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Contraction Mapping Theorem

If $T\colon X\to X$ is a contraction mapping on a complete metric space $(X,d)$ then there is exactly one solution $x\in X$.

Proof:

Let $x_0$ be any point in $X$. We define a sequence by $$x_{n+1}=Tx_n, \qquad \text{for } n\geq 0.$$ Denote the $n$th iterate of $T$ by $T^n$, so that $x_n = T^n x_0$. First, we show that $(x_n)$ is a Cauchy Sequence. If $n\geq m\geq 1$ then $$d(x_n,x_m) = d(T^n x_0, T^m x_0) \leq c^md(T^{n-m} x_0,x_0).$$

This is only some of the proof but my question is about the last inequality in particular the $T^{n-m}$. How is this obtained?

Thank you for any help and comments.

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  • $\begingroup$ Just induct on the condition $d(Tx, Ty) \leq cd(x,y)$ with $x=T^{n-1}x_0, y = T^{m-1}x_0$ $\endgroup$ – Prahlad Vaidyanathan Nov 1 '13 at 3:41
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I assume from context that $c$ is defined to be a constant for which

$$d(Tx, Ty) \le c d(x, y)$$

for all $x$ and $y$ (and in particular, $c < 1$ since $T$ is a contraction). Then

$$d(T^n x_0, T^m x_0) \le c d(T^{n - 1} x_0, T^{m - 1} x_0) \le c^2 d(T^{n - 2} x_0, T^{m - 2} x_0)$$

and so on. Apply the condition a total of $m$ times.

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Officially, it's induction, but do it one step at a time: $$d(T^nx,T^mx)\le c\,d(T^{n-1}x,T^{m-1}x).$$ Now repeat.

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You have

$$[Tx,Ty]\le c[x,y]$$ where I used the bracket for distance. Try to apply $T$ on the rightmost $m$ times and you reach the inequality.

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