§3.4 Further Properties of Holomorphic Functions

3.4 Further Properties of Holomorphic Functions

A useful corollary of Theorem 3.1.10 $Theorem 3.1.10 Theorem 3.1.10$ is the Maximum Modulus Principle.

In preparation for the result, we first introduce the global mean-value property of holomorphic functions: the mean-value property holds on every disk compactly contained in the domain. The adjectives global and local are informal terminology adopted here for clarity; they are not necessarily standard in classical treatments.

Lemma 3.4.1.

Let $𝑈 \subseteq ℂ$ be open and connected, and let $𝑓 : 𝑈 \to ℂ$ be holomorphic. Then $\forall 𝑧 \in 𝑈$ and $\forall 𝜀 > 0$ such that $\overline{𝐷 (𝑧, 𝜀)} \subset 𝑈$ , $𝑓 (𝑧)$ is the average of all $𝑓 (𝜁)$ for $𝜁 \in 𝜕 𝐷 (𝑧, 𝜀)$ . In other words,

𝑓 (𝑧) = \frac{1}{2 π 𝜀} \oint_{𝜕 𝐷 (𝑧, 𝜀)} 𝑓 (𝜁) | d 𝜁 | .

Proof.

By the Cauchy–Goursat Formula (Theorem 3.1.10 $Theorem 3.1.10 Theorem 3.1.10$ ),

𝑓 (𝑧) = \frac{1}{2 π i} \oint_{𝜕 𝐷 (𝑧, 𝜀)} \frac{𝑓 (𝜁)}{𝜁 - 𝑧} d 𝜁 = \frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑧 + 𝜀 e^{i 𝑡}) d 𝑡 .

Observe that

\begin{aligned} 𝑓 (𝑧) = \frac{1}{2 π 𝜀} \oint_{𝜕 𝐷 (𝑧, 𝜀)} 𝑓 (𝜁) | d 𝜁 | & = \frac{1}{2 π 𝜀} \int_{0}^{2 π} 𝑓 (𝑧 + 𝜀 e^{i 𝑡}) | i 𝜀 e^{i 𝑡} d 𝑡 | \\ = \frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑧 + 𝜀 e^{i 𝑡}) d 𝑡, \end{aligned}

and the conclusion follows. ⁠ $□$

Theorem 3.4.2 (Maximum Modulus Principle).

Let

𝑓 (𝑧)

be holomorphic on an open connected region

𝑈 \subseteq ℂ

. If

\exists 𝑧_{0} \in 𝑈

and an open neighborhood

𝑉 \subseteq 𝑈

𝑧_{0}

such that

\forall 𝑧 \in 𝑉

| 𝑓 (𝑧_{0}) | \geq | 𝑓 (𝑧) |

, then

𝑓

is a constant function on

𝑈

Proof.

Assume that $𝑧_{0}$ exists. We will first prove that the set

𝑆 = {𝑧 : 𝑓 (𝑧) = 𝑓 (𝑧_{0}), 𝑧 \in 𝑉}

is all of $𝑉$ . This is equivalent to proving that $𝑆$ is nonempty, open, and closed in $𝑉$ .

Since $𝑧_{0} \in 𝑆$ , the first condition is satisfied (nonemptiness). For any sequence ${𝑧_{𝑛}} \in 𝑆$ converging to some $𝑧_{\infty} \in 𝑉$ , by the continuity of $𝑓$ ,

lim_{𝑛 \to \infty} 𝑓 (𝑧_{𝑛}) = 𝑓 (lim_{𝑛 \to \infty} 𝑧_{𝑛}) = 𝑓 (𝑧_{\infty}) = 𝑓 (𝑧_{0}),

and $𝑧_{\infty} \in 𝑆$ . Thus, $𝑆$ contains all of its accumulation points in $𝑉$ and is therefore closed (if $𝑧_{\infty} \notin 𝑉$ , then it is no longer relevant; we are concerned with its relative closedness in $𝑉$ ).

