Resolvent Set of Bounded Linear Operator equal to Resolvent Set as Densely-Defined Linear Operator

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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:

$\map {\rho_1} T = \map {\rho_2} T$

$\blacksquare$


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