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.)

(In reply to re(2): No Subject by Joel)

With f identically 1,  (1/2)^(n=1) helps a little, but now S(2)=1 and the test to determine S(3) cannot be done because S(2)+1/4=5/4 is not in [0,1]. I can think of other approaches that might work, such as using min{1,S(n)+(1/2)^(n+1)}, with  perhaps also a test for a function value of 1 which if present stops everything and declares 1 to be the fixed point. It looks like not just f identically 1, but f=1 for all x past some point  presents a problem. Your basic idea will still work with suitable modifications, but care will be needed to make sure a sequence is always being defined and that the sequence really does converge to the fixed point.

Posted by Richard on 2006-11-17 21:35:44