Mathematical methods in quantum mechanics


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Mathematical methods in quantum mechanics

  1. 1. Mathematical Methodsin Quantum MechanicsWith Applicationsto Schr¨dinger Operators oGerald TeschlNote: The AMS has granted the permission to post this online edition!This version is for personal online use only! If you like this book and wantto support the idea of online versions, please consider buying this book: Studiesin MathematicsVolume 99 American Mathematical Society Providence, Rhode Island
  2. 2. Editorial Board David Cox (Chair) Steven G. Krants Rafe Mazzeo Martin Scharlemann2000 Mathematics subject classification. 81-01, 81Qxx, 46-01, 34Bxx, 47B25Abstract. This book provides a self-contained introduction to mathematical methods in quan-tum mechanics (spectral theory) with applications to Schr¨dinger operators. The first part cov- oers 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. The second part starts with a detailed study of the free Schr¨dinger operator respectively oposition, momentum and angular momentum operators. Then we develop Weyl–Titchmarsh the-ory for Sturm–Liouville operators and apply it to spherically symmetric problems, in particularto the hydrogen atom. Next we investigate self-adjointness of atomic Schr¨dinger operators and otheir essential spectrum, in particular the HVZ theorem. Finally we have a look at scatteringtheory and prove asymptotic completeness in the short range case. For additional information and updates on this book, visit: by L TEXand Makeindex. Version: February 17, 2009. ALibrary of Congress Cataloging-in-Publication DataTeschl, Gerald, 1970– Mathematical methods in quantum mechanics : with applications to Schr¨dinger operators o/ Gerald Teschl. p. cm. — (Graduate Studies in Mathematics ; v. 99) Includes bibliographical references and index. ISBN 978-0-8218-4660-5 (alk. paper) 1. Schr¨dinger operators. 2. Quantum theory—Mathematics. I. Title. oQC174.17.S3T47 2009 2008045437515’.724–dc22Copying and reprinting. Individual readers of this publication, and nonprofit libraries actingfor them, are permitted to make fair use of the material, such as to copy a chapter for usein teaching or research. Permission is granted to quote brief passages from this publication inreviews, provided the customary acknowledgement of the source is given. Republication, systematic copying, or multiple reproduction of any material in this pub-lication (including abstracts) is permitted only under license from the American MathematicalSociety. Requests for such permissions should be addressed to the Assistant to the Publisher,American Mathematical Society, P.O. Box 6248, Providence, Rhode Island 02940-6248. Requestscan also be made by e-mail to c 2009 by the American Mathematical Society. All rights reserved. The American Mathematical Society retains all rights except those granted too the United States Government.
  3. 3. To Susanne, Simon, and Jakob
  4. 4. ContentsPreface xiPart 0. PreliminariesChapter 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 12 §0.3. The geometry of Hilbert spaces 16 §0.4. Completeness 22 §0.5. Bounded operators 22 §0.6. Lebesgue Lp spaces 25 §0.7. Appendix: The uniform boundedness principle 32Part 1. Mathematical Foundations of Quantum MechanicsChapter 1. Hilbert spaces 37 §1.1. Hilbert spaces 37 §1.2. Orthonormal bases 39 §1.3. The projection theorem and the Riesz lemma 43 §1.4. Orthogonal sums and tensor products 45 §1.5. The C∗ algebra of bounded linear operators 47 §1.6. Weak and strong convergence 49 §1.7. Appendix: The Stone–Weierstraß theorem 51Chapter 2. Self-adjointness and spectrum 55 vii
  5. 5. viii Contents §2.1. Some quantum mechanics 55 §2.2. Self-adjoint operators 58 §2.3. Quadratic forms and the Friedrichs extension 67 §2.4. Resolvents and spectra 73 §2.5. Orthogonal sums of operators 79 §2.6. Self-adjoint extensions 81 §2.7. Appendix: Absolutely continuous functions 84Chapter 3. The spectral theorem 87 §3.1. The spectral theorem 87 §3.2. More on Borel measures 99 §3.3. Spectral types 104 §3.4. Appendix: The Herglotz theorem 107Chapter 4. Applications of the spectral theorem 111 §4.1. Integral formulas 111 §4.2. Commuting operators 115 §4.3. The min-max theorem 117 §4.4. Estimating eigenspaces 119 §4.5. Tensor products of operators 120Chapter 5. Quantum dynamics 123 §5.1. The time evolution and Stone’s theorem 123 §5.2. The RAGE theorem 126 §5.3. The Trotter product formula 131Chapter 6. Perturbation theory for self-adjoint operators 133 §6.1. Relatively bounded operators and the Kato–Rellich theorem 133 §6.2. More on compact operators 136 §6.3. Hilbert–Schmidt and trace class operators 139 §6.4. Relatively compact operators and Weyl’s theorem 145 §6.5. Relatively form bounded operators and the KLMN theorem 149 §6.6. Strong and norm resolvent convergence 153Part 2. Schr¨dinger Operators oChapter 7. The free Schr¨dinger operator o 161 §7.1. The Fourier transform 161 §7.2. The free Schr¨dinger operator o 167
  6. 6. Contents ix §7.3. The time evolution in the free case 169 §7.4. The resolvent and Green’s function 171Chapter 8. Algebraic methods 173 §8.1. Position and momentum 173 §8.2. Angular momentum 175 §8.3. The harmonic oscillator 178 §8.4. Abstract commutation 179Chapter 9. One-dimensional Schr¨dinger operators o 181 §9.1. Sturm–Liouville operators 181 §9.2. Weyl’s limit circle, limit point alternative 187 §9.3. Spectral transformations I 195 §9.4. Inverse spectral theory 202 §9.5. Absolutely continuous spectrum 206 §9.6. Spectral transformations II 209 §9.7. The spectra of one-dimensional Schr¨dinger operators o 214Chapter 10. One-particle Schr¨dinger operators o 221 §10.1. Self-adjointness and spectrum 221 §10.2. The hydrogen atom 222 §10.3. Angular momentum 225 §10.4. The eigenvalues of the hydrogen atom 229 §10.5. Nondegeneracy of the ground state 235Chapter 11. Atomic Schr¨dinger operators o 239 §11.1. Self-adjointness 239 §11.2. The HVZ theorem 242Chapter 12. Scattering theory 247 §12.1. Abstract theory 247 §12.2. Incoming and outgoing states 250 §12.3. Schr¨dinger operators with short range potentials o 253Part 3. AppendixAppendix A. Almost everything about Lebesgue integration 259 §A.1. Borel measures in a nut shell 259 §A.2. Extending a premeasure to a measure 263 §A.3. Measurable functions 268
  7. 7. x Contents §A.4. The Lebesgue integral 270 §A.5. Product measures 275 §A.6. Vague convergence of measures 278 §A.7. Decomposition of measures 280 §A.8. Derivatives of measures 282Bibliographical notes 289Bibliography 293Glossary of notation 297Index 301
  8. 8. PrefaceOverview The present text was written for my course Schr¨dinger Operators held oat the University of Vienna in winter 1999, summer 2002, summer 2005,and winter 2007. It gives a brief but rather self-contained introductionto the mathematical methods of quantum mechanics with a view towardsapplications to Schr¨dinger operators. The applications presented are highly oselective and many important and interesting items are not touched upon. 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.In addition, existence of spectral measures is established via the Herglotztheorem rather than the Riesz representation theorem since this approachpaves the way for an investigation of spectral types via boundary values ofthe resolvent as the spectral parameter approaches the real line. xi
  9. 9. xii Preface Part 2 starts with the free Schr¨dinger equation and computes the ofree resolvent and time evolution. In addition, I discuss position, momen-tum, and angular momentum operators via algebraic methods. This isusually found in any physics textbook on quantum mechanics, with theonly difference 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 thereis 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 [40]–[43] (see also [14]) and by the book by Weidmann [60] (anextended version of which has recently appeared in two volumes [62], [63],however, only in German). Other books with a similar scope are for example[14], [15], [21], [23], [39], [48], and [55]. For those who want to know moreabout the physical aspects, I can recommend the classical book by Thirring[58] and the visual guides by Thaller [56], [57]. Further information can befound in the bibliographical notes at the end.Reader’s guide There is some intentional overlap between 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
