Weak derivative of modulus I am stuck on an introductory problem on Sobolev spaces. Any help would be appreciated!
Suppose $\Omega\subseteq\mathbb{R}^n$ is bounded and open and $u:\Omega\rightarrow\mathbb{R}$ has an $\alpha$-th weak derivative $D^\alpha u$, where $\alpha$ is a multi-index with $|\alpha| = 1$. Can we conclude that $|u|$ also has a weak partial derivative, and if so, how?
Thanks in advance.
 A: Notice that $u \to |u|$ is a Lipschitz map, hence we can apply the chain rule for Sobolev functions to find the weak derivative of $|u|$. (recall that Lipschitz functions are differentiable a.e.). A proof of this fact can be found in any introductory book on Sobolev spaces.
Alternatively, we can prove it directly. Consider 
$$f_{\epsilon}(t) =  
\begin{cases}
\sqrt{t^2 + \epsilon^2} - \epsilon & \text{if}\ t > 0 \\
0 & \text{if}\ t \le 0.
\end{cases}
$$
Notice that for every test function $\varphi \in C^{\infty}_c(\Omega)$ we have $$\int_{\Omega}f_{\epsilon}(u(x))D\varphi(x)\, dx = -\int_{\Omega \cap \{u > 0\}}\frac{u(x)Du(x)}{\sqrt{u(x)^2 + \epsilon^2}}\varphi(x)\, dx.$$ From this we get that $$\int_{\Omega}u^+(x)D\varphi(x)\,dx = \lim_{\epsilon \to 0}\int_{\Omega}f_{\epsilon}(u(x))D\varphi(x)\, dx = -\int_{\Omega \cap \{u > 0\}}Du(x)\varphi(x)\,dx,$$ so that $Du^+ = \chi_{\{u > 0\}}Du$ in the sense of distributions. Similarly $Du^- = -\chi_{\{u < 0\}}Du.$ Notice that if $\{u = 0\}$ has non zero measure we have $Du = 0$ a.e. on this set. Let's put everything together: \begin{align}
\int_{\Omega}|u|(x)D\varphi(x)\,dx = &\ \int_{\Omega}u^+(x)D\varphi(x)\,dx + \int_{\Omega}u^-(x)D\varphi(x)\,dx \\
= &\ -\int_{\Omega}Du^+(x)\varphi(x)\,dx - \int_{\Omega}Du^-(x)\varphi(x)\,dx \\
 = &\ -\int_{\Omega}(Du^+(x) + Du^-(x))\varphi(x)\,dx.
\end{align}
This shows that $|u|$ has a weak gradient and also provides a formula for it. You should be able to fill in all the details where needed.
