# Prove the Contraction Mapping Theorem.

1. Feb 8, 2014

Prove the Contraction Mapping Theorem.

Let $(X,d)$ be a complete metric space and $g : X \rightarrow X$ be a map such that $\forall x,y \in X, d(g(x), g(y)) \le \lambda d(x,y)$ for some $0<\lambda < 1$.Then $g$ has a unique fixed point $x^* \in X$, and it attracts everything, i.e. for any $x_0 \in X$ , the sequence of iterates $x_0, g(x_0), g(g(x_0))$, ... converges to the fixed point $x^* \in X$.

The hint I am given are for existence and convergence - prove that the sequence is Cauchy. For uniqueness, choose two fixed points of $g$ and apply the map to both.

My approach is After we got to $d(x_{n+1}, x_n) \le \lambda^n d(x_1, x_0)$,

then assuming that $m > n$, $d(x_m, x_n) \le d(x_m, x_{m-1}) + d(x_{m-1}, x_{m-2}) + ... + d(x_{n+1}, x_{n})$ since each term of the right hand size is 0 ...so if we add up all the 0 terms, we get 0 on the right hand size. Therefore, $d(x_m, x_n)$ is 0. If this is right, then the question I have here is how do I guarantee that $x_{m-1} > x_n$?

2. Feb 9, 2014

### Dick

Assuming you are trying to prove the sequence is Cauchy and that $x_n=g^n(x_0)$, why do you think the terms on the right side are zero??

3. Feb 9, 2014

I realize my reasoning a little bit off since the difference between m and n could be infinity , is any difference way to argue that the sum of the right hand side are zeros?

4. Feb 9, 2014

### Dick

They aren't zeros. Try to bound it with a geometric series.

5. Feb 9, 2014

You've got $d(x_{n+1}, x_n) \le \lambda^n d(x_1, x_0)$. Use it!