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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 $x_t\to \{x\mid k(x) = 0\}$.

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up vote 6 down vote accepted

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.

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I've give a reference which led me to this question. – Ilya Aug 12 '11 at 13:51
Changing the notation may lead to confusions. I have updated my answer in accordance with the current version of the question. – frog Aug 13 '11 at 8:55
Thank you. I've tried to be consistent the notation in the book to avoid any differences with the reference. I accept your answer. – Ilya Aug 13 '11 at 9:19

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.

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Telescope = tele "far" + skopein "to look or see". By the way, the result you try to prove is false, the problem in your proof being that for some functions $f$, your $a_n-a_1$ cannot be made as large as one wants (see @frog answer). – Did Aug 12 '11 at 16:14
Didier: I was (wrongly) assuming that the condition given was that f'(x) was bounded away from 0 as $x\rightarrow \infty$. My question was more on how the concept of telescope related to this type of series were terms cancel out. – gary Aug 12 '11 at 16:18
This meaning of telescopic probably refers to the object Jack Sparrow and his fellow pirates use to look afar, which is, as you know, designed to be folded up after usage. Likewise, your series may be said to fold up, since the sum of the pieces $a_k-a_{k-1}$ fold up into a unique piece $a_n-a_1$. Maybe... :-) – Did Aug 12 '11 at 16:34
To add to @Didier's explanation, the namer of that concept was referring to this sort of telescope. The collapsing sort to be precise. – J. M. Aug 13 '11 at 8:10

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