Consider the following nonlinear system P:
$$ P = \begin{cases} \dot{x} = f(x,u)\\y = g(x)\end{cases} $$
where $f,g$ are of classes $C^1$, $x \in \mathbb{R}^n$ and $u,y \in \mathbb{R}$.
I am reading a proof about the existence of a unique solution for every $x_0 \in \mathbb{R}^n$ and every $u_0 \in [u_{min},u_{max}]$.
It starts stating the following.
"Let $R>0$ and let $B_R$ be the closed ball of radius $R$ around $x_0 \in \mathbb{R}^n$. Denote $M = \{max||f(x,u)|| s.t. x \in B_R, u \in [u_{min},u_{max}]\}$. Then it is clear from the mean value theorem that for any input $u \in [u_{min},u_{max}]$, the state trajectory of P exists and remains in $B_R$ for all $t \leq R/M$."
I do not understand why the existance follows from the mean value theorem. Is that because the mean value theorem guarantees that the derivative is finite within that interval?
This builds on the existence-and-uniqueness theorem (Picard-Lindelöf, Lipschitz-Cauchy,...) that follows from $u$ constant and $f\in C^1$. As we get existence of a local solution, one can apply the mean value theorem to find that $\|x(t)-x_0\|\le M\,|t-t_0|$ for all $t$ in the domain of the solution.
Now consider the local solution extended to its maximal solution, where at first we know only that the domain contains a neighborhood of $t_0$. The result on when the maximal domain can have finite boundaries tells us that any bounded solution can be extended (at least) to the limits of the bounding condition or of the domain of $f$, whatever comes first. In this case, the restricted domain is $B_R$, this means that the domain of the maximal solution contains the interval where the set defined by the bound stays inside $B_R$, $M\,\!|t-t_0|\le R$.