It is clear that the Gaussian is a fixpoint of the Fourier transform within the space of Schwartz functions S. Until recently I was convinced that this is propably the only fixpoint in S. But now I discovered that the Hyperbolic secant sech(x) = 2/(e^x + e^-x) is another such fixpoint. See F(sech) at https://en.wikipedia.org/wiki/Fourier_transform where a=pi. So, the question arises: Does anyone know other fixpoints of the Fourier transform in S? If yes, is there a general rule to characterize such functions? Thanks for your help!
All Hermite functions with n=4m are fixpoints of the Fourier transform. The subspace of L2(R) generated by these functions gives ALL fixpoints of the Fourier transform in L2. See
https://en.wikipedia.org/wiki/Hermite_polynomials#Hermite_functions_as_eigenfunctions_of_the_Fourier_transform
What happens outside of L2 (say, for Fourier transform in the sense of distributions), I don't know.
Great, thank you very much for this answer. The Hermite functions are of course also members in the Schwartz space S because they are integrable and smooth. They will therefore play an important role not only in S but also in S' and hence in the theory of generalized functions. Very interesting.
so I think you could try with a Laplace operator? I could explain it to you more precisely on the 18 th February at 4 pm at my home? How many you will be?
Regards, Frank
Since I gave my copy of the book of Laurent SCHWARTZ, I cannot check easily, but in his notation exp(-\pi x^{2}) is a fixed point, and he mentioned something about the Hermite polynomials, and the Hermite functions must be exp(-\pi x^{2}) multplied by these polynomials.
I remember (for having read the book around 1970) that he checked that they are eigenfunctions of the Fourier transform for the eigenvalues 1, -1, i, or -i, so that what Vladimir KADETS says suggests that the nth Hermite function corresponds to the eigenvalue i^{n}.
Luc TARTAR
mathematician
Thank you very much for your valuable answers. I'll have a look into Laurent Schwartz' books. So, the Gaussain obviously generates the Hermite functions. But what about the Hyperbolic Secant function? There must be a magic connection between the Gaussian and the Hyperbolic Secant. Physicists send strong laser beams into waveguides where the initial beam is Gaussian distributed. But they end up with Hybolic secant distributed beams ... Interesting, don't you think?
In fact the Fourier invariant functions form a closed subspace of L2 and the orthogonal projection onto this subspace is either the projection on the Hermite functions nr. 0,4,8,12,... (see first answer), but it can be described directly by taking for f simply f+Ff+F^2f+F^3f/4, since F^4f=f it is clear that this produces Fourier invariant functions and that Fourier invariant functions are left invariant by this (self-adjoint and idempotent) operator.
I thank you all for these answers. Especially the idea of having a closed subspace of such functions is most interesting.
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