvalue of a compact linear operator A in a separable Hilbert space is a pole of finite multiplicity of the resolvent

210


Proof. We showed in Theorem 6.6.2 that a compact linear operator A in a separable Hilbert space may be approximated arbitrarily closely by a finite dimensional 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 the numerical series

is convergent; this is so if in this circle is a holomorphic operator-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 of as were, and are holomorphic functions of in the circle For any satisfying the elements are linearly independent, since are linearly independent and is continuously invertible. So, by analogy with (7.4.3), for the solution to (7.4.7) is

7.5 Self-adjoint operators in a Hilbert space

211

We note that and are the same functions of and as and are of and respectively. Thus depends on explicitly as a polynomial of degree no greater than but also depends 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 If is an eigenvalue, then For we may choose so that so that if were not zero, the equation would be soluble for all and so for all which is impossible. This means that the set of eigenvalues of A lying inside any circle coincides with the set of zeros of lying inside this circle.

7.5 Self-adjoint operators in a Hilbert space Many important problems in continuum mechanics may be phrased as problems relating to a self-adjoint linear operator. The theory for such operators is particularly 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) are real, and eigenvectors corresponding to distinct eigenvalues are orthogonal. This, combined with Corollary 7.3.1 shows that a self-adjoint compact linear operator A in a Hilbert space has a spectrum which is a real point spectrum—if it has one at all. We will now show that A has at least one eigenvalue. We start with a definition and a lemma. Definition 7.5.1 A functional on a Hilbert space H is called weakly continuous if it takes weakly convergent sequences into (strongly) convergent (numerical) sequences. Thus if

and

then

Problem 7.5.3 Show that a weakly continuous functional is continuous. Lemma 7.5.1 A real valued weakly continuous functional on a Hilbert space H assumes its maximum and minimum values in any ball Proof. Let sup

There is a sequence The set

such that

and

is bounded and weakly closed by the


212

Corollary to Theorem 4.6.6. Therefore, by Theorem 4.7.1 it is weakly compact. Thus the sequence contains a subsequence converging weakly to some 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 Hilbert space. is a real valued weakly continuous functional on H. Proof. By Problem 7.5.1, to Then

is real valued. Let

be weakly convergent

A is compact and so that i.e. weakly convergent sequence is bounded (Problem 4.6.3) so that, and

On the other hand as

so that and

A

and thus 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 space H has at least one, non-zero, eigenvalue. Proof. By Lemmas 7.5.1, 7.5.2, values on Let these be

assumes its maximum and minimum

then

This must be non-zero because, by homogeneity and Problem 7.5.4,

Therefore there is an

such that

and


213

Now consider the functional

The range of values of Thus

for

coincides with the range of

for

We will show that is an eigenvector of A. Indeed consider where is 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 its minimum 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 space H has a finite or infinite sequence of orthonormal eigenvectors corresponding to non-zero eigenvalues which is complete in the range R(A) of the operator A, i.e. for every the Parseval equality

214


holds. Proof. By Theorem 7.5.1 there is an eigenvector where

with

Rename the Hilbert space the operator as let by and decompose into and The space If then for

This means that we may define a new operator

in

be the space spanned is a Hilbert space.

by

This operator is called the restriction of to it is clearly a self-adjoint compact linear operator in the Hilbert space If this operator is not identically zero we may apply Theorem 7.5.1 to it, and find an eigenvector such that Since

wehave

and

We now continue this process; we let be the space spanned by decompose into and call the restriction of to and find an eigenvector and 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 vectors corresponding to non-zero eigenvalues moreover

and Suppose

and consider


We have so that satisfies (7.5.2) so that with Problem 7.5.4,

215

and hence i.e. Thus

so that

Now consider the case in which the process does not stop. We have an infinite sequence of vectors and a corresponding sequence of non-zero eigenvalues where According to Lemma 7.3.4 we must 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 concerning A; thus we introduce Definition 7.5.2 A self-adjoint continuous linear operator A in a Hilbert space H is called strictly positive if for all and iff

216


For a strictly positive, compact, self-adjoint operator in a Hilbert space the process 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 in an infinite dimensional Hilbert space H. There is an orthonormal system which 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 prove that 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

so that

and

forms a basis for H, and moreover

This theorem shows that one can have a strictly positive compact selfadjoint 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


217

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 is an orthonormal basis for

with

As an example, consider the eigenvalue problem

This has eigenvalues

and eigenfunctions

Let W be the Hilbert space of functions product Remember that Problem 3.6.1 shows that lem (7.5.4) can thus be posed as

with the inner

is a norm for

The prob-

where the operator A is defined as

Thus in the space

which means that for

But

In the language of Problem 7.5.5,

so that

which thus forms a basis for

is an orthonormal basis

218


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. Problem 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 complete in One can have a strictly positive (Definition 7.5.2) compact self-adjoint operator only in a separable Hilbert space. Theorem 7.5.3.

8. Applications to Inverse Problems

As an orthodox mathematician, he believes his formula more than his eyes and common sense, and doesn’t see the incongruity in it. Academician A.N. Krylov, on a formula in an article by Levi-Civita.

8.1 Well-posed and ill-posed problems Most problems in mechanics and physics have the form ‘Find the effect of this cause.’ There are numerous examples: Find how this structure is deformed when these forces are applied to it. Find how heat diffuses through a body when a heat source is applied to a boundary. Find how waves are bent, or absorbed, as they pass through a nonhomogeneous medium. At the turn of this century the French mathematician Jacques Salomon Hadamard (1865–1963) identified three characteristics of what he called a wellposed 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 the effect. Much of the research in theoretical mechanics and physics during this century has been devoted to showing that, under specified conditions, the traditional problems in these fields do in fact possess these properties. Traditional cause and effect problems, with attendant studies of the accuracy and stability of approximate solutions still dominate mechanics. However, during the last three or four decades (since about 1960) there has been a growing recognition that there are important problems which fail to have some or all of the defining properties of a well-posed problem; they are called ill-posed problems. Many, but not all, are concerned with questions of the form ‘what is the cause of this effect?’ Since, in many cases, each problem in this subclass may be associated with a direct problem ‘what is the effect of this cause?’

220


they are, somewhat loosely, called inverse problems. It should be noted that a problem is called an inverse one only because of its relation to another that we call direct; in some cases the choice of which to call direct and which to call inverse is arbitrary. Also, not all inverse problems are ill-posed, nor are all ill-posed problems, inverse problems. It may be shown that many ill-posed and/or inverse problems may be reduced, perhaps after some linearization, to the operator equation

where belong to normed linear spaces X, Y, and A is an operator from X into Y. The simplest, and most common form that this equation takes is the Fredholm 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 found so 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 metric space Y. The Definition 2.7.3 is a straightforward generalization of the definition of a continuous function of a real variable. In § 2.7 we concentrated on contraction mappings (Definition 2.7.4), but at the end of the section we showed that 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 a normed linear space Y. Thereafter all the operators that we have considered have been linear. The important Problem 2.9.3 shows that a linear operator is continuous iff it is continuous at 0, and consequently that an operator is continuous iff it has a bounded norm, in the sense of equation (2.9.2).


