Bounded type of meromorphic functions omitting three half-plane values
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Must a meromorphic function omitting three values in a half-plane be of bounded type there? The counterexample target is a real meromorphic function whose 0-, 1- and ∞-points are all real and infinite, while every other value’s divisor fails the Blaschke condition in both half-planes.
Why it matters and what counts as progress
Why it matters: It tests a proposed link between value omission and bounded-type growth, with a stronger divisor obstruction for all other values.
What counts as progress: A meromorphic function violating the proposed half-plane bounded-type implication, with the omitted values and divisor conditions proved exactly.
How the problem developed
- 1925 · Theorem and question [1]
Rolf Nevanlinna Developed the value-distribution framework behind the half-plane problem of whether omission of three values forces bounded type.
- 25 Aug 2026 · Counterexample [3]
Yixin He and Teng Zhang Constructed an entire meromorphic function whose preimages of 0, 1, and infinity lie on the real axis, while its restriction to the upper half-plane is not of bounded type, answering the question negatively.
- 26 Aug 2026 · Partial result [2]
Tang and coauthors, with GPT-5.6 Sol Ultra Proved bounded-type criteria for specified half-plane domains and constructed examples showing failure under weaker boundary-divisor conditions.
AI contribution
Three omitted values and non-Blaschke point divisors in half-planes
Tang et al.’s Theorem 1.2 constructs a real meromorphic F on C with all 0-, 1- and ∞-points on the real axis, each infinite, and for every other spherical value a, the a-point divisor in each half-plane fails the Blaschke condition. Consequently F is not of bounded type in either half-plane despite omitting three prescribed values there; Möbius postcomposition and affine precomposition extend the example to any triple and Euclidean half-plane.
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AI-origin meromorphic example defeats half-plane bounded type