Rellich's theorem from compact resolvent

Question

On a compact Riemannian manifold, we know that the Laplacian $\Delta$ has compact resolvent. In proving this, one typical way is to use Rellich's theorem about the compact embedding of $H^1(M)$ into $L^2(M)$. I was wondering if it is possible to reverse the roles, i.e., use the compactness of the resolvent $(1 - \Delta)^{-1}$ as a known fact, and use this to prove Rellich's embedding theorem. Any help would be appreciated. A reference would be nice too. Thanks!

Answer 1

If we assume a bit more, namely that $1-\Delta$ is (essentially) self-adjoint, and we denote by $A$ its self-adjoint extension, then we can define the Sobolev space $H^s(M)$, $s\geq 0$, as the domain of $A^{s/2}$ equipped with the graph norm. The compactness of the embeddings $H^s\to H^t$, $s>t$, follows from the compactness the operators $A^{-r}$, $r>0$.

This is a special case of the interpolation construction and I refer to Lions-Peetre's famous paper

Sur une classe d'espaces d'interpolation, IHES Publ. Math., 19(1964), 5-68.

Answer 2

It follows immediately from the condition on the resolvent that $\Delta$ has a discrete spectrum with eigenvalues which go to infinity. One can then use the onb of the eigen vectors to display $L^2$ as an $\ell^2$-space and $H^1$ as a weighted version thereof from which the compactness follows immediately.