Point Spectrum of Symmetric Densely-Defined Linear Operator is Real

Theorem

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

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

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

Then:

$\map {\sigma_p} T \subseteq \R$

Proof

If $\map {\sigma_p} T = \O$, the result is immediate.

Let $\lambda \in \map {\sigma_p} T$.

Then, from Point Spectrum of Densely-Defined Linear Operator consists of its Eigenvalues, there exists $x \in \HH \setminus \set 0$ such that:

$T x = \lambda x$

Then, we have:

$\innerprod {T x} x = \innerprod {\lambda x} x = \lambda \innerprod x x$

and:

\(\ds \innerprod {T x} x\)

\(=\)

\(\ds \innerprod x {T x}\)

Definition of Symmetric Densely-Defined Linear Operator

\(\ds \)

\(=\)

\(\ds \innerprod x {\lambda x}\)

\(\ds \)

\(=\)

\(\ds \overline \lambda \innerprod x x\)

Inner Product is Sesquilinear

So, we have:

$\paren {\lambda - \overline \lambda} \innerprod x x = 0$

Since $x \ne 0$, we have $\innerprod x x \ne 0$, so:

$\lambda = \overline \lambda$

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

$\lambda \in \R$

So:

$\map {\sigma_p} T \subseteq \R$

from the definition of set inclusion.

$\blacksquare$

Sources

• 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $25.3$: The Spectrum of Closed Unbounded Self-Adjoint Operators