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Constructing an analytic repelling Abel function
#1
Hey, everyone. This question is pretty simple, but I'm not sure if the answer is positive or not. My gut is saying it should be, but I'm in no way certain.

Let's take an abel function \(\alpha(b,z)\), such that \(b \in \mathcal{S}\) the Shell-Thron region. Additionally we ask that it is a repelling iteration, so that:

$$
\begin{align}
\alpha(b,\log_b(z)) &= \alpha(b,z) + 1\\
\alpha(b,F(b)) &= \infty\\
\end{align}
$$

Where \(\log|F(b)| > 1\), and \(F(b)\) is the repelling fixed point of the exponential \(b^z\)--\(b = F(b)^{1/F(b)}\); attracting fixed point of \(\log_b(z)\).

Now, this Abel function is always solvable, essentially just take:

$$
\alpha(b,z) = \frac{\log \Psi_b(z)}{-\log \log F(b)}
$$

For the Schroder function \(\Psi_b(z)\) about the fixedpoint \(F(b)\) of the logarithm \(\log_b(z)\). So that,

$$
\begin{align}
\alpha(b,\log_b(z)) &= \frac{\log \Psi_b(\log_b(z))}{-\log\log F(b)}\\
&= \frac{\log \left(\log F(b)^{-1}\Psi_b(z)\right)}{-\log \log F(b)}\\
&= \frac{\log\Psi_b(z)-\log \log F(b)}{-\log \log F(b)}\\
&= \alpha(b,z) + 1\\
\end{align}
$$

The question then becomes simple. Does the limit \(b \to \partial S\) retain holomorphy in \(z\)?

So as I let \(\log F(b) \to e^{i\theta}\) for some \(\theta \in [0,2\pi)\), does this iteration converge uniformly in \(z\)?

I'd be hard pressed if this weren't the case, but I've seen crazier things happen studying tetration.

Any help is greatly appreciated.

This is essentially, does the repelling iteration for \(b \in (1,\eta)\) converge to the "cheta" iteration about \(\eta\)? I know it happens in the attracting case, but I don't know about the repelling case.
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