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The following theorem appears in "Princeton Lectures in Analysis, Complex Analysis," by Stein & Shakarchi:

Suppose $\{F_n\}$ is a sequence of holomorphic functions on the open set $\Omega$.
If there exists constants $c_n \gt 0$ s.t. $\sum c_n < \infty$ and $|F_n(z) - 1| \le c_n$, for all $z \in \Omega$, then
The product $\prod_{n=1}^\infty F_n(z)$ converges uniformly in $\Omega$ to a holomorphic function $F(z)$

The idea is that we can write $\prod_{n=1}^N F_n(z)$ as $\prod_{n=1}^Ne^{\text{log}F_n(z)} = e^{\sum_{n=1}^N\text{log}F_n(z)}$
Using the fact that $|\text{log} f| \le 2|f-1|$ for $f$ close enough to $1$, together with $|F_n(z) - 1| \le c_n$ and $\sum c_n < \infty$, it follows that $e^{\sum_{n=1}^N\text{log}F_n(z)}$ converges to some function $F(z)$,
but how do we know that it converges uniformly?

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up vote 1 down vote accepted

Hint: bound $\left|\sum_{n=M}^N \log F_n(z)\right|$

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I may have some clue at what you're getting at. So I want to write $e^{\sum_{n=1}^{M-1}\text{log}F_n}e^{\sum_{n=M}^{N}\text{log}F_n}$ and then I can show that the second exponential term is close to $1$. But that only allows me to get within a multiplicative constant of my limit function, not an additive constant, right? – Mark Sep 20 '12 at 1:52
If the function is bounded, a multiplicative factor is as good as an additive one. – Robert Israel Sep 20 '12 at 7:46
That's a great point! – Mark Oct 5 '12 at 2:24

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