On analytic families of conformal maps
Let Λ be a domain in C and let fλ(z) = z + a0(λ) + a1(λ)z −1 + ... be meromorphic in D∗ := {z ∈ C : |z| 1} ∪ {∞}. We assume that fλ(z) is holomorphic in λ ∈ Λ for fixed z.The main theorem states: Let Λ0 be a subdomain of Λ such that fλ is univalent in D∗ for λ ∈ Λ0. If fλ0 has a quasiconformal exten...
- Autores:
-
Becker, Jochen
Pommerenke, Christian
- Tipo de recurso:
- Article of journal
- Fecha de publicación:
- 2017
- Institución:
- Universidad Nacional de Colombia
- Repositorio:
- Universidad Nacional de Colombia
- Idioma:
- spa
- OAI Identifier:
- oai:repositorio.unal.edu.co:unal/66436
- Acceso en línea:
- https://repositorio.unal.edu.co/handle/unal/66436
http://bdigital.unal.edu.co/67464/
- Palabra clave:
- 51 Matemáticas / Mathematics
Funciones univalentes
extensión cuasiconforme
parámetro analítico
desigualdad de Grunsky
Univalent function
quasiconformal extension
analytic parameter
Grunsky inequality
- Rights
- openAccess
- License
- Atribución-NoComercial 4.0 Internacional
Summary: | Let Λ be a domain in C and let fλ(z) = z + a0(λ) + a1(λ)z −1 + ... be meromorphic in D∗ := {z ∈ C : |z| 1} ∪ {∞}. We assume that fλ(z) is holomorphic in λ ∈ Λ for fixed z.The main theorem states: Let Λ0 be a subdomain of Λ such that fλ is univalent in D∗ for λ ∈ Λ0. If fλ0 has a quasiconformal extension to the closure of D∗ for one λ0 ∈ Λ0 then fλ has a quasiconformal extension for all λ ∈ Λ0.This result is related to a theorem of Mañé, Sad and Sullivan (1983) where the assumptions are however different. The main tool of our proof is the Grunsky inequality for univalent functions. |
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