§14.2.1 Classifying Growth of Meromorphic Functions

14.2.1 Classifying Growth of Meromorphic Functions

Theorem 14.2.1.1.

Let $𝑓$ be holomorphic on $\overline{𝐷 (0, 𝑅)}$ for $𝑅 > 0$ and define $𝑀 (𝑟, 𝑓)$ to be ${max}_{| 𝑧 | \leq 𝑟} | 𝑓 (𝑧) | = {max}_{| 𝑧 | = 𝑟} | 𝑓 (𝑧) |$ for $0 < 𝑟 < 𝑅$ ; then

𝑇 (𝑟, 𝑓) \leq {log}^{+} 𝑀 (𝑟, 𝑓) \leq \frac{𝑅 + 𝑟}{𝑅 - 𝑟} 𝑇 (𝑅, 𝑓) .

Proof.

Since $𝑓$ is holomorphic in $𝐷 (0, 𝑟)$ , $𝑁 (𝑟, 𝑓) \equiv 0$ and hence

𝑇 (𝑟, 𝑓) = 𝑚 (𝑟, 𝑓) = \frac{1}{2 π} \int_{0}^{2 π} {log}^{+} | 𝑓 (𝑟 e^{i 𝜃}) | d 𝜃 \leq \frac{1}{2 π} \int_{0}^{2 π} {log}^{+} 𝑀 (𝑟, 𝑓) d 𝜃,

which proves the first inequality. Since $𝑇 \geq 0$ trivially, if $𝑀 \leq 1$ the second assertion holds trivially. Hence, assume $𝑀 > 1$ ; by the Poisson–Jensen formula (Theorem 4.4.4.3 $Theorem 4.4.4.3 Theorem 4.4.4.3$ ), letting $𝑧 \in 𝜕 𝐷 (0, 𝑟)$ be where $| 𝑓 |$ attains $𝑀$ , we have

\begin{aligned} {log}^{+} 𝑀 (𝑟, 𝑓) & = \frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (𝑅 e^{i 𝜃}) | \frac{𝑅^{2} - 𝑟^{2}}{{| 𝑅 e^{i 𝜃} - 𝑧 |}^{2}} d 𝜃 \\ + \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} | \\ \leq \frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (𝑅 e^{i 𝜃}) | \frac{𝑅^{2} - 𝑟^{2}}{{| 𝑅 e^{i 𝜃} - 𝑧 |}^{2}} d 𝜃 \\ \leq \frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (𝑅 e^{i 𝜃}) | \frac{𝑅^{2} - 𝑟^{2}}{{(𝑅 - 𝑟)}^{2}} d 𝜃 \\ \leq \frac{1}{2 π} \int_{0}^{2 π} {log}^{+} | 𝑓 (𝑅 e^{i 𝜃}) | \frac{𝑅 + 𝑟}{𝑅 - 𝑟} d 𝜃 = \frac{𝑅 + 𝑟}{𝑅 - 𝑟} 𝑇 (𝑅, 𝑓) . \end{aligned}

(The Möbius transformation-like terms of the zeros

\leq 0

since they map to the unit disk, and the second summation vanishes since there are no

𝑏_{𝑘}

by holomorphy). ⁠

□

Definition 14.2.1.2.

The order of a meromorphic function $𝑓 : ℂ \to \hat{ℂ}$ , denoted $𝜌 = 𝜌 (𝑓)$ , is given by

𝜌 = \underset{𝑟 \to \infty}{lim sup} \frac{{log}^{+} 𝑇 (𝑟, 𝑓)}{log 𝑟},

equivalent to the condition that

𝑇 (𝑟, 𝑓) = 𝒪︀ (𝑟^{𝜌 + 𝜀})

for any

𝜀 > 0

but not for

𝜀 < 0

Proof of equivalent definitions.

For any prescribed $𝜀 > 0$ there exists $𝑟^{'} > 0$ such that for any $𝑟 > 𝑟^{'}$ ,

\frac{{log}^{+} 𝑇 (𝑟, 𝑓)}{log 𝑟} \leq 𝜌 + 𝜀 ⟹ 𝑇 (𝑟, 𝑓) \leq 𝑟^{𝜌 + 𝜀} .

