TITLE: Are Fourier transforms of L^p stable under diffeomorphisms?
QUESTION [15 upvotes]: 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 $L^p$?

REPLY [8 votes]: 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, $p<\infty$.