19. From a resolution of the identity to operators - and bac

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1 19. From a resolution of the identity to operators - and back matkt@imf.au.dk Institut for Matematiske Fag Det Naturvidenskabelige Fakultet Aarhus Universitet December 2005

2 We read Sections in Rudins book - with some changes.

3 Approximation by simple functions - the statement Let Ω be a set, and M a σ-algebra of sets in Ω. Let H be a Hilbert space and E : M B(H) a resolution of the identity. Recall that an M-measurable function g : Ω C is simple when g(ω) is finite. Note that every simple M-measurable function is essentially bounded. Lemma Let f L (E). For every ǫ > 0 there is a simple M-measurable function s : Ω C such that f s ǫ.

4 Approximation by simple functions - the proof Proof. Let M 1,M 2,...,M N be disjoint non-empty Borel sets in C such that i) λ µ ǫ when λ,µ M i,i = 1,2,...,N, ii) N i=1 M i = {λ C : λ f }. Choose an element µ i M i for each i, and g = N µ i 1 ωi, i=1 where ω i = f 1 (M i ) M. Note that g is simple. Then f (x) g(x) ǫ for all x Ω\f 1 ({λ C : λ > f }). Since (f g) 1 ({λ C : λ > ǫ}) f 1 ({λ C : λ > f }) we find that E ( (f g) 1 ({λ C : λ > ǫ}) ) = 0. Hence f g ǫ.

5 L (E) is a C -algebra Lemma L (E) is a C -algebra in the norm and with involution f (x) = f (x). Proof. We leave the reader to check that fg f g. (1) (To get some inspiration (if needed), the reader should look at the proof of the triangle-inequality for.) It remains then only to check the C -identity. Observe that the involution is clearly isometric: f = f. (2)

6 L (E) is a C -algebra Proof. On the other hand {x Ω : f (x) > } ff = { } x Ω : f (x)f (x) > ff = {x Ω : ff (x) > ff }, so that E ({x Ω : f (x) > }) ff = 0. Hence f ff, or f 2 ff. Combined with (1) and (2) we find that f 2 ff f f = f 2.

7 From a resolution of the identity to operators - Theorem Theorem There is an isometric -homomorphism Ψ : L (E) B(H) such that Ψ(f )x,x = f de x (3) for all x H. An operator T B(H) commutes with every operator in Ψ (L (E)) if and only if T commutes with E(ω) for all ω M. Proof. Let f L (E). For x,y H, set L(x,y) = i k k=1 Ω Ω f de x+i k y. (4)

8 From a resolution of the identity to operators - proof of Theorem Proof. We check first that L is sesquilinear. By Lemma 1 we can find a sequence {g n } of simple functions such that lim n f g n = 0. Since Ω f de x+i k y Ω g n de x+i k y Ω f g n de x+i k y f g n E x+i k y (Ω) f g n x + i k y 2 for all x,y,k, we conclude that L(x,y) = lim L n(x,y), n where L n (x,y) = k=1 ik Ω g n de x+i k y. To prove that L is sesqui-linear we may assume that f is simple, say f = N α j 1 ωj. j=1

9 From a resolution of the identity to operators - proof of Theorem Then L(x,y) = 1 4 N 4 k=1 ik Ω i=1 α j1 ωj de x+i k y = N i=1 α j k=1 ik Ω 1 ω j de x+i k y = N i=1 α j k=1 ik E x+i k y (ω j) = N i=1 α j k=1 ik E(ω j )(x + i k y),x + i k y. We calculate 4 i k E(ω j )(x + i k y),x + i k y = k=1 E(ω j )x,x + E(ω j )y,y + E(ω j )y,x + E(ω j )x,y E(ω j )x,x E(ω j )y,y + E(ω j )y,x + E(ω j )x,y + i E(ω j )x,x + i E(ω j )y,y + ii E(ω j )y,x + i( i) E(ω j )x,y i E(ω j )x,x i E(ω j )y,y i( i) E(ω j )y,x ii E(ω j )x,y = 4 E(ω j )x,y (5)

