TITLE: Do elliptic operators on Riemannian manifolds have a regularizing effect?
QUESTION [7 upvotes]: 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.

REPLY [5 votes]: Yes. Write your equation in local coordinates and then use the usual elliptic theory there.
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.
There isn't really a standard reference for this, although it probably appears as a remark buried in most PDE books.