I found myself trying to prove the following, but I had to compute everything explicitly.


It is well known that if $u:\mathbb{R}^n\to\mathbb{R}$ is an harmonic function on $\mathbb{R}^n$, then the so-called Kelvin transform of $u$


$$(Ku)(x):=\frac{1}{|x|^{n-2}}u\biggl(\frac{x}{|x|^2}\biggr)$$


is harmonic in $\mathbb{R}^n\setminus\{0\}$.


Question: Is there a way to prove that $Ku$ is harmonic in $\mathbb{R}^n\setminus\{0\}$ WITHOUT making the explicit computations?


I know that there are tricks to soften the calculations (for example noticing that $1/|x|^{n-2}$ is harmonic in $\mathbb{R}^n\setminus\{0\}$), however I wasn't able to find a "clever" way to do it. I noticed that $x\mapsto x/|x|^2$ is a special transformation but I wasn't able to go anywhere, not even using the mean value property.



Why is $\frac{1}{|x|^{n-2}}u(\frac{x}{|x|^2})$ harmonic if $u$ is harmonic?
An explanation can be the following. Take a harmonic function of the form $u(x)=r^\alpha P(\omega)$, with $r=|x|$, $|\omega|=1$ and $P$ a polynomial. Then $0=\Delta u=r^{\alpha-2}\left (\alpha(N-2+\alpha)P(\omega)+\Delta\_S P(\omega)\right)$ with $\Delta\_S$ the Laplace-Beltrami on the unit sphere $S$. Then $\Delta\_S P=-\alpha(N-2+\alpha)P$. However we know that the eigenvalues of $\Delta\_S$ are given by $-k(N-2+k)$ with $k$ a nonegative integer but the equation $\alpha(N-2+\alpha)=k(N-2+k)$ has another root $\alpha=2-N-k$ which corresponds to the Kelvin transform $Ku(x)=|x|^{2-N}u(x/|x|^2)$. This shows that $Ku$ is harmonic whenever $u$ is an harmonic polynomial and then (approximating locally) for any harmonic function $u$.


