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.

  • $\begingroup$ For future references, several nice observations about similar weak convergences can be found in these notes: uio.no/studier/emner/matnat/math/MAT4380/v06/… $\endgroup$
    – Voliar
    Mar 10, 2022 at 11:29
  • $\begingroup$ The question title is misleading. $\endgroup$
    – daw
    Mar 10, 2022 at 12:21

1 Answer 1


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 $v_n(x)=\int_0^x f_n(s)ds$.


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