I'm no expert on these things (and this may not be cutting edge research level; it's really motivated by this [MSE question](https://math.stackexchange.com/questions/2332292/do-flat-functions-with-flat-fourier-transform-exist/2353043#2353043)), 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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Function and Fourier transform vanish on an interval
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 $f\in\mathcal S$.


