an analysis companion - heil

8/12/2019 An Analysis Companion - Heil

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Christopher Heil

An Analysis Companion

January 23, 2010

c 2010 by Christopher Heil


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Contents

General Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

A Metrics, Norms, Inner Products, and Topology . . . . . . . . . . . . 3A.1 Metrics and Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3A.2 Norms and Seminorms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

A.2.1 Innite Series in Normed Spaces . . . . . . . . . . . . . . . . . . . . . 5A.2.2 Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

A.3 Examples of Banach Spaces: p, C b, C 0 , C mb . . . . . . . . . . . . . . . . 7A.3.1 The p Spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7A.3.2 Some Spaces of Continuous and Differentiable Functions 9

A.4 Inner Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11A.5 Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

A.5.1 Product Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17A.6 Convergence and Continuity in Topological Spaces . . . . . . . . . . . 17

A.6.1 Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17A.6.2 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20A.6.3 Equivalent Norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

A.7 Closed and Dense Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23A.8 Compact Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25A.9 Complete Sequences and a First Look at Schauder Bases . . . . . 28

A.9.1 Span and Closed Span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28A.9.2 Hamel Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30A.9.3 Introduction to Schauder Bases . . . . . . . . . . . . . . . . . . . . . 31

A.10 Unconditional Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32A.11 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

A.11.1 Orthogonality and the Pythagorean Theorem . . . . . . . . . 34A.11.2 Orthogonal Direct Sums. . . . . . . . . . . . . . . . . . . . . . . . . . . . 35A.11.3 Orthogonal Projections and Orthogonal Complements . 35

A.12 Orthogonality and Complete Sequences . . . . . . . . . . . . . . . . . . . . 36A.13 Urysohns Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39


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B Lebesgue Measure and Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41B.1 Exterior Lebesgue Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41B.2 Lebesgue Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

B.2.1 Denition and Basic Properties . . . . . . . . . . . . . . . . . . . . . 43

B.2.2 Equivalent Formulations of Measurability . . . . . . . . . . . . . 44B.2.3 Almost Everywhere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46B.3 Measurable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47B.4 Convergence in Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48B.5 The Lebesgue Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

B.5.1 Integration of Nonnegative Simple Functions . . . . . . . . . . 49B.5.2 Integration of Nonnegative Functions . . . . . . . . . . . . . . . . 50B.5.3 Integration of Real-Valued and Complex-Valued

Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53B.5.4 The Lebesgue Dominated Convergence Theorem . . . . . . 53B.5.5 Relation to the Riemann Integral . . . . . . . . . . . . . . . . . . . . 54

B.6 The L p Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55B.6.1 Norm and Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . 55B.6.2 On Abuses of Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56B.6.3 Convergence in L p(E ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58B.6.4 Local Integrability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

B.7 Repeated Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61B.8 Functions of Bounded Variation . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

B.8.1 Denition and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 64B.8.2 The Jordan Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . 66B.8.3 Differentiability of Functions of Bounded Variation . . . . 67

B.9 Absolutely Continuous and Singular Functions . . . . . . . . . . . . . . 68B.9.1 Singular Functions on the Real Line . . . . . . . . . . . . . . . . . 68B.9.2 Absolutely Continuous Functions on the Real Line . . . . . 70B.9.3 Preparation for the BanachZarecki Theorem . . . . . . . . . 73B.9.4 The BanachZarecki Theorem and its Implications . . . . 75

B.10 Holder Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

C Functional Analysis and Operator Theory . . . . . . . . . . . . . . . . . 81C.1 Linear Operators on Normed Spaces . . . . . . . . . . . . . . . . . . . . . . . 81

C.1.1 Equivalence of Boundedness and Continuity . . . . . . . . . . 83C.1.2 Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84C.1.3 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . 85

C.2 Some Useful Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86C.2.1 Orthogonal Projections . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86C.2.2 Multiplication Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . 87C.2.3 Integral Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87C.2.4 Convolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

C.3 The Space

B (X, Y ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

C.4 Banach Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93C.5 Some Dual Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94


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C.5.1 The Dual of a Hilbert Space . . . . . . . . . . . . . . . . . . . . . . . . 95C.5.2 The Dual of L p(E ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96C.5.3 The Relation between L p

(E ) and L p(E ) . . . . . . . . . . . . 97C.6 Adjoints for Operators on Hilbert Spaces . . . . . . . . . . . . . . . . . . . 99

C.6.1 Adjoints of Bounded Operators . . . . . . . . . . . . . . . . . . . . . 99C.6.2 Adjoints of Unbounded Operators . . . . . . . . . . . . . . . . . . . 101C.6.3 Bounded Self-Adjoint Operators on Hilbert Spaces . . . . 101C.6.4 Positive and Positive Denite Operators on Hilbert

Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103C.7 Compact Operators on Hilbert Spaces. . . . . . . . . . . . . . . . . . . . . . 105

C.7.1 Denition and Basic Properties . . . . . . . . . . . . . . . . . . . . . 105C.7.2 Finite-Rank Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107C.7.3 Integral Operators with Square-Integrable Kernels . . . . . 109

C.8 The Spectral Theorem for Compact Self-Adjoint Operators . . . 110C.8.1 Existence of an Eigenvalue . . . . . . . . . . . . . . . . . . . . . . . . . 111C.8.2 The Spectral Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

C.9 HilbertSchmidt Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115C.9.1 Denition and Basic Properties . . . . . . . . . . . . . . . . . . . . . 115C.9.2 Singular Numbers and Schatten Classes . . . . . . . . . . . . . . 116C.9.3 HilbertSchmidt Integral Operators . . . . . . . . . . . . . . . . . . 119C.9.4 Trace-Class Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

C.10 The HahnBanach Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122C.10.1 Abstract Statement of the HahnBanach Theorem . . . . . 122C.10.2 Implications of the HahnBanach Theorem . . . . . . . . . . . 123C.10.3 Orthogonal Complements in Normed Spaces . . . . . . . . . . 125C.10.4 X and Reexivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125C.10.5 Adjoints of Operators on Banach Spaces . . . . . . . . . . . . . 126

C.11 The Baire Category Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127C.12 The Uniform Boundedness Principle . . . . . . . . . . . . . . . . . . . . . . . 129C.13 The Open Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130C.14 The Closed Graph Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131C.15 Schauder Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

C.15.1 Continuity of the Coefficient Functionals . . . . . . . . . . . . . 132C.15.2 Minimal Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134C.15.3 A Characterization of Schauder Bases . . . . . . . . . . . . . . . . 135C.15.4 Unconditional Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

C.16 Weak and Weak* Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

D Borel and Radon Measures on the Real Line . . . . . . . . . . . . . . 139D.1 -Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139D.2 Signed Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141

D.2.1 The Jordan Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . 143D.3 Positive Measures and Integration . . . . . . . . . . . . . . . . . . . . . . . . . 145

D.3.1 Basic Properties of Positive Measures . . . . . . . . . . . . . . . . 145D.3.2 Borel Measurable Functions . . . . . . . . . . . . . . . . . . . . . . . . 146


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D.3.3 Integration of Nonnegative Functions . . . . . . . . . . . . . . . . 146D.3.4 Integration of Arbitrary Functions . . . . . . . . . . . . . . . . . . . 148

D.4 Signed Measures and Integration . . . . . . . . . . . . . . . . . . . . . . . . . . 150D.5 Complex Measures. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

D.6 Fubini and Tonelli for Borel Measures . . . . . . . . . . . . . . . . . . . . . . 156D.7 Radon Measures. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158D.8 The Riesz Representation Theorem for Positive Linear

Functionals on C c(R) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159D.8.1 Topologies on C c (R) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160D.8.2 Positive Linear Functionals on C c (R) . . . . . . . . . . . . . . . . 162

D.9 The Relation Between Radon and Borel Measures on R . . . . . . 163D.10 The Dual of C 0(R) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

E Topological Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169E.1 Motivation and Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169E.2 Topological Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

E.2.1 Base for a Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174

E.2.2 Topological Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 174E.3 Topologies Induced by Families of Seminorms . . . . . . . . . . . . . . . 176E.3.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176E.3.2 The Topology Associated with a Family of Seminorms . 176E.3.3 The Convergence Criterion . . . . . . . . . . . . . . . . . . . . . . . . . 178E.3.4 Continuity of the Vector Space Operations . . . . . . . . . . . . 180E.3.5 Continuity Equals Boundedness . . . . . . . . . . . . . . . . . . . . . 180

E.4 Topologies Induced by Countable Families of Seminorms . . . . . 182E.4.1 Metrizing the Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182E.4.2 Tempered and Compactly Supported Distributions . . . . 184

E.5 C c (R) and its Dual Space D (R) . . . . . . . . . . . . . . . . . . . . . . . . . . 186E.5.1 The Topology on C c (R) . . . . . . . . . . . . . . . . . . . . . . . . . . . 186E.5.2 The Space of Distributions . . . . . . . . . . . . . . . . . . . . . . . . . 188

E.6 The Weak and Weak* Topologies on a Normed Linear Space . . 189E.6.1 The Weak Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189E.6.2 The Weak* Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190

E.7 Alaoglus Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192E.7.1 Product Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192E.7.2 Statement and Proof of Alaoglus Theorem . . . . . . . . . . . 193E.7.3 Implications for Separable Spaces. . . . . . . . . . . . . . . . . . . . 194

F Complex Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197F.1 Analytic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197F.2 Power Series and Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198F.3 Dirichlet Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199F.4 Trigonometric Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200F.5 Interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201


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G Zorns Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

Hints and Solution Sketches for Exercises and AdditionalProblems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225

Index of Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227

Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231


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General Notation

We review here some of the notational conventions that will be used through-out these notes.

