Fixed point question: $|f'(x)|=1$ case

Question

I have a fixed point question. Let's say we have the following recursive equation: $x_{n+1}=f(x_{n})$. According to Wikipedia, $x_n$ converges to a fixed point $x_0$ as long as $|f'(x)|<1$ in an open neighbourhood of $x_0$.

However, I've read on the Internet that the convergence is not guaranteed for $|f'(x)|=1$.

Do you know if there exists additional conditions that guarantee the convergence for $|f'(x)|=1$? Thanks!

Answer 1

What follows refers to rational functions on the Riemann sphere, i.e. complex dynamics is implied.

Without loss of generality, let the fixed-point $z_0=0$ so that $f(0)=0$ and we are interested in the case $f'(0)=\lambda$ with $|\lambda|=1$, i.e. $$\lambda = \exp(2\pi i\alpha),\qquad \alpha\in[0,1]$$ There are two cases:

• If $\alpha\in\Bbb Q$ is rational, then the fixed-point is called "rationally indifferent" or "parabolic".
• If $\alpha\in\Bbb R\backslash \Bbb Q$ is irrational then the fixed-point is called "irrationally indifferent".

The first result was already known to Julia and Fatou:

• If $z_0$ is a parabolic periodic point of $f$, then $z_0\in J_f$. Where $J_f$ denotes the Julia set of $f$.

• Parabolic case: If $\lambda^n=1$ and $\lambda^k\neq1$ for $0<k<n$, then either $f^n$ is the identity, or there exists a homeomorphism $h$ in a neighborhood of $0$ with $h(0)=0$ and$$h\circ f\circ h^{-1}(z)=\lambda z(1+z^{kn})\quad\text{ for some } k\geqslant 1.$$

The irrational case is more complicated. $f(0)=0$ is called stable, if for any neighborhood $U$ of $0$ there exists a neighborhood $V$ of $0$ such that $V\subset U$ and $f^k(V)\subset U$ for any $k\geqslant 1$.

Attractive fixed-points are obviously stable. For indifferent fixed-points there is a result from Moser and Siegel:

• Let $f(z) = \lambda z+a_2z^2+a_3z^3+\cdots$ be the power series of $f$ and $\lambda^n\neq1$ for all $n\in\Bbb N$. Then $0$ is a stable fixed-point iff the functional equation (Schröder's equation) $$\phi(\lambda z) = f(\phi(z))\tag1$$has an analytic solution in a neighborhood of $0$. This means $f$ is locally equivalent to a rotation by $2\pi\alpha$ if such $\phi$ exists.

Solvability of Schröder's equation is not trivial and characterization involves how $\alpha$ behaves under Diophantine approximation.

There are numbers for which Schröder's equation does not have a solution, for example (Cremer) for $$\{\lambda:|\lambda|=1\text{ and }\liminf_{n\in\Bbb N} |\lambda^n-1|^{1/n}=0\}.$$

If a solution $\phi$ of $(1)$ exists, then $f$ is called linearizable at 0 for obvious reasons, and the maximal domain containing $f(0)=0$ in which $(1)$ holds is called Siegel disc.