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Veral Versions of the Spectral Theorem for Normal Operators in Hilbert Spaces

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Veral Versions of the Spectral Theorem for Normal Operators in Hilbert Spaces 10. The Spectral Theorem The big moment has arrived, and we are now ready to prove se- veral versions of the spectral theorem for normal operators in Hilbert spaces. Throughout this chapter, it should be helpful to compare our results with the more familiar special case when the Hilbert space is finite-dimensional. In this setting, the spectral theorem says that every normal matrix T 2 Cn×n can be diagonalized by a unitary transfor- mation. This can be rephrased as follows: There are numbers zj 2 C n (the eigenvalues) and orthogonal projections Pj 2 B(C ) such that Pm T = j=1 zjPj. The subspaces R(Pj) are orthogonal to each other. From this representation of T , it is then also clear that Pj is the pro- jection onto the eigenspace belonging to zj. In fact, we have already proved one version of the (general) spec- tral theorem: The Gelfand theory of the commutative C∗-algebra A ⊆ B(H) that is generated by a normal operator T 2 B(H) provides a functional calculus: We can define f(T ), for f 2 C(σ(T )) in such a way that the map C(σ(T )) ! A, f 7! f(T ) is an isometric ∗-isomorphism between C∗-algebras, and this is the spectral theorem in one of its ma- ny disguises! See Theorem 9.13 and the discussion that follows. As a warm-up, let us use this material to give a quick proof of the result about normal matrices T 2 Cn×n that was stated above. Consider the C∗-algebra A ⊆ Cn×n that is generated by T . Since T is normal, A is commutative. By Theorem 9.13, A ∼= C(σ(T )) = C(fz1; : : : ; zmg), where z1; : : : ; zm are the eigenvalues of T . We also use the fact that by Theorem 9.16, σA(T ) = σB(H)(T ). All subsets of the discrete space fz1; : : : ; zmg are open, and thus all functions f : fz1; : : : ; zmg ! C are continuous. We will make use of the functional calculus notation: f(T ) 2 A will denote the operator that corresponds to the function f under the isometric ∗-isomorphism that sends the identity function id(z) = z to T 2 A. Write fj = χfzj g 2 ∗ and let Pj = fj(T ). Since fj = fj and fj = fj, we also have Pj = Pj 2 and Pj = Pj, so each Pj is an orthogonal projection by Theorem 6.5. Furthermore, fjfk = 0 if j 6= k, so PjPk = 0, and thus hPjx; Pkyi = hx; PjPkyi = 0 for all x; y 2 H if j 6= k. This says that R(Pj) ? R(Pk) for j 6= k. Also, P1 + ::: + Pm = 1 because we have the same identity for the fj's. It Lm n Pm follows that j=1 R(Pj) = H = C . Finally, since id = j=1 zjfj, we Pm obtain the representation T = j=1 zjPj, as asserted. On infinite-dimensional Hilbert spaces, we have a continuous analog of this representation: every normal T 2 B(H) can be written as T = 105 106 Christian Remling R z dP (z). We first need to address the question of how such an integral can be meaningfully defined. We will also switch to the more common symbol E (rather than P ) for these \measures" (if that's what they are). Definition 10.1. Let M be a σ-algebra on a set Ω, and let H be a Hilbert space. A resolution of the identity (or spectral resolution) on (Ω; M) is a map E : M! B(H) with the following properties: (1) Each E(!)(! 2 M) is a projection; (2) E(;) = 0, E(Ω) = 1; (3) E(!1 \ !2) = E(!1)E(!2)(!1;!2 2 M); (4) For all x; y 2 H, the set function µx;y(!) = hx; E(!)yi is a complex measure on (Ω; M). If Ω is a locally compact Hausdorff space and M = B is the Borel σ-algebra, then we also demand that every µx;y is a regular (Borel) measure. We can think of E as a projection valued measure (of sorts) on (Ω; M): the \measure" E(!) of a set ! 