2
$\begingroup$

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

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ 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
    Commented 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} $ $\endgroup$
    – Paul
    Commented Jun 16, 2020 at 23:38
  • 2
    $\begingroup$ What precisely does "such that $\hat F$ exists" mean? $\endgroup$
    – Jochen Glueck
    Commented Jun 17, 2020 at 0:04
  • 2
    $\begingroup$ 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? $\endgroup$
    – Jochen Glueck
    Commented Jun 17, 2020 at 0:16
  • 1
    $\begingroup$ @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. $\endgroup$
    – Paul
    Commented Jun 17, 2020 at 9:11

2 Answers 2

Reset to default
4
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ We wrote the same proof! $\endgroup$
    – Giorgio Metafune
    Commented Jun 17, 2020 at 16:31
  • $\begingroup$ @ Christian Remling Thank you very very very much $\endgroup$
    – Paul
    Commented Jun 17, 2020 at 18:58
1
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ @ Giorgio Metafune Thank's so much $\endgroup$
    – Paul
    Commented 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 $\endgroup$
    – Paul
    Commented 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. $\endgroup$
    – Giorgio Metafune
    Commented Jun 19, 2020 at 12:56
  • $\begingroup$ thank's so much $\endgroup$
    – Paul
    Commented Jun 19, 2020 at 21:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .