03/26/2014, 03:34 PM

(03/26/2014, 01:23 PM)tommy1729 Wrote: Considering the threadSorry but you dragged me where I can not swim... I don't know what entire, linear independence, analyticity and pseudounivalent functions are...

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

I need alot of time to understand this..

following your post and the definition of univalent function whe have that an u. function is injective (and holomorphic from wiki) thus invertible.

Anyways every invertible function it is the superfunction of another functions and then it is twice a superfunction too since it defines in an unique way the iterates of its subfunction.

So back to my poor knowledge:

I don't know if your weaker assumption (pseudounivalent) for a function is enough to ensure the existences of 2 ("sub")functions that commute (or other) and to prove that f is twice a superfunction of them.

MathStackExchange account:MphLee