LANDAU-SIEGEL ZEROS AND ZEROS OF THE DERIVATIVE OF THE RIEMANN ZETA FUNCTION DAVID W. FARMER AND HASEO KI

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1 LANDAU-SIEGEL ZEROS AND ZEROS OF HE DERIVAIVE OF HE RIEMANN ZEA FUNCION DAVID W. FARMER AND HASEO KI Abstract. We show that if the derivative of the Riemann zeta function has sufficiently many zeros close to the critical line, then the zeta function has many closely spaced zeros. his gives a condition on the zeros of the derivative of the zeta function which implies a lower bound of the class numbers of imaginary quadratic fields.. Introduction he spacing between zeros of the Riemann zeta-function and the location of zeros of the derivative of the zeta-function are closely related problems which have connections to other topics in number theory. For example, if the zeta-function had a large number of pairs of zeros that were separated by less than half their average spacing, one would obtain an effective lower bound on the class numbers of imaginary quadratic fields [, ]. Also, Speiser proved that the Riemann hypothesis is equivalent to the assertion that the nontrivial zeros of the derivative of the zeta-function, ζ, are to the right of the critical line [5]. here is a quantitative version of Speiser s theorem [9] which is the basis for Levinson s method [8]. In Levinson s method there is a loss caused by the zeros of ζ which are close to the critical line, so it would be helpful to understand the horizontal distribution of zeros of ζ. he intuition is that the spacing of zeros of the zeta-function should determine the horizontal distribution of zeros of the derivative. Specifically, a pair of closely spaced zeros of ζs) gives rise to a zero of ζ s) close to the critical line. Our main result is a partial converse, showing that sufficiently many zeros of ζ s) close to the -line implies the existence of many closely spaced zeros of ζs). See heorem.3. We assume the Riemann hypothesis and write the zeros of ζ as ρ j = + iγ j and the zeros of ζ as β j + iγ j, where in both cases we list the zeros by increasing imaginary part, repeated according to their multiplicity. Recall the zero-counting functions [6, ].) N ) := #{j : 0 < γ j < } = π log πe + Olog ) and.) N ) := #{j : 0 < γ j < } = π log + Olog ). 4πe We consider the normalized gaps between zeros of ζ and the normalized distance of ρ j to the right of the critical line, given by λ j = γ j+ γ j ) log γ j he second named author was supported by Mid-career Researcher Program through NRF grant funded by the MES

2 DAVID W. FARMER AND HASEO KI.3) λ j = β j ) log γ j. We are interested in how small the normalized gaps can be, and how small the normalized distance to the critical line can be, so we set.4).5) λ = lim inf j λ j λ = lim inf j λ j. We also consider the cumulative densities of λ j and λ j, given by mν) = lim inf J J #{ j J : λ j ν} m.6) ν) = lim inf J J #{ j J : λ j ν}. Soundararajan s [3] Conjecture B states that λ = 0 if and only if λ = 0. his amounts to conjecturing that zeros of ζ s) close to the -line can only arise from a pair of closely spaced zeros of ζs). Zhang [8] showed that on RH) λ = 0 implies λ = 0. hus, Soundararajan s conjecture is almost certainly true because λ = 0 follows from standard conjectures on the zeros of the zeta-function, based on random matrix theory. However, the second author[7] showed that λ = 0 and λ = 0 are not logically equivalent. Specifically, Ki[7] proved heorem. Haseo Ki [7]). Assuming RH, λ > 0 is equivalent to.7) Mγ j ) := = Olog γ j ). γ j γ n 0< γ j γ n < Note that the theorem implies Zhang s result that λ = 0 implies λ = 0), because if λ = 0 then for some j the sum in.7) will be large because an individual term in the sum is large. But that is not the only way for Mγ j ) to be large. It is possible that there could be an imbalance in the distribution of zeros, such as a very large gap between neighboring zeros, which makes the sum large because many small terms have the same sign. For example, suppose there were consecutive zeros of the zeta function with a gap of size, followed by c log zeros equally spaced this cannot happen, but we are illustrating a point). hen Mγ) would be log log log. hat possibility is the reason attempts to prove λ = 0 implies λ = 0 have been unsuccessful. For example, Garaev and Yıldırım [5] required the stronger assumption λ J log log γ J ) = o) in order to conclude λ J = o). he discussion in the previous paragraph shows that, without detailed knowledge of the distribution of zero spacings, one requires Mγ) C log log log for any C > 0 in order to conclude λ = 0. It is possible that this could be improved by proving results about the rigidity of the spacing between zeros of the zeta function. Random matrix theory could give a clue about the limits of this approach. his would involve finding the expected maximum of the random matrix analogue of the sum.8) log γ j < γ j γ n < γ j γ n. Unfortunately, the necessary random matrix calculation may be quite difficult because a lower bound on γ j γ n requires the exclusion of a varying number of intervening zeros, so the combinatorics of the random matrix calculation may be intricate.

