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Given a function $f$ and an invertible matrix $A$, we have the following relation for the Fourier transforms: $$ \widehat{f \circ A}(\xi) = |\det A|^{-1} \widehat f(A^{-T}\xi). $$ In particular, $\widehat f(\xi) = O(|\xi|^{-\alpha})$ at infinity if and only if $\widehat{f \circ A}(\xi) = O(|\xi|^{-\alpha})$.

Now, given a $C^\infty$-diffeomorphism $g$ which differs from identity only on some compact set $K$ we can only write (see this answer) $$ \widehat{f \circ g}(\xi) = \int f(y)|\det g'(y)|^{-1}\exp(ig^{-1}(y)\xi) dy. $$ Is it still possible to say that $\widehat f(\xi) = O(|\xi|^{-\alpha})$ at infinity if and only if $\widehat{f \circ g}(\xi) = O(|\xi|^{-\alpha})$?

Joonas Ilmavirta
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Appliqué
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1 Answers1

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Let $F(x)=f(g(x))-f(x)$. Now $F$ is supported in the compact set $K$ outside which $g$ is the identity, so by the Paley–Wiener theorem its Fourier transform decays faster than any power of $\xi$. We have $\hat F=\widehat{f \circ g}-\hat f$, so indeed $\widehat f(\xi) = O(|\xi|^{-\alpha})$ if and only if $\widehat{f \circ g}(\xi) = O(|\xi|^{-\alpha})$ for any $\alpha$, since the difference between the two is $o(|\xi|^{-\alpha})$. This allows comparing any decay rates that are slower than what the Paley–Wiener theorem produces.

This was under the assumption that $f$ and $g$ are smooth. If you have continuous differentiability of some finite order, you will get the decay estimate only up to that order. For general distributions there is no decay for $F$. For example, if $f=\delta_a$ and $g(a)=b$, then $F=\delta_b-\delta_a$ and $\hat F(\xi)$ is the difference of two plane waves. However, this does not disprove your claim since neither $\hat f$ nor $\widehat{f\circ g}$ has polynomial decay.

Joonas Ilmavirta
  • 25,809
  • The Paley-Wiener theorem states that the Fourier transform of $F$ is analytic. But how do you get the decay? Your $F$ is not necessarily smooth in $K$ – Appliqué Apr 23 '17 at 23:04
  • @Appliqué There are unfortunately many versions of the Paley-Wiener theorem. The Fourier transform of a compactly supported smooth function has good decay: see this question. What regularity do you assume for $f$? It was not explicitly mentioned in the question. If $f$ and $g$ are smooth, so is $F$. – Joonas Ilmavirta Apr 23 '17 at 23:22
  • $f$ is not $C^\infty$ (otherwise the question does not make sense), typically it continuous and its derivative may have discontinuities along hypersurfaces, for example – Appliqué Apr 23 '17 at 23:53
  • @Appliqué On a second thought I realize that the question indeed only makes sense for non-smooth $f$. (It would have helped to make this explicit in the question.) My intuition is that the answer is yes, but my method of proof is insufficient. I'll try to see if I can figure out a different way to approach it. – Joonas Ilmavirta Apr 24 '17 at 00:02