nonobvious limit commuting

**choovuck** · August 30th 2008, 06:14 PM

hey all,

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

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

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

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

**choovuck** · August 30th 2008, 06:48 PM

ok, this is trivial -- easy application of maximum principle

sorry for bothering

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