Fixed Point Theorem

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 $.

Average: 5 (2 votes)

Back to top