# Showing that there exists a function with a certain property.

I want to prove the following:

$\forall \epsilon > 0: \exists \text{ function }f:\mathbf{R}_{> 0} \rightarrow \mathbf{R}_{> 0}: \forall a,b\in \mathbf{R}_{>0}: a < b: \epsilon \cdot f(b) > f(a).$

$f$ should be continous and strictly monotonous.

My intuition tells me that such a function cannot exist, because basically for any chosen function the numbers $a$ and $b$ can lie so close to each other, that $f$ cannot make their difference be greater than a factor of $\epsilon$.

What do you think? Am I right?

Take $\epsilon = 1/2$, and suppose there is a continuous function $f$. Call $f(1) = A, f(2) = B$. So $B > 2A$. Call $B/A = D$, and $\lceil\log_2(D)\rceil + 1 = d$. So in particular, $2^d > B/A$.
Choose $d$ points $x_i$ between $1$ and $2$. Then we have that $f(x_2) > 2f(x_1)$ and $f(x_3) > 2f(x_2)$ and so on, so that $f(2) > 2^d f(1)$, or rather $B > 2^d A$. But this is a contradiction on $d$.
So there is no such function $f$. $\diamondsuit$
You are right. Let $\epsilon=\frac{1}{2}$, and let $f(1)=c\gt 0$. Then $f(2)\gt 2c$. By the same reasoning, $f(1+1/2)\gt 2c$, so $f(2)\gt 4c$. By the same reasoning, $f(1+1/3)\gt 2c$ so $f(1+1/2)\gt 4c$, so $f(2)\gt 8c$. And so on.