# How to prove that this integral converges?

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?

• Thank you, there was a mistake, sorry. I edit the question. Feb 8, 2020 at 9:04

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 $$L^{p+1}$$.