As the title indicates, my question is:
Question: Does there exist a nonnegative function $f\in L^1(\mathbb R)$ such that the Fourier transform of $f$ satisfies $$\hat f(\xi)<0$$ for all $|\xi|\ge R$, where $R>0$ is some fixed constant.
It is still of interest if one has the weaker property $$\hat f(\xi)\le 0$$ with $\hat f(\xi)< 0$ occurs along a sequence of $\xi$ tending to infinity.
The second Hermite function is even and (mostly) positive and an eigenvector belonging to the eigenvalue $(-1)$ of the Fourier transform.
Edit: I just realized that being only mostly-positive isn't sufficient. In that case we can take (using the notation of that Wiki article) $$\psi_2(x)+\frac{1}{\sqrt{2}}\psi_0(x) = \frac{4\pi x^2}{\sqrt[4]{2}}e^{-\pi x^2},$$ which is non-negative, and have $$\hat{\psi}_2(\xi)+\frac{1}{\sqrt{2}}\hat{\psi}_0(\xi) = -\psi_2(\xi)+\frac{1}{\sqrt{2}}\psi_0(\xi) = \frac{1}{\sqrt[4]{2}}e^{-\pi\xi^2}(2-4\pi\xi^2).$$