Some bounds and limits in the theory of Riemann's zeta function

SOME BOUNDS AND LIMITS IN THE THEORY OF RIEMANN’S ZETA FUNCTION JUAN ARIAS DE REYNA AND JAN VAN DE LUNE Abstract. For any real a > 0 we determine the supremum of the real σsuch that ζ(σ+it) = afor some real t. For 0 <a<1, a= 1, and a > 1 the results turn out to be quite different. We also determine the supremum Eof the real parts of the ‘turning points’, that is points σ+it where a curve Im ζ(σ+it) = 0 has a vertical tangent. This supremum E(also considered by Titchmarsh) coincides with the supremum of the real σsuch that ζ0(σ+it) = 0 for some real t. We find a surprising connection between the three indicated problems: ζ(s) = 1, ζ0(s) = 0 and turning points of ζ(s). The almost extremal values for these three problems appear to be located at approximately the same height. 1. Introduction. In this paper we study various bounds and limits related to the values of Riemann’s ζ(s) = ζ(σ+it) with sin the half-plane σ > 1. For example, in Titchmarsh [8, Theorem 11.5(C)] it is shown that E:= the supremum of all σsuch that ζ0(σ+it) = 0 for some t∈R, satisfies 2< E < 3. Also, one of us [3] proved that σ0:= the unique solution to the equation Pparcsin(p−σ) = π 2, is the supremum of all σsuch that Re ζ(σ+it)<0 for some t∈Rand Re ζ(σ0+it)>0 for all t∈R. In [4] and [5] we encounter the question of the supremum σ(1) of Re (s) for the solutions of ζ(s) = 1. In Sections 3and 4we will solve this problem and also answer the same question for the solutions of ζ(s) = afor any given a > 0. In Section 5we give a more direct proof of Theorem 11.5(C) of Titchmarsh. Our method is constructive so that it allowed us to find explicit roots of ζ(s) = 1 with σnear the extremal value σ(1) ( by means of the Lenstra–Lenstra–Lov´asz lattice basis reduction algorithm ), and analogously solutions of ζ0(s) = 0 with Re (s) near E. We also found a 1 arXiv:1107.5134v1 [math.NT] 26 Jul 2011 2 JUAN ARIAS DE REYNA AND JAN VAN DE LUNE relation between the two problems: Near every almost-extremal solution for ζ(s) = 1 there is one for ζ0(ρ) = 0 with ρ−s≈E−σ(1) ( see Section 6for a more precise formulation ). In Section 7we will discuss some similar aspects of general Dirichlet functions L(s, χ). There are two types of curves Im ζ(σ+it) = 0. One kind (the I1 curves) is crossing the halfplane σ > 0 more or less horizontally whereas the other kind (the I2curves) has the form of a loop. These loops do not stick out arbitrarily far to the right. In Section 9we determine exactly the limit of the I2curves Im ζ(σ+it) = 0. This problem was also mentioned in [3]. The somewhat surprising fact is that this limit of the I2curves is equal to the limit Eof the zeros of ζ0(s) considered in Theorem 11.5 (C) of Titchmarsh. 