Radial extensions in fractional Sobolev spaces
Abstract
Given $f:\partial (1,1)^n\to{\mathbb R}$, consider its radial extension $Tf(X):=f(X/\X\_{\infty})$, $\forall\, X\in [1,1]^n\setminus\{0\}$. In "On some questions of topology for $S^1$valued fractional Sobolev spaces" (RACSAM 2001), the first two authors (HB and PM) stated the following auxiliary result (Lemma D.1). If $0<s<1$, $1< p<\infty$ and $n\ge 2$ are such that $1<sp<n$, then $f\mapsto Tf$ is a bounded linear operator from $W^{s,p}(\partial (1,1)^n)$ into $W^{s,p}((1,1)^n)$. The proof of this result contained a flaw detected by the third author (IS). We present a correct proof. We also establish a variant of this result involving higher order derivatives and more general radial extension operators. More specifically, let $B$ be the unit ball for the standard Euclidean norm $\ $ in ${\mathbb R}^n$, and set $U_af(X):=X^a\, f(X/X)$, $\forall\, X\in \overline B\setminus\{0\}$, ${\forall\,} f:\partial B\to{\mathbb R}$. Let $a\in{\mathbb R}$, $s>0$, $1\le p<\infty$ and $n\ge 2$ be such that $(sa)p<n$. Then $f\mapsto U_af$ is a bounded linear operator from $W^{s,p}(\partial B)$ into $W^{s,p}(B)$.
 Publication:

arXiv eprints
 Pub Date:
 March 2018
 arXiv:
 arXiv:1803.00241
 Bibcode:
 2018arXiv180300241B
 Keywords:

 Mathematics  Functional Analysis;
 46E35
 EPrint:
 Brezis, H., Mironescu, P. &