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Let $\Omega\subseteq \mathbb{C}$ be a region and $f\in H(\mathbb{C})$ ($f$ is analytic on $\mathbb{C}$). If $|f|$ has in $\Omega$ a local maximum, then $f$ is constant.

Proof. Let $ D(a,\ r)\subseteq\Omega$ be a disk with $|f(z)|\leq|f(a)|$ for all $z\in D(a,\ r)$ ($a$ is therefore a local maximum for $f$). Then we have for every $\rho\in(0,\ r)$ with $\gamma(t)=a+\rho e^{it} \ (t\ \in[0,2\pi])$: $$ |f(a)|=\frac{1}{2\pi}\left|\int_{\gamma}\frac{f(\xi)}{\xi-a}\mathrm{d}\xi\right|\ \leq\frac{1}{2\pi}\int_{0}^{2\pi}|f(a+\rho e^{it})|\mathrm{d}t\ \leq|f(a)| \ \ \ (1) $$

$\color{green}{\text{I understand the inequality in (1), but what is the point of showing: } |f(a)|\leq|f(a)|?}$ $\color{green}{\text{Why is this mentioned?}}$

From $|f(a+\rho e^{it})|\leq|f(a)$ for $t \in[0,2\pi]$ and $\displaystyle \frac{1}{2\pi}\int_{0}^{2\pi}|f(a+\rho e^{it})|\mathrm{d}t =|f(a)|$ (2) we have: $$ |f(a+\rho e^{it})|=|f(a)|\ (t\ \in[0,2\pi]).\ \ \ (3) $$

$\color{green}{\text{What is the role of the integral in (2), in showing the equality (3)?}}$

Therefore we have: $$ |f(z)|=|f(a)|\ (z\in D(a,r))\ (*) $$ If $f$ would not be constant, then (because $\Omega$ is a region) also $f:D(a, r)\rightarrow \mathbb{C}$ would not be constant and $f(D(a,\ r))$ would be open, which is a contradiction to $(*)$. $ \square $


I still don't get something:

So we know that:

$|f(a+\rho e^{it})|\leq|f(a)$ for $t \in[0,2\pi]$ and

$\displaystyle \frac{1}{2\pi}\int_{0}^{2\pi}|f(a+\rho e^{it})|\mathrm{d}t =|f(a)|$

I still don't see why:

$f(a+\rho e^{it})|\mathrm{d}t =|f(a)|$ follows.

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For your second question: you have the inequality $|f(a+\rho e^{it})|\leq|f(a)|$, and then the integral shows that on average, your function $|f(a+\rho e^{it})|$ is equal to $|f(a)|$, so it is equal everywhere. –  M Turgeon Jun 6 '12 at 16:39
the point is the if $x\leq y\leq z \leq x$ then $x=y=z$ since if instead of $\leq$ you had $<$ you had $x<x$ –  Belgi Jun 6 '12 at 16:40
I still don't get (2), see the edited post. –  Chris Jun 16 '12 at 3:28

2 Answers 2

up vote 2 down vote accepted

The point of inequality (1) is that we have equality two. Define $g\colon [0,2\pi]\to\Bbb R$ by $g(t)=|f(a)|-|f(a+\rho e^{it})|$. This function is non-negative, continuous and its integral is $0$ hence $g$ is identically $0$ (that's why we used integral).

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I notice that you had just asked about the open mapping theorem for holomorphic function in a previous post. If you assume this result, you can also get a very quick proof of Maximum Principle:

$\Omega$ is an open set. By the open mapping theorem, if $f$ is a not constant, then $f(\Omega)$ is an open set. Suppose $|f|$ takes a maximum on $a$. That is $|f(a)| \geq |f(x)$ for all $x \in \Omega$. However $f(\Omega)$ is open, so there should be a small neighborhood of $f(a)$ in $f(\Omega)$. So there must exists a $x$ such that $|f(x)| > |f(a)|$. Contradiction.

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