I'd like some assistant on the proof of the following Lemma:
If $X_n$ is $AN(\mu,\sigma_n^2)$, then also $X_n$ is $AN(\overline\mu,\overline\sigma_n^2)$ if and only if $\frac{\overline\sigma_n}{\sigma_n}\rightarrow1$, $\frac{\overline\mu_n-\mu_n}{\sigma_n}\rightarrow0$.
The hint says to use Pólya's Theorem, which relates weak convergence and uniform convergence, but I'm not sure about how to use it. Thanks in advance.
By Pólya's theorem, $X_n$ is $AN(\mu_n,\sigma_n^2)$ is equivalent to $$ \sup_{t\in\mathbb R}\left\lvert\Pr\left(\frac{X_n-\mu_n}{\sigma_n }\leqslant t\right)-\Phi(t) \right\rvert\to 0, $$ where $\Phi$ denotes the cumulative distribution function of a standard normal random variable. Changing $t$ to $u:= t\sigma_n +\mu_n$ in the supremum, this is also equivalent to $$ \sup_{u\in\mathbb R}\left\lvert\Pr\left(X_n\leqslant u\right)-\Phi\left(\frac{u-\mu_n}{\sigma_n }\right) \right\rvert\to 0. $$ Now, if $X_n$ is both $AN(\mu_n,\sigma_n^2)$ and $AN(\overline\mu,\overline\sigma_n^2)$, we derive that $$ \sup_{u\in\mathbb R}\left\lvert\Phi\left(\frac{u-\mu_n}{\sigma_n }\right) -\Phi\left(\frac{u-\bar{\mu}_n}{\bar{\sigma}_n}\right) \right\rvert\to 0. $$ Doing the change $v:= (u-\mu_n)/\sigma_n $ gives $$ \sup_{v\in\mathbb R}\left\lvert\Phi\left(v\right) -\Phi\left(\frac{v\sigma_n +\mu_n-\bar{\mu}_n}{\bar{\sigma}_n}\right) \right\rvert\to 0. $$