There are $u$ in $W^{1,p}(D)$ and a weakly converging subsequence $\left\{ u_{m_{k}}\right\} $ to $u$.

Question

Let $D$ be a bounded open subset with smooth boundary in $\mathbb{R}^n$, $p \in (1,\infty)$, and {$u_m$} be a bounded sequence in $W^{1,p}(D)$. Then there are $u$ in $W^{1,p}(D)$ and a subsequence {$u_{m_{k}}$} such that {$u_{m_{k}}$} weakly converges to $u$.

Help me some hints to start.

Answer 1

In general, if the sequence $\{u_n\}_{n\in\mathbb N}\subset X$ is bounded, where $X$ is a separable normed space, then $\{u_n\}_{n\in\mathbb N}$ possesses a weakly converging subsequence to a $u\in X$, with $\|u\|\le\sup_{n\in\mathbb N}\|u_n\|$, as the closed unit ball in $X$ is weak$^*$ sequentially compact, due to Banach-Alaoglou Theorem.

Answer 2

Follow @Quickbeam2k1 's comment, use Proposition $8.1$ and Theorem $3.18$ in the book Haim Brezis - Functional Analysis, Sobolev Spaces and Partial Differential Equations, we are done.