Let $g\in \mathcal{S}(\mathbb{R})$ be a Schwartz class function and denote its Fourier transform by $\hat{g}(x)\equiv \int_{-\infty}^\infty g(y)e^{-2\pi i xy}dy$. Let $t > 0, x\in\mathbb{R}$ be a fixed and arbitrary. I am trying to show that
$$\sum_{n=-\infty}^{\infty}(1+|n|)^{-t}|\hat{g}(x-n)|\leq A(1+|x|)^{-t}$$
for some constant $A > 0$. I am about half-way done with the proof but need help with the finishing touch.
I have been given a hint to consider partitioning the sum to regions
1.) $n\in\mathbb{Z}:|x - n| \leq |x|/2$
2.) $n\in\mathbb{Z}:|x - n| > |x|/2$
With this partitioning we can estimate 1.) with $(1 + |n|)^{-t}\leq (1 + ||n - x| - |x||)^{-t}\leq (1 + |x|/2)^{-t}$. I wasn't quite sure how to then transform $(1 + |x|/2)^{-t}$ into $(1 + |x|)^{-t}$, so I paritioned the original sum into
i.) $n\in\mathbb{Z}:|x - n| \leq 2|x|$
ii.) $n\in\mathbb{Z}:|x - n| > 2|x|$
If I am not mistaken both partitions will work. Now applying the triangle inequality to ii.) will give the $(1 + |x|)^{-t}$.
But how should I estimate i.)? In i.) the sum will behave differently depending on whether $\frac{|x|}{2} \leq |x - n| \leq \frac{3|x|}{2}$ or $|x - n|\in \left[\frac{|x|}{2}, \frac{3|x|}{2}\right]^c\cap[0,2|x|]$. How should this be done?
$\hat{g}$ is a schwarz function so you can bound $|\hat{g}(x)|\le C(1+|x|)^{-t-2}$ ,this will suffice. $$\sum (1+|n|)^{-t}|\hat{g}(x-n)|\le C\sum (1+|n|)^{-t}(1+|x-n|)^{-t}(1+|x-n|)^{-2}$$ $$\le C(1+|x|)^{-t} \sum (1+|x-n|)^{-2}< 100C(1+|x|)^{-t}$$