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A note on a conjecture of Gonek

We derive a lower bound for a second moment of the reciprocal of the derivative of the Riemann zeta-function averaged over the zeros of the zeta-function that is half the size of the conjectured value. Our result is conditional upon the assumption of the Riemann Hypothesis and the conjecture that the zeros of the zeta-function are simple.


Introduction
Let ζ(s) denote the Riemann zeta-function. Using a heuristic method similar to Montgomery's study [13] of the pair-correlation of the imaginary parts of the non-trivial zeros of ζ(s), Gonek has made the following conjecture [7,8].
Conjecture. Assume the Riemann Hypothesis and that the zeros of ζ(s) are simple. Then, as T → ∞, where the sum runs over the non-trivial zeros ρ = 1 2 +iγ of ζ(s).
The assumption on the simplicity of the zeros of the zeta-function in the above conjecture is so that the sum over zeros on the right-hand side of (1.1) is well defined. While the details of Gonek's method have never been published, he announced his conjecture in [5]. More recently, a different heuristic method of Hughes, Keating, and O'Connell [10] based upon modeling the Riemann zeta-function and its derivative using the characteristic polynomials of random matrices has led to the same conjecture. Through the work of Ingham [11], Titchmarsh (Chapter 14 of [21]), Odlyzko and te Riele [17], Gonek (unpublished), and Ng [15], it is known that the behavior of this and related sums are intimately connected to the distribution of the summatory function where µ(·), the Möbius function, is defined by µ(1) = 1, µ(n) = (−1) k if n is divisible by k distinct primes, and µ(n) = 0 if n > 1 is not square-free. See also [9] and [20] for connections between similar sums and other arithmetic problems.
MBM is supported in part by a University of Mississippi College of Liberal Arts summer research grant. NN is supported in part by an NSERC Discovery grant. 1 In support of his conjecture, Gonek [5] has shown, assuming the Riemann Hypothesis and the simplicity of the zeros of ζ(s), that for some constant C > 0 and T sufficiently large. In this note, we show that the inequality in (1.2) holds for any constant C < 3 2π 3 .
Theorem. Assume the Riemann Hypothesis and that the zeros of ζ(s) are simple. Then, for any fixed ε > 0, for T sufficiently large.
While our result differs from the conjectural lower bound by a factor of 2, any improvements in the strength of this lower bound have, thus far, eluded us. It would be interesting to investigate whether for k > 0 there is a constant C k > 0 such that for T sufficiently large. However, a lower bound of this form is probably not of the correct order of magnitude for all k. This is because it is expected that for each ε > 0 there are infinitely many zeros ρ = 1 2 + iγ of ζ(s) satisfying |ζ ′ (ρ)| −1 ≫ |γ| 1/3−ε . If such a sequence were to exist, it would then follow that 0<γ≤T 1 |ζ ′ (ρ)| 2k = Ω T 2k/3−ε and the lower bound in (1.4) would be significantly weaker than this Ω-result when k > 3 2 .

