Actually I already found the flaw.
I was looking at it on the complex plane and found out that \(\arctan(c\tan(z))\) has branch points at \(\pm i\text{artanh}(1/c)+\pi k\) for \(c>1\) and \(\pm i\text{artanh}( c ) + \frac{\pi}{2} + \pi k\) for \(0< c< 1\). (Takes to long for just quickly explaining it, you have to look at the conformal map of tan, and then you see that i/c is mapped (by cz) to the branch point of arctan at i. So there the trouble begins.)
This means these branch points come close to the real line (assume c>1) at \(x_{2k}=\frac{\pi}{2} 2k\) by \(\text{artanh}(c^{-t})\) for \(t\to \infty\) and at \(x_{2k+1}=\frac{\pi}{2}(2k+1)\) by \(\text{artanh}( c^{t} )\) for \(t\to -\infty\).
So for two consecutive fixed points \(x_{2k}\) and \(x_{2k+1}\) one of them can not be contained in an analytic vicinity independent of t (and for two nonconsecutive fixed point the belt it tightened between them).
This time though not because a pole moves in, but a branch point moves in when \(t \to \pm\infty\).
Mhm, so there seems something true about the condition of a domain independent of t.
Though I wonder whether for any non-entire function there can be two fixed points in a domain D such that \(f^{\circ n}\) is holomorphic on D for all \(n\in\mathbb{N}\) (nothing to do with non-integer iterations).
Maybe you have a proof for this - as a first step. (I mean if there is such domain D then the function is already entire)
Looks like branch points and poles are "sucked in" by the fixed point and hence come arbitrarily near.
I was looking at it on the complex plane and found out that \(\arctan(c\tan(z))\) has branch points at \(\pm i\text{artanh}(1/c)+\pi k\) for \(c>1\) and \(\pm i\text{artanh}( c ) + \frac{\pi}{2} + \pi k\) for \(0< c< 1\). (Takes to long for just quickly explaining it, you have to look at the conformal map of tan, and then you see that i/c is mapped (by cz) to the branch point of arctan at i. So there the trouble begins.)
This means these branch points come close to the real line (assume c>1) at \(x_{2k}=\frac{\pi}{2} 2k\) by \(\text{artanh}(c^{-t})\) for \(t\to \infty\) and at \(x_{2k+1}=\frac{\pi}{2}(2k+1)\) by \(\text{artanh}( c^{t} )\) for \(t\to -\infty\).
So for two consecutive fixed points \(x_{2k}\) and \(x_{2k+1}\) one of them can not be contained in an analytic vicinity independent of t (and for two nonconsecutive fixed point the belt it tightened between them).
This time though not because a pole moves in, but a branch point moves in when \(t \to \pm\infty\).
Mhm, so there seems something true about the condition of a domain independent of t.
Though I wonder whether for any non-entire function there can be two fixed points in a domain D such that \(f^{\circ n}\) is holomorphic on D for all \(n\in\mathbb{N}\) (nothing to do with non-integer iterations).
Maybe you have a proof for this - as a first step. (I mean if there is such domain D then the function is already entire)
Looks like branch points and poles are "sucked in" by the fixed point and hence come arbitrarily near.
