Let $f=\{a_n\}$ be a function on $\mathbb{Z}$. We call $f$ rapidly decay (also known as Schwartz space) if for any $k$, we have $$ \sum_{n\in \mathbb{Z}}(1+|n|)^k|a_n|^2<\infty. $$.
The convolution production of functions $f=\{a_n\}$ and $g=\{b_n\}$ is given by $$ (f*g)_n=\sum_{s+t=n}a_sb_t. $$
We know that rapidly decay functions is closed under convolution product. The easiest way to see it is via Fourier transform, in which rapidly decay functions on $\mathbb{Z}$ corresponds to smooth functions on $S^1$, and convolution product corresponds to ordinary product.
Now I want to prove the claim directly without involving Fourier transform. Could we prove it by some estimations?
If $|f(n)| \le A (1+|n|)^{-k}, |g(n)| \le B (1+|n|)^{-k}$ then $$|f \ast g(n)| = |\sum_m g(m) f(n-m)| \le AB\sum_m (1+|m|)^{-k} (1+|n-m|)^{-k}\\ \le AB\sum_m \min((1+|m|)^{-k},(1+|n-m|)^{-k}) \le AB \sum_m (1+|n|/2 + |m|/2)^{-k} \le ABC (1+|n|)^{1-k}$$