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Thread: find function

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    Post 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.
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    Quote Originally Posted by kamaksh_ice View Post
    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
    Last edited by topsquark; Sep 13th 2007 at 06:21 AM.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by kamaksh_ice View Post
    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
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