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Take a function $f$ then and use the following definition of Fourier transform $\mathcal{F}$: $$\mathcal{F}(f)(\xi) = \hat{f}(\xi)=\int_{-\infty}^{\infty} f(x) e^{-i \xi x} d x, \quad \forall \xi \in \mathbb{R} $$ One can prove that taking the Fourier transform twice leads to: $$\mathcal{F}^2(f)(x) = 2 \pi f(-x)$$ a proof of this fact can be found at this post Fourier transform of the Fourier transform.

However now I am interested in Fourier transforms of probability measures. We know that if we let $\mu$ be a probability measure on $(\mathbb{R}, \mathcal{B})$. Its characteristic function $\phi: \mathbb{R} \rightarrow \mathbb{C}$ is defined by $$ \phi(u)=\int_{\mathbb{R}} e^{\mathrm{i} u x} \mu(\mathrm{d} x) $$

Now if we let the $f$ from above to be the density of $\mu$ then we see the equality: $$\phi(u)=\hat{f}(-u)$$ Then we can play the same game as above and see that if we take again the Fourier we obtain again the density $f$ times $2 \pi$. But what happens if $\mu$ does not admit density?

Q: What object do we get if we take twice the Fourier transform of a probability measure that does not admit density?

Grandes Jorasses
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  • What happens if you do a simple example? Say $\mu$ is the point-mass at $1$, then what is $\phi$? And what is $\widehat{\phi}$? I think you find that $\phi$ is not in the domain of $\mathcal F$. (Or, perhaps you say: $\widehat{\phi}$ is not a function, but a distrubution.) – GEdgar Oct 16 '22 at 18:34
  • Can you explain more? A distribution is a function in my knowledge. Do you mean the double transform is again a measure? – Grandes Jorasses Oct 16 '22 at 18:45
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    I mean a Schwartz distribution (a generalized function). I temporarily forgot the conflict in terminology with probability distributions. If $\mu$ is the point-mass at $1$, then $\phi(u) = e^{iu}$, and the integral $\int_{-\infty}^{\infty} \phi(u) e^{-i u x} ,d u$ does not exist (as a function); but does exist (in the sense of Schwartz distributions) as $2\pi\delta(x-1)$ where $\delta$ is the Dirac delta. – GEdgar Oct 16 '22 at 18:57
  • Nice, thank you! I would like to award you some point for your answer. – Grandes Jorasses Oct 16 '22 at 19:34
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    @GrandesJorasses: The Fourier transform, in the context of your posting, is maps measures to continuous functions. As such it does not make sense to apply the Fourier transform twice, unless you extend the definition somewhat, as GEDgar proposes. – Mittens Oct 16 '22 at 19:49
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    Thank you all. I see the issue now. If any of you wants to leave here below an answer which synthesizes both of your comments, it might be useful for future generations. – Grandes Jorasses Oct 16 '22 at 21:00

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