Iteration with two analytic fixed points

[bo198214]

To begin, since \(E\) is simply connected, it is safe to assume that \(E = \mathbb{D}\) the unit disk (conjugate with a riemann mapping). A mapping from the unit disk to itself always has a fixed point, and so again, without loss of generality, we can assume one of the fixed points is \(0\). By schwarz's lemma, this means that \(|f'(0)| \le 1\) and that \(|f(z)| \le |z|\). Where \(|f(z)| = |z|\) only when \(|f'(0)| = 1\). This means that \(f(z) = e^{i\theta}z\) for some \(\theta \in [0,2\pi)\).

So, if \(|f'(0)| < 1\), then \(|f(z)| < |z|\) and there are no more fixed points other than \(0\) by construction. So all that's left is if \(f\) has a neutral fixed point at \(0\). Wlog we can assume that \(f'(0) = 1\). Well then \(f(z) = z\), which is only fixed points, and is therefore the trivial iteration. And can be discarded. Or you can just observe that \(e^{i\theta}z\) has no other fixed points than zero.
....

James, I like that you took initiative, BUT:
Your proof has nothing to do with iteration (all your tediously made definition about local iteration find no application in your proof!), it basically states that a holomorphic function \(f\colon E\to E\) can only have one fixed point.
But the assumption \(E\to E\) is already the problem.
One interesting application would be the two fixed points for base sqrt(2).
The theorem should explain why one can not have one iteration holomorphic at both fixed points.
But I have already difficulties to find a domain \(E\) containing both fixed points such that \(b^z\colon E\to E\) (not even started with iteration here) because one of the fixed points is repelling so it tends to map stuff more far away than it came from.
Except perhaps I choose \(E=\mathbb{C}\) (which is simply connected) but then your proof would state that \(b^z\) can only have one fixed point on \(\mathbb{C}\), which is not true - maybe there is problem with "it is safe to assume ... (conjugate Riemann mapping)".

Correct me if I totally misunderstood your statements!

[JmsNxn]

Oh yes, I'm well aware that I didn't even use the iteration. I'm trying to find the more advanced proof that I had before, but I remembered this one quickly to get started. And also, I apologize for the simply connected domain part, I mean simply connected and biholomorphic to \(\mathbb{D}\). There are only two kinds of simple connectivity, all of \(\mathbb{C}\) or \(\mathbb{D}\). So that works.

For the case \(b^z\), just note that no simply connected domain (bi holomorphic to \(\mathbb{D}\)) can have a repelling fixed point. So by nature, there's no simply connected domain that has any repelling fixed points. This part was crucial to my argument, and I'm trying to remember all the details.

The second part of the proof (which I haven't written yet), is a bit more complicated, and I'm working on remembering how it goes. The iteration part does come into play. I'll see if I can remember. But essentially the rest of the argument uses that if \(f^{\circ t}(z) : D \times A \to A\) where \(A\) is the immediate basin about the fixed point, then that is its maximal domain. This stuff is buried in old notebooks though, so I'm racking my brain trying to remember how to derive it. But it isn't very far off from Milnor, where the Schroder coordinate change has a maximal domain of convergence.

[JmsNxn]

Okay, YES! I think I got it now. I only have part of it though, but the proof went something like this. This is only for the Schroder case, I can't remember what to do for the parabolic case--but that should take care of itself--if we stick to local iteration. There exists no local iteration about a neutral fixed points.

Assume that \(f\) is holomorphic on a domain \(H\) (not simply connected), and let's assume there are two fixed points \(p_0,p_1 \in H\), such that \(p_0\) is geometrically attracting. Let us additionally assume that \(f^{\circ t}(z)\) is holomorphic on all of \(H\), including the fixed point. Define \(\lambda = f'(p_0)\). We can define the domains:


\[
B(z) = \{q \in H\,|\,|\lambda^t| < 1,\,q = f^{\circ t}(z),\,\text{for some}\,z \in H\}\\
\]

This is a simply connected domain. And additionally, as \(t \to \infty\), we converge to \(p_0\), by the nature of the Schroder iteration. Now, we know that \(f:B(z) \to B(z)\) because \(f\) is a contraction mapping here. Therefore, for all \(z \in H\) and \(z \neq p_1\), \(p_1 \not\in B(z)\)--because it would be a simply connected domain with two fixed points. But the union of all \(B(z)\) is precisely \(H\), so \(f^{\circ t}(z)\) can't be holomorphic about all of \(H\).

