Spectrum of Bounded Linear Operator equal to Spectrum 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 $\map {\sigma_1} T$ 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:
- $\map {\sigma_1} T = \map {\sigma_2} T$
$\blacksquare$