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I am trying to read the following paper by Granville-Harper-Soundararajan and I had a few questions regarding the paper. They prove, for $S(x):=\sum_{n\leq x}f(n)$ and $F_x(s):=\prod_{p\leq x}\left(1+\prod_{p\leq x}\frac{f(p^k)}{p^{ks}}\right)$, the following theorem:

Suppose that $f$ is a multiplicative function with $|f(n)|\leq 1$, $$|S(x)|\ll x\frac{L(x)}{\log x}\log\left(100 \frac{\log x}{L(x)}\right)+x\frac{\log\log x}{\log x}\tag{1}$$

One of my questions is: why is this error term in $(1)$ better than the trivial bound $x$?

The second error term on the right, $x\frac{\log\log x}{\log x}$ is clearly better than $x$. But what about the left one? It seems, as the authors state, that $L(x)\leq 6 \log x$ for large $x$. However, that means that $x\frac{L(x)}{\log x}\log\left(100 \frac{\log x}{L(x)}\right)\ll x\log\left(100 \frac{\log x}{L(x)}\right)$. Call $L(x)/\log x$, $a(x)$. Then our error term is clearly $\ll x(\log 100-\log a(x))$, which to me does not seem to be better than $x$.

My second question is: One of the inequalities in the paper is: $$\sum_{x/pq\leq n \leq 2x/pq}\min\left(1, \frac{x}{T|x-pqn|}\right)\ll \left(1+\frac{x}{Tpq}\log T\right)\tag{2}$$ Here $p$ and $q$ are primes such that $p\in \mathcal{P_k}$ (as defined in the paper) and $q\leq 2x/p$. Further, $n$ is a natural number whose largest prime factor $\leq x$.

I was wondering how one could prove inequality $(2)$.

Maybe, it is something standard. Thank you for your time.

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    $\begingroup$ It's not always better than $x$ but it sometimes is (and that's when it's useful). $\endgroup$ Commented Dec 14, 2020 at 10:07
  • $\begingroup$ and for the inequality you should think that "$x/pq-n$ varies over the said range as does $n$ when running from $-x/pq$ to $x/pq$" (you'll need the first term in the minimum for the "zero term") $\endgroup$
    – tomos
    Commented Dec 14, 2020 at 22:56

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Let h(x)=log(100 log x/L(x)) which is >3 for large x. Then the first term in (1) is << x h(x)/e^{h(x)}. This is evidently << x always, and significantly better than that the larger h(x) gets

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  • $\begingroup$ Dear Andrew, you can use TeX on this site. Put your formulae inside dollar symbols etc. as in a normal TeX file. $\endgroup$
    – GH from MO
    Commented Mar 23, 2021 at 9:14

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