Assume there exists some $𝜀 > 0$ such that

𝑇 (𝑟, 𝑓) = ℴ︀ (𝑟^{𝜌 - 𝜀}) .

Then there exist finite $𝑐, 𝑟^{'}$ such that $\forall 𝑟 > 𝑟^{'}$ ,

\begin{aligned} 𝑇 (𝑟, 𝑓) \leq 𝑐 𝑟^{𝜌 - 𝜀} & ⟹ {log}^{+} 𝑇 (𝑟, 𝑓) \leq 𝒪︀ (1) + (𝜌 - 𝜀) log 𝑟 \\ ⟹ \frac{{log}^{+} 𝑇 (𝑟, 𝑓)}{log 𝑟} \leq 𝜌 - 𝜀, \end{aligned}

which is a contradiction. The converse follows similarly to the case of entire functions in Section 4.4.4

Section 4.4.4 Section 4.4.4

. ⁠

□

Theorem 14.2.1.3.

Let

𝜌_{𝑀}

be the order of a non-constant entire function

𝑓

as defined in Section 4.4.4

Section 4.4.4 Section 4.4.4

and let

𝜌_{𝑇}

be the order as defined in Definition 14.2.1.2

Definition 14.2.1.2 Definition 14.2.1.2

. Then

𝜌_{𝑀} = 𝜌_{𝑇}

Proof.

Since $𝑓$ is non-constant, for sufficiently large $𝑟$ , ${log}^{+} 𝑀 (𝑟, 𝑓) = log 𝑀 (𝑟, 𝑓)$ . For the remainder of this proof we will assume the two are equivalent. From Theorem 14.2.1.1 $Theorem 14.2.1.1 Theorem 14.2.1.1$ it is apparent that (under $𝑅 = 2 𝑟$ for sufficiently large $𝑟$ )

\frac{{log}^{+} 𝑇 (𝑟, 𝑓)}{log 𝑟} \leq \frac{{log}^{+} log 𝑀 (𝑟, 𝑓)}{log 𝑟} \leq \frac{log 3}{log 𝑟} + \frac{{log}^{+} 𝑇 (2 𝑟, 𝑓)}{log (2 𝑟)} \cdot \frac{log (2 𝑟)}{log 𝑟} .

Letting $𝑟 \to \infty$ in the limit superior gives that

𝜌_{𝑇} \leq 𝜌_{𝑀} \leq 𝜌_{𝑇} ⟹ 𝜌_{𝑇} = 𝜌_{𝑀} . □

Proposition 14.2.1.4.

Let $𝑓 : ℂ \to \hat{ℂ}$ be a meromorphic function of finite order $𝜌$ . Then for every $𝜀 > 0$ and every $𝑎 \in \hat{ℂ}$ ,

$𝑚 (𝑟, 𝑎, 𝑓) = 𝒪︀ (𝑟^{𝜌 + 𝜀})$ .
$𝑁 (𝑟, 𝑎, 𝑓) = 𝒪︀ (𝑟^{𝜌 + 𝜀})$ .
$𝑛 (𝑟, 𝑎, 𝑓) = 𝒪︀ (𝑟^{𝜌 + 𝜀})$ .

Proof.

By the First Fundamental Theorem of Nevanlinna theory (Theorem 14.2.4 $Theorem 14.2.4 Theorem 14.2.4$ ),

𝑚 (𝑟, 𝑎) = 𝑇 (𝑟, 𝑓) + 𝒪︀ (1) - 𝑁 (𝑟, 𝑎)

and since $𝑚 (𝑟, 𝑎) \geq 0$ and $𝑁 (𝑟, 𝑎) \geq 0$ , we obtain

𝑚 (𝑟, 𝑎) \leq 𝑇 (𝑟, 𝑓) + 𝒪︀ (1) and 𝑁 (𝑟, 𝑎) \leq 𝑇 (𝑟, 𝑓) + 𝒪︀ (1) .

