If a map on a complete metric space has a contraction property, it has a unique fixed point

Question

I am stuck on the following problem:

Prove that if $(X, d)$ is a complete metric space and $f : X\rightarrow X$ is a function with the property that there is a number $A < 1$ such that $d(f(x),f(y))\leq Ad(x,y)$ for all $x$ and $y$ in $X$ then $f$ has a unique fixed point.

I know:

• Since $(X,d)$ is a complete metric space then every Cauchy sequence in $X$ converges.

• I can pick $N$ such that for all $n,m > N$ then $d(x_n,x_m) < E$ for all $E > 0$.

To solve this I need to show that there exists an $x^*$ in $X$ such that $f(x^*) = x^*$. Any hints would be great!

Answer 1

Look here http://en.wikipedia.org/wiki/Banach_fixed-point_theorem

Or here is a hint:

look at the sequence $a_n=T(a_{n-1})$ with arbitrary $a_0$. Prove, that it is Cauchy. Prove that its limit is a fixed point.