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?

Thank you in advance!


1 Answer 1


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.


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