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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 $n \in \mathbb Z$.

How to prove that for some constants $M >0$ and $t> 0$ the following holds: $$ |c_n| \leq M (1+n^2)^t \textrm{ for } n\in \mathbb Z. $$

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  • $\begingroup$ Why did an attempt introducing for example $k^l$ and Cauchy-Schwarz inequality failed? $\endgroup$ Commented Jul 7, 2013 at 20:38

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Define $C_l:=\sup_{n\in\mathbb Z}|n|^l|a_n|$. Then for each $k$, $$|a_k|\cdot |b_{n-k}|\leqslant C_l|k|^{-l}\frac{|b_{n-k}|}{(1+|n-k|^2)^{s/2}}(1+|n-k|)^{s/2}.$$ There is a constant $K$ such that for each $k$, $n$, we have $$(1+|n-k|)^{s/2}\leqslant K(1+|k|)^{s/2}\cdot (1+|n|)^{s/2}.$$ Choosing $l$ such that the sequence $(|k|^{-2l}(1+|k|)^{s/2})$ is summable, we are done by Cauchy-Schwarz inequality.

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