2. The key lemmas. We will use the following Lemma 2.1. There exists a sequence of real numbers (tk)such that lim k→∞ ζ(s+itk) = 2s−1 2s+ 1ζ(s) uniformly on compact sets of the half plane σ > 1. Proof. Since the numbers log pnare linearly independent over Q, there are ( by Kronecker’s theorem [1, Theorem 7.9, p. 150] ) for each positive integer Nand any η > 0 a real number tand integers g1, . . . , gNsuch that | − tlog 2 −π+ 2πg1|< η, | − tlog pj+ 2πgj|< η, 2≤j≤N where pndenotes the n-th prime number. Taking ηsmall enough we may obtain in this way a real tsuch that |2−it + 1|< ε, |p−it j−1|< ε, 2≤j≤N. Repeating this construction we obtain a sequence of real numbers (tk) such that lim k→∞ 2−itk=−1,lim k→∞ p−itk= 1 for any odd prime p. Now we prove that any such sequence satisfies the Lemma. For any natural number nlet ν(n) be the exponent of 2 in the prime factorization of n. Let n= 2ν(n)qa1 1· · · qar rbe the prime factorization of n. Then SOME BOUNDS IN THE THEORY OF RIEMANN’S ZETA FUNCTION 3 we will have n−itk→(−1)ν(n), and as we will show lim k→∞ ζ(s+itk) = ∞ X n=1 (−1)ν(n) nsuniformly for σ≥a > 1. Given a > 1 and ε > 0 we first determine Nsuch that ∞ X n=N+1 1 na< ε. For 1 ≤n≤Nwe then have n−itk→(−1)ν(n), so that there exists a Ksuch that |n−itk−(−1)ν(n)|<ε N,1≤n≤N, k ≥K. For Re (s) = σ≥aand k > K we will then have   ζ(s+itk)−∞ X n=1 (−1)ν(n) ns  ≤   ∞ X n=1 n−itk−(−1)ν(n) ns  ≤ ≤ N X n=1 |n−itk−(−1)ν(n)|n−a+ 2 ∞ X n=N+1 n−a≤3ε. Finally we check whether ∞ X n=1 (−1)ν(n) ns=∞ X j=0 (−1)j 2js  ∞ X k=0 1 (2k+ 1)s= =1 + 1 2s−1Y p≥3 1−1 ps−1=1−1 2s 1 + 1 2s ζ(s) = 2s−1 2s+ 1ζ(s).  Lemma 2.2. There exists a sequence of real numbers (tk)such that lim k→∞ ζ(s+itk) = ζ(2s) ζ(s) uniformly on compact sets of the half plane σ > 1. Proof. The proof is similar to that of the previous Lemma. Applying Kronecker’s theorem we get a sequence of real numbers (tk) such that lim k→∞ p−itk=−1 for all primes p. Similarly as in the proof of Lemma 2.1 we obtain lim k→∞ ζ(s+itk) = ∞ X n=1 (−1)Ω(n) nsuniformly for σ≥σ0>1 4 JUAN ARIAS DE REYNA AND JAN VAN DE LUNE where Ω(n) is the total number of prime factors of ncounting multiplicities. It is well known that this series is equal to ζ(2s) ζ(s)(see Titchmarsh [8, formula (1.2.11)]).  To apply these lemmas we will use a theorem of Hurwitz (see [7, Theorem 3.45, p. 119] or [2, Theorem 4.10d and Corollary 4.10e, p. 282– 283]). We will use it in the following form: Theorem 2.3 ((Hurwitz)).Assume that a sequence (fn)of holomorphic functions on a region Ωconverges uniformly on compact sets of Ω to the function fwhich has an isolated zero a∈Ω. Then for n≥n0 the functions fnhave a zero an∈Ωsuch that limnan=a. 3. The bound for ζ(s) = a(>0) with a6= 1. For a positive real number alet σ(a) denote the supremum of all real σsuch that ζ(σ+it) = afor some t∈R. Theorem 3.1. Let abe >0but 6= 1. If a > 1then σ(a)is the unique solution of ζ(σ) = awith σ > 1. If 0< a < 1then σ(a)is the unique solution of ζ(2σ) ζ(σ)=awith σ > 1. Proof. It will be convenient to define σaas the ( unique ) solution of the equations considered in the theorem. The case a > 1. It is easily seen that in this case we have σ(a) = σa. In the case 0 < a < 1 we consider a solution to ζ(s) = a. Then a=|ζ(s)|=Y p 1   1−1 ps   ≥Y p 1 1 + 1 pσ =ζ(2σ) ζ(σ),(σ > 1). It is clear from the last equality that ζ(2σ) ζ(σ)is strictly increasing ( for σ > 1 ) from 0 to 1. Hence, there exists a unique solution σato the equation a=ζ(2σ) ζ(σ). The inequality a≥ζ(2σ) ζ(σ)is then equivalent to σ≤σa. Taking the supremum of σfor all solutions of ζ(s) = awe obtain σ(a)≤σa. To prove the converse we apply Lemma 2.2: There exists a sequence of real numbers (tk) such that ζ(s+itk)−aconverges uniformly on compact sets of σ > 1 to the function ζ(2s) ζ(s)−a. The limit function has a zero at s=σa. So, by Hurwitz’s theorem σais a limit point of zeros bk(k≥k0) of ζ(s+itk)−a. Therefore ζ(bk+itk)−a= 0 and limkbk=σa. For sk:= bk+itkwe have ζ(sk) = aand lim kRe (sk) = lim kRe (bk) = Re (lim kbk) = σa. SOME BOUNDS IN THE THEORY OF RIEMANN’S ZETA FUNCTION 5 It follows that σ(a) = sup{σ:ζ(s) = a} ≥ lim kRe (sk) = σa. Therefore σ(a) = σa, proving our theorem.  4. The bound for ζ(s) = 1. Theorem 4.1. The supremum σ(1) of all real σsuch that ζ(σ+it) = 1 for some value of t∈R, is equal to the unique solution σ > 1of the equation (1) ζ(σ) = 2σ+ 1 2σ−1. Numerically we have σ(1) = 1.94010 16837 43625 28601 74693 90525 54887 82302 47607 . . . Proof. Assume that ζ(s) = 1 with Re (s) = σ > 1. Then by the Euler product formula 1−1 2s=Y p≥3 1−1 ps−1=∞ X k=1 1 (2k−1)s or −1 = ∞ X k=2 2 2k−1s. Therefore 1 =    ∞ X k=2 2 2k−1s  ≤∞ X k=2 2 2k−1σ. Since the right hand side is decreasing in σ, it follows that there is a unique solution σ1of the equation (2) 1 = ∞ X k=2 2 2k−1σ= (2σ−1)ζ(σ)−2σ and that σ≤σ1. Now observe that (2) is equivalent to (1). Therefore, ζ(s) = 1 implies σ≤σ1which is by definition the solution of equation (1). Taking the sup over all solutions of ζ(s) = 1 we get σ(1) ≤σ1. For the converse inequality we apply Lemma 2.1 to get a sequence of real numbers (tk) such that lim k{ζ(s+itk)−1}=2s−1 2s+ 1ζ(s)−1 uniformly on compact sets of σ > 1. By definition σ1is a zero of the limit function 2s−1 2s+1 ζ(s)−1, so that there exists a natural number n0 6 JUAN ARIAS DE REYNA AND JAN VAN DE LUNE and a sequence of complex numbers (zk) such that ζ(zk+itk)−1 = 0 and limkzk=σ1. For sk:= zk+itkwe then have ζ(sk) = 1 and limkσk=σ1( with σk:= Re (sk) ). It follows that σ(1) = supζ(s)=1 Re s≥σ1, proving the theorem.  5. The bound for ζ0(s) = 0. A new proof of Titchmarsh’s Theorem 11.5(C). Theorem 11.5(C) in Titchmarsh [8] says that there exists a constant Ebetween 2 and 3, such that ζ0(s)6= 0 for σ > E, while ζ0(s) has an infinity of zeros in every strip between σ= 1 and σ=E. In this section we give a more direct proof of this theorem and determine the precise value of E. Theorem 5.1. Let Ebe the unique solution of the equation (3) 2σ+1 4σ−1log 2 = −ζ0(σ) ζ(σ),(σ > 1). Then ζ0(s)6= 0 for σ > E, while ζ0(s)has a sequence of zeros (sk)with limkRe (sk) = E. The value of this constant is E= 2.81301 40202 52898 36752 72554 01216 68696 38461 40560 . . . Proof. Assuming that ζ0(s) = 0 ( for σ > 1 ) we have ζ0(s) ζ(s)=d ds log ζ(s) = d ds X p −log1−1 ps=−X p log p ps−1 so