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I'm failling to see how a function being globally Lipschitz on a variable in a set is any different from saying that the function has bounded continuous (partial) derivatives on that variable.

Could someone give me some intuition about what means for a function to be Lipschitz?

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  • $\begingroup$ Can you be more specific? Do you want to know the relation between the two properties or something else? $\endgroup$ – Megadeth Nov 5 '15 at 2:36
  • $\begingroup$ Try to imagine a function that is Lipschitz but not differentiable, at least at some points. $\endgroup$ – hardmath Nov 5 '15 at 2:39
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    $\begingroup$ A first difference is that a Lipschitz function has no reason to have partial derivatives. A classic example is $f(x)=|x|$, which is $1$-Lipschitz but not differentiable at $0$. $\endgroup$ – Taladris Nov 5 '15 at 2:39
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Your intuition is correct...almost everywhere. This is due to Rademacher's theorem. However, as mentioned in the comments, there are plenty of Lipschitz functions that are not differentiable, like $f(x) = \vert x\vert$. In fact, there are even functions with infinitely many nondifferentiable points that are still Lipschitz!

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If $f$ is differentiable on an interval in one variable, then $f$ is Lipschitz $1$ on that domain iff $f'$ is bounded there. In several variables the situation is different: For example, $f(x,y) = \arg (x+iy)$ on $\{1< x^2 + y^2<2\} \setminus (-\infty,0],$ where $\arg z$ the principal value argument.

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