Resolvent Set of Bounded Linear Operator is Open

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