Let be $f : [0, T] \times \mathbb{R} \to \mathbb{R}$ partially differentiable for the both variables and :
\begin{equation*} \forall (t, x) \in [0, T] \times \mathbb{R}, \left \lvert \dfrac{\partial f}{\partial x}(t, x) \right \rvert \leq C \end{equation*}
If I take a look to the classical ODE: $y'(t) = f(t, y(t))$ and $y(0) = y_0$, I'd like to know if there is a unique solution defined over $[0, T]$ to this problem.
I want the solution to be continuous.
So here's what I've done so far:
First, this hypothesis give me a uniform Lipschitz constant for the second variable.
This gives me a way to find almost everywhere (i.e. discontinuous over a null set) continuous over a neighborhood of $0$ using Caratheodory's theorem, plus, this solution is unique, again by Caratheodory. Plus, $y$ would be absolutely continuous, thus, continuous.
My question, can I get it over $[0, T]$ completely for any arbitrary $T$?
What would be awesome is, could I even go further and get the fact that $f$ is continuous?
My idea is the following, if I can use the fact that $f(t, \cdot)$ is uniformly $C$-Lipschitz for all $t$. By Uniform Lipshitz continuity implies Continuous Differentiability I'd get that $f(t, \cdot)$ is continuously differentiable, thus: the partial derivative is continuous.
But I'm suspicious, the result seems to indicate that the derivative is $0$-Holder, i.e. just bounded. A bounded real derivative is not really continuous, is it? But let's do this assumption that I have the fact that this particular partial derivative is continuous (I might find another way to prove it.) Except if I use something else than a constant polynomial, maybe, I could have a degree 2 polynomial, but I have no idea what I could use, the second derivative is not guaranteed to exist. It could be there is some property I could use I don't see or I could just resort to weaker derivatives (?).
But, by Partial derivatives of $f$ exist, but only $n-1$ of them are continuous, implies differentiability ; I'd get the fact that $f$ is differentiable, thus continuous.
As a result, I get the solution of the ODE for free using classical Cauchy-Lipschitz. I might be wrong though with my reasoning.
