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I'm working with a Hamiltonian system where I have $\dot{x}=f(x,u(x))$ with $u(x)$ being a control piecewise continuous function and differentiable almost everywhere. I know that $f$ is Locally Lipschitz in $x$ for $u$ fixed and in $u$ for $x$ fixed but I need to determine if $u(x)$ is locally Lipschitz. I would like to know, given what I know about $u(x)$, what extra set of requirements are needed for it to be Lipschitz.

Rademacher Theorem says that a Lipschitz function is differentiable almost everywhere. I know the converse put simply is not true, but maybe the converse with some extra condition will make it true.

Edited: And what if $u(x)$ is continuous but still differentiable almost everywhere?

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A function $u$ is Lipschitz if and only if all of the following hold:

  1. $u$ is absolutely continuous
  2. $u$ is differentiable almost everywhere
  3. There is a constant $M$ such that $|u'|\le M$ almost everywhere.

I included item 2 because you mentioned it, but it is actually redundant: it follows from absolute continuity.

Unfortunately, there is no way around the requirement of absolute continuity. Without it, you can look at the a.e. derivative all day and still know next to nothing about the function.

Sufficient condition: if $u$ is piecewise smooth with bounded derivative, then it's Lipschitz.

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  • $\begingroup$ Thank u very much...the sufficient condition u mention will be usefull to me $\endgroup$ – The Mighty Algernon Jun 5 '14 at 10:53
  • $\begingroup$ Thank u very much...the sufficient condition u mention will be useful to me.... $\endgroup$ – The Mighty Algernon Jun 5 '14 at 10:54
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    $\begingroup$ Just for the record, the Cantor function: en.wikipedia.org/wiki/Cantor_function is a nice example of an a.e.-differentiable function that's not abolutely continuous, thus not Lipschitz. $\endgroup$ – fonini Feb 10 '15 at 20:01

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