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Suppose that $u$ is harmonic on the unit disk $D$, $f$ is a continuous bijection from $\bar \Omega$ to $\bar D$, where $\Omega$ is an open connected set in $\mathbb C$, and $f$ sends the boundary of $\Omega$ to the boundary of $D$, and $f(\Omega) = D$. Is $u\circ f$ harmonic on $\Omega$?

I am trying to show that $u\circ f$ satisfies the mean value property.

Given $z_0 \in \Omega$, $f(z_0) \in D$, there exists $\delta > 0$, such that $u(f(z_0)) = \int^{2\pi}_0u(f(z_0)+re^{i\theta})\ d\theta \ \forall\ 0<r <\delta$. Since $u$ is harmonic.

How do I conclude from this that $u\circ f$ satisfies the mean value property? I have the sense that if $f$ is holomorphic this would be easy since $u\circ f$ will be the real part of a holomorphic function.

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  • $\begingroup$ Do you mean $f$ is analytic? Otherwise talking about derivative of $f$ makes no sense $\endgroup$ – user160738 Nov 19 '16 at 20:47
  • $\begingroup$ I am not taking derivative of $f$. $\endgroup$ – koch Nov 19 '16 at 20:54
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If $f$ is not holomorphic, it is false in general. For instance, take $u(z) = z$ and $f : D \to \Omega = D$ some homeomorphism of $D$ which is not harmonic (there are many of those).

If $f$ is holomorphic, you can compute the Laplacian of $f \circ u$ directly using the chain rule and use the Cauchy-Riemann equations to conclude the result. If you want to check the mean value property like you want, the problem is that the image of a circle by $f$ is not necessarily a circle. You could try to get past this difficulty by first generalizing the mean value property to any loop (say of class $C^1$) rather than just circles, by saying that $u$ is the real part of a holomorphic function and using the Cauchy integral formula. This proof would work but I think in retrospect it will look a bit artificial.

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