Assume that a sequence of probability measures $\mu_n$ converges weakly to $\mu$. Let $\phi_n$ and $\phi$ denote respetively the characteristic function of $\mu_n$ and $\mu$. Prove that $\phi_n$ converges uniformly to $\phi$ on any bounded interval.
My thoughts:By assumption, for any bounded continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$, we have $\int f(x)\mu_n(dx)\rightarrow \int f(x)\mu(dx)$. So $$|\int e^{itx}\mu_n(dx)-\int e^{itx}\mu(dx)|=|\int \cos tx\mu_n(dx)-\int \cos tx\mu(dx)+i(\int \sin tx\mu_n(dx)-\int \sin tx\mu(dx))|$$. Then how to show this convergence is independent of $t$ for any $t$ in bounded interval?
Best Answer
The pointwise convergence of the characteristic functions follows directly from the definition of weak convergence. Indeed, since $f(x) := e^{\imath \, x \xi}$ is for each $\xi \in \mathbb{R}$ continuous and bounded, we have
$$\phi_n(\xi) := \int e^{\imath \, x \xi} \, d\mu_n(x) \to \int e^{\imath \, x \xi} \, d\mu(x) =: \phi(\xi).$$
The uniform convergence on compact intervals is more delicate.