  10. 10. Preface xiiichapters and you should study them in detail (except for Section 2.6 whichcan be skipped on first reading). Chapter 4 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. Chapter 5contains 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, Chapter 6 is againof central importance and should be studied in detail. The chapters in the second part are mostly independent of each otherexcept for Chapter 7, which is a prerequisite for all others except for Chap-ter 9. If you are interested in one-dimensional models (Sturm–Liouville equa-tions), Chapter 9 is all you need. If you are interested in atoms, read Chapter 7, Chapter 10, and Chap-ter 11. In particular, you can skip the separation of variables (Sections 10.3and 10.4, which require Chapter 9) 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 Chapter 10, and Chapter 12. Chapter 5 is one of the key prereq-uisites in this case.Updates The AMS is hosting a web page for this book at updates, corrections, and other material may be found, including alink to material on my own web site: I would like to thank Volker Enß for making his lecture notes [18] avail-able to me. Many colleagues and students have made useful suggestions andpointed out mistakes in earlier drafts of this book, in particular: KerstinAmmann, J¨rg Arnberger, Chris Davis, Fritz Gesztesy, Maria Hoffmann- oOstenhof, Zhenyou Huang, Helge Kr¨ger, Katrin Grunert, Wang Lanning, uDaniel Lenz, Christine Pfeuffer, Roland M¨ws, Arnold L. Neidhardt, Harald o
  11. 11. xiv PrefaceRindler, Johannes Temme, Karl Unterkofler, Joachim Weidmann, and RudiWeikard. 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 muchof this writing, most recently under grant Y330. Gerald TeschlVienna, AustriaJanuary 2009Gerald TeschlFakult¨t f¨r Mathematik a uNordbergstraße 15Universit¨t Wien a1090 Wien, AustriaE-mail:
  12. 12. Part 0Preliminaries
  13. 13. Chapter 0A first look at Banachand Hilbert spacesI assume that the reader has some basic familiarity with measure theory and func-tional analysis. For convenience, some facts needed from Banach and Lp spaces arereviewed in this chapter. A crash course in measure theory can be found in theAppendix A. If you feel comfortable with terms like Lebesgue Lp spaces, Banachspace, or bounded linear operator, you can skip this entire chapter. However, youmight want to at least browse through it to refresh your memory.0.1. Warm up: Metric and topological spacesBefore 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 [26]. 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 semi-metric. Moreover, it is straight-forward to see the inverse triangle inequality (Problem 0.1) |d(x, y) − d(z, y)| ≤ d(x, z). (0.1) 3
  14. 14. 4 0. A first look at Banach and Hilbert spacesExample. Euclidean space Rn together with d(x, y) = ( n (xk − yk )2 )1/2 k=1is a metric space and so is Cn together with d(x, y) = ( n |xk −yk |2 )1/2 . k=1 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 setU is called an interior point of U if U contains some ball around x. If x isan interior point of U , then U is also called a neighborhood of x. A pointx is called a limit point of U if (Br (x){x}) ∩ U = ∅ for every ball aroundx. Note that a limit point x need not lie in U , but U must contain pointsarbitrarily close to x.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 are limit pointsof 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 implies α Oα ∈ 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 usually notvery interesting examples are the trivial topology O = {∅, X} and thediscrete topology O = P(X) (the powerset of X). Given two topologiesO1 and O2 on X, O1 is called weaker (or coarser) than O2 if and only ifO1 ⊆ 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 n ˜ d(x, y) = |xk − yk |. (0.3) k=1Then n n n 1 √ |xk | ≤ |xk |2 ≤ |xk | (0.4) n k=1 k=1 k=1
  15. 15. 0.1. Warm up: Metric and topological spaces 5 ˜ ˜shows Br/√n (x) ⊆ Br (x) ⊆ Br (x), where B, B are balls computed using d,˜d, respectively.Example. We can always replace a metric d by the bounded metric ˜ d(x, y) d(x, y) = (0.5) 1 + d(x, y)without changing the topology. 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 such that ˜O = O ∩ Y (induced topology).Example. The set (0, 1] ⊆ R is not open in the topology of X = R, but it isopen in the induced 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 eachx 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 be ˜written as O = O⊇O∈B O and we have ˜Lemma 0.1. If B ⊆ O is a base for the topology, then every open set canbe 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 (and 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 semi-metric space will not be Hausdorff). The complement of an open set is called a closed set. It follows fromde Morgan’s rules that the family of closed sets C satisfies (i) ∅, X ∈ C, (ii) C1 , C2 ∈ C implies C1 ∪ C2 ∈ C, (iii) {Cα } ⊆ C implies α Cα ∈ C.That is, closed sets are closed under finite unions and arbitrary intersections. The smallest closed set containing a given set U is called the closure U= C, (0.6) C∈C,U ⊆C
  16. 16. 6 0. A first look at Banach and Hilbert spacesand the largest open set contained in a given set U is called the interior U◦ = O. (0.7) O∈O,O⊆U We can define interior and limit points as before by replacing the wordball by open set. Then it is straightforward to checkLemma 0.2. Let X be a topological space. Then the interior of U is theset of all interior points of U and the closure of U is the union of U withall limit points of U . A sequence (xn )∞ ⊆ X is said to converge to some point x ∈ X if n=1d(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 semi-metric). Every convergent sequence is a Cauchy sequence; that is, for everyε > 0 there is some N ∈ N such that d(xn , xm ) ≤ ε, n, m ≥ N. (0.8)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. A point x is clearly a limit point of U if and only if there is some sequencexn ∈ U {x} converging to x. HenceLemma 0.3. A closed subset of a complete metric space is again a completemetric space. Note that convergence can also be equivalently formulated in terms oftopological terms: A sequence xn converges to x if and only if for everyneighborhood U of x there is some N ∈ N such that xn ∈ U for n ≥ N . Ina Hausdorff space 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. Notethat X is separable if and only if it is second countable as a topologicalspace.Lemma 0.4. Let X be a separable metric space. Every subset of X is againseparable.Proof. Let A = {xn }n∈N be a dense set in X. The only problem is thatA ∩ Y might contain no elements at all. However, some elements of A mustbe at least arbitrarily close: Let J ⊆ N2 be the set of all pairs (n, m) forwhich B1/m (xn ) ∩ Y = ∅ 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 is
  17. 17. 0.1. Warm up: Metric and topological spaces 7dense, 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. A function 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. (0.9)If f is continuous at every point, it is called continuous.Lemma 0.5. Let X, Y be metric spaces and f : X → Y . The following areequivalent: (i) f is continuous at x (i.e, (0.9) 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 isa neighborhood V of f (x) such that Bδ (x) ⊆ f −1 (V ) for every δ. Hencewe can choose a sequence xn ∈ B1/n (x) such that f (xn ) ∈ f −1 (V ). Thusxn → x but f (xn ) → 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 inverse imageof every open (closed) set is again open (closed). Note: In a topological space, (iii) is used as the definition for continuity.However, 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. 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) = 0}. (0.10) If X and Y are metric spaces, then X × Y together with d((x1 , y1 ), (x2 , y2 )) = dX (x1 , x2 ) + dY (y1 , y2 ) (0.11)is a metric space. A sequence (xn , yn ) converges to (x, y) if and only ifxn → x and yn → y. In particular, the projections onto the first (x, y) → x,respectively, onto the second (x, y) → y, coordinate are continuous. In particular, by the inverse triangle inequality (0.1), |d(xn , yn ) − d(x, y)| ≤ d(xn , x) + d(yn , y), (0.12)we see that d : X × X → R is continuous.
