Let $\{x_n\}$ be a bounded sequence of distinct real numbers such that $|x_{n+1}-x_n|<|x_n-x_{n-1}|,\forall n\in \mathbb N$ , then is it true that $\{x_n\}$ converges ?

The motivation for this comes from the fixed point theorem that if $X$ is compact metric space and $f:X\to X$ is a function such that $d(f(x),f(y))<d(x,y),\forall x,y \in X , x\ne y$ then $f$ has a fixed point.


marked as duplicate by user223391, Claude Leibovici, Did real-analysis Oct 26 '16 at 7:50

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    $\begingroup$ Why the downvotes? $\endgroup$ – Darío G Oct 26 '16 at 4:44
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    $\begingroup$ It should be straightforward to adapt this answer for the strict inequality. $\endgroup$ – dxiv Oct 26 '16 at 5:07
  • $\begingroup$ Voting to reopen. This is not a duplicate $-$ the condition $|x_{n+1}-x_n|<|x_n-x_{n-1}|$ is not at all the same as $|x_n-x_{n+1}|\rightarrow 0$. $\endgroup$ – TonyK Oct 31 '16 at 12:08

Following @dxiv's comment: Consider the sequence defined recursively by putting $x_1=1,x_2=\frac{1}{2}$ and

$$x_{n+1}=\begin{cases}x_n+\frac{1}{n+1} &\text{if $\left(x_{n-1}<x_n \text{ and } x_{n+1}+\frac{1}{n+1}\leq 1\right)$ or $\left( x_{n-1}>x_n \text{ and }x_n-\frac{1}{n+1}<0\right)$}\\ x_n-\frac{1}{n+1} &\text{otherwise}\end{cases}$$

The idea is that this sequence represents the walk of a person who make the $n$-th step of size $\frac{1}{n}$, and changing direction if there is not enough room for the next step.

For every $n\geq 1$, we have $|x_{n+1}-x_n|=\frac{1}{n+1}$, the sequence is bounded by $0$ and $1$, but it is not convergent since it has subsequences converging to $0$ and $1$ (namely, those points when the person change direction).




It does not converge : Consider a sequence $r_n$ s.t. it is monotone strictly decreasing and it converges to $\frac{1}{2}$

Then define $$ x_1=0,\ x_2=r_1,\ x_3=r_1-r_2,\ x_4=r_1-r_2+r_3,\ \cdots $$

That is $$ x_n=r_1-r_2+\cdots + (-1)^{n}r_{n-1} $$

Note that this sequence has two subsequences which converges to different limit