Let $\Omega$ be an open bounded subset of $\mathbb{R}^n$ and $p>1$. Let $(u_n)_n, u$ in $W_0^{1, p}(\Omega)$ be such that $$u_n\rightharpoonup 0 \quad \mbox{ in } W_0^{1, p}(\Omega).$$ Then, it is true that $$\lim_{n\to +\infty}\int_{\Omega} \nabla (u_n - u) dx =0.$$

It was on my notes, so I guess it is true, but I don't understand why.

Could anyone help, please?

Thank you in advance.

  • $\begingroup$ What is $u?\,\,$ $\endgroup$ Jun 17, 2020 at 9:57

1 Answer 1


Observe that $u_n\rightharpoonup u$ in $W_0^{1,p}(\Omega)$ if and only if $u_n\rightharpoonup u$ in $L^p(\Omega)$ and $\nabla u_n\rightharpoonup \nabla u$ in $L^p(\Omega)$ (this is a slight abuse of notation, as I'm really talking about each component). Since $1\in L^p(\Omega)$ ($\Omega$ is bounded), we actually get (in particular) that $\nabla u_n\rightarrow \nabla u$ in $L^1(\Omega)$ (again, same abuse of notation).


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