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 self-adjoint densely defined linear operator.
Then the residual spectrum $\map {\sigma_r} T$ is empty.
Proof
Let $\struct {\map D {T^\ast}, T^\ast}$ be the adjoint of $\struct {\map D T, T}$.
Since $\struct {\map D T, T}$ is self-adjoint, we have:
- $\struct {\map D {T^\ast}, T^\ast} = \struct {\map D T, T}$
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:
\(\ds 0\) | \(=\) | \(\ds \innerprod {T x} y - \lambda \innerprod x y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \innerprod x {T^\ast y} - \innerprod x {\overline \lambda y}\) | Inner Product is Sesquilinear, Definition of Adjoint of Densely-Defined Linear Operator | |||||||||||
\(\ds \) | \(=\) | \(\ds \innerprod x {T y} - \innerprod x {\overline \lambda y}\) | Definition of Self-Adjoint Densely-Defined Linear Operator | |||||||||||
\(\ds \) | \(=\) | \(\ds \innerprod x {T y - \overline \lambda y}\) | Inner Product is Sesquilinear |
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 $\map {\sigma_r} T = \O$.
$\blacksquare$
Sources
- 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $25.3$: The Spectrum of Closed Unbounded Self-Adjoint Operators