A note on Hardy s theorem

Transcription

1 A noe on Hardy s heorem Usha K. Sangale o cie his version: Usha K. Sangale. A noe on Hardy s heorem. Hardy-Ramanujan Journal, Hardy-Ramanujan Sociey, 207, Volume 39, pp <hal > HAL Id: hal hps://hal.archives-ouveres.fr/hal Submied on 3 Jan 207 HAL is a muli-disciplinary open access archive for he deposi and disseminaion of scienific research documens, wheher hey are published or no. he documens may come from eaching and research insiuions in France or abroad, or from public or privae research ceners. L archive ouvere pluridisciplinaire HAL, es desinée au dépô e à la diffusion de documens scienifiques de niveau recherche, publiés ou non, émanan des éablissemens d enseignemen e de recherche français ou érangers, des laboraoires publics ou privés.

2 Hardy-Ramanujan Journal ), submied 3/05/206, acceped 3/05/206, revised 3/05/206 A noe on Hardy s heorem Usha K. Sangale Absrac. Hardy s heorem for he Riemann zea-funcion ζs) says ha i admis infiniely many complex zeros on he line Rs) =. In his noe, we give a simple proof of his saemen which, o he bes of our knowledge, is new. 2 Keywords. Riemann zea-funcion, Hardy s heorem, Hardy s Z-funcion. 200 Mahemaics Subjec Classificaion. E45 primary); M4 secondary).. Inroducion he Riemann Hypohesis RH) is perhaps one of he mos challenging problems in mahemaics o his dae. In 859, Riemann assered ha all he complex zeros of he Riemann zea funcion ζs), s = σ + i lie on he criical line σ = /2. In oher words, RH assers ha ζs) 0 in σ > /2..) So far a proof or disproof of his saemen has eluded he bes minds. he sharpes known resul in his direcion is by I. M. Vinogradov [Vi36] who proved ha here exiss a posiive consan c such ha ζs) 0 in σ > clog ) 2/3 log log ) /3..2) Riemann zea-funcion was invened by Euler o sudy he disribuion of primes. here is a deep connecion beween he zero-free region of ζs) and he error erm in he prime number heorem PN) see [Iv85, Chap. 2]). Le πx) coun he number of primes upo x, hen he prime number heorem says ha here is an absolue consan C > 0 such ha πx) = x 2 { log u du + O x exp Clog x) 3/5 log log x) /5)}. he above error erm is a consequence of.2), which is he bes known uncondiional resul whereas he error on RH is Ox /2+ε ). I was G. H. Hardy who in 94 [Ha4] showed ha ζs) has infiniely many zeros on he criical line σ = /2. his is generally known as a firs sep owards RH! he mehod of Hardy, simplified furher by Hardy and Lilewood [HaLi8] paved way o esablish analogous resuls for more general Dirichle Series see [MRS08]). ichmarsh s famous book on Riemann zea-funcion [i86, Chap. X] discusses several mehods of proving Hardy s heorem. he basic idea behind he proof of Hardy s heorem is very simple. Suppose fx) C [a, b], and fx) 0 in a x b, hen b b fx) dx = fx) dx..3) a herefore, if.3) is violaed, hen i shows ha fx) has a zero of odd order in [a, b]. he proof of Hardy s heorem involves esimaing a cerain inegral boh from above and below and arriving a a conradicion by comparing hese esimaes. he esimaion from below is sraighforward for he Riemann zea funcion and no oo difficul in he case of a general Dirichle series hanks o a beauiful heorem of Ramachandra see [MRS08]). However, obaining good upper bounds, which will lead o a conradicion, is a difficul problem! We hank episciences.org for providing open access hosing of he elecronic journal Hardy-Ramanujan Journal a

