Definition 34.7.2 (Continuous Functional Calculus).label Let $A$ be a unital $C^{*}$-algebra and $x \in A$ be normal, then there exists a unique continuous unital *-homomorphism
such that:
- (1)
$\one(x) = 1_{A}$.
- (2)
$\text{Id}(x) = x$.
- (3)
$\overline{\text{Id}}(x) = x^{*}$.
Moreover,
- (4)
Under the identification $\Omega(A[x]) = \sigma_{A}(x)$, for each $f \in C(\sigma_{A}(x); \complex)$, $f(x) = \Gamma^{-1}_{A[x]}(f)$.
- (5)
The mapping $f \mapsto f(x)$ is a unital *-isomorphism.
- (6)
For each $U \subset \complex$ open and $f \in C(U; \complex)$, the mapping
\[\bracs{x \in A|\sigma_A(x) \subset U}\to A \quad x \mapsto f(x)\]is continuous.
- (7)
Let $B$ be a unital $C^{*}$-algebra and $\Phi: A \to B$ be a unital *-homomorphism, then for any $x \in A$ and $f \in C(\sigma_{A}(x); \complex)$, $\Phi(f(x)) = f(\Phi(x))$.
Proof. (1), (4), (5): By the Spectral Theorem, $\Omega(A[x])$ and $\sigma_{A}(x)$ may be identified. For each $f \in C(\sigma_{A}(x); \complex)$, define $f(x) = \Gamma^{-1}_{A[x]}(f)$, then the mapping $f \mapsto f(x)$ is a unital *-isomorphism by the Gelfand-Naimark Theorem.
(2): The identification
given by the Spectral Theorem implies that $\Gamma_{A[x]}(x) = \text{Id}$.
(3): The mapping $f \mapsto f(x)$ is a *-homomorphism.
(6): Fix $x \in A$ and let $K \in \cn_{U}(\sigma_{A}(x); \complex)$ be compact. By Proposition 33.5.10, there exists $r > 0$ such that $\sigma_{A}(y) \subset K$ for all $y \in B_{A}(x, r)$.
Let $\eps > 0$. By the Stone-Weierstrass Theorem, there exists $p \in \complex[z, \ol z]$ such that $\sup_{\lambda \in K}|p(\lambda) - f(\lambda)| < \eps$. In which case,
for all $y \in B_{A}(x, r)$. Since $p$ is a continuous map on $A$, there exists $\delta \in (0, r)$ such that $\norm{p(x) - p(y)}_{A} < \eps$ for all $y \in B_{A}(x, \delta)$. Therefore $\norm{f(x) - f(y)}_{A} < 3\eps$ for all $y \in B_{A}(x, \delta)$.
(Uniqueness): By the Stone-Weierstrass Theorem.$\square$
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