I have just solved a problem:

Let $f:[0,+\infty)\rightarrow \mathbb{R}$ be continuous on $[0,+\infty)$ and differentiable on $(0,+\infty)$. If $\displaystyle \lim_{x\rightarrow +\infty}f'(x)=0$, prove that $\displaystyle \lim_{x\rightarrow +\infty}\frac{f(x)}{x}=0$

My questions is about the inverse: if $\displaystyle \lim_{x\rightarrow +\infty}\frac{f(x)}{x}=0$, which hypothesis can be added (if needed) so that we can conclude $\displaystyle \lim_{x\rightarrow +\infty}f'(x)=0$ ?

Actually, I tried to solve the following one, but I still cannot solve it:

If $f:[0,+\infty)\rightarrow [0,+\infty)$ such that its second derivative is continuous, $f'\le 0$ and $|f''|\le M$ for some $M$ for all $x\ge 0$, then $\displaystyle \lim_{x\rightarrow +\infty}f'(x)=0$.

From the hypothesis, $f$ decreases and bounded below, so $f$ has a limit when $x$ tends to infinity. Thus $\displaystyle \lim_{x\rightarrow +\infty}\frac{f(x)}{x}=0$.

My questions seems to be in a wider range than the second problem above. Any help would be appreciated.

  • $\begingroup$ Possibly realated $\endgroup$ – Winther Apr 4 '15 at 16:35
  • $\begingroup$ For the 1st problem (which you have already solved) L'Hospital's is the shortest solution. Its converse is not true in general because of the simple reason that a function may be bounded without the derivative being bounded. For the second problem note that $f$ is non-negative and decreasing and hence has a finite limit $L$ as $x \to \infty$. Now the solution is available at math.stackexchange.com/q/1196439/72031 $\endgroup$ – Paramanand Singh Apr 4 '15 at 22:05

The second statement you list is true; in particular, it follows from the fact that bounding the second derivative above means that if $f'(x)$ drops below some $-\varepsilon$, then there must be a corresponding drop in the value of $f(x)$, since the derivative cannot return to $0$ arbitrarily quickly.

In particular, suppose, for some $x$, that $f'(x)<-\varepsilon$. It follows that if we take a function $g$ such that $g(x)=f(x)$ and $g'(x)=-\varepsilon$ but $g''(x)=M$, then $f(y)\leq g(y)$ for all $y\geq x$. However, $g(y)$ can be explicitly found as: $$g(x)-g(x+t)=\varepsilon t - \frac{M}2t^2$$ and so $g(x)-g(x+t)$ has a maximum at $t=\frac{\varepsilon}{M}$ with a value of $\frac{\varepsilon^2}{2M}$. Thus, if $f'(x)<-\varepsilon$, the function $f$ is bound to drop by at least $\frac{\varepsilon^2}{2M}$. However, given that $\lim_{x\rightarrow\infty}f(x)=c$ exists and is a lower bound for $f$, it follows that if $c\leq f(x)<c+\frac{\varepsilon^2}{2m}$, which is true for all large enough $x$, then $f'(x)\geq -\varepsilon$. One can easily from here establish that $\lim_{x\rightarrow\infty}f'(x)=0$.

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