Let us denote the Poisson kernel on $B_r$ by $$ P(x,\zeta)=\frac{r^2-\vert x \vert^2}{r\omega_{n-1}\vert x-\zeta \vert^n}$$ where $x\in B_r$ and $\zeta\in\partial B_r$. Given a boundary function $f$ on $\partial B_r$, the Poisson integral formula is given by $$P[f](x)=\int_{\partial B_r} f(\zeta)P(x,\zeta)\,d\sigma(\zeta)$$ where $d\sigma(\zeta)$ denotes the surface integral.
Fix a point $c\in\mathbb{R}^n$ such that $\vert c \vert^2=r^2+R^2$ and consider the inversion with respect to the sphere whose center is $c$ and radius is $R$. Denoting this inversion map by $T$, it is easy to check that $T$ maps $\partial B_r$ to itself. Define a new function $g$ on $\partial B_r$ by the equation $$g(\zeta):=f(T\zeta)$$ for each $\zeta\in\partial B_r$. My question is whether or not the following is true:
For every $x\in B_r$ satisfying $\vert x-c \vert^2=R^2$, we have $P[g](x)=P[f](x)$.
I came up with the question while trying to generalize the Schwarz reflection principle; I wanted to 'reflect' with respect to a sphere, not a plane. If the answer to my question is positive, the generalization of the Schwarz reflection principle follows. This really is a generalization since a plane can be thought as a sphere of inifinite radius.
I tried to calculate $P[g](x)$ directly, but couldn't think of a way to handle $d\sigma(T\zeta)$. For maps between open subsets of Euclidean spaces, the change of variables formula indicates $d(Ty)=J_T(y)\,dy$ where $J_T$ is the Jacobian of $T$. But here, $T$ is a map between spheres, so I don't have any 'change of variables' formula available. Also, I'm starting to think that there must be a simple, geometric explanation for this invariance. Perhaps we can take a suitable conformal map, which takes a plane to a sphere, and argue that the invariance of the integral is `transfered'. But I have zero knowledge on conformal maps, so that's probably too hopeful.