# 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 $f$ with respect to $z$.

## Proof 1

For $a \in \map \rho T$, define:

$R_a = \paren {T - a I}^{-1}$

Then we have:

 $\ds \frac {\norm {\map f {z + h} - \map f z - \paren {T - z I}^{-2} h }_*} {\size h}$ $=$ $\ds \frac {\norm {R_{z + h} - R_z - R_z^2 h}_*} {\size h}$ $\ds$ $=$ $\ds \frac {\norm {h R_{z + h} R_z - R_z^2 h }_*} {\size h}$ Resolvent Identity $\ds$ $=$ $\ds \frac {\size h \norm {R_{z + h} R_z - R_z^2 }_*} {\size h}$ Operator Norm is Norm $\ds$ $=$ $\ds \norm {R_{z + h} R_z - R_z^2 }_*$

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.

$\blacksquare$

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

Then for each $z \in \C$:

 $\ds T - z I$ $=$ $\ds \paren {T - a I - \paren {z - a} I}$ $\ds$ $=$ $\ds \paren {T - a I} \paren {I - \paren {z - a} \paren {T - a I}^{-1} }$ $\text {(1)}: \quad$ $\ds$ $=$ $\ds \paren {T - a I} \paren {I - \paren {z - a} \map f a}$

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:

 $\ds \paren {I - \paren {z - a} \map f a}^{-1}$ $=$ $\ds \sum_{n \mathop \ge 0} \paren {\paren {z - a} \map f a}^n$ Neumann Series, as $\norm {\paren {z - a} \map f a} < 1$ $\text {(2)}: \quad$ $\ds$ $=$ $\ds \sum_{n \mathop \ge 0} \paren {\map f a}^n \paren {z - a}^n$

Therefore:

 $\ds \paren {T - zI}^{-1}$ $=$ $\ds \paren {\paren {T - a I} \paren {I - \paren {z - a} \map f a} }^{-1}$ by $(1)$ $\ds$ $=$ $\ds \paren {I - \paren {z - a} \map f a}^{-1} \paren {T - a I}^{-1}$ Inverse of Product $\ds$ $=$ $\ds \paren {I - \paren {z - a} \map f a}^{-1} \map f a$ $\ds$ $=$ $\ds \paren {\sum_{n \mathop \ge 0} \paren {\map f a}^n \paren {z - a}^n} \map f a$ by $(2)$ $\ds$ $=$ $\ds \sum_{n \mathop \ge 0} \paren {\map f a}^{n+1} \paren {z - a}^n$

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:

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

$\blacksquare$