  18. 18. 8 0. A first look at Banach and Hilbert spacesExample. If we consider R × R, we do not get the Euclidean distance ofR2 unless we modify (0.11) as follows: ˜ d((x1 , y1 ), (x2 , y2 )) = dX (x1 , x2 )2 + dY (y1 , y2 )2 . (0.13)As noted in our previous example, the topology (and thus also conver-gence/continuity) is independent of this choice. If X and Y are just topological spaces, the product topology is definedby 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 the case ofmetric spaces this clearly agrees with the topology defined via the productmetric (0.11). A cover of a set Y ⊆ X is a family of sets {Uα } such that Y ⊆ α Uα .A cover is called open if all Uα are open. Any subset of {Uα } which stillcovers Y is called a subcover.Lemma 0.6 (Lindel¨f). If X is second countable, then every open cover ohas a countable subcover.Proof. Let {Uα } be an open cover for Y and let B be a countable base.Since every Uα can be written as a union of elements from B, the set of allB ∈ B which satisfy B ⊆ Uα for some α form a countable open cover for Y .Moreover, for every Bn in this set we can find an αn such that Bn ⊆ Uαn .By construction {Uαn } is a countable subcover. A subset K ⊂ X is called compact if every open cover has a finitesubcover.Lemma 0.7. A topological space is compact if and only if it has the finiteintersection property: The intersection of a family of closed sets is emptyif and only if the intersection of some finite subfamily is empty.Proof. By taking complements, to every family of open sets there is a cor-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-tion. A subset K ⊂ X is called sequentially compact if every sequence hasa convergent subsequence.Lemma 0.8. 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, any compact set is closed.
  19. 19. 0.1. Warm up: Metric and topological spaces 9 (iv) The product of finitely many compact sets is compact. (v) A compact set is also sequentially compact.Proof. (i) Observe that if {Oα } is an open cover for f (Y ), then {f −1 (Oα )}is one for Y . (ii) Let {Oα } be an open cover for the closed subset Y . Then {Oα } ∪{XY } is an open cover for X. (iii) Let Y ⊆ X be compact. We show that XY is open. Fix x ∈ XY(if Y = X, there is nothing to do). By the definition of Hausdorff, forevery y ∈ Y there are disjoint neighborhoods V (y) of y and Uy (x) of x. Bycompactness of Y , there are y1 , . . . , yn such that the V (yj ) cover Y . Butthen U (x) = n Uyj (x) is a neighborhood of x which does not intersect j=1Y. (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) = k U (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) Let xn be a sequence which has no convergent subsequence. ThenK = {xn } has no limit points and is hence compact by (ii). For every nthere is a ball Bεn (xn ) which contains only finitely many elements of K.However, finitely many suffice to cover K, a contradiction. In a metric space compact and sequentially compact are equivalent.Lemma 0.9. Let X be a metric space. Then a subset is compact if and onlyif it is sequentially compact.Proof. By item (v) of the previous lemma it suffices to show that X iscompact if it is sequentially compact. First of all note that every cover of open balls with fixed radius ε > 0has a finite subcover since if this were false we could construct a sequencexn ∈ X n−1 Bε (xm ) such that d(xn , xm ) > ε for m < n. m=1 In particular, we are done if we can show that for every open cover{Oα } there is some ε > 0 such that for every x we have Bε (x) ⊆ Oα forsome α = α(x). Indeed, choosing {xk }n such that Bε (xk ) is a cover, we k=1have that Oα(xk ) is a cover as well. So it remains to show that there is such an ε. If there were none, forevery ε > 0 there must be an x such that Bε (x) ⊆ Oα for every α. Choose
  20. 20. 10 0. A first look at Banach and Hilbert spaces 1ε = n and pick a corresponding xn . Since X is sequentially compact, it is norestriction to assume xn converges (after maybe passing to a subsequence).Let x = lim xn . Then x lies in some Oα and hence Bε (x) ⊆ Oα . But choosing 1 ε εn so large that n < 2 and d(xn , x) < 2 , we have B1/n (xn ) ⊆ Bε (x) ⊆ Oα ,contradicting our assumption. Please also recall the Heine–Borel theorem:Theorem 0.10 (Heine–Borel). In Rn (or Cn ) a set is compact if and onlyif it is bounded and closed.Proof. By Lemma 0.8 (ii) and (iii) it suffices to show that a closed intervalin I ⊆ R is compact. Moreover, by Lemma 0.9 it suffices to show thatevery sequence in I = [a, b] has a convergent subsequence. Let xn be oursequence and divide I = [a, a+b ] ∪ [ a+b , b]. Then at least one of these two 2 2intervals, call it I1 , contains infinitely many elements of our sequence. Lety1 = xn1 be the first one. Subdivide I1 and pick y2 = xn2 , 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 for m ≥ n). n A topological space is called locally compact if every point has a com-pact neighborhood.Example. Rn is locally compact. The distance between a point x ∈ X and a subset Y ⊆ X is dist(x, Y ) = inf d(x, y). (0.14) y∈YNote that x is a limit point of Y if and only if dist(x, Y ) = 0.Lemma 0.11. Let X be a metric space. Then | dist(x, Y ) − dist(z, Y )| ≤ d(x, z). (0.15)In particular, x → 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 0.12 (Urysohn). Suppose C1 and C2 are disjoint closed subsets ofa metric space X. Then there is a continuous function f : X → [0, 1] suchthat f is zero on C1 and one on C2 . If X is locally compact and C1 is compact, one can choose f with compactsupport.
  21. 21. 0.1. Warm up: Metric and topological spaces 11 dist(x,C2 )Proof. To prove the first claim, set f (x) = dist(x,C1 )+dist(x,C2 ) . For thesecond claim, observe that there is an open set O such that O is compactand C1 ⊂ O ⊂ O ⊂ XC2 . In fact, for every x, there is a ball Bε (x) suchthat Bε (x) is compact and Bε (x) ⊂ XC2 . Since C1 is compact, finitelymany of them cover C1 and we can choose the union of those balls to be O.Now replace C2 by XO. Note that Urysohn’s lemma implies that a metric space is normal; thatis, for any two disjoint closed sets C1 and C2 , there are disjoint open setsO1 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 0.13. Let X be a locally compact metric space. Suppose K isa compact set and {Oj }n an open cover. Then there is a partition of j=1unity for K subordinate to this cover; that is, there are continuous functionshj : X → [0, 1] such that hj has compact support contained in Oj and n hj (x) ≤ 1 (0.16) j=1with 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 arefunctions gj : X → [0, 1] such that gj = 1 on Kj and gj = 0 on XOj . Nowset j−1 hj = gj (1 − gk ). (0.17) k=1Then hj : X → [0, 1] has compact support contained in Oj and n n hj (x) = 1 − (1 − gj (x)) (0.18) j=1 j=1shows that the sum is one for x ∈ K, since x ∈ Kj for some j impliesgj (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(x , y )| ≤d(x, x ) + d(y, y ).Problem 0.3. Let X be some space together with a sequence of distancefunctions dn , n ∈ N. Show that ∞ 1 dn (x, y) d(x, y) = 2n 1 + dn (x, y) n=1
  22. 22. 12 0. A first look at Banach and Hilbert spacesis again a distance function.Problem 0.4. Show that the closure satisfies U = U .Problem 0.5. Let U ⊆ V be subsets of a metric space X. Show that if Uis dense in V and V is dense in X, then U is dense in X.Problem 0.6. Show that any open set O ⊆ R can be written as a countableunion of disjoint intervals. (Hint: Let {Iα } be the set of all maximal subin-tervals of O; that is, Iα ⊆ O and there is no other subinterval of O whichcontains Iα . Then this is a cover of disjoint intervals which has a countablesubcover.)0.2. The Banach space of continuous functionsNow 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 max-imum norm: f (x) − g(x) ∞ = max |f (x) − g(x)|. (0.19) x∈IIt is not hard to see that with this definition C(I) becomes a normed linearspace: A normed linear space X is a vector space X over C (or R) with areal-valued function (the norm) . such that • f ≥ 0 for all f ∈ X and f = 0 if and only if f = 0, • α f = |α| f for all α ∈ C and f ∈ X, and • f + g ≤ f + g for all f, g ∈ X (triangle inequality). From the triangle inequality we also get the inverse triangle inequal-ity (Problem 0.7) | f − g |≤ f −g . (0.20) Once we have a norm, we have a distance d(f, g) = f −g and hence weknow when a sequence of vectors fn converges to a vector f . We will writefn → f or limn→∞ fn = f , as usual, in this case. Moreover, a mappingF : X → Y between two normed spaces is called continuous if fn → fimplies F (fn ) → F (f ). In fact, it is not hard to see that the norm, vectoraddition, and multiplication by scalars are continuous (Problem 0.8). In addition to the concept of convergence we have also the concept ofa Cauchy sequence and hence the concept of completeness: A normed
