Is there a probability distribution $\mu$ (with reasonably nice density $f$ on $\mathbb{R}$) such that the Fourier transform (aka. characteristic function) $\psi\_\mu(t) = \int\_{\mathbb{R}} e^{itx} \, d\mu(x) = \int\_{\mathbb{R}} e^{itx} f(x) \, dx$ converges to zero for $t \rightarrow \pm\infty$ faster than $\exp(-ct^2)$ for arbitrarily large $c>0$?


The rate $\exp(-ct^2)$ is achieved by the normal distribution $\mathcal{N}(0,\sigma^2)$ for $\sigma^2 \geq 2c$, since $\psi\_{\mathcal{N}(0,\sigma^2)}(t) = e^{-\frac{\sigma^2t^2}{2}}$.


Any help is much apprechiated!




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This question is somewhat related to a previous question ([A kernel 'more analytic' than $\exp(-x^2)$](https://mathoverflow.net/questions/451657/a-kernel-more-analytic-than-exp-x2)). Due to the negative result given to that question, I am now shifting my focus to kernels on $\mathbb{S}\_1 \subset \mathbb{C}$.



A probability distribution, with Fourier transform smaller than $C \exp(-ct^2)$
Yes. E.g., if $f(x)=\dfrac{1-\cos x}{\pi x^2}$ for real $x\ne0$, then $\psi\_\mu(t)=\max(0,1-|t|)$ for real $t$ (which latter $=0$ if $|t|\ge1$).