3 SIEGEL ZEROS AND ZEROS OF HE DERIVAIVE 3 In this paper we consider not λ and λ, but the density functions mν) and m ν). In the next section we illustrate this with the example described above, and then we state our main result... Examples with equally spaced zeros. We illustrate heorem. with examples which can help build intuition for why λ = 0 does not imply λ = 0. Our example involves degree N polynomials with all zeros on the unit circle. In other words, characteristic polynomials of matrices in the unitary group UN). In these examples. λ > 0 but λ = 0, where λ and λ refer respectively to the large N limits of the normalized gap between zeros, and the rescaled distance between zeros of the derivative and the unit circle. his is the random matrix analogue of λ and λ for the zeta function. Figure. illustrates the case of 6 zeros in the interval{e iθ : 0 θ π/}. he plot on the left shows the zeros of the polynomial and its derivative. he figure on the right is the same plot unrolled : the horizontal axis is the argument, and the vertical axis is the distance from the unit circle, rescaled by a constant factor Figure.. On the left, the zeros and the zeros of the derivative of a degree 6 polynomial having all zeros in of the unit circle. On the right, the image of 4 those zeros under the mapping re iθ θ, π 6 r)). Zeros of the function are shown as small squares, and zeros of the derivative as small dots. Figure. is the analogue of the plot on the right side of Figure., for 0 zeros and 50 zeros. Note that in these examples λ π/. In Figure. the vertical scales are stretched by a factor of πn r) where N = 0 and 50, respectively. Figures. and. illustrate that, with this unrolling and rescaling, the zeros of the derivative approach a circle. We see that even though λ > 0 we have λ = 0, but furthermore, since the zeros lie on a rescaled) circle, we have m ν) ν as ν 0. hus, we can have m ν) > 0 for all ν > 0, yet mν) = 0 for ν sufficiently small. We believe that the above example is the limit of this behavior, and we make the following conjecture, which we view as a refinement of Soundararajan s conjecture. Conjecture.. If m ν) ν α for some α <, then mν) > 0 for all ν > 0.

4 4 DAVID W. FARMER AND HASEO KI Figure.. Unrolled and rescaled zeros of the derivative of a polynomial with zeros equally spaced along the arc {e iθ : 0 θ π/}. he polynomial has degree 0 left) and 50 right). 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. For applications to lower bounds of class numbers [, ] one does not actually need mν) > 0 for ν < π; it is sufficient to show that a relatively small number of gaps between zeros of the zeta function are small. Our main result, heorem.3, obtains such bounds from estimates on the zeros of the derivative of the zeta function. Denote log ) t = log log t. heorem.3. Assume RH. Suppose that for all ν > 0,.9) #{0 < γ < : β ) log γ ν} e Cν) log, π as, where Cν) > 0 for ν > 0 with lim ν 0 + νcν) = 0 and limν 0 + Cν) =. hen.0) lim inf for all ν > 0. #{0 < γ n : γ n+ γ n ) log γ n ν} log / log ) he conclusion of the theorem is weaker than mν) > 0 for ν > 0, but only by a factor of log ). hus, it is more than sufficient for applying the results of Conrey and Iwaniec []. In particular, heorem.3 shows that it is possible to obtain lower bounds for class numbers of imaginary quadratic fields from knowledge of the density of zeros of the derivative of the Riemann zeta function. here is an apparent discrepancy between Conjecture. and heorem.3 which we wish to clarify. In heorem.3 we allow exponential decrease of m ν) as ν 0. While the conclusion of the theorem is weaker than mν) > 0 by a factor of log ), it may seem curious that the condition in Conjecture. requires m ν) to be relatively large as ν 0. Indeed, the examples in Section. show that the condition in Conjecture. cannot be improved for general functions. he reason for the apparent inconsistency is that, as described in Section.3, our method relies on a bound on the moments of the logarithmic derivative. For the Riemann zeta > 0