  23. 23. 0.2. The Banach space of continuous functions 13space 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 sequences a = (aj )∞ for which the norm j=1 ∞ a 1 = |aj | (0.21) j=1is finite is a Banach space. To show this, we need to verify three things: (i) 1 (N) is a vector spacethat is closed under addition and scalar multiplication, (ii) . 1 satisfies thethree requirements for a norm, and (iii) 1 (N) is complete. First of all observe k k k |aj + bj | ≤ |aj | + |bj | ≤ a 1 + b 1 (0.22) j=1 j=1 j=1for any finite k. Letting k → ∞, we conclude that 1 (N) is closed underaddition and that the triangle inequality holds. That 1 (N) is closed underscalar multiplication and the two other properties of a norm are straight-forward. It remains to show that 1 (N) is complete. Let an = (an )∞ be j j=1a Cauchy sequence; that is, for given ε > 0 we can find an Nε such that am − an 1 ≤ ε for m, n ≥ Nε . This implies in particular |am − an | ≤ ε for j jany fixed j. Thus an is a Cauchy sequence for fixed j and by completeness jof C has a limit: limn→∞ an = aj . Now consider j k |am − an | ≤ ε j j (0.23) j=1and take m → ∞: k |aj − an | ≤ ε. j (0.24) j=1Since this holds for any finite k, we even have a−an 1 ≤ ε. Hence (a−an ) ∈ 1 (N) and since a ∈ 1 (N), we finally conclude a = a + (a − a ) ∈ 1 (N). n n nExample. The space ∞ (N) of all bounded sequences a = (aj )∞ together j=1with the norm a ∞ = sup |aj | (0.25) j∈Nis a Banach space (Problem 0.10). Now what about convergence in the space C(I)? A sequence of functionsfn (x) converges to f if and only if lim f − fn = lim sup |fn (x) − f (x)| = 0. (0.26) n→∞ n→∞ x∈I
  24. 24. 14 0. A first look at Banach and Hilbert spacesThat is, in the language of real analysis, fn converges uniformly to f . Nowlet us look at the case where fn is only a Cauchy sequence. Then fn (x) isclearly a Cauchy sequence of real numbers for any fixed x ∈ I. In particular,by completeness of C, there is a limit f (x) for each x. Thus we get a limitingfunction f (x). Moreover, letting m → ∞ in |fm (x) − fn (x)| ≤ ε ∀m, n > Nε , x ∈ I, (0.27)we see |f (x) − fn (x)| ≤ ε ∀n > Nε , x ∈ I; (0.28)that is, fn (x) converges uniformly to f (x). However, up to this point wedo not know whether it is in our vector space C(I) or not, that is, whetherit is continuous or not. Fortunately, there is a well-known result from realanalysis which tells us that the uniform limit of continuous functions is againcontinuous. Hence f (x) ∈ C(I) and thus every Cauchy sequence in C(I)converges. Or, in other wordsTheorem 0.14. C(I) with the maximum norm is a Banach space. Next we want to know if there is a countable basis for C(I). We will calla set of vectors {un } ⊂ X linearly independent if every finite subset is andwe will call a countable set of linearly independent vectors {un }N ⊂ X n=1a Schauder basis if every element f ∈ X can be uniquely written as acountable linear combination of the basis elements: N f= cn un , cn = cn (f ) ∈ C, (0.29) n=1where the sum has to be understood as a limit if N = ∞. In this case thespan span{un } (the set of all finite linear combinations) of {un } is dense inX. 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 linear space containing acountable dense set is called separable.Example. The Banach space 1 (N) is separable. In fact, the set of vectorsδ n , with δn = 1 and δm = 0, n = m, is total: Let a = (aj )∞ ∈ 1 (N) be n n j=1 n = ngiven and set a j=1 aj δ j . Then ∞ n a−a 1 = |aj | → 0 (0.30) j=n+1since an = aj for 1 ≤ j ≤ n and an = 0 for j > n. j j Luckily this is also the case for C(I):Theorem 0.15 (Weierstraß). Let I be a compact interval. Then the set ofpolynomials is dense in C(I).
  25. 25. 0.2. The Banach space of continuous functions 15Proof. Let f (x) ∈ C(I) be given. By considering f (x) − f (a) + (f (b) −f (a))(x − b) it is no loss to assume that f vanishes at the boundary points.Moreover, without restriction we only consider I = [ −1 , 1 ] (why?). 2 2 Now the claim follows from the lemma below using 1 un (x) = (1 − x2 )n , Inwhere 1 1 n In = (1 − x2 )n dx = (1 − x)n−1 (1 + x)n+1 dx −1 n+1 −1 n! n! = ··· = 22n+1 = 1 1 (n + 1) · · · (2n + 1) 2 ( 2 + 1) · · · ( 1 + n) 2 √ Γ(1 + n) π 1 = π 3 = (1 + O( )). Γ( 2 + n) n n √In the last step we have used Γ( 1 ) = π [1, (6.1.8)] and the asymptotics 2follow from Stirling’s formula [1, (6.1.37)].Lemma 0.16 (Smoothing). Let un (x) be a sequence of nonnegative contin-uous functions on [−1, 1] such that un (x)dx = 1 and un (x)dx → 0, δ > 0. (0.31) |x|≤1 δ≤|x|≤1(In other words, un has mass one and concentrates near x = 0 as n → ∞.) 1 Then for every f ∈ C[− 2 , 1 ] which vanishes at the endpoints, f (− 2 ) = 2 1f ( 1 ) = 0, we have that 2 1/2 fn (x) = un (x − y)f (y)dy (0.32) −1/2converges uniformly to f (x).Proof. Since f is uniformly continuous, for given ε we can find a δ (indepen-dent of x) such that |f (x)−f (y)| ≤ ε whenever |x−y| ≤ δ. Moreover, we canchoose n such that δ≤|y|≤1 un (y)dy ≤ ε. Now abbreviate M = max{1, |f |}and note 1/2 1/2 |f (x) − un (x − y)f (x)dy| = |f (x)| |1 − un (x − y)dy| ≤ M ε. −1/2 −1/2In fact, either the distance of x to one of the boundary points ± 1 is smaller 2than δ and hence |f (x)| ≤ ε or otherwise the difference between one and theintegral is smaller than ε.
  26. 26. 16 0. A first look at Banach and Hilbert spaces Using this, we have 1/2 |fn (x) − f (x)| ≤ un (x − y)|f (y) − f (x)|dy + M ε −1/2 ≤ un (x − y)|f (y) − f (x)|dy |y|≤1/2,|x−y|≤δ + un (x − y)|f (y) − f (x)|dy + M ε |y|≤1/2,|x−y|≥δ =ε + 2M ε + M ε = (1 + 3M )ε, (0.33)which proves the claim. Note that fn will be as smooth as un , hence the title smoothing lemma.The same idea is used to approximate noncontinuous functions by smoothones (of course the convergence will no longer be uniform in this case).Corollary 0.17. C(I) is separable.However, ∞ (N) is not separable (Problem 0.11)!Problem 0.7. Show that | f − g | ≤ f − g .Problem 0.8. Let X be a Banach space. Show that the norm, vector ad-dition, and multiplication by scalars are continuous. That is, if fn → f ,gn → g, and αn → α, then fn → f , fn + gn → f + g, and αn gn → αg. ∞Problem 0.9. Let X be a Banach space. Show that j=1 fj < ∞ impliesthat ∞ n fj = lim fj n→∞ j=1 j=1exists. The series is called absolutely convergent in this case.Problem 0.10. Show that ∞ (N) is a Banach space.Problem 0.11. Show that ∞ (N) is not separable. (Hint: Consider se-quences which take only the value one and zero. How many are there? Whatis the distance between two such sequences?)0.3. The geometry of Hilbert spacesSo 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:
  27. 27. 0.3. The geometry of Hilbert spaces 17 Suppose H is a vector space. A map ., .. : H × H → C is called asesquilinear form if it is conjugate linear in the first argument and linearin the second; that is, α1 f1 + α2 f2 , g ∗ ∗ = α1 f1 , g + α2 f2 , g , α1 , α2 ∈ C, (0.34) f, α1 g1 + α2 g2 = α1 f, g1 + α2 f, g2 ,where ‘∗’ denotes complex conjugation. A sesquilinear form satisfying therequirements (i) f, f > 0 for f = 0 (positive definite), (ii) f, g = g, f ∗ (symmetry)is called an inner product or scalar product. Associated with everyscalar product is a norm f = f, f . (0.35)The pair (H, ., .. ) is called an inner product space. If H is complete, itis called a Hilbert space.Example. Clearly Cn with the usual scalar product n a, b = a∗ bj j (0.36) j=1is a (finite dimensional) Hilbert space.Example. A somewhat more interesting example is the Hilbert space 2 (N),that is, the set of all sequences ∞ (aj )∞ j=1 |aj |2 < ∞ (0.37) j=1with scalar product ∞ a, b = a∗ bj . j (0.38) j=1(Show that this is in fact a separable Hilbert space — Problem 0.13.) Of course I still owe you a proof for the claim that f, f is indeed anorm. Only the triangle inequality is nontrivial, which will follow from theCauchy–Schwarz inequality below. A vector f ∈ H is called normalized or a unit vector if f = 1.Two vectors f, g ∈ H are called orthogonal or perpendicular (f ⊥ g) if f, g = 0 and parallel if one is a multiple of the other. If f and g are orthogonal, we have the Pythagorean theorem: 2 2 f +g = f + g 2, f ⊥ g, (0.39)which is one line of computation.
