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Let $\Omega \subset \mathbb{R}^{N}$ and $W^{1,p}(\Omega)$ be Sobolev Space. Then we let $u\in W^{1,p}(\Omega)$ and define
(i) $u^{+} := \max\{0,u\}$
(ii) $u^{-} := \max\{0,-u\}$
Then, we claim that $u^{+}, u^{-}\in W^{1,p}(\Omega)$ and we have \begin{equation}\tag{1} \nabla(u^{+}) = \begin{cases} \nabla u &\text{ if }u>0 \\ 0 &\text { if }u\leq0 \end{cases} \end{equation} and \begin{equation}\tag{2} \nabla(u^{-}) = \begin{cases} \nabla u &\text{ if }u<0 \\ 0 &\text { if }u\geq0 \end{cases} \end{equation} almost everywhere in $\Omega$.

How to show that $u^{+},u^{-}\in W^{1,p}(\Omega)$? I know that $||u^{+}||_{p} \leq ||u||_{p}$ and $||u^{-}||_{p} \leq ||u||_{p}$ but I do not know how to show that $||\nabla u^{+}||_{p}$ and $||\nabla u^{-}||_{p}$ are bounded. Also how to show that (1) and (2) hold true? Any hint is much appreciated

Thank you very much!

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It is easy to see that $u^+$ or $u^-$ have as much weak derivatives as $u$, just set

$$\Omega^+:=\{x\in\Omega: u(x)\ge 0\},\quad\Omega^-:=\{x\in\Omega: u(x)\le 0\}$$

Hence

$$\int_{\Omega} u^+(x)\partial_j\varphi(x)\, dx=\int_{\Omega} u(x)\partial_j\varphi(x)\chi_{\Omega^+}(x)\, dx=-\int_{\Omega}\partial_j u(x)\chi_{\Omega^+}(x)\varphi(x)\, dx$$

where $\partial_j u$ is the $j$-weak partial derivative of $u$, and $\varphi$ is any test function (it holds because we can choose any test function with compact support in $\Omega^+$, hence its easy to see that it holds for any test function).

Thus $\partial_j u\chi_{\Omega^+}$ is the $j$-weak partial derivative of $u^+$, hence it follows that $\nabla u^+=\nabla u\chi_{\Omega^+}$.

Thus clearly $\|\nabla u^+\|_p\le\|\nabla u\|_p$, just note that

$$|\nabla u^+(x)|^p=|\nabla u(x)|^p|\chi_{\Omega^+}(x)|^p\le |\nabla u(x)|^p$$

for all $x\in\Omega$ and all $p\ge 0$.

The same can be shown for the other case.

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  • $\begingroup$ I dont think $\partial_{j}(u(x))\chi_{\Omega^{+}}(x)= \partial_{j} (u(x)\chi_{\Omega^{+}}(x))$ $\endgroup$ – Evan William Chandra Jun 14 at 3:41
  • $\begingroup$ @EvanWilliamChandra me neither, I didn't said that. Note that $$\int_\Omega f\chi_A=\int_{\Omega\cap A} f$$ $\endgroup$ – Masacroso Jun 14 at 3:42
  • $\begingroup$ Now I get it! Thank you very much! $\endgroup$ – Evan William Chandra Jun 14 at 3:44
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    $\begingroup$ @EvanWilliamChandra I edited to add some more explanation to see clearly why the identity with the integrals holds. $\endgroup$ – Masacroso Jun 14 at 3:51

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