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Homework Help: Holder Continuity question

  1. Sep 24, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove that if f(x) is Holder continuous, i.e,

    [tex] \sup_{a<x , y<b} \frac{\abs{f(x) - f(y)}}{\abs{x-y}^\alpha} = K^f_\alpha<\inf [/tex]
    with [tex] \alpha > 1 [/tex], then f(x) is a constant function

    2. Relevant equations



    3. The attempt at a solution

    I've been staring at this for a while, but I'm unsure of where to start. I'm guessing that I'm supposed to show that the derivative of something is zero. I've seen hints elsewhere about applying Taylor's theorem, but I am unsure on how to apply it.
     
  2. jcsd
  3. Sep 25, 2007 #2

    morphism

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    First let's fix up your LaTeX: f(x) is Holder continuous (on (a,b)) if

    [tex]\sup_{a < x < y < b} \frac{\left| f(x) - f(y) \right|}{\left| x-y \right|^{\alpha}} < \infty,[/tex]

    where we're assuming that [itex]\alpha>1[/itex]. We can restate this as: there exists a fixed C such that for all x & y in (a,b)

    [tex]\left| f(x) - f(y) \right| \leq C \left| x-y \right|^{\alpha}.[/tex]

    Your idea of proving that f'(x)=0 is a good one, but we don't know a priori that f is differentiable. So let's resort to the epsilon-delta definition. Given e>0, can we find a d>0 such that if |h|<d then

    [tex]\left| \frac{f(x+h) - f(x)}{h} \right| < \epsilon?[/tex]

    Try to use the Holder continuity inequality to find a suitable one.
     
    Last edited: Sep 25, 2007
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