Take the 2-minute tour ×   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.   share|improve this
question The resolvent of the Laplacian is compact if $(M,g)$ is
compact. In general, the spectrum of the Laplacian will not be
discrete. Consider the classical example $-y^2( \partial^2_x +
\partial^2_y)$ on $SL_2(\mathbb{Z}) \backslash \mathbb{H}$ for
example. The spectrum has a continuous part. –  Marc Palm Apr 4
'13 at 7:01 Assuming a self-adjoint realization of the
Laplacian: Yes. The resolvent $R$ of a self-adjoint operator
always satisfies the estimate $\|R(z)\|\leq |\mathrm{Im}
z|^{-1}$ if $z$ is not real. –  Sönke Hansen Apr 4 '13 at 7:02
Just to be sure that I am not c