§4.4.4 Classifying Growth of Entire Functions

4.4.4 Classifying Growth of Entire Functions

Lemma 4.4.4.1.

Let $𝑓 : \overline{𝐷 (0, 𝑟)} \to ℂ^{*}$ (where $𝑟 > 0$ ) be a nowhere-vanishing holomorphic function. It follows that

log | 𝑓 (0) | = \frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (𝑟 e^{i 𝜃}) | d 𝜃 .

Proof.

Without loss of generality, assume $𝑟 = 1$ . Since $𝑓$ is non-vanishing and $\overline{𝔻}$ is simply connected, we may define the holomorphic logarithm as

log (𝑓 (𝑧)) = \int_{𝛾} \frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)} d 𝑧 + log (𝑓 (𝑧_{0}))

for any fixed $𝑧_{0} \in \overline{𝔻}$ and all $𝑧 \in \overline{𝔻}$ , where $𝛾 \subset \overline{𝔻}$ is any piecewise smooth curve from $𝑧_{0}$ to $𝑧$ .

Hence, $log | 𝑓 (𝑧) | = ℜ𝔢 [log (𝑓 (𝑧))]$ and is therefore harmonic. The assertion then follows from the mean-value property. ⁠ $□$

Theorem 4.4.4.2 (Jensen’s Formula).

Let $𝑓 : \overline{𝐷 (0, 𝑟)} \to ℂ$ be meromorphic such that $𝑓 (0) \neq 0$ . If $𝑎_{1}, \dots, 𝑎_{𝑚}$ , $𝑏_{1}, \dots, 𝑏_{𝑛}$ are the zeros and poles of $𝑓$ in $\overline{𝐷 (0, 𝑟)}$ , counted with multiplicities and orders, respectively, then

log | 𝑓 (0) | = \frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (𝑟 e^{i 𝜃}) | d 𝜃 + \sum_{𝑘 = 1}^{𝑚} log | \frac{𝑎_{𝑘}}{𝑟} | - \sum_{𝑘 = 1}^{𝑛} log | \frac{𝑏_{𝑘}}{𝑟} | .

Proof.

For simplicity, assume $𝑎_{1}, \dots, 𝑎_{𝑚_{0}}$ are the zeros in $𝐷 (0, 𝑟)$ and $𝑎_{𝑚_{0} + 1}, \dots, 𝑎_{𝑚}$ are the zeros on $𝜕 𝐷 (0, 𝑟)$ . Similarly, let $𝑏_{1}, \dots, 𝑏_{𝑛_{0}}$ be the poles in $𝐷 (0, 𝑟)$ and $𝑏_{𝑛_{0} + 1}, \dots, 𝑏_{𝑛}$ be the poles on $𝜕 𝐷 (0, 𝑟)$ . Let

𝑓 (𝑧) = 𝑔 (𝑧) \frac{\prod_{𝑗 = 1}^{𝑚} (𝑧 - 𝑎_{𝑗})}{\prod_{𝑘 = 1}^{𝑛} (𝑧 - 𝑏_{𝑘})},

where $𝑔$ is holomorphic and non-vanishing on $\overline{𝐷 (0, 𝑟)}$ . Since

log | 𝑓 (0) | = log | 𝑔 (0) | + \sum_{𝑗 = 1}^{𝑚} log | 𝑎_{𝑗} | - \sum_{𝑘 = 1}^{𝑛} log | 𝑏_{𝑘} |,

by Lemma 4.4.4.1 $Lemma 4.4.4.1 Lemma 4.4.4.1$ on $𝑔$ ,

\begin{aligned} log | 𝑓 (0) | & = \frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (𝑟 e^{i 𝜃}) \frac{\prod_{𝑘 = 1}^{𝑛} (𝑟 e^{i 𝜃} - 𝑏_{𝑘})}{\prod_{𝑗 = 1}^{𝑚} (𝑟 e^{i 𝜃} - 𝑎_{𝑗})} | d 𝜃 \\ \begin{matrix}  \end{matrix} + \sum_{𝑗 = 1}^{𝑚} log | 𝑎_{𝑗} | - \sum_{𝑘 = 1}^{𝑛} log | 𝑏_{𝑘} | \\ = \frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (𝑟 e^{i 𝜃}) | d 𝜃 + \sum_{𝑗 = 1}^{𝑚} log | 𝑎_{𝑗} | - \sum_{𝑘 = 1}^{𝑛} log | 𝑏_{𝑘} | \\ \begin{matrix}  \end{matrix} + \frac{1}{2 π} \int_{0}^{2 π} [\sum_{𝑘 = 1}^{𝑛} (log 𝑟 + log | 1 - \frac{𝑏_{𝑘}}{𝑟 e^{i 𝜃}} |) \\ \begin{matrix}  \end{matrix} - \sum_{𝑗 = 1}^{𝑚} (log 𝑟 + log | 1 - \frac{𝑎_{𝑗}}{𝑟 e^{i 𝜃}} |)] d 𝜃 \\ = \frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (𝑟 e^{i 𝜃}) | d 𝜃 + \sum_{𝑗 = 1}^{𝑚} log | \frac{𝑎_{𝑗}}{𝑟} | - \sum_{𝑘 = 1}^{𝑛} log | \frac{𝑏_{𝑘}}{𝑟} | \\ \begin{matrix}  \end{matrix} + ℜ𝔢 \frac{1}{2 π i} [(\sum_{𝑗 = 1}^{𝑚} ∲_{𝜕 𝐷 (0, | 𝑎_{𝑗} / 𝑟 |)} - \sum_{𝑘 = 1}^{𝑛} ∲_{𝜕 𝐷 (0, | 𝑏_{𝑘} / 𝑟 |)}) \frac{Log (1 - 𝑧) d 𝑧}{𝑧}] \\ = \frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (𝑟 e^{i 𝜃}) | d 𝜃 + \sum_{𝑗 = 1}^{𝑚} log | \frac{𝑎_{𝑗}}{𝑟} | - \sum_{𝑘 = 1}^{𝑛} log | \frac{𝑏_{𝑘}}{𝑟} | \\ \begin{matrix}  \end{matrix} + ℜ𝔢 \frac{1}{2 π i} (\sum_{𝑘 = 𝑛_{0} + 1}^{𝑛} - \sum_{𝑗 = 𝑚_{0} + 1}^{𝑚}) ∳_{𝜕 𝔻} \frac{Log (1 - 𝑧) d 𝑧}{𝑧} \end{aligned}

where $𝑧 = 𝑎_{𝑗} / (𝑟 e^{i 𝜃}), 𝑏_{𝑘} / (𝑟 e^{i 𝜃})$ , $d 𝜃 = i d 𝑧 / 𝑧$ , and the leftover integrals (up until $𝑘 = 𝑛_{0}$ and $𝑗 = 𝑚_{0}$ ) for interior points vanish by Cauchy–Goursat (Theorem 3.1.9 $Theorem 3.1.9 Theorem 3.1.9$ ), since $\frac{Log (1 - 𝑧)}{𝑧}$ has a removable singularity at $𝑧 = 0$ .

We now are left to prove that the remaining integral $𝐼$ vanishes as well, which is not as immediate since the integrand does not extend continuously to the boundary. Let $𝑧 = e^{i 𝜃}$ , $d 𝑧 = i e^{i 𝜃}$ , then (by $𝜓 = \frac{𝜃}{2}$ )

\begin{aligned} 𝐼 & = ∳_{𝜕 𝔻} \frac{Log (1 - 𝑧) d 𝑧}{𝑧} = \int_{0}^{2 π} log | 1 - e^{i 𝜃} | d 𝜃 = 2 \int_{0}^{π} log | e^{- i 𝜓} - e^{i 𝜓} | d 𝜓 \\ = 2 π log 2 + 2 \int_{0}^{π} log | sin 𝜓 | d 𝜓 = 2 π log 2 + 4 \int_{0}^{π / 2} log | sin 𝜓 | d 𝜓 \\ = 2 π log 2 + 4 𝐽 . \end{aligned}

Splitting at $π / 4$ and using $cos$ with a substitution for the second integral then yields

𝐽 = \int_{0}^{π / 4} log | sin 𝜓 | d 𝜓 + \int_{0}^{π / 4} log | cos 𝜓 | d 𝜓 = \int_{0}^{π / 4} log | \frac{1}{2} sin 2 𝜓 | d 𝜓 .

