If the series $\sum f(n) n^{-s}$ converges absolutely for $\sigma > \sigma_a$ and has sum $F(s),$ prove that $F(s) \neq 0$

Question

I'm solving Apostol's introduction to analytic number theory, and got stuck on a problem exercise 11.12

The problem states the following:

Let $f$ be a completely multiplicative function, such that $f(p) = f(p)^2$ for each prime p. If the series $\sum f(n) n^{-s}$ converges absolutely for $\sigma > \sigma_a$ and has sum $F(s),$ prove that $F(s) \neq 0$ and that $\sum_{n=1}^{\infty} \frac{f(n) \lambda(n)}{n^s} = \frac{F(2s)}{F(s)}$ if $\sigma > \sigma_a$.

Here is my attempt of proving it: It is given that f is completely multiplicative, so, for $\sigma > \sigma_a$, I have shown that $\left( \sum_{n=1} ^{\infty} \frac{f(n)}{n^s} \right) \left( \sum_{n=1} ^ {\infty} \frac{f(n) \mu(n)}{n^s} \right) = \sum_{n=1} ^{\infty} \frac{f(n) I(n)}{n^s} = f(1) = 1$, which proves $F(s) \neq 0$.

As for the second part, I don't really have a clue, I thought of using the fact that $f(n) \lambda(n)$ is completely multiplicative, but I have no idea how to keep going from here. Could you give me a hint?

Thanks.

Answer 1

So I have come up with a solution.

We know that $f(n) \lambda (n)$ is completely multiplicative.

Then, by theorem 11.7 it can be said that $$\sum_{n = 0} ^{\infty} \frac{f(n) \lambda (n)}{n^s}$$ $$=\prod_{p} \frac{1}{1 - f(p)\lambda(p)p^{-s}}$$ $$=\prod_{p} \frac{1}{1 + f(p)p^{-s}}$$ $$=\prod_{p} \frac{1}{1 + f(p)p^{-s}} \cdot\frac{F(s)}{F(s)}$$ $$\prod_{p} \frac{1 - f(p) p^{-s}}{1 - (f(p))^2 p^{-2s}}$$ $$=\frac{F(2s)}{F(s)} \square$$