3 U.K. Sangale, A noe on Hardy s heorem 39 In his aricle, we shall firs discuss he classical approach of Hardy s heorem as given in [i86, 0.5]. hen in he nex secion we shall give an alernaive proof of his heorem, which is simpler and seems o be new! 2. Hardy s heorem for he Riemann zea-funcion he Riemann zea funcion, denoed by ζs), s = σ + i, is defined by he series ζs) = I saisfies he following analyic properies: n= n s, σ >. i) Analyic Coninuaion: s )ζs) is an enire funcion of order. ii) Funcional Equaion: π s/2 Γs/2)ζs) = π s)/2 Γ s)/2)ζ s). 2.4) iii) Euler Produc: ζs) = p p s ), σ >, where he produc is over all primes. I is well known ha he gamma funcion Γs) is analyic in σ > 0 and has no zeros. I has simple poles a s = 0,, 2, 3,.... his fac, along wih he funcional equaion 2.4) implies ha ζs) has zeros a s = 2, 4, 6,.... hese zeros on he real line are called rivial zeros of ζs). From he heory of enire funcions of finie order, i can be derived ha all he complex zeros of ζs) have real pars 0 σ. I is also well known ha ζ + i) 0 for 0 and herefore ζi) 0 by 2.4). Furher, i is an elemenary exercise o show ha ζs) 0 for 0 < s <. hus, all he complex zeros of ζs) lie in he verical srip 0 < σ <. hese zeros are called non-rivial zeros and he Riemann Hypohesis assers ha all he non-rivial zeros are, in fac, on he line σ = /2. Now, he heorem of Hardy for he Riemann zea-funcion is he following: Hardy s heorem. ζs) has infiniely many zeros on he line σ = /2. Proof. he funcional equaion 2.4) can be rewrien as where χs) = π s /2 Γ 2 s))/γ 2 s). ζs) = χs)ζ s), 2.5) Observe ha χs)χ s) = or χ 2 + i)χ 2 i) =, i.e., χ 2 + i) =. From 2.5), we obain χ 2 + i) /2 ζ 2 + i) = χ 2 i) /2 ζ 2 i). hus he funcion Z) := χ 2 + i) /2 ζ 2 + i) is real for real values of 0. Moreover, Z) = ζ 2 + i). he funcion Z is popularly called Hardy s Z-funcion. From he above discussion, i is clear ha he zeros of Z) correspond o he zeros of ζs) on he criical line. We shall apply he basic idea elucidaed in he beginning by aking f = Z in.3). o begin wih, we shall assume ha Z) has finiely many zeros or no zeros, hen we can find a real number 0 > 0 sufficienly large such ha for all 0, we have I := 2 Z)d = 2 Z) d. 2.6)

4 40 2. Hardy s heorem for he Riemann zea-funcion Now, we need o esimae he inegrals in 2.6) boh from below and above and arrive a a conradicion o our assumpion. Esimaion from below: his is he easy sep. We have 2 Z) d = Now, by a change of variable we ge 2 2 ζ 2 + i) d ζ. 2 + i)d 2.7) 2 /2+2i i ζ 2 + i)d = ζs)ds. /2+i Moving he line of inegraion o σ = 2 and using Cauchy s heorem, we obain /2+2i /2+i 2+i ζs)ds = /2+i Now, from he convexiy bound 2.6), i follows ha 2+i /2+i ζs)ds = 2 /2 2+2i /2+2i ) + + ζs)ds. 2.8) 2+i 2+2i ζσ + i )dσ = O /4+ε). 2.9) On he oher hand, 2+2i 2+i ζs)ds = 2+2i 2+i = i + i = i + O + ) n s ds n=2 2 n 2 e i log n d n=2 n=2 n 2 log n ) = i + O). 2.0) herefore, from 2.7), 2.8), 2.9) and 2.0), we obain where A > 0 is an absolue consan. 2 Z) d > A. 2.) he main problem, as remarked before, is o find an upper bound of he ype I α. 2.2) If 0 < α <, his will conradic 2.) and hus esablishes he heorem. Remark. As χ 2 + i) =, aking rivial esimae we see ha 2 Z)d = 2 2 χ 2 + i) /2 ζ 2 + i)d ζ 2 + i) d. Even on Lindelöf hypohesis which is a consequence of RH and which says ha ζσ + i) ε for σ 2 ) he above is only O +ε ), no enough o conradic 2.).

