most recent 30 from http://mathoverflow.net 2013-05-24T01:05:05Z http://mathoverflow.net/feeds/question/85926 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85926/subspaces-of-a-sobolev-space McParson 2012-01-17T20:32:30Z 2012-01-18T12:35:41Z

<p>For $a \in \mathbb{R}^N\setminus{0}, N \ge 2$, and $\lambda \in \mathbb{R}$ let <code>$$ X_{\lambda,a}=\{u(\cdot+\lambda a):\, u(x)=u(|x|) \in W^{1,2}(\mathbb{R}^N)\}. $$</code> Denote by $X_a$ the closure of the direct sum: <code>$$ \bigoplus_{\lambda \in \mathbb{R}}X_{\lambda,a}. $$</code> Question: Is $X_a$ a proper subspace of $W^{1,2}(\mathbb{R}^N)$? </p>

http://mathoverflow.net/questions/85926/subspaces-of-a-sobolev-space/85929#85929 Pietro Majer 2012-01-17T21:25:32Z 2012-01-18T12:35:41Z

<p>Yes, all functions in $X_a$ are still symmetric wrto the line generated by $a$. And all traces of these functions on the affine hyperplanes orthogonal to $a$ are radially symmetric in dimension $N-1$.</p> <p><strong>edit.</strong> Consider any pair $x$ and $x'$ in $\mathbb{R}^N$ with $|x|=|x'|=1$ and $a\cdot x=a\cdot x'$. Then $|x-\lambda a |=|x'-\lambda a|\ ,$ so $u(x)=u(x')$ for all $\lambda$ and all $u\in X_{\lambda,a}$. And this symmetry is preserved taking the linear span and the closure: $u=u\circ R$ holds for any $u\in X_a$ for any orthogonal $R$ that fixes $a\ .$</p>