Assume $1 \le p \le \infty$ and $U$ is bounded.
(a) Prove that if $u \in W^{1,p}(U)$, then $|u| \in W^{1,p}(U)$.
(b) Prove $u \in W^{1,p}(U)$ implies $u^+,u^- \in W^{1,p}(U)$, and \begin{align} Du^+ &= \begin{cases} Du & \, \, \, \text{a.e. on } \{u>0\} \\ 0 & \, \, \, \text{a.e. on }\{u \le 0\}, \end{cases} \\ Du^- &= \begin{cases} 0 & \text{a.e. on } \{u \ge 0 \} \\ -Du & \text{a.e. on } \{u < 0 \}. \end{cases} \end{align} (Hint: $u^+ = \lim_{\epsilon \to 0} F_\epsilon(u)$, for $$F_\epsilon(z) := \begin{cases} (z^2+\epsilon^2)^{1/2}-\epsilon & \text{if }z \ge 0 \\ 0 & \text{if }z < 0.) \end{cases}$$
(c) Prove that if $u \in W^{1,p}(U)$, then $$Du = 0 \text{ a.e.} \, \text{ on the set} \{u=0\}.$$
PDE by Evans, 2nd edition: Chapter 5, Exercise 18
Here is what I tried to do so far:
(a) If $u \in W^{1,p}(U)$, then $Du \in L^p(U)$ exists in the weak sense. How can I show that $D|u|$ exists as the weak derivative of $|u|$? Do I also need to show that $D|u|$ belongs to $L^p(U)$?
(b) Part (a) implies $(u^2)^{1/2} = |u| \in W^{1,p}(U)$. Since $W^{1,p}$ is a vector space, $W^{1,p}$ is closed under addition. Clearly, $\epsilon, \epsilon^2 \in W^{1,p}(U)$. So can I conclude that, for $u \ge 0$, we have $F_\epsilon(u) \in W^{1,p}$; therefore, $u^+ \in W^{1,p}$? And is the case for showing $u^- \in W^{1,p}(U)$ similar or nearly identical to showing $u^+ \in W^{1,p}(U)$?
(c) Suppose $u \in W^{1,p}(U)$. Then $u^+,u^- \in W^{1,p}(U)$ from part (b). If $u \le 0$, then $\|u^+\|=\sum_{|\alpha| \le k} \|D^\alpha u^+ \|_{L^p(U)} = 0$, and if $u \le 0$, then $\|u^-\|=\sum_{|\alpha| \le k} \|D^\alpha u^- \|_{L^p(U)} = 0$. If what I said is fine so far, then can I conclude that $Du=0$ a.e. on $\{u=0\}$?
This exercise is the result of Chain rule:
Recall that if $f\in C^1(\mathbb R)$ with $f'\in L^\infty(\mathbb R)$ then we have $f(u)\in W^{1,p}(U)$ if $U$ is bounded, and we have $\partial_i f(u)=f'(u)\partial_i u$ in weak sense.
Hence, for part $(a)$, the function $f(x)=|x|$ has derivative $-1$ or $1$ and is (pise-wise) $C^1$, and hence the chain rule applied.
The part $b$ works in the same way. Function $F_\epsilon$ satisfies the condition of chain and hence $F_\epsilon(u)\in W^{1,p}(U)$ and the chain rule applied. What you need is push $\epsilon\to 0$. Notice that $F_\epsilon(x)\to x^+$ for $\epsilon\to 0$.
For part $c$, notice that $u^+=u^-=0$ on the set $\{u\equiv0\}$. And from part $b$, you actually have \begin{cases} Du^+=Du &x\in \,spt(u^+)\\ Du^+=0 & x\notin\,spt(u^+) \end{cases}