Spectral Theory in Hilbert Space

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16 3. HILBERT SPACE Given a normed space one may of course ask whether there is a scalar product on the space which gives rise to the given norm in the usual way. Here is a simple criterion. Lemma 3.3. (parallelogram identity) If u and v are elements of H, then u + v 2 + u − v 2 = 2u 2 + 2v 2 . Proof. A simple calculation gives u±v 2 = 〈u±v, u±v〉 = u 2 ± (〈u, v〉 + 〈v, u〉) + v 2 . Adding this for the two signs the parallelogram identity follows. The name parallelogram identity comes from the fact that the lemma can be interpreted geometrically, as saying that the sum of the squares of the lengths of the sides in a parallelogram equals the sum of the squares of the lengths of the diagonals. This is a theorem that can be found in Euclid’s Elements. Given a normed space, Lemma 3.3 shows that a necessary condition for the norm to be associated with a scalar product is that the parallelogram identity holds for all vectors in the space. It was proved by von Neumann in 1929 that this is also sufficient (Exercise 3.3). We shall soon have another use for the parallelogram identity. In practice it is quite common that one has a space with scalar product which is not complete (such a space is often called a pre-Hilbert space). In order to use Hilbert space theory, one must then embed the space in a larger space which is complete. The process is called completion and is fully analogous to the extension of the rational numbers to the reals, which is also done to make the Cauchy convergence principle valid. In very brief outline the process is as follows. Starting with a (not complete) normed linear space L let Lc be the set of all Cauchy sequences in L. The set Lc is made into a linear space in the obvious way. We may embed L in Lc by identifying u ∈ L with the sequence (u, u, u, . . . ). In Lc we may introduce a semi-norm · (i.e., a norm except that there may be non-zero elements u in the space for which u = 0) by setting (u1, u2, . . . ) = limuj. Now let Nc be the subspace of Lc consisting of all elements with semi-norm 0, and put H = Lc/Nc, i.e., elements in Lc are identified whenever the distance between them is 0. One may now prove that · induces a norm on H under which H is complete, and that through the identification above we may consider the original space L as a dense subset of H. If the original norm came from a scalar product, then so will the norm of H. We leave to the reader to verify the details, using the hints provided (Exercise 3.4). The process above is satisfactory in that it shows that any normed space may be ‘completed’ (in fact, the same process works in any metric space). Equivalence classes of Cauchy sequences are of course rather abstract objects, but in concrete cases one can often identify the elements
3. HILBERT SPACE 17 of the completion of a given space with more concrete objects. So, for example, one may view L 2 (Ω, µ) as the completion, in the appropriate norm, of the linear space C0(Ω) of functions which are continuous in Ω and 0 outside a compact subset of Ω. In the sequel H is always assumed to be a Hilbert space. There are two properties which make Hilbert spaces far more convenient to deal with than more general spaces. The first is that any closed, linear subspace has a topological complement which can be chosen in a canonical way (Theorem 3.7). The second, a Hilbert space can be identified with its topological dual (Theorem 3.8). Both these properties are actually true even if the space is not assumed separable (and of course if the space is finite-dimensional), as our proofs will show. To prove them we start with the following definition. Definition 3.4. A set M is called convex if it contains all linesegments connecting two elements of the set, i.e., if u and v ∈ M, then tu + (1 − t)v ∈ M for all t ∈ [0, 1]. A subset of a metric space is of course called closed if all limits of convergent sequences contained in the subset are themselves in the subset. It is easily seen that this is equivalent to the complement of the subset being open, in the sense that it is a neighborhood of all its points (check this!). Lemma 3.5. Any closed, convex subset K of H has a unique element of smallest norm. Proof. Put d = inf{u | u ∈ K}. Let u1, u2, . . . be a minimizing sequence, i.e., uj ∈ K and uj → d. By the parallelogram identity we then have uj − uk 2 = 2uj 2 + 2uk 2 − 4(uj + uk)/2 2 . On the right hand side the two first terms both tend to 2d 2 as j, k → ∞. By convexity (uj + uk)/2 ∈ K so the last term is ≥ 4d 2 . Therefore u1, u2, . . . is a Cauchy sequence, and has a limit u which obviously has norm d and is in K, since K is closed. If u and v are both minimizing elements, replacing uj by u and uk by v in the calculation above immediately shows that u = v, so the minimizing element is unique. Lemma 3.6. Suppose M is a proper ( i.e., M = H) closed, linear subspace of H. Then there is a non-trivial normal to M, i.e., an element u = 0 in H such that 〈u, v〉 = 0 for all v ∈ M. Proof. Let w /∈ M and put K = w + M. Then K is obviously closed and convex so it has a smallest element u which is non-zero since 0 /∈ K. Let v = 0 be in M so that u + av ∈ K for any scalar a. Hence u 2 ≤ u + av 2 = u 2 + 2 Re(a〈v, u〉) + |a| 2 v 2 . Setting a = −〈u, v〉/v 2 we obtain −(|〈u, v〉|/v) 2 ≥ 0 so that 〈u, v〉 = 0.
