# How to directly prove rapidly decay functions on $\mathbb{Z}$ is closed under convolution?

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 $$|f \ast g(n)| = |\sum_m g(m) f(n-m)| \le AB\sum_m (1+|m|)^{-k} (1+|n-m|)^{-k}\\ \le AB\sum_m \min((1+|m|)^{-k},(1+|n-m|)^{-k}) \le AB \sum_m (1+|n|/2 + |m|/2)^{-k} \le ABC (1+|n|)^{1-k}$$