Given the function $f:\mathbb{R^n\setminus{0}}\rightarrow \mathbb{R},\:f(x)=||x||$, $||*||$ being the euclidean distance, show that $\Delta(f^{2-n}) = 0$ for $n>0$
What is the correct way to approach this problem? Calculating everything to n=4 hasn't yielded any insights.
Let $w$ be a unit vector.
Note that by a Taylor expansion
$|w+u|^{2-n}=1+L_w(u)+\frac{n(n-2)}{2}(w \cdot u)^2 +\frac{2-n}{2}|u|^2+o(|u|^2)$ as $u$ goes to zero, where $L_w$ is linear: this gives us $\langle 2D^2f(w)u,\,u \rangle$ for all $u$.
Note that if $w$ is any unit vector, the average of $\langle w,\,u\rangle^2$ over unit vectors $u$ doesn’t depend on $w$: by taking an ONB of $\mathbb{R}^n$ one finds this average to be $\frac{1}{n}$.
As a consequence, if $S$ is a symmetric real matrix, the average of $\langle Su,\,u\rangle$ over unit vectors $u$ is $\frac{tr{S}}{n}$.
In particular, $\Delta f(w)$ is $n/2$ times the average over unit vectors $u$ of $\frac{n(n-2)}{2}(w \cdot u)^2 +\frac{2-n}{2}|u|^2$. From what was written above, this average is $-\frac{n-2}{2}+\frac{n-2}{2}=0$, which proves that $\Delta f$ vanishes on the sphere.
A similar scheme of proof works in general.