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!


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

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


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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.