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David Corfield asked the following questions yesterday: Is the n-dimensional Fourier transform of exp(-||x||) always non-negative, where ||.|| is the Euclidean norm on R^n? What is its support?

I want to ask a more general question: what happens when ||.|| is the p-norm, for arbitrary p in [1, 2]?

David's question, and Josh Shadlen's helpful answer, are here: Is the Fourier transform of exp(-||x||) non-negative?

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Here's an update. Mark Lewko notes that my question is closely related to the positive definiteness of the function exp(-||x||). In fact, I posed this question precisely because I want to know whether exp(-||x||) is strictly positive definite. Without "strictly", I can prove it. But with "strictly", I can't. So that's the challenge! – Tom Leinster Oct 21 '09 at 22:17
There's now a write-up and discussion of this problem at the n-Category Cafe:… – Tom Leinster Nov 4 '09 at 8:26
up vote 8 down vote accepted

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 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 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 Berstein to reduce to the case p=2.)

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Thanks Mark! That looks promising. I'll try going through your argument tomorrow. (Incidentally, our library doesn't have Koldobsky's book, but I did eventually find it on the web, after a long trawl through various sites of increasing dodginess.) – Tom Leinster Oct 23 '09 at 23:53
There are probably more-canonical sources for this than Koldobsky's book; that was just the one book on my shelf that I knew had a proof of that lemma. – Mark Meckes Oct 24 '09 at 12:34
Mark, Koldobsky's book was exactly what came to my mind when I saw this question. I think it is the right place for this. – Deane Yang Oct 24 '09 at 14:11
Deane, that's because you're a convex geometer. I think most probabilists would go for a book on infinitely divisible distributions for a proof of the lemma. I'm enough of a probabilist to think that's what a canonical source ought to be, but not enough of one actually to be familiar with those sources. In any case, lest I give anyone the wrong impression, Koldobsky's book is very well-written and makes an excellent reference for everything it has in it. – Mark Meckes Oct 24 '09 at 15:54

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:

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Thanks again. But I'm still not sold. A version of Bochner's Theorem is that a continuous L^1 function is strictly positive definite iff it is bounded and its Fourier transform is non-negative and not identically zero. So, my question is equivalent to asking whether e^{-||x||_p} is strictly positive definite - and in fact, that's the question I really want answered. The question about Fourier transforms was a reformulation that I thought people would find easier to answer. The results of Schoenberg show that e^{-||x||_p} is pos def. But my difficulty is: why is it strictly pos def? – Tom Leinster Oct 20 '09 at 14:19
Just a note on terminology, because it's somewhat treacherous. What many analysts call "positive definite" is what one might these days prefer to call "positive semidefinite": the assertion is that some quantity is \geq 0, with no conditions on when equality is attained. What they call "strictly positive definite" is what one might these days call "positive definite". I'm using the analysts' convention. – Tom Leinster Oct 20 '09 at 14:21

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 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.

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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: Note that, in general, the function is e^(-||x||_{p}) is p-radial.

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Thanks, Mark. I'm having trouble seeing this argument through, because the Wikipedia article only seems to cover the single-variable case, whereas we're doing R^n with n > 1. I'm not currently seeing anything else useful on the web. Could you give me a pointer or two? – Tom Leinster Oct 18 '09 at 6:44

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