Mathematical methods in quantum mechanics

If you feel comfortable with terms like Lebesgue L p spaces, Banach space, or bounded linear operator, you can skip this entire chapter.. In general, a space X together with a family of

American Mathematical Society

Providence, Rhode Island

Rafe Mazzeo (Chair) Gigliola Staffilani

2010 Mathematics subject classification 81-01, 81Qxx, 46-01, 34Bxx, 47B25

Abstract This book provides a self-contained introduction to mathematical methods in tum mechanics (spectral theory) with applications to Schr¨ odinger operators The first part cov- ers mathematical foundations of quantum mechanics from self-adjointness, the spectral theorem, quantum dynamics (including Stone’s and the RAGE theorem) to perturbation theory for self- adjoint operators.

quan-The second part starts with a detailed study of the free Schr¨ odinger operator respectively position, momentum and angular momentum operators Then we develop Weyl–Titchmarsh the- ory for Sturm–Liouville operators and apply it to spherically symmetric problems, in particular

to the hydrogen atom Next we investigate self-adjointness of atomic Schr¨ odinger operators and their essential spectrum, in particular the HVZ theorem Finally we have a look at scattering theory and prove asymptotic completeness in the short range case.

For additional information and updates on this book, visit:

http://www.ams.org/bookpages/gsm-157/

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Library of Congress Cataloging-in-Publication Data

Teschl, Gerald, 1970–

Mathematical methods in quantum mechanics : with applications to Schr¨ odinger operators / Gerald Teschl.– Second edition

p cm — (Graduate Studies in Mathematics ; volume 157)

Includes bibliographical references and index.

ISBN 978-1-4704-1704-8 (alk paper)

1 Schr¨ odinger operator 2 Quantum theory—Mathematics I Title.

pub-Send requests for translation rights and licensed reprints to reprint-permission@ams.org c

The American Mathematical Society retains all rights except those granted to the United States Government.

To Susanne, Simon, and Jakob

Part 0 Preliminaries

Chapter 0 A first look at Banach and Hilbert spaces 3

§0.1 Warm up: Metric and topological spaces 3

§0.2 The Banach space of continuous functions 14

§0.3 The geometry of Hilbert spaces 21

§0.5 Bounded operators 27

§0.6 Lebesgue Lp spaces 30

§0.7 Appendix: The uniform boundedness principle 38

Part 1 Mathematical Foundations of Quantum Mechanics

Chapter 1 Hilbert spaces 43

§1.2 Orthonormal bases 45

§1.3 The projection theorem and the Riesz lemma 49

§1.4 Orthogonal sums and tensor products 52

§1.5 The C∗ algebra of bounded linear operators 54

§1.6 Weak and strong convergence 55

§1.7 Appendix: The Stone–Weierstraß theorem 59

Chapter 2 Self-adjointness and spectrum 63

vii

§2.1 Some quantum mechanics 63

§2.2 Self-adjoint operators 66

§2.3 Quadratic forms and the Friedrichs extension 76

§2.4 Resolvents and spectra 83

§2.5 Orthogonal sums of operators 89

§2.6 Self-adjoint extensions 91

§2.7 Appendix: Absolutely continuous functions 95

Chapter 3 The spectral theorem 99

§3.1 The spectral theorem 99

§3.2 More on Borel measures 111

§3.4 Appendix: Herglotz–Nevanlinna functions 119

Chapter 4 Applications of the spectral theorem 131

§4.6 Tensor products of operators 143

Chapter 5 Quantum dynamics 145

§5.1 The time evolution and Stone’s theorem 145

§5.2 The RAGE theorem 150

§5.3 The Trotter product formula 155

Chapter 6 Perturbation theory for self-adjoint operators 157

§6.1 Relatively bounded operators and the Kato–Rellich theorem 157

§6.2 More on compact operators 160

§6.3 Hilbert–Schmidt and trace class operators 163

§6.4 Relatively compact operators and Weyl’s theorem 170

§6.5 Relatively form-bounded operators and the KLMN theorem 174

§6.6 Strong and norm resolvent convergence 179

Part 2 Schr¨odinger Operators

Chapter 7 The free Schr¨odinger operator 187

§7.1 The Fourier transform 187

Contents ix

§7.3 The free Schr¨odinger operator 197

§7.4 The time evolution in the free case 199

§7.5 The resolvent and Green’s function 201

Chapter 8 Algebraic methods 207

§8.1 Position and momentum 207

§9.4 Inverse spectral theory 238

§9.5 Absolutely continuous spectrum 241

§9.6 Spectral transformations II 244

§9.7 The spectra of one-dimensional Schr¨odinger operators 250

Chapter 10 One-particle Schr¨odinger operators 257

§10.1 Self-adjointness and spectrum 257

§10.2 The hydrogen atom 258

§10.3 Angular momentum 261

§10.4 The eigenvalues of the hydrogen atom 265

§10.5 Nondegeneracy of the ground state 272

Chapter 11 Atomic Schr¨odinger operators 275

§11.1 Self-adjointness 275

§11.2 The HVZ theorem 278

Chapter 12 Scattering theory 283

§12.1 Abstract theory 283

§12.2 Incoming and outgoing states 286

§12.3 Schr¨odinger operators with short range potentials 289

Part 3 Appendix

Appendix A Almost everything about Lebesgue integration 295

§A.1 Borel measures in a nutshell 295

§A.2 Extending a premeasure to a measure 303

§A.3 Measurable functions 307

§A.4 How wild are measurable objects? 309

§A.5 Integration — Sum me up, Henri 312

§A.6 Product measures 319

§A.7 Transformation of measures and integrals 322

§A.8 Vague convergence of measures 328

§A.9 Decomposition of measures 331

§A.10 Derivatives of measures 334

Overview

The present text was written for my course Schr¨odinger Operators held

at the University of Vienna in winter 1999, summer 2002, summer 2005,and winter 2007 It gives a brief but rather self-contained introduction

to the mathematical methods of quantum mechanics with a view towardsapplications to Schr¨odinger operators The applications presented are highlyselective; as a result, many important and interesting items are not touchedupon

Part 1 is a stripped-down introduction to spectral theory of unboundedoperators where I try to introduce only those topics which are needed forthe applications later on This has the advantage that you will (hopefully)not get drowned in results which are never used again before you get tothe applications In particular, I am not trying to present an encyclopedicreference Nevertheless I still feel that the first part should provide a solidbackground covering many important results which are usually taken forgranted in more advanced books and research papers

