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I am really confused by the above proof as the following argument seems a lot more simple:


(i) If $\prod f_n$ is normally convergent in $G$, then $\prod f_n$ is compactly convergent.

(ii) By definition, $\prod f_n$ is compactly convergent if for all compact subset $K \subseteq X$ exists index $m=m(K)$ such that $$ p_{m,n} = f_m \cdots f_n $$ converges uniformly on $K$ as $n \rightarrow \infty$ to $\hat{f}_m$ that is nonvanishing on $K$.


If we fix $K$, then $\hat{f}_n = \frac{\hat{f}_m}{p_{m,n-1}} \rightarrow 1 $ uniformly on $K$, hence compactly in $G$. Also we know that $\hat{f}_n \in \mathcal{O}(G)$ as $\hat{f}_n = \prod _{k \ge n } f_k $ is compactly convergent. What is wrong with this argument?


Definitions:

  1. A product $\prod f_n$ with $f_n= 1+ g_n \in C(X)$ is called normally convergent in $X$ if the series $\sum g_n$ converges normally in $X$.

1.' A series $\sum g_n $ converges normally in $X$ if for all compact subsets $K \subseteq X$, $\sum |g_n|_K < \infty$ where $|g_n|_K$ is sup norm on set $K$.

  1. If $f_n \in C(X)$, the $\prod f_n$ is called compactly convergent in $X$, if for all compact subset $K \subseteq X$ exists index $m=m(K)$ such that $$ p_{m,n} = f_m \cdots f_n $$ converges uniformly on $K$ as $n \rightarrow \infty$ to $\hat{f}_m$ that is \textit{non vanishing} on $K$. We let $f = \prod f_n$ and on $K$, $f|K = (f_0|K) \cdots (f_{m-1} | K) \hat{f}_m $.
  2. $\mathcal{O}(G)$ is the ring of complex analytic functions $f: G \rightarrow \mathbb{C}$
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  • $\begingroup$ What is $O(G)$, what is normally convergent, what is compactly convergent? $\endgroup$ – mathworker21 Jul 6 '17 at 7:36
  • $\begingroup$ Sorry, I have added them now. Thanks! $\endgroup$ – CL. Jul 6 '17 at 7:58
  • $\begingroup$ In complex analysis we only consider uniformly convergent sequences of analytic functions on some open $U$. In that case the limit is analytic on $U$. When there are some poles at $z_k \in U$, the sequence won't converge uniformly on $U$, but it will be on $U \cap \{|z-z_k| > \epsilon\}$. This leads to the idea of being locally-uniformly-convergent. $\endgroup$ – reuns Jul 6 '17 at 8:05
  • $\begingroup$ @user1952009, the confusion is, in the proof provided, we could have completed avoid the case of a sequence of analytic functions with poles by making the a choice of nonvanishing $\hat{f}_m$ (unless this is not possible?) $\endgroup$ – CL. Jul 6 '17 at 8:14

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