Prove: Let f be a nondecreasing, but not necessarily continuous, mapping of the closed interval [0,1] into itself. Then f has a fixed point, i.e., there is some x in [0,1] such that f(x)=x.

(Just to avoid any misunderstanding, [0,1] is the set of all real numbers between 0 and 1, with both 0 and 1 also included.)

Give me such a function f.  I will define a sequence S whose limit is such a fixed point x.

S(0) = 0
S(n+1) = S(n) if f(S(n)+(1/2)^n) < S(n)+(1/2)^n
           = S(n)+(1/2)^n otherwise.

The limit of this sequence is some x and the limit of f(y) as y approaches x=x.  Since f is monotonic, defined at x and the limit (from both directions) = x then f(x)=x

(note that the limit may not exist in both directions if x = 0 or 1 but if the limit from below as y->1 is 1 then f(1)=1 and similarly for f(0)=0.

Posted by Joel on 2006-11-17 13:37:39