221

We returned to the theory of continuous (i.e. bounded) linear operators in Chapter 5. In Theorem 5.2.1 we showed that a linear operator which is bounded on a domain which is dense in a normed space X, and whose range lies in a Banach space Y, can be extended, without increasing its norm, to the whole space X. This means that if R(A) lies in a Banach space, there is no loss of generality in assuming that D(A) is closed (If it is not, then Theorem 5.2.1 shows that we can extend A to its closure Further, if X is a Banach space then being a closed subspace of a Banach space, is itself a Banach space. In this case there is no loss of generality in supposing that A is defined on X, i.e. D(A) = X, for D(A) being a closed subspace of a Banach space, is a Banach space; it is this Banach space that we call X. In § 5.3, we considered whether a continuous linear operator i.e. on X into Y, had an inverse. A necessary and sufficient condition for to exist is (Problem 5.3.2) that there should not be two distinct and such that the null space N(A) must be empty. In Theorem 5.3.1 we proved that the operator is a continuous linear operator iff

The most important results concerning were derived from Banach’s open mapping theorem (Theorem 5.3.3), that if X, Y are Banach spaces and A is a continuous linear operator on X onto Y, then A maps open sets of X onto open sets of Y. (This is a much deeper result than the straightforward Theorem 2.7.2) Note that, as we said in the previous paragraph, since Y is a Banach space, there is no loss of generality in assuming that A is defined on X, i.e. D(A) = X. By contrast, it is a restriction, and a prerequisite of the theorem, that A be an operator from X onto a complete space Y. As we pointed out earlier, after the proof of Theorem 5.3.4, when Y is complete, we can replace ‘R(A) = Y ’ by ‘R(A) is closed’. (The old Y is replaced by the complete space R(A) From the open mapping theorem we derived the fundamental Theorem 5.3.4. This states that if X,Y are Banach spaces and A is a one-to-one continuous linear operator on X onto Y, then A has a continuous inverse on Y onto X. This suggests that the equation (8.1.1) should present no difficulty: the solution is simply However, if we accept this suggestion readily it is because we have underestimated the power of the restriction ‘onto Y’ i.e. R(A) is closed. The difficulties start to become apparent 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 bounded sets onto precompact sets. In § 6.5 we showed that we may consider the integral operator A in equation (8.1.2) either in C[0,1] or in in both cases it is compact. The fundamental result concerning compact operators is Theorem 6.5.4: if a compact operator has a bounded inverse then X must be finite dimensional. There are two important corollaries of this results:

222


Corollary 8.2.1 If X, Y are normed linear spaces and X is infinite dimensional, a compact linear operator A from X onto Y cannot have a bounded inverse. This agrees with what we found for equation (6.5.4): is infinite dimensional and A does not have a bounded inverse. In § 5.3 we showed, on the basis of the open mapping theorem, that a oneto-one continuous linear operator A from a Banach space X onto a Banach space Y does have a bounded inverse. From this we can deduce that if indeed it is onto Y, then the only way it can fail to satisfy Theorem 5.3.4 is that it is not 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, and A is a one-to-one compact linear operator from X into Y, then R(A) cannot be Y, and indeed cannot be closed, for then R(A) being a closed subspace of a 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 from the infinite dimensional Banach space into the Banach space Corollary 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 expressed in the form

i.e. there is a

which cannot be

for any In other words the solution of (8.1.2) cannot be unique and exist for all one or other of uniqueness and existence must fail, 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 not stable; 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 can take sufficiently large that

223

we

is so small that

We describe this situation by saying that small amplitude high frequency noise in may cause large errors in the solution. We conclude that the operator equation faces us with three difficulties: R(A) does not exhaust Y, i.e. there are which are not in the range 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. There are various ways in which one or more of these difficulties may be overcome, as we 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 oneto-one operator from X into Y. Let S be a compact subspace of X and let be the restriction of A to S, then is continuous. Note that this does nothing to the difficulty that R(A) is strictly contained in Y, and it assumes that A is one-to-one; it simply ensures that the inverse operator, which will have a range within R(A), has a continuous inverse. Proof. According to Theorem 5.3.1 we must show that if a constant such that

then there is

Suppose this were not so. Then we could find a sequence such that and Since and S is compact, there is a subsequence converging to and

Thus but (8.2.3) holds and

so that A is not one-to-one as we assumed. Therefore is continuous.

To apply this theorem to the integral equation (8.1.2) we must restrict the function to a 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.


224

Tikhonov’s theorem deals with the continuity of the inverse by restricting the domain, and hence also the range of A. We now consider how we can enlarge the range of A and also deal with the fact that A may not be one-to-one. To do so we will assume that the operator A is a continuous linear operator on a Hilbert space into a Hilbert space The closure of the range of A is a closed subspace of Theorem 4.3.2 states that may be decomposed into and its orthogonal complement (because the orthogonal complement is always closed). Thus the closure of is or in other words, the subspace of is dense in We show how we can extend the inverse operator from R(A) to Suppose then its projection onto is actually in R(A). This means that there is an such that

This being the projection of onto closest to , i.e. according to Theorem 4.3.1

is the element of

The decomposition Theorem 4.3.2 states that any

Here that

By saying that

which is

may be written

we state that there is an

such

Any such is called a least squares solution of the equation, because it minimizes the norm But Problem 5.5.3 states that (Remember that a null space and an orthogonal complement are both automatically 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 that Suppose 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). This subset is closed and convex. (It is convex because

and imply We may apply Theorem 4.3.1 to M. This shows that there is a unique which minimizes on M. We take this to be the generalized solution of equation (8.2.1); it gives a unique solution for which is a dense subspace of This solution


225

is called the least squares solution of minimum norm. (Of course we can use Theorem 4.3.1 to find the least squares solution which is closest, in the norm of to some other element ) The mapping from into D(A) which associates to 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 choosing the of minimum norm. Another way of proceeding is to restrict to This however gives exactly the same solution, as shown by Problem 8.2.1 Let A be a continuous linear operator from to Show that is the unique least squares solution in Problem 8.2.2 If show that for any projection of on

Suppose

represents the restriction of A to where

is the

What have we achieved so far? We started with a continuous linear operator which could have a null space (i.e. need not be one-to-one) and which could have a range R(A) which was not dense in We have constructed a generalized inverse of A which is defined on a dense subspace of and which yields a unique for any The critical question is whether this generalized inverse is a continuous operator. In general it is not; it is merely a closed operator, as discussed in § 5.4. We prove Theorem 8.2.2 Let be Hilbert spaces, and A be a continuous linear operator from into The generalized inverse from into is a closed operator. It is continuous iff R(A) is closed. Proof. We recall Definition 5.4.1, and reword it for our case. the three statements

is closed iff

together imply Problem 8.2.1 states that is the unique solution of in Thus and is closed, imply Also and imply But implies so that and Thus and is a solution of Again 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

226


continuous and

is closed

so that and Therefore R(A) is closed. Now suppose R(A) is closed, then is a closed linear operator on into continuous.

On the other hand

Thus

and

so that so that, by Theorem 5.4.1, it is

Note that Theorem 8.2.2 leaves us with a most unsatisfying result if, as often the case, A is a compact operator. For we showed in Theorem 6.6.2 that if A is compact then its range will be closed iff it is finite dimensional. For the Fredholm integral equation this means that the equation must be degenerate. We still have not achieved the construction of a stable ‘inverse’ for the general non-degenerate integral equation.