that we may write the equation ζ0(s) = 0 as X p log p ps−1= 0 or −log 2 2s−1=X p≥3 log p ps−1. So, we must necessarily have log 2 2σ+ 1 ≤  −log 2 2s−1  =  X p≥3 log p ps−1  ≤X p≥3 log p pσ−1 and we may write this inequality as log 2 ≤X p≥3 (2σ+ 1)1 pσ+1 p2σ+1 p3σ+· · · log p. SOME BOUNDS IN THE THEORY OF RIEMANN’S ZETA FUNCTION 7 Since the right hand side is strictly decreasing in σthis is equivalent to σ≤E:= the unique solution of the equation log 2 2σ+ 1 +log 2 2σ−1=X p≥2 log p pσ−1 which is equivalent to (3). This proves that there is no zero of ζ0(s) with σ > E. Now we must find a sequence of complex numbers (sk) with ζ0(sk)=0 and limkRe (sk) = E. By Lemma 2.1 ζ0(s+itk) converges uniformly on compact sets of σ > 1 to the function d ds 2s−1 2s+ 1ζ(s) = 2s+1 4s−1log 2 + ζ0(s) ζ(s)·2s−1 2s+ 1ζ(s). This function has a zero at s=E(see equation (3)), so that by Hurwitz’s theorem, there exist for k≥k0numbers zksuch that zk→E and ζ0(zk+itk) = 0. Taking sk=zk+itkwe will have ζ0(sk) = 0 and lim kRe (sk) = lim kRe (zk+itk) = lim kRe (zk) = E as we wanted to show. With Mathematica we found that the solution to equation (3) is approximately the number given in the theorem.  6. The connection between ζ(s) = 1 and ζ0(s) = 0. We have seen that to get points with ζ(s) = 1 and σnear σ(1), and points ρwith ζ0(ρ) = 0 and Re ρnear E, we have applied in both cases Lemma 2.1. The limit function f(s) := 2s−1 2s+1ζ(s) satisfies f(σ(1)) = 1 and f0(E) = 0. Hence, from the approximate function ζ(s+itk) we may obtain simultaneously points sand ρwith ζ(s) = 1 and ζ0(ρ) = 0 and more or less to the same height tk. We will say that a sequence of complex numbers (sn) is almost extremal for ζ(s) = 1 if ζ(sn) = 1 and limnRe (sn) = σ(1). Analogously (ρn) is said to be almost extremal for ζ0(s) = 0 if ζ0(ρn) = 0 and limnRe (ρn) = E. First we prove that an almost extremal sequence is related to the situation of Lemma 2.1. Theorem 6.1. (a) If (sn)is an almost extremal sequence for ζ(s)=1, then tn:= Im (sn)satisfies (4) lim n→∞ 2−itn=−1,lim n→∞ p−itn= 1 for every odd prime p. 8 JUAN ARIAS DE REYNA AND JAN VAN DE LUNE (b) If (ρn)is an almost extremal sequence for ζ0(s)=0, then tn:= Im (ρn)also satisfies (4). Proof. (a) Let sn=σn+itn. Since limnσn=σ(1) >1 we may assume that σn>1 for all n. As in the proof of Theorem 4.1 the equation ζ(sn) = 1 may be written as −1 = ∞ X k=2 2 2k−1σn+itn. Since limnσn=σ(1), we see that σnconverges to the unique solution to the equation 1 = ∞ X k=2 2 2k−1σ. Therefore ∞ X k=2 2 2k−1σn+itn=−∞ X k=2 2 2k−1σ(1) so that, for all n∈Nwe have (5) ∞ X k=2 2 2k−1σ(1)1 + 2 2k−1σn−σ(1)+itn= 0. We now prove that for each k≥2 we must have (6) lim n2 2k−1itn=−1. We proceed by contradiction and assume that (6) is not true for some k0. Since the absolute value of 2 2k−1itnis 1, there must exist a subsequence njsuch that lim j2 2k0−1itnj=ak06=−1,|ak0|= 1. By a diagonal argument we may assume that for this subsequence we also have the limits lim j2 2k−1itnj=ak,|ak|= 1, k 6=k0. Now consider the equation (5) for n=njand take the limit for j→ ∞. Interchanging limit and sum we then obtain ∞ X k=2 2 2k−1σ(1)(1 + ak) = 0. Now take real parts in this equation. Since Re (1 + ak)≥0 but Re (1+ ak0)>0 we get a contradiction, proving (6). SOME BOUNDS IN THE THEORY OF RIEMANN’S ZETA FUNCTION 9 Hence, for any kwe have (6). Now if pis an odd prime we have p= 2k+ 1 and p2= 2m+ 1 so that lim n2 pitn=−1,lim n2 p2itn=−1. Hence lim npitn=2 pitn·2 p2−itn= 1 so that lim n2itn= lim n2 pitnpitn=−1. (b) Assume now that (ρn) is an almost extremal sequence for ζ0(s) = 0. Let ρn=σn+itn. Since limnσn=E > 1 we may assume that σn>1 for all n. As in the proof of Theorem 5.1 we will have log 2 2σn+ 1 ≤  −log 2 2ρn−1  =  X p≥3 log p pρn−1  ≤X p≥3 log p pσn−1. Since limnσn=Eand Esatisfies equation (3) we have lim n→∞ log 2 2σn+ 1 = lim n→∞ X p≥3 log p pσn−1 so that (7) lim n→∞  −log 2 2ρn−1  =log 2 2E+ 1 =X p≥3 log p pE−1= lim n→∞  X p≥3 log p pρn−1  . The first equality in (7) implies that limn|1−2σn+itn|= 1 + 2E. Let abe a limit point of the sequence (2itn). We may choose a sequence (nk) such that limk2itnk=a. Then limk|1−2σnk+itnk|=|1−2Ea|= 1+2E. Since |a|= 1 this is possible only if a=−1. Therefore, (2itn), beeing a bounded sequence with a unique limit point, is convergent and limn2itn=−1. For each odd prime pthe sequence (pitn) has 1 as unique limit point. Indeed, if not, then there is an odd prime qand a sequence (nk) with lim kqitnk=aq6= 1. By a diagonal argument we may assume that the limits limkpitnk=ap exist for each prime p. We will always have |ap|= 1. Taking limits in 16 JUAN ARIAS DE REYNA AND JAN VAN DE LUNE Now we introduce some notations. We may write equations (8) in the form (10) −∞ X k=1 1 k sin(kt log 2) 2kσ =X p≥3 ∞ X k=1 1 k sin(kt log p) pkσ −∞ X k=1 cos(kt log 2) 2kσ log 2 = X p≥3 ∞ X k=1 cos(kt log p) pkσ log p. For σ > 0 and t∈Rwe now define (11) f(σ, t) := ∞ X k=1 1 k sin(kt log 2) 2kσ and (12) g(σ, t) := ∂ ∂tf(σ, t) = ∞ X k=1 cos(kt log 2) 2kσ log 2. Note that fand gare periodic functions of twith period 2π/ log 2. So, a turning point σ+it must satisfy −f(σ, t) = X p≥3 ∞ X k=1 1 k sin(kt log p) pkσ and −g(σ, t) = X p≥3 ∞ X k=1 cos(kt log p) pkσ log p. We now consider the function U(σ, t) := 22σf(σ, t)2+2σ log 22g(σ, t)2 the choice of the coefficients 22σand (2σ/log 2)2being motivated by ( use (11) and (12) ) lim σ→+∞U(σ, t) = lim σ→+∞n22σ∞ X k=1 1 k sin(kt log 2) 2kσ 2+2σ log 22∞ X k=1 cos(kt log 2) 2kσ log 22o = sin2(tlog 2) + cos2(tlog 2) = 1. Lemma 9.4. Let aand bbe arbitrary real numbers. Then there exist real numbers xand ysuch that ax +by = (a2+b2)1/2and x2+y2= 1. SOME BOUNDS IN THE THEORY OF RIEMANN’S ZETA FUNCTION 17 Proof. If a2+b2= 0 then a=b= 0 and we need only take xand y such that x2+y2= 1. If a2+b26= 0 then we can take x=a √a2+b2and y=b √a2+b2. Lemma 9.5. If σ+it is a turning point of ζ(s)with σ > A, then U(σ, t)<2σ log 22X p≥3 ∞ X k=1 log p pks 2. Proof. We apply Lemma 9.4 to a=−2σf(σ, t) and b=−2σ log 2g(σ, t) to get n22σf(σ, t)2+2σ log 22g(σ, t)2o1/2=−2σxf(σ, t)−2σ log 2yg(σ, t) which, by (10), may be written as U(σ, t)1/2= 2σX p≥3 ∞ X k=1 x k sin(kt log p) pkσ + +2σ log 2 X p≥3 ∞ X k=1 ycos(kt log p) pkσ log p= = 2σX p≥3 ∞ X k=1 x k sin(kt log p) pkσ +ylog p log 2 cos(kt log p) pkσ . Applying the Cauchy-Schwarz inequality to the right hand side we obtain the condition (13) U(σ, t)1/2≤ ≤2σX p≥3 ∞ X k=1 x2 k2+y2log2p log221/2sin2(kt log p) + cos2(kt log p) p2kσ 1/2. Now observe that in (13)1 k2<log2p log22so that x2 k2+y2log2p log22<(x2+y2) log2p log22≤log2p log22. Using this we thus obtain the condition U(σ, t)1/2<2σX p≥3 ∞ X k=1 log p log 2 1 pkσ 18 JUAN ARIAS DE REYNA AND JAN VAN DE LUNE or U(σ, t)<2σ log 22X p≥3 ∞ X k=1 log p pkσ 2 as we wanted to show.  For σ > 1 we define (14) H(σ) := 2σ log 22X p≥3 ∞ X k=1 log p pkσ 2. Lemma 9.6. For each t∈Rthere exists a largest solution u(t)to the equation in σ (15) U(σ, t) = H(σ) and U(σ, t)> H(σ),(σ > u(t)). Proof. By (14) it is easily seen that H(σ) is continuous and strictly decreasing for σ > 1 from +∞to 0. In particular lim σ→∞ H(σ)=0. Since U(σ, t) is continuous for σ > 0 and t∈R, and lim σ→+∞U(σ, t) = 1 we see that for every tthe infimum u(t) of the asuch that U(σ, t)> H(σ) for σ > a exists and is larger than 1. From this it is clear that u(t) must be a solution of equation (15) in σ. Lemma 9.7. We have the closed formulas f(σ, t) = arctan sin(tlog 2) 2σ−cos(tlog 2), g(σ, t) = −(1 −2σcos(tlog 2)) log 2 1+4σ−21+σcos(tlog 2). Proof. The first follows from the identity f(σ, t) = Im (log(1 −2−s)), and the second by differentiation.  Lemma 9.8. We have u(π/ log 2) = E. Proof. We have X p≥3 ∞ X k=1 log p pkσ =X p ∞ X k=1 log p pkσ −∞ X k=1 log 2 2kσ =−ζ0(σ) ζ(σ)−log 2 2σ−1 SOME BOUNDS IN THE THEORY OF RIEMANN’S ZETA FUNCTION 19 so that H(σ) = 2σ log 22ζ0(σ) ζ(σ)+log 2 2σ−12. By its definition u(t) is the largest solution of the equation U(σ, t) = H(σ). For t=π/ log 2 we have f(σ, t) = ∞ X k=1 1 k sin(kt log 2) 2kσ =∞ X k=1 1 k sin(kπ) 2kσ = 0 and g(σ, t) = ∞ X k=1 cos(kt log 2) 2kσ log 2 = ∞ X k=1 cos(kπ) 2kσ log 2 = =∞ X k=1 (−1)k 2kσ log 2 = −log 2 2σ+ 1 so that u(π/2) satisfies the equation 2σ log 22log 2 2σ+ 12=2σ log 22ζ0(σ) ζ(σ)+log 2 2σ−12. Since ζ0(σ) ζ(σ)+log 2 2σ−1<0 this is equivalent to log 2 2σ+ 1 =−ζ0(σ) ζ(σ)−log 2 2σ−1 or 2σ+1 4σ−1log 2 = −ζ0(σ) ζ(σ). But Eis the unique solution of this equation for σ > 1 ( see Theorem 5.1 ). Hence u(π/ log 2) = E. Lemma 9.9. For all σ > 1and all t∈Rwe have (16) U(σ, t)≥U(σ, π/ log 2). Proof. We have computed U(σ, π/ log 2) in the proof of Lemma 9.8. Substituting this value and the definition of U(σ, t), (16) may be written 22σf(σ, t)2+2σ log 22g(σ, t)2≥2σ log 22log 2 2σ+ 12. 