# Compact set - A non-constant holomorphic function - Maximum modulus principle

Why a compact, non-constant holomorphic function always reaches a maximum module on the edge of the set, never in its interior?

Maximum modulus principle : Let $D \subset \mathbb{C}$ a domain and $f : D \to \mathbb{C}$ a holomorphic function, non-constant. Then the module $|f|$ of $f$ doesn't have local maximum or local minimum in $D$ except at the points where $f$ vanishes.

Solution :

Intuitively it is true because an interior point $z_0$ cannot have a local maximum of the modulus. If the function is not constant, then developing it as a power series around $z_0$ will yield: $$f(z) = a_0 + a_n(z-z_0)^n + o((z-z_0)^n)$$ for some $n\ge 1$ and some $a_n\ne 0$, in some neighborhood of $z_0$. Thus, no matter what $a_n$ and $n$ are, we can find some point in the neighborhood of $z_0$ where the modulus is larger than $|a_0|$. Therefore there is no maximum of the modulus at $z_0$ when $z_0$ is an interior point.

On the other hand $|f(z)|$ is a continuous function of $z$ and must attain its maximum somewhere on a compact set. Since that somewhere cannot be an interior point, it can only be on the boundary.

Someone gave an intuitive explanation to the question. Does someone could thoroughly explain the part in bold?

• The key is the open mapping theorem. Mar 21, 2016 at 4:18
• Could you explain the reasoning behind rigorously?
– user320554
Mar 21, 2016 at 4:20
• en.wikipedia.org/wiki/… Mar 21, 2016 at 4:28
• note that at first this is a property of polynomials, and it is transferred to holomorphic functions which are locally very much like polynomials Mar 21, 2016 at 4:34

## 2 Answers

Any non-constant holomorphic function $f(z)$ defined on a domain $D$ is an open map, so in particular $f(D)$ is open. Therefore for any $z_0\in D$ there some $\varepsilon>0$ such that $B_{\varepsilon}(f(z_0))\subset f(D)$. $B_{\varepsilon}(f(z_0))$ will certainly have elements of larger modulus than $f(z_0)$, which shows that $f$ cannot have a local maximum at $z_0$.

• Can you explain the last sentence?
– user320554
Mar 21, 2016 at 4:45
• Any open disk will contain points which have larger modulus than the center. Since in this case the disk is contained in $f(D)$, this shows that there is some $w\in D$ such that $|f(w)|>|f(z_0)|$. Mar 21, 2016 at 5:40

If $z$ runs around $z_0$ in an $\varepsilon$-circle, then $z - z_0$ runs through the $\varepsilon$-circle around zero. In turn, $(z-z_0)^n$ runs through the $\varepsilon^n$-circle around zero. In particular, it attains the value $\frac{|a_n|a_0}{a_n|a_0|}\varepsilon^n$ at some $z$. So, $f(z)\approx a_0 + a_n\frac{|a_n|a_0}{a_n|a_0|}\varepsilon^n = (1 + |\frac{a_n}{a_0}|\varepsilon^n)a_0$, which has a larger modulus than $a_0$.