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Recall that the extension of function from $u:\mathbb{R}^n\to \mathbb{R}$ can be defined using the Poisson Kernel as follows: $$u^{\mathrm{e}}(\mathbf{x}):=\gamma_{n} \int_{\mathbb{R}^{n}} \frac{x_{n+1} u(y)}{\left(|x-y|^{2}+x_{n+1}^{2}\right)^{\frac{n+1}{2}}} \mathrm{~d} y \quad \text { for } \mathbf{x}=\left(x, x_{n+1}\right) \in \mathbb{R}_{+}^{n+1}$$ where $\gamma_n>0$ is some dimension dependent constant. This is also called the Harmonic Extension since $\Delta u^e = u$ on the half-space and is equal to the function $u$ on the boundary.

I came across a paper where the author mentioned that the Fourier transform of the above kernel is $\exp(-2\pi |\xi| x_{n+1})$. So I am wondering what happens in the fractional case, that is when

\begin{align*} u^e(x,x_{n+1}) = c_{n,s} \int \frac{x_{n+1}^{2s}u(y)}{(|x-y|^2+x_{n+1}^2)^{\frac{n+2s}{2}}}dy \end{align*}

for $s\in (0,1)$. Here $\operatorname{div}(x_{n+1}^{1-2s}\nabla u^e)=0$ on the half space. Is there an explicit expression (or $L^2$ and $L^\infty$ estimates) known for the Fourier Transform of the kernel $$P=\frac{x_{n+1}^{2s}}{(|x|^2+|x_{n+1}|^2)^{n/2+s}}?$$

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  • $\begingroup$ don't you want $P=x^{2s}_{n+1}/(|\mathbf{x}|^2+x^2_{n+1})^{s+n/2}$ , and then Fourier transform with respect to $\mathbf{x}$ ? (otherwise, how will you avoid the pole at $\mathbf{x}=0$?) $\endgroup$ Commented Mar 24, 2022 at 15:10
  • $\begingroup$ Aah yes, let me fix that. $\endgroup$
    – Student
    Commented Mar 24, 2022 at 17:02

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Yes indeed, for fractional $s$ it is related to Modified Bessel Functions of 2nd Kind (Macdonald Function). For $r=|\mathbf{x}|,\ a=|x_{n+1}|$ $$P(r,a)=\frac{a^{2s}}{(r^2+a^2)^{n/2+s}}$$ The Fourier Transform of an n-dimensional radial function $f(r)$ is a radial function in the trasformed space $\mathcal{F}(|\xi|)$. It is given by the following Hankel Transform. $$\mathcal{F}(|\xi|)=\frac{1}{|\xi|^{\frac{n}{2}-1}}\int_0^\infty r^{\frac{n}{2}-1}\left[r\,J_{\frac{n}{2}-1}(|\xi|\cdot r)\right]\cdot f(r)\,dr$$ See, for instance, Sec 11 pg. 63-65 in Sneddon, Ian N., Fourier transforms, New York: McGraw-Hill Book Company, Inc. (1950). ZBL0038.26801. for a proof.

Also, Sneddon, Ian N., A note on some relations between Fourier and Hankel transforms, Bull. Acad. Pol. Sci., Sér. Sci. Math. Astron. Phys. 9, 799-806 (1961). ZBL0100.31501. Therefore $$\mathcal{F}_P(|\xi|,a)=\frac{a^{2s}}{|\xi|^{\frac{n}{2}-1}}\int_0^\infty J_{\frac{n}{2}-1}(|\xi|\cdot r)\cdot \frac{r^{\frac{n}{2}}\,dr}{(r^2+a^2)^{\frac{n}{2}+s}}$$ But integral is formula 6.565.4 in 7-th Edition of

Gradshteyn, I. S.; Ryzhik, I. M., Table of integrals, series, and products. Ed. by Alan Jeffrey. CD-ROM version 1. 0 for PC, MAC, and UNIX computers., San Diego, CA: Academic Press. (1996). ZBL0918.65001.

It reads (adapted to our notation), for $s>0$ $$\int_0^\infty J_{\frac{n}{2}-1}(|\xi|\cdot r)\cdot \frac{r^{\frac{n}{2}}\,dr}{(r^2+a^2)^{\frac{n}{2}+s}}=\left(\frac{|\xi|}{2}\right)^{\frac{n}{2}+s-1}\cdot\frac{K_s(a\,|\xi|)}{a^s\,\Gamma(\frac{n}{2}+s)}$$ Being $K_s(\cdot)$ the Modified Bessel Function of 2nd Kind. Therefore, $$\mathcal{F}_P(|\xi|,a)=\frac{(a\,|\xi|)^sK_s(a\,|\xi|)}{2^{\frac{n}{2}+s-1}\Gamma(\frac{n}{2}+s)}$$ Since $K_\frac{1}{2}(z)=\sqrt{\frac{\pi}{2z}}\cdot e^{-z}$, for $s=\frac{1}{2}$ we get $$\mathcal{F}_P(|\xi|,a)=\frac{\sqrt{\pi}}{2^\frac{n}{2}\Gamma\left(\frac{n+1}{2}\right)}\cdot e^{-a\,|\xi|}$$ Note: Dimension constant and $2\pi$ in exponent are matter of Fourier Transform's normalization and convention used respectively. By 'Rosetta Table' Formulae 504 at the end of this Wiki page this can be fitted accordingly.

Using question convention -FT unitary, ordinary frequency, $\mathcal{F}^*$- and notation we get,

$$P(|x|,x_{n+1})=\frac{2^{s-\frac{1}{2}}\Gamma\left(\frac{n}{2}+s\right)}{\pi^{\frac{n+1}{2}}}\cdot\frac{x_{n+1}^{2s}}{(|x|^2+|x_{n+1}|^2)^{n/2+s}}$$ producing this amazingly simple expression $$\mathcal{F}_P^*(|\xi|,x_{n+1})=\sqrt{\frac{2}{\pi}}\cdot \left(2\pi\,x_{n+1}\,|\xi|\right)^s\cdot K_s\left(2\pi\, x_{n+1}\,|\xi|\right)$$

This gives $\exp(-2\pi |\xi| x_{n+1})$ for $s=1/2$.

Being so simple, this result should be added as new Formulae 505 in Wiki's Table Fourier Transforms of n-dimensional functions.

I leave this for further reading, a nice article on Fourier Transforms of Fractional Poison Kernels and Fractional Laplacian (See Section 2.8) at

Kwaśnicki, Mateusz, Ten equivalent definitions of the fractional Laplace operator, Fract. Calc. Appl. Anal. 20, No. 1, 7-51 (2017). ZBL1375.47038.

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