The zeros of the derivative of the Riemann zeta function near the critical line

0.15 MB·English

by  Haseo Ki

The zeros of the derivative of the Riemann zeta
7 function near the critical line
0
0
2

n
Haseo Ki
a
J

5
2 Department of Mathematics, Yonsei University, Seoul 120–749, Korea

]
T
haseoyonsei.ac.kr
N
.
h
t
a February 2, 2008
m
[

1
v Abstract
6
2
7 We study the horizontal distribution of zeros of ζ′(s) which are denoted as ρ′ =
1
β′ +iγ′. We assume the Riemann hypothesis which implies β′ > 1/2 for any non-real
0
7 zero ρ′, equality being possible only at a multiple zero of ζ(s). In this paper we prove
0 that liminf(β′ 1/2)logγ′ = 0 if and only if for any c > 0 and s = σ + it with
/ − 6
h σ 1/2 < c/logt(t > 10)
t | − |
a
m ζ′ 1
(s)= +O(logt),
: ζ s ρ
v −
i
X where ρ = 1/2+iγ isthe closest zero of ζ(s)to sand the origin. Wealso show that if
r liminf(β′ 1/2)logγ′ = 0,thenforanyc > 0ands = σ+it(t > 10),wehave
a − 6
(logt)2−2σ
logζ(s)= O
loglogt
(cid:18) (cid:19)
uniformly for1/2+c/logt 6 σ 6 σ < 1.
1
1 Introduction
TheRiemannhypothesis(RH)statesthattherealpartofanynonrealzerooftheRiemannzeta
function ζ(s) is 1/2. A. Speiser [13] has a theorem which says that RH is equivalent to the
nonexistenceofnonreal zeros ofζ′(s) inRe(s) < 1/2.
1
2
We let ρ′ = β′ +iγ′ denote a zero of ζ′(s) where a sum over ρ′ is repeated according to
multiplicity. N.LevinsonandH.L.Montgomery[8]provedthatfornonrealzerosofζ′(s)the
averagevalueofβ′ with0 < γ′ 6 T is1/2+loglogT/logT. But itislikelythatβ′ 1/2is
−
usuallyoftheorder1/logT rather thanloglogT/logT.
By aresultofB. C. Berndt [1], thenumberofzeros withordinateless thanT is
T T
1 = log +O(logT).
2π 4πe
0<γ′6T
X
UnderRH,K.Soundararajan[12]demonstratedthepresenceofapositiveproportionofzeros
of ζ′(s) in the region σ < 1/2 + ν/logT for all ν > 2.6. ‘A positive proportion of zeros’
means
1
liminf # ρ′ : β′ 6 1/2+ν/logT,0 < γ′ 6 T > 0.
T→∞ T logT { }
2π
In his remarkable work, Y. Zhang [15] showed that not only unconditionally there exists a
ν > 0such thatapositiveproportionofzeros ofζ′(s) are intheregion σ 1/2 < ν/logT,
| − |
but also assuming RH and a strong hypothesis of the distribution of zeros of ζ(s), ν can be
arbitrary small. Recently, Feng [5] proved the Zhang’s second theorem only assuming the
stronghypothesis. Thusitis veryprobablethat
liminf(β′ 1/2)logγ′ = 0.
−
AssumingthetruthofRH,K.Soundararajan[12]conjecturedthatthefollowingtwostate-
mentsare equivalent:
(i)liminf(β′ 1/2)logγ′ = 0;
−
(ii)liminf(γ+ γ)logγ = 0whereγ+ istheleastordinateofazeroofζ(s)withγ+ > γ.
−
Y. Zhang[15]has shownthat (ii)implies(i)as follows.
Theorem A. AssumeRH.Let α andα bepositiveconstantssatisfyingα < 2π and
1 2 1
−1
α
1
α > α 1 .
2 1
− 2π
(cid:18) r (cid:19)
If ρ = 1/2+ iγ is a zero of ζ(s) such that γ is sufficiently large and γ+ γ < α (logγ)−1,
1
−
thenthereexistsa zeroρ′ of ζ′(s) suchthat
ρ′ ρ < α (logγ)−1.
2
| − |
In this paper, we consider the converse of Theorem A. Namely is it true that (i) implies
(ii)? Concerning thisproblem,wehavethefollowing.
3
Theorem 1. AssumeRH.Then thefollowingareequivalent:
(1)liminf(β′ 1/2)logγ′ = 0;
− 6
(2)Foranyc > 0and s = σ +it with σ 1/2 < c/logt(t > 10)
| − |
ζ′ 1
(s) = +O(logt),
ζ s ρ
−
whereρ = 1/2+iγ is theclosest zeroofζ(s)tos andtheorigin;
Corollary1. AssumeRHandliminf(β′ 1/2)logγ′ = 0. Then,foranyc > 0ands = σ+it
− 6
(t > 10),wehave
ζ′
(s) = O (logt)2−2σ
ζ
uniformlyfor1/2+c/logt 6 σ 6 σ < 1. (cid:0) (cid:1)
1
Based on ourtheoremsandSoundararajan’s conjecture, wespeculateas follows.
Conjecture. AssumeRH. Then thefollowingareequivalent:
(i)′ Foranyc > 0 ands = σ +it (t > 10),we have
ζ′
(s) = O (logt)2−2σ
ζ
(cid:0) (cid:1)
uniformlyfor1/2+c/logt 6 σ 6 σ < 1;
1
(ii)′ Thenegationof(ii), i.e.,liminf(γ+ γ)logγ = 0.
− 6
We briefly introduce why (2) in Theorem 1 doesn’t seem possible. We let s = σ +it for
real numbers, σ,t. It is known in [3, p. 99] and [14, Theorem 9.6(A)] that for t > 10 and
1 6 σ 6 2wehave
− ζ′ 1
(s) = +O(logt), (1.1)
ζ s ρ
|γ−t|61 −
X
where the sum is over the ordinates of the complex zeros ρ = β + iγ of ζ(s). However,
assumingRH, wemay expectastrongerresult. UnderRH wehave
ζ′ 1
(s) = +O(logt) (1.2)
ζ s ρ
|γ−t|61/loglogt −
X
fort > 10. Forthiswereferto[14,p. 357(14.15.2)]. Accordingtotheformula(1.2),wecan
seethattheformula(2)inTheorem1isveryunrealistic,because‘ σ 1/2 < c/logt’in(2)
| − |
doesn’tseem probableinsteadof‘ γ t 6 1/loglogt’in (1.2).
| − |
4
Concerning Corollary1, itis worthnotingthatunderRH
ζ′
(s) = O (logt)2−2σ (1.3)
ζ
(cid:0) (cid:1)
holds uniformly for 1/2+c/loglogt 6 σ 6 σ < 1, where t > 10, c > 0 and ‘O’ depends
1
uponc andσ . Fortheproofofit,weapply thefact [14, (14.14.5)]to theformula
1
ζ′(s) 1 logζ(z)
= dz.
ζ(s) 2πi (s z)2
Z|s−z|=2logclogt −
Then we immediately get (1.3). In proving (1.3), one cannot relax the condition ‘1/2 +
c/loglogt’as
1 c
+ (t ).
2 logt → ∞
Withthisinformation,itisvery likelythaton Re(s) = 1/2+c/logt (t > 10),
ζ′
(s) = O(logt).
ζ 6
Ontheotherhand,Corollary1followsfromaPhragme´n-Lindelo¨fargument,providedthat
wehave
ζ′
(s) = O(logt)
ζ
on Re(s) = 1/2+c/logt (t > 10). Thus we find the behavior of the logarithmicderivative
of the Riemann zeta function near Re(s) = 1/2 subtle and so we need a deep observation
abouttheRiemannzetafunctionnearthecriticallinetoestablishliminf(β′ 1/2)logγ′ = 0.
−
In fact we will see from Theorem 4 in Section 2 that the behavior of ζ′(s)/ζ(s) on Re(s) =
1/2+c/logtisvery muchrelated to
1
,
γ γ˜
0<|γ−γ˜|<1 −
X
where1/2+iγ,1/2+iγ˜ are complexzeros ofζ(s)and thesumis overγ˜.
From Corollary1, wecan demonstratethefollowing.
Corollary2. AssumeRHandliminf(β′ 1/2)logγ′ = 0. Then,Foranyc > 0ands = σ+it
− 6
(t > 10),wehave
(logt)2−2σ
logζ(s) = O
loglogt
(cid:18) (cid:19)
uniformlyfor1/2+c/logt 6 σ 6 σ < 1.
1
5
Assuming RH, it is known in [14, p. 355 (14.14.5)] and [14, Theorem 14.14(B)] that for
anyc > 0 ands = σ +it(t > 10), wehave
(logt)2−2σ
logζ(s) = O (1.4)
loglogt
(cid:18) (cid:19)
holds uniformly for 1/2 + c/loglogt 6 σ 6 σ < 1, and there exists an absolute constant
1
c∗ > 0 dependingonc such that
logt 2 logt
c∗ log < log ζ(s) < c∗ (1.5)
− loglogt σ 1 loglogt | | loglogt
!
