# the derivative for a Lipschitz function

For the solution of the ODE $dy/dt=f(t,y)$, $f$ has to be Lipschitz for all $t$. So, if $f$ is a function that is not differentiable with respect to $y$ but Lipschitz, what can I say about $f_y$? Can I estimate some norm of it? I can't say $||f_y||_C$ because $f$ is not $C^1$, however, I can say that $f_y$ is defined in a weak sense and then this weak derivative is bounded (by the Lipschitz property) and I can estimate $||f_y||_{\mathbb{L}^{\infty}}$ and have it bounded. Is that correct?

-
In the ODE $dy/dt=f(t,y)$ the function $f$ is understood to be given, while $y$ is a solution we don't know yet. It seems that your question has nothing to do with the function $y(t)$, hence it's not really a question about the differential equation. Is this correct? –  user31373 Jul 11 '12 at 20:30
I just thought what does it mean to be Lipshitz in case of non differentiability. I understand that we don't know range for $y$ and thus can't say what the constant that bounds derivative is, but we just know it exists. –  Medan Jul 12 '12 at 1:00

Yes, every Lipschitz function (on an open subset $\Omega\subset \mathbb R^n$) has a weak derivative which belongs to $L^\infty(\Omega)$. The $L^\infty$ norm of the derivative is bounded by the Lipschitz constant of the function.
Under appropriate assumptions on the geometry of $\Omega$ one can reverse this implication and obtain the global Lipschitz condition from the boundedness of the derivative.