# Limit of a sequence of holomorphic functions

Let $f_n$ be a sequence of holomorphic functions on a domain $D \subset \mathbb{C}$ converging to a function $f$, and also converging uniformly on compact subsets. Suppose each function has at most $m$ zeros (counted with multiplicity) for some fixed $m \in \{0, 1, 2, ... \}$. Show that either $f$ is exactly zero on $D$ or has at most $m$ zeros on $D$.

I'm not sure how to approach this problem, but a hint to get my started would be much appreciated.

Context: I'm studying for a qual.

• Calculating a path integral commutes with taking a uniformly convergent limit on the compact set (=the path). So... Argument principle? Hmm. May be problematic if zeros converge towards the boundary. If $f$ has $m+1$ zeros in some compact set contained in $D$ and with a nice boundary curve, then... – Jyrki Lahtonen Aug 26 '15 at 20:10
• Do we know that the zeros converge somewhere? My approach was to assume that there were at least $m+1$ zeros and then show it must be exactly zero, is this reasonable? – user19817 Aug 26 '15 at 20:19
• I don't know. I am hoping that if $f$ has $m+1$ zeros, then over some nice path $C$ we have $$2\pi i(m+1)=\int_C\frac{f'(z)}{f(z)}\,dz=\lim_{n\to\infty}\int_C\frac{f'_n(z)}{f_n(z)}\,dz.$$ Implying that for all large enough $n$ the function $f_n$ also has at least $m+1$ zeros inside $C$. But it's been a while since I did anything like this, so it is not clear to me how to justify all of the above. – Jyrki Lahtonen Aug 26 '15 at 20:22
• Nice! And the right side is at most $2\Pi*i(m)$, giving a contradiction. Thanks for your comments. – user19817 Aug 26 '15 at 20:26

Let $\Gamma\subset D$ be a curve that $f(z)\ne 0$ for any $z\in\Gamma$. Denote $$m = \min_{z\in\Gamma}|f(z)|$$ Since we have convergence on compact sets, concretely on the compact enclosed by $\Gamma$, then $n\ge n_0$ we have $$|f_n(z)-f(z)|<m\le f(z).$$
It follews from the Rouché's theorem that the functions $f$ and $f_n=(f_n-f)+f$ have the same number of zeroes inside $\Gamma.$