Proof. (1): Let $\lambda \in \complex \setminus f(\sigma_{A}(x))$, then $1/(\lambda - f) \in H(\sigma_{A}(x); \complex)$. Since the holomorphic functional calculus is a homomorphism, $[1/(\lambda - f)](x) = (\lambda - f(x))^{-1}$. Thus $\sigma_{A}(f(x)) \subset f(\sigma_{A}(x))$.

On the other hand, let $\lambda \in \sigma_{A}(x)$, then the function $h(\mu) = [f(\mu) - f(\lambda)]/(\mu - \lambda)$ extends to an element of $H(\sigma_{A}(x); \complex)$. Therefore by the homomorphism property,

Since $(x - \lambda) \not\in G(A)$, $f(x) - f(\lambda) \not\in G(A)$ as well. Therefore $\sigma_{A}(f(x)) \supset f(\sigma_{A}(x))$.

(2): If additionally $f, g \in \complex(z)$, then the theorem holds by (1) and (2) of Definition 29.5.1.

Now suppose that $f \in H(\sigma_{A}(x); \complex)$ is arbitrary and $g \in \complex(z)$. For each $\eps > 0$, by Proposition 29.4.10 and continuity of the holomorphic functional calculus, there exists $U \in \cn_{H(\sigma_A(x); \complex)}(f)$ such that for each $h \in U$,

1. (a)

   $g \in H(\sigma_{A}(h(x)); \complex)$.

2. (b)

   $\norm{g(f(x)) - g(h(x))}_{A} < \eps$.

By Runge’s Theorem, there exists $h \in U \cap \complex(z)$. Thus $\norm{g(f(x)) - (g \circ f)(x)}_{A} < \eps$ as well. As this holds for all $\eps > 0$, $g(f(x)) = (g \circ f)(x)$ for any $f \in H(\sigma_{A}(x); \complex)$ and $g \in \complex(x) \cap H(f(\sigma_{A}(x)); \complex)$.

By Runge’s Theorem again and continuity of the holomorphic functional calculus, the above also holds for all $f \in H(\sigma_{A}(x); \complex)$ and $g \in H(f(\sigma_{A}(x)); \complex)$.$\square$