Resolvent Set of Bounded Linear Operator equal to Resolvent Set as Densely-Defined Linear Operator
Theorem
Let $\HH$ be a Hilbert space over $\C$.
Let $T : \HH \to \HH$ be a bounded linear operator.
Let $\map {\rho_1} T$ be the resolvent set of $T$ as a bounded linear operator.
Let $\map {\rho_2} T$ be the resolvent set of $T$ as a densely-defined linear operator $\struct {\HH, T}$.
Then:
- $\map {\rho_1} T = \map {\rho_2} T$
Proof
Let $\lambda \in \map {\rho_1} T$.
Then $T - \lambda I$ is invertible in the sense of a bounded linear transformation.
That is, $T - \lambda I$ is bijective and $\paren {T - \lambda I}^{-1}$ is bounded.
From Underlying Set of Topological Space is Everywhere Dense, we have that $\HH$ is everywhere dense in $\HH$.
So, $T - \lambda I$ is injective, $\map {\paren {T - \lambda I} } {\map D T}$ is everywhere dense in $\HH$, and $\paren {T - \lambda I}^{-1}$ is bounded.
So $\lambda \in \map {\rho_2} T$.
We therefore have:
- $\map {\rho_1} T \subseteq \map {\rho_2} T$
by the definition of set inclusion.
Now let $\lambda \in \map {\rho_2} T$.
Then:
- $T - \lambda I$ is injective, $\map {\paren {T - \lambda I} } {\map D T}$ is everywhere dense in $\HH$, and $\paren {T - \lambda I}^{-1}$ is bounded.
To show that $\lambda \in \map {\rho_1} T$, we just need to show that:
- $\map {\paren {T - \lambda I} } {\map D T} = \HH$.
Let $y \in \HH$.
For brevity, let $S_\lambda = \paren {T - \lambda I}^{-1}$.
From Point in Closure of Subset of Metric Space iff Limit of Sequence, there exists a sequence in $\map {\paren {T - \lambda I} } {\map D T}$ with:
- $y_n \to y$
Define:
- $x_n = \map {S_\lambda^{-1} } {y_n}$
for each $n \in \N$.
We show that $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy.
Since $\HH$ is a Hilbert space, we will then have that $\sequence {x_n}_{n \mathop \in \N}$ converges.
We have:
\(\ds \norm {x_n - x_m}\) | \(=\) | \(\ds \norm {\map {S_\lambda^{-1} } {y_n} - \map {S_\lambda^{-1} } {y_m} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {\map {S_\lambda^{-1} } {y_n - y_m} }\) | Inverse of Linear Transformation is Linear Transformation | |||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {S_\lambda^{-1} } \norm {y_n - y_m}\) | Definition of Norm on Bounded Linear Transformation |
From Convergent Sequence in Normed Vector Space is Cauchy Sequence, we have:
- $\sequence {y_n}_{n \mathop \in \N}$ is Cauchy.
So, for each $\epsilon > 0$ there exists $N \in \N$ such that:
- $\ds \norm {y_n - y_m} < \frac \epsilon {\norm {S_\lambda^{-1} } }$ for all $n \ge N$.
Then:
- $\norm {x_n - x_m} < \epsilon$ for all $n \ge N$.
So $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy, and so converges to a limit $x$.
We then have:
\(\ds S_\lambda x\) | \(=\) | \(\ds \map {S_\lambda} {\lim_{n \mathop \to \infty} x_n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \to \infty} S_\lambda x_n\) | Continuity of Linear Transformation between Normed Vector Spaces, Continuous Mappings preserve Convergent Sequences | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{n \to \infty} y_n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds y\) |
That is:
- $y \in \map {\paren {T - \lambda I} } {\map D T}$
Since $y \in \HH$ was arbitrary, we have:
- $\HH \subseteq \map {\paren {T - \lambda I} } {\map D T}$
from the definition of set inclusion.
We also have:
- $\map {\paren {T - \lambda I} } {\map D T} \subseteq \HH$
by definition and so:
- $\map {\paren {T - \lambda I} } {\map D T} = \HH$
as required.
So $\lambda \in \map {\rho_1} T$.
We therefore have:
- $\map {\rho_2} T \subseteq \map {\rho_1} T$
and can therefore conclude:
- $\map {\rho_1} T = \map {\rho_2} T$
$\blacksquare$
Sources
- 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $25.3$: The Spectrum of Closed Unbounded Self-Adjoint Operators