Weak solution of elliptic differential equation of divergence type

Assume that $$u\in W^{1,2}(B^n,R)$$ is a weak solution to the elliptic pde of type $$\sum_{i,j=1}^n\partial_j \left(a_{ij}(x) \partial_i u(x)\right)=f\in L^p(B^n),$$ where $$n/2, and $$A=(a_{ij}(x))$$ is a smooth elliptic symmetric matrix defined in the unit ball $$B^n$$. Assume also that $$u|_{\partial B^n}=0$$. I need a reference to the following claim $$\|Du\|_{L^{p^*}}\le C(\|f\|_{L^p}+\|Du\|_{L^2}),$$ where $$p^*=np/(n-p)$$.

• If $p^* \leq 2$ the inequality is obvious (omitting $\|f\|_p$ on the RHS. In the other case, since the coefficients are smooth, you can use the estimate $\|u\|_{2,p} \leq C\|f\|_p$ and then Sobolev embedding $\|Du\|_{p^*} \leq C\|u\|_{2,p}$ to obatain the inequality without $Du$ on the right hand side (when $p^* >2$ the $W^{2,p}$ solution is in $H^1$). Jan 10 at 21:17
• @Giorgio Metafune, how do you know $\|u\|_{2,p}\le C\|f\|_p$? Maybe u is not in W^{2,p}??
– Dejv
Jan 10 at 21:42
• Solve in $W^{2,p}$ and you get a solution in $H^1$ which, by uniquesess in $H^1$, coincides with your original solution. This is what I have in mind. Jan 10 at 23:22
• Giorgio Metafune. It seems you are right. Thank you very much.
– Dejv
Jan 11 at 8:27