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

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!

• Dear @user0430 I see that, although you have already asked 18 question in this site and received answers in most of them, you have not mark a best answer in any of them. You can do it so by clicking on the checkmark next to the answer that you think is the one that helped you the most. Please read here for more detail. Jul 2, 2015 at 21:35

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.