# If $f(x)\to 0$ as $x\to\infty$ and $f''$ is bounded, show that $f'(x)\to0$ as $x\to\infty$

Let $f\colon\mathbb R\to\mathbb R$ be twice differentiable with $f(x)\to 0$ as $x\to\infty$ and $f''$ bounded. Show that $f'(x)\to0$ as $x\to\infty$. (This is inspired by a comment/answer to a different question)

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Asked 24 mins ago. Answered 24 mins ago. That was fast. You're in complete autarky. – 1015 May 19 '13 at 22:18
@julien That feature is called self-answer. However, I wanted to wait for (and accepted) a smarter answer than mine. – Hagen von Eitzen May 19 '13 at 22:29
I am aware of that feature and I think it is great one. I was just surprised by the fact that you managed to post the question and the answer within the same minute. With, however, a fairly long answer. There is probably some obvious explanation I can't see due to my lack of imagination. – 1015 May 19 '13 at 22:34
Proof by picture – 40 votes Jul 15 '13 at 19:45

Let $|f''|\le 2M$ on $\mathbb R$ for some $M>0$. By Taylor's expansion, for every $x\in\mathbb R$ and every $\delta>0$, there exists $y\in[x,x+\delta]$, such that $$f(x+\delta)=f(x)+f'(x)\delta+\frac{1}{2}f''(y)\delta^2.$$ It follows that $$|f(x+\delta)-f(x)-f'(x)\delta|\le M\delta^2.\tag{1}$$ Since $\lim_{x\to\infty}f(x)=0$, fixing $\delta>0$ and letting $x\to\infty$ in $(1)$, we have $$\limsup_{x\to\infty}|f'(x)|\le M\delta.$$ Since $\delta>0$ is arbitrary, the conclusion follows.

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How is Eq. (1) valid? Can you really take the absolute value for the inequality? – João Victor Bateli Romão Nov 16 '15 at 16:48

Let me just mention that proposed fact immediately follows from Landau-Kolmogorov inequality which in this particular case reduces to $$\|f'\|^2_{L_{\infty}{\mathbb{(R)}}}\le 4\|f\|_{L_{\infty}{\mathbb{(R)}}}\|f''\|_{L_{\infty}{\mathbb{(R)}}}$$

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Let $M$ be a bound for $f''$. Then $|f'(x+h)-f'(x)|\le M|h|$ for all $x,h$. Let $\epsilon>0$ be given. We have to show that $|f'(x)|<\epsilon$ for all sufficiently big $x$. As $f(x)\to0$, there is an $x_0$ such that $|f(x)|<\frac{\epsilon^2}{4 M}$ for all $x>x_0$. Consider $x>x_0$ and assume $f'(x)> 0$. Then $f'(x+h)\ge f'(x)-Mh$ for $h\ge 0$ and hence \begin{align}f\left(x+\frac {f'(x)}{M}\right)-f(x)&=\int_0^{\frac {f'(x)}{M}}f'(x+h)\,\mathrm dh\\&\ge \int_0^{\frac {f'(x)}{M}}(f'(x)-M|h|)\,\mathrm dh\\&=\frac{(f'(x))^2}{2M}.\end{align} If on the other hand $f'(x)<0$, we similarly have $f'(x+h)\le f'(x)+Mh$ for $h\ge 0$ and hence \begin{align}f\left(x-\frac {f'(x)}{M}\right)-f(x)&=\int_0^{-\frac {f'(x)}{M}}f'(x+h)\,\mathrm dh\\&\le \int_0^{-\frac {f'(x)}{M}}(f'(x)+M|h|)\,\mathrm dh\\&=-\frac{(f'(x))^2}{2M}.\end{align} In both cases we find $$\left|f\left(x+\frac {|f'(x)|}{M}\right)-f(x)\right|\ge \frac{(f'(x))^2}{2M}$$ and as $x+\frac {|f'(x)|}{M}>x_0$, we conclude $\frac{(f'(x))^2}{2M}< 2\cdot \frac{\epsilon^2}{4M}$ and hence $|f'(x)|<\epsilon$ as was to be shown.$_\square$

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Should the first line not read "Then $|f'(x+h)-f^{\color{red}{\prime}}(x)|<\cdots$"? – Pedro Tamaroff May 19 '13 at 22:39
Did you see my comment? – Pedro Tamaroff May 23 '13 at 23:50
@PeterTamaroff Yes, it should. Sorry, I didn't. Thanks. – Hagen von Eitzen Jun 23 '13 at 13:08