# Is $\frac{\sin |\xi|}{|\xi|}$ in range of Fourier Transform for $n \ge 3$?

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 $$n \ge 3$$?

• For $n=3$, see mathoverflow.net/questions/315536
– YCor
Commented Nov 18, 2018 at 16:46
• 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 ${\bf R}^3$. Commented Nov 18, 2018 at 16:52
• 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. Commented Nov 18, 2018 at 17:00
• 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 $(\sin|\xi|)/|\xi|$ Commented Nov 18, 2018 at 17:06
• @LiviuNicolaescu Gelfand and Shilov's book covers the $n = 2m+3$ case. How'd we argue for the even dimensions? Commented Nov 18, 2018 at 17:39

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