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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

$$F(s)=\sum_{1}^{\infty}\frac{f(n)}{n^{s}}.$$

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?

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Can you elaborate on your 'which is equivalent to saying that ... for all multiplicative functions $f$' equation above? –  Stopple Oct 18 '11 at 16:49
    
@ Stopple. Yes- take logs on both sides of ... and divide by $log x$ to get the formula for the corresponding abscissi of convergence. Thanks. –  Kevin Smith Oct 18 '11 at 16:57
    
I'm sorry, but I'm confused. Why isn't any Dirichlet $L$-function attached to a character $\chi$ modulo $d$ a counterexample: $\sigma_c=0$, $\sigma_a=1$. –  Stopple Oct 18 '11 at 17:16
    
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 $l$. –  Kevin Smith Oct 18 '11 at 18:06
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1 Answer

up vote 4 down vote accepted

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.

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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. –  Kevin Smith Oct 18 '11 at 23:17
    
The obvious typo being that i mean $1/2$ not $2$. –  Kevin Smith Oct 18 '11 at 23:29
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