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Let $f:(0,\infty)\rightarrow\mathbb{R}$ differentiable and $F$ one of its primitives. Prove that if $f'$ is bounded and $\lim\limits_{x\rightarrow\infty}F(x)=0$, then $\lim\limits_{x\rightarrow\infty}f(x)=0$. I found this problem too on mathstack but I can't find it. However, I wrote down the lemma that solves it, but I didn't understood it's proof. Can you give a proof for: Lemma: For $g:(a,\infty)\rightarrow\mathbb{R}$ double differentiable on $(a,\infty)$ with it's second derivative $g''$ bounded on $(a,\infty)$, if $\lim\limits_{x\rightarrow\infty}g(x)=L$ then $\lim\limits_{x\rightarrow\infty}g'(x)=0$. For $F=g$ this solves our problem, but I don't know a proof for this lemma to the level of highschool maths.

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Indeed, $\lim_{x\to\infty}F(x)$ does not have to be $0$, it is okay with any $L=\lim_{x\to\infty} F(x)$. Let $\delta>0$ be arbitrary and $|f'|\le M$. By mean value theorem, there is $s\in (0,1)$ and $s'\in (0,1)$ such that $$ \left|\frac{F(x+\delta)-F(x)}{\delta}-f(x)\right|=\left|f(x+s\delta)-f(x)\right|=s\delta|f'(x+s'\delta)|\le M\delta. $$ Let $x$ tend to $\infty$ to obtain $$\begin{align*} \limsup_{x\to\infty}\left|\frac{F(x+\delta)-F(x)}{\delta}-f(x)\right|&=\limsup_{x\to\infty}\left|f(x)\right|\le M\delta. \end{align*}$$ Since $\delta>0$ is arbitrary, we get that $ \lim_{x\to\infty} f(x)=0. $

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  • $\begingroup$ Should your $f''$ not be $f'$? $\endgroup$ – K.Power Mar 12 at 19:11
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    $\begingroup$ Ah, you're absolutely right. Thank you for the correction ! $\endgroup$ – Song Mar 12 at 23:39
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Hint: Show that \begin{align} \max|g'(t)|^2 \leq 4\left(\max|g(t)|\right) \left( \max|g''(t)|\right). \end{align} for any twice differentiable function on any interval $[a, b]$.

Additional Hint to the Hint: Observe \begin{align} g(t+2h) = g(t)+ 2g'(t)h+2g''(\xi)h^2 \end{align} for some $\xi \in (t, t+2h)$. Hence it follows \begin{align} g'(t) = \frac{g(t+2h)-g(t)}{2h}-g''(\xi)h \end{align} which means \begin{align} |g'(t)| \leq \frac{\max |g(t)|}{h}+\max| g''(t)| h. \end{align} Choose an appropriate $h$.

Response to @MarkViola: Consider the interval $(-1, \infty)$ and the function \begin{align} g(x) = \begin{cases} 2x^2-1 & \text{ if } (-1<x<0),\\ \frac{x^2-1}{x^2+1} & \text{ if } (0\leq x<\infty) \end{cases} \end{align} which is twice differentiable. Observe \begin{align} \sup |g(t)| = 1, \sup |g'(t)|=4, \text{ and } \sup |g''(t)|=4. \end{align} I stole this example from Baby Rudin Exercise 5.15.

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  • $\begingroup$ Ops. I left the 4. Now it's correct. $\endgroup$ – Jacky Chong Mar 13 at 3:37
  • $\begingroup$ In THIS ANSWER, I showed that $|g'|\le \sqrt{2||g||_\infty\,||g''||_\infty}$. $\endgroup$ – Mark Viola Mar 13 at 4:14
  • $\begingroup$ Yes. I left the square yes well. $\endgroup$ – Jacky Chong Mar 13 at 4:15
  • $\begingroup$ The $4$ is overkill. $\endgroup$ – Mark Viola Mar 13 at 4:16
  • $\begingroup$ @MarkViola I have updated my post. $\endgroup$ – Jacky Chong Mar 13 at 4:27

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