1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Holder Continuity question
  1. Continuity question (Replies: 3)

  2. Continuity question (Replies: 3)

  3. Continuity question (Replies: 7)

Loading...