My question is the following: Provided that $u_n \to u$ in $W^{1,p} (\Omega)$, where $\Omega$ is a smooth bounded set in $\mathbb R^n$ and $p > 1$, does $u_n^+ \to u^+$ in the strong sense?
It is easy to show that $u_n^+ \rightharpoonup u^+$ in $W^{1,p}$, since one can easily show that $u_n^+$ is bounded in $W^{1,p}$, and hence characterize the limit in $L^p (\Omega)$. Nonetheless, the problem remains, is the convergence also strong?
Thanks. Cheers D
Define $$ \theta(s)=\left\{\begin{array}{ll}s, \text{ if }s>0,\\0, \text{ else.}\end{array}\right.$$ Then first show $Du^+=\theta'(u)Du$. Hence $$ \int_{\Omega}|u_n^+-u^+|^pdx=\int_{\Omega}|\theta(u_n)-\theta(u)|^pdx\le\sup|\theta'|\int_{\Omega}|u_n-u|^pdx\to0\text{ as }n\to\infty.$$ By using $(a+b)^p\le 2^p(a^p+b^p)$, one has \begin{eqnarray} &&\int_{\Omega}|Du_n^+-Du^+|^pdx\\ &=&\int_{\Omega}|\theta'(u_n)Du_n-\theta'(u)Du|^pdx\\ &=&\int_{\Omega}|(\theta'(u_n)-\theta'(u))Du_n+\theta'(u)(Du_n-Du)|^pdx\\ &\le&2^p\left(\sup|\theta'|\int_{\Omega}|Du_n-Du|^pdx+\int_{\Omega}|(\theta'(u_n)-\theta'(u))Du_n|^pdx\right). \end{eqnarray} Since $u_n\to u$ in $L^p(\Omega)$, there is a subsequence, still defined by $\{u_n\}$, of $\{u_n\}$ such that $u_n\to u$ a.e. on $\Omega$. Since $\theta'$ is continuous a.e, $\theta'(u_n)\to\theta'(u)$ a.e. on $\Omega$. Also since $u_n\to u$ in $W^{1,p}(\Omega)$, $\{Du_n\}$ is bounded in $L^p(\Omega)$. By the DCT, the last integral goes to $0$. So $$ \int_{\Omega}|Du_n^+-Du^+|^pdx\to0\text{ as }n\to\infty.$$ Thus $u_n^+\to u^+$ in $W^{1,p}(\Omega)$.