§14.4.1 The Estimation of $𝑆 (𝑟, 𝑓)$

14.4.1 The Estimation of $𝑆 (𝑟, 𝑓)$

The second part of the Second Fundamental Theorem is typically a suitable estimate on $𝑆 (𝑟, 𝑓)$ .

Theorem 14.4.1.5 (Second Fundamental Theorem of Nevanlinna Theory, Part 2).

If $𝑅_{0} = \infty$ and $𝑓$ has finite order, then Let $𝑓 : 𝐷 (0, 𝑅_{0}) \to \hat{ℂ}$ ( $0 < 𝑅_{0} \leq \infty$ ) be meromorphic and non-constant. Define $𝑆 (𝑟, 𝑓)$ as in (14.4.1) $(14.4.1) (14.4.1)$ . Then
$𝑆 (𝑟, 𝑓) = 𝒪︀ (log 𝑇 (𝑟, 𝑓)) + 𝒪︀ (log 𝑟) (†)$
as $𝑟 \to \infty$ through all values.
If $𝑅_{0} = \infty$ and $𝑓$ has infinite order, then $(†)$ holds as $𝑟 \to \infty$ through all values of $𝑟$ outside a set $𝐸$ of finite linear measure.
If $𝑅_{0} < \infty$ , then
$𝑆 (𝑟, 𝑓) = 𝒪︀ ({log}^{+} 𝑇 (𝑟, 𝑓) + log \frac{1}{𝑅_{0} - 𝑟}) (‡)$
as $𝑟 \to 𝑅_{0}^{-}$ outside a set $𝐸$ satisfying
$\int_{𝐸} \frac{d 𝑟}{𝑅_{0} - 𝑟} < \infty .$
Moreover, for any $𝜌, 𝜌^{'} \in (0, 𝑅_{0})$ such that $𝑅_{0} - 𝜌^{'} < e^{- 2} (𝑅_{0} - 𝜌)$ (implying that $𝜌 < 𝜌^{'}$ ), $\exists 𝑟 \in (𝜌, 𝜌^{'})$ such that $𝑟 \notin 𝐸$ .

Modern results vastly improve said estimations, and the search for a sharp estimation is one of the most challenging questions in Nevanlinna Theory.

Lemma 14.4.1.1.

Let $𝑧 \in ℂ$ be arbitrary and fixed and suppose $0 < 𝑟 < \infty$ . Let

𝐸_{𝑘} = {𝜃 : - π < 𝜃 < π \land | 𝑧 - 𝑟 e^{i 𝜃} | < 𝑘 𝑟}

for each $0 < 𝑘 \leq 1$ . It follows that

𝐼 = \int_{𝐸_{𝑘}} log | \frac{𝑟}{𝑧 - 𝑟 e^{i 𝜃}} | d 𝜃 < π 𝑘 (1 - log 𝑘) .

Proof.

Let $𝑧 = 𝜌 e^{i 𝜓}$ . Performing a substitution of $𝜗 = 𝜃 - 𝜓$ , we have $𝐼 = \int_{𝐸_{𝑘}^{'}} log | \frac{𝑟}{𝜌 - 𝑟 e^{i 𝜗}} | d 𝜗$ , where $𝐸_{𝑘}^{'} = {𝜗 : - π - 𝜓 < 𝜗 < π - 𝜓 \land | 𝜌 - 𝑟 e^{i 𝜗} | < 𝑘 𝑟}$ .

Because the exponential is periodic every $2 π i$ , we may assume that $𝐸_{𝑘}^{'}$ is restricted to $- π < 𝜗 < π$ instead of $- π - 𝜓 < 𝜗 < π - 𝜓$ . The expression $| 𝜌 - 𝑟 e^{i 𝜗} | = 𝑘 𝑟$ gives at most two symmetric solutions $\pm 𝜃_{0}$ (let $𝜃_{0} \geq 0$ ).

Evidently, for $𝜗 > \frac{π}{2}$ or $- 𝜗 < - \frac{π}{2}$ ,

| 𝜌 - 𝑟 e^{i 𝜗} | = \sqrt{𝜌^{2} + 𝑟^{2} - 2 𝑟 𝜌 cos (𝜗)} \geq 𝑟 \geq 𝑘 𝑟, (Law of Cosines)

so it follows that $𝜃_{0} \leq \frac{π}{2}$ , and furthermore, all $- 𝜃_{0} < 𝜗 < 𝜃_{0}$ lie in $𝐸_{𝑘}^{'}$ by the geometry of a circle. Since $| 𝜌 - 𝑟 e^{i 𝜗} | \geq | ℑ𝔪 (𝜌 - 𝑟 e^{i 𝜗}) | = 𝑟 | sin (𝜗) |$ , assuming that $𝐸_{𝑘}^{'} \neq ⌀$ , it follows that

\begin{aligned} 𝐼 & = 2 \int_{0}^{𝜃_{0}} log | \frac{𝑟}{𝜌 - 𝑟 e^{i 𝜗}} | d 𝜗 \leq 2 \int_{0}^{𝜃_{0}} log csc (𝜗) d 𝜗 < 2 \int_{0}^{𝜃_{0}} log (\frac{π}{2 𝜗}) d 𝜗 \\ = 2 𝜃_{0} log (\frac{π}{2}) - 2 \int_{0}^{𝜃_{0}} log (𝜗) d 𝜗 = 2 𝜃_{0} log (\frac{π}{2}) - 2 (𝜃_{0} log (𝜃_{0}) - 𝜃_{0}) \\ = 2 𝜃_{0} (log (\frac{π}{2 𝜃_{0}}) + 1) . \end{aligned}

Since $\frac{π}{2} sin 𝑥 \geq 𝑥$ for $𝑥 \in [0, \frac{π}{2}]$ , it follows that $𝜃_{0} \leq \frac{π}{2} sin 𝜃_{0} \leq \frac{π 𝑘}{2}$ . Moreover, $𝜃 \mapsto 2 𝜃 (log (\frac{π}{2 𝜃}) + 1)$ is an increasing function of $0 \leq 𝜃 \leq \frac{π}{2}$ ; thus we may replace all instances of $𝜃_{0}$ with $\frac{π 𝑘}{2}$ , proving the lemma. ⁠ $□$

Lemma 14.4.1.2.

Let ${𝑧_{𝜈}}_{𝜈 = 1}^{𝑛}$ be $𝑛 \geq 1$ (possibly indistinct) complex numbers. Define $𝛿 (𝑧) = min {| 𝑧 - 𝑧_{𝜈} | : 𝜈 \in ℕ_{\leq 𝑛}}$ . It follows that for any $𝑟 > 0$ ,

\frac{1}{2 π} \int_{0}^{2 π} {log}^{+} \frac{𝑟}{𝛿 (𝑟 e^{i 𝜃})} d 𝜃 \leq 2 log 𝑛 + \frac{1}{2}

Proof.

