1
$\begingroup$

If we have a differentiable function $ f:\mathbb{R}^n \to \mathbb{R}^n $, does it have to be locally Lipschitz? It's obviously true for continuously differentiable functions, but what happens without that assumption?

$\endgroup$
  • 5
    $\begingroup$ What about stuff like $x^2\sin(1/x^2)$? $\endgroup$ – zhw. Apr 17 '15 at 21:52
1
$\begingroup$

You can refer to another page, about Cantor function, which is differentiable but not locally Lipschitz. This function is not even weakly differentiable.

Showing the Cantor function is not Lipschitz.

$\endgroup$
  • 2
    $\begingroup$ The Cantor function is only a.e. differentiable, so this is irrelevant to the question. $\endgroup$ – Ian Apr 18 '15 at 1:40
  • 1
    $\begingroup$ In fact it is a simpler problem. It is hard to find a everywhere differentiable function that is no-where loc Lipschitz continuous. It suffices to get a point x in the domain, that f is not locally Lipschitz near x. You can just take the example on this page. math.stackexchange.com/questions/303695/… $\endgroup$ – Tony Low Apr 18 '15 at 6:39

Not the answer you're looking for? Browse other questions tagged or ask your own question.