$ \vert x_m - x_n\vert \le \frac{K^{n-1}}{1-K}\vert x_2-x_1\vert $

Question

Prove that for all m > n $\ge$ 1,

$ \vert x_m - x_n\vert \le \frac{K^{n-1}}{1-K}\vert x_2-x_1\vert $

Where K is a contraction factor of f, such that K $\in$ [0,1) and $\vert f(x)-f(y) \vert \le K\vert x-y \vert $

$(x_n)$ is the sequence defined as $x_1 = \alpha \in R$ and $x_n = f(x_{n-1}) $ for all $n\ge2$

I have already shown that $\forall n \ge 1$, $\vert x_{n+1}-x_n\vert \le K^{n-1}\vert x_2-x_1 \vert$,

so that can be used without proof.

I tried using induction by showing m=n+1 holds, assuming m=n+k holds, but couldn't show that m=n+k+1 holds.

I don't know if you're supposed to use induction but that's the only method I could think of.

Answer 1

If you've shown that $ \left(\forall n\geq 1\right),\ \left|x_{n+1}-x_{n}\right|\leq K^{n-1}\left|x_{2}-x_{1}\right| $ then we'll be using it directly :

Let $ n,m $ be positive integers such that $ m\geq n $, we have : \begin{aligned}\left|x_{m}-x_{n}\right|&=\left|\sum_{k=n}^{m-1}{\left(x_{k+1}-x_{k}\right)}\right|\\ &\leq\sum_{k=n}^{m-1}{\left|x_{k+1}-x_{k}\right|}\\ &\leq\sum_{k=n}^{m-1}{K^{k-1}\left|x_{2}-x_{1}\right|}=\frac{K^{n-1}-K^{m-1}}{1-K}\left|x_{2}-x_{1}\right|\leq\frac{K^{n-1}}{1-K}\left|x_{2}-x_{1}\right|\end{aligned}