The Fourier transform $\hat u$ is defined on the Schwartz space $\mathscr S(\mathbb R^n)$ by $ \hat u(\xi)=\int e^{-2iπ x\cdot \xi} u(x) dx. $ It is an isomorphism of $\mathscr S(\mathbb R^n)$ and the inversion formula is $ u(x)=\int e^{2iπ x\cdot \xi} \hat u(\xi) d\xi. $ The Fourier transformation can be extended to the tempered distributions $\mathscr S'(\mathbb R^n)$ with the formula $$ \langle \hat T,\phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)} = \langle T, \hat \phi\rangle_{\mathscr S'(\mathbb R^n), \mathscr S(\mathbb R^n)}. $$ Below we note $\mathcal F$ the Fourier transformation on $\mathscr S'(\mathbb R^n)$. We find easily that $\mathcal F^4=Id$, so that if $T\in\mathscr S'(\mathbb R^n)$ is such that $\mathcal F T=\lambda T$, then $\lambda$ is a fourth root of unity.

Question: Are all the tempered distributions $T$ such that $\mathcal FT=T$ known? Two examples are very classical: first the Gaussians $e^{-π\vert x\vert^2}, e^{-π\langle Ax,x\rangle } $ where $A$ is a positive definite matrix with determinant 1, second the case $$ T_0=\sum_{k\in \mathbb Z^n}\delta_k, $$ where the equality $\mathcal FT_0=T_0$ is the Poisson summation formula.

I think that, thanks to your answers and a reference, I found the answer: using What are fixed points of the Fourier Transform which deals with the $L^2$ case, one may guess that the fixed points of $\mathcal F$ in $\mathscr S'(\mathbb R^n)$ are $$ \Bigl\{S+\mathcal F S+\mathcal F^2 S+\mathcal F^3 S \Bigr\}_{S\in \mathscr S'(\mathbb R^n)}=(Id+\mathcal F+\mathcal F^2+\mathcal F^3)(\mathscr S'(\mathbb R^n)). $$ Since $\mathcal F^4=Id$ on $\mathscr S'(\mathbb R^n)$, the above distributions are indeed fixed points and if $\mathcal F T=T$, then $$ T=\frac14\bigl(T+\mathcal F T+\mathcal F^2 T+\mathcal F^3 T\bigr), $$ conluding the proof.

  • $\begingroup$ I am most grateful for all the answers below. However I am looking for all the tempered distributions solutions $T$ such that $\mathcal F T=T$. $\endgroup$ – Bazin May 10 '16 at 17:24
  • 3
    $\begingroup$ The answer in your edit has almost nothing to do with Fourier transformation and uses only linearity and the periodicity. Are you really satisfied by this characterization? $\endgroup$ – Jochen Wengenroth May 11 '16 at 12:38
  • $\begingroup$ Not claiming to answer your question, but in $L^2(\mathbb{R})$, an orthonormal basis that diagonalizes the Fourier transform is given by the Hermite functions $H_n(x)\,e^{-x^2/2}$. The closed span of those for $n$ multiple of $4$ gives a convenient description of $L^2$ functions equal to their Fourier transform. $\endgroup$ – Gro-Tsen May 11 '16 at 12:39
  • $\begingroup$ Possible duplicate of What are fixed points of the Fourier Transform $\endgroup$ – Nemo Dec 20 '17 at 7:01
  • $\begingroup$ Did you notice that in $\mathbb{R}^2$, the distribution $\langle T,\phi\rangle = \int \phi(x,0)\,dx+\int\phi(0,y)\,dy$ is an eigenvalue? There are other similar examples. $\endgroup$ – user90189 Jan 5 '18 at 22:36

Given a square-integrable, positive semi-definite function $f$, with its Fourier transform $\hat{f}$, then the function


with $\star$ the convolution, is its own Fourier transform: $\hat{F}=F$.

If we require that $F$ is a probability density (absolutely integrable and positive semi-definite), then any $F$ with $\hat{F}=F$ is of this form, see A. Nosratinia, Self-characteristic distributions.

The decomposition $F=f^2+\hat{f}\star\hat{f}$ for a given probability density $F=\hat{F}$ is not unique, one realization is $f=\sqrt{F/2}$.

  • $\begingroup$ Are there results on the size of the equivalence classes of $f$s that will yield the same $F$s? Can $f$ be recovered from $F$, or at least some member of the class? $\endgroup$ – Emilio Pisanty May 10 '16 at 19:46
  • $\begingroup$ @EmilioPisanty -- one member of the equivalence class can indeed be recovered immediately, $f=\sqrt{F/2}$; for a particular example Nosratinia constructs a one-parameter family of solutions to $F=f^2+\hat{f}\star\hat{f}$, but there may be more. $\endgroup$ – Carlo Beenakker May 10 '16 at 20:17
  • $\begingroup$ Well, indeed it would. That's pretty clever, thanks for that. $\endgroup$ – Emilio Pisanty May 10 '16 at 21:02
  • $\begingroup$ It needs to be pointed out that $f$ should be an even function. In the Nosratinia paper it is assumed "positive semi-definite", and hence even a.e. $\endgroup$ – davyjones May 11 '16 at 12:39
  • $\begingroup$ @davyjones -- condition on $f$ added, thanks for the pointer $\endgroup$ – Carlo Beenakker May 11 '16 at 12:45

There is a whole chapter in Titchmarsh, Fourier transform, describing all eigenfunctions in great detail. He calls them "self-reciprocal" functions. A more difficult question is about the eigenvectors of discrete Fourier transform, but this one you do not ask.

Edit. Since you are asking now about discrete transform too, it has 4 eigenspaces corresponding to eigenvalues $1,-1,i,-i$ and the question is how to choose a convenient basis in each. Of course the choice is very non-unique, so many bases were proposed. A survey can be found here: http://www.cs.princeton.edu/~ken/Eigenvectors82.pdf. Here is a newer paper: http://www.sciencedirect.com/science/article/pii/S0165168408001783. More can be found by typing "Eigenvectors of the discrete Fourier transform" on Google.

  • $\begingroup$ ...but I would like to ask :) $\endgroup$ – Delio Mugnolo May 10 '16 at 21:03
  • $\begingroup$ Primitive Dirichlet characters are eigenvectors of the DFT (up to complex conjugaison, so identifying $\mathbb{C}^n$ with $\mathbb{R}^{2n}$) $\endgroup$ – reuns Jan 4 '18 at 5:10

For any even function, $\hat{f}(x) + f(x)$ is a fixed point of the Fourier transform.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.