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I am in the middle of the proof of the maximum principle for harmonic functions.

Given a harmonic function $u$ on the complex plane and $M_0\in \mathbb{C}$. Take $r>0$ and suppose there is an open arc $\ell$ contained in the circle $\{M_0+re^{it}\colon t\in [0,2\pi)\}$ such that $$u(M) Does it follow from this that $$u(M_0)\neq \frac{1}{2\pi}\int_0^{2\pi}u(M_0+re^{it})dt?$$

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    When $u$ only has to be continuous it is impossible to tell anything about $u(M_0)$ using only data referring to $u$ on the circle of radius $r>0$ around $M_0$.2012-08-18
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    I have edited my post.2012-08-18
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    As stated, your question has a simple answer: no, it does not. An inequality valid on a part of the domain of integration does not give enough information about the value of the integral to make such a conclusion2012-08-18
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    What if we assume that $M_0=\max_{z\in \mathbb{C}} u(z)$.2012-08-18
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    Well, since $M_0$ is a point in the complex plane and $u$ is a real-valued functions, making such an assumption leads us nowhere fast.2012-08-20

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