Resolvent Mapping is Analytic/Bounded Linear Operator/Proof 1

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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 domain
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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.

$\blacksquare$

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