Changing back to $𝜃 = 2 𝜓$ , we have

𝐽 = \frac{𝐽}{2} - \frac{π}{4} log 2 ⟹ 4 𝐽 = - 2 π log 2 ⟹ 𝐼 = 0 . □

As an immediate consequence, we have:

Corollary 4.4.4.2.1 (Jensen’s Inequality).

Let $𝑓$ be holomorphic on $\overline{𝐷 (0, 𝑟)}$ such that $𝑓 ≢ 0$ and $𝑓 (0) \neq 0$ . It follows that

log | 𝑓 (0) | \leq \frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (𝑟 e^{i 𝜃}) | d 𝜃 .

Theorem 4.4.4.3 (Poisson–Jensen Formula).

Suppose $𝑓$ is a meromorphic function on $\overline{𝐷 (0, 𝑟)}$ such that $𝑓 ≢ 0$ on $𝐷 (0, 𝑟)$ and is non-vanishing and non-infinity on $𝜕 𝐷 (0, 𝑟)$ . Let $𝑎_{1}, \dots, 𝑎_{𝑚}$ and $𝑏_{1}, \dots, 𝑏_{𝑛}$ be the zeros and poles of $𝑓$ in $𝐷 (0, 𝑟)$ , counted with multiplicity and order, respectively (multiplicities and orders count as multiple zeros or poles). Then it follows that

\begin{aligned} log | 𝑓 (𝑧) | & = \int_{0}^{2 π} log | 𝑓 (𝜁) | 𝑃 (𝜁, 𝑧) d 𝜃 \\ \begin{matrix}  \end{matrix} + \sum_{𝑗 = 1}^{𝑚} log | \frac{𝑟 (𝑧 - 𝑎_{𝑗})}{𝑟^{2} - \overline{𝑎_{𝑗}} 𝑧} | - \sum_{𝑘 = 1}^{𝑛} log | \frac{𝑟 (𝑧 - 𝑏_{𝑘})}{𝑟^{2} - \overline{𝑏_{𝑘}} 𝑧} |, (4.4.4 . 1) \end{aligned}

where

𝜁 = 𝑟 e^{i 𝜃}

𝑧 \in 𝐷 (0, 𝑟) ∖ ({𝑎_{𝑗}}_{𝑗 = 1}^{𝑚} \cup {𝑏_{𝑘}}_{𝑗 = 1}^{𝑛})

, and

𝑃 (𝜁, 𝑧)

is the Poisson kernel in (3.6.6)

(3.6.6) (3.6.6)

Proof.

For fixed $𝑧 \in 𝐷 (0, 𝑟)$ not at zeros or poles, let

𝑔_{𝑧} (𝜁) = 𝑓 (𝑟 𝜑_{- \frac{𝑧}{𝑟}} (\frac{𝜁}{𝑟}))

where $𝜑_{- 𝑎} = {(𝜑_{𝑎})}^{- 1}$ is the unit disk automorphism sending 0 to $𝑎$ . Then $𝑔_{𝑧}$ maps 0 to $𝑓 (𝑧)$ , and has zeros at $𝑟 𝜑_{- \frac{𝑧}{𝑟}} (\frac{𝜁}{𝑟}) = 𝑎_{𝑘}$ or $𝜁 = 𝑟 𝜑_{\frac{𝑧}{𝑟}} (\frac{𝑎_{𝑘}}{𝑟})$ and poles at $𝑟 𝜑_{- \frac{𝑧}{𝑟}} (\frac{𝜁}{𝑟}) = 𝑏_{𝑘}$ or $𝜁 = 𝑟 𝜑_{\frac{𝑧}{𝑟}} (\frac{𝑏_{𝑘}}{𝑟})$ . By Jensen’s formula (Theorem 4.4.4.2 $Theorem 4.4.4.2 Theorem 4.4.4.2$ ),

\begin{aligned} log | 𝑔_{𝑧} (0) | & = log | 𝑓 (𝑧) | = \frac{1}{2 π} \int_{0}^{2 π} log | 𝑔_{𝑧} (𝑟 e^{i 𝜃}) | d 𝜃 \\ \begin{matrix}  \end{matrix} + \sum_{𝑘 = 1}^{𝑚} | 𝜑_{\frac{𝑧}{𝑟}} (\frac{𝑎_{𝑘}}{𝑟}) | - \sum_{𝑘 = 1}^{𝑛} | 𝜑_{\frac{𝑧}{𝑟}} (\frac{𝑏_{𝑘}}{𝑟}) | . □ \end{aligned}

Lemma 4.4.4.4.

Let $𝑓 : 𝔻 \to ℂ$ be a non-constant bounded holomorphic function whose zeros are $𝑎_{1}, 𝑎_{2}, \dots$ , counted according to their multiplicities, ordered such that $| 𝑎_{𝑛} | \leq | 𝑎_{𝑛 + 1} |$ for all $𝑛 \in ℕ$ . Then,

\sum_{𝑛 = 1}^{\infty} (1 - | 𝑎_{𝑛} |)

is convergent.

Proof.

First assume $𝑓 (0) \neq 0$ and choose $𝑀$ such that $| 𝑓 | \leq 𝑀$ on $𝔻$ . Let $𝑛 (𝑟, 0, 𝑓)$ count the number of zeros of $𝑓$ , according to multiplicities, inside $\overline{𝐷 (0, 𝑟)}$ . By Jensen’s Formula (Theorem 4.4.4.2 $Theorem 4.4.4.2 Theorem 4.4.4.2$ ), we have

\begin{aligned} log | 𝑓 (0) | & = \frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (𝑟 e^{i 𝜃}) | d 𝜃 \\ \begin{matrix}  \end{matrix} + \sum_{𝑘 = 1}^{𝑛 (𝑟, 0, 𝑓)} log | \frac{𝑎_{𝑘}}{𝑟} | \leq log (𝑀) + \sum_{𝑘 = 1}^{𝑛 (𝑟, 0, 𝑓)} log | \frac{𝑎_{𝑘}}{𝑟} | . \end{aligned}

For any fixed positive integer $𝑘$ , choose $𝑟$ such that $| 𝑎_{𝑘} | < 𝑟 < 1$ . Then $𝑛 (𝑟, 0, 𝑓) \geq 𝑘$ and

\sum_{𝑗 = 1}^{𝑛 (𝑟, 0, 𝑓)} log | \frac{𝑎_{𝑗}}{𝑟} | \leq \sum_{𝑗 = 1}^{𝑘} log | \frac{𝑎_{𝑗}}{𝑟} |,

since each $log | \frac{𝑎_{𝑗}}{𝑟} | < 0$ for $𝑗 \leq 𝑛 (𝑟, 0, 𝑓)$ . Therefore,

log | 𝑓 (0) | \leq log 𝑀 + \sum_{𝑗 = 1}^{𝑘} log | \frac{𝑎_{𝑗}}{𝑟} | = log 𝑀 + \sum_{𝑗 = 1}^{𝑘} log | 𝑎_{𝑗} | - 𝑘 log 𝑟 .

Rearranging,

\sum_{𝑗 = 1}^{𝑘} log | 𝑎_{𝑗} | \geq log | 𝑓 (0) | - log 𝑀 + 𝑘 log 𝑟 .

Now let $𝑟 \to 1^{-}$ with $𝑟 > | 𝑎_{𝑘} |$ . Since $𝑘 log 𝑟 \to 0$ , it follows that

\sum_{𝑗 = 1}^{𝑘} log | 𝑎_{𝑗} | \geq log | 𝑓 (0) | - log 𝑀 .

