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Math Help - If f is holder continuous with a > 1, then f is constant?

  1. #1
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    If f is holder continuous with a > 1, then f is constant?

    Hello

    I'm trying to prove that if f satisfies the holder condition, i.e.
    there exists a constan K such that for all x,y in R : |f(x) - f(y)| <= K(|x - y|^a)
    and a > 1
    then f is a constant function.

    I assume that the right direction is proving that the derivative of f is constantly zero, but I don't see why f has to be differentiable in the first place...and even assuming that f is differentiable I don't see why the derivative must be zero..
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  2. #2
    Super Member girdav's Avatar
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    Let x_0\in\mathbb R. We have \left|\frac{f(x)-f(x_0)}{x-x_0}\right|\leq K|x-x_0| if x\neq x_0. Now you can see what the derivative in x_0 is and conclude.
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  3. #3
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    So would it be correct to take the limit of both sides of the inequality as X goes to Xo, and then we get that the absolute value of the derivative at Xo is smaller or equal to zero?
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  4. #4
    Super Member girdav's Avatar
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    Yes, it's correct.
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