Let $U = \mathbf{R}^n - { 0 }$, $n > 2$ and consider for a function $f \in C^2(U)$ the Kelvin transform
$$f^\star(x) = r^{2-n} f\left(\frac{x}{r^2}\right),$$
where $r = \lvert x \rvert$. One can verify by explicit computation that,
$$r^2 \Delta(f^\star(x))= (r^2\Delta f)^{\star}(x),$$
where $\Delta$ denotes the usual Laplace operator. In particular the Kelvin transform maps harmonic functions to harmonic functions. Is there a way to see this last fact without resorting to explicit computation? I suspect it has to do with how the group of conformal transformations acts.