Now, particularly, the assumption that might be troubling you is that we converge to \(p_0\) as \(t \to \infty\)--what if it converges somewhere else?

Well to play this out, we have to allow \(z \to J\) where \(J\) is the julia set of \(f\). The schroder coordinate change is no longer valid here, and if we could extend this value here, we've incidentally analytically continued \(\Psi^{-1}\) to a domain larger than its maximal domain; which is erroneous to say the least. I'm going to think about this more though, to further justify it for you. There should be something technical which will save me

I'll keep working on polishing this, I know it's something close to this.

And again, this says nothing about your iteration, because \(B(z)\) is no longer a domain in your case (there are poles).

---

Also! for a bit more intuition. Every regular iteration that is a local iteration that is a schroder iteration is periodic. So say it has period \(\ell\), then the arc \(\gamma\) connecting \(\ell\) and \(0\), induces a simply connected domain through the jordan curve \(\mu = f^{\gamma}\) about any \(z\), so there can only be one fixed point contained within the center of this jordan curve. I can't see how we can move \(z\) to include neither fixed points/the other fixed points... I'll have to think about this more.

[bo198214]

I think its not yet as mature as you wish, but keep going.
In the mean time I was constructing another counter example, this time without linear fractional mappings.
The idea is the following, we know that the Schröder iteration is given by:
\[ f^{\mathfrak{R} t}(x) = \chi^{-1}(c^t \chi(x)) \]
Instead for looking for functions \(f\) I just took \(\chi\) as my start of research, how does \(\chi\) need to look, such that \(f^{\mathfrak{R}t}\) is still analytic at a second fixed point.
For a second fixed point \(z_2\) it must be valid that:
\[ c^t \chi(z_2) = \chi(z_2) \]
which is basically satisfied for \(\chi(z_2)=0\) or \(\chi(z_2)=\pm\infty\).
So I took a function that satisfies all 3 values  : the tangent! And constructed the following function
\[ f(x) = \arctan(c\tan(x)) \] with the iteration \( f^{\mathfrak{R}t}(x) = \arctan(c^t\tan(x)) \). One has to be a bit careful with the choosen branch, so we choose:
\[ \arctan_0(c^t\tan(x)) + \pi\left\lfloor \frac{x+\frac{\pi}{2}}{\pi}\right\rfloor \]
This looks like:
   
So the main question is: Is this function analytic at \(\frac{\pi}{2} + k\pi \)?
Expressing tan with sin:
\[ \tan(x)=\frac{\sin(x)}{\sqrt{1-\sin(x)^2}}, \quad\arctan(x) = \arcsin\left(\frac{x}{\sqrt{1+x^2}}\right) \]
For brevity, write \(s\) for \(\sin(x)\)
\[\arctan(c\tan(x)) = \arcsin\left(\frac{\frac{cs}{\sqrt{1-s^2}}}{\sqrt{1+\frac{c^2s^2}{1-s^2}}}\right) = 
\arcsin\left(\frac{cs}{\sqrt{1-s^2+c^2s^2}}\right) = \arcsin\left(\frac{c\sin(x)}{\sqrt{1+\sin(x)^2(-1+c^2)}}\right)\]
So we see: nothing scary happens around \(x\approx \frac{\pi}{2}\), \(\sin(x)\approx 1\) just combination of analytic functions.
Hence \(f\) and \(f^{\mathfrak{R}t}\) is analytic on whole \(\mathbb{R}\) at infintely many fixed points.

[tommy1729]

I think its not yet as mature as you wish, but keep going.
In the mean time I was constructing another counter example, this time without linear fractional mappings.
The idea is the following, we know that the Schröder iteration is given by:
\[ f^{\mathfrak{R} t}(x) = \chi^{-1}(c^t \chi(x)) \]
Instead for looking for functions \(f\) I just took \(\chi\) as my start of research, how does \(\chi\) need to look, such that \(f^{\mathfrak{R}t}\) is still analytic at a second fixed point.
For a second fixed point \(z_2\) it must be valid that:
\[ c^t \chi(z_2) = \chi(z_2) \]
which is basically satisfied for \(\chi(z_2)=0\) or \(\chi(z_2)=\pm\infty\).
So I took a function that satisfies all 3 values  : the tangent! And constructed the following function
\[ f(x) = \arctan(c\tan(x)) \] with the iteration \( f^{\mathfrak{R}t}(x) = \arctan(c^t\tan(x)) \). One has to be a bit careful with the choosen branch, so we choose:
\[ \arctan_0(c^t\tan(x)) + \pi\left\lfloor \frac{x+\frac{\pi}{2}}{\pi}\right\rfloor \]
This looks like:

