Obviously, if $f(x) \geq 0 $ on $(a,b)$, the above claim is trivial. We also see that if $a = -\infty, b = \infty$, above is claiming that $|\hat{f}(z)| \leq |\hat{f}(0)|$ for all real $z$, where $\hat{f}(z)$ is the Fourier Transform of $f$.
My question is whether anyone has an idea of how nontrivial the question in the title is? Is it exceptionally "case by case"? Are there any really powerful results concerning this question when $f$ is fairly general and not just nonnegative? Obviously there are very subtle ways to tweak the $f \geq 0 $ requirement. For example, require $f\geq 0 $ a.e. or $f<0$ on a set with very small measure, etc.
This idea has shown up in something I'm working on and I am curious how challenging of a topic this is.
Thank you.
