6
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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
12
$\begingroup$

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.
$\endgroup$
  • $\begingroup$ Thank you so much! The proof is so nice. $\endgroup$ – Shine Oct 25 '14 at 13:19
  • 2
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.