Since $𝑆$ is a subset of $𝑉$ and $𝑉$ is open, $\forall 𝑧 \in 𝑆$ , $\exists 𝜆 > 0$ such that $𝐷 (𝑧, 𝜆) \subseteq 𝑉$ . By Lemma 3.4.1 $Lemma 3.4.1 Lemma 3.4.1$ , $\forall 0 < 𝜀 < 𝜆$ ,

\begin{aligned} | 𝑓 (𝑧) | & = | \frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑧 + 𝜀 e^{i 𝑡}) d 𝑡 | \leq \frac{1}{2 π} \int_{0}^{2 π} | 𝑓 (𝑧 + 𝜀 e^{i 𝑡}) | d 𝑡 \\ \leq \frac{1}{2 π} \int_{0}^{2 π} | 𝑓 (𝑧) | d 𝑡 = | 𝑓 (𝑧) | . \end{aligned}

It follows that all inequalities above are equalities, or that

\begin{aligned} | 𝑓 (𝑧) | & = | \frac{1}{2 π} \int_{0}^{2 π} 𝑓 (𝑧 + 𝜀 e^{i 𝑡}) d 𝑡 | = \frac{1}{2 π} \int_{0}^{2 π} | 𝑓 (𝑧 + 𝜀 e^{i 𝑡}) | d 𝑡 \\ = \frac{1}{2 π} \int_{0}^{2 π} | 𝑓 (𝑧) | d 𝑡 = | 𝑓 (𝑧) | . \end{aligned}

From the equality of the last two integrals,

\int_{0}^{2 π} [| 𝑓 (𝑧) | - | 𝑓 (𝑧 + 𝜀 e^{i 𝑡}) |] d 𝑡 = 0 .

Since this integrand is strictly non-negative and continuous, we have equality everywhere. Indeed, if not, there exists a $𝑡_{0}$ at which the integrand is positive. Then the integrand is bounded below by a positive constant on a neighborhood of $𝑡_{0}$ by continuity, giving a positive integral.

Thus we the integrand must have $| 𝑓 (𝑧) | = | 𝑓 (𝑧 + 𝜀 e^{i 𝑡}) |$ for all $𝑡$ .

Thus, $\forall 𝑧 \in 𝑆$ , $\exists 𝜆 > 0$ such that $𝐷 (𝑧, 𝜆) \subseteq 𝑆$ . In other words, every $𝑧 \in 𝑆$ has an open neighborhood that also lies in $𝑆$ . Therefore, $𝑆$ is open and $𝑆 = 𝑉$ as it is a nonempty clopen subset (Theorem 3.2.1.19 $Theorem 3.2.1.19 Theorem 3.2.1.19$ ). Since $𝑉$ is nonempty and open, it has an accumulation point in $𝑈$ . It follows that $𝑓 (𝑧) \equiv 𝑓 (𝑧_{0})$ over $𝑈$ by the Identity Theorem (Theorem 3.3.3 $Theorem 3.3.3 Theorem 3.3.3$ ). ⁠ $□$

Remark.

𝑓

is holomorphic and non-constant on an open region

𝑈 \subseteq ℂ

, then for any compact set

𝐾 ⋐ 𝑈

, the maximum of

𝑓

𝐾

lies on

𝜕 𝐾

. Otherwise,

𝑓

would attain a maximum at some

𝑧 \in \overset{⚬}{𝐾}

, and contradict the statement of Theorem 3.4.2

Theorem 3.4.2 Theorem 3.4.2

under the assumption of being non-constant.

By nature of the proof, it follows that a similar theorem holds for any continuous function that satisfies the global mean-value property.

Theorem 3.4.3.

Let $𝑈 \subseteq ℂ$ be open and connected and let $𝑓 : 𝑈 \to ℝ$ be continuous such that $\forall 𝑧 \in 𝑈$ and $\forall 𝜀 > 0$ such that $\overline{𝐷 (𝑧, 𝜀)} \subset 𝑈$ ,

𝑓 (𝑧) = \frac{1}{2 π 𝜀} \oint_{𝜕 𝐷 (𝑧, 𝜀)} 𝑓 (𝜁) | d 𝜁 |

is true. Suppose that $\exists 𝑧_{0} \in 𝑈$ and a neighborhood $𝑉 \subseteq 𝑈$ of $𝑧_{0}$ such that

𝑓 (𝑧_{0}) \geq 𝑓 (𝑧) \forall 𝑧 \in 𝑉 .

Then

𝑓

is constant on

𝑈

Theorem 3.4.4.

