It is known that, for $n \ge 3, 2 < p< 2^\*$, the imbedding $H^1(\mathbb{R}^n) \hookrightarrow L^p(\mathbb{R}^n)$ is not compact. Let $G=O(n\_1) \times O(n\_2)\times\cdots\times O(n\_k)$, with
$n\_1+n\_2+\cdots+n\_k=n, n\_i \ge 2$, and $k \ge 1$. Define an action of $G$ on $H^1(\mathbb{R}^n)$ by $g.u=u\circ g^{-1}$, and denote by $H^1\_G(\mathbb{R}^n)$ the subspace of $H^1(\mathbb{R}^n)$ which consists of the fixed points of that action, i.e. $g.u=u$ for all $g \in G$. Then the imbedding $H^1\_G(\mathbb{R}^n) \hookrightarrow L^p(\mathbb{R}^n)$ is compact. The question is whether there exists a space $E \subsetneq H^1(\mathbb{R}^n)$, with $H^1\_G(\mathbb{R}^n) \subsetneq E$, that is compactly imbedded into $L^p(\mathbb{R}^n)$ ? 



Sobolev imbedding
Of course yes, basically you achieve compactness with $H^1\_r$ because you have local regularity plus decay at infinity (pointwise decay like $|x|^{(1-n)/2}$ to be precise, by Strauss-type inequalities). If I'm not mistaken, any weighted $H^1$ space with norm $\|\langle x\rangle^\epsilon u\|\_{H^1}$ , $\epsilon>0$, should do the trick.