5 function one expects.) SIEGEL ZEROS AND ZEROS OF HE DERIVAIVE 5 ζ ζ + ) k log + it dt k log k. he bound.) should follow by the method of Selberg [], although we give a conditional proof that allows us to explicitly determine the implied constant. Such a bound, for one fixed k, would establish a weaker version of heorem.3 that required m ν) ν k. However, more general functions like the polynomials in Section. do not satisfy an analogous bound to.). In fact, they are very large on the unit circle and do not satisfy the analogue of the Lindelöf hypothesis. Conjecture. is intended to cover those more general cases, while stronger statements should be true for the zeta function. It is interesting to speculate on the precise nature of the function m ν) for the Riemann zeta function. Dueñez et. al. [3] give a detailed analysis of the relationship between small gaps between zeros of the zeta function and analogously for zeros of the characteristic polynomial of a random unitary matrix) and the zeros of the derivative which arise from the small gaps. For the case of the Riemann zeta function they indicate that the random matrix conjectures for the zeros of the zeta function should imply.) m ζν) 8 9π ν 3, as conjectured by Mezzadri [0]. hat calculation is based on a more general result which suggests that if mν) κν β then m ν) κ ν β/ where.3) κ = π κ ) β. β π he factor of π comes from a different normalization used in [3] and here we work with the cumulative distribution functions m and m, while in [3] they use density functions. hat derivation assumed that zeros of ζ close to the -line only arise from closely spaced zeros of the zeta-function. he discussion above shows that, without further knowledge of the zeros, this is not a valid assumption. But, as indicated in our Conjecture., if β < 4 then we believe that the almost all zeros close to the -line do arise in such a manner. he random matrix prediction for the neighbor spacing of zeros of the zeta-function has κ = π/6 and β = 3, which is covered by Conjecture.. So our results support the analysis of Dueñez et. al. [3]. he remainder of this paper is devoted to the proof of heorem.3.. Proof of heorem.3 heorem.3 says that sufficiently many zeros of ζ close to the -line can only arise from closely spaced zeros of the zeta-function. If ρ = β + iγ is a zero of ζ, then we denote by ρ c = + iγ c the zero of the zeta-function which is closest to ρ. hus, we must show that if there are many β very close to, then often there is another zero of the zeta-function close to γ c. Our approach involves a study of the quantity.) M γc = 0< γ γ c Xγ c) γ γ,

6 6 DAVID W. FARMER AND HASEO KI where the range in the sum, Xγ c ), turns out to be a limiting factor in our method. We will define Xγ c ) in.0) in Section.. By analogy to a similar quantity studied in [7], we expect that M γc should be large if and only if β is small. And just like in [7], there are two ways that M γc can be large. here could be an individual term which is large. hat would happen if γ was near two γs that are very close together. Or there could be a large imbalance in the the distribution of the γs, for example if there was an unusually large gap between γ c and one of the adjacent zeros. We must show that the second possibility cannot occur too often. his is accomplished by showing that an imbalance in the distribution of zeros causes the zeta function to be large, and bounds on moments of the zeta function show that this cannot happen too often. he proof involves two steps. Assume the zeros of the zeta function rarely get close together. First we show that if β is small then M γ c is large. Second, we show that if M γc is large then usually ζ s) is large near + ζ iγ, subject to our assumption that the zeros of the zeta function rarely get close together. Standard bounds for the moments of ζ σ + it) ζ let us conclude that β cannot be small too often, which is what we wanted to prove. he relationship between M γc and ζ /ζ relies on an estimate for ζ /ζ in terms of a short sum over zeros. Suppose we have.) ζ ζ s) = γ t <Xt) + Olog t). s ρ On RH, with Xt) = / log ) t the above holds for all t [6]. Using this, instead of our.3) below, leads to a weaker version of heorem.3, where the log ) in the denominator of.0) is replaced by log. We prove the following strengthening of.), but only near almost all γ. Proposition.. Assume RH. Let m 0 be a positive integer. If C > is sufficiently large, then the number of 0 < γ n < such that ζ.3) ζ s) = s ρ + Olog γ n) γ t C log ) γ log γ for s = σ + it with s / iγ n A/ log γ n is.4) π log ) πe + O log ) m 0 as. he proof of Proposition. is in Section Restricting to zeros with special properties. We begin the proof of heorem.3. he lemmas in this section show that, in the context of the proof of heorem.3, we only have to deal with zeros that are well spaced. Suppose, for the purposes of contradiction, that there exists ϵ > 0 so that.5) lim inf #{0 < γ n : γ n+ γ n ϵ/ log γ n } log / log ) = 0. hen, we can find a sequence l such that is sufficiently large, l and.6) #{0 < γ n l : γ n+ γ n ϵ/ log γ n } = o l log l / log ) l )

