Absolute value of functions in Sobolev space

Solution 1:

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.

Solution 2:

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 $L^p$.