An excerpt from Introduction to Analytic Number Theory by Tom M. Apostol.
I have three main concerns regarding this proof:
- What is the abscissa of convergence for $\frac{1}{F(s)}$ - why can we take the product of the two series. I only know this is true if the two series are absolutely convergent.
- Why can we do term by term integration? I believe its suffices to show that the series converges locally uniformly - BUT is this true?
- How does the proof show that $G(s)$ converges uniformly?
Concerning the abscissa of absolute convergence of the Dirichlet series of $1/F(s)$, it seems that is at least a hole in Apostol's proof. I took a quick look at the chapter and didn't see anything about the abscissa of (absolute) convergence of $\sum \frac{f^{-1}(n)}{n^s}$. If that series always converges absolutely for $\sigma > \sigma_0$ (where $\sigma_0$ is as in the theorem), then - unless I overlooked a relevant theorem/note - it's just a hole in that Apostol uses something true he neither proves, nor refers to a proof elsewhere. Mistakes like that happen. If, however, there are arithmetic functions $f$ with $f(1) \neq 0$ and $\sigma_a(f) < +\infty$ such that $\sigma_a(f^{-1}) > \sigma_0$, then it might be that the theorem asserts more than actually holds. But even then, everything works if we replace $\sigma_0$ with $\max \: \{ \sigma_0, \sigma_a(f^{-1})\}$, so it's not a huge error.
Thus let's suppose $\sigma_0$ were subject to the additional constraint that $\sigma_0 \geqslant \sigma_a(f^{-1})$.
Per the note after theorem 11.12, the Dirichlet series of $F'$ has the same abscissa of absolute convergence as the series of $F$. (Apostol doesn't explicitly prove that, but the proof is easy.) Hence the series
$$- \sum_{n = 2}^{\infty} \frac{(f' \ast f^{-1})(n)}{n^s}\tag{1}$$
is absolutely convergent for $\sigma > \sigma_0$ (since both factors are absolutely convergent there), and it represents $\frac{F'(s)}{F(s)}$ in that half plane. Let me note that by the identity theorem it follows that
$$\frac{F'(s)}{F(s)} = - \sum_{n = 2}^{\infty} \frac{(f' \ast f^{-1})(n)}{n^s}\tag{2}$$
holds wherever the right hand side converges and the left hand side is defined. Thus $(2)$ can also hold for (some) $s$ with $\operatorname{Re} s < \sigma_0$, although for the justification of the rearrangement we needed the absolute convergence of both factors.
We come to your second point,
Yes. Apostol proves that in theorem 11.11, the convergence of a Dirichlet series is locally uniform on its half plane of convergence. Since $\mathbb{C}$ is locally compact, locally uniform convergence and "uniform convergence on every compact subset (of an open $U \subset \mathbb{C}$)" are the same thing. [Locally uniform convergence always implies uniform convergence on every compact subset of the domain, the local compactness is needed for the other direction.] Thus we can integrate (and differentiate) Dirichlet series term by term on their half plane of convergence.
With regard to
there seems to be a misreading or a typo on your part. Apostol doesn't claim uniform convergence, but absolute convergence of $G(s)$. This follows from absolute convergence of $(1)$ - which is guaranteed on every half plane where $F'(s)$ and $\frac{1}{F(s)}$ converge absolutely, but may hold on a larger half plane. Of course as for every Dirichlet series the convergence is locally uniform on the half plane of convergence, and it is uniform on the half plane $\sigma \geqslant \sigma_a(G) + \varepsilon$ for every $\varepsilon > 0$.
Let's return to the question of the abscissa of (absolute) convergence of $1/F$. If $F$ has a zero at $s$, then $1/F$ has a pole at $s$, and hence $\sigma_c(f^{-1}) \geqslant \operatorname{Re} s$. This shows that $\sigma_c(f^{-1})$ and $\sigma_a(f^{-1})$ have no simple relation to $\sigma_c(f)$ and/or $\sigma_a(f)$, for if we take $\tilde{f}(n) = f(n)$ except for e.g. $n = 37$ so that the modified series has a zero at some $s$ of our choice, we have $\sigma_c(\tilde{f}) = \sigma_c(f)$ and $\sigma_a(\tilde{f}) = \sigma_a(f)$, but if $\operatorname{Re} s > \sigma_a(f^{-1})$, then $\sigma_c(\tilde{f}^{-1}) > \sigma_a(f^{-1})$.
I don't know if $\sigma_0 \geqslant \sigma_a(f)$ and $F(s) \neq 0$ for $\operatorname{Re} s > \sigma_0$ is sufficient to have $\sigma_a(f^{-1}) \leqslant \sigma_0$. A sufficient condition for $\sigma_a(f^{-1}) \leqslant \sigma_1$ is that
$$\sum_{n = 2}^{\infty} \frac{\lvert f(n)\rvert}{n^{\sigma}} < \lvert f(1)\rvert \tag{$\ast$}$$
for $\sigma > \sigma_1$. Then for $\operatorname{Re} s > \sigma_1$ we can expand the geometric series
$$\frac{1}{f(1) + \sum_{n = 2}^{\infty} f(n)n^{-s}} = \frac{1}{f(1)}\cdot \frac{1}{1 + f(1)^{-1}\sum_{n = 2}^{\infty} f(n)n^{-s}} = \sum_{k = 0}^{\infty} \frac{(-1)^k}{f(1)^{k+1}}\Biggl(\sum_{n = 2}^{\infty} \frac{f(n)}{n^s}\Biggr)^k,$$
and since $(\ast)$ guarantees that
$$\sum_{k = 0}^{\infty} \frac{1}{\lvert f(1)\rvert^{k+1}}\Biggl(\sum_{n = 2}^{\infty} \frac{\lvert f(n)\rvert}{n^{\operatorname{Re} s}}\Biggr)^k < +\infty,$$
we can rearrange and group as we please, so we find a Dirichlet series
$$\sum_{n = 1}^{\infty} \frac{g(n)}{n^s}$$
that represents $1/F$ and is absolutely convergent for $\sigma > \sigma_1$. By the uniqueness theorem for Dirichlet series or by analysing the expression for $g(n)$, we then find $g(n) = f^{-1}(n)$ for all $n$.
Note that the Taylor series of $\log (1+z)$ gives another argument that $G(s) = \log F(s)$ has an absolutely convergent Dirichlet series for $\sigma > \sigma_1$.
However, in general we must expect
$$\sigma_1 > \inf \: \{ \sigma > \sigma_a(f) : F(s) \neq 0 \text{ if } \operatorname{Re} s > \sigma \},$$
so there remains a gap between what I can prove now and what Apostol's theorem asserts.