TITLE: Resolvent of Laplacian
QUESTION [7 upvotes]: Hello!
Let $(M,g)$ be a Riemannian manifold and $-\Delta$ the Laplacian on M (acting on smooth functions). Then the resolvent $R(\xi)$ of $-\Delta$ is a compact operator.
Is it possible to find for every $\epsilon>0$, a point in the resolvent set $\xi$, s.t. $\Vert R(\xi) \Vert\leq \epsilon$? 
Maybe it is very easy to prove, but I'm not so familiar with spectral theory. I hope you can help me.
Regards.

REPLY [7 votes]: This is mostly enhancing Nik Weaver's comments.  Suppose  that $M$ is compact of dimension $m$. If $m\geq 2$, then for a generic metric $g$ on $M$ the eigenvalues $\lambda_k$  of the Laplacian $\Delta_g$ are   simple. In general, for any $m$,  Weyl's spectral estimates  imply that
$$\lambda_k \sim C_m \left(\frac{k}{{\rm vol}_g(M)}\right)^{\frac{2}{m}}\;\;\mbox{as $k\to\infty$},  $$
where $C_m$ is an explicit universal constant that depends only on $m$. (Hat-tip to Marc Palm!)  In particular this shows that for $m\geq 2$ and a generic metric  we have
$$0<\lambda_{k+1}-\lambda_k =O(1). $$
Now Nik Weaver's argument shows that there exists $r_0>0$ such   for any $\xi\in [0,\infty)\setminus {\rm spec}\;(\Delta)$  we have  $\Vert(R(\xi)\Vert\geq r_0$.