Resolvent Mapping is Analytic

Theorem
Let $B$ be a Banach space.

Let $\map \LL {B, B}$ be the set of bounded linear operators from $B$ to itself.

Let $T \in \map \LL {B, B}$.

Let $\map \rho T$ be the resolvent set of $T$ in the complex plane.

Then the resolvent mapping $f : \map \rho T \to \map \LL {B, B}$ given by $\map f z = \paren {T - z I}^{-1}$ is analytic, and:


 * $\map {f'} z = \paren {T - z I}^{-2}$

where $f'$ denotes the derivative of $f$ $z$.

Proof
For $a \in \map \rho T$, define:
 * $R_a = \paren {T - a I}^{-1}$

Then we have:

From Resolvent Mapping is Continuous we have:
 * $R_{z + h} \to R_z$ as $h \to 0$

Taking limits of both sides and using Norm is Continuous, we get:


 * $\ds \lim_{h \mathop \to 0} \dfrac {\norm {\map f {z + h} - \map f z - \paren {T - z I}^{-2} h }_*} {\size h} = \norm {R_z^2 - R_z^2}_* = 0$

which is the result.