Assume you are given a non-negative, continuous, radial function $f\in L^q(\mathbb{R}^3)$ (for any $q\geq 1$).
Are there any conditions which would guarantee that the Fourier transform of $f$, that is $\hat{f}(p)$, is also non-negative?
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2 Answers
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the three-dimensional Fourier transform $F(\vec{p})$, of the radial function $f(r)$ has Fourier transform
$$F(\vec{p})=\int_0^\infty dr\int_0^\pi d\theta \int_0^{2\pi}d\phi\;e^{ ipr\cos\theta}f(r) r^2\sin\theta$$ $$\qquad=\frac{4\pi}{p}\int_0^\infty rf(r)\sin(pr)\,dr,\;\;{\rm with}\;\;p=|\vec{p}|.$$
so you're asking when the Fourier-sine-transform $S(p)$ of $rf(r)$ will be (pointwise) positive for $p>0$. A sufficient condition is that $rf(r)$ is a decreasing function of $r>0$, see On positivity of Fourier transforms.
For a related question, see this MO post.
For the connection to Bochner's theorem: the OP's question amounts to finding a function that is both positive and positive-definite, since positive functions have positive-definite Fourier transforms.
answered Jan 16, 2015 at 21:25
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$\begingroup$ I am somewhat puzzled. My understanding is that the OP has a radial function (which happens to be positive). By Bochner, he needs to check whether this is positive definite, and this is exactly what is done in the reference in my answer. Presumably, checking that a function is positive is not hard (well, it can be, of course...) FINDING a function which is positive and positive definite is not rocket science since the Gaussian density works. $\endgroup$ Commented Jan 17, 2015 at 4:32
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$\begingroup$ @IgorRivin --- indeed, theorem 2.4.1 cited in your answer, applied to $\mathbb{R}^3$, gives the same condition as above: $\int_0^\infty rf(r)sin(pr)dr>0$ for $p>0$. The OP's question, as I understand it, is: what are the requirements on $f(r)>0$ such that this integral condition holds; I only have one sufficient requirement ($rf(r)$ decreasing) --- can this be improved? $\endgroup$ Commented Jan 17, 2015 at 13:37
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$\begingroup$ Well, it is not entirely clear what the OP's question is. I agree that a characterization of radial functions such that both the function and its fourier transform are positive and positive definite sounds tractable (with enough regularity thrown in...) $\endgroup$ Commented Jan 17, 2015 at 14:52
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$\begingroup$ @CarloBeenakker $f(r)>0$, $rf(r)$ is $0$ at $0$, it will never be decreasing for all $r$ $\endgroup$ Commented Jul 14, 2020 at 16:28
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$\begingroup$ @TanyaVladi --- try $f(r)=1/r^2$ $\endgroup$ Commented Jul 14, 2020 at 17:37
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The magic words are "Bochner's theorem" (which says that the Fourier transform is positive for positive definite functions). For more on radial functions see Theorem 2.4.1 in this book. (Greg Fasshauer)
answered Jan 16, 2015 at 20:43
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1$\begingroup$ Yeah, but the fact that a function is positive definite does not imply it is pointwise positive, doesn't it? $\endgroup$– scouserCommented Jan 16, 2015 at 20:59
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1$\begingroup$ @scouser what does that have to do with anything? Your function is positive by assumption, and pos. def. since the fourier transform is positive. $\endgroup$ Commented Jan 16, 2015 at 21:04
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$\begingroup$ maybe I expressed myself incorrect in the comment. My point is that if I treat $f$ as the density of my measure, then the theorem states that $\hat{f}(p)$ is positive definite, but this does not imply that $\hat{f}(p)\geq 0$ pointwise. This is what I meant by non-negative in the question. $\endgroup$– scouserCommented Jan 16, 2015 at 21:13
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1$\begingroup$ @scouser think inverse fourier transform. FT is positive -> YOUR FUNCTION is p.d. $\endgroup$ Commented Jan 16, 2015 at 21:25