  28. 28. 18 0. A first look at Banach and Hilbert spaces Suppose u is a unit vector. Then the projection of f in the direction ofu is given by f = u, f u (0.40)and f⊥ defined via f⊥ = f − u, f u (0.41)is perpendicular to u since u, f⊥ = u, f − u, f u = u, f − u, f u, u =0. f  w  f f   f f⊥   f   I f     f u I  Taking any other vector parallel to u, it is easy to see 2 2 2 f − αu = f⊥ + (f − αu) = f⊥ + | u, f − α|2 (0.42)and hence f = u, f u is the unique vector parallel to u which is closest tof. As a first consequence we obtain the Cauchy–Schwarz–Bunjakowskiinequality:Theorem 0.18 (Cauchy–Schwarz–Bunjakowski). Let H0 be an inner prod-uct space. Then for every f, g ∈ H0 we have | f, g | ≤ f g (0.43)with equality if and only if f and g are parallel.Proof. It suffices to prove the case g = 1. But then the claim followsfrom f 2 = | g, f |2 + f⊥ 2 . Note that the Cauchy–Schwarz inequality entails that the scalar productis continuous in both variables; that is, if fn → f and gn → g, we have fn , gn → f, g . As another consequence we infer that the map . is indeed a norm. Infact, 2 2 2 f +g = f + f, g + g, f + g ≤ ( f + g )2 . (0.44) 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(Problem 0.17).
  29. 29. 0.3. The geometry of Hilbert spaces 19Theorem 0.19 (Jordan–von Neumann). A norm is associated with a scalarproduct if and only if the parallelogram law 2 2 2 2 f +g + f −g =2 f +2 g (0.45)holds. In this case the scalar product can be recovered from its norm by virtueof the polarization identity 1 2 2 2 2 f, g = f +g − f −g + i f − ig − i f + ig . (0.46) 4Proof. If an inner product space is given, verification of the parallelogramlaw and the polarization identity is straightforward (Problem 0.14). To show the converse, we define 1 2 2 2 2 s(f, g) = f +g − f −g + i f − ig − i f + ig . 4Then s(f, f ) = f 2 and s(f, g) = s(g, f )∗ are straightforward to check.Moreover, another straightforward computation using the parallelogram lawshows g+h s(f, g) + s(f, h) = 2s(f, ). 2Now choosing h = 0 (and using s(f, 0) = 0) shows s(f, g) = 2s(f, g ) and 2thus s(f, g) + s(f, h) = s(f, g + h). Furthermore, by induction we inferm m2n s(f, g) = s(f, 2n g); that is, α s(f, g) = s(f, αg) for every positive rationalα. By continuity (check this!) this holds for all α 0 and s(f, −g) =−s(f, g), respectively, s(f, ig) = i s(f, g), finishes the proof. Note that the parallelogram law and the polarization identity even holdfor sesquilinear forms (Problem 0.14). But how do we define a scalar product on C(I)? One possibility is b f, g = f ∗ (x)g(x)dx. (0.47) aThe corresponding inner product space is denoted by L2 (I). Note that contwe have f ≤ |b − a| f ∞ (0.48)and hence the maximum norm is stronger than the L2 norm. cont Suppose we have two norms . 1 and . 2 on a space X. Then . 2 issaid to be stronger than . 1 if there is a constant m 0 such that f 1 ≤m f 2. (0.49)It is straightforward to check the following.
  30. 30. 20 0. A first look at Banach and Hilbert spacesLemma 0.20. If . 2 is stronger than . 1 , then any . 2 Cauchy sequenceis also a . 1 Cauchy sequence. Hence if a function F : X → Y is continuous in (X, . 1 ), it is alsocontinuous in (X, . 2 ) and if a set is dense in (X, . 2 ), it is also dense in(X, . 1 ). In particular, L2 is separable. But is it also complete? Unfortunately contthe answer is no:Example. Take I = [0, 2] and define  1 0,  0 ≤ x ≤ 1 − n, fn (x) = 1 + n(x − 1), 1 1 − n ≤ x ≤ 1, (0.50)  1, 1 ≤ x ≤ 2. Then fn (x) is a Cauchy sequence in L2 , but there is no limit in L2 ! cont contClearly the limit should be the step function which is 0 for 0 ≤ x 1 and 1for 1 ≤ x ≤ 2, but this step function is discontinuous (Problem 0.18)! This shows that in infinite dimensional spaces different norms will giverise to different convergent sequences! In fact, the key to solving problems ininfinite dimensional spaces is often finding the right norm! This is somethingwhich cannot happen in the finite dimensional case.Theorem 0.21. If X is a finite dimensional space, then all norms are equiv-alent. That is, for any two given norms . 1 and . 2 , there are constantsm1 and m2 such that 1 f 1 ≤ f 2 ≤ m1 f 1 . (0.51) m2Proof. Clearly we can choose a basis uj , 1 ≤ j ≤ n, and assume that . 2 2 2is the usual Euclidean norm, j αj uj 2 = j |αj | . Let f = j αj uj .Then by the triangle and Cauchy–Schwartz inequalities f ≤ |αj | uj ≤ uj 2 f 1 1 1 2 j jand we can choose m2 = j uj 1. In particular, if fn is convergent with respect to . 2 , it is also convergentwith respect to . 1 . Thus . 1 is continuous with respect to . 2 and attainsits minimum m 0 on the unit sphere (which is compact by the Heine-Boreltheorem). Now choose m1 = 1/m.Problem 0.12. Show that the norm in a Hilbert space satisfies f + g = f + g if and only if f = αg, α ≥ 0, or g = 0.
  31. 31. 0.3. The geometry of Hilbert spaces 21Problem 0.13. Show that 2 (N) is a separable Hilbert space.Problem 0.14. 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) (0.52)and the polarization identity 1 s(f, g) = (q(f + g) − q(f − g) + i q(f − ig) − i q(f + ig)) . (0.53) 4 Conversely, show that any quadratic form q(f ) : Q → R satisfyingq(αf ) = |α|2 q(f ) and the parallelogram law gives rise to a sesquilinear formvia the polarization identity.Problem 0.15. A sesquilinear form is called bounded if s = sup |s(f, g)| f = g =1is finite. Similarly, the associated quadratic form q is bounded if q = sup |q(f )| f =1is finite. Show q ≤ s ≤2 q .(Hint: Use the parallelogram law and the polarization identity from the pre-vious problem.)Problem 0.16. 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. Show that theCauchy–Schwarz inequality |s(f, g)| ≤ q(f )1/2 q(g)1/2 (0.54)holds if q(f ) ≥ 0. (Hint: Consider 0 ≤ q(f + αg) = q(f ) + 2 Re(α s(f, g)) + |α|2 q(g) andchoose α = t s(f, g)∗ /|s(f, g)| with t ∈ R.)Problem 0.17. Show that the maximum norm (on C[0, 1]) does not satisfythe parallelogram law.Problem 0.18. Prove the claims made about fn , defined in (0.50), in thelast example.
