Multiplicative arithmetic function on the unit disk

Question

Suppose $f$ is a multiplicative arithmetic function that takes values inside the unit disk, and let Re$(s)>1$. We define $F(s) = \sum_{n\ge1}^{}\dfrac{f(n)}{n^s}$. I want to show that $$\text{log } F(s) = \sum_{p \text{ prime}} \dfrac{f(p)}{p^s} + O(1)$$ Now I have tried to expand $\text{log } F(s)$ into its Euler product (i.e. writing it in terms of prime powers) and then trying to solve this, but I'm getting stuck after this. I have no idea how to proceed. I am guessing I need to approximate $\text{log} (1+\dfrac{f(p)}{p^s}+\dfrac{f(p^2)}{p^{2s}}+ \dots)$ somehow, but I don't know what to do exactly. Any help on this?

Answer 1

Since by Euler's product $$\sum_{n\geq 1}\frac{f(n)}{n^s} = \prod_{p}\left(1+\frac{f(p)}{p^s}+\frac{f(p^2)}{p^{2s}}+\ldots\right) $$ and $\|f\|\leq 1$ we have $$ \log F(s) = \sum_{p}\log\left(1+\frac{f(p)}{p^s}+O\left(\frac{1}{p^{2s}}\right)\right)=\sum_{p}\frac{f(p)}{p^s}+O\left(\sum_{p}\frac{1}{p^{2s}}\right)=\sum_{p}\frac{f(p)}{p^{s}}+O(1) $$ since in a neighbourhood of the origin $\log(1+z)=z+o(z)$ and for any $\eta>1$ the series $\sum_{p}\frac{1}{p^{\eta}}$ is clearly convergent.