What are fixed points of the Fourier Transform - MathOverflow

What are fixed points of the Fourier Transform

2010-01-16T23:46:46Z

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 by Andy Putman for What are fixed points of the Fourier Transform

2010-01-16T23:58:36Z

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 by Yemon Choi for What are fixed points of the Fourier Transform

2010-01-17T00:04:43Z

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

Answer by Darsh Ranjan for What are fixed points of the Fourier Transform

2010-09-18T01:47:20Z

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)