Given $x_n$ uniformly convergent, is $\log |x_n|$ uniformly convergent? Suppose the sequence $x_n(z)$ satisfies:
a) uniformly convergence, and
b) $x_n$ is real-valued, and
c) $0 < N \le |x_n| \le M$, and
d) $x_n$ is analytic
on a compact set $K \subset \mathbb{C}$. Is it true that $g_n = \log |x_n|$ is uniformly convergent on $K$? If it's not, how about the subsequence of $g_n$?
I think it is true, but I'm getting stuck at fining an estimate for $\log$ functions.
Thank you.
 A: Note that $x_j \to x$ uniformly on $K$ implies $|x_j| \to |x|$ uniformly on $K.$ That follows from $||b|-|a|| \le |b-a|.$
Because each $|x_j|$ maps $K$ into $[M,N],$ the same is true of $|x|.$ Now since $0 < M <N,$ $\ln t$ is continuous on $[M,N].$ Since $[M,N]$ is compact, $\ln t$ is uniformly continuous on $[M,N].$
So let $\epsilon>0.$ Then there exists $\delta > 0$ such that $s,t\in [M,N], |s-t| < \delta,$ implies $|\ln s - \ln t| < \epsilon.$ Choose $J$ such that $j\ge J$ implies $||x_j(z)|-|x(z)|| < \delta$ for all $z\in K.$ Then for $j\ge J,$ $|\ln |x_j(z)|-\ln |x(z)|| <\epsilon$ for all $z\in K.$ That gives the desired uniform convergence.
Note that we didn't need the $x_n$ to be analytic, or even continuous, or much of anything really, to get the conclusion. Also, we didn't need $K$ to be compact. And there is nothing special about the logarithm here. Any function continuous on a compact set containing $\cup_j x_j(K)$ would have done as well.
A: Since $0 < N \leq |x_n| \leq M$, $\log |x_n|$ just transforms $|x_n|$ by at most $\sup_{y\in [N,M]} y-\log y$, which exists. Can you use this to help your uniform estimate?
