Let $\xi$ be a compactly supported distribution on $\mathbb R^n$ and assume that its Fourier transform is in $L^p$. Let $\phi:\mathbb R^n\to\mathbb R^n$ be a diffeomorphism. Does the Fourier transform of $\phi^*(\xi)$ always lie in $L^p$?
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$\begingroup$ tided up spelling errors and some grammar $\endgroup$– Yemon ChoiCommented Aug 18, 2016 at 17:21
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1$\begingroup$ @Denis Serre - the question does state that $\xi$ is compactly supported. If $\xi$ is an $L^2$ function of compact support, then $\phi\xi$ is in $L^2$ for any diffeomorphism $\phi$, unless I'm missing something? $\endgroup$– Pablo ShmerkinCommented Aug 19, 2016 at 15:53
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$\begingroup$ Oh ! I see. I missed the compact support of $\xi$. $\endgroup$– Denis SerreCommented Aug 19, 2016 at 18:37
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2 Answers
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Don't think so. Consider the surface measure on a compact surface (e.g. a sphere). Its Fourier transform has a rate of decay which depends on the curvature of the surface; more precisely, on the order of degeneracy of the surface. By a change of variables it is easy to change the order of degeneracy locally. If you are familiar with these topics it should be trivial to construct an explicit counterexample.
answered Aug 18, 2016 at 23:10
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$\begingroup$ How does the decay rate of the Fourier transform depend on the curvature of the surface? $\endgroup$ Commented Aug 19, 2016 at 13:30
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1$\begingroup$ @IgorKhavkine: pretty complicated for the degenerate cases (I am not sure if the sharp answer is known in the degenerate case). But in the non-degenerate case the decay corresponding to a hypersurface is $|x|^{-(d-1)/2}$. In the fully degenerate case there is no decay (think delta function of a hyperplane). See e.g. terrytao.wordpress.com/2010/12/28/… for an overview. $\endgroup$ Commented Aug 19, 2016 at 15:08
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$\begingroup$ Is it true that the first work on this topic is due to Strichartz ? $\endgroup$ Commented Aug 19, 2016 at 15:34
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$\begingroup$ @DenisSerre: doesn't Tomas-Stein predate Strichartz? $\endgroup$ Commented Aug 19, 2016 at 15:46
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1$\begingroup$ @WillieWong, thanks! This discussion seems to suggest that Sobolev-type wave front sets (which use Sobolev/$L^p$ norms to measure the decay of the Fourier transform of a distribution after localization by a bump function) do not pull-back/push-forward in a clean way under diffeomorphisms. $\endgroup$ Commented Aug 19, 2016 at 21:33
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This is false in all dimensions, even if $\phi$ is real-analytic. In dimensions $n\ge 2$ Piero D'Ancona already explained why: the Fourier transform of surface measure on a spherical cap in $\mathbb{R}^n$ is bounded by $O(1)|\xi|^{-(n-1)/2}$ (what matters here is curvature), while surface measure on a piece of hyperplane does not decay at all in the direction orthogonal to the hyperplane, so it is in no $L^p$ with finite $p$ (see edit below for explanation).
In the line, let $\mu$ be the uniform (Cantor-Lebesgue) measure on the ternary Cantor set. It is well known that $\widehat{\mu}(\xi)$ does not tend to $0$ as $\xi\to\infty$ (since $\widehat{\mu}(3^k\xi)=\widehat{\mu}(\xi)$), so it is in no $L^p$ space with $p<\infty$. However, Kaufman proved that there is $\delta>0$ such that for any $C^2$ map $\phi$ with $\phi''>0$ everywhere, the push down measure $\phi\mu$ satisfies that $$ |\widehat{\phi\mu}(\xi)| \le O(1)|\xi|^{-\delta}, $$ so that $\widehat{\phi\mu}$ is in $L^p$ for sufficiently large $p=p(\delta)$.
I believe it is even possible to construct examples of measures $\mu$ and diffeomorphisms $\phi$ such that $\widehat{\mu}\notin L^p$ for any finite $p$ and $\widehat{\phi\mu}\in L^{2+\varepsilon}$ where $\varepsilon>0$ is arbitrarily small.
Edit: The Fourier transform of compactly supported probability measures is uniformly continuous (in fact Lipschitz). Then, if $\widehat{\mu}$ does not tend to $0$ at infinity, there is $c>0$ such that $|\widehat{\mu}(\xi)|>c$ has infinite measure, so $\widehat{\mu}$ can be in no $L^p$ space, $p<\infty$.
answered Aug 19, 2016 at 15:24
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$\begingroup$ I don't understand: one can take $\phi$ the identity, and then $\widehat{\phi^*\mu}=\hat\mu$ does not decay. $\endgroup$ Commented Aug 19, 2016 at 15:35
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$\begingroup$ @Denis Serre - thanks, it was a typo, $\phi$ has to be strictly convex, not strictly increasing. $\endgroup$ Commented Aug 19, 2016 at 15:37
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$\begingroup$ @ChristianRemling - $\widehat{\mu}$ is Lipschitz, so in this case it is true that if it doesn't go to $0$ then it is in no $L^p$, added edit explaining this. $\endgroup$ Commented Aug 20, 2016 at 13:16
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$\begingroup$ @PabloShmerkin: Yes, you're right of course, that was obvious in fact. $\endgroup$ Commented Aug 20, 2016 at 15:36