# Prove that if $f'$ is bounded and $\lim\limits_{x\rightarrow\infty}F(x)=0$, then $\lim\limits_{x\rightarrow\infty}f(x)=0$.

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.

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 $$\lim_{x\to\infty} f(x)=0.$$

• Should your $f''$ not be $f'$? Mar 12, 2019 at 19:11
• Ah, you're absolutely right. Thank you for the correction ! Mar 12, 2019 at 23:39

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 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.

• Ops. I left the 4. Now it's correct. Mar 13, 2019 at 3:37
• In THIS ANSWER, I showed that $|g'|\le \sqrt{2||g||_\infty\,||g''||_\infty}$. Mar 13, 2019 at 4:14
• Yes. I left the square yes well. Mar 13, 2019 at 4:15
• The $4$ is overkill. Mar 13, 2019 at 4:16
• @MarkViola I have updated my post. Mar 13, 2019 at 4:27