This holds for every $𝑘$ . Since $log | 𝑎_{𝑗} | < 0$ for all $𝑗$ , the partial sums $\sum_{𝑗 = 1}^{𝑘} log | 𝑎_{𝑗} |$ are decreasing and bounded below by $log | 𝑓 (0) | - log 𝑀$ , hence converge to some finite limit, and

\sum_{𝑗 = 1}^{\infty} log | 𝑎_{𝑗} | \geq log | 𝑓 (0) | - log 𝑀,

or equivalently,

log | 𝑓 (0) | \leq log | 𝑀 | + \sum_{𝑘 = 1}^{\infty} log | 𝑎_{𝑘} | .

For any $0 < 𝑎 < 1$ , we have $- log (𝑎) = 1 - 𝑎 + \sum_{𝑛 = 2}^{\infty} {(1 - 𝑎)}^{𝑛} 𝑎^{𝑛} > 1 - 𝑎$ . Hence,

0 \leq \sum_{𝑘 = 1}^{\infty} (1 - | 𝑎_{𝑘} |) < - \sum_{𝑘 = 1}^{\infty} log | 𝑎_{𝑘} | \leq log | 𝑀 | - log | 𝑓 (0) | .

𝑓

has a zero of multiplicity

𝑚

0

, then the argument applies to

𝑧 \mapsto \frac{𝑓 (𝑧)}{𝑧^{𝑚}}

. ⁠

□

Theorem 4.4.4.5 (Blaschke Product).

Let ${𝑎_{𝑘}}_{𝑘 \in ℕ} \subset 𝔻^{*} = 𝔻 ∖ {0}$ be a sequence such that the series $\sum_{𝑘 = 1}^{\infty} (1 - | 𝑎_{𝑘} |)$ is convergent (known as the Blaschke condition). Then the Blaschke product, defined by

𝐵 (𝑧) = \prod_{𝑘 = 1}^{\infty} [- \frac{| 𝑎_{𝑘} |}{𝑎_{𝑘}} 𝜑_{𝑎_{𝑘}} (𝑧)], (4.4.4 . 2)

(where

𝜑_{𝑎} (𝑧)

is a Möbius transformation in the form of (3.5.1)

(3.5.1) (3.5.1)

), locally uniformly converges to an analytic function on

𝔻

such that

| 𝐵 | \leq 1

𝔻

, and its only zeros are precisely at each of

{𝑎_{𝑘}}_{𝑘 \in ℕ}

, counted according to multiplicities.

Proof.

If it can be shown that

\sum_{𝑘 = 1}^{\infty} | \frac{| 𝑎_{𝑘} |}{𝑎_{𝑘}} \frac{𝑎_{𝑘} - 𝑧}{1 - \overline{𝑎_{𝑘}} 𝑧} - 1 |

locally uniformly converges, we can use Lemma 4.4.3.3 $Lemma 4.4.3.3 Lemma 4.4.3.3$ to show that the infinite product converges uniformly on compact subsets of $𝔻$ . Let $\overline{𝐷 (0, 𝑟)} \subset 𝔻$ be a compact subset. The summand can be bounded with

\begin{aligned} | \frac{| 𝑎_{𝑘} |}{𝑎_{𝑘}} \frac{𝑎_{𝑘} - 𝑧}{1 - \overline{𝑎_{𝑘}} 𝑧} - 1 | & = | \frac{\overline{𝑎_{𝑘}}}{| 𝑎_{𝑘} |} \frac{𝑎_{𝑘} - 𝑧}{1 - \overline{𝑎_{𝑘}} 𝑧} - 1 | = | \frac{{| 𝑎_{𝑘} |}^{2} - \overline{𝑎_{𝑘}} 𝑧}{| 𝑎_{𝑘} | (1 - \overline{𝑎_{𝑘}} 𝑧)} - 1 | \\ = | \frac{{| 𝑎_{𝑘} |}^{2} - \overline{𝑎_{𝑘}} 𝑧 - | 𝑎_{𝑘} | + | 𝑎_{𝑘} | \overline{𝑎_{𝑘}} 𝑧}{| 𝑎_{𝑘} | (1 - \overline{𝑎_{𝑘}} 𝑧)} | \\ = | \frac{\overline{𝑎_{𝑘}} 𝑧 (| 𝑎_{𝑘} | - 1) + | 𝑎_{𝑘} | (| 𝑎_{𝑘} | - 1)}{| 𝑎_{𝑘} | (1 - \overline{𝑎_{𝑘}} 𝑧)} | \\ = | \frac{(\overline{𝑎_{𝑘}} 𝑧 + | 𝑎_{𝑘} |) (1 - | 𝑎_{𝑘} |)}{| 𝑎_{𝑘} | (1 - \overline{𝑎_{𝑘}} 𝑧)} | \\ \leq (1 - | 𝑎_{𝑘} |) \frac{| \overline{𝑎_{𝑘}} | (1 + 𝑟)}{| 𝑎_{𝑘} | (1 - | 𝑎_{𝑘} | 𝑟)} < (1 - | 𝑎_{𝑘} |) \frac{1 + 𝑟}{1 - 𝑟} . \end{aligned}

Since

\sum_{𝑘 = 1}^{\infty} | \frac{| 𝑎_{𝑘} |}{𝑎_{𝑘}} \frac{𝑎_{𝑘} - 𝑧}{1 - \overline{𝑎_{𝑘}} 𝑧} - 1 | < \frac{1 + 𝑟}{1 - 𝑟} \sum_{𝑘 = 1}^{\infty} (1 - | 𝑎_{𝑘} |)

is convergent (Blaschke condition), by the Weierstrass $𝑀$ -Test (Theorem 2.3.3 $Theorem 2.3.3 Theorem 2.3.3$ ), $\sum_{𝑘 = 1}^{\infty} | \frac{| 𝑎_{𝑘} |}{𝑎_{𝑘}} \frac{𝑎_{𝑘} - 𝑧}{1 - \overline{𝑎_{𝑘}} 𝑧} - 1 |$ converges uniformly on $\overline{𝐷 (0, 𝑟)}$ . By Lemma 4.4.3.3 $Lemma 4.4.3.3 Lemma 4.4.3.3$ , the infinite product in (4.4.4.2) $(4.4.4.13) (4.4.4.2)$ converges uniformly on compact subsets of $𝔻$ . The properties of its zeros follow from the lemma.

Lastly, since $| 𝜑_{𝑎_{𝑘}} | \leq 1$ and each partial product is bounded by 1, it follows that $| 𝐵 (𝑧) | \leq 1$ on $𝔻$ . ⁠ $□$

Remark.

A more general Blaschke product has an additional factor of

𝑧^{𝑚}

to account for a zero at the origin, similar to the case of the Weierstrass product.

Corollary 4.4.4.5.1.

Let $𝑓 : 𝔻 \to ℂ$ be bounded and holomorphic whose multiplicity of the zero at 0 is $𝑚$ (if $𝑓$ does not vanish at 0, then $𝑚 = 0$ ). If ${𝑎_{𝑛}}_{𝑛 \in ℕ}$ are its zeros in $𝔻^{*}$ , counting multiplicities, then

𝑓 (𝑧) = 𝐹 (𝑧) 𝑧^{𝑚} \prod_{𝑛 = 1}^{\infty} [- \frac{| 𝑎_{𝑛} |}{𝑎_{𝑛}} 𝜑_{𝑎_{𝑛}} (𝑧)],

where $𝐹$ is bounded, holomorphic, and non-vanishing on $𝔻$ . Moreover,

sup_{𝑧 \in 𝔻} | 𝑓 (𝑧) | = sup_{𝑧 \in 𝔻} | 𝐹 (𝑧) | .

Proof.

