If $u$ is a solution of $\Delta_p u=0$ weakly, then $u^{+}$ is also satisfies $\Delta_p v=0$ weakly. To prove this result, I need to prove that $$ \int_{\Omega}|\nabla v|^{p-2}\nabla v.\nabla \phi\,dx=0 $$ where $v=u^{+}$ for every $\phi\in W_{0}^{1,p}(\Omega)$, whereas I know that $$ \int_{\Omega}|\nabla u|^{p-2}\nabla u.\nabla \phi\,dx=0\:\text{ for every }\:\phi\in W_{0}^{1,p}(\Omega).$$

Can you please help me with this one?

  • $\begingroup$ I don't believe this is true. Consider the 1D-case, where $\Delta_p u = 0$ just means that $u$ is linear. $\endgroup$ – Michał Miśkiewicz Nov 8 '18 at 14:46
  • $\begingroup$ Thanks, but what about if $\Omega\subset\mathbb{R}^N$ for $N\geq 2$? $\endgroup$ – Mathlover Nov 9 '18 at 19:46
  • $\begingroup$ The same. If $u$ is a counterexample in 1D, then $v(x_1,\ldots,x_n) = u(x_1)$ is a counterexample in $n$ dimensions. $\endgroup$ – Michał Miśkiewicz Nov 10 '18 at 10:11

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