Definition 37.4.6 ($L^{\infty}$ Functional Calculus).label Let $H$ be a complex Hilbert space, $T \in B(H)$ be normal, and $A \subset B(H)$ be the smallest von Neumann algebra acting on $H$ containing $T$ and $I$, then
- (1)
There exists a unique spectral measure $E: \cb_{\sigma_{B(H)}(T)}\to A$ such that
\[T = \int_{\sigma_{B(H)}(T)}\lambda E(d\lambda) \quad T^{*} = \int_{\sigma_{B(H)}(T)}\ol \lambda E(d\lambda)\] - (2)
Let $\mathscr{E}\subset M_{R}(\sigma_{B(H)}(T); \complex)$ be the closed subspace generated by $\bracsn{E_{x, y}|x, y \in H}$, and $A \subset B(H)$ be the von Neumann algebra generated by $I$ and $T$, then
\[I_{E}: \mathscr{E}^{*} \to A \quad \phi \mapsto \phi(T) := \int_{\sigma_{B(H)}(T)}\phi dE\]is a *-isomorphism.
- (3)
$I_{E}: \mathscr{E}^{*} \to A$ is the unique weak* to weak-operator continuous unital *-homomorphism such that $I_{E}(\text{Id}) = T$.
The spectral measure $E$ is the resolution of the identity for $T$, and the mapping $f \mapsto f(T)$ on $\mathscr{E}^{*}$ is the $L^{\infty}$-functional calculus of $T$.
Proof. (1), (2): By Spectral Theorem I applied to $B(H)[T]$, there exists a unique spectral measure $E$ on $\sigma_{B(H)}(T)$ such that the mapping
is a *-isomorphism that extends the inverse Gelfand transform $\Gamma_{B(H)[T]}^{-1}: C(\sigma_{B(H)}(T); \complex) \to B(H)[T]$.
For each $\phi \in \mathscr{E}^{*}$, let $\phi(T) = \int_{\sigma_{B(H)}(T)}\phi dE$, then the mapping $\phi \mapsto \phi(T)$ is continuous from the weak* topology on $\mathscr{E}^{*}$ to the weak operator topology on $A$ by Definition 37.4.3.
(3): By uniqueness of the continuous functional calculus, Goldstine’s Theorem, and (2), the mapping $\phi \mapsto \phi(T)$ is unique.$\square$
Post a Comment