Unless otherwise specied, all vector spaces are taken over the complexeld C. In particular, functions whose domain is Rd (or a subset of Rd) aregenerally allowed to take values in the complex plane C .

Integrals with unspecied limits are taken over either the real line or Rd ,according to context. In particular, if f : R C , then we take

f (x) dx = f (x) dx.The extended real line is R{, }= [, ]. We use the conventionsthat 1 / 0 = , 1/ = 0 , and 0 = 0 .If 1 p is given, then its dual index or dual exponent is the extendedreal number p that satises

1 p +

1 p = 1 .

Explicitly, p =

p p1

.

The dual index lies in the range 1 p , and we have 1 = , 2 = 2 ,and = 1 .The Kronecker delta is ij =

1, i = j,0, i = j.

If S is a subspace of a vector space V, then the nite linear span of S is

denoted by span( S ). If V is also a metric space, then the closed linear spanspan( V ) is dened to be the closure of span( S ) in V. We say that S is complete if span( S ) = V.


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2 Contents

We use the symbol to denote the end of a proof, and the symbol todenote the end of a denition, remark, example, or exercise, or the end of thestatement of a theorem whose proof will be omitted.


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A

Metrics, Norms, Inner Products, and Topology

These notes provide mini-courses on analysis, Lebesgue measure, functionalanalysis, Borel measures, topological vector spaces, and other related material.They were originally written as background notes for my text Introduction toHarmonic Analysis. Each appendix gives a substantial, though not exhaus-tive, introduction to and review of its respective subject. Topics that are nottypically part of standard beginning mathematics graduate courses are givenmore detailed attention, while other results are either formulated as exercises(often with hints) or stated without proof. Sources for additional informationon the material of this and most of the other appendices include Follands realanalysis text [Fol99], Conways functional analysis text [Con90], and the theoperator theory/Hilbert space text [GG01] by Gohberg and Goldberg. Hintsand some solution sketches for exercises and problems are given at the end of the notes.

A.1 Metrics and ConvergenceA metric determines a notion of distance between points in a set.

Denition A.1 (Metric Space). Let X be a set. A metric on X is a functiond: X X R such that for all f, g, h X we have:(a) d( f, g ) 0,(b) d( f, g ) = 0 if and only if f = g,(c) d( f, g ) = d( g, f ), and(d) the Triangle Inequality : d(f, h ) d(f, g ) + d( g, h).In this case, X is a called a metric space . The value d( f, g ) is the distance from f to g.

If we need to explicitly identify the metric we write let X be a metricspace with metric d or let ( X, d) be a metric space.


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4 A Metrics, Norms, Inner Products, and Topology

A metric space need not be a vector space, although this will be true of most of the metric spaces encountered in this volume.

Once we have a notion of distance, we have a corresponding notion of convergence.

Denition A.2 (Convergent and Cauchy Sequences). Let X be a met-ric space with metric d , and let {f n }nN be a sequence of elements of X.(a) We say that {f n }nN converges to f X if limn d(f n , f ) = 0 , i.e., if

> 0, N > 0, n N, d(f n , f ) < .In this case, we write lim

n f n = f or f n f.

(b) We say that {f n }nN is Cauchy if > 0, N > 0, m, n N, d(f m , f n ) < .

Exercise A.3. Let X be a metric space.

(a) Every convergent sequence in X is Cauchy.(b) The limit of a convergent sequence is unique.

In general, however, a Cauchy sequence need not be convergent (Exer-cises A.63A.61).

Denition A.4 (Complete Metric Space). If every Cauchy sequence in ametric space X has the property that it converges to an element of X, thenX is said to be complete .

Beware that the term complete is heavily overused and has a number of distinct mathematical meanings.

Notation A.5. Let X be a metric space. Given f

X and r > 0, the open

ball in X of radius r centered at f is

B r (f ) = g X : d(f, g ) < r . (A.1)

A.2 Norms and Seminorms

A norm provides a notion of the length of a vector in a vector space.

Denition A.6 (Seminorms and Norms). Let X be a vector space overthe eld C of complex scalars. A seminorm on X is a function : X Rsuch that for all f , g X and all scalars c C we have:(a)

f

0,

(b) cf = |c| f , and(c) the Triangle Inequality : f + g f + g .


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A.2 Norms and Seminorms 5

A seminorm is a norm if we also have:

(d) f = 0 if and only if f = 0 .A vector space X together with a norm is called a normed linear space

or simply a normed space . If the norm is not clear from context, we may write(X, ) to denote that is the norm on X. If S is a subspace of a normed space X, then S is itself a normed space

with respect to the norm on X (restricted to S ).

Exercise A.7. If X is a normed space, then d( f, g ) = f g denes a metricon X, called the induced metric . The Schwartz space (Example E.4) is an example of a metric space whose

metric is not induced from any norm; another is p with 0 < p < 1 (seeExercise A.18).

Exercise A.8. Show that if X is a normed linear space, then the followingstatements hold.

(a) Reverse Triangle Inequality: f g f g .(b) Continuity of the norm: f n f = f n f .(c) Continuity of vector addition: f n f and gn g = f n + gn f + g.(d) Continuity of scalar multiplication: f n f and n = n f n f.(e) Boundedness of convergent sequences: If {f n }nN converges then we havesup f n < .(f) Boundedness of Cauchy sequences: If {f n }nN is Cauchy then we havesup f n < . Denition A.9 (Banach Space). A normed linear space X is called a Ba-nach space if it is complete, i.e., if every Cauchy sequence is convergent.

Thus, the terms Banach space and complete normed space are inter-changeable.

An important fact that we will assume without proof is that the complexplane C under absolute value is a Banach space.

A.2.1 Innite Series in Normed Spaces

Since a normed space has both an operation of vector addition and a notionof convergence, we can consider innite series.

Denition A.10 (Convergent Series). Let {f n }nN be a sequence in anormed linear space X. Then the series n =1 f n converges and equals f X if the partial sums sN =

N n =1 f n converge to f , i.e., if

limN

f sN = limN f N

n =1f n = 0 .


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Note that the ordering of a series may be important! If we reorder a series,or in other words consider a new series n =1 f (n ) where : N N is abijection, there is no guarantee that this reordered series will still converge.These issues are addressed in more detail in Section A.10.

Denition A.11 (Absolutely Convergent Series). Let X be a normedspace and let {f n }nN be a sequence of elements of X. If

n =1f n < ,

then we say that the series n =1 f n is absolutely convergent in X. The denition of absolute convergence does not require that the series

f n converge in X. This will always be the case if X is a Banach space, andindeed this property is an equivalent characterization of completeness.

Exercise A.12. Let X be a normed space. Prove that X is a Banach spaceif and only if every absolutely convergent series in X converges in X. A.2.2 Convexity

Denition A.13 (Convex Set). If X is a vector space and K X, then K is convex if x, y K, 0 t 1 = tx + (1 t)y K.

Thus, the entire line segment between x and y is contained in K (includingthe midpoint 12 x +

12 y in particular).

Every subspace of a vector space is convex by denition. The fact thatballs in a normed space are convex is an important property.

Exercise A.14. Show that if X is a normed linear space, then each open ballB r (f ) in X is convex. Additional Problems

A.1. If {f n }nN is a Cauchy sequence in a normed space X and there existsa subsequence {f n k }kN that converges to f X, then f n f.A.2. If {f n }nN is a Cauchy sequence in a normed space X, then there existsa subsequence {f n k }kN such that f n k +1 f n k < 2 k for all k N.A.3. Let X be a normed space. Show that if f n X satisfy f n +1 f n < 2 nfor every n, then {f n }nN is Cauchy.


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A.3 Examples of Banach Spaces: p , C b , C 0 , C mb 7

A.4. Let {f n }nN be a sequence in a normed space X, and let f X bexed. Suppose that every subsequence {gn }nN of {f n }nN has a subsequence{hn }nN of {gn }nN such that hn f. Show that f n f.

A.5. Let X be a normed space. Extend the denition of convergence to fam-ilies indexed by a real parameter by declaring that if f X and f t X fort R, then f t f as t 0 if for every > 0 there exists a > 0 suchthat f f t < whenever |t| < . Show that f t f as t 0 if and only if f t k f for every sequence of real numbers {tk}kN such that tk 0.

