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Transcendental characteristic dominates logarithmic growth

Statement

Let $f$ be a nonconstant meromorphic function on $C$ . If $f$ is transcendental, then $\frac{T (r, f)}{\log r} ⟶ + \infty (r \to \infty) .$ Consequently, off the exceptional set belonging to any occurrence of $S (r, f)$ , the right-hand side $C (\log^{+} T (r, f) + \log r)$ is $o (T (r, f))$ .

If $f$ is rational of degree $d \geq 1$ , then $T (r, f) = d \log r + O (1)$ , and this $O (1)$ term is $o (T (r, f))$ .

Facts & Assumptions

Given: A nonconstant meromorphic function $f$ on $C$ , with chordal characteristic $T (r, f)$ as in Counting, chordal proximity and characteristic.

[F1]

Ahlfors–Shimizu: $T (r, f) = T_{A S} (r, f) + C_{\infty} (f)$ for a constant $C_{\infty} (f)$ , where $T_{A S}$ is finite and nondecreasing in $r$ and convex as a function of $\log r$ (Ahlfors–Shimizu area form of the characteristic).

[F2]

$f$ is rational if and only if $T (r, f) = O (\log r)$ ; more precisely, if $f$ is rational of degree $d \geq 1$ , then $T (r, f) = d \log r + O (1)$ (Rational functions are exactly those with logarithmic characteristic).

[F3]

$T$ is nondecreasing, and $T (r, f) > 1$ for all sufficiently large $r$ ; the characteristic is finite for every $r > 0$ (Counting, chordal proximity and characteristic).

Proof

technique · shift the convex Ahlfors–Shimizu potential to the logarithmic variable; a bounded ratio on an unbounded sequence and the convexity chord bound force $T=O(\log r)$, which the rational characterization converts into a contradiction with transcendence

1.1F1F3construct

Put $u (x) : = T (e^{x}, f)$ for $x \in R$ . By [F1], $u (x) - C_{\infty} (f)$ is convex and nondecreasing in $x$ ; hence $u$ itself is convex and nondecreasing. By [F3], $u (x) > 1$ for all sufficiently large $x$ .

1.2F2F3algebra

Since $T$ is nondecreasing by [F3], the limit $L : = \lim_{r \to \infty} T (r, f)$ exists in $(0, + \infty]$ . If $L < \infty$ , then $T (r, f) = O (1) = O (\log r)$ , so [F2] would make $f$ rational, contrary to transcendence; hence $L = + \infty$ , that is, $T (r, f) \to \infty$ .

2.1step 1.1algebra

(Convexity chord bound) Put $x_{0} : = 1$ . By convexity of $u$ from step 1.1, for all $x_{0} \leq x \leq y$ , $u (x) \leq u (x_{0}) + \frac{x - x_{0}}{y - x_{0}} (u (y) - u (x_{0})) .$

3.1step 1.1step 2.1algebra

Assume for contradiction that ${lim inf}_{r \to \infty} T (r, f) / \log r < \infty$ . Then there are a constant $C < \infty$ and a sequence $x_{n} \to + \infty$ with $u (x_{n}) \leq C x_{n}$ for every $n$ . Discard finitely many terms so that $x_{n} \geq 2 x_{0}$ for all $n$ ; applying the chord bound of step 2.1 with $y = x_{n}$ to each $x \in [x_{0}, x_{n}]$ gives $u (x) \leq u (x_{0}) + C x_{n} \frac{x - x_{0}}{x_{n} - x_{0}} \leq u (x_{0}) + 2 C (x - x_{0}),$ because $x_{n} \geq 2 x_{0}$ gives $x_{n} / (x_{n} - x_{0}) \leq 2$ . Thus $T (e^{x}, f) = u (x) \leq C^{'} x$ for all $x \geq 1$ , i.e. $T (r, f) = O (\log r)$ .

4.1F2step 3.1discharge-contradiction

By [F2] the bound $T (r, f) = O (\log r)$ makes $f$ rational, contradicting the hypothesis. Therefore ${lim inf}_{r \to \infty} T (r, f) / \log r = + \infty$ , which for a nonnegative function is the assertion $T (r, f) / \log r \to + \infty$ .

5.1step 1.2step 4.1algebra

By step 1.2, $T (r, f) \to \infty$ , so $\log^{+} T (r, f) = o (T (r, f))$ ; by step 4.1, $\log r = o (T (r, f))$ . Hence $\log^{+} T (r, f) + \log r = o (T (r, f))$ : for every $ε > 0$ the inequality $C (\log^{+} T (r, f) + \log r) \leq ε T (r, f)$ holds for all sufficiently large $r$ . In particular this applies to the right-hand side of any occurrence of $S (r, f)$ at every nonexceptional large radius.

6.1F2algebra∎

If $f$ is rational of degree $d \geq 1$ , then [F2] gives $T (r, f) = d \log r + O (1) \geq \frac{d}{2} \log r$ for all large $r$ , so $T (r, f) \to \infty$ and $O (1) = o (T (r, f))$ .

Depends on

Used by

Little and Great Picard consequences of Nevanlinna theory Corollary
Nevanlinna deficiency and ramification index Definition
Nevanlinna deficiency and ramification defect relations Theorem
Nevanlinna five-value uniqueness theorem Theorem

Dependency tree · two levels

15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Alexandre Eremenko, Lectures on Nevanlinna Theory, §§4–6 (standard reference, not scraped)
Goldberg–Ostrovskii, Value Distribution of Meromorphic Functions (standard reference, not scraped)