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Suppose $|a_{n}| \leq 1$ completely multiplicative function assuming real values. Suppose further that, $ L(s)=\sum_{n} \frac{a_{n}}{n^s} $ may be continued analytically to the left of $s=1$ a bit (say $Re(s)>1-\epsilon$)

((This is to be done by writing $L(s)=exp(\sum_{p} \frac{a_{p}}{p^s})G(s)$, where $G(s)$ is a harmless function and by partial summation together with the bound, $\sum_{p \leq x} a_{p} \ll x^{1-\epsilon}$ )) .
Can we conclude that $L(s)$ converges for $Re(s)>1-\frac{\epsilon}{2}$ ?

Regards.

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  • $\begingroup$ Do you have an actual example in mind, or are you trying to find another proof for a known example? $\endgroup$
    – KConrad
    Commented Jan 26, 2014 at 18:40
  • $\begingroup$ As a matter of the fact: I know that how to extend the series $\sum_{p } \frac{a_{p}}{p^s}$, to the left of s=1 and make sure that the "prime" series converges there. The question turns out to be how one can conclude conclude results on convergence of $\sum \frac{a_{n}}{n^s} $? $\endgroup$
    – Yıldırım A.
    Commented Jan 26, 2014 at 18:44
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    $\begingroup$ @user4486 You presumably know very well, but I say it just in case, that when the coefficients $a_n$ are positive this is true, by a result of Landau (which is in chapter V of Serre's course in arithmetic). I believe it is false without the positivity hypothesis, but don't have a counter-example at hand. Good question. $\endgroup$
    – Joël
    Commented Jan 26, 2014 at 22:06
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    $\begingroup$ @user4486: I had asked if you are looking at something that is not a Dirichlet $L$-function. Could you tell us exactly what your example is? Without the condition that the coefficients are bounded there is an easy counterexample: let $\chi$ be a nontrivial quadratic character and set $a_n = \chi(n)n$. Then $L(s) = L(s-1,\chi)$, which extends analytically to the whole plane but converges nowhere on the half-plane ${\rm Re}(s) < 1$ since it would imply convergence of $L(1) = \sum \chi(n)$. $\endgroup$
    – KConrad
    Commented Jan 26, 2014 at 23:23
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    $\begingroup$ If your analytic continuation of $L$ comes with some kind of polynomial bounds for $|L(s)|$ in the region Re$(s)>1-\delta$, then you could shift contours and prove that $\sum_{n\le x} a(n) \ll x^{1-c\delta}$ for some constant $c>0$ (depending on your polynomial type bound for $L(s)$). That'll give the desired convergence to the right of $1-c\delta$. $\endgroup$
    – Lucia
    Commented Jan 27, 2014 at 5:06

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