On a necessary and sufficient condition for $\sum_{k\in\mathbb{Z}}a_k=L$ ($a_k\in\mathbb{C}$)

Question

Following this question and more particularly this question where the proof of the equivalence of two notions was left as an exercise, here is my attempt to solve that exercise:

Questions:

1. Is the definition below the correct particular case of the general one linked?

2. Is my proof of the proposition below correct? If not, what mistakes did I make?

3. In case my proof is incorrect: what would be a valid one?

4. In case my proof is correct (statistically improbable ^^), is there a "neater" one? Or one which takes Riemann rearrangement theorem for granted but not Steinitz's theorem?

In the following $a_k\in\mathbb{C}$ ($k\in\mathbb{Z})$.

Definition: $\displaystyle\sum_{k\in\Bbb{Z}} a_k$ is the (unique) limit of the net $ \left\{\displaystyle\sum_{k \in F}a_k:F\subset\Bbb{Z}\text{ and $F$ is finite}\right\}$ as defined in here. That is, $\displaystyle\sum_{k\in\mathbb{Z}}a_k=L$ if, and only if, $$ \forall\epsilon>0,\exists F_{\epsilon}\subset\Bbb{Z}\text{ finite},\forall \Bbb{Z}\supset F\supseteq F_{\epsilon}\text{ finite}:\left|\sum_{k\in F}a_k-L\right|<\epsilon\tag{1} $$ Proposition: $$ \sum_{k\in\mathbb{Z}}a_k=L\iff\sum_{k=-\infty}^{\infty}a_k\text{ converges absolutely and }\sum_{k=-\infty}^{\infty}a_k=L $$ Proof: $[\Longrightarrow]$ (i) Let $\epsilon>0$ and $F_{\epsilon}$ be as in $(1)$. Let $N:=\max\{|a|:a\in F_{\epsilon}\}$. Then for all $m,n>N$, the subset $F_{m,n}:=[-m,n]$ contains $F_{\epsilon}$ hence $$ \left|\sum_{k=-m}^na_k-L\right|=\left|\sum_{k\in F_{m,n}}a_k-L\right|<\epsilon $$ So, $\displaystyle\sum_{k=-\infty}^{\infty}a_k=L$.

(ii) Suppose that absolute convergence doesn't hold. Then, WLOG, $\displaystyle\sum_{k=0}^{\infty}a_k$ converges conditionally. By Steinitz's theorem, there exists $z_1,z_2$, $z_1\neq z_2$ and permutations $\sigma_1,\sigma_2$ on $\mathbb{Z}_{\geq0}$ such that $$ \sum_{k=0}^{\infty}a_{\sigma_1(k)}=z_1\text{ and }\sum_{k=0}^{\infty}a_{\sigma_2(k)}=z_2 $$ By (i), $\displaystyle\sum_{k=1}^{\infty}a_{-k}$ converges, say to $\alpha$.

Given $\epsilon>0$, a standard argument then shows that there exists a $N$ such that $$ \left|\left(\sum_{k=1}^ma_{-k}+\sum_{k=0}^na_{\sigma_1(k)}\right)-(\alpha+z_1)\right|<\epsilon $$ whenever $m,n>N$.

Now, for any $\epsilon>0$, it is (clearly) possible to choose $F$ in $(1)$ such that $$ \sum_{k=1}^ma_{-k}+\sum_{k=0}^na_{\sigma_1(k)}=\sum_{k\in F}a_k $$ and $$ \left|\left(\sum_{k=1}^ma_{-k}+\sum_{k=0}^na_{\sigma_1(k)}\right)-(\alpha+z_1)\right|<\epsilon\\ \left|\sum_{k\in F}a_k-L\right|<\epsilon $$ simultaneously. This implies $L=\alpha+z_1$. Similarly, we must have $L=\alpha+z_2$. This is a contradiction because $z_1\neq z_2$.

Hence $\displaystyle\sum_{k=-\infty}^{\infty}a_k$ converges absolutely.

