# Tetration Forum

Full Version: Universal uniqueness criterion II
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Lets summarize what we have so far:

Proposition. Let $S$ be a vertical strip somewhat wider than $1$, i.e. $S=\{z\in\mathbb{C}: x_1-\epsilon<\Re(z) for some $x_1\in\mathbb{R}$ and $\epsilon>0$.
Let $D=\mathbb{C}\setminus (-\infty,x_0]$ for some $x_0 and let $G\subseteq G'$ be two domains (open and connected) for values, and let $F$ be holomorphic on $G'$. Then there is at most one function $f$ that satisifies
(1) $f$ is holomorphic on $D$ and $f(S)\subseteq G\subseteq f(D)=G'$
(2) $f$ is real and strictly increasing on $\mathbb{R}\cap S$
(3) $f(z+1)=F(f(z))$ for all $z\in D$ and $f(x_1)=y_1$
(4) There exists an inverse holomorphic function $f^{-1}$ on $G$, i.e. a holomorphic function such that $f(f^{-1}(z))=z=f^{-1}(f(z))$ for all $z\in G$.

Proof. Let $g,h$ be two function that satisfy the above conditions. Then the function $\delta(z)=g^{-1}(h(z))$ is holomorphic on $S$ (because $h(S)\subseteq G$ and (4)) and satisfies $h(z)=g(\delta(z))$. By (3) and (4)
$\delta(z+1)=g^{-1}(F(h(z)))=g^{-1}(F(g(\delta(z))))=g^{-1}(g(\delta(z)+1))=\delta(z)+1$ and $\delta(0)=0$.
So $\delta$ can be continued from $S$ to an entire function and is real and strictly increasing on the real axis.

$\delta(\mathbb{C})=\mathbb{C}$ by our previous considerations.
By Big Picard every real value of $\delta$ is taken on infinitely often if $\delta$ is not a polynomial, but every real value is only taken on once on the real axis, thatswhy still $\delta(\mathbb{C}\setminus\mathbb{R})=\mathbb{C}$. But this is in contradiction to $g^{-1}:G\to D=\mathbb{C}\setminus [x_0,\infty)$. So $\delta$ must be a polynomial that takes on every real value at most once. This is only possible for $\delta(x)=x+c$ with $c=0$ because $\delta(0)=0$.$\boxdot$.

In the case of tetration one surely would chose $x_0=-2$ and $x_1=0$ or $x_1=-1$. However I am not sure about the domain $G$ which must contain $f(S)$ and hence give some bijection $f:G\leftrightarrow S'$, with some $S'\supseteq S$.

Of course in the simplest case one just chooses $G=f(S)$ if one has some function $f$ in mind already. However then we can have a different function $f_2$ with $f(S)\neq f_2(S)$ but our intention was to have a criterion that singles out other solutions.
So we need an area $G$ on which every slog should be defined at least and satisfy $\text{sexp}(\text{slog})=\text{id}=\text{slog}(\text{sexp})$ as well as $\text{sexp}(S)\subseteq G$.
I just see that we can essentially simplify our conditions:

Proposition. Let $S$ be a vertical strip somewhat wider than $1$, i.e. $S=\{z\in\mathbb{C}: x_1-\epsilon<\Re(z) for some $x_1\in\mathbb{R}$ and $\epsilon>0$.
Let $D$, $G$, $G'$ be three domains (open and connected) such that $S\subseteq D$, $G\subseteq G'$ and let $F$ be holomorphic on $G'$, let $y_1\in G$. Then there exist at most one function $f$ that satisifies
(1) $f$ is holomorphic on $D$ and $f(S)\subseteq G\subseteq f(D)=G'$
(2) $f(z+1)=F(f(z))$ for all $z\in D$ and $f(x_1)=y_1$
(3) There is a $D'\subseteq D$ such that $f: D' \leftrightarrow G$ is biholomorphic.

Proof. Let $g,h$ be two function that satisfy the above conditions. Then the function $\delta(z)=g^{-1}(h(z))$ is holomorphic on $S$ (because $h(S)\subseteq G$ and (3)) and satisfies $h(z)=g(\delta(z))$. By (3) and (4)
$\delta(z+1)=g^{-1}(F(h(z)))=g^{-1}(F(g(\delta(z))))=g^{-1}(g(\delta(z)+1))=\delta(z)+1$ and $\delta(0)=0$.
So $\delta$ can be continued from $S$ to an entire function.
But the same is also true for $\delta_2=h^{-1}(g(z))$ by the same reasoning. But as $\delta\circ\delta_2=g^{-1}\circ h\circ h^{-1}\circ g =\text{id}$ and $\delta_2\circ \delta = \text{id}$, we see that $\delta:\mathbb{C}\leftrightarrow\mathbb{C}$ is a bijection!
Again with Picard's big theorem we conclude that $\delta(x)=x+c$ with $c=0$.$\boxdot$
bo198214 Wrote:...Let $D=\mathbb{C}\setminus [x_0,\infty)$ for some $x_0...
Henryk: For tetration, I would define
$D=\mathbb{C}\setminus (-\infty, x_0]$ for some $x_0...

