How to prove that this integral converges?

Question

EDITED. Let $p > 2$ and $\Omega \subset R^n$ an open bounded subset. Let $(u_n)_n, v\in W_0^{1, p}(\Omega)$ such that \begin{align*} \int_{\Omega} \vert\nabla u_n \vert^p dx \rightarrow 0 \quad \mbox{ and } \quad \Vert v\Vert_{W_0^{1,p}}\leq 1. \end{align*} I want to prove that \begin{align*} \int_{\Omega} \vert\nabla u_n\vert^{p - 1} \vert \nabla u_n\cdot\nabla v\vert dx \rightarrow 0. \end{align*} I am proceeding in this way (by using Cauchy - Schwartz inequality), but I'm not sure that this is right: \begin{align*} \int_{\Omega} \vert\nabla u_n\vert^{p - 1} \vert \nabla u_n\cdot\nabla v\vert dx \leq \int_{\Omega} \vert\nabla u_n\cdot\nabla v\vert^p dx \left(\int_{\Omega}\vert\nabla u_n\vert^p dx\right)^{\frac{p - 1}{p}} \rightarrow 0. \end{align*} Could anyone help?

Answer 1

This is false! Take $u=v$. Then you are asking if $\|\nabla u\|_{L^{p}} \to 0$ implies $\|\nabla u\|_{L^{p+1}} \to 0$. But there are functions in $L^p$ that are not in $L^{p+1}$.