4
$\begingroup$

Suppose $(f_n(s))_n$ and $(g_n(s))_n$ are two sequences of Dirichlet series with positive coefficients such that $\exists \alpha\in\mathbb R$ such that for all $s\in\mathbb C$ with $\Re(s)>\alpha$ and for all $n$, the series $f_n(s)$ and $g_n(s)$ converge. Call the sequences $(f_n(s))_n$ and $(g_n(s))_n$ equivalent if there exists some $C>0$ such that for any $s\in\mathbb R$ such that $f_n(s)$ and $g_n(s)$ converge for all $n$ we have $$C^{-1-s}g_n(s)\le f_n(s)\le C^{1+s}g_n(s),\quad \forall n\in\mathbb N.$$

Put $F(s)=\prod_n(1+f_n(s))$ and $G(s)=\prod_n(1+g_n(s))$.

It is not too hard to check that if $(f_n(s))$ and $(g_n(s))$ are equivalent than for any $s\in\mathbb C$, $F(s)$ converges iff $G(s)$ converges.

My question is this: Suppose that $(f_n(s))$ and $(g_n(s))$ are equivalent and let $\alpha_0$ be the infimum $\alpha$ such that $F(s)$ (and hence $G(s)$ as well) converges on the half-plain $\Re(s)>\alpha$ (i.e. the abscissa of convergence of $F(s)$). Assume that there exists some $\delta>0$ such that $F(s)$ can be meromorphically continued to the half plain $\Re(s)>\alpha_0-\delta$.

Does it follow that $G(s)$ can be meromorphically continued to the half plain $\Re(s)>\alpha_0-\delta'$ for some $\delta'>0$?

Intuition: I have to say that my intuition says that the answer to this question is no, but I haven't managed to find a counter-example yet. The reason I believe it is false is since while the abscissa of convergence of a Dirichlet series is determined by the behavior of the series along the real-line- the question of meromorphic continuation is more subtle and require to understand the behavior of $F(s)$ along the line $\Re(s)=\alpha_0$.

But it is very possible that I am lacking knowledge on Dirichlet series an in fact such an equivalence does determine whether a function can be continued along the line $\Re(s)=\alpha_0$.

What I have done do far: So my attempts of finding a counter example reduce to trying to work with known sequences $(f_n(s))$ such that the product $F(s)=\prod_n(1+f_n(s))$ has a natural bound at imaginary line through the abscissa, and hence can not be continued to any half plain $\Re(s)>\alpha-\delta$.

One such example appears here, and is given by $f_p(s)=\frac{p^{-1-s}}{1-p^{-s}}=\sum_n p^{-1-(n+1)s}$, whenever $n=p$ is prime, and $f_n(s)=0$ otherwise. It can be checked that the product $F(s)=\prod_{p\:prime}1+\frac{p^{-1-s}}{1-p^{-s}}$ converges whenever $s>0$ and that $F(s)$ vanishes along the line $\Re(s)=0$ and hence can not be merom. continued past it.

It tried to an easy sequence $g_n(s)$ for which the product $G(s)=\prod_n(1+g_n(s))$ is known to have a merom. continuation past $0$ and such that $(g_n(s))$ and $f_n(s))$ are equivalent. For example $g_p(s)=\frac{p^{-1-s}}{1-p^{-1-s}}$ for $n=p$ prime and $g_n(s)=0$ otherwise, for which $G(s)=\zeta(s+1)$ is the shifted Riemann zeta function, and hence can be merom. continued. Unfortunately, for this $g_n$ it only holds that for any $s>0$ $$g_n(s)=g_n(s)\cdot 1^{-1-s}<f_n(s)$$ and there exists no $C>0$ such that $f_n(s)<C^{1+s}g_n(s)$ for all $n$'s. This follows simply from the fact that $f_n(s)/g_n(s)$ tends to infinity as $s\to 0$ for any $n$ prime.

In any case, if someone here has an idea about how to make this example into a counter-example, or maybe knows of another counter example, or otherwise knows of a reason why the answer to my question should actually be yes, I would appreciate a response very much.

Thank you, Shai

Remark- I did not write in any information for why this question interests me, as I thought this would be irrelevant. If anyone else thinks otherwise, I can definitely fill in this gap.

$\endgroup$
4
  • 1
    $\begingroup$ Why do you think (at the end) that it would be irrelevant to explain why the question is worthwhile? $\endgroup$
    – KConrad
    Mar 13, 2015 at 20:30
  • $\begingroup$ When you write "the abscissa of convergence of a Dirichlet series is determined by the behavior of the series along the real line [but] the question of meromorphic continuation is more subtle and requires understanding the behavior of $F(s)$ along the line ${\rm Re}(s) = \alpha_0$," that is strictly speaking not true. A meromorphic function is determined completely by its values on any interval of the real line larger than a point, so in principle everything about a meromorphic function is encoded in its values on $[3,5]$. That doesn't mean extracting information from $[3,5]$ is practical. $\endgroup$
    – KConrad
    Mar 13, 2015 at 20:32
  • 1
    $\begingroup$ Regarding your second comment- you are of course right. What I was trying to say is that convergence is a question which can be answered for Dirichlet series by just observing the real line, so this of equivalence makes sense. It somehow seems possible that two functions would be 'close' (in the sense of equivalent) and yet there will be some point with large imaginary value such that they are 'far apart enough at that point' F(s) cannot be merom. continued around this point, while G(s) can. Of course, I'm being incredibly informal here, and i admit to lack a lot of knowledge in the subject. $\endgroup$
    – kneidell
    Mar 13, 2015 at 20:39
  • 2
    $\begingroup$ @KConrad, regarding your first comment: the question comes up from the study representation zeta functions. I was reading an article by N. Avni (arxiv.org/abs/0803.1331) where he proves a result on the abscissa of convergence of one such product by equating it with another product for which it is somehow easier to 'understand' the abscissa, using model theoretic tools. I was just trying to get a grasp of 'how far' he got from proving merom. continuation of the initial function. $\endgroup$
    – kneidell
    Mar 13, 2015 at 20:42

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.