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Assume that we have two sequences $(a_n)_{n \in \mathbb Z}, (b_n)_{n \in \mathbb Z}$ such that

  • for each $l\in \mathbb N $ the sequence $\left(|n|^l a_n\right)_{n \in \mathbb Z}$ is bounded,
  • there exists $s \in \mathbb N$ such that $\displaystyle\sum_{n \in \mathbb Z} \frac{|b_n|^2}{(1+n^2)^s}< \infty$.

Let $\displaystyle c_n=\sum_{k \in \mathbb Z} a_k b_{n-k}$ for $n \in \mathbb Z$. Is it then true that $$\sum_{n \in \mathbb Z} |c_n|^2 <\infty ?$$

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up vote 1 down vote accepted

No, an easy counterexample is given by the sequences $a_0 = 1$, $a_n=0$ for $n \ne 0$, $b_n=1$ for all $n$. In that case the first condition is trivially satisfied since $|n|^l a_n = 0$ for all $n$, and the second condition is satisfied for any $s \in \mathbb{N}$ since $\sum\limits_{n \in \mathbb{Z}} (1+n^2)^{-s}$ converges for every $s \ge 1$. Also $c_n = b_n = 1$ for all $n$ in this example, so $\sum\limits_{n \in \mathbb{Z}} |c_n|^2 = \infty$.

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