Is the $n$dimensional Fourier transform of $\exp(\x\)$ always nonnegative, where $\\cdot\$ is the Euclidean norm on $\mathbb{R}^n$? What is its support?
This Fourier transform is positive, supported everywhere, and has polynomial decay. It is the Poisson kernel evaluated at time 1, up to some rescaling.

$\begingroup$ Could you clarify? According to wikipedia, the Poisson kernel is supported on the unit disc, and there is no mention of a time parameter. $\endgroup$ Oct 16 '09 at 16:11

$\begingroup$ Never mind, I found it. Check the last section of the article, entitled "On the upper halfspace". Thanks, Josh! $\endgroup$ Oct 16 '09 at 16:14
These questions are closely related to the socalled 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 integrableotherwise the 'universal' Central Limit Theorem would hold. The cauchy distribution is only weakly integrable.

$\begingroup$ 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)? $\endgroup$ Oct 31 '09 at 4:22

$\begingroup$ 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. $\endgroup$ Oct 31 '09 at 14:41