# If $f_n\rightarrow f$ uniformly on compact subsets $\Omega$ and $f$ is not constant, prove $f(\Omega)\subseteq \Omega$

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?

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.)
• So the argument principle states that if $f$ is meromorphic with roots and poles not on the contour $\gamma$, we have $\frac{1}{2\pi i}\int_{\gamma}\frac{f'}{f}=$ number of zeros in $\gamma$ - number of poles in $\gamma$. How do we know that $\int_{\gamma}\frac{f'}{f}\geq1$? I guess I just need some more explanation on your reasoning. – Ya G Dec 19 '18 at 19:22
• As $f_n \rightarrow f$ uniformly on compact sets, $f_n$ holomorphic, we know by Morerra Theorem and uniform convergence that $f$ is holomorphic. So it has no poles, but $f(z_0) = 0$, so it has at least one zero in $\gamma$ namely $z_0$. So the integral is greater than or equal to 1. – Story123 Dec 19 '18 at 19:26
• Got it! So you are assuming that there is a root inside the contour such that the root itself is not in $\Omega$. But since it's inside the contour, the argument principle suggests that there exists at least one zero in the contour. – Ya G Dec 19 '18 at 19:31
• Yes, you can assume WLOG it's a root since if it isn't consider $g(z) := f(z) - w_0$, then $g'(z) = f'(z)$ and you can apply the argument principle to $g$. – Story123 Dec 19 '18 at 19:36
• What if $z_0\in \partial \Omega?$ Then every nbhd of $z_0$ intersects $\Omega$ – Matematleta Dec 19 '18 at 22:06