10 From a resolution of the identity to operators - proof of Theorem It follows that L(x,y) = N i=1 α j E(ω j )x,y, which is clearly sesqui-linear. When x 1 and y 1, we find that L(x,y) k=1 Ω f de x+i k y k=1 f E x+i k y (Ω) 1 4 f 4 k=1 x + i k y f 4 k=1 22 = 4 f. It follows now from Theorem 12.8 in Rudin that there is an operator Ψ(f ) B(H) such that Ψ(f ) 4 f and L(x,y) = Ψ(f )x,y (6) for all x,y H. To prove that Ψ is linear we calculate Ψ(f + zg)x,y = k=1 ik Ω f + zg de x+i k y = k=1 ik Ω f de x+i k y + z k=1 ik Ω g de x+i k y = Ψ(f )x,y + z Ψ(g)x,y = (Ψ(f ) + zψ(g)) x,y. Since this holds for all x, y H we conclude that Ψ(f ) + zψ(g) = Ψ(f + zg), proving the linearity of Ψ.

11 From a resolution of the identity to operators - proof of Theorem By using (5) we find that Ψ (1 ω ) x,x = k=1 ik Ω 1 ω de x+i k x = k=1 ik E(ω)(x + i k x),x + i k x = E(ω)x,x. Since this holds for all x H we conclude from the corollary to Theorem 12.7 that Ψ (1 ω ) = E(ω), ω M. (7) Thus, when f = N j=1 α j1 ωj is a simple function we find that Ψ (f )x,x = N j=1 α ( ) j Ψ 1ωj x,x = N j=1 α j E(ω j )x,x = Ω f de x. To prove that this equality holds for arbitrary f L (E), choose a sequence {g n } of simple functions such that lim n f g n = 0. By linearity and continuity of Ψ and Ω be find that Ψ (f ) x,x = lim n Ψ (g n )x,x = lim n Ω g n de x = Ω f de x. Thus (3) holds.

12 From a resolution of the identity to operators - proof of Theorem Note that Ψ (1 ω ) Ψ (1 ω ) = E(ω)E(ω ) = E(ω ω ) = Ψ (1 ω ω ) = Ψ (1 ω 1 ω ) for all ω,ω. It follows by linearity that Ψ(fg) = Ψ(f )Ψ(g) for all simple functions f,g. It follows by continuity (and Lemma 1) that Ψ(fg) = Ψ(f )Ψ(g) for all f,g L (E). When f = N j=1 α j1 ωj is a simple function we find that ( N ) Ψ(f ) = Ψ j=1 α j1 ωj = N j=1 α jψ ( 1 ωj ) = Ψ(f ). It follows by continuity (and Lemma 1) that Ψ(f ) = Ψ(f ) for all f L (E). We have shown that Ψ is a -homomorphism.

13 From a resolution of the identity to operators - proof of Theorem Let f L (E),x H. Then Ψ(f )x 2 = Ψ(f )x,ψ(f )x = Ψ(f ) Ψ(f )x,x = Ψ(f f )x,x = Ω f 2 de x f 2 E x(ω) f 2 x 2. It follows that Ψ(f ) f. To show that we actually have equality here, consider first a simple function f = N j=1 α j1 ωj with ω i ω j = when i j. Then f = α j0, where α j0 = max { α j : E (ω j ) 0}. Let y be a unit vector in the subspace E (ω j0 ) H ) of H. Then Ψ(f )y = Ψ(f )E (ω j0 )y = Ψ (f 1 ωj0 y = ) ) Ψ (α j0 1 ωj0 y = α j0 Ψ (1 ωj0 y = α j0 E (ω j0 ) y = α j0 y = α j0. Hence Ψ(f ) = f when f is a simple function.

14 From a resolution of the identity to operators - proof of Theorem That equality holds generally follows by approximation: Choose a sequence {g n } of simple functions such that lim n f g n = 0. Then f = lim n g n and Ψ(f ) = lim n Ψ(g n ). Since Ψ(g n ) = g n for all n, we conclude that Ψ(f ) = f. It remains now only to prove the statements about commotators: If T B(H) commutes with all E(ω), we see immediately that T commutes with Ψ(g) when is simple. The general case follows by an obvious approximation argument; viz. TΨ(f ) = lim n TΨ(g n) = lim n Ψ(g n)t = Ψ(f )T. The converse is trivial: If T commutes with all Ψ(f ), it commutes in particular with E(ω) = Ψ (1 ω ).

