# Fourier transform of $f_a(x)= a^{-2}\exp(-|x|^a)$, $a \in (0,2)$, is decreasing in $a$

Can one show that Fourier transform of $$f_a(x) = a^{-2} \exp(-|x|^a), \qquad a \in (0,2)$$ is decreasing in $$a$$?

I have a solution for $$a \in (0,1]$$ which cannot be used for $$a\in (1,2)$$.

• @user44191: Doesn't look like a homework to me. Commented May 21, 2020 at 22:13
• First of all it is not a homework. If you think it is a trivial question -please just refer me to a textbook. There is no closed form expression for Fourier for exp(-|x|^a) for any a in (0,2], just for a=1 and a=2. Context I am researching positive definite radial functions. I have solved the problem for a \in (0,1) and x\in R^d by using results from theory of k-times monotone functions. Commented May 21, 2020 at 22:20
• @TanyaApanasovich: There must be a typo in your question: $f_a(0) = a^2$ is increasing in $a$, and hence the integral of the Fourier transform of $f_a$ is increasing in $a$. In fact, the Fourier transform of $f_a$ seems to be increasing in $a$ near the origin, but of course not in a neighbourhood of infinity (because tails become lighter as $a$ grows). Commented May 21, 2020 at 22:33
• @IosifPinelis I study positive definite functions. Positive definite function has a non-negarive Fourier transform. There is a sufficient condition, called multiple monotonicity. I can exchange integral and derivative - so I need to show that f(x)=\exp(-x^a)(x^a log(x) +2/a) is positive definite. Using multiple monotonicity, for d=1, x \in R^1 , I just need to show that f(x) is positive, non-increasing and convex. I take two derivatives and show that. This sufficient condition does not work for a>1. Commented May 22, 2020 at 15:55
• @TanyaVladi I believe I can prove it when the test point is sufficiently far from the origin (distances around 20 and up should be fine). Would you be interested in such partial result or you need "all or nothing"? Commented Jul 9, 2020 at 13:24

Too long for a comment. We have $$\widehat{f_a}(\xi)=\int_{\mathbb R} a^{-2} e^{-\vert x\vert ^a-ix \xi} dx= 2\int_0^{+\infty} a^{-2} e^{-x^a} \cos(x\vert\xi\vert)dx,$$ so that $$J_a(\xi)=-\frac{a^3}2\partial_a\widehat{f_a}(\xi)= \int_0^{+\infty} \bigl( 2 + x^a \ln (x^a)\bigr)e^{-x^a}\cos(x\vert\xi\vert)dx,\quad\text{and}$$ $$aJ_a(\xi)=\int_0^{+\infty} t^{\frac 1a-1}\bigl( 2 + t \ln t\bigr)e^{-t}\cos(t^{1/a}\vert\xi\vert)dt.$$ We note that $$\frac d{dt}(t\ln t)=\ln t+1$$ which is positive iff $$t>1/e$$ so that $$2+t\ln t\ge 2-\frac{1}{e}>0,$$ proving that $$J_a(0)>0$$.
Remark. Positive values of $$\xi$$ remain to be checked. Note that for $$\chi_0$$ smooth equal to 1 near 0 and vanishing outside of $$[0,1]$$, we have $$t^{\frac 1a-1}\bigl( 2 + t \ln t\bigr)e^{-t}=t^{\frac 1a-1}\bigl( 2 + t \ln t\bigr)e^{-t}\chi_0(t) +\psi(t),$$ where $$\psi$$ belongs to the Schwartz space, thus as well as its cosine transform. Somehow the main part of the integral is located near $$t=0$$ and $$\int_0^{+\infty} \cos(2π s\vert\xi\vert) ds 2πa=π a\delta_0(\xi),$$ which is indeed positive for $$a>0$$.
• I am struggling to understand why Fourier transform of $\psi$ is positive. Can you please give me a reference to that decomposition for $\chi$ and $\psi$ you have used in the proof. Thank you Commented Jun 16, 2020 at 21:08
• "ln𝑡+1 which is positive iff 𝑡>1/𝑒 " . and it seems the only condition for positivity of that Fourier transform of the derivative . So following your arguments I can replace the power in $1/\alpha^{2}$ from 2 to a smaller number as soon as it is bigger than $\exp(-1)$. However a simple numerical analysis shows that if the power is $0.4$ and $\alpha=1.2$, the Fourier transform can be negative for some $\omega$ Commented Jun 16, 2020 at 21:51
• @Tanya Vladi I did not write that the Fourier transform of $\psi$ is positive. I said that the $\partial_a\hat{f_a}(\xi=0)<0$. On the other hand it is indeed tempting to consider the positive function $\mathbb R_+\ni t\mapsto t^{1/a-1}(2+t\ln t)e^{-t}$ which is in $L^1$ and somewhat concentrated (in fact singular) at 0. I guess that for $\phi$ positive decreasing (and smooth), it is possible to decide the sign of the cosine transform, if $\phi(0)$ is large enough, but this case is indeed a bit different. Commented Jun 17, 2020 at 10:06
• About $\chi_0$ and $\psi$, I simply claim that, given a positive $T$, I can find two smooth functions on $[0,+\infty)$ such that $\chi_0(t)=1$ on $[0,T]$ and $\chi_0(t)=0$ on $[2T,+\infty)$ so that with $\psi=1-\chi_0$, you get that $\psi$ vanishes near 0. Commented Jun 17, 2020 at 10:19
• I guess I do not understand your proof at all. However a simple example, Fourier transform of $-{\partial\over \partial \alpha }[\exp(-x^{\alpha})\alpha^{-0.4}]$, for $\alpha$ can be negative for some $\omega$ (just from plotting using MATLAB), but according to your arguments it should be nonnegative Commented Jun 17, 2020 at 15:11