What are fixed points of the Fourier Transform - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T21:41:13Z http://mathoverflow.net/feeds/question/12045 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12045/what-are-fixed-points-of-the-fourier-transform What are fixed points of the Fourier Transform pavpanchekha 2010-01-16T23:46:46Z 2010-09-18T01:47:20Z <p>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?</p> http://mathoverflow.net/questions/12045/what-are-fixed-points-of-the-fourier-transform/12047#12047 Answer by Andy Putman for What are fixed points of the Fourier Transform Andy Putman 2010-01-16T23:58:36Z 2010-01-17T02:31:23Z <p>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).</p> <p>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!</p> <p>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:</p> <p>$f = f_1 + f_{-1} + f_{i} + f_{-i}$</p> <p>$T(f) = f_1 - f_{-1} +i f_{i} -i f_{-i}$</p> <p>$T^2(f) = f_1 + f_{-1} - f_{i} - f_{-i}$</p> <p>$T^3(f) = f_1 - f_{-1} -i f_{i} +i f_{-i}$</p> <p>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 $T$.</p> http://mathoverflow.net/questions/12045/what-are-fixed-points-of-the-fourier-transform/12050#12050 Answer by Yemon Choi for What are fixed points of the Fourier Transform Yemon Choi 2010-01-17T00:04:43Z 2010-01-17T00:04:43Z <p>Following on a little from <a href="http://mathoverflow.net/questions/12045/what-are-fixed-points-of-the-fourier-transform/12047#12047" rel="nofollow">Andy's comment</a>, <a href="http://en.wikipedia.org/wiki/Hermite%5Fpolynomials#Hermite%5Ffunctions%5Fas%5Feigenfunctions%5Fof%5Fthe%5FFourier%5Ftransform" rel="nofollow">Hermite polynomials</a> (multiplied by a Gaussian factor) give a basis of eigenvectors for the FT as an operator on $L^2({\mathbb R})$ </p> http://mathoverflow.net/questions/12045/what-are-fixed-points-of-the-fourier-transform/39187#39187 Answer by Darsh Ranjan for What are fixed points of the Fourier Transform Darsh Ranjan 2010-09-18T01:47:20Z 2010-09-18T01:47:20Z <p>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. </p> <p>(I wrote an argument explaining why $D$ should be its own Fourier transform in an answer to another question: <a href="http://mathoverflow.net/questions/14568/truth-of-the-poisson-summation-formula/14580#14580" rel="nofollow">http://mathoverflow.net/questions/14568/truth-of-the-poisson-summation-formula/14580#14580</a>)</p>