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 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 that also has and as non-inclusive boundaries.

(1) Since is biholomorphic on , will still be biholomorphic on the curve , because if is biholomorphic, so is .

(2) We can do this for every curve in that has and as boundaries, to extend the domain of

(3) We can repeat (4.2) as many times as needed to get to . (we can do the above the other way around, using to get from to , because f is BIholomorphic) Then is biholomorphic where 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 where not only the biholomorphism that maps d to c exists, but also

Corollary.

There exists a unique superlogarithm for 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 .

*Now I don't know if in . It must have something to do with the condition . 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 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 that also has and as non-inclusive boundaries.

(1) Since is biholomorphic on , will still be biholomorphic on the curve , because if is biholomorphic, so is .

(2) We can do this for every curve in that has and as boundaries, to extend the domain of

(3) We can repeat (4.2) as many times as needed to get to . (we can do the above the other way around, using to get from to , because f is BIholomorphic) Then is biholomorphic where 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 where not only the biholomorphism that maps d to c exists, but also

Corollary.

There exists a unique superlogarithm for 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 .

*Now I don't know if in . It must have something to do with the condition . For your theorem to apply, I need to prove that both A1 and A2 are equal in some neighborhood of d.