Let $\omega \subset \mathbb{R}^N$ with $\vert \omega \vert < \infty$ and let $\mathcal{H} \subset C(\overline{\omega})$ have compact compact closure in $C(\overline{\omega})$. Then, $\mathcal{H}$ also has compact closure in $L^p(\omega)$? Here, $C(\overline{\omega})$, the continous functions on $\overline{\omega}$, is equipped with the supreme norm.
Definitely, $C(\overline{\omega})\subset L^p(\omega)$, but if that is to work, the supreme norm would have to be equivalent to the $L^p$ norm... Any help would be appreciated!