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.

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:


 * $\map \sigma T \subseteq \set {\lambda \in \C : \cmod \lambda \le \norm T_{\map \BB X} }$