# How to prove this?

Suppose $\Omega\subset\mathbb{C}$ is a region, $f_n\in H(\Omega)$, for $n=1,2,3,...,$ none of the functions ${f_n}$ has a zero in $\Omega$, and $\{f_n\}$ converges to $f$ uniformly on compact subsets of $\Omega$. Prove that:

(1)either f has no zero in $\Omega$ or $f(z)=0$ for all $z\in\Omega$

(2)if $\Omega'$ is a region that contains every $f_n(\Omega)$, and if $f$ is not constant, then $f(\Omega)\subset\Omega'$

Luckily, I can prove the first question, using Rouche's Theorem. However, I really have no idea how to prove the second one, because I don't know what "$f_n(\Omega)\subset\Omega'$" implies. It seems that I have few tools to deal with that, based on theorems I learnt from my textbook.

PS: I cannot see why $\Omega'$ should be a region.

Any hints will be very appreciated. Actually, I just want some hints. I believe that this question is very easy.

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Have you seen Montel's theorem and normal families? –  gary Sep 13 '11 at 10:40
ah....no, I haven't learnt so far.Actually, I've been teaching myself using the Big Rudin –  Y. Fan Sep 13 '11 at 12:35

Perhaps you should try to use question 1.) to solve question 2.). Pick a point $z_0 \in \mathbb{C}\setminus \Omega'$ and look at the sequence of functions $g_n$ defined by .... (I guess this is a pretty huge hint, hope I didnt give away to much).

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Thank you very much!!!!!!! I can see your point in your hint!Thanks! –  Y. Fan Sep 13 '11 at 12:42

Lemma if $f$ is holomorphic on an open set containing the closed disc $B(c, r]$, and $|f(z)| > |f(c)|$ for every $z\in\partial B(c, r]$, then $f(w) = 0$ for some $w \in B(c, r[$.

Proof Clearly if $|f(c)|=0$ there is nothing to prove. So we will assume $|f(c)|>0$. Suppose the thesis is false. Then the open set $D\backslash Z(f)$, where $g(z) = 1/f(z)$ is defined and holomorphic, contains $B(c, r]$, so that the first Cauchy estimate can be applied to $g$ on $B(c, r]$, giving $|g(c)| \leq \|g\|_r$. But $|g(c)| = 1/|f(c)|$ and $\|g\|_r = 1/|f(z)|$ for some $z \in \partial B(c, r]$, so that $|f(c)| \geq |f(z)|$ for this $z \in \partial B(c, r]$, contradicting the hypothesis.

Using this lemma:

a) Assume that $f(c) = 0$ for some $c \in D$, with $f$ not identically zero in $D$; then we can pick a disc $B(c, r] \subseteq D$ such that $Z_D(f) \cap B(c, r] = {c}$; then $\min\{|f(z)| : z ∈ \partial B(c, r]\} = μ > 0$ Take $n_0 \in \mathbb N$ such that $\|f − f_n\|_{B(c,r]} \leq \mu/3$ for $n\geq n_0$. Then $|f_n(c)| \leq \mu/3 < 2\mu/3 \leq \min\{|f_n(z)| : z \in \partial B(c, r]\}$, so that $f_n$ has a zero in $B(c, r[$, a contradiction.

b) Pick any $w\in \mathbb C\backslash \Omega'$. Then $f_n-w$ is zero free in $\Omega$ and converges compactly on $\Omega$ to the function $f-w$, which, being non constant, is then by a) zero free in $\Omega$. Hence $f(\Omega)\subseteq \Omega'$.

In fact, there's no need for $\Omega'$ to be a region, while on $\Omega$ it is necessary since otherwise you may not have any disc contained in it. I think it was just a choice of your book, there are no deep reasoning.

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Is your lemma equivalent to the minimum modulus theorem? –  Y. Fan Sep 14 '11 at 2:35