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Let $\Omega\subset\mathbb{R}^n$ be an open connected domain, and let $u\in C^2(\Omega)$ be a harmonic function on $\Omega$.

Then for every ball $B_R(x)=\{y\in\Omega:|x-y|<R\}$ in $\Omega$ we have $$u(x)=\frac 1{|\partial B_R(x)|}\int_{\partial B_R(x)} u(y) dy=\frac 1{n\omega_nR^{n-1}}\int_{\partial B_R(x)} u(y) dy$$ Where $\omega_n$ is the volume of the unit $n$ dimensional ball.

I saw some proofs, but they were always longer then mine. So I deduce I have a mistake. I saw a similar proof to mine, but with a change of variables to $\alpha=\frac y{||y||}$ which I didn't see the need for. What is my mistake?

My proof:

Let $r\in[0,R]$. By Gauss' theorem and using the fact $u$ is harmonic and $r$ is constant on $\partial B_r(x)$, $$0=\int_{B_r(x)}\Delta udy=\int_{\partial B_r(x)} \frac \partial {\partial r}u(y)ds_y=\frac \partial {\partial r}\int_{\partial B_r(x)} u(y)ds_y$$ It follows that $$\int_{\partial B_r(x)} u(y)ds_y=const.\xrightarrow{|\partial B_r(x)|=const.} \frac 1 {|\partial B_r(x)|}\int_{\partial B_r(x)} u(y)ds_y=const.$$ So $$\frac 1 {|\partial B_R(x)|}\int_{\partial B_R(x)} u(y)ds_y=\lim_{r\rightarrow 0}\frac 1 {|\partial B_r(x)|}\int_{\partial B_r(x)} u(y)ds_y=u(x)$$

Q.E.D

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1 Answer 1

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The claim $$\int_{\partial B_r(x)} \frac \partial {\partial r}u(y)ds_y=\frac \partial {\partial r}\int_{\partial B_r(x)} u(y)ds_y$$

is not true since the domain of integration depends on $r$.

You also later claim that $|\partial B_r(x)|$ is constant, when it is in fact proportional to $r^{n-1}$.

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