The obvious ones are 0 and $e^{x^2}$ (with annoying factors), and someone I know suggested hyperbolic secant. What other fixed points (or even eigenfunctions) of the Fourier transform are there?

$\begingroup$ math.stackexchange.com/questions/118078/… $\endgroup$– Tom CopelandMar 11, 2016 at 23:46
4 Answers
The following is discussed in a little more detail on pages 337339 of Frank Jones's book "Lebesgue Integration on Euclidean Space" (and many other places as well).
Normalize the Fourier transform so that it is a unitary operator $T$ on $L^2(\mathbb{R})$. One can then check that $T^4=1$. The eigenvalues are thus $1$, $i$, $1$, and $i$. For $a$ one of these eigenvalues, denote by $M_a$ the corresponding eigenspace. It turns out then that $L^2(\mathbb{R})$ is the direct sum of these $4$ eigenspaces!
In fact, this is easy linear algebra. Consider $f \in L^2(\mathbb{R})$. We want to find $f_a \in M_a$ for each of the eigenvalues such that $f = f_1 + f_{1} + f_{i} + f_{i}$. Using the fact that $T^4 = 1$, we obtain the following 4 equations in 4 unknowns:
$f = f_1 + f_{1} + f_{i} + f_{i}$
$T(f) = f_1  f_{1} +i f_{i} i f_{i}$
$T^2(f) = f_1 + f_{1}  f_{i}  f_{i}$
$T^3(f) = f_1  f_{1} i f_{i} +i f_{i}$
Solving these four equations yields the corresponding projection operators. As an example, for $f \in L^2(\mathbb{R})$, we get that $\frac{1}{4}(f + T(f) + T^2(f) + T^3(f))$ is a fixed point for $T$.

23$\begingroup$ To add a little detail: The four eigenspaces are the closed linear spans of concrete functions, called Hermite functions, that are of the form (Hermite polynomial)e^{x^2}. So you get a lot of fixpoints of the Fourier transform, namely everything that is the limit in mean square of linear combinations of those of the Hermite functions that belong to the eigenvalue 1. $\endgroup$ Jan 17, 2010 at 0:14

$\begingroup$ @engelbrekt: did our answers cross? I think we must have commented at about the same time $\endgroup$ Jan 17, 2010 at 4:21

