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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?

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Yes we can. But why do you need? – user17762 Mar 9 '12 at 2:42
The key here was the phrase "fixed point" or "eigenfunction", then Google performed the rest of the work.… and – dls Mar 9 '12 at 2:49
Also, have a look at the answers here. – kuch nahi Mar 9 '12 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, . You might follow the references in this paper, particularly those of Titchmarsh– Tom Copeland 1 min ago – Tom Copeland Apr 7 '12 at 22:30

2 Answers 2

Well, the Dirac delta function, strictly speaking, is not a 'function' but it serves your purpose.

The Fourier transform of a Dirac comb (or an impulse train, as it is called in Electrical Engineering) is another Dirac comb.

$$ \sum_{n=-\infty}^{\infty} \delta(t-nT_o) = \frac {2\pi} {T_o} \sum_{m=-\infty}^{\infty} \delta(\omega - m\omega_o) $$

For the proof, refer

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Welcome to Maths.SE ! Thanks for your answer. Please note that on this site it is recomended to include excerpts of your work, providing a link is not enough to reach this site's standards. – Tom-Tom Sep 29 at 19:38

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).

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