# Does converge (uniformly) absolutely imply normal convergence in an open subset of $\mathbb{C}$

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]

• 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 $\Omega$? Aug 26, 2021 at 7:45
• This definition is defined by Wikipedia Aug 26, 2021 at 7:49
• Check again. I think you missed a word in the definition. Aug 26, 2021 at 7:51
• 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 :) Aug 26, 2021 at 7:51
• No, what you mention only deal with the case of a single compact set. Aug 26, 2021 at 15:49

(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. $$\square$$