37.4 The Borel Functional Calculus

Proof. Firstly, let $x, y \in H$, $\seqf{B_j}\subset \cb_{X}$ be disjoint Borel sets, and $B = \bigsqcup_{j = 1}^{n} B_{j}$, then for each $1 \le i < j \le n$, $E(B_{i})(H) \perp E(B_{j})(H)$, so by the Cauchy-Schwarz inequality and the Pythagorean Theorem,

As the above holds for all finite sequences of disjoint Borel sets, $\norm{E_{x, y}}_{C(X; \complex)^*}\le \norm{x}_{H} \norm{y}_{H}$. Thus for any $\phi \in C(X; \complex)^{**}$,

Since the above holds for all $x, y \in H$, $I_{E}(\phi) \in B(H)$ with $\norm{I_E(\phi)}_{B(H)}\le \norm{\phi}_{C(X; \complex)^{**}}$.

(1): For each $x, y \in H$, $E_{x, y}\in C(X; \complex)^{*}$. Since $\angles{\int \phi dE \cdot x, y}_{H}= \dpn{E_{x, y}, \phi}{C(X; \complex)^*}$ for every $\phi \in C(X; \complex)^{**}$, $I_{E}$ is continuous from the weak* topology on $C(X; \complex)^{**}$ to the weak operator topology on $B(H)$.

(2): By Lemma 26.6.4, the simple functions $\Sigma(X; \complex)$ are uniformly dense in the bounded Borel functions $B^{\infty}(X; \complex)$. Since

1. (i)

   $I_{E}$ restricted to $\Sigma(X; \complex)$ is a *-homomorphism.

2. (ii)

   Multiplication and conjugation are continuous in the uniform norm on $B^{\infty}(X; \complex)$

3. (iii)

   Composition and transposition are continuous in the operator norm on $B(H)$

the map $I_{E}$ restricted to $B^{\infty}(X; \complex)$ is a *-homomorphism by continuity. By Goldstine’s Theorem, $C(X; \complex) \subset B^{\infty}(X; \complex)$ is weak*-dense in $C(X; \complex)^{**}$. So as

1. (i)

   $I_{E}$ restricted to $B^{\infty}(X; \complex)$ is a *-homomorphism.

2. (ii)

   The involution $\phi \mapsto \ol \phi$ is weak*-continuous on $C(X; \complex)^{**}$.

3. (iii)

   The transpose $T \mapsto T^{*}$ is weak-operator continuous on $B(H)$.

4. (iv)

   The product $(\phi, \psi) \mapsto \phi \psi$ is separately weak*-continuous on $C(X; \complex)^{**}$.

5. (v)

   The composition $(S, T) \mapsto ST$ is separately weak-operator continuous on $B(H)$.

the map $I_{E}$ is a *-homomorphism by (1). Finally, since $E(X) = I_{B(H)}$, $I_{E}$ is a unital *-homomorphism.$\square$

Proof, [Theorem 20.2, Zhu93]. (1): By the Gelfand-Naimark Theorem, $\Gamma_{A}: A \to C(\Omega(A); \complex)$ is a unital *-isomorphism. For each $x, y \in H$, $\Gamma_{A}^{-1}$ induces a mapping

which, by the Riesz Representation Theorem, takes the form of a complex Radon measure on $\Omega(A)$. Thus by the uniqueness part of the Riesz Representation Theorem, such a spectral measure must be unique if it exists.

Since $\norm{E_{x, y}}_{C(\Omega(A); \complex)^*}\le \norm{x}_{H}\norm{y}_{H}$ for all $x, y \in H$, $\bracsn{E_{x, y}|x, y \in H}$ induces a bounded linear map

with $I_{E}(f) = \Gamma_{A}^{-1}(f)$ for all $f \in C(\Omega(A); \complex)$.

For any $C \in \cb_{\Omega(A)}$, $\one_{C}$ is a projection in $B^{\infty}(\Omega(A); \complex)$. So to see that

defines a spectral measure, it is sufficient to show that $I_{E}|_{B^\infty(\Omega(A); \complex)}$ is a *-homomorphism.

Let $x, y \in H$, then as $\Gamma_{A}$ is a *-isomorphism, for any $f, g \in C(\Omega(A); \complex)$,

As the above holds for all $g \in C(\Omega(A); \complex)$, $fE_{x, y}= E_{x, I_E(f)^*y}$. Now, fix $\phi \in B^{\infty}(\Omega(A); \complex)$, then for every $f \in C(\Omega(A); \complex)$,

so $\phi E_{x, y}= E_{I_E(\phi)x, y}$ for all $\phi \in B^{\infty}(\Omega(A); \complex)$. Thus for any $\phi, \psi \in B^{\infty}(\Omega(A); \complex)$,

and $I_{E}|_{B^\infty(\Omega(A); \complex)}$ is a homomorphism^[1].

Finally, let $f \in C(\Omega(A); \real)$, then since $\Gamma_{A}$ is a *-isomorphism, $I_{E}(f) = \Gamma_{A}^{-1}(f)$ is self-adjoint. As such, for any $x \in H$, $\dpn{f, E_{x, x}}{C(\Omega(A); \complex)}= \dpn{I_E(f)x, x}{H}\in \real$, so $E_{x, x}$ is real-valued. Thus for any $\phi \in B^{\infty}(\Omega(A); \real)$ and $x \in H$, $\dpn{I_E(\phi)x, x}{H}= \dpn{E_{x, x}, \phi}{C(\Omega(A); \complex)^*}\in \real$ as well. Therefore $I_{E}(\phi)$ is self-adjoint, and $I_{E}|_{B^\infty(\Omega(A); \complex)}$ is a *-homomorphism.

(2): By Goldstine’s Theorem, $C(\Omega(A); \complex)$ is weak*-dense in $C(\Omega(A); \complex)^{**}$. Since the mapping $\phi \mapsto \int_{\Omega(A)}\phi dE$ is continuous from the weak* topology on $C(\Omega(A); \complex)^{**}$ to the weak operator topology on $B(H)$,

by Proposition 5.5.3.

On the other hand, by the Banach-Alaoglu Theorem, $\ol{B_{C(\Omega(A); \complex)^{**}}(0, 1)}$ is weak*-compact, so

is weak-operator compact. As $\Gamma_{A}: A \to C(\Omega(A); \complex)$ is an isometric isomorphism, $S \supset \ol{B_A(0, 1)}$. By the Kaplansky Density Theorem, $S \supset B_{B}(0, 1)$, and

$\square$

Proof. By the Spectral Theorem applied to $B(H)[T]$, there exists a unique spectral measure $E$ on $\sigma_{B(H)}(T)$ such that the mapping

extends the inverse Gelfand transform $\Gamma_{B(H)[T]}^{-1}: C(\sigma_{B(H)}(T); \complex) \to B(H)[T]$.

For each $\phi \in C(\sigma_{B(H)}(T); \complex)^{**}$, let $\phi(T) = \int_{\sigma_{B(H)}(T)}\phi dE$, then the mapping $\phi \mapsto \phi(T)$ is continuous from the weak* topology on $C(\sigma_{B(H)}(T); \complex)^{**}$ to the weak operator topology on $A$ by Definition 37.4.2.

Finally, by uniqueness of the continuous functional calculus, Goldstine’s Theorem, and (2), the mapping $\phi \mapsto \phi(T)$ is unique.$\square$