I'm trying to prove the following: If $L$ is a uniformly elliptic PDE, without costant or linear terms and with essentialy bounded coefficents $a^{ij}$. If $u\in H^1(B_2)$ satisfies $Lu = f$, then $$||u||_{L^{\infty}(B_1)} \le C(||u||_{L^2(B_2)} + ||f||_{L^{\frac{n}{2}}(B_2)}),$$ at least as long $||f||_{L^{\frac{n}{2}}}$ is small enough.
I have seen a similar result for the Laplacian and I believe it should work but I'm having trouble:
The proof for the case of the Laplacian uses estimates that come from the inequality $$\int\phi^2u^{p - 2}|\nabla u|^2 \le (\dfrac{2}{p - 1})^2 \int|\nabla \phi|^2 u^p - \dfrac{2}{p - 1}\int\phi^2u^{p - 1}\Delta u,$$
where $\phi$ is some cut off function. This inequality uses explicitly the laplacian and is proved via the divergence theorem, but in our case $u\in H^1$ so the Laplacian is not defined. I do not see how to change this inequality in some way that $L$ apears instead of $\nabla$, and without $L$ I do not know how to make $f$ appear suddenly.
In the cases I've seen for the Laplacian what they actually prove is that $$||u||_{L^{\infty}(B_1)} \le C(||u||_{L^2(B_2)}) $$ if $||f||_{L^{\frac{n}{2}}(B_2)}$ is small, so $f$ does not appear in the bounds. Can the inequality that I'm trying to prove be improved to an inequality of this form or there is a reason why $f$ has to appear in the right hand side?
If the inequality with the Laplacian can't be fixed is there any suggestion on how to make $L$ actually appear. Sobolev/Poincare inequalities seem to only make the gradient appear, not L (hence f), so I do not really know what to do.
Thanks!
For constant coefficents, I recalled you can rotate the coordinate to get it into the standard form, then apply your result.