7 as l. We set SIEGEL ZEROS AND ZEROS OF HE DERIVAIVE 7.7) = l. he following lemma shows that we can restrict our attention to those zeros whose immediate neighbors are well spaced. Lemma.. Let K = 4C [ log ) ]. Under assumption.5) we have.8) #{0 < γ n < : 0 < m K, γ n+m γ n+m Proof. For each m = ±, ±,..., let.9) A m = {0 < γ n < : γ n+m γ n+m ϵ } = log + o)). log γ n π ϵ log γ n }. Here, we exclude the case n + m. By assumption.5) together with.) we have ).0) #A m ) = π log + o log log ) for 0 < m log. We see that.) #.) 0< m K A m = 0< m K #A m ) K m<k m 0 K π log + o K log log ) # A m ) m<l K l 0 A l K ) log + OK log ) π.3) = log + o log ). π he next Proposition shows that we can restrict to intervals where the number of zeros is close to its average. Fix C >, let l and l be integers, and for + iγ a zero of the zeta function set.4) Nγ, l, l ) = N γ + l C ) log ) γ N γ + l C ) log ) γ l l )C log ) γ. log γ log γ π Using an argument in [6], we get the following. Proposition.3. Let m 0 > 0. here exists C > 0 such that the number of 0 < γ n < with.5) Nγ n, l, l ) C log ) for all l, l log /C log ) ) and 0 < l l log /C log ) ), is.6) π log ) πe + O ). log ) m 0 he proof of Proposition.3 is in Section 3..

8 8 DAVID W. FARMER AND HASEO KI.. Lower bound for M γc. Let β + iγ be a zero of ζ, and assuming RH) let + iγ c be the zero of the zeta function which is closest to iγ. If there are two closest zeros, choose the one nearer to the origin. We will use the above Propositions and Lemma to give a lower bound for M γc, assuming β is small. Let Z ) be the set of 0 < γ c < which satisfy the following three conditions: ϵ.7) γ c {0 < γ n < : 0 < m K, γ n+m γ n+m }; log γ n.8).9) Nγ c, l, l ) C log ) ζ ζ s) = γ t C log ) γ log γ log C log ) l < l s ρ + Olog γ c), log C log ) where s = σ + it with s / iγ c A/ log γ c. By the Propositions and Lemma in the previous section, as the set Z ) contains log elements. π In the definition.) of M γc, let.0) Xγ c ) = C log ) γ c log γ c, Lemma.4. If ν > 0 and k is a positive integer then.) e Cν) log k ν k A k + O ν)) k π log log γ β ) log γ ν γ c Z ) M γc k. Proof. hroughout the proof we assume γ c Z ). From equation..7) in [6] and standard properties of the Γ-function, we get ζ.) ζ s) = log t + O) + s ρ ), ρ ρ uniformly for t 0 and σ. Let β + iγ be a zero of ζ s) where 0 < γ < is sufficiently large. aking the real part.) with s = β + iγ, we have.3) log γ + O) = β ) β + γ γ c ) + β γ γ β c ) + γ γ). We consider three cases, with Case 3 requiring some manipulations which will be used to complete the proof of the lemma. Case. β / > γ γ c. hen, by.3), we get.4) hus, we have β / / log γ. log γ + O) β Case. β / γ γ c and γ γ c > δϵ)/ log γ, where δϵ) = 8/ϵ. ). ) ;

