Showing $\varphi(t)\neq 0$ when $\varphi$ is a characteristic function of an infinitely divisible distribution

Question

Let $\varphi$ be a characteristic function of an infinitely divisible random variable. Show that $\varphi(t) \neq 0$ for all $t$.

Sorry, I have no clue how to do it, because if the exponential is not real, then it can turn around at the origin.

Answer 1

Since $\varphi$ is the characteristic function of an infinitely divisible distribution we have that $$ \varphi(t)=\varphi_n(t)^n,\quad n\in\mathbb{N},\tag{1} $$ for a sequence of characteristic functions $\varphi_n$. Now we use that $|\varphi_n|^2$ is also a characteristic function for each $n$, and thus by $(1)$ we have that $|\varphi|^{2/n}$ is a characteristic function for each $n$. Define $\psi$ by $\psi(t)=\lim\limits_{n\to \infty}|\varphi(t)|^{2/n}$, then $$ \psi(t)= \begin{cases} 1,\quad &\text{if }\varphi(t)\neq 0,\\ 0,&\text{if }\varphi(t)=0. \end{cases} $$

Since $\psi$ is continuous at $0$ we know that $\psi$ is also a characteristic function and hence it is continuous. Using that $\psi(0)=1$ and that $\psi$ only takes on the values $0$ and $1$ we must have that $\psi(t)=1$ for all $t$ meaning that $\varphi(t)\neq 0$ for all $t$.

Answer 2

This is response to 'Why is continuous at 0?'

Since is a characteristic function, it is continuous. In particular, since (0)=1, it is non-zero near 0. So, by its construction =1 near 0. Therefore, is continuous at t=0 and hence by Levy's continuity theorem should be a characteristic function of a probability measure. Hence, (t)=1 for all t.