proof of Banach fixed point theorem

Let (X,d) be a non-empty, complete metric space, and let T be a contraction mapping on (X,d) with constant q. Pick an arbitrary x0X, and define the sequence (xn)n=0 by xn:=Tnx0. Let a:=d(Tx0,x0). We first show by inductionMathworldPlanetmath that for any n0,


For n=0, this is obvious. For any n1, suppose that d(Tn-1x0,x0)1-qn-11-qa. Then

d(Tnx0,x0) d(Tnx0,Tn-1x0)+d(x0,Tn-1x0)
= qn-1-qn1-qa+1-qn-11-qa
= 1-qn1-qa

by the triangle inequalityMathworldMathworldPlanetmath and repeated application of the property d(Tx,Ty)qd(x,y) of T. By induction, the inequalityMathworldPlanetmath holds for all n0.

Given any ϵ>0, it is possible to choose a natural numberMathworldPlanetmath N such that qn1-qa<ϵ for all nN, because qn1-qa0 as n. Now, for any m,nN (we may assume that mn),

d(xm,xn) = d(Tmx0,Tnx0)
< qn1-qa<ϵ,

so the sequence (xn) is a Cauchy sequencePlanetmathPlanetmath. Because (X,d) is completePlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath, this implies that the sequence has a limit in (X,d); define x* to be this limit. We now prove that x* is a fixed pointPlanetmathPlanetmath of T. Suppose it is not, then δ:=d(Tx*,x*)>0. However, because (xn) convergesPlanetmathPlanetmath to x*, there is a natural number N such that d(xn,x*)<δ/2 for all nN. Then

d(Tx*,x*) d(Tx*,xN+1)+d(x*,xN+1)
< δ/2+δ/2=δ,

contradictionMathworldPlanetmathPlanetmath. So x* is a fixed point of T. It is also unique. Suppose there is another fixed point x of T; because xx*, d(x,x*)>0. But then


contradiction. Therefore, x* is the unique fixed point of T.

Title proof of Banach fixed point theorem
Canonical name ProofOfBanachFixedPointTheorem
Date of creation 2013-03-22 13:08:34
Last modified on 2013-03-22 13:08:34
Owner asteroid (17536)
Last modified by asteroid (17536)
Numerical id 5
Author asteroid (17536)
Entry type Proof
Classification msc 54A20
Classification msc 47H10
Classification msc 54H25