# Do elliptic operators on Riemannian manifolds have a regularizing effect?

I'm working on my master thesis and need to handle some spectral theory of the Laplace operator on compact Riemannian manifolds and especially on the sphere. While investigating essential self-adjointness I stumbled on the following problem.

*Problem*$\quad$ In a compact Riemannian manifold $M$ let $$\Delta=\operatorname{div}\operatorname{grad}$$ and let $f\in L^2(M)$ be such that $(f, u-\Delta u)=0$ for every $u \in C^{\infty}(M)$. Prove that $f=0$.

I believe that the claim is true, because the condition $(f, u-\Delta u)=0$ means exactly that $f$ is a distributional solution of the elliptic equation $-\Delta f + f=0$, and so I expect it to be a $H^2_{\text{loc}}$ function (see Theorem 2.1 of Berezin - Shubin's book). Since $M$ is compact this must imply that $f\in H^1(M)$ so that integrating by parts we get $\lVert f \rVert_{H^1}^2=(f, f)+(\operatorname{grad}f, \operatorname{grad}f)=0$.

Unfortunately Theorem 2.1 above is set in an open subset of the Euclidean space and I don't know if it is applicable verbatim in a Riemannian manifold. Can you point me to some reference on this?

Thank you.

-
I don't know the answer to this but would first try to look in Chavel's book 'Eigenvalues in Riemannian Geometry'. – user20266 Jun 28 '12 at 18:07

Here is an example. The equation $$\Delta_g u = h$$ in local coordinates is $$g^{ij}\frac{\partial^2u}{\partial x^i\partial x^j} - \frac{1}{\sqrt g} \frac{\partial}{\partial x^i}(\sqrt g g^{ij})\frac{\partial u}{\partial x^j} = h$$
Notice that the operator on the LHS is still an elliptic operator on $\mathbb R^n$ in the given local coordinates, due to the fact that the Riemannian metric is positive definite. Therefore all of the standard elliptic regularity theorems you know for operators on $\mathbb R^n$ still apply.