# Gaussian fixed point Fourier transform

We know that $\text{exp}(-\alpha |x|^2)$ is a fixed point for the unitary Fourier transform if $\text{Re } \alpha > 0$.

I know many arguments to show this (contour-integration and differentiation).

Is there a not an elegant way where we can exploit that fact that the Gaussian is rotationally symmetric? A sketch would be fine.

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## 1 Answer

Here's a different argument. Take a zero-mean Gaussian random variable $X$. We know that the sum of $n$ copies of $X$, scaled down by a factor of $\sqrt{n}$, is distributed the same as $X$. On the other hand, by the characteristic function argument (which can be used to prove the central limit theorem) we know that the Fourier transform of $(X_1+\cdots+X_n)/\sqrt{n}$ converges to $\exp(- \sigma^2x^2/2)$.

Edit: an even simpler way. Again take $X$ to be a zero-mean Gaussian. We know that $(X+X)/\sqrt{2}$ is equidistributed with $X$. We immediately deduce that all higher-order cummulants are nil, and since the second cummulant is the variance, we get that the Fourier transform is $\exp(-\sigma^2x^2/2)$.

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Nice! I like stochastic arguments in analysis. –  Jonas Teuwen Jan 14 '11 at 23:56
Well, the usual way is rather to prove that the characteristic function of a Gaussian random variable is a Gaussian function and use this to prove the central limit theorem. –  Rasmus Jan 15 '11 at 1:12
A better way to think about the usual proof is that we find the characteristic function of the limiting random variable, and then we verify this c.f. corresponds to a Gaussian. Alternatively, we could find the Gaussian distribution by applying an inverse Fourier transform. –  Yuval Filmus Jan 15 '11 at 1:16
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