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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 $L^p(\mathbb{R}^n)$ ?

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up vote 6 down vote accepted

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.

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McPerso: you edited the original question but the argument holds the same, the underlying reason for compactness is decay at infinity of the functions with additional symmetry. – Piero D'Ancona Apr 30 '11 at 16:14
    
sorry that I modified my question, but thanks for the hint! – Mercy King Apr 30 '11 at 19:12

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