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