Spectrum of Self-Adjoint Bounded Linear Operator is Real and Closed

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Theorem

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.

Let $T : \HH \to \HH$ be a bounded self-adjoint operator.

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


Then:

$\map \sigma T \subseteq \R$


Proof 1

This follows from:

Spectrum of Self-Adjoint Densely-Defined Linear Operator is Real and Closed
Spectrum of Bounded Linear Operator equal to Spectrum as Densely-Defined Linear Operator

$\blacksquare$


Proof 2

For all $\phi \in \HH$ and $\lambda := a + i b \in \C$:

\(\ds \norm {\paren {A - \lambda I} \phi}_\HH^2\) \(=\) \(\ds \norm {\paren {A - a I} \phi}_\HH^2 + b^2 \norm {\phi}_\HH^2\)
\(\text {(1)}: \quad\) \(\ds \) \(\ge\) \(\ds b^2 \norm {\phi}_\HH^2\)

Let $b \ne 0$.

In view of $(1)$, both $A - \lambda I$ and $A - \overline \lambda I$ are injective.

Moreover:

\(\ds \paren {\Img {A - \lambda I} }^\perp\) \(=\) \(\ds \paren {\Img {A - \overline \lambda I}^\ast }^\perp\) as $A - \lambda I = \paren {A - \overline \lambda I}^\ast$
\(\ds \) \(=\) \(\ds \ker \paren {A - \overline \lambda I}\) Kernel of Linear Transformation is Orthocomplement of Image of Adjoint
\(\ds \) \(=\) \(\ds \set 0\) as $A - \overline \lambda I$ is injective

By Linear Subspace Dense iff Zero Orthocomplement, $\Img {A - \lambda I}$ is dense in $\HH$.

Now applying $(1)$ again, we can conclude that $A - \lambda I$ is invertible.

That is, $\lambda \not \in \map \sigma T$.