Apropos "fix"point: are the fractional iterations from there "fix" as well?

[MphLee]

Typos and errors fixed.

I hope everyone forgives me, but I need to put all of this on solid, formal grounds. I need to prove every step.
Now I'm doing it for James' post... but I've done only for the first half of it.

So, this is actually a defining property of the standard Schroeder iteration. But it's a little difficult to fully flesh out why.  Now, to begin I'll construct an arbitrary iteration which has a constant noodle, and show there are many of them.
If you iterate locally about a fixed point, and your solution satisfies \(f^t(p) = p\), then the iteration is expressible via Shroeder iteration (provided that \(|f'(p)| \neq 0,1\)).  For convenience, assume that \(|f'(p)| < 1\).
You can actually prove this pretty fast. Assume that \(f^t(x)\) is a super function in \(t\) about a fixed point \(p\), and \(x\) is in the neighborhood of \(p\). Assume that \(f^t(p) = p\).  Well then:
\[
\Psi(f^{t+1}(x)) = \lambda \Psi(f^t(x))\\
\]
So that:

\[
\theta(t) = \frac{\Psi(f^t(x))}{\lambda^t}\\
\]

And we know that there must be some 1-periodic function \(\theta(t)\), such that:
\[
f^t(x) = \Psi^{-1}\left(\lambda^{t}\theta(t) \Psi(x)\right)\\
\]
In fact, any periodic function will work fine here, and will have a constant noodle, will be a super function, but will not be Schroeder iteration.

Now let's add one more constraint, let's say that \(f^{t}(f^{s}(x)) = f^{t+s}(x)\). Well then, we have that \(\theta\) must be constant. By which, we are guaranteed that it's a Schroeder iteration...

So to clarify. Any fractional iteration with a constant noodle is a Schroeder iteration. But there are plenty of superfunctions of \(f\) which have a constant noodle, but in turn, they aren't fractional iterations then (don't satisfy the semi-group law).

I'm trying to scan the deep meaning behind your moves... what is happening in the background. I'd like a feedback so I can go on to the next step and then to Tommy's observations.

When you ask for superfunction about a fixed point \(p\) you, in fact, are actually taking a complex object. You are not just taking A superfunction but an entire family of superfunctions \({\varphi}_{x\in U}:\mathbb R^+\to X\), superfunctions of \(f:X\to X\), parametrized by a small \(f\)-stable neighborhood of the fixed point \(p\in U\subseteq X\). In other words you asking for a \(\varphi:\mathbb R\times U\to U\), i.e. a family \(\{\varphi_x\}_{x\in U}:\mathbb R^+\to X\) of superfunctions of \(f\) restricted to the neighborhood of the fixed point \(p\) (that should be the local part about it).

Notice that, as recent posts by me, James and Tommy are making clear... being a superfunction is something weaker than being an iteration, like being an interpolation is weaker than being a superfunction. In other words we just ask for the family \(\varphi_x\) to satisfy the mild conditions \[\varphi_x(t+1)=f(\varphi_x(t))\quad \quad \varphi_x(0)=x\]

No other coherence conditions, like the semigroup property... a property that is very strong. Note that what I write as \(\varphi_x(t)\) we could write as \(\varphi(t,x)\) and James denotes it as \(\varphi_x(t)=f^t(x)\). I use \(\varphi_x\) notation to emphasize that it is a FAMILY OF POSSIBLE SOLUTIONS of the equation \(\varphi\circ  S=f\circ \varphi\). We can think of this as an assignment \(\varphi_{-}:U\to [S,f]\) where \([S,f]\) is the space of all the superfunctions of \(f\)...

Now... we want to know if \(\varphi_p(t)\) is a constant path or... if it loops around \(p\in X\) infinitely many times. The answer James gives is... it is constant if the family \(\varphi_x\) satisfies the stronger condition of being an iteration: i.e. \[\varphi_{\varphi_x(t)}(s)=\varphi(s,\varphi(t,x))=\varphi(s+t,x)=\varphi_x(s+t)\]

The reasoning, subject to multipliers conditions to ensure convergence of Schroeder functions, seems to begin in the following way.
Given a family of superfunctions of \(f\) locally about a fixed point \(p\in U\), i.e. given a family of superfunctions \(\varphi_{x\in U}\) and given a function \(\Psi\in [f,{\rm mul}_\lambda]\), a solution of the Schroeder eq. of \(f\) we define another family
\[A_x=\Psi\circ \varphi_x\]
this time this family lies in the space \([S,{\rm mul}_\lambda]\) that is, each \(A_x\) is a superfunction of multiplication \({\rm mul}_\lambda\) by \(\lambda\).
\[A_x(t+1)=\lambda A_x(t)\]
this is what James denotes by \(A_x(t)=\Psi(f^t(x))\).

Remember that we have a family of functions in the space  \([S,{\rm mul}_\lambda]\) and remember, in this space there is a special solution: exponentiation \(\exp_\lambda \in [S,{\rm mul}_\lambda]\). What James does now is measuring the difference between the \(A_x\) and \(\exp_\lambda\). They differ by a periodic function.
\[\theta_x(t)\lambda^t=A_x(t)\]
If the periodic function is constant \(\theta_x(t)=K\) then \(K\lambda^t=\Psi(\varphi_x(t))\) and \(\varphi_x(t)=\Psi^{-1}(K\lambda^t)\) is purely obtained by the Schroeder function. James calls it the Schroeder iteration.

---

At this level of detail each \(\theta_x\) can be a different periodic function. What I managed to prove is that... if we ask the family \(\phi_x\) tho satisfy the additional condition \(\varphi_{x}(t+n)=\varphi_{f^n(x)}(t)\), weaker than being an iteration but implied by it we obtain that if \(x\sim_f y\) then \(\theta_x=\theta_y\)

Proposition. let \(\varphi_{-}:U\to [S;f|_U]\) be a family of superfunctions of \(f|_U\) for \(U\subseteq X\) an \(f\) stable subset, let \(\Psi \in [f, \lambda ]\) and define the family of 1-periodic function \[\theta_x(t)=\frac{\Psi(\varphi_x(t))}{\lambda^t}\] we prove that

• if for each \(x\in U\) we have \(\varphi_x(0)=x\) (a section); This condition is not needed
  
• for each \(t\) \(\varphi_{x}(t+n)=\varphi_{f^n(x)}(t)\) (weakly iterative)
  

then for \(x,y\in U\) are \(f\) connected, i.e. \(f^n(x)=f^m(y)\) for some \(m,n\in\mathbb N\), \[\theta_x(t)=\theta_y(t)\]

Proof. assume \(\varphi_{x}(t+n)=\varphi_{f^n(x)}(t)\). We prove that \(\theta_x=\theta_{f^n(x)}\). Compute \(\theta_{f^n(x)}(t)=\frac{\Psi(\varphi_{f^n(x)}(t))}{\lambda^t}=\frac{\Psi(\varphi_{x}(t+n))}{\lambda^t}=\theta_x(t+n)\) by assumption. \(\theta_{f^n(x)}(t)=\theta_{x}(t)\) by periodicity. If \(f^n(x)=f^m(y)\) then \(\theta_x(t)=\theta_y(t)\) by transitivity of the identity. \(\square\)

At this point I can't wait to translate the rest of the argument...
Maybe \(\phi_x\) satisfies semigroup, is an iteration, iff \(\theta_x\) is constant. If this was the case, as Tommy seemed to suggest elsewhere... this is something like a criterion to select among the space of superfunctions the special ones... something like an uniqueness criterion.