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)
 A: Let me just mention that proposed fact immediately follows from Landau-Kolmogorov inequality which in this particular case reduces to 
$$\|f'\|^2_{L_{\infty}{\mathbb{(R)}}}\le 4\|f\|_{L_{\infty}{\mathbb{(R)}}}\|f''\|_{L_{\infty}{\mathbb{(R)}}}$$
A: 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.$_\square$
A: 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.
A: This is inspired by the proof by picture mentioned in the comment.
Geometrically, if $f'(x)$ is large, then because $f''$ is bounded, $f'(x)$ won't vary too much in a short period, therefore $f(x)$ cannot be Cauchy.
Given $\epsilon>0$, for sufficiently large $a$, due to the convergence of $f(x)$ we have $$|f(a+\epsilon)-f(a)| = |\int_a^{a+\epsilon} f'(x)dx|<\epsilon^2$$
By the mean value theorem, $f'(x) = f'(a) + f''(b)(x-a)$ for some $b$, therefore $$|\int_a^{a+\epsilon} f'(x)dx-\int_a^{a+\epsilon}f'(a)dx|\le M|\int_a^{a+\epsilon}(x-a)dx|$$ Then
$$|\int_a^{a+\epsilon}f'(a)dx|\le |\int_a^{a+\epsilon}f'(x)dx|  + M|\int_a^{a+\epsilon}(x-a)dx|\le \epsilon^2 + \frac{M\epsilon^2}{2}$$
Finally $$ |f'(a)|\le (1+\frac{M}{2})\epsilon$$
Note that we only need $f(x)$ converges (not necessarily to $0$) and $|f''(x)|<M$.
