Consider a function $f:(0,\infty)\rightarrow \mathbb{N}$ with argument $\epsilon$. Suppose $f$ is decreasing in $\epsilon$. Let $0<b<1$, $K>0$, $d \in \mathbb{N}$, $\delta>0$. Assume $$ 1\leq f(\epsilon* b)\leq K*\Big(\frac{2\delta}{\epsilon}\Big)^d $$ $\forall$ $0<\epsilon<2\delta$.
This implies that $$ 0 \leq \log(f(\epsilon*b))\leq \log(K)+d* \log(2)+d*\log\Big(\frac{2\delta}{\epsilon}\Big) $$ $\forall$ $0<\epsilon<2\delta$.
Is it true that $$ \exists \text{ }M>0 \text{ s.t. } \log(K)+d* \log(2)+d*\log\Big(\frac{2\delta}{\epsilon}\Big) \leq M*\log\Big(\frac{\delta}{\epsilon}\Big) $$ $\forall$ $0<\epsilon<\delta b$?
If no, could you give a counterexample? If yes, could you explain why? Maybe, it holds only for sufficiently small $\delta$?
I can see that $\log(K)+d* \log(2)+d*\log\Big(\frac{2\delta}{\epsilon}\Big)=O\Big(\log\Big(\frac{\delta}{\epsilon}\Big)\Big)$ as $\epsilon$ goes to $0$. This means that $\exists M>0, \eta>0$ s.t. $\log(K)+d* \log(2)+d*\log\Big(\frac{2\delta}{\epsilon}\Big)\leq M*\log\Big(\frac{\delta}{\epsilon}\Big) $ $\forall \epsilon<\eta$. I don't know how to show that $\eta\geq \delta b$.
The inequality is equivalent to $\log K+2 d \log 2\leq (M-d)\log \delta /\epsilon.$ Since $\epsilon\in (0,\delta b)\implies \delta /\epsilon >1/b>1\implies \log \delta /\epsilon >\log 1/b >0$ , the inequality holds for all $\epsilon \in (0,\delta b)$ if $(M-d)\log 1/b \geq \log K+2 d \log 2,$ equivalently, if $M\geq d+(\log 4^d K)/(\log 1/b).$ This is satisfied if $M>0$ and if $M$ is sufficiently large.
What does this have to do with $f$?.