2 M is a projection. The E(!) are also called spectral projections. Let's start out with some quick observations. For every x 2 H, we have 2 2 µx;x(!) = hx; E(!)xi = hx; E(!) xi = hE(!)x; E(!)xi = kE(!)xk ; 2 so µx;x is a finite positive measure with µx;x(Ω) = kxk . Property (3) implies that any two spectral projections E(!1), E(!2) commute. Mo- reover, if !1 ⊆ !2, then R(E(!1)) ⊆ R(E(!2)). If !1 \ !2 = ;, then R(E(!1)) ? R(E(!1)), as the following calculation shows: hE(!1)x; E(!2)yi = hx; E(!1)E(!2)yi = hx; E(!1 \ !2)yi = 0 for arbitrary x; y 2 H. E is finitely additive: If !1;:::;!n 2 M are disjoint sets, then Sn Pn E j=1 !j = j=1 E(!j). To prove this, notice that (4) implies that n ! n ! n [ [ X hx; E !j yi = µx;y !j = µx;y(!j) j=1 j=1 j=1 n n X X = hx; E(!j)yi = hx; E(!j)yi j=1 j=1 for arbitrary x; y 2 H, and this gives the claim. Is E also σ-additive (as it ought to be, if we are serious about inter- preting E as a new sort of measure)? In other words, if !n 2 M are The Spectral Theorem 107 disjoint sets, does it follow that ! 1 [ X (10.1) E !n = E(!n): n2N n=1 The answer to this question depends on how one defines the right- hand side of (10.1). We observe that if E(!n) 6= 0 for infinitely many n, then this series can never be convergent in operator norm. Indeed, kE(!n)k = 1 if E(!n) 6= 0, and thus the partial sums do not form a Cauchy sequence. However, (10.1) will hold if we are satisfied with strong operator convergence: We say that Tn 2 B(H) converges strongly s to T 2 B(H) (notation: Tn −! T ) if Tnx ! T x for all x 2 H. To prove that (10.1) holds in this interpretation, fix x 2 H and use the fact that the E(!n)x form an orthogonal system (because the ranges of the projections are orthogonal subspaces for disjoint sets). We normalize the non-zero vectors: let y = E(!n)x if E(! )x 6= 0. n kE(!n)xk n Then the yn form an ONS, and thus, by Theorem 5.15, the series P P hyn; xiyn = E(!n)x converges. Now if y 2 H is arbitrary, then the continuity of the scalar product and the fact that µx;y is a complex measure show that 1 1 ! X X [ hy; E(!n)xi = hy; E(!n)xi = hy; E !n xi: n=1 n=1 n2N P1 Since this holds for every y 2 H, it follows that n=1 E(!n)x = E S ! x, and this is (10.1), with the series interpreted as a strong n2N n operator limit. Definition 10.2. A set N 2 M with E(N) = 0 is called an E-null set. We define L1(Ω;E) as the set of equivalence classes of measurable, essentially bounded functions f :Ω ! C. Here, f ∼ g if f and g agree off an E-null set. Also, as usual, we say that f is essentially bounded if jf(x)j ≤ M (x 2 Ω n N) for some M ≥ 0 and some E-null set N ⊆ Ω. Exercise 10.1. Prove that a countable union of E-null sets is an E-null set. Recall that for an arbitrary positive measure µ on X, the space L1(X; µ) only depends on what the µ-null sets are and not on the specific choice of the measure µ. For this reason and because of Exercise 10.1, we can also, and in fact without any difficulty, introduce L1 spaces that are based on resolutions of the identity. These spaces have the same basic properties: L1(Ω;E) with the essential supremum of jfj as the norm and the involution f ∗(x) = f(x) is a commutative 108 Christian Remling C∗-algebra. The spectrum of a function f 2 L1(Ω;E) is its essential range. Exercise 10.2. Write down precise definitions of the essential supremum and the essential range of a function f 2 L1(Ω;E). R 1 We would like to define an integral Ω f(t) dE(t) for f 2 L (Ω;E). This integral should be an operator from B(H), and it also seems reasonable to demand that Z Z hx; f(t) dE(t) yi = f(t) dµx;y(t) Ω Ω for all x; y 2 H. It is clear that this condition already suffices to un- R iquely determine Ω f(t) dE(t), should such an operator indeed exist. As for existence, we have the following result; it will actually turn out that the integral with respect to a resolution of the identity has many other desirable properties, too. Theorem 10.3. Let E be a resolution of the identity. Then there exists a unique map Ψ: L1(Ω;E) ! A onto a C∗-subalgebra A ⊆ B(H), such that Z (10.2) hx; Ψ(f)yi = f(t) dµx;y(t) Ω for all f 2 L1(Ω;E), x; y 2 H.
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