Added continuity of homomorphisms.
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@@ -41,6 +41,17 @@
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Let $A, B$ be unital $C^*$-algebras and $\Phi: A \to B$ be a unital *-homomorphism, then $\Phi(A)$ is closed.
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\end{theorem}
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\begin{proof}[Proof, {{\cite[Theorem 11.1]{Zhu}}}. ]
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Let $y \in \ol{\Phi(A)} \cap B_{sa}$, then there exists $x \in A_{sa}$ such that $\norm{y - \Phi(x)}_B \le \norm{y}_B/2$. Let
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\[
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f: \complex \to \complex \quad z \mapsto \begin{cases}
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z & |z| \le 2\norm{y}_F \\
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2\norm{y}_F \cdot \sgn z = 2\norm{y}_F \cdot \frac{z}{|z|} & |z| \ge 2\norm{y}_F
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\end{cases}
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\]
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then $f \in C(\complex; \complex)$. Since $\norm{\Phi(x)}_B \le \norm{y}_B + \norm{y - \Phi(x)}_B \le 2\norm{y}_B$, $\sigma_B(\Phi(x)) \subset \ol{B_\complex(0, 2\norm{y}_B)}$, and $f|_{\sigma_B(\Phi(x))}$ is the identity. Thus by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $\Phi(x) = f(\Phi(x)) = \Phi(f(x))$. By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-continuous}, $\sigma_A(f(x)) = f(\sigma_A(x))$. By \autoref{theorem:c-star-normal-spectral-radius}, $\norm{f(x)}_A = [f(x)]_{sp} \le \norm{f}_u = 2\norm{y}_F$.
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The above setup implies that for every $y \in \ol{\Phi(A)} \cap B_{sa}$, there exists $z \in A_{sa}$ such that $\norm{y - \Phi(z)}_{B} \le \norm{y}_B/2$, and $\norm{z}_A \le 2\norm{y}_B$. By the \hyperref[method of successive approximations]{theorem:successive-approximation}, $\phi(A_{sa}) = \ol{\Phi(A)} \cap B_{sa}$. Therefore $\Phi(A) = \ol{\Phi(A)}$.
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\end{proof}
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