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.



Resolvent of Laplacian
Sure, since $-\Delta$ is a positive operator, by the spectral theorem it can be realized as multiplication by a positive function $f(x)$ on some $L^2(X)$ space. Then $R(\xi)$ is multiplication by $\frac{1}{\xi - f(x)}$ and its operator norm is the sup norm of this function. If $\xi < 0$ then the function $\frac{1}{\xi - f(x)}$ is bounded by $\frac{1}{|\xi|}$ in absolute value, so as $\xi \to -\infty$ we have $\|R(\xi)\| \to 0$.


The point is that $\|R(\xi)\|$ equals one over the distance from $\xi$ to ${\rm spec}(-\Delta)$.


If you restrict $\xi$ to be positive the question is more interesting. On the unit circle ${\bf T}^1$ the eigenvalues of $-\Delta$ are the square integers, and for any $\epsilon > 0$ we can find $\xi > 0$ such that $|\xi - n^2| > \frac{1}{\epsilon}$ for every $n \in {\bf Z}$, so we can still ensure that $\|R(\xi)\| \to 0$. In other words, the eigenvalues are spaced farther and farther apart so $\xi > 0$ can be chosen arbitrarily far from the spectrum of $-\Delta$. But on ${\bf T}^4$ the eigenvalues of $-\Delta$ are of the form $a^2 + b^2 + c^2 + d^2$ for $a, b, c, d \in {\bf Z}$, which means that every positive integer is an eigenvalue. Thus any $\xi > 0$ is at most $\frac{1}{2}$ units away from an eigenvalue, and therefore $\|R(\xi)\| \geq 2$ for every $\xi > 0$.


