Decay of Fourier transform of function composition 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 $\widehat{f \circ g}(\xi) = O(|\xi|^{-\alpha})$?
 A: 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.
