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Let $f\in L^m(\Omega)$ for some $m>1$, where $\Omega$ is a smooth bounded domain in $\mathbb{R}^N$ ($N\geq 2$). Consider the equation, $$ \Delta_p u=f(x) $$for $p=N$, then $u$ is bounded in $\Omega$. Moreover, $u$ is continuous upto the boundary. Can anyone help me with the solution of this one. I have got this question while going through the paper : Lemma 3.7 of the paper below https://link.springer.com/content/pdf/10.1007%2Fs00030-016-0361-6.pdf

Thank you very much.

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  • $\begingroup$ What is $p$, $m$ and $N$? $\endgroup$ Commented Nov 10, 2018 at 23:09

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I'm not sure what your assumptions are, but I assume $p$ is equal to the dimension of the domain.

In the special case $f \equiv 0$ the continuity of $u$ can be easily shown. Look up Peter Lindqvist's Notes on the $p$-Laplace equation, section 3.2. The proof can be adjusted to work in the case of sufficiently regular nonzero right-hand side $f$.


As for the continuity-up-to-the boundary, this obviously depends on the boundary data. If one solves the equation with non-continuous boundary data (chosen in the trace space for $W^{1,n}$), then the solution is not in $C(\overline{\Omega})$.

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  • $\begingroup$ I am extremely sorry for your confusion. Here $p>1$. $N\geq 2$ is the dimension of $\mathbb{R}^N$ where I assumed $p=N$. Moreover suppose $f\in L^m(\Omega)$ is nonnegative for some $m>1$. Then I want to prove that any weak solution $u\in W_0^{1,p}(\Omega)$ of the equation $-\Delta_p u=f$ in $\Omega$ belong to $C_0(\overline\Omega)$. $\endgroup$
    – Mathlover
    Commented Nov 11, 2018 at 4:28
  • $\begingroup$ May I kindly request you to have a look at the following question. $\endgroup$
    – Mathlover
    Commented Nov 11, 2018 at 4:33
  • $\begingroup$ math.stackexchange.com/questions/2992644/… $\endgroup$
    – Mathlover
    Commented Nov 11, 2018 at 4:33
  • $\begingroup$ You can (and should) edit your question so that it's clear for everyone (there might be someone else willing to answer your question, so why confuse him too). $\endgroup$ Commented Nov 11, 2018 at 8:37
  • $\begingroup$ And $f \in L^m$ with some $m > 1$ should be enough to adjust the proof I linked. If you show your efforts, I can comment on possible difficulties. $\endgroup$ Commented Nov 11, 2018 at 8:38

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