Theorem 1 [1,] 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 $g:[-1,1]\to \sR$ defined by $g(x)=f(x)-x$ . It satisfies \begin{eqnarray*} g(+1) &<& 0,\\ g(-1) &>& 0, \end{eqnarray*}so by the intermediate value theorem, there is a point $x$ such that $g(x)=0$ , i.e., $f(x)=x$ .

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 $a,b\in[-1,1]$ , $$|f(b)-f(a)|\le C |b-a|.$$

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.

Bibliography

1

A. Mukherjea, K. Pothoven, Real and Functional analysis, Plenum press, 1978.