Complex Analysics/Maximum Principle

Introduction

The maximum principle is a statement about holomorphic functions from the Complex Analysis. The magnitude $|f|$ of a holomorphic function $f:G\to \mathbb {C}$ cannot attain any strict local maxima within the domain of definition. Specifically, it asserts the following statement.

Statement

Let $U\subseteq \mathbb {C}$ be a domain, and let $f:U\to \mathbb {C}$ be holomorphic. If $|f|$ has a local maximum in $U$ , then $f$ is constant. If $U$ is bounded and $f$ can be continuously extended to ${\bar {U}}$ , then $f$ attains its maximum on $\partial U$ .

To prove this, we require a lemma that locally implies the conclusion.

Lemma

Let $G\subseteq \mathbb {C}$ be open, and $f:G\to \mathbb {C}$ be holomorphic. Let $z_{0}\in G$ be a local maximum point of $|f|$ . Then $f$ is constant in a neighborhood of $z_{0}$ .

Proof of Lemma 1

Let $r>0$ be chosen such that $|f(z)|\leq |f(z_{0})|$ for all $z\in {\bar {D}}r(z_{0})\subseteq G$ . The Cauchy's integral formula'gives, for all $\varepsilon \leq r$ :

f(z_{0})={\frac {1}{2\pi i}}\int {\partial D_{\varepsilon }(z_{0})}{\frac {f(z)}{z-z_{0}}},dz

This allows us to establish the following estimation:

Proof of Lemma 2

We derive the following estimation:

{\begin{array}{rl}|f(z_{0})|&={\frac {1}{2\pi }}\left|\int _{\partial D_{\varepsilon }(z_{0})}{\frac {f(z)}{z-z_{0}}}\,dz\right|\\&={\frac {1}{2\pi }}\left|\int _{0}^{2\pi }{\frac {f(z_{0}+\varepsilon e^{it})}{\varepsilon \cdot e^{it}}}\varepsilon \cdot i\cdot e^{it}\,dt\right|\\&\leq {\frac {1}{2\pi }}\int _{0}^{2\pi }|f(z_{0}+\varepsilon \cdot e^{it})|\,dt\\&\leq \sup _{t\in [0,2\pi ]}|f(z_{0}+\varepsilon \cdot e^{it})|\\&\leq |f(z_{0})|\end{array}}

Proof of Lemma 3

It follows that the inequality $\leq$ must be an equality chain, implying

{\begin{array}{rl}|f(z_{0})|=&\int _{0}^{2\pi }|f(z_{0})|\cdot {\frac {1}{2\pi }}dt={\frac {1}{2\pi }}\int _{0}^{2\pi }|f(z_{0}+\varepsilon \cdot e^{it})|\,dt\\\Rightarrow &0=\int _{0}^{2\pi }(|f(z)|-|f(z_{0})|)dt\\\Rightarrow &|f(z)|=|f(z_{0})|\end{array}}

.

Proof of Lemma 4

Thus, we establish the constancy of $|f|$ using the property:

|f(z)|=|f(z_{0})|

for all

z\in {\bar {D}}_{r}(z_{0})

,

i.e., $|f|$ is constant on $D_{r}(z_{0})$ .

Proof of Lemma 5

If $|f|={\sqrt {{\mathfrak {Re}}(f)^{2}+{\mathfrak {Im}}(f)^{2}}}$ is constant on $D_{r}(z_{0})$ , then ${\mathfrak {Re}}(f)^{2}+{\mathfrak {Im}}(f)^{2}=c$ must also be constant, where $c\in \mathbb {R}$ is a constant.

Proof of Lemma 6

Since $f$ is holomorphic on $D_{r}(z_{0})$ , the Cauchy-Riemann-Differential equation'apply:

u:={\mathfrak {Re}}(f),\ v:={\mathfrak {Im}}(f),\ {\text{and}}\ f(z):=u(x,y)+i\cdot v(x,y)

,

and the following holds:

{\frac {\partial u(x,y)^{2}+v(x,y)^{2}}{\partial x}}=0\ {\text{and}}\ {\frac {\partial u(x,y)^{2}+v(x,y)^{2}}{\partial y}}=0

.

Proof of Lemma 7

Let $u_{x}={\frac {\partial u}{\partial x}}$ and $u_{y}={\frac {\partial u}{\partial y}}$ . Applying the chain rule to the partial derivatives, we obtain:

0=2uu_{x}+2vv_{x}

and

0=2uu_{y}+2vv_{y}

.

Using the Cauchy-Riemann-Differential equation', replace the partial derivatives of $v$ with those of $u$ :

u_{x}=v_{y}

and

u_{y}=-v_{x}

, leading to:

0=2\cdot (uu_{x}-vu_{y})

and

0=2\cdot (uu_{y}+vu_{x})

.

Proof of Lemma 8

Squaring the above equations yields:

0=(uu_{x}-vu_{y})^{2}=u^{2}u_{x}^{2}-2uu_{x}\cdot vu_{y}+v^{2}u_{y}^{2}

,

0=(uu_{y}+vu_{x})^{2}=u^{2}u_{y}^{2}+2uu_{y}\cdot vu_{x}+v^{2}u_{x}^{2}

.

Adding these equations gives:

0=u^{2}u_{x}^{2}+v^{2}u_{y}^{2}+u^{2}u_{y}^{2}+v^{2}u_{x}^{2}

.

Proof of Lemma 9

Factoring out $u^{2}$ and $v^{2}$ :

0=u^{2}(u_{x}^{2}+u_{y}^{2})+v^{2}(u_{x}^{2}+u_{y}^{2})=(u^{2}+v^{2})\cdot (u_{x}^{2}+u_{y}^{2})

.

Thus,

0=u^{2}+v^{2}

or

0=u_{x}^{2}+u_{y}^{2}

.

Proof of Lemma 10

With $u^{2}=-v^{2}$ , it follows that $u=v=0$ since $u(x,y)$ and $v(x,y)$ are real-valued, implying $f=u+iv=0$ .

If $u_{x}^{2}=-u_{y}^{2}$ , then $u_{x}^{2}=u_{y}^{2}=0$ , and $u_{x}=u_{y}=0$ . By the Cauchy-Riemann-Differential equation, $f'=0$ . Thus, $f$ is constant on $D_{r}(z_{0})$ .

Proof

Let $z_{0}\in G$ be a local maximum point of $|f|$ in the domain $G$ . Define $V:={z\in G:f(z)=f(z_{0})}$ as the set of all $z\in G$ mapped to $w:=f(z_{0})\in \mathbb {C}$ (level set).

Proof 1: V is closed

Since $f$ is continuous, preimages of open sets are open, and preimages of closed sets are closed (in the relative topology of $G$ ). Thus, $V=f^{-1}({w})$ is closed in $G$ .

Proof 2: V is open

Using the lemma, $V$ can also be represented as a union of open disks, and unions of open sets are open.

Proof 3: Connectivity

Thus, $V=G$ due to the connectivity of $U$ , i.e., $f$ is constant.

Proof 4: G is bounded

If $G$ is bounded, then ${\bar {G}}$ is compact. Therefore, the continuous function $f$ attains its maximum on ${\bar {G}}$ , say at $z_{0}\in {\bar {G}}$ . If $z_{0}\in G$ , then $f$ is constant on $G$ (by the lemma) and hence on ${\bar {G}}$ , so $f$ also attains its maximum on $\partial G$ . Otherwise, $z_{0}\in \partial G$ , completing the proof.

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Date: 12/26/2024

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