## Is the Fourier transform of exp(-||x||) non-negative?

Is the n-dimensional Fourier transform of exp(-||x||) always non-negative, where ||.|| is the Euclidean norm on R^n ? What is its support?

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 Very useful. Thanks Josh. – David Corfield Oct 17 2009 at 11:52

## 3 Answers

This Fourier transform is positive, supported everywhere, and has polynomial decay. It is the Poisson kernel evaluated at time 1, up to some rescaling.

http://en.wikipedia.org/wiki/Poisson_kernel

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 Could you clarify? According to wikipedia, the Poisson kernel is supported on the unit disc, and there is no mention of a time parameter. – David Speyer Oct 16 2009 at 16:11 Never mind, I found it. Check the last section of the article, entitled "On the upper half-space". Thanks, Josh! – David Speyer Oct 16 2009 at 16:14

These questions are closely related to the so-called stable distributions. In particular, the cauchy distribution on the real line has the characteristic function e^{-|x|}.

Go to the wikipedia page, and in the definition section set: mu=0 (this is the drift parameter) alpha=0 (this is the skewness parameter)

To get the same thing in higher dimensions, take independent copies in each coordinate.

Take note: These distributions are not square integrable--otherwise the 'universal' Central Limit Theorem would hold. The cauchy distribution is only weakly integrable.

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 Agreed that this works in one dimension, but for higher dimensions I think taking the product density doesn't always give the right formula (depending on the value of p)? – Yemon Choi Oct 31 2009 at 4:22 Mark Lewko and I had a discussion about this over here: mathoverflow.net/questions/959/… . I couldn't see how to make this strategy work in dimension greater than 1. – Tom Leinster Oct 31 2009 at 14:41

Thanks, Josh. (I wanted to know too.)

Do you know the answer when ||.|| is the p-norm for arbitrary p in [1, 2]?

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 You should probably post this as its own question (it's a lot more likely to get noticed that way). – Ben Webster♦ Oct 17 2009 at 22:01 OK, thanks. Will do. – Tom Leinster Oct 17 2009 at 23:44