User talk:Dmitrii Kouznetsov/Analytic Tetration

Henryk Trappmann 's theorems

This is approach to the Second part of the Theorem 0, which is still absent in the main text.

Copypast from http://math.eretrandre.org/tetrationforum/showthread.php?tid=165&pid=2458#pid2458

Theorem T1. (about Gamma function)

Let $~F~$ be holomorphic on the right half plane let $~F(z+1)=zF(z)~$ for all $~z~$ such that $~\Re (z)>0~$ .
Let $~F(1)=1~$ .
Let $F$ be bounded on the strip $~1\leq \Re (z)<2~$ .
Then $~F~$ is the gamma function.

Proof see in Reinhold Remmert, "Funktionnentheorie", Springer, 1995. As reference is given: H. Wielandt 1939. (Mein Gott, so old reference!)

Consider function $~v=F-\Gamma ~$ on the right half plane, it also satisfies equation $~v(z+1)~=~z~v(z)~$ Hence, $~v~$ has a meromorphic continuation to $~\mathbb {C} ~$ ; and the poles are allowed only at non–positive integer values of the argument.

While $~v(1)=0~$ , we have $~\lim _{z\rightarrow 0}z~v(z)=0~$ , hence, $~v~$ has a holomorphic continuation to 0 and also to each $~-n~$ , $~n\in \mathbb {N}$ by $~v(z+1)=z~v(z)~$ .

In the range $~1\leq \Re (z)<2~$ , $~v(z)~<math>ispounded.Itisbecausefunction<math>~\Gamma ~$ is bounded there.

Then $~v(z)~$ is also restricted on $~\mathbb {S} ~$ , because $~v(z)!$ and $~v(1-z)!$ have the same values on $~\mathbb {S} ~$ . Now $~q(z+1)=-q(z)~$ , hence $~q~$ is bounded on whole $~\mathbb {C} ~$ , and by the Liouville Theorem, $~q(z)=q(1)=0$ . Hence, $~v=0~$ and $~F=\Gamma ~$ .

(end of proof)

Theorem T2 (about exponential)

Let $~E~$ be solution of $~E(z+1)=bE(x)~$ , $~E(1)=b~$ , bounded in the strip $~0\leq \Re (z)<1~$ .

Then $~E~$ is exponential on base $~b~$ , id est, $~E=\exp _{b}~$ .

Proof. We know that every other solution must be of the form $~g(z)=f(z+p(z))~$ where $~p~$ is a 1-periodic holomorphic function. This can roughly be seen by showing periodicity of $~h(z)=f^{-1}(g(z))-z~$ .