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 otherwisethere 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 theBanach fixed point theorem. In one dimension it states the following:

Theorem 2 Suppose $f:[-1,1]\to [-1,1]$ is a function that satisfies thefollowing 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 existsone and only one point $x\in [-1,1]$ such that $f(x)=x$.

RemarksThe 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 detailsare given on the entry for the Banach fixed point theorem.A function that satisfies thecondition in Theorem 2 is called a contraction mapping. Such mappings also satisfy theLipschitz condition (http://planetmath.org/LipschitzCondition).