I'm so stuck right now. I feel like I lost all my analysis skills. Assume I have a continuous map $s:[0,\infty) \to [0,\infty)$ with bounded Dini derivative for all $t_0 > 0$, in particular $$\limsup\limits_{t\nearrow t_0} \frac{s(t_0)-s(t)}{t_0-t} \le L \cdot s(t_0)$$ for a constant $L>0$ independent of $t_0$. Assume further that $s(0) = 0$.
Show that $s\le0$.
Of course I tried assuming there was a $t>0$ with $s(t) > 0$, but I were not able to get a contradiction.
Any help is greatly appreciated.
In this answer I will, for simplicity, be writing $f'(t_0) = \limsup\limits_{t\nearrow t_0} \frac{f(t_0)-f(t)}{t_0-t}$ for functions $f:[0,\infty) \to [0,\infty)$. So by assumption we know that $$s'(t_0) \le L \cdot s(t_0)$$ for $t_0 > 0 $. Define $g:[0,\infty)\to[0,\infty)$ by $g(t) := s(t) \cdot e^{-Lt}$. Since $s$ is continuous, the same is true for $g$.
We now want to apply the product rule in the case of Dini derivatives. The same exact proof as for the normal product rule goes through, but here we use $\limsup (a_n + b_n) \le (\limsup a_n) + (\limsup b_n)$ and therefore get the "$\le$" instead of "$=$". Also, the Dini derivative of a differentiable function (here $e^{-Lt}$) is of course just its normal derivative.
So we get $$g'(t_0) \le s'(t_0)\cdot e^{-Lt_0} + s(t) \cdot (-L)e^{-Lt_0} \le L \cdot s(t_0) \cdot e^{-Lt_0} + s(t) \cdot (-L)e^{-Lt_0} = 0$$ for all $t_0 >0$. Since $s(0)=0$, we also have $g(0) = 0$.
From $$g'(t_0) = \limsup_{t \nearrow t_0} \frac{g(t_0)-g(t)}{t_0-t} \le 0$$ one immediately gets that $g(t_0)-g(t)\le 0$ for all $t\le t_0$ close to $t_0$, since $t_0-t > 0$ (otherwise we would find a sequence $(t_n)_n$ converging to $t$ from below with $\frac{g(t_0)-g(t)}{t_0-t} > 0$). In particular, we can conclude that $g(t) \ge g(t_0)$ for $t\le t_0$ close to $t_0$. In other words, $g$ has to decrease infinitesimally. Using $g(0)=0$ and continuity of $g$, it follows that $g\le0$.
Finally, we can also conclude that $s\le0$, since $e^{-Lt} > 0$ for all $t$.