Self-Adjoint Densely-Defined Linear Operator has Empty Residual Spectrum

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

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

Then the residual spectrum $\map {\sigma_r} T$ is empty.

Proof
Let $\struct {\map D {T^\ast}, T^\ast}$ be the adjoint of $\struct {\map D T, T}$.

Suppose that $\map {\sigma_r} T$ is non-empty.

Then there exists $\lambda \in \map {\sigma_r} T$.

That is, there exists $\lambda \in \C$ such that $T - \lambda I$ is injective, but such that:


 * $\map {\paren {T - \lambda I} } {\map D T}$ is not everywhere dense in $\HH$.

Let:


 * $R = \overline {\map {\paren {T - \lambda I} } {\map D T} }$

Then from Linear Subspace Dense iff Zero Orthocomplement, we have:


 * $R^\bot \ne \set 0$

where $R^\bot$ is the orthocomplement of $R$.

Then there exists some $y \in R^\bot$ such that $y \ne 0$.

Then:


 * $\innerprod {\paren {T - \lambda I} x} y = 0$ for all $x \in \map D T$

from the definition of the orthocomplement.

From Inner Product is Sesquilinear, we have:


 * $\innerprod {T x} y = \lambda \innerprod x y$ for each $x \in \map D T$.

As shown in Riesz Representation Theorem (Hilbert Spaces):


 * $x \mapsto \innerprod x y$ is a bounded linear functional.

So:


 * $x \mapsto \innerprod {T x} y$ is a bounded linear functional.

So from the definition of the adjoint, we have:


 * $y \in \map D {T^\ast}$

We therefore have:

Since $\map D T$ is everywhere dense in $H$, there exists a sequence in $\map D T$ with:


 * $x_n \to T y - \overline \lambda y$

Then:


 * $\innerprod {x_n} {T y - \overline \lambda y}$ for each $n \in \N$.

So, taking $n \to \infty$ and using Inner Product is Continuous:


 * $\innerprod {T y - \overline \lambda y} {T y - \overline \lambda y} = 0$

So from the definition of the inner product, we have:


 * $T y = \overline \lambda y$

So $\lambda$ is in the point spectrum $\map {\sigma_p} T$ of $T$.

From Point Spectrum of Symmetric Densely-Defined Linear Operator is Real, we have:


 * $\overline \lambda \in \R$

so from Complex Number equals Conjugate iff Wholly Real:


 * $\overline \lambda = \lambda$

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

But then:


 * $\paren {T - \lambda I} y = \map {\paren {T - \lambda I} } 0$

with $y \ne 0$.

So $T - \lambda I$ is not injective.

So we have a contradiction, and so $\map {\sigma_r} T = \O$.