Define the set

𝐸_{𝜈} = {𝜃 \in (- π, π) : | 𝑧_{𝜈} - 𝑟 e^{i 𝜃} | < \frac{𝑟}{𝑛}}

and let $𝐸 = ⋃_{𝜈 = 1}^{𝑛} 𝐸_{𝜈}$ . For $𝜃 \in 𝐸$ , $𝛿 (𝑟 e^{i 𝜃}) < \frac{𝑟}{𝑛}$ , implying that $\frac{𝑟}{𝛿 (𝑟 e^{i 𝜃})} > 𝑛$ . Define

{log}_{0} 𝑥 = {\begin{cases} log 𝑥 & if 𝑥 \geq 𝑛 \\ 0 & otherwise \end{cases} (\geq 0),

so that for $𝜃$ in the prescribed range, we have

\begin{aligned} {log}^{+} \frac{𝑟}{𝛿 (𝑟 e^{i 𝜃})} & = {log}_{0} (\frac{𝑟}{𝛿 (𝑟 e^{i 𝜃})}) = {log}_{0} (\frac{𝑟}{{min}_{𝜈 \in ℕ_{\leq 𝑛}} | 𝑟 e^{i 𝜃} - 𝑧_{𝜈} |}) \\ \leq \sum_{𝜈 = 1}^{𝑛} {log}_{0} (\frac{𝑟}{| 𝑟 e^{i 𝜃} - 𝑧_{𝜈} |}) . \end{aligned}

Since $0 < \frac{1}{𝑛} \leq 1$ , by virtue of Lemma 14.4.1.1 $Lemma 14.4.1.1 Lemma 14.4.1.1$ for $𝑘 = \frac{1}{𝑛}$ ,

\begin{aligned} \frac{1}{2 π} \int_{𝐸} {log}^{+} \frac{𝑟}{𝛿 (𝑟 e^{i 𝜃})} d 𝜃 & \leq \frac{1}{2 π} \sum_{𝜈 = 1}^{𝑛} \int_{0}^{2 π} {log}_{0} (\frac{𝑟}{| 𝑟 e^{i 𝜃} - 𝑧_{𝜈} |}) d 𝜃 \\ = \frac{1}{2 π} \sum_{𝜈 = 1}^{𝑛} \int_{0}^{2 π} log (\frac{𝑟}{| 𝑟 e^{i 𝜃} - 𝑧_{𝜈} |}) d 𝜃 \\ < \frac{1}{2 π} \sum_{𝜈 = 1}^{𝑛} \frac{π}{𝑛} (1 + log 𝑛) = \frac{1}{2} + \frac{log 𝑛}{2} . (14.4.1 . 1) \end{aligned}

For $𝜃 \notin 𝐸$ , for each $𝜈$ we have $| 𝑧_{𝜈} - 𝑟 e^{i 𝜃} | \geq \frac{𝑟}{𝑛}$ and thus $𝛿 (𝑟 e^{i 𝜃}) \geq \frac{𝑟}{𝑛}$ . It follows that

\frac{1}{2 π} \int_{𝜃 \notin 𝐸} {log}^{+} \frac{𝑟}{𝛿 (𝑟 e^{i 𝜃})} d 𝜃 \leq \frac{1}{2 π} \int_{0}^{2 π} {log}^{+} 𝑛 d 𝜃 \leq log 𝑛 . (14.4.1 . 2)

Combining (14.4.1.1)

(14.4.1.4) (14.4.1.1)

, (14.4.1.2)

(14.4.1.5) (14.4.1.2)

gives the desired result. ⁠

□

Proposition 14.4.1.3.

If $𝑓$ is meromorphic on $𝐷 (0, 𝑅)$ , then for $0 < 𝑟 < 𝑅$ , letting $𝑐$ be the first nonzero coefficient of the innermost Laurent expansion of $𝑓$ about the origin,

\begin{aligned} 𝑚 (𝑟, \frac{𝑓^{'}}{𝑓}) & < 4 {log}^{+} 𝑇 (𝑅, 𝑓) + 4 {log}^{+} {log}^{+} | \frac{1}{𝑐} | \\ \begin{matrix}  \end{matrix} + 5 {log}^{+} 𝑅 + 6 {log}^{+} \frac{1}{𝑅 - 𝑟} + {log}^{+} \frac{1}{𝑟} + 14 . \end{aligned}

Proof.

By the Poisson–Jensen Formula (Theorem 4.4.4.3 $Theorem 4.4.4.3 Theorem 4.4.4.3$ ), letting $𝜌 = \frac{1}{2} (𝑅 + 𝑟)$ , for each non-singular point $𝑧 \in 𝜕 𝐷 (0, 𝑟)$ ,

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

where ${𝑎_{𝑘}}$ and ${𝑏_{𝑗}}$ are the respective zeros and poles in the disk. By (4.4.4.4) $(4.4.4.15) (4.4.4.4)$ , we have $\frac{𝜕}{𝜕 𝑧} (ℜ𝔢 𝑔 (𝑧)) = \frac{1}{2} 𝑔^{'} (𝑧)$ for all holomorphic $𝑔$ . Applying this with differentiation under the integral sign, we have that

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

Define

𝛿 (𝑧) = min ({| 𝑧 - 𝑎_{𝑘} | : 𝑘 \in ℕ_{\leq 𝑛 (𝜌, 0, 𝑓)}} \cup {| 𝑧 - 𝑏_{𝑗} | : 𝑗 \in ℕ_{\leq 𝑛 (𝜌, 𝑓)}}) .

Evidently the number of zeros and poles is given by

𝑛 = 𝑛 (𝜌, 𝑓) + 𝑛 (𝜌, 0, 𝑓) .

Since

| 𝜌^{2} - \overline{𝑎_{𝑘}} 𝑧 | \geq 𝜌^{2} - 𝑟 | 𝑎_{𝑘} | \geq 𝜌^{2} - 𝑟 𝜌, | 𝜌^{2} - \overline{𝑏_{𝑗}} 𝑧 | \geq 𝜌^{2} - 𝑟 𝜌,

it follows that

| \frac{\overline{𝑎_{𝑘}}}{𝜌^{2} - \overline{𝑎_{𝑘}} 𝑧} |, | \frac{\overline{𝑏_{𝑗}}}{𝜌^{2} - \overline{𝑏_{𝑗}} 𝑧} | \leq \frac{𝜌}{𝜌^{2} - 𝑟 𝜌} = \frac{1}{𝜌 - 𝑟}, (14.4.1 . 3)

and additionally,

| \frac{1}{𝑏_{𝑗} - 𝑧} | \leq \frac{1}{𝛿 (𝑧)}, | \frac{1}{𝑎_{𝑘} - 𝑧} | \leq \frac{1}{𝛿 (𝑧)} . (14.4.1 . 4)