So the main question is: Is this function analytic at \(\frac{\pi}{2} + k\pi \)?
Expressing tan with sin:
\[ \tan(x)=\frac{\sin(x)}{\sqrt{1-\sin(x)^2}}, \quad\arctan(x) = \arcsin\left(\frac{x}{\sqrt{1+x^2}}\right) \]
For brevity, write \(s\) for \(\sin(x)\)
\[\arctan(c\tan(x)) = \arcsin\left(\frac{\frac{cs}{\sqrt{1-s^2}}}{\sqrt{1+\frac{c^2s^2}{1-s^2}}}\right) = 
\arcsin\left(\frac{cs}{\sqrt{1-s^2+c^2s^2}}\right) = \arcsin\left(\frac{c\sin(x)}{\sqrt{1+\sin(x)^2(-1+c^2)}}\right)\]
So we see: nothing scary happens around \(x\approx \frac{\pi}{2}\), \(\sin(x)\approx 1\) just combination of analytic functions.
Hence \(f\) and \(f^{\mathfrak{R}t}\) is analytic on whole \(\mathbb{R}\) at infintely many fixed points.

Epic answer Bo !

This reminds me I posted a question about iterations of periodic functions + identity. 

But my main question is this :

How about polynomial cases ?

I somewhat considered a degree 2 case with 2 fixpoints and its entire superfunction. 
Not quite the same as here but in the same spirit , the fermat superfunction.

https://math.eretrandre.org/tetrationfor...hp?tid=809&

Which bring me to the following " psychological " analysis ;

when we consider super( abel(x) + c ) 

we tend to jump to conclusions ...
(demonstration ; )

if x is a fixpoint then super(v) = x only at v around infinity.

in the limit to v we get super(v) = x.

reaching a limit at infinity means having derivative going to zero.

so the abel at x must have the derivative 1/0 , hence a log , pole or singularity.

so super( abel(x) + c) for nonzero c has the singularity of abel(x) , unless in special cases where the super cancels it 

( aka such as exp( ln(x) + c ) but not in say cube(cuberoot(x) + 1)  )

since the singularity or pole was not removed 

super ( abel(x) + c) ) for x a fixpoint must also have a singularity or pole.

we then believe 

super ( abel(fix) + 1/2 ) has a pole or singularity for fix.

and even if it is not at one fixpoints we believe it must be at another fix.

this psychological or rational failure might even happen unconsciously.

It is a very subtle mistake !!

Second " psychological " analysis :

just because the regalar iterations ( koenigs function ) suggest a periodic superfunction , this is not true !!

it is pseudoperiodic near the fixpoint used for the regular iteration.

but away from the fixpoint it might become less and less speudoperiodic or even take ANOTHER pseudoperiodic when we get closer to the other fixpoint.

An easy proofs follows from using better asymtotics than linear for the regular itertion near a fixpoint ; then suddenly it is no longer periodic.

***

it would be nice to see a degree 4 polynomial where they agree on the half-iterate at 2 fixpoints and are analytic at the fixpoints too.
with an analytic path from one fixpoint to the other ( repel and attract ).
and ofcourse the half-iterate not simply being a degree 2 polynomial.

Or a proof that it is not possible.

Or other degrees ofcourse.

I assume degree 4 is the smallest ( lower bound ) possible.



regards

tommy1729

[bo198214]

How about polynomial cases ?

Depends on what you mean by polynomial case. If the Schröder/Kœnigs function is a polynomial we can do the same as with tangent, but not quite:
What one needs to create two fixed points is two values that are either 0,\(\pm\infty\). However not all combinations do work.
Best combination is \(\{0,\infty\}\) or \(\{0,-\infty\}\) and the function being analytic at \(x_0\), and (I call) quasi-analytic at \(x_\infty\) i.e. its real valued left and right of \(x_\infty\) and its holomorphic on a vicinity \(V\setminus\{x_\infty\}\) (or can be continued from left to right avoiding \(x_\infty\) and the function is strictly increasing/decreasing.
I think if you have these ingredients then the resulting function iteration is analytic at both fixed points for all t.
Similar consideration for \(\{0,-\infty\}\), the case \(\{\pm\infty,\mp\infty\}\) is a combination of the previous cases as the function needs to pass through 0.

