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Locally Lipschitz

  1. Jan 20, 2010 #1

    I am struggling with the concept of "locally Lipschitz". I have read the formal definition but i cannot see how that differs from saying something like: "A function is locally Lipschitz in x on domain D if the function doesn't blow up anywhere on D"? It seems that, when talking about local Lipschitz, you can always pick the domain small enough or L large enough to satisfy the condition; unless it blows up.

    Maybe an example, that I am struggling with, may help. Is |x| locally Lipschitz, and why?
    I, by looking at the derivative which is sgn(x) am convinced that this satisfies the Lipschitz condition, the way I interpret it, but still, i would not bet my life on it. :)

  2. jcsd
  3. Jan 22, 2010 #2
    |x| is Lipschitz continuous since its derivative is bounded (by -1 and 1). Since it's Lipschitz continuous it's also locally Lipschitz continuous.

    For an example of function that isn't locally Lipschitz continuous consider [itex]|x|^{1/2}[/itex].

    This is the general idea of what it means to be locally Lipschitz continuous and local Lipschitz continuity could in some sense be considered a formal way of stating that the function never "blow up". Actually all [itex]C^1[/itex] functions are locally Lipschitz.
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