# Is a polynomial decay sufficient for a smooth function to be in $\mathcal{F}(L^1)$?

Background: I have a function $g(\omega)\in C^{\infty}(\mathbb{R})$, which vanishes like $O(|\omega|^{-\beta})$ at infinity for some $\beta>0$.

This answer states that functions that decays "too slowly" failes to be in $\mathcal{F}L^1(\mathbb{R})$, and exercise VI.1.7 of Kaznelson's "An Introduction to Harmonic Analysis" states that $C^{2}_{c}(\mathbb{R})\subset\mathcal{F}L^1(\mathbb{R})$.

Problem: Can we deduce $g\in \mathcal{F}L^1(\mathbb{R})$ from these conditions? Heuristically, the decay condition means that $\hat{g}$ should not have singularities of "order $\geq1$" , and therefore $\hat{g}$ should be locally integrable. Can this intuition be made precise?

This does not follow. There are compactly supported, finite, purely singular measures $\mu$ whose Fourier transform has power decay: $|\widehat{\mu}(x)|\lesssim (1+|x|)^{-\beta}$; for more background, you can search for Fourier dimension, for example on this site.
This function $g=\widehat{\mu}$ satisfies all your conditions, but its Fourier transform $\mu$ is not an integrable function.