# $\sum |f_n|$ converges uniformly implies $\sum f_n$ converges normally

Is the following proposition true?

Let $f_n \colon U\subset\mathbb{C}\longrightarrow \mathbb{C}$ be a sequence of continuous (or holomorphic) functions such that:

$\sum |f_n|$ converges uniformly on every compact $K\subset U$

Then $\sum f_n$ converges normally on every compact $K\subset U$

i.e. $\sum ||f_n||_K < \infty$ where $||f_n||_K =\sup_{z\in K} |f_n(z)|$

Any help would be appreciated.

• What you're trying to prove is that if $\sum |f_\nu(z)|$ converges uniformly on every compact $K$, then $\sum |f_\nu|_K<+\infty$ on every compact $K$. Can you see that? – Pedro Tamaroff Dec 5 '14 at 20:37
• Yes, I think a counterexample is needed, the converse is true. – felipeuni Dec 5 '14 at 20:45
• I will answer this in a bit, even though you have an answer, just to make sure I can do it! – Alec Teal Dec 20 '14 at 23:56

Let $U$ be the unit disk and $K$ be the closed disk of radius $1/2$. The series of constant functions $g_n = 1/n^2$ converges uniformly on $U$. Use a continuous partition of unity to write $$g_n = h_{n,1}+\dots + h_{n,n}$$ where $0\le h_{n,k}\le 1/n^2$ and $\max_K h_{n,k}=1/n^2$ for every $k$. One way to do this is to pick $n$ distinct points $z_k\in K$, let $\delta$ be the smallest distance between them, and define $$\begin{split} h_{n,k}(z) &= \frac{1}{n^2}(1-2\delta^{-1}|z-z_k|)^+,\quad k=1,\dots,n-1 \\ h_{n,n} &= g_n-\sum_{k=1}^{n-1} h_{n,k}\end{split}$$ The series $$h_{1,1}+h_{2,1}+h_{2,2}+h_{3,1}+h_{3,2}+h_{3,3}+h_{4,1}+\cdots$$ converges uniformly on $U$, since every partial sum is sandwiched between two partial sums of $\sum g_n$. On the other hand, the sum of suprema of its terms on $K$ is $$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots$$ which diverges.