Question

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?

Answer 1

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 $T$.

Answer 2

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 $L^2({\mathbb R})$