Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $$u(M_0)\neq \frac{1}{2\pi}\int_0^{2\pi}u(M_0+re^{it})dt?$$

share|cite|improve this question
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

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 $f'(s)=u(M_0+re^{is})$.

share|cite|improve this answer
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

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 $z\in\gamma$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.