− 2
holdsfor1/2 < σ 6 1/2+c/log(cid:0)logt; (cid:1)
logt
argζ(s) = O (1.6)
loglogt
(cid:18) (cid:19)
holds uniformly for 1/2 6 σ 6 1/2 + c/loglogt. However, as in the proof of (1.3), one
cannot relax the condition ‘1/2 + c/loglogt’ of (1.4) as 1/2 + c/logt (t ). On the
→ ∞
otherhand,wehaveΩ-theoremsrelatedtologζ(s)nearthecriticalline. Forthese,wereferto
Montgomery’sresults[10]and[14,p. 209]. Inparticular,Montgomeryshowedthatassuming
RH, for1/2 6 σ < 1 and anyreal θ, thereis atwithT1/6 6 t 6 T such that
1
Re e−iθlogζ(s) > (logT)1−σ(loglogT)−σ.
20
See [10, p. 512]. Recent(cid:0)ly, from rando(cid:1)m matrix theory, in the case that θ = 0 in the above
Ω-result,it isconjectured in[4]that wehavethefollowing:
1 1
max ζ +it = exp (1+o(1)) logT loglogT .
t∈[0,T](cid:12) (cid:18)2 (cid:19)(cid:12) r2 !
(cid:12) (cid:12)
Concerningnegativev(cid:12)aluesoflog(cid:12)ζ(s) ,weobservethatby(1.5)and(1.6),itispossiblethat
(cid:12) (cid:12)
| |
wehave
1 1 logt
logζ + +it > ψ(t)
2 logt loglogt
(cid:12) (cid:18) (cid:19)(cid:12)
for some arbitrarily large v(cid:12)alues of t, where ψ(t) (cid:12) as t . A sharp Ω-result like this
(cid:12) (cid:12)
impliesliminf(β′ 1/2)l(cid:12)ogγ′ = 0. Namely,we→n(cid:12)ote∞thatit→wil∞lfollowifweshowthat(1.4)
−
does not hold uniformly for 1/2 + 1/logt 6 σ 6 1/2 + 1/loglogt. Thus, Corollary 2 is
useful in investigating the horizontal behavior of zeros of the derivative of the Riemann zeta
function.
We apply our theorem to mean values of the logarithmic derivative of the Riemann zeta
function:
T ζ′ 2
(σ +it) dt
ζ
Z0 (cid:12) (cid:12)
forσ = 1/2+a/logT. (cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
Wemay getthefollowingfromaresult ofA.Selberg [11,equation(1.2)].
6
Theorem B. AssumeRH. Then
T ζ′ 2 1
(σ +it) dt T log2T
ζ ∼ 4a2
Z0 (cid:12) (cid:12)
(cid:12) (cid:12)
holdswhere σ = 1/2+a/logT (cid:12)anda (cid:12), a = o(logT).
(cid:12) (cid:12)
→ ∞
Weintroducemorestudiesonmeanvaluesofthelogarithmicderivativeofζ(s). Following
Montgomery[9], let
1
F(α,T) = Tiα(γ−γ˜)w(γ γ˜),
T logT −
2π 0<γ,γ˜6T
X
where β + iγ and β˜+ iγ˜ are zeros of ζ(s) and w(u) = 4 . Montgomery conjectured that
4+u2
foranyfixed A > 1,
(MH) F(α,T) 1 uniformlyfor1 6 α 6 A.
∼
UnderRH, D. A.Goldston,S. M.Gonek andH. L. Montgomery[7]considered
T ζ′ 2
(σ +it) dt
ζ
Z0 (cid:12) (cid:12)
(cid:12) (cid:12)
forσ = 1/2+a/logT as a 0 and pro(cid:12)vedthefoll(cid:12)owing,providedthat (MH)is valid.
(cid:12) (cid:12)
→
Theorem C. AssumeRHand (MH). Then
T ζ′ 2 1
(σ +it) dt T log2T
ζ ∼ 2a
Z0 (cid:12) (cid:12)
(cid:12) (cid:12)
holdswhere σ = 1/2+a/logT a(cid:12)nda = a(T(cid:12)) 0 (sufficientlyslowly)asT .
(cid:12) (cid:12)
→ → ∞
Wehavethefollowingas inTheorem 1.
