Theorem 39.5.4 (Spectral Theorem I).label Let $H$ be a complex Hilbert space and $A \subset B(H)$ be a commutative $C^{*}$-subalgebra with $I \in A$, then:
- (1)
There exists a unique spectral measure $E: \cb_{\Omega(A)}\to B(H)$ such that[1]
\[T = \int_{\Omega(A)}\Gamma_{A} T dE \quad \forall T \in A\] - (2)
Let $B \subset B(H)$ be the strong-operator closure of $A$ in $B(H)$, and $\mathscr{E}\subset M_{R}(\Omega(A); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, then
\[I_{E}: \mathscr{E}^{*} \to B \quad \phi \mapsto \int_{\Omega(A)}\phi dE\]is a *-isomorphism.
The measure $E$ is the spectral measure associated with $A$, and the homomorphism $I_{E}$ is the extended inverse Gelfand transform of $A$.
Proof, [Theorem 20.2, Zhu93]. (1): By the Gelfand-Naimark Theorem, $\Gamma_{A}: A \to C(\Omega(A); \complex)$ is a *-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 $J_{E}(f) = \Gamma_{A}^{-1}(f)$ 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 $J_{E}$ 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 $g \in C(\Omega(A); \complex)$, $fE_{x, y}= E_{x, J_E(f)^*y}$. Now, fix $\phi \in B^{\infty}(\Omega(A); \complex)$, then for every $f \in C(\Omega(A); \complex)$,
so $\phi E_{x, y}= E_{J_E(\phi)x, y}$ for all $\phi \in B^{\infty}(\Omega(A); \complex)$. Thus for any $\phi, \psi \in B^{\infty}(\Omega(A); \complex)$,
and $J_{E}$ is a homomorphism.
Finally, let $f \in C(\Omega(A); \real)$, then since $\Gamma_{A}$ is a *-isomorphism, $J_{E}(f) = \Gamma_{A}^{-1}(f)$ is self-adjoint. As such, for any $x \in H$, $\dpn{f, E_{x, x}}{C(\Omega(A); \complex)}= \dpn{J_E(f)x, x}{H}\in \real$, so $E_{x, x}$ is real-valued. Thus for any $\phi \in B^{\infty}(\Omega(A); \real)$ and $x \in H$, $\dpn{J_E(\phi)x, x}{H}= \dpn{E_{x, x}, \phi}{C(\Omega(A); \complex)^*}\in \real$ as well. Therefore $J_{E}(\phi)$ is self-adjoint, and $J_{E}$ is a *-homomorphism.
(2): By Definition 39.5.3, $I_{E}$ is an injective unital *-homomorphism, so it is sufficient to show that $I_{E}(\mathscr{E}^{*}) = B$.
Let $J: C(\Omega(A); \complex) \to \mathscr{E}^{*}$ be defined by $\dpn{\mu, J(f)}{\mathscr{E}}= \int_{\Omega(A)}f d\mu$ for each $\mu \in \mathscr{E}$ and $f \in C(\Omega(A); \complex)$. By Goldstine’s Theorem, $C(\Omega(A); \complex)$ is weak*-dense in $C(\Omega(A); \complex)^{**}$, so $J(C(\Omega(A); \complex))$ is weak*-dense in $\mathscr{E}^{*}$. Since $I_{E}$ is continuous from the weak* topology on $\mathscr{E}^{*}$ to the weak operator topology on $B(H)$, $I_{E}(\mathscr{E}^{*}) \subset B$ by Proposition 5.5.3.
On the other hand, by the Banach-Alaoglu Theorem, $\ol{B_{\mathscr{E}^*}(0, 1)}$ is weak*-compact, so $I_{E}(\ol{B_{\mathscr{E}^*}(0, 1)})$ is weak-operator compact. As $\Gamma_{A}: A \to C(\Omega(A); \complex)$ is an isometric isomorphism, $I_{E}(\ol{B_{\mathscr{E}^*}(0, 1)}) \supset \ol{B_A(0, 1)}$. By the Kaplansky Density Theorem, $I_{E}(\ol{B_{\mathscr{E}^*}(0, 1)}) \supset \ol{B_B(0, 1)}$, and $I_{E}(\mathscr{E}^{*}) = B$.$\square$
- Omitting the natural map $C(\Omega(A); \complex) \to \mathscr{E}^{*}$. keyboard_return
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