Brouwer fixed point in one dimension

Theorem 1 [1, adams] Suppose f is a continuous functionMathworldPlanetmath f:[-1,1][-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] 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.

Assuming slightly more of the function f yields the Banach fixed point theoremMathworldPlanetmath. In one dimensionMathworldPlanetmathPlanetmath it states the following:

Theorem 2 Suppose f:[-1,1][-1,1] is a function that satisfies the following condition:

  • for some constant C[0,1), we have for each a,b[-1,1],


Then f has a unique fixed point in [-1,1]. In other words, there exists one and only one point x[-1,1] such that f(x)=x.

Remarks The fixed point in Theorem 2 can be found by iteration from any s[-1,1] as follows: first choose some s[-1,1]. Then form s1=f(s), then s2=f(s1), and generally sn=f(sn-1). As n, sn 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 conditionMathworldPlanetmath (


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