compact inclusion of domains of unbounded operators Let $L$ be a positive self-adjoint operator defined densely on $L^2(M)$ where $M$ is a compact manifold.
Also, let $\mathcal{D}(L) \subset H^1(M)$. It is known that $\mathcal{D}(L) \subset \mathcal{D}((L)^{1/2})$ as a continuous inclusion. I am trying to see whether this inclusion is also compact. I suspect that this has something to do with Rellich's theorem and interpolation spaces, but cannot work out a rigorous proof. Any help would be appreciated.
Edit: I can also work out that $L$ has compact resolvent.
Further edit: Please take a look at my proposed answer below.
 A: The mention of $M$ and $H^1(M)$ is irrelevant.  The information on the resolvent implies that the spectrum of  $L$ is discrete with a sequence of eigenvalues which increases to infinity.  By the spectral theorem, the underlying Hilbert space can then be identified with $\ell^2$ and the other two spaces with weighted versions thereof.  It is then transparent that the  inclusion is compact as required.
A: I think I have the idea for an answer, but I would really appreciate people's opinion on this. Here is what I think: let $R(\lambda)$ denote the resolvent $(\lambda + L)^{-1}$. Then, $R(\lambda)^{1/2} : \mathcal{D}(L^{1/2}) \to \mathcal{D}(L)$ is continuous, hence maps bounded sets in $\mathcal{D}(L^{1/2})$ to bounded sets in $\mathcal{D}(L)$. Also, $R(\lambda)^{1/2} : \mathcal{D}(L^{1/2}) \to \mathcal{D}(L^{1/2})$ is compact, as $R(\lambda) : \mathcal{D}(L^{1/2}) \to \mathcal{D}(L^{1/2})$ is compact, and fractional powers of compact operators are compact. So bounded sets in $\mathcal{D}(L)$ are precompact in $\mathcal{D}(L^{1/2})$.
