Definition

set up $X$ Is the distance space ,$T:X\to X$ Is a mapping , If exist $0\le \lambda <1$, bring
$$d(Tx,Ty)\le \lambda d(x,y),\forall x,y\in X$$
said $T$ It's compressed

lemma

The contractive mapping is continuous

if $x_n\to x$, be ：$T{x}_n\to Tx$

prove ：
$$d(T{x}_n,Tx)\le \lambda d(x_n,x)\to 0$$

Contraction mapping theorem

There is a unique fixed point for contractive mappings on complete metric spaces

prove ：

set up $X$ Is an arbitrary complete distance space ,$T$ yes $X$ Compression mapping on . Take whatever you like $x_0\in X$, The following proves that the sequence of numbers { x n } \{x_{n}\} { xn​} :
$$x_n=T{x}_{n-1}$$
The limits of $x=\underset{n\to \infty }{\lim }{x}_n$ There is , And is a fixed point

(1) First of all, prove that { x n } \{x_{n}\} { xn​} yes Cauchy Sequence ：
$$d(x_{n+1},x_n)=d(T{x}_n,T{x}_{n-1})\le \lambda d(x_n,x_{n-1})$$
, in turn,
$$d(x_{n+1},x_n)\le {\lambda }^{n}d(x_1,x_0)$$
For any positive integer $n$ and $p$ ：
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So for $\forall ϵ>0$,$\exists N$, Properly $n>N$ when , Yes $\forall p$ Satisfy ：
$$|d(x_{n+p},x_n)|\le ϵ$$

(2) Because of space $X$ It's complete , therefore { x n } \{x_{n}\} { xn​} convergence

(3) , in turn, $x=\underset{n\to \infty }{\lim }{x}_n=\underset{n\to \infty }{\lim }T{x}_{n-1}=T\underset{n\to \infty }{\lim }{x}_{n-1}=Tx$