When a meromorphic function that omits three values has a bounded type

TL;DR

This paper investigates conditions under which a meromorphic function omitting three values in a specific domain is of bounded type, linking this property to the convergence of a logarithmic integral of a domain-defining function.

Contribution

It establishes a criterion based on the convergence of a logarithmic integral of m for the boundedness of meromorphic functions omitting three values in a domain.

Findings

If the logarithmic integral of m converges, the function is of bounded type.

If the logarithmic integral of m diverges, there exists an unbounded type function omitting three values.

The results relate to a longstanding question posed by Rolf Nevanlinna.

Abstract

Suppose that a function $F$ is meromorphic in the domain $\mathbb H(-m) = \{ z : \mathrm{Im}\, z > -m(\mathrm{Re}\, z) \}$, where $m$ is an even, positive, and continuous function that does not increase on $\mathbb R_{\ge 0}$, and suppose that $F$ omits there three distinct values. Then $F$ is of bounded type in the upper half-plane (i.e., is represented there as a quotient of two bounded analytic functions), provided that the logarithmic integral of the function $m$ is convergent. On the other hand, if the logarithmic integral of $m$ diverges, there exists a function $F$ meromorphic in $\mathbb H(-m)$, that omits there three distinct values, and which is of unbounded type in the upper half-plane. This result is motivated by a century old question originating with Rolf Nevanlinna, which remains unsettled.