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.)

Define M:={ x in [0,1] | t<=x => f(t)>=t }, s:=supremum(M) and d:=f(s)-s. We prove d=0 by contradiction.

d>0: Then f(s+e)>s+e for all e<d, which means s is not the supremum of M.

d<0: Then f has a decreasing jump discontinuity at x=s with the jump being at least |d|, which means f is not monotonic.

Interestingly f is also continuous at x=s although continuity has not been assumed.

Posted by JLo on 2006-11-19 05:51:03