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I am working right now with "Independent and Stationary Sequences of Random Variables" from Ibragimov 1971.

I am trying to understand the proof of the following Lemma (18.2.4):

$h: \mathbb N \rightarrow \mathbb R$ is a slowly varying function, i.e. for all $a > 0$ $ \lim_{n\to\infty}\frac{h(an)}{h(n)}=1. $

For all sufficiently small $c$ and all sufficiently large $n$, $ \frac{h(cn)}{h(n)} < c^{-\frac 1 2}. $

We remark that this inequality holds for all $c where $c_0$ does not depend on $n$.

Proof: From what has been proved about $h(n)$, [I come to this later] \begin{align*} \log \frac{h(cn)}{h(n)} & = \sum_{k=0}^{\lfloor-\frac{\log c}{\log 2}\rfloor} \log h(\lfloor2^{-k-1}n\rfloor) - \log h(\lfloor2^{-k}n\rfloor) + \log h(cn) - \log h(\lfloor2^{-\lfloor \frac{\log c}{\log 2}\rfloor} n\rfloor) \\ & < \frac 1 2 \log c^{-1}. \end{align*}

Thats what is in the book. The first "=" is a telescoping sum, but the last term of the equation should be $\log h(\lfloor2^{\lfloor \frac{\log c}{\log 2}\rfloor} n\rfloor)$ because for all $x < 0, x \notin \mathbb Z$ $ -\lfloor -x \rfloor - 1 = \lfloor x \rfloor. $ Note that $2^{\frac{\log c}{\log 2}} = c$ and every term of the series goes to zero, since $\lim_{n\to\infty}\frac{h(2n)}{h(n)}=1$.

But I dont understand the "<". Do you have an idea?

Here is what has been proved about $h$ before this lemma:(I couldn't use these information) $ \text{for all } \varepsilon > 0: \\ \lim n^\varepsilon h(n) = \infty \\ \lim n^{-\varepsilon} h(n) = 0. $ $ \text{If $n$ is sufficiently large, then } \\ \sup_{n\le r\le 2n} \frac{h(n)}{h(r)} \le 4. $

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