Exponential decay of resolvent kernel

For the integral kernel of the Laplacian $$\Delta$$ on $$\mathbb{R}^n$$, consider the resolvent $$R(\lambda) := (\lambda - \Delta)^{-1}$$ and let $$R(\lambda; x, y)$$ be its kernel, which is a smooth function away from the diagonal $$\{x=y\}$$. If I calculated correctly, we have estimates of the form $$|R(\lambda;x, y)| \leq C e^{-\delta(\lambda)|x-y|},$$ whenever $$|x-y| \geq 1$$, where $$\delta(\lambda)$$ is a certain positive constant, depending (quite explicitly) on the distance of $$\lambda$$ to the spectrum of $$\Delta$$.

Q: On general Riemannian manifolds, do we still have estimates like this on the resolvent kernel of the Laplacian (or more general operators)?

More specifically, if $$M$$ is a complete Riemannian manifold, say with bounded geometry, and $$\Delta$$ is the Laplace-Beltrami operator, is it true that for any $$\varepsilon>0$$, there exist constants $$C, \delta>0$$ such that we have $$|R(\lambda;x, y)| \leq C e^{-\delta\,d(x, y)}$$ whenever $$d(x, y) \geq 1$$ and $$\lambda$$ has distance at least $$\varepsilon$$ from the spectrum of $$\Delta$$? Here $$d(x, y)$$ is the Riemannian distance function.

• In the slightly different setting of a Schrodinger operator $-\Delta+V(x)$ on $L^2(\mathbb R^n)$ such estimates are well known and go by the name Combes-Thomas bounds. Maybe searching for this will give something for your case also. – Christian Remling Nov 12 '18 at 20:00