# Proving the Maximum Modulus Principle using the Open Mapping Theorem

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?

• 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.) Aug 10, 2011 at 23:53

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.

• As Qiaochu Yuan mentioned, this is not true if $\frak{U}$ contains the origin.
– lhf
Aug 11, 2011 at 10:47
• Plus the conclusion is that $f$ is constant, not a contradiction.
– lhf
Aug 11, 2011 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, 2011 at 11:19
• @lhf: why does the origin matter? Aug 11, 2011 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, 2011 at 17:42

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 $$f= a_0 + a_1(z-z_0)+....$$

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 $$z_0 is a maximum point for $$|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)