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If $u$ is harmonic function on disk with radius $R$ around the origion, and non constant in it.

why is it true that $u$ cannot be constant in any sub-Disk (i.e disk with radius less than $R$)

thanks

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  • $\begingroup$ Do you know how to represent a harmonic function on a disk as a Poisson integral? Do you now the maximum principle for harmonc harmonics? What kinds of things have you studied so far. $\endgroup$ – Disintegrating By Parts Jun 18 '15 at 18:16
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Because than $u$ is equal to a constant function on an open set (the sub disk) which implies by the identity theorem for harmonic function that $u$ is constant on the disk with radius $R$.

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  • $\begingroup$ I am not familiar with complex analysis yet. is it possible without this Thm? $\endgroup$ – Lin Jun 18 '15 at 15:32
  • $\begingroup$ It's harmonic function, so it has to do something with complex analysis.. $\endgroup$ – Snufsan Jun 18 '15 at 15:34
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    $\begingroup$ Well now, people certainly study harmonic functions in for example $R^n$, which can't have anything to do with complex analysis. Of course complex analysis often does give the simplest approach when $n=2$. $\endgroup$ – David C. Ullrich Jun 18 '15 at 16:14

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