The remaining cases (0,0) and \((\pm\infty,\pm\infty)\) don't work well typically, because the function is not injective (has a min/max).
Maybe its possible but I doubt.

So in case of a polynomial we have a lot of zeros but only one \(\infty\) at \(\infty\). So you don't have the case \(\{0,\pm\infty\}\).
If you however would allow rational functions, one can quite do a similar construction as with the tangent.

       
And this linear fractional mapping stuff works like this (these functions always have a pole and zero and hence the case \(\{0,\pm\infty\}\) is paramount).

But if you mean: iterating a polynomial at one of its fixed points - well this method of taking \(\chi^{-1}(c^t\chi(x))\) does mostly not yield polynomials. So this approach would not much help here.

I somewhat considered a degree 2 case with 2 fixpoints and its entire superfunction. 
Not quite the same as here but in the same spirit , the fermat superfunction.

https://math.eretrandre.org/tetrationfor...hp?tid=809&

Yes, I also stumbled over a similar case as f(x) = (x-1)^2 + 1 recently.
The regular iteration is \(f^{\mathfrak{R}t}(x) = (x-1)^{2^t} + 1\), though at its left fixed point (at 1) it is not analytic
which has to do with its Schröder/Koenigs function \(\log(x-1)\) is not real valued left of 1 (see my previous preconditions).

Second " psychological " analysis :

just because the regalar iterations ( koenigs function ) suggest a periodic superfunction , this is not true !!

Is it not??? The super function is \(z\mapsto\chi^{-1}(c^z \cdot \chi(z_0))\) and \(c^z\) is periodic with period \(\frac{2\pi i}{\log( c)}\) ... or do you mean that the choice of the branch of \(\chi^{-1}\) can make it non-periodic (like the tangent construction, but there the choice of the branch depended on \(z_0\)).

it would be nice to see a degree 4 polynomial where they agree on the half-iterate at 2 fixpoints and are analytic at the fixpoints too.

Well as I said, the applied method would not work for this case ... but let the hunt be open!

[bo198214]

Just to be complete, the same procedure as with the Schröder-Koenigs iteration can also done with the Abel function tan.
When using Abel functions though the fixed points need to be \(\pm\infty\) not 0 anymore as with the Schröder iteration, because:
\[ \alpha(z_0) = t + \alpha(z_0) \]
If one now takes \(\alpha(x) = \tan(x) \) then we get a parabolic iteration that is analytic at infinitely many fixed points:

   

[bo198214]

Now that we have so ample evidence for iteration analytic at two fixed points I was wondering, what's the difference: What Schröder functions create multi-fp-analytic iterations and which don't? My guess is: if the Schröder function is meromorphic (forget about my "quasi-analytic"  ) then the iteration is analytic at all fixed points, otherwise it can't.
But JmsNxn knows better about this meromorphic business ...

[bo198214]

I think it is as easy as this: if \(f\) has a pole at \(z_0\) then \(\frac{1}{f}=:g\) is analytic there, and then we have
\[ f^{-1}(c^t f(z)) = g^{-1}\left(\frac{1}{c^t\frac{1}{g(z)}}\right) = g^{-1}(c^{-t} g(z)) \]
and hence it is analytic at the fixed point.
A similar reasoning works with the Abel function
\[ f^{-1}(t+ f(z)) = g^{-1}\left(\frac{1}{t+\frac{1}{g(z)}}\right) = g^{-1}\left(\frac{ g(z)}{tg(z)+1}\right) \]

[JmsNxn]

I think its not yet as mature as you wish, but keep going.
In the mean time I was constructing another counter example, this time without linear fractional mappings.
The idea is the following, we know that the Schröder iteration is given by:
\[ f^{\mathfrak{R} t}(x) = \chi^{-1}(c^t \chi(x)) \]
Instead for looking for functions \(f\) I just took \(\chi\) as my start of research, how does \(\chi\) need to look, such that \(f^{\mathfrak{R}t}\) is still analytic at a second fixed point.
For a second fixed point \(z_2\) it must be valid that:
\[ c^t \chi(z_2) = \chi(z_2) \]
which is basically satisfied for \(\chi(z_2)=0\) or \(\chi(z_2)=\pm\infty\).
So I took a function that satisfies all 3 values  : the tangent! And constructed the following function
\[ f(x) = \arctan(c\tan(x)) \] with the iteration \( f^{\mathfrak{R}t}(x) = \arctan(c^t\tan(x)) \). One has to be a bit careful with the choosen branch, so we choose:
\[ \arctan_0(c^t\tan(x)) + \pi\left\lfloor \frac{x+\frac{\pi}{2}}{\pi}\right\rfloor \]
This looks like:

