How Fourier transform behaves if we kills the oscillation?

Let $a, b \in \mathbb R$ such that $ab> 1$ ; put

$$L^{1}_{a}(\mathbb R)= \{ f:\mathbb R\to \mathbb C \ \text {measurable} : ||(1+|x|)^{a}f||_{L^{1}(\mathbb R)}< \infty \},$$

and

$$FL^{1}_{b}(\mathbb R)= \{ f:\mathbb R \to \mathbb C \ \text {measurable} : ||(1+|w|)^{b} \hat {f}||_{L^{1}(\mathbb R)}< \infty \}.$$

We consider a Foureier-Lebsgue space, $$FL:=L^{1}_{a}(\mathbb R) \cap FL^{1}_{b}(\mathbb R).$$

My question is: Does there exist a $f\in FL$ such that (the Fourier transform of $|f|$) $\widehat{|f|} \notin L^{1}(\mathbb R)$ ?

(A bit roughly speaking, the question is about the investigation of the fourier transform of $|f|$ is in $L^{1}(\mathbb R)$ or not; once we killed the oscillation, in some sense)

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