inte-By studying the convergence under the supremum norm, that is, by asking if agiven function can be approximated uniformly on the entire interval by staircasefunctions, we are led to
Trang 1SinhVienZone.Com
Trang 2Joachim Escher
Analysis II
Translated from the German
by Silvio Levy and Matthew Cargo
Birkhäuser
Basel · Boston · Berlin
SinhVienZone.Com
Trang 3Originally published in German under the same title by Birkhäuser Verlag, Switzerland
© 1999 by Birkhäuser Verlag
2000 Mathematics Subject Classification: 26-01, 26A42, 26Bxx, 30-01
Library of Congress Control Number: 2008926303
Bibliographic information published by Die Deutsche Bibliothek
Die Deutsche Bibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data is available in the Internet at <http://dnb.ddb.de>.
ISBN 3-7643-7472-3 Birkhäuser Verlag, Basel – Boston – Berlin
This work is subject to copyright All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microfilms or in other ways, and storage in data banks For any kind of use permission of the copyright owner must be obtained
© 2008 Birkhäuser Verlag AG
Basel · Boston · Berlin
P.O Box 133, CH-4010 Basel, Switzerland
Part of Springer Science+Business Media
Printed on acid-free paper produced of chlorine-free pulp TCF d
Printed in Germany
Institut für Mathematik Institut für Angewandte MathematikUniversität Zürich Universität Hannover
Trang 4As with the first, the second volume contains substantially more material than can
be covered in a one-semester course Such courses may omit many beautiful andwell-grounded applications which connect broadly to many areas of mathematics
We of course hope that students will pursue this material independently; teachersmay find it useful for undergraduate seminars
For an overview of the material presented, consult the table of contents andthe chapter introductions As before, we stress that doing the numerous exercises
is indispensable for understanding the subject matter, and they also round outand amplify the main text
In writing this volume, we are indebted to the help of many We especiallythank our friends and colleages Pavol Quittner and Gieri Simonett They havenot only meticulously reviewed the entire manuscript and assisted in weeding outerrors but also, through their valuable suggestions for improvement, contributedessentially to the final version We also extend great thanks to our sta2 for theircareful perusal of the entire manuscript and for tracking errata and inaccuracies.Our most heartfelt thank extends again to our “typesetting perfectionist”,without whose tireless e2ort this book would not look nearly so nice.1 We alsothank Andreas for helping resolve hardware and software problems
Finally, we extend thanks to Thomas Hintermann and to Birkh¨auser for thegood working relationship and their understanding of our desired deadlines
1 The text was set in L A TEX, and the figures were created with CorelDRAW! and Maple.
SinhVienZone.Com
Trang 5Foreword to the second edition
In this version, we have corrected errors, resolved imprecisions, and simplifiedseveral proofs These areas for improvement were brought to our attention byreaders To them and to our colleagues H Crauel, A Ilchmann and G Prokert,
we extend heartfelt thanks
Z¨urich and Hannover, December 2003 H Amann and J Escher
Foreword to the English translation
It is a pleasure to express our gratitude to Silvio Levy and Matt Cargo for theircareful and insightful translation of the original German text into English Theireffective and pleasant cooperation during the process of translation is highly ap-preciated
SinhVienZone.Com
Trang 6Foreword v
Chapter VI Integral calculus in one variable 1 Jump continuous functions . 4
Staircase and jump continuous functions 4
A characterization of jump continuous functions 6
The Banach space of jump continuous functions 7
2 Continuous extensions . 10
The extension of uniformly continuous functions 10
Bounded linear operators 12
The continuous extension of bounded linear operators 15
3 The Cauchy–Riemann Integral . 17
The integral of staircase functions 17
The integral of jump continuous functions 19
Riemann sums 20
4 Properties of integrals . 25
Integration of sequences of functions 25
The oriented integral 26
Positivity and monotony of integrals 27
Componentwise integration 30
The first fundamental theorem of calculus 30
The indefinite integral 32
The mean value theorem for integrals 33
5 The technique of integration . 38
Variable substitution 38
Integration by parts 40
The integrals of rational functions 43
SinhVienZone.Com
Trang 76 Sums and integrals . 50
The Bernoulli numbers 50
Recursion formulas 52
The Bernoulli polynomials 53
The Euler–Maclaurin sum formula 54
Power sums 56
Asymptotic equivalence 57
The Riemann 2 function 59
The trapezoid rule 64
7 Fourier series . 67
The L2scalar product 67
Approximating in the quadratic mean 69
Orthonormal systems 71
Integrating periodic functions 72
Fourier coe3cients 73
Classical Fourier series 74
Bessel’s inequality 77
Complete orthonormal systems 79
Piecewise continuously di2erentiable functions 82
Uniform convergence 83
8 Improper integrals . 90
Admissible functions 90
Improper integrals 90
The integral comparison test for series 93
Absolutely convergent integrals 94
The majorant criterion 95
9 The gamma function . 98
Euler’s integral representation 98
The gamma function onC\(−N) 99
Gauss’s representation formula 100
The reflection formula 104
The logarithmic convexity of the gamma function 105
Stirling’s formula 108
The Euler beta integral 110
SinhVienZone.Com
Trang 8Chapter VII Multivariable differential calculus
1 Continuous linear maps 118
The completeness ofL(E, F ) 118
Finite-dimensional Banach spaces 119
Matrix representations 122
The exponential map 125
Linear di2erential equations 128
Gronwall’s lemma 129
The variation of constants formula 131
Determinants and eigenvalues 133
Fundamental matrices 136
Second order linear di2erential equations 140
2 Differentiability 149
The definition 149
The derivative 150
Directional derivatives 152
Partial derivatives 153
The Jacobi matrix 155
A differentiability criterion 156
The Riesz representation theorem 158
The gradient 159
Complex differentiability 162
3 Multivariable di2erentiation rules 166
Linearity 166
The chain rule 166
The product rule 169
The mean value theorem 169
The differentiability of limits of sequences of functions 171
Necessary condition for local extrema 171
4 Multilinear maps 173
Continuous multilinear maps 173
The canonical isomorphism 175
Symmetric multilinear maps 176
The derivative of multilinear maps 177
5 Higher derivatives 180
Definitions 180
Higher order partial derivatives 183
The chain rule 185
Taylor’s formula 185
SinhVienZone.Com
Trang 9Functions of m variables 186
Su3cient criterion for local extrema 188
6 Nemytskii operators and the calculus of variations 195
Nemytskii operators 195
The continuity of Nemytskii operators 195
The differentiability of Nemytskii operators 197
The differentiability of parameter-dependent integrals 200
Variational problems 202
The Euler–Lagrange equation 204
Classical mechanics 207
7 Inverse maps 212
The derivative of the inverse of linear maps 212
The inverse function theorem 214
Di2eomorphisms 217
The solvability of nonlinear systems of equations 218
8 Implicit functions 221
Di2erentiable maps on product spaces 221
The implicit function theorem 223
Regular values 226
Ordinary di2erential equations 226
Separation of variables 229
Lipschitz continuity and uniqueness 233
