# Help with Theorem - Local Contraction Mapping

• Aug 5th 2010, 09:27 PM
Mollier
Help with Theorem - Local Contraction Mapping
Hi,
I am having some problems understanding the proof of the following theorem.

Theorem:
Suppose that g is a real-valued function, defined and continuous on a bounded closed interval $[a,b]$ of the real line, and assume that $g(x)\in [a,b]$ for all $x\in [a,b]$.
Let $\xi = g(\xi)\in [a,b]$ be a fixed point of $g$, and assume that $g$ has a continuous derivative in some neighborhood of $\xi$ with $|g'(x)|<1$. Then the sequence $(x_k)$ defined by $x_{k+1}=g(x_k),k\geq 0$, converges to $\xi$ as $k\rightarrow\infty$, provided that $x_0$ is sufficiently close to $\xi$.

Proof:
By hypothesis, there exists $h>0$ such that $g'$ is continuous in the interval
$[\xi-h, \xi+h]$. Since $|g'(\xi)|<1$ we can find a smaller interval
$I_{\delta}=[\xi-\delta,\xi+\delta]$, where $0<\delta\leq h$, such that $|g'(x)\leq L|$ in this interval, with $L<1$.
To do so, take $L=\frac{1}{2}(1+|g'(\xi)|)$ and then choose $\delta\leq h$ such that,
$|g'(x)-g'(\xi)|\leq \frac{1}{2}(1-|g'(\xi)|)$
for all $x$ in $I_{\delta}$; this is possible since $g'$ is continuous at $\xi$.

I will stop there as already I am not sure what's going on.
Why should I take $L=\frac{1}{2}(1+|g'(\xi)|)$?
By the way, in this book L is used to denote the "contraction factor" such that;
$|g(x)-g(y)|\leq L|x-y|$ for all $x,y\in[a,b]$.

Thanks.