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

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


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

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

My question is: Let $f\in FL$. Can we expect $|f|\in FL$ ; Or we get a counter example ?

Thanks to math fraternity;-)

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The answer is no. Take, for example for example if $a=1,b=2$. The function $$f(x)=\frac{\sin\pi x}{x(x^2-1)}$$ has Fourier transform with compact support, by Paley-Wiener, or by direct computation, so it belongs to your space. However $|f|$ is not differentiable at some points, therefore it cannot belong to your space (your condition on the Fourier transform with $b=2$ permits to differentiate $f$).

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