Timeline for Does there exist a continuous function of compact support with Fourier transform outside L^1?
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
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| Nov 6, 2009 at 17:48 | comment | added | Julián Aguirre | The Fourier transform of (1-x^2)_+^a (a>0) in one dimension is given by a 0F1 hypergeometric function (Mathematica computes it exactly). It's behaviour at ∞ is y^{-(1+a)}sin(π a/2-y), so that it is integrable. Maybe one should look at (1-log(1-x^2)_+)^{-1}, or (1-log(1+log(1-x^2)))^{-1}, or ... | |
| Nov 5, 2009 at 9:56 | comment | added | Gian Maria Dall'Ara | I don't know if you can compute it explicitly, but trying to substitute x=Sin(y) in the Fourier integral gives you the oscillatory integral \int Cos^2(y) e^{-i\pi z Sin(y)}dy. The general principles of oscillatory integrals would say to look at the critical points of the phase Sin(y). Since there Sin vanishes to first order one would expect a decay z^{-1/2}, but unfortunately the coefficient Cos^2(y) vanishes (to second order), accelerating the decay for large values of z. | |
| Nov 4, 2009 at 16:54 | comment | added | Darsh Ranjan | Is sqrt(1-x^2) a counterexample in one dimension? (I tried computing it explicitly, but without success.) | |
| Nov 4, 2009 at 16:11 | history | answered | Gian Maria Dall'Ara | CC BY-SA 2.5 |