The Picard–Lindel¨of theorem 235
9 Manifolds 242
Submanifolds ofRn 242
Graphs 243
The regular value theorem 243
The immersion theorem 244
Embeddings 247
Local charts and parametrizations 252
Change of charts 255
10 Tangents and normals 260
The tangential inRn 260
The tangential space 261
Characterization of the tangential space 265
Di2erentiable maps 266
The di2erential and the gradient 269
Normals 271
Constrained extrema 272
Applications of Lagrange multipliers 273
SinhVienZone.Com
Trang 10Chapter VIII Line integrals
1 Curves and their lengths 281
The total variation 281
Rectifiable paths 282
Di2erentiable curves 284
Rectifiable curves 286
2 Curves inRn 292
Unit tangent vectors 292
Parametrization by arc length 293
Oriented bases 294
The Frenet n-frame 295
Curvature of plane curves 298
Identifying lines and circles 300
Instantaneous circles along curves 300
The vector product 302
The curvature and torsion of space curves 303
3 Pfa2 forms 308
Vector fields and Pfa2 forms 308
The canonical basis 310
Exact forms and gradient fields 312
The Poincar´e lemma 314
Dual operators 316
Transformation rules 317
Modules 321
4 Line integrals 326
The definition 326
Elementary properties 328
The fundamental theorem of line integrals 330
Simply connected sets 332
The homotopy invariance of line integrals 333
5 Holomorphic functions 339
Complex line integrals 339
Holomorphism 342
The Cauchy integral theorem 343
The orientation of circles 344
The Cauchy integral formula 345
Analytic functions 346
Liouville’s theorem 348
The Fresnel integral 349
The maximum principle 350
SinhVienZone.Com
Trang 11Harmonic functions 351
Goursat’s theorem 353
The Weierstrass convergence theorem 356
6 Meromorphic functions 360
The Laurent expansion 360
Removable singularities 364
Isolated singularities 365
Simple poles 368
The winding number 370
The continuity of the winding number 374
The generalized Cauchy integral theorem 376
The residue theorem 378
Fourier integrals 379
References 387
Index 389
SinhVienZone.Com
Trang 12Integral calculus in one variable
Integration was invented for finding the area of shapes This, of course, is anancient problem, and the basic strategy for solving it is equally old: divide theshape into rectangles and add up their areas
A mathematically satisfactory realization of this clear, intuitive idea is ingly subtle We note in particular that is a vast number of ways a given shapecan be approximated by a union of rectangles It is not at all self-evident they alllead to the same result For this reason, we will not develop the rigorous theory
amaz-of measures until Volume III
In this chapter, we will consider only the simpler case of determining the areabetween the graph of a sufficiently regular function of one variable and its axis
By laying the groundwork for approximating a function by a juxtaposed series ofrectangles, we will see that this boils down to approaching the function by a series
of staircase functions, that is, functions that are piecewise constant We will showthat this idea for approximations is extremely flexible and is independent of itsoriginal geometric motivation, and we will arrive at a concept of integration thatapplies to a large class of vector-valued functions of a real variable
To determine precisely the class of functions to which we can assign an gral, we must examine which functions can be approximated by staircase functions
inte-By studying the convergence under the supremum norm, that is, by asking if agiven function can be approximated uniformly on the entire interval by staircasefunctions, we are led to the class of jump continuous functions Section 1 is devoted
to studying this class
There, we will see that an integral is a linear map on the vector space ofstaircase functions There is then the problem of extending integration to thespace of jump continuous functions; the extension should preserve the elementaryproperties of this map, particularly linearity This exercise turns out to be a specialcase of the general problem of uniquely extending continuous maps Because theextension problem is of great importance and enters many areas of mathematics, we
SinhVienZone.Com
Trang 13will discuss it at length in Section 2 From the fundamental extension theorem foruniformly continuous maps, we will derive the theorem of continuous extensions
of continuous linear maps This will give us an opportunity to introduce theimportant concepts of bounded linear operators and the operator norm, whichplay a fundamental role in modern analysis
After this groundwork, we will introduce in Section 3 the integral of jumpcontinuous functions This, the Cauchy–Riemann integral, extends the elemen-tary integral of staircase functions In the sections following, we will derive itsfundamental properties Of great importance (and you can tell by the name) isthe fundamental theorem of calculus, which, to oversimplify, says that integrationreverses differentiation Through this theorem, we will be able to explicitly calcu-late a great many integrals and develop a flexible technique for integration Thiswill happen in Section 5
In the remaining sections — except for the eighth — we will explore tions of the so-far developed di2erential and integral calculus Since these are notessential for the overall structure of analysis, they can be skipped or merely sam-pled on first reading However, they do contain many of the beautiful results ofclassical mathematics, which are needed not only for one’s general mathematicalliteracy but also for numerous applications, both inside and outside of mathemat-ics
applica-Section 6 will explore the connection between integrals and sums We derivethe Euler–Maclaurin sum formula and point out some of its consequences Specialmention goes to the proof of the formulas of de Moivre and Sterling, which describethe asymptotic behavior of the factorial function, and also to the derivation of
several fundamental properties of the famous Riemann 2 function The latter is
important in connection to the asymptotic behavior of the distribution of primenumbers, which, of course, we can go into only very briefly
In Section 7, we will revive the problem — mentioned at the end of ter V — of representing periodic functions by trigonometric series With help fromthe integral calculus, we can specify a complete solution of this problem for alarge class of functions We place the corresponding theory of Fourier series inthe general framework of the theory of orthogonality and inner product spaces.Thereby we achieve not only clarity and simplicity but also lay the foundation for
Chap-a number of concrete Chap-applicChap-ations, mChap-any of which you cChap-an expect see elsewhere.Naturally, we will also calculate some classical Fourier series explicitly, leading
to some surprising results Among these is the formula of Euler, which gives an
explicit expression for the 2 function at even arguments; another is an interesting
expression for the sine as an infinite product
Up to this point, we have will have concentrated on the integration of jumpcontinuous functions on compact intervals In Section 8, we will further extendthe domain of integration to cover functions that are defined (and integrated)
on infinite intervals or are not bounded We content ourselves here with simplebut important results which will be needed for other applications in this volume
SinhVienZone.Com
Trang 14because, in Volume III, we will develop an even broader and more flexible type ofintegral, the Lebesgue integral.
