Smooth function on $\mathbb R$ whose small increments are not controlled by the first derivative at infinity I need some help in finding a (as simple as possible) smooth function $f:\mathbb R \rightarrow \mathbb R$ which does NOT satisfy the following:
There exist a constant $C>0$, a compact $K\subset\mathbb R$ and $h_0>0$ such that for every $|h| \leq h_0$ and every $x\in\mathbb R\setminus K$  
$|h|^{-1}|f(x+h) - f (x)|  \leq C |f'(x)|$ 
EDIT: and there exists a $\tilde C>0$ such that $|f'(x)|>\tilde C$ for $x\in\mathbb R\setminus K$. 
EDIT 2: my intuition is that such an $f$ may look like this: the first derivative stays always positive and oscillates (around g(x)=|x| for example), the oscillations becoming both faster and larger in amplitude when $x$C goes to infinity.   
Many thanks. 

Try e.g. $f(x) = \cos(x)$.  All you need is that $f$ is not constant on any interval and $f'$ has arbitrarily large zeros.
EDIT: With the new condition, take $$f(x) = \int_0^x (1 + t^2 \cos^2(t^2))\ dt = x+\frac{x \sin \left( 2\,{x}^{2} \right)}{8}-\frac{\sqrt {\pi }}{16}{\rm 
FresnelS} \left( 2\,{\frac {x}{\sqrt {\pi }}} \right) +\frac{{x}^{3}}{6}
$$
Note that for $x > 0$
$$\eqalign{\frac{f(x+h) -f(x)}{h} &= \frac{1}{h} \int_x^{x+h} (1 + t^2 \cos^2(t^2))\ dt\cr
&> \frac{x^2}{h} \int_x^{x+h} \cos^2(t^2)\ dt \cr
&= \frac{x^2}{h} \left(\frac{h}{2} + \frac{\sqrt{\pi}}{4} \left({\rm FresnelC}\left(
\frac{2(x+h)}{\sqrt{\pi}}\right) - {\rm FresnelC}\left(
\frac{2x}{\sqrt{\pi}}\right) \right)\right)
\cr
&=  \frac{x^2}{2} + o(x^2)}$$
as $x \to \infty$ for any fixed $h$ (since ${\rm FresnelC}(t) \to 1/2$ as $t \to \infty$), while $f'(\sqrt{(n+1/2)\pi})=1$ for positive integers $n$.
