I am wondering if the following assertion is true: 


Let $f,g:\mathbb{R}\_+\rightarrow [0,1]$ be completely monotone functions on $\mathbb{R}\_+^\*$, that is, $(-1)^n f^{(n)}(x)\geq 0$ and $(-1)^n g^{(n)}(x)\geq 0$ for any $x>0$ and any $n\in \mathbb{N}$. Assume that $f(0)=g(0)$ and $\lim\_{x\rightarrow \infty} f(x) = \lim\_{x\rightarrow \infty} g(x)=0$, and that there exist $0<a<A$ such that $f\leq g$ on $[0,a]$ and $f\leq g$ on $[A,\infty)$.


Then $f\leq g$ on $\mathbb{R}\_+$ (?).


If not, does anyone have a counter-example?


N.B.: 1) one can think of $f$ and $g$ as two Laplace transforms of positive measures.


2) Removing any of the assumptions seems to lead to a counter-example.


Thank you for your help!



Order between two completely monotone functions?
No. Let's take $g(x)=\epsilon e^{-\epsilon x}+(1-\epsilon)e^{-\alpha x}$,
$$
f(x) = \int\_{\epsilon}^{1+\epsilon} e^{-tx}\, dt = \frac{1}{x}e^{-\epsilon x}(1-e^{-x}) .
$$
Then clearly $f\le g$ near infinity, and near zero, $f(x)\simeq 1-(1/2+\epsilon)x$, $g(x)\simeq 1-(\epsilon^2+(1-\epsilon)\alpha)x$. I also want $f\le g$ near zero, and this gives a condition on $\alpha$ (given $\epsilon$). For small $\epsilon$, I can take $\alpha\approx 1/2$.


On the other hand, as $\epsilon\to 0$, $\alpha\to 1/2$, we have that $f(1)\to 1-e^{-1}$, $g(1)\to e^{-1/2}$, and now a calculator will tell us that $f(1)>g(1)$ for sufficiently small $\epsilon$. (There might be easier counterexamples.)


