Let $I$ be a bounded open interval, assume for simplicity that $I=(0,1)$. Let $(u_n)_n \subset W_0^{1,p}(I)\ \ \ (1<p<\infty)\ $ be a sequence such that the sequence of weak derivatives $(u_n') \subset L_p(I)$ is bounded and $u_n(t)\to u(t)$ for all $t \in I$. How can one show that $u_n\to u$ $\textbf{uniformly}$ on $\overline I$?
My attempt: By Poincare's inequality, we have that $(u_n)_n$ is bounded and since $W_0^{1,p} \hookrightarrow C(\overline I) \ \textbf{compactly}$, we see that $(u_n)_n$ is relatively compact and thus we can find a uniformly convergent sequence: $$u_{n_k} \to \hat u \ \ \text{ in } \ C(\overline I) .$$ Since $u_n(t) \to u(t)$ for all $t \in I$, we infer that $u=\hat u$. Can we somehow show that $u_n \to u$ in $C(\overline I)$?
Thanks in advance!
You can use a subsequence-subsequence argument:
Assume that $u_n$ does not converge towards $u$ in $C(\bar I)$. Then, there is a subsequence $u_{n_k}$ with $\| u_{n_k} \ge u\|_C \ge \varepsilon > 0$. But your argument shows that this subsequence has a subsequence which converges towards $u$ in $C(\bar I)$. This is a contradiction.