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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.

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  • $\begingroup$ 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
    Commented May 10, 2016 at 17:24
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    $\begingroup$ 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
    Commented May 11, 2016 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
    Commented May 11, 2016 at 12:39
  • $\begingroup$ Possible duplicate of What are fixed points of the Fourier Transform $\endgroup$
    – Nemo
    Commented Dec 20, 2017 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
    Commented Jan 5, 2018 at 22:36

3 Answers 3

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

$$F=f^2+\hat{f}\star\hat{f},$$

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 $f=\sqrt{F/2}$.

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  • $\begingroup$ 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
    Commented May 10, 2016 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
    Commented May 10, 2016 at 20:17
  • $\begingroup$ Well, indeed it would. That's pretty clever, thanks for that. $\endgroup$
    – Emilio Pisanty
    Commented May 10, 2016 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
    Commented May 11, 2016 at 12:39
  • $\begingroup$ @davyjones -- condition on $f$ added, thanks for the pointer $\endgroup$
    – Carlo Beenakker
    Commented May 11, 2016 at 12:45
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For any even function, $\hat{f}(x) + f(x)$ is a fixed point of the Fourier transform.

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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.

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  • $\begingroup$ ...but I would like to ask :) $\endgroup$
    – Delio Mugnolo
    Commented May 10, 2016 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
    Commented Jan 4, 2018 at 5:10

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