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$f(x)$ is a 1D bump function which real, even and compactly supported in the interval $[-a,a]$, and strictly positive within that interval.

Are there any guarantees on the Fourier transform of $f(x)$,

$$ \hat{f}(s) = \int_{-\infty}^{\infty} f(x) \exp(-2 \pi i x s) dx $$

having at least one root in the interval $[-\frac{1}{a},\frac{1}{a}]$? Given that $f(x)$ is real and even, $|\hat{f}(s)|$ will also be real, and my intuition leads me to believe the above is true but I didn't find any theorem related to it.

I've moved the followup question to a new page so I could mark the answer to the first one here.

Filipe Maia
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  • By the Paley-Wiener theorem you also know that $\hat f$ is entire, but I do not know if that helps here – MSDG Jun 16 '18 at 12:21
  • Actually $\hat f(s)$ will be real and even, if that helps. – Andy Walls Jun 16 '18 at 13:11
  • I don't quite know what you mean by bump function, but if $f(x) = \delta(x)$, the transform will not have any zeros. In general, the narrower the time domain pulse, the wider the transform in the frequency domain. – Andy Walls Jun 16 '18 at 13:14
  • By bump function I mean a function which is smooth and has continuous derivatives of all orders. https://en.wikipedia.org/wiki/Bump_function. So that would rule out the dirac delta function. – Filipe Maia Jun 16 '18 at 13:16
  • Right. However, a sequence of ever narrowing, unit area bump functions are a $\delta()$ in the limit, as the width goes to zero. My intuition tells me you are not guaranteed a zero in the transform domain, as ever narrowing bump functions will have ever widening transforms. – Andy Walls Jun 16 '18 at 13:55
  • But the function is required to be strictly positive in the interval so you can't really even approach $\delta()$ – Filipe Maia Jun 16 '18 at 15:00

1 Answers1

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The function

$$ f(x) = \begin{cases} (1 + \cos x)^2 & -\pi < x < \pi, \\ 0 & \text{otherwise} \end{cases} $$

is real, even, compactly supported, strictly positive on the interior of its support, and has three continuous derivatives.

Its Fourier transform,

$$ \hat f(s) = \frac{3\sin(2\pi^2 s)}{2\pi s(1-\pi^2 s^2)(1-4\pi^2 s^2)}, $$

is strictly positive on $[-1/\pi,1/\pi]$.

It seems to me that we can find a $C^\infty$ bump function arbitrarily close to $f$, and certainly close enough so that its Fourier transform is strictly positive on $[-1/\pi,1/\pi]$.

Antonio Vargas
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  • Thanks, this is good enough for me! One can even iteratively square that function to get wider and wider regions without zeros, by moving the "power" of the $f(x)$ closer and closer to the origin.

    I've moved my followup question to a new page, and will now try to figure out how to accept this answer (I'm new to stackexchange).

     – Filipe Maia Jun 17 '18 at 09:09