Let $U$ be a complex domain, and $(f_n)_{n\in \mathbb{N}}$ be a sequence on one-to-one analytic functions defined on $U$. Suppose that $f_n$ converges to $f$ uniformly on every compact subset of $U$. Prove that $f$ is either constant or one-to-one on $U$.
Here's my proof. I would appreciate if you guys point out any possible mistakes, or maybe give a different proof that uses different techniques.
We may assume that the zeros of $f$ have no accumulation point in $U$, otherwise, $f$ is identically zero.
Now we proceed by contradiction; suppose, WLOG, that $f$ has two zeros in $U$; say $f(a)=f(b)=0$. There exists $\delta >0$ such that $a$ and $b$ are the only zeroes of $f$ in $B(a,\delta)$ and $B(b,\delta)$, respectively. Moreover, the Maximum Modulus Principle implies that $|f(z)|>0$ on the boundaries of the two balls above; say $$|f(z)|>m>0\, \text{ for all }\, z\in \partial B(a,\delta)\cup \partial B(b,\delta).$$
Since $f_n \rightarrow f$ uniformly on compact sets, then for $n$ large enough, we have $$|f_n (z)-f(z)|<m<|f(z)| \, \text { for all } \,z\in \partial B(a,\delta)\cup \partial B(b,\delta).$$
By Rouche's Theorem, $f_n$ and $f$ have the same number of zeroes in $B(a,\delta)$ and $B(b,\delta)$, namely one in each ball (not counting multiplicities). But this contradicts the assumption that $f_n$ is one-to-one.
Best Answer
May I do it in this way?
Let $a \in U$, there is a compact subset $\overline{B_R(a)}\subset U$. Then $f_n$ converges to a analytic $f$ uniformly on $\overline{B_R(a)}$. Define $g_n=f_n'\neq 0$ ($f_n$ is one to one) which is also a analytic sequence. Suppose $g_n$ converges to a analytic $g$ uniformly on $\overline{B_R(a)}$. By uniform convergence of $f_n$ and Cauchy Integral formula $$ g(a)=\lim g_n(a)=\dfrac{1}{2\pi} \lim \int_{B_{R/2}(a)} \dfrac{f_n}{(z-a)^2}=\dfrac{1}{2\pi} \int_{B_{R/2}(a)} \dfrac{\lim f_n}{(z-a)^2}=\dfrac{1}{2\pi} \int_{B_{R/2}(a)} \dfrac{ f}{(z-a)^2}=f'(a)$$ By Corollary of Hurwitz's theorem, $g=f'$ is non-vanishing or identically zero.