Fixed Point Theorem

Posted November 11th, 2007 by Structure

Let (X,d) a complete metric space and f:X--->X a contraction (i.e. a function for which there is a real c, $0\leq c<1$ such as $\forall x,y \in X, d(f(x),f(y))\leq c d(x,y)$ Then there is a unique $\psi$ such as $f(\psi)=\psi$ . Such a point $\psi$ is called a fixed point for f.

Proof.
Unicity. Let $\psi_1\neq\psi_2$ two different fixed points.Then $d(\psi_1,\psi_2)>0$ . We have
$0<d(\psi_1,\psi_2)=d(f(\psi_1),f(\psi_2))\leq c d(\psi_1,\psi_2)<d(\psi_1,\psi_2)$ contradiction!
Existence.
Let $x_0\in X$ an arbitrary point and $x_1=f(x_0)$ If $x_1=x_0$ then $x_0$ is the unique fixed point.
Otherwise, let $x_{n+1}=f(x_n)$ We shall prove that sequence $(x_n)_{n\in N}$ is a Cauchy sequence and as X is a complete metric space , the sequence $(x_n)_{n\in N}$ will be a convergent sequence.
$\lim_{n\to\infty}x_n=\psi$
We have

$d(x_{n},x_{n+1})=d(f(x_{n-1}),f(x_{n}))\leq cd(x_{n-1},x_{n})=cd(f(x_{n-2}),f(x_{n-1}))\leq c^2d(x_{n-2},x_{n-1})\leq...\leq c^{n}d(x_0,x_1)$

Then we also have

$d(x_{n},x_{n+p})\leq d(x_{n},x_{n+1})+d(x_{n+1},x_{n+2})+...d(x_{n+p-1},x_{n+p})\leq c^{n}d(x_0,x_1)+c^{n+1}d(x_0,x_1)+...+c^{n+p-1}d(x_0,x_1)=c^n d(x_0,x_1)(1+c+c^2+...+c^{p-1})=c^n d(x_0,x_1)\frac{1-c^p}{1-c}<\frac{c^n}{1-c}d(x_0,x_1)$

.
Now let $\epsilon>0$ .
$d(x_{n},x_{n+p})<\frac{c^n}{1-c}d(x_0,x_1)<\epsilon$ if n is big enough as $\lim_{n\to\infty}\frac{c^n}{1-c}d(x_0,x_1)=0$
As any contraction is a continuous function from $x_{n+1}=f(x_n)$ we have $\psi=\lim_{n\to +\infty}x_{n+1}=\lim_{n\to +\infty}f(x_n)=f(\lim_{n\to +\infty}x_n)=f(\psi)$
From $d(x_{n},x_{n+p})<\frac{c^n}{1-c}d(x_0,x_1)<\epsilon$ taking limit as p tends to $+\infty$ we have also
$d(x_{n},\psi)\leq\frac{c^n}{1-c}d(x_0,x_1)$ which gives an estimation of the error made by replacing $\psi$ by $x_n$ .

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