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I'm reading an article which claims the following result (p.9): if $f : \mathbb{R}^{2} \to \mathbb{R}$ is of the form $f(x_1,x_2) = \sin (N x_{1}) h (g^{-1}(x))$, where $g$ is a diffeomorphism and $h$ is $C^{\infty}$ and compactly supported on $[0,1]^2$ , then if we use the notation $\langle u\rangle = \sqrt{1 + |u|^{2}}$ and $N_{1} = \pi* N *(1,0)$, then the Fourier Transform of $f$ is such that, for all $M>0$ there exists $C_M$ such that $\widehat{f}(\omega) \leq C_{M}. \big( \langle\omega - N_{1} \rangle^{-M} + \langle\omega + N_{1} \rangle^{-M} \big)$.

Knowing that the Fourier transform is defined as $\widehat{f}(\omega ) = \int e^{-i\langle x,w\rangle}f(x)dx$, I was wondering which mathematical result could justify this estimation. Is it the Payley-Wiener theorem?

Thank you for your help.

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  • $\begingroup$ I've improved the formatting a bit; the question is still the same. $\endgroup$
    – Tobias Fritz
    Commented Apr 12, 2016 at 1:27
  • $\begingroup$ I don't understand what is $N_{1}$ (is that even a number? I'll take that as a constant number), and notice that $C=C(M,f)$ (this is written implicitly in your estimate), therefore the asymptotic expression is true because $f$ is smooth. As the lower modes are controllable by the integral of $f$, you can use this implicit $C_{M,f}$ to have control via the inequality also in the lower modes. Anyhow this is far from being a research question, and should be posted to mathstack. $\endgroup$
    – Asaf
    Commented Apr 12, 2016 at 8:31
  • $\begingroup$ @Asaf: $N_1$ is a vector in $\mathbb{R}^2$, so is $\omega$, the frequency. And also, I think implicitly the constant depends on $h$ and $g$, but not on $N$; otherwise as you indicated the estimate is entirely trivial. $\endgroup$
    – Willie Wong
    Commented Apr 12, 2016 at 13:10
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    $\begingroup$ @OP: Do you know how to prove it for $N = 0$? (As Asaf said, this is just the standard decay estimate for Fourier transforms of smooth functions.) For $N \neq 0$, you get it by the formula that frequency modulation in physical space equals translation in Fourier space. E.g. en.wikipedia.org/wiki/Fourier_transform#Basic_properties (So in the end I also agree with Asaf that this should be asked on Math.SE instead.) $\endgroup$
    – Willie Wong
    Commented Apr 12, 2016 at 13:14

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$$2if(x_1,x_2)=(e^{i\langle N_1,x\rangle}-e^{-i\langle N_1,x\rangle})h(g^{-1}(x))$$ so $$2i\hat f(\omega)=\widehat{h\circ g^{-1}}(\omega-N_1)-\widehat{h\circ g^{-1}}(\omega+N_1).$$ Since $h$ is smooth and compactly supported, so is $h\circ g^{-1}$. In particular it is of Schwarz class, for which we have $$\widehat{h\circ g^{-1}}(\omega)\le C_M\langle\omega\rangle^{-M}.$$

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