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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 $$u(M)<u(M_0)\mbox{ for each }M\in \ell.$$ 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 $M_0$. –  Christian Blatter Aug 18 '12 at 15:58
    
I have edited my post. –  SiwyLigr Aug 18 '12 at 16:12
    
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 conclusion –  user31373 Aug 18 '12 at 17:43
    
What if we assume that $M_0=\max_{z\in \mathbb{C}} u(z)$. –  SiwyLigr Aug 18 '12 at 17:52
    
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. –  user31373 Aug 20 '12 at 2:10
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2 Answers

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? –  SiwyLigr Aug 18 '12 at 16:44
    
Maybe mean value theorem or something similar? –  timur Aug 18 '12 at 16:49
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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 $z\in\gamma$.

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