Let $(\mu_n)_{n\in\Bbb{N}}$ be a sequence of probability measure on $\Bbb{R}$ with characteristic functions $(\phi_n)_{n\in\Bbb{N}}$. Assume that $\lim_{n\rightarrow\infty}\phi_n(t)=1$ for all $t\in[-\delta,\delta]$ for some $\delta>0.$ Prove that $\mu_n$ converges weakly to $\delta_0$, and hence $\phi_n$ converges uniformly to 1 on any bounded interval.

My thought: I try to prove the property of tightness of $(\mu_n)$. Then if subsequence of $(\mu_n)$ can be proved to weakly converge to $\delta_0$. By the uniqueness, we are done.


marked as duplicate by Davide Giraudo, Claude Leibovici, Joonas Ilmavirta, user147263, drhab Oct 31 '14 at 11:31

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  1. Deduce from the truncation inequality, $$\mu_n(B(0,R)^c) \leq 7R \int_0^{\frac{1}{R}} (1-\text{Re} \, \phi_n(t)) \, dt,$$ and $\lim_{n \to \infty} \phi_n(t)=1$ for $|t|< \delta$ that $$\lim_{n \to \infty} \mu_n(B(0,R)^c) = 0$$ for $R$ sufficiently large. Here $B(0,R) = \{x; |x|<R\}$ denotes the open ball centered at $0$ of radius $R$.
  2. Using the first step, show that $(\mu_n)_{n \in \mathbb{N}}$ is tight.
  3. By tightness, there exists a weakly convergent subsequence. Show that the limit of the subsequence equals $\delta_0$.
  4. Conclude from the subsequence principle that $\mu_n$ converges weakly to $\delta_0$.
  • $\begingroup$ How to show the limt of subsequence equals to $\delta_0$? $\endgroup$ – Shine Nov 3 '14 at 1:29
  • $\begingroup$ @Shine There are several possibilities. One way is to prove that the characteristic function of the limit is analytic (see DavideGaurido's answer). Another one is to apply some known results, e.g. the following: "Let $\varphi$ be an characteristic function such that $|\varphi(\xi_1)| = |\varphi(\xi_2)|=1$ for $\xi_1,\xi_2 \in \mathbb{R} \backslash \{0\}$ satisfying $\xi_1/\xi_2 \notin \mathbb{Q}$. Then $|\varphi(\xi)|=1$ for all $\xi \in \mathbb{R}$." $\endgroup$ – saz Nov 3 '14 at 10:36

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