Theorem 2. AssumeRHand liminf(β′ 1/2)logγ′ = 0. Then
− 6
T ζ′ 2 1 log(2πe) 1 1
(σ +it) dt = T log2T 1 +O alog +O
ζ 2a − logT a T
Z0 (cid:12) (cid:12) (cid:18) (cid:18) (cid:19) (cid:18) (cid:19)(cid:19)
(cid:12) (cid:12)
holdswher(cid:12)e σ = 1/2+(cid:12) a/logT anda 0. Here‘O’doesn’tdependupon aandT.
(cid:12) (cid:12)
→
(MH)impliesMontgomery’spaircorrelation conjecture. Thatis,
T β sinπu 2
1 logT 1 du.
∼ 2π − πu
0<γ,γ˜6T Z0 (cid:18) (cid:19)
X
0<γ−γ˜62πβ/logT
7
Clearly (MH)implies
liminf(γ+ γ)logγ = 0.
−
AssumingRH, TheoremA says that
liminf(β′ 1/2)logγ′ = 0implies liminf(γ+ γ)logγ = 0.
− 6 − 6
ThustheassumptionsofTheoremCandTheorem2arecontradictorytoeachother. However,
we have the similar conclusion in Theorem C and Theorem 2. Further, a theorem of D. A.
Goldston,S. M.Gonek and H.L. Montgomery[7, Theorem 3]says thefollowing.
Theorem D. AssumeRH.Then, forσ = 1/2+a/logT andanyfixed a > 0,
T ζ′ 2 1 e−2a
(σ +it) dt − T log2T
ζ ∼ 4a2
Z0 (cid:12) (cid:12)
(cid:12) (cid:12)
holdsas T ifand onlyif(cid:12)thepairco(cid:12)rrelationconjectureis true.
(cid:12) (cid:12)
→ ∞
CombiningTheorem Band Theorem2, weimmediatelyhavethefollowing.
Theorem 3. AssumeRHand liminf(β′ 1/2)logγ′ = 0. Then, forσ = 1/2+a/logT,
− 6
T ζ′ 2 1 e−2a
(σ +it) dt − T log2T
ζ ∼ 4a2
Z0 (cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
holdswhere a 0 ora anda = o(logT).
(cid:12) (cid:12)
→ → ∞
Apparently, Theorem D and Theorem 3 are similar. However the conclusion of Theorem
D says a much stronger statement than that of Theorem 3. We noticethat thepair correlation
conjecture implies liminf(β′ 1/2)logγ′ = 0. Thus the conclusion of Theorem 3 under
−
liminf(β′ 1/2)logγ′ = 0isinstructivetounderstandthebehaviorofthemeanvalueofthe
− 6
logarithmicderivativeoftheRiemann zetafunction.
2 Proof of Theorem 1 and Corollaries 1, 2
We arrange the zeros of the Riemann zeta function ζ(s) on the upper half-plane as ρ ,ρ ,...
1 2
withρ = β +iγ and
n n n
0 < γ 6 γ 6 ...,
1 2
where it appears precisely m times consecutivelyin the above sequence, if a zero is multiple
withmultiplicitym. RH isthat β = 1/2foranyn = 1,2,3,....
n
Westatebasicfacts.
8
Proposition2.1. Let T > 0. Then, we have:
(1)Thenumber ofzerosofζ(s)in0 < Im(s) 6 T is
T T
1 = log +O(logT);
2π 2πe
1<Xγn6T
(2)Thenumber ofzerosofζ(s)inT 6 Im(s) 6 T +1is O(logT).
For Proposition 2.1(1), see [3, p. 98] and [14, Theorem 9.4]. Proposition 2.1(2) immedi-
atelyfollowsfrom (1).
Westart withthefollowingtheorem.
Theorem 4. Assume RH and liminf(γ γ )logγ > 0. Then the following three state-
n+1 n n
−
mentsareequivalent:
(A)liminf(β′ 1/2)logγ′ > 0;
−
(B) Let c > 0 ands = σ +it. Forsufficientlylargen, we have
ζ′ 1
(s) = +O(logt),
ζ s ρ
n
−
where γn−1+γn < t 6 γn+1+γn and σ 1/2 < c/logγ ;
2 2 | − | n
(C) limsup M /logγ < ,where
n n
| | ∞
1
M = .
n
γ γ
n m
0<|γmX−γn|61 −
ProofofTheorem 4. We may assume that all but finitely many nontrivial zeros of ζ(s) are
simpleand onRe(s) = 1/2,becauseweassumeRH andliminf(γ γ )logγ > 0.
n+1 n n
−
(C) (B). Werecall that
⇒
ζ′ 1
(s) = O(logt)+
ζ s ρ
|γ−t|61 −
X
for 1 6 Re(s) 6 2and t > 2.
