For a harmonic function $u(x)$, on domain $\Omega$ where $x \in \Omega \subset \Bbb R^n $, how to show that
$$ u(x) = \frac{1}{\omega_n R^{n-1}}\int_{\partial B_R(x)} u(\sigma) d\sigma$$
where $\omega_n$ is the area of the unit sphere $\partial B_1(x)$.
I am looking for simple case $n=2$ centered at $0$. An argument in my note uses the definition
$\displaystyle g(r) \overset{(def)}{=} \frac{1}{2\pi r }\int_{\partial B_r(0)} u(\sigma) d\sigma$ and shows that $g'(r) = 0$. I don't see how does it prove above? I think it proves $g(r) = \text{constant}$.
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Onil90
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Mula Ko Saag
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You are right, but what is that constant? – Oct 30 '13 at 05:08
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@John is it supposed to be $u(0)$ ?? – Mula Ko Saag Oct 30 '13 at 05:09
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Note that $g(r)$ is the average of $g$ over the circle of radius $r$. If $g$ is continuous, $g(r)$ should goes to $g(0)$. Of course one needs to prove that, though the proof is not difficult. – Oct 30 '13 at 05:53
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Hint: consider $\displaystyle\lim_{r\to 0}g(r)$.
detnvvp
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could you be a bit explicit in you answer?? I am not getting it :(( – Mula Ko Saag Oct 30 '13 at 05:14
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Ok, since $g'=0$, you have that $g$ is a constant. Therefore, $\displaystyle g(r)=\lim_{s\to 0}g(s)$. All you have to do is prove that $\displaystyle\lim_{s\to 0}g(s)=u(0)$. – detnvvp Oct 30 '13 at 05:16
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