= Solution
<Weak convergence> $X_k\Rightarrow X$ means that $\mathbb E f(X_k)\to\mathbb E f(X)$ for every bounded continuous real function $f$ on $\mathbb R^d$; it is a statement about their distributions, not their coupling.
The <central limit theorem> says that for independent identically distributed real variables of finite mean $\mu$ and variance $\sigma^2$, $\sqrt n(\bar X_n-\mu)\Rightarrow N(0,\sigma^2)$. For $\sigma>0$ this is equivalent to convergence of $(\sum X_j-n\mu)/(\sigma\sqrt n)$ to the standard <normal distribution>. For $\sigma=0$, every variable equals $\mu$ almost surely and the limit is the point mass at zero.
For the proof when $\sigma>0$, put $Y=(X-\mu)/\sigma$. Its <characteristic function> has expansion $\varphi_Y(t)=1-t^2/2+o(t^2)$. To justify this using only a second moment, subtract the first two Taylor terms of $e^{itY}$ and use <dominated convergence theorem>, with the second-order remainder bounded by a constant times $t^2Y^2$. Independence gives the characteristic function of the normalized sum as $[\varphi_Y(t/\sqrt n)]^n\to e^{-t^2/2}$. The <Lévy continuity theorem> says that pointwise convergence of characteristic functions to a characteristic function continuous at zero implies <weak convergence>. Since $e^{-t^2/2}$ is the characteristic function of $N(0,1)$, the result follows.
Back to article page
