## What are fixed points of the Fourier Transform

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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?

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## 3 Answers

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 $T$.

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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. – engelbrekt Jan 17 2010 at 0:14
@engelbrekt: did our answers cross? I think we must have commented at about the same time – Yemon Choi Jan 17 2010 at 4:21
@Choi: Yes, they crossed. I remember noticing that. – engelbrekt Jan 17 2010 at 7:15

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

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A very important fixed point of the Fourier transform that isn't in $L^2$ is the Dirac comb distribution, informally $$D(x) = \sum_{n\in Z} \delta(x-n),$$ 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: http://mathoverflow.net/questions/14568/truth-of-the-poisson-summation-formula/14580#14580)

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These are tempered distributions, and Andy's argument carries over verbatim to these. – Robin Chapman Sep 18 2010 at 6:44
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