Let $D$ be an open connected set and $\{f_n\}$ be a sequence of holomorphic functions on $D$ that converge uniformly to a holomorphic function $f$ on all compact subsets of $D$. Suppose that $f_n(D) \subset U$ for all $n$. Show that either $f(D) \subset U$ or there is a $c \in \partial U$ such that $f(z) = c$ for all $z \in D$.
First I supposed that $f$ was not a constant function. Then I suppose that there is a $w \in D$ such that $f(w) \notin \overline{U}$ By the open mapping theorem, I can take a small ball around this point. Consider the closure of this ball. Then the pullback of this ball under $f$ is a closed set in $D$. Then I hope to use uniform convergence on compact sets to show that for $n$ large enough $f_n(D) \not\subset U$ which would be a contradiction.
I don't feel confident with this approach. Also, this homework was given in a problem set that uses the Schwarz Reflection Principle as well as Caratheodory's Theorem so perhaps I should make use of these theorems.
Any tips?
Best Answer
Assume that there is $p\in D$ such that $f(p)\notin U$. Since $f(p)$ is the limit of $f_n(p)\in U$ you must have that $f(p)\in\overline{U}$.
If $f\not\equiv f(p)$, then by Hurwitz's theorem (which proof is just an exercise using the argument principle), if we take a neighborhood of $p$ inside $D$, then for all large enough $n$ the functions $f_n(z)-f(p)$ will have solutions in that neighborhood. But this is not possible, since $f_n(z)\in U$ and $f(p)\notin U$.
Therefore $f\equiv f(p)$.