Let

𝐹 (𝑧) = \frac{𝑓 (𝑧)}{𝑧^{𝑚} \prod_{𝑛 = 1}^{\infty} [- \frac{| 𝑎_{𝑘} |}{𝑎_{𝑘}} 𝜑_{𝑎_{𝑛}} (𝑧)]},

which is well-defined by Lemma 4.4.4.4 $Lemma 4.4.4.4 Lemma 4.4.4.4$ and Theorem 4.4.4.5 $Theorem 4.4.4.5 Theorem 4.4.4.5$ . By construction, $𝐹$ extends to its removable singularities to a holomorphic function that does not vanish. Because

sup_{𝑧 \in 𝔻} | 𝑧^{𝑚} \prod_{𝑛 = 1}^{\infty} [- \frac{| 𝑎_{𝑛} |}{𝑎_{𝑛}} 𝜑_{𝑎_{𝑛}} (𝑧)] | \leq 1,

it follows that

sup_{𝑧 \in 𝔻} | 𝐹 (𝑧) | \geq sup_{𝑧 \in 𝔻} | 𝑓 (𝑧) | . (4.4.4 . 3)

The partial products

𝐵_{𝑛} (𝑧) = \prod_{𝑘 = 1}^{𝑛} [- \frac{| 𝑎_{𝑘} |}{𝑎_{𝑘}} 𝜑_{𝑎_{𝑘}} (𝑧)]

give for fixed $𝜃 \in ℝ$ , $𝜀 > 0$ , the existence of $0 < 𝑟^{'} < 1$ such that $𝑟^{'} < 𝑟 < 1$ implies

| 𝐵_{𝑛} (𝑟 e^{i 𝜃}) | > 1 - 𝜀 .

Then by the Maximum Modulus Principle (Theorem 3.4.2 $Theorem 3.4.2 Theorem 3.4.2$ ),

\begin{aligned} sup_{𝑧 \in 𝔻} | \frac{𝑓 (𝑧)}{𝑧^{𝑚} 𝐵_{𝑛} (𝑧)} | & = sup_{𝑧 \in 𝔻 ∖ \overline{𝐷 (0, 𝑟)}} | \frac{𝑓 (𝑧)}{𝑧^{𝑚} 𝐵_{𝑛} (𝑧)} | \\ \leq \frac{1}{𝑟^{𝑚} (1 - 𝜀)} sup_{𝑧 \in 𝔻} | 𝑓 (𝑧) | \to \frac{1}{1 - 𝜀} sup_{𝑧 \in 𝔻} | 𝑓 (𝑧) | \end{aligned}

as $𝑟 \to 1^{-}$ . Letting $𝜀 \to 0^{+}$ , $𝑛 \to \infty$ gives

sup_{𝑧 \in 𝔻} | 𝐹 (𝑧) | \leq sup_{𝑧 \in 𝔻} | 𝑓 (𝑧) |,

which in conjunction with (4.4.4.3)

(4.4.4.14) (4.4.4.3)

, completes the final assertion. ⁠

□

From the results above, a recurring theme in complex analysis is hinted at; the rate of growth of functions provides insight towards the distribution of its zeros.

The subjects to be discussed here are relevant and preliminary to Nevanlinna theory, or the study of holomorphic value distribution.

For an entire function $𝑓$ , let $𝑀 (𝑟, 𝑓) = {sup}_{| 𝑧 | = 𝑟} | 𝑓 (𝑧) | = {sup}_{| 𝑧 | \leq 𝑟} | 𝑓 (𝑧) |$ (by the Maximum Modulus Principle in Theorem 3.4.2 $Theorem 3.4.2 Theorem 3.4.2$ ).

Definition 4.4.4.6 (Growth Order of Entire Functions).

An entire function $𝑓$ is said to be of finite order if there exists $𝛼, 𝑟_{𝛼} > 0$ such that

𝑀 (𝑟, 𝑓) \leq exp (𝑟^{𝛼}), \forall 𝑟 > 𝑟_{𝛼},

or in loose terms,

𝑓

is of finite order if it grows at most exponentially for large

𝑧

. The order of

𝑓

, or

𝜌 (𝑓)

is defined to be the infimum of all

𝛼

satisfying the previous condition.

Proposition 4.4.4.7.

Let $𝑓$ be entire; then if there exist $𝑎, 𝑏, 𝛼, 𝑟_{𝛼, 𝑏} > 0$ such that

𝑀 (𝑟, 𝑓) \leq exp (𝑎 𝑟^{𝛼} + 𝑏), \forall 𝑟 > 𝑟_{𝛼, 𝑏},

then

𝜌 (𝑓) \leq 𝛼

Proof.

For $𝜀 > 0$ , since $𝑟^{𝜀} \to \infty$ as $𝑟 \to \infty$ , for any $𝜀 > 0$ , there exists $𝑟_{𝜀}$ such that

𝑟^{𝜀} \geq 2 𝑎 ⟹ \frac{1}{2} 𝑟^{𝛼 + 𝜀} \geq 𝑎 𝑟^{𝛼}

for $𝑟 > 𝑟_{𝜀}$ . There exists $𝑟_{𝜀}^{'} > 0$ such that

𝑟 > 𝑟_{𝜀}^{'} ⟹ \frac{1}{2} 𝑟^{𝛼 + 𝜀} \geq 𝑏 .

For simplicity, let the value $max {𝑟_{𝜀}, 𝑟_{𝜀}^{'}}$ be denoted by $𝑟_{𝜀}$ . Then

𝑟 > 𝑟_{𝜀} ⟹ 𝑎 𝑟^{𝛼} + 𝑏 \leq \frac{1}{2} 𝑟^{𝛼 + 𝜀} + \frac{1}{2} 𝑟^{𝛼 + 𝜀} = 𝑟^{𝛼 + 𝜀} .

By assumption, we have

𝑀 (𝑟, 𝑓) \leq exp (𝑎 𝑟^{𝛼} + 𝑏) \leq exp (𝑟^{𝛼 + 𝜀}) ⟹ 𝛼 + 𝜀 \geq 𝜌 (𝑓) .

Letting

𝜀 \to 0^{+}

, the assertion follows. ⁠

□

Theorem 4.4.4.8.

The order of a non-constant entire function $𝑓$ may be explicitly given by

𝜌 (𝑓) = \underset{𝑟 \to \infty}{lim sup} \frac{log (log 𝑀 (𝑟, 𝑓))}{log 𝑟} .

Proof.

By definition, $𝜌 (𝑓)$ is the infimum of all $𝛼 > 0$ for which there exists $𝑟_{𝛼} > 0$ such that

𝑀 (𝑟, 𝑓) \leq exp (𝑟^{𝛼}), \forall 𝑟 > 𝑟_{𝛼} .

Let

𝐿 = \underset{𝑟 \to \infty}{lim sup} \frac{log (log 𝑀 (𝑟, 𝑓))}{log 𝑟} .

First suppose $𝜌 (𝑓) < \infty$ . Fix $𝜀 > 0$ . Since $𝜌 (𝑓)$ is the infimum, there is an $𝛼$ satisfying this condition such that $𝜌 (𝑓) \leq 𝛼 < 𝜌 (𝑓) + 𝜀$ .

Hence, for all sufficiently large $𝑟$ ,

𝑀 (𝑟, 𝑓) \leq exp (𝑟^{𝛼}) \leq exp (𝑟^{𝜌 (𝑓) + 𝜀}),

and hence $𝐿 \leq 𝜌 (𝑓) + 𝜀$ . Letting $𝜀 \to 0^{+}$ gives $𝐿 \leq 𝜌 (𝑓)$ .

Conversely, suppose $0 < 𝜀 < 𝜌 (𝑓)$ and let $𝛽 = 𝜌 (𝑓) - 𝜀$ . Since $𝛽 < 𝜌 (𝑓)$ , this condition fails for $𝛽$ . Thus, for every $𝑅 > 0$ , there exists $𝑟 > 𝑅$ such that

𝑀 (𝑟, 𝑓) > exp (𝑟^{𝛽}) = exp (𝑟^{𝜌 (𝑓) - 𝜀}) .

