# Does a self-adjoint operator with discrete spectrum have compact resolvent?

I know that if one has a densely-defined operator $T$ with compact resolvent $(T-\lambda)^{-1}$ (for some $\lambda\in\mathbb{C}$) then its spectrum $\sigma(T)$ comprises a countable set of isolated eigen-values of finite multiplicity, accumulating at no finite point.

What I don't know (and can't seem to find anywhere) is whether there is some sort of converse implication in the case of $T$ being a self-adjoint operator on a Hilbert space. Can anybody offer an answer to the title question? If positively, please may I have a reference?

• What about $T=\Bbb1$, this has discrete spectrum but the resolvent is never a compact operator. Apr 17 '18 at 10:34
• The eigen-value $1$ for your $T$ has infinite multiplicity, since the eigen-space is the entire Hilbert space. I'm more interested in the situation where all the eigen-values of $T$ have finite multiplicities. Apr 17 '18 at 14:01
• What would be the type of converse statement you want? Apr 17 '18 at 17:09
• For example, the statement "Let $T$ be self-adjoint and suppose $\sigma(T)$ is discrete (countable, isolated eigen-values of finite multiplicity) and accumulates only at $\pm\infty$. Then $(T-\lambda)^{-1}$ is compact for any $\lambda\in\rho(T)$." would suffice for my purposes! :-) Apr 20 '18 at 13:11
• I have the feeling it's somehow a totally obvious consequence of the spectral theorem, but I can't tease it out from any of the statements of the theorem. Apr 20 '18 at 13:12

For $\lambda$ in the resolvent set, $(T-\lambda)^{-1}$ is the limit in the operator norm topology of the following finite rank operators: $$K_n = \int_{|\sigma(T)| \le n} \dfrac{1}{\lambda-t} \, dP(t)$$ where $P$ is the projection valued measure such that $T = \int_{\sigma(T)} t \, dP(t)$. These are finite rank since the range of $K_n$ is just the union of finitely many finite-dimensional eigenspaces. $$(T-\lambda)^{-1}-K_n = \int_{|\sigma(T)|>n} \dfrac{1}{\lambda-t} \, dP(t)$$ which goes to $0$ as $n \to \infty$.