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Universal uniqueness criterion II - bo198214 - 11/16/2008
Lets summarize what we have so far:
Proposition. Let \( S \) be a vertical strip somewhat wider than \( 1 \), i.e. \( S=\{z\in\mathbb{C}: x_1-\epsilon<\Re(z)<x_1+1\} \) for some \( x_1\in\mathbb{R} \) and \( \epsilon>0 \).
Let \( D=\mathbb{C}\setminus (-\infty,x_0] \) for some \( x_0<x_1 \) 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 \( \text{sexp}(S)\subseteq G \).
Simplified Universal Uniqueness Criterion - bo198214 - 11/16/2008
I just see that we can essentially simplify our conditions:
Proposition. Let \( S \) be a vertical strip somewhat wider than \( 1 \), i.e. \( S=\{z\in\mathbb{C}: x_1-\epsilon<\Re(z)<x_1+1\} \) 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 \( c=0 \).\( \boxdot \)
RE: Universal uniqueness criterion II - Kouznetsov - 11/17/2008
bo198214 Wrote:...Let \( D=\mathbb{C}\setminus [x_0,\infty) \) for some \( x_0<x_1 \)...Henryk: For tetration, I would define
\( D=\mathbb{C}\setminus (-\infty, x_0] \) for some \( x_0<x_1 \)...
then I like your proof.
I include below the small part of
http://math.eretrandre.org/tetrationforum/showthread.php?tid=214&pid=2477#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 \( \mathrm{sexp}(S): \leftrightarrow S \).
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 \( 1<b<\exp(1/{\mathrm e}) \) at once).
RE: Universal uniqueness criterion II - bo198214 - 11/17/2008
Kouznetsov Wrote:bo198214 Wrote:...Let \( D=\mathbb{C}\setminus [x_0,\infty) \) for some \( x_0<x_1 \)...Henryk: For tetration, I would define
\( D=\mathbb{C}\setminus (-\infty, x_0] \) for some \( x_0<x_1 \)...
then I like your proof.
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/tetrationforum/showthread.php?tid=147&pid=2477#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 \( f^{-1} \).
RE: Universal uniqueness criterion II - Kouznetsov - 11/18/2008
bo198214 Wrote: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:2. Can we claim, that some of singularities of a modified tetration are atDmitrii, 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 \( f^{-1} \).
\( \Re(z)>-2 \) ?
RE: Universal uniqueness criterion II - bo198214 - 11/18/2008
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?
RE: Universal uniqueness criterion II - Kouznetsov - 11/19/2008
bo198214 Wrote:ok, I do know. Conjectiure: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..
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)
RE: Simplified Universal Uniqueness Criterion - Kouznetsov - 11/19/2008
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 \( f \)?
RE: Simplified Universal Uniqueness Criterion - bo198214 - 11/19/2008
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."