Let $u_\epsilon \in W^{1,p}(\Omega)$ be a sequence with $\Omega \subset \mathbb{R}^n$ bounded. Let $u_\epsilon \rightharpoonup^* u$ weakly* in $L^\infty (\Omega)$ (for a subsequence) with $u \in W^{1,p}(\Omega)$ and assume $\sup_\epsilon \vert \nabla u_\epsilon\vert_{L^p (\Omega; \mathbb{R}^n)} < C$.
Proof that $\nabla u_\epsilon \rightharpoonup \nabla u$ weakly in $L^p (\Omega; \mathbb{R}^n)$ (for a subsequence).
My idea so far:
$\nabla u_\epsilon$ is bounded, therefore there exists a weak limit (for a subsequence).
$u_\epsilon \rightharpoonup^* u$ weakly* in $L^\infty (\Omega)$, therefore $u_\epsilon \rightharpoonup u$ weakly in $L^p (\Omega)$, which means that $~\int_\Omega \phi \partial_x u_\epsilon \to \int_\Omega \phi \partial_x u ~$ for all $~\phi \in W^{1,p^*}(\Omega)$.
If I am correct, the weak convergence in $L^p (\Omega; \mathbb{R}^n)~$ is $~\int_\Omega \Phi \cdot \nabla u_\epsilon \to \int_\Omega \Phi \nabla u~$ for all $~\Phi \in L^{p^*} (\Omega; \mathbb{R}^n)$, but I can't seem to get that from the lines above.
Any help is appreciated.