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Math Help - analysis

  1. #1
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    analysis

    Problem Statement:Suppose that f : \Re \rightarrow \Re is such that, for some \epsilon > 0, if x,y \in \re, then |f(x)-f(y)|<|x-y|^{1+\epsilon}. Prove that f is constant.


    I dont see how this is true. for instance, f(x)=x/100. Is there something I am missing?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    I think it is something with the derivative... f'(x)=|f(x)-f(y)|/|x-y|=0 when epsilon is very small.

    f'(x)=0 for all x ==> f(x) constant for all x.


    Hmmm...
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    I think it is something with the derivative... f'(x)=|f(x)-f(y)|/|x-y|=0 when epsilon is very small.

    f'(x)=0 for all x ==> f(x) constant for all x.


    Hmmm...
    You could state this a little bit more rigorously.

    1) Let x_0\in\mathbb{R} then the above says that \left|\frac{f(x_0)-f(y)}{x_0-y}\right|\leqslant |x_0-y|^{\varepsilon}. Thus, since \varepsilon>0 we have that \displaystyle \lim_{y\to x_0}|x_0-y|^{\varepsilon}=|x_0-x_0|^{\varepsilon}=0 (by continuity of f(y)=|x_0-y|^{\varepsilon}) and thus by the squeeze theorem f is differentiable at x_0 and \displaystyle f'(x_0)=\lim_{y\to x_0}\frac{f(x_0)-f(y)}{x_0-y}=0. Since x_0 was arbitrary it follows that f'(x)=0 for all x\in\mathbb{R}

    2) Since \mathbb{R} is an interval we may conclude that f(x)=c.

    Just a comment though.
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