Is the Function |x| Locally Lipschitz?

[zdenko]

Hi,

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. :)

Thanks

 

[rasmhop]

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. :)

|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 |x|^{1/2}.

"A function is locally Lipschitz in x on domain D if the function doesn't blow up anywhere on D"?

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 C^1 functions are locally Lipschitz.