Partition of Spectrum of Densely-Defined Linear Operator

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Theorem

Let $\HH$ be a Hilbert space over $\C$.

Let $\struct {\map D T, T}$ be a densely-defined linear operator.

Let $\map \sigma T$ be the spectrum of $T$.


Then:

$\map \sigma T = \map {\sigma_p} T \cup \map {\sigma_s} T \cup \map {\sigma_r} T$

where:

$\map {\sigma_p} T$ is the point spectrum of $T$
$\map {\sigma_s} T$ is the continuous spectrum of $T$.
$\map {\sigma_r} T$ is the residual spectrum of $T$


Proof

Let $\lambda \in \map \sigma T$.

Then, from the definition of the resolvent set of $T$, at least one of the following is false:

$(1) \quad$ $T - \lambda I$ is injective
$(2) \quad$ $\map {\paren {T - \lambda I} } {\map D T}$ is everywhere dense in $\HH$
$(3) \quad$ $\paren {T - \lambda I}^{-1}$ is bounded.

Note that either $(1)$ is true and one of $(2)$ or $(3)$ is false, or $(1)$ is false.

So we have three cases:

$(\text A) \quad$ $(1)$ is false
$(\text B) \quad$ $(1)$ is true, $(2)$ is true, $(3)$ is false
$(\text C) \quad$ $(1)$ is true, $(2)$ is false

Note that the last case includes both the case that $(3)$ is true and $(3)$ is false.

Clearly only one of $(\text A)$, $(\text B)$ or $(\text C)$ can be true.

$(\text A)$ is precisely the definition of the point spectrum, $(\text B)$ is precisely the definition of the continuous spectrum and $(\text C)$ is precisely the definition of the residual spectrum.

So we have the result.

$\blacksquare$


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