8.3 Singular value decomposition In this section we suppose that A is a compact linear operator on a Hilbert space into a Hilbert space As we showed in § 6.6, this means and are compact self-adjoint linear operators in and respectively. We consider the eigenvalues and eigenvectors of these operators. Both operators are nonnegative in the sense that and for and Thus their eigenvalues will be non-negative. Note that in these equations, as elsewhere in this section and the remainder of the chapter, we use the same symbol to denote 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 suppose that is an eigenvector of corresponding to then so that Then so that is an eigenvector of corresponding to and similarly vice versa. We may now use Theorem 7.5.2. This states that since it is selfadjoint, has a finite or infinite sequence of orthonormal eigenvectors corresponding to positive eigenvalues and that the are complete in the closure of the range of We note that Theorem 5.5.2 shows that and that for if then on the other hand if then so that ) Let and then

and

8.3 Singular value decomposition

227

so that Thus the form an orthonormal set of eigenvectors for and Theorem 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 we may write

where

on N(A):

Since the

are complete in

Hence

This is called the singular value decomposition (SVD) of the operator A. We now return to the equation

If complete in

then this equation has a solution. We recall that the (Theorem 5.5.1). Thus if

then

This imposes a restriction on

Conversely, if

and

for it implies

are

228


then any element

where is a 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 existence criterion, after Charles Emile Picard (1856–1941). We note that will either have a finite number of eigenvalues, or an infinite sequence of eigenvalues. Since the eigenvectors span the former case will hold only when R(A) is finite dimensional, i.e. A is degenerate. 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 than We showed earlier that when gives the unique element in which satisfies (8.2.5). Equation (8.3.9) shows that when A is 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 inverse we do two things: replace by its projection on find the unique such that Equation (8.3.10) shows that if there are an infinity of singular values then is unbounded because, for example, while

In order to obtain an approximation to we may truncate the expansion (8.3.10) and take the approximation as

then as However, the question arises as to how many terms 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 evaluate it for some nearby such that We will obtain a bound for the difference 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 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 (8.3.12) implies that in choosing an say corresponding to a given data error we must do so in such a way that

Thus there are two conflicting requirements on it must be large enough to make 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 generalized inverse gives a ‘solution’ of (8.3.6) for all a dense subspace of which satisfy Picard’s criterion. However is not continuous unless R(A) is finite dimensional (and then closed). The unboundedness of arises because the tend to zero, and this in turn can be attributed to the fact that does not have a bounded inverse. The operator arose in equation (8.2.7); this in turn followed from (8.2.6) and (8.2.5). To find we first found the closest to in the sense of (8.2.5), then, if there were more than 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

for

To do this we set up a new Hilbert space with elements where We define the inner product in this space by

230


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. a complete inner product space. We can now imitate for this new space H what we did with in § 8.2. For the continuous linear operator A which takes into induces another continuous linear operator, which we will call which takes into in H. It is continuous, because

The range, that

of this new operator is the set of those and it is a closed subspace of H.

Problem 8.4.2 Show that if a sequence in the norm (8.4.3), then i.e. that

such

converges to is closed.

The Hilbert space H may be decomposed into and its orthogonal complement According to Theorem 4.3.1 this means that for any there is an such that is the element of R(A) which is closest to i.e.

The decomposition Theorem 4.3.2 applied to H states that any may be written

Here m is the projection of such that

onto so that

Since

there is an

In § 8.2 we used the result, proved in Problem 5.5.3, that To use this we must first define the adjoint of the operator from to H. We note that for any i.e. for any the functional

is a continuous linear functional on the Hilbert space Therefore, by Riesz’s representation theorem (Theorem 4.3.3), there is an element of which we call such that

8.4 Regularization

231

Thus

so that Since this holds for all

we have

Now we return to equation (8.4.5) and use the result that to 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 to for those (satisfying Picard’s condition) for which exists. We note that

But

so that so that we may write

But we showed that the

span

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 operator which gives in terms of is bounded. We note that since the are positive and tend to zero we can find such that

232


Thus when

while if

so that if

we may write

Now we show the convergence of to existence criterion holds. We note that

for those

for which Picard’s

so that

Choose Since the series in (8.4.12) converges, the sum from to must tend to zero as N tends to infinity. Therefore we can find N such that that 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, for those for which exists, converges to as Now suppose that the data, is subject to error. This means that instead of solving equation (8.4.7) for we are actually solving it for some nearby such that We wish to obtain a bound for the difference between the formed from which we will call and the ‘true’ formed from we wish to estimate We have

8.4 Regularization

233

so that, by proceeding as in (8.4.10), (8.4.11), we have

where Since the series converges, we may, for any given find 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 properties 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 error we must do so in such a way that

When we choose so that (8.4.15) holds, the difference between satisfies the inequality

and we have already shown that the second term tends to zero with of such that

and

A choice

is called a regular scheme for approximating The 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 that due to using rather than the limit as It is theoretically attractive to ask whether we can choose the way in which depends on i.e. so that both error terms are of the same order.

234


To bound the second term we return to the inequality (8.4.13). This holds for 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 other hand if we take then and will both be of order so that

8.5 Morozov’s discrepancy principle We continue to assume that A is a compact linear operator from to The choice is theoretically attractive, but difficult to apply. Morozov (1984) put forward a discrepancy principle in which the choice of a is made so that the error in the prediction of i.e. is equal to the error in the 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 means that, in computing there is no loss of generality in assuming that i.e. that Therefore we assume that the error in the data is less than or equal to and that the signal to noise ratio is greater than unity:

Decompose into and Theorem 5.5.1 states that and equation (8.3.3) states that the are complete in Thus

where find

because

is the projection of on Equation (8.4.10) gives we may apply A term by term to get

Thus

To

8.5 Morozov’s discrepancy principle

235

and

This equation shows that is a monotonically increasing function 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 a unique which we call such that Since we have shown 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

while

from which we conclude that

so that

236


i.e. the are bounded. Now suppose that is a sequence converging to and that Each such pair will determine an and a corresponding which we will call We now show that there is a subsequence of for which the converge to The sequence lies in the closed ball with center O and radius in The 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 subsequence of which converges weakly to some i.e. such that Equation (8.4.9) shows that is a closed subspace of and therefore also weakly closed, by Problem 4.6.6. Thus Let be the pair corresponding to then

Lemma 5.5.2 states that a continuous linear operator in a Hilbert space is weakly continuous, so that according to Definition 5.5.3, it maps a weakly convergent sequence into a weakly convergent sequence. Therefore A maps which converges weakly to into a sequence which converges weakly to But converges strongly to and therefore weakly to The weak limit is unique (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 there is a subsequence of which converges strongly to According to Theorem 4.6.2, in order to show that it is sufficient to show that We know that so that lies in the compact set of Therefore there is a subsequence of such that and On the other hand, since we have

so that Therefore and hence and in fact and We conclude that Morozov’s discrepancy principle does provide a regular scheme for solving equation (8.3.6) when A is a compact linear operator from into


237

Conclusion

A work of fiction usually has an ending: the murderer is unmasked, the prince and princess live happily ever after, or Romeo and Juliet lie dead. Sometimes, however, the writer purposely leaves the reader in suspense: the hero lifts up 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 is left for the reader to continue. The theory described in this book has already been applied to numerous problems, but there are countless more possible applications and extensions which have been described elsewhere, and many more extensions and applications remain to be discovered. May happiness attend your search.

