Natural question about weak convergence. Let $u_k, u \in H^{1}(\Omega)$ such that $u_k \rightharpoonup u$ (weak convergence) in $H^{1}(\Omega)$. Is true that $u_{k}^{+}\rightharpoonup u^{+}$ in $\{u\geqslant 0\}$? You can do hypothesis on $\Omega$ if you need.

I get the idea from Richard's answer.
Let $\Omega:=(0,2\pi)$ and $u_k(x):=\frac{\cos(kx)}{k+1}$. Then $\{u_k\}$ converges weakly to $0$ in $H^1(\Omega)$ (as it's bounded in $H^1(\Omega)$, and $\int_{\Omega}(u_k\varphi+u'_k\varphi')dx\to 0$ for all test function $\varphi$). 
Assume that $\{u_k^+\}$ converges weakly to $0$ in $H^1(\Omega)$. Up to a subsequence, using the fact that $L^2(\Omega)$ is a Hilbert space, we can assume that $(v_k^+)'$ and $v_k^+$ converge weakly in $L^2(\Omega)$, where $v_k:=u_{n_k}$, respectively to $f$ and $g$. It's due to the fact that these sequences are bounded in $L^2$.
Then writing the definition $g'=f$. We also have $f=0$, hence (by connectedness of $\Omega$), $g$ is constant and should be equal to $0$.
But $$\int_{\Omega}|\sin(kx)|dx\geqslant \int_{\Omega}\sin^2(kx)dx=\pi,$$
and we should have that $2u_k^+-u=2u^+_k-u_k^++u_k^-=|u_k|$ weakly converge to $0$. 