Section 9 is devoted to the theory of the gamma function This is one ofthe most important nonelementary functions, and it comes up in many areas ofmathematics Thus we have tried to collect all the essential results, and we hopeyou will find them of value later This section will show in a particularly nice waythe strength of the methods developed so far
SinhVienZone.Com
Trang 151 Jump continuous functions
In many concrete situations, particularly in the integral calculus, the constraint ofcontinuity turns out to be too restrictive Discontinuous functions emerge natu-rally in many applications, although the discontinuity is generally not very patho-logical In this section, we will learn about a simple class of maps which containsthe continuous functions and is especially useful in the integral calculus in oneindependent variable However, we will see later that the space of jump continu-ous functions is still too restrictive for a flexible theory of integration, and, in thecontext of multidimensional integration, we will have to extend the theory into aneven broader class containing the continuous functions
In the following, suppose
• E := (E, ·2 ) is a Banach space;
I := [α, 4 ] is a compact perfect interval.
Staircase and jump continuous functions
We callZ := (30, , 3 n ) a partition of I, if n 3 N × and
3 = 30< 31< · · · < 3 n = β
If{30, , 3 n } is a subset of the partition Z := (40, , 4 k),Z is called a refinement
ofZ, and we write Z 4 Z.
The function f : I 5 E is called a staircase function on I if I has a partition
Z := (30, , 3 n ) such that f is constant on every (open) interval (3 j−1 , 3 j) Then
we sayZ is a partition for f, or we say f is a staircase function on the partition Z.
A staircase function
If f : I 5 E is such that the limits f(3 + 0), f(4 − 0), and
f (x ± 0) := lim
y2 x±0 y3=x
f (y)
exist for all x 3 ˚ I, we call f jump continuous.1 A jump continuous function is
piecewise continuousif it has only finitely many discontinuities (“jumps”) Finally,
1Note that, in general, f (x + 0) and f (x − 0) may di2er from f(x).
SinhVienZone.Com
Trang 16we denote by
T (I, E), S(I, E), SC(I, E)
the sets of all functions f : I 5 E that are staircase, jump continuous, and
piece-wise continuous, respectively.2
A piecewise continuous function Not a jump continuous function
1.1 Remarks (a) Given partitionsZ := (30, , 3 n) and Z := (40, , 4 m ) of I,
the union{30, , 3 n } 6{ 40, , 4 m } will naturally define another partition Z 7 Z
of I Obviously, Z 4 Z 7 Z and Z 4 Z 7 Z In fact, 4 is an ordering on the set of
partitions of I, and Z 7 Z is the largest from {Z, Z}.
(b) If f is a staircase function on a partition Z, every refinement of Z is also a
partition for f
(c) If f : I 5 E is jump continuous, neither f(x + 0) nor f(x − 0) need equal f(x)
for x 3 I.
(d) S(I, E) is a vector subspace of B(I, E).
Proof The linearity of one-sided limits implies immediately that S(I, E) is a vector
space If f ∈ S(I, E)\B(I, E), we find a sequence (x n ) in I with
3 f(x n) 4 n for n 2 N (1.1) Because I is compact, there is a subsequence (x n k ) of (x n ) and x 2 I such that x n k 5 x
as k → 6 By choosing a suitable subsequence of (x n k ), we find a sequence (y n),
that converges monotonically to x.3 If f is jump continuous, there is a v 2 E with
lim f (y n ) = v and thus lim 3 f(y n)3 = 3 v3 (compare with Example III.1.3(j)) Because
every convergent sequence is bounded, we have contradicted (1.1) ThereforeS(I, E) 7 B(I, E).2
(e) We have sequences of vector subspaces
T (I, E) ⊂ SC(I, E) ⊂ S(I, E) and C(I, E) ⊂ SC(I, E)
(f ) Every monotone function f : I 5 R is jump continuous.
2 We usually abbreviateT (I) := T (I, K) etc, if the context makes clear which of the fields R
or C we are dealing with.
3 Compare with Exercise II.6.3.
SinhVienZone.Com
Trang 17Proof This follows from Proposition III.5.3.2
(g) If f belongs to T (I, E), S(I, E), or SC(I, E), and J is a compact perfect
subinterval of I, then f |J belongs to T (J, E), S(J, E), or SC(J, E).
(h) If f belongs to T (I, E), S(I, E), or SC(I, E), then 2 f2 belongs to T (I, R), S(I, R), SC(I, R).2
A characterization of jump continuous functions
1.2 Theorem A function f : I 5 E is jump continuous if and only if there is a sequence of staircase functions that converges uniformly to it.
