GAPS BETWEEN ZEROS OF ζ(s) AND THE DISTRIBUTION OF ZEROS OF ζ (s)

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1 GAPS BETWEEN ZEROS OF ζs AND THE DISTRIBUTION OF ZEROS OF ζ s MAKSYM RADZIWI L L Abstract. We settle a conjecture of Farmer and Ki in a stronger form. Roughly speaking we show that there is a positive proportion of small gaps between consecutive zeros of the zeta-function ζs if and only if there is a positive proportion of zeros of ζ s lying very closely to the half-line. Our work has applications to the Siegel zero problem. We provide a criterion for the non-existence of the Siegel zero, solely in terms of the distribution of the zeros of ζ s. Finally on the Riemann Hypothesis and the Pair Correlation Conjecture we obtain near optimal bounds for the number of zeros of ζ s lying very closely to the halfline. Such bounds are relevant to a deeper understanding of Levinson s method, allowing us to place one-third of the zeros of the Riemann zeta-function on the half-line. 1. Introduction. The inter-relation between the horizontal distribution of zeros of ζs denoted ρ = β +iγ and the horizontal distribution of the zeros of ζ s denoted ρ = β + iγ is the basis of Levinson s method [1] allowing us to place one third of the zeros of ζs on the critical line. Recently it has been understood that the horizontal distribution of the zeros of ζ s is also related to the vertical distribution of zeros of ζs. As an first attempt at capturing such a relationship we have the following conjecture of Soundararajan [16]. Note: Throughout we assume the Riemann Hypothesis. We recall that β 1 for all non-trivial zeros of ζ s see [18] and that this is equivalent to the Riemann Hypothesis. Conjecture 1 Soundararajan [16]. We have A lim inf γ γ+ γ log γ = 0 with γ + the ordinate suceeding γ, if and only if B lim inf γ β 1 log γ = 0 Zhang [19] shows that A = B see also [8] for a partial converse. Ki [11] obtained a necessary and sufficient condition for the negation of B. Ki s result shows that zeros ρ with β 1 log γ = o1 arise not only from small gaps between zeros of ζs but also, for example, from clusters of regularly spaced zeros of ζs. Therefore given our current knowledge about the zeros of ζs it is possible for B and the negation of A to co-exist. The assertion A is arithmetically very interesting, since, following an idea of Montgomery made explicit by Conrey and Iwaniec in [] if there are many small gaps between consecutive zeros of ζs then the class number of Q d is large and there are no Siegel zeros Mathematics Subject Classification. Primary: 11M06, Secondary: 11M6. The author is partially supported by a NSERC PGS-D award. 1

2 A more recent attempt at capturing the relation between the distribution of zeros of ζs and ζ s is due to Farmer and Ki [4]. Let wx be the indicator function of the unit interval. Following Farmer and Ki we introduce two distribution functions, β 1 m ε := lim inf T mε := lim inf T π T π T T γ T T γ T w ε γ + γ w ε These are indeed distribution functions, since in a rectangle of length T, both ζs and ζ s have asymptotically NT T/π zeros see [1], and it is conjectured that m v 1 as v, whereas it is known that mv 1 as v see [16], [7]. Zhang shows in [19] that if mε > 0 for all ε > 0, then m ε > 0. An analogue of Soundararajan s conjecture would assert that mε > 0 for all ε > 0 if and only if m ε > 0 for all ε > 0. As explained by Farmer and Ki in [4] if for example the zeros are wellspaced with sporadic large gaps, something we cannot rule out at present, then in principle m ε > 0 is not enough to imply mε > 0. Farmer and Ki propose the following alternative conjecture. Conjecture Farmer and Ki [4]. If m ε ε v with a v < as ε 0 then mε > 0 for all ε > 0. This is a realistic conjecture since we expect that m ε 8/9πε 3/ as ε 0 see [3]. Farmer and Ki comment we intend this as a general conjecture, applying to the Riemann zeta-function but also to other cases such as a sequence of polynomials with all zeros on the unit circle and that stronger statements should be true for the zeta function. Our main result is a proof of Conjecture in a stronger and quantitative form for the Riemann zeta-function. Main Theorem. Let A, δ > 0 be given. If m ε ε A as ε 0 then mε 1/ ε A+δ for all ε 1. If mε 1/ ε A as ε 0 then m ε ε A+δ for all ε 1. We conjecture that m ε mε 1/ provided that one of mε or m ε is ε A for some A > 0. This is consistent with the expectation that mε π/6ε 3 and m ε 8/9πε 3/ as ε 0 see [3]. Our Main Theorem could be restated as saying that log mε log m ε 1/ as ε 0 provided that one of mε or m ε is greater than ε A. As a consequence of the Main Theorem we obtain estimates for m ε assuming the Pair Correlation Conjecture. Corollary 1. Assume the Pair Correlation Conjecture. Let δ > 0. Then as ε 0. ε 3/+δ m ε ε 3/ δ An assumption on the zero distribution in Corollary 1 is inevitable, since m ε 0 implies that almost all the zeros of ζs are simple. Corollary 1 allows one to quantify the loss in Levinson s method coming from the zeros of ζ s lying closely to the half-line. Unfortunately.

