Spectral Theory in Hilbert Space

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118 A. FUNCTIONAL ANALYSIS Proposition A.2. A subset of a meager set is meager, a countable union of meager sets is meager, and no meager set has an interior point. Proof. The first two claims are left as exercises for the reader to verify; the third claim is Baire’s theorem. The following theorem is one of the cornerstones of functional analysis. Theorem A.3 (Banach). Suppose B1 and B2 are Banach spaces and T : B1 → B2 a bounded, injective (one-to-one) linear map. If the range of T is not meager, in particular if it is all of B2, then T has a bounded inverse, and the range is all of B2. Proof. We denote the norm in Bj by ·j. Let An = {T x | x1 ≤ n} be the image of the closed ball with radius n, centered at 0 in B1. The balls expand to all of B1 as n → ∞, so the range of T is ∪ ∞ n=1An ⊂ ∪ ∞ n=1An. The range not being meager, at least one An must have an interior point y0. Thus we can find r > 0 so that {y0 + y | y2 < r} ⊂ An. Since An is symmetric with respect to the origin, also −y0+y ∈ An if y2 < r. Furthermore, An is convex, as the closure of (the linear image of) a convex set. It follows that y = 1 2 ((y0 +y)+(−y0 +y)) ∈ An. Thus 0 is an interior point of An. Since all An are similar (An = nA1), 0 is also an interior point of A1. This means that there is a number C > 0, such that any y ∈ B2 for which y2 ≤ C is in A1. For such y we may therefore find x ∈ B1 with x1 ≤ 1, such that T x is arbitrarily close to y. For example, we may find x ∈ B1 with x1 ≤ 1 such that y − T x2 ≤ 1 2 C. For arbitrary non-zero y ∈ B2 we set ˜y = C y2 y, and then have ˜y2 = C, so we can find ˜x with ˜x1 ≤ 1 and ˜y − T ˜x2 ≤ 1 2 C. Setting x = y2 C ˜x we obtain (A.1) x1 ≤ 1 C y2 and y − T x2 ≤ 1 2y2. Thus, to any y ∈ B2 we may find x ∈ B1 so that (A.1) holds (for y = 0, take x = 0). We now construct two sequences {xj} ∞ j=0 and {yj} ∞ j=0, in B1 respectively B2, by first setting y0 = y. If yn is already defined, we define xn and yn+1 so that xn1 ≤ 1 C yn2, yn+1 = yn − T xn, and yn+12 ≤ 1 2yn2. We obtain yn2 ≤ 2−ny2 and xn1 ≤ 1 C 2−ny2 from this. Furthermore, T xn = yn+1 − yn, so adding we obtain T ( n j=0 xj) = y − yn+1 → y as n → ∞. But the series ∞ j=0xj1 converges, since it is dominated by 1 C y2 ∞ j=0 2−j = 2 C y2. Since B1 is complete, the series ∞ j=0 xj therefore converges to some x ∈ B1 satisfying x1 ≤ 2 C y2, and since T is continuous we also obtain T x = y. In
A. FUNCTIONAL ANALYSIS 119 other words, we can solve T x = y for any y ∈ B2, so the inverse of T is defined everywhere, and the inverse is bounded by 2 , so it is C continuous. The proof is complete. In these notes we do not actually use Banach’s theorem, but the following simple corollary (which is actually equivalent to Banach’s theorem). Recall that a linear map T : B1 → B2 is called closed if the graph {(u, T u) | u ∈ D(T )} is a closed subset of B1 ⊕B2. Equivalently, if uj → u in B1 and T uj → v in B2 implies that u ∈ D(T ) and T u = v. Corollary A.4 (Closed graph theorem). Suppose T is a closed linear operator T : B1 → B2, defined on all of B1. Then T is bounded. Proof. The graph {(u, T u) | u ∈ B1} is by assumption a Banach space with norm (u, T u) = u1 + T u2, where ·j denotes the norm of Bj. The map (u, T u) ↦→ u is linear, defined in this Banach space, with range equal to B1, and it has norm ≤ 1. It is obviously injective, so by Banach’s theorem the inverse is bounded, i.e., there is a constant so that (u, T u) ≤ Cu1. Hence also T u2 ≤ Cu1, so that T is bounded. In Chapter 3 we used the Banach-Steinhaus theorem, Theorem 3.10. Since no extra effort is involved, we prove the following slightly more general theorem. Theorem A.5 (Banach-Steinhaus; uniform boundedness principle). Suppose B is a Banach space, L a normed linear space, and M a subset of the set L(B, L) of all bounded, linear maps from B into L. Suppose M is pointwise bounded, i.e., for each x ∈ B there exists a constant Cx such that T xB ≤ Cx for every T ∈ M. Then M is uniformly bounded, i.e., there is a constant C