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I was reading on Wikipedia that

"The maximum modulus principle can be viewed as a special case of the open mapping theorem, which states that a nonconstant holomorphic function maps open sets to open sets. If $|f|$ attains a local maximum at $a$, then the image of a sufficiently small open neighborhood of $a$ cannot be open. Therefore, $f$ is constant".

Could someone expand upon that? I don't follow why the image of a open neighborhood of a would not be open?

Davide Giraudo
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    Because it can't contain an open neighborhood of $|f(a)|$ by maximality. (Here you need to observe that the absolute value function is also open away from the origin, but this is not hard to see.) – Qiaochu Yuan Aug 10 '11 at 23:53

2 Answers2

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If $\frak{U}$ is an open set in $\mathbb{C}$, then $|\frak{U}|$ can have no greatest element. Thus, $|f(\frak{U})|$ is cannot have a greatest element since $f(\frak{U})$ is open.

Suppose that $f$ attains its maximum in $\frak{U}$; this means that for some $z_0\in\frak{U}$, $|f(z)|\le|f(z_0)|$ for all $z\in\frak{U}$. Thus, $|f(z_0)|$ is the greatest element of $|f(\frak{U})|$. This means that $f(\frak{U})$ is not open. Therefore, $f$ is constant.

robjohn
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    As Qiaochu Yuan mentioned, this is not true if $\frak{U}$ contains the origin. – lhf Aug 11 '11 at 10:47
  • Plus the conclusion is that $f$ is constant, not a contradiction. – lhf Aug 11 '11 at 10:49
  • @lhf: Thanks! I forgot about the origin. I had assumed $f$ was non-constant, but never wrote it down. It does read more like the maximum modulus principle to conclude that $f$ is constant, so rather than introduce my hidden assumption, I have concluded that $f$ is constant. – robjohn Aug 11 '11 at 11:19
  • @lhf: why does the origin matter? – user1736 Aug 11 '11 at 15:23
  • You have now phrased your answer differently but the original version had ${\frak{U}} \mbox{ open in } \mathbb C \Rightarrow |{\frak{U}}| \mbox{ open in } \mathbb R$ and this is not true if $\frak{U}$ contains the origin. – lhf Aug 11 '11 at 17:42
  • @user1736: if $\frak{U}$ contains the origin, then $|\frak{U}|$ is not open (it has $0$ as a minimum element). What you can say is that if $\frak{U}$ is open, then $|\frak{U}|$ has no greatest element. – robjohn Aug 11 '11 at 17:55
  • Why is f(U) open? – gen May 30 '17 at 08:27
  • @gen: We are trying to prove that if $|f|$ attains its maximum in $\frak{U}$, then $f$ is constant. If $f$ is constant, there is nothing to prove, so we assume that $f$ is non-constant. The open mapping theorem says that $f(\frak{U})$ is open. – robjohn May 30 '17 at 11:38
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Proposition 1

Let f be analytic on an open set U. Let $z_0$ a member of $U$ be a maximum for $|f|$, that is, $|f(z)|\geq|f(z_0)$ , for all $z$ members of $U$. Then $f$ is locally constant at $z_0$

Proof:

Since $f$ is analytic we have $f= a_0 + a_1(z-z_0)+....$

We proof this through contradiction.

If $f$ is not constant $a_0=f(z_0)$ then by the Open Mapping Theorem we know $f$ is an oppen mapping thus the image of $f$ contains a disc $D(a_0,s)$. Hence the set of numbers $|f(z)|$, for z in a neighbourhood of $z_0$, contains an open interval around $a_0$ s.t. $f(z)>f(z_0)$. But thats a contradiction hence $f$ is locally constant at $z_0$.

Proposition 2

Let $f,g,$ be analytic on $U$. Let $S$ be a set of points in $U$ which is not discrete. Assume that $f(z)=g(z)$ for all $z$ in $S$. Then $f=g=$ on $U$.

Proof:

The proof is found in many textbooks (such as Serge Lang's Complex Analysis) and I think is known to OP but if needed I can provide it.

Maximum Modulus Principle Statement:

Let $U$ be a connected open set, and let $f$ be an analytic function on $U$. If $z_0<U$ is a maximum point for $|f|$, that is $|f(z_0)|\geq|f(z)|$ for all $z∈U$, then $f$ is constant on $U$

Proof Now for proving the maximum modulus principle, by Proposition 1 we have $f$ locally constant at $z_0$. Then by Proposition 2 $f$ is constant on $U$ (compare $f$ with the constant function)

EDIT: corrected mathematical notation

Lucius Aelius Seianus
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