Elliptic theory on compact manifolds 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.
 A: Previous answer replaced by new one:
Fix a finite relatively open
cover $U_1, \dots, U_m$ of $M$, where
each $U_i$ is diffeomorphic to either the open ball 
or the half-ball obtained by intersecting the open ball with a closed
half-space. Let $\phi_1, \dots, \phi_N$ be a partition of unity
subordinate to this cover.
If $u$ satisfies
$$
\Delta u = f\text{ and }u = 0\text{ on }\partial M,
$$
then
$$
\Delta(\phi_iu) = \phi_if + \Delta\phi_iu + 2\nabla\phi\cdot\nabla u.
$$
If the support of $\phi_i$ is contained in an open ball, then
$\phi_iu$ vanishes on the boundary. If the support of $\phi_i$ is
contained in the half-ball, then $\phi_i u$ vanishes on
the boundary of the half-ball, because $u$ is assumed to be zero on
$\partial M$. Therefore, by the estimate for the Dirichlet problem and
the Gagliardo-Nirenberg inequality,
\begin{align*}
\|\phi_iu\|_{2,p}
&\le C_1(\|f\|_p  + \|u\|_p + \|\nabla u\|_p)\\
&\le C_2(\|f\|_p + \|u\|_p +
2\|\nabla^2u\|_p^{1/2}\|u\|_p^{1/2})\\
&\le C_2(\|f\|_p + \|u\|_p +
\epsilon\|\nabla^2u\|_p + \epsilon^{-1}\|u\|_p).
\end{align*}
Set $\epsilon = 1/(2NC_2)$. Adding everything up, we get
\begin{align*}
\|u\|_{2,p} &=\|\sum_i \phi_iu\|_{2,p} \le \sum_i \|\phi_iu\|_{2,p}\\
&\le C_2N(\|f\|_p + (1+2NC_2)\|u\|_p) + \frac{1}{2}\|u\|_{2,p}.
\end{align*}
Therefore,
$$
\|u\|_{2,p} \le C(\|f\|_p + \|u\|_p).
$$
A: For the specific question of getting rid of the low-order term $\|u\|_p$, what you really need is the compact embedding of $W^{2,p}$ in $L^p$ (Rellich–Kondrachov theorem). Indeed, suppose there is no bound $\|u\|_{2,p}\le C\|f\|_p$, then there is a sequence $u_n$ such that $\|u_n\|_{2,p}=1$ and $\|f_n\|_p\to 0$, where $f_n=\Delta u_n$. By the compact embedding, we can pass to a subsequence so that $u_n$ converges in $L^p$. Combining this with $f_n\to 0$ in $L^p$, we know $u_n$ converges in $W^{2,p}$. Let $u$ be the limit, then $u$ would be a nonzero function whose Laplacian is zero, which is impossible by your boundary conditions.
A: 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 $p$.
