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 $a\in (1,2)$.
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 $a\in (1,2)$.
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 $a>0$.