The contraction mapping principle converts a distance-shrinking condition into existence and uniqueness of a fixed point. The focus keyword contraction mapping is fundamental in metric space theory, functional analysis, differential equations, and numerical methods. The theorem also gives convergence of successive approximations. :::definition[Contraction Mapping] Let $(X,d)$ be a metric space. A mapping $T:X\to X$ is called a contraction if there exists $\alpha$ with $0<\alpha<1$ such that $d(Tx,Ty)\leq\alpha d(x,y)$ for all $x,y\in X$. ::: :::remark Every contraction is uniformly continuous. Given $\epsilon>0$, choose $\delta=\epsilon/\alpha$. Then $d(x,y)<\delta$ implies $d(Tx,Ty)<\epsilon$. ::: :::example[Differentiable Self-Map of an Interval] If $f:[a,b]\to[a,b]$ is differentiable and $|f'(x)|\leq K<1$ on $(a,b)$, then $f$ is a contraction by the mean value theorem. ::: :::definition[Fixed Point] Let $T:X\to X$. A point $x\in X$ is called a fixed point of $T$ if $Tx=x$. ::: :::theorem[Contraction Mapping Principle] Let $(X,d)$ be complete. If $T:X\to X$ is a contraction, then $T$ has a unique fixed point. ::: :::proof Given that $(X,d)$ is complete and $T$ is a contraction with constant $\alpha$. To prove existence and uniqueness of a fixed point. Choose $x_0\in X$ and define $x_{n+1}=Tx_n$. Then $d(x_{n+1},x_n)\leq\alpha^n d(x_1,x_0)$. For $m>n$, $d(x_m,x_n)\leq\sum_{k=n}^{m-1}\alpha^k d(x_1,x_0)\leq\alpha^n d(x_1,x_0)/(1-\alpha)$. Hence $\{x_n\}$ is Cauchy and converges to some $x\in X$. Continuity of $T$ gives $Tx=x$. If $x$ and $y$ are fixed points, then $d(x,y)=d(Tx,Ty)\leq\alpha d(x,y)$, so $d(x,y)=0$ and $x=y$. $\square$ ::: :::remark The strict inequality $d(Tx,Ty)<d(x,y)$ for $x\neq y$ does not by itself guarantee a fixed point. For example, $T(x)=x+1/x$ on $[1,\infty)$ has no fixed point. ::: :::theorem[Convergence from Arithmetic Subsequences] Let $\{x_n\}$ be a sequence in a metric space, and let $k\in\mathbb N$. If each subsequence $\{x_{mk+j}\}$, $j=0,1,\ldots,k-1$, converges to the same point $x$, then $x_n\to x$. ::: :::proof Given that each arithmetic subsequence converges to $x$. To prove that $x_n\to x$. For each residue class choose an index after which the corresponding subsequence lies within $\epsilon$ of $x$. Taking the maximum of finitely many such indices works for all sufficiently large $n$. Hence, $x_n\to x$. $\square$ ::: :::corollary Let $T:X\to X$ be continuous on a complete metric space. If $T^k$ is a contraction for some positive integer $k$, then $T$ has a unique fixed point, and $T^n x$ converges to it for every $x\in X$. ::: :::proof The contraction mapping principle gives a unique fixed point $x_0$ of $T^k$. Since $T^k(Tx_0)=T(T^kx_0)=Tx_0$, the point $Tx_0$ is also a fixed point of $T^k$, so $Tx_0=x_0$. Thus $x_0$ is a fixed point of $T$. Uniqueness and convergence follow from the same argument and convergence of arithmetic subsequences. $\square$ :::