Furthermore,

\begin{matrix} | \frac{1}{2 π} \int_{0}^{2 π} log | 𝑓 (𝜌 e^{i 𝜃}) | \frac{2 𝜌 e^{i 𝜃} d 𝜃}{{(𝜌 e^{i 𝜃} - 𝑧)}^{2}} | \leq \frac{1}{2 π} \frac{2 𝜌}{{(𝜌 - 𝑟)}^{2}} \int_{0}^{2 π} | log | 𝑓 (𝜌 e^{i 𝜃}) | | d 𝜃 \\ = \frac{2 𝜌}{{(𝜌 - 𝑟)}^{2}} [𝑚 (𝜌, 𝑓) + 𝑚 (𝜌, \frac{1}{𝑓})] . (14.4.1 . 5) \end{matrix}

by Part 4 $Part 4 Part 4$ of Proposition 14.1.2 $Proposition 14.1.2 Proposition 14.1.2$ . Combining (14.4.1.3) $(14.4.1.6) (14.4.1.3)$ , (14.4.1.4) $(14.4.1.7) (14.4.1.4)$ , (14.4.1.5) $(14.4.1.8) (14.4.1.5)$ , we have

\begin{aligned} | \frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)} | & \leq \frac{2 𝜌}{{(𝜌 - 𝑟)}^{2}} [𝑚 (𝜌, 𝑓) + 𝑚 (𝜌, \frac{1}{𝑓})] \\ \begin{matrix}  \end{matrix} + \sum_{𝑘 = 1}^{𝑛 (𝜌, 0, 𝑓)} (\frac{1}{𝜌 - 𝑟} + \frac{1}{𝛿 (𝑧)}) + \sum_{𝑗 = 1}^{𝑛 (𝜌, 𝑓)} (\frac{1}{𝜌 - 𝑟} + \frac{1}{𝛿 (𝑧)}) \\ = \frac{2 𝜌}{{(𝜌 - 𝑟)}^{2}} [𝑚 (𝜌, 𝑓) + 𝑚 (𝜌, \frac{1}{𝑓})] + 𝑛 (\frac{1}{𝜌 - 𝑟} + \frac{1}{𝛿 (𝑧)}) . \end{aligned}

By Proposition 14.1.5 $Proposition 14.1.5 Proposition 14.1.5$ , we have $𝑇 (𝜌, 0, 𝑓) = 𝑇 (𝜌, 𝑓) - log | 𝑐 |$ , where $𝑐$ is the first nonzero coefficient of the innermost Laurent expansion of $𝑓$ about the origin. It follows that $𝑚 (𝜌, \frac{1}{𝑓}) = 𝑇 (𝜌, 𝑓) - log | 𝑐 | - 𝑁 (𝜌, 0, 𝑓)$ and $𝑚 (𝜌, 𝑓) = 𝑇 (𝜌, 𝑓) - 𝑁 (𝜌, 𝑓)$ , thus

\begin{aligned} 𝑚 (𝜌, \frac{1}{𝑓}) + 𝑚 (𝜌, 𝑓) & = 2 𝑇 (𝜌, 𝑓) - log | 𝑐 | - 𝑁 (𝜌, \frac{1}{𝑓}) \\ \begin{matrix}  \end{matrix} - 𝑁 (𝜌, 𝑓) \leq 2 𝑇 (𝜌, 𝑓) + 2 {log}^{+} | \frac{1}{𝑐} |, \end{aligned}

and

| \frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)} | \leq \frac{4 𝜌}{{(𝜌 - 𝑟)}^{2}} [𝑇 (𝜌, 𝑓) + {log}^{+} | \frac{1}{𝑐} |] + \frac{𝑛}{𝑟} (\frac{𝑟}{𝜌 - 𝑟} + \frac{𝑟}{𝛿 (𝑧)}) .

By the subadditive properties of Proposition 14.1.2 $Proposition 14.1.2 Proposition 14.1.2$ , we have

\begin{aligned} {log}^{+} | \frac{𝑓^{'} (𝑧)}{𝑓 (𝑧)} | & \leq {log}^{+} [\frac{4 𝜌}{{(𝜌 - 𝑟)}^{2}} [𝑇 (𝜌, 𝑓) + {log}^{+} | \frac{1}{𝑐} |]] \\ \begin{matrix}  \end{matrix} + {log}^{+} [\frac{𝑛}{𝑟} (\frac{𝑟}{𝜌 - 𝑟} + \frac{𝑟}{𝛿 (𝑧)})] + log 2 \\ \leq {log}^{+} 𝜌 + {log}^{+} [{(\frac{2}{𝜌 - 𝑟})}^{2}] + {log}^{+} [𝑇 (𝜌, 𝑓) + {log}^{+} | \frac{1}{𝑐} |] \\ \begin{matrix}  \end{matrix} + {log}^{+} \frac{𝑛}{𝑟} + {log}^{+} (\frac{𝑟}{𝜌 - 𝑟} + \frac{𝑟}{𝛿 (𝑧)}) + log 2 \\ \leq {log}^{+} 𝜌 + 2 {log}^{+} \frac{1}{𝜌 - 𝑟} + {log}^{+} 𝑇 (𝜌, 𝑓) + {log}^{+} {log}^{+} | \frac{1}{𝑐} | \\ \begin{matrix}  \end{matrix} + {log}^{+} \frac{𝑛}{𝑟} + {log}^{+} \frac{𝑟}{𝜌 - 𝑟} + {log}^{+} \frac{𝑟}{𝛿 (𝑧)} + 5 log 2 . \end{aligned}

Integrating on $| 𝑧 | = 𝑟$ with Lemma 14.4.1.2 $Lemma 14.4.1.2 Lemma 14.4.1.2$ gives

\begin{aligned} 𝑚 (𝑟, \frac{𝑓^{'}}{𝑓}) & \leq {log}^{+} 𝜌 + 2 {log}^{+} \frac{1}{𝜌 - 𝑟} + {log}^{+} 𝑇 (𝜌, 𝑓) + {log}^{+} {log}^{+} | \frac{1}{𝑐} | \\ \begin{matrix}  \end{matrix} + {log}^{+} \frac{𝑛}{𝑟} + {log}^{+} \frac{𝑟}{𝜌 - 𝑟} + \frac{1}{2 π} \int_{0}^{2 π} {log}^{+} \frac{𝑟}{𝛿 (𝑟 e^{i 𝜃})} d 𝜃 + 5 log 2 \\ \leq {log}^{+} 𝜌 + 3 {log}^{+} \frac{1}{𝜌 - 𝑟} + {log}^{+} 𝑇 (𝜌, 𝑓) + {log}^{+} {log}^{+} | \frac{1}{𝑐} | \\ \begin{matrix}  \end{matrix} + 3 {log}^{+} 𝑛 + {log}^{+} \frac{1}{𝑟} + {log}^{+} 𝑟 + \frac{1}{2} + 5 log 2 . (14.4.1 . 6) \end{aligned}

To derive a more useful estimate for $𝑛$ , we note that

\begin{matrix} 𝑁 (𝑅, 𝑓) \geq \int_{𝜌}^{𝑅} \frac{𝑛 (𝑥, 𝑓) d 𝑥}{𝑥} \geq \int_{𝜌}^{𝑅} \frac{𝑛 (𝜌, 𝑓) d 𝑥}{𝑅} = 𝑛 (𝜌, 𝑓) \frac{𝑅 - 𝜌}{𝑅}, \\ 𝑁 (𝑅, 0, 𝑓) \geq 𝑛 (𝜌, 0, 𝑓) \frac{𝑅 - 𝜌}{𝑅} . \end{matrix}