5 U.K. Sangale, A noe on Hardy s heorem 4 hus we need o esimae I nonrivially, which we do below. Esimaion from above: Our aim is o esimae he inegral I = /2+2i /2+i χs) /2 ζs)ds. We move he line of inegraion o he righ of he line σ = so ha ζs) can be replaced by is series represenaion and hen evaluae he inegral using Cauchy s heorem. Accordingly, le us consider he recangle R wih he verices a /2 + i, + δ + i, + δ + 2i, /2 + 2i for some δ > 0 o be chosen laer. As χs) 0 in σ > 0, χs) /2 is single-valued analyic funcion in σ /2, herefore, by Cauchy s heorem χs) /2 ζs)ds = 0. 2πi R hus, /2+2i χs) /2 ζs)ds = +δ+i +δ+2i + + 2πi /2+i 2πi /2+i +δ+i /2+2i +δ+2i ) χs) /2 ζs)ds = I + I 2 + I 3 say). 2.3) As I and I 3 are of he same order of magniude, i is enough o esimae I only. We require growh esimaes for he inegrand in he region σ /2. he well-known Sirling s formula for Γs) says ha in any fixed srip α σ β, we have, Γσ + i) = σ+i 2 e 2 π i+ 2 iπσ 2 ) 2π) 2 + O ) ), as. his gives as. herefore, χs) = /2π) /2 σ+i) e i+π/4) + O ) ), 2.4) χs) /2 = O σ/2 /4), 0 σ + δ. 2.5) We also require an upper bound for ζs) in he criical srip which is uniform in σ. his is obained by using convexiy principle. Firs, one shows ha ζs) = Olog ) for σ. hen 2.5) combined wih 2.4) gives ζ0 + i) = O /2 log ). Finally he Phragmen-Lindelöf convexiy principle implies ) ζs) = O σ)/2 log ), uniformly in 0 σ. 2.6) Now, combining 2.5) and 2.6), we obain { χs) /2 O /4 log ) ), 0 σ ζs) = O /4+δ/2 ), σ + δ. hus, I = +δ /2 χσ + i ) /2 ζσ + i ) dσ = O Now comes he difficul par, he esimaion of I 2, which we carry ou below. /4+δ/2). 2.7)

6 42 2. Hardy s heorem for he Riemann zea-funcion We have I 2 = 2πi = 2π By Sirling s formula 2.4), we have herefore, he above inegral is +δ+2i +δ+i 2 χs) /2 ζs)ds χ + δ + i) /2 ζ + δ + i))d. χ + δ + i) /2 = 2π ) /4+δ/2+i/2 e i/2+π/8) { + O/)}. = n= 2 ) /4+δ/2+i/2 n +δ 2π e i/2+π/8) n i { + O/)} d. he O-erm is O 2 3/4+δ/2 d) = O /4+δ/2 ) = o ) provided δ < 3/2. he main erm is a consan muliple of = n= n= 2 ) /4+δ/2+i/2 n +δ 2π e i/2 i log n d 2 ) /4+δ/2 i n +δ 2π e 2 log/2πen2) d. 2.8) Observe ha a rivial esimaion of he inegral in 2.8) will no yield he desired upper bound, hus we need o exploi he possible cancellaions arising from he exponenial erm in he inegrand and hope ha he resuling esimae will be small! he following resuls are sandard ools in he heory of exponenial inegral echniques for insance see [i86, pages 7 72]). Lemma 2.. Le F x) be a real-valued differeniable funcion such ha F x) is monoonic, and F x) m > 0, or F x) m < 0, hroughou he inerval [a, b]. hen b a e if x) dx 4 m Lemma 2.2. Le F x)be wice differeniable funcion and F x) r > 0 or F x) r < 0, hroughou he inerval [a, b] and le Gx)/F x) be monoonic wih Gx) M, hen b a Gx)e if x) dx 8M r Now, aking F n ) = 2 log ), we see ha F 2πen 2 n ) = /2 /4, 2. hus from he Lemma 2.2 we obain 2 ) /4+δ/2e i 2 log/2πen2) 2π d = O /4+δ/2 /2) = O 3/4+δ/2). 2.9) Now combining 2.3), 2.7), 2.8) and 2.9), we obain 2 Z)d = O 3/4+δ/2 ) = o ) provided δ < / )

