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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?

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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.)

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  • $\begingroup$ 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. $\endgroup$ – Ya G Dec 19 '18 at 19:22
  • $\begingroup$ 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. $\endgroup$ – Story123 Dec 19 '18 at 19:26
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    $\begingroup$ 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. $\endgroup$ – Ya G Dec 19 '18 at 19:31
  • $\begingroup$ 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 $g$. $\endgroup$ – Story123 Dec 19 '18 at 19:36
  • $\begingroup$ What if $z_0\in \partial \Omega?$ Then every nbhd of $z_0$ intersects $\Omega$ $\endgroup$ – Matematleta Dec 19 '18 at 22:06

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