We wish to show that the terms of xn form a Cauchy sequence, which suffices to demonstrate they converge in a complete space. Take m,n∈N+, and WLOG m<n. Then we know from the definition of contraction that d(xm,xn)≤qm⋅d(x0,xn−m). This converges to 0 as m increases, so the sequence is Cauchy.
It’s easy to see that this makes the rate of convergence between terms of the Cauchy sequence exponentially quick. Intuitively that seems like it ought to make the sequence converge to its limit with the same speed, but I don’t think that can be made rigorous without more steps.
Q2
Take a sequence {xn=fn(x0)}. This converges to some L. Suppose L was not a fixed point. Then choose an ϵ=(L−f(L))/10 . A sequence {xn} which converges to a limit has, for every ϵ, some N such that ∀n>=N:|xn−L|<ϵ. Then we know that d(xN,L)<ϵ but d(f(xN),f(L))>ϵ , contradicting the contraction condition. So there is at least one fixed point, L.
Suppose there are two fixed points, f(x)=x, f(y)=y for distinct x and y. If so, d(f(x),f(y))=d(x,y), which again contradicts the contraction condition. So there is at most one fixed point.
Q3
Take as the space {n∈R+:n≥1}, with the usual metric. Define f(x)=x2+1x. This is a weak contraction (toward infinity) and has no fixed points within this space.
Q1
We wish to show that the terms of xn form a Cauchy sequence, which suffices to demonstrate they converge in a complete space. Take m,n∈N+, and WLOG m<n. Then we know from the definition of contraction that d(xm,xn)≤qm⋅d(x0,xn−m). This converges to 0 as m increases, so the sequence is Cauchy.
It’s easy to see that this makes the rate of convergence between terms of the Cauchy sequence exponentially quick. Intuitively that seems like it ought to make the sequence converge to its limit with the same speed, but I don’t think that can be made rigorous without more steps.
Q2
Take a sequence {xn=fn(x0)}. This converges to some L. Suppose L was not a fixed point. Then choose an ϵ=(L−f(L))/10 . A sequence {xn} which converges to a limit has, for every ϵ, some N such that ∀n>=N:|xn−L|<ϵ. Then we know that d(xN,L)<ϵ but d(f(xN),f(L))>ϵ , contradicting the contraction condition. So there is at least one fixed point, L.
Suppose there are two fixed points, f(x)=x, f(y)=y for distinct x and y. If so, d(f(x),f(y))=d(x,y), which again contradicts the contraction condition. So there is at most one fixed point.
Q3
Take as the space {n∈R+:n≥1}, with the usual metric. Define f(x)=x2+1x. This is a weak contraction (toward infinity) and has no fixed points within this space.