Definition 29.4.3 (Resolvent).label Let $A$ be a unital Banach algebra and $x \in A$, then
\[R_{x}: \complex \setminus \sigma_{A}(x) \to A \quad \lambda \mapsto \frac{1}{\lambda - x}\]
is the resolvent function of $x$, which is holomorphic on $\complex \setminus \sigma_{A}(x)$.
Proof. By Proposition 29.2.3, the mapping $y \mapsto y^{-1}$ is smooth on $G(A)$. Hence $R_{x}: \complex \setminus \sigma_{A}(x) \to A$ is holomorphic.$\square$
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