Thus,

\begin{aligned} 𝑛 & = 𝑛 (𝜌, 𝑓) + 𝑛 (𝜌, 0, 𝑓) \leq \frac{𝑅}{𝑅 - 𝜌} [𝑁 (𝑅, 𝑓) + 𝑁 (𝑅, 0, 𝑓)] \\ \leq \frac{𝑅}{𝑅 - 𝜌} [𝑇 (𝑅, 𝑓) + 𝑇 (𝑅, 0, 𝑓) - 𝑚 (𝑅, 𝑓) - 𝑚 (𝑅, 0, 𝑓)] \\ \leq \frac{𝑅}{𝑅 - 𝜌} [2 𝑇 (𝑅, 𝑓) + log | \frac{1}{𝑐} |] \leq \frac{2 𝑅}{𝑅 - 𝜌} [𝑇 (𝑅, 𝑓) + {log}^{+} | \frac{1}{𝑐} |] . \end{aligned}

Using the ${log}^{+}$ subadditivity properties (as with earlier), we have

\begin{aligned} {log}^{+} 𝑛 & \leq {log}^{+} \frac{2 𝑅}{𝑅 - 𝜌} + {log}^{+} 𝑇 (𝑅, 𝑓) + {log}^{+} {log}^{+} | \frac{1}{𝑐} | + log 2 \\ \leq {log}^{+} 𝑅 + {log}^{+} \frac{1}{𝑅 - 𝜌} + {log}^{+} 𝑇 (𝑅, 𝑓) + {log}^{+} {log}^{+} | \frac{1}{𝑐} | + 2 log 2 \end{aligned}

Substituting this into (14.4.1.6) $(14.4.1.9) (14.4.1.6)$ , and using $𝑅 - 𝜌 = 𝜌 - 𝑟 = \frac{𝑅 - 𝑟}{2}$ we have that (recognizing ${log}^{+} 𝑟, {log}^{+} 𝜌 < {log}^{+} 𝑅$ )

\begin{aligned} 𝑚 (𝑟, \frac{𝑓^{'}}{𝑓}) & \leq {log}^{+} 𝜌 + 3 {log}^{+} \frac{2}{𝑅 - 𝑟} + {log}^{+} 𝑇 (𝜌, 𝑓) + {log}^{+} {log}^{+} | \frac{1}{𝑐} | + {log}^{+} \frac{1}{𝑟} \\ \begin{matrix}  \end{matrix} + {log}^{+} 𝑟 + \frac{1}{2} + 5 log 2 + 3 {log}^{+} 𝑅 + 3 {log}^{+} \frac{2}{𝑅 - 𝑟} \\ \begin{matrix}  \end{matrix} + 3 {log}^{+} 𝑇 (𝑅, 𝑓) + 3 {log}^{+} {log}^{+} | \frac{1}{𝑐} | + 6 log 2 \\ < 5 {log}^{+} 𝑅 + 17 log 2 + 6 {log}^{+} \frac{1}{𝑅 - 𝑟} + 4 {log}^{+} 𝑇 (𝜌, 𝑓) \\ \begin{matrix}  \end{matrix} + 4 {log}^{+} {log}^{+} | \frac{1}{𝑐} | + {log}^{+} \frac{1}{𝑟} + \frac{1}{2} \\ \leq 5 {log}^{+} 𝑅 + 17 log 2 + 6 {log}^{+} \frac{1}{𝑅 - 𝑟} + 4 {log}^{+} 𝑇 (𝑅, 𝑓) \\ \begin{matrix}  \end{matrix} + 4 {log}^{+} {log}^{+} | \frac{1}{𝑐} | + {log}^{+} \frac{1}{𝑟} + \frac{1}{2}, \end{aligned}

since

𝑇 (𝜌, 𝑓) \leq 𝑇 (𝑅, 𝑓)

(Theorem 14.2.3

Theorem 14.2.3 Theorem 14.2.3

). This proves the proposition. ⁠

□

Lemma 14.4.1.4 (Borel).

Let $𝑇 : [𝑟_{0}, 𝑅_{0}) \to ℝ$ be an increasing, continuous function satisfying $𝑇 > 1$ on all of its domain (where $𝑟_{0} < 𝑅_{0} \leq \infty$ ).

If $𝑅_{0} = \infty$ , for $𝑟 \geq 𝑟_{0}$ outside a set $𝐸_{0} \subseteq ℝ_{\geq 𝑟_{0}}$ whose linear measure does not exceed 2 (satisfying $\int_{𝐸_{0}} d 𝑟 \leq 2$ with respect to the Lebesgue measure¹),
$𝑇 (𝑟 + \frac{1}{𝑇 (𝑟)}) < 2 𝑇 (𝑟) (14.4.1 . 7)$
If $𝑅_{0} < \infty$ , then $\exists 𝐸_{0} \subseteq [𝑟_{0}, 𝑅_{0})$ satisfying $\int_{𝐸_{0}} \frac{d 𝑟}{𝑅_{0} - 𝑟} \leq 2$ such that for $𝑟 \in [𝑟_{0}, 𝑅_{0}) ∖ 𝐸_{0}$ ,
$𝑇 (𝑟 + \frac{𝑅_{0} - 𝑟}{e 𝑇 (𝑟)}) < 2 𝑇 (𝑟) (14.4.1 . 8)$
Moreover, for any $𝑅, 𝜌, 𝜌^{'} \in (0, 𝑅_{0})$ such that $𝑅 - 𝜌^{'} < e^{- 2} (𝑅 - 𝜌)$ (implying that $𝜌 < 𝜌^{'}$ ), $\exists 𝑟 \in (𝜌, 𝜌^{'})$ such that $𝑟 \notin 𝐸_{0}$ .

Proof.

We observe the two cases separately.