3 Corollary 1 is a conditional result, and as such it cannot be used to put a greater proportion of the zeros of ζs on the half-line see [5] for related work. A final consequence of our work is a criterion for the non-existence of the Siegel zero in terms of the zeros of ζ s. We state it only for completeness since a stronger result has been obtained by Farmer and Ki [4]. Corollary. Let A > 0. If m ε ε A, for all ε > 0, then for primitive characters χ modulo q, L1; χ > log q 18. for all q sufficiently large. Proof. If m ε ε A for every ε > 0 then m1/4 > 0 by our Main Theorem, hence L1; χ > log q 18 for all q sufficiently large by Theorem 1.1 of Conrey-Iwaniec, []. With some care it is possible to turn the above Corollary into an effective result. By Dirichlet s formula Corollary also implies that the class number of Q d is at least as large as c dlog d 18 with c constant. Farmer and Ki show that if m ε exp ε 1/+δ as ε 0, for some δ > 0, then there are NT / log ordinates of zeros of ζs lying in [T ; T ] and such that γ + γ log γ = o1 as T. Using the result of Conrey and Iwaniec [] this is enough to rule out the existence of Siegel zeros. It is an interesting question to determine whether, given the current technology, one can increase the exponent 1 in Farmer and Ki s assumption m ε exp ε 1/+δ and still guarantee the non-existence of Siegel zeros.. Main ideas The first part of our Main Theorem follows from the stronger Theorem 1 below. Theorem 1. Let A, δ > 0. There is a constant C = Cδ, A such that if 0 < ε < C and m ε cε A then mε 1/ δ c/8ε A. The approximate value of Cδ, A is Bδ/A 3A/δ with B an absolute constant. Theorem 1 follows from two technical Propositions which we now describe. Given a zero ρ = β + iγ of ζ s we denote by ρ c = 1 + iγ c the zero of ζs lying closest to ρ. If there are two choices of ρ c then we pick the one lying closer to the origin. For any ordinate γ of a zero of ζs we denote by γ + the ordinate suceeding γ and by γ the ordinate preceeding γ. We denote by γ ± the ordinate closest to γ. Theorem 1 follows quickly from the following Proposition. Proposition 1. Let 0 < δ, ε < 1. Let S ε,δ T be a set of zeros ρ = β + iγ of ζ s such that T γ T, β 1 ε/ and γ c γ ± c > ε 1/ δ / There is a C = Cδ, A such that if 0 < ε < C then S ε,δ T ε A T. The proof of Proposition 1 rests on a Proposition describing the structure the roots of ζ s lying close to the half-line. The Proposition which we are about to state complements with a corresponding upper bound the classical lower bound, ρ β 1 ρ c 3

