# Resolvent Mapping is Analytic/Bounded Linear Operator/Proof 1

## Theorem

### Bounded Linear Operator

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 $z$.

### Banach Algebra

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:

- $\map {R'} \lambda = -\paren {\lambda {\mathbf 1}_A - x}^{-2}$

Although this article appears correct, it's inelegant. There has to be a better way of doing it.In particular: Derivative is not defined for complex domainYou can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Improve}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## Proof

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.

$\blacksquare$