The contraction lemma: Suppose that X is a complete metric space and that f:X→X is a contraction mapping on X. Then there is a unique z∈Xsuch that f(z)=z. Furthermore, if x0 is any point in X,then fn(x0)→z as n→∞.Proof. Let x0∈X. We will show that the sequence {xR=fn(x0)}n−1∞ is a Cauchysequence in X. Let D=d(x0,f(x0))and let 0<c<1 be the contraction constant for f. By the definition of contraction mapping d(f(x0),f2(x0))≤c.d(x0,f(x0))=c.D. By induction one can establish that d(fn(x0),fn+1(x0))≤cn.D. Thus,d(x0,fn+1(x0))≤k=0∑nck.D<1−cDSimilarly, d(fn(x0),fm(x0))≤D.cn.k=0∑m−n−1ck<1−ccn.D for all n<m. So, to seethat the sequence is Cauchy, let ϵ>0 and choose N such that 1−ccN.D<ϵ. then for N≤n≤m,d(fn(x0),fm(x0))<ϵ. So the sequence is Cauchy.since {xR=fn(x0)}n−1∞ is Cauchy,it converges to a point z∈X. But for this zn→∞limfn(x0)=z=n→∞limfn+1(x0)=f(z). So, z is a fixed point for f. On the other hand, if there were another fixed point z′=z,then d(f(z),f(z′))=d(z,z′)>c.d(z,z′). This last inequality contradicts the assumption that fis a contraction mapping.So, there is only one fixed point.
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