Number of roots of a sequence of a uniformly convergent holomorphic functions implies an upper bound for the number of roots of their limit

Question

Let $G$ be an open, simply connected region in $\mathbb{C}$. We define a sequence of holomorphic functions $(f_n)_{n \in \mathbb{N}}, f_n: G \to \mathbb{C}$ as almost uniformly convergent iff $(f_n)$ converges uniformly on all compact subsets of $G$.

I now want to show that, if $f_n \to f$ for a non-constant function $f: G \to \mathbb{C}$, and none of the $f_n$ has more than $m \in \mathbb{N}$ roots, then $f$ also has not more than $m$ roots. (If we count roots with their multiplicities.)

I already know that the limit of an almost uniformly convergent sequence of holomorphic functions is also holomorphic, and (although I don't think that's helpful here), I also know that $(f_n')$ then converges almost uniformly against $f'$. So I only have to show this statement about the roots.

I thought about using Rouché's Theorem, mostly because I couldn't think about any other Theorem I know that concretly talks about the actual number of roots of different functions. But I don't know how exactly I can apply Rouché here: what would I choose as the functions $f$ and $g$ that Rouché's Theorem demands, in order to show the inequality in the Theorem?

Answer 1

It is a bit easier to prove the other way around:

If $f$ has $m$ roots in $G$ (counted with multiplicity) then there exists an $n_0 \in \Bbb N$ that for all $n \ge n_0$, $f_n$ has at least $m$ roots (counted with multiplicity).

Since the roots of $f$ can be separated into disjoint circles, it suffices to show:

If $D$ is a circle such that $D \subset G$ and $f$ has a single root $z_1$ in $D$ with multiplicity $m_1$, then there exists an $n_0 \in \Bbb N$ that for all $n \ge n_0$, $f_n$ has $m_1$ roots in $D$ (counted with multiplicity).

Here one can use Rouché's theorem. Without loss of generality we can assume that $\overline D \subset G$ and $f$ is not zero on the boundary of $D$ (otherwise make the circle smaller). Then $$ \varepsilon := \frac 12 \min \{ |f(z)| : z \in \partial D \} $$ is strictly positive, and $f_n \to f$ uniformly on $\partial D$. Therefore there exists an $n_0 \in \Bbb N$ such that $$ | f_n(z) - f(z) | \le \varepsilon < | f(z) | $$ for all $z \in \partial D$ and $n \ge n_0$. It follows that $f_n$ and $f$ have the same number of roots in $D$ for $n \ge n_0$.