Characteristic function of an infinitely divisible distribution

Question

I need to prove that for random variable $\xi$ which comes from infinitely divisible distribution characteristic function has no zeros, i.e. $\phi_{\xi}(u) \neq 0\: \: \forall \: \: u \in \mathbb{R}$

Answer 1

Hint:

Look, for each $n\geq 1$, at $\phi_n$, the characteristic function of the $n$ i.i.d. random variables guaranteed by the hypothesis that $\chi$ is infinitely divisible. Letting $\phi$ be the characteristic function of $\chi$, you get for all $t\in \mathbb{R}$ $$ \chi_n(t) = \chi(t)^{\frac{1}{n}}$$ Now, show that

1. for each $t\in\mathbb{R}$, $\lim_{n\to\infty}\lvert \chi_n(t) \rvert$ exists, and is either $0$ or $1$ depending on whether $\chi(t)=0$ or not.
2. using this, that for each $t\in\mathbb{R}$, $\phi^{(\infty)}(t)\stackrel{def}{=}\lim_{n\to\infty} \chi_n(t) $ exists, and is either $0$ or $1$ depending on whether $\chi(t)=0$ or not.
3. that $\phi^{(\infty)}$ is a characteristic function;
4. that combining 2. and 3., this implies that $\phi^{(\infty)}$ is identically equal to $1$.
5. Conclude using 4. and 2. that $\phi$ cannot cancel.