1. ## find function

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.

2. Originally Posted by kamaksh_ice
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.
Why can't $\displaystyle f(x) = \frac{1}{3}x$? Seems to me that one works as well.

Edit: Ah wait! That doesn't work for negative x, does it?

-Dan

3. Originally Posted by kamaksh_ice
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.
Wait a minute.

If $\displaystyle f(x) = x$ then $\displaystyle f(x + y) = x + y \not \leq x ~ \forall y$. So $\displaystyle f(x) = x$ 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 $\displaystyle f(x)$ is an absolute maximum for the function for all x, which is a contradiction, unless we want to use the trivial $\displaystyle f(x) = \text{constant}$.

-Dan