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Cauchy product with both extremes infinites

  1. Sep 17, 2014 #1
    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. jcsd
  3. Sep 18, 2014 #2

    mathman

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    ##\{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.

    I don't understand your problem.
     
  4. Sep 19, 2014 #3
    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
  5. Sep 19, 2014 #4

    mathman

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    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})##
     
  6. Sep 20, 2014 #5
    ##\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!
     
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