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Lipschitz function and Baire Category Theorem

  1. Sep 19, 2011 #1
    hey,
    I need to show, using Baire Category Theorem, that there exits a continuous function
    f: [0,1] to R , that isn't Lipschitz on the interval [r,s] for every 0<=r<s<=1 .

    I defined the set A(r,s) to be all the continuous functions that are lipschitz on the interval [r,s]. I showed that A(r,s) is closed , but i'm having trouble showing it's nowhere dense.

    help please! :)
     
  2. jcsd
  3. Sep 19, 2011 #2

    micromass

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    So, you must show that A(r,s) has empty interior. So take an f in A(r,s). Try to find a g arbitrary close to f such that [itex]g\notin A(r,s)[/itex]. This will show that f is not in the interior of A(r,s).
    Note that our g will always be uniform continuous (since it is continuous on a compact domain). So you must somehow use a function that is uniform continuous but not Lipschitz. Do you know such a function?
     
  4. Sep 19, 2011 #3
    sqrt(x), for example, is continuous but not lipschitz on [0,1], but i don't know how to continue from here.
     
  5. Sep 19, 2011 #4

    micromass

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    Do you see why it is not Lipschitz?? Can you do something similar on [r,s]??
    And can you modify the function such that it is close to f?
     
  6. Sep 20, 2011 #5
    it's not Lipschitz because it's first derivative isn't bounded close to 0.
    o.k., so given f and a>0, define g as follows: g=sqrt[x-r] in the interval [r,a^2/4+r], g=0 for x<=r and g=a/2 for x>=a^2/4+r. g is not Lipschitz on [r,s]. define h=f+g.
    h isn't Lipschitz because g isn't Lipschitz and d(f,h)<a.
     
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