Resolvent Mapping is Analytic/Banach Algebra

Theorem

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 You 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

Let $\lambda, \mu \in \map {\rho_A} x$ be such that $\lambda \ne \mu$.

Then, we have:

\(\ds \frac {\paren {\mu {\mathbf 1}_A - x}^{-1} - \paren {\lambda {\mathbf 1}_A - x}^{-1} } {\mu - \lambda}\)

\(=\)

\(\ds \frac {\paren {\mu {\mathbf 1}_A - x}^{-1} \paren { {\mathbf 1}_A - \paren {\mu {\mathbf 1}_A - x} \paren {\lambda {\mathbf 1}_A - x}^{-1} } } {\mu - \lambda}\)

\(\ds \)

\(=\)

\(\ds \frac {\paren {\mu {\mathbf 1}_A - x}^{-1} \paren { \paren {\lambda {\mathbf 1}_A - x} - \paren {\mu {\mathbf 1}_A - x} } \paren {\lambda {\mathbf 1}_A - x}^{-1} } {\mu - \lambda}\)

\(\ds \)

\(=\)

\(\ds \frac {\paren {\mu {\mathbf 1}_A - x}^{-1} \paren {\lambda - \mu} \paren {\lambda {\mathbf 1}_A - x}^{-1} } {\mu - \lambda}\)

\(\ds \)

\(=\)

\(\ds -\paren {\mu {\mathbf 1}_A - x}^{-1} \paren {\lambda {\mathbf 1}_A - x}^{-1}\)

From Resolvent Mapping is Continuous: Banach Algebra, we have:

$\paren {\mu {\mathbf 1}_A - x}^{-1} \to \paren {\lambda {\mathbf 1}_A - x}^{-1}$ as $\mu \to \lambda$ in $\map {\rho_A} x$.

So, from Product Rule for Sequence in Normed Algebra, we have:

$-\paren {\mu {\mathbf 1}_A - x}^{-1} \paren {\lambda {\mathbf 1}_A - x}^{-1} \to -\paren {\lambda {\mathbf 1}_A - x}^{-2}$ as $\mu \to \lambda$.

So, we have that:

$\ds \lim_{\mu \mathop \to \lambda} \frac {\paren {\mu {\mathbf 1}_A - x}^{-1} - \paren {\lambda {\mathbf 1}_A - x}^{-1} } {\mu - \lambda}$ exists

and:

$\ds \lim_{\mu \mathop \to \lambda} \frac {\map R \mu - \map R \lambda} {\mu - \lambda} = \lim_{\mu \mathop \to \lambda} \frac {\paren {\mu {\mathbf 1}_A - x}^{-1} - \paren {\lambda {\mathbf 1}_A - x}^{-1} } {\mu - \lambda} = -\paren {\lambda {\mathbf 1}_A - x}^{-2}$

So, we obtain:

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

$\blacksquare$