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 $g:[-1,1]\to ℝ$ 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 $a,b\in [-1,1]$,

  $$|f(b)-f(a)|\le C|b-a|.$$

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 $s_n=f(s_{n-1})$. 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).

Title	Brouwer fixed point in one dimension
Canonical name	BrouwerFixedPointInOneDimension
Date of creation	2013-03-22 13:46:25
Last modified on	2013-03-22 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