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Let $s=α+iβ$ be a complex number. Consider the Dirichlet series of the form $$f(s)=∑_{n=1}^{∞}(a_{n})/n^{s}$$ where $(a_{n})_{n≥1}$ is a real sequence.

We consider the class of Dirichlet series satisfying the following conditions:

(1) The real sequence $(a_{n})_{n≥1}$ is not identically zero.

(2) The function $f(s)$ is analytic for all $s=α+iβ∈ℂ$ such that $α>0$.

(3) The first term of $f(s)$ is not zero, i.e., $a₁≠0$.

(4) There exist a complex zero $a=1/2+iβ₁$ of $f $ in the critical strip: $s=α+iβ∈ℂ$ with $0<α<1$.

(5) $f $ has infinitely many non trivial zeros.

(6) For all $α∈(0,1)$ with $α≠1/2$ we have $f(α+iβ₁)≠0$.

My question is: Does there exist a known Dirichlet series verifying all these conditions and have non trivial zeros off the critical line.

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    $\begingroup$ What's your definition of a "trivial zero" for general Dirichlet series? $\endgroup$ – Mark Lewko Jan 23 '17 at 4:42
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What's wrong with the most obvious choice for a Dirichlet series that converges for $\text{Re}(s)>0$?
$$ f(s)=(1-2^{1-s})\zeta(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} $$ converges for real $s>0$ by Dirichlet's test and in the half plane by standard facts about the convergence of Dirichlet series. Any nontrivial zero of $\zeta(s)$ is a zero of $f(s)$. The only non obvious fact is your 6) above, but this could be verified for the zero at $1/2+i\cdot 14.1347\ldots$ via the Argument Principle and a numerical estimate of the integral. Since it's an integer, careful bounds on the error will give you a rigorous answer.

The term $(1-2^{1-s})$ is zero for $(1-s)\log(2)=2\pi i\cdot k$.

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Pursuing @Stopple's line, Landau had already observed that things like $\zeta(2s)\zeta(2s-1)$ are positive Dirichlet series, have Euler products, nice functional equations (and analytic continuations), but have no zeros at all on the critical line, and many off. One can create minor variations such as $\zeta(2s)\zeta(2s-1)(1-p^{s-1/2})$... (and given the incommensurability of logs of distinct primes, we can arrange sufficiently large $p$ to avoid zeros of zetas near a given height).

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  • $\begingroup$ gareett: Can you give a referrence about this. Also, these examples do not verify condition (4). $\endgroup$ – Safwane Jan 23 '17 at 9:18
  • $\begingroup$ The $1-p^{s-1/2}$ term gives infinitely-many zeros on the critical line. A way that such things arise is in zeta functions of quaternion (division) algebras, which appear in Weil's "Basic Number Theory", and also as examples in his "Adeles and Algebraic Groups". But in any case once one notices the possibility, it is not hard to verify. $\endgroup$ – paul garrett Jan 23 '17 at 14:37

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