9 By.3),.7), and.8), we have log γ β ).5) log γ + and so again we have SIEGEL ZEROS AND ZEROS OF HE DERIVAIVE 9 β ) log γ, β ) + mϵ log γ m=.6) β log γ. Here the implied constants depend only on ϵ. Case 3. β / γ γ c and γ γ c δϵ)/ log γ. Using.3),.7), and.8), as in Case, we get.7) and.8) hus we have ) β log) γ ) log) γ ) m log) γ log γ + log γ m=0 log γ + O) β γ γ c ) log γ β γ γ c ) + β ) log γ β γ γ c )..9) γ γ c ) log γ β γ γ c ) log γ. Here the implied constants depend only on ϵ. By.9) and the conditions of Case 3 we have γ c γ.30) ) β + γ γ c ) = Olog γ ). γ c γ aking the imaginary part of.9) with s = β + iγ and using the conditions in Case 3 we get γ γ.3) ) β + γ γ) + γ c γ β ) + γ γ c ) = O log γ ). 0< γ γ c C log ) γc log γc By.7), recalling that K log ), we have γ γ β ) + γ γ) M β γ c = O ).3) 0< γ γ c C log ) γc log γc k= = Olog γ ). By combining.30),.3),.3), and.9), we have.33) Olog γ ) = M γc + log γ = M γ c γ γc + A γc β, where A γc, with the implied constants depending only on ϵ. ) 3 ϵk log γ

10 0 DAVID W. FARMER AND HASEO KI hat ends our consideration of Case 3. Now we complete the proof of the lemma. Let ν be a positive number. Suppose that.34) β ) log γ ν. hen, for sufficiently small ν, we see that only Case 3 is possible for sufficiently large γ, namely we have log γ.35) M γc + A γc i.e., β = Olog γ ), log γ.36) M γc A + O ν)). ν By this, the assumption in heorem.3, and the fact that #Z ) log, we have π e Cν) π log log γ β ) log γ ν γ c Z ) log γ β ) log γ ν γ c Z ) M γc log γ A ν + O ν)) ) k. he last inequality gives.). In the next section we describe upper bounds for the moments of M γc. his will contradict.) and complete the proof of heorem Bounding the moments of M γc. We obtain an upper bound on M γc from a bound on moments of the logarithmic derivative of the zeta function. his makes use of that fact that, assuming the zeros of the zeta function do not get close together, the logarithmic derivative can be approximated either by a short sum over zeros, or by a short Dirichlet series. Lemma.5. Assume RH. Let / log < γ < such that γ γ c δϵ)/ log γ and assume.7).9). hen.37) for t γ A/ log γ. M γc i + ζ ζ + ) log + it = O ϵ log ),

11 Proof. By the assumptions we have M γc + ζ i ζ + ) log + it = M γ c i.38) SIEGEL ZEROS AND ZEROS OF HE DERIVAIVE = + 0< γ γ c C log ) γc log γc 0< γ γ c C log ) γc log γc m= +δϵ) log γ = O ϵ log ). log + Olog ) + it γ) + it log γ ) iγ γ ) log ) + Olog ) mϵ log γ + it γ)) + Olog ) Lemma.6. Assume RH and.7).8). Let s = + + it with t, and let log x = /00k. hen if γ γ c δϵ)/ log γ and t γ ϵ/ log γ, we have.39) ζ ζ s) = Λ x n) + O n s ϵ k log ), n<x where { Λn) n x.40) Λ x n) = Λn) log x n ) x n x. log x Proof. By [6], heorem 4.0,.4) ζ ζ s) = Λ x n) + x s) x s n s s) log x n<x + log x q= x q s x q+s) + x ρ s x ρ s). q + s) log x s ρ) ρ he assumptions on the zero spacings give the claimed bound on the terms involving the zeros. Lemma.7 Soundararajan, Lemma 3 of [4]). Let be large, and let x. Let k be a natural number such that x k / log. For any complex numbers ap) we have k ) ap) k ap) dt k!, p where the sum is over the primes. p x p +it Lemma.8 Soundararajan, Proposition 6 of [3]). Suppose /+iγ and /+iγ are two consecutive zeros of ζs) with γ < γ for large. Suppose the disc s / + iγ ) is devoid of exceptions to the Riemann hypothesis. hen the box p x {s = σ + it : / < σ < / + / log, γ t γ }

