The motivation for this enquiry is to understand something about the impact of multiplicativity for $f:\mathbb{N}\rightarrow\mathbb{C}$ on the conditional convergence of Dirichlet series


The existence of an abscissa of absolute convergence $\sigma_a$ naturally implies the existence of an abscissa of conditional convergence $$\sigma_c\leq\sigma_a$$ yet, in many cases of interest to number theorists, improvements on this upper bound for $\sigma_c$ are presently unavailable.

I want to know about the impact of multiplicativity on lower bounds for $\sigma_c$ in the same "functional" setting, i.e. in relation to $\sigma_a$. For example, although one can easily find non-multiplicative $f$ for which $\sigma_c/\sigma_a=0$, one is tempted to suggest that multiplicativity for $f$ implies that $$ l\leq \sigma_c/\sigma_a $$

for some $0<l\leq 1,$ which is equivalent to saying that $$\limsup_{x\rightarrow\infty}\sum_{n\leq x}|f(n)|\leq\limsup_{x\rightarrow\infty} \left|\sum_{n\leq x}f(n)\right|^{1/l} $$ for all (or perhaps some class of) multiplicative arithmetic functions $f$.

Specifically, I would like to ask if the assumption of multiplicativity (perhaps with additional hypotheses) leads to such (or similar) functional inequalities?

  • $\begingroup$ Can you elaborate on your 'which is equivalent to saying that ... for all multiplicative functions $f$' equation above? $\endgroup$
    – Stopple
    Oct 18 '11 at 16:49
  • $\begingroup$ @ Stopple. Yes- take logs on both sides of ... and divide by $log x$ to get the formula for the corresponding abscissi of convergence. Thanks. $\endgroup$ Oct 18 '11 at 16:57
  • $\begingroup$ I'm sorry, but I'm confused. Why isn't any Dirichlet $L$-function attached to a character $\chi$ modulo $d$ a counterexample: $\sigma_c=0$, $\sigma_a=1$. $\endgroup$
    – Stopple
    Oct 18 '11 at 17:16
  • $\begingroup$ Ok, thanks Stopple. You needn't be confused- those characters are counter-examples, so I should allow for $l=0$ too. I will edit my question appropriately as I am trying to get at additional hypotheses that give a non-zero value for $l$. $\endgroup$ Oct 18 '11 at 18:06

If $f(n)$ is a multiplicative function then so is $f(n)n^{-w}$ for any fixed complex number $w$. In particular, find a multiplicative $f(n)$ for which $\sigma_c$ is strictly smaller than $\sigma_a$, and take $w=\sigma_c-\epsilon$; the modified function $\tilde f(n) = f(n)n^{-w}$ will have its $\tilde\sigma_c = \epsilon$ and its $\tilde\sigma_a$ bounded below by the positive number $\sigma_a-\sigma_c$. This example shows that $\tilde\sigma_c/\tilde\sigma_a$ can be arbitrarily close to 0.

Although it doesn't directly address your question: perhaps you already know that the inequalities $\sigma_c \le \sigma_a \le \sigma_c+1$ are true and optimal, in the class of Dirichlet series as a whole (see for example Chapter 1 of Montgomery/Vaughan). Because of the ability to translate Dirichlet series as in the above construction, it seems to be that the difference between the two abscissae is more important than the quotient.

  • $\begingroup$ Thanks for this explanation, Greg. The kind of inequality I am getting at certainly requires further hypotheses on $f$- I think I should perhaps restrict $f(p)$. I am not really interested in translations or periodicities. For example, there may be a good reason why $$\limsup_{x\rightarrow\infty}|M(x)|\geq \frac{x^2}{\zeta^2(2)}$$ or better, trivial as it may be when one has the Mellin transform. You have answered my question as it stands though so I accept you answer. $\endgroup$ Oct 18 '11 at 23:17
  • $\begingroup$ The obvious typo being that i mean $1/2$ not $2$. $\endgroup$ Oct 18 '11 at 23:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.