TITLE: Can someone help me obtain a contradiction for this statement in 1-D real analysis
QUESTION [6 upvotes]: I am trying to prove the following is not possible:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a strictly increasing function such that $f(0)=0$. Fix $\alpha >1$ and suppose that for $h>0$

$\lim_{h \to 0} \frac{f(h)}{h} = \infty$ and
$\lim_{h \to 0} \frac{f(h)}{f(\alpha h)} =0$


I am quite sure it is not possible for these two limits to be both true and I've been trying to prove it from first definitions but I can't seem to see it.

REPLY [3 votes]: Assume the second condition holds, and let $\epsilon>0$ be sufficiently small so that 
$$
\frac{f(h)}{f(\alpha h)}<\frac{1}{2\alpha}
$$
whenever $0<h\leq \epsilon$.
It follows by induction that $f(h/\alpha^n)<(2\alpha)^{-n} f(h)$ for every positive integer $n$ and $h\leq \epsilon$.
If we also assume the first condition, then we get a contradiction for $h<\epsilon$:
$$
\infty=\lim_{x\to 0}\frac{f(x)}{x}=\lim_{n\to\infty}\frac{f(h/\alpha^n)}{h/\alpha^n}\leq \limsup_{n\to\infty}\frac{(2\alpha)^{-n}f(h)}{h/\alpha^n}=\frac{f(h)}{h}\limsup_{n\to\infty}2^{-n}=0.
$$