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?

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    $\begingroup$ You don't seem to be on the right track, IMHO. I guess that it is simpler than that, just apply the definition of characteristic function. $\endgroup$ – Giuseppe Negro Oct 25 '14 at 8:59
  • $\begingroup$ Yes. Thank you. I got it. $\endgroup$ – Shine Oct 25 '14 at 9:04
  • $\begingroup$ @Shine Why not write an answer to your question? $\endgroup$ – saz Oct 25 '14 at 9:06
  • $\begingroup$ OK. I am typying. But Actually, I got stuck at one step. $\endgroup$ – Shine Oct 25 '14 at 9:22
  • $\begingroup$ @GiuseppeNegro, Can you find a way to prove the statement above? $\endgroup$ – Shine Oct 25 '14 at 12:29

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.

  • Step 1: The family of measure $\{\mu_n; n \in \mathbb{N}\}$ is tight, i.e. for any $\varepsilon>0$ there exists a compact set $K$ such that $$\mu_n(K^c) \leq \varepsilon.$$ Proof: Choose $r>0$ such that $\mu(B[0,r])< \varepsilon$ and set $K := B[0,2r]$. Pick a continuous function $\chi$ satisfying $1_{B[0,r]} \leq \chi \leq 1_{K}$. In particular, we have $1-\chi(x)=1$ for any $x \in K^c$ and $1-\chi(x)=0$ for any $x \in B[0,r]$. Hence, $$\mu_n(K^c) \leq \int (1-\chi) \, d\mu_n \stackrel{n \to \infty}{\to} \int (1-\chi) \, d\mu \leq \mu(B[0,r]^c) < \varepsilon.$$ This shows that for $n \geq N$ sufficiently large, $\mu_n(K^c) \leq \varepsilon$. Enlarging $K$ yields $$\mu_n(K^c) \leq \varepsilon$$ for all $n \in \mathbb{N}$.
  • Step 2: $(\varphi_n)_{n \in \mathbb{N}}$ is uniformly equicontinuous, i.e. for any $\varepsilon>0$ there exists $\delta>0$ such that for any $|\xi-\eta| \leq \delta$ and $n \in \mathbb{N}$, we have $$|\phi_n(\xi)-\phi_n(\eta)| \leq \varepsilon. \tag{1}$$ Proof: Let $\varepsilon>0$ and $K$ as in step 1. Since the mapping $\xi \mapsto e^{\imath \, \xi}$ is continuous, we can pick $\delta>0$ such that $$|1-e^{\imath \, x (\xi-\eta)}| \leq \varepsilon$$ for any $|\xi-\eta| \leq \delta$ and $x \in K$. Consequently, $$\begin{align*}|\phi_n(\eta)-\phi_n(\xi)| &\leq \int_K \underbrace{|e^{\imath \, \xi x}-e^{\imath \, \eta x}|}_{|1-e^{\imath \, x (\xi -\eta)}| \leq \varepsilon} \, d\mu_n(x) + \int_{K^c} \underbrace{|e^{\imath \, \xi x}-e^{\imath \, \eta x}|}_{\leq 2} \, d\mu_n(x) \\ &\leq \varepsilon + 2 \mu_n(K^c) \leq 3 \varepsilon. \end{align*}$$
  • Step 3: Fix $\xi \in \mathbb{R}$ and $\varepsilon>0$. Choose $\delta$ as in step 2. Since $\phi$ is continuous, we may assume that $$|\phi(\xi)-\phi(\eta)| \leq \varepsilon \tag{2}$$ for any $\eta \in B(\xi,\delta)$. Hence, $$\begin{align*} |\phi_n(\eta)-\phi(\eta)| \leq \underbrace{|\phi_n(\eta)-\phi_n(\xi)}_{\stackrel{(1)}{\leq} \varepsilon} + \underbrace{|\phi_n(\xi)-\phi(\xi)}_{\stackrel{n \to \infty}{\to 0}} + \underbrace{|\phi(\xi)-\phi(\eta)|}_{\stackrel{(2)}{\leq} \varepsilon}. \end{align*}$$ Letting $n \to \infty$ and $\varepsilon \to 0$ shows local uniform convergence. Since local uniform convergence is equivalent to uniform convergence on compact sets, this finishes the proof.
  • $\begingroup$ Thank you so much! The proof is so nice. $\endgroup$ – Shine Oct 25 '14 at 13:19
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    $\begingroup$ @Shine Then why not accept the answer? (I don't care for the reputation; I'm just curious.) $\endgroup$ – saz Oct 25 '14 at 15:38

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