1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.