My approach is built around the spectral theorem as the central object.Hence I try to get to it as quickly as possible Moreover, I do not take thedetour over bounded operators but I go straight for the unbounded case Inaddition, existence of spectral measures is established via the Herglotz ratherthan the Riesz representation theorem since this approach paves the way for

an investigation of spectral types via boundary values of the resolvent as thespectral parameter approaches the real line

xi

Part 2 starts with the free Schr¨odinger equation and computes the freeresolvent and time evolution In addition, I discuss position, momentum,and angular momentum operators via algebraic methods This is usu-ally found in any physics textbook on quantum mechanics, with the onlydifference being that I include some technical details which are typicallynot found there Then there is an introduction to one-dimensional mod-els (Sturm–Liouville operators) including generalized eigenfunction expan-sions (Weyl–Titchmarsh theory) and subordinacy theory from Gilbert andPearson These results are applied to compute the spectrum of the hy-drogen atom, where again I try to provide some mathematical details notfound in physics textbooks Further topics are nondegeneracy of the groundstate, spectra of atoms (the HVZ theorem), and scattering theory (the Enßmethod).

Prerequisites

I assume some previous experience with Hilbert spaces and boundedlinear operators which should be covered in any basic course on functionalanalysis However, while this assumption is reasonable for mathematicsstudents, it might not always be for physics students For this reason there

is a preliminary chapter reviewing all necessary results (including proofs)

In addition, there is an appendix (again with proofs) providing all necessaryresults from measure theory

Literature

The present book is highly influenced by the four volumes of Reed andSimon [49]–[52] (see also [16]) and by the book by Weidmann [70] (an ex-tended version of which has recently appeared in two volumes [72], [73],however, only in German) Other books with a similar scope are, for ex-ample, [16], [17], [21], [26], [28], [30], [48], [57], [63], and [65] For thosewho want to know more about the physical aspects, I can recommend theclassical book by Thirring [68] and the visual guides by Thaller [66], [67].Further information can be found in the bibliographical notes at the end

Reader’s guide

There is some intentional overlap among Chapter 0, Chapter 1, andChapter 2 Hence, provided you have the necessary background, you canstart reading in Chapter 1 or even Chapter 2 Chapters 2 and 3 are key

Preface xiii

chapters, and you should study them in detail (except for Section2.6whichcan be skipped on first reading) Chapter4 should give you an idea of howthe spectral theorem is used You should have a look at (e.g.) the firstsection, and you can come back to the remaining ones as needed Chapter5

contains two key results from quantum dynamics: Stone’s theorem and theRAGE theorem In particular, the RAGE theorem shows the connectionsbetween long-time behavior and spectral types Finally, Chapter6 is again

of central importance and should be studied in detail

The chapters in the second part are mostly independent of each otherexcept for Chapter7, which is a prerequisite for all others except for Chap-ter 9

If you are interested in one-dimensional models (Sturm–Liouville tions), Chapter 9is all you need

equa-If you are interested in atoms, read Chapter7, Chapter 10, and ter11 In particular, you can skip the separation of variables (Sections 10.3

Chap-and10.4, which require Chapter9) method for computing the eigenvalues ofthe hydrogen atom, if you are happy with the fact that there are countablymany which accumulate at the bottom of the continuous spectrum

If you are interested in scattering theory, read Chapter 7, the first twosections of Chapter10, and Chapter 12 Chapter5is one of the key prereq-uisites in this case

2nd edition

Several people have sent me valuable feedback and pointed out misprintssince the appearance of the first edition All of these comments are of coursetaken into account Moreover, numerous small improvements were madethroughout Chapter 3 has been reworked, and I hope that it is now moreaccessible to beginners Also some proofs in Section9.4have been simplified(giving slightly better results at the same time) Finally, the appendix onmeasure theory has also grown a bit: I have added several examples andsome material around the change of variables formula and integration ofradial functions

Updates

The AMS is hosting a web page for this book at

http://www.ams.org/bookpages/gsm-157/

where updates, corrections, and other material may be found, including alink to material on my own web site:

http://www.mat.univie.ac.at/~gerald/ftp/book-schroe/

Acknowledgments

I would like to thank Volker Enß for making his lecture notes [20] able to me Many colleagues and students have made useful suggestions andpointed out mistakes in earlier drafts of this book, in particular: KerstinAmmann, J¨org Arnberger, Chris Davis, Fritz Gesztesy, Maria Hoffmann-Ostenhof, Zhenyou Huang, Helge Kr¨uger, Katrin Grunert, Wang Lanning,Daniel Lenz, Christine Pfeuffer, Roland M¨ows, Arnold L Neidhardt, SergeRichard, Harald Rindler, Alexander Sakhnovich, Robert Stadler, JohannesTemme, Karl Unterkofler, Joachim Weidmann, Rudi Weikard, and DavidWimmesberger

avail-My thanks for pointing out mistakes in the first edition go to: ErikMakino Bakken, Alexander Beigl, Stephan Bogend¨orfer, Søren Fournais,Semra Demirel-Frank, Katrin Grunert, Jason Jo, Helge Kr¨uger, Oliver Lein-gang, Serge Richard, Gerardo Gonz´alez Robert, Bob Sims, Oliver Skocek,Robert Stadler, Fernando Torres-Torija, Gerhard Tulzer, Hendrik Vogt, andDavid Wimmesberger

If you also find an error or if you have comments or suggestions(no matter how small), please let me know

I have been supported by the Austrian Science Fund (FWF) during much

of this writing, most recently under grant Y330

Part 0

Preliminaries

Chapter 0

A first look at Banach

and Hilbert spaces

I assume that the reader has some basic familiarity with measure theory and tional analysis For convenience, some facts needed from Banach and L p spaces are reviewed in this chapter A crash course in measure theory can be found in Appendix A If you feel comfortable with terms like Lebesgue L p spaces, Banach space, or bounded linear operator, you can skip this entire chapter However, you might want to at least browse through it to refresh your memory.

func-0.1 Warm up: Metric and topological spaces

Before we begin, I want to recall some basic facts from metric and topologicalspaces I presume that you are familiar with these topics from your calculuscourse As a general reference I can warmly recommend Kelly’s classicalbook [33]

A metric space is a space X together with a distance function d :

X × X → R such that

(i) d(x, y) ≥ 0,

(ii) d(x, y) = 0 if and only if x = y,

(iii) d(x, y) = d(y, x),

(iv) d(x, z) ≤ d(x, y) + d(y, z) (triangle inequality)

If (ii) does not hold, d is called a pseudometric Moreover, it is forward to see the inverse triangle inequality (Problem0.1)

straight-|d(x, y) − d(z, y)| ≤ d(x, z) (0.1)

