I don't understand a paragraph in Conway's complex analysis at the beginning of Chapter VI page 128 (Maximum Modulus Theorem).

He says: "Note that in Theorem 1.2 we did not assume that $G$ is connected as in Theorem 1.1. Do you understand how Theorem 1.1 puts the finishing touches on the proof of Theorem 1.2? Or could the assumption of connectedness in Theorem 1.1 be dropped?"

For reference I included both theorems here: 1.1 Maximum Modulus Theorem, First Version. If $f$ is analytic in a region $G$ and $a$ is a point in $G$ with $|f(a)| \geq |f(z)|$ for all $z$ in $G$, then $f$ must be constant. (it is proved with the open mapping theorem).

1.2 Maximum Modulus Theorem, Second Version. Let $G$ be a bounded open set in $\mathbb{C}$ and suppose $f$ is continuous on the closure of $G$ and analytic in $G$. Then the maximum is attained on $\partial G$.

I was thinking that I understood the maximum modulus but since I can understand Conway's little paragraph I must be missing something important. If someone has the book, thanks for any help.

  • 8
    $\begingroup$ In Theorem 1.1, you can drop the connected assumption on $G$, but then you can no longer say that $f$ is constant on all of $G$. Think of a simple example of a $G$ that is not connected, and try to imagine what happens in that setting. $\endgroup$
    – Braindead
    Aug 3, 2011 at 7:24
  • 1
    $\begingroup$ Also, is there a part of the proof of Theorem 1.2 that you don't understand? I'm not exactly sure what you are really confused about. $\endgroup$
    – Braindead
    Aug 3, 2011 at 7:29
  • $\begingroup$ For the proof of 1.2 since $G$ is bounded there is $a$ in the closure of $G$ such that $|f(a)| \geq |f(z)|$ for every $z$ in the closure of $G$. Now if $f$ is not constant then by 1.1 for every $a$ there is a $z$ such that $|f(a)|<|f(z)|$ which is a contradiction. I think I understand the proof of 1.2. I just didn't know if I could drop the connectedness assumption. Let me try to think about your hint for a counterexample. $\endgroup$
    – user786
    Aug 3, 2011 at 7:51
  • $\begingroup$ You don't need $G$ connected for Theorem 1.2. $\endgroup$
    – Braindead
    Aug 3, 2011 at 15:02
  • 6
    $\begingroup$ The characterization of Conway's paragraph as obscure seems undue. $\endgroup$
    – Did
    Aug 4, 2011 at 17:11

1 Answer 1


I think this answer may help to this question (even though the question is too old):

In class I was given the proof of 1.2 in the following way (Answering to Conway's question "do you understand(...)":

Let $M=\max\{|f(z)|: z\in \overline{G}\}$ and $N=\max\{|f(z)|: z\in \partial G\}$. Suppose that $M>N$ so there exists $a\in G$ such that $M=|f(a)|$. Let $G_a$ be the connected component of $G$ which contains $a$. By Maximum Modulus (the first version) $f(z)=f(a)$ $\forall z\in G_a$ and by continuity $f(z)=f(a)$ $\forall z\in \partial G_a$. As $\mathbb{C}$ is locally connected, it is true that $\partial G_a\subset \partial G$ so there exists $z\in\partial G$ such that $f(z)=f(a)$, which implies $M=N$, absurd.


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