Thus this inequality holds along a sequence $𝑟_{𝑗} \to \infty$ , so $𝐿 \geq 𝜌 (𝑓) - 𝜀$ . Letting $𝜀 \to 0^{+}$ gives $𝐿 \geq 𝜌 (𝑓)$ .

If $𝜌 (𝑓) = 0$ , then $𝐿 \geq 0$ because $𝑀 (𝑟, 𝑓) \to \infty$ for a non-constant entire function, while the first half gives $𝐿 \leq 0$ . Therefore $𝐿 = 𝜌 (𝑓)$ .

Finally, if $𝜌 (𝑓) = \infty$ , then this condition fails for every $𝛽 > 0$ . The same argument gives $𝐿 \geq 𝛽$ for every $𝛽 > 0$ , and hence $𝐿 = \infty = 𝜌 (𝑓)$ . ⁠ $□$

Example 4.4.4.9.

The function

sin

is of order 1, while

exp ⚬ exp

is not of finite order.

Solution to Example 4.4.4.9 $Example 4.4.4.9 Example 4.4.4.9$ .

We consider the two examples separately:

Observe that
$sup_{| 𝑧 | = 𝑟} | sin (𝑧) | \leq sup_{| 𝑧 | = 𝑟} \frac{| e^{i 𝑧} | + | e^{- i 𝑧} |}{2} = sup_{| 𝑧 | = 𝑟} \frac{e^{| 𝑦 |} + e^{- | 𝑦 |}}{2} \leq sup_{| 𝑧 | = 𝑟} e^{| 𝑦 |} = e^{𝑟} .$
For $𝑟 > 1$ , we have $e^{- 𝑟} < 1 < \frac{1}{2} e^{𝑟}$ , and hence for $𝑧 = i 𝑟$ , we have
$| sin (𝑧) | = \frac{e^{𝑟} - e^{- 𝑟}}{2} > \frac{1}{4} e^{𝑟} .$
Therefore,
$\frac{1}{4} e^{𝑟} < sup_{| 𝑧 | = 𝑟} | sin (𝑧) | \leq e^{𝑟} ⟹ 𝜌 (𝑓) = \underset{𝑟 \to \infty}{lim sup} \frac{log (𝑟 + 𝒪︀ (1))}{log 𝑟} = 1 .$
Let $𝑧 = 𝑟$ , then
$\begin{aligned} sup_{| 𝑧 | = 𝑟} | exp ⚬ exp | \geq exp (exp (𝑟)) & ⟹ log ⚬ log sup_{| 𝑧 | = 𝑟} | 𝑓 (𝑧) | \geq 𝑟 \\ ⟹ 𝜌 (𝑓) \geq \underset{𝑟 \to \infty}{lim sup} \frac{𝑟}{log 𝑟} = \infty . □ \end{aligned}$

The utility of $𝜌$ is that it gives implications on the rate of which the zeros of an entire function tend to $\infty$ . The order for meromorphic functions is more general and is pertinent in Nevanlinna Theory (Section 14 $Section 14 Section 14$ ). This is quantified technically by the convergence range of the sum given by

\sum_{𝑛 = 1}^{\infty} \frac{1}{{| 𝑎_{𝑛} |}^{𝑘 + 1}},

where each $𝑎_{𝑛}$ is a zero. Specifically, the infimum of all such $𝑘$ under which the prescribed sum converges correlates to this right. For example, let $𝑎_{𝑛} = 𝑛$ for each $𝑛$ . Then for any $𝑘 > 0$ , the integral test gives the convergence of the series, while if $𝑎_{𝑛} = \sqrt{𝑛}$ (corresponding to a slower approach to $\infty$ ), the series converges for $𝑘 > 1$ .

For the following discussions, let $𝑛 (𝑟, 0, 𝑓)$ count the zeros of $𝑓$ in $𝐷 (0, 𝑟)$ according to multiplicity.

Lemma 4.4.4.10.

If $𝑓$ is entire with $𝑓 (0) = 1$ , then

log 2 \cdot 𝑛 (𝑟, 0, 𝑓) \leq log 𝑀 (2 𝑟, 𝑓) .

Proof.

By Jensen’s formula (Theorem 4.4.4.2 $Theorem 4.4.4.2 Theorem 4.4.4.2$ ), for $𝑟 > 0$ , we have

\sum_{𝑘 = 1}^{𝑛 (2 𝑟, 0, 𝑓)} log | \frac{2 𝑟}{𝑎_{𝑘}} | = \frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (2 𝑟 e^{i 𝜃}) | d 𝜃,

where $𝑎_{1}, \dots, 𝑎_{𝑛 (2 𝑟, 0, 𝑓)}$ are the zeros of $𝑓$ in $𝐷 (0, 2 𝑟)$ , ordered such that each $| 𝑎_{𝑘} | \leq | 𝑎_{𝑘 + 1} |$ . Then

\begin{aligned} \sum_{𝑘 = 1}^{𝑛 (𝑟, 0, 𝑓)} log 2 & \leq \sum_{𝑘 = 1}^{𝑛 (𝑟, 0, 𝑓)} log | \frac{2 𝑟}{𝑎_{𝑘}} | \leq \sum_{𝑘 = 1}^{𝑛 (2 𝑟, 0, 𝑓)} log | \frac{2 𝑟}{𝑎_{𝑘}} | \\ = \frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (2 𝑟 e^{i 𝜃}) | d 𝜃 \leq log 𝑀 (2 𝑟, 𝑓) . □ \end{aligned}

Theorem 4.4.4.11.

For a nonzero complex sequence ${𝑎_{𝑘}}_{𝑘 \in ℕ}$ counting multiplicities (such that $| 𝑎_{1} | \leq | 𝑎_{2} |$ , etc.), the sum

\sum_{𝑘 = 1}^{\infty} \frac{1}{{| 𝑎_{𝑘} |}^{𝜎}}

converges for any

𝜎 > \underset{𝑟 \to \infty}{lim sup} \frac{log 𝑛 (𝑟)}{log 𝑟}

where

𝑛 (𝑟)

counts

𝑎_{𝑘}

in the closed disk of radius

𝑟

Proof.

Choose $𝜎^{'}$ such that

𝜎 > 𝜎^{'} > \underset{𝑟 \to \infty}{lim sup} \frac{log 𝑛 (𝑟)}{log 𝑟} .

For sufficiently large $𝑟$ ,

\frac{log 𝑛 (𝑟)}{log 𝑟} < 𝜎^{'} ⟹ 𝑛 (𝑟) \leq 𝑟^{𝜎^{'}} .

For sufficiently large $𝑘 \in ℕ$ , by the ordering of zeros, it follows that

𝑘 \leq 𝑛 (| 𝑎_{𝑘} | + 𝛿) \leq {(| 𝑎_{𝑘} | + 𝛿)}^{𝜎^{'}}

for sufficiently small $𝛿$ . As $𝛿 \to 0^{+}$ , we have

𝑘 \leq {| 𝑎_{𝑘} |}^{𝜎^{'}} ⟹ \frac{1}{𝑘} \geq \frac{1}{{| 𝑎_{𝑘} |}^{𝜎^{'}}} ⟹ \frac{1}{𝑘^{𝜎 / 𝜎^{'}}} \geq \frac{1}{{| 𝑎_{𝑘} |}^{𝜎}} .

By the comparison test, we then have the convergence of

\sum_{𝑘 = 1}^{\infty} \frac{1}{{| 𝑎_{𝑘} |}^{𝜎}} . □

Theorem 4.4.4.12.

For an entire function $𝑓$ ( $𝑓 (0) = 1$ ) of finite order $𝜌 (𝑓)$ whose zeros are at ${𝑎_{𝑘}}_{𝑘 \in ℕ}$ counting multiplicities (such that $| 𝑎_{1} | \leq | 𝑎_{2} |$ , etc.), the sum

\sum_{𝑘 = 1}^{\infty} \frac{1}{{| 𝑎_{𝑘} |}^{𝜌 (𝑓) + 𝜂}}

converges for any

𝜂 > 0

Proof.