3

Example Euclidean space Rntogether with d(x, y) = (Pn

k=1(xk− yk)2)1/2

is a metric space and so is Cntogether with d(x, y) = (Pn

k=1|xk−yk|2)1/2 The set

Br(x) = {y ∈ X|d(x, y) < r} (0.2)

is called an open ball around x with radius r > 0 A point x of some set

U is called an interior point of U if U contains some ball around x If x

is an interior point of U , then U is also called a neighborhood of x Apoint x is called a limit point of U (also accumulation or cluster point)

if (Br(x)\{x}) ∩ U 6= ∅ for every ball around x Note that a limit point

x need not lie in U , but U must contain points arbitrarily close to x Apoint x is called an isolated point of U if there exists a neighborhood of xnot containing any other points of U A set which consists only of isolatedpoints is called a discrete set If any neighborhood of x contains at leastone point in U and at least one point not in U , then x is called a boundarypoint of U The set of all boundary points of U is called the boundary of

U and denoted by ∂U

Example Consider R with the usual metric and let U = (−1, 1) Thenevery point x ∈ U is an interior point of U The points [−1, 1] are limitpoints of U , and the points {−1, +1} are boundary points of U 

A set consisting only of interior points is called open The family ofopen sets O satisfies the properties

(i) ∅, X ∈ O,

(ii) O1, O2∈ O implies O1∩ O2∈ O,

(iii) {Oα} ⊆ O impliesS

αOα ∈ O

That is, O is closed under finite intersections and arbitrary unions

In general, a space X together with a family of sets O, the open sets,satisfying (i)–(iii), is called a topological space The notions of interiorpoint, limit point, and neighborhood carry over to topological spaces if wereplace open ball by open set

There are usually different choices for the topology Two not too esting examples are the trivial topology O = {∅, X} and the discretetopology O = P(X) (the powerset of X) Given two topologies O1 and O2

inter-on X, O1 is called weaker (or coarser) than O2 if and only if O1 ⊆ O2.Example Note that different metrics can give rise to the same topology.For example, we can equip Rn (or Cn) with the Euclidean distance d(x, y)

as before or we could also use

˜d(x, y) =

n

X

k=1

|xk− yk| (0.3)

0.1 Warm up: Metric and topological spaces 5

Then

1

√n

n(x) ⊆ ˜Br(x) ⊆ Br(x), where B, ˜B are balls computed using d,

˜

Example We can always replace a metric d by the bounded metric

˜d(x, y) = d(x, y)

without changing the topology (since the family of open balls does not

Every subspace Y of a topological space X becomes a topological space

of its own if we call O ⊆ Y open if there is some open set ˜O ⊆ X suchthat O = ˜O ∩ Y This natural topology O ∩ Y is known as the relativetopology (also subspace, trace or induced topology)

Example The set (0, 1] ⊆ R is not open in the topology of X = R, but it isopen in the relative topology when considered as a subset of Y = [−1, 1] 

A family of open sets B ⊆ O is called a base for the topology if for each

x and each neighborhood U (x), there is some set O ∈ B with x ∈ O ⊆ U (x).Since an open set O is a neighborhood of every one of its points, it can bewritten as O =S

O⊇ ˜ O∈BO and we have˜Lemma 0.1 If B ⊆ O is a base for the topology, then every open set can

be written as a union of elements from B

If there exists a countable base, then X is called second countable.Example By construction, the open balls B1/n(x) are a base for the topol-ogy in a metric space In the case of Rn(or Cn) it even suffices to take ballswith rational center, and hence Rn(as well as Cn) is second countable 

A topological space is called a Hausdorff space if for two differentpoints there are always two disjoint neighborhoods

Example Any metric space is a Hausdorff space: Given two differentpoints x and y, the balls Bd/2(x) and Bd/2(y), where d = d(x, y) > 0, aredisjoint neighborhoods (a pseudometric space will not be Hausdorff) The complement of an open set is called a closed set It follows from

de Morgan’s rules that the family of closed sets C satisfies

(i) ∅, X ∈ C,

(ii) C1, C2 ∈ C implies C1∪ C2 ∈ C,

Lemma 0.2 Let X be a topological space Then the interior of U is the set

of all interior points of U , and the closure of U is the union of U with alllimit points of U

Example The closed ball

A sequence (xn)∞n=1 ⊆ X is said to converge to some point x ∈ X ifd(x, xn) → 0 We write limn→∞xn = x as usual in this case Clearly thelimit is unique if it exists (this is not true for a pseudometric)

Every convergent sequence is a Cauchy sequence; that is, for every

ε > 0 there is some N ∈ N such that

d(xn, xm) ≤ ε, n, m ≥ N (0.10)

0.1 Warm up: Metric and topological spaces 7

If the converse is also true, that is, if every Cauchy sequence has a limit,then X is called complete

Example Both Rn and Cn are complete metric spaces Note that in a metric space x is a limit point of U if and only if thereexists a sequence (xn)∞n=1⊆ U \{x} with limn→∞xn= x Hence U is closed

if and only if for every convergent sequence the limit is in U In particular,Lemma 0.3 A closed subset of a complete metric space is again a completemetric space

Note that convergence can also be equivalently formulated in topologicalterms: A sequence xn converges to x if and only if for every neighborhood

U of x there is some N ∈ N such that xn ∈ U for n ≥ N In a Hausdorffspace the limit is unique

A set U is called dense if its closure is all of X, that is, if U = X Ametric space is called separable if it contains a countable dense set.Lemma 0.4 A metric space is separable if and only if it is second countable

as a topological space

Proof From every dense set we get a countable base by considering openballs with rational radii and centers in the dense set Conversely, from everycountable base we obtain a dense set by choosing an element from each

Lemma 0.5 Let X be a separable metric space Every subset Y of X isagain separable

Proof Let A = {xn}n∈N be a dense set in X The only problem is that

A ∩ Y might contain no elements at all However, some elements of A must

be at least arbitrarily close: Let J ⊆ N2 be the set of all pairs (n, m) forwhich B1/m(xn) ∩ Y 6= ∅ and choose some yn,m ∈ B1/m(xn) ∩ Y for all(n, m) ∈ J Then B = {yn,m}(n,m)∈J ⊆ Y is countable To see that B isdense, choose y ∈ Y Then there is some sequence xnk with d(xnk, y) < 1/k.Hence (nk, k) ∈ J and d(ynk,k, y) ≤ d(ynk,k, xnk) + d(xnk, y) ≤ 2/k → 0 Next, we come to functions f : X → Y , x 7→ f (x) We use the usualconventions f (U ) = {f (x)|x ∈ U } for U ⊆ X and f−1(V ) = {x|f (x) ∈ V }for V ⊆ Y The set Ran(f ) = f (X) is called the range of f , and X is calledthe domain of f A function f is called injective if for each y ∈ Y there

is at most one x ∈ X with f (x) = y (i.e., f−1({y}) contains at most onepoint) and surjective or onto if Ran(f ) = Y A function f which is bothinjective and surjective is called bijective

