Show that the Kelvin-transform is harmonic First, I have to give you our definitions:
Consider $\Omega:=\mathbb{R}^n\setminus\overline{B}_R(0)$ with $R>0$ and $n>1$. The function $\phi\colon\Omega\to G:=B_R(0)\setminus\left\{0\right\}$ with
$$
y=\phi(x):=\frac{R^2}{\| x\|^2}x\text{  for  }\| x\|>R
$$
is called Kelvin-transformation (relating to the sphere $S_R(0)$). Let $u\colon\Omega\to\mathbb{R}$. The function defined by
$$
v(y):=\frac{\| x\|^{n-2}}{R^{n-2}}u(x)\text{  for  }\lVert x\rVert >R~~~(*)
$$
is called Kelvin-transform of u.
Alternatively, the Kelvin-transform of $u$ can be definied by
$$
v(z):=\frac{R^{n-2}}{\lVert z\rVert^{n-2}}u\left(\frac{R^2}{\lVert z\rVert^2}z\right)\text{  for  }\lVert z\rVert <R.
$$


Let $u$ be harmonic in $\Omega$. Then its Kelvin-transform is harmonic in $G$. Show that for the case $n=2$.


To be honest, I do not know, what is to show here, because if I take the alternative definition of the Kelvin-transform, for $n=2$ it is
$$
v(z)=u\left(\frac{R^2}{\lVert z\rVert^2}z\right), z\in G
$$
and then
$$
\Delta v(z)=\Delta u\left(\frac{R^2}{\lVert z\rVert^2}z\right)=0,
$$
because $x:=\frac{R^2}{\lVert z\rVert^2}z$ is in $\Omega$ and $u$ is harmonic in $\Omega$.
So I only had to use that the inverse of the Kelvin-transformation is given by
$$
\phi^{-1}\colon G\to\Omega, y\longmapsto\frac{R^2}{\lVert y\rVert^2}y.
$$
Is that the proof? It seems TOO easy...
 A: Generally, to compute the partial derivatives of a composite function, you use the chain rule,
$$\frac{\partial (u\circ \phi)}{\partial z_i}(z) = \sum_{k=1}^n \bigl(\partial_k u\bigr)(\phi(z))\cdot \frac{\partial \phi_k}{\partial z_i}(z).$$
To compute the Laplacian, you need second order derivatives,
$$\frac{\partial^2 (u\circ\phi)}{\partial z_i^2}(z) = \sum_{k,m = 1}^n \bigl(\partial_m\partial_k u\bigr)(\phi(z))\cdot \frac{\partial \phi_k}{\partial z_i}(z)\cdot\frac{\partial \phi_m}{\partial z_i}(z) + \sum_{k=1}^n \bigl(\partial_k u\bigr)(\phi(z))\cdot \frac{\partial^2 \phi_k}{\partial z_i^2}(z),$$
which can result in a fair amount of computation. If we collect terms, we can write somewhat more succinctly
$$\Delta (u\circ\phi)(z) = \sum_{k,m=1}^n \bigl(\partial_m\partial_k u\bigr)(\phi(z))\langle \nabla \phi_k(z)\mspace{-3mu}\mid\mspace{-3mu} \nabla \phi_m(z)\rangle + \sum_{k=1}^n \bigl(\partial_k u\bigr)(\phi(z))\bigl(\Delta \phi_k\bigr)(z).\tag{1}$$
For the general case $n \neq 2$, you additionally have a factor $k_n(z) = \dfrac{R^{n-2}}{\lVert z\rVert^{n-2}}$, which by the product rule leads to
$$\begin{align}
\Delta v(z) &= \Delta \bigl( k_n(z)\cdot (u\circ\phi)(z)\bigr)\\
&= \bigl(\Delta k_n\bigr)(z)\cdot (u\circ\phi)(z) + 2 \langle \nabla k_n(z)\mspace{-3mu}\mid\mspace{-3mu} \nabla (u\circ\phi)(z)\rangle + k_n(z)\cdot \bigl(\Delta (u\circ\phi)\bigr)(z).
\end{align}$$
The properties of $\phi$ and $k_n$ (the latter is harmonic in $\mathbb{R}^n\setminus\{0\}$, by the way) make the computation less ugly than it looks in abstract form, when organised in the right way, but it's still offering ample opportunities to make a mistake.
