David Corfield asked the following questions yesterday: Is the $n$-dimensional Fourier transform of $\exp(-\|x\|)$ always non-negative, where $\|\cdot\|$ is the Euclidean norm on $\Bbb R^n$? What is its support?
I want to ask a more general question: what happens when $\|\cdot\|$ is the $p$-norm, for arbitrary $p\in [1, 2]$?
David's question is here: Is the Fourier transform of $\exp(-\|x\|)$ non-negative?
Okay, I think I do have an answer now. I'm borrowing arguments from the proof of Lemma 2.27 in the book "Fourier Analysis in Convex Geometry" by A. Koldobsky (apparently not available online at all). That lemma states that the Fourier transform of the function (on $\Bbb R$) $\exp(-|x|^p)$ is positive everywhere for $p \in (0,2]$.
The central tool is a theorem of Berstein, which in particular implies that if $s$ is in $(0,1]$ then $\exp(-z^s)$ is the Laplace transform of some finite positive measure $\mu$ on $[0,\infty)$; that is, $$ \exp(-z^s) = \int \exp(-uz) d\mu(u). $$ Applying this with $s=1/p$ and $z=\|x\|_p^p$ yields $$ \exp(-\|x\|_p) = \int \exp(-u \|x\|_p^p) d\mu(u). $$ Now calculate the Fourier transform on $\Bbb R^n$ of this. Using Fubini you get an integral wrt $\mu$ of a product of Fourier transforms of $\exp(-|x|^p)$, and you can now apply the one-dimensional lemma. (The one-dimensional lemma is proved by using the same theorem of Bernstein to reduce to the case $p=2$.)
If the function was $e^{-\|x|\|_{p}^{p}}$, the result would follow from the discussion in wikipedia article and a product argument. This is what I was thinking when I wrote the comment above. Of course, I now realize that this doesn't quite work for $e^{-\|x\|_{p}}$. In general, by Bochner's theorem, your problem is equivalent to asking when $e^{-\|x\|_{p}}$ is a positive definite function. This is a special case of a more general problem of Schoenberg which asks for which values of p and b is the function $e^{-\|x\|_{p}^{b}}$ be positive definite. I believe this problem is now solved. See: http://arxiv.org/abs/math/9210207.
I don't have an actual answer for you, but I'll give you some possibly helpful buzzwords to try to google your way to an answer: stable random vectors. As I understand it (at least in the case $p=2$), the existence of 1-stable random vectors is essentially the positive definiteness of $\exp(-\|x\|)$. So you want to know if a $1$-stable random vector in $\Bbb R^n$ has a strictly positive density. There are probably probabilists who know this, but I'm not one of them and don't have the time to hunt it down right now.
The Fourier transform of $e^{-\|x\|_{p}}$ is non-negative for $p\in [1,2]$, and takes negative values for $p>2$. For $p \in [1,2]$ the function is the characteristic function of a Lévy stable distribution (which implies that its Fourier transform is non-negative). For more information, look at: http://en.wikipedia.org/wiki/Stable_distribution. Note that, in general, the function is $e^{-\|x\|_{p}}$ is $p$-radial.
