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Does there exist $f \in L^1(\mathbb{R}^n)$ s.t., $\displaystyle \widehat{f}(\xi) = \frac{\sin |\xi|}{|\xi|}$ in case of dimension $n \ge 3$?

It is known that for $n = 2$, the function $\displaystyle f(x) = \frac{\chi_{\{|x| < 1\}}}{\sqrt{1-|x|^2}}$ curiously has some constant multiple of $\dfrac{\sin |\xi|}{|\xi|}$ as Fourier transform. Or at least is there a way of finding the Fourier transform $\frac{\sin |\xi|}{|\xi|}$ in sense of tempered distributions for $n \ge 3$?

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    $\begingroup$ For $n=3$, see mathoverflow.net/questions/315536 $\endgroup$
    – YCor
    Commented Nov 18, 2018 at 16:46
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    $\begingroup$ Just to expand very slightly on @YCor's comment/answer: because the Fourier transform is injective from $M({\bf R}^n)$ to $C_b({\bf R}^n)$, there is no $f\in L^1({\bf R}^3)$ with the properties that you desire, but instead you need to take the Fourier transform of a certain probability measure that is singular w.r.t. Lebesgue measure on ${\bf R}^3$. $\endgroup$
    – Yemon Choi
    Commented Nov 18, 2018 at 16:52
  • $\begingroup$ There is of course a (radial) distribution that has the desired FT, by Fourier inversion. The only meaningful question you can ask along these lines is how much regularity this distribution has. $\endgroup$
    – Christian Remling
    Commented Nov 18, 2018 at 17:00
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    $\begingroup$ Check Volume 1 of Gelfand and Shilov's book Generalized functions. You will find and explicit description of a distribution, concentrated along the unit sphere, whose Fourier transform is $(\sin|\xi|)/|\xi|$ $\endgroup$
    – Liviu Nicolaescu
    Commented Nov 18, 2018 at 17:06
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    $\begingroup$ @LiviuNicolaescu Gelfand and Shilov's book covers the $n = 2m+3$ case. How'd we argue for the even dimensions? $\endgroup$
    – sciona
    Commented Nov 18, 2018 at 17:39

1 Answer 1

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Here's a formula from Duistermaat and Kolk book Distributions: Theory and Applications, Chapter 17, Eq, (17.13). We denote by $\newcommand{\eF}{\mathscr{F}}$ $\eF$ the Fourier transform. Then $\newcommand{\ii}{\boldsymbol{i}}$ $\newcommand{\ve}{\varepsilon}$

$$ \eF^{-1}\left(\frac{e^{\ii t\Vert \xi\Vert}}{\Vert\xi\Vert}\right)=p_n\lim_{\ve\searrow 0} \Big(\; \Vert x\Vert^2-(t+\ve\ii)^2\;\Big)^{-\frac{n-1}{2}}, $$ where $p_n$ is a certain universal constant,

$$p_n=\frac{\Gamma\big(\frac{n+1}{2}\big)}{(n-1)\pi^{(n+1)/2}}. $$

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  • $\begingroup$ Thank you very much for the reference. $\endgroup$
    – sciona
    Commented Nov 18, 2018 at 19:26

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