Maybe this is silly.


On a bounded set $\Omega\subset\mathbb{R}^n$ consider the equation


$$ \Delta u=f \quad\text{ in $\Omega$}$$
$$ u=0\quad\text{ on $\partial\Omega$}.$$


One has the following elliptic estimates


$$ \| u\|\_{W^{2,p}}\le C\|f\|\_{L^p}. $$


Does one can have the same result if instead of $\Omega\subset\mathbb{R}^n$ one consider the same problem in a compact Riemannian manifold with boundary?
I know that one has


$$\|u\|\_{W^{2,p}}\le C(\|f\|\_{L^p}+\|u\|\_{L^p}) $$


for the general case.. is it true that one can get rid of $\| u\|\_{L^p}$ considering zero boundary conditions?


This should be true using a covering argument and the result in $\mathbb{R}^n$ but I cannot find anything in literature. Any help will be appreciated.



Elliptic theory on compact manifolds
Consider the case of a sphere. The boundary is empty. A constant $u$ satisfies the equation with $f=0$. You cannot get rid of the $u$ on the right hand side of the inequality in this case. 


EDIT: I think this example demonstrates the difficulty of trying to use only local estimates pieced together with partitions of unity to eliminate $u$ on the right hand side of the inequality.


On the other hand, a possible strategy for proving the inequality without $u$ in the right hand side could be the following:


* Note that the Laplace-Beltrami operator with Dirichlet boundary conditions is (essentially) self-adjoint on, say the domain of $u$ in $W^{2,2}$ vanishing at the boundary, and the spectrum is discrete (by compactnesss)
* A maximum principle shows that if the boundary is nonempty (and manifold connected), then $0$ is not an eigenvalue. This is the essential global ingredient.
* The resolvent $\Delta^{-1}$ at $0$ is bounded in $L^2$, which means that we can eliminate $u$ in the right hand side of the inequality for $p=2$.
* We might be able to proceed by interpolation to general $p$.