I include below the small part of
http://math.eretrandre.org/tetrationforu...77#pid2477
which is picture of slog(S), assuming, $x_1=-1$ and that $\epsilon$ is small and not seen. Vertical lines correspond to
$\Re(\mathrm{sexp}(z))=0$ and
$\Re(\mathrm{sexp}(z))=1$
Horisontal lines correspond to
$\Im(\mathrm{sexp}(z))=0$ and
$\Im(\mathrm{sexp}(z))=1$
The curvilinear mesh is produced by images of lines
$\Re(z)= 0.2, 0.4, 0.6, 0.8, 1 ; \Im(z)>0$ and
$\Im(z)= 0.2, 0.4, 0.6, 0.8, 1, 1.2, 1.4, 1.6, 1.8, 2, 2.2, 2.4, 2.6, 2.8, 4; 0\le /Re(z) \le 1$.
The pink cutline corresponds to $\Re(z)= -2$.
The images of lines with integer $\Im(z)$ are a little bit extended.
Also, images of lines $\Re(z)=2$ and $\Re(z)=3$are shown.
There is biholomorphizm $\mathrm{sexp}(S): \leftrightarrow S$.
[attachment=414]

Henrik,
1. Do you plan to polish this proof or I may include this into the paper?
2. Can we claim, that some of singularities of a modified tetration are at
$\Re(z)>-2$ ? (In this case, we can include the case $1 at once).
Kouznetsov Wrote:
bo198214 Wrote:...Let $D=\mathbb{C}\setminus [x_0,\infty)$ for some $x_0...
Henryk: For tetration, I would define
$D=\mathbb{C}\setminus (-\infty, x_0]$ for some $x_0...

Yes that was a mistype, it should read $\mathbb{C}\setminus(-\infty,x_0]$. I change that in the original post.

Quote:I include below the small part of
http://math.eretrandre.org/tetrationforu...77#pid2477
which is picture of slog(S),

thanks, for the illustration.

Quote:1. Do you plan to polish this proof or I may include this into the paper?
Its not yet finished, we still have no universal domain $G$ for the slog, which we need to have uniqueness independent of the specific domain $\text{sexp}(S)$.

Quote:2. Can we claim, that some of singularities of a modified tetration are at
$\Re(z)>-2$ ?

Dmitrii, modified tetration is not all, if you want to consider $g(z)=f(J(z))$ then you need to have a $J$ first, a $J=f^{-1}\circ g$ that is holomorphic on $S$ which in turn imposes conditions on the cut, set in the domain of definition of $f^{-1}$.
bo198214 Wrote:
Kouznetsov Wrote:2. Can we claim, that some of singularities of a modified tetration are at
$\Re(z)>-2$ ?
Dmitrii, modified tetration is not all, if you want to consider $g(z)=f(J(z))$ then you need to have a $J$ first, a $J=f^{-1}\circ g$ that is holomorphic on $S$ which in turn imposes conditions on the cut, set in the domain of definition of $f^{-1}$.
Sorry, I try again: If $h$ is entire and 1-periodic and $J(z)=z+h(z)$, then can we claim that $\exists z \in \mathbb{C} ~:~ \Re(z)>-2 ~,~ J(z)\!=\!-2~.~$ ?
Kouznetsov Wrote:I try again: If $h$ is entire and 1-periodic and $J(z)=z+h(z)$, then can we claim that $\exists z \in \mathbb{C} ~:~ \Re(z)>-2 ~,~ J(z)\!=\!-2~.~$ ?

Hm, sorry, I dont know. Wherefore do you need such a statement?
bo198214 Wrote:
Kouznetsov Wrote:I try again: If $h$ is entire and 1-periodic and $J(z)=z+h(z)$, then can we claim that $\exists z \in \mathbb{C} ~:~ \Re(z)>-2 ~,~ J(z)\!=\!-2~.~$ ?
Hm, sorry, I dont know..
ok, I do know. Conjectiure:
Let
$h$ be entire 1-periodic non constant function,
$h(0)=0$,
$|h'(z)| <1$ for all $z$ in some vicinity of the real axis
$J(z)=z+h(z)$ for all complex $z$
Then there exist $z \in \mathbb{C}$ such that $\Re(z)>0$ and $J(z)=-2$

bo198214 Wrote:Wherefore do you need such a statement?
Yes, I do. Then we have beautiful and general proof of uniqueness of tetration as it is defined at http://en.citizendium.org/wiki/Tetration

P.S. For other participants, I repeat here the essense from one of our previous discussions.
Many times I tried to build-up an example to negate the conjecture above.
Therefore I claim this conjecture.
I agree with you, that it is not sufficient reason for such a claim.
(Although I never met a simple condition for an existing function, such that I could not provide an example.)
It would be interesting to construct the sufficient reason. (then the conjecture becomes Theorem)
bo198214 Wrote:Proof. Let $g,h$ be two function that satisfy the above conditions. Then the function $\delta(z)=g^{-1}(h(z))$ is holomorphic on $S$ (because $h(S)\subseteq G$ and (3)) and satisfies $h(z)=g(\delta(z))$...
Why $h$? There was no $h$ above. Should not be $f$?
Kouznetsov Wrote:
bo198214 Wrote:Proof. Let $g,h$ be two function that satisfy the above conditions. Then the function $\delta(z)=g^{-1}(h(z))$ is holomorphic on $S$ (because $h(S)\subseteq G$ and (3)) and satisfies $h(z)=g(\delta(z))$...
Why $h$? There was no $h$ above. Should not be $f$?

$h$ and $g$ are two functions that satisfy the above conditions, as I wrote. Perhaps write better: "Let $f=g$ and $f=h$ be two solutions of the above conditions."