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Are there any real-valued strictly positive functions besides the Gaussian function whose cosine transform is also strictly positive?

For instance with $F_c(e^{-x^2}) = \frac{e^{-\frac{\omega^2}{4}}}{\sqrt{2}}$, both sides are strictly positive, whereas with $f(x) = \begin{cases} 1, & -1<x<1 \\0, & \text{otherwise}\end{cases}$ which is strictly positive we get $F_c(f(x)) = \frac{\sqrt{\frac{2}{\pi}} \sin(\omega)}{\omega}$, which is not strictly positive.

Edit: by "positive" I actually meant non-negative, as in zero values are welcome too.

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    $\begingroup$ There is $g(x) = (1-|x|) 1_{|x| < 1}$ which is compactly supported and whose Fourier transform is non-negative, and $f(x) = g(x)+g(\pi x)$ whose FT is strictly positive $\endgroup$ – reuns Apr 25 '17 at 17:33
  • $\begingroup$ Sorry for having to show my lack of education, but what does $1_{|x| < 1}$ mean? $\endgroup$ – Michel Rouzic Apr 25 '17 at 17:46
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    $\begingroup$ The function that is $1$ when $|x|<1$ and zero otherwise, also known as an indicator function $\endgroup$ – Chappers Apr 25 '17 at 17:47
  • $\begingroup$ Otherwise, take $g$ compactly supported and let $f(x)= \int_{-\infty}^\infty \overline{g(-y)} g(x-y)dy$. Then $\hat{f}(\xi) = |g(\xi)|^2$ $\endgroup$ – reuns Apr 25 '17 at 17:53
  • $\begingroup$ Oh I see, so that's a triangular function. I always knew its Fourier transform had ripples but never realised they weren't negative! The FT of the triangular function could possibly be the basis for an interesting filter (in signal/image processing). By the way clearly adding triangles of different widths starts giving you something a Gaussian function if their height decreases with the increased width, but what if you don't scale them differently like you did with $g(x)+g(\pi x)$? Because it seems that any function made from sums of triangles would be positive and have a positive FT too! $\endgroup$ – Michel Rouzic Apr 25 '17 at 18:13
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How about $f(x)=\exp(-|x|)$?

We have $$\int_{-\infty}^\infty f(x)\cos tx= \int_0^\infty (e^{itx}+e^{-itx})e^{-x}=\frac1{1-it}+\frac1{1+it} =\frac2{1+t^2}.$$

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  • $\begingroup$ Yep looks like a good one. Does this function or its cosine transform $\frac{\sqrt{\frac{2}{π}}}{ω^2 + 1}$ have a name? $\endgroup$ – Michel Rouzic Apr 25 '17 at 17:36
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By Bochner's theorem, a function $F$ is the Fourier transform of a positive function if and only if it is positive-definite, i.e. $$ \sum_{1\leq i,j \leq n} \xi_i \bar{\xi}_j F(a_i-a_j) \geq 0 $$ for any $n$, any set of reals $a_i$ and any complex numbers $\xi_i$. One certainly needs $f(0)>0$ and $\lvert f(x) \rvert \leq f(0)$ for this to happen (from the $n=1,2$ cases). Invoking this condition on both $F$ and its Fourier transform implies that $F$ and $\tilde{F}$ are both positive if and only if they are both positive-definite; hence the positive positive-definite functions are closed under Fourier transforms.

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  • $\begingroup$ If we look at semi positive-definite functions, then they are of the form $f(x)= \int_{-\infty}^\infty \overline{g(-y)} g(x-y)dy$ $\endgroup$ – reuns Apr 25 '17 at 17:55
  • $\begingroup$ By applying the convolution theorem to $|f(k)|^2$, eh? $\endgroup$ – Chappers Apr 25 '17 at 18:05
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This might be a little late.. but I just want to put this out here anyway:

All convex functions have positive Fourier-Cosine transform. This was proven in by E.O Tucker

Here is the link to the paper: On positivity of Fourier transforms

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