# Cauchy product $\sum_{n=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}a_{n-k}b_k$

I have been told that, if $\{a_n\}_{n\in\mathbb{N}}$, $\{a_{-n}\}_{n\in\mathbb{N}^+}$, $\{b_n\}_{n\in\mathbb{N}}$ and $\{b_{-n}\}_{n\in\mathbb{N}^+}$ are absolutely summable complex sequences, then$$\Bigg(\sum_{n=-\infty}^{\infty}a_n\Bigg)\Bigg(\sum_{n=-\infty}^{\infty}b_n\Bigg)=\sum_{n=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}a_{n-k}b_k$$where $\sum_{n=-\infty}^{\infty}a_n=\sum_{n=0}^{\infty}a_n+\sum_{n=1}^{\infty}a_{-n}$.

I know that if $\{a_n\}_{n\in\mathbb{N}}$ or $\{b_n\}_{n\in\mathbb{N}}$ is absolutely summable, then $(\sum_{n=0}^{\infty}a_n)(\sum_{n=0}^{\infty}b_n)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_{n-k}b_k=\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}a_{n-k}b_k$, i.e. the proposition is true if $\forall n\leq-1\quad a_n=0=b_n$, and have tried to use that to prove the general case, but I get nothing. I have not studied any theory of measure yet.

I $+\infty$-ly thank you for any help!!!

EDIT: my question had been considered as a duplicate to a question already asked, but that is about the case $(\sum_{n=0}^{\infty}a_n)(\sum_{n=0}^{\infty}b_n)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_{n-k}b_k$, which I do know, as I had written. Though, I cannot generalise that to show that $(\sum_{n=-\infty}^{\infty}a_n)(\sum_{n=-\infty}^{\infty}b_n)$, which I hope I correctly write as $$(\sum_{k=0}^{\infty}a_k+\sum_{k=1}^{\infty}a_{-k}) (\sum_{k=0}^{\infty}b_k+\sum_{k=1}^{\infty}b_{-k})$$ $$=\sum_{n=0}^{\infty}\sum_{k=0}^{n} a_{n-k}b_k +\sum_{n=0}^{\infty}\sum_{k=0}^{n} a_{n-k}b_{-k-1}+\sum_{n=0}^{\infty}\sum_{k=0}^{n} a_{-n+k-1}b_{k}+\sum_{n=1}^{\infty}\sum_{k=1}^{n} a_{-n+k-1}b_{-k},$$

is equal to $\sum_{n=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}a_{n-k}b_k$. The people who regarded this question of mine as a duplicate to that were enough to have this question closed for a while, therefore, I suppose that it may well be trivial that the equality $(\sum_{n=-\infty}^{\infty}a_n)(\sum_{n=-\infty}^{\infty}b_n)=\sum_{n=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}a_{n-k}b_k$ derives from the equality, known by me, $(\sum_{n=0}^{\infty}a_n)(\sum_{n=0}^{\infty}b_n)=\sum_{n=0}^{\infty}\sum_{k=0}^{n}a_{n-k}b_k$. If that is trivial, I do not realise that fact: could anybody, either from those considering the question identical to that or anybody else, please, show me how to derive the equality with the indices $n$ from $-\infty$ to $+\infty$? I... $+\infty$-ly thank you! ;-)

• This is essentially Fubini's theorem on the measure space $\Bbb Z^2$ with the counting measure. The proof of this case will differ from the proof in the question you linked to only in the bookkeeping of indices. – Greg Martin Sep 18 '14 at 16:11
• @GregMartin If both are summable [absolutely convergent, here], then it is indeed straightforward. If one of the series is only conditionally convergent, it is not so straightforward (and I'm not even 100% sure it works then). – Daniel Fischer Sep 18 '14 at 16:27
• Thank you both!!! I've edited to cover the both absolutely convergent case only. I have tried to apply the proof of linked question to this case, but I have had serious difficulties, since we have four series to handle here: $U=u_0,+u_1+u_2+...$ $+u_{-1},+u_{-2}+...$, $V=v_0,+v_1+v_2+...$ $+v_{-1},+v_{-2}+...$, whose product I wish to prove to be... – Self-teaching worker Sep 18 '14 at 19:26
• $UV=\sum_{n=0}^{+\infty}(\sum_{k=0}^{+\infty}u_{n-k}v_k+\sum_{k=-\infty}^{-1}u_{n-k}v_k)+\sum_{n=-\infty}^{-1}(\sum_{k=0}^{+\infty}u_{n-k}v_k+\sum_{k=-\infty}^{-1}u_{n-k}v_k)=(u_0v_0+u_{-1}v_1+u_{-2}v_2+...+u_1v_0+u_0v_1++u_{-1}v_{2}+...)+(u_1v_{-1}+u_2v_{-2}+u_3v_{-3}+...+u_2v_0+u_3v_{-1}+u_4v_{-2}+...)+(u_{-1}v_0+u_{-2}v_1+u_{-3}v_2+...u_{-2}v_0+u_{-3}v_1+u_{-4}v_2+...)+(u_0v_{-1}+u_1v_{-2}+u_2v_{-3}+...+u_{-1}v_{-1}+u_0v_{-2}+u_1v_{-3}+...)$. – Self-teaching worker Sep 18 '14 at 19:38
• You really ought to reformat this question - it is really hard to read. Too many complex $\sum$ operations in the middle of paragraphs. – Thomas Andrews Sep 19 '14 at 18:12

