Is it possible to show that the Gaussian is a fixed point of the Fourier transform using a fixed point theorem?

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

Is it possible to show this using a fixed point theorem?

• It's only a fixed point for one value of $\alpha$ (which depends on which Fourier transform you use). – Daniel Fischer Sep 24 '13 at 13:00

The metric fixed point theorem is somewhat constructive: one could try to find the limit of the sequence $x_{n+1}=f(x_n)$, which is the unique fixed point provided that $f$ has Lipschitz constant less than $1$. However, this method is not going to work with the Fourier transform $\mathcal F$, since $\mathcal F^4=I$ (up to a constant). Iteration of $\mathcal F$ does not lead anywhere.