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 $sgeq 1$ and $(u_n)_n$ be a sequence such that
$$|u_n|^s u_nrightharpoonup|u|^s uquadmbox{ in } W_0^{1,p}(Omega).$$
If we suppose $u=0$, it is true that
$$int_{Omega} A(x) nabla(|u_n|^s u_n) dxto 0?$$

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

Thank you in advance!