Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

2 Answers 2

up vote 1 down vote accepted

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).

share|improve this answer
    
Thank you very much!!!!!!! I can see your point in your hint!Thanks! –  Y. Fan Sep 13 '11 at 12:42
add comment

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.

share|improve this answer
    
Is your lemma equivalent to the minimum modulus theorem? –  Y. Fan Sep 14 '11 at 2:35
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.