Proof, [Proposition I.2.7, Tak01]. (Definition): Let $U, V \in \cn_{\complex}(\sigma_{A}(x))$ such that $\ol V$ is a compact subset of $U$, then by Proposition 27.6.1, there exists closed rectifiable curves $\seqf{\gamma_j}$ on $V \setminus \sigma_{A}(x)$ such that

1. (a)

   For all $f \in H(V; \complex)$ and $z_{0} \in \sigma_{A}(x)$,

   \[f(z_{0}) = \frac{1}{2\pi i}\sum_{j = 1}^{n} \int_{\gamma_j}\frac{f(z)}{z - z_{0}}dz\]

   
   
2. (b)

   For all $f \in H(U; \complex)$ and $z_{0} \in U \setminus V$,

   \[\frac{1}{2\pi i}\sum_{j = 1}^{n} \int_{\gamma_j}\frac{f(z)}{z - z_{0}}dz = 0\]

   
   

For each $f \in H(U; \complex)$, define

then by Cauchy’s Theorem, the above definition is independent of the choice of curves satisfying (a).

(Linearity): By Proposition 13.4.1, the mapping $f \mapsto f(z)$ is a continuous linear map from $H(\sigma_{A}(x); \complex)$ to $A$.

(Homomorphism): Let $T$ be the union of the image of $\seq{\gamma_j}$. By Proposition 27.6.1, there exists closed rectifiable curves $\seqf[m]{\mu_j}$ on $U \setminus \ol V$ such that $g(z_{0}) = \frac{1}{2\pi i}\sum_{j = 1}^{m} \int_{\mu_j}g(z)/(z - z_{0})dz$ for all $f \in H(U; \complex)$ and $z_{0} \in \ol V$. Now,

For each $z, w \in U \setminus \sigma_{A}(x)$ with $z \ne w$, by the resolvent equation,

By assumptions on $\seqf[m]{\mu_j}$, for each $1 \le j \le n$,

By assumption (b) and Fubini’s Theorem, for each $1 \le k \le m$,

Therefore

and the mapping $f \mapsto f(x)$ is a homomorphism.

(1): Since the constant $1$ function is the identity in $H(\sigma_{A}(x); \complex)$, $1(x) = 1$ by the homomorphism property.

(2): Let $R > 0$ such that $\sigma_{A}(x) \subset B_{\complex}(0, R)$, then by Proposition 13.4.2 and Cauchy’s Integral Formula,

and

(Uniqueness): By (2), the homomorphism extends uniquely to $\complex(z) \cap H(\sigma_{A}(x); \complex)$. By Runge’s Theorem, it extends uniquely to $H(\sigma_{A}(x); \complex)$ by continuity.$\square$