hey all,

basically I need to justify $\displaystyle \lim_{n\to\infty}\lim_{k\to\infty}f_n(z_k)=\lim_{k \to\infty}\lim_{n\to\infty}f_n(z_k)$ in this very particular setup

let $\displaystyle f_n(z)$ be analytic functions on some fixed annulus $\displaystyle \{z:r<|z|<R\}$ where $\displaystyle r<1<R$. It is given that $\displaystyle f_n(z)$ tend to the constant function 1 uniformly on compact sets of $\displaystyle \{z:r<|z|<1\}$ and of $\displaystyle \{z:1<|z|<R\}$ (so everywhere except of the unit circle). Prove that $\displaystyle f_n(z)$ tend to 1 for $\displaystyle z$ on the unit circle as well.

WLOG take $\displaystyle z=1$. we can choose $\displaystyle z_k\to 1$, $\displaystyle |z_k|\ne 1$, and so $\displaystyle \lim_{n\to\infty}\lim_{k\to\infty}f_n(z_k)=\lim_{k \to\infty}\lim_{n\to\infty}f_n(z_k)$ is what we need

it's clearly crucial that $\displaystyle f_n(z)$ are analytic and converge both inside and outside the unit circle (as the counterexample $\displaystyle 1+z^n$ suggests), but I can't find a way to use it