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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?

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If f is in S, then |f| is in $H^1$. This is enough to guarantee that its Fourier transform is in $L^1$.

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    $\begingroup$ $g\in H^1\Longleftrightarrow (1+|\xi|^2)^{\frac{1}{2}}\widehat{g}(\xi)\in L^2$. Why does the last condition imply that $\widehat{g}\in L^1$? $\endgroup$ Commented Nov 16, 2013 at 16:28
  • $\begingroup$ Because $(1+\xi ^2)^{-\frac{1}{2} }$ is in $L^2$, and the product of two functions of $L^2$ is in $L^1$. $\endgroup$
    – abx
    Commented Nov 16, 2013 at 16:54
  • $\begingroup$ Let $f\in S(\mathbb R)$ such that $|f|\not \in S(\mathbb R)$; (so $\widehat {|f|} \in L^{2}(\mathbb R)$ by Plancherel theorem). But why $(1+|\xi|^{2})^{\frac {1}{2}}\widehat {|f|} \in \mathbb L^{2}(\mathbb R) ?$ $\endgroup$ Commented Nov 17, 2013 at 8:20
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    $\begingroup$ It is more elementary than you think. If f is in S, it is Lipschitz continuous. It is then easy to prove that |f| is also Lipschitz continuous, and therefore in $H^1$. $\endgroup$ Commented Nov 17, 2013 at 16:51
  • $\begingroup$ Oops! I forgot that $f$ is a one-variable function. $(1+|\xi|^2)^{-1/2}$ is not in $L^2(\mathbb{R}^n)$ if $n\geq 2$. $\endgroup$ Commented Nov 18, 2013 at 13:10

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