238


Synopsis of Chapter 8: Inverse Problems

Well-posed problems: existence, uniqueness, stability The operator equation: If X is infinite dimensional a compact operator A from X onto Y cannot have a bounded inverse. Corollary 8.2.1 Tikhonov’s Theorem 8.2.1 If 5 is compact then 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) The effect of error (8.4.14)

Morozov’s discrepancy principle

(8.4.13)

is continuous.


239

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 uses the methods of functional analysis, and has many instructive examples from the theory of Fredholm integral equations. The reader who has studied the present book will have more than sufficient background knowledge in functional analysis to understand it. Classical treatment of the abstract theory of ill-posed problems is to be found in 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 have studied in this chapter is C.W. Groetsch, Inverse Problems in the Mathematical Sciences, Vieweg, Braunschweig, 1993. This motivates the study of inverse problems by many examples taken from different areas of mathematics, physics and engineering. It provides a very brief summary of functional analysis and then applies it to the inverse problem stated as a Fredholm integral equation of the first kind, or more generally as the equation The principal aim of Chapter 8 has been to expand on Groetsch’s treatment, trying to fill in some of the steps which he left to the reader. The book has a valuable guide to the literature.

Index

a-priori estimate, 129 absolute convergence of series of operators, 142 accumulation point, 25 approximation in a Hilbert space, 106 Faedo–Galerkin, 135 in a normed space, 103 Ritz, 128 Arzelà, Cesare, 176 Arzelà–Ascoli theorem, 176 Ascoli, Guilio, 176 axiom triangle, 20 axioms inner product, 58 metric, 20 norm, 41

set of real numbers, 5 uniformly, 175 bounded above, 7 bounded below, 7 Buniakowski, Victor Yakovlevich, 59 Cantor’s theorem, 100 Cantor, Georg, 100 Cauchy problem, 179 for N equations, 181 Cauchy sequence, 3, 30 weak, 120 stationary, 4 Cauchy, Augustin-Louis, 3 Cauchy–Schwarz inequality, 59 class equivalence, 32 null, 52 closed interval, 5 set, 26 set of real numbers, 5 closed system, 117 closure, 25 of a set of real numbers, 6 compact sequentially, 168 set, 170 set of real numbers, 6 compact linear operator, 183 limit of, 185 compact linear operators in a separable Hilbert space, 190 product of, 184 compact operator resolvent of, 208 compact support, 10 complete system, 114

ball open, 9, 25 Banach space, 44 Banach’s fixed point theorem, 36 Banach’s open mapping theorem, 151, 221 Banach, Stefan, 36 Banach–Steinhaus theorem, 144, 145 basis for normed linear space, 113 Bernoulli, Daniel, 74 Bernstein polynomial, 15 Bernstein, Serge, 15 Bessel’s inequality, 116 Bessel, Friedrich Wilhelm, 116 Bolzano, Bernard, 6 Bolzano–Weierstrass theorem, 6 Borel, Emile, 168 bounded 241

242

Index

completeness of IR, 5 completion of a metric space, 32 cone property, 96 contact point, 25 continuation of operator, 144 continuity of a function, 10 of an operator, 46 of inverse operator, 147 continuous functional, 174 continuous function, 10 contraction mapping, 36 contraction operator, 36 convergence in a metric space, 29 of linear operators, 142 pointwise, 143 strong, 143 uniform, 143 weak, 120 convergent sequence, 2, 29 convex, 105 correspondence isometric, 32 countable, 99 countable dense subset, 101 cover, 169 d’Alembert, Jean le Rond, 68 degenerate, 148 derivative generalized, 67, 157 direct problem, 219 Dirichlet, Gustave Peter Lejeune, 133 distance Euclidean, 19 domain, 9 of operator, 35 dual space, 110 eigenvalue, 196 Einstein’s double suffix summation convention, 83 Einstein’s summation convention, 85 Einstein, Albert, 83

elastic body with free boundary, 88 energy space, 25 for clamped membrane, 78 for elastic body, 86 separability of, 103 equal almost everywhere, 53 equicontinuous, 176 equivalence class, 32 of Cauchy sequences, 4 representative of a, 4 stationary, 32 equivalent norms, 41 equivalent Cauchy sequences, 3 equivalent metrics, 20 equivalent sequence, 32 Euclidean distance, 9, 19 Euler’s method justification of, 181 Euler, Leonhard, 74 Euler–Bernoulli beam, 74 evolution problems, 132 extension of a function, 12 of an operator, 155 of operator, 144 external forces work of, 68, 84 Faedo, 135 Faedo–Galerkin approximation, 135 family of functionals, 175 equicontinuous, 175 uniformly bounded, 175 finite set of real numbers, 5 fixed point, 36 Fourier expansion, 21 Fourier coefficients, 116 Fourier series, 116 Fourier, Jean Baptiste, 115 Fredholm alternative, 159, 206 Fredholm integral equation, 220 Fredholm integral operator, 162, 187 Fredholm operator, 61 Fredholm, Ivar, 148

Index Friedrichs’ inequality, 78 Friedrichs, Kurt Otto, 78 function of compact support, 10 continuous, 10 definition of, 9 extension of, 12 support of, 9 tent, 52 uniformly continuous, 11 functional, 35 complex, 35 continuous, 174 real, 35 uniformly continuous, 175 weakly continuous, 211 work, 68, 76, 84 Galerkin, Boris Grigor’evich, 135 generalized solution, 224 for eigenvalue problem, 82 for free vibration of a membrane, 113 for Neumann problem, 82 for plate, 84 for the rod, 71 Generalized solutions for evolution problems, 132 Gram, Jórgen Pedersen, 115 Gram–Schmidt process, 115 graph of an operator, 153 Hölder condition, 91 Hölder continuous, 91 Hölder’s inequality, 48 for integrals, 55 Hölder, Ludwig Otto, 48 Hadamard, Jacques Salomon, 219 Hausdorff, Felix, 171 heat transfer equation, 132 Heine, Heinrich Eduard, 168 Hermite, Charles, 162 Hilbert identity, 200 Hilbert space, 60 orthogonal decomposition of, 108 separable, 115, 144, 190 Hilbert space, approximation in, 106 Hilbert, David, 60

ill-posed problem, 219 image, 35 imbedding, 55, 67, 75 induced inner product, 60 metric, 26 normed, 41 inequality Friedrichs’, 78 Poincaré’s, 81 Bessel’s, 116 Cauchy–Schwarz, 59 Friedrich’s, 87 Hölder’s, 48 Hölder’s integral, 55 Jensen’s, 50 Korn’s, 86, 87 Minkowski’s, 49 Minkowski’s integral, 51 Poincaré’s, 89 Schwarz, 59 triangle, 20 infimum, 7 inner product induced, 60 inner product space, 58 integral Lebesgue, 54 Riemann, 51 integral equation Fredholm, 220 integral operator compact, 185 interior point, 25 inverse operator continuity of, 147 inverse problem, 220 isometric, 32 Jensen’s inequality, 50 Jensen, Valdemar, 50 kernel, 109 degenerate, 148 kinetic energy of membrane, 82 Korn, 86 Korn’s inequality, 86

