Bounded sequence in $W^{1,p}(I)$ that has no convergent sub-sequence in $L^{\infty}$

Question

Let $I \subset \mathbb{R}$ be an open interval and $p \geqslant 1$. Is there a bounded sequence in $W^{1,p}(I)$ that has no convergent sub-sequence in $L^{\infty}(I)$?

Answer 1

For $p=1$, yes. Take $I = (0,1)$ and let $f_n(x) = \begin{cases} nx, & x < 1/n \\ 1, & x \ge 1/n \end{cases}$.

For $p > 1$, no. This is a standard Sobolev embedding theorem, but in this case you can prove it with Arzela-Ascoli. Recall that every function in $W^{1,p}(I)$ has a continuous version, so we work with those. Now use Holder's inequality to see that $$\begin{align*}|f(x) - f(y)| &= \left|\int_x^y f'(t)\,dt \right| \\ &\le \left(\int_x^y 1^q\,dt\right)^{1/q} \left(\int_x^y |f'(t)|^p\,dt\right)^{1/p}, && \frac{1}{p} + \frac{1}{q}=1 \\ &\le (y-x)^{1/q} \|f\|_{W^{1,p}} \end{align*}$$ If $\{f_n\}$ is a sequence bounded in $W^{1,p}$ norm, you can use this to show that $\{f_n\}$ is equicontinuous.