Let $(X,d)$ be a metric space. A map $f: X \longrightarrow X$ is called contracting if there exists a $\lambda < 1$ such that for any $x, y \in X$

$$d(f(x),f(y)) \leq \lambda d(x,y)$$

It is well known that a contracting map $f$ on a complete metric space has a unique fixed point to which all points converge under interation by $f$.

In Katok's and Hasselblatt's Introduction to the Modern Theory of Dynamical Systems, there is an exercise designed to show the importance of the uniform contraction condition in the definition of contracting maps, that is, to show the importance of the existence of a $\lambda$ strictly less than $1$ that works for all $x, y \in X$. The question is as follows:

Let $(X,d)$ be a complete metric space. Construct an example of a map $f:X \longrightarrow X$ such that $d(f(x),f(y)) < d(x,y)$ for $x \neq y$, $f$ has no fixed point, and $d(f^n(x),f^n(y))$ does not converge to zero for some $x,y \in X$.

I've been having a lot of trouble keeping everything in check: $X$ complete, no fixed point, and also the existence of some pair $x,y$ for which the distance between iterates does not go to $0$. Are any examples of this highly pathological or am I missing something simple?

Thank you in advance.


surface $r = 1 + e^{-z}$ in cylindrical coordinates. Map is just $z \mapsto 1 + z,$ that is $$ ( ( 1 + e^{-z}) \cos \theta, ( 1 + e^{-z}) \sin \theta, z ) \mapsto ( ( 1 + e^{-z -1}) \cos \theta, ( 1 + e^{-z-1}) \sin \theta, z +1 ) $$

Or, intersection of the same thing with the $xz$ plane, easier, and the proofs become easier. No apparent requirement for connectivity.

  • $\begingroup$ This makes perfect sense, and sheds some light into the process behind it. Thank you. $\endgroup$ – Fimpellizieri Sep 22 '14 at 3:42

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