Assume $𝑅_{0} = \infty$ . Then define $𝑟_{0}^{'} = 𝑟_{0} + \frac{1}{𝑇 (𝑟_{0})}$ , and
$𝑟_{1} = inf {𝑟 > 𝑟_{0}^{'} : (14.4.1.7) (14.4.1.10) (14.4.1.7) fails}, 𝑟_{1}^{'} = 𝑟_{1} + \frac{1}{𝑇 (𝑟_{1})} > 𝑟_{1},$
and let
$𝑟_{2} = inf {𝑟 > 𝑟_{1}^{'} : (14.4.1.7) (14.4.1.10) (14.4.1.7) fails}, 𝑟_{2}^{'} = 𝑟_{2} + \frac{1}{𝑇 (𝑟_{2})} > 𝑟_{2},$
repeating this process recursively by
$𝑟_{𝑛} = inf {𝑟 > 𝑟_{𝑛 - 1}^{'} : (14.4.1.7) (14.4.1.10) (14.4.1.7) fails}, 𝑟_{𝑛}^{'} = 𝑟_{𝑛} + \frac{1}{𝑇 (𝑟_{𝑛})} > 𝑟_{𝑛} .$
It follows that $𝑟_{𝑛} \geq 𝑟_{𝑛 - 1}^{'} > 𝑟_{𝑛 - 1}$ , and $𝑟_{𝑛} - 𝑟_{𝑛 - 1} \geq 𝑟_{𝑛 - 1}^{'} - 𝑟_{𝑛 - 1}$ .
Let $𝑁 \in ℕ \cup {\infty}$ be the largest $𝑛$ for which the infimum in $𝑟_{𝑛}$ is defined. Observe that by the definition of the infimum, for each $𝑛$ , (14.4.1.7) $(14.4.1.10) (14.4.1.7)$ must be satisfied throughout each $(𝑟_{𝑛 - 1}^{'}, 𝑟_{𝑛})$ . Indeed, if there exists some value within this interval for which (14.4.1.7) $(14.4.1.10) (14.4.1.7)$ fails, then $𝑟_{𝑛}$ , which was defined by an infimum, would then be at most this new value, contradicting the assertion that it lay within the interval. It follows that the exceptional set must lie outside of the disjoint union; in other words,
$𝐸_{0} \subseteq ℝ_{\geq 𝑟_{0}} ∖ ⋃_{𝑛 = 1}^{𝑁} (𝑟_{𝑛 - 1}^{'}, 𝑟_{𝑛}) = ⋃_{𝑛 = 0}^{𝑁} [𝑟_{𝑛}, 𝑟_{𝑛}^{'}] .$
We state without proof that subadditivity holds:
$\int_{𝐸_{0}} d 𝑟 \leq \sum_{𝑛 = 0}^{𝑁} (𝑟_{𝑛}^{'} - 𝑟_{𝑛}) .$
By construction, for points sufficiently close to $𝑟_{𝑛}$ or at $𝑟_{𝑛}$ itself,
$𝑇 (𝑟 + \frac{1}{𝑇 (𝑟)}) \geq 2 𝑇 (𝑟),$
continuity allows (in either case) this inequality to be extended to $𝑟 = 𝑟_{𝑛}$ . Then for each $𝑛$ ,
$\begin{matrix} 𝑇 (𝑟_{𝑛}^{'}) = 𝑇 (𝑟_{𝑛} + \frac{1}{𝑇 (𝑟_{𝑛})}) \geq 2 𝑇 (𝑟_{𝑛}), \\ 𝑇 (𝑟_{𝑛}^{'}) \leq 𝑇 (𝑟_{𝑛 + 1}) ⟹ 𝑇 (𝑟_{𝑛 + 1}) \geq 2 𝑇 (𝑟_{𝑛}) . \end{matrix}$
Using induction, $𝑇 (𝑟_{𝑛}) \geq 2^{𝑛} 𝑇 (𝑟_{0})$ , implying that
$\begin{matrix} 𝑟_{𝑛}^{'} - 𝑟_{𝑛} = \frac{1}{𝑇 (𝑟_{𝑛})} \leq \frac{1}{2^{𝑛} 𝑇 (𝑟_{0})} \\ \sum_{𝑛 = 0}^{𝑁} (𝑟_{𝑛}^{'} - 𝑟_{𝑛}) \leq \frac{1}{𝑇 (𝑟_{0})} \sum_{𝑛 = 0}^{\infty} \frac{1}{2^{𝑛}} = \frac{2}{𝑇 (𝑟_{0})} \leq 2 . \end{matrix}$
Suppose $𝑅_{0} < \infty$ in the theorem statement. Let $𝑇_{0} (𝜌) = 𝑇 (𝑅_{0} - e^{- 𝜌})$ , where $𝜌_{0} = log \frac{1}{𝑅_{0} - 𝑟_{0}} \leq 𝜌 < \infty$ , so that $𝑇_{0}$ is an increasing function on $[𝜌_{0}, \infty)$ . By the result above, $\exists 𝐸_{0}^{'} \subseteq [𝜌_{0}, \infty)$ such that
$\int_{𝐸_{0^{'}}} d 𝜌 \leq 2, 𝑇_{0} (𝜌 + \frac{1}{𝑇_{0} (𝜌)}) < 2 𝑇_{0} (𝜌) \forall 𝜌 \notin 𝐸_{0}^{'} .$
Substitute $𝜌 = log \frac{1}{𝑅_{0} - 𝑟}$ , $𝑟 = 𝑅_{0} - e^{- 𝜌}$ , $𝐸_{0} = (𝜌 \mapsto 𝑅_{0} - e^{- 𝜌}) (𝐸_{0}^{'})$ , so that $𝑇_{0} (𝜌) = 𝑇 (𝑟)$ , and
$\int_{𝐸_{0}} \frac{d 𝑟}{𝑅_{0} - 𝑟} \leq 2, 𝑇 (𝑅_{0} - (𝑅_{0} - 𝑟) e^{- \frac{1}{𝑇 (𝑟)}}) < 2 𝑇 (𝑟) \forall 𝑟 \notin 𝐸_{0} .$
We are left to bound the inequality to be in the form of (14.4.1.8) $(14.4.1.11) (14.4.1.8)$ . For $𝑥 \in (0, 1)$ ,
$e^{- 𝑥} < 1 - \frac{e - 1}{e} 𝑥 < 1 - \frac{𝑥}{e},$
thus for $𝑟$ within $[𝑟_{0}, 𝑅_{0})$ outside $𝐸_{0}$ , we have from $e^{- \frac{1}{𝑇 (𝑟)}} < 1 - \frac{1}{e 𝑇 (𝑟)}$ . Hence,
$\begin{aligned} 𝑇 (𝑅_{0} - (𝑅_{0} - 𝑟) e^{- \frac{1}{𝑇 (𝑟)}}) & \geq 𝑇 (𝑅_{0} - (𝑅_{0} - 𝑟) (1 - \frac{1}{e 𝑇 (𝑟)})) \\ = 𝑇 (𝑟 + \frac{𝑅_{0} - 𝑟}{e 𝑇 (𝑟)}) . \end{aligned}$
If $𝜌, 𝜌^{'}$ are chosen such that $𝑅_{0} - 𝜌^{'} < e^{- 2} (𝑅_{0} - 𝜌)$ , then
$\int_{𝜌}^{𝜌^{'}} \frac{d 𝑟}{𝑅_{0} - 𝑟} = log (\frac{𝑅_{0} - 𝜌}{𝑅_{0} - 𝜌^{'}}) > log (e^{2}) = 2 .$
Since $\int_{𝐸_{0}} \frac{d 𝑟}{𝑟_{0} - 𝑟} \leq 2$ , there then $(𝜌, 𝜌^{'}) \subset . \neq 𝐸_{0}$ . Thus, $\exists 𝑟 \in (𝜌, 𝜌^{'})$ such that $𝑟 \notin 𝐸_{0}$ . $□$

Theorem 14.4.1.5 (Second Fundamental Theorem of Nevanlinna Theory, Part 2).