12 DAVID W. FARMER AND HASEO KI contains at most one zero counted with multiplicity) of ζ s). We assemble the above lemmas to bound the moments of M γc. By Lemma.5 and Lemma.6, with A = A ϵ a constant depending only on ϵ, which may be different in each inequality, we have M γc k A k log k + k ζ ζ + ) k log + it k A k k k log k + k Λ x n) n<x n + log +it k A k k k log k + k Λ x p).4) log +it, p<x p + where x = /00k, for t γ ϵ/ log γ and γ γ c δϵ)/ log γ, provided γ c satisfies.7).9). hat is, provided γ c Z ). Recall that for sufficiently small ν with β /) log γ ν, only Case 3 is possible, i.e., β γ γ c and γ γ c δϵ) log γ for sufficiently large γ. hus, using the property.7) of Z ), Lemma.8 and.4), we get k ϵ.43) M γc k A k k k log k + k Λ x p) log log γ log p<x p + log +it dt β ) log γ ν γ c Z ) By this and Lemma.7.44) ϵ log log γ β ) log γ ν γ c Z ) M γc k A k k k log k + k k! Λ x p) p<x A k k k log k. he last step used Λ x p) Λp) and the fact that.45) Λp) log x, p p x which is a weak form of the prime number theorem. Rearranging the above inequality and combining with.), we have p + log k.46) e Cν) ν k + O ν)) k log k+ A k k k log k+,

13 which rearranges to give SIEGEL ZEROS AND ZEROS OF HE DERIVAIVE 3.47) + O ν)) k A k k k ν k e Cν). Letting k = [/ A eν], we have a contradiction if νcν) 0 as as ν 0. his completes the proof of heorem Proofs of technical results In this section we provide the proofs of Proposition. and Proposition Proof of Proposition.3. Assume that γ n satisfies /log ) m 0+3 < γ n <. We recall 3.) +H St + h) St) k dt = Hk)! π ) k k! logk + h log ) )) + O Hck) k k + log k / + h log ) uniformly for a < H, a > /, 0 < h < and any positive integer k, where c is a positive constant and St) = arg ζ/ + it). For this, see [7, heorem 4]. hus we have π 3.) 0 St + h) St) k dt Ak) k, where log + h log ) k. Let St, h) = St + h) St). We note that 3.3) St, h) = Nt + h) Nt) h π log t π + O where Nt) is the number of zeros of ζs) in 0 < Is < t. For 0 h log ), we have log h + t ), 3.4) Nγ n, l, l ) Nγ n + l C log ) log l l )C log ) π Sγ n + l C ) log ) log + h) Nγ n + l C ) log ) log + olog ) ) + h, l l )C + ) log ) log + h) ) + log ).

14 4 DAVID W. FARMER AND HASEO KI Using this, we have 3.5) log ) m 0 +3 <γn< Nγ n,l,l )C log ) C log ) ) k log 4 log ) ) k + log 4 log ) ) k + log log ) log ) log ) Nγ n, l, l ) log ) m 0 + <γn< 0<γ n< log ) 4 log ) ) k + Ak) k ) 0 S γ n+ l C log ) log γ n+ l C ) log ) log k S t, l l )C + ) log ) log t, l l )C ) + ) log ) k dt log ) k dt for any sufficiently large and any l, l log /C log ) ) with 0 < l l log /C log ) ). We put 3.6) k = [log ) ] and C = e m 0+ A + 5). By these and the last inequality, we have 3.7) l, l log C log ) 0<l l log C log ) Since N ) = O log ) m ) log ) m 0 +3 <γ n< Nγ n,l,l )C log ) log ) m 0 l, l log C log ) 0<l l log C log ) ), we have hat completes the proof of Proposition Proof of Proposition.. We recall 3.9) log )4 C log ) ) k C log) ) k log ) m 0. 0<γ n < Nγ n,l,l )C log ) ζ s) = Olog t) + ζ γ t log ) m 0. s ρ holds uniformly for t > and Res. For this, see [6, heorem 9.6 A)]. Using the last formula, it suffices to show that for any γ n [/log ) m0+, ] satisfying the condition in Proposition.3, 3.0) C log ) log < γ n γ m γ n γ m = Olog )

