Uniqueness of analytic tetration, scheme of the proof.
Notations.
Let \( \mathbb{C} \) be set of complex numbers.
Let \( \mathbb{R} \) be set of real numbers.
Let \( \mathbb{N} \) be set of integer numbers.
Let \( \mathbb{C}^{\prime}=\mathbb{C} \backslash \{ z| z \in \mathbb{R}, z\le 2 \} \)
Let \( b>1 \) be base of tetration.
Assume that there exist analytic tetration \( F \) on base \( b \), id est,
(0) \( F \) is analytic at \( \mathbb{C}^{\prime} \)
(1) for all \( z\in \mathbb{C}^{\prime} \), the relation \( F(z+1)=\exp_b(F(z)) \) holds
(2) \( F(0)=1 \)
(3) \( F \) is real increasing function at \( \{z \in \mathbb{R}| z>-2 \} \).
Properties.
From assumption (1) and (2) it follows, that function \( F \) has singularity at -2, at -3 and so on.
Consider following Assumption:
There exist entire 1-periodic function \( h \) such that
\( G(z)=F(z+h(z)) \) is also analytic tetration on base \( b \).
Then, function \( I(z)=z+h(z) \) is not allowed to take values -2, -3, ..
being evaluated at elements of \( \mathbb{C}^{\prime} \).
This means that function
\( J(z)=I(z) \frac{z+2}{I(z)+2} =(z+h(z)) \frac{z+2}{z+h(z)+2} \)
is entire function.
(Weak statement which seems to be true)
Function \( J \) cannot grow faster than linear function at infinity in any direction.
Therefore, it is linear function. Therefore, \( h=0 \)
Therefore, there exist only one analytic tetration.
Notations.
Let \( \mathbb{C} \) be set of complex numbers.
Let \( \mathbb{R} \) be set of real numbers.
Let \( \mathbb{N} \) be set of integer numbers.
Let \( \mathbb{C}^{\prime}=\mathbb{C} \backslash \{ z| z \in \mathbb{R}, z\le 2 \} \)
Let \( b>1 \) be base of tetration.
Assume that there exist analytic tetration \( F \) on base \( b \), id est,
(0) \( F \) is analytic at \( \mathbb{C}^{\prime} \)
(1) for all \( z\in \mathbb{C}^{\prime} \), the relation \( F(z+1)=\exp_b(F(z)) \) holds
(2) \( F(0)=1 \)
(3) \( F \) is real increasing function at \( \{z \in \mathbb{R}| z>-2 \} \).
Properties.
From assumption (1) and (2) it follows, that function \( F \) has singularity at -2, at -3 and so on.
Consider following Assumption:
There exist entire 1-periodic function \( h \) such that
\( G(z)=F(z+h(z)) \) is also analytic tetration on base \( b \).
Then, function \( I(z)=z+h(z) \) is not allowed to take values -2, -3, ..
being evaluated at elements of \( \mathbb{C}^{\prime} \).
This means that function
\( J(z)=I(z) \frac{z+2}{I(z)+2} =(z+h(z)) \frac{z+2}{z+h(z)+2} \)
is entire function.
(Weak statement which seems to be true)
Function \( J \) cannot grow faster than linear function at infinity in any direction.
Therefore, it is linear function. Therefore, \( h=0 \)
Therefore, there exist only one analytic tetration.