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Let $f:\mathbb R \to \mathbb R$ such that $$f(x)= \frac{\sin \pi x}{x (x^{2}-1)}$$ for $x\in \mathbb R - \{ 0, -1, 1 \}$ and $f(x):= \pi $ for $x=0$ and $f(x)=-\frac{\pi}{2}$ for $x= -1, 1$.

Clearly, $f\in C_{0}(\mathbb R) $ and note that $\int_{\mathbb R} |f(x)| (1+|x|) dx < \infty.$

My questions are:

(1) How to compute the Fourier transform $\hat{f}$ of the $f$ (as defined above), that is, for $\xi \in \mathbb R$, $\hat {f}(\xi)= \int_{\mathbb R} f(x) e^{-2\pi i x \xi} dx = $ ? Or, even without computation can we conclude that that, $\int_{\mathbb R} (1+|\xi|)^{2} |\hat{f}(\xi)| d\xi < \infty $ ?

(2) Define $g:\mathbb R \to \mathbb R$ such that $g(x):= |f(x)|$ , for $x\in \mathbb R$. Is it true that $\int_{\mathbb R} (1+|\xi|)^{2} |\hat{g}(\xi)| d\xi < \infty $ ?

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

$$L^{1}_{a}(\mathbb R)= \{ f:\mathbb R\to \mathbb C \ \text {measurable} : ||(1+|x|)^{a}f||_{L^{1}(\mathbb R)}< \infty \},$$ and $$FL^{1}_{b}(\mathbb R)= \{ f:\mathbb R \to \mathbb C \ \text {measurable} : ||(1+|w|)^{b} \hat {f}||_{L^{1}(\mathbb R)}< \infty \}.$$ We consider a Foureier-Lebsgue space, $$FL:=L^{1}_{a}(\mathbb R) \cap FL^{1}_{b}(\mathbb R).$$ The question is: If $f\in FL$. Can we expect $|f|\in FL$ ?; so, my motivation here is that above example serve here as a counter example. )


Thanks to math fraternity;-)

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2 Answers 2

Use partial fractions, and the fact that you know the Fourier transform of $\frac{\sin(\pi x)}x$, and also if you know the Fourier transform of $f(x)$, then it is easy to obtain the Fourier transforms of $f(x+1)$ and $f(x-1)$. I think you will get something compactly supported with Lipschitz derivative.

$g = |f|$ is harder. We will outline a proof $\int_{\mathbb R} (1+|\xi|)^{2} |\hat{g}(\xi)| d\xi = \infty $. Since the Fourier transform of $f$ is very nice, we will instead work with $h = \frac12(f+g) = f_+$. It is sufficient to show $\int_{\mathbb R} (1+|\xi|)^{2} |\hat{h}(\xi)| d\xi = \infty $.

Find a smooth bump function $\phi$, supported on $[-3,3]$, such that $\phi(x) h(x) = \sin(\pi x) I_{|x|<2}$. We can compute $\widehat{\phi h}(\xi)$ explicitly, seeing that it is something like an oscillating function that decays like $|\xi|^{-2}$. Hence $\int_{\mathbb R} (1+|\xi|)^{2} |\widehat{\phi h}(\xi)| d\xi = \infty $.

However, $\widehat{\phi h} = \hat\phi*\hat h$, where $*$ denotes convolution. Since $\hat\phi$ is localized, we can make some approximations when $\xi$ is large, and see that $\int_{\mathbb R} (1+|\xi|)^{2} |\widehat{\phi h}(\xi)| d\xi \le C \int_{\mathbb R} (1+|\xi|)^{2} |\hat{h}(\xi)| d\xi $. I hope that you are satisfied with proving this last bit yourself, because it is probably going to be a bit fiddly.

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As for the Fourier transform of $|f|$, that looks quite a bit harder. My wife is calling me to bed, so maybe I'll try it tomorrow. –  Stephen Montgomery-Smith Jan 4 at 7:32
    
Thanks a lot prof. S M, ; and hope to see you tomorrow; –  Inquisitive Jan 4 at 8:22

Note that $$\frac{\sin(\pi\,x)}{x(1-x^2)}=\frac{\sin(\pi\,x)}{x}+\frac{\sin(\pi(x+1))}{2(x+1)}+\frac{\sin(\pi(x-1))}{2(x-1)}.$$

The first summand is a multiple of the sinc function and its Fourier transform is given in the link. Use shift the shift property of the Fourier transform to compute the transform of the other two summands.

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