Merging with respect to bounded uniformly continuous functions in terms of characteristic functions

characteristic-functionsprobabilityweak-convergence

I would like to know if there are any results, where merging
of probability measures in $R^n$ with respect to bounded uniformly continuous functions is deduced from some conditions on characteristic functions?

In partucular, is it true that if for all t $f_n(t)-g_n(t)$ converges to zero, when $n$ goes to infinity, where $f_n(t)=\int e^{itx} dP_n(x)$ and $g_n(t)=\int e^{itx} dQ_n(x)$, then sequences of measures $P_n$ and $Q_n$ merge wrt bounded uniformly continuous functions?

(sequences $P_n$ and $Q_n$ of probability measures merge wrt bounded uniformly continuous functions if for every bounded uniformly continuous function $f$ $(\int f dP_n)-(\int f dQ_n)$ converges to zero when n goes to infinity)

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The counterexample is given in R.M.Dudley, "Real analysis and probability"(proposition 11.7.6). Take $P_n$ with density $\frac{1}{x\cdot log(n)}$ on the interval $[1,n]$. Take $dQ_n(x)=dP_n(-x)$. And take $f(x)=max(0,min(x,1))$, for example.

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[Math] uniform convergence of characteristic functions

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.

The convergence of the random variables in probability implies weak convergence of the corresponding probability measures

Let $S$ be a metric space with distance $\mathrm d$, and $\xi_n$, $\xi\in S$ such that $\lim_{n\rightarrow\infty}\mathbb P(|\xi_n-\xi|\ge\varepsilon)=0$ for all $\varepsilon>0$. Let $f:S\rightarrow\mathbb R$ be bounded and $L$-Lipschitz. Then we have \begin{aligned} |\mathbb E[f(\xi_n)]-\mathbb E[f(\xi)]| &\le\mathbb E[|f(\xi_n)-f(\xi)|]\\ &=\mathbb E[\unicode{120793}\{|X-X_n|\le\varepsilon\}|f(\xi_n)-f(\xi)|] +\mathbb E[\unicode{120793}\{|X-X_n|>\varepsilon\}|f(\xi_n)-f(\xi)|]\\ &\le \mathbb E[\unicode{120793}\{|X-X_n|\le\varepsilon\}|f(\xi_n)-f(\xi)|]+2\|f\|_\infty\mathbb P(|X-X_n|>\varepsilon)\\ &\le \mathbb E[\unicode{120793}\{|X-X_n|\le\varepsilon\}L|\xi_n-\xi|]+2\|f\|_\infty\mathbb P(|X-X_n|>\varepsilon)\\ &\le L\varepsilon\mathbb P(|X-X_n|>\varepsilon)+2\|f\|_\infty\mathbb P(|X-X_n|>\varepsilon). \end{aligned} using Jensen's inequality with $|\cdot|$, the triangle inequality to obtain $2\|f\|_\infty$, Lipschitz continuity, and the $\varepsilon$-bound from the event. Taking the limit and using convergence in probability gives \begin{aligned} \lim_{n\rightarrow\infty}|\mathbb E[f(\xi_n)]-\mathbb E[f(\xi)]|\le L\varepsilon. \end{aligned} Taking $\varepsilon$ to $0$ completes the proof.

EDIT: We provide a few more details. The triangle inequality yields $\mathbb E[\unicode{120793}\{|X-X_n|>\varepsilon\}|f(X_n)-f(X)|]\le\mathbb E[\unicode{120793}\{|X-X_n|>\varepsilon\}|f(X_n)|+|f(X)|]$, where we recall that $|x-y|=|x-0+0-y|\le|x-0|+|y-0|=|x|+|y|$. Next, we use $|f(x)|\le\sup_{y}|f(y)|=\|f\|_\infty$, i.e.~the absolute of the function value can be upper bounded by the (uniform or) $\infty$-norm. Finally, we have \begin{aligned} \lim_{n\rightarrow\infty}|\mathbb E[f(\xi_n)]-\mathbb E[f(\xi)]| &\le\lim_{n\rightarrow\infty} (L\varepsilon\mathbb P(|X-X_n|>\varepsilon)+2\|f\|_\infty\mathbb P(|X-X_n|>\varepsilon))\\ &= L\varepsilon(1-\lim_{n\rightarrow\infty}\mathbb P(|X-X_n|>\varepsilon))+2\|f\|_\infty\lim_{n\rightarrow\infty}\mathbb P(|X-X_n|>\varepsilon)\\ &=L\varepsilon \end{aligned} by the assumption. Since we could choose any $\varepsilon>0$ and the left hand side does not depend on $\varepsilon$, we make take the limit on both sides to obtain \begin{aligned} \lim_{n\rightarrow\infty}|\mathbb E[f(\xi_n)]-\mathbb E[f(\xi)]| =\lim_{\varepsilon\rightarrow 0} \lim_{n\rightarrow\infty}|\mathbb E[f(\xi_n)]-\mathbb E[f(\xi)]| \le\lim_{\varepsilon\rightarrow 0}L\varepsilon=0. \end{aligned} This implies that $\lim_{n\rightarrow\infty}\mathbb E[f(\xi_n)]=\mathbb E[f(\xi)]$. Since this holds for all bounded Lipschitz continuous functions $f$, we obtain convergence in distribution (for the random variables and their laws).

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[Math] uniform convergence of characteristic functions

The convergence of the random variables in probability implies weak convergence of the corresponding probability measures

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