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Lipschitz Condition and Differentiability

  1. Feb 21, 2009 #1
    Let K>0 and a>0. The function f is said to satisfy the Lipschitz condition if
    |f(x)-f(y)|<= K |x-y|a ..

    I am given a problem where I must prove that f is differentiability if a>1.

    I know I need to show that limx->c(f(x)-f(c))/ (x-c) exists. I am having quite a hard time. Any hints?
     
  2. jcsd
  3. Feb 21, 2009 #2
    I hope I remember my analysis, but that's usually called Hölder condition. Lipschitz we generally reserve for case when a=1. if a>1, the function is not just differentiable, but constant.

    Proving that might be easier.
     
  4. Feb 22, 2009 #3

    HallsofIvy

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    I have often seen "Lipschitz of order a" for that. I notice that Planet Math gives both:
    http://planetmath.org/encyclopedia/HolderContinuous.html [Broken]

    MathewSmith2, look at
    [tex]\lim_{x\rightarrow y}\frac{|f(x)- f(y)|}{|x-y|}= \lim_{x\rightarrow y}\frac{K|x-y|^a}{|x-y|}[/tex]

    How is that related to the derivative and what happens on the right when a> 1?

    Yes, it is true that such a function is constant, but I think proving that involves proving the derivative is 0 which requires first proving that the derivative exists.
     
    Last edited by a moderator: May 4, 2017
  5. Apr 30, 2011 #4
    I'm interested in this, too. The definition of locally Lipschitz is something like this: [itex]f[/itex] is locally Lipschitz at [itex]x_0[/itex] if there exists [itex]M > 0[/itex] and [itex]\epsilon > 0[/itex] such that

    [tex]
    |f(x) - f(x_0)| \leq M|x-y| \quad \text{whenever} \quad |x-x_0| < \epsilon.
    [/tex]

    Doesn't differentiability at [itex]x_0[/itex] imply this? I mean, if [itex]f[/itex] isn't locally Lipschitz at [itex]x_0[/itex], it's not differentiable there, right?
     
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