- #1
DavideGenoa
- 155
- 5
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
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!
##(\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!