Metric and Hilbert Spaces

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 4 November 2014

Lecture 20: Banach fixed point theorem

Let $(X,d)$ be a metric space.

A contraction mapping is a function $f:X\to X$ such that there exists $\alpha \in (0,1)$ such that if $x,y\in X$ then $d(f\left(x\right),f\left(y\right))\le \alpha d(x,y)\text{.}$

A fixed point of $f:X\to X$ is an element $x\in X$ such that $f\left(x\right)=x\text{.}$

(Banach fixed point theorem) Let $(X,d)$ be a complete metric space and let $f:X\to X$ be a contraction mapping. Let $x\in X$ and let $x_1,x_2,\dots$ be the sequence $$x_1=f\left(x\right),x_2=f\left(f\left(x\right)\right),x_3=f\left(f\left(f\left(x\right)\right)\right),\dots$$ Then $x_1,x_2,\dots$ converges and $p=\underset{n\to \infty }{\text{lim}}{x}_n$ is the unique fixed point of $f\text{.}$

		Proof.
	To show: (a) $x_1,x_2,\dots$ is a Cauchy sequence. (b) $p=\underset{n\to \infty }{\text{lim}}{x}_n$ exists. (c) $f\left(p\right)=p\text{.}$ (d) If $q$ is a fixed point of $f$ then $q=p\text{.}$ (a) To show: If $\epsilon \in {ℝ}_{>0}$ then there exists $N\in {\mathbb{Z}}_{>0}$ such that if $m,n\in {\mathbb{Z}}_{>0}$ and $m>N$ and $n>N$ then $d(x_m,x_n)<\epsilon \text{.}$ Assume $\epsilon \in {ℝ}_{>0}\text{.}$ Let $N$ be the smallest integer in ${\mathbb{Z}}_{>0}$ such that $$\frac{{\alpha }^{N}d(f\left(x\right),x)}{1-\alpha }<\epsilon (\text{so}{\alpha }^N<\frac{\epsilon (1-\alpha )}{d(f\left(x\right),x)})\text{.}$$ To show: If $m,n\in {\mathbb{Z}}_{>0}$ and $m>N$ and $n>N$ then $d(x_m,x_n)<\epsilon \text{.}$ Assume $m,n\in {\mathbb{Z}}_{>0}$ and $m>N$ and $n>N\text{.}$ Assume $n<m\text{.}$ To show: $d(x_m,x_n)<\epsilon \text{.}$ Since $$\begin{array}{rcl}{\displaystyle d(x_2,x_1)}& =& {\displaystyle d(f^2\left(x\right),f\left(x\right))\le \alpha d(f\left(x\right),x),}\\ {\displaystyle d(x_3,x_2)}& =& {\displaystyle d(f^3\left(x\right),f^2\left(x\right))\le \alpha d(f^2\left(x\right),f\left(x\right))\le {\alpha }^{2}d(f\left(x\right),x),}\\ {\displaystyle d(x_4,x_3)}& =& {\displaystyle d(f^4\left(x\right),f^3\left(x\right))\le \alpha d(f^3\left(x\right),f^2\left(x\right))\le {\alpha }^{3}d(f\left(x\right),x),}\\ {\displaystyle }& ⋮& {\displaystyle }\end{array}$$ then $$\begin{array}{rcl}{\displaystyle d(x_n,x_m)}& \le & {\displaystyle d(x_n,x_{n+1})+d(x_{n+1},x_{d+2})+\cdots +d(x_{m-1},x_m)}\\ & \le & {\displaystyle {\alpha }^{n}d(f\left(x\right),x)+{\alpha }^{n+1}d(f\left(x\right),x)+\cdots +{\alpha }^{m-1}d(f\left(x\right),x)}\\ & =& {\displaystyle ({\alpha }^n,{\alpha }^{n+1}+\cdots +{\alpha }^{m-1})d(f\left(x\right),x)}\\ & \le & {\displaystyle {\alpha }^n(1+\alpha +{\alpha }^2+\cdots )d(f\left(x\right),x)}\\ & =& {\displaystyle \frac{{\alpha }^n}{1-\alpha }d(f\left(x\right),x)}\\ & <& {\displaystyle \epsilon \text{.}}\end{array}$$ So $x_1,x_2,\dots$ is a Cauchy sequence in $X\text{.}$ (b) Since $(X,d)$ is complete and $x_1,x_2,\dots$ is a Cauchy sequence in $X$ then $x_1,x_2,\dots$ converges in $X\text{.}$ So there exists $p\in X$ with $p=\underset{n\to \infty }{\text{lim}}{x}_n\text{.}$ (c) To show: $f\left(p\right)=p\text{.}$ To show: $d(f\left(p\right),p)=0\text{.}$ To show: If $\epsilon \in {ℝ}_{>0}$ then $d(f\left(p\right),p)<\epsilon \text{.}$ Assume $\epsilon \in {ℝ}_{>0}\text{.}$ Since $\underset{n\to \infty }{\text{lim}}{x}_n=p,$ there exists $N\in {\mathbb{Z}}_{>0}$ such that if $n\in {\mathbb{Z}}_{>0}$ and $n\ge N$ then $d(x_n,p)<\frac{\epsilon }{2}\text{.}$ Then $$\begin{array}{rcl}{\displaystyle d(f\left(p\right),p)}& \le & {\displaystyle d(f\left(p\right),x_{N+1})+d(x_{N+1},p)}\\ & \le & {\displaystyle \alpha d(p,x_N)+d(x_{N+1},p)\text{(since}\hspace{0.17em}{x}_{N+1}=f\left(x_N\right)\text{)}}\\ & <& {\displaystyle \alpha \frac{\epsilon }{2}+\frac{\epsilon }{2}}\\ & <& {\displaystyle \epsilon \text{.}}\end{array}$$ So $d(f\left(p\right),p)=0\text{.}$ So $f\left(p\right)=p\text{.}$ (d) To show: If $q$ is a fixed point of $f$ then $q=p\text{.}$ To show: If $q\in X$ and $f\left(q\right)=q$ then $q=p\text{.}$ Assume $q\in X$ and $f\left(q\right)=q\text{.}$ To show: $q=p\text{.}$ To show: $d(q,p)=0\text{.}$ $$\begin{array}{rcl}{\displaystyle d(q,p)}& =& {\displaystyle d(f\left(q\right),f\left(p\right))\text{(since}\hspace{0.17em}q=f\left(q\right)\hspace{0.17em}\text{and}\hspace{0.17em}p=f\left(p\right)\text{)}}\\ & \le & {\displaystyle \alpha d(q,p)\text{(since}\hspace{0.17em}f\hspace{0.17em}\text{is contractive).}}\end{array}$$ So $(1-\alpha )d(q,p)\le 0\text{.}$ Since $(1-\alpha )d(q,p)\ge 0$ and $(1-\alpha )d(q,p)\le 0$ then $(1-\alpha )d(q,p)=0\text{.}$ Since $\alpha \in (0,1)$ then $(1-\alpha )\ne 0,$ and so $d(q,p)=0\text{.}$ So $p=q\text{.}$ $\square$

Notes and References

These are a typed copy of Lecture 20 from a series of handwritten lecture notes for the class Metric and Hilbert Spaces given on August 29, 2014.

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