Spectrum of Bounded Linear Operator is Compact
Theorem
Let $X$ be a Banach space over $\C$.
Let $T : X \to X$ be a bounded linear operator.
Let $\map \sigma T$ be the spectrum of $T$.
Then $\map \sigma T$ is compact, and:
- $\map \sigma T \subseteq \set {\lambda \in \C : \cmod \lambda \le \norm T_{\map \BB X} }$
Proof
From Spectrum of Bounded Linear Operator is Closed, $\map \sigma T$ is closed in $\C$.
It therefore suffices to show that $\map \sigma T$ is bounded.
Hence it will suffice to show that:
- $\map \sigma T \subseteq \set {\lambda \in \C : \cmod \lambda \le \norm T_{\map \BB X} }$
Let $\lambda \in \C$ be such that $\cmod \lambda > \norm T_{\map \BB X}$.
Then, we have:
- $\ds \norm {\frac 1 \lambda T}_{\map \BB X} = \frac 1 {\cmod \lambda} \norm T_{\map \BB X} < 1$
from Norm Axiom $\text N 2$: Positive Homogeneity.
From Invertibility of Identity Minus Operator:
- $\ds I - \frac 1 \lambda T$ is invertible as a bounded linear operator.
Since $\map \BB X$ is a vector space, we have:
- $\ds -\frac 1 \lambda \paren {I - \frac 1 \lambda T} = T - \lambda I$ is invertible as a bounded linear operator.
That is $\lambda \in \map \rho T$, where $\map \rho T$ is the resolvent set of $T$.
So if $\cmod \lambda > \norm T_{\map \BB X}$, we have $\lambda \in \C \setminus \map \sigma T$.
So if $\lambda \in \map \sigma T$, we have $\cmod \lambda \le \norm T_{\map \BB X}$.
That is:
- $\map \sigma T \subseteq \set {\lambda \in \C : \cmod \lambda \le \norm T_{\map \BB X} }$
$\blacksquare$
Sources
- 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $14.1$: The Resolvent and Spectrum