# Is there any Banach space $X$ that $L^2(\Omega)$ is compactly embedded into?

Let $\Omega \subset \mathbb{R}^n$. Is there a good (*) Banach space $X$ that $L^2(\Omega)$ is compactly embedded into: $$L^2(\Omega) \subset\!\subset X$$?

If not compactly embedding, I at least would like "any bounded sequence in $L^2(\Omega)$ has a convergent subsequence in some space $X$.

(*) good in the sense that I can use such compact embedding to prove existence of solution of PDE using Faedo--Galerkin method.

• @mookid Sorry I made a mistake. I meant the compact embedding the other way round. – maximumtag Mar 7 '14 at 13:32
• Well, if you have a PDE, I doubt that the "good" space to work in is larger than $L^2$. Usually you definitely need some assumption on derivatives... – Siminore Mar 7 '14 at 13:48

Think of $L^2$ as the space of bounded functionals on $L^2$. In particular, your bounded sequence is a bounded sequence of linear functionals $f_n: L^2\to \mathbb R$. If you have a space $Z$ compactly embedded into $L^2$, then the restrictions of $f_n$ to the unit ball of $Z$ have a convergent subsequence (by the Arzelà–Ascoli theorem). Also, the restriction is an injective operator provided that $Z$ is dense in $L^2$. So, in this situation $L^2$ compactly embeds into $Z^*$.
A concrete example would be $Z=H_0^1$, for which $Z^*=H^{-1}$.