1
$\begingroup$

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?

$\endgroup$
0
1
$\begingroup$

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 $f$. $\diamondsuit$

$\endgroup$
1
$\begingroup$

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.

Note that continuity, monotonicity are irrelevant.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.