Let $u_n \rightharpoonup u$ in $W^{2,p}(\Omega)$ for sufficiently large $p$. Does $\Delta u_n^+ \rightharpoonup \Delta u^+$ in $L^2(\Omega)$?

Question

On a smooth and bounded domain, let $u_n \rightharpoonup u$ in $W^{2,p}(\Omega)$ for sufficiently large $p$. Does $\Delta u_n^+ \rightharpoonup \Delta u^+$ in $L^2(\Omega)$?

Here $u^+$ means the positive part of $u$.

I think it is no, since we only have weak convergence in $L^2$ of the Laplacians of $u_n$.. but do I miss anything?

Answer 1

For example, let $u_n(x) = n^{-2} \sin nx$ (the domain being a one-dimensional interval such as $[0, 1]$.). Then $u_n'' = -\sin nx$ which converges weakly (in any $L^p$ with $p<\infty$) to zero. So, $u_n\to 0$ weakly in $W^{2, p}$.

On the other hand, $(u_n'')^+ $ does not converge weakly to zero, because $$ \int_0^1 (u_n'')^+ \to \frac{1}{\pi} $$ (the average value of $(\sin x)^+$ is $1/\pi$).