Maximum Principle

Introduction

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

Statement

Let $U\subseteq ℂ$ be a domain, and let $f:U\to ℂ$ 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 $\overline{U}$, then $f$ attains its maximum on $\partial U$.

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

Lemma

Let $G\subseteq ℂ$ be open, and $f:G\to ℂ$ 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 $|f(z)|\le |f(z_0)|$ for all $z\in \overline{D}r(z_0)\subseteq G$. The Cauchy's integral formula'gives, for all $\epsilon \le r$:

$f(z_0)=\frac{1}{2\pi i}\int \partial D_{\epsilon }(z_0)\frac{f(z)}{z-z_0},dz$

This allows us to establish the following estimation:

Proof of Lemma 2

We derive the following estimation:

$\begin{array}{cc}|f(z_0)|& =\frac{1}{2\pi }\left|\underset{\partial D_{\epsilon }(z_0)}{\int }\frac{f(z)}{z-z_0}dz\right|\\ & =\frac{1}{2\pi }|\underset{0}{\overset{2\pi }{\int }}\frac{f(z_0+\epsilon e^{it})}{\epsilon \cdot e^{it}}\epsilon \cdot i\cdot e^{it}dt|\\ & \le \frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}|f(z_0+\epsilon \cdot e^{it})|dt\\ & \le \underset{t\in [0,2\pi ]}{\sup }|f(z_0+\epsilon \cdot e^{it})|\\ & \le |f(z_0)|\end{array}$

Proof of Lemma 3

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

$\begin{array}{cc}|f(z_0)|=& \underset{0}{\overset{2\pi }{\int }}|f(z_0)|\cdot \frac{1}{2\pi }dt=\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}|f(z_0+\epsilon \cdot e^{it})|dt\\ \Rightarrow & 0=\underset{0}{\overset{2\pi }{\int }}(|f(z)|-|f(z_0)|)dt\\ \Rightarrow & |f(z)|=|f(z_0)|\end{array}$.

Proof of Lemma 4

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

$|f(z)|=|f(z_0)|$ for all $z\in {\overline{D}}_r(z_0)$,

i.e., $|f|$ is constant on $D_r(z_0)$.

Proof of Lemma 5

If $|f|=\sqrt{\Re 𝔢(f{)}^2+\Im 𝔪(f{)}^2}$ is constant on $D_r(z_0)$, then $\Re 𝔢(f{)}^2+\Im 𝔪(f{)}^2=c$ must also be constant, where $c\in ℝ$ is a constant.

Proof of Lemma 6

Since $f$ is holomorphic on $D_r(z_0)$, the Cauchy-Riemann-Differential equation'apply:

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

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:

$0=2u{u}_x+2v{v}_x$ and $0=2u{u}_y+2v{v}_y$.

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:

$0=2\cdot (u{u}_x-v{u}_y)$ and $0=2\cdot (u{u}_y+v{u}_x)$.

Proof of Lemma 8

Squaring the above equations yields:

$0=(u{u}_x-v{u}_y{)}^2=u^2{u}_x^2-2u{u}_x\cdot v{u}_y+v^2{u}_y^2$,

$0=(u{u}_y+v{u}_x{)}^2=u^2{u}_y^2+2u{u}_y\cdot v{u}_x+v^2{u}_x^2$.

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, $f^{\prime }=0$. 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 $w:=f(z_0)\in ℂ$ (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 $\overline{G}$ is compact. Therefore, the continuous function $f$ attains its maximum on $\overline{G}$, say at $z_0\in \overline{G}$. If $z_0\in G$, then $f$ is constant on $G$ (by the lemma) and hence on $\overline{G}$, so $f$ also attains its maximum on $\partial G$. Otherwise, $z_0\in \partial G$, completing the proof.

See Also

• Cauchy-Riemann-Differential equation'
• Application of Cauchy-Riemann Equations

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