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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)

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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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Assume the contrary and consider the function $$ f(s)=\int_0^{s}u(M_0+re^{it})dt, $$ that has the properties $f(0)=0$, $f(2\pi)=2\pi u(M_0)$, and $f'(s)=u(M_0+re^{is})$.

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    Sure, but what is your claim?2012-08-18
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    Maybe mean value theorem or something similar?2012-08-18
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When $u$ is a harmonic function then it has the mean value property for circles. Therefore in any case $$u(M_0)={1\over2\pi}\int_0^{2\pi} u\bigl(M_0+r e^{it}\bigr)\ dt\ .$$ Now about small boundary values on part of a circle: Consider the harmonic function $$u(z):={\rm Re}(z^2)=x^2-y^2\ .$$ Then $u(i)=-1<0=u(0)$, and by continuity there is a rather large arc $\gamma$ with midpoint $i$ on the unit circle such that $u(z)<-{1\over2}$ for all $z\in\gamma$.