# 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 $f(+1)<1$, 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

 $\displaystyle g(+1)$ $\displaystyle<$ $\displaystyle 0,$ $\displaystyle g(-1)$ $\displaystyle>$ $\displaystyle 0,$

so by the intermediate value theorem, there is a point $x$ such that $g(x)=0$, i.e., $f(x)=x$. $\Box$

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 $a,b\in[-1,1]$,

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

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 $n\to\infty$, $s_{n}$ 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

Title Brouwer fixed point in one dimension BrouwerFixedPointInOneDimension 2013-03-22 13:46:25 2013-03-22 13:46:25 mathcam (2727) mathcam (2727) 10 mathcam (2727) Proof msc 47H10 msc 54H25 msc 55M20 LipschitzCondition