# Weak convergence in $W_0^{1,p}(\Omega)$ implies weak convergence in $(L^2(\Omega))^N$?

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.

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

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