Spectrum of Compact Linear Operator on Infinite-Dimensional Banach Space contains Zero

Theorem

Let $X$ be an infinite-dimensional Banach space over $\C$.

Let $T : X \to X$ be a compact linear operator.

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

Then $0 \in \map \sigma T$.

That is, $T$ is not invertible as a bounded linear operator.

Proof

Suppose $0 \not \in \map \sigma T$.

Then $T$ is invertible as a bounded linear operator with bounded inverse $T^{-1}$ so that:

$T T^{-1} = I$

From Left Composition of Compact Linear Transformation with Bounded Linear Transformation is Compact, this implies that $I$ is compact.

From Identity Operator is Compact iff Finite-Dimensional Normed Vector Space, we have that $X$ is finite-dimensional.

This is a contradiction, so we have $0 \in \map \sigma T$.

$\blacksquare$

Sources

• 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $15.3$: Compact Linear Operators