A.3 Examples of Banach Spaces: p , C b , C 0 , C mb

In this section we give a few examples of Banach and other spaces.

A.3.1 The p Spaces

Denition A.15. Let I be a nite or countably innite index sequence.(a) If 0 < p < , then p(I ) consists of all sequences of scalars x = ( xk )kI such that

x p = (xk )kI p =kI

|xk | p1/p

< .

(b) For p = , the space (I ) consists of all sequences of scalars x = ( xk )kI such thatx = (xk )kI = supkI |xk | < .

If I = N, then we write p instead of p(N).

If I = {1, . . . , d}, then p

(I ) = Cd

, and in this case we refer to p

(I ) as Cd

under the p norm. The 2 norm on C d is called the Euclidean norm . Now we prove a fundamental inequality for the p spaces.

Theorem A.16 (H olders Inequality). Let I be a nite or countable index set. Given 1 p , if x = ( xk )kI p(I ) and y = ( yk )kI p

(I ), then xy = ( xk yk )kI 1(I ) and

xy 1 x p y p .For 1 < p < , this inequality is

kI |xk yk | kI |xk | p

1/p

kI |yk | p

1/p

.


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Proof. The cases p = 1 and p = are straightforward exercises. Assume1 < p < . The key to the proof is a special case of an inequality dueto Young for continuous, strictly increasing functions. Namely, since x p 1 iscontinuous and strictly increasing and its inverse function is y

1p 1 , we have

for all a, b 0 thatab a0 x p 1 dx + b0 y 1p 1 dy = a p p + b p p

(see the proof by picture in Figure A.1, or Problem A.10).

a

b

Fig. A.1. Illustration of Youngs Inequality. Area of the vertically hatched regionis R a0 xp

1 dx ; area of the horizontally hatched region is R b0 y1

p 1 dy ; area of therectangle is ab.

Consequently, if x

p(I ) and y

p

(I ) satisfy

x

p = 1 =

y

p , then

xy 1 =kI

|xk yk | kI

|xk | p p

+ |yk | p

p =

1 p

+ 1 p

= 1 . (A.2)

For general nonzero x, y , we apply equation (A.2) to the normalized vectorsx/ x p and y/ y p to obtain

xy 1x p y p

= xx p

yy p 1 1.

The next exercise shows that if p 1 then p is a norm on p(I ).The Triangle Inequality on p (often called Minkowskis Inequality ) is easy toprove for p = 1 and p = , but more difficult for 1 < p < . A hint for usingHolders Inequality to prove Minkowskis Inequality is given in the solutionssection at the end of the text.


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A.3 Examples of Banach Spaces: p , C b , C 0 , C mb 9

Exercise A.17. Let I be a nite or countable index set. Show that if 1 p , then p is a norm on p(I ), and p(I ) is a Banach space with respectto this norm.

On the other hand, if p < 1 then p fails the Triangle Inequality andhence is not a norm. Still, we can modify the distance function so that p is acomplete metric space, though this metric is not induced from any norm.

Exercise A.18. Let I be a nite or countably innite index set. Show that if 0 < p < 1, then x + y p p x p p + y p p. Consequently, p(I ) is a vector spaceand d( x, y ) = x y p p is a metric on p(I ). Show that p(I ) is complete withrespect to this metric. However, if I contains more than one element, then theunit ball B 1(0) is not convex, and hence this metric is not induced from anynorm (compare Exercise A.14).

We can also dene p(I ) when the index set I is uncountable. In this case,for p < we dene p(I ) to be the space of all sequences x = ( xk )kI withat most countably many terms nonzero such that |x

k | p

< . With thisdenition, p(I ) is again a Banach space.

A.3.2 Some Spaces of Continuous and Differentiable Functions

We now give some examples of normed spaces of functions.

Denition A.19. The support of a function f : R C is the closure of theset of points where f is nonzero:supp( f ) = {xR : f (x) = 0 }.

Since the support of a function is a closed set, a function on R has compact

support if and only if it is zero outside of a nite interval.Exercise A.20. Let C b(R) denote the space of continuous, bounded functionsf : R C. Show that C b(R) is a Banach space with respect to the uniform norm

f = suptR |f (t)|.

Show that the subspace

C 0(R) = f C b(R) : lim| t | f (t) = 0is also a Banach space with respect to the uniform norm, but the subspace

C c(R

) = f C b(R

) : supp( f ) is compact (A.3)is not complete with respect to the uniform norm.


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Beware, some authors use the symbols C 0 to denote the space that werefer to as C c .

Exercise A.20 has an extension to m-times differentiable functions, as fol-lows.

Exercise A.21. Let C mb (R) be the space of all m-times differentiable func-tions on R each of whose derivatives is bounded and continuous, i.e.,

C mb (R) = f C b(R) : f , f , . . . , f (m ) C b(R) .Show that C mb (R) is a Banach space with respect to the norm

f C mb = f + f + + f (m ) ,and

C m0 (R) = f C 0(R) : f, f , . . . , f (m )C 0(R)is a subspace of C mb (R) that is also a Banach space with respect to the same

norm. However,C mc (R) = f C c(R) : f, f , . . . , f (m )C c(R)

is not a Banach space with respect to this norm. Although they are not normed spaces, it is often important to consider

the space of functions that are continuous or m-times differentiable but notbounded. We denote these by:

C (R) = f : R C : f is continuous ,C m (R) = f C (R) : f, f , . . . , f (m ) C (R) .

Additionally, we sometimes need to consider spaces of innitely differentiablefunctions, including the following:

C (R) = f C (R) : f, f , . . .C (R) ,C b (R) = f C b(R) : f, f , . . . C b(R) .C 0 (R) = f C 0 (R) : f, f , . . . C 0(R) .C c (R) = f C c(R) : f, f , . . .C c(R) .

The space C c (R) will be especially important to us in Appendix E. Althoughnot a normed space, it is topological vector space, and is often denoted by

D(R) = C c (R).


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A.4 Inner Products 11

Additional Problems

A.6. Fix 0 < p . Let {xn }nN be a sequence of vectors in p(I ), and x avector in p(I ). Write the components of xn and x as xn = ( xn (1) , xn (2) , . . . )and x = ( x(1) , x(2) , . . . ), and prove the following statements.(a) If xn x in p(I ), then xn converges componentwise to x, i.e., foreach xed k we have limn xn (k) = x(k).

(b) If I is nite then componentwise convergence implies convergence withrespect to the norm p.

(c) If I is innite then componentwise convergence need not imply conver-gence in the norm of p(I ).

A.7. Show that if 1 p < q , then p q , and x q x p for allx p.A.8. Show that if x q(I ) for some nite q then x p x as p ,but this can fail if x /

q(I ) for any nite q.

A.9. Let I be a nite or countable index set, and let w : I (0, ) be xed.Given a sequence of scalars x = ( xk )kI , set

x p,w = kI |xk | p w(k) p

1/p

, 0 < p < ,supkI |xk |w(k), p = ,

and dene the weighted p space pw (I ) = x : x p,w < . Show that pw (I )is a Banach space for each 1 p .A.10. (a) Show that if 0 < < 1, then t

t + (1

) for t > 0, with

equality if and only if t = 1 .(b) Suppose that 1 < p < and a, b 0. Apply part (a) with t = a pb p

and = 1 /p to show that ab a p/p + b p

/p , with equality if and only if b = a p 1 .

A.11. Show that equality holds in H olders Inequality (Theorem A.16) if andonly if there exist scalars , , not both zero, such that |xk | p = |yk | p

foreach k I.

A.4 Inner Products

While a norm on a vector space provides a notion of the length of a vector,an inner product provides us with a notion of the angle between vectors.


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Denition A.22 (Semi-Inner Product, Inner Product). If H is a vectorspace over the complex eld C , then a semi-inner product on H is a function

, : H H C such that for all f, g, h H and scalars , C we have:(a)

f, f

0,

(b) f, g = g, f ,(c) Linearity in the rst variable: f + g,h = f, h + g, h .If a semi-inner product , also satises:

(d) f, f = 0 if and only if f = 0 ,then it is called an inner product on H. In this case, H is called a inner product space or a pre-Hilbert space .

There are many different standard notations for semi-inner products, in-cluding [f, g ], (f, g ), or f |g , in addition to our preferred notation f, g .Exercise A.23. If , is a semi-inner product on a vector space H, showthat the following statements hold.(a) Antilinearity in the second variable: f,g + h = f, g +

f, h .(b) f, 0 = 0 = 0, f .

(c) 0, 0 = 0 . A function of two variables that is linear in the rst variable and antilinear

(also called conjugate linear ) in the second variable conjugate linear function is referred to as a sesquilinear form . Thus each semi-inner product , is anexample of a sesquilinear form. 1 Sometimes an inner product is required tobe antilinear in the rst variable and linear in the second; this is common inthe physics literature.

Every subspace S of an inner product space H is itself an inner productspace (using the inner product on H restricted to S ).

