Which function's Fourier transform is the function itself?

We know that the Fourier transform of a Gaussian function is Gaussian function itself. Can anyone give one or more functions which have themselves as Fourier transform?

• Yes we can. But why do you need?
– user17762
Mar 9, 2012 at 2:42
• The key here was the phrase "fixed point" or "eigenfunction", then Google performed the rest of the work. mathoverflow.net/questions/12045/… and en.wikipedia.org/wiki/Fourier_transform#Eigenfunctions
– dls
Mar 9, 2012 at 2:49
• Also, have a look at the answers here. Mar 9, 2012 at 2:59
• In the old days the "fixed-point functions" were called self-reciprocal functions and investigated by the likes of Hardy and Titchmarsh. See "On self-reciprocal functions under a class of integral transforms" by Kurt Wolf, fis.unam.mx/~bwolf/Articles/28.pdf . You might follow the references in this paper, particularly those of Titchmarsh– Tom Copeland 1 min ago Apr 7, 2012 at 22:30
• See this MO post.
– Nemo
Nov 24, 2017 at 10:02

Pick a function $f$ that is reasonable enough for the inversion formula to hold (e.g. take $f$ in Schwartz space, which contains the Gaussian among other functions). If $\mathcal{F}$ denotes the linear transformation which takes $f$ to its Fourier transform, then it's easy to check that $\mathcal{F}^{4}$ is the identity map. In particular, by playing some games you find that $$g \ = \ f + \mathcal{F}(f) + \mathcal{F}^{2}(f) + \mathcal{F}^{3}(f)$$ is fixed by $\mathcal{F}$. So $g$ is its own Fourier transform.
This argument doesn't produce a concrete function, but it at least shows you that the Gaussian is far from the only function that is equal to its own Fourier transform. If you want a more specific example, you can show that $(\cosh \pi x)^{-1}$ is its own Fourier transform (use contour integration and the residue theorem).
$$\sum_{n=-\infty}^{\infty} \delta(t-nT_o) = \frac {2\pi} {T_o} \sum_{m=-\infty}^{\infty} \delta(\omega - m\omega_o)$$