Let xi+1=f(xi) for arbitrary x0. Call c=d(x0,x1). Then by induction (i<j)d(xi,xj)≤∑k=j−1k=id(xk,xk+1)≤∑k=j−1k=icqk≤cqi1−q (power series simplification)
Therefore ∀δ>0:∃n∈N∀i>n,j>i:d(i,j)<cqn1−q<δ ie xi is a cauchy sequence. However (X,d) is said to be complete, which by definition means any cauchy sequence is convergent. So xn→y and d(xi,y)≤sup∞j=id(xi,xj)≤cqi1−q So xn converges exponentially quickly
Answer to question 2.
From part 1, as f is continuous, y=limn→∞(f(xn+1))=limn→∞(f(xn))=f(limn→∞(xn))=f(y) So y is a fixed point. Suppose x and y are both fixed points of f(x) a contraction map. Then f(x)=x and f(y)=y so d(f(x),f(y))≤qd(x,y)=qd(f(x),f(y)) therefore d(x,y)=0 so x=y. Thus f has a unique fixed point.
Answer to question 3.
(R,d(x,y)=|x−y|) is a metric space. Its the real line with normal distance. Let f(x)=√1+x2 . Then f is a contraction map because f is differentiable and f′(x)=x√1+x2has the property ∀x:|f′(x)|<1. However no fixed point exists as ∀x:f(x)>x. This works because the sequence xi generated from repeated applications of f will tend to infinity, despite successive terms becoming ever closer.
Answer to question 1.
Let xi+1=f(xi) for arbitrary x0. Call c=d(x0,x1). Then by induction (i<j)d(xi,xj)≤∑k=j−1k=id(xk,xk+1)≤∑k=j−1k=icqk≤cqi1−q (power series simplification)
Therefore ∀δ>0:∃n∈N∀i>n,j>i:d(i,j)<cqn1−q<δ ie xi is a cauchy sequence. However (X,d) is said to be complete, which by definition means any cauchy sequence is convergent. So xn→y and d(xi,y)≤sup∞j=id(xi,xj)≤cqi1−q So xn converges exponentially quickly
Answer to question 2.
From part 1, as f is continuous, y=limn→∞(f(xn+1))=limn→∞(f(xn))=f(limn→∞(xn))=f(y) So y is a fixed point. Suppose x and y are both fixed points of f(x) a contraction map. Then f(x)=x and f(y)=y so d(f(x),f(y))≤qd(x,y)=qd(f(x),f(y)) therefore d(x,y)=0 so x=y. Thus f has a unique fixed point.
Answer to question 3.
(R,d(x,y)=|x−y|) is a metric space. Its the real line with normal distance. Let f(x)=√1+x2 . Then f is a contraction map because f is differentiable and f′(x)=x√1+x2has the property ∀x:|f′(x)|<1. However no fixed point exists as ∀x:f(x)>x. This works because the sequence xi generated from repeated applications of f will tend to infinity, despite successive terms becoming ever closer.
For Q2, I believe you aren’t done:
You have established that there is at most one fixed point, but not that a fixed point exists.
For question 2
you haven’t proven f is continuous
For question 3 you say
f is a contraction map because f is differentiable and … ∀x:|f′(x)|<1
I would think proving this is part of what is asked for.