# Theorem 11.14, Apostol, pg 238 - need explanation

An excerpt from Introduction to Analytic Number Theory by Tom M. Apostol.

I have three main concerns regarding this proof:

1. 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.
2. Why can we do term by term integration? I believe its suffices to show that the series converges locally uniformly - BUT is this true?
3. How does the proof show that $G(s)$ converges uniformly?
• Hasn't Apostol proven any stability results concerning abcissas of convergence (i.e. if $F$ has certain $\sigma_c$ then $F'$ and $F^{-1}$ have this other...?) – Pedro Tamaroff Jul 26 '17 at 13:03
• @PedroTamaroff $F'$ has the same abscissa of convergence and of absolute convergence as $F$, but $F^{-1}$ can have a wildly different abscissa of convergence. I suppose Apostol has proven that $F^{-1}$ always is convergent when $F$ is, though. – Daniel Fischer Jul 26 '17 at 13:08
• Related: math.stackexchange.com/questions/1043825 In Cohen, Number Theory: Volume II: Analytic and Modern Tools exercise 9 p. 259 it says: In particular, if a Dirichlet series $f$ is invertible, show that the abscissa of absolute convergence of its inverse is equal to that of $f$. Not sure what is meant by an invertible dirichlet series. If it means nonzero, it's exactly the result we're looking for. --- In the special case where $f$ is multiplicative, they have the same $\sigma_a$ (by the Euler product). – punctured dusk Jul 26 '17 at 13:10
• @DanielFischer (I know that, just trying to lead the OP and ponder about it.) – Pedro Tamaroff Jul 26 '17 at 13:11
• @barto It's a mistake in Cohen's book, it seems. (Might be true for multiplicative $f$, but not in general.) – Daniel Fischer Jul 26 '17 at 13:14

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,

Why can we do term by term integration? I believe its suffices to show that the series converges locally uniformly - BUT is this true?

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

How does the proof show that $G(s)$ converges uniformly?

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.

• Thank you so much Daniel for your detailed answer. By the way, what was your proof for absolute convergence of $\sum \frac{ f(n) \log n }{n^s}$? I do it by splitting $\sigma - \alpha > \sigma_a$ where $\alpha > 0$, so that $\frac{ \log n }{n^{s- \alpha} } \rightarrow 0$ as $n \rightarrow \infty$ and the sum is bounded by $C\sum \frac{|f(n)| }{n^{s - \alpha}}$ for some constant $C$. – CL. Jul 28 '17 at 9:36
• Also, why is $\log f(1)$ well -defined, or do we pick any branch of logarithm that contans both $f(1)$ and the positive real line. – CL. Jul 28 '17 at 10:20
• Yes, we have $\sigma - \sigma_a > 0$, and we take a bit of that difference to compensate the logarithm. Alternatively, one can use that $\sum_{n \leqslant x} \lvert f(n)\rvert\log n \leqslant \log x \sum_{n\leqslant x} \lvert f(n)\rvert$, and the formula for the abscissa of absolute convergence gives $\sigma_a(f') \leqslant \sigma_a(f)$. Since $\sigma_a(f) \leqslant \sigma_a(f')$ follows from $\log n \to +\infty$, we have equality. It doesn't matter which logarithm of $f(1)$ we use, the resulting $G$s only differ by a constant ($2k\pi i$), every choice works. – Daniel Fischer Jul 28 '17 at 12:33
• To me Apostol statement is false. – reuns Aug 31 '17 at 23:16
• @reuns I suspect that too, but so far I haven't found an example where $\sigma_a(f^{-1}) > \sigma_0$. Nor have I found a theorem giving the abscissa of absolute convergence of the inverse. Have you an idea where one should look? – Daniel Fischer Sep 1 '17 at 10:56