The next exercise gives the prototypical example of an inner product.

Exercise A.24. Given an index set I and x = ( xk )kI , (yk )kI 2(I ), setx, y =

kI

xk yk .

Show that , is an inner product on 2(I ). Our next goal is to show that every semi-inner product induces a seminorm

on H, and every inner product induces a norm.

Notation A.25. If , is a semi-inner product on a vector space H, then wewritef = f, f 1/ 2 , f H.

Prejudicing the issue, we refer to

as the seminorm induced by

,

. If

,

is an inner product, then we call the norm induced by , . 1 The prex sesqui- means one and a half.


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A.4 Inner Products 13

Before showing that actually is a seminorm or norm, we derive someof its basic properties.Exercise A.26. Given a semi-inner product , on a vector space H, showthat the following statements hold for all f , g H.(a) Polar Identity: f + g 2 = f 2 + 2Re f, g + g 2 .(b) Parallelogram Law: f + g 2 + f g 2 = 2 f 2 + g 2 .

Now we prove an important inequality; this should be compared toHolders Inequality (Theorem A.16) for p = 2 . This inequality is variouslyknown as the Schwarz , CauchySchwarz , or CauchyBunyakowskiSchwarz Inequality .

Theorem A.27 (CauchyBunyakowskiSchwarz Inequality). If , is a semi-inner product on a vector space H, then f,g,H, | f, g | f g .

Proof. If f = 0 or g = 0 then there is nothing to prove, so suppose both arenonzero. Write f, g = | f, g | where C and | | = 1 . Then for tR wehave by the Polar Identity that

0 f tg 2 = f 2 2Ret f, g + t2 g 2= f 2 2t | f, g |+ t2 g 2 .

This is a real-valued quadratic polynomial in the variable t. In order for itto be nonnegative, it can have at most one real root. This requires that thediscriminant be at most zero, so 2 | f, g |

2

4 f 2 g 2 0. The desiredinequality then follows upon rearranging. When f, g = 0 , we say that f and g are orthogonal . More details onorthogonality appear in Section A.11.Finally, the CauchyBunyakowskiSchwarz Inequality can be used to show

that is indeed a seminorm or norm on H.Exercise A.28. Given a semi-inner product , on a vector space H, showthat is a seminorm on H, and if , is an inner product then is anorm on H.

Thus, all inner product spaces are normed linear spaces. The followingexercise shows that there are normed space whose norm is not induced fromany inner product on the space.

Exercise A.29. Let I be an index set containing at least two elements. Showthat p does not satisfy the Parallelogram Law if 1 p and p = 2 .Therefore, there is no inner product on p(I ) whose induced norm is p.


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On the other hand, it is certainly possible for p(I ) to be an inner productspace with respect to some inner product. For example, if I is nite or if p < 2,then p(I )

2(I ), so in this case p(I ) is an inner product space with respectto the inner product on 2(I ) restricted to the subspace p(I ). However, the

norm induced from this inner product is 2 and not p. Further, if I isinnite then p(I ) is not complete with respect to the induced norm 2 ,whereas it is complete with respect to the norm p.Denition A.30 (Hilbert Space). An inner product space H is called aHilbert space if it is complete with respect to the induced norm, i.e., if everyCauchy sequence is convergent.

Thus, a Hilbert space is an inner product space that is a Banach spacewith respect to the induced norm. In particular, 2(I ) is a Hilbert space.

Additional Problems

A.12. Show that , is an inner product on Cd

if and only if there exists apositive denite matrix A such that x, y = Ax y, where x y = x1 y1 + +xd yd denotes the usual dot product on C d .A.13. Continuity of the inner product: If H is an inner product space andf n f, gn g in H, then f n , gn f, g .A.14. Let H be an inner product space. Show that if the series n =1 f nconverges in H, then for any g H,

n =1f n , g =

n =1f n , g .

Note that this is not merely a consequence of the linearity of the inner product

in the rst variable the continuity of the inner product is also needed.A.15. Show that equality holds in the CauchyBunyakowskiSchwarz In-equality if and only if there exist scalars , C , not both zero, such thatf + g = 0 . In particular, if is a norm, then either f = cg or g = cf for some scalar c.A.16. Justify the following statement: The angle between two vectors f , g ina Hilbert space H is the value of that satises Re f, g = f g cos .

A.5 Topology

Now we consider topologies, especially on metric and normed spaces. Generalbackground references on topology include Munkres [Mun75] and Singer andThorpe [ST76].


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A.5 Topology 15

Denition A.31 (Topology). A topology on a set X is a family T of subsetsof X such that the following statements hold.(a) , X T .(b) Closure under arbitrary unions: If I is any index set and U i

T for i

I ,

then iI U i T .(c) Closure under nite intersections: If U, V T , then U V T .If these hold then T is called a topology on X and X is called a topological space . The elements of T are called the open subsets of X. The complementsof the open subsets are the closed subsets of X. A neighborhood of a pointx X is any set A such that there exists an open set U with x U A.In particular, an open neighborhood of x is any open set U that contains x. If the topology on X is not clear from context, we may write ( X, T ) to denotethat T is the topology on X.

The following is a convenient criterion for testing for openness.

Exercise A.32. Let X be a topological space. Prove that V

X is open if

and only if

x V, open U V such that x U. If a space X has two topologies T 1 , T 2 and if T 1 T 2 , i.e., every set thatis open with respect to T 1 is also open with respect to T 2 , then we say that

T 1 is weaker than T 2 and T 2 is stronger than T 1 . These terms should not beconfused with the weak topology or the strong topology on a space. The strongtopology on a normed space is dened below (see Denition A.34) and theweak topology is dened in Example E.7.

The most familiar topologies are those associated with metric spaces.

Exercise A.33. Let (X, d) be a metric space. Declare a subset U X to beopen if

f U, r > 0 such that B r (f )U,where B r (f ) is the open ball centered at f with radius r. Let T be the collec-tion of all subsets of X that are open according to this denition. Show thatT is a topology on X.

Thus every metric space has a natural topology associated with it, andconsequently so does every normed space.

Denition A.34. (a) If ( X, d) is a metric space, then the topology T denedin Exercise A.33 is called the topology on X induced from the metric d, orsimply the induced topology on!X.

(b) If ( X,

) is a normed linear space, then the topology

T induced from the

metric d( f, g ) = f g is called the norm topology , the strong topology ,or the topology induced from .


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(c) Let ( X, T ) be a topological space. If there exists a metric d on X whoseinduced topology is exactly T , then the topology T on X is said to bemetrizable .(d) Let X be a vector space with a topology

T . If there exists a norm

on X whose induced topology is exactly T , then the topology T on X issaid to be normable . The Schwartz space S (R) and the space C (R) are important examples of topological spaces that are metrizable but not normable (see Example E.28).The topology induced by a metric has the following special property.

Denition A.35. A topological space X is Hausdorff if

x = y X, disjoint U, V T such that x U, y V. Exercise A.36. Every metric space is Hausdorff.

Every subset of a topological space X inherits a topology from X.

Exercise A.37. Let X be a topological space. Given Y

X, show that

T Y = {U Y : U is open in X }is a topology on Y , called the topology on Y relative to X, the topology on Y inherited from X, the topology on X restricted to Y , etc.

One way to generate a topology on a set X is to begin with a collectionof sets that we want to be open, and then to create a topology that includesthose particular sets. There will be many such topologies in general, but thefollowing exercise shows that there is a smallest topology that includes thosechosen sets.

Exercise A.38. Let E be a collection of subsets of a set X. Show that

T (

E ) =

T :

T is a topology on X with

E T is a topology on X. We call T (E ) the topology generated by E . The collectionE is sometimes called a subbase for the topology T (E ).

Note that if T 1 , T 2 are topologies, then T 1 T 2 is not formed by intersectingthe elements of T 1 with those of T 2 . Rather, it is the collection of all sets thatare common to both T 1 and T 2 . Thus, if T is any topology that contains E thenwe will have T (E ) T , which explains why T (E ) is the smallest topologythat contains E . In particular, the topology induced by a metric d on a metricspace X is the topology generated by the set of open balls in X.We can characterize the generated topology T (E ) as follows.

Exercise A.39. If E is a collection of subsets of a set X whose union is X,then T (E ) is set of all unions of nite intersections of elements of E :

T (E ) = iI n

j =1E ij : I arbitrary, n N, E ij E .


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A.6 Convergence and Continuity in Topological Spaces 17

A.5.1 Product Topologies

As an application of generated topologies, we show how to construct a nat-ural topology on the Cartesian product of two topological spaces. A product

topology on an innite collection of topological spaces can also be dened butrequires a little more care, see Denition E.44.

