Let $φ_n$ be a sequence of characteristic functions . We know that they converge at every point in a nonempty open interval around 0 to a function $φ$ , which is also continuous at 0.
Can it be said that the sequence of the distribution functions corresponding to $φ_n$’s are tight.
I am aware of the proof of a different version where the the convergence is given on the entire real line . But what to do in this case ?
This is true. Proposition 8.29 in Breiman's book says $P(|X|>\frac 1 u) \leq \frac {\alpha} u\int_0^{u}[1- Re(f(v))]dv$ for any random variable $X$ with characteristic function $f$. Here $\alpha$ is an absolute constant). If $\phi_n \to \phi$ in some interval around $0$ and $\phi$ is continuous we can choose $u$ such that $\frac {\alpha} u\int_0^{u}[1- Re(\phi (v))]dv <\epsilon$. It the follows that $P(|X_n|>\frac 1 u) \leq \frac {\alpha} u\int_0^{u}[1- Re(\phi_n(v))]dv <\epsilon$ for $n$ sufficiently large from which tightness follows. [Here $X_n$ is a random variable with characteristic function $\phi_n$].