Resolvent Mapping is Continuous/Banach Algebra/Lemma

Lemma

Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra over $\C$.

Let $x \in A$.

Define $S : \C \to A$ by:

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

Then $S$ is continuous.

Proof

We have, for $\lambda, \mu \in \C$:

\(\ds \norm {\map S \lambda - \map S \mu}\)

\(=\)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \cmod {\lambda - \mu} \norm { {\mathbf 1}_A}\)

Norm Axiom $\text N 2$: Positive Homogeneity

\(\ds \)

\(=\)

\(\ds \cmod {\lambda - \mu}\)

Definition of Unital Banach Algebra

Let $\epsilon > 0$ and $\lambda, \mu \in \C$ be such that:

$\cmod {\lambda - \mu} < \epsilon$

Then, we have:

$\norm {\map S \lambda - \map S \mu} < \epsilon$

Hence $S$ is continuous.

$\blacksquare$