Full Problem
Assume $1 \leq p \leq \infty$, $U \subseteq \mathbb{R}^N$ bounded. Show that $u \in W^{1,p} (U)$ implies $u^{+}, u^{-} \in W^{1,p}(U)$, and
$Du^+ = \begin{cases} Du, \ \text{a.e. on } \{ u>0 \} \\ 0, \ \text{a.e. on } \{ u \leq 0 \} \end{cases}$
$Du^- = \begin{cases} 0, \ \text{a.e. on } \{ u \geq 0 \} \\ Du, \ \text{a.e. on } \{ u < 0 \} \end{cases}$
Hint: Define $F_\epsilon (z) = \begin{cases} (z^2 + \epsilon^2)^{1/2} - \epsilon, \ z \geq 0 \\ 0, \ z < 0 \end{cases}$ and note that $u^+ = \lim_{\epsilon \rightarrow 0} F_\epsilon (u) $.
My Question
It is clear that $F_\epsilon (u) \in W^{1,p}(U)$, and that $F_\epsilon (u(x))$ tends to $u^+ (x)$ pointwise. I believe that in order to complete the proof that $u^+ \in W^{1,p}(U)$, however, we need convergence in $W^{1,p}(U)$.
I'm not sure how to show convergence in the $W^{1,p}(U)$-norm directly, but if we take the squares of each function, we clearly have
$||F_\epsilon (u)^2 - (u^+)^2||_{W^{1,p}(U)} = || u^2 + \epsilon ^2 - 2(u^2 + \epsilon ^2)^{1/2}\epsilon + \epsilon ^2 + u^2 ||_{W^{1,p}(U)}$ $= 2\epsilon||\epsilon - (u^2 + \epsilon ^2)^{1/2}||_{W^{1,p}(U)}$, on the set $\{ u > 0 \}$, which tends to $0$ as $\epsilon \rightarrow 0$, since $u$ bounded in $W^{1,p}(U)$. Also clearly we trivially have $0$ on the set $\{ u \leq 0 \}$.
Since we are looking at $u^+$, which is already always non-negative, is taking the square like this ok for proving convergence? Or is some other method necessary? Thank you.
Note that $$ \nabla F_\epsilon (u)(x) = \begin{cases}\frac{u(x)}{\sqrt{u(x)^2 + \epsilon^2}}\nabla u(x) & u(x)\ge0\\ 0 & u(x)<0\end{cases} $$ converges pointwise to $\nabla u(x)$ if $u(x)\ge0$ or $0$ if $u(x)<0$. In addition, $|\nabla F_\epsilon (u)(x)|\le |\nabla u(x)|$. Then you can use dominated convergence to show that $\nabla F_\epsilon (u)$ converges in $L^p$ to $\nabla u^+$.
You can also use dominated convergence to proof $F_\epsilon(u)$ to $u^+$ in $L^p$.