Short question cornerning some lectures notes in my current calculus of variation class:
Let $\Omega \subset \mathbb{R^n}$ be open and bounded. It is now stated that if $(\phi_j)_{j \in \mathbb{N}} \subset W^{1, \infty}(\Omega, \mathbb{R^N})$ convergences uniformly to zero and futhermore $\| \nabla\phi_j \|_{L^p} \leq C$ for some constant $C$. Then $\phi_j$ convergences weakly to $0$ in $W^{1, p}(\Omega, \mathbb{R^N})$. I don't really see why this holds, I just know that there needs to be some convergent subsequence, but can't even tell why this needs to converge to $0$. Can someone help me?
You have $\phi_j \to 0$ in $L^\infty(\Omega)$. In addition, $\phi_{j_k} \rightharpoonup \phi$ in $W^{1,p}(\Omega)$ for a subsequence. In particular, $\phi_{j,k}\rightharpoonup \phi$ in $L^p(\Omega)$.
If $\Omega$ is bounded, then $\phi_j\to0$ in $L^q(\Omega)$ for all $q\in [1,+\infty]$, in particular for $q=p$. By uniqueness of limits, $\phi=0$.