Let $\Omega$ be a compact subset of $\Bbb R^3$, $\mu(\Omega)<\infty$. I want to show that $$ \int_{\Omega} |u_m-u|^2 (|\nabla u_m|+|\nabla u|)^4 d\mu \to 0 $$ as $m\to\infty$. It is given that $u,u_m\in L^{\infty}(\Omega)$ while $\nabla u,\nabla u_m\in L^4(\Omega)$. We also have $u_m\to u$ in $L^2$ (and almost everywhere) and $\nabla u_m\to\nabla u$ in $L^2$. Moreover, $\nabla u_m\overset{w}\to \nabla u$ weakly in $L^4$ so $||\nabla u_m||_{L^4}<C$ uniformly for some $C>0$.
It appears that this follow somehow from Vitali's theorem but I don't see how. Any help is extremely appreciated.
PS. The probability tag is from the fact that I often see Vitali's theorem used there, and this question can be interpreted in term of random variables if one wish.
Yes, indeed one can prove your claim by Vitali's theorem, however a direct estimate already gives the claim if one uses the compact Sobolev embedding $W^{1,4}(\Omega) \hookrightarrow L^\infty(\Omega)$ for a bounded domain $\Omega$ of dimension $d \leq 3$: $$\begin{align} &\quad\int_\Omega |u_m - u|^2 (|\nabla u_m| + |\nabla u|)^4 d\mu \\&\leq C \|u_m - u\|_{L^\infty}^2 \cdot \int_{\Omega}(|\nabla u_m|^4 + |\nabla u|^4)d\mu \\ &\leq C' \|u_m - u\|_{L^\infty}^2 \quad\longrightarrow 0 \quad\text{for}\quad m\to\infty, \end{align} $$ where $C,C'>0$ denote constants independent of $m$.