Cauchy sequence & Fixed point

[kingwinner]

Cauchy sequence & "Fixed" point

Homework Statement

Suppose that f: R^d->R^d and there is a constant c E (0,1) such that
||f(x)-f(y)|| ≤ c||x-y|| for all x, y E R^d. Let x_o E R^d be an arbitrary point in R^d, let x_n+1=f(x_n). Prove that
a) f is continuous everywhere.
b) (x_n) is Cauchy.
c) (x_n) 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.

Homework Equations

N/A

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!

 

[netheril96]

Isn't (c) obvious? Since x_n is Cauchy and R^d is complete,then x_n 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 $$\[<br /> \left. {\left\| {f(y) - f(z)} \right.} \right\| = \left\| {\left. {y - z} \right\|} \right. \le c\left\| {\left. {y - z} \right\|} \right.<br /> \]$$
But since 0<c<1,we conclude that $$\[<br /> \left\| {\left. {y - z} \right\|} \right. = 0<br /> \]$$
Thus y=z