Resolvent Mapping is Analytic/Bounded Linear Operator

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.

Proof 2
If $B=\set {\mathbf 0_B}$, the statement is trivial, since $\map f z = \mathbf 0_{\map \LL {B, B}}$ for all $z \in \C$.

We assume that $B \ne \set {\mathbf 0_B}$.

Especially, for all $z \in \map \rho T$:
 * $\map f z \ne \mathbf 0_{\map \LL {B, B}}$

since:
 * $\paren {T - z I} \map f z = I \ne \mathbf 0_{\map \LL {B, B}}$

Let $a \in \map \rho T$.

That is, $\paren {T - a I}^{-1}$ exists.

Thus for each $z \in \C$:

Recall that $\map B {c, r}$ denotes the open disc of center $c \in \C$ and radius $r>0$.

For all $z \in \map B {a, \frac{1}{\norm {\map f a} } }$, we have:

Therefore:

That is, in a neighborhood of each $a \in \map \rho T$, $f$ can be written as:
 * $\ds \map f z = \sum_{n \mathop \ge 0} \paren {\map f a}^{n+1} \paren {z - a}^n$

This means that $f$ is analytic on $\map \rho T$.

Finally, by Derivative of Power Series, in the neighborhood of each $a \in \map \rho T$:
 * $\ds \map {f'} z = \sum_{n \mathop \ge 1} \paren {\map f a}^{n+1} n \paren {z - a}^{n-1}$

Choosing $z = a$, we obtain:
 * $\map {f'} a = \paren {\map f a}^2 = \paren {T - aI}^{-2}$