I am having a hard time trying to study complex analysis for one of my exams. I have some problems with an exercise that I am supposed to solve. The setting is the following.
Let $D$ be a domain (open and connected subset) in $\mathbb{C}$, $n_{0}\in\mathbb{N}$ and $(f_{n})_{n\geq n_{0}}$ be a sequence in $\mathcal{O}(D)$, the set of holomorphic functions $D\to\mathbb{C}$. Assume that, for each $n\geq n_{0}$, $f_{n}\neq 0$ and $\prod_{n\geq n_{0}} f_{n}$ converges normally in $\mathcal{O}(D)$. Then I know that $(\prod_{n=n_{0}}^{j} f_{n})_{j\geq n_{0}}$ converges to some $f\in \mathcal{O}(D)$ in $\mathcal{O}(D)$.
(Here $\mathcal{O}(D)$ has the topology induced by that on $\mathcal{C}(D):=\{f\colon D\to \mathbb{C}:\ \text{f is continuous}\}$ which is given, for any compact $K\subseteq D$, any $\epsilon >0$ and any $f\in\mathcal{C}(D)$, by the balls $B_{K,\epsilon}(f)=\{g\in\mathcal{C}(D):\ \vert\vert g -f\vert\vert_{K}:=\max_{z\in K} \vert g(z)-f(z)\vert <\epsilon\}$).
Now I am asked to prove the following statement.
Let $D':=\{z\in D:\ \forall n\geq n_{0} (f_{n}(z)\neq 0)\}$. Prove that $\prod_{n\geq n_{0}} 1/f_{n}$ converges normally to $1/f$ in $\mathcal{O}(D')$.
I don't know where to start in showing this. I suppose that I should prove first the normal convergence of $\prod_{n\geq n_{0}} 1/f_{n}$ and then show that it converges precisely to $1/f$. As far as the first part is concerned, a so to say standard approach might be that of fixing a compact $K\subseteq D'$ and show that $g_{n}:= 1/f_{n}-1$ satisfies $\vert\vert g_{n}\vert\vert_{K}\leq m_{n}$, for some $m_{n}\in\ \mathbb{R}$ such that $\sum_{n\geq n_{0}} m_{n}$ is known to converge. Still, I don't go too far from this (also because I fear I am not so much confident with all of this stuff).
Any help or reference for the problem would be greately appreciated. Thanks.
Let $g_j = \prod_{n=n_0}^j f_n$, and focus on the following facts: $g_j\to f$ uniformly on compact subsets of $D'$, and $g_j\ne 0$ for all $j$. By Hurwitz's theorem either $f \equiv 0$ in $D'$, or $f\ne 0$ in $D'$.
If $f\equiv 0$, then $g_j$ are uniformly small on compact subsets of $D'$, which implies $1/g_j\to \infty$ uniformly on compact subsets. But I suspect the case $f\equiv 0$ is ruled out because an infinite product that tends to $0$ is not considered converging.
If $f\ne 0$, then $f$ is bounded away from $0$ on every compact subset $K\subset D'$. Hence $g_j$ are also bounded away from $0$ on $K$. This allows you to conclude that $$ \left|\frac{1}{g_j} - \frac{1}{f}\right| \le \frac{|f-g_j|}{|g_j||f|} \to 0 $$ uniformly on $K$.