$[\Longleftarrow]$ Let $\epsilon>0$. Since $\displaystyle\sum_{k=-\infty}^{\infty}a_k$ converges absolutely, both $\displaystyle\sum_{k=0}^{\infty}a_k$ and $\displaystyle\sum_{k=1}^{\infty}a_{-k}$ converge absolutely. Consequently, we may take $N$ sufficiently large so that $$ \sum_{k=N+1}^{\infty}|a_k|<\frac{\epsilon}{2}\text{ and }\sum_{k=N+1}^{\infty}|a_{-k}|<\frac{\epsilon}{2} $$ Taking $F_{\epsilon}:=[-N,N]$, if $F\supseteq F_{\epsilon}$ then \begin{align} \left|\sum_{k\in F}a_k-L\right|&=\left|\sum_{k\in F}a_k-\left(\sum_{k=0}^{\infty}a_k+\sum_{k=1}^{\infty}a_{-k}\right)\right|\\ &=\left|\sum_{\substack{k\geq N+1\\k\not\in F}}a_{k}+\sum_{\substack{k\geq N+1\\k\not\in F}}a_{-k}\right|\\ &\leq\sum_{k=N+1}^{\infty}|a_{k}|+\sum_{k=N+1}^{\infty}|a_{-k}|\\ &<\epsilon \end{align} This shows that $\displaystyle\sum_{k\in\Bbb{Z}}a_k=L$. $\square$

Answer 1

I'll just answer your question 4. I am not saying the proof is neater, but it does not use anything deep. I will prove that if your limit (1) holds, then there is a constant $c>0$ such that $$ \sum_{k\in F}\left\vert a_{k}\right\vert \leq c $$ for every finite set $F\subseteq\mathbb{Z}$. I will use real numbers and not complex (you can just consider the real and immaginary sums separately).

Step 1: Let $F\subseteq\mathbb{Z}$ be a finite set. We claim that there exists a subset $G\subseteq F$ such that $$ \sum_{k\in F}\left\vert a_{k}\right\vert \leq2\left\vert \sum_{k\in G} a_{k}\right\vert . $$ Let $s:=\sum_{k\in F}\left\vert a_{k}\right\vert $. If $s=0$, take $G:=\emptyset$. If $s>0$, then writing $$ s:=\sum_{k\in F}\left\vert a_{k}\right\vert =\sum_{k\in F}a_{k}^{+}+\sum_{k\in F}a_{k}^{-}, $$ where $x^{+}:=\max\left\{ x,0\right\} $ and $x^{-}:=\max\left\{ -x,0\right\} $, we have that $\sum_{k\in F}a_{k}^{+}\geq\frac{s}{2}$ or $\sum_{k\in F}a_{k}^{-}\geq\frac{s}{2}$. Assume that $\sum_{k\in F}a_{k} ^{+}\geq\frac{s}{2}$ (the other case is similar). Let $G=\{k\in F:$ $a_{k}>0\}$. Then $$ \left\vert \sum_{k\in G}a_{k}\right\vert =\sum_{k\in G}a_{k}=\sum_{k\in F}a_{k}^{+}\geq\frac{s}{2}=\frac{1}{2}\sum_{k\in F}\left\vert a_{k}\right\vert . $$ Step 2: We claim that there exists $C>0$ such that $$ \left\vert \sum_{k\in F}a_{k}\right\vert \leq C $$ for every finite set $F\subseteq\mathbb{Z}$. Take $\varepsilon=1$ and let $E_1$ be the set given in (1). If $F\subseteq\mathbb{Z}$ then \begin{align} \left\vert\sum_{k\in F} a_{k}\right\vert= \left\vert\sum_{k\in (F\cup E_1)} a_{k}-\sum_{k\in (E_1\setminus F )} a_{k}\right\vert\le \left\vert\sum_{k\in (F\cup E_1)} a_{k}\right\vert+\sum_{k\in E_1} |a_{k}| \\ \le 1+|L|+ +\sum_{k\in E_1} |a_{k}|=:C \end{align}

Step 3: We claim that $$ \sum_{k\in F}\left\vert a_{k}\right\vert \leq2C $$ for every finite set $F\subseteq\mathbb{Z}$. This follows from Step 2 by applying Step 1.