Theorem 34.7.3 (Spectral Mapping Theorem (Continuous)).label Let $A$ be a unital $C^{*}$-algebra and $x \in A$ be normal, then:
- (1)
For every $f \in C(\sigma_{A}(x); \complex)$, $\sigma_{A}(f(x)) = f(\sigma_{A}(x))$.
- (2)
For every $f \in C(\sigma_{A}(x); \complex)$ and $g \in C(\sigma_{A}(f(x)); \complex)$, $(f \circ g)(x) = f(g(x))$.
Proof. (1): Since $f(x) \in A[x]$, by Proposition 33.8.2 and definition of the continuous functional calculus,
(2): Firstly, the theorem holds directly if $f, g \in \complex[z, \ol z]$ are polynomials in $z$ and $\ol z$.
Suppose that $g \in \complex[z, \ol z]$ but $f$ is arbitrary. By the Stone-Weierstrass Theorem, there exists $\seq{p_n}\subset \complex[z, \ol z]$ such that $p_{n} \to f$ uniformly on $\sigma_{A}(x)$. By property (6) of the continuous functional calculus,
Finally, suppose that both $f$ and $g$ are arbitary. By the Stone-Weierstrass Theorem, there exists $\seq{p_n}\subset \complex[z, \ol z]$ such that $p_{n} \to g$ uniformly on $\sigma_{A}(f(x))$. By continuity of the continuous functional calculus,
$\square$
Post a Comment