uniform convergence of characteristic functions

Question

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?

Answer 1

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.

Answer 2

Take any bounded interval $I$. We want to show that $\sup_{t\in I}|\phi_n(t)-\phi(t)|\to 0$. Since this supremum does not decrease if we widen the interval $I$, it suffices to prove the claim for $I=[a,b]$.

Let us assume to the contrary that $s_n := \sup_{t\in [a,b]}|\phi_n(t)-\phi(t)|\not\to 0$. Then there is $\varepsilon>0$ and a subsequence $\phi_{n_k}$ such that $s_{n_k} > \varepsilon$ for all $k$. From definition of supremum, it follows that for each $k$, there is some $t_k\in [a,b]$ such that $|\phi_{n_k}(t_k) - \phi(t_k)| > \varepsilon$. Now $\{t_k\}_{k\ge 1}$ being a bounded sequence, must have a convergent subsequence, say $t_{k_m} \to x$.

To summarize, we have a sequence $\varphi_m := \phi_{n_{k_m}}$ of characteristic functions converging pointwise to a characteristic function $\phi$, and a sequence $x_m := t_{k_m}$ converging to a point $x$ such that $|\varphi_m(x_m)- \phi(x_m)| = |\phi_{n_{k_m}}(t_{k_m}) - \phi(t_{k_m})| > \varepsilon$ for every $m\ge 1$.

On the other hand, $\varphi_m\to \phi$ and $x_m\to x$ implies $\varphi_m(x_m)\to \phi(x)$ (see below for a proof). Moreover, the continuity of $\phi$ tells us that $x_m\to x$ implies $\phi(x_m)\to \phi(x)$. Hence $\varphi_m(x_m) - \phi(x_m)\to 0$ as $m\to\infty$. But this contradicts the fact that $|\varphi_m(x_m)-\phi(x_m)|>\varepsilon$ for every $m\ge 1$. This completes the proof.

Proof of the fact that $\varphi_m(x_m)\to \phi(x)$: Suppose that $\varphi_m$ is the characteristic function of a random variable $Y_m$ and $\phi$ is the characteristic function of $Y$. Invoking Levy's continuity theorem, $\varphi_m \to \phi$ implies that $Y_m \stackrel{d}{\rightarrow} Y$. Since $x_m \to x$, we can apply Slutsky's theorem to say that $x_m Y_m \stackrel{d}{\rightarrow} xY$. Appealing to Levy's continuity theorem once again, we can say that the characteristic function of $x_m Y_m$ evaluated at $ 1$ converges to the characteristic function of $xY$ evaluated at $ 1$, i.e., $$\varphi_m(x_m) = E\left(e^{ix_mY_m}\right)\to E\left(e^{ixY}\right)=\phi(x).$$