Global Version of the maximum modulus principle. Proof by Serge Lang

Question

I have 3 problems with Theorem 1.3.

Problem 1) I don't see how Theorem 7.1 applies here. I have provided the statement and the proof given by Serge Lang and all the other proofs used for Theorem 1.3.

Theorem 1.3 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:

By Theorem 7.1 we have $f$ locally constant at $z_0$. Then by Theorem 1.2 ii) (compare with the constant function and $f$) $f $ is constant on $U$

Theorem 7.1 Statement:

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

Theorem 1.2 ii) Statement:

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 $U$.

Problem 2) Is the difference between Theorem 1.3 (Maximum Modulus Principle) and Theorem 7.1(Local Maximum Modulus Principle) that in 1.3 we are talking about any open set but in Theorem 7.1 we are specifically talking about connected open set? Also if it is, I can't intietively see why in the local case the maximum can exist whereas in the connected case the maximum can not exist and the function must be constant

Problem 3) In the case of regular calculus it is obvious that Theorem 1.3 doesn't hold but shouldn't it since we have the set of real numbers beeing an open set that is connected?

Any help would be appreciated. If you wish for me to clarify something from the book please do so and I will respond as quickly as I can.

Answer 1

To understand the difference between Theorem 7.1 and Theorem 1.3, look at the following.

Let $z_0$ be a maximum for $f$, and set $f(z)=a$.

Consider now the set $$V:= \{ z \in U : f(x)=a \}$$

Now, Theorem 7.1 says that $V$ is an open subset of $U$.

But, since $V= f^{-1}(\{ a \})$ and $\{a \}$ is a closed set, we also get that $V$ is closed in $U$.

Therefore, $V$ is open and closed in $U$, and this is the gist of Theorem 7.1.

Now, if you know that $U$ is connected (the extra assumption in Theorem 1.3), then any non-empty subset which is open and closed must be the entire set, thus $V=U$.

P.S. If $U$ is not connected, lets say $U=B_1(0) \cup B_1(2019)$ is the disjoint union of two balls of radius one lets say, then you can have $f(z)=z$ for $z \in B_1(0)$ and $f(z)=2019$ for $z \in B_1(2019)$. Note that this function has a maximum of $2019$, is locally constant around this maximum, but not constant....

Answer 2

It seems just a matter of understanding the difference between a connected open set and just an open set. Recall, a connected open set is an open set which cannot be represented as a union of two non-empty non-intersecting open sets. Such a union $U=U_1\cup U_2$ of connected open sets is a simple example of a disconnected open set (and the most general "example" is an at most countable union). It is where Theorem 7.1 is applicable (giving that if, say, $z_0\in U_2$, then $f$ is constant on $U_2$, but it can be whatever one wants on $U_1$), but Theorem 1.3 is not. So:

1. A connected open set is open, so Theorem 7.1 is applicable, and we know that $f$ is locally constant at $z_0$, which means (by definition) that there is an open neighborhood $S$ of $z_0$ such that $f$ is constant on $S$. Now Theorem 1.2 ii) is applied with this $S$ (and with $g$ being that same constant).

2. Yes, exactly. In fact, the outcome of Theorem 7.1 is that $f$ is constant on the whole connected component of $U$ containing $z_0$ (i.e., the union of all open connected subsets of $U$ containing $z_0$). I wonder why Lang didn't go this way in 1. (perhaps to avoid too much topology?..).

3. The set $\mathbb{R}$ of real numbers is not open if considered as a subset of $\mathbb{C}$ (with its topology).