Proof, [Theorem 11.1, Zhu93]. 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

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 continuous functional calculus, $\Phi(x) = f(\Phi(x)) = \Phi(f(x))$. By the Spectral Mapping Theorem, $\sigma_{A}(f(x)) = f(\sigma_{A}(x))$. By Theorem 34.4.3, $\norm{f(x)}_{A} = [f(x)]_{sp}\le \norm{f}_{u} = 2\norm{y}_{F}$.

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 method of successive approximations, $\phi(A_{sa}) = \ol{\Phi(A)}\cap B_{sa}$. Therefore $\Phi(A) = \ol{\Phi(A)}$.$\square$