06/20/2009, 07:27 PM
(This post was last modified: 07/06/2009, 12:08 AM by Base-Acid Tetration.)
Theorem.
Let \( f \) be a function that is biholomorphic on both of the initial regions \( D_1 \) and \( D_2 \) that share as boundaries the same conjugate pair of fixed points of f. Also let f have no other fixed points. Let there be two Abel functions of f, \( A_1 \) and \( A_2 \), that are biholomorphic on these initial regions and satisfy A(d) = c.
For each f, there exists exactly one biholomorphism \( A_{cont}(z) \) on a single simply connected open set \( C \supseteq D_1,\,D_2 \) such that \( \forall z \in D_i\, A_{cont}(z) = A_i (z). \) (i is an index that can be 1 or 2) i.e. there is an analytic continuation, and it's unique.
Proof.
1.
Let \( D_1, D_2 \) be disjoint, simply connected domains that have as boundaries:
(1) \( \partial_1 D,\, \partial_2 D \not \subset D \), disjoint curves which are homeomorphic to (0,1);
(2) \( L \) and \( \bar{L} \), which are boundaries of, but not contained in, \( \partial _1 D_1, \, \partial_2 D_1,\, \partial_1 D_2, \,\partial_2 D_2 \).
2.
Let \( f \) be a biholomorphism on \( S := \lbrace z:|\Im(z)| < \Im(L) + \epsilon i \rbrace \), where \( \epsilon>0 \), (tried to make the domain of biholomorphism into an open set) that:
(1) bijects \( \partial_1 D \) to \( \partial_2 D \);
(2) has a conjugate pair of fixed points \( L \) and \( \bar{L} \);
(3) has no other fixed points in the domain of biholomorphy.
3.
(1) Let \( A_1 \), a biholomorphism on \( D_1 \), and \( A_2 \), a biholomorphism on \( D_2 \), both satisfy \( A(f(z))=A(z)+1 \) for all applicable z. (for all z such that A(f(z)) is defined)
(2) Let \( A(d)=c \) for some \( d \in S \).
to be continued
Let \( f \) be a function that is biholomorphic on both of the initial regions \( D_1 \) and \( D_2 \) that share as boundaries the same conjugate pair of fixed points of f. Also let f have no other fixed points. Let there be two Abel functions of f, \( A_1 \) and \( A_2 \), that are biholomorphic on these initial regions and satisfy A(d) = c.
For each f, there exists exactly one biholomorphism \( A_{cont}(z) \) on a single simply connected open set \( C \supseteq D_1,\,D_2 \) such that \( \forall z \in D_i\, A_{cont}(z) = A_i (z). \) (i is an index that can be 1 or 2) i.e. there is an analytic continuation, and it's unique.
Proof.
1.
Let \( D_1, D_2 \) be disjoint, simply connected domains that have as boundaries:
(1) \( \partial_1 D,\, \partial_2 D \not \subset D \), disjoint curves which are homeomorphic to (0,1);
(2) \( L \) and \( \bar{L} \), which are boundaries of, but not contained in, \( \partial _1 D_1, \, \partial_2 D_1,\, \partial_1 D_2, \,\partial_2 D_2 \).
2.
Let \( f \) be a biholomorphism on \( S := \lbrace z:|\Im(z)| < \Im(L) + \epsilon i \rbrace \), where \( \epsilon>0 \), (tried to make the domain of biholomorphism into an open set) that:
(1) bijects \( \partial_1 D \) to \( \partial_2 D \);
(2) has a conjugate pair of fixed points \( L \) and \( \bar{L} \);
(3) has no other fixed points in the domain of biholomorphy.
3.
(1) Let \( A_1 \), a biholomorphism on \( D_1 \), and \( A_2 \), a biholomorphism on \( D_2 \), both satisfy \( A(f(z))=A(z)+1 \) for all applicable z. (for all z such that A(f(z)) is defined)
(2) Let \( A(d)=c \) for some \( d \in S \).
to be continued