A function f between metric spaces X and Y is called continuous at apoint x ∈ X if for every ε > 0 we can find a δ > 0 such that

dY(f (x), f (y)) ≤ ε if dX(x, y) < δ (0.11)

If f is continuous at every point, it is called continuous

Lemma 0.6 Let X be a metric space The following are equivalent:(i) f is continuous at x (i.e., (0.11) holds)

(ii) f (xn) → f (x) whenever xn→ x

(iii) For every neighborhood V of f (x), f−1(V ) is a neighborhood of x.Proof (i) ⇒ (ii) is obvious (ii) ⇒ (iii): If (iii) does not hold, there is

a neighborhood V of f (x) such that Bδ(x) 6⊆ f−1(V ) for every δ Hence

we can choose a sequence xn ∈ B1/n(x) such that f (xn) 6∈ f−1(V ) Thus

xn → x but f (xn) 6→ f (x) (iii) ⇒ (i): Choose V = Bε(f (x)) and observethat by (iii), Bδ(x) ⊆ f−1(V ) for some δ The last item implies that f is continuous if and only if the inverseimage of every open set is again open (equivalently, the inverse image ofevery closed set is closed) If the image of every open set is open, then f

is called open A bijection f is called a homeomorphism if both f andits inverse f−1 are continuous Note that if f is a bijection, then f−1 iscontinuous if and only if f is open

In a topological space, (iii) is used as the definition for continuity ever, in general (ii) and (iii) will no longer be equivalent unless one usesgeneralized sequences, so-called nets, where the index set N is replaced byarbitrary directed sets

How-The support of a function f : X → Cn is the closure of all points x forwhich f (x) does not vanish; that is,

supp(f ) = {x ∈ X|f (x) 6= 0} (0.12)

If X and Y are metric spaces, then X × Y together with

d((x1, y1), (x2, y2)) = dX(x1, x2) + dY(y1, y2) (0.13)

is a metric space A sequence (xn, yn) converges to (x, y) if and only if

xn→ x and yn→ y In particular, the projections onto the first (x, y) 7→ x,respectively, onto the second (x, y) 7→ y, coordinate are continuous More-over, if X and Y are complete, so is X × Y

In particular, by the inverse triangle inequality (0.1),

|d(xn, yn) − d(x, y)| ≤ d(xn, x) + d(yn, y), (0.14)

we see that d : X × X → R is continuous

0.1 Warm up: Metric and topological spaces 9

Example If we consider R × R, we do not get the Euclidean distance of

R2 unless we modify (0.13) as follows:

˜

d((x1, y1), (x2, y2)) =pdX(x1, x2)2+ dY(y1, y2)2 (0.15)

As noted in our previous example, the topology (and thus also gence/continuity) is independent of this choice 

conver-If X and Y are just topological spaces, the product topology is defined

by calling O ⊆ X × Y open if for every point (x, y) ∈ O there are openneighborhoods U of x and V of y such that U × V ⊆ O In other words, theproducts of open sets form a basis of the product topology In the case ofmetric spaces this clearly agrees with the topology defined via the productmetric (0.13)

A cover of a set Y ⊆ X is a family of sets {Uα} such that Y ⊆S

By construction, {Uαn} is a countable subcover 

A subset K ⊂ X is called compact if every open cover has a finitesubcover A set is called relatively compact if its closure is compact.Lemma 0.8 A topological space is compact if and only if it has the finiteintersection property: The intersection of a family of closed sets is empty

if and only if the intersection of some finite subfamily is empty

Proof By taking complements, to every family of open sets there is a responding family of closed sets and vice versa Moreover, the open setsare a cover if and only if the corresponding closed sets have empty intersec-

Lemma 0.9 Let X be a topological space

(i) The continuous image of a compact set is compact

(ii) Every closed subset of a compact set is compact

(iii) If X is Hausdorff, every compact set is closed

(iv) The product of finitely many compact sets is compact

(v) The finite union of compact sets is again compact

(vi) If X is Hausdorff, any intersection of compact sets is again pact.

com-Proof (i) Observe that if {Oα} is an open cover for f (Y ), then {f−1(Oα)}

j=1Uyj(x) is a neighborhood of x which does not intersect

Y

(iv) Let {Oα} be an open cover for X × Y For every (x, y) ∈ X × Ythere is some α(x, y) such that (x, y) ∈ Oα(x,y) By definition of the producttopology there is some open rectangle U (x, y) × V (x, y) ⊆ Oα(x,y) Hence forfixed x, {V (x, y)}y∈Y is an open cover of Y Hence there are finitely manypoints yk(x) such that the V (x, yk(x)) cover Y Set U (x) =T

kU (x, yk(x)).Since finite intersections of open sets are open, {U (x)}x∈X is an open coverand there are finitely many points xj such that the U (xj) cover X Byconstruction, the U (xj) × V (xj, yk(xj)) ⊆ Oα(xj,yk(xj)) cover X × Y (v) Note that a cover of the union is a cover for each individual set andthe union of the individual subcovers is the subcover we are looking for.(vi) Follows from (ii) and (iii) since an intersection of closed sets is

As a consequence we obtain a simple criterion when a continuous tion is a homeomorphism

func-Corollary 0.10 Let X and Y be topological spaces with X compact and

Y Hausdorff Then every continuous bijection f : X → Y is a phism

homeomor-Proof It suffices to show that f maps closed sets to closed sets By (ii)every closed set is compact, by (i) its image is also compact, and by (iii) it

A subset K ⊂ X is called sequentially compact if every sequencefrom K has a convergent subsequence In a metric space, compact andsequentially compact are equivalent

0.1 Warm up: Metric and topological spaces 11

Lemma 0.11 Let X be a metric space Then a subset is compact if andonly if it is sequentially compact

Proof Suppose X is compact and let xnbe a sequence which has no gent subsequence Then K = {xn} has no limit points and is hence compact

conver-by Lemma0.9 (ii) For every n there is a ball Bεn(xn) which contains onlyfinitely many elements of K However, finitely many suffice to cover K, acontradiction

