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?



What are fixed points of the Fourier Transform
The following is discussed in a little more detail on pages 337-339 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 $T$.


