# Weak convergence in Sobolev Space: does this integral converge to $0$?

Let $$\Omega$$ be an open bounded domain in $$\mathbb{R^N}$$ and $$A(x)\in L^{\infty}(\Omega)$$. Let $$s\geq 1$$ and $$(u_n)_n$$ be a sequence such that $$|u_n|^s u_n\rightharpoonup|u|^s u\quad\mbox{ in } W_0^{1,p}(\Omega).$$ If we suppose $$u=0$$, it is true that $$\int_{\Omega} A(x) \nabla(|u_n|^s u_n) dx\to 0?$$

About me the answer is yes, I reasoned in this way. Since $$|u_n|^s u_n\rightharpoonup|u|^s u$$ in $$W_0^{1, p}(\Omega)$$ and $$u=0$$, thus $$|u_n|^s u_n\rightharpoonup 0$$ in $$L^p(\Omega)$$ and $$\nabla(|u_n|^s u_n)\rightharpoonup 0$$ in $$L^p(\Omega)$$. Moreover, since $$1\in L^p(\Omega)$$ and since $$A(x)\in L^{\infty}(\Omega)$$, a constant $$c\geq 0$$ exists such that $$\int_{\Omega} |A(x)| \nabla(|u_n|^s u_n) dx\leq c \int_{\Omega} \nabla(|u_n|^s u_n) dx\to 0.$$ It is true or am I missing something? Could anyone please help?

Your idea is right and your proof is almost correct. However your last estimate $$\int_{\Omega} |A(x)| \nabla(|u_n|^s u_n) \,\mathrm{d}x\leq c \int_{\Omega} \nabla(|u_n|^s u_n) \,\mathrm{d}x\to 0$$ does not imply that $$\int_{\Omega} A(x) \nabla(|u_n|^s u_n) \,\mathrm{d}x$$ vanishes in the limit, as these quantities can be signed (note $$\nabla(|u_n|^s u_n)$$ is not necessarily non-negative!).
Instead observe that since $$\Omega$$ is bounded and $$A \in L^{\infty}(\Omega),$$ we have $$A \in L^{p'}(\Omega) \cong L^p(\Omega)^*.$$ Hence as $$\nabla(|u_n|^su_n) \rightharpoonup 0$$ in $$L^p(\Omega),$$ we have
$$\lim_{n \to \infty} \int_{\Omega} A(x) \nabla(|u_n|^su_n)\,\mathrm{d} x = \int_{\Omega} A(x) \cdot 0 \,\mathrm{d} x = 0$$ as required.