Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\liminf_{t\to\infty} f'(t) = 0$?

share|improve this question
4  
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

1 Answer 1

up vote 3 down vote accepted

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, $\lim f'=0$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.