Resolvent Set of Bounded Linear Operator is Open
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Theorem
Let $X$ be a Banach space over $\C$.
Let $T : X \to X$ be a bounded linear operator.
Let $\map \rho T$ be the resolvent set of $T$.
Then $\map \rho T$ is open.
Proof
Let $\lambda \in \map \rho T$.
Then $T - \lambda I$ is invertible as a bounded linear operator.
Let $\delta > 0$ be such that:
- $\cmod \delta < \norm {\paren {T - \lambda I}^{-1} }_{\map \BB X}^{-1}$
Then, we have:
- $\norm {\delta I}_{\map \BB X} \norm {\paren {T - \lambda I}^{-1} } < 1$
From Invertibility of Identity Minus Operator: Corollary, we have:
- $T - \paren {\lambda + \delta} I$ is invertible as a bounded linear operator.
That is, if:
- $\ds \cmod {\mu - \lambda} < \norm {\paren {T - \lambda I}^{-1} }_{\map \BB X}^{-1}$
we have:
- $T - \mu I$ is invertible as a bounded linear operator
so that $\mu \in \map \rho T$.
So $\map \rho T$ contains an open neighborhood of each of its points.
So $\map \rho T$ is open.
$\blacksquare$
Sources
- 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $14.1$: The Resolvent and Spectrum