I am not so convinced whether \( f \) is invertible (it is usually not injective on the complex plane) without getting trouble. So here is the next revision/attempt:
(U1) \( f \) is bounded on the strip \( S:=\{ x+iy| 0< x\le 1, y\in\mathbb{R}\} \).
(U2) \( \lim_{n\to\infty}\exp_b^{\circ n}(z)= \infty \) for all \( z\in f(S) \). (\( \lim_{n\to\infty} z_n=\infty \) is just an abbreviation for the more save \( \lim_{n\to\infty}1/z_n=0 \).)
Proof attempt:
Let \( g \) be some function that also satisfies (1), (2), (3),(4) and (U1), (U2).
Then \( g(z)=f(\delta(z)) \), where \( \delta(z)-z \) is a 1-periodic function (This is easily provable for bijective \( f \) and \( g \) but I am not 100% sure yet for arbitrary \( f \) and \( g \), but it should go through.)
As in the previous proof we just need to show that \( \delta \) has bounded real part on \( S \).
We know already that \( g(S)=f(\delta(S)) \) is bounded.
We can write
\( f(\delta(z))=\exp_b^{{\delta_Z(z)}} (f({\delta_S(z)}) \)
where \( \delta_Z(z)\in\mathbb{Z} \) is the next lower integer (floor) of the real part of \( \delta(z) \) and \( \delta_S(z):=\delta(z)-\delta_Z(z)\in S \).
\( \delta_S(S)\subseteq S \) hence \( f(\delta_S(S))\subseteq f(S) \) is bounded by (U1) and \( \lim_{n\to\infty}\exp_b^{\circ n}(z)=\infty \) for every \( z\in f(\delta_S(S)) \) by (U2).
If the real part of \( \delta \) would be unbounded then also \( \delta_Z \) would be unbounded on \( S \). We choose a sequence \( z_n \) such that \( \delta_Z(z_n) \) is strictly increasing (and hence unbounded).
The sequence \( f(z_n) \) is bounded and hence has a converging subsequence. Let \( v_n \) be a subsequence of \( z_n \) such that \( \lim_{n\to\infty} f(v_n)=d \).
The sequence \( f(\delta_S(v_n)) \) is bounded and hence has a converging subsequence. Let \( w_n \) be a subsequence of \( v_n \) such that \( c_n:= f(\delta_S(w_n))\in f(S) \) converges, let \( c:=\lim_{n\to\infty}c_n \).
We still have \( w_n\in S \), \( \lim_{n\to\infty} f(w_n)=d \) and \( \delta_Z(w_n) \) is strictly increasing.
\( c \) is element of the closure \( \overline{f(S)} \). (U2) holds also on \( \overline{f(S)}=f(\overline{S}) \) by (3).
Thatswhy
\( \begin{align*}
\infty=\lim_{m\to\infty} \exp_b^{\circ \delta_Z(w_m)}( c)=\lim_{m\to\infty} \lim_{n\to\infty} \exp_b^{\circ \delta_Z(w_m)}(c_n)=\lim_{n\to\infty} \exp_b^{\circ \delta_Z(w_n)}(c_n)= \lim_{n\to\infty} f(w_n)=d
\end{align*} \)
in contradiction to \( d \) being finite.
bo198214 Wrote:We have the usual conditionsThe following criterion should pick a unique function out of the possible solutions for the above 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.
(U1) \( f \) is bounded on the strip \( S:=\{ x+iy| 0< x\le 1, y\in\mathbb{R}\} \).
(U2) \( \lim_{n\to\infty}\exp_b^{\circ n}(z)= \infty \) for all \( z\in f(S) \). (\( \lim_{n\to\infty} z_n=\infty \) is just an abbreviation for the more save \( \lim_{n\to\infty}1/z_n=0 \).)
Proof attempt:
Let \( g \) be some function that also satisfies (1), (2), (3),(4) and (U1), (U2).
Then \( g(z)=f(\delta(z)) \), where \( \delta(z)-z \) is a 1-periodic function (This is easily provable for bijective \( f \) and \( g \) but I am not 100% sure yet for arbitrary \( f \) and \( g \), but it should go through.)
As in the previous proof we just need to show that \( \delta \) has bounded real part on \( S \).
We know already that \( g(S)=f(\delta(S)) \) is bounded.
We can write
\( f(\delta(z))=\exp_b^{{\delta_Z(z)}} (f({\delta_S(z)}) \)
where \( \delta_Z(z)\in\mathbb{Z} \) is the next lower integer (floor) of the real part of \( \delta(z) \) and \( \delta_S(z):=\delta(z)-\delta_Z(z)\in S \).
\( \delta_S(S)\subseteq S \) hence \( f(\delta_S(S))\subseteq f(S) \) is bounded by (U1) and \( \lim_{n\to\infty}\exp_b^{\circ n}(z)=\infty \) for every \( z\in f(\delta_S(S)) \) by (U2).
If the real part of \( \delta \) would be unbounded then also \( \delta_Z \) would be unbounded on \( S \). We choose a sequence \( z_n \) such that \( \delta_Z(z_n) \) is strictly increasing (and hence unbounded).
The sequence \( f(z_n) \) is bounded and hence has a converging subsequence. Let \( v_n \) be a subsequence of \( z_n \) such that \( \lim_{n\to\infty} f(v_n)=d \).
The sequence \( f(\delta_S(v_n)) \) is bounded and hence has a converging subsequence. Let \( w_n \) be a subsequence of \( v_n \) such that \( c_n:= f(\delta_S(w_n))\in f(S) \) converges, let \( c:=\lim_{n\to\infty}c_n \).
We still have \( w_n\in S \), \( \lim_{n\to\infty} f(w_n)=d \) and \( \delta_Z(w_n) \) is strictly increasing.
\( c \) is element of the closure \( \overline{f(S)} \). (U2) holds also on \( \overline{f(S)}=f(\overline{S}) \) by (3).
Thatswhy
\( \begin{align*}
\infty=\lim_{m\to\infty} \exp_b^{\circ \delta_Z(w_m)}( c)=\lim_{m\to\infty} \lim_{n\to\infty} \exp_b^{\circ \delta_Z(w_m)}(c_n)=\lim_{n\to\infty} \exp_b^{\circ \delta_Z(w_n)}(c_n)= \lim_{n\to\infty} f(w_n)=d
\end{align*} \)
in contradiction to \( d \) being finite.