  32. 32. 22 0. A first look at Banach and Hilbert spaces0.4. CompletenessSince L2cont is not complete, how can we obtain a Hilbert space from it?Well, the answer is simple: take the completion. If X is an (incomplete) normed space, consider the set of all Cauchy ˜sequences X. Call two Cauchy sequences equivalent if their difference con- ¯verges to zero and denote by X the set of all equivalence classes. It is easyto see that X¯ (and X) inherit the vector space structure from X. Moreover, ˜Lemma 0.22. If xn is a Cauchy sequence, then xn converges. Consequently the norm of a Cauchy sequence (xn )∞ can be defined by n=1 (xn )∞ n=1 = limn→∞ xn and is independent of the equivalence class (show ¯ ˜this!). Thus X is a normed space (X is not! Why?). ¯Theorem 0.23. X is a Banach space containing X as a dense subspace ifwe identify x ∈ X with the equivalence class of all sequences converging tox. ¯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=1is its limit. ¯ Let me remark that the completion X is unique. More precisely anyother complete space which contains X as a dense subset is isomorphic to ¯X. This can for example be seen by showing that the identity map on X ¯has a unique extension to X (compare Theorem 0.26 below). In particular it is no restriction to assume that a normed linear spaceor an inner product space is complete. However, in the important caseof L2cont it is somewhat inconvenient to work with equivalence classes ofCauchy sequences and hence we will give a different characterization usingthe Lebesgue integral later.0.5. Bounded operatorsA linear map A between two normed spaces X and Y will be called a (lin-ear) operator A : D(A) ⊆ X → Y. (0.55)The linear subspace D(A) on which A is defined is called the domain of Aand is usually required to be dense. The kernel Ker(A) = {f ∈ D(A)|Af = 0} (0.56)and range Ran(A) = {Af |f ∈ D(A)} = AD(A) (0.57)
  33. 33. 0.5. Bounded operators 23are defined as usual. The operator A is called bounded if the operatornorm A = sup Af Y (0.58) f X =1is finite. The set of all bounded linear operators from X to Y is denoted byL(X, Y ). If X = Y , we write L(X, X) = L(X).Theorem 0.24. The space L(X, Y ) together with the operator norm (0.58)is a normed space. It is a Banach space if Y is.Proof. That (0.58) is indeed a norm is straightforward. If Y is complete andAn is a Cauchy sequence of operators, then An f converges to an elementg for every f . Define a new operator A via Af = g. By continuity ofthe vector operations, A is linear and by continuity of the norm Af =limn→∞ An f ≤ (limn→∞ An ) f , it is bounded. Furthermore, givenε 0 there is some N such that An − Am ≤ ε for n, m ≥ N and thus An f −Am f ≤ ε f . Taking the limit m → ∞, we see An f −Af ≤ ε f ;that is, An → A. By construction, a bounded operator is Lipschitz continuous, Af Y ≤ A f X, (0.59)and hence continuous. The converse is also trueTheorem 0.25. 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 1of unit vectors un such that Aun ≥ n. Then fn = n un converges to 0 but Afn ≥ 1 does not converge to 0. Moreover, if A is bounded and densely defined, it is no restriction toassume that it is defined on all of X.Theorem 0.26 (B.L.T. theorem). Let A ∈ L(X, Y ) and let Y be a Banachspace. If D(A) is dense, there is a unique (continuous) extension of A to Xwhich has the same norm.Proof. Since a bounded operator maps Cauchy sequences to Cauchy se-quences, this extension can only be given by Af = lim Afn , fn ∈ D(A), f ∈ X. n→∞To show that this definition is independent of the sequence fn → f , letgn → f be a second sequence and observe Afn − Agn = A(fn − gn ) ≤ A fn − gn → 0.
  34. 34. 24 0. A first look at Banach and Hilbert spacesFrom continuity of vector addition and scalar multiplication it follows thatour extension is linear. Finally, from continuity of the norm we concludethat the norm does not increase. An operator in L(X, C) is called a bounded linear functional and thespace X ∗ = L(X, C) is called the dual space of X. A sequence fn is said toconverge weakly, fn f , if (fn ) → (f ) for every ∈ X ∗ . The Banach space of bounded linear operators L(X) even has a multi-plication given by composition. Clearly this multiplication satisfies (A + B)C = AC + BC, A(B + C) = AB + BC, A, B, C ∈ L(X) (0.60)and (AB)C = A(BC), α (AB) = (αA)B = A (αB), α ∈ C. (0.61)Moreover, it is easy to see that we have AB ≤ A B . (0.62)However, note that our multiplication is not commutative (unless X is one-dimensional). We even have an identity, the identity operator I satisfying I = 1. A Banach space together with a multiplication satisfying the above re-quirements is called a Banach algebra. In particular, note that (0.62)ensures that multiplication is continuous (Problem 0.22).Problem 0.19. Show that the integral operator 1 (Kf )(x) = K(x, y)f (y)dy, 0where K(x, y) ∈ C([0, 1] × [0, 1]), defined on D(K) = C[0, 1] is a boundedoperator both in X = C[0, 1] (max norm) and X = L2 (0, 1). cont dProblem 0.20. Show that the differential operator A = dx defined onD(A) = C 1 [0, 1] ⊂ C[0, 1] is an unbounded operator.Problem 0.21. Show that AB ≤ A B for every A, B ∈ L(X).Problem 0.22. Show that the multiplication in a Banach algebra X is con-tinuous: xn → x and yn → y imply xn yn → xy.Problem 0.23. Let ∞ f (z) = fj z j , |z| R, j=0
  35. 35. 0.6. Lebesgue Lp spaces 25be a convergent power series with convergence radius R 0. Suppose A isa bounded operator with A R. Show that ∞ f (A) = fj Aj j=0exists and defines a bounded linear operator (cf. Problem 0.9).0.6. Lebesgue Lp spacesFor this section some basic facts about the Lebesgue integral are required.The necessary background can be found in Appendix A. To begin with,Sections A.1, A.3, and A.4 will be sufficient. We fix some σ-finite measure space (X, Σ, µ) and denote by Lp (X, dµ),1 ≤ p, the set of all complex-valued measurable functions for which 1/p f p = |f |p dµ (0.63) Xis finite. First of all note that Lp (X, dµ) is a linear space, since |f + g|p ≤2p max(|f |, |g|)p ≤ 2p max(|f |p , |g|p ) ≤ 2p (|f |p + |g|p ). Of course our hopeis that Lp (X, dµ) is a Banach space. However, there is a small technicalproblem (recall that a property is said to hold almost everywhere if the setwhere it fails to hold is contained in a set of measure zero):Lemma 0.27. Let f be measurable. Then |f |p dµ = 0 (0.64) Xif and only if f (x) = 0 almost everywhere with respect to µ.Proof. Observe that we have A = {x|f (x) = 0} = n An , where An = 1{x| |f (x)| n }. If |f |p dµ = 0, we must have µ(An ) = 0 for every n andhence µ(A) = limn→∞ µ(An ) = 0. The converse is obvious. Note that the proof also shows that if f is not 0 almost everywhere,there is an ε 0 such that µ({x| |f (x)| ≥ ε}) 0.Example. Let λ be the Lebesgue measure on R. Then the characteristicfunction of the rationals χQ is zero a.e. (with respect to λ). Let Θ be the Dirac measure centered at 0. Then f (x) = 0 a.e. (withrespect to Θ) if and only if f (0) = 0. Thus f p = 0 only implies f (x) = 0 for almost every x, but not for all!Hence . p is not a norm on Lp (X, dµ). The way out of this misery is toidentify functions which are equal almost everywhere: Let N (X, dµ) = {f |f (x) = 0 µ-almost everywhere}. (0.65)
  36. 36. 26 0. A first look at Banach and Hilbert spacesThen N (X, dµ) is a linear subspace of Lp (X, dµ) and we can consider thequotient space Lp (X, dµ) = Lp (X, dµ)/N (X, dµ). (0.66)If dµ is the Lebesgue measure on X ⊆ Rn , we simply write Lp (X). Observethat f p is well-defined on Lp (X, dµ). Even though the elements of Lp (X, dµ) are, strictly speaking, equiva-lence classes of functions, we will still call them functions for notationalconvenience. However, note that for f ∈ Lp (X, dµ) the value f (x) is notwell-defined (unless there is a continuous representative and different con-tinuous functions are in different equivalence classes, e.g., in the case ofLebesgue measure). With this modification we are back in business since Lp (X, dµ) turnsout to be a Banach space. We will show this in the following sections. But before that let us also define L∞ (X, dµ). It should be the set ofbounded measurable functions B(X) together with the sup norm. The onlyproblem is that if we want to identify functions equal almost everywhere, thesupremum is no longer independent of the equivalence class. The solutionis the essential supremum f ∞ = inf{C | µ({x| |f (x)| C}) = 0}. (0.67)That is, C is an essential bound if |f (x)| ≤ C almost everywhere and theessential supremum is the infimum over all essential bounds.Example. If λ is the Lebesgue measure, then the essential sup of χQ withrespect to λ is 0. If Θ is the Dirac measure centered at 0, then the essentialsup of χQ with respect to Θ is 1 (since χQ (0) = 1, and x = 0 is the onlypoint which counts for Θ). As before we set L∞ (X, dµ) = B(X)/N (X, dµ) (0.68)and observe that f ∞ is independent of the equivalence class. If you wonder where the ∞ comes from, have a look at Problem 0.24. As a preparation for proving that Lp is a Banach space, we will needH¨lder’s inequality, which plays a central role in the theory of Lp spaces. oIn particular, it will imply Minkowski’s inequality, which is just the triangleinequality for Lp .Theorem 0.28 (H¨lder’s inequality). Let p and q be dual indices; that is, o 1 1 + =1 (0.69) p q
  37. 37. 0.6. Lebesgue Lp spaces 27with 1 ≤ p ≤ ∞. If f ∈ Lp (X, dµ) and g ∈ Lq (X, dµ), then f g ∈ L1 (X, dµ)and f g 1 ≤ f p g q. (0.70)Proof. The case p = 1, q = ∞ (respectively p = ∞, q = 1) follows directlyfrom the properties of the integral and hence it remains to consider 1 p, q ∞. First of all it is no restriction to assume f p = g q = 1. Then, usingthe elementary inequality (Problem 0.25) 1 1 a1/p b1/q ≤ a + b, a, b ≥ 0, (0.71) p qwith a = |f |p and b = |g|q and integrating over X gives 1 1 |f g|dµ ≤ |f |p dµ + |g|q dµ = 1 X p X q Xand finishes the proof. As a consequence we also getTheorem 0.29 (Minkowski’s inequality). Let f, g ∈ Lp (X, dµ). Then f +g p ≤ f p + g p. (0.72)Proof. Since the cases p = 1, ∞ are straightforward, we only consider 1 p ∞. Using |f + g|p ≤ |f | |f + g|p−1 + |g| |f + g|p−1 , we obtain fromH¨lder’s inequality (note (p − 1)q = p) o p f +g p ≤ f p (f + g)p−1 q + g p (f + g)p−1 q p−1 =( f p + g p ) (f + g) p . This shows that Lp (X, dµ) is a normed linear space. Finally it remainsto show that Lp (X, dµ) is complete.Theorem 0.30. The space Lp (X, dµ) is a Banach space.Proof. Suppose fn is a Cauchy sequence. It suffices to show that somesubsequence converges (show this). Hence we can drop some terms suchthat 1 fn+1 − fn p ≤ n . 2Now consider gn = fn − fn−1 (set f0 = 0). Then ∞ G(x) = |gk (x)| k=1
  38. 38. 28 0. A first look at Banach and Hilbert spacesis in Lp . This follows from n n 1 |gk | ≤ gk (x) p ≤ f1 p + p 2 k=1 k=1using the monotone convergence theorem. In particular, G(x) ∞ almosteverywhere and the sum ∞ gn (x) = lim fn (x) n→∞ n=1is absolutely convergent for those x. Now let f (x) be this limit. Since|f (x) − fn (x)|p converges to zero almost everywhere and |f (x) − fn (x)|p ≤2p G(x)p ∈ L1 , dominated convergence shows f − fn p → 0. In particular, in the proof of the last theorem we have seen:Corollary 0.31. If fn − f p → 0, then there is a subsequence which con-verges pointwise almost everywhere. Note that the statement is not true in general without passing to asubsequence (Problem 0.28). Using H¨lder’s inequality, we can also identify a class of bounded oper- oators in Lp .Lemma 0.32 (Schur criterion). Consider Lp (X, dµ) and Lq (X, dν) with1 1p + q = 1. Suppose that K(x, y) is measurable and there are measurablefunctions K1 (x, y), K2 (x, y) such that |K(x, y)| ≤ K1 (x, y)K2 (x, y) and K1 (x, .) q ≤ C1 , K2 (., y) p ≤ C2 (0.73)for µ-almost every x, respectively, for ν-almost every y. Then the operatorK : Lp (X, dµ) → Lp (X, dµ) defined by (Kf )(x) = K(x, y)f (y)dν(y) (0.74) Yfor µ-almost every x is bounded with K ≤ C1 C2 .Proof. Choose f ∈ Lp (X, dµ). By Fubini’s theorem Y |K(x, y)f (y)|dν(y)is measurable and by H¨lder’s inequality we have o |K(x, y)f (y)|dν(y) ≤ K1 (x, y)K2 (x, y)|f (y)|dν(y) Y Y 1/q 1/p ≤ K1 (x, y)q dν(y) |K2 (x, y)f (y)|p dν(y) Y Y 1/p ≤ C1 |K2 (x, y)f (y)|p dν(y) Y
  39. 39. 0.6. Lebesgue Lp spaces 29(if K2 (x, .)f (.) ∈ Lp (X, dν), the inequality is trivially true). Now take thisinequality to the p’th power and integrate with respect to x using Fubini p p |K(x, y)f (y)|dν(y) dµ(x) ≤ C1 |K2 (x, y)f (y)|p dν(y)dµ(x) X Y X Y p p p = C1 |K2 (x, y)f (y)|p dµ(x)dµ(y) ≤ C1 C2 f p p. Y XHence Y |K(x, y)f (y)|dν(y) ∈ Lp (X, dµ) and in particular it is finite forµ-almost every x. Thus K(x, .)f (.) is ν integrable for µ-almost every x and Y K(x, y)f (y)dν(y) is measurable. It even turns out that Lp is separable.Lemma 0.33. Suppose X is a second countable topological space (i.e., ithas a countable basis) and µ is a regular Borel measure. Then Lp (X, dµ),1 ≤ p ∞, is separable.Proof. The set of all characteristic functions χA (x) with A ∈ Σ and µ(A) ∞ is total by construction of the integral. Now our strategy is as follows:Using outer regularity, we can restrict A to open sets and using the existenceof a countable base, we can restrict A to open sets from this base. Fix A. By outer regularity, there is a decreasing sequence of open setsOn such that µ(On ) → µ(A). Since µ(A) ∞, it is no restriction to assumeµ(On ) ∞, and thus µ(On A) = µ(On ) − µ(A) → 0. Now dominatedconvergence implies χA − χOn p → 0. Thus the set of all characteristicfunctions χO (x) with O open and µ(O) ∞ is total. Finally let B be acountable basis for the topology. Then, every open set O can be written asO = ∞ Oj with Oj ∈ B. Moreover, by considering the set of all finite j=1 ˜ ˜unions of elements from B, it is no restriction to assume n Oj ∈ B. Hence j=1 ˜there is an increasing sequence On˜ ˜ O with On ∈ B. By monotone con-vergence, χO − χOn p → 0 and hence the set of all characteristic functions ˜χO with O ˜ ˜ ∈ B is total. To finish this chapter, let us show that continuous functions are densein Lp .Theorem 0.34. Let X be a locally compact metric space and let µ be aσ-finite regular Borel measure. Then the set Cc (X) of continuous functionswith compact support is dense in Lp (X, dµ), 1 ≤ p ∞.Proof. As in the previous proof the set of all characteristic functions χK (x)with K compact is total (using inner regularity). Hence it suffices to showthat χK (x) can be approximated by continuous functions. By outer regu-larity there is an open set O ⊃ K such that µ(OK) ≤ ε. By Urysohn’s
