My question is: Let $ (u_n)_n $ is a sequence of the Sobolev space $ W^{1,\infty}( \Omega) $ and $ u \in W^{1,\infty}( \Omega) $ be such that $ u_n \to u $ in $ W^{1,\infty}( \Omega) $ where $ \Omega $ is some open set of $ \mathbb{R}^N,\ N \geq 1. $ Is it true that $ u_n^+ \to u^+ $ in $ W^{1,\infty}( \Omega)? $ (Here $ u_n^+ = \max(u_n,0) $).
Clearly, $ \vert u_n^+ -u^+ \vert \leq \vert u_n -u \vert. $ Thus, $$ \vert u_n^+ -u^+ \vert_{L^{\infty}( \Omega)} \leq \vert u_n -u \vert_{L^{\infty}( \Omega)} \to 0,\ n \to + \infty. $$ The problem is with the gradient. In fact, we know that $ Du_n^+ = g(u_n) Du_n $ where $ g(s) = \chi_{(0, + \infty)}(s). $ I cannot see very well how to prove that $$ \vert g(u_n) Du_n - g(u) Du \vert_{L^{\infty}(\Omega)} \to 0,\ n \to + \infty. $$
If we write $$ g(u_n) Du_n - g(u) Du = g(u_n)(Du_n -Du) + (g(u_n)-g(u))Du, $$ we can easily see that $$ \vert g(u_n)(Du_n -Du) \vert_{L^{\infty}(\Omega)} \leq \vert (Du_n -Du) \vert_{L^{\infty}(\Omega)} \to 0,\ n \to + \infty. $$ But the convergence of the other term is not clear.