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If my memory serves I have heard something like "the less smooth your function $f$ is, the worse its Fourier transform $\hat{f}$ decay because its Fourier transform $\hat{f}$ needs more waves of high frequency". I now would like to formulate the claim above properly.

I am interested in relation of smoothness and decay property of Fourier transformation. I think this can be rigorously shown by the fundamental properties of Fourier transformation:

$\widehat{\partial_{x}^n f}(\xi)=(i\xi)^n\hat{f}(\xi)$ and $\widehat{(i\xi)^n f}(\xi)=\partial_{x}^n\hat{f}(\xi)$

From these equalities how can one conclude smoothness and decay property of Fourier transformation?

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up vote 3 down vote accepted

If $f$ has an integrable (weak) derivative of order $n$, then the map $\xi\mapsto \xi^n\widehat f(\xi)$ is in $C_0$, the space of continuous functions which go to $0$ at infinity. This is because we expressed $\xi^n\widehat f(\xi)$ as a Fourier transform of an integrable function.

We deduce that $\widehat f$ decays at least like $|\xi|^{-n}$.

Conversely, a good decay of Fourier transform used with inverse Fourier transform and differentiation under the integral gives smoothness of the function. Furthermore, we can "read" on the Fourier transform how smooth is the function.

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Sorry for the late reply. I thought I had commented the post. Your answer made mu understanding much better. Many thanks! – M. K. Oct 11 '12 at 4:01

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