I showed this to my teacher this morning, and he remarked that the argument was a bit strange...but that it might be correct; he recommended using the ball mean property (which I will do, but I'm curious about this path)...
So, the question is: Does the following argument work? Is it a valid proof?
Partial Differential Equations, Jurgen Jost, pg 31, problem 1.7:
Let $\Omega \subset R^3 \backslash ${$0$}$,u:\Omega \rightarrow R$ be harmonic. Show that $v(x_1,x_2,x_3):=\frac{1}{|x|}u(\frac{x_1}{|x|^2},\frac{x_2}{|x|^2},\frac{x_3}{|x|^2})$ is harmonic in the region $\Omega ' := \{ x \in R^3:(\frac{x_1}{|x|^2},\frac{x_2}{|x|^2},\frac{x_3}{|x|^2})\in \Omega\}$.
Proposed solution:
$x=(x_1,x_2,x_3)\\\mu = \frac{x}{|x|}=(\frac{x_1}{|x|},\frac{x_2}{|x|},\frac{x_3}{|x|})\\\bar{\mu}=\frac{\mu}{|x|}=(\frac{x_1}{|x|^2},\frac{x_2}{|x|^2},\frac{x_3}{|x|^2})\\\frac{d\bar{\mu}}{d\mu}=\frac{1}{|x|}\\\bar{u}=u(\bar{\mu})\\\frac{\partial \bar{u}}{\partial \mu}=\frac{1}{|x|}*\frac{\partial\bar{u}}{\partial\bar{\mu}}\\v(x)=\frac{\bar{u}}{|x|}$
Since the region $\Omega ' \subset \Omega$, $\Delta \bar{u} = 0$.
Let $\Omega '' \subset \Omega '$ be $\Omega '' := \{ x\in R^3:|x| \gt \epsilon : \epsilon > 0\}$
Green's Formula writes out as:
$\int_{\Omega ''}\{v\Delta \bar{u}-\bar{u}\Delta v\}dv=\int_{\partial\Omega ''}\{v\frac{\partial \bar{u}}{\partial \mu} - \bar{u}\frac{\partial v}{\partial \mu}\}da$
Which rewrites as:
$\int_{\Omega ''}\{\frac{\bar{u}}{|x|}\Delta \bar{u}-\bar{u}\Delta \frac{\bar{u}}{|x|}\}dv=\int_{\partial\Omega ''}\{\frac{\bar{u}}{|x|^2}\frac{\partial \bar{u}}{\partial \bar{\mu}} - \frac{\bar{u}}{|x|^2}\frac{\partial \bar{u}}{\partial \bar{\mu}}\}da$
Noting that the right side equals zero and the first term on the left side equals zero, it simplifies to:
$-\int_{\Omega ''} \bar{u}\Delta\frac{\bar{u}}{|x|}dv = 0 $
And the non-trivial solution is $\Delta\frac{\bar{u}}{|x|}=0: x \in \Omega ''$.
And, to qualify the last step (triviality): Either $u$ is identically equal to zero, in which case $\Delta v = 0$, or $u\ne 0$, in which case $\Delta v = 0$ anyways.
Then, since $\Omega ' = lim_{\epsilon \rightarrow 0} \Omega '' : x\in R^3 : \frac{x}{|x|^2} \gt \epsilon \gt 0$, $\Delta v(x)=0\forall x \in \Omega'$.
I realize the above notation is sketchy, I just don't know how to apply a limit and an $\epsilon$/$\delta$ type-argument to sets; it is supposed to prevent the integration over the singularity at $\{0\}$ in order to satisfy the divergence theorem.
$\Omega ''$ and the step above was an attempt to incorporate Tomas's suggestions...but I still don't know if all the steps are valid.