Radial extensions in fractional Sobolev spaces

Haïm Brezis (1, 2) , Petru Mironescu (3) , Itai Shafrir (2)

1 RU - Rutgers, The State University of New Jersey [New Brunswick]
2 TECHNION - Department of Mathematics
3 EDPA - Équations aux dérivées partielles, analyse

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 $(s-a)p$<$n$. Then $f\mapsto U_af$ is a bounded linear operator from $W^{s,p}(\partial B)$ into $W^{s,p}(B)$.

Mots clés: Sobolev spaces; fractional Sobolev spaces; radial extensions

Domaines: Analyse classique [math.CA]

Haïm Brezis, Petru Mironescu, Itai Shafrir. Radial extensions in fractional Sobolev spaces. RACSAM. Real Academia de Ciencias. Serie A, Matemáticas - Revista de la Real Academia de Ciencias Exactas, Físicas y Naturales. Serie A, Matemáticas, 2019, 113 (2), pp.707-714. ⟨10.1007/s13398-018-0510-3⟩. ⟨hal-01713818v2⟩