Suppose $f \ \text{and} \ f|f|\in L^{1}(\mathbb R).$ Then, clearly, $|f|\in L^{2}(\mathbb R)$ and therefore by Plancheral theorem, we get, $\widehat{|f|} \in L^{2}(\mathbb R).$ Also, assume, $f, |f|f, \hat{f}, \widehat{f|f|} \in L^{1}(\mathbb R) \cap C_{0}(\mathbb R).$
My Question: Can we expect $|f|\in H_{1}(\mathbb R)=\{ f\in \mathcal{S'}(\mathbb R): (1+|\xi|^{2})^{1/2}\hat{f}\in L^{2}(\mathbb R) \}$ (Sobolev spaces ), Or, we we can produce a counter example ?
(If we put, additional condition, $|(\widehat{|f|}(\xi))| \leq \frac{1}{\sqrt{1 + |\xi|^{2}}}, \ (\xi \in \mathbb R)$, then, certainly, by definition of sobolev space, it will follows; so is this condition necessary ?)
Motivation: Actually, I want to show: $\widehat{|f|}\in L^{1}(\mathbb R);$ so it sufficient to show, $|f|\in H_{1}$. If there is other way, I would like to have your suggestions.
Thanks,