Consider a continuous time point process $\eta(t)$ representing the number of children in the interval $[0,t]$. Then $\eta(\infty)$ gives the total number of children of an individual. We define $\mu(t):= \mathbb{E}[\eta(t)]$ as the expected number of children in the interval $[0,t]$. In particular, we define $\mathcal{L}\mu(z):= \int_0^{\infty}e^{-zx} \mu(dx)$ for $z\in\mathbb{C}$.
The Malthusian parameter is defined as the solution $\alpha$ of $\mathcal{L}\mu(\alpha)=1$.
If we assume that the Malthusian Parameter exists, then it follows $\mu(\infty)>1$. I ask myself if the converse is true. So if we have $\mu(\infty)>1$, thus their a Malthusian Parameter exsists. Therefore should be $\mathcal{L}\mu(z)$ continuous, but i am not sure if this is true.
You have to use the fact that $k=\log \mathcal {L}$ is a strictly convex function on $(0,\infty)$ (unless than $\mu$ is proportional to a Dirac measure.) We have $k(0)=\log \mu(\infty)$ and $k(\alpha)=0.$ If $k(0)<0$ then $\alpha$ exists and is unique. If $k(0)\geq 0$ there are $0,1,2$ solutions to the equation $k(\alpha)=0.$