# Does weak convergence in a Sobolev space implies uniform convergence?

Let $\Omega$ be a nice open, bounded domain in $\mathbb{R}^n$. (e.g suppose $\bar \Omega$ is a smooth manifold with boundary).

Let $f_n \in W^{1,p}(\Omega),p>n$ and suppose that $f_n \rightharpoonup f$ in $W^{1,p}(\Omega)$.

Question: Is it true that $f_n$ converges uniformly to $f$?

(Since $p>n$ the $f_n$ are continuous, so this is well-defined).

I know two things:

1. If $f_n$ converges uniformly to some function $g$, then $g=f$.

2. There is always a subsequence of $f_n$ which converges uniformly to $f$. Indeed, any weakly convergent sequence is bounded in $W^{1,p}$, hence by the Rellich-Kondrachov theorem it has a subsequence which converges in $C(\Omega)$. By point $(1)$ the limit must be $f$.

So, the question is can we can conclude directly that the original sequence was uniformly convergent, without passing to a subsequence.

(This is the case when we have strong Sobolev convergence, by the way).

If $\{f_n\}$ does not converge to $f$ in $C(\Omega)$, then there is $\epsilon_0>0$ and a subsequence $\{f_{n_k}\}$ so that $\|f_{n_k} - f\|_{C^0} \ge \epsilon_0$.
But by taking further subsequence if necessary, $f_{n_k} \to g$ in $C^0$ for some $g$. Then $g$ has to be $f$. This is a contradiction.