$\begingroup$ @Choi: Yes, they crossed. I remember noticing that. $\endgroup$ Jan 17, 2010 at 7:15
$\bf{1.}$ A more complete list of particular selfreciprocal Fourier functions of the first kind, i.e. eigenfunctions of the cosine Fourier transform $\sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\cos ax dx=f(a)$:
$1.$ $\displaystyle e^{x^2/2}$ (more generally $e^{x^2/2}H_{2n}(x)$, $H_n$ is Hermite polynomial)
$2.$ $\displaystyle \frac{1}{\sqrt{x}}$ $\qquad$ $3.$ $\displaystyle\frac{1}{\cosh\sqrt{\frac{\pi}{2}}x}$ $\qquad$ $4.$ $\displaystyle \frac{\cosh \frac{\sqrt{\pi}x}{2}}{\cosh \sqrt{\pi}x}$ $\qquad$$5.$ $\displaystyle\frac{1}{1+2\cosh \left(\sqrt{\frac{2\pi}{3}}x\right)}$
$6.$ $\displaystyle \frac{\cosh\frac{\sqrt{3\pi}x}{2}}{2\cosh \left( 2\sqrt{\frac{\pi}{3}} x\right)1}$ $\qquad$ $7.$ $\displaystyle \frac{\cosh\left(\sqrt{\frac{3\pi}{2}}x\right)}{\cosh (\sqrt{2\pi}x)\cos(\sqrt{3}\pi)}$ $\qquad$ $8.$ $\displaystyle \cos\left(\frac{x^2}{2}\frac{\pi}{8}\right) $
$9.$ $\displaystyle\frac{\cos \frac{x^2}{2}+\sin \frac{x^2}{2}}{\cosh\sqrt{\frac{\pi}{2}}x}$ $\qquad$ $10.$ $\displaystyle \sqrt{x}J_{\frac{1}{4}}\left(\frac{x^2}{2}\right)$ $\qquad$ $11.$ $\displaystyle \frac{\sqrt[4]{a}\ K_{\frac{1}{4}}\left(a\sqrt{x^2+a^2}\right)}{(x^2+a^2)^{\frac{1}{8}}}$
$12.$ $\displaystyle \frac{x e^{\beta\sqrt{x^2+\beta^2}}}{\sqrt{x^2+\beta^2}\sqrt{\sqrt{x^2+\beta^2}\beta}}$$\qquad$ $13.$ $\displaystyle \psi\left(1+\frac{x}{\sqrt{2\pi}}\right)\ln\frac{x}{\sqrt{2\pi}}$, $\ \psi$ is digamma function.
Examples $15,810$ are from the chapter about selfreciprocal functions in Titschmarsh's book "Introduction to the theory of Fourier transform". Examples $11$ and $12$ can be found in Gradsteyn and Ryzhik. Examples $6$ and $7$ are from this question What are all functions of the form $\frac{\cosh(\alpha x)}{\cosh x+c}$ selfreciprocal under Fourier transform?. Some other selfreciprocal functions composed of hyperbolic functions are given in Bryden Cais's paper On the transformation of infinite series. Discussion of $13$ can be found in Berndt's article.
$\bf{2.}$ Selfreciprocal Fourier functions of the second kind, i.e. eigenfunctions of the sine Fourier transform $\sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\sin ax dx=f(a)$:
$1.$ $\displaystyle \frac{1}{\sqrt{x}}$ $\qquad$ $2.$ $\displaystyle xe^{x^2/2}$ (and more generally $e^{x^2/2}H_{2n+1}(x)$)
$3.$ $\displaystyle \frac{1}{e^{\sqrt{2\pi}x}1}\frac{1}{\sqrt{2\pi}x}$ $\qquad$ $4.$ $\displaystyle \frac{\sinh \frac{\sqrt{\pi}x}{2}}{\cosh \sqrt{\pi}x}$ $\qquad$ $5.$ $\displaystyle \frac{\sinh\sqrt{\frac{\pi}{6}}x}{2\cosh \left(\sqrt{\frac{2\pi}{3}}x\right)1}$
$6.$ $\displaystyle \frac{\sinh(\sqrt{\pi}x)}{\cosh \sqrt{2\pi} x\cos(\sqrt{2}\pi)}$ $\qquad$ $7.$ $\displaystyle \frac{\sin \frac{x^2}{2}}{\sinh\sqrt{\frac{\pi}{2}}x}$ $\qquad$ $8.$ $\displaystyle \frac{xK_{\frac{3}{4}}\left(a\sqrt{x^2+a^2}\right)}{(x^2+a^2)^{\frac{3}{8}}}$
$9.$ $\displaystyle \frac{x e^{\beta\sqrt{x^2+\beta^2}}}{\sqrt{x^2+\beta^2}\sqrt{\sqrt{x^2+\beta^2}+\beta}}$$\qquad$ $10.$ $\displaystyle \sqrt{x}J_{\frac{1}{4}}\left(\frac{x^2}{2}\right)$$\qquad$ $11.$ $\displaystyle e^{\frac{x^2}{4}}I_{0}\left(\frac{x^2}{4}\right)$
$12.$ $\displaystyle \sin\left(\frac{3\pi}{8}+\frac{x^2}{4}\right)J_{0}\left(\frac{x^2}{4}\right) $$\qquad$ $13.$ $\displaystyle \frac{\sinh \sqrt{\frac{2\pi}{3}}x}{\cosh \sqrt{\frac{3\pi}{2}}x}$
Examples $15,7$ can be found in Titschmarsh's book cited above. $812$ can be found in Gradsteyn and Ryzhik. $13$ is from Bryden Cais, On the transformation of infinite series, where more functions of this kind are given.

$\begingroup$ it's brilliant @Nemo , where do you find all this ? i'm just learning Fourier analysis , can you please tell me how and where can i find these techniques and learn them properly $\endgroup$– Zeno SanJul 15, 2017 at 15:25

$\begingroup$ thanks much appreciated ! "Introduction to the theory of Fourier integral" you are talking about this one , right ? @Nemo $\endgroup$– Zeno SanJul 15, 2017 at 18:02
A very important fixed point of the Fourier transform that isn't in $L^2$ is the Dirac comb distribution, informally $$D(x) = \sum_{n\in Z} \delta(xn),$$ or more properly, defined by its pairing on smooth functions of sufficient decay by $$\langle D, f\rangle = \sum_{n\in Z} f(n).$$ The fact that $D$ is equal to its Fourier transform is really just the Poisson summation formula.
(I wrote an argument explaining why $D$ should be its own Fourier transform in an answer to another question: Truth of the Poisson summation formula)

3$\begingroup$ These are tempered distributions, and Andy's argument carries over verbatim to these. $\endgroup$ Sep 18, 2010 at 6:44
Following on a little from Andy's comment, Hermite polynomials (multiplied by a Gaussian factor) give a basis of eigenvectors for the FT as an operator on $L^2({\mathbb R})$