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.
Best 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 $\exp(-\sigma^2x^2/2)$.