Proof of Theorem
The method we use to prove our theorem is based on a recent idea of Rudnick and Soundararajan [18]. Let where µ is the Möbius function. Assuming the Riemann Hypothesis, for any non-trivial zero . From this observation and Cauchy's inequality it follows that Our Theorem is a consequence of the following proposition.
Proposition. Assume the Riemann Hypothesis and let 0 < ϑ < 1 be fixed. Then If we further assume that the zeros of ζ(s) are all simple, then there exists a sequence T := {τ n } ∞ n=3 such that n < τ n ≤ n + 1 and for T ∈ T we have We now deduce our theorem from the above proposition.
Proof of the Theorem. Let T ≥ 4 and choose τ n to satisfy T − 1 ≤ τ n < T . Combining (2.2), (2.4), and (2.3) we see that under the assumption of the Riemann Hypothesis and the simplicity of the zeros of ζ(s). From (2.5), our theorem follows by letting ϑ → 1 − .
We could have just as easily estimated the sums M 1 and M 2 using a Dirichlet polynomial n≤ξ a n n −s for a large class of coefficients a n in place of M ξ (s). In the special case where a n = µ(n)P log ξ/n log ξ for polynomials P , we can show that the choice P = 1 is optimal in the sense that it leads to largest lower bound in (1.3). We prove the above proposition in the next two sections; the sum M 1 is estimated in section 3 and the sum M 2 is estimated in section 4. The evaluation of sums like M 1 dates back to Ingham's [11] important work on M(x) in which he considered sums of the form is a Dirichlet polynomial with ξ ≤ T . Such sums have played an important role in various applications. For instance, results concerning the distribution of consecutive zeros of ζ(s) and discrete mean values of the zeta-function and its derivatives are proven in [1,2,3,6,12,16,19]. In each of these articles, the evaluation of the discrete mean (2.6) either makes use of the Guinand-Weil explicit formula or of Gonek's uniform version [6] of Landau's formula is an explicit error function uniform in x and T . A novel aspect of our approach is that it does not require the use of the Guinand-Weil explicit formula or of the Landau-Gonek explicit formula (2.7). Instead we evaluate M 2 using the residue theorem and a version of Montgomery and Vaughan's mean value theorem for Dirichlet polynomials [14]. Our approach is simpler and it is likely that it can be extended to evaluate the discrete mean (2.6) for a large class of coefficients a n with ξ ≤ T .

The estimation of M 1
To estimate M 1 , we require the following version of Montgomery and Vaughan's mean value theorem for Dirichlet polynomials.
Lemma. Let {a n } and {b n } be two sequences of complex numbers. For any real number Proof. This is Lemma 1 of Tsang [22]. The special case where b n = a n , is originally due to Montgomery and Vaughan [14]. It turns out, as shown by Tsang, that this special case is equivalent to the more general case stated in the lemma.
Let T ≥ 4 and set c = 1 + (log T ) −1 . It is well known (see Theorem 14.16 of Titchmarsh [21]) that assuming the Riemann Hypothesis there exists a sequence T = {τ n } ∞ n=3 , n < τ n ≤ n + 1, and a fixed constant A > 0 such that (3.2) ζ(σ+iτ n ) −1 ≪ exp A log τ n log log τ n uniformly for 1 2 ≤ σ ≤ 2. We now prove the estimate (2.4) assuming that T ∈ T. Recall that |γ| > 1 for every non-trivial zero ρ = 1 2 + iγ of ζ(s). Thus, assuming that all the zeros of ζ(s) are simple, the residue theorem implies that say. Here we are using the fact that the residue of the function 1/ζ(s) at s = ρ equals 1/ζ ′ (ρ) if ρ is a simple zero of ζ(s). 4 The main contribution to M 1 comes from the integral I 1 ; the remainder of the integrals contribute an error term. Observe that By (3.1) with a m = µ(m)m −c and b n = µ(n)n −1+c it follows that we conclude that To estimate the contribution from the integral I 2 , we recall the functional equation for the Riemann zeta-function which says that Stirling's asymptotic formula for the Gamma-function can be used to show that uniformly for −1 ≤ σ ≤ 2 and |t| ≥ 1. Combining this estimate and (3.2), it follows that, for T ∈ T, In addition, we have the trivial bound Thus, estimating the integral I 2 trivially, we find that To bound the contribution from the integral I 3 , we notice that the functional equation for ζ(s) combined with the estimate in (3.5) implies that, for 1 ≤ |t| ≤ T , It therefore follows that Finally, since 1/ζ(s) and M ξ (1−s) are bounded on the interval [1 − c + i, c + i], we find that I 4 ≪ 1. Hence, our combined estimates for I 1 , I 2 , I 3 , and I 4 imply that From this and (2.1), the estimate in (2.4) follows.

The estimation of M 2
We now turn our attention to estimating the sum M 2 . As before, let T ≥ 4 and c = 1 + (log T ) −1 . Assuming the Riemann Hypothesis, we notice that Therefore, by the residue theorem, we see that  In each interval of length one such a T exists. This well-known argument may be found in [4], page 108. Applying (3.6) we find