I'm trying to find a continuous increasing function $f$ in $[1,\infty)$ such that
$1-\frac{f(x)}{f(2x)} = O(1/\log^c(x))$ for some constant $c>1$, and $\lim_{x\to \infty} f(x) = \infty$.
Note if $c=1$, $f(x)=\log(x)$ works well, when as soon as $c>1$, I can't come up with any nice function that works. It seems iterating log $k$ times is not enough.
Interestingly, such function doesn't exist. The point is, that $$ \sum_{n= 1}^{\infty}\frac{1}{n^c} $$ converges for $c > 1$. Assuming otherwise, condition implies existence of constant $C \in \mathbb{R}$, such that for any $x$, $$ 1-\frac{f(x)}{f(2x)} < \frac{C}{\log^{c}(x)} $$ Now apply this for numbers $b_n = 2^{n}$ to get $$ 1-\frac{f(2^n)}{f(2^{n+1})} < \frac{C}{\log^{c}(2^{n})} = \frac{C}{n^c\log^{c}(2)}. $$ Now, summing this inequality for all positive integers ($f(b_n) = a_{n})$, we get $$ \sum_{n = 1}^{\infty}\left(1-\frac{a_{n}}{a_{n+1}}\right) < \frac{C}{\log^{c}(2)}\sum_{n= 1}^{\infty}\frac{1}{n^c} < \infty, $$ but since as $n \to \infty$, $2^{n} \to \infty $ and so $a_{n} = f(2^{n}) \to \infty $, according to Prove $\sum^{\infty}_{n=1} \frac{a_{n}-a_{n-1}}{a_{n}}=\infty$ left-hand side diverges, a contradiction.
