Summability
Given a Banach space $E$.
Consider sums: $$\sum_{\lambda\in\Lambda}\varphi_\lambda:=\bigg\{\sum_{\lambda\in\Lambda_0}\varphi_\lambda\bigg\}_{\Lambda_0\subseteq\Lambda:\#\Lambda_0<\infty}$$
Ordered by inclusion: $$\Lambda_0,\Lambda_0'\subseteq\Lambda:\quad \Lambda_0\leq\Lambda_0':\iff\Lambda_0\subseteq\Lambda_0'$$
(Distinct to series!)
Problem
Given the complex plane $\mathbb{C}$.
Then for sums: $$\sum_\lambda a_\lambda\in\mathbb{C}\iff\sum_\lambda|a_\lambda|\in\mathbb{R}_+$$
How can I prove this?
(Elegant proof?)
Attempt
By Lipschitz and completeness: $$\sum_\lambda|a_\lambda|\in\mathbb{R}_+\implies\sum_\lambda a_\lambda\in\mathbb{C}$$
By continuity and linearity: $$\sum_\lambda a_\lambda\in\mathbb{C}\implies\sum_\lambda\Re a_\lambda\in\mathbb{R}$$ $$\sum_\lambda a_\lambda\in\mathbb{C}\implies\sum_\lambda\Im a_\lambda\in\mathbb{R}$$
But how to proceed?
To clarify notation that wasn't clear to me at first: Say $f:X\to\mathbb C$. We say $$\sum_{x\in X}f(x)=s$$if for every $\epsilon>0$ there exists a finite set $F\subset X$ such that if $E$ is finite and $F\subset E\subset X$ then $$\left|s-\sum_{x\in E}f(x)\right|<\epsilon.$$
Lemma: Suppose $f:X\to\mathbb C$ and $\sum_{x\in X}f(x)$ exists. Suppose $A\subset X$. Then $\sum_{x\in A}f(x)$ exists. (Exercise.)
Lemma: Obvious lemma about the case $0\le g\le f$. (Exercise.)
Lemma: Say $f=g+ih$ where $g$ and $h$ are real-valued. Then $\sum f(x)$ exists if and only if $\sum g(x)$ and $\sum h(x)$ both exist. (Exercise.)
Suppose now that $\sum_{x\in X}f(x)$ exists. Say $f=g+ih$ as in the lemma. So $\sum g$ and $\sum h$ exist. Write $g=g^+-g^-$, $|g|=g^++g^-$. Let $A$ be the set of $x$ with $g(x)\ge 0$. Then $\sum_Xg^+=\sum_Ag$ exists. Similarly $\sum g^-$ exists. Since $|g|=g^++g^-$ this shows that $\sum_X|g|$ exists. Similarly $\sum|h|$ exists. And so $\sum|f|$ exists, because $0\le|f|\le|g|+|h|$.