Conversely, suppose X is sequentially compact and let {Oα} be someopen cover which has no finite subcover For every x ∈ X we can choosesome α(x) such that if Br(x) is the largest ball contained in Oα(x), theneither r ≥ 1 or there is no β with B2r(x) ⊂ Oβ (show that this is possible).Now choose a sequence xn such that xn 6∈ S

m<nOα(xm) Note that byconstruction the distance d = d(xm, xn) to every successor of xm is eitherlarger than 1 or the ball B2d(xm) will not fit into any of the Oα

Now let y be the limit of some convergent subsequence and fix some r ∈(0, 1) such that Br(y) ⊆ Oα(y) Then this subsequence must eventually be in

Br/5(y), but this is impossible since if d = d(xn1, xn2) is the distance betweentwo consecutive elements of this subsequence, then B2d(xn 1) cannot fit into

Oα(y) by construction whereas on the other hand B2d(xn 1) ⊆ B4r/5(a) ⊆

In a metric space, a set is called bounded if it is contained inside someball Note that compact sets are always bounded since Cauchy sequencesare bounded (show this!) In Rn(or Cn) the converse also holds

Theorem 0.12 (Heine–Borel) In Rn (or Cn) a set is compact if and only

if it is bounded and closed

Proof By Lemma0.9(ii) and (iii) it suffices to show that a closed interval

in I ⊆ R is compact Moreover, by Lemma 0.11, it suffices to show thatevery sequence in I = [a, b] has a convergent subsequence Let xn be oursequence and divide I = [a,a+b2 ] ∪ [a+b2 , b] Then at least one of these twointervals, call it I1, contains infinitely many elements of our sequence Let

y1 = xn 1 be the first one Subdivide I1 and pick y2 = xn 2, with n2 > n1 asbefore Proceeding like this, we obtain a Cauchy sequence yn (note that byconstruction In+1⊆ In and hence |yn− ym| ≤ b−a

n for m ≥ n) 

By Lemma0.11 this is equivalent to

Theorem 0.13 (Bolzano–Weierstraß) Every bounded infinite subset of Rn

(or Cn) has at least one limit point

Combining Theorem 0.12 with Lemma 0.9 (i) we also obtain the treme value theorem

ex-Theorem 0.14 (Weierstraß) Let X be compact Every continuous function

f : X → R attains its maximum and minimum

A metric space for which the Heine–Borel theorem holds is called proper.Lemma0.9(ii) shows that X is proper if and only if every closed ball is com-pact Note that a proper metric space must be complete (since every Cauchysequence is bounded) A topological space is called locally compact if ev-ery point has a compact neighborhood Clearly a proper metric space islocally compact

The distance between a point x ∈ X and a subset Y ⊆ X is

dist(x, Y ) = inf

y∈Yd(x, y) (0.16)Note that x is a limit point of Y if and only if dist(x, Y ) = 0

Lemma 0.15 Let X be a metric space Then

| dist(x, Y ) − dist(z, Y )| ≤ d(x, z) (0.17)

In particular, x 7→ dist(x, Y ) is continuous

Proof Taking the infimum in the triangle inequality d(x, y) ≤ d(x, z) +d(z, y) shows dist(x, Y ) ≤ d(x, z)+dist(z, Y ) Hence dist(x, Y )−dist(z, Y ) ≤d(x, z) Interchanging x and z shows dist(z, Y ) − dist(x, Y ) ≤ d(x, z) Lemma 0.16 (Urysohn) Suppose C1 and C2 are disjoint closed subsets of

a metric space X Then there is a continuous function f : X → [0, 1] suchthat f is zero on C2 and one on C1

If X is locally compact and C1is compact, one can choose f with compactsupport

Proof To prove the first claim, set f (x) = dist(x,C2 )

dist(x,C 1 )+dist(x,C 2 ) For thesecond claim, observe that there is an open set O such that O is compactand C1 ⊂ O ⊂ O ⊂ X\C2 In fact, for every x ∈ C1, there is a ball Bε(x)such that Bε(x) is compact and Bε(x) ⊂ X\C2 Since C1 is compact, finitelymany of them cover C1 and we can choose the union of those balls to be O

Note that Urysohn’s lemma implies that a metric space is normal; that

is, for any two disjoint closed sets C1 and C2, there are disjoint open sets

O1 and O2 such that Cj ⊆ Oj, j = 1, 2 In fact, choose f as in Urysohn’slemma and set O1= f−1([0, 1/2)), respectively, O2= f−1((1/2, 1])

Lemma 0.17 Let X be a locally compact metric space Suppose K is acompact set and {Oj}n

j=1 is an open cover Then there is a partition of

0.1 Warm up: Metric and topological spaces 13

unity for K subordinate to this cover; that is, there are continuous functions

hj : X → [0, 1] such that hj has compact support contained in Oj and

n

X

j=1

with equality for x ∈ K

Proof For every x ∈ K there is some ε and some j such that Bε(x) ⊆ Oj

By compactness of K, finitely many of these balls cover K Let Kj be theunion of those balls which lie inside Oj By Urysohn’s lemma there arecontinuous functions gj : X → [0, 1] such that gj = 1 on Kj and gj = 0 onX\Oj Now set

shows that the sum is one for x ∈ K, since x ∈ Kj for some j implies

gj(x) = 1 and causes the product to vanish Problem 0.1 Show that |d(x, y) − d(z, y)| ≤ d(x, z)

Problem 0.2 Show the quadrangle inequality |d(x, y) − d(x0, y0)| ≤d(x, x0) + d(y, y0)

Problem 0.3 Show that the closure satisfies the Kuratowski closure axioms.Problem 0.4 Show that the closure and interior operators are dual in thesense that

X\A = (X\A)◦ and X\A◦ = (X\A)

(Hint: De Morgan’s laws.)

Problem 0.5 Let U ⊆ V be subsets of a metric space X Show that if U

is dense in V and V is dense in X, then U is dense in X

Problem 0.6 Show that every open set O ⊆ R can be written as a countableunion of disjoint intervals (Hint: Let {Iα} be the set of all maximal opensubintervals of O; that is, Iα ⊆ O and there is no other subinterval of Owhich contains Iα Then this is a cover of disjoint open intervals which has

a countable subcover.)

Problem 0.7 Show that the boundary of A is given by ∂A = A\A◦

0.2 The Banach space of continuous functions

Now let us have a first look at Banach spaces by investigating the set ofcontinuous functions C(I) on a compact interval I = [a, b] ⊂ R Since wewant to handle complex models, we will always consider complex-valuedfunctions!

