TITLE: Showing $\varphi(t)\neq 0$ when $\varphi$ is a characteristic function of an infinitely divisible distribution
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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.