Is $H^2(\Omega)\cap H_0^1(\Omega)$ compactly embedded on $H_0^1(\Omega)$?

Question

Considering $\Omega$ bounded and $\partial \Omega$ smooth. I already know that $H^2(\Omega)\cap H_0^1(\Omega)$ is continuously embedded on $H_0^1(\Omega)$, thus if I take a bounded sequence in $H^2(\Omega)\cap H_0^1(\Omega)$ it is also bounded on $H_0^1(\Omega)$, and it has a weakly convergent subsequence, but I didn't succeed with that approach. I also tried to use Rellich Theorem that gives me a strongly convergent subsequence $u_{k_j}$ on $ L^{2}(\Omega)$ when I take $ \{ u_k\}\subset H^2(\Omega)\cap H_0^1(\Omega) : \|u_k\|_{H^2(\Omega)}\leq M $. Because $u_k$ is also bounded on $H_0^1(\Omega)$. But then I don't know how to make $$\int\limits_\Omega |Du_{k_j}|^2$$ convergent. Can anyone please help me with that? Thanks in advance.

Answer 1

As Ian suggested, the statement follows by applying the Rellich-Kondrachov theorem to the gradients.

Generally, any time you apply the Sobolev embedding on a bounded domain and give up a bit in the exponent of the target space, the embedding is compact. An intuitive reason is that when exponents are related in this way, small scale features affect the norm of the source space much more than they affect the norm of the target space. So, with a uniform norm bound you have a collection of functions that all live in the same bounded domain and whose features are limited in scale from below. There are only so many substantially different photographs of given size and given resolution.