Convergence of sequence of function for a bounded sequence in the Sobolev space

Question

Let $u_n$ be a bounded sequence in $W_{0}^{1,p}(\Omega)$. Then upto a subsequence one has $$ u_n\to u \mbox{ weakly in}\,W_{0}^{1,p}(\Omega). $$ How the following statement is true? $$ \int_{\Omega}|\nabla u_n|^{p-2}\nabla u_n\cdot\nabla\phi\,dx\to\int_{\Omega}|\nabla u|^{p-2}\nabla u\cdot\nabla\phi\,dx\,\,\forall\,\phi\in{C_c^{\infty}(\Omega)}. $$

Answer 1

I don't think that this is true. This could be a counterexample: Let $\Omega = (0,1)$ and $$ f_n(x) := \begin{cases} 0 & x < \frac12 \text{ and } \sin(4 \, \pi \, n\, x) > 0 ,\\ 2 & x < \frac12 \text{ and } \sin(4 \, \pi \, n\, x) \le 0, \\ -1 & x \ge \frac12. \end{cases} $$ We set $u_n(x) = \int_0^x f(t) \mathrm{d}t$. One can check that $u_n \in W_0^{1,p}(\Omega)$ for all $p \in (1,\infty)$ and $u_n \rightharpoonup u$ for $$u(x) = \begin{cases} x & x \le \frac12, \\ 1-x & x \ge \frac12.\end{cases}$$ However, $$ \int_0^1 |\nabla u_n|^{p-2} \nabla u_n \cdot\nabla\varphi \, \mathrm{d}x \to 2^{p-2} \, \int_0^{\frac12} \nabla\varphi \,\mathrm{d}x - \int_{\frac12}^1 \nabla\varphi\,\mathrm{d}x = (2^{p-2} + 1)\, \varphi(\frac12)\\ \ne 2 \, \varphi(\frac12) = \int_0^1 |\nabla u|^{p-2} \nabla u \cdot\nabla\varphi \, \mathrm{d}x $$ unless $p = 2$.