Resolvent Mapping is Analytic/Bounded Linear Operator/Proof 2

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

Then 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:
 * $\map {f'} a = \paren {\map f a}^2 = \paren {T - aI}^{-2}$