# Counterexample for a weakened definition of contraction mapping

I was trying to solve this exercise form a book on dynamical systems. There, after introducing the notion of contraction mapping, it is shown that, given a complete metric space $$(X,d)$$ and a contraction mapping $$f:X\to X$$, there exists a unique point $$p\in X$$ s.t. $$f(p)=p$$ and $$\lim_{n\to+\infty}f^n(x)=p$$ for all $$x\in X$$ (here $$f^n$$ represents the $$n-$$th iterate of the map).

After that, a problem ask the reader to find essentially a counterexample of this result in case the definition of contraction mapping is weakened. Specifically

Construct an example of a map $$f:X\to X$$ (where $$(X,d)$$ is a complete metric space) such that

• $$d(f(x),f(y)) for $$x\neq y$$
• $$f$$ has no fixed points
• $$d(f^n(x),f^n(y))$$ does not converge to zero for some $$x,y$$.

I tried to construct an example in $$(X,d)=(\mathbb{R},|\cdot|)$$ and I managed to satisfy the first two conditions, but not the third.

My attempt was the function $$f:\mathbb{R}\to\mathbb{R}$$ $$f(x)=\sqrt{1+x^2}.$$ Since $$f(x)>x$$ for all $$x\in\mathbb{R}$$, there are no fixed points, and since we have that $$d(f(x),f(y))=\left|\sqrt{1+x^2}-\sqrt{1+y^2}\right|=\left|\dfrac{x^2-y^2}{\sqrt{1+x^2}+\sqrt{1+y^2}}\right|<\left|\dfrac{x^2-y^2}{x+y}\right|=|x-y|=d(x,y)\ .$$ However, since its iterates are $$f^n(x)=\sqrt{n+x^2}\ ,$$ $$\lim_{n\to+\infty}(f^n(x)-f^n(y))=\lim_{n\to+\infty}\left(\sqrt{n+x^2}-\sqrt{n+y^2}\right)=0$$ for all $$x,y\in\mathbb{R}$$.

I would like to see how to construct a function satisfying all three conditions.

Thanks!

• One answer is given here. Jun 24, 2021 at 18:35
• As a modification to the answer given in the above link: take $$X = \{(x,y) \in \Bbb R^2 : xy = 1, \quad x > 0\}.$$ Define $f:X \to X$ by $$f(x,\frac 1x) = (x+1,\frac 1{x+1}).$$ Equivalently, $$f(x,y) = (x+1,\frac y{y+1}).$$ Jun 24, 2021 at 18:45