4 valid for all ρ = β + iγ see [16]. It might be of independent interest. Proposition. Let 0 < δ < 1, 0 < ε < c with c > 0 an absolute constant. Let T be large and Z := Z ε,δ T be a set of δ/ well-spaced ordinates of zeros ρ such that ρ ρ, β 1 ε/ and T γ T. If Z ε,δ T ε A T then, for any given κ > 0, all but κ Z elements ρ Z satisfy the inequality, β 1 ρ ρ c β 1 A logεκδ 1. The proof of the converse part of our Main Theorem builds on ideas of Zhang, and follows from the following more precise statement valid for any fixed ε > 0. Theorem. Let A, δ > 0. There is a C = Cδ, A such that if 0 < ε < C and mε 1/ cε A then m ε c/4ε A+δ. The paper is organized as follows. Most of the paper, all the way until section 7, is devoted to the proof of the propositions above and the deduction of Theorem 1 from them. Following section 7 we prove Theorem and Corollary 1. with Define, 3. Lemma on Dirichlet polynomials W N n = The lemma below is due to Selberg. A N s := n N ΛnW N n n s { 1 for 1 n N 1/ logn/n/ for N 1/ < n N Lemma 1. Let σ = 1 + /, with N T. Then for T t T, Proof. This is equation. in [15]. ρ σ 1 σ 1 + t γ A Ns + Using the explicit formula we obtain an upper bound for the number of zeros in a small window [t πk/ log t; t + πk/ log t], in terms of the Dirichlet polynomial B N s := Λn 1 log n. n s n N We have the following lemma. Lemma. For T t T and N T, N t + π π N t + B N 1 + it 4

5 Proof. Let sin π v F v = π v be the Fejer kernel. The Fourier transform of F v is for x < F x := F te πitx dx = 1 1 x and F x = 0 for x >. By the explicit formula see Lemma 1 in [9], 3 F γ t = Oe π / t + 1/ + 1 F u t R Γ 1 π γ Γ 4 + iu 1 Λn log n F π n π n= du + F log n π The integral over u is bounded by log t/. On the other hand the prime sum is bounded by, 1 Λn π n 1 log n 1/+it π n e π Finally π/ γ F γ t is an upper bound for the number of zeros in the interval going from t 1/ to t + 1/. If T t T we choose π = and we are done. In order to understand the average behavior of the Dirichlet polynomials A N s and B N s we use a version of the large sieve. Lemma 3. Let As be a Dirichlet polynomial with positive coefficients and of length x. Let s r = σ r + it r be points with T t r T and 0 σ r 1 ε/ for some small ε > 0. Suppose that t i t j δ/ for i j, with 100ε < δ < 1. Then, for x k T, As r k s r 0 δ T T A 1 + it k dt Proof. Let Ds = As k. For any s we have, with C a circle of radius δ/ around s, Ds 4 Dx + iy dxdy πδ Summing over all s = s r, since the circles are disjoint we obtain, s r Ds r 4 πδ C 1 +δ/ 1 δ/ T +1 T 1 Dσ + it dtdσ Since the coefficients of Ds are positive, and D is of length at most T, by a majorant principle see Chapter 3, Theorem 3 in [13], the inner integral is bounded by 3e δ T T D 1 + it dt Since in addition δ < 1, the claim follows we obtain a constant of 8e /π < 0. 5

6 Combining the above lemma with Chebyschev s inequality allows us to understand the average size of the Dirichlet polynomials A N s and B N s. Lemma 4. Let s r = σ r + it r be a set of well-spaced points as appearing in the statement of Lemma 3. Suppose that N k T/. The number of points s r for which we have is bounded above by e k /δt. A N s r > k/e or B N s r > k/e Proof. Let L N s be either A N s or B N s. Let D N s = n N Λn n s. By a majorant principle see Chapter 3, Theorem 3 in [13] we have, T T L N 1 + it k dt 3 T T D N 1 + it k dt By Soundararajan s lemma 3 in [17], for N k T/ we have, T T D N 1 + it k dt k!t k Therefore, for N k T/, by the previous lemma, LN s r k k! δ T k It follows that for N k T/, the number of points s r for which L N s r > B is less than, k k T/δ B Choosing B = k/e we conclude that the number of points for which L N s r > k/e is bounded by e k /δt as desired. Lemma 5. Let 0 < c < 1. Uniformly in T t T and N T, ζ ζ s = 1 s ρ + O E T,N s c where s ρ <c/ E T,N s := 1 A N s + B N 1 + it +. Furthermore, if s r is a set of well-spaced points as in Lemma 3, and N k T/, then s r E T,N s r k k k /δt. 6