So the main question is: Is this function analytic at \(\frac{\pi}{2} + k\pi \)?
Expressing tan with sin:
\[ \tan(x)=\frac{\sin(x)}{\sqrt{1-\sin(x)^2}}, \quad\arctan(x) = \arcsin\left(\frac{x}{\sqrt{1+x^2}}\right) \]
For brevity, write \(s\) for \(\sin(x)\)
\[\arctan(c\tan(x)) = \arcsin\left(\frac{\frac{cs}{\sqrt{1-s^2}}}{\sqrt{1+\frac{c^2s^2}{1-s^2}}}\right) = 
\arcsin\left(\frac{cs}{\sqrt{1-s^2+c^2s^2}}\right) = \arcsin\left(\frac{c\sin(x)}{\sqrt{1+\sin(x)^2(-1+c^2)}}\right)\]
So we see: nothing scary happens around \(x\approx \frac{\pi}{2}\), \(\sin(x)\approx 1\) just combination of analytic functions.
Hence \(f\) and \(f^{\mathfrak{R}t}\) is analytic on whole \(\mathbb{R}\) at infintely many fixed points.

BO! You are a beautiful soul! That is fucking unreal! But please hear me, when I say that is not a counter example!

You are using \(\tan\) which is a MEROMORPHIC function. This is something I'm finding far deeper in what I have. I swear, my next argument will be as solid as your last few posts. You've just gotta let me ruminate. We need that these are EUCLIDEAN maps, as Milnor calls them. This means, they are maps from the complex plane to itself. The function \(\tan\) is a map from the complex plane TO THE RIEMANN SPHERE. I know this sounds like a cop out on my part. But this is no different than the LFT results. There's something going on that is very different here. The exponential, and any transcendental entire function (Euclidean mapping--as Milnor calls it) won't suffer these results.

You again, primarily solve on the real line, where poles appear as you add imaginary argument. This forces what won't be a local iteration; it'll be a whole nother beast entirely. I just want to show that.

\[
f : \mathbb{C} \to \mathbb{C}\\
\]

\[
f^{\circ t}(z) : D \times H \to H\\
\]

Then \(H\) can't contain two fixed points of \(f\). And what you have shown to me is that EUCLIDEAN is necessary. Also, recall that \(D\) is a semi group under addition. So this is a semi group translation. And it is not \(\mathbb{R}\), it is a DOMAIN in \(\mathbb{C}\). These restrictions I think are the exact amounts.

I really want to get this write bo, and I promise I'll try my fucking hardest. But you still ain't provided a true true counter example to what I mean. I've been rereading milnor again to make sure I can argue from scratch. lol

I mean, with all due respect, not trying to be contrarian:

\[
\arctan(c\tan(x))\\
\]

Is not Euclidean. You tell me what world that's an entire function.

I know I'm adding these points now, but they were always my assumptions, lol.

So you are absolutely right, we can iterate about multiple fixed points. My point is that this fails for Euclidean cases. Which is again entire functions. Which is a specific branch of complex dynamics which relates to Tetration the most. Meromorphic dynamics is far more whacky. And as I see it, this is meromorphic dynamics.

Again, not trying to be snarky. Just saying how I see it. I love being proven wrong though--because that would make all this shit so much simpler.

I hope I don't seem like I'm moving the goal posts. Like I'm trying not to be proven wrong. You've definitely proven me wrong on my general statement. But that has helped me hone it a lot more. And I'm really fuckign sure there is a way using Euclidean maps \(\mathbb{C} \to \mathbb{C}\) to explain why you can't iterate about two fixed points. The moment you allow for the compact domain \(\widehat{\mathbb{C}}\) to enter the picture; or branching in some manner; everything becomes far more of a free for all, and multiple fixed point iteration exists.

But for \(\mathbb{C} \to \mathbb{C}\), I don't think it's possible. And I am at least 75% there in a much better argument. Just let me ruminate and think and jot down some shit; reread milnor a couple times, lol.