Resolvent Mapping is Analytic/Banach Algebra

Theorem
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra over $\C$.

Let ${\mathbf 1}_A$ be the identity element of $A$.

Let $x \in A$.

Let $\map {\rho_A} x$ be the resolvent set of $x$ in $A$.

Define $R : \map {\rho_A} x \to A$ by:
 * $\map R \lambda = \paren {\lambda {\mathbf 1}_A - x}^{-1}$

Then $R$ is analytic with derivative:
 * $\map {R'} \lambda = -\paren {\lambda {\mathbf 1}_A - x}^{-2}$

Proof
Let $\lambda, \mu \in \map {\rho_A} x$ be such that $\lambda \ne \mu$.

Then, we have:

From Resolvent Mapping is Continuous: Banach Algebra, we have:
 * $\paren {\mu {\mathbf 1}_A - x}^{-1} \to \paren {\lambda {\mathbf 1}_A - x}^{-1}$ as $\mu \to \lambda$ in $\map {\rho_A} x$.

So, from Product Rule for Sequence in Normed Algebra, we have:
 * $-\paren {\mu {\mathbf 1}_A - x}^{-1} \paren {\lambda {\mathbf 1}_A - x}^{-1} \to -\paren {\lambda {\mathbf 1}_A - x}^{-2}$ as $\mu \to \lambda$.

So, we have that:
 * $\ds \lim_{\mu \mathop \to \lambda} \frac {\paren {\mu {\mathbf 1}_A - x}^{-1} - \paren {\lambda {\mathbf 1}_A - x}^{-1} } {\mu - \lambda}$ exists

and:
 * $\ds \lim_{\mu \mathop \to \lambda} \frac {\map R \mu - \map R \lambda} {\mu - \lambda} = \lim_{\mu \mathop \to \lambda} \frac {\paren {\mu {\mathbf 1}_A - x}^{-1} - \paren {\lambda {\mathbf 1}_A - x}^{-1} } {\mu - \lambda} = -\paren {\lambda {\mathbf 1}_A - x}^{-2}$

So, we obtain:
 * $\map {R'} \lambda = -\paren {\lambda {\mathbf 1}_A - x}^{-2}$