Sobolev imbedding - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T19:27:42Z http://mathoverflow.net/feeds/question/63532 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/63532/sobolev-imbedding Sobolev imbedding McParson 2011-04-30T15:25:22Z 2011-04-30T16:02:12Z <p>It is known that, for $n \ge 3, 2 &lt; p&lt; 2^*$, 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 <code>$n_1+n_2+\cdots+n_k=n, n_i \ge 2$</code>, 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 $L^p(\mathbb{R}^n)$ ? </p> http://mathoverflow.net/questions/63532/sobolev-imbedding/63536#63536 Answer by Piero D'Ancona for Sobolev imbedding Piero D'Ancona 2011-04-30T15:50:44Z 2011-04-30T15:50:44Z <p>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.</p>