Absolute value of a function in $W^{1,2}_0$?

Question

Let $u\in W^{1,2}_0(\Omega)$. Can we conclude that $|u|\in W^{1,2}_0(\Omega)$? Here $\Omega$ is an arbitrary area in $\mathbb{R}^n$.

This should hold, can you help me proving this? I have issues with the not-necessary bounded domain and Lebesgue's Dominated Convergence Theorem.. Thanks

Answer 1

I believe that this is the case. Suppose that $\Omega$ is bounded and $p \in [1,\infty[$.

First of all, note that if $u,v \in W_0^{1,p}(\Omega)$, then $\max(u,v), \min(u,v) \in W_0^{1,p}(\Omega)$. In particular, we conclude that if $u \in W_0^{1,p}(\Omega)$, then $u^+:=\max(u,0),u^-:=\max(0,-u) \in W_0^{1,p}(\Omega)$.

It suffices to note that $|u|=u^+-u^-$ in order to conclude.