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Thread: find function

  1. #1
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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 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 f(x) = x then f(x + y) = x + y \not \leq x ~ \forall y. So 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 f(x) is an absolute maximum for the function for all x, which is a contradiction, unless we want to use the trivial f(x) = \text{constant}.

    -Dan
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