7 Proof. Selberg shows in [14] see equation 14 on page 4 that ζ ζ s = 1 s ρ + O A Ns + s ρ < 1 It suffices to notice that the contribution of the zeros ρ with c 1 < s ρ < 1 is bounded above by N t + π π N t c c + 1 B N 1 + it Combining the above two equations we obtain the first part of the lemma. Now it remains to estimate the moments of E T,N. We have, k C E T,N s r k A N s r k + B N s r k + Ck k /δt s r s r with C > 0 an absolute constant. Using Lemma 3 and proceeding as in Lemma 4 we find that the k-th moments of the Dirichlet polynomials A N and B N is bounded above by k!/δt k. Hence we conclude that the k-th moment of E N,T is bounded above by Ck k /δt 4. Proof of Proposition The proof of Proposition rests on the following classical lemma. Lemma 6. If ρ ρ then, 1 log γ = ρ Proof. See Zhang [19], Lemma 3. β 1 β 1 + γ γ + O1. We will show that on average the zero ρ = ρ c dominates, the claim then follows shortly. In order to simplify the notation we define, as in the previous section, A N s := n N s r ΛnW N n n s with W N n the same smoothing as defined in the previous section. We also define B N s := Λn 1 log n. n s n N On average both Dirichlet polynomials are of size. Proof of Proposition. Let N T to be fixed later. In the formula 4 1 log γ = ρ β 1 β 1 + γ γ + O1 7

8 The contribution of the ρ s for which γ γ < π 1 is bounded above by 5 N γ + π N γ π β 1 ρ c ρ + 1 B N 1 + iγ β 1 ρ c ρ. by Lemma. On the other hand, to bound the contribution of the ρ s for which γ γ > π 1 we notice that if γ γ > π 1 then β 1 + γ γ / + γ γ. Therefore the contribution of the ρ s with γ γ > π 1 to 4 is bounded above by 6 β 1 / / + γ γ ρ β log 1 T + 1 A N iγ by Lemma 1. Combining 4, 5 and 6 we conclude that B N 1 + β 1 iγ ρ c ρ + + β log 1 T + A N iγ Suppose that N k T/ with a k to be fixed later and N the largest integer such that N k < T/. By Lemma 4 the number of ρ Z ε,δ for which B N 1 + iγ > k/e is bounded above by ce k /δt with c a constant. Similarly the number of ρ Z ε,δ for which A N 1 + 1/ + iγ > k/e is also bounded by above by ce k /δt. Choose k so that ce k /δt κ/ Z ε,δ. Since Z ε,δ c 1 ε A T we can take k to be the closest integer to c A logκεδ 1 with c an absolute constant. Choose N to be the largest integer such that N k T/. With this choice of k and N it follows that for at most κ Z ε,δ elements ρ Z ε,δ we have B N 1 + iγ k/e or A N 1 + iγ k/e. It follows that for all but at most κ Z ε,δ of the ρ Z ε,δ we have, c k β 1 ρ c ρ + 1 k β 1 with c > 0 an absolute constant. If ε is choosen so that ε < c/ then since β 1 we obtain < ε/ c/ k β 1 ρ c ρ hence ρ c ρ kβ 1/ 1/ which gives the desired bound for all but at most κ Z ε,δ elements ρ Z ε,δ. Recall that k A logεδκ 1. 8

