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?
Best Answer
A function $u$ is Lipschitz if and only if all of the following hold:
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.