If a function f:R->R satisfies the inequalities f(x) <=x, f(x+y)<=f(x), for all real numbers x and y, show that f(x)=x.
If then . So isn't correct.
I'm beginning to think there is no such function because of the second condition. If I'm seeing this correctly it implies that is an absolute maximum for the function for all x, which is a contradiction, unless we want to use the trivial .