Proof “=⇒” Suppose f ∈ S(I, E) and n 3 N × Then for every x 3 I, there are
numbers 3 (x) and 4 (x) such that 3 (x) < x < 4 (x) and
50 := 3 , 5 j+1 := x j for j = 0, , m, and 5 m+2 := 4 , we let Z0= (50, , 5 m+2)
be a partition of I Now we select a refinementZ1= (60, , 6 k) ofZ0 with
Trang 18“⇐=” Suppose there is a sequence (f n) inT (I, E) that converges uniformly to
f The sequence also converges to f in B(I, E) Let ε > 0 Then there is an n 3 N
such that 2 f(x) − f n (x) 2 < 7/ 2 for all x 3 I In addition, for every x 3 (α, 4 ]
there is an 3 5 3 [α, x) such that f n (s) = f n (t) for s, t 3 (3 5 , x) Consequently,
2 f(s) − f(t) 4 2 f(s) − f n (s) 2 + 2 f n (s) − f n (t) 2 + 2 f n (t) − f(t)2 < 7 (1.2) for s, t 3 (3 5 , x).
Suppose now (s j ) is a sequence in I that converges from the left to x Then there is an N 3 N such that s j 3 (3 5 , x) for j ≥ N, and (1.2) implies
2 f(s j)− f(s k)2 < 7 for j, k ≥ N
Therefore 2
f (s j)3
j6 N is a Cauchy sequence in the Banach space E, and there is
an e 3 E with lim j f (s j ) = e If (t k ) is another sequence in I that converges from the left to x, then we can repeat the argument to show there is an e 5 3 E such
that limk f (t k ) = e 5 Also, there is an M ≥ N such that t k 3 (3 5 , x) for k ≥ M.
Consequently, (1.2) gives
2 f(s j)− f(t k)2 < 7 for j, k ≥ M
After taking the limits j → ∞ and k → ∞, we find 2 e − e 5 4 7 Now e and e 5
agree, because ε > 0 was arbitrary Therefore we have proved that lim y2 x−0 f (y)
exists By swapping left and right, we show that for x 3 [α, 4 ) the right-sided
limits limy2 x+0 f (y) exist as well Consequently f is jump continuous.2
1.3 Remark If the function f ∈ S(I, R) is nonnegative, the first part of the above
proof shows there is a sequence of nonnegative staircase functions that converges
uniformly to f 2
The Banach space of jump continuous functions
1.4 Theorem The set of jump continuous functions S(I, E) is a closed vector subspace of B(I, E) and is itself a Banach space; T (I, E) is dense in S(I, E).
Proof From Remark 1.1(d) and (e), we have the inclusions
T (I, E) ⊂ S(I, E) ⊂ B(I, E)
According to Theorem 1.2, we have
T (I, E) = S(I, E) , (1.3) when the closure is formed in B(I, E) Therefore S(I, E) is closed in B(I, E) by
Proposition III.2.12 The last part of the theorem follows from (1.3).2
SinhVienZone.Com
Trang 193 Prove or disprove thatSC(I, E) is a closed vector subspace of S(I, E).
4 Show these statements are equivalent for f : I 5 E:
(i) f ∈ S(I, E);
(ii) ∃ (f n) inT (I, E) such thatn 3 f n 3 2 < 6 and f =2
Trang 209 Suppose E j , j = 0, , n are normed vector spaces and
f = (f0, , f n ) : I 5 E := E0× · · · × E n
Show
f ∈ S(I, E) ⇐⇒ f j ∈ S(I, E j ) , j = 0, , n
10 Suppose E and F are normed vector spaces and that f ∈ S(I, E) and 2 : E 5 F
are uniformly continuous Show that 2 ◦ f ∈ S(I, F ).
11 Suppose f, g ∈ S(I, R) and im(g) 7 I Prove or disprove that f ◦ g ∈ S(I, R).
SinhVienZone.Com
Trang 21to understand the results well.
The extension of uniformly continuous functions
2.1 Theorem (extension theorem) Suppose Y and Z are metric spaces, and Z is
complete Also suppose X is a dense subset of Y , and f : X 5 Z is uniformly continuous.1 Then f has a uniquely determined extension f : Y 5 Z given by
f (y) = lim
x2 y x6 X
f (x) for y 3 Y ,
and f is also uniformly continuous.
Proof (i) We first verify uniqueness Assume g, h 3 C(Y, Z) are extensions of f.
Because X is dense in Y , there is for every y 3 Y a sequence (x n ) in X such that
x n 5 y in Y The continuity of g and h implies
g(y) = lim
n g(x n) = lim
n f (x n) = lim
n h(x n ) = h(y) Consequently, g = h.
(ii) If f is uniformly continuous, there is, for every ε > 0, a δ = δ(7 ) > 0 such
Trang 22f (x j)3
is a Cauchy sequence in Z Because Z is complete, we can find
a z 3 Z such that f(x j)5 z If (x 5
k ) is another sequence in X such that x 5 k 5 y,
we reason as before that there exists a z 5 3 Z such that f(x 5
k)5 z 5 Moreover, we
can find M ≥ N such that d(x 5
k , y) < δ/2 for k ≥ M This, together with (2.2),
true every positive 7 , we have z = z 5 These considerations show that the map
f : Y 5 Z , y 5 lim
x2 y x6 X
f (x)
is well defined, that is, it is independent of the chosen sequence
For x 3 X, we set x j := x for j 3 N and find f(x) = lim j f (x j ) = f (x) Therefore f is an extension of f
(iii) It remains to show that f is uniformly continuous Let ε > 0, and choose
δ > 0 satisfying (2.1) Also choose y, z 3 Y such that d(y, z) < δ/3 Then there
are series (y n ) and (z n ) in X such that y n 5 y and z n 5 z Therefore, there is
an N 3 N such that d(y n , y) < δ/3 and d(z n , z) < δ/3 for n ≥ N In particular,
Therefore f is uniformly continuous.2
2.2 Application Suppose X is a bounded subset ofKn Then the restriction2
T : C(X) 5 BUC(X) , u 5 u|X (2.4)
2BUC(X) denotes the Banach space of bounded and uniformly continuous functions on X,
as established by the supremum norm Compare to Exercise V.2.1.