15 - and back: A resolution of the identity from an abelian C -subalgebra of B(H) When E is a resolution of the identity and f L (E), we write in the followinf f de for the operator Ψ(f ) constructed above. Theorem Ω (Theorem 12.22) Let A be an abelian C -subalgebra of B(H), and let be the Gelfand spectrum of A (i.e. the character space of A). (a) There is a resolution of the identity E defined on the Borel σ-algebra of such that T = T de, where T C() is the Gelfand transform of T.

16 Theorem Theorem (b) E (ω) 0 for every non-empty open set ω. (c) An operator S B(H) commutes with every T A if and only if S commutes with every E(ω). Proof of Theorem 12.22: Let Φ : C() A B(H) be the inverse of the Gelfand transform. For every x H we obtain a linear and positive functional l x : C() C such that l x (f ) = Φ(f )x,x. By the Riesz representation theorem there is a meause µ x defined on the Borel σ-algebra B of such that l x (f ) = Φ(f )x,x = f dµ x.

17 Theorem Let ψ : C be a bounded Borel function. Define L ψ : H H C such that L ψ (x,y) = 1 4 i k ψ dµ 4 x+i k y. k=1 We check that L ψ is sesqui-linear. For this we will use some of properties of the measures µ x which were emphasized in the statement of the Riesz representation theorem - in the following abbreviated to RRT. - The proof of sesqui-linearity is based on the following Assertion A: Let x 1,x 2,...,x m be a finite set of vectors in H, and let f : C be a bounded Borel function. There is then a sequence {g n } C() such that lim n for all i = 1,2,...,m. g n dµ xi = f dµ xi

18 Theorem Let K be a compact subset and let ǫ > 0. It follows from RRT that there are open sets U i,i = 1,2,...,m, in such that K U = m i=1 U i and µ xi (U) µ xi (K) + ǫ for all i = 1,2,...,m. By Urysohn s lemma there is a g C() such that K g U. Then 0 g 1 K dµ xi 1 U 1 K dµ xi = µ xi (U) µ xi (K) ǫ (8) for all i = 1,2,...,m. Let then B be a Borel subset. Since µ xi (B) µ xi () = l xi (1) = x i 2 <, it follows from RRT that there is a compact subset K = K 1 K 2 K m B such that µ xi (B) ǫ µ xi (K i ) µ xi (K) for all i. If now g C() satisfies (8) we find that 1 B g dµ xi 1 B 1 K dµ xi + 1 K g dµ xi 1 K 1 B dµ xi + ǫ µ xi (B) µ xi (K) + ǫ 2ǫ for all i = 1,2,...,m.

19 Theorem Let now f = N i=1 α j1 Bj be a simple Borel function. It follows from the above that we can find continuous functions g 1,g 2,...,g N such that 1 B j dµ xi g j dµ xi 2ǫ for all i,j. Then for all i. f dµ xi N α j g j dµ xi j=1 N α j 2ǫ Since ǫ > 0 was arbitrary we conclude that Assertion A is true when f is simple Borel function. It follows then from Lemma 1 that Assertion A is true for all bounded Borel functions: Let f : C be a bounded Borel function. Apply the version of Lemma 1 for measures instead of resolutions of the identity to get a sequence {g n } of simple Borel functions such that the essential supremum g n f with respect to the measure m i=1 µ x i converges to 0 as n. j=1

20 Theorem Since g n dµ xi f dµ x i gn f µ xi () g n f x i 2, we find that lim n g n dµ xi f dµ xi = 0. This establishes Assertion A. Assertion A gives the sesqui-linearity of L ψ in the following way: Let x,y,z H and α C, we use Assertion A to get a sequence {g n } in C() such that and lim n lim n lim n g n dµ (x+αy)+i k z = g n dµ x+i k z = g n dµ y+i k z = f dµ (x+αy)+i k z, f dµ x+i k z f dµ y+i k z.