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Let $(X,d)$ be a complete metric space and $f$ a mapping of $X$ into itself. Let $\{f^n(x)\}=\{x_n\}$ be the sequence of iterated transforms.

Suppose $f$ satisfies that for each $\varepsilon >0$,there exists $\delta>0$ such that for all $x,y\in X$, $$\varepsilon\leq d(x,y)<\varepsilon+\delta\implies d(f(x),f(y))<\varepsilon.$$

Is $\{x_n\}$ a Cauchy sequence ?

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    $\begingroup$ Suppose we assume the weaker condition (1) that $d(f(x), f(y)) < d(x,y)$, then is it obviously false? Or if we started with the stronger assumption (2) that $d(f(x), g(x)) < d(x, y) / 1.001$, then is it obviously true? $\endgroup$
    – Pat Devlin
    Commented Dec 21, 2016 at 5:07
  • $\begingroup$ Answering my own question, (2) is true, but it gives me an idea that might show (1) is false. $\endgroup$
    – Pat Devlin
    Commented Dec 21, 2016 at 5:09
  • $\begingroup$ Just for me to understand, $\{x_n\}$ is a sequence of points in $X$ given an arbitrary $x\in X$? $\endgroup$
    – Amir Sagiv
    Commented Dec 21, 2016 at 5:34
  • $\begingroup$ $x$ is an arbitrary but fixed element in $X$ such that : ${x_n}={f^n(x)}$ $\endgroup$
    – Isra El
    Commented Dec 22, 2016 at 3:30
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    $\begingroup$ @Pat Devlin (1) is vacuous (the hypothesis fails in every nonempty metric space: take $x=y$); but if you require this only for $x\neq y$ then yes it's false: let $X$ consist of the set of real numbers $a_n = n + 1/2^n$ and let $f(a_n) = a_{n+1}$. $\endgroup$
    – Noam D. Elkies
    Commented Dec 22, 2016 at 22:15

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$\let\eps\varepsilon$Yes, that's true, and the argument is somewhat similar to that for your previous question. Surely, completeness is not needed, since you wish to obtain a Cauchy sequence, not a convergent one.

Denote $d_i=d(x_i,x_{i+1})$. Applying the condition for $\eps=d_i$, we get that $d_i$ strictly decrease, so there exists $D=\lim_{i\to\infty} d_i$. If $D>0$, we get a contradiction by setting $\eps=D$; so $D=0$.

Now let us show that for every $\eps>0$ there exists $N$ such that for all $k\geq n>N$ we have $d(x_k,x_n)<2\eps$; this is exactly what we need. Choose $\delta$ for our $\eps$ from the condition; we may assume that $\delta<\eps$. There exists $N$ such that $d_n<\delta$ for all $n>N$. We claim that this $N$ fits.

Indeed, let us show that for all $k\geq n>N$ we have $d(x_k,x_n)<\eps+\delta$; this is clearly sufficient. Induction on $k$. If $k=n$ then there is nothing to prove. Assume now that $k>n$. [ADDED] By the induction hypothesis, we have $d(x_{k-1},x_n)<\eps+\delta$. Now consider the following two cases.[/ADDED]

If $d(x_{k-1},x_n)<\eps$ then $$ d(x_k,x_n)\leq d_{k-1}+d(x_{k-1},x_n)<\delta+\eps. $$ Otherwise, we have $\eps \leq d(x_{k-1},x_n)<\eps+\delta$, so by the condition $d(x_k,x_{n+1})<\eps$, and $$ d(x_k,x_n)\leq d(x_k,x_{n+1})+d_n\leq \eps+\delta. $$ The claim is proved.

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  • $\begingroup$ Thank you Ilya Bogdanov ;but a have some problems with the choosing of $\delta$. $\endgroup$
    – Isra El
    Commented Dec 22, 2016 at 3:33
  • $\begingroup$ May you specify them? $\endgroup$
    – Ilya Bogdanov
    Commented Dec 22, 2016 at 9:04
  • $\begingroup$ why we have this: $\varepsilon\leq d(x_{k-1},x_n)<\varepsilon+\delta$ ? $\endgroup$
    – Isra El
    Commented Dec 22, 2016 at 21:24
  • $\begingroup$ Sorry, I missed to say a few words on this; now they're in. Basically, this is by the induction hypothesis. $\endgroup$
    – Ilya Bogdanov
    Commented Dec 22, 2016 at 21:44

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