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Thread: analysis

  1. #1
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    analysis

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


    I dont see how this is true. for instance, $\displaystyle f(x)=x/100$. 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 $\displaystyle x_0\in\mathbb{R}$ then the above says that $\displaystyle \left|\frac{f(x_0)-f(y)}{x_0-y}\right|\leqslant |x_0-y|^{\varepsilon}$. Thus, since $\displaystyle \varepsilon>0$ we have that $\displaystyle \displaystyle \lim_{y\to x_0}|x_0-y|^{\varepsilon}=|x_0-x_0|^{\varepsilon}=0$ (by continuity of $\displaystyle f(y)=|x_0-y|^{\varepsilon}$) and thus by the squeeze theorem $\displaystyle f$ is differentiable at $\displaystyle x_0$ and $\displaystyle \displaystyle f'(x_0)=\lim_{y\to x_0}\frac{f(x_0)-f(y)}{x_0-y}=0$. Since $\displaystyle x_0$ was arbitrary it follows that $\displaystyle f'(x)=0$ for all $\displaystyle x\in\mathbb{R}$

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

    Just a comment though.
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