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I'm no expert on these things (and this may not be cutting edge research level; it's really motivated by this MSE question), but it seems that there are non-zero measures (and also functions (?), I would think) on $\mathbb R$ such that both the function and its Fourier transform vanish on an interval. This is usually mentioned in the context of Beurling's work on spectral gaps.

(1) Is there a nice snappy way to see this? (2) How regular can such an $f$ be? Could it be a Schwartz function?

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  • $\begingroup$ That's a very good MSE question. I am also very curious after reading the MSE question if there are 'nice' functions that do this (not something crazy like the Cantor function or what have you that isn't 'nice' per se). Personally, I would think it's impossible for a Schwartz function to satisfy this, but I could be way off. If so, I'd like to see this Schwartz function! $\endgroup$
    – user78249
    Jul 10, 2017 at 2:52
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    $\begingroup$ I am not expert either but according to this paper of Kargaev--Volberg (jstor.org/stable/24897037) there are nonzero functions vanishing on an interval and having the spectral gap. Following references therein actually more peculiar results are obtained in this direction. $\endgroup$
    – Paata Ivanishvili
    Jul 10, 2017 at 5:02
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    $\begingroup$ Also see Section 2 of Kolountzakis and Lev's paper: arxiv.org/pdf/1505.06833.pdf. $\endgroup$
    – Mark Lewko
    Jul 10, 2017 at 5:06

2 Answers 2

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A distribution for which both the distribution and its Fourier transform vanish on some interval is the Shah function aka Dirac comb
$$ Ш = \sum_{k=-\infty}^\infty \delta_k $$ which is its own Fourier transform (if the right scaling and normalization is used), i.e. $\hat Ш = Ш$.

Mollfying $Ш$ gives a function that vanishes on some interval and the Fourier transform (still not a function) does so, too.

To get a Schwartz function with a Schwartz function as Fourier transform, we could take two compactly supported mollifiers $\phi$ and $\psi$ and set $f = \hat\psi\cdot (\phi * Ш)$ which is a blurred and damped version of the Shah function (and actually a Schwartz function). Its Fourier transform is $\hat f = \psi * (\hat\phi\cdot Ш)$ which is a damped and blurred version of the Shah function. Choosing the support of $\phi$ and $\psi$ narrow enough will lead to $f$ and $\hat f$ which are zero on intervals of positive length. By construction, both functions are Schwartz functions. This is basically the same construction as in Christian Remling's answer and the adaption to the previous version that made this approach work is due to reuns comment below.

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  • $\begingroup$ It's easy to produce an explicit bump function that is unity on (-a,a) and zero outside (-b,b) for a < b. Dubrovin, Fomenko, and Novikov part II does this explicitly in their discussion of partitions of unity. $\endgroup$
    – Steve Huntsman
    Jul 10, 2017 at 20:00
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    $\begingroup$ Look at $f = \hat{\varphi}\, (\phi \ast \Pi\text{I}), \hat{f} = \varphi_- \ast (\hat{\phi} \, \Pi\text{I})$ where $\phi,\varphi$ are $C^\infty_c([a,a+1-\epsilon])$ (this is equivalent to the answer I posted on MSE) @ChristianRemling $\endgroup$
    – reuns
    Jul 10, 2017 at 21:07
  • $\begingroup$ Please keep the t in Laurent Schwartz's name, otherwise people might confuse him with the author of Cauchy-Schwarz inequality :) (as he once predicted). $\endgroup$
    – Jean Duchon
    Jul 11, 2017 at 13:10
  • $\begingroup$ Oh boy, how could I forget that t... $\endgroup$
    – Dirk
    Jul 11, 2017 at 13:41
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Over at MSE (see linked question), user1952009 (= reuns on MO) has given a very elegant construction of a Schwartz function with this property. I am reproducing his/her answer here.

Let $g,h\in C_0^{\infty}$ with support contained in $[1/3,1/2]$, and write $\widehat{h}_n=\int_0^1 h(x)e^{-2\pi inx}\, dx$ for the discrete Fourier coefficients of $h$, viewed as a $1$-periodic function $h_p$. Let $$ f(x)=\sum_{n=-\infty}^{\infty} \widehat{h}_n g(x-n) . $$ Then $$ \widehat{f}(\xi)= \sum \widehat{h}_n e^{2\pi in\xi}\widehat{g}(\xi) = h_p(\xi)\widehat{g}(\xi) . $$ Thus $f$ is as desired; this calculation also confirms that $f\in\mathcal S$.

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