Harmonic function composed with conformal map is harmonic (in $\mathbb{R}^n$) Here's the setup: Let $U,V$ open $\subset \mathbb{R}^n$, and let $u:V\rightarrow \mathbb{R}$ be harmonic, and $v:U\rightarrow V$ be conformal, i.e. $v$ is $C^1$ and the Jacobian $J_v(x)$ is a scalar multiple of an orthogonal transformation for all [; x\in U ;].  
I'm trying to prove $u\circ v$ is harmonic. [I've seen this stated as a fact in a few places without reference, namely here: http://en.wikipedia.org/wiki/Conformal_map#Uses , but maybe my hypotheses are slightly different and this is not true at all]
I've seen a proof that if $u$ is $C^2$ and $T$ is an orthogonal transformation then 
$\Delta (u \circ T) = \Delta(u) \circ T$.
So I'm thinking that to show $u\circ v$ is harmonic, we can use the fact that $v$ acts locally as its Jacobian, which is an orthogonal transformation, and move the Laplacian onto $u$ and conclude $u\circ v$ is harmonic.
However, I'm having trouble making this idea precise.  After glancing at my copy of baby Rudin, my hunch is to use the inverse function theorem or constant rank theorem, but I'm unsure how to apply those.  Any suggestions?
 A: Though this is an old question, it seems worthwhile to set this matter straight: 


*

*The proof by @alext87 is flawed: it does not account the dependency of $x_\epsilon$ on $x$. 

*The statement is false in dimensions $n>2$: the composition of a harmonic function with a Möbius transformation is not harmonic in general. Example in 3 dimensions: let $u(x)=x_1$, which is harmonic. Let $v(x)=x/|x|^2$ be the inversion in the unit sphere, which is conformal. Now $(u\circ v)(x)=x_1/|x|^2$, and computation yields $\Delta(u\circ v)=-2x_1/|x|^4$.

*The proof for $n=2$ can be found in many places, e.g., in Composition of a harmonic function with a holomorphic function 

*One can retain some of the connection between conformality and harmonicity in higher dimension by replacing the Laplacian $\Delta u=\mathrm{div}\,\nabla u$ with the $n$-Laplacian $\Delta_n u=\mathrm{div}\,(|\nabla u|^{n-2}\nabla u)$. The composition of an $n$-harmonic function with a Möbius transformation is $n$-harmonic.
A: Here is an attempt at a solution: 
Let $x_0\in U$, since $U$ is open there is a $\delta>0$ such that $B(x_0,\delta)\subset U$ and by Taylor's theorem (using $v\in\mathcal{C}^1$):
$v(x) = v(x_0)+\nabla v(x_{\epsilon})^T(x-x_0)$  for some $x_{\epsilon}\in B(x_0,\delta)$
We want to show that $\Delta u(v(x_0)+\nabla v(x_{\epsilon})^T(x-x_0))=0$ for $\forall x\in B(x_0,\delta)$. 
From the fact that $\nabla v(x_{\epsilon})^T$ is orthogonal and the theorem stated in the question we have 
$\Delta u(\nabla v(x_{\epsilon})^T(x-x_0))=\Delta (u)\nabla v(x_{\epsilon})^T(x-x_0)=0$  $\forall x\in B(x_0,\delta)$. 
If $v(x_0)$ is the zero vector then we are done. Else $v(x_0)+\nabla v(x_{\epsilon})^T(x-x_0)$ is a translation of $\nabla v(x_{\epsilon})^T(x-x_0)$ by a constant vector $v(x_0)$ which is an orthogonal transformation. Let this translation be denoted as the linear orthogonal operator $T_{v(x_0)}$ then we have 
$\Delta u(T_{v(x_0)}[\nabla v(x_{\epsilon})^T(x-x_0)]) = \Delta (u\circ T_{v(x_0)})\nabla v(x_{\epsilon})^T(x-x_0)$
as $v(x_{\epsilon})^T$ is orthogonal and
$\Delta (u\circ T_{v(x_0)})\nabla v(x_{\epsilon})^T(x-x_0)= \Delta (u)\circ T_{v(x_0)}(\nabla v(x_{\epsilon})^T(x-x_0))=0$
as $T_{v(x_0)}$ is orthogonal. 
Since $x_0$ was any point in $U$ we have showed that $\Delta(u\circ v)=0$ in $U$ and thus is a harmonic function. 
