# Properties of Functions in $W^{1,p}(\Omega)$ and Their Weak Derivatives

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 $$$$\tag{1} \nabla(u^{+}) = \begin{cases} \nabla u &\text{ if }u>0 \\ 0 &\text { if }u\leq0 \end{cases}$$$$ and $$$$\tag{2} \nabla(u^{-}) = \begin{cases} \nabla u &\text{ if }u<0 \\ 0 &\text { if }u\geq0 \end{cases}$$$$ almost everywhere in $$\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

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

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.

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