Functional Analysis
SOLID MECHANICS AND ITS APPLICATIONS Volume 100 Series Editor:
G.M.L. GLADWELL Department of Civi...
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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
1.1 Real and complex numbers
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
1.1 Real and complex numbers
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
1.2 Theory of functions
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:
1.2 Theory of functions
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
2. Introduction to Metric Spaces
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
2. Introduction to Metric Spaces
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
2. Introduction to Metric Spaces
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
2. Introduction to Metric Spaces
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
2. Introduction to Metric Spaces
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
2. Introduction to Metric Spaces
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
2. Introduction to Metric Spaces
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
2. Introduction to Metric Spaces
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
2. Introduction to Metric Spaces
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
2.7 An introduction to operators
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
2. Introduction to Metric Spaces
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
2.7 An introduction to operators
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.
2. Introduction to Metric Spaces
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.
2. Introduction to Metric Spaces
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
2. Introduction to Metric Spaces
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
2. Introduction to Metric Spaces
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
2. Introduction to Metric Spaces
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
2. Introduction to Metric Spaces
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
2. Introduction to Metric Spaces
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,
2.11 Lebesgue spaces
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
2. Introduction to Metric Spaces
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
2.11 Lebesgue spaces
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
2. Introduction to Metric Spaces
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
2.11 Lebesgue spaces
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.
2. Introduction to Metric Spaces
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
2. Introduction to Metric Spaces
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
2.12 Inner product spaces
Problem 2.12.6 What are the Schwarz inequalities for
61
and the real
Problem 2.12.7 Show that
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
2. Introduction to Metric Spaces
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
2.12 Inner product spaces
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
3. Energy Spaces and Generalized Solutions
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
3. Energy Spaces and Generalized Solutions
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
3. Energy Spaces and Generalized Solutions
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
3. Energy Spaces and Generalized Solutions
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
3. Energy Spaces and Generalized Solutions
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
3. Energy Spaces and Generalized Solutions
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
3. Energy Spaces and Generalized Solutions
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
3. Energy Spaces and Generalized Solutions
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
3. Energy Spaces and Generalized Solutions
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
3. Energy Spaces and Generalized Solutions
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
3. Energy Spaces and Generalized Solutions
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.
3. Energy Spaces and Generalized Solutions
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
3. Energy Spaces and Generalized Solutions
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
Problem 3.6.1 Show that
is a norm which is equivalent to
on
3.7 Some imbedding theorems In these first three chapters we have introduced a number of function spaces, and at various times, e.g. in § 2.11, § 3.1, § 3.2, we have shown that one 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
3. Energy Spaces and Generalized Solutions
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
3.7 Some imbedding theorems
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
3. Energy Spaces and Generalized Solutions
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
3.7 Some imbedding theorems
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
3. Energy Spaces and Generalized Solutions
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
3.7 Some imbedding theorems
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
3. Energy Spaces and Generalized Solutions
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
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4. Approximation in a Normed Linear Space
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
4. Approximation in a Normed Linear Space
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.
4. Approximation in a Normed Linear Space
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
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4. Approximation in a Normed Linear Space
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
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4. Approximation in a Normed Linear Space
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
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4. Approximation in a Normed Linear Space
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
4. Approximation in a Normed Linear Space
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
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4. Approximation in a Normed Linear Space
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:
4.5 Bases and complete systems
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
4. Approximation in a Normed Linear Space
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),
4.5 Bases and complete systems
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
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4. Approximation in a Normed Linear Space
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
4.5 Bases and complete systems
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
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4. Approximation in a Normed Linear Space
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.
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4. Approximation in a Normed Linear Space
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
4.6 Weak convergence in a Hilbert space
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
4. Approximation in a Normed Linear Space
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
4.6 Weak convergence in a Hilbert space
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
4. Approximation in a Normed Linear Space
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
4. Approximation in a Normed Linear Space
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.
4. Approximation in a Normed Linear Space
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
4. Approximation in a Normed Linear Space
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
4. Approximation in a Normed Linear Space
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
4.9 Generalized solutions of evolution problems
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
4. Approximation in a Normed Linear Space
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
4.9 Generalized solutions of evolution problems
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
4. Approximation in a Normed Linear Space
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.
4.9 Generalized solutions of evolution problems
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
5. Elements of the Theory of Linear Operators
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
5. Elements of the Theory of Linear Operators
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
5. Elements of the Theory of Linear Operators
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
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5. Elements of the Theory of Linear Operators
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
5.3 The inverse operator
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
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5. Elements of the Theory of Linear Operators
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
5.3 The inverse operator
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
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5. Elements of the Theory of Linear Operators
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
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5. Elements of the Theory of Linear Operators
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
5. Elements of the Theory of Linear Operators
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
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5. Elements of the Theory of Linear Operators
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
is orthogonal to every
it is in
is the
5.5 The adjoint operator
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
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5. Elements of the Theory of Linear Operators
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
5.5 The adjoint operator
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
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5. Elements of the Theory of Linear Operators
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
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5. Elements of the Theory of Linear Operators
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
5. Elements of the Theory of Linear Operators
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
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6. Compactness and Its Consequences
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
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6. Compactness and Its Consequences
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
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6. Compactness and Its Consequences
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
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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.
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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
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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
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6. Compactness and Its Consequences
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.
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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
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6. Compactness and Its Consequences
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
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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
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6. Compactness and Its Consequences
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
6. Compactness and Its Consequences
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
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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).
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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.
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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.
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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
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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
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6. Compactness and Its Consequences
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-
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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
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6. Compactness and Its Consequences
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
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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
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6. Compactness and Its Consequences
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.
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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:
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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.
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7. Spectral Theory of Linear Operators
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
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7. Spectral Theory of Linear Operators
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
7. Spectral Theory of Linear Operators
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
7.3 The spectrum of a compact linear operator in a Hilbert space
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
is orthogonal to every
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
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7. Spectral Theory of Linear Operators
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
7.3 The spectrum of a compact linear operator in a Hilbert space
Thus
and
are orthogonal and
which means that
cannot contain a Cauchy sequence.
Theorem 7.3.2
iff
Note that, since
we know that
iff
Proof. Suppose that but Take 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
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7. Spectral Theory of Linear Operators
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:
7.3 The spectrum of a compact linear operator in a Hilbert space
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
7. Spectral Theory of Linear Operators
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
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7. Spectral Theory of Linear Operators
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
7. Spectral Theory of Linear Operators
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
7.5 Self-adjoint operators in a Hilbert space
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
7. Spectral Theory of Linear Operators
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
7.5 Self-adjoint operators in a Hilbert space
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
7. Spectral Theory of Linear Operators
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
7.5 Self-adjoint operators in a Hilbert space
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
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7. Spectral Theory of Linear Operators
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
8. Applications to Inverse Problems
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).
8.2 The operator equation
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:
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8. Applications to Inverse Problems
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
8.2 The operator equation
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.
8. Applications to Inverse Problems
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
8.2 The operator equation
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
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8. Applications to Inverse Problems
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
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8. Applications to Inverse Problems
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
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8. Applications to Inverse Problems
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
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8. Applications to Inverse Problems
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.
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8. Applications to Inverse Problems
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
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8. Applications to Inverse Problems
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
8.5 Morozov’s discrepancy principle
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.
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8. Applications to Inverse Problems
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.
8.5 Morozov’s discrepancy principle
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
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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.
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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
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P. Pedersen and M.P. Bendsøe (eds.): IUTAM Symposium on Synthesis in Bio Solid Mechanics. Proceedings of the IUTAM Symposium held in Copenhagen, Denmark. 1999 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.
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