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A reverse Denjoy theorem

A reverse Denjoy theorem Suppose that C1 and C2 are two simple curves joining 0 to ∞, non‐intersecting in the finite plane except at 0 and enclosing a domain D which is such that, for all large r, {θ:reiθ∈D¯} has measure at most 2α, where 0 < α < π. Suppose also that u is a non‐constant subharmonic function in the plane such that u(z) = B(|z|, u) for all large z ∈ C1 ∪ C2. Let AD(r, u) = inf { u(z):z ∈ D and | z | = r }. It is shown that if AD(r, u) = O(1) (or AD(r, u) = o(B(r, u))), then limr → ∞ B(r, u)/rπ/2α > 0 (or limr→∞ log B(r, u)/log r ⩾ π/2α). http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Bulletin of the London Mathematical Society Wiley

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References (4)

Publisher
Wiley
Copyright
© London Mathematical Society
ISSN
0024-6093
eISSN
1469-2120
DOI
10.1112/blms/bdn098
Publisher site
See Article on Publisher Site

Abstract

Suppose that C1 and C2 are two simple curves joining 0 to ∞, non‐intersecting in the finite plane except at 0 and enclosing a domain D which is such that, for all large r, {θ:reiθ∈D¯} has measure at most 2α, where 0 < α < π. Suppose also that u is a non‐constant subharmonic function in the plane such that u(z) = B(|z|, u) for all large z ∈ C1 ∪ C2. Let AD(r, u) = inf { u(z):z ∈ D and | z | = r }. It is shown that if AD(r, u) = O(1) (or AD(r, u) = o(B(r, u))), then limr → ∞ B(r, u)/rπ/2α > 0 (or limr→∞ log B(r, u)/log r ⩾ π/2α).

Journal

Bulletin of the London Mathematical SocietyWiley

Published: Feb 1, 2009

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