Banach fixed point theorem

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[edit] Theorem

Let $(M, d)$ be a complete metric space and $f$ a contraction on $M$ with Lipschitz constant $k$ . Then there exists a unique $x\in M$ such that $f (x) = x$ .

Let us define a sequence $(x n)$ of elements in $M$ in the following way

and show that this sequence is a Cauchy sequence.
Since $f$ is a contraction, we have that, for $n\in \mathbb N_0$

which results in the formula

As such, we obtain, for any $n,m\in \mathbb N_0$ ( $n\leq m$ )

$d(x_n,x_m)\leq \sum_{i=n}^{m-1} d(x_i,x_{i+1})\leq d(x_0,x_1)\sum_{i=n}^{m-1}k^i\leq d(x_0,x_1)\frac{k^n}{1-k}$

by basic results on geometric series.
Since $k\in [0,1)$ , $(d(x_0,x_1)\frac{k^n}{1-k})$ is a sequence converging to $0$ , and hence

$\underset{\epsilon >0}{\forall} \; \underset{n\in \mathbb N}{\exists} \; \underset{m>n}{\forall} \; d(x_0,x_1)\frac{k^m}{1-k}<\epsilon$

which results in

$\underset{\epsilon >0}{\forall} \; \underset{p\in \mathbb N}{\exists}\; \underset{m,n>p}{\forall} \; d(x_n,x_m)<\epsilon$ ,

or, in other words, $(x n)$ is a Cauchy sequence.
$M$ being complete, $(x n)$ converges to a point which we will call $x$ . Finally, since $f$ is a contraction, it is continuous, and hence, the following holds

$x=\lim_{n\to \infty} x_n=\lim_{n\to \infty} f(x_n)=f(\lim_{n\to \infty} x_n)=f(x)$

which states that $x$ is a fixed point of $f$ .
To prove uniqueness, suppose $y$ $(\not =x)$ is also a fixed point of $f$ . Then

and we would also obtain

resulting in a contradiction.

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