One way of declaring a distance, well-known from calculus, is the imum norm:

• kf k > 0 for f 6= 0 (positive definiteness),

• kα f k = |α| kf k for all α ∈ C, f ∈ X (positive homogeneity),and

• kf + gk ≤ kf k + kgk for all f, g ∈ X (triangle inequality)

If positive definiteness is dropped from the requirements, one calls k.k aseminorm

From the triangle inequality we also get the inverse triangle ity (Problem 0.8)

inequal-|kf k − kgk| ≤ kf − gk (0.20)Once we have a norm, we have a distance d(f, g) = kf − gk and hence

we know when a sequence of vectors fn converges to a vector f Wewill write fn → f or limn→∞fn = f , as usual, in this case Moreover, amapping F : X → Y between two normed spaces is called continuous if

fn → f implies F (fn) → F (f ) In fact, the norm, vector addition, andmultiplication by scalars are continuous (Problem0.9)

In addition to the concept of convergence we have also the concept of

a Cauchy sequence and hence the concept of completeness: A normedspace is called complete if every Cauchy sequence has a limit A completenormed space is called a Banach space

Example The space `1(N) of all complex-valued sequences a = (aj)∞j=1forwhich the norm

0.2 The Banach space of continuous functions 15

To show this, we need to verify three things: (i) `1(N) is a vector spacethat is closed under addition and scalar multiplication, (ii) k.k1 satisfies thethree requirements for a norm, and (iii) `1(N) is complete

First of all, observe

be a Cauchy sequence; that is, for given ε > 0 we can find an Nε such that

kam− ank1 ≤ ε for m, n ≥ Nε This implies in particular |amj − an

j| ≤ ε forevery fixed j Thus anj is a Cauchy sequence for fixed j and, by completeness

of C, it has a limit: limn→∞anj = aj Now consider

`1(N) and since an ∈ `1(N), we finally conclude a = an+ (a − an) ∈ `1(N)

By our estimate ka − ank1≤ ε, our candidate a is indeed the limit of an Example The previous example can be generalized by considering thespace `p(N) of all complex-valued sequences a = (aj)∞j=1 for which the norm

is finite By |aj+ bj|p≤ 2pmax(|aj|, |bj|)p = 2pmax(|aj|p, |bj|p) ≤ 2p(|aj|p+

|bj|p) it is a vector space, but the triangle inequality is only easy to see inthe case p = 1 (It is also not hard to see that it fails for p < 1, whichexplains our requirement p ≥ 1 See also Problem0.17.)

To prove it we need the elementary inequality (Problem 0.12)

kabk1 ≤ kakpkbkq (0.27)

for x ∈ `p(N), y ∈ `q(N) In fact, by homogeneity of the norm it suffices toprove the case kakp= kbkq= 1 But this case follows by choosing α = |aj|p

and β = |bj|q in (0.26) and summing over all j

Now using |aj+ bj|p ≤ |aj| |aj+ bj|p−1+ |bj| |aj+ bj|p−1, we obtain fromH¨older’s inequality (note (p − 1)q = p)

ka + bkpp ≤ kakpk(a + b)p−1kq+ kbkpk(a + b)p−1kq

= (kakp+ kbkp)k(a + b)kp−1p Hence `p is a normed space That it is complete can be shown as in the case

a closed subspace In fact, if a ∈ `∞(N) \ c0(N), then lim infj→∞|aj| ≥ ε > 0and thus ka − bk∞≥ ε for every b ∈ c0(N) Now what about convergence in the space C(I)? A sequence of functions

fn(x) converges to f if and only if

is clearly a Cauchy sequence of real numbers for every fixed x ∈ I Inparticular, by completeness of C, there is a limit f (x) for each x Thus weget a limiting function f (x) Moreover, letting m → ∞ in

|fm(x) − fn(x)| ≤ ε ∀m, n > Nε, x ∈ I, (0.30)

we see

|f (x) − fn(x)| ≤ ε ∀n > Nε, x ∈ I; (0.31)that is, fn(x) converges uniformly to f (x) However, up to this point we donot know whether it is in our vector space C(I), that is, whether it is con-tinuous Fortunately, there is a well-known result from real analysis whichtells us that the uniform limit of continuous functions is again continuous:Fix x ∈ I and ε > 0 To show that f is continuous we need to find a δ such

0.2 The Banach space of continuous functions 17

that |x − y| < δ implies |f (x) − f (y)| < ε Pick n so that kfn− f k∞ < ε/3and δ so that |x − y| < δ implies |fn(x) − fn(y)| < ε/3 Then |x − y| < δimplies

|f (x)−f (y)| ≤ |f (x)−fn(x)|+|fn(x)−fn(y)|+|fn(y)−f (y)| < ε

Theorem 0.18 C(I) with the maximum norm is a Banach space

Next we want to look at countable bases To this end we introduce a fewdefinitions first

The set of all finite linear combinations of a set of vectors {un} ⊂ X iscalled the span of {un} and denoted by span{un} A set of vectors {un} ⊂ X

is called linearly independent if every finite subset is If {un}N

n=1 ⊂ X,

N ∈ N ∪ {∞}, is countable, we can throw away all elements which can beexpressed as linear combinations of the previous ones to obtain a subset oflinearly independent vectors which have the same span

We will call a countable set of vectors {un}N

n=1 ⊂ X a Schauder sis if every element f ∈ X can be uniquely written as a countable linearcombination of the basis elements:

Example The set of vectors δn, with δn

n = 1 and δn

m = 0, n 6= m, is aSchauder basis for the Banach space `1(N)

Let a = (aj)∞j=1∈ `1(N) be given and set an=Pn

A set whose span is dense is called total, and if we have a countable totalset, we also have a countable dense set (consider only linear combinationswith rational coefficients — show this) A normed vector space containing

a countable dense set is called separable

Example Every Schauder basis is total and thus every Banach space with

a Schauder basis is separable (the converse is not true) In particular, the

While we will not give a Schauder basis for C(I), we will at least showthat it is separable In order to prove this, we need a lemma first

Lemma 0.19 (Smoothing) Let un(x) be a sequence of nonnegative uous functions on [−1, 1] such that

M = maxx∈[−1/2,1/2]{1, |f (x)|} and note

0.2 The Banach space of continuous functions 19

Using this, we have

Note that fn will be as smooth as un, hence the title smoothing lemma.Moreover, fnwill be a polynomial if unis The same idea is used to approx-imate noncontinuous functions by smooth ones (of course the convergencewill no longer be uniform in this case)

Now we are ready to show:

Theorem 0.20 (Weierstraß) Let I be a compact interval Then the set ofpolynomials is dense in C(I)

Proof Let f (x) ∈ C(I) be given By considering f (x)−f (a)−f (b)−f (a)b−a (x−a) it is no loss to assume that f vanishes at the boundary points Moreover,without restriction, we only consider I = [−12,12] (why?)

