Let $\{u_n\} \subset W_0^{1,p}(\Omega)$, where $\Omega \subset \mathbb{R}^N$ is a bounded domain and $p>1$.
Assume that there exists $C \in (0,\infty)$ such that
$$
\tag{1}
\int_\Omega |\nabla u_n|^p \,dx < C,
\quad \forall n \in \mathbb{N},
$$
that is, $\{u_n\}$ is bounded in $W_0^{1,p}(\Omega)$.
Then $u_n \to u$ weakly in $W_0^{1,p}(\Omega)$ to some $u \in W_0^{1,p}(\Omega)$, up to a subsequence.
The boundedness (1) can be also interpreted as the boundedness of $\{|\nabla u_n|^\frac{p-2}{2}\nabla u_n\}$ in $(L^2(\Omega))^N$.
(Or, more generally, as the boundedness of $\{|\nabla u_n|^q\nabla u_n\}$ in $(L^\frac{p}{q+1}(\Omega))^N$).
Thus, $|\nabla u_n|^\frac{p-2}{2}\nabla u_n \to z$ weakly in $(L^2(\Omega))^N$ to some $z \in (L^2(\Omega))^N$, up to a subsequence.
Is it true that $z = |\nabla u|^\frac{p-2}{2}\nabla u$?
This question appeared in the discussion of the related question.
Best Answer
Here is a counterexample. Let $\Omega=(0,1)$. Define $$ f(x):= \begin{cases} -1 & \mbox{ if } t\in (0,2/3)\\ 2& \mbox{ if } t\in (2/3,1)\end{cases} $$ and extend $f$ $1$-periodically to $\mathbb R$. Define $f_n:=f(nx)$. Then $f_n \rightharpoonup 0$ in $L^2(\Omega)$. In addition, $$ |f|^{\frac{p-2}2}f = \begin{cases} -1 & \mbox{ if } t\in (0,2/3)\\ 2^{p/2}& \mbox{ if } t\in (2/3,1)\end{cases}, $$ so $|f_n|^{\frac{p-2}2}f_n \rightharpoonup \frac23(1-2^{p/2})\ne 0$, where the latter value is the integral mean of $|f|^{\frac{p-2}2}f $.
Now define $v_n(x)=\int_0^x f_n(s)ds$.