Let $f\in L^{1} (\mathbb R) := \{f:\mathbb R \rightarrow \mathbb C \ \text {measurable functions} : \int\_{\mathbb R} | f(x)| dx < \infty \}$ and the Fourier transform of $f$, 
$\hat{f} (y) : = \int \_ {\mathbb R} f(x) e^{-2\pi i x\cdot y} dx ; y \in \mathbb R $ 
and $\widehat{|f|} (y) : = \int \_ {\mathbb R} |f(x)| e^{-2\pi i x\cdot y} dx ; y \in \mathbb R $.


The Schwartz space, $S(\mathbb R): = \{f\in C^{\infty}(\mathbb R): \sup\_{x\in \mathbb R} |x^{\alpha} D^{\beta}f(x)|< \infty , \forall \alpha, \beta \in \mathbb N \cup \{0\} \}$.


Theorem: The Fourier transform is a linear isomorphism $F:S(\mathbb R)\rightarrow S(\mathbb R) \ni f\mapsto \hat {f}$.


My question is: 
Let $f\in S(\mathbb R)$ such that $|f|\not \in S(\mathbb R)$. 
Can we expect $\widehat{|f|} \in L^{1} (\mathbb R)$? 
Or can we produce a counter-example?



Let $f \in S(\mathbb R)$. Can we say $\widehat{|f|} \in L^{1}(\mathbb R)$?
If f is in S, then |f| is in $H^1$. This is enough to guarantee that its Fourier transform is in $L^1$.


