Convergence of characteristic functions to $1$ on a neighborhood of $0$ and weak convergence Prove the following statement: 
$ X_n \Rightarrow 0 $ (convergence in distribution) if and only if $ (\exists\; \epsilon>0: |t|<\epsilon) \;\; \phi_n(t) \rightarrow 1  $, where $\phi_n(t)$ is the characteristic function of the random variable $X_n$.
We could write $X_n \rightarrow 0 $ in probability since convergence in probability to a constant and convergence in distribution to a constant are the same concept.
I guess Lévy's continuity theorem implies the 'only if' part.
Thank you very much for your help in advance!
 A: If $X_n\to 0$ in distribution, then any $\varepsilon$ does the job. 
The converse is harder. Here it is the proof of Levy's continuity theorem which will be used rather than the result in itself. Denoting by $\varphi_n$ the characteristic function of $X_n$ and $\mu_n$ its distribution, we indeed have the equality 
$$\tag{1}\varepsilon^{-1}\int_{-\varepsilon}^\varepsilon(1-\varphi_n(t))\mathrm dt=2\int_\Omega\left(1-\frac{\sin(x\varepsilon)}{x\varepsilon}\right)\mathrm d\mu_n(x)$$
which admits the lower bound
$$\tag{2}2\int_{\{|x|\geqslant 2/\varepsilon\}}\left(1-\frac1{|\varepsilon x|}\right)\mathrm d\mu_n(x)\geqslant \mu\{|X_n|\geqslant 2/\varepsilon\}.$$
The LHS in (1) goes to $0$ as $n$ goes to infinity because of the assumption and dominated convergence. Hence, by (2), 
$$\tag{3}\lim_{n\to\infty}\mu\{|X_n|\geqslant 2/\varepsilon\}=0.$$
This gives in particular tightness of the sequence $(X_n)_{n\geqslant 1}$. Now we have to prove that the only limit we can have for a subsequence is necessarily $0$. 
Assume that a subsequence $(Y_k)_{k\geqslant 1}$ of $(X_n)_{n\geqslant 1}$ converges in distribution to some $Y$. Then $\varphi_Y(t)=1$ if $|t|\lt \varepsilon$ and $Y$ is bounded. Indeed, we take $t_0$ a continuity point of the cumulative distribution function of $|Y|$ such that $t_0\geqslant 2/\varepsilon$. Then by (3),
$$0=\lim_{n\to \infty}\mu\{|X_n|\gt t_0\}=\mu\{|Y|\gt t_0\}.$$
The characteristic function is then analytic, hence equal to $1$ on the whole real line.