243

244

Index

Lebesgue integral, 54 Lebesgue space, 51 separability of, 102 Lebesgue, Henri Léon, 54 limit, 29 limit point, 25 linear elasticity, 85 linear functional kernel of, 109 Linear operator space of, 141 linear operator, 45 bounded, 47 continuous, 35 domain of, 45 norm of, 46, 141 linear operators product of, 143 linearly independent, 42 Lipschitz continuous, 91 Lipschitz property, 96 Lipschitz, Rudolf Otto Sigismund, 91 mapping contraction, 36 matrix Hermitian, 162 matrix operator infinite dimensional, 186 maximum metric, 27 maximum value, 11 membrane, 78 clamped, 78 metric, 20 axioms, 20 equivalent, 20 induced, 26 maximum, 27 uniform, 27 metric space complete, 30 completion of, 32 incomplete, 30 separable, 101 minimum value, 11 Minkowski’s inequality, 49 Minkowski, Hermann, 49 Moore, Eliakim Hastings, 225

Moore–Penrose generalized inverse, 225 Morozov, 234 Morozov’s discrepancy principle, 234 natural boundary condition, 73, 81 natural end condition, 70 natural end conditions, 77 natural frequencies of clamped membrane, 82 neighborhood, 25 25 norm, 41 of an operator, 46 axioms, 41 equivalent, 43, 152 induced, 41 Sobolev, 89 normed linear space, 41 basis for, 113 strictly normed, 105 norms equivalent, 41 null class, 52 null sequence, 53 null space, 158 open ball, 25 interval, 5 set, 25 set of real numbers, 5 open ball, 9 operator, 35 compact, 183 continuation of, 144 integral, 162 adjoint, 157 bounded, 47 closed extension of, 155 closed linear, 152, 154 continuous, 35 continuously invertible, 148 contraction, 36 coordinate, 198 domain of, 35 eigenvalue of, 196 extension of, 144 fixed point of, 36 Fredholm, 61

Index Fredholm integral, 162, 187 graph of, 153 imbedding, 56, 67, 76 integral, 61, 148 inverse, 147 linear, 45 matrix, 162 norm of, 46 null space of, 158 projection, 144 range of, 35 residual spectrum of, 196 resolvent set of, 195 self-adjoint, 160 spectrum of, 195 strictly positive, 215 weakly continuous, 161 orthogonal, 60 mutually, 108 orthogonal decomposition, 108 orthogonal system, 115 orthonormal, 115 parallelogram law, 59 Parseval’s equality, 117 Parseval, Marc Antoine, 117 Peano’s local existence theorem, 179 Peano, Giuseppe, 179 Penrose, Roger, 225 Picard’s existence criterion, 228 Picard, Charles Emile, 228 plate, 83 stability of, 163 Poincaré’s inequality, 81, 89 Poincaré, Jules Henri, 81 point accumulation, 25 contact, 25 fixed, 36 interior, 25 isolated, 26 limit, 25 principle of uniform boundedness, 146 Principle of Minimum Energy, 71 Principle of Virtual Work, 68, 76 problem direct, 219

245

ill-posed, 219 inverse, 220 well posed, 219 product scalar, 40 product space, 153 range of operator, 35 real number, 4 regular points, 195 representative sequence, 32 resolvent operator, 199 resolvent set, 195 Riemann integral, 51 Riemann, Georg Friedrich Bernhard, 51 Riemann–Lebesgue lemma, 119 Riesz’s representation theorem, 106, 110 Riesz, Frédéric, 106 Ritz method, 128 Ritz, Walter, 128 rod cantilever, 65 free, 73 Saint Venant, 163 Schauder, Pavel Julius, 201 Schmidt, Erhard, 115 Schwarz inequality, 59 Schwarz, Hermand Amandus, 59 semi-norm, 88 sequence Cauchy, 30 Cauchy, 3 convergent, 2, 29 null, 53 representative, 32 set closed, 26, 60 closure of, 25 compact, 170 convex, 105 countable, 99 cover of, 169 dense, 26, 31 linearly dependent, 42 of measure zero, 53 open, 25 separable, 174 sequentially compact, 168

246

Index

weakly closed, 125 set of real numbers bounded, 5 closed, 5 compact, 6 finite, 5 infimum of, 7 open, 5 supremum of, 7 singular system, 227 singular value decomposition, 227 singular values, 227 Sobolev norm, 89 Sobolev space, 88 separability of, 102 Sobolev’s imbedding theorem, 95 Sobolev, Sergei L’vovich, 88 solution generalize, 224 least squares, 225 minimum norm, 225 space infinite dimensional, 42 separable, 101, 174 Sobolev, 88 Banach, 44 complete, 30 dual, 110 energy, 25 finite dimensional, 42 Hilbert, 60 incomplete, 30 inner product, 58 Lebesgue, 51 linear, 40 product, 153 real inner product, 58 separable, 99 strictly normed linear, 105 weakly complete, 125 space inner product, 58 spectrum residual, 196 continuous, 196 of compact linear operator, 201, 208 of coordinate operator, 198 of differential operators, 196

of linear operators, 195 of membrane equation, 196 stationary Cauchy sequence, 4 Steinhaus, Hugo Dyonis, 145 strain energy of linearly elastic body, 85 of membrane, 78, 82 of plate, 83 of rod, 65 of string, 24 subspace closed, 42, 60 dimension of, 42 finite dimensional, 42 linear, 45 of a metric space, 26 subspaces mutually orthogonal, 108 supremum, 7 system closed, 117 complete, 114 orthogonal, 115 orthonormal, 117 tent function, 52 Theorem Weierstrass’ polynomial approximation, 14 Weierstrass’ uniform convergence, 13 Banach’s open mapping, 221 theorem Banach’s fixed point, 36 Weierstrass’ polynomial approximation, 31 Arzelà–Ascoli, 176 Banach’s open mapping, 151 Banach–Steinhaus, 144, 145 Bolzano–Weierstrass, 6 Cantor’s, 100 closed graph, 154, 155 contraction mapping, 36 Heine–Borel, 169 imbedding, 55 Riesz’s representation, 106, 109, 110 Sobolev’s imbedding, 95 Tikhonov, 223 Tikhonov’s theorem, 223

Index Tikhonov, Andrei Nikolaevich, 223 total energy of rod, 71 triangle axiom, 20 triangle inequality, 20 uniform boundedness principle of, 123 uniform convergence of sequence of operators, 143 uniform metric, 27 uniformly bounded, 175 uniformly continuous functional, 175 variational formulation for rod, 71 virtual displacement, 68 Virtual Work Principle of, 68, 76 weak Cauchy sequence, 120 Convergence, 120 weakly closed, 125 weakly complete, 125 Weierstrass’ polynomial approximation theorem, 14 uniform convergence theorem, 13 Weierstrass’ polynomial approximation theorem, 31 Weierstrass, Karl Theodor Wilhelm, 6 well posed problem, 219 work of external forces, 68, 84 Young, Thomas, 65 zero almost everywhere, 53

