Suppose that $f:\mathbb R \to \mathbb R$ is a smooth compactly supported function, specifically, $f=1$ on $[-1/2, 1/2)$ and the support of $f$ is contained in $[-1,1].$
Suppose that $g$ is smooth and compactly supported function on $[0, 2\pi).$
For $n\ne m \in \mathbb Z,$ put $F(\xi)=f(\xi-n)\cdot \widehat{g}(\xi-m), \xi \in \mathbb R.$. (Here $\widehat{\cdot}$ denote the Fourier transform)
My question: Can we say that $\int_{\mathbb R}|F(\xi)|^p d\xi \leq C \frac{1}{\big(1+ (n-m)^2\big)^{\frac{s+2}{2p}}}$ for $s\geq 0$?
My attempts: I think both $f$ and $\widehat{g} $ are both Schwartz class function and its product as well. So it is $L^p$ integrable. How can I get the decay?
As you mentioned, $\hat{g}$ is a Schwartz function. In particular, for every $s \geq 0$, there exists $C_s > 0$ such that: $$ \forall \xi \in \mathbb{R}, \quad |\hat{g}(\xi)| \leq (1+|\xi|^2)^{-\frac s 2}.$$ Since $f$ is compactly supported in $[-1,1]$, $F = 0$ on $(-\infty,n-1] \cup [n+1,\infty)$. Hence, for $p \geq 1$, $$ \int_{\mathbb{R}} |F(\xi)|^p \mathrm{d}\xi = \int_{n-1}^{n+1} |F(\xi)|^p \mathrm{d}\xi \leq 2 \|f\|_{L^\infty(-1,1)}^p \int_{-1}^{+1} |\hat{g}((n-m)+\xi')|^p \mathrm{d}\xi'. $$ Combining both bounds readily implies that $$ \forall s \geq 0, \exists C_s' > 0, \quad \| F \|_{L^p} \leq \frac{C_s'}{(1+(n-m)^2)^{\frac{s}{2}}}, $$ which implies the bound you are looking for (since this one is valid for all $s \geq 0$).