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Jan 12, 2023 at 15:00 answer added Giuseppe Di Fazio timeline score: 1
Jan 11, 2023 at 8:27 comment added Dejv Giorgio Metafune. It seems you are right. Thank you very much.
Jan 10, 2023 at 23:22 comment added Giorgio Metafune 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, 2023 at 21:42 comment added Dejv @Giorgio Metafune, how do you know $\|u\|_{2,p}\le C\|f\|_p$? Maybe u is not in W^{2,p}??
Jan 10, 2023 at 21:17 comment added Giorgio Metafune 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, 2023 at 18:07 history asked Dejv CC BY-SA 4.0