Now the claim follows from Lemma 0.19using

un(x) = 1

In(1 − x

2)n,where

n!

1

2(12 + 1) · · · (12 + n).Indeed, the first part of (0.33) holds by construction, and the second partfollows from the elementary estimate

22n + 1 ≤ In< 2.

Corollary 0.21 C(I) is separable

However, `∞(N) is not separable (Problem0.15)!

Problem 0.8 Show that |kf k − kgk| ≤ kf − gk

Problem 0.9 Let X be a Banach space Show that the norm, vector dition, and multiplication by scalars are continuous That is, if fn → f ,

ad-gn→ g, and αn→ α, then kfnk → kf k, fn+ gn→ f + g, and αngn→ αg.Problem 0.10 Let X be a Banach space Show that P∞

j=1kfjk < ∞implies that

Problem 0.11 While `1(N) is separable, it still has room for an able set of linearly independent vectors Show this by considering vectors ofthe form

uncount-aα= (1, α, α2, ), α ∈ (0, 1)

(Hint: Take n such vectors and cut them off after n + 1 terms If the off vectors are linearly independent, so are the original ones Recall theVandermonde determinant.)

cut-Problem 0.12 Prove (0.26) (Hint: Take logarithms on both sides.)Problem 0.13 Show that `p(N) is a separable Banach space

Problem 0.14 Show that `∞(N) is a Banach space

Problem 0.15 Show that `∞(N) is not separable (Hint: Consider quences which take only the value one and zero How many are there? What

se-is the dse-istance between two such sequences?)

Problem 0.16 Show that if a ∈ `p 0(N) for some p0∈ [1, ∞), then a ∈ `p(N)for p ≥ p0 and

lim

p→∞kakp= kak∞.Problem 0.17 Formally extend the definition of `p(N) to p ∈ (0, 1) Showthat k.kp does not satisfy the triangle inequality However, show that it is

a quasinormed space; that is, it satisfies all requirements for a normedspace except for the triangle inequality which is replaced by

ka + bk ≤ K(kak + kbk)with some constant K ≥ 1 Show, in fact,

ka + bkp≤ 21/p−1(kakp+ kbkp), p ∈ (0, 1)

Moreover, show that k.kpp satisfies the triangle inequality in this case, but

of course it is no longer homogeneous (but at least you can get an honestmetric d(a, b) = ka−bkpp which gives rise to the same topology) (Hint: Show

α + β ≤ (αp+ βp)1/p ≤ 21/p−1(α + β) for 0 < p < 1 and α, β ≥ 0.)

0.3 The geometry of Hilbert spaces 21

0.3 The geometry of Hilbert spaces

So it looks like C(I) has all the properties we want However, there isstill one thing missing: How should we define orthogonality in C(I)? InEuclidean space, two vectors are called orthogonal if their scalar productvanishes, so we would need a scalar product:

Suppose H is a vector space A map h., i : H × H → C is called asesquilinear form if it is conjugate linear in the first argument and linear

in the second; that is,

hα1f1+ α2f2, gi = α∗1hf1, gi + α∗2hf2, gi,

hf, α1g1+ α2g2i = α1hf, g1i + α2hf, g2i, α1, α2 ∈ C, (0.36)where ‘∗’ denotes complex conjugation A sesquilinear form satisfying therequirements

(i) hf, f i > 0 for f 6= 0 (positive definite),

The pair (H, h., i) is called an inner product space If H is complete(with respect to the norm (0.37)), it is called a Hilbert space

Example Clearly Cn with the usual scalar product

n(aj)∞j=1

∞

X

j=1

|aj|2 < ∞o (0.39)with scalar product

A vector f ∈ H is called normalized or a unit vector if kf k = 1.Two vectors f, g ∈ H are called orthogonal or perpendicular (f ⊥ g) if

hf, gi = 0 and parallel if one is a multiple of the other

If f and g are orthogonal, we have the Pythagorean theorem:

kf + gk2 = kf k2+ kgk2, f ⊥ g, (0.41)which is one line of computation (do it!)

Suppose u is a unit vector Then the projection of f in the direction of



Taking any other vector parallel to u, we obtain from (0.41)

kf − αuk2= kf⊥+ (fk− αu)k2 = kf⊥k2+ |hu, f i − α|2 (0.44)and hence fk = hu, f iu is the unique vector parallel to u which is closest to

Proof It suffices to prove the case kgk = 1 But then the claim followsfrom kf k2= |hg, f i|2+ kf⊥k2 Note that the Cauchy–Schwarz inequality implies that the scalar product

is continuous in both variables; that is, if fn → f and gn → g, we have

hfn, gni → hf, gi

0.3 The geometry of Hilbert spaces 23

As another consequence we infer that the map k.k is indeed a norm Infact,

kf + gk2 = kf k2+ hf, gi + hg, f i + kgk2 ≤ (kf k + kgk)2 (0.46)But let us return to C(I) Can we find a scalar product which has themaximum norm as associated norm? Unfortunately the answer is no! Thereason is that the maximum norm does not satisfy the parallelogram law(Problem0.20)

Theorem 0.23 (Jordan–von Neumann) A norm is associated with a scalarproduct if and only if the parallelogram law

kf + gk2+ kf − gk2 = 2kf k2+ 2kgk2 (0.47)holds

In this case the scalar product can be recovered from its norm by virtue

of the polarization identity

hf, gi = 1

4 kf + gk

2− kf − gk2+ ikf − igk2− ikf + igk2
(0.48)

Proof If an inner product space is given, verification of the parallelogramlaw and the polarization identity is straightforward (Problem0.22)

To show the converse, we define

s(f, g) = 1

4 kf + gk

2− kf − gk2+ ikf − igk2− ikf + igk2
Then s(f, f ) = kf k2 and s(f, g) = s(g, f )∗ are straightforward to check.Moreover, another straightforward computation using the parallelogram lawshows

2 ns(f, g) = s(f,2mng); that is, α s(f, g) = s(f, αg) for every positive rational

α By continuity (which follows from the triangle inequality for k.k) thisholds for all α > 0 and s(f, −g) = −s(f, g), respectively, s(f, ig) = i s(f, g),