Page 1 and 2: Spectral Theory in Hilbert Space Le
Page 3: Preface The aim of these notes is t
Page 6 and 7: iv CONTENTS Exercises for Chapter 1
Page 8 and 9: 2 0. INTRODUCTION cos( √ λ t), i
Page 10 and 11: 4 0. INTRODUCTION • Can the equat
Page 12 and 13: 6 1. LINEAR SPACES of u so one may
Page 14 and 15: 8 1. LINEAR SPACES Exercises for Ch
Page 16 and 17: 10 2. SPACES WITH SCALAR PRODUCT on
Page 18 and 19: 12 2. SPACES WITH SCALAR PRODUCT Pr
Page 20 and 21: 14 2. SPACES WITH SCALAR PRODUCT Ex
Page 24 and 25: 18 3. HILBERT SPACE Two subspaces M
Page 26 and 27: 20 3. HILBERT SPACE ˆvn = limj→
Page 29 and 30: CHAPTER 4 Operators A bounded linea
Page 31 and 32: 4. OPERATORS 25 Proposition 4.2. A
Page 33 and 34: 4. OPERATORS 27 In the rest of this
Page 35 and 36: 4. OPERATORS 29 must both be 0. Hen
Page 37 and 38: CHAPTER 5 Resolvents We now conside
Page 39: EXERCISES FOR CHAPTER 5 33 Analytic
Page 42 and 43: 36 6. NEVANLINNA FUNCTIONS However,
Page 44 and 45: 38 6. NEVANLINNA FUNCTIONS Proof. B
Page 46 and 47: 40 7. THE SPECTRAL THEOREM function
Page 48 and 49: 42 7. THE SPECTRAL THEOREM as The r
Page 50 and 51: 44 7. THE SPECTRAL THEOREM as a clo
Page 52 and 53: 46 8. COMPACTNESS weakly convergent
Page 54 and 55: 48 8. COMPACTNESS only if g ∈ L 2
Page 57 and 58: CHAPTER 9 Extension theory We will
Page 59 and 60: 2. SYMMETRIC RELATIONS 53 We will n
Page 61 and 62: 2. SYMMETRIC RELATIONS 55 then it i
Page 63: EXERCISES FOR CHAPTER 9 57 Exercise
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66 10. BOUNDARY CONDITIONS uniforml
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68 10. BOUNDARY CONDITIONS Hint: If
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70 11. STURM-LIOUVILLE EQUATIONS fu
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72 11. STURM-LIOUVILLE EQUATIONS ob
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74 11. STURM-LIOUVILLE EQUATIONS Ta
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76 11. STURM-LIOUVILLE EQUATIONS Pr
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78 11. STURM-LIOUVILLE EQUATIONS No
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80 11. STURM-LIOUVILLE EQUATIONS Th
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CHAPTER 12 Inverse spectral theory
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1. ASYMPTOTICS OF THE m-FUNCTION 85
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2. UNIQUENESS THEOREMS 87 hand, the
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2. UNIQUENESS THEOREMS 89 the bound
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CHAPTER 13 First order systems We s
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u(c) = 0 is given by (13.2) u(x) =
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13. FIRST ORDER SYSTEMS 95 and (u2,
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13. FIRST ORDER SYSTEMS 97 a non-ze
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EXERCISES FOR CHAPTER 13 99 Exercis
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102 14. EIGENFUNCTION EXPANSIONS We
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104 14. EIGENFUNCTION EXPANSIONS Ex
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106 15. SINGULAR PROBLEMS We obtain
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108 15. SINGULAR PROBLEMS We will g
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110 15. SINGULAR PROBLEMS Proof. Le
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112 15. SINGULAR PROBLEMS in the pr
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114 15. SINGULAR PROBLEMS Since vK
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APPENDIX A Functional analysis In t
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A. FUNCTIONAL ANALYSIS 119 other wo
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122 B. STIELTJES INTEGRALS (4) C (5
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124 B. STIELTJES INTEGRALS Definiti
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126 B. STIELTJES INTEGRALS We obtai
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APPENDIX C Linear first order syste
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C. LINEAR FIRST ORDER SYSTEMS 131 s
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Spectral Theory in Hilbert Space

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