Theorem 37.4.3 (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
\[T = \int_{\Omega(A)}\Gamma_{A} T dE \quad \forall T \in A\] - (2)
Let $B \subset B(H)$ be the strong-operator closure of $A$ in $B(H)$, then
\[B = \bracs{\int_{\Omega(A)} \phi dE \bigg | \phi \in C(\Omega(A); \complex)^{**}}\]
The mapping $C(\Omega(A); \complex)^{**}\to B$ defined by $\phi \mapsto \int \phi dE$ 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 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 $B(H)$,
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
$\square$
- With the same amount of writing and considerably more mental gymnastics, it can be shown that $I_{E}$ is a *-homomorphism on the full space $C(\Omega(A); \complex)^{**}$. However, it is not needed to show that $E$ is a spectral measure, and the homomorphism property falls out at the end anyways.keyboard_return
Post a Comment