Note that the parallelogram law and the polarization identity even holdfor sesquilinear forms (Problem0.22)

But how do we define a scalar product on C(I)? One possibility is

hf, gi =

Z b

a

f∗(x)g(x)dx (0.49)

The corresponding inner product space is denoted by L2cont(I) Note that

we have

kf k ≤p|b − a|kfk∞ (0.50)and hence the maximum norm is stronger than the L2

cont norm

Suppose we have two norms k.k1 and k.k2 on a vector space X Thenk.k2 is said to be stronger than k.k1 if there is a constant m > 0 such that

kf k1≤ mkf k2 (0.51)

It is straightforward to check the following

Lemma 0.24 If k.k2is stronger than k.k1, then every k.k2Cauchy sequence

is also a k.k1 Cauchy sequence

Hence if a function F : X → Y is continuous in (X, k.k1), it is alsocontinuous in (X, k.k2), and if a set is dense in (X, k.k2), it is also dense in(X, k.k1)

In particular, L2cont is separable But is it also complete? Unfortunatelythe answer is no:

Example Take I = [0, 2] and define

is something which cannot happen in the finite dimensional case

Theorem 0.25 If X is a finite dimensional vector space, then all normsare equivalent That is, for any two given norms k.k1 and k.k2, there arepositive constants m1 and m2 such that

1

m2kf k1 ≤ kf k2 ≤ m1kf k1. (0.53)Proof Since equivalence of norms is an equivalence relation (check this!) wecan assume that k.k2 is the usual Euclidean norm Moreover, we can choose

an orthogonal basis uj, 1 ≤ j ≤ n, such that kP

jαjujk2

2 = P

j|αj|2 Let

0.3 The geometry of Hilbert spaces 25

Problem 0.20 Show that the maximum norm on C[0, 1] does not satisfythe parallelogram law

Problem 0.21 In a Banach space, the unit ball is convex by the triangleinequality A Banach space X is called uniformly convex if for every

ε > 0 there is some δ such that kxk ≤ 1, kyk ≤ 1, and kx+y2 k ≥ 1 − δ imply

kx − yk ≤ ε

Geometrically this implies that if the average of two vectors inside theclosed unit ball is close to the boundary, then they must be close to eachother

Show that a Hilbert space is uniformly convex and that one can chooseδ(ε) = 1 −

q

1 −ε42 Draw the unit ball for R2 for the norms kxk1 =

|x1| + |x2|, kxk2 = p|x1|2+ |x2|2, and kxk∞ = max(|x1|, |x2|) Which ofthese norms makes R2 uniformly convex?

(Hint: For the first part, use the parallelogram law.)

Problem 0.22 Suppose Q is a vector space Let s(f, g) be a sesquilinearform on Q and q(f ) = s(f, f ) the associated quadratic form Prove theparallelogram law

q(f + g) + q(f − g) = 2q(f ) + 2q(g) (0.54)and the polarization identity

s(f, g) = 1

4(q(f + g) − q(f − g) + i q(f − ig) − i q(f + ig)) (0.55)

Show that s(f, g) is symmetric if and only if q(f ) is real-valued.

Problem 0.23 A sesquilinear form is called bounded if

|s(f, g)| ≤ q(f )1/2q(g)1/2 (0.56)holds if q(f ) ≥ 0

(Hint: Consider 0 ≤ q(f + αg) = q(f ) + 2 Re(α s(f, g)) + |α|2q(g) andchoose α = t s(f, g)∗/|s(f, g)| with t ∈ R.)

Problem 0.25 Prove the claims made about fn, defined in (0.52), in thelast example

to see that ¯X (and ˜X) inherit the vector space structure from X Moreover,Lemma 0.26 If xn is a Cauchy sequence, then kxnk converges

Consequently, the norm of a Cauchy sequence (xn)∞n=1 can be defined

by k(xn)∞n=1k = limn→∞kxnk and is independent of the equivalence class(show this!) Thus ¯X is a normed space ( ˜X is not! Why?)

Theorem 0.27 ¯X is a Banach space containing X as a dense subspace if

we identify x ∈ X with the equivalence class of all sequences converging tox

0.5 Bounded operators 27

Proof (Outline) It remains to show that ¯X is complete Let ξn= [(xn,j)∞j=1]

be a Cauchy sequence in ¯X Then it is not hard to see that ξ = [(xj,j)∞j=1]

In particular, it is no restriction to assume that a normed vector space

or an inner product space is complete However, in the important case

of L2cont, it is somewhat inconvenient to work with equivalence classes ofCauchy sequences and hence we will give a different characterization usingthe Lebesgue integral later

Problem 0.26 Provide a detailed proof of Theorem 0.27

Ran(A) = {Af |f ∈ D(A)} = AD(A) ⊆ Y (0.59)are defined as usual The operator A is called bounded if the operatornorm

kAk = sup

f ∈D(A),kf kX=1

is finite

By construction, a bounded operator is Lipschitz continuous,

kAf kY ≤ kAkkf kX, f ∈ D(A), (0.61)and hence continuous The converse is also true:

Theorem 0.28 An operator A is bounded if and only if it is continuous.Proof Suppose A is continuous but not bounded Then there is a sequence

of unit vectors un such that kAunk ≥ n Then fn= n1unconverges to 0 but

In particular, if X is finite dimensional, then every operator is bounded.Note that in general one and the same operation might be bounded (i.e.continuous) or unbounded, depending on the norm chosen.

Example Consider the vector space of differentiable functions X = C1[0, 1]and equip it with the norm (cf Problem 0.29)

However, if we consider A = dxd : D(A) ⊆ Y → Y defined on D(A) =

C1[0, 1], then we have an unbounded operator Indeed, choose

un(x) = sin(nπx)which is normalized, kunk∞= 1, and observe that

Af = lim

n→∞Afn, fn∈ D(A), f ∈ X

To show that this definition is independent of the sequence fn → f , let

gn→ f be a second sequence and observe

kAfn− Agnk = kA(fn− gn)k ≤ kAkkfn− gnk → 0

Since for f ∈ D(A) we can choose fn= f , we see that Af = Af in this case;that is, A is indeed an extension From continuity of vector addition andscalar multiplication it follows that A is linear Finally, from continuity ofthe norm we conclude that the operator norm does not increase 

In particular,... no:

Example Take I = [0, 2] and define

is something which cannot happen in the finite dimensional case

Theorem 0.25 If X is a finite dimensional vector space, then all normsare... class="page_container" data-page="40">

In particular, if X is finite dimensional, then every operator is bounded.Note that in general one and the same operation might be bounded (i.e.continuous) or