Brouwer fixed point in one dimension
Theorem 1 [1, adams] Suppose $f$ is a continuous function^{} $f:[1,1]\to [1,1]$. Then $f$ has a fixed point, i.e., there is a $x$ such that $f(x)=x$.
Proof (Following [1]) We can assume that $f(1)>1$ and $$, since otherwise there is nothing to prove. Then, consider the function $g:[1,1]\to \mathbb{R}$ defined by $g(x)=f(x)x$. It satisfies
$g(+1)$  $$  $0,$  
$g(1)$  $>$  $0,$ 
so by the intermediate value theorem, there is a point $x$ such that $g(x)=0$, i.e., $f(x)=x$. $\mathrm{\square}$
Assuming slightly more of the function $f$ yields the Banach fixed point theorem^{}. In one dimension^{} it states the following:
Theorem 2 Suppose $f:[1,1]\to [1,1]$ is a function that satisfies the following condition:

for some constant $C\in [0,1)$, we have for each $a,b\in [1,1]$,
$$f(b)f(a)\le Cba.$$
Then $f$ has a unique fixed point in $[1,1]$. In other words, there exists one and only one point $x\in [1,1]$ such that $f(x)=x$.
Remarks The fixed point in Theorem 2 can be found by iteration from any $s\in [1,1]$ as follows: first choose some $s\in [1,1]$. Then form ${s}_{1}=f(s)$, then ${s}_{2}=f({s}_{1})$, and generally ${s}_{n}=f({s}_{n1})$. As $n\to \mathrm{\infty}$, ${s}_{n}$ approaches the fixed point for $f$. More details are given on the entry for the Banach fixed point theorem. A function that satisfies the condition in Theorem 2 is called a contraction mapping. Such mappings also satisfy the Lipschitz condition^{} (http://planetmath.org/LipschitzCondition).
References
 1 A. Mukherjea, K. Pothoven, Real and Functional analysis^{}, Plenum press, 1978.
Title  Brouwer fixed point in one dimension 

Canonical name  BrouwerFixedPointInOneDimension 
Date of creation  20130322 13:46:25 
Last modified on  20130322 13:46:25 
Owner  mathcam (2727) 
Last modified by  mathcam (2727) 
Numerical id  10 
Author  mathcam (2727) 
Entry type  Proof 
Classification  msc 47H10 
Classification  msc 54H25 
Classification  msc 55M20 
Related topic  LipschitzCondition 