The high level question is: Just as the Fourier transform of a Gaussian is a Gaussian, is the Fourier Transform of a sub-Gaussian also a sub-Gaussian?


Let $x \in \mathbf{R}^n$ denote some sub-Gaussian random vector, i.e.
there exists $b>0$ such that for any $t\in \mathbf{R}^n$ its Laplace
transform is upper-bounded as follows:
$$
\mathbf{E}[ e^{t \cdot x} ] \leq e^{b^2 \|t\|^2}.
$$


What is the Fourier Transform of the PDF of $x$, is it also sub-Gaussian?


It seems that just substituting $t = 2\pi i$ would immediately imply that, but is this substitution justified?



Fourier Transform of sub-Gaussian distributions
No, the Fourier Transform of a sub-Gaussian is not necessarily sub-Gaussian. By common wisdom, the decay properties of the Fourier transform $\hat f$ of an $ f \in L^1({\bf R}) $ are related to the smoothness of $f$ (and vice versa). To obtain examples take a gaussian $g(x)=Ce^{-x^2}$ and a bounded function $h(x)$ with slowly decaying FT (e.g. characteristic function of [0,1]) and set
$$ f(x) := g(x)h(x). $$
Then $f$ is clearly sub-gaussian, but the decay of $\hat f \propto \hat g \star \hat h$ is essentially given by the slow decay of $\hat h$. 


(However by the uncertainty principle, if $f$ and $\hat f$ are both sub-gaussian, then the product of widths is subject to a lower bound, cf. <https://en.wikipedia.org/wiki/Uncertainty_principle#Hardy.27s_uncertainty_principle> or <https://terrytao.wordpress.com/2009/02/18/hardys-uncertainty-principle/> )


