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Let $\gamma$ be defined on $\mathbb R^n$ by $\gamma (x)=e^{-π x^2}$. With $\mathcal F$ standing for the Fourier transformation defined on the Schwartz space by $$ (\mathcal F u)(\xi)=\int e^{-2iπ x\cdot \xi} u(x) dx, $$ we have $ \mathcal F\gamma =\gamma. $ We can also define $\mathcal F$ for the tempered distribution ($\mathscr S'$) with the duality formula $$ \langle \widehat T,\phi\rangle_{\mathscr S',\mathscr S}=\langle T,\widehat{\phi}\rangle_{\mathscr S',\mathscr S}. $$ For instance, the Poisson summation formula is $\widehat S=S$ with $S=\sum_{k\in \mathbb Z^n}\delta_k$. Finally the question: determine all the tempered distributions $T$ such that $$ \mathcal F T=T. $$

Ricardo Andrade
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Bazin
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    Related: http://mathoverflow.net/questions/12045/what-are-fixed-points-of-the-fourier-transform – Christian Remling Sep 15 '14 at 18:21
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    The question is not Hilbertian, but on $\mathcal S'$: I want also to include the Poisson summation formula. – Bazin Sep 15 '14 at 18:23
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    I don't see why this makes any difference. Voting to close as a duplicate. – Michael Renardy Sep 15 '14 at 18:30
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    A tempered distribution satisfies $T=\hat T$ iff $\langle T,\hat\phi-\phi\rangle=0$ for all Schwartz functions $\phi$. If the Schwartz space is decomposed into eigenspaces of the Fourier transform as $E_1\oplus E_{-1}\oplus E_i\oplus E_{-i}$, then tempered distributions with eigenvalue one correspond to the dual space of $E_1$ (by extension by zero to the Schwartz space). – Joonas Ilmavirta Sep 15 '14 at 18:43
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    That is, you want all the tempered distributions that annihilate $E_{-1}$, $E_i$, and $E_{-i}$, and equivalently the test functions $H_j(x) \exp(-x^2/2)$ for $j \not\equiv 0 \mod 4$ where $H_j$ are the Hermite polynomials. – Robert Israel Sep 15 '14 at 18:50
  • @Bazin: I understand, but I still thought the linked question was worth pointing out. – Christian Remling Sep 15 '14 at 19:41
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    @Robert Israel: yes, but I would like a somewhat more explicit description yielding in particular the Poisson summation formula. The latter formula is not exactly a triviality and to provide an algebraic proof would be interesting. – Bazin Sep 15 '14 at 19:44
  • @Christian Remling: yes, of course, thanks. – Bazin Sep 15 '14 at 19:46
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    A bit more explicit: $T = X + \mathcal F(X) + \mathcal F^2(X) + \mathcal F^3(X)$ for some tempered distribution $X$. Note that $\mathcal F^2$ is the reflection $\mathcal F^2 T(x) = T(-x)$. – Robert Israel Sep 15 '14 at 23:51
  • Possible duplicate of What are fixed points of the Fourier Transform – Martin Nicholson Dec 20 '17 at 07:01

2 Answers2

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In case the specific distribution-theoretic argument is not clear... as it hadn't really been overtly mentioned in comments or answers, and is not really suggested by the classical argument as in Titchmarsh and such:

E.g., for Poisson summation on $\mathbb R$: observe that the distribution "sum over integers" is annihilated by multiplication by $e^{2\pi ix}-1$, and is translation invariant. Observe that these two conditions are interchanged by Fourier transform. Show that the space of such distributions is one-dimensional: the multiplication annihilation shows that any such distribution is of order $0$ and supported at integers. By classification of distributions supported at a point, it is a sum of Dirac deltas at integers. By translation-invariance, it is (a scalar multiple of) sum-over-evaluation-at-integers. A just-slightly-more-complicated version applies in $n$-dimensions.

paul garrett
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This question is addressed in great detail in the book of Titchmarsh, Introduction to the theory of Fourier integrals. Such functions are called self-reciprocal, and there is a separate chapter about them. Of course Titchmarsh did not use the language of distributions, but it is easy to translate. A recent paper on the related subject is http://arxiv.org/abs/1203.2427.

Alexandre Eremenko
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