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5.8 Lemma Let (X,d) be a compact metric space and let $\{g_n\}$ be a sequence of continuous functions from X into $\mathbb{C}$ such that $ \sum g_n(x)$ converges absolutely and uniformly for x in X. Then the product $f(x)=\prod(1+g_n(x))$ converges absolutely and uniformly for x in X.

Proof. Since $\sum g_n(x)$ converges uniformly for x in X there is an integer $n_0$ such that $|g_n(x)|<\frac{1}{2}$ for all x in X and $n>n_0$. This implies that $Re[1+g_n(x)]>0$ and also, according to inewuality (5.3), $|log(1+g_n(x))| \leq \frac{3}{2}|g_n(x)|$ for all $n>n_0$ and x in X. Thus \begin{equation} h(x)=\sum_{n=n_0+1}^\infty log(1+g_n(x)) \end{equation} converges uniformly for x in X.

First of all it's not clear to me whether "converges absolutely and uniformly" means the uniform convergence of the series $\sum |g_n(x)|$ or just the pointwise convergece of the series $\sum |g_n(x)|$ and the uniform convergence of the series $\sum g_n(x)$. In the latter case I cant understand how the implication "Thus $ h(x)=\sum_{n=n_0+1}^\infty log(1+g_n(x))$ converges uniformly..." is justified.

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  • $\begingroup$ I would guess that "converges absolutely and uniformly" means that $\sum |g_n(x)|$ converges pointwise and that $\sum g_n(x)$ converges uniformly. $\endgroup$ – Arthur Oct 25 '15 at 18:37
  • $\begingroup$ yes. It's just an abbreviated way of saying the same thing. $\endgroup$ – DanielWainfleet Oct 25 '15 at 21:22

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