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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

  • $d(f(x),f(y))<d(x,y)$ 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!

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    $\begingroup$ One answer is given here. $\endgroup$ Jun 24, 2021 at 18:35
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    $\begingroup$ 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, $$ f(x,y) = (x+1,\frac y{y+1}). $$ $\endgroup$ Jun 24, 2021 at 18:45

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