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, $\map {\paren {T - \lambda I} } \HH$ 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, $\map {\paren {T - \lambda I} } {\HH}$ 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:


 * $\map {\paren {T - \lambda I} } \HH = \HH$.

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 $\map {\paren {T - \lambda I} } \HH$ 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:

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:

That is:


 * $y \in \map {\paren {T - \lambda I} } \HH$

Since $y \in \HH$ was arbitrary, we have:


 * $\HH \subseteq \map {\paren {T - \lambda I} } \HH$

from the definition of set inclusion.

We also have:


 * $\map {\paren {T - \lambda I} } \HH \subseteq \HH$

by definition and so:


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

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:


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