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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?

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  • $\begingroup$ Substituting $t=i\xi$ in the inequality is clearly not justified, if you mean $\mathbb E[e^{i\xi \cdot x}]\leq e^{-b^2|\xi|^2}$ (consider a Gaussian, and increase $b$). But for n=1, see en.wikipedia.org/wiki/Sub-Gaussian_random_variable $\endgroup$
    – Jean Duchon
    Aug 1, 2016 at 8:31
  • $\begingroup$ In view of alphanum's answer (upvoted and accepted!), I think the question was not clear enough. You define "sub-Gaussian" for a random variable (or its distribution, OK) by an inequality on the Laplace transform, then ask if the Fourier transform (which is not a probability measure) is itself sub-Gaussian, suggesting it means "dominated by a Gaussian" if I understood the last sentence. The example given in the answer doesn't use the same definition of "sub-Gaussian"... $\endgroup$
    – Jean Duchon
    Aug 1, 2016 at 15:31

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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/ )

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