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

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

• @metamorphy thanks for the response. I just have a couple of more questions. 1) I still can't see how theorem 1.2 implies that $f$ is constant on all of $U$. IF you could explicitly clarify that for me I would be very grateful 2) I can't understand why is it then not the same case for theorem 7.1 where a maximum is allowed to exist. Could you please explain the intuition between distinguishing the local and the global versions of the Maximum Modulus Principle? – Maths Wizzard Sep 8 '19 at 20:54
• Moved everything to an answer. – metamorphy Sep 8 '19 at 21:35

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....

• Thank you very much for the clear explanation. I just have one question. Why did Serge Lang need Theorem 1.2 ii) if a proof can be written, like you did, without the use of Theorem 1.2 ii)? – Maths Wizzard Sep 8 '19 at 22:37
• @KanyeWest I don't know, probably because the argument is shorter: You simply observe that $f(z)=a$ for some open set $a \in V \subset U$ and then, you apply Theorem 1.2 ii) to $f(z)$ and $g(z)=a$, the constant function. This works because $V$ has accumulation points. – N. S. Sep 8 '19 at 22:40

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).