03/26/2014, 01:23 PM
Considering the thread
http://math.eretrandre.org/tetrationforu...hp?tid=844
It seems to follow that from the (probably) fact that sexp is not pseudounivalent then a superfunction cannot be twice a superfunction on a half-plane.
The proof sketch is as follows : Let f be a nonspeudounivalent superfunction.
Lef f(a)= b.
Then there must exist for most such a and b :
f(a2)=f(a3)=f(a)=b
where a2 and a3 are lineair indep. (real lineair indep of course )
then from the 2 functional equations that make up twice the superfunction it follows that f must be double periodic.
( the superfunctions equations are in the directions a-a2 and a-a3 )
Since double periodic functions are not entire , then the superfunction cannot be analytic on a halfplane.
regards
tommy1729
http://math.eretrandre.org/tetrationforu...hp?tid=844
It seems to follow that from the (probably) fact that sexp is not pseudounivalent then a superfunction cannot be twice a superfunction on a half-plane.
The proof sketch is as follows : Let f be a nonspeudounivalent superfunction.
Lef f(a)= b.
Then there must exist for most such a and b :
f(a2)=f(a3)=f(a)=b
where a2 and a3 are lineair indep. (real lineair indep of course )
then from the 2 functional equations that make up twice the superfunction it follows that f must be double periodic.
( the superfunctions equations are in the directions a-a2 and a-a3 )
Since double periodic functions are not entire , then the superfunction cannot be analytic on a halfplane.
regards
tommy1729