09/18/2008, 01:42 PM
We have the usual conditions
(1) \( f \) is a holomorphic function defined on \( C'=\mathbb{C}\setminus (-\infty,-2] \)
(2) \( f(0)=1 \)
(3) \( f(z+1)=b^{f(z)} \), \( b>1 \).
(4) \( f \) is real on \( (-2,\infty) \) and strictly increasing.
The following criterion should pick a unique function out of the possible solutions for the above conditions:
(U1) \( f \) is bounded on the strip \( S:=\{ x+iy| 0< x\le 1, y\in\mathbb{R}\} \).
(U2) The values of \( f \) fill the complex plane: \( f(D)=\mathbb{C} \), \( D:=\{x+iy| -2<x, y\in\mathbb{R}\} \).
Proof attempt:
Let \( g \) be some function that also satisfies (1), (2), (3),(4) and (U1), (U2). We put appropriate cuts on the domain of \( f \) such that we have an inverse \( f^{-1} \) which always maps reals to reals.
By some considerations previously made already on this forum
\( f^{-1}(g(z))-z \) is a holomorphic and 1-periodic function. We know that if a 1-periodic holomorphic function is bounded on \( S \) then it is already a constant.
So if we could show that \( \delta(z):=f^{-1}(g(z)) \) has a bounded real part on \( S \) then would
\( e^{\delta(z)-z}=e^{\Re(\delta(z))+i\Im(\delta(z))+\Re(z)+i\Im(z)}=e^{\Re(\delta(z))+\Re(z)}e^{i\left(...\right)} \)
be bounded and 1-periodic on \( S \), hence \( e^{\delta(z)-z}=c \) hence \( \delta(z)-z=d \) hence \( \delta(z)=z+d \) for some \( d\in\mathbb{C} \). But of course \( d=0 \) because \( 0=f^{-1}(1)=f^{-1}(g(0))=\delta(0)=d \).
So we still need show that \( f^{-1}(g(z)) \) has bounded real part on \( S \). We know already that \( g(S) \) is bounded.
By condition (U2) we divide the area \( g(S) \) into images of \( f \) of \( S_n = \{ x+iy| n<x\le n+1, y\in\mathbb{R}\} \), i.e.
\( g(S)\subseteq \mathbb{C}= f(D) =\bigcup_{n=-2}^{\infty} f(S_n) \).
Because \( g(S) \) is bounded, it is a subset of a compact set, which can be covered by merely finitely many \( f(S_n) \):
\( g(S)\subseteq f(S_{n_1})\cup f(S_{n_2})\cup \dots \cup f(S_{n_k}) \)
so \( f^{-1}(g(S))\subseteq S_{n_1}\cup \dots\cup S_{n_k} \) which has bounded real part.
(1) \( f \) is a holomorphic function defined on \( C'=\mathbb{C}\setminus (-\infty,-2] \)
(2) \( f(0)=1 \)
(3) \( f(z+1)=b^{f(z)} \), \( b>1 \).
(4) \( f \) is real on \( (-2,\infty) \) and strictly increasing.
The following criterion should pick a unique function out of the possible solutions for the above conditions:
(U1) \( f \) is bounded on the strip \( S:=\{ x+iy| 0< x\le 1, y\in\mathbb{R}\} \).
(U2) The values of \( f \) fill the complex plane: \( f(D)=\mathbb{C} \), \( D:=\{x+iy| -2<x, y\in\mathbb{R}\} \).
Proof attempt:
Let \( g \) be some function that also satisfies (1), (2), (3),(4) and (U1), (U2). We put appropriate cuts on the domain of \( f \) such that we have an inverse \( f^{-1} \) which always maps reals to reals.
By some considerations previously made already on this forum
\( f^{-1}(g(z))-z \) is a holomorphic and 1-periodic function. We know that if a 1-periodic holomorphic function is bounded on \( S \) then it is already a constant.
So if we could show that \( \delta(z):=f^{-1}(g(z)) \) has a bounded real part on \( S \) then would
\( e^{\delta(z)-z}=e^{\Re(\delta(z))+i\Im(\delta(z))+\Re(z)+i\Im(z)}=e^{\Re(\delta(z))+\Re(z)}e^{i\left(...\right)} \)
be bounded and 1-periodic on \( S \), hence \( e^{\delta(z)-z}=c \) hence \( \delta(z)-z=d \) hence \( \delta(z)=z+d \) for some \( d\in\mathbb{C} \). But of course \( d=0 \) because \( 0=f^{-1}(1)=f^{-1}(g(0))=\delta(0)=d \).
So we still need show that \( f^{-1}(g(z)) \) has bounded real part on \( S \). We know already that \( g(S) \) is bounded.
By condition (U2) we divide the area \( g(S) \) into images of \( f \) of \( S_n = \{ x+iy| n<x\le n+1, y\in\mathbb{R}\} \), i.e.
\( g(S)\subseteq \mathbb{C}= f(D) =\bigcup_{n=-2}^{\infty} f(S_n) \).
Because \( g(S) \) is bounded, it is a subset of a compact set, which can be covered by merely finitely many \( f(S_n) \):
\( g(S)\subseteq f(S_{n_1})\cup f(S_{n_2})\cup \dots \cup f(S_{n_k}) \)
so \( f^{-1}(g(S))\subseteq S_{n_1}\cup \dots\cup S_{n_k} \) which has bounded real part.