SinhVienZone.Com
Trang 23is an isometric isomorphism.
Proof (i) Suppose u 2 C(X) Then, because X is compact by the Heine–Borel theorem,
Corollary III.3.7 and Theorem III.3.13 imply that u — and therefore also T u = u |X — is
bounded and uniformly continuous Therefore T is well defined Obviously T is linear (ii) Suppose v 2 BUC(X) Because X is dense in X, there is from Theorem 2.1 a
uniquely determined u 2 C(X) such that u|X = v Therefore T : C(X) 5 BUC(X) is
a vector space isomorphism
(iii) For u 2 C(X), we have
On the other hand, there is from Corollary III.3.8 a y 2 X such that 3 u3 2 =|u(y)| We
choose a sequence (x n ) in X such that x n 5 y and find
3 u3 2 =|u(y)| = | lim u(x n)| = | lim T u(x n)| ≤ sup
x 3 X |T u(x)| = 3 T u3 2 .
This shows that T is an isometry.2
Convention If X is a bounded open subset of K n , we always identify BUC(X) with C(X) through the isomorphism (2.4).
Bounded linear operators
Theorem 2.1 becomes particularly important for linear maps We therefore compilefirst a few properties of linear operators
Suppose E and F are normed vector spaces, and A : E 5 F is linear We
call A bounded3 if there is an 3 ≥ 0 such that
2 Ax 4 3 2 x2 for x 3 E (2.5)
We define
L(E, F ) :=4A 3 Hom(E, F ) ; A is bounded5 .
For every A ∈ L(E, F ), there is an 3 ≥ 0 for which (2.5) holds Therefore
2 A2 := inf{ 3 ≥ 0 ; 2 Ax 4 3 2 x2 , x 3 E }
is well defined We call2 A2 L(E,F ):=2 A2 the operator norm of A.
3For historical reasons, we accept here a certain inconsistency in the nomenclature: if F is
not the null vector space, there is (except for the zero operator)no bounded linear operator that
is a bounded map in the terms of Section II.3 (compare Exercise II.3.15) Here, a bounded linear operator maps bounded sets to bounded sets (compare Conclusion 2.4(c)).
SinhVienZone.Com
Trang 242.3 Proposition For A ∈ L(E, F ) we have4
sup
7x ≤1 2 Ax 4 sup
7y 7=1 2 Ay2
Thus we have shown the theorem’s first three equalities
For the last, let a := sup x6 B E 2 Ax2 and y 3 B := B E Then 0 < λ < 1 means
λy is in B Thus the estimate 2 Ay2 = 2 A(λy)2 /λ 4 a/λ holds for 0 < λ < 1, as
2 Ay 4 a shows Therefore
Proof If x, y 2 E, then 3 Ax − Ay3 = 3 A(x − y) ≤ 3 A 3 x − y3 Thus A is Lipschitz
continuous with Lipschitz constant3 A3 2
4Here and in similar settings, we will implicitly assume E 2= {0}.
SinhVienZone.Com
Trang 25(c) Let A 3 Hom(E, F ) Then A belongs to L(E, F ) if and only if A maps bounded
sets to bounded sets
Proof “=⇒” Suppose A ∈ L(E, F ) and B is bounded in E Then there is a β > 0 such
that3 x ≤ 3 for all x 2 B It follows that
3 Ax ≤ 3 A 3 x ≤ 3 A3 3 for x 2 B
Therefore the image of B by A is bounded in F
“⇐=” Since the closed unit ball B E is closed in E, there is by assumption an α > 0
such that3 Ax ≤ 4 for x 2 B E Because y/ 3 y 2 B E for all y 2 E\{0}, it follows that
3 Ay ≤ 4 3 y3 for all y 2 E.2
(d) L(E, F ) is a vector subspace of Hom(E, F ).
Proof Let A, B ∈ L(E, F ) and 5 2 K For every x 2 E, we have the estimates
3 (A + B)x3 = 3 Ax + Bx ≤ 3 Ax3 + 3 Bx ≤23 A3 + 3 B3 33 x3 (2.6)
and
3 λAx3 = |5 | 3 Ax ≤ |5 | 3 A 3 x3 (2.7) Therefore A + B and λA also belong to L(E, F ).2
(e) The map
that3 x ≤ 1 Then it follows from (2.6) and (2.7) that
3 (A + B)x ≤ 3 A3 + 3 B3 and 3 λAx3 = |5 | 3 Ax3 ,
and taking the supremum of x verifies the remaining parts of the definition. 2
(f ) Suppose G is a normed vector space and B ∈ L(E, F ) and also A ∈ L(F, G).
Then we have
AB ∈ L(E, G) and 2 AB 4 2 A 2 B2
Proof This follows from
3 ABx ≤ 3 A 3 Bx ≤ 3 A 3 B 3 x3 for x 2 E ,
and the definition of the operator norm.2
(g) L(E) := L(E, E) is a normed algebra5 with unity, that is,L(E) is an algebra
and2 1 E 2 = 1 and also
2 AB 4 2 A 2 B2 for A, B ∈ L(E)
Proof The claim follows easily from Example I.12.11(c) and (f).2
5 Compare with the definition of a Banach algebra in Section V.4.
SinhVienZone.Com
Trang 26Convention In the following, we will always endow L(E, F ) with the operator
norm Therefore
L(E, F ) :=2L(E, F ), ·2 3
is a normed vector space with·2 := ·2 L(E,F )
The next theorem shows that a linear map is bounded if and only if it is continuous
2.5 Theorem Let A 3 Hom(E, F ) These statements are equivalent:
(i) A is continuous.
(ii) A is continuous at 0.
(iii) A ∈ L(E, F ).
Proof “(i)=⇒(ii)” is obvious, and “(iii)=⇒(i)” was shown in Conclusions 2.4(b).
