# What if the cauchy product of two series in $\mathbf{Z}$ is null

I have a problem I do not find a solution. Given two series $\left(a_n\right)_{n \in \mathbf{Z}}$ and $\left(b_n\right)_{n \in \mathbf{Z}}$ which have a cauchy product $\left(c_n\right)_{n \in \mathbf{Z}} = \left(\sum_{i \in \mathbf{Z}}{a_i b_{n-i}}\right)_{n \in \mathbf{Z}}$ and $\forall n,\sum_{i\in\mathbf{Z}}{\left|a_ib_{n-i}\right|}<\infty$.

Do $\left(c_n\right)_{n \in \mathbf{Z}}=0$ implies $\left(a_n\right)_{n \in \mathbf{Z}}=0$ or $\left(b_n\right)_{n \in \mathbf{Z}}=0$?

It is true if $\exists n_0, \forall n<n_0, a_n=0$ (like in traditional formal Laurent series). I can not find a demonstration or a counterexample for the general case.

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Your sum over $i\in \mathbb Z$ need not be well defined without further restrictions - in what order are the terms to be added? –  Mark Bennet May 31 '13 at 20:16
I supposed implicitely that $\forall n, \sum_{i\in\mathbf{Z}}{\left|a_ib_{n-i}\right|}$ is defined. I add it to the question. –  oao May 31 '13 at 20:20
What if $a_n=1$ for all $n$ and $b_0=1, b_1=-1$ and $b_n=0$ otherwise? –  Hagen von Eitzen May 31 '13 at 20:25
It is a good idea, but $\sum{\left|a_i b_{n-i}\right|}$ does not converge. –  oao May 31 '13 at 20:29

Just look at some Fourier series (two functions with product zero), and use their coefficients. Example: $$a_0 = \frac{1}{2},\qquad a_{n} = \frac{1}{\pi i n}, n \text{ odd}, \qquad a_n = 0, \text{otherwise}$$

$$b_0 = \frac{1}{2},\qquad b_{n} = \frac{-1}{\pi i n}, n \text{ odd}, \qquad b_n = 0, \text{otherwise}$$

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I am not sure to understand. $c_0 = \sum_j{a_j b_{-j}} = 1/4 + \sum_{j \neq 0}{\frac{1}{-(2j+1)^2}\frac{1}{-\pi}} = 0$, but $c_1 = 0 + \sum_{i \neq -1,1}{\frac{1}{(i+1)(i-1)}\frac{1}{-\pi}}\neq 0$. –  oao May 31 '13 at 22:30
The last term is of course c_1 = 0 + \sum_{i \neq -1,1}{\frac{1}{(2i+1)(2i-1)}\frac{1}{-\pi}}\neq 0. Too late to edit. –  oao May 31 '13 at 22:40
The sum of any two odd number is even, but $1$ is odd. So $c_1 = a_0b_1+a_1b_0$. –  GEdgar Jun 1 '13 at 12:14

Proof by induction:

$n=0$. If $a_0b_0=0$, one of them is zero. If both are zero, repeat the problem with sums from $1$ to $\infty$ until the first non-zero one is found. If all are zero, we are done.

We assume $a_0 = 0$ and $b_0 \ne 0$.

$n > 0$. Suppose $c_n = \sum_{i=0}^n a_i b_{n-i} = 0$ and $a_i = 0$ for $i$ from $0$ to $n-1$. Then $a_i b_{n-i} = 0$ for $i$ from $0$ to $n-1$, so $0 = c_n = a_n b_0$. Since $b_0 \ne 0$, $a_n = 0$.

Therefore all the $a_n = 0$.

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It is not possible to do that, the sum $c_n=\sum_{i \in\mathbf Z}{a_i b_{n-1}}$ are on an infinite number of items. –  oao May 31 '13 at 22:53
This is the solution for index set $\mathbb N$. And (essentially) what the OP means by "traditional formal Laurent series". –  GEdgar Jun 1 '13 at 14:40