9 5. Proof of Proposition 1 The lemma below is critical, in that it allows us to produce a sufficiently dense well-spaced sequence of zeros of ζ s. Lemma 7 Soundararajan [16]. Suppose that ρ 1 = 1 + iγ 1 and ρ = 1 + iγ are two consecutive zeros of ζs with T γ 1 < γ T for large T. Then the box, {s = σ + it : 1 σ < 1 + 1/, γ 1 < t < γ } contains at most one zero counted with multiplicity of ζ s. Proof. The only way that ρ can lie on the critical line is if ρ = ρ. Since γ 1 < t < γ this possibility is excluded. As for the box 1 < σ < 1 + 1/ we know by Soundararajan s work [16] see Proposition 6 that the box 1 < σ < 1 + 1/ with t in [γ 1, γ ] can contain at most on zero of ζ s, counted with multiplicity. We are now ready to prove Proposition 1. Proof of Proposition 1. Suppose that S = S ε,δ T > ε A T. We will show that this leads to a contradiction when 0 < ε < Cδ, A with Cδ, A some explicit constant depending only on δ and A for example we could take Cδ, A = cδ/a 3A/δ with c > 0 an absolute constant. Since each ρ S satisfies γc γ γ c + and γ c γ c ± > ε 1/ δ / by the above lemma for each ρ S there is at most one zero of ζ s in [γc, γ c ] and at most one zero of ζ s in [γ c, γ c + ]. We construct a subset S of S by skipping every second element in S. This produces a subset of at least 1/ S elements, with the property that the ordinates of elements of S are ε 1/ δ / well-spaced, because γ c γ c ± ε 1/ δ / for each ρ S. By Proposition, we have for at least half of the ρ S, 8 γ γ c ρ ρ c C Aε logε 1. with C > 0 an absolute constant. We call S the subset of S satisfying the above inequality. Since γ c ± γ c > ε 1/ δ / for each ρ S the interval γ t ε 1/ δ / contains exactly one ordinate of a zero of ζs namely γ c once ε is choosen so small so as to make the right-hand side of 8 less than ε 1/ δ / for example ε < δ/ca /δ would suffice. Using Lemma 5, we have at s = ρ S, 1 s ρ E T,N s c s ρ <c/ Choose s = ρ S, c = ε 1/ δ and N the largest integer such that N k T/ with a k to be fixed later ultimately k = A + 1/δ. By our previous remark the left-hand side of the above expression consists of only one term ρ ρ c 1. Raising the above expression to the k-th power and then summing over all ρ S we obtain 1 9 ρ ρ ρ S c k ε k+kδ C k E T,N ρ k ρ S 10 ε k+kδ Ck k /ε 1/ δ T k+1 9

10 by Lemma 5, with C > 0 an absolute constant not necessarily the same in each occurence. Since for each ρ S we have, Aε logε ρ 1 ρ c the left-hand side of 9 is at least 1 11 ρ ρ ρ S c k S C/A k ε k logε 1 k k 1 C/A k ε A k logε 1 k T k+1 since S ε A T. Combining the upper bound 9 and the lower bound 11 we get ε A k logε 1 k ε k 1/+k+1δ CAk k with C > 0 an absolute constant. The above inequality simplifies to Using the inequality log x x δ /δ we obtain ε A+1/ CAk k ε k+1δ logε 1 k. ε A+1/ CAk/δ k ε kδ Choosing k to be the smallest integers with kδ > A + 1 we obtain a contradiction once ε < Cδ /A 16A/δ with C an absolute constant. Note: We have certainly not tried to optimize the constant Cδ, A. 6. Proof of Theorem 1. Let T be large. By assumption each interval [T ; T ] contains at least cε A NT ordinates T γ T with β 1 < ε/. If ρ = ρ for more than half of these zeros of ζ s, then we have c/ε A NT zeros ρ with γ + = γ and so we are done. Thus we can assume that there are c/ε A NT zeros ρ with T γ T, ρ ρ and β 1 < ε/. We call the set of such ρ by S. By Lemma 7 between any two consecutive zeros of ζs there is at most one ρ S. For each ρ S consider two possibilities 1 γ ± c γ c ε 1/ δ / γ ± c γ c > ε 1/ δ / Call S the subset of S for which the second possibility holds. If the second possibility holds for at least one half of the elements in S then S c/ε A T. But this is impossible by Proposition 1 once ε is less than c/4cδ, A + 1, with Cδ, A as in the statement of Proposition 1. Therefore the second possibility can hold for at most one half of the elements in S. Hence the first possibility holds for at least a half of the elements in S. Call S 1 the subset of S for which the first possibility holds. By Lemma 7, there are no two ρ S 1 lying between the same tuple of consecutive zeros of ζs. Every ρ S 1 lies either between [γ c, γ c ] or [γ c ; γ + c ] and moreover one of these intervals is of length ε 1/ δ / log γ c. Skipping every second ρ S 1 we make sure that no two ρ 1 S 1 and ρ S 1 lie between the same set of consecutive zeros. Therefore every second ρ S 1 gives rise to one new zero γ namely γ c or γ c with γ + γ log γ ε 1/ δ. Thus we have at least 1/ S 1 c/8ε A T zeros T γ T such that γ + γ log γ ε 1/ δ. 10

