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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?

Michael Albanese
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David Corfield
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2 Answers2

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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

thel
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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 E 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 E Speyer Oct 16 '09 at 16:14
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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.

michael lacey
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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 04:22
  • Mark Lewko and I had a discussion about this over here: http://mathoverflow.net/questions/959/fourier-transform-of-exp-xp-more-general-question . I couldn't see how to make this strategy work in dimension greater than 1. – Tom Leinster Oct 31 '09 at 14:41