247

Mechanics SOLID 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 of this series is to provide lucid accounts written by authoritative researchers giving vision and insight in answering these questions on the subject of mechanics as it relates to solids. The scope of the series covers the entire spectrum of solid mechanics. Thus it includes the foundation of mechanics; variational formulations; computational mechanics; statics, kinematics and dynamics of rigid and elastic bodies; vibrations of solids and structures; dynamical systems and chaos; the theories of elasticity, plasticity and viscoelasticity; composite materials; rods, beams, shells and membranes; structural control and stability; soils, rocks and geomechanics; fracture; tribology; experimental mechanics; 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., 1990 ISBN 0-7923-0608-2 J.J. Kalker: Three-Dimensional Elastic Bodies in Rolling Contact. 1990 ISBN 0-7923-0712-7 P. Karasudhi: Foundations of Solid Mechanics. 1991 ISBN 0-7923-0772-0 Not published Not published. J.F. Doyle: Static and Dynamic Analysis of Structures. With an Emphasis on Mechanics and Computer Matrix Methods. 1991 ISBN 0-7923-1124-8; Pb 0-7923-1208-2 O.O. Ochoa and J.N. Reddy: Finite Element Analysis of Composite Laminates. ISBN 0-7923-1125-6 M.H. Aliabadi and D.P. Rooke: Numerical Fracture Mechanics. ISBN 0-7923-1175-2

J. Angeles and C.S. López-Cajún: Optimization of Cam Mechanisms. 1991 ISBN 0-7923-1355-0 D.E. Grierson, A. Franchi and P. Riva (eds.): Progress in Structural Engineering. 1991 ISBN 0-7923-1396-8 R.T. Haftka and Z. Gürdal: Elements of Structural Optimization. 3rd rev. and exp. ed. 1992 ISBN 0-7923-1504-9; Pb 0-7923-1505-7 J.R. Barber: Elasticity. 1992 ISBN 0-7923-1609-6; Pb 0-7923-1610-X H.S. Tzou and G.L. Anderson (eds.): Intelligent Structural Systems. 1992 ISBN 0-7923-1920-6 E.E. Gdoutos: Fracture Mechanics. An Introduction. 1993 ISBN 0-7923-1932-X J.P. Ward: Solid Mechanics. An Introduction. 1992 ISBN 0-7923-1949-4 M. Farshad: Design and Analysis of Shell Structures. 1992 ISBN 0-7923-1950-8 H.S. Tzou and T. Fukuda (eds.): Precision Sensors, Actuators and Systems. 1992 ISBN 0-7923-2015-8 J.R. Vinson: The Behavior of Shells Composed of Isotropic and Composite Materials. 1993 ISBN 0-7923-2113-8 H.S. Tzou: Piezoelectric Shells. Distributed Sensing and Control of Continua. 1993 ISBN 0-7923-2186-3 W. Schiehlen (ed.): Advanced Multibody System Dynamics. Simulation and Software Tools. 1993 ISBN 0-7923-2192-8 C.-W. Lee: Vibration Analysis of Rotors. 1993 ISBN 0-7923-2300-9 D.R. Smith: An Introduction to Continuum Mechanics. 1993 ISBN 0-7923-2454-4 G.M.L. Gladwell: Inverse Problems in Scattering. An Introduction. 1993 ISBN 0-7923-2478-1

Mechanics SOLID MECHANICS AND ITS APPLICATIONS Series Editor. G.M.L. Gladwell 24. 25. 26. 27. 28. 29. 30. 31. 32.

33. 34. 35. 36. 37.

38. 39.

40. 41. 42. 43.

44. 45. 46.

47.

48.

G. Prathap: The Finite Element Method in Structural Mechanics. 1993 ISBN 0-7923-2492-7 ISBN 0-7923-2510-9 J. Herskovits (ed.): Advances in Structural Optimization. 1995 ISBN 0-7923-2536-2 M.A. González-Palacios and J. Angeles: Cam Synthesis. 1993 ISBN 0-7923-2580-X W.S. Hall: The Boundary Element Method. 1993 J. Angeles, G. Hommel and P. Kovács (eds.): Computational Kinematics. 1993 ISBN 0-7923-2585-0 ISBN 0-7923-2761-6 A. Curnier: Computational Methods in Solid Mechanics. 1994 ISBN 0-7923-2866-3 D.A. Hills and D. Nowell: Mechanics of Fretting Fatigue. 1994 B. Tabarrok and F.P.J. Rimrott: Variational Methods and Complementary Formulations in ISBN 0-7923-2923-6 Dynamics. 1994 E.H. Dowell (ed.), E.F. Crawley, H.C. Curtiss Jr., D.A. Peters, R. H. Scanlan and F. Sisto: A Modern Course in Aeroelasticity. Third Revised and Enlarged Edition. 1995 ISBN 0-7923-2788-8; Pb: 0-7923-2789-6 ISBN 0-7923-3036-6 A. Preumont: Random Vibration and Spectral Analysis. 1994 J.N. Reddy (ed.): Mechanics of Composite Materials. Selected works of Nicholas J. Pagano. ISBN 0-7923-3041-2 1994 ISBN 0-7923-3329-2 A.P.S. Selvadurai (ed.): Mechanics of Poroelastic Media. 1996 Z. Mróz, D. Weichert, S. Dorosz (eds.): Inelastic Behaviour of Structures under Variable ISBN 0-7923-3397-7 Loads. 1995 R. Pyrz (ed.): IUTAM Symposium on Microstructure-Property Interactions in Composite Materials. Proceedings of the IUTAM Symposium held in Aalborg, Denmark. 1995 ISBN 0-7923-3427-2 M.I. Friswell and J.E. Mottershead: Finite Element Model Updating in Structural Dynamics. ISBN 0-7923-3431-0 1995 D.F. Parker and A.H. England (eds.): IUTAM Symposium on Anisotropy, Inhomogeneity and Nonlinearity in Solid Mechanics. Proceedings of the IUTAM Symposium held in Nottingham, ISBN 0-7923-3594-5 U.K. 1995 J.-P. Merlet and B. Ravani (eds.): Computational Kinematics ’95. 1995 ISBN 0-7923-3673-9 L.P. Lebedev, I.I. Vorovich and G.M.L. Gladwell: Functional Analysis. Applications in Mechanics and Inverse Problems. 1996 ISBN 0-7923-3849-9 Mechanics of Components with Treated or Coated Surfaces. 1996 ISBN 0-7923-3700-X D. Bestle and W. Schiehlen (eds.): IUTAM Symposium on Optimization of Mechanical Systems. Proceedings of the IUTAM Symposium held in Stuttgart, Germany. 1996 ISBN 0-7923-3830-8 D.A. Hills, P.A. Kelly, D.N. Dai and A.M. Korsunsky: Solution of Crack Problems. The Distributed Dislocation Technique. 1996 ISBN 0-7923-3848-0 V.A. Squire, R.J. Hosking, A.D. Kerr and P.J. Langhorne: Moving Loads on Ice Plates. 1996 ISBN 0-7923-3953-3 A. Pineau and A. Zaoui (eds.): IUTAM Symposium on Micromechanics of Plasticity and Damage of Multiphase Materials. Proceedings of the IUTAM Symposium held in Sèvres, Paris, France. 1996 ISBN 0-7923-4188-0 A. Naess and S. Krenk (eds.): IUTAM Symposium on Advances in Nonlinear Stochastic Mechanics. Proceedings of the IUTAM Symposium held in Trondheim, Norway. 1996 ISBN 0-7923-4193-7 D. and A. Scalia: Thermoelastic Deformations. 1996 ISBN 0-7923-4230-5

Mechanics SOLID MECHANICS AND ITS APPLICATIONS Series Editor. G.M.L. Gladwell 49. 50. 51. 52.

