When $X$ is Banach, the spectrum of $T$ is either finite or countably infinite.

Question

Consider the claim:

"Let $X$ be an infinite dimensional Banach space and suppose that $T\in\mathcal B(X)$ is compact. Then the spectrum of $T$ is either a finite set or a sequence converging to zero."

In the sources and references I have available, I have only been able to find the proof of this fact for when we are on an infinite dimensional Hilbert space, $\mathcal H$, and not only a Banach space. How does the proof change when one only has a Banach space structure? And how does one get around the caveats left by reducing the structure from Hilbert to Banach?

It seems to me that the proof in a Hilbert space relies on the use of an orthonormal sequence $\{e_n\}$; is this all that we rely on in the Hilbert space setting? And if it is, how do we rectify it's absence for the Banach space scenario?

Answer 1

Here is a sketch of the ingredients of a proof in the Banach space setting. Throughout $T \in \mathcal{B}(X)$ will be a compact operator on an infinite dimensional Banach space.

• Lemma: If $\lambda \in \mathbb{C} \setminus \{0\}$ is not an eigenvalue then there is a $c > 0$ such that $\|(T-\lambda)x\| \geq c \|x\|$ for any $x \in X$. In particular, $\sigma(T) \setminus \{0\}$ consists of eigenvalues of $T$.
• Lemma: If $\lambda \in \mathbb{C} \setminus \{0\}$ is an eigenvalue of $T$ then $\dim \ker(T - \lambda) < \infty$.
• Lemma: If $r>0$ then there are only finitely many eigenvalues $\lambda$ of $T$ with $|\lambda| > r$.

It is easy to see that these results will imply your desired result. The first two are fairly straightforward exercises.

The idea for the last lemma is to assume that $\lambda_j$ is an infinite sequence of eigenvalues with $|\lambda_j| > r > 0$. Then for $V_k := \operatorname{span} \bigg( \bigcup_{j = 1}^k \ker(T-\lambda_j) \bigg )$ inductively find a sequence $x_j$ such that $\|x_j\| = 1, x_j \in V_j$ and $\operatorname{dist}(x_j, V_{j-1}) = 1$. This is possible since $\ker(T-\lambda_j) \not \subset V_{j-1}$ and $\dim V_j < \infty$. To conclude, check that $Tx_j$ has no Cauchy subsequence which gives a contradiction.