Let $s\in (0, 1)$. By Corollary 7.2 in the paper https://arxiv.org/pdf/1104.4345.pdf, it is know that, if $s<1/2$, then $H^s(\mathbb R)$ is pre-compact in $L^p_{loc}(\mathbb R)$ for any $p\in [1, 2^*_s)$, being $2^*_s$ the fractional Sobolev exponent.
What happens if $s\in (\frac12, 1)$, instead?
There is any $p$ for which we can say that $H^s(\mathbb R)$ is pre-compact in $L^p_{loc}(\mathbb R)$?
It seems to me that I read it should be so for $p\in [1, +\infty]$, but I can't find confirmation anywhere.
($s=1/2$ case.) Pick your favourite $p\in [1,\infty)$ and let $s_0<1/2$ be such that $2^*_{s_0} > p$. Then note $$H^{1/2} \subset H^{s_0} \Subset L^p_{\text{loc}}.$$
As for $p=\infty$, there are counterexamples; for example, from this MO page, you may recall that if $\phi$ is a cutoff near the origin in $\mathbb R^2$, then $\phi(x) \log\log|x| \in H^1(\mathbb R^2)$. Taking the trace along the $x$-axis, we obtain a function unbounded near 0 in $H^{1/2}(\mathbb R)$. In particular, $H^{1/2}\not\subset L^\infty_{\text{loc}}$.
($s>1/2$ case.) In this case $H^s\subset C^{\alpha}$ for an appropriate $\alpha\in(0,1)$ by Morrey's inequality, and $C^\alpha\Subset L^\infty_{\text{loc}}$. So the result holds for all $p\in[1,\infty]$.