Spectrum of Bounded Linear Operator contains Point Spectrum
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Theorem
Let $X$ be a Banach space over $\C$.
Let $T : X \to X$ be a bounded linear operator.
Let $\map {\sigma_p} T$ be the point spectrum of $T$.
Let $\map \sigma T$ be the spectrum of $T$.
Then $\map {\sigma_p} T \subseteq \map \sigma T$.
Proof
Let $\lambda \in \map {\sigma_p} T$.
Then there exists $x \ne \mathbf 0_X$ such that $T x = \lambda x$.
So $\paren {T - \lambda I} x = 0$ for some $x \ne \mathbf 0_X$.
So $\ker T \ne \set {\mathbf 0_X}$.
So from Linear Transformation is Injective iff Kernel Contains Only Zero, $T$ is not injective.
So $T$ cannot be invertible as a bounded linear operator.
So $\lambda \in \map \sigma T$.
$\blacksquare$