Weak Convergence If and Only If (Pointwise) Convergence of Characteristic Function

This is actually a theorem from lecture notes, with the corresponding proof. Unfortunately, it doesn't prove the last bit, or mention it at all (!), and I have a question about the penultimate bit. This is the definition of weak convergence being used:

Let $\mu$ be a Borel probability measure on $\Bbb R^d$ and let $(\mu_n)$ be a sequence of Borel probability measures on $\Bbb R^d$.

We say that $(\mu_n)$ converges weakly to $\mu$, written $\mu_n \Rightarrow \mu$, if $\mu_n(f) \rightarrow \mu(f)$ as $n \rightarrow \infty$ for all bounded, continuous functions $f:\Bbb R^d \rightarrow \Bbb R$.

We say that the sequence $(X_n)$ of random variables on $\Bbb R^d$ converges weakly to $X$ if $\mu_{X_n} \Rightarrow \mu_X$.

Firstly, consider the penultimate claim (If a sequence...). What I don't understand is why this example is not a counter-example to the claim. ($\phi$ is the characteristic function.)

Secondly, consider the final claim (Conversely, if...). No proof was given in the lecture notes. I think I have a solution, but I'm not sure; if someone could look over my answer (given as an answer below), then I'd be most appreciative. Thanks!

• Regarding the first, $\phi_{X_n}(t) \to \phi_X(t)$ says, not merely that CFs converges pointwise to some function, but that it convergest to the CF of the random variable $X$, hence $\phi_X(t)$ is continuous at $t=0$. This is not true in the counterexample. – leonbloy Jan 8 '15 at 15:34
• Ahh, I see. I did notice that the limit function in the other answer wasn't continuous, but forgetting that the characteristic function needs to be continuous (not so for the density function (pdf) yes?). – Sam T Jan 8 '15 at 15:43

Suppose $X_n \Rightarrow X$, ie $\mu_{X_n} \Rightarrow \mu_X$. We desire to show that $\phi_{X_n}(\xi) \rightarrow \phi_{X_n}$ as $n \to \infty$ $\forall \xi \in \Bbb R^d.$ Write $\mu_n = \mu_{X_n}$ and $\mu = \mu_X$.
Since, by definition, $\mu_n \Rightarrow \mu$, in particular we have that, for each (fixed) $\xi \in \Bbb R^d$, $\mu_n(e^{i x \cdot \xi}) \rightarrow \mu(e^{i x \cdot \xi})$. Thus, \begin{align} |\phi_{X_n} - \phi_X| & = |\hat \mu_{X_n}(\xi) - \hat \mu_X(\xi) | \\ & = | \int_{\Bbb R^d} e^{i x \cdot \xi} \, d\mu_{X_n}(x) - \int_{\Bbb R^d} e^{i x \cdot \xi} \, d\mu_{X_n}(x) | \\ & = | \mu_n(e^{i x \cdot \xi}) - \mu(e^{i x \cdot \xi}) | \\ & \rightarrow 0. \end{align} as $n \to \infty$. (No need for the dominated convergence theorem.)
Also, this is for fixed $\xi$, so I have shown pointwise, not uniform convergence.
• No, your proof is not valid. In general, we cannot expect uniform convergence. First of all: $$\hat{\mu}_{X_n}(\xi) \neq \int \mu_{X_n}(x) e^{\imath \, x \xi} \, dx.$$ The expression $\mu_{X_n}(x)$ doesn't even make sense. Instead it should read $$\hat{\mu}_{X_n}(\xi) = \int e^{\imath \, x \xi} \, d\mu_{X_n}(x).$$ (Note that this is an integral with respect to the distribution $\mu_{X_n}$ of the random variable $X_n$.) – saz Jan 8 '15 at 16:06
• Yes I see. This is exactly what I thought the issue would be. What do I then do with $|\hat\mu_{X_n}(\xi)-\hat\mu_X(\xi)|$? I realise that we won't necessarily get uniform convergence, but we should get pointwise. How would I rectify my error? – Sam T Jan 8 '15 at 17:00
• I see. Then the hint goes like that: Note that for each fixed $\xi \in \mathbb{R}^d$ the function $f(x) := e^{\imath \, x \cdot \xi}$ is bounded and continuous. – saz Jan 8 '15 at 17:21