Negative discrete moments of the derivative of the Riemann zeta-function

We obtain conditional upper bounds for negative discrete moments of the derivative of the Riemann zeta-function averaged over a subfamily of zeros of the zeta function which is expected to have full density inside the set of all zeros. For $k\leq 1/2$, our bounds for the $2k$-th moments are expected to be almost optimal. Assuming a conjecture about the maximum size of the argument of the zeta function on the critical line, we obtain upper bounds for these negative moments of the same strength while summing over a larger subfamily of zeta zeros.


Introduction
Let ζ(s) denote the Riemann zeta-function.In this paper, we study negative discrete moments of the derivative ζ(s) of the form where k ∈ R and ρ = β + iγ runs over the non-trivial zeros of ζ(s).We will be interested in the case k > 0, where we note that J −k (T ) is only defined if the zeros ρ are all simple.
Gonek [8] and Hejhal [12] independently conjectured that as T → ∞ and for every real number k.Using a random matrix theory model for ζ(s), Hughes, Keating, and O'Connell [13] have refined this conjecture to predict that for a specific constant C k , when k ∈ C and ℜ(k) < 3/2.This conjecture was recovered by Bui, Gonek, and Milinovich [3] using a hybrid Euler-Hadamard product for ζ(s).The random matrix model for ζ(s), as well as unpublished work of Gonek, suggests that there are infinitely many zeros ρ such that |ζ ′ (ρ)| −1 ≫ |γ| 1/3−ε , for any ε > 0. This, in turn, would imply that for k > 0, J −k (T ) = Ω(T 2k/3−ε ), suggesting that (1.1)only holds for k < 3/2.However, it is noted in [13] that while (1.1) is not expected to hold for k ≥ 3/2, it is plausible that (1.1) could hold when summing over a subfamily of zeros of full density inside the set of zeros with ordinates between T and 2T .Indeed, if one were to redefine J −k (T ) to exclude a set of rare points where |ζ ′ (ρ)| is very close to 0, then (1.1) should still predict the behavior of the redefined sum J −k (T ).In Theorems 1.1 and 1.2, below, we make the first progress towards this idea, by obtaining upper bounds for J −k (T ) over certain subfamilies of the nontrivial zeros of ζ(s) that expected to have full density inside the set of all nontrivial zeros of the zeta function.
While asymptotic formulas are known only in the cases mentioned above, progress has been made towards upper and lower bounds in certain cases.When k < 0 (i.e., for positive discrete moments), Ng and Milinovich [18] obtained sharp lower bounds consistent with (1.1) for integer k under the Generalized Riemann Hypothesis (GRH) for Dirichlet L-functions.Milinovich [16] obtained almost sharp upper bounds for positive discrete moments when k is an integer under the RH, and Kirila [14] extended this result to all k < 0, obtaining sharp upper bounds of the order conjectured in (1.1).These results settle, on GRH, conjecture (1.1) for negative integers k.
When k > 0 (i.e., in the case of negative discrete moments), less is known.Heap, Li, and J. Zhao [11] obtained lower bounds of the size in (1.1) for all fractional k under RH and the assumption of simple zeros, while recently Gao and L. Zhao [6] extended this result to all k > 0 under the same assumptions.As mentioned above, these lower bounds are expected to be sharp for k < 3/2, but not for k ≥ 3/2.We expect F to have full density inside the set of zeros in (T, 2T ].Indeed, for any fixed β > 0, Montgomery's Pair Correlation Conjecture [19] implies that if γ + denotes the next largest ordinate of a nontrivial zero of ζ(s) above γ, then where N (T ) denotes the number of nontrivial zeros of ζ(s) with ordinates γ ∈ (0, T ].Therefore, the set of excluded zeros whose ordinates do not belong to the family F conjecturally has measure 0. We will prove the following theorem.
Theorem 1.1.Assume the Riemann hypothesis and let ε > 0.Then, for any δ > 0, we have that 2) The proof of Theorem , for any other ordinate γ ′ . (1.4) Then we will prove the following theorem.
Theorem 1.2.Let ε > 0, and assume both the Riemann hypothesis and the conjecture in (1.3).Then, for any δ > 0, we have that Remark.Note that if we instead assume a conjectural bound for S(t) of the form for some function h(t) such that h(t) = o( log t log log t ), then we can define a family of zeros , for any other ordinate γ ′ , where, as before, f (t) is any function slowly going to infinity.Then, modifying our proof of Theorems 1.1 and 1.2 in a straightforward manner, we can similarly prove that (1.2) holds when summing over the zeros with γ ∈ F ′ .1.2.Bounds for J −k (T ) assuming the Weak Mertens Conjecture.If one were to sum over the full family of zeros of ζ(s), the problem of obtaining upper bounds for the negative discrete moments becomes more delicate.In order to obtain an upper bound, it seems that simply assuming RH and the simplicity of zeros is not enough.Indeed, from equation (2.1) in the proof of Lemma 2.1, one sees that log |ζ ′ (ρ)| is closely connected to a sum over the zeros ρ ′ = ρ, and in order to control log |ζ ′ (ρ)| we need to understand the small differences γ ′ − γ.Hence, one would need to assume a lower bound on the size of the smallest difference between consecutive zeros of ζ(s).