If $𝑅_{0} = \infty$ and $𝑓$ has finite order, then Let $𝑓 : 𝐷 (0, 𝑅_{0}) \to \hat{ℂ}$ ( $0 < 𝑅_{0} \leq \infty$ ) be meromorphic and non-constant. Define $𝑆 (𝑟, 𝑓)$ as in (14.4.1) $(14.4.1) (14.4.1)$ . Then
$𝑆 (𝑟, 𝑓) = 𝒪︀ (log 𝑇 (𝑟, 𝑓)) + 𝒪︀ (log 𝑟) (†)$
as $𝑟 \to \infty$ through all values.
If $𝑅_{0} = \infty$ and $𝑓$ has infinite order, then $(†)$ holds as $𝑟 \to \infty$ through all values of $𝑟$ outside a set $𝐸$ of finite linear measure.
If $𝑅_{0} < \infty$ , then
$𝑆 (𝑟, 𝑓) = 𝒪︀ ({log}^{+} 𝑇 (𝑟, 𝑓) + log \frac{1}{𝑅_{0} - 𝑟}) (‡)$
as $𝑟 \to 𝑅_{0}^{-}$ outside a set $𝐸$ satisfying
$\int_{𝐸} \frac{d 𝑟}{𝑅_{0} - 𝑟} < \infty .$
Moreover, for any $𝜌, 𝜌^{'} \in (0, 𝑅_{0})$ such that $𝑅_{0} - 𝜌^{'} < e^{- 2} (𝑅_{0} - 𝜌)$ (implying that $𝜌 < 𝜌^{'}$ ), $\exists 𝑟 \in (𝜌, 𝜌^{'})$ such that $𝑟 \notin 𝐸$ .

Proof.

As a function of $𝑟$ ,

𝑆 (𝑟, 𝑓) = 𝑚 (𝑟, \frac{𝑓^{'}}{𝑓}) + 𝑚 (𝑟, \frac{𝜙^{'}}{𝜙}) + constant,

where $𝜙 (𝑧) = \prod_{𝜈 = 1}^{𝑞} (𝑓 (𝑧) - 𝑎_{𝜈})$ , where the constant term does not depend on $𝑟$ . In all three perceivable cases,

\begin{matrix} 𝑇 (𝑟, 𝜙) \leq \sum_{𝜈 = 1}^{𝑞} 𝑇 (𝑟, 𝑓 - 𝑎_{𝜈}) = 𝑞 𝑇 (𝑟, 𝑓) + 𝒪︀ (1), \\ {log}^{+} 𝑇 (𝑟, 𝜙) \leq {log}^{+} 𝑇 (𝑟, 𝑓) + 𝒪︀ (1) (14.4.1 . 9) \end{matrix}

by Proposition 14.2.5 $Proposition 14.2.5 Proposition 14.2.5$ (product-subadditivity), Theorem 14.2.4 $Theorem 14.2.4 Theorem 14.2.4$ (first fundamental theorem). In the first case, where $𝑅_{0} = \infty$ , by Proposition 14.4.1.3 $Proposition 14.4.1.3 Proposition 14.4.1.3$ with $𝑅 = 2 𝑟$ ,

\begin{aligned} 𝑚 (𝑟, \frac{𝑓^{'}}{𝑓}) & < 4 {log}^{+} 𝑇 (𝑅, 𝑓) + 𝒪︀ (1) + 5 {log}^{+} (2 𝑟) \\ = 𝒪︀ ({log}^{+} 𝑇 (2 𝑟, 𝑓)) + 𝒪︀ ({log}^{+} 𝑟) . \\ 𝑚 (𝑟, \frac{𝜙^{'}}{𝜙}) & < 4 {log}^{+} 𝑇 (𝑅, 𝜙) + 𝒪︀ (1) + 5 {log}^{+} (2 𝑟) \\ = 𝒪︀ ({log}^{+} 𝑇 (2 𝑟, 𝜙)) + 𝒪︀ ({log}^{+} 𝑟) . \end{aligned}

Since $𝑓$ has finite order, $𝑇 (2 𝑟, 𝑓) = 𝒪︀ ({(2 𝑟)}^{𝜌 + 𝜀}) = 𝒪︀ (𝑟^{𝜚})$ , so ${log}^{+} 𝑇 (2 𝑟, 𝑓) = 𝒪︀ ({log}^{+} 𝑟)$ , thus as $𝑟 \to \infty$ , using (14.4.1.9) $(14.4.1.12) (14.4.1.9)$ we have

\begin{matrix} 𝑚 (𝑟, \frac{𝑓^{'}}{𝑓}) = 𝒪︀ ({log}^{+} 𝑇 (2 𝑟, 𝑓)) + 𝒪︀ ({log}^{+} 𝑟) = 𝒪︀ (log 𝑟), \\ 𝑚 (𝑟, \frac{𝜙^{'}}{𝜙}) = 𝒪︀ ({log}^{+} 𝑇 (2 𝑟, 𝜙)) + 𝒪︀ ({log}^{+} 𝑟) = 𝒪︀ (log 𝑟) . \end{matrix}

Then $𝑆 (𝑟, 𝑓) = 𝒪︀ (log 𝑟)$ , clearly satisfying $(†)$ .

Now assume $𝑓$ is not necessarily of finite order and $𝑅_{0} = \infty$ . Then by Proposition 14.4.1.3 $Proposition 14.4.1.3 Proposition 14.4.1.3$ with $𝑅 = 𝑟 + \frac{1}{𝑇 (𝑟)}$ ,

\begin{matrix} 𝑚 (𝑟, \frac{𝑓^{'}}{𝑓}) < 𝒪︀ ({log}^{+} 𝑇 (𝑟 + \frac{1}{𝑇 (𝑟, 𝑓)}, 𝑓)) + 𝒪︀ (1) + 𝒪︀ ({log}^{+} 𝑟) \\ 𝑚 (𝑟, \frac{𝜙^{'}}{𝜙}) < 𝒪︀ ({log}^{+} 𝑇 (𝑟 + \frac{1}{𝑇 (𝑟, 𝜙)}, 𝜙)) + 𝒪︀ (1) + 𝒪︀ ({log}^{+} 𝑟) \end{matrix}

By Lemma 14.4.1.4 $Lemma 14.4.1.4 Lemma 14.4.1.4$ , there exist some exceptional sets $𝐸_{1}, 𝐸_{2}$ each of linear measure not exceeding 2 such that for $𝑟 \notin 𝐸_{1} \cup 𝐸_{2}$ ,

\begin{matrix} {log}^{+} 𝑇 (𝑟 + \frac{1}{𝑇 (𝑟, 𝑓)}, 𝑓) \leq {log}^{+} 𝑇 (𝑟, 𝑓) + 𝒪︀ (1) \\ {log}^{+} 𝑇 (𝑟 + \frac{1}{𝑇 (𝑟, 𝜙)}, 𝜙) \leq {log}^{+} 𝑇 (𝑟, 𝜙) + 𝒪︀ (1) \end{matrix}

thus for $𝑟 \to \infty$ with $𝑟 \notin 𝐸_{0}$ ,

𝑆 (𝑟, 𝑓) = 𝒪︀ ({log}^{+} 𝑇 (𝑟, 𝑓)) + 𝒪︀ ({log}^{+} 𝑟) .

