# Spectrum of Self-Adjoint Densely-Defined Linear Operator is Real and Closed

## Theorem

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

Let $\struct {\map D T, T}$ be a self-adjoint densely-defined linear operator.

Let $\map \sigma T$ be the spectrum of $\struct {\map D T, T}$.

Then $\map \sigma T$ is a closed subset of $\C$ and:

$\map \sigma T \subseteq \R$

## Proof

Let $\lambda \in \map \sigma T$.

We show that $\lambda \in \R$.

From Element of Spectrum of Self-Adjoint Densely-Defined Linear Operator is Approximate Eigenvalue, there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\map D T$ with:

$\paren {T - \lambda I} x_n \to 0$

with $\norm {x_n} = 1$ for each $n \in \N$.

Then, we have:

 $\ds \cmod {\innerprod {\paren {T - \lambda I} x_n} {x_n} }$ $\le$ $\ds \norm {\paren {T - \lambda I} {x_n} } \norm {x_n}$ Cauchy-Bunyakovsky-Schwarz Inequality $\ds$ $=$ $\ds \norm {\paren {T - \lambda I} {x_n} }$ since $\norm {x_n} = 1$ $\ds$ $\to$ $\ds 0$ Modulus of Limit: Normed Vector Space

So:

$\innerprod {\paren {T - \lambda I} x_n} {x_n} \to 0$

We have:

 $\ds \innerprod {\paren {T - \lambda I} {x_n} } {x_n}$ $=$ $\ds \innerprod {T x_n - \lambda x_n} {x_n}$ $\ds$ $=$ $\ds \innerprod {T x_n} {x_n} - \lambda \innerprod {x_n} {x_n}$ $\ds$ $=$ $\ds \innerprod {T x_n} {x_n} - \lambda \norm {x_n}^2$ Definition of Inner Product Norm $\ds$ $=$ $\ds \innerprod {T x_n} {x_n} - \lambda$

So, from Sum Rule for Complex Sequences, we have:

$\innerprod {T x_n} {x_n} \to \lambda$
$\overline {\innerprod {T x_n} {x_n} } \to \overline \lambda$

We then have:

 $\ds \overline {\innerprod {T x_n} {x_n} }$ $=$ $\ds \innerprod {x_n} {T x_n}$ conjugate symmetry of the inner product $\ds$ $=$ $\ds \innerprod {T x_n} {x_n}$ Definition of Self-Adjoint Densely-Defined Linear Operator

So:

$\innerprod {T x_n} {x_n} \to \overline \lambda$

From Convergent Complex Sequence has Unique Limit, we have:

$\lambda = \overline \lambda$

From Complex Number equals Conjugate iff Wholly Real, we then have:

$\lambda \in \R$

$\Box$

We now show that $\map \sigma T$ is closed.

We show that $\C \setminus \map \sigma T$ is open.

That is, we show that the resolvent set of $T$, $\map \rho T$, is open.

Let $\lambda \in \map \rho T$.

Then $T - \lambda I$ is injective, has everywhere dense image and $\paren {T - \lambda I}^{-1}$ is bounded.

So there exists a real number $C > 0$ such that:

$\norm {\paren {T - \lambda I}^{-1} y} \le C \norm y$ for each $y \in \map {\paren {T - \lambda I} } \HH$.

Each $y \in \map {\paren {T - \lambda I} } {\map D T}$ can be written $y = {\paren {T - \lambda I} } x$ for $x \in \map D T$ and conversely $y$ of this form is contained in $\map {\paren {T - \lambda I} } {\map D T}$.

Then we have:

$\norm {\paren {T - \lambda I}^{-1} \paren {T - \lambda I} x} \le C \norm {\paren {T - \lambda I} x}$ for each $x \in \map D T$

That is:

$\ds \norm {\paren {T - \lambda I} x} \ge \frac 1 C \norm x$ for $x \in \map D T$.

We show that $\map \rho T$ contains an open neighborhood of $\lambda$.

Let $\mu \in \C$ have:

$\ds \cmod {\lambda - \mu} < \frac 1 {2 C}$

Then, for each $x \in \map D T$ we have:

 $\ds \norm {\paren {T - \mu I} x}$ $=$ $\ds \norm {\paren {T - \lambda I} x - \paren {\lambda - \mu} x}$ $\ds$ $\ge$ $\ds \norm {\paren {T - \lambda I} x} - \norm {\paren {\lambda - \mu} x}$ Reverse Triangle Inequality $\ds$ $\ge$ $\ds \frac 1 C \norm x - \cmod {\lambda - \mu} \norm x$ $\ds$ $\ge$ $\ds \frac 1 C \norm x - \frac 1 {2 C} \norm x$ $\ds$ $=$ $\ds \frac 1 {2 C} \norm x$

Then, if $\paren {T - \mu I} x = 0$ for some $x \in \map D T$, we have $\norm x = 0$, so $x = 0$.

So $T - \mu I$ is injective.

Suppose that $\mu \in \map \sigma T$.

From Partition of Spectrum of Densely-Defined Linear Operator, we have that $\mu$ is contained in either the continuous spectrum of $T$ or the residual spectrum of $T$.

From Self-Adjoint Densely-Defined Linear Operator has Empty Residual Spectrum, the residual spectrum of $T$ is empty, so if $\map \sigma T$, we must have that $\mu$ is in the continuous spectrum of $T$.

Then $\paren {T - \mu I}^{-1}$ is not bounded, but we have, for each $y \in \map {\paren {T - \mu I} } {\map D T} = \map D {\paren {T - \mu I}^{-1} }$ we have:

 $\ds \norm y$ $=$ $\ds \norm {\paren {T - \mu I} \paren {T - \mu}^{-1} y}$ setting $x = \paren {T - \mu I}^{-1} y$ in the previous computation $\ds$ $\ge$ $\ds \frac 1 {2 C} \norm {\paren {T - \mu I}^{-1} y}$

so:

$\norm {\paren {T - \mu I}^{-1} y} \le 2 C \norm y$

for all $y \in \map D {\paren {T - \mu I}^{-1} }$.

So, we have that $\paren {T - \mu I}^{-1}$ is bounded, a contradiction.

So we have $\mu \not \in \map \sigma T$, so $\mu \in \map \rho T$.

So for all $\mu$ with:

$\ds \cmod {\lambda - \mu} < \frac 1 {2 C}$

we have $\mu \in \map \rho T$.

So $\map \rho T$ is open.

So $\map \sigma T$ is closed.

$\blacksquare$