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Global and local Lipschitz proof

  1. Sep 2, 2012 #1
    1. The problem statement, all variables and given/known data
    f(x)={0 for x<0, [itex]\sqrt{x}[/itex] else}

    a) Is f(x) globally Lipschitz? Explain

    b) Find the area for which f(x) is locally Lipschitz.

    2. Relevant equations



    3. The attempt at a solution
    a) f(x) is not globally Lipschitz in x on [a,b]xRn since there is a discontinuity at x=0.

    b) I would think that f(x) is locally Lipschitz for all real positive numbers, R+, but how do I go about proving it?
     
  2. jcsd
  3. Sep 3, 2012 #2
    Ok, f(x) and f'(x) are continuous throughout R+. This can be proven by limits. So therefore f(x) is by definition locally Lipschitz in x on R+.
     
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