15 C log ) log < γ n γ m SIEGEL ZEROS AND ZEROS OF HE DERIVAIVE 5 holds, because log γ n = + o)) log and for s with s / iγ n A/ log, we have s + iγ m) iγ n γ m ) = O log ) ) log m log) = Olog ). We recall that the condition in Proposition.3 is ) 3.) N γ n + lc log ) = Nγ n ) + lc log ) + O log log π ) ) for any integer l with l log /C log ) ). Let γ m be the least one in [γ n, γ n C log ) / log ) and γ m the greatest one in γ n + C log ) / log, γ n + ]. Let N be the smallest positive integer such that γ n C log ) / log γ m +N and γ m N γ n + C log ) / log. By taking C large enough, we have γ m +N < γ n < γ m N. Since the error term in 3.) is uniform for any integers l log /C log ) ), we have 3.) max 0 k N γ n γ m +k γ m k = O m= log) log Using this and the fact [6, heorems 9.3 and 4.3] that the number of zeros between t and t + is ) log t 3.3) π + O log t as t ), log ) t we have 3.4) C log ) log < γ n γ m γ n γ m = 0 k N = O log ) log = Olog ). hus, we complete the proof of Proposition.. References ). log γ n γ m +k γ m k γ n γ m +k)γ n γ m k) + Olog ) k= log ) ) k log) log + Olog ) [] B.C. Berndt, he number of zeros for ζ k) s), J. London Math. Soc.) 970), [] J.B. Conrey and H. Iwaniec, Spacing of zeros of Hecke L-functions and the class number problem, Acta Arith ), no. 3, [3] E. Dueñez, D.W. Farmer, S. Froehlich, C. P. Hughes, F. Mezzadri, and. Phan, Roots of the derivative of the Riemann zeta function and of characteristic polynomials, Nonlinearity 3 00), no. 0, [4] D.W. Farmer and R. Rhoades, Differentiation evens out zero spacing, rans. Amer. Math. Soc ), no. 9, [5] M.Z. Garaev and C.Y. Yıldırım, On small distances between ordinates of zeros of ζs) and ζ s), Int. Math. Res. Notices ), Art. ID rnm09, 4 pp.

16 6 DAVID W. FARMER AND HASEO KI [6] A. Ivıć, On small values of the Riemann zeta-function on the critical line and gaps between zeros, Lietuvos Mat. Rinkinys 4 00), no., 3 45; translation in Lithuanian Math. J. 4 00), no., [7] H. Ki, he zeros of the derivative of the Riemann zeta function near the critical line, Int Math Res Notices ), Art. ID rnn064, 3 pp. [8] N. Levinson More than one third of zeros of Riemann s zeta function are on σ = /, Adv. Math ), [9] N. Levinson and H.L. Montgomery, Zeros of the derivatives of the Riemann zeta-function, Acta Math ), [0] F. Mezzadri, Random matrix theory and the zeros of ζ s), J. Phys. A: Math. Gen ), no., [] H. Montgomery, he pair correlation of zeros of the zeta function, Proc. Sympos. Pure Math. 4, Amer. Math. Soc., Providence, R. I. 973), [] A. Selberg, On the normal density of primes in small intervals, and the difference between consecutive primes, Arch. Math. Naturvid ), no. 6, [3] K. Soundararajan, he horizontal distribution of zeros of ζ s), Duke Math. J ), no., [4], Moments of the Riemann zeta-function, Ann. of Math. ) ), no., [5] A. Speiser, Geometrisches zur Riemannschen Zetafunktion, Math. Ann ), [6] E.C. itchmarsh, he heory of the Riemann Zeta-function, nd ed., revised by D. R. Heath-Brown, Oxford University Press, Oxford, 986. [7] K.-M. sang, Some Ω-theorems for the Riemann zeta-function, Acta Arith ), no. 4, [8] Y. Zhang, On the zeros of ζ s) near the critical line, Duke Math. J. 0 00), no. 3, American Institute of Mathematics, 360 Portage Ave Palo Alto, CA address: farmer@aimath.org Department of Mathematics, Yonsei University, Seoul, 0-749, Korea address: haseo@yonsei.ac.kr