For every $𝜀 > 0$ and all sufficiently large $𝑟$ , the definition of order gives

𝑀 (2 𝑟, 𝑓) \leq exp ({(2 𝑟)}^{𝜌 (𝑓) + 𝜀}) .

Also, Lemma 4.4.4.10 $Lemma 4.4.4.10 Lemma 4.4.4.10$ gives

log 2 \cdot 𝑛 (𝑟, 0, 𝑓) \leq log 𝑀 (2 𝑟, 𝑓) .

Hence,

log 2 \cdot 𝑛 (𝑟, 0, 𝑓) \leq {(2 𝑟)}^{𝜌 (𝑓) + 𝜀} ⟺ \frac{𝑛 (𝑟, 0, 𝑓)}{𝑟^{𝜌 (𝑓) + 2 𝜀}} \leq \frac{1}{log 2} 2^{𝜌 (𝑓) + 𝜀} 𝑟^{- 𝜀} \to 0^{+}

as $𝑟 \to \infty$ . Hence, for sufficiently large $𝑟$ ,

𝑛 (𝑟, 0, 𝑓) \leq 𝑟^{𝜌 (𝑓) + 2 𝜀} .

For sufficiently large $𝑘 \in ℕ$ , by the ordering of zeros, it follows that

𝑘 \leq 𝑛 (| 𝑎_{𝑘} | + 𝛿, 0, 𝑓) \leq {(| 𝑎_{𝑘} | + 𝛿)}^{𝜌 (𝑓) + 2 𝜀}

for sufficiently small $𝛿$ . As $𝛿 \to 0^{+}$ , we have

𝑘 \leq {| 𝑎_{𝑘} |}^{𝜌 (𝑓) + 2 𝜀} ⟹ \frac{1}{𝑘} \geq \frac{1}{{| 𝑎_{𝑘} |}^{𝜌 (𝑓) + 2 𝜀}} ⟹ \frac{1}{𝑘^{(𝜌 (𝑓) + 𝜂) / (𝜌 (𝑓) + 2 𝜀)}} \geq \frac{1}{{| 𝑎_{𝑘} |}^{𝜌 (𝑓) + 𝜂}} .

The left-hand side as a summation is convergent for $2 𝜀 < 𝜂$ or lower, and hence we have the convergence of

\sum_{𝑘 = 1}^{\infty} \frac{1}{{| 𝑎_{𝑘} |}^{𝜌 (𝑓) + 𝜂}} . □

Therefore, for any $𝑟 > 0$ , the series

\sum_{𝑘 = 1}^{\infty} {| \frac{𝑟}{𝑎_{𝑘}} |}^{⌊ 𝜌 (𝑓) ⌋ + 1}

converges, because $⌊ 𝜌 (𝑓) ⌋ + 1 > 𝜌 (𝑓)$ and Theorem 4.4.4.12 $Theorem 4.4.4.12 Theorem 4.4.4.12$ applies with $𝜂 = ⌊ 𝜌 (𝑓) ⌋ + 1 - 𝜌 (𝑓)$ . Then by the Weierstrass Factorization Theorem (Theorem 4.4.3.5 $Theorem 4.4.3.5 Theorem 4.4.3.5$ ),

𝑓 (𝑧) = 𝑧^{𝑚} e^{𝜑 (𝑧)} \prod_{𝑘 = 1}^{\infty} 𝐸_{⌊ 𝜌 (𝑓) ⌋} (\frac{𝑧}{𝑎_{𝑘}})

locally uniformly converges on $ℂ$ , where $𝜑$ is entire.

Definition 4.4.4.13.

The rank of an entire function is the smallest $𝑝 \in ℤ_{\geq 0}$ for which the associated sum

\sum_{𝑘 = 1}^{\infty} \frac{1}{{| 𝑎_{𝑘} |}^{𝑝 + 1}}

converges, where

{𝑎_{𝑘}}_{𝑘}

are its zeros in

ℂ^{*}

The conclusion of Theorem 4.4.4.12 $Theorem 4.4.4.12 Theorem 4.4.4.12$ is that the rank of an entire function with finite order is finite. Moreover, the rank $\leq ⌊ 𝜌 ⌋$ .

Definition 4.4.4.14 (Weierstrass Canonical Factorization).

Let $𝑓$ be entire of finite rank $𝑝$ . Using this rank in the Weierstrass Factorization theorem (Theorem 4.4.3.5 $Theorem 4.4.3.5 Theorem 4.4.3.5$ ),

𝑓 (𝑧) = 𝑧^{𝑚} e^{𝜑 (𝑧)} \prod_{𝑘 = 1}^{\infty} 𝐸_{𝑝} (\frac{𝑧}{𝑎_{𝑘}}) .

𝜑

is a polynomial of degree

𝑞

, then

𝑓

is said to be of finite genus

𝜇 = max {𝑝, 𝑞}

This particular Weierstrass factorization (using the rank $𝑝$ ) is the Weierstrass canonical factorization of $𝑓$ (the portion corresponding to the product of elementary factors itself is the Weierstrass canonical product). Now that we have indulged in the implications of $𝜌 (𝑓)$ to its zero distribution, we now turn to the function $𝜑$ in the exponential.

Lemma 4.4.4.15.

Let $𝑓$ be entire with finite order such that $𝑓 (0) = 1$ . Let ${𝑎_{𝑘}}_{𝑘 \in ℕ}$ be the zeros of $𝑓$ , listed with multiplicities, such that $| 𝑎_{1} | \leq | 𝑎_{2} | \leq | 𝑎_{3} | \leq \dots$ . Suppose $𝑝 > 𝜌 (𝑓) - 1$ ; then for any $𝑧 \in ℂ$ ,

lim_{𝑟 \to \infty} \sum_{𝑘 = 1}^{𝑛 (𝑟, 0, 𝑓)} {\overline{𝑎_{𝑘}}}^{𝑝 + 1} {(𝑟^{2} - \overline{𝑎_{𝑘}} 𝑧)}^{- 𝑝 - 1} = 0 .

Proof.

For fixed $𝑧$ , let $𝑟 > 2 | 𝑧 |$ such that $𝑎_{1}, \dots, 𝑎_{𝑛 (𝑟, 0, 𝑓)}$ lie in $𝐷 (0, 𝑟)$ . For each $𝑘$ we obtain

\begin{matrix} | 𝑟^{2} - \overline{𝑎_{𝑘}} 𝑧 | \geq 𝑟^{2} - | 𝑎_{𝑘} | | 𝑧 | > 𝑟^{2} - 𝑟 \cdot \frac{𝑟}{2} = \frac{𝑟^{2}}{2} \\ ⟹ {| 𝑎_{𝑘} |}^{𝑝 + 1} {| 𝑟^{2} - \overline{𝑎_{𝑘}} 𝑧 |}^{- 𝑝 - 1} < {(\frac{2}{𝑟})}^{𝑝 + 1} \end{matrix}

since $𝜌 (𝑓) \geq 0$ by the logarithm formula. Now by definition of $𝜌 (𝑓)$ , Lemma 4.4.4.10 $Lemma 4.4.4.10 Lemma 4.4.4.10$ gives the estimate for sufficiently large $𝑟$ and arbitrarily small $𝜀 > 0$ :

𝑛 (𝑟, 0, 𝑓) 𝑟^{- 𝑝 - 1} \leq \frac{log 𝑀 (2 𝑟, 𝑓)}{log 2} 𝑟^{- 𝑝 - 1} \leq \frac{{(2 𝑟)}^{𝜌 (𝑓) + 𝜀} 𝑟^{- 𝑝 - 1}}{log 2} .

Thus,

| \sum_{𝑘 = 1}^{𝑛 (𝑟, 0, 𝑓)} {\overline{𝑎_{𝑘}}}^{𝑝 + 1} {(𝑟^{2} - \overline{𝑎_{𝑘}} 𝑧)}^{- 𝑝 - 1} | \leq 𝑛 (𝑟, 0, 𝑓) {(\frac{2}{𝑟})}^{𝑝 + 1} \leq \frac{𝑟^{𝜌 (𝑓) + 𝜀 - 𝑝 - 1} 2^{𝜌 (𝑓) + 𝜀 + 𝑝 + 1}}{log 2} .

