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.
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.