Lastly, assume that $𝑅_{0} < \infty$ . Now apply Proposition 14.4.1.3 $Proposition 14.4.1.3 Proposition 14.4.1.3$ with $𝑅 = 𝑟 + \frac{𝑅_{0} - 𝑟}{e 𝑇 (𝑟, 𝑓)}$ and $𝑅 = 𝑟 + \frac{𝑅_{0} - 𝑟}{e 𝑇 (𝑟, 𝜙)}$ , so that for $𝑟$ outside some exceptional set satisfying $\int_{𝐸} \frac{d 𝑟}{𝑅_{0} - 𝑟} \leq 4 < \infty$ , by Lemma 14.4.1.4 $Lemma 14.4.1.4 Lemma 14.4.1.4$ ,

\begin{aligned} 𝑚 (𝑟, \frac{𝑓^{'}}{𝑓}) & < 𝒪︀ ({log}^{+} 𝑇 (𝑟 + \frac{𝑅_{0} - 𝑟}{e 𝑇 (𝑟, 𝑓)}, 𝑓)) + 𝒪︀ (1) \\ \begin{matrix}  \end{matrix} + 𝒪︀ ({log}^{+} (𝑟 + \frac{𝑅_{0} - 𝑟}{e 𝑇 (𝑟, 𝑓)})) + 𝒪︀ ({log}^{+} \frac{e 𝑇 (𝑟, 𝑓)}{𝑅_{0} - 𝑟}), \\ < 𝒪︀ ({log}^{+} 𝑇 (𝑟, 𝑓)) + 𝒪︀ (1) + 𝒪︀ ({log}^{+} \frac{e 𝑇 (𝑟, 𝑓)}{𝑅_{0} - 𝑟}) \\ = 𝒪︀ ({log}^{+} 𝑇 (𝑟, 𝑓) + log \frac{1}{𝑅_{0} - 𝑟}), \end{aligned}

where the $𝒪︀ ({log}^{+} (𝑟 + \frac{𝑅_{0} - 𝑟}{e 𝑇 (𝑟, 𝑓)}))$ term has been coalesced to $𝒪︀ (1)$ as $𝑟 + \frac{𝑅_{0} - 𝑟}{e 𝑇 (𝑟, 𝑓)}$ is bounded.

\begin{aligned} 𝑚 (𝑟, \frac{𝜙^{'}}{𝜙}) & = 𝒪︀ ({log}^{+} 𝑇 (𝑟, 𝜙) + log \frac{1}{𝑅_{0} - 𝑟}) \\ \leq 𝒪︀ ({log}^{+} 𝑇 (𝑟, 𝑓) + log \frac{1}{𝑅_{0} - 𝑟}) . (by (14.4.1.9) (14.4.1.12) (14.4.1.9)) \end{aligned}

Therefore,

𝑆 (𝑟, 𝑓) = 𝒪︀ ({log}^{+} 𝑇 (𝑟, 𝑓) + log \frac{1}{𝑅_{0} - 𝑟}) . □

Proposition 14.4.1.6.

𝑓 : ℂ \to \hat{ℂ}

is meromorphic and

𝑇 (𝑟, 𝑓) = 𝒪︀ (log 𝑟)

, then

𝑓

is a rational function.

Proof.

We must have (for any $𝑎 \in \hat{ℂ}$ ) $𝑁 (𝑟, 𝑎, 𝑓) = 𝒪︀ (log 𝑟)$ . We assert that $𝑓$ attains each value finitely many times. Otherwise, for each $𝑘 \in ℕ$ , there exists $𝑟^{'}$ such that $\forall 𝑟 > 𝑟^{'}$ , $𝑛 (𝑟, 𝑎, 𝑓) > 𝑘$ . Then

𝑁 (𝑟, 𝑎, 𝑓) = 𝑐 + \int_{𝑟^{'}}^{𝑟} \frac{𝑛 (𝑥, 𝑎, 𝑓) - 𝑛 (0, 𝑎, 𝑓)}{𝑥} d 𝑥 + 𝑛 (0, 𝑎, 𝑓) log 𝑟 > 𝑐 + 𝑘 log 𝑟 .

Thus, $𝑁$ eventually grows faster than $𝑘 log 𝑟$ for any $𝑘$ . It follows that each $𝑎$ is attained finitely many times by contradiction. Construct $𝑔 = 𝒪︀ ({| 𝑧 |}^{𝑑})$ to be a polynomial with its zeros at each pole (finitely many), counting orders, of $𝑓$ .

Then $ℎ = 𝑓 𝑔$ is entire, and furthermore, $𝑇 (𝑟, ℎ) \leq 𝑇 (𝑟, 𝑓) + 𝑇 (𝑟, 𝑔)$ . Since $𝑇 (𝑟, 𝑓) = 𝒪︀ (log 𝑟)$ and

𝑇 (𝑟, 𝑔) = 𝑚 (𝑟, 𝑔) \leq {log}^{+} 𝑀 (𝑟, 𝑔) = 𝒪︀ (log 𝑟),

then

log 𝑀 (𝑟, ℎ) \leq 3 𝑇 (2 𝑟, ℎ) = 𝒪︀ (log 𝑟) . (by Theorem 14.2.1.1 Theorem 14.2.1.1 Theorem 14.2.1.1 with 𝑅 = 2 𝑟)

Then

𝑀 (𝑟, ℎ) \leq 𝑎 𝑟^{𝑏} + 𝑐 \forall 𝑟 > 0,

where $𝑎, 𝑏, 𝑐$ are constants. Thus the growth of $ℎ$ may not exceed that of a polynomial. By Cauchy’s Estimate (Theorem 3.2.3 $Theorem 3.2.3 Theorem 3.2.3$ ), for any natural number $𝑛$ and positive $𝑟$ ,

| ℎ^{(𝑛)} (0) | \leq \frac{𝑛! (𝑎 𝑟^{𝑏} + 𝑐)}{𝑟^{𝑛}} = 𝒪︀ (𝑟^{𝑏 - 𝑛}) .

By letting $𝑟 \to \infty$ , it follows that $ℎ^{(𝑛)} (0) = 0$ for any $𝑛 > 𝑏$ . Then there are finitely many terms in the Taylor expansion of $ℎ$ , and therefore $ℎ$ is a polynomial. Consequently $𝑓$ is the quotient of two polynomials. ⁠ $□$

Definition 14.4.1.7.

Let $𝑓 : 𝐷 (0, 𝑅_{0}) \to \hat{ℂ}$ be meromorphic for $𝑅_{0} \in ℝ_{> 0} \cup {\infty}$ . Then $𝑓$ is said to be admissible if either of the following are true:

$𝑅_{0} = \infty$ and $𝑓$ is non-constant.
$𝑅_{0} < \infty$ and ${lim sup}_{𝑟 \to 𝑅_{0}^{-}} \frac{𝑇 (𝑟, 𝑓)}{log \frac{1}{𝑅_{0} - 𝑟}} = \infty$ .