−
Proposition2.2. Let δ > 0. Supposethatliminf(γ γ )logγ > δ. Then we have
n+1 n n
−
1 log2γ
n
= O
(γ γ )2 δ2
m6=n n − m (cid:18) (cid:19)
X
forsufficientlylargen.
9
ProofofProposition2.2. Wewrite
1 1 1
= +
(γ γ )2 (γ γ )2 (γ γ )2
m n m n m n (2.1)
mX6=n − |γmX−γn|>1 − 0<|γmX−γn|61 −
=I +II.
ByProposition2.1(2),thenumberofγ ’sbetweentandt+1isO(logt)fort > 1. Then,for
n
someC > 0, weget
∞ ∞
1 1
I = +
(γ γ )2 (γ γ )2
m n m n
Xk=1γn+k<γXm6γn+k+1 − Xk=1γn−k−1X6γm<γn−k −
∞
Clog(γ +k) Clog(γ k)
6 n + n − (2.2)
k2 k2
Xk=1 γnX−k>1
∞ ∞
logγ +logk logγ
6C n +C n = O(logγ )
k2 k2 n
k=1 k=1
X X
By theassumptionofProposition2.2, thereexistsapositiveintegern such that
1
δ
γ γ > for allm > n .
m+1 m 1
− logγ
m
Usingthis,forsufficientlylargen, wehave
k δ k δ
γ γ > | | > | |
n+k n
| − | log(γ +1) 2logγ
n n
for γ γ 6 1and k > 1. Herethereexistsaa > 0 suchthat1 6 k 6 alogγ . Thus
n+k n n
| − | | | | |
weget
1 ∞ 2 log2γ
II 6 < = O n .
2 2 δ2
16|k|X6alogγn 2lokgδγn Xk=1 2lokgδγn (cid:18) (cid:19)
Weapplythisinequalityand (2.2(cid:16))to(2.1(cid:17))and then(cid:16)weobt(cid:17)ain
1 log2γ
n
= O
(γ γ )2 δ2
m6=n m − n (cid:18) (cid:19)
X
forsufficientlylarge n. ThisprovesProposition2.2.
Proposition2.3. Choosec > 0suchthatliminf(γ γ )logγ > 2c . DefineM (t) by
1 n+1 n n 1 n
−
1
M (t) =
n
t γ
m
0<|γmX−γn|61 −
forγ +c /logγ 6 t 6 γ c /logγ . Then we have
n−1 1 n n+1 1 n
−
M (t) = O(logt),
n
wheretheimpliedconstantis absolute.
10
ProofofProposition2.3. Since
1
M′(t) = − < 0,
n (t γ )2
m
0<|γmX−γn|<1 −
M (t) is decreasing. Thus it suffices to consider the endpoints for the proof. By Proposition
n
2.2and ourassumption(C), wehave
c 1
1
M γ =
n n+1 − logγ γ γ c1
(cid:18) n(cid:19) 0<|γmX−γn|61 n+1 − m − logγn
1
= +O(logγ )
γ γ c1 n
0<|γmX−γn+1|61 n+1 − m − logγn
1 1
= +
γ γ c1 − γ γ
0<|γmX−γn+1|61 n+1 − m − logγn n+1 − m
M +O(logγ )
n+1 n
1 1
=O +O(logγ )
logγ (γ γ )2 n+1
n n+1 m
0<|γmX−γn+1|<1 −
=O(logγ ). 
n
Similarly,wehave
c
1
M γ + = O(logγ ).
n n−1 n
logγ
(cid:18) n(cid:19)
Proposition2.3follows.
Using Proposition 2.1(2), Proposition 2.3 and (C), for ρ = 1/2 + iγ and s = σ + it
n n
(t > 10),weget
ζ′ 1 1
(s) = + +O(logt)
ζ s ρ s ρ
n
− ρX6=ρn −
0<|γ−t|61
1 1 1
= + + +O(logt)
s ρ s ρ s ρ
n m m
− 0<|γmX−γn|61 − |γmX−γn|>1 −
0<|γm−t|61
1 1
= + +O 1 +O(logt)
s ρ s ρ  
n m
− 0<|γmX−γn|61 − 0<|Xγ−t|61
1 1 1 M(t)
n
= + + +O(logt)
s ρ s ρ − i(t γ ) i
n m m
− 0<|γmX−γn|61 − −
1 σ 1
= +O − 2 +O(logt)
s ρ (γ γ )2
n m n !
− m6=n −
X