I am looking for the construction of a smooth bump function, $f$, mapping the real line to itself which has two special properties:
(1) $f$ is constant on some interval in its support (for instance $f=1$ on $(-1,1)$ and $f=0$ outside $(-2,2)$.)
(2) The Fourier transform of $f$ is non-negative and rapidly decaying.
Thank you
Best Answer
Such a function (at least with the additional properties you name) cannot exist:
If the Fourier transform of $f$ is non-negative, then (by Fourier inversion) $$ |f(x)| = |\mathcal{F}^{-1}(\hat{f})(x)| = |\int \hat{f}(\xi) e^{2\pi i \langle x,\xi\rangle}\, d\xi | \leq \int |\hat{f}(\xi)|\, d\xi = \int \hat{f}(\xi)\, d\xi = f(0), $$ where the step before the last used $\hat{f}\geq 0$.
Equality in the above estimate can only hold if $\hat{f}\cdot e^{2\pi i \langle x,\xi \rangle}$ has constant (complex) sign/direction for (almost) all $\xi$. But for $x \neq 0$, this is impossible.
Hence, $f$ can not be constant in a neighborhood of $0$ and $f$ needs to attain its (unique) maximal absolute value at $0$.
A general construction, which will probably yield something interesting for you is to take
$$ f = g \ast g^\ast, $$
where $g^\ast (x) = \overline{g(-x)}$ (or something similar). By the convolution theorem (and some other easy properties of the Fourier transform) this yields
$$ \hat{f}= \hat{g}\cdot \overline{\hat{g}}=|\hat{g}|^2, $$
so that the Fourier transform is nonnegative.
Compact support and smoothness of $f$ (which implies smoothness and fast decay of the Fourier transform) is also easy to achieve. But as the considerations above imply, you will have problems with obtaining a constant value around $0$.