53.

54. 55. 56. 57. 58. 59. 60.

61. 62.

63.

64. 65. 66.

67. 68.

J.R. Willis (ed): IUTAM Symposium on Nonlinear Analysis of Fracture. Proceedings of the ISBN 0-7923-4378-6 IUTAM Symposium held in Cambridge, U.K. 1997 A. Preumont: Vibration Control of Active Structures. An Introduction. 1997 ISBN 0-7923-4392-1 G.P. Cherepanov: Methods of Fracture Mechanics: Solid Matter Physics. 1997 ISBN 0-7923-4408-1 D.H. van Campen (ed.): IUTAM Symposium on Interaction between Dynamics and Control in Advanced Mechanical Systems. Proceedings of the IUTAM Symposium held in Eindhoven, ISBN 0-7923-4429-4 The Netherlands. 1997 N. A. Fleck and A.C.F. Cocks (eds.): IUTAM Symposium on Mechanics of Granular and Porous Materials. Proceedings of the IUTAM Symposium held in Cambridge, U.K. 1997 ISBN 0-7923-4553-3 J. Roorda and N.K. Srivastava (eds.): Trends in Structural Mechanics. Theory, Practice, Education. 1997 ISBN 0-7923-4603-3 Yu. A. Mitropolskii and N. Van Dao: Applied Asymptotic Methods in Nonlinear Oscillations. ISBN 0-7923-4605-X 1997 C. Guedes Soares (ed.): Probabilistic Methods for Structural Design. 1997 ISBN 0-7923-4670-X D. A. Pineau and A. Zaoui: Mechanical Behaviour of Materials. Volume I: Elasticity ISBN 0-7923-4894-X and Plasticity. 1998 D. A. Pineau and A. Zaoui: Mechanical Behaviour of Materials. Volume II: ViscoISBN 0-7923-4895-8 plasticity, Damage, Fracture and Contact Mechanics. 1998 L.T. Tenek and J. Argyris: Finite Element Analysis for Composite Structures. 1998 ISBN 0-7923-4899-0 Y.A. Bahei-El-Din and G.J. Dvorak (eds.): IUTAM Symposium on Transformation Problems in Composite and Active Materials. Proceedings of the IUTAM Symposium held in Cairo, ISBN 0-7923-5122-3 Egypt. 1998 ISBN 0-7923-5257-2 I.G. Goryacheva: Contact Mechanics in Tribology. 1998 O.T. Bruhns and E. Stein (eds.): IUTAM Symposium on Micro- and Macrostructural Aspects of Thermoplasticity. Proceedings of the IUTAM Symposium held in Bochum, Germany. 1999 ISBN 0-7923-5265-3 F.C. Moon: IUTAM Symposium on New Applications of Nonlinear and Chaotic Dynamics in Mechanics. Proceedings of the IUTAM Symposium held in Ithaca, NY, USA. 1998 ISBN 0-7923-5276-9 R. Wang: IUTAM Symposium on Rheology of Bodies with Defects. Proceedings of the IUTAM ISBN 0-7923-5297-1 Symposium held in Beijing, China. 1999 Yu.I. Dimitrienko: Thermomechanics of Composites under High Temperatures. 1999 ISBN 0-7923-4899-0 P. Argoul, M. Frémond and Q.S. Nguyen (eds.): IUTAM Symposium on Variations of Domains and Free-Boundary Problems in Solid Mechanics. Proceedings of the IUTAM Symposium ISBN 0-7923-5450-8 held in Paris, France. 1999 F. J. Fahy and W.G. Price (eds.): IUTAM Symposium on Statistical Energy Analysis. Proceedings ISBN 0-7923-5457-5 of the IUTAM Symposium held in Southampton, U.K. 1999 H.A. Mang and F.G. Rammerstorfer (eds.): IUTAM Symposium on Discretization Methods in Structural Mechanics. Proceedings of the IUTAM Symposium held in Vienna, Austria. 1999 ISBN 0-7923-5591-1

Mechanics SOLID MECHANICS AND ITS APPLICATIONS Series Editor. G.M.L. Gladwell 69.

70. 71. 72. 73.

74. 75. 76. 77. 78. 79. 80.

81.

82.

83. 84. 85.

86. 87.

88. 89.

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 ISBN 0-7923-5615-2 S.K. Agrawal and B.C. Fabien: Optimization of Dynamic Systems. 1999 ISBN 0-7923-5681-0 A. Carpinteri: Nonlinear Crack Models for Nonmetallic Materials. 1999 ISBN 0-7923-5750-7 F. Pfeifer (ed.): IUTAM Symposium on Unilateral Multibody Contacts. Proceedings of the ISBN 0-7923-6030-3 IUTAM Symposium held in Munich, Germany. 1999 E. Lavendelis and M. Zakrzhevsky (eds.): IUTAM/IFToMM Symposium on Synthesis of Nonlinear Dynamical Systems. Proceedings of the IUTAM/IFToMM Symposium held in Riga, ISBN 0-7923-6106-7 Latvia. 2000 ISBN 0-7923-6308-6 J.-P. Merlet: Parallel Robots. 2000 J.T. Pindera: Techniques of Tomographic Isodyne Stress Analysis. 2000 ISBN 0-7923-6388-4 G.A. Maugin, R. Drouot and F. Sidoroff (eds.): Continuum Thermomechanics. The Art and ISBN 0-7923-6407-4 Science of Modelling Material Behaviour. 2000 N. Van Dao and E.J. Kreuzer (eds.): IUTAM Symposium on Recent Developments in Non-linear ISBN 0-7923-6470-8 Oscillations of Mechanical Systems. 2000 S.D. Akbarov and A.N. Guz: Mechanics of Curved Composites. 2000 ISBN 0-7923-6477-5 ISBN 0-7923-6489-9 M.B. Rubin: Cosserat Theories: Shells, Rods and Points. 2000 S.Pellegrino and S.D. Guest (eds.): IUTAM-IASS Symposium on Deployable Structures: Theory and Applications. Proceedings of the IUTAM-IASS Symposium held in Cambridge, U.K., 6–9 ISBN 0-7923-6516-X September 1998. 2000 A.D. Rosato and D.L. Blackmore (eds.): IUTAM Symposium on Segregation in Granular Flows. Proceedings of the IUTAM Symposium held in Cape May, NJ, U.S.A., June 5–10, ISBN 0-7923-6547-X 1999. 2000 A. Lagarde (ed.): IUTAM Symposium on Advanced Optical Methods and Applications in Solid Mechanics. Proceedings of the IUTAM Symposium held in Futuroscope, Poitiers, France, ISBN 0-7923-6604-2 August 31–September 4, 1998. 2000 D. Weichert and G. Maier (eds.): Inelastic Analysis of Structures under Variable Loads. Theory ISBN 0-7923-6645-X and Engineering Applications. 2000 T.-J. Chuang and J.W. Rudnicki (eds.): Multiscale Deformation and Fracture in Materials and ISBN 0-7923-6718-9 Structures. The James R. Rice 60th Anniversary Volume. 2001 S. Narayanan and R.N. lyengar (eds.): IUTAM Symposium on Nonlinearity and Stochastic Structural Dynamics. Proceedings of the IUTAM Symposium held in Madras, Chennai, India, ISBN 0-7923-6733-2 4–8 January 1999 S. Murakami and N. Ohno (eds.): IUTAM Symposium on Creep in Structures. Proceedings of the IUTAM Symposium held in Nagoya, Japan, 3-7 April 2000. 2001 ISBN 0-7923-6737-5 W. Ehlers (ed.): IUTAM Symposium on Theoretical and Numerical Methods in Continuum Mechanics of Porous Materials. Proceedings of the IUTAM Symposium held at the University ISBN 0-7923-6766-9 of Stuttgart, Germany, September 5-10, 1999. 2001 D. Durban, D. Givoli and J.G. Simmonds (eds.): Advances in the Mechanis of Plates and Shells ISBN 0-7923-6785-5 The Avinoam Libai Anniversary Volume. 2001 U. Gabbert and H.-S. Tzou (eds.): IUTAM Symposium on Smart Structures and Structonic Systems. Proceedings of the IUTAM Symposium held in Magdeburg, Germany, 26–29 September 2000. 2001 ISBN 0-7923-6968-8

