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?](https://mathoverflow.net/questions/723/is-the-fourier-transform-of-exp-x-non-negative)



Fourier transform of $\exp(-\|x\|_p)$: more general question
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 $p=2$.)