  40. 40. 30 0. A first look at Banach and Hilbert spaceslemma (Lemma 0.12) there is a continuous function fε which is 1 on K and0 outside O. Since |χK − fε |p dµ = |fε |p dµ ≤ µ(OK) ≤ ε, X OKwe have fε − χK → 0 and we are done. If X is some subset of Rn , we can do even better. A nonnegative function ∞u ∈ Cc (Rn ) is called a mollifier if u(x)dx = 1. (0.75) Rn 1The standard mollifier is u(x) = exp( |x|2 −1 ) for |x| 1 and u(x) = 0otherwise. If we scale a mollifier according to uk (x) = k n u(k x) such that its mass ispreserved ( uk 1 = 1) and it concentrates more and more around the origin, T u k Ewe have the following result (Problem 0.29):Lemma 0.35. Let u be a mollifier in Rn and set uk (x) = k n u(k x). Thenfor any (uniformly) continuous function f : Rn → C we have that fk (x) = uk (x − y)f (y)dy (0.76) Rnis in C ∞ (Rn ) and converges to f (uniformly). Now we are ready to proveTheorem 0.36. If X ⊆ Rn and µ is a regular Borel measure, then the set ∞Cc (X) of all smooth functions with compact support is dense in Lp (X, dµ),1 ≤ p ∞.Proof. By our previous result it suffices to show that any continuous func-tion f (x) with compact support can be approximated by smooth ones. Bysetting f (x) = 0 for x ∈ X, it is no restriction to assume X = Rn . Nowchoose a mollifier u and observe that fk has compact support (since f
  41. 41. 0.6. Lebesgue Lp spaces 31has). Moreover, since f has compact support, it is uniformly continuousand fk → f uniformly. But this implies fk → f in Lp . We say that f ∈ Lp (X) if f ∈ Lp (K) for any compact subset K ⊂ X. locLemma 0.37. Suppose f ∈ L1 (Rn ). Then loc ∞ ϕ(x)f (x)dx = 0, ∀ϕ ∈ Cc (Rn ), (0.77) Rnif and only if f (x) = 0 (a.e.).Proof. First of all we claim that for any bounded function g with compact ∞support K, there is a sequence of functions ϕn ∈ Cc (Rn ) with support inK which converges pointwise to g such that ϕn ∞ ≤ g ∞ . To see this, take a sequence of continuous functions ϕn with supportin K which converges to g in L1 . To make sure that ϕn ∞ ≤ g ∞ , justset it equal to g ∞ whenever ϕn g ∞ and equal to − g ∞ wheneverϕn g ∞ (show that the resulting sequence still converges). Finally use(0.76) to make ϕn smooth (note that this operation does not change therange) and extract a pointwise convergent subsequence. Now let K be some compact set and choose g = sign(f )χK . Then |f |dx = f sign(f )dx = lim f χn dx = 0, K K n→∞which shows f = 0 for a.e. x ∈ K. Since K is arbitrary, we are done.Problem 0.24. Suppose µ(X) ∞. Show that lim f p = f ∞ p→∞for any bounded measurable function.Problem 0.25. Prove (0.71). (Hint: Take logarithms on both sides.)Problem 0.26. Show the following generalization of H¨lder’s inequality: o fg r ≤ f p g q, (0.78) 1 1where p + q = 1. rProblem 0.27 (Lyapunov inequality). Let 0 θ 1. Show that if f ∈Lp1 ∩ Lp2 , then f ∈ Lp and θ 1−θ f p ≤ f p1 f p2 , (0.79) 1 θ 1−θwhere p = p1 + p2 .
  42. 42. 32 0. A first look at Banach and Hilbert spacesProblem 0.28. Find a sequence fn which converges to 0 in Lp ([0, 1], dx) butfor which fn (x) → 0 for a.e. x ∈ [0, 1] does not hold. (Hint: Every n ∈ N canbe uniquely written as n = 2m +k with 0 ≤ m and 0 ≤ k 2m . Now considerthe characteristic functions of the intervals Im,k = [k2−m , (k + 1)2−m ].)Problem 0.29. Prove Lemma 0.35. (Hint: To show that fk is smooth, useProblems A.7 and A.8.)Problem 0.30. Construct a function f ∈ Lp (0, 1) which has a singularity atevery rational number in [0, 1]. (Hint: Start with the function f0 (x) = |x|−αwhich has a single pole at 0. Then fj (x) = f0 (x − xj ) has a pole at xj .)0.7. Appendix: The uniform boundedness principleRecall that the interior of a set is the largest open subset (that is, the unionof all open subsets). A set is called nowhere dense if its closure has emptyinterior. The key to several important theorems about Banach spaces is theobservation that a Banach space cannot be the countable union of nowheredense sets.Theorem 0.38 (Baire category theorem). Let X be a complete metric space.Then X cannot be the countable union of nowhere dense sets.Proof. Suppose X = ∞ Xn . We can assume that the sets Xn are closed n=1and none of them contains a ball; that is, XXn is open and nonempty forevery n. We will construct a Cauchy sequence xn which stays away from allXn . Since XX1 is open and nonempty, there is a closed ball Br1 (x1 ) ⊆XX1 . Reducing r1 a little, we can even assume Br1 (x1 ) ⊆ XX1 . More-over, since X2 cannot contain Br1 (x1 ), there is some x2 ∈ Br1 (x1 ) that isnot in X2 . Since Br1 (x1 ) ∩ (XX2 ) is open, there is a closed ball Br2 (x2 ) ⊆Br1 (x1 ) ∩ (XX2 ). Proceeding by induction, we obtain a sequence of ballssuch that Brn (xn ) ⊆ Brn−1 (xn−1 ) ∩ (XXn ).Now observe that in every step we can choose rn as small as we please; hencewithout loss of generality rn → 0. Since by construction xn ∈ BrN (xN ) forn ≥ N , we conclude that xn is Cauchy and converges to some point x ∈ X.But x ∈ Brn (xn ) ⊆ XXn for every n, contradicting our assumption thatthe Xn cover X. (Sets which can be written as the countable union of nowhere dense setsare said to be of first category. All other sets are second category. Hencewe have the name category theorem.)
  43. 43. 0.7. Appendix: The uniform boundedness principle 33 In other words, if Xn ⊆ X is a sequence of closed subsets which coverX, at least one Xn contains a ball of radius ε 0. Now we come to the first important consequence, the uniform bound-edness principle.Theorem 0.39 (Banach–Steinhaus). Let X be a Banach space and Y somenormed linear space. Let {Aα } ⊆ L(X, Y ) be a family of bounded operators.Suppose Aα x ≤ C(x) is bounded for fixed x ∈ X. Then Aα ≤ C isuniformly bounded.Proof. Let Xn = {x| Aα x ≤ n for all α} = {x| Aα x ≤ n}. αThen n Xn = X by assumption. Moreover, by continuity of Aα and thenorm, each Xn is an intersection of closed sets and hence closed. By Baire’stheorem at least one contains a ball of positive radius: Bε (x0 ) ⊂ Xn . Nowobserve Aα y ≤ Aα (y + x0 ) + Aα x0 ≤ n + Aα x0 xfor y ε. Setting y = ε x , we obtain n + C(x0 ) Aα x ≤ x εfor any x.
  44. 44. Part 1MathematicalFoundations ofQuantum Mechanics
  45. 45. Chapter 1Hilbert spacesThe phase space in classical mechanics is the Euclidean space R2n (for the nposition and n momentum coordinates). In quantum mechanics the phasespace is always a Hilbert space H. Hence the geometry of Hilbert spacesstands at the outset of our investigations.1.1. Hilbert spacesSuppose H is a vector space. A map ., .. : H×H → C is called a sesquilinearform if it is conjugate linear in the first argument and linear in the second.A positive definite sesquilinear form is called an inner product or scalarproduct. Associated with every scalar product is a norm ψ = ψ, ψ . (1.1)The triangle inequality follows from the Cauchy–Schwarz–Bunjakowskiinequality: | ψ, ϕ | ≤ ψ ϕ (1.2)with equality if and only if ψ and ϕ are parallel. If H is complete with respect to the above norm, it is called a Hilbertspace. It is no restriction to assume that H is complete since one can easilyreplace it by its completion.Example. The space L2 (M, dµ) is a Hilbert space with scalar product givenby f, g = f (x)∗ g(x)dµ(x). (1.3) M 37