Mechanics SOLID MECHANICS AND ITS APPLICATIONS Series Editor. G.M.L. Gladwell Y. Ivanov, V. Cheshkov and M. Natova: Polymer Composite Materials - Interface Phenomena & Processes. 2001 ISBN 0-7923-7008-2 91. R.C. McPhedran, L.C. Botten and N. A. Nicorovici (eds.): IUTAM Symposium on Mechanical and Electromagnetic Waves in Structured Media. Proceedings of the IUTAM Symposium held in Sydney, NSW, Australia, 18-22 Januari 1999. 2001 ISBN 0-7923-7038-4 92. D.A. Sotiropoulos (ed.): IUTAM Symposium on Mechanical Waves for Composite Structures Characterization. Proceedings of the IUTAM Symposium held in Chania, Crete, Greece, June 14-17, 2000. 2001 ISBN 0-7923-7164-X 93. V.M. Alexandrov and D.A. Pozharskii: Three-Dimensional Contact Problems. 2001 ISBN 0-7923-7165-8 94. J.P. Dempsey and H.H. Shen (eds.): IUTAM Symposium on Scaling Laws in Ice Mechanics and Ice Dynamics. Proceedings of the IUTAM Symposium held in Fairbanks, Alaska, U.S.A., 13-16 June 2000. 2001 ISBN 1-4020-0171-1 95. U. Kirsch: Design-Oriented Analysis of Structures. A Unified Approach. 2002 ISBN 1-4020-0443-5 96. A. Preumont: Vibration Control of Active Structures. An Introduction ( Edition). 2002 ISBN 1-4020-0496-6 97. B.L. Karihaloo (ed.): IUTAM Symposium on Analytical and Computational Fracture Mechanics of Non-Homogeneous Materials. Proceedings of the IUTAM Symposium held in Cardiff, U.K., 18-22 June 2001. 2002 ISBN 1-4020-0510-5 98. S.M. Han and H. Benaroya: Nonlinear and Stochastic Dynamics of Compliant Offshore Structures. 2002 ISBN 1-4020-0573-3 99. A.M. Linkov: Boundary Integral Equations in Elasticity Theory. 2002 ISBN 1-4020-0574-1 100. L.P. Lebedev, I.I. Vorovich and G.M.L. Gladwell: Functional Analysis. Applications in Mechanics and Inverse Problems ( Edition). 2002 ISBN 1-4020-0667-5; Pb 1-4020-0756-6 90.

Kluwer Academic Publishers – Dordrecht / Boston / London

ICASE/LaRC Interdisciplinary Series in Science and Engineering 1. 2.

3.

4. 5. 6. 7.

J. Buckmaster, T.L. Jackson and A. Kumar (eds.): Combustion in High-Speed Flows. 1994 ISBN 0-7923-2086-X M.Y. Hussaini, T.B. Gatski and T.L. Jackson (eds.): Transition, Turbulence and Combustion. Volume I: Transition. 1994 ISBN 0-7923-3084-6; set 0-7923-3086-2 M.Y. Hussaini, T.B. Gatski and T.L. Jackson (eds.): Transition, Turbulence and Combustion. Volume II: Turbulence and Combustion. 1994 ISBN 0-7923-3085-4; set 0-7923-3086-2 D.E. Keyes, A. Sameh and V. Venkatakrishnan (eds): Parallel Numerical Algorithms. 1997 ISBN 0-7923-4282-8 T.G. Campbell, R.A. Nicolaides and M.D. Salas (eds.): Computational ElectromagISBN 0-7923-4733-1 netics and Its Applications. 1997 V. Venkatakrishnan, M.D. Salas and S.R. Chakravarthy (eds.): Barriers and Challenges in Computational Fluid Dynamics. 1998 ISBN 0-7923-4855-9 M.D. Salas, J.N. Hefner and L. Sakell (eds.): Modeling Complex Turbulent Flows. 1999 ISBN 0-7923-5590-3

KLUWER ACADEMIC PUBLISHERS – DORDRECHT / BOSTON / LONDON

Functional Analysis: Applications in Mechanics and Inverse Problems (Solid Mechanics and Its Applications)

Functional Analysis: Applications in Mechanics and Inverse Problems

Functional analysis: Applications in mechanics and inverse problems

Fracture Mechanics: Inverse Problems and Solutions (Solid Mechanics and Its Applications)

Elasticity of Transversely Isotropic Materials (Solid Mechanics and Its Applications) (Solid Mechanics and Its Applications)

Elasticity (Solid Mechanics and Its Applications)

Elasticity (Solid Mechanics and Its Applications)

Non-Linear Mechanics of Materials (Solid Mechanics and Its Applications)

Fracture Mechanics: An Introduction (Solid Mechanics and Its Applications)

Design-Oriented Analysis of Structures: A Unified Approach (Solid Mechanics and Its Applications, Volume 95) (Solid Mechanics and Its Applications)

Wave Mechanics and Its Applications

Functional Analysis and its Applications

Functional analysis and its applications

Applications of Methods of Functional Analysis to Problems in Mechanics

Applications of Methods of Functional Analysis to Problems in Mechanics

Continuum and solid mechanics : concepts and applications

A Modern Course in Aeroelasticity (Solid Mechanics and Its Applications)

Inverse problems in engineering mechanics

Inverse Problems in Engineering Mechanics

Functional analysis in mechanics

Functional Analysis in Mechanics

Functional Analysis in Mechanics

Functional Analysis in Mechanics

Functional Analysis in Mechanics

Functional analysis in mechanics

Functional Analysis in Mechanics

Functional analysis in mechanics

Vibration Control of Active Structures (Solid Mechanics and Its Applications)

Elasticity of Transversely Isotropic Materials (Solid Mechanics and Its Applications)

An introduction to structural optimization (Solid Mechanics and Its Applications)

Flexible Multibody Dynamics (Solid Mechanics and Its Applications 176)

Functional Analysis: Applications in Mechanics and Inverse Problems (Solid Mechanics and Its Applications)