“(ii)=⇒(iii)” By assumption, there is a δ > 0 such that 2 Ay2 = 2 Ay−A02 < 1
for all y 3 B(0, δ) From this, it follows that
and therefore A is closed.2
The continuous extension of bounded linear operators
2.6 Theorem Suppose E is a normed vector space, X is a dense vector subspace
in E, and F is a Banach space Then, for every A ∈ L(X, F ) there is a uniquely determined extension A ∈ L(E, F ) defined through
Ae = lim
x2 e x6 X
and 2 A2 L(E,F )=2 A2 L(X,F )
Proof (i) According to Conclusions 2.4(b), A is uniformly continuous Defining
f := A, Y := E, and Z := F , it follows from Theorem 2.1 that there exists a
uniquely determined extension of A 3 C(E, F ) of A, which is given through (2.8).
(ii) We now show that A is linear For that, suppose e, e 5 3 E, λ 3 K, and
(x n ) and (x 5 n ) are sequences in X such that x n 5 e and x 5 n 5 e 5 in E From the
linearity of A and the linearity of limits, we have
A(e + λe 5) = lim
n A(x n + λx 5 n) = lim
n Ax n + λ lim
n Ax 5 n = Ae + λAe 5
Therefore A : E 5 F is linear That A is continuous follows from Theorem 2.5,
because A belongs to L(E, F ).
SinhVienZone.Com
Trang 27(iii) Finally we prove that A and A have the same operator norm From the
continuity of the norm (see Example III.1.3(j)) and from Conclusions 2.4(a), itfollows that
2 Ae2 = 2 lim Ax n 2 = lim 2 Ax n 4 lim 2 A 2 x n 2 = 2 A 2 lim x n 2 = 2 A 2 e2
for every e 3 E and every sequence (x n ) in X such that x n 5 e in E Consequently,
2 A ≤ 2 A2 Because A extends A, Proposition 2.3 implies
2 A2 = sup
7y 7< 1 2 Ay ≥ sup
7x 7< 1 x6 X
2 Ax2 = sup
7x 7< 1 x6 X
2 Ax2 = 2 A2 ,
and we find2 A2 L(E,F )=2 A2 L(X,F ).2
Exercises
In the following exercises, E, E j , F , and F jare normed vector spaces
1 Suppose A 2 Hom(E, F ) is surjective Show that
A −1 ∈ L(F, E) ⇐⇒ ∃ α > 0 : 4 3 x ≤ 3 Ax3 , x 2 E
If A also belongs to L(E, F ), show that 3 A −1 ≥ 3 A3 −1.
2 Suppose E and F are finite-dimensional and A ∈ L(E, F ) is bijective with a
contin-uous inverse6A −1 ∈ L(F, E) Show that if B ∈ L(E, F ) satisfies
with j = 1, 2, 6 , verify for A ∈ L(E j) that
(i) 3 A3 L(E1)= maxk
6 Suppose (A n) is a sequence inL(E, F ) converging to A, and suppose (x n) is a sequence
in E whose limit is x Prove that (A n x n ) in F converges to Ax.
7 Show that ker(A) of A ∈ L(E, F ) is a closed vector subspace of E.
6If E is a finite-dimensional normed vector space, then Hom(E, F ) = L(E, F ) (see
Theo-rem VII.1.6) Consequently, we need not assume that A, A −1 , and B are continuous.
SinhVienZone.Com
Trang 283 The Cauchy–Riemann Integral
Determining the area of plane geometrical shapes is one of the oldest and mostprominent projects in mathematics To compute the areas under graphs of realfunctions, it is necessary to simplify and formalize the problem It will help to gain
an intuition for integrals To that end, we will first explain, as simply as possible,how to integrate staircase functions; we will then extend the idea to jump con-tinuous functions These constructions are based essentially on the results in thefirst two sections of this chapter These
present the idea that the area of a graph
is made up of the sum of areas of
rectan-gles that are themselves determined by
choosing the best approximation of the
graph to a staircase function By the
“unlimited refinement” of the width of
the rectangles, we anticipate the sum of
their areas will converge to the area of
the figure
In the following we denote by
• E := (E, |·|), a Banach space;
I := [α, 4 ], a compact perfect interval.
The integral of staircase functions
Suppose f : I 5 E is a staircase function and Z := (30, , 3 n ) is a partition of I for f Define e j through
f (x) = e j for x 3 (3 j−1 , 3 j ) and j = 1, , n , that is, e j is the value of f on the interval (3 j−1 , 3 j) We call
( Z)f :=
n
j=1
e j (3 j − 3 j−1)
the integral of f with respect to the partition Z Obviously(Z)f is an element
of E We note that the integral does not depend on the values of f at its jump discontinuities In the case that E = R, we can interpret |e j | (3 j − 3 j−1) as thearea of a rectangle with sides of length|e j | and (3 j −3 j−1) Thus
( Z)f represents
a weighted sum of rectangular areas, where the area|e j | (3 j − 3 j−1) is weighted
by the sign of e j In other words, those rectangles that rise above the x-axis are
SinhVienZone.Com
Trang 29weighted by 1, whereas those below are given−1.
