Absolute value of functions in Sobolev space I'm just trying to show that if $u\in W^{1,p}(U)$ (with $1\leq p<\infty$) then $|u|\in W^{1,p}(U)$. Here $U$ is bounded.
I originally took smooth functions $a_n$, equal to $|x|$ whenever $|x|\geq 1/n$ and which converged uniformly to $|x|$; this implies that $a_n(u)$ is Cauchy in $L^p$, so all I needed to do was show that $Da_n(u)$ was Cauchy, but then I ran into problems.
Namely, since no smooth functions approach the sign function uniformly, I can't seem to bound $||D_{x_i}(a_n(u)-a_m(u))||=||(a_n'(u)-a_m'(u))u_{x_i}||$.
Anyone know a way around this, or an easier approach? (If you're going to quote mollifiers, please be specific because I feel I've exhausted all approaches that use them!)
 A: A characterization of $W^{1,p}(U)$ is that $u\in W^{1,p}(U)$ if and only if $u$ is absolutely continuous in almost every horizontal and almost every vertical line, and the partial derivatives are in $L^p(U)$.
We assume that $u\in W^{1,p}(U)$. Note that $\left|\frac{|u(x+he_i)|-|u(x)|}{h}\right| \leq \frac{|u(x+he_i)-u(x)|}{|h|}$
so if the partial derivatives of $|u|$ exist, they will also lie in $L^p(U)$.
Furthermore, note that for almost every horizontal or vertical line $\gamma$ and for all points $x,y\in \gamma$ we have
$$||u(x)|- |u(y)|| \leq |u(x)-u(y)| = \left| \int_{\gamma|_{[x,y]}} \nabla u ds \right|  \leq \int_{\gamma |_{[x,y]}} |\nabla u| ds$$
where $\gamma|_{[x,y]}$ is the subsegment of $\gamma$ from $x$ to $y$. 
The above inequality shows that $|u|$ is absolutely continuous in almost every horizontal and almost every vertical line, and in particular the partial derivatives exist, as desired.
A: You don't need $a_n'$ to converge uniformly to the sign function; it's enough to have them uniformly bounded and converging pointwise.   (If your chosen $a_n$ functions don't do this, then pick other ones that do.)  Then you can use the dominated convergence theorem to show that $a_n'(u) u_{x_i} \to \operatorname{sgn}(u) u_{x_i}$ in $L^p$.
