I have a "simple" question but I'm not able to solve it.
Suppose that $u$ is a harmonic function on $\mathbb{R}^3$. Prove that the function $H(x) := |x|^{-1} u(x/|x|^2) $ is harmonic on $\mathbb{R}^3 \setminus\{0\}$
I tried brute force evaluating $\Delta H$ but even there I had some problems.. Any hint or partial solution is welcomed.
My attempt:
$$D_iH = \frac{x_i}{|x|^3}u+\frac 1{|x|}\sum_{j=1}^3D_ju\cdot\left(\frac {\delta_{ij}}{|x|^2}-\frac{2x_ix_j}{|x|^4}\right) $$ and \begin{align*} D_{ii}H & = \left(\frac{1}{|x|^3}-\frac{3x_i^2}{|x|^5}\right)u +\frac{x_i}{|x|^3}\sum_{j=1}^3D_ju\cdot\left(\frac {\delta_{ij}}{|x|^2}-\frac{2x_ix_j}{|x|^4}\right)\\ & -\frac{x_i}{|x|^3}\sum_{j=1}^3D_ju\cdot\left(\frac {\delta_{ij}}{|x|^2}-\frac{2x_ix_j}{|x|^4}\right)\\ & + \frac 1{|x|}\cdot\frac{D_{ii} u}{|x|^2}+\frac 1{|x|}\frac{2D_iu\cdot x_i}{|x|^4}\\ &-\frac{1}{|x|}\sum_{j=1}^3\left(D_{ij}u\cdot \frac{2x_ix_j\delta_{ij}}{|x|^4} + D_j u\cdot \left(\frac{2x_j}{|x|^4} + \frac{2x_i\delta_{ij}}{|x|^4} + \frac{4x_i^2x_j}{|x|^6} \right)\right)\\ & = \frac{1}{|x|^3}-\frac{3x_i^2}{|x|^5} + \frac{D_{ii}}{|x|^3}. \end{align*} Using the harmonicity of $u,$ we get $\Delta H=0$ if I don't miscalculate anything (when cancelling some terms).
