Given a Dirichlet series $$\phi(s)=\sum_{n\ge1}\frac{a_n}{n^s}$$
let $\sigma_{\text{conv}}\in\bar{\mathbb{R}}$ its abscissa of convergence, then we know that $\phi(s)$ is holomorphic on the half-plan $\mathrm{Re}(s)>\sigma_{\text{conv}},$ then if we denot by $\sigma_{\text{hol}}$ the abscissa of holomorphy of $\phi(s)$ we have $$\sigma_{\text{hol}}\le \sigma_{\text{conv}}.$$ My question is : it is in general true that $\sigma_{\text{hol}}= \sigma_{\text{conv}}$?

Edit : $\sigma_{\text{hol}}:=\inf\{\sigma\in\mathbb{R}\; ; \; \phi(s)\;\text{has a holomorphic prolongation on}\; \mathrm{Re}(s)>\sigma \}$

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    $\begingroup$ I've never heard of the notion of "abscissa of holomorphy". Note that for example $L(s, \chi)$ can be continued analytically to an entire function, while $\sigma_{\text{conv}} = 0$, so that would mean that the answer to your question is no (but you didn't define abscissa of holomorphy so I can't know). Note also that there is usually a distinction between the abscissa of absolute and conditional convergence. You would need to specify which one you mean. $\endgroup$ – COnstructor Jan 11 '17 at 0:29
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    $\begingroup$ @COnstructor, Thanks for your comment I'll add the definition of "abscissa of holomorphy" $\endgroup$ – Adam Jan 11 '17 at 0:35
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    $\begingroup$ The simplest counter-example is $\eta(s) = \sum_{n=1}^\infty (-1)^{n+1} n^{-s}$ converging for $Re(s) > 0$ but holomorphic on the whole complex plane. $\endgroup$ – reuns Jan 11 '17 at 10:14

The answer to your question is no, the counterexample is $L(s, \chi) = \sum_{n = 1}^{\infty} \chi(n) n^{-s}$ with $\chi$ any non-principal character. $L(s,\chi)$ can be continued to an entire function thus $\sigma_{\text{hol}} = -\infty$, while $\sigma_{\text{abs}} = 1$ (abscissa of absolute convergence) and $\sigma_{\text{cond}} = 0$ (abscissa of conditional convergence).

  • $\begingroup$ I have another question if $\phi(s)$ is holomorphic at $s_0$ it is necessary that $\phi(s)$ convergente at $s_0$ ? $\endgroup$ – Adam Jan 11 '17 at 1:09
  • $\begingroup$ No, it could be holomorphic without being convergent, for example $L(-1, \chi)$ is well-defined but the series is not convergent there $\endgroup$ – COnstructor Jan 11 '17 at 1:50
  • $\begingroup$ @Adam No, otherwise it would imply that $\sigma_{conv} \le \sigma_{hol}$ $\endgroup$ – reuns Jan 11 '17 at 10:15

In general the answer is no, but if you assume that the $a_n$ are non-negative, then Landau's theorem tells you that $\phi$ has a singularity at $\sigma_{\mathrm{conv}}$, in particular $\sigma_{\mathrm{conv}}=\sigma_{\mathrm{hol}}$. See http://math.mit.edu/~ebelmont/229-notes.pdf , Theorem 3.5. There is a variety of results which weaken the non-negativity assumption, e.g. non-negative partial sums, so for a concrete application you should check the proof of Landau's theorem and try to adapt it to your problem.

  • $\begingroup$ Thank you very much, that's exactly what I am looking for. $\endgroup$ – Adam Jan 11 '17 at 13:11
  • $\begingroup$ @Adam In the non-negative case $a_n \ge 0$. If $\sum_{n=1}^\infty a_n n^{-s}$ diverges at $s_0$ then it diverges at $Re(s_0)$ and so $\lim_{x \to Re(s_0)}\sum_{n=1}^\infty a_n n^{-x} = + \infty$ and hence $F(s)$ has a singularity at some $\sigma \ge Re(s_0)$. Together with $\sigma_{conv} \ge \sigma_{hol}$ it proves that $\sigma_{conv} = \sigma_{hol}$ in that case $\endgroup$ – reuns Jan 11 '17 at 16:35
  • $\begingroup$ @user1952009, thank you very much for your comment. $\endgroup$ – Adam Jan 11 '17 at 20:44

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