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

Question

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.

Answer 1

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

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

Answer 2

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 $a_n = 0$.