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Math Help - Lipschitz continuity

  1. #1
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    [SOLVED] Lipschitz continuity

    I want to find out if a function is lipschitz continuous.
    Take for example the function f(x) = \sqrt{|x|} on (-1,1)
    I know it's not lipschitz continuous since the graph's slope goes to +/-infinity when x approaches 0, but is this argument sufficient?
    Last edited by madflame991; December 14th 2011 at 05:54 AM.
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  2. #2
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    Re: Lipschitz continuity

    Quote Originally Posted by madflame991 View Post
    Take for example the function f(x) = \sqrt{|x|} on (-1,1)
    I know it's not lipschitz continuous since the graph's slope goes to +/-infinity when x approaches 0, but is this argument sufficient?
    According to this theorem, yes. We have that \mathbb{R} is a Banach space and A=(0,1) is a convex open subset of \mathbb{R}. Also, \sqrt{|x|}=\sqrt{x} is continuous on the closure \bar{A}=[0,1] of A and differentiable on A. Therefore, \sqrt{x} is Lipschitz continuous on \bar{A} iff (\sqrt{x})' is bounded on A, which it is not.

    However, it is easiest to show that \sqrt{x} is not Lipschitz continuous on [0,1) by definition since (\sqrt{x}-\sqrt{0})/(x-0) is not bounded on that interval.
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