Suppose $\nu$ is a probability measure on the positive integers $\{1,2,3\dots,\}$, and let $\pi_{a}$ and $\pi_b$ be Poisson distributions on the positive integers with parameter $a,b$ respectively, where $a \le b$.
We know that for any $x \ge 0$, the generating function $$G_a(x):=\sum_{n = 0}^\infty x^n \pi_a(n) = e^{-a}\sum_{n = 0}^\infty x^n \frac{a^n}{n!} = \exp\left(ax - a \right)$$ is finite for all $x\ge 0$.
Suppose now I know that the generating function of $\nu$, call it $G_\nu$ satisfies the following for all $x \in [0,1]$: $$ G_{b}(x) \le G_\nu(x) \le G_{a}(x). $$ I want to know if $G_\nu(x)$ exists for $x >1$, that is, if we have $G_\nu(x)<\infty$ for $x >1$.
I believe $G_\nu(x)< \infty$ for all $x \ge 0$, and here is the reason: the inequality $G_a(x) \le G_\nu(x)$ means, at intuitive level, that $\nu(n)$ has too be small for large $n$, i.e. the tail of $\nu$ has to be comparable to that of $\pi_a$.
I've tried: the only tail estimates might work is $\int_0^\infty 1-e^{-tx}\nu(dx) \ge \frac{1}{2} \nu(t^{-1},\infty)$, but this estimate is too rough and it does not even give the existence of first moment.
Any help will be predicated.
No, it does not necessarily exist. Let $\mu$ be a probability measure on the positive integers whose generating function $G_\mu$ diverges at some $x_0 > 1$ but converges on $[0,x_0)$, and consider $\nu = (1-\epsilon) (\pi_a + \pi_b)/2 + \epsilon \mu$ where $\epsilon > 0$ is small. Of course, $G_\nu(x) = (1-\epsilon)(G_a(x) + G_b(x))/2 + \epsilon G_\mu(x)$. For sufficiently small $\epsilon$, $G_b(x) < G_\nu(x) < G_a(x)$ for $0 < x < 1$ while $G_b(1) = G_\nu(1) = G_a(1) = 1$, and $G_\nu$ diverges at $x = x_0$.