Propert of convergence of sequences in Sobolev space

Question

Let $ (u_n)_n \subset W^{1,p}( \Omega) $ where $ 1 \leq p \leq + \infty $ and $ \Omega $ an open set of $ \mathbb{R}^N,\ N \geq 2. $ Assume that $ u_n \to u $ in $ W^{1,p} ( \Omega). $ We want to prove that $ u_n^+ = \max\left\{u_n,0\right\} \to u^+ $ in $ W^{1,p}(\Omega). $ The convergence of $ (u_n^+)_n $ to $ u^+ $ in $ L^p( \Omega) $ is trivial. We know that $ \nabla u_n^+ = g(u_n) \nabla u_n $ where $ g(s) = \chi_{(0, + \infty)}(s). $ Thus, $$ \left|\nabla u_n^+ - \nabla u^+\right|_{L^p( \Omega)} \leq \left|\nabla u_n - \nabla u\right|_{L^p( \Omega)} + \left|(g(u_n) -g(u)) \nabla u\right|_{L^p( \Omega)}. $$ The problem is with the convergence a.e. of the sequence $ g(u_n) -g(u). $ There is problem with points $ x \in \Omega $ such that $ u(x) = 0 $ and $ u_n(x) > 0. $ Any help is welcome.

Answer 1

As your above proof, we suffice to prove that $$|(g(u_k)-g(u))\nabla u|_{L^p(\Omega)}\to0\label{1}\tag{1}$$ as $k\to\infty.$

Without loss of generality, we may assume that $\Omega\subset\mathbf R^n$ is bounded; otherwise, for any $\epsilon>0,$ there exists a large $R>0$ such that $|\nabla u|_{L^p(\Omega_R)}<\epsilon,$ we then prove \eqref{1} in $\Omega_R,$ here $\Omega_R=\Omega\cap B_R.$ Then, we can obtain \eqref{1} by sending $\epsilon\to0.$

Since $\{u_k\}$ converges to $u$ in $L^p(\Omega)$, we obtain that $\{u_k\}$ converges to $u$ with respect to the Lebesgue measure in $\Omega$, and thus there is a subsequence, which is still denoted by $\{u_k\},$ such that $$u_k\to u\quad\hbox{a.e. in $\Omega$.}$$ By the absolute continuity of integral, there exists $\delta>0$ such that $$|\nabla u|_{L^p(V)}<\epsilon$$ while $\mathcal L^n(V)<\delta.$ By Egorov’s theorem, there exists a measurable subset $U\subset\Omega$ such that $\mathcal L^n(\Omega\setminus U)<\delta$ and $u_k$ converges to $u$ uniformly on $U.$ Note that $\nabla u=0$ a.e. on $\{u=0\}.$ Thus, \begin{align*}|(g(u_k)-g(u))\nabla u|_{L^p(\Omega)}=&|(g(u_k)-g(u))\nabla u|_{L^p(\Omega\cap\{u\neq0\})}\\ \leq&|(g(u_k)-g(u))\nabla u|_{L^p(U\cap\{u>0\})}+\\ &|(g(u_k)-g(u))\nabla u|_{L^p(U\cap\{u<0\})}+\\ &2|\nabla u|_{L^p(\Omega\setminus U)}\\ \leq&|(g(u_k)-g(u))\nabla u|_{L^p(U\cap\{u>0\})}+\\ &|(g(u_k)-g(u))\nabla u|_{L^p(U\cap\{u<0\})}+2\epsilon.\end{align*} Then, by the uniform convergence, we get $$\limsup_{k\to\infty}|(g(u_k)-g(u))\nabla u|_{L^p(\Omega)}\leq2\epsilon,$$ which implies \eqref{1} by sending $\epsilon\to0.$