Denition A.40 (Product Topology). Let X and Y be topological spaces,and set

B = {U V : open U X, open V Y }.The product topology on X Y is the topology T (B ) generated by B . Exercise A.41. (a) Show that B as given above is a base for T (B ), whichmeans that if W T (B ), then there exist sets U open in X and V openin Y such that W = (U V ).(b) Show that if W X Y is open with respect to the product topology, thenfor each x

X the restriction W x =

{y

Y : (x, y )

W

} is open in Y,

and likewise for each y Y the restriction W y = {x X : (x, y ) W } isopen in X.

A.6 Convergence and Continuity in Topological Spaces

In large part, the importance of topologies in this volume is that they providenotions of convergence and continuity. Indeed, a basic philosophy that wewill expand upon in this section is that topologies and convergence criteriaare equivalent. Thus, though many of the spaces that we will encounter aredened in terms of norms or families of seminorms (i.e., convergence criteria),this is equivalent to dening them in terms of a topology.

A.6.1 Convergence

In metric spaces, convergence is dened with respect to sequences indexed bythe natural numbers (Denition A.2). In a general topological space, conver-gence must be formulated in terms of nets instead of countable sequences.

Denition A.42 (Directed Sets, Nets). A directed set is a set I togetherwith a relation on I such that:(a) is reexive: i i for all i I,(b) is transitive: i j and j k implies i k, and(c) for any i, j I, there exists k I such that i k and j k.

A net in a set X is a sequence {xi}iI of elements of X indexed by adirected set ( I, ).


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Remark A.43. By denition, a sequence {xi}iI is shorthand for the functionx : I X dened by x(i) = xi for i I . In particular, unlike a set, a sequenceallows repetitions of the xi . Technically, we should be careful to distinguishbetween a sequence {x i}iI and a set {x i : i I }, but it is usually clear fromcontext whether a sequence or a set is meant.

The set of natural numbers I = N under the usual ordering is one exampleof a directed set, and hence every ordinary sequence indexed by the naturalnumbers is a net. Another typical example is I = P (X ), the power set of X,ordered by reverse inclusion , i.e.,

U V V U.Denition A.44 (Convergence of a Net). Let X be a topological space,let {xi}iI be a net in X, and let x X be given. Then we say that {xi}iI converges to x (with respect to the directed set I ), and write xi x, if forany open neighborhood U of x there exists i0 I such that

i i0 = xi U. Next, we will dene the notion of accumulation points of a subset of a

generic topological space and see how this denition can be reformulated interms of nets. We will also see that topologies induced from a metric havethe advantage that we only need to use convergence of ordinary sequencesindexed by N instead of general nets.

Denition A.45 (Accumulation Point). Let E be a subset of a topologicalspace X. Then a point x X is an accumulation point of E if every openneighborhood of x contains a point of E other than x itself, i.e.,

U open and x U = E (U \{x}) = . Lemma A.46. If E is a subset of a topological space X and x X, then the following statements are equivalent.(a) x is an accumulation point of E.(b) There exists a net {xi}iI contained in E \{x} such that x i x.

If X is a metric space, then these statements are also equivalent to the following.(c) There exists a sequence {xn }nN contained in E \{x} such that xn x.Proof. (a) (b). Assume that x is an accumulation point of E . Dene

I = U X : U is open and x U .

Exercise: Show that I is a directed set when ordered by reverse inclusion.For each U I, by denition of accumulation point there exists a pointxU E (U \{x}). Then {xU }U I is a net in E \{x}, and we claim that


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xU x. To see this, x any open neighborhood V of x. Set U 0 = V, andsuppose that U U 0 . Then, by denition, U I and U U 0 . Hence xU U U 0 = V. Therefore xU x.(b)

(a). Suppose that

{xi

}iI is a net in E

\{x

} and x i

x. Let U be

any open neighborhood of x. Then there exists an i0 such that x i U for alli i0 . Since x i = x, this implies that x i E (U \{x}) for all i i0 .(a) (c), assuming X is metric. Suppose that x is an accumulation pointof E. For each n N, the open ball B1/n (x) is an open neighborhood of x,and hence there must exist some xn E (B1/n (x)\{x}). Therefore {xn }nNis a sequence in E \{x}, and since d( x, x n ) < 1/n, we have xn x.(c) (b), assuming X is metric. This follows from the fact that everycountable sequence {xn }nN is a net. We can now give an equivalent formulation of closed sets in terms of nets

and accumulation points.

Exercise A.47. Given a subset E of a topological space, prove that the fol-

lowing statements are equivalent.(a) E is closed, i.e., X \E is open.(b) If x is an accumulation point of E, then x E.(c) If {xi}iI is a net in E and x i x X, then x E.

If X is a metric space, show that these are also equivalent to the followingstatement.(d) If {xn }nN is a sequence in E and xn x X, then x E.

Now we can quantify the philosophy that topologies and convergence cri-teria are equivalent. For arbitrary topologies, this requires that we use con-vergence with respect to nets, but for topologies induced from a metric we areable to use convergence of ordinary sequences indexed by the natural numbers.

Exercise A.48. Given two topologies T 1 , T 2 on a set X, prove that the fol-lowing statements are equivalent.(a) T 1 T 2 , i.e.,

U is open with respect to T 1 = U is open with respect to T 2 .(b) If {xi}iI is a net in X and x X, then

xi x with respect to T 2 = xi x with respect to T 1 .If T 1 is induced from a metric d 1 on X, and T 2 is induced from a metric d 2on X, show that these statements are also equivalent to the following.

(c) If

{xn

}nN is a sequence in X and x

X, then

limn

d2 (xn , x) = 0 = limn d1(xn , x) = 0 .


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Interchanging the roles of T 1 and T 2 in Exercise A.48, we see that T 1 = T 2if and only if T 1 and T 2 dene exactly the same convergence criterion.Example A.49. An example of a topological space where it is important todistinguish between convergence of ordinary sequences and convergence withrespect to nets is the sequence space 1 under the weak topology. This topologywill be dened precisely in Section E.6, but the important point for us at themoment is that it can be shown that if {xn }nN is a sequence in 1 and xn xwith respect to the weak topology, then xn x in norm, i.e., x xn 1 0(see [Con90, Prop. V.5.2]). However, the weak topology on 1 is not the sameas the topology induced by the norm 1 . The moral is that when discussingconvergence in a topological space that is not a metric space, it is importantto consider nets instead of ordinary sequences. A.6.2 Continuity

Our next goal is to reformulate continuity of a function in terms of convergence

of nets or sequences. Recall that if f : X Y and V Y, then the preimageof V is f 1(V ) = {x X : f (x) V }.Denition A.50 (Continuity). Let X, Y be topological spaces. Then afunction f : X Y is continuous if

V is open in Y = f 1(V ) is open in X.

We say that f is a topological isomorphism or a homeomorphism if f is abijection and both f and f 1 are continuous.

It will be convenient to restate continuity in terms of continuity at a point.

Denition A.51 (Continuity at a Point). Let X, Y be topological spacesand let x

X be given. Then a function f : X

Y is continuous at x if for

each open neighborhood V of f (x) in Y, there exists an open neighborhoodU of x in X such that U f 1(V ). Exercise A.52. Prove that f is continuous if and only if f is continuous ateach x X.

Now we can formulate continuity in terms of preservation of convergence of nets. For the case of a metric space, this reduces to preservation of convergenceof sequences.

Lemma A.53. If X, Y be topological spaces, f : X Y, and x X are given,then the following statements are equivalent.(a) f is continuous at x.

(b) For any net {xi}iI in X,x i x in X = f (x i ) f (x) in Y.


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If X is a metric space, then these are also equivalent to the following.(c) For any sequence {xn }nN in X,

xn x in X = f (xn ) f (x) in Y.Proof. (a) (b). Assume that f is continuous at x X. Let {xi}iI be anynet in X such that xi x. Let V be any open neighborhood of f (x). Thenby denition of continuity at a point, there exists an open neighborhood U of x that is contained in f 1(V ). Hence, by denition of x i x, there existsan i0 I such that xi U f 1(V ) for all i i0 . Hence f (x i ) V for alli i0 , which means that f (xi ) f (x).

(b) (a). Suppose that statement (a) fails, i.e., f is not continuous atx X. Then by denition there exists an open neighborhood V of f (x) suchthat no open neighborhood U of x can be contained in f 1(V ). Thereforeeach open neighborhood U of x must contain some point xU U \f 1(V ).Now let

I = U : U is an open neighborhood of x .

Then I is a directed set when ordered by reverse inclusion, so {xU }U I is anet in X. Exercise: Show that xU x. However, f (xU ) does not converge tof (x) because V is an open neighborhood of f (x) but V contains no pointsf (xU ). Hence statement (b) fails.

The remaining implications are exercises. A.6.3 Equivalent Norms

Next we consider the equivalence of convergence criteria and topologies forthe case of normed spaces.

