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Let $f\in L^1(\mathbb{R})$ and continuous on $\mathbb{R}$ such that its Fourier transform $\hat f$ equals zero in a neighborhood of zero.

Let $F$ be function such that $\hat F$ exists and

$$\hat f(x) =x\hat F(x),\quad \forall x\in \mathbb{R}$$

Prove that $F\in L^1(\mathbb{R})$.

Any hints on how to prove that?

I already asked this question on MSE, I hope it will have a chance here

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    $\begingroup$ Could you say more about where this question/problem arose, and why it is supposed to have a positive answer? That is: does this come from a paper you are reading? A course you are taking? etc $\endgroup$
    – Yemon Choi
    Jun 16, 2020 at 23:16
  • $\begingroup$ it is a question of a professor without indications. If F is a primitive of f (F '= f) we see that $ \hat f (x) = i x \hat F (x), \forall x \in \mathbb {R} $ $\endgroup$
    – Pascal
    Jun 16, 2020 at 23:38
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    $\begingroup$ What precisely does "such that $\hat F$ exists" mean? $\endgroup$ Jun 17, 2020 at 0:04
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    $\begingroup$ As for your previous question I changed the title to make it a bit more informative, and I changed the format of the link. May I kindly suggest for future questions that you try to (i) choose titles that contain a bit more concrete information about the question (phrases such as "Prove that" don't really add much information to the title) and (ii) use urls within links instead of writing down the entire url in plain text? $\endgroup$ Jun 17, 2020 at 0:16
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    $\begingroup$ @Jochen Glueck thanks for the formatting (title and link) For your question, I am a little embarrassed because the question does not specify the meaning. I believe in the sence of moderate distributions. If the problem is badly posed I will wait seven september to see the professor. $\endgroup$
    – Pascal
    Jun 17, 2020 at 9:11

2 Answers 2

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We have $\widehat{f}=0$ near zero, so we can write $\widehat{F}=\widehat{g}\widehat{f}$ with a function $\widehat{g}\in C^{\infty}$, $\widehat{g}(x)=1/x$ for $|x|\ge a>0$.

Since $\widehat{g}'' = -(t^2 g)\,\widehat{}\in L^1$, we have $|g(t)|\lesssim 1/t^2$. Moreover, $g\in L^2\subseteq L^1_{\textrm{loc}}$. So $g\in L^1$, and thus also $F=g*f\in L^1$, since $f\in L^1$ by assumption.

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    $\begingroup$ We wrote the same proof! $\endgroup$ Jun 17, 2020 at 16:31
  • $\begingroup$ @ Christian Remling Thank you very very very much $\endgroup$
    – Pascal
    Jun 17, 2020 at 18:58
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Assume that $\hat f$ vanishes in $[-2a,2a]$ and take $\phi$ odd, vanishing in $[-a,a]$ and equal to $1/x$ if $|x| \ge 2a$. Then $\hat F=\phi \hat f$ and $F=\psi*f$ where $\psi$ is the inverse Fourier transform of $\phi$ $$\psi (\xi)=2 \int_a^\infty \phi(x)\sin (\xi x)\, dx.$$ Then $\psi$ is bounded for $|\xi| \le 1$ and, integrating by parts twice, decays at least as $1/\xi^2$ at infinity. This shows that $\psi \in L^1$ and $F \in L^1$, too.

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  • $\begingroup$ @ Giorgio Metafune Thank's so much $\endgroup$
    – Pascal
    Jun 17, 2020 at 18:59
  • $\begingroup$ @ Giorgio Metafune According to my teacher, we can only express a fourier transform or its inverse by an integral if the function is in L ^ 1, here the function is in L ^ 2, the inversion formula $\psi (\xi)=\int_0^\infty \phi(x)e^{(i\xi x)}\, dx.=2 \int_a^\infty \phi(x)\sin (\xi x)\, dx.$ is not justified $\endgroup$
    – Pascal
    Jun 19, 2020 at 11:46
  • $\begingroup$ If $\phi_R(x)=\phi \chi_{(0,R)}$, then $\phi_R \to \phi$ and $\hat{\phi_R} \to \hat{\phi}$ in $L^2$. But $\hat{\phi_R}=\int_0^R \phi (x) \sin(\xi x)dx \to \psi$ pointwise. $\endgroup$ Jun 19, 2020 at 12:56
  • $\begingroup$ thank's so much $\endgroup$
    – Pascal
    Jun 19, 2020 at 21:03

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