7 U.K. Sangale, A noe on Hardy s heorem 43 herefore, he desired conradicion is arrived from comparing 2.) and 2.20), by choosing δ < /2. Remark. Anoher mehod of esimaing he upper bound for I uses he following weak version of he well-known Riemann-Siegel formula ) /2π) Z) = 2 n /2 cos log n 2 π + O /4). 2.2) 8 n /2π) he inegral I is hen evaluaed using Lemma 2.2 o obain I = O 3/4 ) see [Iv3, Chap. 2] for deails). 3. A new approach using firs approximaion formula for ζs) In his mehod, he inegral I = 2 χ 2 + i) /2 ζ 2 + i)d 3.22) is evaluaed by replacing he zea funcion wih a Dirichle polynomial having a good error erm. As menioned in he remark above, a weak version of Riemann-Siegel formula can be used for he upper bound esimaion. We will use he well-known firs order approximaion of ζs) see [i86, page 77]) o esimae I boh from below and above. he esimaion from above seems o be new. Lemma 3.. We have ζs) = n x uniformly for σ σ 0 > 0, < 2πx/C when C is a given consan >. n s + x s s + Ox σ ), 3.23) aking x = C/π in he above Lemma and noing ha 2, we ge ζ 2 + i) = ) n 2 i /2 + O + O /2 ). 3.24) n C/π For he lower bound, recall ha we have 2 herefore, from 3.24) we have 2 Z) d = where A > 0 is an absolue consan. herefore, we have 2 2 ζ 2 + i)d = { + 2 ζ 2 + i) d ζ. 2 + i)d 3.25) 2 n C /π n 2 i }d + O /2 ) 3.26) = + O /2 ) > A, 3.27) 2 Z) d A. 3.28) On he oher hand, for esimaion of upper bound, we have from 3.24), ) Z) = χ 2 + i) /2 n 2 i /2 + O + O /2 ). 3.29) n C/π

8 44 References he O-erms conribues O /2 ) o he inegral I. hus, by Sirling s formula for χs), we have I = 2 n C/π n 2 i ) i/2 2π e i/2+π/8) { + O )} d + O /2 ). 3.30) Noe ha he error erm in he above expression is O /2 ). hus he main erm is a consan muliple of 2 e if n,) d, n /2 n C/π 3.3) where F n, ) = 2 log ) 2πen. Observe ha 2 F n, ) = 2 log ) 2πn, F n, ) = 2 2 > ) As F n, ) = 0 when = 2πn 2 or when n = /2π, we break he sum over n ino wo pars viz., n 3 /π and 3 /π < n C/π. he inegral in he firs sum is evaluaed using Lemma 2.2 and he inegral in he second sum is evaluaed using Lemma 2.. hus from Lemma 2.2 and 3.32) we obain n 3 /π 2 n /2 e if n,) d = O n 3 /π n /2 /2 = O o evaluae he second sum, we use he elemenary inequaliy log n m n m n ) ) log 2πn 2 2πn 2 8 2πn 2 9, 3/4). 3.33) for n > m. his gives as 2 and 3 /π < n C/π. hus F n, ) = 2 log ) 2πn 4/9. Now applying 2 Lemma 2., we ge 2 n /2 e if n,) d = O n /2 = O /2). 3.34) 3 /π<n C/π 3 /π<n C/π hus, i follows ha 2 Z)d = O 3/4 ). 3.35) Equaion 3.28) and 3.35) leads o a conradicion o our assumpion and hus esablishes Hardy s heorem. Acknowledgemens. I wish o hank Prof. K. Srinivas for many useful discussions. I also hank Prof. A. Ivić for his advice and encouragemen. I express my graiude o he Insiue of Mahemaical Sciences for providing congenial environmen which enabled me o carry ou his projec. his work consiues par of my PhD hesis. I sincerely hank Prof. C. S. Aravinda for he promp handling of he manuscrip and for suggesing some changes in he layou of aricle.

9 U.K. Sangale, A noe on Hardy s heorem 45 References [Ha4] G.H. Hardy, Sur les zeros de la foncion ζs), Comp. Rend. Acad. Sci., 58 94), [HaLi8] [Iv85] G.H. Hardy and J.E. Lilewood, Conribuions o he heory of he Riemann zea-funcion and he disribuion of primes, Aca Mah, 4 98), A. Ivić, he Riemann zea-funcion,he heory of he Riemann zea-funcion wih applicaions, A Wiley-Inerscience Publicaion. John Wiley & Sons, Inc., New York, 985. [Iv3] A. Ivić, he heory of Hardy s Z-funcion, Cambridge Universiy Press, UK, 203. [MRS08] A. Mukhopadhayay, K. Rajkumar and K. Srinivas, On he zeros of funcions in he Selberg class, Func. Approx. Commen. Mah. XXXVIII, ), [Vi36] I. M. Vinogradov, A new mehod of esimaion of rigonomerical sums, Ma. Sbornik ), ), [i86] E. C. ichmarsh, he heory of he Riemann zea-funcion, Second ediion. Edied and wih a preface by D.R. Heah-Brown, he Clarendon Press, Oxford Universiy Press, New York, 986. Usha K. Sangale SRM Universiy Nanded Maharashra, INDIA. ushas073@gmail.com