The following lemma shows that
( Z)f depends only on f and not on the choice
of partitionZ
3.1 Lemma If f ∈ T (I, E), and Z and Z 5 are partitions for f , then
( Z)f =
( Z3)f
Proof (i) We first treat the case that Z5 := (30, , 3
k , γ, 3 k+1 , , 3 n) hasexactly one more partition point thanZ := (30, , 3 n) Then we compute
(iii) Finally, ifZ and Z5 are arbitrary partitions of f , then Z 7 Z 5, according
to Remark 1.1(a) and (b), is a refinement of bothZ and Z5 and is therefore also a
partition for f Then it follows from (ii) that
( Z)
f =
(Z∨Z 3)f =
( Z3)f 2
Using Lemma 3.1, we can define for f ∈ T (I, E) the integral of f over I by
SinhVienZone.Com
Trang 30using an arbitrary partition Z for f Obviously, the integral induces a map from
|f| 4 2 f2 4 (4 − 3 ) Also3
4 belongs to L2T (I, E), E3, and 2 3
4 2 = 4 − 3
Proof The first statement is clear To prove the inequality, suppose f ∈ T (I, E)
and (30, , 3 n ) is a partition for f Then
Con-4 belongs to L2T (I, E), E3, and we have 2 3
4 4 4 − 3 For the
constant function1 ∈ T (I, R) with value 1, we have 3
4 1 = 4 − 3 , and the last
claim follows.2
The integral of jump continuous functions
From Theorem 1.4, we know that the space of jump continuous functions S(I, E)
is a Banach space when it is endowed with the supremum norm and thatT (I, E) is
a dense vector subspace ofS(I, E) (see Remark 1.1(e)) If follows from Lemma 3.2
that the integral3
4 is a continuous, linear map fromT (I, E) to E We can apply
the extension Theorem 2.6 to get a unique continuous linear extension of3
4 intothe Banach spaceS(I, E) We denote this extension using the same notation, so
Trang 31where (f n) is an arbitrary sequence of staircase functions that converges uniformly
to f The element3
4 f of E is called the (Cauchy–Riemann) integral of f or the
integral of f over I or the integral of f from 2 to 3 We call f the integrand.
3.3 Remarks Suppose f ∈ S(I, E).
(a) According to Theorem 2.6 3
4 f is well defined, that is, this element of E is
independent of the approximating sequence of staircase functions In the special
4 f with this oriented area.
(b) There are a number of other notations for3
the mesh ofZ, and we call any element 6 j 3 [3 j−1 , 3 j] a between point With this
terminology, we now prove an approximation result for3
Trang 32for every partition Z := (30, , 3 n ) of I of mesh Z < δ and for every value of the between point 6 j
Proof (i) We treat first the staircase functions Suppose then that f ∈ T (I, E)
and ε > 0 Also let Z := (30, , 3 n ) be a partition of f , and let e j = f |(3 j−1 , 3 j)for 14 j 4 n We set δ := 74n 2 f2 4 and choose a partitionZ := (30, , 3 n) of
I such that Z< δ We also choose between points 6 j 3 [3 j−1 , 3 j] for 14 j 4 n.
k can only hold on the partition points{30, , 3 n } Therefore
the last sum in (3.1) has at most 2n terms that do not vanish For each of these,
(ii) Suppose now that f ∈ S(I, E) and ε > 0 Then, according to
Theo-rem 1.4, there is a g ∈ T (I, E) such that
Z < δ and every value 6 j 3 [3 j−1 , 3 j], we have
Trang 33g(6 j)− f(6 j)3
(3 j − 3 j−1) < 7 ,and the claim is proved.2
3.5 Remarks (a) A function f : I 5 E is said to Riemann integrable if there is
an e 3 E with the property that for every ε > 0 there is a δ > 0, such that
If f is Riemann integrable, the value e is called the Riemann integral of f
over I, and we will again denote it using the notation3
4 f dx Consequently,
The-orem 3.4 may be stated as, Every jump continuous function is Riemann integrable,
and the Cauchy–Riemann integral coincides with the Riemann integral.
(b) One can showS(I, E) is a proper subset of the Riemann integrable functions.1Consequently, the Riemann integral is a generalization of the Cauchy–Riemannintegral
We content ourselves here with the Cauchy–Riemann integral and will notseek to augment the set of integrable functions, as there is no particular need for
it now However, in Volume III, we will introduce an even more general integral,the Lebesgue integral, which satisfies all the needs of rigorous analysis
(c) Suppose f : I 5 E, and Z := (30, , 3 n ) is a partition of I with between points 6 j 3 [3 j−1 , 3 j] We call
Trang 34the Riemann sum If f is Riemann integrable, then
[nx2]
n2 dx , (iii)
1 0
[nx2]
n dx , (iv)
2 3
sign x dx
2 Compute1
−1 f for the function f of Exercise 1.2 and also
1
0 f for the f of Exercise 1.7.
3 Suppose F is a Banach space and A ∈ L(E, F ) Then show for f ∈ S(I, E) that
Trang 358 Suppose f 2 B(I, R) and Z 4 := (3
0, , 3 m) is a refinement of Z := (40, , 4 n).Show that
S(f, I, Z) − S(f, I, Z 4)≤ 2(m − n) 3 f3 2 3Z , S(f, I,Z4)− S(f, I, Z) ≤ 2(m − n) 3 f3 2 3Z .
9 Let f 2 B(I, R) From Exercise 7(ii), we know the following exist in R:
(ii) for every ε > 0 there is a δ > 0 such that for every partition Z of I with 3Z < 6 ,
we have the inequalities
In addition, Exercise 7(i) gives S(f, I, Z ∨ Z 4)≤ S(f, I, Z 4).)
10 Show these statements are equivalent for f 2 B(I, R):
(i) f is Riemann integrable.
(Hint: “(ii)⇐⇒(iii)” follows from Exercise 9.
“(i)=⇒(iii)” Suppose ε > 0 and e :=I f Then there is a δ > 0 such that
Trang 364 Properties of integrals
In this section, we delve into the most important properties of integrals and, inparticular, prove the fundamental theorem of calculus This theorem says thatevery continuous function on an interval has a primitive function (called the anti-derivative) which gives another form of its integral This antiderivative is known
in many cases and, when so, allows for the easy, explicit evaluation of the integral
As in the preceding sections, we suppose
• E = (E, |·|) is a Banach space; I = [α, 4 ] is a compact perfect interval.
Integration of sequences of functions
The definition of integrals leads easily to the following convergence theorem, whosetheoretical and practical meaning will become clear
4.1 Proposition (of the integration of sequences and sums of functions) Suppose
(f n ) is a sequence of jump continuous functions.
