# 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 Nov 18 '18 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 ${\bf R}^3$. – Yemon Choi Nov 18 '18 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. – Christian Remling Nov 18 '18 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 $(\sin|\xi|)/|\xi|$ – Liviu Nicolaescu Nov 18 '18 at 17:06
• @LiviuNicolaescu Gelfand and Shilov's book covers the $n = 2m+3$ case. How'd we argue for the even dimensions? – sciona Nov 18 '18 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,
$$p_n=\frac{\Gamma\big(\frac{n+1}{2}\big)}{(n-1)\pi^{(n+1)/2}}.$$