11 7. Lemma: Zeros of the Riemann zeta-function In this section we collect a few facts concerning the zeros of the Riemann zeta-function. They will be used in the proof of Theorem and Corollary 1. We first need Gonek s lemma. Lemma 8 Gonek [10]. If x = a/b 1 and a, b N, then, x iγ N log T T γ T T γ T Proof. As noted by Ford and Zaharescu Lemma 1, [6], it follows from Gonek s work that, x 1/+iγ = Λn x e it logx/nx 1 + O x log xt +. π i logx/n x log x Since x is not an integer we have x n x. Therefore the closest that x/n x = a/bn x can be to 1 is when bn x is equal to a ± 1. This shows that logx/n x a 1 N 1. Therefore the main term in the above equation is bounded by N, This gives a bound of T γ T xiγ N/ x + x log T for x > 1. For x < 1 this bound is reversed to xn + log T/ x. In either case the final bound is N log T because N 1 x N. An quick consequence of the above lemma is a bound for Dirichlet polynomials. Lemma 9. Let B N s be as in Lemma. If N k T then, B N 1 + iγ k Ck k T k T γ T for some absolute constant C > 0. Proof. First notice that for T t T log p 1 p log pk k/+kit p k N k>1 = 1 πi +i i ζ N s/ ds s it ζ s log N + O1 = N it t + ζ 1 + it + O 1 + ζ N 1/8 and that the above expression is less than log by a classical estimate for the size of ζ /ζ on the Riemann Hypothesis. Therefore, B N 1 + iγ k C k log p p 1 log p k + T C log k. 1/+iγ T γ T T γ T p N with C > 0 some absolute constant. We denote the coefficients of the Dirichlet polynomial over primes by ap. We have, k app iγ = iγ p1... p k ap 1... ap k aq 1... aq k q 1... q k T γ T p N p 1,...,p k N q 1,...,q k N 11 T γ T

12 The diagonal terms p 1... p k = q 1... q k contribute at most T k! k ap Ck k T k p N because given q 1,..., q k all the solutions to the equation p 1... p k = q 1... q k are obtained by pairing together each prime p i with some other prime q j, and there is at most k! such pairings. To bound the off-diagonal terms p 1... p k q 1... q k we notice that p 1... p k N k T and similarly that q 1... q k N k T. Therefore by Gonek s lemma iγ p1... p k T log T. T γ T q 1... q k Since p N ap N it follows that the off-diagonal terms contribute at most C k N k T log T C k T log T, which is less than the main term as soon as k > 0 An immediate consequence of the above lemma is the following. Lemma 10. Let T t T. Then, N γ + π π k N γ Ck k T. T γ T with C > 0 an absolute constant. Proof. Let N be the largest integer such that N k T. We have N γ + π π π π N γ N γ + N γ + B N 1 + iγ by Lemma. Raising the above expression to the k-th power and then summing over all T γ T we obtain N γ + π π k N γ Ck k T + Ck B k N 1 +iγ k T γ T T γ T with C > 0 an absolute constant. By the previous lemma the sum over T γ T is Ck k T k and so the claim follows. Corollary 3. Let A > 0 and δ > 0 be given. If 0 < ε < Cδ, A, with Cδ, A depending only on δ and A, then, { # T γ T : N γ + π } π N γ > ε δ ε A+1 T. Proof. By the previous lemma we have for k > 1, N γ + π π k N γ Ck k T T γ T 1

