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.)

If f:[0,1]->[0,1] is non-increasing, then there is some x such that f(f(x))=x.

Simple, now that we know Richard's little fix point theorem, but who would have thought that before?

I wonder if there is also a funny criterion for a function to have some x with f(f(f(x)))=x...

Edited on November 19, 2006, 7:18 am

Posted by JLo on 2006-11-19 05:54:54