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

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

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Very useful. Thanks Josh. –  David Corfield Oct 17 '09 at 11:52

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 '09 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 '09 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 '09 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 '09 at 14:41