Iteration with two analytic fixed points

[JmsNxn]

So this seems to be a dead end, it is as frustrating for me.
Ok, try it in a different way. You say:

can we agree that if \(f^{\circ t}(z) : D \times H \to H\) for two domains \(D\subset \mathbb{C}\) and \(H\subset \mathbb{C}\)--where \(D\) is closed under addition (is a semigroup under \(\{+\}\)); such that:

\[
\begin{align}
f^{\circ t}(f^{\circ s}(z)) &= f^{\circ t+s}(z)\\
f^{\circ 1}(z) &= f(z)\\
\end{align}
\]

Then there can only be one point \(p \in H\) such that \(f(p) = p\).

But for me this has nothing to do with non-integer iteration. Your proposition would already be valid for just (positive) integer iteration, isn't it? (I rephrase your proposition to see what I mean)

if \(f^{\circ n}(z) : \mathbb{N} \times H \to H\) for a domain \(H\subset \mathbb{C}\):

Then there can only be one point \(p \in H\) such that \(f(p) = p\).

I didn't think deeply about it. It could be wrong. If you have a counter-example please put it out.

But if this is true, then the preconditions of your statement are so restrictive that just no function with two fixed points can fit.
And then you can not call it a statement about non-integer iteration.

I'm sorry bo, but for you to say I am the one repeating myself isn't correct.

This is already the second time that you claim I said something that I didn't say (and also didnt mean).
Please be more careful what you allege people have said.
I don't know how one can misread this:

I thought I emphasized that too much already, and now you ask me whether I understand that?

Ya I apologize bo. I was in a bad mood when I made that post and it was late and I was grumpy and tired, lol.

The only difference, which I would say that is different from what you are saying; is that yes you are right if \(H\) is simply connected (and mappable to \(\mathbb{D}\)) you are correct. It can be reduced to the action \(\mathbb{N} \times H \to H\). If \(H\) is not simply connected this isn't necessary. But If you require that the function is a semi-group action; this will force only one fixed point, for an arbitrary domain \(H\). And then we're pretty much back to your scenario of \(\mathbb{N}\)-actions. I see that as a strength though (which means we are reducing the fractional iteration to the natural action).

Another point, is that these iterations are always periodic if the fixed point is on the interior: (It's mapping a s.c. domain to a s.c. domain and has a fixed point on its interior--and therefore is geometric). Therefore since it's periodic, you can make "disks" about the fixed point by drawing Jordan curves like:

\[
\partial\mathcal{B}(z) = \{f^{\circ it}(z)\,|\, 0 \le t < \ell\}\\
\]

(where \(i \ell\) is the period of \(f^{\circ t}\)).

Which the interior of this jordan curve, has the fixed point \(p\), and additionally is invariant under maps. The interior looks like:

\[
\mathcal{B}(z) = \{w \in \mathbb{C}\,|\, |w - p| < |f^{\circ it}(z) - p|\}\\
\]

where \(f : \mathcal{B} \to \mathcal{B}\) for arbitrary \(z\).

This crucially uses the semi-group action--which only happens about a fixed point if its contracting (\(\mathcal{B}\) in this case, is a simply connected domain). To ask that you have \(\mathcal{B}(z) \cap \mathcal{B}(z') = \emptyset\) can't happen, unless we are no longer univalent in the strip \(0 < \Im(t) < \ell\) & \(\Re(t) > 0\), but every Schroder iteration is univalent here.

The only exception to this rule, would be if the semi-group action fails; we cannot construct \(B(z)\) because for some \(t\), the value \(f^{\circ t}(z)\) diverges, or somewhere within its interior. This would equate to your observations involving a fixed point with mult \(\lambda\) and nearby fixed point with mult \(1/\lambda\).

Where about the second fixed point we have a semigroup action for \(\Re(t) < 0\), and the first fixed point still has the semi group action for \(\Re(t)> 0\); And additionally these will have different domains in \(z\). We would have an \(A^+\) and an \(A^-\) attracting basins for the separate cases. Then if \(A^+\) and \(A^-\) intersect non trivially; we must have that the Schroder functions about both don't agree (otherwise they would create a non permissable function \(f^{\circ t}(z) : \mathbb{C} \times A^+ \cap A^- \to A^+ \cup A^-\)--they're not the same function. Because any entire function can at most omit a single point. Here we  crucially assume that \(f\) is euclidean; by which there are infinite periodic points, where therefore \(A^+ \cup A^- \subset \mathbb{C}\), so there can't be an entire function which sends there--\(A^+ \cup A^-\) omits more than one point.

Which is again seen by your iterations about two fixed points; in that as we let \(t \to -\infty\) or we grow \(t\)'s domain, we start entering in \(z\)'s domain being dependent on \(t\)'s domain.

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But it seems we have hit a stand still. I rather understand my definition is very restrictive. But you get A LOT of cool shit for free when you are this restrictive, which is why I like it. And why I like to see if iterations can be reduced to this case. It can extend to parabolic iterations too, but there also it is very restrictive. Just swap \(f(0) = 0\) in \(f: \mathbb{D} \to \mathbb{D}\) to \(f(1) = 1\).

Either way, it seems it's a large divergence of our interests. As this is needed when we start talking about group actions on arbitrary integrals, rather than integral equations of the form \(y '(x) = p(x)g(y(x))\) (which generates a semi-group). So it interests me. Where as you just want to have a god damned iteration--which I can respect.

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Again, I'd like to apologize for my tone in the last post; it was a little rude of me. Being in a bad mood is no excuse to be rude.