It is possible to obtain upper bounds on the negative discrete moments J −k (T ) when summing over the full family of zeros of ζ(s) assuming a well-known conjecture for the distribution of partial sums of the Möbius function.Defining M (x) = n≤x µ(n), the Weak Mertens Conjecture (WMC) states that Titchmarsh [22,Chapter 14] has shown that the WMC implies RH, that the zeros of ζ(s) are all simple, the fact that sum over all nontrivial zeros of ζ(s) is convergent.From these facts, we now observe that WMC implies that Though these estimates apply to sums over all zeros, we note that both of these bounds are weaker than the bounds obtained in Theorem 1.1 for zeros restricted to the set F.
If k ≥ 1, then using Cauchy's inequality and WMC, we have as T → ∞, where we used the fact that the first sum on the right-hand side of the inequality in the first line is o(1), which follows from (1.5) and the fact that k ≥ 1.This implies that J −k (T ) = o T 2k for k ≥ 1.On the other hand, if k ∈ (0, 1), then using Hölder's inequality, we have that as T → ∞, which follows by using the bound J −1 (T ) = o(T 2 ) derived in (1.6) and the fact that N (2T ) − N (T ) ≪ T log T .1.3.Overview of the paper.To prove Theorems 1.1 and 1.2, we start by relating log |ζ ′ (ρ)| to log |ζ(ρ + 1/ log T )| and a sum over the zeros ρ ′ = ρ, which we then bound using the lower bound on the difference between consecutive zeros provided by the families F and F enl .The problem of bounding the discrete negative moments of the derivative of ζ(s) then reduces to bounding negative discrete shifted moments of ζ(s).To do that, we build on work of the first two authors [1] on upper bounds for the negative continuous moments of ζ(s), which in turn builds on work of Harper [10] and Soundararajan [23] on upper bounds for the positive moments of ζ(s) on the critical line.Our analysis uses a discrete mean-value theorem for the mean-square of Dirichlet polynomials averaged over zeros of ζ(s), similar in flavor to the Montgomery-Vaughan's mean-value theorem for continuous averages Dirichlet polynomials [20,Corollary 3].The proof of this discrete mean-value estimate relies on the Landau-Gonek explicit formula [9].
In Section 2, we relate log |ζ ′ (ρ)| to log |ζ(ρ + 1/ log T )| building upon previous observations of Kirila [14].Then we gather a few facts that we will need from [1] and state the main propositions which evaluate moments of certain Dirichlet polynomials.In Section 3.1, we prove our discrete mean-value theorem for Dirichlet polynomials averaged over zeros of ζ(s), and in Section 4 we prove the main propositions stated in Section 2. We finally prove Theorems 1.1 and 1.2 in Section 5.
Lemma 2.1.Assume RH.For γ ∈ F, we have that Proof.Equation 3.3 in Kirila [14] states that for any σ > 1/2.We will obtain an upper bound for the sum over ρ ′ .Let σ − 1/2 = α.We write Let M 1 (γ) denote the first term on the right-hand side, and let M 2 (γ) denote the second.We pick We first bound M 1 (γ).We have where we used the fact that γ ∈ F, and that Standard estimates imply that Hence for γ ∈ F, using (2.2) and (2.3), we have that This completes the proof of the lemma.
We similarly obtain the following estimate.
Lemma 2.2.For γ ∈ F enl , we have that where f is the function appearing in (1.4).
Proof.Upon noticing that the proof is very similar to the proof of Lemma 2.1.
Throughout the paper, we rely on the following lemma proved in [1, Lemma 2.1 and (3.2)].
Lemma 2.3.Assume RH.Then we have that where the coefficients b(p; ∆) can be written down explicitly, and Moreover, the coefficients b(p; ∆) satisfy the bound where, for any ε > 0, (2.5) Now we introduce some of the notation in [1].Let for a sequence β 0 , . . ., β K to be chosen later such that β j+1 = rβ j , for some r > 1.Also, let ℓ j be even parameters which we will also choose later on.Let s j be integers.For now, we can think of s j β j ≍ 1, and K h=0 ℓ h β h ≪ 1.We let T β j = e 2π∆ j for every 0 ≤ j ≤ K and let and we extend b(p; ∆) to a completely multiplicative function in the first variable.Let Denote the set of γ such that γ ∈ T u for all u ≤ K by T ′ .For 0 ≤ j ≤ K − 1, let S j denote the subset of γ ∈ (T, 2T ] such that γ ∈ T h for all h ≤ j, but γ / ∈ T j+1 .Now for ℓ an integer, and for z ∈ C, let If z ∈ R, then for ℓ even, E ℓ (z) > 0. We will often use the fact that for ℓ an even integer and z a complex number such that |z| ≤ ℓ/e 2 , we have Let ν(n) be the multiplicative function given by for p a prime and any integer j.We will frequently use the following fact: For any interval I, s ∈ N and a(n) a completely multiplicative function, we have where Ω(n) denotes the number of prime factors of n, counting multiplicity.We now state the following result, which is similar to [1, Lemma 3.1].
We also have the following propositions, whose proofs we postpone to Section 4.

