Let $\Omega$ be open and connected, and let $\{f_n\}$ be a sequence of holomorphic functions on $\Omega$ such that $f_n(\Omega)\subseteq \Omega$. If $f_n\rightarrow f$ uniformly on compact subsets $\Omega$ and $f$ is not constant, prove $f(\Omega)\subseteq \Omega$
My attempt:
Let $z_0\in \Omega$ and let $g(z)=f(z)-f(z_0)$ and $h(z)=f_n(z)-f(z_0)$. Clearly, $g(z)$ has a zero in a small neighborhood of $z_0$, which is $z_0$.
If I'm able to show that $h(z)$ also has a zero in that neighborhood, say $z_1$, this implies that $f(z_0)=f_n(z_1)\in \Omega$ and this completes the proof.
But I'm having trouble showing that $h(z)$ has a zero in the neighborhood of $z_0$. I'm trying to use Rouche's theorem by showing if $|g(z)-h(z)|<|g(z)|$ in the boundary of some neighborhood of $z_0$, $g(z)$ and $h(z)$ has the same number of roots in the neighborhood, hence, there exists some $z_1$ that satisfies my claim.
Can anyone show me how to get the inequality?
Best Answer
Suppose for the sake of contradiction there exists a point $z_0$ such that $z_0 \notin \Omega$, such that $f(w_0) = z_0$. WLOG $z_0 = 0$, then by the argument principle \begin{equation} \int_{\gamma} \frac{f'}{f} \geq 1 \end{equation} where $\gamma$ is a small contour around zero so that it doesn't intersect $\Omega$; this is possible since $\Omega$ is open. But observe \begin{equation} \int_{\gamma} \frac{f_n'}{f_n} = 0 \end{equation} for any $n$ since $f_n(\Omega) \subset \Omega$, so $f_n \neq 0$ for any $z \in \Omega$. So by uniform convergence we get \begin{equation} \int_{\gamma} \frac{f'}{f} = 0 \end{equation} which is our contradiction. In general uniform convergence + argument principle gives you many strong results such as if $f_n$ is $1-1$ and converges uniformly to $f$, then either $f$ is constant or $1-1$ and similarly if $f_n \neq 0$ for any $z$. (Uniform convergence allows you to pass limit to integrals, and this integral tells you how many zeros there are.)