# Cauchy product with both extremes infinites

1. Sep 17, 2014

### DavideGenoa

Dear friends, 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 -maybe even if only one i between $\{a_n\}_{n\in\mathbb{Z}}$ and $\{b_n\}_{n\in\mathbb{Z}}$ is- then

$(\sum_{n=-\infty}^{\infty}a_n)(\sum_{n=-\infty}^{\infty}b_n)=\sum_{n=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}a_{n-k}b_k$​

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 $(\sum_{n=0}^{\infty}a_n)(\sum_{n=0}^{ \infty}b_n)=\sum_{n=0}^{ \infty}\sum_{k=0}^{ \infty}a_{n-k}b_k$, 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 $\sum_{k=0}^{M}a_k\sum_{k=0}^{N}b_k+\sum_{k=1}^{P}a_{-k}\sum_{k=0}^{Q}b_k+\sum_{K=0}^{M}a_k\sum_{k=1}^{R}b_{-k}+\sum_{k=1}^{P}a_{-k}\sum_{k=1}^{R}b_{-k}$, with $M,N,P,Q\to+\infty$, without being able to handle the indices to get $\sum_{n=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}a_{n-k}b_k$.
Thank you so much for your help!!!

2. Sep 18, 2014

### mathman

$\{a_n\}_{n\in\mathbb{N}}$, $\{a_{-n}\}_{n\in\mathbb{N}^+}$, $\{b_n\}_{n\in\mathbb{N}}$ and $\{b_{-n}\}_{n\in\mathbb{N}^+}$ absolutely summable complex sequences imply $\{a_n\}_{n\in\mathbb{N}}$ and $\{b_n\}_{n\in\mathbb{N}}$ are absolutely summable.

3. Sep 19, 2014

### DavideGenoa

I know, but I am not able to see that $(\sum_{n=-\infty}^{+\infty}a_n)(\sum_{n=-\infty}^{+\infty}b_n):=(\sum_{n=0}^{+\infty}a_n+\sum_{n=1}^{+\infty}a_{-n})(\sum_{n=0}^{+\infty}b_n+\sum_{n=1}^{+\infty}b_{-n})$ is equal to $\sum_{n=-\infty}^{+\infty}\sum_{k=-\infty}^{+\infty}a_{n-k}b_k:=\sum_{n=0}^{+\infty}(\sum_{k=0}^{+\infty}a_{n-k}b_k+\sum_{k=1}^{+\infty}a_{n+k}b_{-k})+\sum_{n=1}^{+\infty}(\sum_{k=0}^{+\infty}a_{-n-k}b_k+\sum_{k=1}^{+\infty}a_{-n+k}b_{-k})$.

I have also tried this road, which I hope not to lack mathematical rigour: I think we can say that $(\sum_{k=-\infty}^{+\infty}a_k)(\sum_{k=-\infty}^{+\infty}b_k)=(\lim_{p\to+\infty}\sum_{k=-p}^{+\infty}a_k)(\lim_{q\to+\infty}\sum_{k=-q}^{+\infty}b_k)$. Therefore I guess we could work (knowing that under the hypothesis of absolute summability $(\sum_{n=0}^{+\infty}a_n)(\sum_{n=0}^{+\infty}b_n)=\sum_{n=0}^{+ \infty}\sum_{k=0}^{n}a_{n-k}b_k$, which is the case I know) with $(\sum_{k=-p}^{+\infty}a_k)(\sum_{k=-q}^{+\infty}b_k)=\sum_{n=0}^{+\infty}\sum_{k=0}^{n}a_{n-p-k}b_{k-q}=\sum_{n=-p}^{+\infty}\sum_{k=0}^{n}a_{n-k}b_{k-q}$, but, here, I cannot arrange the more internal sum in order to get extremal indices approaching $\pm\infty$...
Any idea?
Thank you so much again!!!

Last edited: Sep 19, 2014
4. Sep 19, 2014

### mathman

It is much easier if you don't split the sums into two domains. Simply changes indices: a sum, n -> k, b sum, n -> m-k.

$(\sum_{n=-\infty}^{+\infty}a_n)(\sum_{n=-\infty}^{+\infty}b_n)=(\sum_{k=-\infty}^{+\infty}a_k)(\sum_{m=-\infty}^{+\infty}b_{m-k})$

5. Sep 20, 2014

### DavideGenoa

$\sum_{n=-\infty}^{+\infty}$ thank$_n$! ;) It seems to me that it perfectly works even with only one series absolutely convergent: am I right? Thank you so much again!