Letting $𝜀 = \frac{𝑝 + 1 - 𝜌 (𝑓)}{2}$ (positive by theorem assumption), we obtain

| \sum_{𝑘 = 1}^{𝑛 (𝑟, 0, 𝑓)} {\overline{𝑎_{𝑘}}}^{𝑝 + 1} {(𝑟^{2} - \overline{𝑎_{𝑘}} 𝑧)}^{- 𝑝 - 1} | \leq \frac{2^{(3 𝑝 + 3 - 𝜌 (𝑓)) / 2} 𝑟^{- 𝜀}}{log 2} \to 0 as 𝑟 \to \infty . □

Theorem 4.4.4.16.

Let $𝑓$ be entire with $𝑓 (0) = 1$ . Then for $𝑝 > 𝜌 (𝑓) - 1$ ( $𝑝$ integer) and $𝑧 \in ℂ$ ,

lim_{𝑟 \to \infty} \int_{0}^{2 π} \frac{𝑟 e^{i 𝜃} log | 𝑓 (𝑟 e^{i 𝜃}) |}{{(𝑟 e^{i 𝜃} - 𝑧)}^{𝑝 + 2}} d 𝜃 = 0 .

Proof.

For fixed $𝑧$ , $𝑟 > 2 | 𝑧 |$ , we have

\int_{0}^{2 π} \frac{i 𝑟 e^{i 𝜃} d 𝜃}{{(𝑟 e^{i 𝜃} - 𝑧)}^{𝑝 + 2}} = ∳_{𝜕 𝐷 (0, 𝑟)} \frac{d 𝑤}{{(𝑤 - 𝑧)}^{𝑝 + 2}} = 2 π i \underset{𝑤 = 𝑧}{Res} \frac{1}{{(𝑤 - 𝑧)}^{𝑝 + 2}}

by the Residue Theorem (Theorem 4.5.2 $Theorem 4.5.2 Theorem 4.5.2$ ). Since $𝑝 + 2 > 𝜌 (𝑓) + 1 \geq 1$ where $𝑝$ is an integer, we must have $𝑝 + 2 \geq 2$ and thus

\int_{0}^{2 π} \frac{𝑟 e^{i 𝜃} d 𝜃}{{(𝑟 e^{i 𝜃} - 𝑧)}^{𝑝 + 2}} = 0 .

Therefore,

\begin{aligned} | \int_{0}^{2 π} \frac{𝑟 e^{i 𝜃} log | 𝑓 (𝑟 e^{i 𝜃}) |}{{(𝑟 e^{i 𝜃} - 𝑧)}^{𝑝 + 2}} d 𝜃 | \\ \begin{matrix}  \end{matrix} = | \int_{0}^{2 π} \frac{𝑟 e^{i 𝜃}}{{(𝑟 e^{i 𝜃} - 𝑧)}^{𝑝 + 2}} [log | 𝑓 (𝑟 e^{i 𝜃}) | - log 𝑀 (𝑟, 𝑓)] d 𝜃 | \\ \begin{matrix}  \end{matrix} \leq \int_{0}^{2 π} \frac{𝑟}{{(𝑟 / 2)}^{𝑝 + 2}} [log 𝑀 (𝑟, 𝑓) - log | 𝑓 (𝑟 e^{i 𝜃}) |] d 𝜃 \\ \begin{matrix}  \end{matrix} = 2^{𝑝 + 3} 𝑟^{- 𝑝 - 1} [2 π log 𝑀 (𝑟, 𝑓) - \int_{0}^{2 π} log | 𝑓 (𝑟 e^{i 𝜃}) | d 𝜃] \\ \begin{matrix}  \end{matrix} \leq 2^{𝑝 + 4} 𝑟^{- 𝑝 - 1} π log 𝑀 (𝑟, 𝑓), \end{aligned}

where the last expression uses the inequality derived from Jensen’s formula (Corollary 4.4.4.2.1 $Corollary 4.4.4.2.1 Corollary 4.4.4.2.1$ ) on the remaining integral.

Now by assumption, we have

log 𝑀 (𝑟, 𝑓) \leq 𝑟^{𝜌 (𝑓) + 𝜀}

for any $𝜀 > 0$ and sufficiently large $𝑟$ . Hence,

| \int_{0}^{2 π} \frac{𝑟 e^{i 𝜃} log | 𝑓 (𝑟 e^{i 𝜃}) |}{{(𝑟 e^{i 𝜃} - 𝑧)}^{𝑝 + 2}} d 𝜃 | \leq 2^{𝑝 + 4} 𝑟^{𝜌 (𝑓) + 𝜀 - 𝑝 - 1} π = 2^{𝑝 + 4} 𝑟^{(𝜌 (𝑓) - 𝑝 - 1) / 2} π

at $𝜀 = \frac{𝑝 + 1 - 𝜌 (𝑓)}{2}$ . Then since $\frac{𝜌 (𝑓) - 𝑝 - 1}{2} < 0$ , the expression vanishes as $𝑟 \to \infty$ . ⁠ $□$

Proposition 4.4.4.17.

Let $𝑓$ be entire, non-constant, and of finite order such that $𝑓 (0) = 1$ . Let ${𝑎_{𝑘}}_{𝑘 \in ℕ}$ be the zeros of $𝑓$ counted according to multiplicities such that $| 𝑎_{1} | \leq | 𝑎_{2} | \leq | 𝑎_{3} | \leq \dots$ . If $𝑝 > 𝜌 (𝑓) - 1$ is an integer, then

\frac{d^{𝑝}}{d 𝑧^{𝑝}} (\frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)}) \equiv - \sum_{𝑘 = 1}^{\infty} \frac{𝑝!}{{(𝑎_{𝑘} - 𝑧)}^{𝑝 + 1}}

for all

𝑧 \in ℂ ∖ {𝑎_{1}, 𝑎_{2}, \dots}

Proof.

Let $𝑟 > 2 | 𝑧 |$ . By the Poisson–Jensen Formula (Theorem 4.4.4.3 $Theorem 4.4.4.3 Theorem 4.4.4.3$ ), at each non-singular point, we have (the kernel representation derived in (3.6.6) $(3.6.6) (3.6.6)$ )

\begin{aligned} ℜ𝔢 log 𝑓 (𝑧) & = \frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (𝑟 e^{i 𝜃}) | ℜ𝔢 (\frac{𝑟 e^{i 𝜃} + 𝑧}{𝑟 e^{i 𝜃} - 𝑧}) d 𝜃 \\ \begin{matrix}  \end{matrix} + \sum_{𝑘 = 1}^{𝑛 (𝑟, 0, 𝑓)} ℜ𝔢 log (\frac{𝑟 (𝑧 - 𝑎_{𝑘})}{𝑟^{2} - \overline{𝑎_{𝑘}} 𝑧}) . \end{aligned}

For any holomorphic $𝑔 = 𝑢 + i 𝑣$ , we have

\begin{aligned} \frac{𝜕 (ℜ𝔢 𝑔 (𝑧))}{𝜕 𝑧} & = \frac{1}{2} (\frac{𝜕 𝑢 (𝑧)}{𝜕 𝑥} - i \frac{𝜕 𝑢 (𝑧)}{𝜕 𝑦}) \\ = \frac{1}{2} (\frac{𝜕 𝑢 (𝑧)}{𝜕 𝑥} + i \frac{𝜕 𝑣}{𝜕 𝑥}) = \frac{1}{2} \frac{𝜕 𝑔 (𝑧)}{𝜕 𝑥} = \frac{𝑔^{'} (𝑧)}{2} . (4.4.4 . 4) \end{aligned}

