User:Caliburn/s/fa/Spectrum of Linear Operator is Bounded

Theorem

Let $\struct {X, \norm \cdot_X}$ be a Banach space over $\C$.

Let $A : X \to X$ be a bounded linear operator.

Let $\map \sigma A$ be the spectrum of $A$.

Then $\map \sigma A$ is a bounded subset of $\C$.

In particular:

$\cmod \lambda \le \norm A$

for all $\lambda \in \map \sigma A$, where $\norm A$ denotes the norm of $A$.

Proof

Let $\lambda \in \C$ be such that:

$\cmod \lambda > \norm A$

We show that:

$\lambda \not \in \map \sigma A$

Let $\map \rho A$ be the resolvent set of $A$.

By the definition of spectrum, we have:

$\map \sigma A = \C \setminus \map \rho A$

So, we aim to prove that:

$\lambda \in \map \rho A$

We have:

$\norm {\lambda^{-1} A} < 1$

So, from User:Caliburn/s/fa/2, we have that:

$I - \lambda^{-1} A$ is invertible with bounded inverse.

So:

$-\lambda \paren {I - \lambda^{-1} A} = A - \lambda I$

is invertible with bounded inverse.

So by the definition of the resolvent set of $A$, we have:

$\lambda \in \map \rho A$

so:

$\lambda \not \in \map \sigma A$

as required.

By Proof by Contraposition, we therefore have that:

if $\lambda \in \map \sigma A$ then $\cmod \lambda \le \norm A$

So:

$\map \sigma A$ is bounded.