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 $n/2<p<n$, 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 $p^*=np/(n-p)$.
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$\begingroup$ 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 $H^1$). $\endgroup$– Giorgio MetafuneCommented Jan 10, 2023 at 21:17
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$\begingroup$ @Giorgio Metafune, how do you know $\|u\|_{2,p}\le C\|f\|_p$? Maybe u is not in W^{2,p}?? $\endgroup$– DejvCommented Jan 10, 2023 at 21:42
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$\begingroup$ 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. $\endgroup$– Giorgio MetafuneCommented Jan 10, 2023 at 23:22
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1$\begingroup$ Giorgio Metafune. It seems you are right. Thank you very much. $\endgroup$– DejvCommented Jan 11, 2023 at 8:27
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you could also give a look to the paper G.Di Fazio -L^p estimates for divergence form elliptic equations with discontinuous coefficients. Boll. Un. Mat. Ital. A (7) 10 (1996), no. 2, 409–420 where similar estimates are shown.
