Uniqueness of solution ODE I am working on a problem and kind of stuck.
Let $T>0$ and $F:[0,T] \times \mathbb{R}^n \to \mathbb{R}^n$ be continuous with 
$$|t| \ |F(t,v) - F(t,w)| \leq |v-w| \quad \forall t\in [0,T], \ v,w\in \mathbb{R}^n$$
Show that 
$$\begin{cases} \frac{du}{dt} = F(t,u) & t\in [0,T], \\
u(0) = u_0 \end{cases}$$
for every $u_0\in \mathbb{R}^n$ has at most one solution $u\in C^1([0,T], \mathbb{R}^n)$. 
Hint: Show that for $D(t):= \frac{|v(t)-w(t)|}{t}$ for $t\in (0,T]$ $\lim_{t\to 0} D(t) = 0$ and compare the maximum $D(t_{max})$ to $\frac{1}{t_{max}} \int_0^{t_{max}} D(t) dt$.
I have shown that $\lim_{t\to 0} D(t) = 0$ and that $D(t_{max}) \geq \frac{1}{t_{max}} \int_0^{t_{max}} D(t) dt$. However, I don't know how to stick the pieces together and make sense of the last part. I also have:
$$|v(t)-w(t)|\leq \int_0^t |F(\tau, v(\tau))- F(\tau, w(\tau))| d\tau \leq \int_0^t D(\tau) d\tau$$
Thanks in advance for any hints.
 A: Since $v'$ and $w'$ are continuous, one has
$$
\lim\limits_{t\to 0^+} \frac{1}{t} \int\limits_{0}^{t} (v'(\tau) - w'(\tau))\, d\tau = v'(0) - w'(0) = F(0, u_0) - F(0, u_0) =0,
$$
that is, $\lim\limits_{t \to 0^+} D(t) = 0$.  So the function $D$ can be extended in a continuous way to $0$ (denote the extension by $D$, too). For $t \in (0, T]$ the right-hand side of the equation is Lipschitz continuous in $u$, with the Lipschitz constant bounded for $t$ in compact subsets of $(0, T]$.  Consequently, both $v$ and $w$ are defined on $[0, T]$.  We estimate
\begin{align*}
\lvert v(t) - w(t) \rvert & = \left\lvert \int\limits_{0}^{t} (v'(\tau) - w'(\tau))\, d\tau \right\rvert  \le \int\limits_{0}^{t} \lvert v'(\tau) - w'(\tau) \rvert \, d\tau
\\
& = \int\limits_{0}^{t} \lvert F(\tau, v(\tau) - F(\tau, w(\tau))\rvert \, d\tau \le 
\int\limits_{0}^{t} \frac{\lvert v(\tau) - w(\tau) \rvert}{\tau} d\tau \qquad t \in (0, T],
\end{align*}
that is, 
$$
\tag{*}
D(t) \le \frac{1}{t} \int\limits_{0}^{t} D(\tau) \, d\tau, \qquad t \in (0, T].
$$
Let $t_{\mathrm{max}} \in (0, T]$ be such that $D(t_{\mathrm{max}}) = \max\{D(\tau): \tau \in [0, T]\}$. According to $(*)$ we have
$$
D(t_{\mathrm{max}}) \le \frac{1}{t_{\mathrm{max}}} \int\limits_{0}^{t_{\mathrm{max}}} D(\tau) \, d\tau.
$$
But, by the properties of the mean value we have
$$
\frac{1}{t_{\mathrm{max}}} \int\limits_{0}^{t_{\mathrm{max}}} D(\tau) \, d\tau \le \max\{D(\tau): \tau \in [0, T]\},
$$
from which we obtain that the continuous function $D$ is constant on $[0, T]$, hence, since $D(0) = 0$, $D$ is constantly equal to zero, which gives the uniqueness of the solution of the IVP. 
The condition 
$$
\ \lvert F(t,v) - F(t,w) \rvert \leq \frac{\lvert v-w \rvert}{t} \quad t\in {\color{red}(}0,T], \ v,w\in \mathbb{R}^n
$$
is an example of the so-called Osgood condition.
