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I want to prove that $\prod f_n (z)$ uniformly converges for every compact subset of complex plane.

The textbook goes like this: It is enough to show that we can find a convergent series $ \sum M_n$ $ |Log f_n (z)| \le M_n$ on each $|z| \le R$. But why is it enough?

Or more generally, is there something like this: $\prod f_n (z)$ uniformly converges on $|z| \le R$ iff $\sum Log f_n (z)$ uniformly converges on $|z| \le R$ (or some other $R'$)?

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Do you mean $|\log f_n(z)| \leq M_n$ (instead of $\sum \ldots \leq M_n$)? – saz Nov 18 '12 at 11:42
@saz Sorry, I edited it. – Gobi Nov 20 '12 at 9:36
up vote 1 down vote accepted

$$\left|\prod_{j=n}^N f_n(z)\right| = \left|\prod_{j=n}^N \exp(\log(f_n(z)) \right| = \left| \exp \left( \sum_{j=n}^N \log f_n(z) \right) \right|\\ \stackrel{\ast}{\leq} \exp \left( \sum_{j=n}^N |\log f_n(z)| \right) \leq \exp \left( \sum_{j=n}^N M_n \right) \to 1 \quad (n,N \to \infty)$$

In $(\ast)$ we used

$$|e^z| = |e^{\text{Re} \, z}| \leq e^{|z|}$$

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