# Average orders of multiplicative functions in Iwaniec-Kowalski

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?

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.