Denition A.54. Suppose that X is a normed linear space with respect to

a norm a and also with respect to another norm b. Then we say thatthese norms are equivalent if there exist constants C 1 , C 2 > 0 such thatf X, C 1 f a f b C 2 f a . (A.4)

We write a b to denote that a and b are equivalent norms. Theorem A.55. Let a and b be two norms on a vector space X. Then the following statements are equivalent.(a) a and b are equivalent norms.(b) a and b induce the same topologies on X.(c) a and b dene the same convergence criterion. That is, if {xn }nNis a sequence in X and x

X, then

limn

x xn a = 0 limn x xn b = 0 .


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Proof. (b) (a). Assume that statement (b) holds. Let Bar (x) and Bbr (x)denote the open balls of radius r centered at x X with respect to a andb, respectively. Since B a1 (0) is open with respect to a , statement (b)implies that B a1 (0) is open with respect to b. Therefore, since 0 B a1 (0) ,

there must exist some r > 0 such that Bbr (0) B

a1 (0) .Now choose any x X and any > 0. Then

(r )x b

x Bbr (0) B a1 (0) ,so

(r ) xx b a

< 1.

Rearranging, this implies ( r ) x a < x b. Since this is true for every ,we conclude that r x a x b.A symmetric argument, interchanging the roles of the two norms, showsthat there exists an s > 0 such that x b s x a for every x X. Hence thetwo norms are equivalent. Given any nite-dimensional vector space X, we can dene many normson X. In particular, the following norms are analogues of the p norms denedin Section A.3.

Exercise A.56. Let B = {x1 , . . . , x d} be any basis for a nite-dimensionalvector space X, and let x = dk=1 ck (x) xk denote the unique expansion of x X with respect to this basis (the vector [ x]B = ( c1 (x), . . . , cd (x)) is calledthe coordinate vector of x with respect to the basis B ). Show that

x p =

d

k=1|ck (x)| p

1/p

, 1 p < ,max

k |ck (x)|, p = ,are norms on X, and X is complete with respect to each of these norms. Notethat x p is simply the p norm of the coordinate vector [ x]B .

It is not difficult to see that all of the norms dened in Exercise A.56are equivalent. Although we will not prove it, it is an important fact that all norms on a nite-dimensional space are equivalent.

Theorem A.57. If X is a nite-dimensional vector space, then any twonorms on X are equivalent. In particular, if is any norm on X and pis any one of the norms constructed in Exercise A.56, then p.


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A.7 Closed and Dense Sets 23

Additional Problems

A.17. (a) Let X be a Hausdorff topological space. Show that if a net {xi}iI converges in X, then the limit is unique.(b) Show that if X is not Hausdorff, then there exists a net {xi}iI in X that has two distinct limits.

A.18. Let {xi}iI be a net in a Hausdorff topological space X. Show that if xi x in X, then either:(a) there exists an open neighborhood U of x and some i0 I such thatxi = x for all i i0 , or(b) every open neighborhood U of x contains innitely many distinct x i ,

i.e., the set {x i : i I and x i U } is innite.A.19. Show that if ( X, d1) and ( Y, d2 ) are metric spaces, then

d (f 1 , g1), (f 2 , g2) = d 1(f 1 , f 2) + d 2(g1 , g2)

denes a metric on X Y that induces the product topology. Conclude thatconvergence in X Y is componentwise convergence, i.e., ( f n , gn ) (f, g ) inX Y if and only if f n f in X and gn g in Y .A.20. Let X be a vector space with a metric d . We say that the metricis translation-invariant if d(f + h, g + h) = d( f, g ) for every f , g, h X.Show that vector addition is continuous in this case, i.e., ( f, g ) f + g is acontinuous mapping of X X into X.A.21. Let X, Y be topological spaces. Let {(f i , gi )}iI be any net in X Y,and suppose ( f, g ) X Y. Show that ( f i , gi ) (f, g ) with respect to theproduct topology on X Y if and only if f i f in X and gi g in Y.

A.7 Closed and Dense Sets

The smallest closed set that contains a given set is called its closure, denedprecisely as follows.

Denition A.58. If E is a subset of a topological space X, then the closure of E , denoted E, is the smallest closed set in X that contains E :

E = {F X : F is closed and F E }.If E = X, then we say that E is dense in X.

Often it is more convenient to use the following equivalent form of theclosure of a set.


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Exercise A.59. Given a subset E of a topological space X, show that E isthe union of E and all the accumulation points of E .

The typical method for showing that a subset of a metric space is dense isgiven in the next exercise.

Exercise A.60. Let X be a metric space, and let E X be given. Show thatE is dense in X if and only if for each f X there exist a sequence {f n }nNin E such that f n f. The next exercise characterizes the closed subspaces of a Banach space.

Exercise A.61. Let M be a subspace of a Banach space X. Then M is itself aBanach space (using the norm inherited from X ) if and only if M is closed.

Every subspace of a nite-dimensional normed space is closed, but thisneed not be the case in innite dimensions.

Exercise A.62. Show that every nite-dimensional subspace of a normed

linear space is closed. Exercise A.63. (a) Fix 1 p . Denec00 = x = ( x1 , . . . , x N , 0, 0, . . . ) : N > 0, x1 , . . . , x N C .

Prove that c00 is a proper dense subspace of p for each 1 p < , andhence is not closed with respect to p. The vectors in c00 are sometimescalled nite sequences because they contain at most nitely many nonzerocomponents.

(b) Denec0 = x = ( xk )k=1 : limk xk = 0 .

Prove that c0 is a closed subspace of (N), and c0 is the closure of c00in -norm.

(c) Show that C c(R) is a proper dense subspace of C 0 (R), and C 0(R) is theclosure of C c(R) in the uniform norm. We now introduce a denition that in some sense distinguishes between

small and large innite-dimensional spaces.

Denition A.64. A topological space that contains a countable dense subsetis said to be separable . Exercise A.65. (a) Show that if I is a nite or countably innite index set,then p(I ) is separable for 0 < p < .

(b) Show that if I is innite, then (I ) is not separable.

(c) Show that if I is uncountable, then p

(I ) is not separable for any p. Exercise A.66. Show that c0 and C 0 (R) are separable.


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A.8 Compact Sets 25

A.8 Compact Sets

Now we briey review the denition and properties of compact sets.

Denition A.67 (Compact Set). A subset K of a topological space X iscompact if every cover of K by open sets has a nite subcover. That is, K iscompact if it is the case that whenever

K iI U i ,where {U i}iI is any collection of open subsets of X, there exist nitely manyi1 , . . . , i N I such that

K N

k=1U i k .

In a metric space, we can reformulate compactness in several equivalentways. We need the following terminology.

Denition A.68. Let E be a subset of a metric space X.(a) E is sequentially compact if every sequence {f n }nN of points of E containsa convergent subsequence {f n k }kN whose limit belongs to E .(b) E is complete if every Cauchy sequence in E converges to a point in E .(c) E is totally bounded if for every r > 0, we can cover E by nitely many

open balls of radius r, i.e., there exist nitely many x1 , . . . , x N X suchthatE

N

k=1B r (xk ).

Similarly to Exercise A.61, if X is complete then a subset E is completeif and only if it is closed.

Theorem A.69. If E is a subset of a metric space X, then the following statements are equivalent.(a) E is compact.(b) E is sequentially compact.(c) E is complete and totally bounded.

Proof. (a) (b). Suppose there exists a sequence {f n }nN in E that has nosubsequence that converges to an element of E. Choose any f E . If everyopen ball centered at f contained innitely many vectors f n , then there wouldbe a subsequence {f n k }kN that converged to f. Therefore, there must existsome open ball Bf centered at f that contains only nitely many f n . Butthen {B f }f E is an open cover of E that has no nite subcover. Hence E isnot compact.


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(b) (c). Suppose that E is sequentially compact. Exercise: Show that E is complete.Suppose that E was not totally bounded. Then there is an r > 0 such

that E cannot be covered by nitely many open balls of radius r centered at

points of X. Choose any f 1 E. Since E cannot be covered by a single r -ball,E cannot be a subset of B r (f 1). Hence there exists a point f 2 E \B r (f 1).In particular, we have d( f 2 , f 1) r. But E cannot be covered by two r -balls,so there must exist an f 3 E \ B r (f 1) B r (f 2 ) . In particular, we haveboth d( f 3 , f 1) r and d( f 3 , f 2) r. Continuing in this way we obtain asequence of points {f n }nN in E that has no convergent subsequence, whichis a contradiction.(c) (b). Assume that E is complete and totally bounded, and let

{f n }nN be any sequence of points in E . Since E is covered by nitely manyopen balls of radius 12 , one of those balls must contain innitely many f n , say{f

(1)n }nN . Then we have

m, n

N, d f (1)m , f

(1)n < 1.

Since E is covered by nitely many open balls of radius 14 , we can nd asubsequence {f

(2)n }nN of {f

(1)n }nN such that

m, n N, d f (2)m , f (2)n < 12

.

Continuing by induction, for each k > 1 we nd a subsequence {f (k )n }nN of

{f (k 1)n }nN such that d f

(k )m , f (k )n < 1k for all m, n N.

