Conditionally hyperbolic attractor

Question

Motivated by this post and the comment- conversation of this post we consider the following definition:

Assume that $f$ is a map on a neighborhood of $0$ with $f(0)=0$. We say that $0$ is a conditionally hyperbolic attractor if for every $p$ sufficiently close to $0$, the series $\sum f^{n}(p)$ is a convergent series. This concept is invariant under a linear change of coordinate. So it can be defined on any arbitrary Affine manifold.

What is an example of a local homeomorphism with a conditionally hyperbolic attractor at $0$ but $0$ is not a hyperbolic fixed point.

Answer 1

One possibility is that $\sum f^n(p)$ converges while the rate at which $f^n(p) \to 0$ is only polynomial in $n$ and not geometric. I cooked up the example

$$ f(x) = \begin{cases} \frac{x}{x-1} & x < 0 \\ - \frac{x}{x+1} & x > 0 \\ 0 & x = 0 \end{cases} $$

Clearly $f$ is continuous and defined in a neighborhood of $0$, and $f(0) = 0$ while $f'(0) = -1$ (from both sides) so that $f$ is not hyperbolic. It is asymptotically stable however, as one can check by looking at forward iterates of intervals of the form $[0,1/n]$ and $[-1/n,0]$ for $n \geq 2$. Since $f(1/n) = -(1/n)/(1+1/n)= - 1/(n+1)$ we have $$f([0,1/n]) = f([-1/(n+1), 0])$$ and since $f(-1/n) = (-1/n)/(-1-1/n)=1/(n+1)$ we have $$f([-1/n,0]) = [0,1/(n+1)]$$

These arguments imply $\sum f^n(p)$ converges conditionally for $p$ sufficiently close to $0$ as an alternating series.

One could also construct an example sending $1/n^2 \to 1/(n+1)^2$, in which case $\sum f^n(p)$ converges absolutely while $0$ is a nonhyperbolic fixed point.