# 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.

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It can't be oscillating if $f'$ is not allowed to have zeros. Also, do you want the "sufficiently small" uniformly in $x$? Maybe try to rephrase the whole question to make it a little clearer. –  Lukas Geyer Oct 22 '12 at 16:42
How did ${\mathbb R}^n$ get in there? –  Robert Israel Oct 22 '12 at 16:53
@LukasGeyer and Robert Israel. Thanks for your comments, I edited, hope it is clear now (yes the "sufficiently small" should be uniform in $x$). With oscillating I meant not the function itself, but higher derivatives. Anyway I cancelled the comment on the oscillations, maybe it is only confusing. –  Hans Oct 22 '12 at 17:00

Take $$\begin{eqnarray*} f(x) &=& x + e^x + sin(e^x) \\ f'(x) &=& 1 + e^x + e^xcos(e^x) \end{eqnarray*}$$ Then $f'(x) \geq 1 + e^x - e^x \geq 1$

You have $f'(\log(\pi + 2\pi n)) = 1$ and $f'(\log(2\pi n)) = 1 + 2e^x$, i.e. arbitrarily close points with arbitrarily large differences of $f'$. This looks promising. Take now, for example, the differential quotient at $$\begin{eqnarray} x &=& \log(\pi + 2\pi n) \\ h &=& \log(2\pi (n+1)) - \log(\pi + 2\pi n) \end{eqnarray}$$ which is $$\begin{eqnarray} && \frac{h + e^{x+h} - e^{x} + sin(e^{x+h}) - sin(e^{x})}{h} \\ &=& \frac{h}{h} + \frac{2\pi(n+1) - (\pi + 2\pi n)}{h} + \frac{0 - 0}{h} \\ &=& 1+ \frac{\pi}{h} \end{eqnarray}$$

Now, since $h \to 0$ as $n \to \infty$, and since $f'(x) = 1$, the differential quotient at $x$ isn't bound by $f'$ globally.

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Tanks, this is what I was looking for! –  Hans Oct 24 '12 at 15:46

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 $n$.

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ops, I forgot to write that I want also the first derivative bounded away from zero..see edit. sorry. –  Hans Oct 22 '12 at 15:21
I accept the other answer because it is bit simpler, but this is interesting anyway, thanks! –  Hans Oct 24 '12 at 15:45