13 with C > 0 a positive absolute constant. Therefore the number of T γ T for which the interval [γ π/ ; γ + π/ ] contains more than ε δ zeros is bounded above by ε kδ Ck k T. Choose k = A/δ. Then ε kδ Ck k ε A provided that ε ca/δ 4/δ with c > 0 an absolute constant. We will require the following two lemma. 8. Proof of Theorem Lemma 11 Zhang [19]. Let ε < 1. If ρ = 1 + iγ is a zero of ζs such that γ is sufficiently large and γ + γ log γ < ε then there exists a zero ρ of ζ s such that Lemma 1 Soundararajan [16]. We have, We are now ready to prove Theorem. ρ ρ ε log γ. ρ ρ c β 1. log γ Proof of Theorem. Suppose that there are at least cε A T zeros T γ T such that γ + γ log γ ε 1/. Call this set S. If γ + = γ for at least a half of the elements in S then ρ = ρ and hence β = 1 for at least c/εa T zeros. Hence suppose that γ + > γ for at least half of the elements in S and call the subset of such elements S 1. By Corollary 3, the number of T γ T such that the interval [γ π/ ; γ + π/ ] contains more than ε δ zeros is c/4ε A T, provided that ε is small enough with respect to δ and A. Therefore there is a subset S of S 1 of cardinality c/4ε A T with the properties that 0 < γ + γ log γ < ε 1/ and the number of zeros in the interval [γ π/, γ + π/ ] is less than ε δ. By Lemma 10 each ρ S gives rise to a zero ρ such that ρ ρ ε/. By Lemma 11 the zero ρ satisfies β 1 log γ ε. Furthermore the interval t γ < ε/ contains at most ε δ zero. Therefore striking out at most ε δ zeros from S we obtain each time a new and distinct zero ρ of ζ s. It follows that ε δ S is a lower bound for the number of zeros ρ with β 1 log γ ε. Hence m ε c/4ε A+δ, as desired. 9. Proof of Corollary 1 The Pair Correlation Conjecture asserts that the number of zeros T γ 1, γ T for which πα/ < γ 1 γ πβ/ is asymptotically β sinπu NT 1 + δu du πu α with δ denoting Dirac s delta function. Here we derive a simple consequence of the Pair Correlation Conjecture for small gaps between consecutive zeros of the Riemann zeta-function. The lower bound is not optimal but sufficient for our needs. Lemma 13. Assume the Pair Correlation Conjecture. Let δ > 0 be given. Then ε 3+δ mε ε 3 provided that 0 < ε < Cδ with Cδ a constant depending only on δ. 13

14 Proof. The Pair Correlation Conjecture asserts that the number of distinct zeros T γ 1, γ T for which 0 γ 1 γ πα/ is asymptotically NT fα with fα such that fα c α 3 as α 0. The number of T γ T such that γ + γ log γ ε is less than the number of distinct T γ 1, γ T for which 0 γ 1 γ πε/ therefore mε fε ε 3. Now consider the set of T γ 1, γ T for which ε γ 1 γ log γ 1 ε. Call S the set of T γ 1 T for which the interval [γ 1 π/ ; γ 1 + π/ ] contains at most ε δ zeros. By Corollary 3, the zero T γ 1 T with γ 1 / S have cardinality ε A T provided that 0 < ε < Cδ, A we choose A = 100 for example. We have 13 1 = γ 1 S T γ T γ 1 S T γ T T γ 1,γ T ε γ 1 γ log γ 1 ε ε γ 1 γ log γ 1 ε ε γ 1 γ log γ 1 ε Since γ 1 S there can be at most ε δ zeros γ satisfying ε/ γ 1 γ log γ 1 ε. Therefore the first sum is bounded by ε δ ε δ mε T γ 1 S γ + 1 γ 1 log γ 1 ε because for each γ 1 S the inner sum over γ is ε δ if γ 1 + γ 1 log γ 1 ε and is 0 otherwise. On the other hand the second sum is by Cauchy-Schwarz less than, 1/ S 1/ 1 T γ 1 T T γ T ε γ 1 γ log γ 1 ε S 1/ T γ 1 T N γ 1 + π N γ1 π 1/ ε A/ T by Lemma 9. By the Pair Correlation Conjecture the left-hand side of 13 is asymptotically C NT ε 3 for some absolute constant C > 0. Combining the above three equations we get Cε 3 mεε δ + C 1 ε A/ for some absolute constant C, C 1 > 0. Therefore if ε is small enough then ε 3+δ mε. We are now ready to prove Corollary 1. Proof of Corollary 1. By the previous lemma, on the Pair Correlation, we have mε 1/ ε 3/+δ for all Cδ > ε > 0. Therefore by the second part of our Main Theorem we get m ε ε 3/+δ for all Cδ > ε > 0. Now suppose to the contrary that there is a η > 0 and a sequence of ε 0 such that m ε ε 3/ η. Then, by Theorem 1 on the same subsequence of ε 0 we have mε 1/ δ ε 3/ η. However by the Pair Correlation Conjecture we have ε 3/ 3δ mε 1/ δ ε 3/ η. Choosing 0 < δ < 1/3η and letting ε 0 along the subsequence, we obtain a contradiction. 10. Acknowledgments I would like to thank Prof. Farmer, Prof. Ki and Prof. Soundararajan for comments on an early draft of this paper. 14

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