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Carlo Beenakker
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Given a square-integrable, positive semi-definite function $f$, with its Fourier transform $\hat{f}$, then the function

$$F=f^2+\hat{f}\star\hat{f},$$

with $\star$ the convolution, is its own Fourier transform: $\hat{F}=F$.

If we require that $F$ is a probability density (absolutely integrable and positive semi-definite), then any $F$ with $\hat{F}=F$ is of this form, see A. Nosratinia, Self-characteristic distributions.

The decomposition $F=f^2+\hat{f}\star\hat{f}$ for a given probability density $F=\hat{F}$ is not unique, one realization is $f=\sqrt{F/2}$.

Carlo Beenakker
  • 188.1k
  • 18
  • 448
  • 651