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

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? $$\int_{\Omega}|\nabla u_n|^{p-2}\nabla u_n\cdot\nabla\phi\,dx\to\int_{\Omega}|\nabla u|^{p-2}\nabla u\cdot\nabla\phi\,dx\,\,\forall\,\phi\in{C_c^{\infty}(\Omega)}.$$

• Do you have a reference for that? Please edit it into the question. – Fritz May 17 at 19:35
• YES. PLEASE SEE LEMMA 3.9 IN THE ARTICLE : ejde.math.txstate.edu/Volumes/2012/35/belhaj.pdf – Mathlover May 18 at 8:34
• I'm wondering about the same question at the moment, have you been able to find a solution? By the way, the source you mention further requires strong convergence in $L^p$, which is probably the reason why the counterexample in the answer below works. – jason paper Jul 28 at 15:19
• I think I found the solution, its explained on page 10 of the above article. It uses that $u_n$ is defined as solution to the eigenvalue problem (1.6). We can apply the inverse of the p-Laplacian to this equation, so on the left-hand side we get exactly $u_n$. For the right-hand side we know strong $L^\frac{p}{p-1}$ convergence from the strong $L^p$ convergence of $u_n$, so the application of the inverse $p$-Laplacian yields strong $W^{1,p}$-convergence which allows us to identify the limit of the desired integral. – jason paper Jul 28 at 20:58

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 $$p = 2$$.