Spectrum of Bounded Linear Operator equal to Spectrum 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 $\map {\sigma_1} T$ be the spectrum of $T$ as a bounded linear operator.

Let $\map {\sigma_2} T$ be the spectrum of $T$ as a densely-defined linear operator $\struct {\HH, T}$.


Then:

$\map {\sigma_1} T = \map {\sigma_2} T$


Proof

From the definition of the spectrum of $T$ as a bounded linear operator, we have:

$\map {\sigma_1} T = \C \setminus \map {\rho_1} T$

where $\map {\rho_1} T$ is the resolvent set of $T$ as a bounded linear operator.

From the definition of the spectrum of $T$ as a densely-defined linear operator $\struct {\HH, T}$, we have:

$\map {\sigma_2} T = \C \setminus \map {\rho_2} T$

where $\map {\rho_2} T$ is the resolvent set of $T$ as a densely-defined linear operator $\struct {\HH, T}$.

From Resolvent Set of Bounded Linear Operator equal to Resolvent Set as Densely-Defined Linear Operator, we have:

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

So:

$\map {\sigma_1} T = \map {\sigma_2} T$

$\blacksquare$