Locally uniformly convergent productfunctions imply locally uniformly convergent function

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.

• Would you know how to solve the problem if you knew that $g$ is never zero? – Eric Wofsey Feb 2 at 8:20
• @EricWofsey Firstly, thank you very much for you answer! But I sadly have to say: no, even then I would not know how to solve it. I know Hurwitz's Theorem says sth. about the limit-function of a locally uniformly convergent sequence of holomorphic functions given the multiplicity of zeros... but I don't know to apply it here. Could you help me further please? – JoHe Feb 2 at 12:01
• The case where $g$ is never zero requires no complex analysis whatsoever. It's just a basic real analysis argument, using nothing but continuity of the functions. – Eric Wofsey Feb 2 at 16:08
• @EricWofsey Again thank you! But the only thing that I know I can say with certainty is that since $(g_n)$ is locally uniformly convergent $g$ has to be holomorphic (Weierstrass Convergence Theorem) on $\Omega$ and since it is not identically zero $g$ has only isolated zeros (Identity Theorem). But maybe by some other Theorem one can imply that g is never zero. Could you show that argument please? Maybe one can use that $g$ has only isolated singularities or sth. else avoiding the condition that $g$ must not be zero anywhere. Thank you very much! – JoHe Feb 2 at 17:01

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.
• Thank you! The first part makes much sense to me but for the second part I need that $f_ng_n/g_n\subset\mathcal{O}(D)$ but $g_n$ has a zero at the center of $D$ - so I don't meet the conditions of the previous task, do I? – JoHe Feb 2 at 18:01
• $f_ng_n/g_n$ is just $f_n$! (OK, $g_n$ might vanish somewhere, but we don't care since we're actually just talking about the convergence of $f_n$.) – Eric Wofsey Feb 2 at 18:34
• so the argument goes as follows: $f_n g_n/g_n\subset\mathcal{O}(D)$; $f_n g_n/g_n$ converges uniformly on the boundary of $D$ where there are no zeros of $f_n g_n/g_n$ by choice of $D$; by "the previous task" it follows that $f_n g_n/g_n$ is uniformly convergent on $D$; so the point that $g_n$ has no zero is only important in the second step, i.e. when we want to show that $f_n g_n/g_n$ converges uniformly on $\partial D$ and not important when we want to show that $f_n g_n/g_n$ is holomorphic - am I right? – JoHe Feb 2 at 19:01
• Right, you only need to not be dividing by $0$ on $\partial D$. When you're looking at $D$ itself, you can just use the functions $f_n$ which you already know are holomorphic. – Eric Wofsey Feb 2 at 19:09
• Thank you so much @EricWofsey! I really do appreciate your help. Lastly could you tell me some open source, where I can do some reading on the things you meant by "some details to fill in" concerning uniform convergence of quotients of functions (e.g. theorems of real analysis) or is there really nothing special about it if $g_n$ is never zero? Also I am a little bit confused on when do I need to argue that $g_n$ is never zero and when that $g$ is never zero (on the boundary), since in your answer you only talked about $g$ being never zero. – JoHe Feb 2 at 19:21