The case $n = 2$ is a bit simpler. If we have a bit of complex analysis at our disposal, it becomes simpler still. We have
$$\Delta = 4\frac{\partial^2}{\partial w\partial \overline{w}}$$
if we identify $\mathbb{R}^2$ with $\mathbb{C}$, and using that simplifies the computation a lot when the function one composes with is holomorphic - $\dfrac{\partial \psi}{\partial \overline{w}} \equiv 0$ - or antiholomorphic - $\dfrac{\partial\psi}{\partial w}\equiv 0$:
$$\begin{align}
\frac{\partial}{\partial w}\frac{\partial (u\circ\psi)}{\partial\overline{w}}(w) &= \frac{\partial}{\partial w}\left( (\partial u)(\psi(w))\cdot \frac{\partial \psi}{\partial\overline{w}}(w) + (\overline{\partial}u)(\psi(w))\cdot\frac{\partial \overline{\psi}}{\partial \overline{w}}(w)\right)\\
&= \frac{\partial}{\partial w} \left((\overline{\partial} u)(\psi(w))\cdot \overline{\frac{\partial \psi}{\partial w}(w)}\right)\\
&= (\partial \overline{\partial} u)(\psi(w)) \left\lvert \frac{\partial \psi}{\partial w}(w)\right\rvert^2
\end{align}$$
for holomorphic $\psi$, since $\overline{\psi'}$ is antiholomorphic, and similar for antiholomorphic $\psi$. The end result in either case is
$$\Delta (u\circ\psi)(w) = \bigl(\Delta u\bigr)(\psi(w))\cdot \lvert \tilde{\psi}'(w)\rvert^2,$$
where $\tilde{\psi}$ is the holomorphic one of $\psi$ and $\overline{\psi}$, which shows that the composition of a harmonic function with a holomorphic or antiholomorphic function is again harmonic.
In our case, we have - identifying $\mathbb{R}^2$ with $\mathbb{C}$ -
$$\phi(w) = \frac{R^2}{\lvert w\rvert^2}\cdot w = \frac{R^2}{\overline{w}},$$
which is antiholomorphic, and thus the result, $v = u\circ \phi$ is harmonic for harmonic $u$.
Using real methods only, we have
$$\phi(x,y) = \left(\frac{R^2 x}{x^2+y^2},\, \frac{R^2 y}{x^2+y^2}\right),$$
and thus
$$\begin{gather}
\frac{\partial \phi_1}{\partial x}(x,y) = \frac{R^2}{x^2+y^2} - \frac{2R^2x^2}{(x^2+y^2)^2} = \frac{R^2(y^2-x^2)}{(x^2+y^2)^2} = - \frac{\partial\phi_2}{\partial y}(x,y),\\
\frac{\partial \phi_1}{\partial y}(x,y) = -\frac{2R^2xy}{(x^2+y^2)^2} = \frac{\partial \phi_2}{\partial x}(x,y).
\end{gather}$$
From these we obtain
$$\frac{\partial^2 \phi_1}{\partial x^2} = -\frac{\partial}{\partial x}\frac{\partial\phi_2}{\partial y} = - \frac{\partial}{\partial y}\frac{\partial\phi_2}{\partial x} = - \frac{\partial}{\partial y}\frac{\partial \phi_1}{\partial y} = - \frac{\partial^2 \phi_1}{\partial y^2},$$
so $\phi_1$ is harmonic, and similar for $\phi_2$, so the second sum in $(1)$ vanishes. For the first sum, we note that $\lVert \nabla \phi_1\rVert^2 = \lVert \nabla\phi_2\rVert^2$ by symmetry, and
$$\begin{align}
\langle \nabla \phi_1\mspace{-3mu}\mid\mspace{-3mu} \nabla \phi_2\rangle
&= \frac{\partial \phi_1}{\partial x}\frac{\partial \phi_2}{\partial x} + \frac{\partial \phi_1}{\partial y} \frac{\partial \phi_2}{\partial y}\\
&= \frac{\partial\phi_1}{\partial x}\frac{\partial\phi_1}{\partial y} + \frac{\partial\phi_1}{\partial y}\left(-\frac{\partial\phi_1}{\partial x}\right)\\
&= 0,
\end{align}$$
so the first sum simplifies to $\bigl(\Delta u\bigr)(\phi(w))\lVert\phi_1(w)\rVert^2$, and thus the desired conclusion that $v = u\circ \phi$ is harmonic if $u$ is.