Therefore, by differentiation under the integral sign,

\begin{aligned} \frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)} & = \frac{1}{π} \int_{0}^{2 π} log | 𝑓 (𝑟 e^{i 𝜃}) | \frac{𝑟 e^{i 𝜃} d 𝜃}{{(𝑟 e^{i 𝜃} - 𝑧)}^{2}} + \sum_{𝑘 = 1}^{𝑛 (𝑟, 0, 𝑓)} \frac{𝑟^{2} - {| 𝑎_{𝑘} |}^{2}}{(𝑟^{2} - \overline{𝑎_{𝑘}} 𝑧) (𝑧 - 𝑎_{𝑘})} \\ = \frac{1}{π} \int_{0}^{2 π} log | 𝑓 (𝑟 e^{i 𝜃}) | \frac{𝑟 e^{i 𝜃} d 𝜃}{{(𝑟 e^{i 𝜃} - 𝑧)}^{2}} + \sum_{𝑘 = 1}^{𝑛 (𝑟, 0, 𝑓)} \frac{\overline{𝑎_{𝑘}}}{𝑟^{2} - \overline{𝑎_{𝑘}} 𝑧} + \sum_{𝑘 = 1}^{𝑛 (𝑟, 0, 𝑓)} \frac{1}{𝑧 - 𝑎_{𝑘}} . \end{aligned}

Differentiating $𝑝$ times from here gives

\begin{aligned} \frac{d^{𝑝}}{d 𝑧^{𝑝}} (\frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)}) & = \frac{1}{π} \int_{0}^{2 π} log | 𝑓 (𝑟 e^{i 𝜃}) | \frac{𝑟 e^{i 𝜃} (𝑝 + 1)! d 𝜃}{{(𝑟 e^{i 𝜃} - 𝑧)}^{𝑝 + 2}} \\ + \sum_{𝑘 = 1}^{𝑛 (𝑟, 0, 𝑓)} \frac{{\overline{𝑎_{𝑘}}}^{𝑝 + 1} 𝑝!}{{(𝑟^{2} - \overline{𝑎_{𝑘}} 𝑧)}^{𝑝 + 1}} - \sum_{𝑘 = 1}^{𝑛 (𝑟, 0, 𝑓)} \frac{𝑝!}{{(𝑎_{𝑘} - 𝑧)}^{𝑝 + 1}} . \end{aligned}

The first two terms vanish as

𝑟 \to \infty

by Theorem 4.4.4.16

Theorem 4.4.4.16 Theorem 4.4.4.16

and Lemma 4.4.4.15

Lemma 4.4.4.15 Lemma 4.4.4.15

. ⁠

□

Lemma 4.4.4.18 (Logarithmic Factorization).

Let $𝑓$ be entire, non-constant, and of finite order $𝜌$ such that $𝑓 (0) = 1$ . Let

𝑃 (𝑧) = \prod_{𝑘 = 1}^{\infty} 𝐸_{rank 𝑓} (\frac{𝑧}{𝑎_{𝑘}})

be the associated product. If $𝑝 > 𝜌 (𝑓) - 1$ is an integer, then

\frac{d^{𝑝}}{d 𝑧^{𝑝}} (\frac{𝑃^{'} (𝑧)}{𝑃 (𝑧)}) \equiv - \sum_{𝑘 = 1}^{\infty} \frac{𝑝!}{{(𝑎_{𝑘} - 𝑧)}^{𝑝 + 1}}

for all

𝑧 \in ℂ ∖ {𝑎_{1}, 𝑎_{2}, \dots}

Proof.

Let $𝑃_{𝑛}$ be the $𝑛$ -th partial product of $𝑃$ . Then

\frac{𝑃_{𝑛}^{'} (𝑧)}{𝑃_{𝑛} (𝑧)} = \sum_{𝑘 = 1}^{𝑛} \frac{\frac{d}{d 𝑧} 𝐸_{rank 𝑓} (\frac{𝑧}{𝑎_{𝑘}})}{𝐸_{rank 𝑓} (\frac{𝑧}{𝑎_{𝑘}})} = \sum_{𝑘 = 1}^{𝑛} [\frac{1}{𝑧 - 𝑎_{𝑘}} + \sum_{𝑗 = 1}^{rank 𝑓} \frac{𝑧^{𝑗 - 1}}{𝑎_{𝑘}^{𝑗}}],

implying that

\frac{d^{𝑝}}{d 𝑧^{𝑝}} (\frac{𝑃_{𝑛}^{'} (𝑧)}{𝑃_{𝑛} (𝑧)}) = - \sum_{𝑘 = 1}^{𝑛} \frac{𝑝!}{{(𝑎_{𝑘} - 𝑧)}^{𝑝 + 1}} + \frac{d^{𝑝}}{d 𝑧^{𝑝}} \sum_{𝑗 = 1}^{rank 𝑓} \sum_{𝑘 = 1}^{𝑛} \frac{𝑧^{𝑗 - 1}}{𝑎_{𝑘}^{𝑗}} .

Since the polynomial in the rightmost term has degree at most $max 𝑗 - 1 = rank 𝑓 - 1$ , and because $𝑝 > ⌊ 𝜌 ⌋ - 1 \geq rank 𝑓 - 1$ , after $𝑝$ derivatives each term of the expression vanishes. For an arbitrarily chosen compact $𝐾 \subset ℂ$ avoiding the zeros, choose $𝑁 \in ℕ$ so that $| 𝑎_{𝑘} | \geq 2 {max}_{𝑧 \in 𝐾} | 𝑧 |$ for all $𝑘 > 𝑁$ . Then, for $𝑧 \in 𝐾$ and $𝑘 > 𝑁$ ,

| 𝑎_{𝑘} - 𝑧 | \geq | 𝑎_{𝑘} | - | 𝑧 | \geq \frac{| 𝑎_{𝑘} |}{2} .

The convergence of

\sum_{𝑘 = 1}^{\infty} \frac{2^{𝑝 + 1}}{{| 𝑎_{𝑘} |}^{𝑝 + 1}}

from Theorem 4.4.4.12 $Theorem 4.4.4.12 Theorem 4.4.4.12$ implies the absolute convergence of

\sum_{𝑘 = 1}^{\infty} \frac{1}{{(𝑎_{𝑘} - 𝑧)}^{𝑝 + 1}}

uniformly in $𝐾$ . Moreover, the locally uniform convergence $𝑃_{𝑛} \to 𝑃$ implies $𝑃_{𝑛}^{'} \to 𝑃^{'}$ by the Weierstrass Convergence Theorem (Theorem 4.1.1 $Theorem 4.1.1 Theorem 4.1.1$ ). Since $𝑃$ is bounded away from zero on $𝐾$ , it follows that $\frac{𝑃_{𝑛}^{'}}{𝑃_{𝑛}} \to \frac{𝑃^{'}}{𝑃}$ locally uniformly away from the zeros. Applying the Weierstrass Convergence Theorem $𝑝$ more times gives

\begin{aligned} lim_{𝑛 \to \infty} \frac{d^{𝑝}}{d 𝑧^{𝑝}} (\frac{𝑃_{𝑛}^{'} (𝑧)}{𝑃_{𝑛} (𝑧)}) & = \frac{d^{𝑝}}{d 𝑧^{𝑝}} (\frac{𝑃^{'} (𝑧)}{𝑃 (𝑧)}) = - lim_{𝑛 \to \infty} \sum_{𝑘 = 1}^{𝑛} \frac{𝑝!}{{(𝑎_{𝑘} - 𝑧)}^{𝑝 + 1}} \\ = - \sum_{𝑘 = 1}^{\infty} \frac{𝑝!}{{(𝑎_{𝑘} - 𝑧)}^{𝑝 + 1}} . □ \end{aligned}

The two preceding results are similar in conclusion, but Lemma 4.4.4.18 $Lemma 4.4.4.18 Lemma 4.4.4.18$ is not a special case of Proposition 4.4.4.17 $Proposition 4.4.4.17 Proposition 4.4.4.17$ since we have not asserted that the canonical product is of finite order.