Kelvin transform and the fractional Laplacian For $0< \alpha< 2$ the fractional Laplacian of order $\alpha$ is defined up to multiplicative constants by
\begin{equation*}
 (-\Delta)^{\alpha/2}u(x) = \int_{\mathbb R^n}\frac{u(x) - u(y)}{|x - y|^{n + \alpha}}\; dy, 
\end{equation*}
where the integral is understood in the principal value sense and, for my purposes, $u:\mathbb R^n \to \mathbb R$ is as smooth and rapidly decaying as you want. For such $u$, define the Kelvin transform of $u$ to be
s\begin{equation*}
 v(x) = \left(\frac 1{|x|}\right)^{n + \alpha}u(\bar x), 
\end{equation*}
where $\bar x = \frac{x}{|x|^2}$. 
My question is about the relationship between the fractional Laplacian $(-\Delta)^{\alpha/2}$ and the Kelvin transform. Is there a direct proof for the equality 
\begin{equation*}
 \left((-\Delta)^{\alpha/2}v\right)(x) = \left(\frac{1}{|x|}\right)^{n + \alpha}\left((-\Delta)^{\alpha/2}u\right)(\bar x)? 
\end{equation*}
where by ``direct'' I mean only using the above definition of $(-\Delta)^{\alpha/2}$ and possibly a slick change of variable? \
Here is an attempt that is not quite giving me what I want but will hopefully provide a starting point for suggestions. Below I use the change of variable $y = \bar z = \frac z{|z|^2}$, $dy = \frac{1}{|z|^{2n}}dz$ together with the equality 
\begin{equation*}
 |\bar x - \bar z| = \frac{|x - z|}{|x||z|}. 
\end{equation*}
\begin{eqnarray*}
 \left((-\Delta)^{\alpha/2}u\right)(\bar x)
 & = & 
 \int_{\mathbb R^n}\frac{u(\bar x) - u(y)}{|\bar x - y|^{n + \alpha}}dy
 \\
 & = & 
 \int_{\mathbb R^n}\frac{u(\bar x) - u(\bar z)}{|\bar x - \bar z|^{n + \alpha}}\frac 1{|z|^{2n}}dz
 \\
 & = & 
 \int_{\mathbb R^n}\frac{u(\bar x) - u(\bar z)}{|x - z|^{n + \alpha}}\frac{(|x||z|)^{n +\alpha}}{|z|^{2n}}dz
 \\
 & = & 
 |x|^{n + \alpha}
 \int_{\mathbb R^n} \frac{\left(\frac{1}{|z|}\right)^{n - \alpha}u(\bar x) - \left(\frac 1{|z|}\right)^{n - \alpha} u(\bar z)}
 {|x - z|^{n+ \alpha}}dz
 \\
 & = & 
 |x|^{n + \alpha}
 \int_{\mathbb R^n} \frac{\left(\frac{|x|}{|z|}\right)^{n - \alpha}v(x) - v(z)}{|x - z|^{n+ \alpha}}dz. 
\end{eqnarray*}
Edit:
As it turns out  the above computation is just one step away from the desired equality. The function $\Phi(x) = |x|^{n - \alpha}$ is the fundamental solution for $(-\Delta)^{\alpha/2}$ in the sense that (still up to a multiplicative constant)
\begin{equation*}
 (-\Delta)^{\alpha/2}\Phi = \delta_0. 
\end{equation*}
Therefore, continuing the above computation for $x\neq 0$ gives
\begin{eqnarray*}
 \left((-\Delta)^{\alpha/2}u\right)(\bar x)
 & = & 
 |x|^{n + \alpha}\int_{\mathbb R^n} \frac{\left(\frac{|x|}{|z|}\right)^{n - \alpha}v(x) - v(z)}{|x - z|^{n+ \alpha}}dz
 \\
 & = & 
 -v(x)|x|^{2n}\int_{\mathbb R^n} \frac{|x|^{\alpha - n} - |z|^{\alpha -n}}{|x - z|^{n +\alpha}}dz
 \\
 &&
 +
 |x|^{n + \alpha}\int_{\mathbb R^n}\frac{v(x) - v(z)}{|x - z|^{n + \alpha}}dz
 \\
 & = & 
 -v(x)|x|^{2n}\left((-\Delta)^{\alpha/2}\Phi\right)(x) + |x|^{n + \alpha}\left((-\Delta)^{\alpha/2}v\right)(x)
 \\
 & = & 
 |x|^{n + \alpha}\left((-\Delta)^{\alpha/2}v\right)(x),  
\end{eqnarray*}
which is the desired equality. However, the only proof that I know for the fact that $(-\Delta)^{\alpha/2}\Phi = \delta_0$ relies on the Fourier transform of $\Phi$. Thus, this entire computation is still not as ``direct'' as I would like, but I suppose it will do for now. 
