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Let $\Omega$ be a domain in $\mathbb{C}$ and $\{f_n\}$ be a sequence of holomorphic functions on $\Omega$ which converges uniformly on compact subsets of $\Omega$ to a non-constant function $f$. Prove that if $f$ has at least $m$ zeroes in $\Omega$, then all but finitely many of the functions $f_n$ have at least $m$ zeroes in $\Omega$

My try:

Since $\{f_n\}$ converges uniformly on compact subsets of $\Omega$ to a non-constant function $f$, $f$ is holomorphic on $\Omega$. Let $z_0 \in \Omega$ be a zero of $f$. Then there exists $r \gt 0$ such that closure($ {B(z_0,r)}$) $\subset \Omega$. Then $|f(z)|$ attains its minimum on the boundary of $B(z_0,r)$ ( As |f| is continuous and boundary is compact) (say $m$). Then for any $\epsilon \gt 0$ (less than m), there exists a $n_0 \in \mathbb{N}$ such that for all $n \ge n_0$, $|f_n(z)-f(z)|\lt \epsilon \lt|f(z)|$ for all $z \in\{ |z-z_0|= r\}$. By Rouche's Theorem $f_n$ & $f$ have the same no of zeroes in $B(z_0,r)$ for all $n \ge n_0$.Since I can do this for the $m$ zeroes of $f$, We conclude that $f_n$ has at least $m$ zeroes in $\Omega$ for all $n \gt n^{*}$, where $n^{*}$= maximum$(n_0,n_1,..n_{m-1})$, where each $n_i$ corresponds to the zero $z_i$.

There is a little problem in this argument. What if $m$ turns out to be $0$?? For this not to happen I have to ensure that $f(z)$ doesn't vanish anywhere on the boundary. How can I guarantee the existence of such a $r$??

Thanks for the help!!

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  • $\begingroup$ @SrinivasK I wanted to use Rouche's Theorem $\endgroup$ – tattwamasi amrutam Apr 3 '15 at 9:12
  • $\begingroup$ But I don't know much about rouche's theorem. Please point out if there is any error in this argument. Also, please add this point about using rouche's theorem in the question itself. $\endgroup$ – Srinivas K Apr 3 '15 at 9:18
  • $\begingroup$ @SrinivasK..How u will get such a sequence? $\endgroup$ – tattwamasi amrutam Apr 3 '15 at 9:36
  • $\begingroup$ We need to prove that if $f$ has no zeroes, then all but finiteley many $f_n's$ have no zeroes : Suppose that $f$ has no zeroes but infinitely many $f′ns$ have zeroes.So, we can find $n_o$ such that for all $n \geq n_o$ , each $f_n$ has atleast one zero say at $z_n$ So, we can find a sequence$(z_n)$ such that for all $n>n_o,|f_n(z_n)−f(z_n)|=|0−f(z_n)|=|f(z_n)|≤ϵ. $After this i feel we can say that within the compact subset, every sequence has a convergent subsequence and so there must be a point $z$ such that $z_n→z$ and by continuity of$ f ,f(z)=0$ which is a contradiction. $\endgroup$ – Srinivas K Apr 3 '15 at 14:31
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What if $m$ turns out to be $0$?? For this not to happen I have to ensure that $f(z)$ doesn't vanish anywhere on the boundary. How can I guarantee the existence of such a $r$??

You know that $f$ is a non-constant holomorphic function. Hence the zeros of $f$ are isolated. That means there is a $\rho > 0$ such that $z_0$ is the only zero of $f$ in the set $\{ z \in \Omega : \lvert z-z_0\rvert < \rho\}$. Then you just need to pick an $r \in (0,\rho)$ such that $\overline{B(z_0,r)} \subset \Omega$.

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  • $\begingroup$ Yeah Got it. Thanks. Is the solution alright?? Otherwise? $\endgroup$ – tattwamasi amrutam Apr 3 '15 at 9:57
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    $\begingroup$ Yes, it's a good solution. But you need to choose the radii around the zeros of $f$ so small that the disks $B(z_k,r_k)$ don't intersect. $\endgroup$ – Daniel Fischer Apr 3 '15 at 9:58
  • $\begingroup$ yes. They won't by the solution you gave. $\endgroup$ – tattwamasi amrutam Apr 3 '15 at 10:12

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