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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?

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  • $\begingroup$ It's only a fixed point for one value of $\alpha$ (which depends on which Fourier transform you use). $\endgroup$ – Daniel Fischer Sep 24 '13 at 13:00
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No.

Fixed point theorems provide sufficient conditions for the existence, and sometimes uniqueness, of a fixed point. They do not tell you what those points are.

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.

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