# Missing assumption? (Convergence of random variables and characteristic functions)

Here are two exercises from my Probability book ("Probabilidade: um curso em nível intermediário", by Barry James). (I have translated them from Portuguese to English and modified them a bit.)

Prove the following claims:

$1$) If $\phi$ is a characteristic function and there exists $\delta > 0$ such that $\phi (t) = 1$ for all $t$ with $|t| < \delta$, then $\phi (t) = 1 \forall t$.

$2$) Let $X_{1}, X_{2}, \ldots$ be random variables whose characteristic functions $\phi _{1}, \phi_{2}, \ldots$ converge pointwise. If there exists $\delta > 0$ such that $\phi_{n} (t) \rightarrow 1$ for all $t$ with $|t| < \delta$, then $X_{n}$ converges in probability to $0$. Hint: use Claim $1$.

I've proven Claim $1$, and I can prove Claim $2$ if I assume that the limit of the characteristic functions $\phi _{1}, \phi_{2}, \ldots$ is also a characteristic function. My question is: is Claim $2$ still true without this additional assumption?

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Yes, thanks to Lévy's continuity theorem, the pointwise limit is a characteristic function as soon as it is continuous at $0$.
Thanks! We've only seen in class a "weaker" form of this theorem, namely: "$X_n$ converges in distribution to $X$ if and only if $\phi_{X_{n}} (t)$ converges to $\phi_{X} (t)$ for all $t$." I guess my teacher forgot about this when he assigned us this exercise. – TuringMachine Feb 25 '13 at 22:26