Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let's say $f$ is a Dirichlet series which converges on the half-plane $\text{Re }s>\sigma$ to a function $f(s)$. Suppose further that $f(s)$ admits an analytic continuation to an entire function, together with the standard sort of functional equation. Let $g_n$ be a sequence of Dirichlet series, also convergent on $\text{Re }s>\sigma$, which each admit an analytic continuation and functional equation, though their precise FEs may vary. We assume that $g_n$ converges to $f$ in the following sense: for every $m>0$ there exists an $N$ for which the series $g_n$ and $f$ match on every term up to the $m$th, for all $n>N$. Note this implies that $g_n(s)$ converges to $f(s)$ for every $\text{Re }s>\sigma$.

Can it be said that $g_n(s)$ converges to $f(s)$ for any $s$ outside the domain of convergence?

Perhaps that's too much to hope for, and you can't even expect that $g_n(s)$ converges to $f(s)$ even for the point $s=\sigma$. I'd certainly be interested in a counterexample which does this!

share|improve this question
    
Do you mean to say that g_n and f match on every term up to the m-th? –  Matt Young Feb 20 '10 at 2:30
    
Of course. Corrected. –  Jared Weinstein Feb 20 '10 at 2:46
    
If one imposes some conditions on the functional equations of the g_n's (in particular, bounding the conductors to avoid FC's counterexample) then the answer could potentially become "yes." –  Matt Young Feb 20 '10 at 3:59
    
FC- I don't understand why you wrote that as a comment rather than an answer. In particular, it gives a misleading impression on the front page. –  Ben Webster Feb 20 '10 at 4:24
1  
If you don't want to play any reputation games, you could instead make all your new answers "community wiki." This allows the site to function correctly while not accruing any reputation to you. –  Noah Snyder Feb 20 '10 at 8:15

1 Answer 1

up vote 2 down vote accepted

(This answer is a community wiki version of a comment above by FC which answered the question.)

For any integer M, there exists a prime p such that chi_p(n) = (n/p) = 1 for all n = 1...M. This means that the Dirichlet series L(s,V chi_p) (for any representation V) "converges" in your sense to L(s,V). but they do not converge at s = 0. If V is trivial, then we are comparing zeta(0) = -1/2 with L(0,chi_p) which grows without bound by Brauer-Siegel. I think in this class of examples one does get convergence at the critical point.

share|improve this answer
    
The question requires f(s) to be entire, which is excluded if V is trivial -- I've thought of this counterexample already. Can anything be said if V is nontrivial? Still, this example hints at a negative answer to the question, since zeta(s) is regular at 0 whereas the f_n(s) do not converge. So, +1. –  Jared Weinstein Feb 21 '10 at 5:00
    
So it does. Accept. –  Jared Weinstein Feb 21 '10 at 7:08

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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