I have a Lipschitz function $X=X(t)$ with the property that, at all points $t$, the right derivative $\lim\_{\epsilon \downarrow 0} \epsilon^{-1}(X(t+\epsilon)-X(t))$ exists and is given by $f(X)$ for some (discontinuous) function $f$. (Of course, at all regulat points, i.e. almost everywhere, this is then the ordinary derivative.) Is it true that such $X$ is necessarily unique?


Of course, if I just specify that a Lipschitz function satisfies $X'(t) = f(X(t))$ at all regular points, the solution doesn't have to be unique; but I'm requiring that this hold for the right derivative at *all* times, which intuitively seems like it ought to work.



Uniqueness of Lipschitz function satisfying differential equation
I suppose you want uniqueness assuming some initial condition. 


Even when $F$ is continuous and $X$ is $C^1$ you don't have uniqueness. For $t\geq 0$ $X(t)=0$ and $X(t)=t^2$ are both solution of $X'=2\sqrt{X}$.


Maybe with stronger assumptions on $F$ you can do something assuming only that $X$ is Lipschitz but I'm not sure. Without the Lipschitz condition you don't have uniqueness assuming that the differential equation is satisfied a.e., Cantor stair is a counterexample.


Edit : The uniqueness statement you want is true when $F$ is Lipschitz. In this case, with your assumptions, $X$ has a right derivative which is continuous. From that it's not too complicated to show that $X$ has a derivative everywhere which is continuous. And then the classical Cauchy-Lipschitz applies.


