Let $(f_n)\subset\mathcal{O}(\Omega)$ be a sequence and let $(g_n)\subset\mathcal{O}(\Omega)$ be a locally uniformly convergent sequence with limit-function $g$, not being identically zero, such that the sequence of the productfunctions $f_n g_n$ be locally uniformly convergent.
This implies that $f_n$ itself is locally uniformly convergent.
The task is to show the above implication by using a previous task, which was to show a stronger version of the "Weierstrass Convergence Theorem". I've found the same task asked and solved here:
A sequence of holomorphic functions $\{f_n\}$ uniformly convergent on boundary of open set.
I have the following problems:
I sadly have no idea how to use the theorem (i.e the above given link) to proof this implication.
Even worse, I've also failed to prove it otherwise, i.e. trying to make use of the functions that are given as locally uniformly convergent and making use of the triangle inequality and adding zeros.
Any help very much appreciated!
Edit: I guess, since it isn't stated otherwise and because of the stated connection between the previous task and this one, that $\Omega$ is a bounded connected open subset of $\mathbb{C}$ just as in the previous task (see above link).
Edit: Sorry, I was wrong: The implication does not presuppose that $f_n g_n\subset\mathcal{O}(\Omega)$. It simply presupposes that the productfunction $f_n g_n$ converges locally uniformly. I'm sorry for mistaken that.
Best Answer
Here's the idea. If you knew that $g$ was never $0$, the problem would be easy: since $f_ng_n$ converges locally uniformly and $g_n$ converges locally uniformly to a nonzero value $g$, then the quotient $f_ng_n/g_n=f_n$ also converges locally uniformly. (There are some details to fill in to prove that, but there is no major insight needed and in particular no theorems from complex analysis are needed; it's basically just the fact that division is continuous if you stay away from dividing by $0$.)
Unfortunately, $g$ might be $0$ at some points, so that we cannot just divide by $g_n$ freely. Here's where your "previous task" comes in. If you pick a small disk $D$ around a point $a$ such that $g(a)=0$, then $g(z)\neq 0$ for all $z\in D$ except for $z=a$. In particular, $g$ is nonzero on the boundary $\partial D$, and so $f_ng_n/g_n=f_n$ converges uniformly on $\partial D$. But now by your "previous task", this is enough to conclude that $f_n$ converges uniformly on $D$ as well.