Convergence in $L^p$ implies convergence of weak derivative Assume $1 \le p,q < \infty$ and $u \in L^p(\mathbb{R}^N)$ such that $|\nabla u| \in L^q(\mathbb{R}^N)$. Suppose $(u_n)$ is a sequence in $C^{\infty}_c(\mathbb{R}^N)$, which converges to $u$ in $L^p(\mathbb{R}^N)$.
Is it true that $|\nabla u_n|$ converges to $|\nabla u$| in $L^q(\mathbb{R}^N)$?
I can show that $|\nabla u_n|$ weakly converges to $|\nabla u|$, but I'm not sure about convergence in norm. Thanks for your help.

This is not true even for weak convergence. Let $p=q=2$, $N=1$ and consider the functions $u_n(x) = n^{-1}\sin nx$,  restricted to a bounded interval or multiplied by a compactly supported bump function. Then $u_n\to 0$ in the norm, but $|u_n'| = |\cos nx|$ converges to the constant function $2/\pi$ weakly. 
Weak convergence is not really compatible with nonlinear operations such as taking the magnitude of gradient. 