Now consider the diagonal subsequence {f (k )k }kN . Given > 0, let N be

large enough that 1N < . If j k > N , then f ( j )j is one element of the

sequence

{f (k )n

}nN , say f

( j )j = f

(k)n . Hence

d f ( j )j , f (k )k = d f

(k )n , f

(k )k 0 such that if B is any open ball of

radius that intersects E , then there is an J such that B U .To prove the claim, suppose that {U }J was an open cover of E such thatno with the required property existed. Then for each n N, we could nd aopen ball Bn with radius 1n that intersects E but is not contained in any U .


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A.8 Compact Sets 27

Choose any f n Bn E. Since E is sequentially compact, there must be asubsequence {f n k }kN that converges to an element of E , say f n k f E .Since {U }J is a cover of E, we must have f U for some J, andsince U is open there must exist some r > 0 such that Br (f ) U . Nowchoose k large enough that we have both

1nk

< r3

and d f, f n k < r3

.

Then it follows that Bn k B r (f ) U , which is a contradiction. Hence theclaim holds.To nish the proof, we use the fact that E is totally bounded, and therefore

can be covered by nitely many open balls of radius . By the claim, eachof these balls is contained in some U , so E is covered by nitely many of the U .

We will state only a few of the many special properties possessed by func-tions on compact sets.

Exercise A.70. Let X and Y be topological spaces. If f : X Y is contin-uous and K X is compact, then f (K ) is a compact subset of Y. Denition A.71. Let (X, dX ) and ( Y, dY ) be metric spaces. We say that afunction f : X Y is uniformly continuous if for every > 0 there exists a > 0 such that

x, y X, dX (x, y ) < = dY (f (x), f (y)) < . Exercise A.72. Let (X, dX ) and ( Y, dY ) be metric spaces, and let K X be compact. Then any continuous function f : K Y is uniformly continu-ous. Additional Problems

A.22. Let X be a topological space.(a) Prove that every closed subset of a compact subset of X is compact.(b) Show that if the topology on X is Hausdorff, then every compact subset

of X is closed.

A.23. Show that if K is a compact subset of a topological space X, then everyinnite subset of K has an accumulation point.

A.24. Show that if E is a totally bounded subset of a metric space X, then itsclosure E is compact. (A set with compact closure is said to be precompact .)

A.25. Show that a function f : Rd C is uniformly continuous if and only if lima 0 f T a = 0 , where T a f (x) = f (x a) is the translation operator.


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A.26. (a) Show that every compact subset of a normed linear space X is bothclosed and bounded.

(b) Show that every closed and bounded subset of R d or Cd is compact.(c) Use the fact that all norms on a nite-dimensional vector space are

equivalent to extend part (b) to arbitrary nite-dimensional vector spaces.(d) Let H be an innite-dimensional inner product space. Show that the

closed unit ball {xH : x 1} is closed and bounded but not compact.A.27. This problem will extend Problem A.26(d) to an arbitrary normedlinear space X.

(a) Prove F. Rieszs Lemma : If M is a proper, closed subspace of X and > 0, then there exists g X with g = 1 such that

dist( g, M ) = inf f M

g f > 1.(b) Prove that the closed unit ball {x X : x 1} in X is compact if and only if X is nite-dimensional.

A.9 Complete Sequences and a First Look at SchauderBases

In this section we dene complete sequences of vectors in normed spaces.In nite dimensions, these are simply spanning sets. However, in innite di-mensions there are subtle, but important, distinctions between spanning sets,complete sets, and bases. For more details on bases in Banach spaces, we re-fer to the texts by Singer [Sin70], Lindenstrauss and Tzafriri [LT77], or theintroductory text [Hei10].

A.9.1 Span and Closed Span

Denition A.73 (Span). Let A be a subset of a vector space X. The nite linear span of A, denoted span( A), is the set of all nite linear combinationsof elements of A:

span( A) = n

k=1

k f k : n > 0, f k A, k C .We also refer to the nite linear span of A as the nite span , the linear span ,or simply the span of A.

In particular, if A is a countable sequence, say A = {f k}kN , thenspan {f k}kN =

n

k=1

k f k : n > 0, k C .We will use the following vectors to illustrate many of the concepts in this

section.


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A.9 Complete Sequences and a First Look at Schauder Bases 29

Denition A.74 (Standard Basis Vectors). Let

en = ( mn )mN = (0 , . . . , 0, 1, 0, 0, . . . )

denote the sequence that has a 1 in the n th component and zeros elsewhere.We call en the nth standard basis vector , and refer to E = {en }nN as thestandard basis for p ( p nite) or c0 ( p = ).

Of course, this denition begs the question of in what sense the standardbasis vectors form a basis. This is answered in Exercise A.81, but for now letus consider the nite span of the standard basis.

Example A.75. The nite span of {en }nN is span {en }nN = c00 . In particular, the nite span of {en }nN is not p for any p and likewiseis not c0 . The standard basis vectors do not form a basis for p or c0 in the

vector space sense. Instead, in the strict vector space sense of spanning andbeing nitely independent, the standard basis vectors form a vector space

basis for c00 .In a generic vector space, we can form nite linear combinations of vectors,but unless we have a notion of convergence, we cannot take limits or forminnite sums. However, once we impose some extra structure, such as theexistence of a norm, these concepts make sense.

Denition A.76 (Closed Span and Complete Sets). Let A be a subsetof a normed linear space X.(a) The closed nite span of A, denoted span( A), is the closure of the set of

all nite linear combinations of elements of A:

span( A) = span( A) = {z X :yn span( A) such that yn z}.

We also refer to the closed nite span of A as the closed linear span orthe closed span of A.

(b) We say that A is complete in X if span(A) is dense in X, that is, if span( A) = X. There are many other terminologies in use for complete sets, for example,

they are also called total or fundamental .

Example A.77. Let E = {en }nN be the standard basis. Taking the closurewith respect to the norm p, we have span( E ) = p if 1 p < , andspan( E ) = c0 if p = . Beware: The denition of the closed span does NOT imply that

span( A) =

k=1

k f k : f k A, k C This need not hold!


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In particular it is NOT true that an arbitrary element of span( A) can bewritten f = k=1 k f k for some f k A, k C (see Exercise A.83). Instead,to illustrate the meaning of the closed span, consider the case of a countableset A = {f k}kN . Here we have

span {f k}kN = f X : k,n C such thatn

k=1

k,n f k f as n .

That is, an element f lies in the closed span of {f k}kN if there exist k,n Csuch thatn

k=1

k,n f k f as n .

In contrast, to say that f = k=1 k f k means thatnk=1

k f k f as n . (A.5)

In particular, in order for equation (A.5) to hold, the scalars k must beindependent of n.

A.9.2 Hamel Bases

We have already dened the nite span of a set, and now we recall the de-nition of nite linear independence.

Denition A.78 (Finite Independence). A subset A of a vector space X is nitely linearly independent if every nite subset of A is linearly independent.That is, if f 1 , . . . , f n are any choice of distinct vectors from A, then we musthave that

n

k=1

k f k = 0 = 1 = = n = 0 .We often simply say that A is linearly independent or just independent tomean that it is nitely linearly independent.

A Hamel basis is an ordinary vector space basis.

Denition A.79 (Hamel Basis). A subset A of a vector space X is a Hamel basis or a vector space basis for X if it spans and is nitely linearly indepen-dent. That is, A is a Hamel basis if span( A) = X and every nite subset of Ais linearly independent.

For example, the standard basis

E =

{e

n }nN is a Hamel basis for c

00,

and the set of monomials {xn : n 0} is a Hamel basis for the set P of allpolynomials.


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A.9 Complete Sequences and a First Look at Schauder Bases 31

It is shown in Theorem G.3 that the Axiom of Choice implies that everyvector space has a basis (in fact, this statement is equivalent to the Axiomof Choice). However, if X is an innite-dimensional Banach space, then anyHamel basis for X must be uncountable (Exercise C.99). Typically, we cannot

explicitly exhibit a Hamel basis for X, i.e., we usually only know that oneexists because of the Axiom of Choice. Consequently, Hamel bases are usuallynot very useful in the setting of normed spaces, and they make only rareappearances in this volume. One interesting consequence of the existence of Hamel bases is that there exist unbounded linear functionals on any innite-dimensional normed linear space (Problem C.8).

A.9.3 Introduction to Schauder Bases

In a normed space, Hamel bases are unnecessarily restrictive, because theyrequire the nite linear span to be the entire space. In contrast, Schauderbases allow innite linear combinations, and as such are much more usefulin normed spaces. We will be careful to avoid using the term basis without

qualication, because in the setting of an abstract vector space it usuallymeans a Hamel basis , while in the setting of a Banach space it usually meansa Schauder basis .

Denition A.80 (Schauder Basis). A sequence F = {f k}kN in a Banachspace X is a Schauder basis for X if we can write every f X asf =

k=1

k (f ) f k (A.6)

for a unique choice of scalars k (f ), where the series converges in the normof X.

