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A boundary point whose complementary component contains another point is regular
Statement
Let be a bounded complex domain and let . If the connected component of containing also contains a second point, then is regular for .
Facts & Assumptions
Given: A bounded complex domain , a boundary point , and a second point in the same connected component of as .
For a cycle, the index is locally constant off the trace and vanishes on every connected set in the zero-index region that meets infinity (The index of a cycle is locally constant off its trace and vanishes far from it).
A complex domain is homologically simply connected exactly when every cycle with trace in it is null-homologous there, equivalently when every holomorphic nowhere-zero function on it has a holomorphic logarithm (Null-homologous cycles and homologous cycles in an open set, Equivalent characterisations of a homologically simply connected domain).
Harmonicity is preserved by holomorphic changes of coordinate (Plane harmonicity is preserved by holomorphic and antiholomorphic changes of coordinate).
A weak local subharmonic peak function implies regularity (A weak local subharmonic peak function upgrades to regularity).
Proof
Let lie in the same connected component of as , and let [given, construct] This Möbius map sends to , sends to , and maps biholomorphically onto the domain . Its image is a connected subset of containing both and . Put . Then is an open connected neighbourhood of and .
Let be any cycle whose trace lies in . By [L1], the index is locally constant on and vanishes on the unbounded zero-index region. The connected set is disjoint from , and because also contains , the local constancy from [L1] forces for every . Since , this says exactly that is null-homologous in by [L2]. Therefore is homologically simply connected.
Because is homologically simply connected and misses , [L2] gives a holomorphic logarithm of the identity map on , so for . Hence [L2, step 1.1, construct] Because as through and , choose a small neighbourhood of with . On the function is harmonic, negative, and tends to as through , because . So is a weak local harmonic peak function at for the domain .
The composition is harmonic on by [L3], is negative there, and tends to as through . Thus is a weak local subharmonic peak function at . Applying [L4] shows that is regular for .
Depends on
- A weak local subharmonic peak function upgrades to regularity
- The index of a cycle is locally constant off its trace and vanishes far from it
- Null-homologous cycles and homologous cycles in an open set
- Equivalent characterisations of a homologically simply connected domain
- Plane harmonicity is preserved by holomorphic and antiholomorphic changes of coordinate
Used by
Dependency tree · two levels
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Sources
- Harold P. Boas, Class Notes Math 618: Complex Variables II, Spring 2016 (standard reference, not scraped)
- Boris Khoruzhenko, Potential Theory lecture notes (standard reference, not scraped)