Resolvent Mapping is Analytic

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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:

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