Average orders of multiplicative functions in Iwaniec-Kowalski

Question

I don't understand how Corollary 1.2 is derived in Iwaniec-Kowalski. Specifically I start getting lost in the last sentence of page 27. They define a completely multiplicative function $f$ just after Theorem 1.1 and at the bottom of page 27 seem to me to be saying an asymptotic for \[ \sum _{d\leq x}\frac {\mu (d)f(d)}{d}\] follows from Theorem 1.1, and the constants in Equations (1.97) and (1.100) suggest to me that they are applying that theorem to the multiplicative function $F(d):=\mu (d)f(d)/d$. In particular Theorem 1.1 needs Equation (1.88) to be satisfied, so assuming that they do apply the theorem to $F$ then we'd need to know \[ \sum _{n\leq x}\Lambda _F(n)\sim -\kappa \log x.\] Definitions (1.50) and the sentence following it seem to me to say that \[ D_F(s)=\prod _{p}\left (1-\frac {f(p)}{p^{s+1}}\right )\] so that \[ 1/D_F(s)=\prod _p\left (1+\frac {f(p)}{p^{s+1}}+\frac {f(p)^2}{p^{2(s+1)}}+...\right )=\sum _{n=1}^\infty \frac {f(n)}{n^{s+1}}\] so that $\mu _F(n)=f(n)/n$ and so (again using the sentence following (1.50)) \[ \Lambda _F(n)=(\mu _F\star FL)(n)=\sum _{dm=n}\frac {f(d)F(m)\log (m)}{d}\] and now somehow I need to use assumption (1.99) to deduce the asymptotic for $\Lambda _F(n)$. But this is as far as I can get. And I'm not really sure that this is the correct line of reasoning at all and I think I'm misunderstanding something. Can anyone clear this up for me?

Answer 1

Note that $$ \sum _{dm=n}\frac {f(d)F(m)\log (m)}{d} = \sum _{dm=n}\frac {f(d)f(m)\mu(m)\log (m)}{md} = \frac{f(n)}n \sum_{m\mid n} \mu(m)\log m = -\frac{f(n)}n \Lambda(n) $$ since $f$ is completely multiplicative. Then $\sum_{n\le x} -\frac{f(n)}n \Lambda(n)$ is essentially $-\sum_{p\le x} \frac{f(p)\log p}p$ (the prime powers contribute $O(1)$) and the given assumption on $\mathcal P$ finishes it from there.