A mean-value theorem
We now prove the following mean-value theorem, which is of independent interest.Theorem 3.1.Assume RH.Then for x ≥ 2 and T > 1, we have where {a n } is any sequence of complex numbers.

The proof relies on following version of the
Applying Lemma 3.2 with y = m/n, we have say.In Σ 1 , the only nonzero terms occur if n divides m (and m/n is a prime power).Thus, making the substitution m = nk, it follows that Next we estimate Σ 2 .Observe that for any ∆ > 0, we have This implies that upon choosing ∆ = (log x) 1/2 .From this calculation, we see that We next estimate Σ 3 .Again using (3.1), for any ∆ > 0, we have say.Write m = qn + r with − n 2 < r ≤ n 2 .Then m/n = q + r n = |r| n if q is a prime power, and m/n ≥ 1 2 if q is not a prime power or r = 0. Then

Now again writing
if q is a prime power, and that log(m/n) n m/n ≪ log m n if q is not a prime power or r = 0.The number of prime powers dividing n equals Ω(n).Since Ω(n) ≪ log n, it follows that Choosing ∆ = log x in (3.2), we see that Finally, we estimate Σ 4 .Note that if m > n, then log(m/n) = − log 1 − m−n m > m−n m , and it therefore follows that Combining estimates, the theorem follows.
Proof of Proposition 2.5.Using (2.6), we have Then it follows that Using Theorem 3.1 twice for 0 < γ ≤ 2T and 0 < γ ≤ T and then differencing, we have that In the second sum above, note that we must have k = 1, in which case the term vanishes.Then it follows that where we have used the fact that N (2T ) ≪ N (T ).Now using the condition that β 0 s 0 ≤ 1 − log log T / log T and the bound ν(n) 2 ≤ ν(n), we get that where we use the bound (2.4) to derive the estimate in the second line.This completes the proof.
By (2.4) and the work in [1] we get that Now we use the fact that and with the choice (5.1) and (5.6), we get that We also have that , which shows (4.3) in the case m = 0.
and again The two bounds above establish (4.3) when b(∆ j )(log log T ∆ j ) η(∆ j ) k < 1.On the other hand, if b(∆ j )(log log T ∆ j ) η(∆ j ) k ≥ 1, then we have that and then , which follows by rewriting n m → n m p a and noting that p a/ log T ≪ 1. Hence (4.3) also follows in this case.Now from (4.2) and (4.3), it follows that γ∈(T,2T ] . Now using the work in [1] to deal with the product over h ≤ j, we get that .
This completes the proof of the proposition.
The proof of Proposition 2.7 is very similar to the proof of Proposition 2.6, so we leave the details to the interested reader.

Proof of Theorems 1.1 and 1.2
We now prove Theorem 1.1.The proof of Theorem 1.2 will follow in exactly the same way.
Proof of Theorem 1.1.If 2k(1 + ε) ≤ 1, we choose and where we can choose We choose K such that β K ≤ c, (5.4) for c a small constant such that The conditions above ensure that for any j < K, and that If 2k(1 + ε) > 1, we choose the parameters as follows.Let where we pick We choose β K as in (5.4) and (5.5).
We have that Using Lemma 2.1, we have that for some 0 where we used the pointwise bound which is proven in [1, Lemma 2.2].Using Proposition 2.5, we have that the above is Now similarly as in [1], using Stirling's formula, we have that the above is If 2k(1 + ε) ≤ 1 then using the choice of parameters (5.1) and the fact that η(∆ 0 ) = 1, we get that We first consider the contribution from those j for which ∆ j = o(log T ), i.e., those j for which β j → 0, in which case η(∆ j ) = 1.Let R 1 denote this contribution.Note that the first term inside the exponential is negative, so we have We now consider the contribution from those j in (5.13) for which η(∆ j ) = 0, i.e., those j for which β j ≫ 1.It is easy to see that in this case, we have R 2 ≪ N (T )(log T ) O (1) . (5.15)

1. 1 .
Main results.No upper bounds are known for J −k (T ) in the case when k > 0. The main results of this paper are to obtain upper bounds for the negative discrete moments when summing over two different subfamilies of the nontrivial zeros of ζ(s), both of which are expected to have full density inside the set of all zeros.Our first result assumes RH, while our second result assumes RH and a conjectural upper bound on |S(t)|, where S(t) = π −1 arg ζ(1/2 + it) and the argument is obtained by continuous variation along the straight line segments joining the points 2, 2 + it, and 1/2 + it starting with the value arg ζ(2) = 0. Let F = γ ∈ (T, 2T ] : |γ − γ ′ | ≫ 1 log T , for any other ordinate γ ′ .