Non-negative derivative of a bounded function Let $V\in C^1$ and $\dot{V} = k(t)\geq 0$. How to show that $V\leq m$ implies $k(t)\to 0$ with $t\to\infty$? I guess it's simple, but I cannot derive an argument for it.
This question is motivated by Kushner's "Stochastic Stability and Control", discussion on p.34. There he discuss the properties of a Lyapunov (deterministic) function and states that if 
$$
\dot{x} = f(x)
$$
and $\nabla V\cdot f\leq 0$ for a non-negative function $V$ on the bounded set $Q_m:=\{x\mid V(x)<m\}$ then $x_t\to \{x\mid k(x) = 0\}$.
 A: The function $V$ may have an infinite collection of very small intervals where its derivative is bounded away from $0$. It is easy to construct a counterexample if the sum of their lengths is finite.
A: I think I remember similar questions here before. By contraposition: If k(t) is bounded away from 0, i.e.k(t)> r, then, using the MVT, on each interval (a,b),$\frac {f(b)-f(a)}{b-a}>r(b-a)>>0$ Now apply this to an interval (b,c), then an interval (c,d) , etc. and if you do it long-enough, your f values can become indefinitely-large, i.e., unbounded.
I guess this would be more rigorous: we can construct a collection of intervals $(a_1,a_2),(a_2,a_3),...,(a_k,a_{k+1}),.....$, and applying the MVT on each interval:
$\frac {f(a_2)-f(a_1)}{a_2-a_1}>r(a_2-a_1); \frac{f(a_3)-f(a_2)}{a_3-a_2}>r(a_3-a_2))$, so that $f(a_3)-f(a_1)>r(a_3-a_1)$, so that you get a telescope*, and $\frac {f(a_n)-f(a_{n-1})}{a_n-a_{n-1}}>r(a_n-a_1)$. Then choose $(a_n-a_1)>\frac{2M}{r}$, and you can see your function going away to $\infty $
EDIT: I clearly worked under the assumption that the condition given was that f'(x) is bounded away from 0 as $x\rightarrow \infty$. Like the above reply, we need for f to be bounded away from 0 outside of an interval of finite length for the above argument to hold. One of these days I'll actually read the question, I promise.
*always wondered where  that name 'telescope' came from. 
