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.

  • $\begingroup$ Are $\mathcal{D}(L)$, $\mathcal{D}(L^{1/2})$ equipped with their graph norms, or something else? $\endgroup$ – Nate Eldredge Apr 23 '15 at 23:44
  • $\begingroup$ @NateEldredge I think $\mathcal{D}((L)^{s})$ is getting the norm from $H^s$. $\endgroup$ – anonymous Apr 23 '15 at 23:50
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    $\begingroup$ How do we know that $\mathcal{D}(L^{1/2}) \subset H^{1/2}$? It isn't obvious to me. $\endgroup$ – Nate Eldredge Apr 23 '15 at 23:51
  • $\begingroup$ @NateEldredge By interpolation, $\mathcal{D}(L^s) = [L^2, \mathcal{D}(L)]_s$? $\endgroup$ – anonymous Apr 24 '15 at 0:46

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.


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 $\mathcal{D}(L^{1/2})$.


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