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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?

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    I'd try to reduce the statement to the case of the reflection about the unit sphere, and then to check that the statement holds in this particular case. I also suspect that the cases n=2 and n>2 must be treated separately. [The case n=2 is clear. But, if memory serves, for n>2, conformal maps defined on connected open subsets extend to the whole sphere. Thanks for correcting me if I'm wrong.]2010-09-17
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    This extension theorem seems to be due to Liouville: http://en.wikipedia.org/wiki/Conformal_map#Higher-dimensional_Euclidean_space - The whole Wikipedia entry looks interesting.2010-09-17
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    There's a proof-by-computation available on-line, at this blog: http://anhngq.wordpress.com/2010/04/24/conformal-invariant-operators-laplacian-operator/ I've spent a couple minutes trying to find a conceptual proof but the ideas I'm attempting seem to all be a little off the mark.2010-09-18

2 Answers 2

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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.

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    Thanks! This looks good to me. I thought Taylor's theorem might be involved, but I didn't know this formulation of it.2010-09-20
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    Great! I'll leave you to tidy it up a bit!2010-09-20
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Though this is an old question, it seems worthwhile to set this matter straight:

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

  2. 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 $\Delta(u\circ v)=-2x_1/|x|^4$.

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

  4. 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.