# Cauchy sequence & Fixed point

1. Feb 24, 2010

### kingwinner

Cauchy sequence & "Fixed" point

1. The problem statement, all variables and given/known data
Suppose that f: Rd->Rd and there is a constant c E (0,1) such that
||f(x)-f(y)|| ≤ c||x-y|| for all x, y E Rd. Let xo E Rd be an arbitrary point in Rd, let xn+1=f(xn). Prove that
a) f is continuous everywhere.
b) (xn) is Cauchy.
c) (xn) converges to a limit y.
d) Show that y is a fixed point of f ,that is f(y)=y, and that f has exactly one fixed point.

2. Relevant equations
N/A
3. The attempt at a solution
I proved part a & part b, but I have no idea how to prove parts c & d.
Any help is appreciated!

2. Feb 24, 2010

### netheril96

Re: Cauchy sequence & "Fixed" point

Isn't (c) obvious? Since xn is Cauchy and Rd is complete,then xn will converge at some point

The first part of (d):Just limit both sides and you will get y=f(y)
Second part:
Suppose that there is another z such that z=f(z)
Then $$$\left. {\left\| {f(y) - f(z)} \right.} \right\| = \left\| {\left. {y - z} \right\|} \right. \le c\left\| {\left. {y - z} \right\|} \right.$$$
But since 0<c<1,we conclude that $$$\left\| {\left. {y - z} \right\|} \right. = 0$$$
Thus y=z