Because $𝑓$ has order $𝜌$ , we have

𝑇 (𝑟, 𝑓) = 𝒪︀ (𝑟^{𝜌 + 𝜀}),

which proves Part 1 $Part 1 Part 1$ and Part 2 $Part 2 Part 2$ . Since

\begin{aligned} 𝑁 (2 𝑟, 𝑎, 𝑓) - 𝑁 (𝑟, 𝑎, 𝑓) & = \int_{𝑟}^{2 𝑟} \frac{𝑛 (𝑥, 𝑎, 𝑓)}{𝑥} d 𝑥 \\ \geq 𝑛 (𝑟, 𝑎, 𝑓) \int_{𝑟}^{2 𝑟} \frac{d 𝑥}{𝑥} = log 2 \cdot 𝑛 (𝑟, 𝑎, 𝑓), \end{aligned}

it follows from Part 2 $Part 2 Part 2$ that

𝑛 (𝑟, 𝑎, 𝑓) \leq \frac{𝑁 (2 𝑟, 𝑎, 𝑓) - 𝑁 (𝑟, 𝑎, 𝑓)}{log 2} \leq \frac{𝑁 (2 𝑟, 𝑎, 𝑓)}{log 2} = 𝒪︀ ({(2 𝑟)}^{𝜌 + 𝜀}) = 𝒪︀ (𝑟^{𝜌 + 𝜀})

and hence Part 3

Part 3 Part 3

follows. ⁠

□

Theorem 14.2.1.5.

Let $𝑓 : ℂ \to \hat{ℂ}$ be a meromorphic function of finite order $𝜌$ . For $𝑎 \in ℂ$ , let $𝑟_{𝑛} (𝑎)$ denote the moduli of the zeros of $𝑓 (𝑧) - 𝑎$ (counted with multiplicity) in non-decreasing order, and let $𝑟_{𝑛} (\infty)$ denote the moduli of the poles of $𝑓$ in non-decreasing order. Then for every $𝜀 > 0$ and every $𝑎 \in \hat{ℂ}$ ,

\sum_{𝑛} \frac{1}{𝑟_{𝑛} {(𝑎)}^{𝜌 + 𝜀}} < \infty .

Proof.

Observe that

𝜌 + 𝜀 > \underset{𝑟 \to \infty}{lim sup} \frac{{log}^{+} 𝑇 (𝑟, 𝑓)}{log 𝑟} \geq \underset{𝑟 \to \infty}{lim sup} \frac{log 𝑁 (𝑟, 𝑓)}{log 𝑟} .

Since

𝑁 (e 𝑟, 𝑓) - 𝑁 (𝑟, 𝑓) = \int_{𝑟}^{e 𝑟} \frac{𝑛 (𝑥, 𝑓) d 𝑥}{𝑥} \geq \int_{𝑟}^{e 𝑟} \frac{𝑛 (𝑟, 𝑓) d 𝑥}{𝑥} = 𝑛 (𝑟, 𝑓),

it follows that $𝑛 \leq 𝑁$ , and

𝜌 + 𝜀 > \underset{𝑟 \to \infty}{lim sup} \frac{log 𝑛 (𝑟, 𝑓)}{log 𝑟},

and hence the conclusion follows for $𝑎 = \infty$ by Theorem 4.4.4.11 $Theorem 4.4.4.11 Theorem 4.4.4.11$ . Assume $𝑎$ is finite; observe that

𝑇 (𝑟, 𝑓) = 𝑇 (𝑟, 𝑎, 𝑓) + 𝒪︀ (1) = 𝑇 (𝑟, \frac{1}{𝑓 - 𝑎}) + 𝒪︀ (1)

by the First Fundamental Theorem (Theorem 14.2.4

Theorem 14.2.4 Theorem 14.2.4

). It follows that

𝜌

is the order of

\frac{1}{𝑓 - 𝑎}

. Applying the previous result to this function, whose poles are precisely at

{𝑟_{𝑛} (𝑎)}

, the conclusion follows. ⁠

□