$f(x)$ 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.
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$\begingroup$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 $[-1/\pi,1/\pi]$.
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