Proof: Let \(\varphi\) denote the characteristic function of \(\x - \bm\) (which does not require a subscript since the random vectors are iid), and let \(\varphi_n\) denote the characteristic function of \(\sqrt{n}(\bar{\x}-\bm) = \tfrac{1}{\sqrt{n}}\sum_i (\x_i - \bm)\). The numbers in parentheses refer to the numbered properties of characteristic functions.

\[\begin{alignat*}{2} \varphi_n(t) &= \varphi(t/\sqrt{n})^n &\hspace{4em}& (2), (4), \textnormal{ independence} \\ &= [\varphi(\zero) + \nabla\varphi(0) \Tr \t/\sqrt{n} + \tfrac{1}{2n} \t \Tr \nabla^2 \varphi(\t^*/\sqrt{n}) \t ]^n && \href{taylor-series.html}{\textnormal{Taylor series}}; (6); \href{random-vector.html}{\Ex \norm{\x}^2 < \infty} \\ &= [1 + \tfrac{1}{2n} \t \Tr \nabla^2 \varphi(\t^*/\sqrt{n}) \t ]^n && (1), (5), \Ex(\x_i - \bm) = \zero \\ &\to \exp\left\{ -\tfrac{1}{2} \t \Tr \bS \t \right\} && \href{exponential-limit.html}{\textnormal{Exponential limit}}; (6) \end{alignat*}\]

Therefore,

\[\sqrt{n}(\bar{\x} - \bm) \inD \Norm(\zero, \bS)\]

by the continuity theorem and (8).