Can we conclude that $v_{n}\rightarrow v$ in $L^{\infty}\left(\Omega\right)$ if $p>N$ Let $\Omega\subset\mathbb{R}^{N}$
  be a smooth bounded domain, $v_{n}\rightharpoonup v$
  in $W_{0}^{1,p}\left(\Omega\right)$
 , $\left\Vert v_{n}\right\Vert _{W_{0}^{1,p}}=1$
  $\forall n$
 . So we can assume that $v_{n}\rightarrow v$
  a.e, $v_{n}\rightarrow v$
  in $L^{s}\left(\Omega\right)$
  for $1<s<\dfrac{Np}{N-p}$
 . Can we conclude that $v_{n}\rightarrow v$
  in $L^{\infty}\left(\Omega\right)$
  if $p>N$
 ?
 A: The argument is the following: 
Let $v_n$ be a sequence in $W^{1, p}_0(\Omega)$ for some $\Omega \subset \mathbb R^N$ and $p >N$. If $v_n \to v$ weakly in $W^{1,p}_0(\Omega)$, then $||v_n||_{1, p}$ is uniformly bounded. By the Sobolev Embedding (Theorem 7.17) and the fact that $$C^{0, \alpha}(\overline \Omega) \to C(\overline \Omega)$$
is compact, there is $\tilde v\in C(\overline \Omega)$ such that $v_n \to \tilde v$ in $C(\overline \Omega)$
But you know also that $v_n \to v$ strongly in $L^s$ for some $s$, so $\tilde v = v$. 
(We are essentially using the fact that a compact operator sends weakly convergent sequence to strong convergent one in the image, see here)
Remark: Let $ u \in C(\overline\Omega)$. Then by definition of $\|u\|_{C^0}$, 
$$\{ x\in  \Omega : |u(x)| > \|u\|_{C^0}\} = \emptyset \Rightarrow \|u\|_\infty \le \|u\|_{C^0}.$$
On the other hand, we show by contradiction that we do not have strictly inequality. Assume that $\|u\|_\infty < \|u\|_{C^0}$. Then by definition, there is $x\in \Omega$ so that $u(x) > \| u\|_\infty + \epsilon$ for some small epsilon. As $u$ is continuous, there is $\delta>0$ so that $u(y) > \| u\|_\infty + \epsilon/2$ for all $y$ such that $|y-x|<\delta$. In particular, the set 
$$\{ x\in \Omega: u(x) > \|u\|_\infty + \epsilon/3\}$$
has positive measure. But that contradict the definition of $\|u\|_\infty$, which is 
$$\|u\|_\infty := \inf\{ r>0 : \{|u|\ge r\} \text{ has measure zero}\}\}.$$
