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Let $H^s_p(\mathbb{R}^n)$ be the fractional Sobolev space of fractional order $s\in \mathbb{R}$, for $1<p<\infty$, and let $\phi:\mathbb{R}^n\to\mathbb{R}^n$ be a diffeomorphism. Assume that the Jacobian of $\phi$ and $\phi^{-1}$ are bounded everywhere by some constant $c$, and that $\phi$ is in a Hölder class $C^\alpha(\mathbb{R}^n)$ for some $\alpha\geq 1$ (the space of $\lfloor \alpha\rfloor$ times continuously differentiable functions, with $\lfloor \alpha\rfloor$th derivative $(\alpha-\lfloor \alpha\rfloor)-$Hölder continuous), with $\|\phi\|_{C^\alpha}\leq L$.

Is the following assertion true? If $\alpha\geq |s|$, then there exists a constant $C=C(p,n,s,\alpha,c,L)$ such that, for $f\in H^s_p(\mathbb{R}^n)$ (one can further assume that $f$ is supported on the unit ball if necessary) $$\|f\circ \phi\|_{H^s_p} \leq C\|f\|_{H^s_p}.$$

If $s$ is an integer, then straightforward computations show that the proposition holds, but for $s\not\in \mathbb{N}$, I could not find any reference. The textbook "Theory of function spaces" by Triebel and the article "Mappings of Homogeneous Groups and Imbeddings of Functional Spaces" by Vodopyanov (among others) study this type of question, but only for $\alpha$ an integer.

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  • $\begingroup$ I would not expect this to be true. If $\phi$ is in $C^\alpha$ and $f\in H^s$, we can only expect $f\circ \phi$ to be in $H^{\alpha s}$, not $H^s$. I do not see how bounds on just the Jacobian would improve this. $\endgroup$ Feb 10 '20 at 3:21
  • $\begingroup$ Indeed, one need to assume that $\alpha\geq 1$. I have changed the question. $\endgroup$
    – Vincent
    Feb 10 '20 at 8:08
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In the paper [MR3135704 Inci, H.; Kappeler, T.; Topalov, P. On the regularity of the composition of diffeomorphisms. Mem. Amer. Math. Soc. 226 (2013), no. 1062, vi+60 pp. ISBN: 978-0-8218-8741-7] you find the following

Theorem 1.1

Putting $r=0$ and choosing $s$ such that $C^\alpha\subset H^s$ gives a positive answer under slightly more strict conditions on $\Phi$: It should differ from the identity by an $H^s$ map. See also

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With respect to $s$ this result ist best possible: Take $f=Id$ on some precompact subset $U$ where $\Phi$ is not better that $H^s$ on $\Phi^{-1}(U)$, then the same holds for $f\circ \Phi$.

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