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 $\xi > 0$.

`$R$`

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