Does $u \in W^{1,2}_0(\Omega)$ imply $|u| \in W^{1,2}_0(\Omega)$?

Question

Is the mentioned implication valid? On the first look, of course, but I've got no clue how to prove it. I guess we also may assume, that the trace-operator exists. Thanks, FFoDWindow

Answer 1

Apparently there is a chain rule for weak derivatives; someone on this site recommends Evans and Gariepy as a reference: How to prove the chain rule with respect to weak derivatives?

EDIT: The below argument is wrong - see comments.

Define $F_\epsilon(t)=\sqrt{t^2+\epsilon^2}-\epsilon.$ For smooth $u$, the usual chain rule shows that $u_\epsilon=F(u)$ is in $W^{1,2}_0(\Omega)$ with weak derivatives $D_iu_\epsilon=F_\epsilon'(u)D_iu,$ which converge in $L^2$ to $\operatorname{sgn}(u)D_iu$ by dominated convergence (taking $\operatorname{sgn}(0)=0$). Note $\|\operatorname{sgn}(u)D_iu\|_2\leq \|D_iu\|_2.$ (Presumably they're actually equal, though I think that would take more work to show.) Since $C^\infty_0(\Omega)$ is dense in $W^{1,2}_0(\Omega)$, this proves that the map $u\mapsto |u|$ extends to a $1$-Lipschitz operator on $W^{1,2}_0(\Omega).$