Resolvent Mapping is Analytic/Bounded Linear Operator/Proof 2

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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

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:

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

$\blacksquare$

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