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Math Help - Prove that f is a constant.

  1. #1
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    Prove that f is a constant.

    Suppose f : R ------> R, and suppose that \vert f(x) - f(y) \vert \leq (x - y)^2 for all x, y in R.
    Prove that f is a constant.




    I tried solving this problem using the Mean Value Theorem, and showing that f'(xo) is always zero.

    \vert f'(xo) \vert = \vert \frac{f(x) - f(y)}{x -y} \vert \leq \frac{(x-y)^2}{(x-y)} \leq (x-y) \leq \epsilon

    Now, is there an argument I can use to let (x-y) be less than epsilon? Is there something that I'm overlooking?




    Any insight would be welcomed,

    Thanks
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Prove that f is a constant.

    isnt lim_{x \to x_0} \frac{|f(x)-f(x_0)|}{|x-x_0|}} another way of saying f is differentiable at x_0 if this limit exists?
    then lim_{x \to x_0} \frac{|f(x)-f(x_0)|}{|x-x_0|}} \leq lim_{x \to x_0}|x-x_0|. The right side is going to 0. Since x_0 was an arbitrary point.
    It means the function has 0 derivative for all points. Since \mathbb{R} is path connected, for any x_0,y_0, there exists a parametrization g(t), such that g(0) = x_0 and g(1)=y_0

    \frac{|x_0 - y_0|}{1} = 0, which implies x_0=y_0 thus f is constant

    We need to show this path connectivity part because f might be a different constant value at different parts in the domain (f=3 in one neighborhood, f=5 in another so f'(x) = 0 for both but they are not the same right?)
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    Re: Prove that f is a constant.

    Jill90

    follow please the steps below .

    |f(x)-f(y)|<= (x-y)^2 implies |f(x)-f(y)|<=|x-y|^2

    then |(f(x)-f(y))/(x-y)|<=|x-y| and taking into account a well known property of the moduli this imples: -|x-y|=<|(f(x)-f(y))/(x-y)|<= +|x-y|

    Taking the limits when x goes to y the middle part is the derivative of f(x) and both |x-y| become zero.

    as per the squezze theorem this implies that the derivative of f"(x) = 0

    This implies that the function f(x) is CONSTANT .

    MINOAS
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  4. #4
    Senior Member jakncoke's Avatar
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    Re: Prove that f is a constant.

    Quote Originally Posted by MINOANMAN View Post
    Jill90

    follow please the steps below .

    |f(x)-f(y)|<= (x-y)^2 implies |f(x)-f(y)|<=|x-y|^2

    then |(f(x)-f(y))/(x-y)|<=|x-y| and taking into account a well known property of the moduli this imples: -|x-y|=<|(f(x)-f(y))/(x-y)|<= +|x-y|

    Taking the limits when x goes to y the middle part is the derivative of f(x) and both |x-y| become zero.

    as per the squezze theorem this implies that the derivative of f"(x) = 0

    This implies that the function f(x) is CONSTANT .

    MINOAS
    you still have to assume path connectivity of any open set in \mathbb{R}. You have to atleast mention this part because f could be different constants on different open subsets if the open subsets of R are not path connected.
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