(i) If (f n ) converges uniformly to f , then f is jump continuous and
f n converges uniformly, then4
n=0 f n is jump continuous and
Proof From Theorem 1.4 we know that the space S(I, E) is complete when
en-dowed with the supremum norm Both parts of the theorem then follow from thefacts that3
4 is a linear map fromS(I, E) to E and that the uniform convergence
agrees with the convergence inS(I, E).2
4.2 Remark The statement of Proposition 4.1 is false when the sequence (f n) isonly pointwise convergent
Proof We set I = [0, 1], E :=R, and
Trang 37For the first application of Proposition 4.1, we prove that the importantstatement of Lemma 3.2 — about interchanging the norm with the integral — isalso correct for jump continuous functions.
4.3 Proposition For f ∈ S(I, E) we have |f| ∈ S(I, R) and
Proof According to Theorem 1.4, there is a sequence (f n) inT (I, E) that
con-verges uniformly to f Further, it follows from the inverse triangle inequality that
|f n (x) | E − |f(x)| E ≤ |f n (x) − f(x)| E ≤ 2 f n − f2 4 for x 3 I
Therefore|f n | converges uniformly to |f| Because every |f n | belongs to T (I, R),
it follows again from Theorem 1.4 that|f| 3 S (I, R).
From Lemma 3.2, we get
Because (f n ) converges uniformly to f and ( |f n |) converges likewise to |f|, we can,
with the help of Proposition 4.1, pass the limit n → ∞ in (4.1) The desired
inequality follows.2
The oriented integral
If f ∈ S(I, E) and γ, δ 3 I, we define1
5 f “the integral of f from γ to δ” We call γ the lower limit and δ the
upper limit of the integral of f , even when γ > δ According to Remark 1.1(g), the integral of f from γ to δ is well defined through (4.2), and
Trang 384.4 Proposition (of the additivity of integrals) For f ∈ S(I, E) and a, b, c 3 I we have
Proof It suffices to check this for a 4 b 4 c If (f n) is a sequence of staircase
functions that converge uniformly to f and J is a compact perfect subinterval of
Positivity and monotony of integrals
Until now, we have considered the integral of jump continuous functions takingvalues in arbitrary Banach spaces For the following theorem and its corollary, wewill restrict to the real-valued case, where the ordering of the reals implies someadditional properties of the integral
4.5 Theorem For f ∈ S(I, R) such that f(x) ≥ 0 for all x 3 I, we have3
4 f ≥ 0.
Proof According to Remark 1.3, there is a sequence (f n) of nonnegative staircase
functions that converges uniformly to f Obviously then,
4 g, that is, integration is monotone on real-valued functions.
Proof This follows from the linearity of integrals and from Proposition 4.5.2
SinhVienZone.Com
Trang 394.7 Remarks (a) Suppose V is a vector space We call any linear map from V
to its field is a linear form or linear functional.
Therefore in the scalar case, that is, E =K, the integral3
4 is a continuouslinear form onS(I).
(b) Suppose V is a real vector space and P is a nonempty set with
call P a (in fact convex) cone in V (with point at 0) This designation is justified
because P is convex and, for every x, it contains the “half ray”R+
x In addition,
P contains “straight lines” Rx only when x = 0.
We next define4 through
x 4 y :⇐⇒ y − x 3 P , (4.6)
and thus get an ordering on V , which is linear on V (or compatible with the vector
space structure ofV ), that is, for x, y, z 3 V ,
Therefore we call P the positive cone of (V, 4 ).
If, conversely, a linear ordering 4 is imposed on V , then (4.7) defines convex cone in V , and this cone induces the given ordering Hence there is a bijection between the set of convex cones on V and the set of linear orderings on V Thus
we can write either (V, 4 ) or (V, P ) whenever P is a positive cone in V
(c) When (V, P ) is an ordered vector space and is a linear form on V , we call
positiveif (x) ≥ 0 for all x 3 P 2 A linear form on V is then positive if and only
if it is increasing3
Proof Suppose : V 5 R is linear Because (x−y) = (x)− (y), the claim is obvious.2
(d) Suppose E is a normed vector space (or a Banach space) and 4 is a linear
ordering on E Then we say E := (E, 4 ) is an ordered normed vector space (or
an ordered Banach space) if its positive cone is closed.
2 Note that the null form is positive under this definition.
3We may also say increasing forms are monotone linear forms.
SinhVienZone.Com
Trang 40Suppose E is an ordered normed vector space and (x j ) and (y j) are sequences
in E with x j 5 x and y j 5 y Then x j 4 y j for almost all j 3 N, and it follows
that x 4 y.
Proof If P is a positive cone in E, then y j −x j 2 P for almost all j 2 N Then it follows
from Proposition II.2.2 and Remark II.3.1(b) that
y − x = lim y j − lim x j = lim(y j − x j)2 P ,
and hence P is closed.2
(e) Suppose X is a nonempty set The pointwise ordering of Example I.4.4(c),
namely,
f 4 g :⇐⇒ f(x) 4 g(x) for x 3 X
and f, g 3 R XmakesRXan ordered vector space We call this ordering the natural
orderingonRX In turn, it induces on every subset M ofRX the natural ordering
on M (see Example I.4.4(a)) Unless stated otherwise, we shall henceforth always
provide RX and any of its vector subspaces the natural ordering In particular,
B(X,R) is an ordered Banach space with the positive cone
B+(X) := B(X,R+
) := B(X, R) ∩ (R+
)X
Therefore every closed vector subspaceF(X, R) of B(X, R) is an ordered Banach
space whose positive cone is given through
F+
(X) := F(X, R) ∩ B+
(X) = F(X, R) ∩ (R+
)X
Proof It is obvious that B+(X) is closed in B(X,R).2
(f ) From (e) it follows thatS(I, R) is an ordered Banach space with positive cone
S+(I), and Proposition 4.5 says the integral 3
4 is a continuous positive (and therefore monotone) linear form on S(I, R).2
The staircase function f : [0, 2] 5 R defined by
f (x) :=
7
1 , x = 1 ,
0 , otherwiseobviously satisfies
f = 0. Hence there is a nonnegative function that doesnot identically vanish but whose integral does vanish The next theorem gives acriterion for the strict positivity of integrals of nonnegative functions