Condition on characteristic function to converge in distribution to zero

Question

I am trying to show the following, we have a sequence of random variable $Y_n$ with ch.f $\phi_n$.

$\exists \delta > 0$ such that $\forall |t|<\delta, \phi_n(t) \rightarrow 1 \Rightarrow Y_n$ converge in distribution to zero.

I understand that from the condition we can obtain the tightness of the sequence and that the tightness implies that every subsequence of $Y_n$ converge in distribution to a random variable Y but I am not sure how to conclude the proof.

I saw an other post talking about the analytic property of the characteristic function but I don't know this notion.

Answer 1

Here is an other way than in this thread (where analyticity was used) to conclude that the only potential limit is $0$. It was shown that if a subsequence converges to some $Y$, then $Y$ is bounded and its characteristic function $\varphi$ is equal to $1$ on $(-\delta,\delta)$.

Using boundedness, we can prove that $\varphi''\left(0\right)=\mathbb E\left[Y^2\right]$. This follows from applications of the dominated convergence theorem. We first prove that $\varphi'(t)=\mathbb E\left[iY e^{itY} \right]$ for $t\in (-\delta,\delta)$ by going back to the definition of derivative $(\varphi(t+h)-\varphi(t))/h$ and considering all the sequences $\left(h_n\right)_{n\geqslant 1}$ going to zero. Then we do the same for the second derivative.