such that T xB ≤ CxL for all x ∈ B and all T ∈ M. Proof. Put Fn = {x ∈ B | T xB ≤ n for all T ∈ M}. Then Fn is closed, as the intersection of the closed sets which are inverse images of the closed interval [0, n] under a continuous function B1 ∋ x ↦→ T xL ∈ R. The assumption means that ∪ ∞ n=1Fn = B. By Baire’s theorem at least one Fn must have an interior point. Since Fn is convex (if x, y ∈ Fn and 0 ≤ t ≤ 1, then tT x + (1 − t)T yL ≤ tT xL + (1 − t)T yL ≤ n) and symmetric with respect to the origin it follows, like in the proof of Banach’s theorem, that 0 is an interior point in Fn. Thus, for some r > 0 we have T xL ≤ n for all T ∈ M, if xB ≤ r. By homogeneity follows that T xL ≤ n r xB for all T ∈ M and x ∈ B.
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Spectral Theory in Hilbert Space Le
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Preface The aim of these notes is t
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iv CONTENTS Exercises for Chapter 1
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2 0. INTRODUCTION cos( √ λ t), i
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4 0. INTRODUCTION • Can the equat
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6 1. LINEAR SPACES of u so one may
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8 1. LINEAR SPACES Exercises for Ch
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10 2. SPACES WITH SCALAR PRODUCT on
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12 2. SPACES WITH SCALAR PRODUCT Pr
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14 2. SPACES WITH SCALAR PRODUCT Ex
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16 3. HILBERT SPACE Given a normed
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18 3. HILBERT SPACE Two subspaces M
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20 3. HILBERT SPACE ˆvn = limj→
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CHAPTER 4 Operators A bounded linea
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4. OPERATORS 25 Proposition 4.2. A
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4. OPERATORS 27 In the rest of this
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4. OPERATORS 29 must both be 0. Hen
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CHAPTER 5 Resolvents We now conside
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EXERCISES FOR CHAPTER 5 33 Analytic
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36 6. NEVANLINNA FUNCTIONS However,
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38 6. NEVANLINNA FUNCTIONS Proof. B
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40 7. THE SPECTRAL THEOREM function
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42 7. THE SPECTRAL THEOREM as The r
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44 7. THE SPECTRAL THEOREM as a clo
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46 8. COMPACTNESS weakly convergent
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48 8. COMPACTNESS only if g ∈ L 2
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CHAPTER 9 Extension theory We will
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2. SYMMETRIC RELATIONS 53 We will n
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2. SYMMETRIC RELATIONS 55 then it i
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EXERCISES FOR CHAPTER 9 57 Exercise
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60 10. BOUNDARY CONDITIONS the same
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62 10. BOUNDARY CONDITIONS locally
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64 10. BOUNDARY CONDITIONS of a sym
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66 10. BOUNDARY CONDITIONS uniforml
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Page 89 and 90: CHAPTER 12 Inverse spectral theory
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Page 105: EXERCISES FOR CHAPTER 13 99 Exercis
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Page 123: APPENDIX A Functional analysis In t
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