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?