How to show that $\delta_{x_n}\xrightarrow{w}\delta_{x} \iff x_n \to x$

Question

Let $x_n$ be a sequence of reals. Show that $$\delta_{x_n}\xrightarrow{w}\delta_{x} \iff x_n \to x$$

Since the weak convergence is equivalent to pointwise convergence of characteristic functions and since $$\varphi_{\delta_x}(t)=\int e^{its} \ d\delta_x(s) = e^{itx}$$ the "$\Leftarrow$" direction follows with continuity of $e^{it(\cdot)}$.

I am somewhat confused about the "$\Rightarrow$" direction. At first, I was going to do pretty much the same but with $\log$ yet I recalled that the complex log isn't continuous.

What is the proper way to do it?

Answer 1

Let $F,F_n$ denote the CDF's of probability measures $\delta_{x}$ and $\delta_{x_n}$. These are the characteristic functions of the sets $[x,\infty)$ and $[x_n,\infty)$.

Then: $$\delta_{x_{n}}\stackrel{w}{\rightarrow}\delta_{x}\iff\forall y\neq x\; F_{n}\left(y\right)\rightarrow F\left(y\right)$$

This because $F$ is continuous at $y$ if and only if $y\neq x$.

Assume that $x_{n}$ does not converge to $x$.

Then some $\epsilon>0$ can be found such that $\left|x-x_{n}\right|>\epsilon$ for $n$ large enough and a subsequence $\left(x_{n_{k}}\right)$ must exist with $x_{n_{k}}>x+\epsilon$ for each $k$ or $x_{n_{k}}<x-\epsilon$ for each $k$.

In the first case for $y\in\left(x,x+\epsilon\right)$ we have $F\left(y\right)=1$ and $F_{n_k}\left(y\right)=0$ for each $k$.

In the second case for $y\in\left(x-\epsilon,x\right)$ we have $F\left(y\right)=0$ and $F_{n_k}\left(y\right)=1$ for each $k$.

So in both cases we find that $\delta_{x_{n}}\stackrel{w}{\rightarrow}\delta_{x}$ is not true.

Proved is now $\neg x_{n}\rightarrow x\Rightarrow\neg\delta_{x_{n}}\stackrel{w}{\rightarrow}\delta_{x}$ or equivalently $\delta_{x_{n}}\stackrel{w}{\rightarrow}\delta_{x}\Rightarrow x_{n}\rightarrow x$.

Answer 2

I think it is best to use the definition of weak convergence, denoted "$\Rightarrow$": $\mu_n \Rightarrow \mu$ if and only if $\int f d\mu_n \rightarrow \int f d\mu$ for all bounded and continuous $f$.

If $x_n \rightarrow x$ and $f \in C_b(\mathbb{R})$, then \begin{equation} \int f(y) \delta_{x_n}(y) = f(x_n) \rightarrow f(x) = \int f(y) \delta_x(y), \tag{1}\end{equation} so $\delta_{x_n} \Rightarrow \delta_x$. On the other hand, if $x_n \not\rightarrow x$, then $|x_n - x| > \epsilon$ for infinitely many $n$ for some $\epsilon > 0$. Take $f(y) = \left(1 - \frac{d(x,y)}{\epsilon}\right)^+$; note $f(x) = 1$ while $f(x_n) = 0$ whenever $|x - x_n| > \epsilon$, which happens for infinitely many $n$. Thus (1) fails and $\delta_{x_n}\not\Rightarrow \delta_x$.

Here $a^+ = \max(a,0)$.