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Let $X$ be an open subset of $\mathbb{R}^m$ and let $f_n\colon X\to \mathbb{C}$ be complex-valued functions. Then one has the following two notions:

$\textbf{1.}$ The series $\sum\limits_{n=0}^{\infty} f_n$ converges compactly normally on $X$ if for each compact $K\subseteq X$ $$\sum\limits_{n=0}^{\infty} \vert\vert f_n \vert\vert _K < \infty,$$ where $\vert\vert f_n\vert\vert_K:=\sup_{x\in K}\{\vert f_n(x)\vert\} $
$\textbf{2.}$ The series $\sum\limits_{n=0}^{\infty} f_n$ converges locally normally on $X$ if for each point $x\in X$ there exists an open neighborhood $U$ of $x$ in $X$ such that $$ \sum\limits_{n=0}^{\infty} \vert\vert f_n \vert\vert _U < \infty,$$ where $\vert\vert f_n\vert\vert_U:=\sup_{x\in U}\{\vert f_n(x)\vert\}$.

Now, the wikipedia article on normal convergence claims that both notions are equivalent.

The point where I get stuck and doubt the result to be true is the following in $\textbf{2} \Rightarrow$ $\textbf{1}$, in fact I just try to use the usual argument one would use to prove this if the supremum was taken of the whole series and not only its summands: Let $K$ be compact and choose finitely many $U_i$, $i=1,\ldots,l$, such that the union of these cover $K$ and for each $i$ $$\sum\limits_{n=0}^{\infty} \vert\vert f_n \vert\vert _{U_i} < \infty.$$ Letting $U$ be the union of the $U_i$'s I would like to prove that $\textbf{2}$ is true for $U$. For each $\epsilon > 0,$ we certainly find some $N$ such that $\sum\limits_{k=n+1}^{\infty} \vert\vert f_n \vert\vert _{U_i} < \epsilon$ for all $n\geq N$ for each $i$. However, $U$ could now be too big and the problem is really that the supremum is taken over each summand.

$\textbf{Question:}$ Is Wikipedia actually correct? If yes, how would a proof look like?

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Note that if $x\in U,$ then $x\in U_{i_0}$ for some $i_0.$ This implies $|f(x)| \le \|f\|_{U_{i_0}} \le \sum_{i=1}^{l} \sup_{U_i}|f|.$ Therefore

$$\|f_n\|_{U} \le \sum_{i=1}^{l}\|f_n\|_{U_i}.$$

So we have

$$\infty >\sum_{i=1}^{l}\sum_{n=0}^{\infty} \|f_n\|_{U_i} = \sum_{n=0}^{\infty} \sum_{i=1}^{l}\|f_n\|_{U_i} \ge \sum_{n=0}^{\infty} \|f_n\|_{U}.$$

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