contractive sequence

The sequence

$a_0,a_1,a_2,\mathrm{\dots }$(1)

in a metric space $(X,d)$ is called contractive, iff there is a real number$r\in (0,1)$ such that for any positive integer $n$ the inequality

$d(a_n,a_{n+1})\leqq r\cdot d(a_{n-1},a_n)$(2)

is true.

We will prove the

Theorem.  If the sequence (1) is contractive, it isa Cauchy sequence.

Proof.  Suppose that the sequence (1) iscontractive. Let $\epsilon$ be an arbitrary positive number and$m,n$ some positive integers from which e.g. $n$ is greater than$m,$ $n=m+\delta$.

Using repeatedly the triangle inequality we get

$d(a_m,a_n)$$\leqq d(a_m,a_{m+1})+d(a_{m+1},a_{m+\delta })$

$\leqq d(a_m,a_{m+1})+d(a_{m+1},a_{m+2})+d(a_{m+2},a_{m+\delta })$

$\mathrm{\dots }$

$\leqq d(a_m,a_{m+1})+d(a_{m+1},a_{m+2})+d(a_{m+2},a_{m+3})+\mathrm{\dots }+d(a_{n-1},a_n).$

Now the contractiveness gives the inequalities

$$d(a_1,a_2)\leqq rd(a_0,a_1),$$

$$d(a_2,a_3)\leqq rd(a_1,a_2)\leqq r^{2}d(a_0,a_1),$$

$$d(a_3,a_4)\leqq rd(a_2,a_3)\leqq r^{3}d(a_0,a_1),$$

$$\mathrm{\dots }$$

$$d(a_m,a_{m+1})\leqq r^{m}d(a_0,a_1),$$

$$\mathrm{\dots }$$

$$d(a_{n-1},a_n)\leqq r^{n-1}d(a_0,a_1),$$

by which we obtain the estimation

$d(a_m,a_n)$$\leqq d(a_0,a_1)(r^m+r^{m+1}+\mathrm{\dots }+r^{m+\delta -1})$

$=d(a_0,a_1)r^m(1+r+r^2+\mathrm{\dots }+r^{\delta -1})$

$=d(a_0,a_1)r^m{\displaystyle \frac{1-r^{\delta }}{1-r}}$

The last expression tends to zero as $m\to \mathrm{\infty }$.  Thusthere exists a positive number $M$ such that

when $n>m$.  Consequently, (1) is a Cauchy sequence.

Remark.  The assertion of the Theorem cannot bereversed. E.g. in the usual metric of $ℝ$, thesequence  $1,\frac{1}{2},\frac{1}{3},\mathrm{\dots }$  converges to 0and hence is Cauchy, but for it the ratio

$$|a_n-a_{n+1}|:|a_{n-1}-a_n|=\mathrm{\hspace{0.33em}1}-\frac{2}{n+1}$$

tends to 1 as  $n\to \mathrm{\infty }$.

Cf. sequences of bounded variation (http://planetmath.org/SequenceOfBoundedVariation).