06/22/2009, 12:37 AM
(This post was last modified: 06/22/2009, 04:49 AM by Base-Acid Tetration.)
Please go back up to the beginning of the proof;
I made the conditions a lot stronger: changed the holomorphy condition for \( f \) to BIholomorphy on a strip with infinite range of real parts), and added "f has no other fixed points", because I thought other fixed points would mess things up.
Proof continued...
4.
Consider a simple curve \( \gamma \subset D_1 \) that also has \( L \) and \( \bar{L} \) as non-inclusive boundaries.
(1) Since \( A_1 \) is biholomorphic on \( \gamma \), \( A_1 \) will still be biholomorphic on the curve \( f(\gamma) \), because if \( A_1(z) \) is biholomorphic, so is \( A_1(z)+1=A_1(f(z)) \forall z \in D_1 \).
(2) We can do this for every curve in \( D_1 \) that has \( L \) and \( \bar{L} \) as boundaries, to extend the domain of \( A_1 \)
(3) We can repeat (4.2) as many times as needed to get to \( D_2 \). (we can do the above the other way around, using \( f^{-1} \) to get from \( D_2 \) to \( D_1 \), because f is BIholomorphic) Then \( A_1 \) is biholomorphic where \( A_2 \) was defined to be biholomorphic.* Then we know, by the theorem that was proven on the other day, that they are the same biholomorphism. So we know that there exists a single open set C that includes \( D_1 \cup D_2 \) where not only the biholomorphism \( A_{cont} \)that maps d to c exists, but also \( A_1 = A_2 = A_{cont} \forall z \in C. \)
\( \mathcal{Q. E. D.} \)
Corollary.
There exists a unique superlogarithm for \( b>e^{1/e} \) that uniquely bijects holomorphically each simple initial region of arbitrary "width" in an open set C that: (1) contains in its boundary fixed points of exp_b; (2) does not include branch cuts of the superlogarithm; to its resp. vertically infinite strip of arbitrary width in \( A( C ) \).
*Now I don't know if \( A_1 = A_2 \) in \( D_2 \). It must have something to do with the condition \( A(d) = c \). For your theorem to apply, I need to prove that both A1 and A2 are equal in some neighborhood of d.
I made the conditions a lot stronger: changed the holomorphy condition for \( f \) to BIholomorphy on a strip with infinite range of real parts), and added "f has no other fixed points", because I thought other fixed points would mess things up.
Proof continued...
4.
Consider a simple curve \( \gamma \subset D_1 \) that also has \( L \) and \( \bar{L} \) as non-inclusive boundaries.
(1) Since \( A_1 \) is biholomorphic on \( \gamma \), \( A_1 \) will still be biholomorphic on the curve \( f(\gamma) \), because if \( A_1(z) \) is biholomorphic, so is \( A_1(z)+1=A_1(f(z)) \forall z \in D_1 \).
(2) We can do this for every curve in \( D_1 \) that has \( L \) and \( \bar{L} \) as boundaries, to extend the domain of \( A_1 \)
(3) We can repeat (4.2) as many times as needed to get to \( D_2 \). (we can do the above the other way around, using \( f^{-1} \) to get from \( D_2 \) to \( D_1 \), because f is BIholomorphic) Then \( A_1 \) is biholomorphic where \( A_2 \) was defined to be biholomorphic.* Then we know, by the theorem that was proven on the other day, that they are the same biholomorphism. So we know that there exists a single open set C that includes \( D_1 \cup D_2 \) where not only the biholomorphism \( A_{cont} \)that maps d to c exists, but also \( A_1 = A_2 = A_{cont} \forall z \in C. \)
\( \mathcal{Q. E. D.} \)
Corollary.
There exists a unique superlogarithm for \( b>e^{1/e} \) that uniquely bijects holomorphically each simple initial region of arbitrary "width" in an open set C that: (1) contains in its boundary fixed points of exp_b; (2) does not include branch cuts of the superlogarithm; to its resp. vertically infinite strip of arbitrary width in \( A( C ) \).
*Now I don't know if \( A_1 = A_2 \) in \( D_2 \). It must have something to do with the condition \( A(d) = c \). For your theorem to apply, I need to prove that both A1 and A2 are equal in some neighborhood of d.