Limit of the derivative of a nonnegative decreasing function

Let $f:[0,\infty) \to [0,\infty)$ be a $C^2$ and monotone decreasing function. I think it can be shown that this condition implies that $\limsup_{x\to\infty} f'(x) = 0$. However, I don't think that the same is in general true for the $\liminf$.

But suppose that $f'' > M$ for some $M<0$. Is it then possible to show that $\liminf_{t\to\infty} f'(t) = 0$?

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Do you mean $\limsup_{x\to\infty}f'(x)=0$? –  Nick Strehlke Aug 12 '11 at 23:02
Yea, I did. Thanks. –  user1736 Aug 12 '11 at 23:38

Both results are true.

First, the limsup:

If $f$ is nonincreasing, $f'(x)\le0$ for every $x$ hence $\limsup f'\le0$. Assume that $\limsup f'\le-2u$ for a given positive $u$. Then $f'(x)\le-u$ for every $x$ large enough, say for every $x\ge z$. Integrating, one sees that $f(x)\le f(z)-u\cdot(x-z)$ for every $x\ge z$. For $x$ large enough, namely $x>z+f(z)/u$, one gets $f(x)<0$, which contradicts the hypothesis that $f\ge0$ everywhere. Hence $\limsup f'=0$.

Second, the liminf:

We assume furthermore from now on that $f''\ge-K$ for some positive $K$. Assume that $\liminf f'\ne0$, then $\liminf f'<0$ and there exists a positive $u$ and an unbounded sequence $(x_n)$ such that $f'(x_n)\le-2u$ for every $n$. For every $x\le x_n$, $$f'(x_n)-f'(x)\ge(x_n-x)\cdot\inf f''\ge-K\cdot(x_n-x),$$ hence $$f'(x)\le f'(x_n)+K\cdot(x_n-x)\le-2u+K\cdot(x_n-x),$$ and $f'(x)\le-u$ for every $x$ in the interval $I_n=(x_n-u/K,x_n)$. One can assume without loss of generality that the sequence $(x_n)$ was chosen in a way such that the intervals $I_n$ are disjoint. Each $I_n$ has length $u/K$ and $f'(x)\le-u$ uniformly on $x$ in $I_n$ hence $f$ is at least $u^2/K$ smaller at the end of $I_n$ than at the beginning. Since $f$ cannot increase between the intervals $I_n$, after $I_n$ the function $f$ can only take values smaller than $f(0)-n\cdot u^2/K$, which for $n$ large enough is again a contradiction with the fact that $f\ge0$ everywhere.

Finally, $\liminf f'\ne0$ is impossible hence $\liminf f'=0$ and in fact, $\lim f'=0$.

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Sweet! Yeah, so I guess the idea for the $\liminf$ part is to take the sequence of points converging to the $\liminf$, and noticing that in a neighborhood of those points (whose size doesn't depend on the specific point), the derivative can be bounded above by a negative number. Thanks for the quick and very clear proof. –  user1736 Aug 13 '11 at 0:26
to take the sequence of points converging to the lim inf... Hardly! The sequence of points converges to +infty but each of them is chosen such that the derivative of f there is uniformly away from zero (less than -2u in my answer). The rest of your sentence is correct. –  Did Aug 13 '11 at 0:37
Hm, I guess I meant a sequence of points whose images converge to the $\liminf$ of $f'$. That's how you can say that the images of those points are bounded uniformly away from zero right (after a certain point in the sequence)? –  user1736 Aug 13 '11 at 0:50
I see what you mean but not even that is required. The actual liminf may very well be around -100u and still, picking up points where the value is less than -2u is enough to make you happy. To get values uniformly away from zero is crucial--and all you need. –  Did Aug 13 '11 at 0:55
Yeah, I see what you mean. Thanks for following up! –  user1736 Aug 13 '11 at 0:58