# When is the spectrum of a compact operator countable and consists only of eigenvalues?

I'm trying to prove the following statement.

Let $\mathcal H$ be a Hilbert space and $T: D \rightarrow \mathcal H$, $D \subset \mathcal H$, a linear operator. Let $\lambda_0 \in \rho(T)$ such that $(T - \lambda_0)^{-1}$ is a compact operator. Then $(T-\lambda)^{-1}$ is compact for each $\lambda \in \rho(T)$, $\sigma(T)$ consists only of a countable number of eigenvalues and doesn't have a limit point in $\mathbb C$.

It was said that this follows from well known facts about the spectral theory of compact operators.

Now, the notable facts that I recall are: (let $K$ be a compact operator)

• each non-zero spectral value of $K$ is also an eigenvalue,
• for non-zero eigenvalues of $K$, the dimension of their eigenspace is finite,
• $\sigma(K)$ is countable and its only limit point is 0.

Unfortunately, I'm unable to cook up the desired conclusion from those properties alone. So I guess there is some missing link here.

Can someone point me into the right direction?

You didn't say that but I'm assuming you also want $T$ to be densely defined.

Let us write $R_\lambda=(T-\lambda)^{-1}$ for $\lambda\in\rho(T)$ to denote the resolvent. If $R_\lambda$ is compact for some $\lambda$, then $T$ is said to have compact resolvent (just to give you a term to search for more information).

The statements that you make are very well-known facts about operators with compact resolvent.

For your convenience I will prove them below.

Assume that $R_{\lambda_0}$ is compact for some $\lambda_0\in\rho(T)$.

Claim 1. $R_\lambda=(T-\lambda)^{-1}$ is compact for every $\lambda\in\rho(T)$.

Proof. Recall the resolvent relation:

$$R_\lambda-R_{\lambda_0} = (\lambda-\lambda_0)R_\lambda R_{\lambda_0}$$

Adding $R_{\lambda_0}$ on both sides we get

$$R_\lambda = ( (\lambda-\lambda_0)R_\lambda + I) R_{\lambda_0}.$$

Thus, we have written $R_\lambda$ as composition of a bounded operator and a compact operator and therefore it is compact. $\square$

Claim 2. $\sigma(T)$ is a discrete set (countable and no accumulation points) and consists only of eigenvalues.

Proof. Set $S=R_{\lambda_0}$. The assumptions is that $S$ is compact. We claim that this implies that $\sigma(S^{-1})$ is a discrete set of eigenvalues (which immediately implies the same for $\sigma(T)$ since $S^{-1}=T-\lambda$).

Because $S$ is invertible, we have $0\not\in\sigma(S)$, so by the spectral theorem $\sigma(S)$ consists of countably many non-zero eigenvalues with the only possible accumulation point being $0$.

Observe that $\lambda\not=0$ is an eigenvalue of $S$ if and only if $\lambda^{-1}$ is an eigenvalue of $S^{-1}$ (check that!).

Next, observe that $\sigma(S^{-1})$ consists only of eigenvalues. To see this, say $0\not=\lambda$ is not an eigenvalue of $S$. Then $\lambda\in\rho(S)$. So $S-\lambda$ is bounded and invertible. This implies that $S^{-1}-\lambda^{-1}$ is also bounded and invertible: $$S-\lambda = \lambda S(\lambda^{-1}-S^{-1})$$ $$(S^{-1}-\lambda^{-1})^{-1} = -\lambda (S-\lambda)^{-1}S$$ Thus $\lambda^{-1}\in \rho(S^{-1})$.

Putting these observations together we have proved that $$\sigma(S^{-1})=\sigma_p(S^{-1}) = \{\lambda^{-1}\,:\,\lambda\in\sigma(S)\}$$ which is a discrete set. $\square$

• I think, I understand both proofs. Thank you so much! – Amarus Nov 16 '16 at 22:19