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 $\{\nabla u_n\}$ in $(L^p(\Omega))^N$. Thus, $\nabla u_n \to z$ weakly in $(L^p(\Omega))^N$ to some $z \in (L^p(\Omega))^N$, up to a subsequence. Since the map $u \mapsto \nabla u$ is a bounded linear operator from $W_0^{1,p}(\Omega)$ to $(L^p(\Omega))^N$, it is not hard to conclude that $z = \nabla u$.
At the same time, the boundedness (1) can be also interpreted as the boundedness of $\{|\nabla u_n|^q\}$ in $L^{p/q}(\Omega)$ for some $q \in [1,p)$.
Consequently, $|\nabla u_n|^q \to z$ weakly in $L^{p/q}(\Omega)$ to some $z \in L^{p/q}(\Omega)$, up to a subsequence.
Is it true that $z = |\nabla u|^q$?
It is very likely that the answer is affirmative.
Could you, please, help me with finding an argument or a counterexample, or a corresponding reference?
Best Answer
No, this is not true, not even for $q=p$. In general, weak convergence and nonlinear mappings do not go well with each other.
Here is a counterexample. Let $\Omega=(0,1)$. Construct $v_n\in W^{1,p}_0(\Omega)$ such that $$ \nabla v_n(x) = sign(\sin(2n\pi x)). $$ Then $v_n \rightharpoonup 0$ in all $L^p(\Omega)$, $p<\infty$, but $|\nabla v_n| \equiv 1$.