So, I am reading in a introductory numerical analysis book, but I believe my question fits into the calculus forums.

First, the definition of a simple iteration as given in my book:

*Suppose that $\displaystyle g$ is a real-valued function, defined and continuous on a bounded closed interval $\displaystyle [a,b]$ of the real line, and assume that $\displaystyle g(x)\in [a,b]$ for all $\displaystyle x\in [a,b]$. Given that $\displaystyle x_0\in [a,b]$, the recursion defined by *

* $\displaystyle x_{k+1}=g(x_k),\;k=0,1,2,\cdots,$*

* is called a ***simple iteration**.

__Contraction Mapping Theorem:__

*Let g be as in the definition above. Suppose further, that g is a contraction on $\displaystyle [a,b]$. Then, $\displaystyle g$ has a unique fixed point $\displaystyle \xi$ in the interval $\displaystyle [a,b]$. Moreover the sequence $\displaystyle (x_k)$ defined above converges to $\displaystyle \xi$ as $\displaystyle k\rightarrow \infty$ for any starting value $\displaystyle x_0$ in $\displaystyle [a,b]$.*

First of all, I don't really understand the simple iteration. I see that because $\displaystyle g(x)\in [a,b]$ for all $\displaystyle x\in [a,b]$, then there exists a number $\displaystyle \xi\in [a,b]$ such that $\displaystyle g(\xi)=\xi$. It would be great if someone had the time to explain this in more detail.

As for the contraction mapping theorem, I understand how $\displaystyle g$ has a unique fixed point $\displaystyle \xi$ in the interval $\displaystyle [a,b]$. However, I do not understand how it implies that the recursive sequence converges to $\displaystyle \xi$ as $\displaystyle k\rightarrow \infty$.

The proof of this is given in the book and is as follows;

$\displaystyle |x_k-\xi|=|g(x_{k-1})-g(\xi)|\leq L|x_{k-1}-\xi|,\;k\geq 1$

we then deduce by induction that

$\displaystyle |x_k-\xi|\leq L^k|x_0-\xi|,\;k\geq 1$

I do not understand how they come to that last part by induction.

Any help is appreciated. Thanks!