Does tetration take the right half plane to itself?

[JmsNxn]

...  It easily follows from this that \( \log^{\circ n}(\lambda^n z + L) \) is fully monotone, it uniformly converges to the inverse Schroder function, and therefore the inverse Schroder function is a fully monotone function...

Good luck with your paper.  We need more rigorous iterated exponentiation papers.
I just wondered what your approach to show the sequence "uniformly converges".  Is there an easy theorem, or did you want to go with a more complicated approach something like my lemma5?  Just curious.  Also, you probably meant  \( \log_b^{\circ n}(\lambda^n z + L) \) where b is the tetration base.

Yeah, I meantt \( \log_b \). This is actually a well known result on how to represent the inverse Schroder function. Bo used it before, and that's how I first learnt about it. 

Essentially if

\( f(z_0) = z_0 \) and \( 0 < |f'(z_0)| < 1 \), then not only does

\( \lim_{n\to\infty} \frac{f^{\circ n}(z) - z_0}{\lambda^n} \to \Psi(z) \)

for \( |z- z_0|<\delta \), where \( \Psi \) is the Schroder function. 

We also get that, if \( g = f^{-1} \)

\( \lim_{n\to\infty} g^{\circ n}(\lambda^n z + z_0) \to \Psi^{-1}(z) \)

for \( |z-z_0| < \delta \), for \( \delta \) sufficiently small. This is all you really need.