Because the series are absolutely convergent, then so are $a(z)=\sum_{n=-\infty}^{\infty}a_{n}z^{n}$ and $b(z)=\sum_{n=-\infty}^{\infty}b_{n}z^{n}$ absolutely convergent for $|z|=1$. Therefore, the following is also absolutely convergent for $|z|=1$: $$a(z)b(z) = \sum_{n=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}a_{n}z^{n}b_{m}z^{m}.$$ So it is permissible to arbitrarily rearrange the terms of the series, without changing the value of the sum. In particular, one may collect all like powers of $z$ without changing the value of the sum. The coefficient of $z^{k}$ is $\sum_{n=-\infty}^{\infty}a_{n}b_{k-n}$ because $n+(k-n)=k$. Therefore, $$a(z)b(z) = \sum_{k=-\infty}^{\infty}\left[\sum_{n=-\infty}^{\infty}a_{n}b_{k-n}\right]z^{k}.$$ The powers of $z$ were introduced only to illustrate how to collect the terms, and to know that none of the terms were omitted from the final sum. Finally, set $z=1$.
• Considering the multiple non-generalised (from $k=0$ or $k=1$ to $\infty$) series confused me. Enlightening answer. Thank you so much! – Self-teaching worker Sep 22 '14 at 9:17
This is about $L^1({\mathbb Z}^2,\#)$.
When $S$ is a countable set ($S={\mathbb Z}^2$ in our example) a function $f:\ S\to{\mathbb C}$ is summable when $$\sup_{J\subset S,\ J\ {\rm finite}}\sum_{x\in J} |f(x)|<\infty\ .$$ If this is the case then it follows from the completeness of ${\mathbb C}$ that there is a unique $s\in{\mathbb C}$ with the following property: For each $\epsilon>0$ there is a finite set $J_0\subset S$ such that for all finite sets $J$ with $J_0\subset J\subset S$ one has $$\left|\sum_{x\in J}f(x)-s\right|<\epsilon\ .$$ Denote this number $s$ by $\int_Sf$.
Given two absolutely convergent complex series $$\sum_{k\in{\mathbb Z}}a_k=:A, \quad \sum_{k\in{\mathbb Z}}b_k=:B$$ it is easy to check that $$f:\ {\mathbb Z}^2\to{\mathbb C},\qquad(j,k)\mapsto a_j\>b_k$$ is summable and that $$\int_{{\mathbb Z}^2}f=A\cdot B\ .$$ On the other hand, for summable $f$ we have a Fubini theorem: Assume that $f$ is summable over $S$, and that $S=\bigcup_{\iota\in I} S_\iota$ is any (maybe infinite) partition of $S$. Then $$\int_S f=\sum_{\iota\in I} s(\iota)\ ,$$ where $$s(\iota)=\int_{S_\iota} f\ .$$ Now apply this to the partition $${\mathbb Z}^2=\bigcup_{n=-\infty}^\infty\bigl\{(n-k,k)\>|\> k\in{\mathbb Z}\bigr\}\ .$$
• I heartily thank you, dear Professor! A constructive and overwhelmingly interesting answer. I don't know anything about the theory of measure until now, though I plan to study it on Kolmogorov-Fomin's. Could the equality be proven with more elementary methods? As to the fact that "it follows from the completeness of ${\mathbb C}$ that there is a unique $s\in{\mathbb C}$ such that...", how do you know that (I do know some topology and theory of metric spaces)? $\infty$ thanks again!!! – Self-teaching worker Sep 20 '14 at 20:17