the product of two completely monotone functions A function $f:(0,∞)→[0,∞]$ is said to be completely monotonic if its $n$-th derivative exists and $(−1)^nf^{(n)}(x)≥0$, where $f^{(n)}(x)$ is the $n$-th derivative of $f$.
Given two completely monotone functions, would the multiplication product of these two functions be completely monotone?
if $f(x)$ and $g(x)$ are completely monotone, would $f(x)*g(x)$ be completely monotone too?
 A: Hint:
We can observe that $f$ is CM $\iff-f'$ is CM and this generalizes to higher orders by induction.
Then
$$f,g\ge0,f',g'\le0\implies-(fg)'=-fg'-f'g\ge0.$$
A: This paper on arXiv (https://arxiv.org/pdf/1211.0900.pdf ) provides a lot of detail about CM (completely monotone) functions and may be helpful for you.
CM functions can be characterised as the real one-sided Laplace transforms of positive measures on $[0,\infty]$, i.e. given a positive Borel measure $\mu$, 
$$ f \in \mbox{CM} \Longleftrightarrow f(x) = \int_0^{\infty} e^{-xt} \ d\mu(t)$$
(provided, of course that the integral converges).  If $g(x)$ if also CM, so is the Laplace transform of a real positive Borel measure $\nu$ we quickly obtain
$$fg = \int_0^{\infty} e^{-xt} \ d(\mu * \nu)(t) $$
i.e. that multiplication of the functions involves convolution of the measures, and so remains CM.
It may be of interest as well to note that CM functions are related to infintely divisible measures, which in turn are used to define Levy processes, which turn up in financial and biological models.
