Spectrum of Bounded Linear Operator contains Point Spectrum

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

$\blacksquare$