# Fourier cosine transforms of Schwartz functions and the Fejer-Riesz theorem

This question spanned from a previous interesting one.
Let $k$ be a real number greater than $2$ and $$\varphi_k(\xi) = \int_{0}^{+\infty}\cos(\xi x) e^{-x^k}\,dx$$ the Fourier cosine transform of a function in the Schwartz space.

Is is possible to use the Fejér-Riesz theorem or some variation of it,
to prove that $\varphi_k(\xi)<0$ for some $\xi\in\mathbb{R}^+$?

• If the integral over $[0,\pi/2\xi]$ is not smaller than that over $[\pi/2\xi,3\pi/2\xi]$, then $\varphi_k(\xi)$ is never negative. Have you looked at that? – DisintegratingByParts Jul 5 '16 at 17:04
• @TrialAndError: that is clear, but I actually want to prove the opposite, that for some xi the integral is negative, and my conjecture is supported by numerical simulations. – Jack D'Aurizio Jul 5 '16 at 17:34
• Do you believe $\varphi_k(\xi)$ to be negative for all $\xi$, or what did you have in mind for F-R? – DisintegratingByParts Jul 5 '16 at 17:38
• Note the integral we have can be written in term of Fox-Write function \begin{align} \int_0^\infty \cos(tx) e^{-x^k}dx= \Psi_{1,1} \left[ \begin{array}{l} (1/k,2/k) \\ (1/2,1)\end{array} ; -\frac{ \left( t\right)^2}{4}\right] \end{align}. You can find this result here: maths.manchester.ac.uk/~goran/research-reports/psrr18-2009.pdf . So the in this case the quesiont boils down what values of $k$ does Fox-Write function have or does not have zeros. – Boby Aug 3 '16 at 15:58
• I've now posted an answer supplying a proof of the result that @Boby asked about earlier (math.stackexchange.com/questions/1846072: the cosine transform is a decreasing function for $0<k<2$) using the same source. – Noam D. Elkies Aug 31 '16 at 16:43

Instead of Fejér-Riesz, we use the Fourier inversion formula directly. Assume for contradiction that $\phi_k(\xi) \geq 0$ for all $\xi$. Then for all $x$ we would have $$0 \leq \int_0^\infty \phi_k(\xi) \, (1 - \cos (\xi x))^2 \, dx = \frac12 \int_0^\infty \phi_k(\xi) \, (3 - 4 \cos (\xi x) + \cos (2\xi x)) \, dx,$$ which means $$3 - 4 e^{-x^k} + e^{-(2x)^k} \geq 0$$ for all $x$. But for $x$ near $0$ we may write $e^{-x^k} = 1 - \epsilon$ and $e^{-(2x)^k} = (1 - \epsilon)^{2^k} = 1 - 2^k \epsilon + O(\epsilon^2)$, so $$3 - 4 e^{-x^k} + e^{-(2x)^k} = 3 - 4(1-\epsilon) + (1 - 2^k \epsilon + O(\epsilon^2)) = (4 - 2^k) \epsilon + O(\epsilon^2).$$ Thus $3 - 4 e^{-x^k} + e^{-(2x)^k}$ becomes negative for small $x$ once $2^k > 4$, which is equivalent to $k=2$, and we have our desired contradiction.
• That is just brilliant: simple and effective. It reminded me of Hadamard's proof that $\zeta(s)\neq 0$ for $\text{Re}(s)=1$. Good work and many, many thanks. – Jack D'Aurizio Aug 21 '16 at 8:15
• You're welcome, and thank you! Indeed it's the same $(3, -4, +1)$ trick, as I note in my lectures on analytic number theory: see Exercise 2 on the last page of the chapter on the zero-free region, math.harvard.edu/~elkies/M229.15/free.pdf . – Noam D. Elkies Aug 21 '16 at 14:45
• @NoamD.Elkies Also, do you think we can say something about distribution of zeros. For example, if $t_0$ is a zero does it mean that $t_1=t_0+2\pi$ is also a zero. – Boby Aug 23 '16 at 18:03
• Re: "Any way we can find a pdf of your paper . . ." to my surprise, yes, since it appeared long enough ago to be on GDZ: gdz.sub.uni-goettingen.de/dms/load/img/?PID=GDZPPN002108941 That paper also gives an asymptotic expansion for the Fourier transform (Lemma 10 on page 634); there are infinitely many zeros, but they're not spaced as regularly as an arithmetic progression -- the case "$k=\infty$" is misleading. – Noam D. Elkies Aug 23 '16 at 18:25