Conditions for a nonnegative Fourier transform

Question

I was wondering if there was known necessary and sufficient conditions for the result of a Fourier transform to be nonnegative, i.e.:

$$ F(f) \ge 0 $$

where $F$ is the Fourier transform.

Answer 1

Since you ask for any necessary and sufficient condition, I provide one of the top of my head but it might not be incredibly useful.

The (very simple) idea is based on the convolution theorem, i.e., if $f_1,f_2 \in L^1(\mathbb{R})$ then $F(f_1\cdot f_2) = F(f_1) * F(f_2)$, where $*$ denotes the convolution. The same result holds for the inverse Fourier transform $F^{-1}$.

It is clear that if $F(f) \geq 0$, then there exist a real-valued $h$ such that $F(f) = h\cdot h$. By the convolution theorem (here you need $h\in L^1(\mathbb R)$) one can show that $f = g * g$ for $g = F^{-1}(h)$ and $g$ is hermitian i.e., $g(-x) = \overline{g(x)}$. The fact that $g$ is hermitian follows from a simple computation using the explicit formula for the (inverse) Fourier transform.

In other words, if $F(f) \geq 0$ and $\sqrt{F(f)} \in L^1(\mathbb{R})$ then $f = g*g$ for a hermitian function $g$.

This condition is also sufficient under some integrability condition on $g$, i.e., if $f = g*g$ for two hermitian $g \in L^1(\mathbb{R})$. Again by the convolution theorem and a computation showing that the Fourier transform of a hermitian function is real-valued.

Clearly, checking if a function is a convolution of two hermitian functions is not exactly easier than computing its Fourier transform and checking if it is non-negative. So, I am sorry if this condition is not insightful enough. I am not aware of a more accessible condition, though.

Answer 2

Since the Fourier transform is invertible, you are asking for necessary and sufficient condition on the function $f$ so that $$f(t) = \int_{\mathbb{R}}g(s) e^{i t s} ds$$ for some positive function $g$. The function $f$ must be positive (semi) definite, that is, for every $t_1$, $\ldots$, $t_n$ the matrix $$(f(t_i-t_j))_{i,h}$$ is Hermitian and positive semidefinite. See positive-definite functions. Note that in general $g(s) ds$ will be a positive measure ( that can be singular, like a Dirac measure). So one also needs that $f$ comes from an actual function.

The equivalent condition by @P. Pet: is sometimes easier to check if it is true. This one may be easier to check that it does not happen.

Check what happens if $f(t) = e^{a t}$ for some $a\in \mathbb{R}$. $f$ should be a positive "linear combination" of such functions.