Let x0:=x, let D:=d(x0,x1), and let k∈N. Then d(xk,xk+1)=d(f(xk−1,f(xk))≤qd(xk−1,xk)⋯≤qkD. For each ϵ∈R+, we find an n∈N such that qn<ϵD, then d(xn,xn+1)≤qnD<ϵ. This proves that (xk)k∈N is a Cauchy-sequence, which (because (X,d) is complete) means it converges to some point x∗∈X.
Given a any sequence (xk)k∈N in X, it converges to some point x∗, and it’s easy to see that x∗ is a fixed point of f. Let y∗ be a fixed point of X. Then, d(f(x∗),f(y∗))=d(x∗,y∗), hence x=y. (Otherwise, this contradicts the fact that f is a contraction.)
Ex3
Choose X⊊R2 as X:=R×N+. Then X is complete because, given any Cauchy sequence (xk)k∈N, it’s easy to prove that there is an n∈N such that all but finitely many xk are in the subspace R×{n}. However, the map f:X→X given by f(x,n):=f(x+1n,n+1) has no fixed points since it moves each point by at least 1. (And it’s straight-forward to verify that f is a weak contraction.)
Ex1
Let x0:=x, let D:=d(x0,x1), and let k∈N. Then d(xk,xk+1)=d(f(xk−1,f(xk))≤qd(xk−1,xk)⋯≤qkD. For each ϵ∈R+, we find an n∈N such that qn<ϵD, then d(xn,xn+1)≤qnD<ϵ. This proves that (xk)k∈N is a Cauchy-sequence, which (because (X,d) is complete) means it converges to some point x∗∈X.
Furthermore, given a position n∈N, we have
d(xn,x∗)≤∞∑k=nd(xk,xk+1)≤∞∑k=nqkD=qnD∞∑k=0qk=qnD11−q=D⋅qn1−q<D⋅(q1−q)n.
Ex2
Given a any sequence (xk)k∈N in X, it converges to some point x∗, and it’s easy to see that x∗ is a fixed point of f. Let y∗ be a fixed point of X. Then, d(f(x∗),f(y∗))=d(x∗,y∗), hence x=y. (Otherwise, this contradicts the fact that f is a contraction.)
Ex3
Choose X⊊R2 as X:=R×N+. Then X is complete because, given any Cauchy sequence (xk)k∈N, it’s easy to prove that there is an n∈N such that all but finitely many xk are in the subspace R×{n}. However, the map f:X→X given by f(x,n):=f(x+1n,n+1) has no fixed points since it moves each point by at least 1. (And it’s straight-forward to verify that f is a weak contraction.)