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Let $f_n$ be a sequence of holomorphic functions defined on an open set $\Omega\subset \mathbb{C}$.

We say that the series $\sum_n f_n$ is uniformly absolutely-convergent if $\sum_n |f_n|$ converges uniformly on any compact subset $K\subset \Omega$.

We say that the series $\sum_n f_n$ converges normally if for every compact subset $K\subset \Omega, \sum_n \sup_K |f_n|$ converges.

We know if the series converge normally, then it must be uniformly absolutely-convergent. But how about the converse?

Does there exist a counterexample which shows that a series of holomorphic functions may be uniformly absolutely convergent but not normally.

Also, there exist a counterexample if we just consider single compact subset [1]

[1]. Does absolute and uniform convergence imply normal convergence?

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    $\begingroup$ calling the first condition "uniformly absolutely-convergent" is weird, because there is no uniformity there. So, isn't the first condition just absolute convergence at every point of $\Omega$? $\endgroup$
    – peek-a-boo
    Aug 26, 2021 at 7:45
  • $\begingroup$ This definition is defined by Wikipedia $\endgroup$
    – qinxs
    Aug 26, 2021 at 7:49
  • $\begingroup$ Check again. I think you missed a word in the definition. $\endgroup$ Aug 26, 2021 at 7:51
  • $\begingroup$ ok Wikipedia adds the extra condition "$\sum_n|f_n|$ converges uniformly on any compact subset". I probably should have guessed this is what you intended, but I can't read minds :) $\endgroup$
    – peek-a-boo
    Aug 26, 2021 at 7:51
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    $\begingroup$ No, what you mention only deal with the case of a single compact set. $\endgroup$
    – qinxs
    Aug 26, 2021 at 15:49

1 Answer 1

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Theorem Notation as above, TFAE:
(1) The series $\sum_n f_n$ is uniformly absolutely-convergent.
(2) The series $\sum_n f_n$ converges normally.
Proof We just need to prove (1) implies (2). By the continuity, for any $n\in \mathbb{N}^{+}$, we can choose $z_n\in K$ such that $$|f_n(z_n)|=\sup_K|f_n|$$ Consider two cases:
(i) If $\partial \Omega\neq \emptyset$, let $$\delta:=d(K,\partial\Omega)>0,\ L:=\{z\in\Omega|\ d(z,K)\leq \frac{\delta}{2}\},$$ then $L\supset K$ is a compact subset of $\Omega$, and the balls $B(z_n,\delta/2) \subset L$. By [Hor, Thm 1.6.7], $|f_n|$ is subharmonic for any $n$. By [Hor,Thm 1.6.3], we have $$|f_n(z_n)|\int_0^{\delta/2} rdr\leq \frac{1}{2\pi}\int_L |f_n(z)|d\lambda(z)\Longrightarrow |f_n(z_n)\leq \frac{4}{\pi\delta^2}\int_L|f_n(z)|d\lambda(z).$$ where $\lambda$ denotes the Lebesgue measure.
By the convergence, we know the series $\sum_n |f_n|$ is uniformly bounded over $L$, hence $$\sum_n \sup_n |f_n(z)|=\sum_n |f_n(z_n)|\leq \frac{4}{\pi\delta^2}\int_L \sum_n |f_n(z)|d\lambda(z)<\infty.$$ (ii) If $\partial \Omega=\emptyset$, then $\Omega=\mathbb{C}$, this can be proved similary.
[Hor] Lars Hormander. An introduction to several complex variables. $\square$

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