Resolvent Mapping is Continuous/Banach Algebra/Lemma
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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.
$\blacksquare$