 A: SUMMARY.
I am going to prove that
$$\tag 1
(-\Delta)^{\frac\alpha 2}_x u = \lvert \bar x \rvert^{-\alpha-d}(-\Delta)_{\bar x}^\frac\alpha 2\bar u,$$
where $\bar x = \frac x {\lvert x \rvert^2}$ and $\bar u(\bar x)=\lvert \bar x \rvert^{\alpha -d}u(x).$

INTRODUCTION.
I have noticed quite a few times that the fractional Laplacian with positive exponent $(-\Delta)^{\alpha/2}$ is much messier than its inverse
$$
(-\Delta)^{-\alpha/2} v(x)=C\int_{\mathbb R^d} \frac{ v(y) }{ \lvert x- y \rvert^{d-\alpha}}\,dy.$$
So in this answer I am going to work with the latter, which is also known as Riesz potential; see the paper of Kwaśnicki 10 definitions of fractional Laplacian, Theorem 1.1 (i). There, you will also find the exact value of the constant $C>0$, which depends on $d$ and on $\alpha$. After the derivation of the correct transformation formula for $(-\Delta)^{-\alpha /2}$, taking inverses will yield (1).
The Riesz potential is a proper convolution, with a locally integrable kernel. On the other hand,  $(-\Delta)^{\alpha/2}$ is given by a crazy singular integral. This is the reason I give to myself to justify why $(-\Delta)^{-\alpha/2}$ is better behaved than $(-\Delta)^{\alpha/2}$.

PROOF.
We apply the change of variable
$$
\bar x=\frac x {\lvert x\rvert^2}, \quad \bar y=\frac y {\lvert y\rvert^2},\quad v(x)=\bar v (\bar x) W(\bar x), $$
where the weight $W$ is to be determined. Using the Jacobian formula $dy=\lvert \bar y \rvert^{-2d}d\bar y$ and the "chordal distance formula" (terminology mutuated from the stereographic projection)
$$
\lvert \bar x - \bar y \rvert = \frac{ \lvert x - y \rvert}{\lvert x\rvert \lvert y\rvert}, $$
we obtain
$$
(-\Delta)^{-\alpha/2}v(x)= C\int_{\mathbb R^d} \frac{W(\bar y)\bar v (\bar y) \lvert \bar x\rvert^{d-\alpha}}{\lvert \bar y\rvert^{d+\alpha}  \lvert \bar x- \bar y\rvert^{d-\alpha}}\, d\bar y, $$
so letting
$$W(\bar y)=\lvert \bar y\rvert^{d+\alpha}$$
yields
$$\tag 2 
(-\Delta)^{-\alpha/2}_xv=\lvert \bar x \rvert^{d-\alpha} (-\Delta)^{-\alpha/2}_{\bar x}(\lvert \bar x\rvert^{d+\alpha}v) .$$
Now we show that (2) implies (1). We let $u=(-\Delta)^{-\alpha/2}v$, so that $v=(-\Delta)^{\alpha/2}u$ and so (2) reads
$$
u=\lvert \bar x\rvert^{d-\alpha}(-\Delta)^{-\alpha/2}_{\bar x}\left( \lvert \bar x\rvert^{d+\alpha} (-\Delta)^{\alpha/2}_x u\right), $$
from which we conclude
$$
(-\Delta)^{\alpha/2}_{\bar x} \left( \lvert \bar x \rvert^{\alpha - d} u\right) = \lvert \bar x \rvert^{d+\alpha} (-\Delta)^{\alpha/2}_x u, $$
which is (1). The proof is complete.