Note that the uniqueness requirement implies that f k = 0 for every k.Exercise A.81. Show that the standard basis E = {en }nN is a Schauderbasis for p for each 1 p < , and E is a Schauder basis for c0 with respectto -norm.

Every Schauder basis is both complete and nitely linearly independent.However, the next exercise shows that the converse fails in general . For thisexample we will need the following useful theorem on approximation by poly-nomials. Given a < b, we set

C [a, b] = {f : [a, b] C : f is continuous }. (A.7)This is a Banach space under the uniform norm.

Theorem A.82 (Weierstrass Approximation Theorem). If f C [a, b]and > 0, then there exists a polynomial p such that f p = supx[a,b ] |f (x) p(x)| < .


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Exercise A.83. Use the Weierstrass Approximation Theorem to show thatthe set of monomials {xk}k=0 is complete and nitely linearly independentin C [a, b]. Prove that there exist functions f C [a, b] that cannot be writtenas f (x) =

k=0 k x

k with convergence of the series in the uniform norm.

Consequently {xk}k=0 is not a Schauder basis for C [a, b]. A complete discussion of Schauder bases requires the HahnBanach and

Uniform Boundedness Theorems, and therefore we will postpone additionaldiscussion of the relations and distinctions between bases and complete lin-early independent sets until Section C.15. One interesting fact that we willsee there is that the uniqueness requirement in the denition of a Schauderbasis implies that the functionals k appearing in equation (A.6) must becontinuous, and hence belong to the dual space of X.

We end this section with one important exercise.

Exercise A.84. Let X be a Banach space. Show that if there exists a count-able subset {f n }nN in X that is complete, then X is separable.

In particular, a Banach space that has a Schauder basis must be separable.The question of whether every separable Banach space has a basis was alongstanding open problem known as the Basis Problem . It was nally shownby Eno [Enf73] that there exist separable reexive Banach spaces that haveno Schauder bases!

A.10 Unconditional Convergence

Recall that a series n =1 f n in a normed space X converges and equals avector f if the sequence of partial sums N n =1 f n converges to f as N .However, the ordering of the f n can be important, for if we rearrange theseries, it may no longer converge. A series that converges regardless of orderingpossesses important stability properties that we will review in this section.

Denition A.85 (Unconditionally Convergent Series). Let X be anormed linear space and let {f n }nN be a sequence of elements of X. Thenthe series n =1 f n is said to converge unconditionally if every rearrangementof the series converges, i.e., if for each bijection : N N the series

n =1f (n )

converges in X. Thus, a series

n =1f

n converges unconditionally if and only if for each

bijection : N N, the partial sums N n =1 f (n ) converge to some vector as

N . Note that while we do not explicitly require in this denition that


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A.10 Unconditional Convergence 33

the partial sums converge to the same limit for each bijection , Exercise A.90shows that this must in fact happen.

The method of the next example can be used to show that if X is a nite-dimensional normed space, then absolute convergence is equivalent to

unconditional convergence. However, this equivalence fails in any innite di-mensional normed space. That is, if X is innite-dimensional, then there willexist series that converge unconditionally but not absolutely.

Example A.86. To illustrate the importance of unconditional convergence,consider the Banach space X = C. The harmonic series n =1 1n does notconverge, as its partial sums are unbounded. On the other hand, the alter-nating harmonic series n =1 (1)n +1 1n does converge (in fact, its partialsums converge to ln 2). While this latter series converges, it does not convergeabsolutely.

Now consider what happens if we change the order of summation. Forn N, let pn = 12n 1 and q n = 12n , i.e., the pn are the positive terms from thealternating series and the q n are the absolute values of the negative terms.Each series p

n and q n diverges to innity. Hence there must exist an

m1 > 0 such that p1 + + pm 1 > 1.

Also, there must exist an m2 > m 1 such that

p1 + + pm 1 q 1 + pm 1 +1 + + pm 2 > 2.Continuing in this way, we can create a rearrangement

p1 + + pm 1 q 1 + pm 1 +1 + + pm 2 q 2 + of

(1)n 1n that diverges to + . Likewise, we can construct a rearrange-ment that diverges to

, converges to any given real number r, or simply

oscillates without ever converging. A modication of the argument above gives us the following theorem for

series of scalars.

Theorem A.87. A series of scalars converges absolutely if and only if it con-verges unconditionally. That is, given cn C ,

n =1cn converges unconditionally

n =1|cn | < .

One direction of Theorem A.87 generalizes to all Banach spaces.

Exercise A.88. Let X be a Banach space. Show that if a series

n =1f n

converges absolutely in X then it converges unconditionally.


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The converse of Exercise A.88 fails in general. For example, if {en }nN isany orthonormal sequence in a Hilbert space H, then the series 1n en con-verges unconditionally but not absolutely. It can be shown that absolute andunconditional convergence are equivalent only for nite-dimensional spaces

(this is the DvoretzkyRogers Theorem , see [LT77] or [Hei10]).The following theorem, which we state without proof, gives several refor-mulations of unconditional convergence (see [Sin70] or [Hei10] for details). Forthis result, recall the denition of directed set given in Denition A.42, andnote that the set I consisting of all nite subsets of N forms a directed setwith respect to inclusion of sets.

Theorem A.89. Let {f n }nN be a sequence in a Banach space X. Then the following statements are equivalent.(a) n =1 f n converges unconditionally.(b) The net nF f n : F N, F nite converges in X. That is, there exists an f X such that for every > 0, there is a nite F 0 N such that

F 0 F N, F nite = f nF

f n < .

(c) If (cn )nN is any bounded sequence of scalars, then n =1 cn f n converges in X. Exercise A.90. Use Theorem A.89 to show that if a series n =1 f n convergesunconditionally in a Banach space X, then there exists a single f X suchthat for every bijection : N N we have

n =1 f (n ) = f.

A.11 Orthogonality

In this section we review some of the special properties of inner product spaces,especially in relation to orthogonality.

A.11.1 Orthogonality and the Pythagorean Theorem

Denition A.91. Let H be an inner product space, and let I be an arbitraryindex set.(a) Vectors f , g H are orthogonal , denoted f g, if f, g = 0 .(b) A collection of vectors {f i}iI is orthogonal if f i , f j = 0 whenever i = j.(c) A collection of vectors {f i}iI is orthonormal if it is orthogonal and eachvector is a unit vector, i.e., f i , f j = ij .

For example, the standard basis is an orthonormal sequence in 2 .


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A.11 Orthogonality 35

Exercise A.92 (Pythagorean Theorem). Show that if f 1 , . . . , f n H areorthogonal, thenn

k=1

f k2

=n

k=1

f k 2. The existence of a notion of orthogonality gives inner product spaces a

much simpler structure than general Banach spaces. We will derive some of the basic properties of inner product spaces below, most of which are straight-forward consequences of orthogonality. Some (but not all) of these results haveanalogues for general Banach spaces. However, even for the results that dohave analogues, the corresponding proofs in the non-Hilbert space setting areusually much more complicated or are nonconstructive (see the discussion of the HahnBanach Theorem in Section C.10).

A.11.2 Orthogonal Direct Sums

Denition A.93 (Orthogonal Direct Sum). Let M, N be closed sub-spaces of a Hilbert space H.(a) The direct sum of M and N is M + N = {f + g : f M, g N }.(b) We say that M and N are orthogonal subspaces , denoted M N, if f gfor every f M and g N.(c) If M, N are orthogonal subspaces in H, then we call their direct sum the

orthogonal direct sum of M and N, and denote it by M N. Exercise A.94. Show that if M, N are closed, orthogonal subspaces of H,then M N is a closed subspace of H. A.11.3 Orthogonal Projections and Orthogonal Complements

Denition A.95. Let X be a normed linear space and x A H. The dis-tance from a point f H to the set A isdist( f, A ) = inf f g : g A .

Given a closed convex subset K of a Hilbert space H and given any x H,there is a unique point in K that is closest to x.Exercise A.96 (Closest Point Property). If H is a Hilbert space and K is a nonempty closed, convex subset of H, then given any h H there existsa unique point k0 K that is closest to h. That is, there is a unique pointk0 K such that

h k0 = dist( h, K ) = inf h k : k K . Applying the closest point theorem to the particular case of closed sub-

spaces gives us the existence of orthogonal projections in a Hilbert space.


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Denition A.97 (Orthogonal Projection). Let M be a closed subspaceof a Hilbert space H. For each h H, the unique point p M closest to h iscalled the orthogonal projection of h onto M.

Orthogonal projections can be characterized as follows.

Exercise A.98. Let M be a closed subspace of a Hilbert space H. Givenh H, show that the following statements are equivalent.(a) h = p + e where p is the (unique) point in M closest to h.(b) h = p + e where p M and e M (i.e., e f for every f M ). Denition A.99 (Orthogonal Complement). Let A be a subset (not nec-essarily a subspac

an analysis companion - heil

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