Resolvent Mapping is Continuous/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 continuous.

Proof

Lemma

Define $S : \C \to A$ by:

$\map S \lambda = \lambda {\mathbf 1}_A - x$

Then $S$ is continuous.

$\Box$

From the Lemma, we have:

the mapping $S : \C \to A$ defined by:

$\map S \lambda = \lambda {\mathbf 1}_A - x$

for each $\lambda \in \C$, is continuous.

From Restriction of Continuous Mapping is Continuous, $S \restriction_{\map {\rho_A} x}$ is continuous.

From the definition of the resolvent set, we have $S \sqbrk {\map {\rho_A} x} \subseteq \map G A$.

From Inverse Mapping on Group of Units in Unital Banach Algebra is Continuous, the mapping $\phi : \map G A \to \map G A$ defined by:

$\map \phi x = x^{-1}$ for each $x \in \map G A$

is continuous.

From Composite of Continuous Mappings is Continuous, $\phi \circ \paren {S \restriction_{\map {\rho_A} x} } : \map {\rho_A} x \to A$ is continuous.

Since $\phi \circ S = R$, we have that $R$ is continuous.

$\blacksquare$