These definitions ensure that $𝑆 (𝑟, 𝑓)$ is generally sufficiently small:

Theorem 14.4.1.8.

An admissible meromorphic function $𝑓 : 𝐷 (0, 𝑅_{0}) \to \hat{ℂ}$ (given for $𝑅_{0} \in ℝ_{> 0} \cup {\infty}$ ) pertains to one of the following cases:

If $𝑅_{0} = \infty$ and $𝑓$ has finite order, then $𝑆 (𝑟, 𝑓) = ℴ︀ (𝑇 (𝑟, 𝑓))$ for $𝑟 \to 𝑅_{0}^{-}$ through all values.
If $𝑅_{0} = \infty$ and $𝑓$ has infinite order, then $𝑆 (𝑟, 𝑓) = ℴ︀ (𝑇 (𝑟, 𝑓))$ for $𝑟 \to 𝑅_{0}^{-}$ outside a set $𝐸_{0}$ of finite linear measure, which depends on $𝑓$ but not on ${𝑎_{𝜈}}$ and $𝑞$ .
If $𝑅_{0} < \infty$ , then $𝑆 (𝑟, 𝑓) = ℴ︀ (𝑇 (𝑟, 𝑓))$ for $𝑟 \to 𝑅_{0}^{-}$ through any sequence ${𝑟_{𝑛}}_{𝑛}$ dependent on $𝑓$ but not on ${𝑎_{𝜈}}$ and $𝑞$ .

Proof.

Assume the conditions of the theorem.

Assume $𝑅_{0} = \infty$ and $𝑓$ has finite order. Then $𝑇 (𝑟, 𝑓) = 𝒪︀ (𝑟^{𝜌 + 𝜀})$ , $𝑟^{𝜌 - 𝜀} = ℴ︀ (𝑇 (𝑟, 𝑓))$ (assuming that $𝜌 > 0$ ), so $log 𝑟 = ℴ︀ (𝑇 (𝑟, 𝑓))$ . Therefore, by Theorem 14.4.1.5 $Theorem 14.4.1.5 Theorem 14.4.1.5$ , it follows that through all values as $𝑟 \to \infty$ ,
$𝑆 (𝑟, 𝑓) = 𝒪︀ (log 𝑇 (𝑟, 𝑓)) + 𝒪︀ (log 𝑟) = 𝒪︀ (log 𝑇 (𝑟, 𝑓)) = ℴ︀ (𝑇 (𝑟, 𝑓)) .$
If $𝜌 = 0$ , either $log 𝑟 = ℴ︀ (𝑇 (𝑟, 𝑓))$ or $𝑇 (𝑟, 𝑓) = 𝒪︀ (log 𝑟)$ . The former case is the same as above.
In the latter case, Proposition 14.4.1.6 $Proposition 14.4.1.6 Proposition 14.4.1.6$ gives that $𝑓$ is rational. Hence, as $𝑟 \to \infty$ , $𝑀 (𝑟, \frac{𝑓^{'}}{𝑓}), 𝑀 (𝑟, \sum_{𝜈 = 1}^{𝑞} \frac{𝑓^{'}}{𝑓 - 𝑎_{𝜈}}) \to 0$ . Hence for sufficiently large $𝑟$ , $𝑚 (𝑟, \frac{𝑓^{'}}{𝑓})$ and $𝑚 (𝑟, \sum_{𝜈 = 1}^{𝑞} \frac{𝑓^{'}}{𝑓 - 𝑎_{𝜈}})$ vanish and $𝑆 (𝑟, 𝑓) = 𝒪︀ (1)$ , therefore $\frac{𝑆}{𝑇} \to 0$ .
Assume $𝑅_{0} = \infty$ and $𝑓$ has infinite order. Then by Theorem 14.4.1.5 $Theorem 14.4.1.5 Theorem 14.4.1.5$ , it follows that for $𝑟 \to 𝑅_{0}^{-}$ outside a set $𝐸_{0}$ of finite linear measure, $𝑆 (𝑟, 𝑓) = 𝒪︀ (log 𝑇 (𝑟, 𝑓)) + 𝒪︀ (log 𝑟)$ . Since $\frac{log 𝑇 (𝑟, 𝑓)}{log 𝑟} \to \infty$ , then $log 𝑟 = ℴ︀ (𝑇 (𝑟, 𝑓))$ , and thus $𝑆 (𝑟, 𝑓) = ℴ︀ (𝑇 (𝑟, 𝑓))$ outside this exceptional set.
Assume $𝑅_{0} < \infty$ . Construct two sequences ${𝜌_{𝑛}}_{𝑛}, {𝜌_{𝑛}^{'}}_{𝑛}$ such that $𝜌_{𝑛} < 𝜌_{𝑛}^{'}$ , $𝜌_{𝑛}, 𝜌_{𝑛}^{'} \to 𝑅_{0}^{-}$ , and
$lim_{𝑛 \to \infty} \frac{𝑇 (𝜌_{𝑛}, 𝑓)}{log \frac{1}{𝑅_{0} - 𝜌_{𝑛}}} = \infty$
(observe that this is no longer a limit superior) and $𝜌_{𝑛}^{'} = 𝑅_{0} - e^{- 2} \frac{𝑅_{0} - 𝜌_{𝑛}}{2}$ , so that $𝑅_{0} - 𝜌_{𝑛}^{'} < e^{- 2} (𝑅_{0} - 𝜌_{𝑛})$ .
Then by Theorem 14.4.1.5 $Theorem 14.4.1.5 Theorem 14.4.1.5$ , there exists some $𝑟_{𝑛} \in (𝜌_{𝑛}, 𝜌_{𝑛}^{'})$ such that $𝑟_{𝑛} \notin 𝐸_{0}$ . It follows that $𝑅_{0} - 𝜌_{𝑛} > 𝑅_{0} - 𝑟_{𝑛} > 𝑅_{0} - 𝜌_{𝑛}^{'} = \frac{𝑅_{0} - 𝜌_{𝑛}}{2 e^{2}}$ . Then for $𝑛 \to \infty$ ,
$\frac{1}{𝑅_{0} - 𝜌_{𝑛}} = ℴ︀ (𝑇 (𝜌_{𝑛})) = ℴ︀ (𝑇 (𝑟_{𝑛})),$
since $𝑇$ is a non-decreasing function of radius. Then as $𝑛 \to \infty$ ,
$𝑆 (𝑟_{𝑛}, 𝑓) = 𝒪︀ [{log}^{+} 𝑇 (𝑟_{𝑛}, 𝑓) + log ℴ︀ (𝑇 (𝑟_{𝑛}, 𝑓))] = ℴ︀ (𝑇 (𝑟_{𝑛}, 𝑓)) . □$

§14.4.1 The Estimation of 𝑆(𝑟,𝑓)

14.4.1 The Estimation of 𝑆(𝑟,𝑓)

§14.4.1 The Estimation of $𝑆 (𝑟, 𝑓)$

14.4.1 The Estimation of $𝑆 (𝑟, 𝑓)$