$f, \hat{f} \in L^{1}(\mathbb R) \implies \widehat {\text{Re}(f)}, \widehat{\text{Im}(f)} \in L^{1}(\mathbb R)$?

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Let $f:\mathbb R \to \mathbb C$ such that $f(x)= (f_{1}(x), f_{2}(x))$; where $f_{1}(x)=\text{Re}(f(x))=\text{the real part of} \ f $ and $f_{2}(x)=\text{Im}(f(x))= \text{ the imaginary part of} \ f .$ We suppose that, $f\in L^{1}(\mathbb R)$ and its Fourier transform, $\hat{f} \in L^{1}(\mathbb R)$ also.

My questions: (I) Can we expect $\hat{f_{1}}$ and $\hat{f_{2}}$ are in $L^{1}(\mathbb R)$ ? (II) Can we expect $\hat{f_{1}}\in L^{1}(\mathbb R)$ ? (III) Can we expect $\hat{f_{2}} \in L^{1}(\mathbb R) ?$ (IV) Or, we get a counter examples ?

Tahnks,

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Since we have

$$\hat{f} = \widehat{f_1 + if_2} = \hat{f}_1 + i\hat{f}_2,$$

we have, under the assumption that $f,\hat{f}\in L^1(\mathbb{R})$,

$$\hat{f}_1 \in L^1(\mathbb{R}) \iff \hat{f}_2\in L^1(\mathbb{R}),$$

so $(I) \iff (II) \iff (III)$.

Now, we have

$$\hat{\overline{f}}(\omega) = \frac{1}{\sqrt{2\pi}}\int_\mathbb{R} \overline{f}(x) e^{-ix\omega}\,dx = \frac{1}{\sqrt{2\pi}} \overline{\int_\mathbb{R} f(x) e^{ix\omega}\,dx} = \overline{\hat{f}(-\omega)},$$

and $g\in L^1(\mathbb{R}) \iff (\omega\mapsto \overline{g(-\omega)})\in L^1(\mathbb{R})$ is easily seen, so we also have $\overline{f},\hat{\overline{f}} \in L^1(\mathbb{R})$, and therefore

$$\hat{f_1} = \frac{1}{2}\widehat{f+\overline{f}} = \frac{1}{2}\left(\hat{f} + \hat{\overline{f}}\right)\in L^1(\mathbb{R}),$$

so the answer to the first three is yes, to the fourth, no.