Let

𝑈 \subseteq ℂ

be an open disk centered at

𝑧_{0}

. If

𝑢 : 𝑈 \to ℝ

is harmonic, then there exists a holomorphic function

𝑓 : 𝑈 \to ℂ

such that

𝑢 \equiv ℜ𝔢 𝑓

Proof.

Let $𝑢$ be harmonic on the open disk $𝑈$ . Then define a complex-valued function $𝑔 : 𝑈 \to ℂ$ by

𝑔 \equiv \frac{𝜕 𝑢}{𝜕 𝑥} - i \frac{𝜕 𝑢}{𝜕 𝑦} .

Then

\frac{𝜕 𝑔}{𝜕 \overline{𝑧}} \equiv \frac{1}{2} [\frac{𝜕}{𝜕 𝑥} (\frac{𝜕 𝑢}{𝜕 𝑥} - i \frac{𝜕 𝑢}{𝜕 𝑦}) + i \frac{𝜕}{𝜕 𝑦} (\frac{𝜕 𝑢}{𝜕 𝑥} - i \frac{𝜕 𝑢}{𝜕 𝑦})] \equiv 0

because $𝑢$ is harmonic. Therefore, $𝑔$ is holomorphic. Since $𝑈$ is simply connected, $𝑔$ has a holomorphic antiderivative. Then let

𝑓 (𝑧) = \int_{𝑧_{0}}^{𝑧} 𝑔 (𝜁) d 𝜁 + 𝑢 (𝑧_{0}) .

Then for all $𝑧 \in 𝑈$ , letting $𝜁 = 𝜉 + i 𝜂$ , $𝑧_{0} = 𝑥_{0} + i 𝑦_{0}$ , $𝑧 = 𝑥 + i 𝑦$ , and integrating on the curve $𝑧_{0} \to 𝑥 + i 𝑦_{0} \to 𝑧$ , we have

\begin{aligned} ℜ𝔢 𝑓 (𝑧) & = ℜ𝔢 \int_{𝑧_{0}}^{𝑧} (\frac{𝜕 𝑢}{𝜕 𝜉} - i \frac{𝜕 𝑢}{𝜕 𝜂}) (d 𝜉 + i d 𝜂) + 𝑢 (𝑧_{0}) \\ = \int_{𝑥_{0}}^{𝑥} \frac{𝜕 𝑢 (𝜉 + i 𝑦_{0})}{𝜕 𝜉} d 𝜉 + \int_{𝑦_{0}}^{𝑦} \frac{𝜕 𝑢 (𝑥 + i 𝜂)}{𝜕 𝜂} d 𝜂 + 𝑢 (𝑧_{0}) \\ = 𝑢 (𝑥 + i 𝑦_{0}) - 𝑢 (𝑥_{0} + i 𝑦_{0}) + 𝑢 (𝑥 + i 𝑦) - 𝑢 (𝑥 + i 𝑦_{0}) + 𝑢 (𝑧_{0}) \\ = 𝑢 (𝑧) \end{aligned}

by the Fundamental Theorem of Calculus. ⁠

□

Therefore, for any real harmonic function $𝑢$ on an open region $𝑈$ , in each small disk $𝐷 \subseteq 𝑈$ , there exists a holomorphic function $𝑓 : 𝐷 \to ℂ$ such that $𝑢 \equiv ℜ𝔢 𝑓$ . Then taking the real part of both sides in Lemma 3.4.1 $Lemma 3.4.1 Lemma 3.4.1$ on $𝑓$ , we find that $𝑢$ satisfies the global mean-value property as well.

Because harmonic functions satisfy the global mean-value property, Theorem 3.4.3 $Theorem 3.4.3 Theorem 3.4.3$ is equivalent to:

Theorem 3.4.5 (Maximum Principle for Harmonic Functions).

Let $𝑈 \subseteq ℂ$ be open and connected and let $𝑢 : 𝑈 \to ℝ$ be harmonic. Suppose that $\exists 𝑧_{0} \in 𝑈$ and a neighborhood $𝑉 \subseteq 𝑈$ of $𝑧_{0}$ such that

𝑢 (𝑧) \leq 𝑢 (𝑧_{0}) \forall 𝑧 \in 𝑉 .

Then

𝑢

is constant on

𝑈