I had to prove a proposition. Let $f$ be a characteristic function(ch.f.). There exists a sequence of ch.f. $\{\phi_{n_{k}}\}$ and $n_k$ is a sequence of positive integers tending to infinity. $f=(\phi_k)^{n_k}$. Prove $f$ is infinitely divisible.
I think this means that the infinitely divisible condition is satisfied on some integers. So I try to find a function sequence and let it tend to $f^{\frac{1}{n}}$. Then I use the Levy continuity theorem to try to prove that $f^{\frac{1}{n}}$ is a characteristic function.
I'm sorry I didn't succeed, I hope to get the correct answer.
Best Answer
Let us fix a choice of $n$. Note that we may choose a sequence of positive integers $m_k$, such that $m_k/n_k\to 1/n.$ Our goal is to show that there is a subsequence of $j_k$, such that $\phi_{n_{j_k}}^{m_{j_k}}\to \phi$ uniformly on compact subsets to some $\phi$ with $\phi^n=f$. By Levy's Theorem, this is sufficent.
Now, by continuity of $f$, and the fact that $f(0)=1$, there is an choice of $\epsilon>0$ such that $\text{Re}(f(x))>0$ for $x\in (-\epsilon,\epsilon)$. In this region, we may find continous functions $\theta(x)\in (-\pi,\pi),$ and $r(x)>0$ such that $$f(x)=r(x)\exp(i\theta(x)).$$ As $\phi_{n_k}(0)=1$, by continuity, we see that for $x\in (-\epsilon,\epsilon)$ $$\phi_{n_k}(x)=r(x)^{1/n_k}\exp(i\theta(x)/n_k).$$
It is clear from this for $x\in (-\epsilon,\epsilon)$, we have that $\phi_{n_k}^{m_k}$ are equicontinous around $(-\epsilon,\epsilon)$.
Now note that the modulus of continuity of a positive definite function $f$, is controlled by that around the original. This follows, for example, the answer to this implies that $$(2-\text{Re}(f(h)))\ge |f(t+h)-f(t)|^2.$$ Thus the sequence $\phi_{n_k}(x)^{m_k}$ are uniformly equicontinous. Thus by the Arzela-Ascoli theorem, there is a subsequence $j_k$, such that $\phi_{n_{j_k}}^{m_{j_k}}\to \phi$ uniformly for some continous $\phi$.
To last fact follows from the following easy determinsitic exercize: For any $a\in \mathbb{C}$, and any choices, $a_{n_k}\in \mathbb{C}$, such that $a_{n_k}^{n_k}=a$, we have that $(a^{m_k}_{n_k})^n\to a$