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I want to prove that if $f \in C^{1}(\mathbb{R})$ is compactly supported then its Fourier transform is integrable. I was able to prove the result for $f \in C^{2}(\mathbb{R})$ and compactly supported. I used the fact that if $f \in C^{2}(\mathbb{R})$, then $\hat{f}$ is bounded by $\frac{c}{1+{|x|}^{2}}$. So it is integrable. I failed to prove it if $f \in C^{1}(\mathbb{R})$

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This is probably false: there must exist compactly supported $C^1$-functions whose Fourier transform is not integrable. I'll look for an example. –  Denis Serre Apr 21 '11 at 6:13
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Does page 4 of the following link help? math.unc.edu/Faculty/met/s14.pdf –  Suvrit Apr 21 '11 at 7:35
    
No, but page 6 does :), thanks a lot,... I'm trying to see now if this result can be upgraded to $\mathbb{R}^{n}$ –  jessica Apr 21 '11 at 9:07

1 Answer 1

Here is a link to the paper about necessary conditions for the integrability of the Fourier transform:

http://www.heldermann-verlag.de/gmj/gmj16/gmj16043.pdf

It is stated in that paper that sufficient conditions for the integrability of the Fourier transform are given in the book

R. M. Trigub and E. S. Bellinsky, Fourier analysis and approximation of functions. Kluwer Academic Publishers, Dordrecht, 2004.

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