If I have a function that is always larger than zero, $g(t)>0$, and I take the fourier transform of this function.
Is there a constraint I can place on the fourier transformed function?
I have checked to see if the transform is also positive, but it is not.
$Fg(t)>0$ is not true.
Is there any constraint or equation that can be written for $Fg(t)$?
Thanks Paul
Usually people look at functions with positive Fourier transform, thus switching the roles of $g$ and $Fg$ in your situation. (I will write $\widehat g$ instead of $Fg$ below). But this amounts merely to the change of sign in the Fourier transform convention, so the results still apply. From Bochner's theorem we conclude that $\widehat g$ is a positive-definite function, in the sense that for any integer $n$ and any $\xi_1,\dots,\xi_n\in\mathbb R$ the matrix with entries $a_{ij}=\widehat g(\xi_i-\xi_j)$ is positive definite. In particular, it is Hermitian, which amounts to saying that $\widehat g(-\xi)=\overline{\widehat g(\xi)}$. Of course, the latter is easy to see directly from the definition of $\widehat g$ and the fact that $g$ is real-valued.
Bochner's theorem has the (more difficult) converse direction, which asserts that all continuous positive-definite functions are Fourier transforms of positive measures. This tells you not to expect other relations or equations for $\widehat g$.
