Qs on extension of continuous iterations from analytic functs to non-analytic

[MphLee]

I'd thought about the partitions, and your partitions worked out very well, em, I don't think the functions have to distinguish each self in \(\mathcal{C}^0,\mathcal{C}^1\) or etc.,  maybe globally \(\mathcal{C}^0\) is what I meant mostly, (thus it won't allow someone to approximate any iteration by fixed point or other analytic methods)

I believed the point was to obtain a non \(\mathcal{C}^\omega\) (analytic) \(f:X\to X\) s.t. we know how to continuously iterate it \(f^t\).
My method was just to chop \(X\) into a bunch of \(X_i\) and define on every piece a function \(f_i:X_i\to X_i\) that we know how to \(\mathbb R\)-iterate. What I was saying is that given all the \(f_i\) the pasting \(f:=\coprod_{i\in I}f_i\) needs not to have any nice properties, even if all the \(f_i\) are, say, \(\mathcal C^n\) for some \(n\in \mathbb N\cup \{\infty,\omega\}\).
What I mean is that if we want the gluing \(f:=\coprod_{i\in I}f_i\)  to carry some continuity/analytical property of the \(f_i\) we need the bundle \(X\to I\), to have some additional structure, e.g. a way to measure how the fibers \(X_i\) sits in \(X\) as \(i\in I\) varies in a continuous way.

This inspires me more,
For \(\mathcal{C}^0\), one can cut the \(\mathbb{C}\) with parallel lines and the function can be for each slice, move it in the direction of these parellel lines, which is an action of addition, or \(f_k^t(z)=z+tk_{a*Im(z)+b*Re(z)}\)

I don't think that the global function you are describing is \(\mathcal{C}^0\). Let me restate abstractly the underlying concept.

Assume we want to present \(\mathbb C\) as a bundle of lines, each of the same shape of the real line and all parallel. This reduces to a decomposition of \(\mathbb C\) as a real vector space along its two dimensions. So given a base for \(\mathbb C\), say \(\{v,w\}\), we can express every complex \(z\) as \(kv+jw\) for \(k,j\in\mathbb R\). This means that \[\mathbb C\simeq \mathbb R\times\mathbb R\]

Now this defines a partition of the complex plane into a bundle of parallel lines as \(\mathbb C=\bigcup_{p\in \mathbb Rv}p+\mathbb R w\), where \(\{v,w\}\) is a base, where \(p+\mathbb Rw=\{z\,|\,\exists k\in \mathbb R,\, z=p+kw\}\) are all the lines parallel to the line \(\mathbb R w\) and \(p=jv\). Each complex number can be expressed in this base using the change of coordinate functions associated with that base \(z=\varphi_0(z)v+\varphi_1(z)w\).
The bundle we are using is \(\varphi_0:\mathbb C=\bigcup_{p\in \mathbb Rv}p+\mathbb R w\to \mathbb R\).

The classes are sets of the form \(jv+\mathbb R w\). On each of these we will move by sliding, as you were proposing. So the function on the piece \(f_j:jv+\mathbb R w\to jv+\mathbb R w\) is just addittion by some element of \(\mathbb R w=\{kw\,|\, k\in\mathbb R\}\) that varies on \(j\).

\[f_j(z)=f_j(jv+kw)=(jv+kw)+ \alpha_j w=z+\alpha_j w\]

Thus the pasting looks as \[f(z)=f_{\varphi_0(z)}(z)=(\varphi_0(z)v+\varphi_1(z)w)+ \alpha_{\varphi_1(z)} w=\varphi_0(z)v+(\varphi_1(z)+ \alpha_{\varphi_1(z)})w\]
And its iteration is \[f^t(z)=\varphi_0(z)v+(\varphi_1(z)+ t\alpha_{\varphi_1(z)})w\]

(In)Sanity check: is this a group action?
Identity: \(f^0(z)=\varphi_0(z)v+(\varphi_1(z)+ 0\alpha_{\varphi_1(z)})w=\varphi_0(z)v+\varphi_1(z)w=z\);

homomorphism proposition: unwind the definition \(f^t(f^s(z))=f^t(f_{\varphi_0(z)}^s(z))=f_{\varphi_0(f_{\varphi_0(z)}^s(z))}^t   (f_{\varphi_0(z)}^s(z)) \). Now assume wlog that \(\varphi_0(z)=j\), then by definition also \(\varphi_0(   f_{\varphi_0(z)}^s(z)  )=\varphi_0 ( jv+(\varphi_1(z)+s\alpha_j )w  )=j\)
We conclude that \(f^t(f^s(z))= f_j^t(f_j^s(z)) \) that, by initial assumption satisfies \( f_j^{t+s}(z) \). \(\square\)

Why this is not continuous? Yes, you analysis guys can find the prof easily, but I'd observe the structural reason for it. Since each \(\alpha_j w\) may be oriented in both ways, and they are totally arbitrary when \(j\) varies, we have no way to assume that "nearness" in the neighboring fibers get preserved after the transformation. I.e. the various given a neighborhood \(U\subseteq \mathbb R\) of \(j\) the action \(f_u\) on the \(z'\) near \(z=jv+kw\) could scatter badly all the \(f_u(z')\) very far from \(f_j(z)\).

however I don't quite understand all the terms (spectral decomposition of linear operators to fiber bundles especially) in topology, but I'll learn them as soon as possible.

I was sloppy, sorry. I meant that this relates many topics like spectral decomposition of linear operators and fiber bundles.

But partition won't work out for more cases, like abs(z), sgn(z), arg(z), log(abs(arg(z)+sgn(z))) and more wierder functions,

That's quite the problem!
Given an arbitrary function \( f:X\to X\) can we partition \(X\) is pieces \(X_i\) were we know how to iterate the restrictions \(f|_{X_i}\)? If we can do this then it's like diagonalizing, and the set of indices of the partition is much like the spectrum of a matrix. I think of this like abstract eigentheory.
Some function will be non abstractly diagonalizable, obviously.

The key here is, whenever this partition technique can work, is to find the "optimal" components of \(X\) that are closed under \(X\). This is not the precise statement. I believe that given a \(f:X\to X\) one has to start from the partition on \(X\) generated by the relation \(x\sim_f y\) iff \(f^m(x)=f^n(y)\) for some \(m,n \in\mathbb N\). But even this, is not sufficient. We must merge into this a study of group and monoid actions in order to find the right orbit...
and even then it could fail.
That's only my opinion btw.

I'm still stuck about the conj(z), as its transformation cannot be described by countably infinitely many partitions \(X_i\) and each has an \(f:X_i\to X_i\).
I wonder how one can get a continuous iteration without endomorphism, since it breaks the symmetry \(f:X_i\to X_i\), or even a more complex question, like their iterations to complex heights, since then you can't use eigen decomposition or linear operators, like the cases \(k\in \mathbb{C},f_k^t(z)=k_{\arg(z)}z\)

Anyway thank u

I think we can adapt the partition technique to conj since conjugation respects the concentric circle-partition you were describing. Idk how, I have not time to investigate this but... take a circle of radius \(r\), i.e. consider a function \(f_r:rS^1\to rS^1\). We need it to be idempotent and reflect each point vertically, as conjugation does. But I believe you can interpolate it as you like, almost. Maybe playing with \(f_r^t(re^{i\theta})=re^{iK_r(t,\theta)}\) as a function of the angle on the circle where \(K_r(t, -):{}S^1\to S^1\) are an endofunctions of the circle that satisfy \(K_r(1,\theta)=-\theta\) and \(K_r(s+t,\theta)=K_r(s,K_r(t,\theta))\).