20 JUAN ARIAS DE REYNA AND JAN VAN DE LUNE In view of Lemma 9.7 we thus need to prove (17) arctan2sin(tlog 2) 2σ−cos(tlog 2)+(1 −2σcos(tlog 2)) 1+4σ−21+σcos(tlog 2)2≥ ≥1 2σ+ 12. We change notations putting tlog 2 = ϕand 2σ=x−1, so that we have to prove for 0 < x < 1 and 0 <ϕ<2π (18) u(x, ϕ) := arctan2xsin ϕ 1−xcos ϕ+x(x−cos ϕ) 1 + x2−2xcos ϕ2≥x 1 + x2. The right hand side is the value for ϕ=πof the left hand side. So, we want to prove that u(x, ϕ) has an absolute minimum at ϕ=π. It is easy to show that u(x, π −θ) = u(x, π +θ). So, we only have to prove inequality (18) for 0 < ϕ < π. We will split the proof in two cases. (1) Proof of (18) for π 2< ϕ < π. If we differentiate u(x, ϕ) with respect to ϕand simplify we arrive at (19) uϕ(x, ϕ) = 2x(x−cos ϕ) (1 + x2−2xcos ϕ)3n−arctanxsin ϕ 1−xcos ϕ× ×(1 + x2−2xcos ϕ)2+x(1 −x2) sin ϕo. We will show that uϕ(x, ϕ)<0 for π 2< ϕ < π, so that (18) will follow. In this interval cos ϕ < 0 and sin ϕ > 0. The first factor in the right hand side of (19) is positive, and we will show that the second is negative. That is we will show that (20) x(1 −x2) sin ϕ≤arctanxsin ϕ 1−xcos ϕ(1 + x2−2xcos ϕ)2. Let (21) α= arctanxsin ϕ 1−xcos ϕ,tan α=xsin ϕ 1−xcos ϕ, 1 cos2α= 1 + xsin ϕ 1−xcos ϕ2=1 + x2−2xcos ϕ (1 −xcos ϕ)2, cos2α=(1 −xcos ϕ)2 1 + x2−2xcos ϕ, SOME BOUNDS IN THE THEORY OF RIEMANN’S ZETA FUNCTION 21 (22) sin2α= 1−(1 −xcos ϕ)2 1 + x2−2xcos ϕ=x2−x2cos2ϕ 1 + x2−2xcos ϕ=x2sin2ϕ 1 + x2−2xcos ϕ so that sin α=xsin ϕ p1 + x2−2xcos ϕ (with the sign + since certainly α∈(0, π/2), since tan α > 0). Now we have 1−x2<1<(1 + x2−2xcos ϕ)3/2 so that x(1 −x2) sin ϕ≤xsin ϕ(1 + x2−2xcos ϕ)3/2 and x(1 −x2) sin ϕ≤sin α(1 + x2−2xcos ϕ)2≤α(1 + x2−2xcos ϕ)2 which is equivalent to (20). (2) Proof of (18) for 0 <ϕ<π 2. Defining αas in (21), sin2αis still given by (22). Although in this case we do not know the sign of sin α, inequality (18) will still follow from (23) x2sin2ϕ 1 + x2−2xcos ϕ+x(x−cos ϕ) 1 + x2−2xcos ϕ2≥x 1 + x2 since sin2α < α2. To prove (23) we consider two cases. (2a) Proof of (23) when 1 + x2−2xcos ϕ > 1. Then (1 + x2−2xcos ϕ)2>1 + x2−2xcos ϕ, so that sin2ϕ 1 + x2−2xcos ϕ+(x−cos ϕ)2 (1 + x2−2xcos ϕ)2≥ ≥sin2ϕ (1 + x2−2xcos ϕ)2+(x−cos ϕ)2 (1 + x2−2xcos ϕ)2= =1 + x2−2xcos ϕ (1 + x2−2xcos ϕ)2=1 1 + x2−2xcos ϕ. Recall that 0 <ϕ<π 2. Then −2xcos ϕ < 2x, so that 1+x2−2xcos ϕ < 1 + x2+ 2x= (1 + x)2, and we obtain sin2ϕ 1 + x2−2xcos ϕ+(x−cos ϕ)2 (1 + x2−2xcos ϕ)2>1 (1 + x)2. 22 JUAN ARIAS DE REYNA AND JAN VAN DE LUNE (2b) Proof of (23) when 1 + x2−2xcos ϕ≤1. In this case (1 + x2− 2xcos ϕ)2≤1 + x2−2xcos ϕso that sin2ϕ 1 + x2−2xcos ϕ+(x−cos ϕ)2 (1 + x2−2xcos ϕ)2≥ ≥sin2ϕ (1 + x2−2xcos ϕ)+(x−cos ϕ)2 (1 + x2−2xcos ϕ)= =1 + x2−2xcos ϕ 1 + x2−2xcos ϕ= 1 >1 (1 + x)2.  Lemma 9.10. For each t∈Rwe have u(t)≤u(π/ log 2). Proof. By Lemma 9.6 U(σ, π/ log 2) > H(σ) for σ > u(π/ log 2) and by Lemma 9.9 U(σ, t)≥U(σ, π/ log 2). It follows that U(σ, t)> H(σ),(σ > u(π/ log 2)). By definition U(σ, t)> H(σ) is not true for σ=u(t), and it follows that u(t)≤u(π/ log 2).  Proof of the first half of Theorem 9.1.Let σ+it be a turning point for ζ(s). It is clear that σ≤A= 1.192 . . . implies σ < E = 2.813 . . . . For σ > A, by Lemma 9.5 we will have U(σ, t)< H(σ) so that Lemma 9.6 implies that σ < u(t). By Lemma 9.10 u(t)≤u(π/ log 2) and by Lemma 9.8 u(π/ log 2) = E. It follows that σ < E. Therefore, the supremum Tof the real parts of the turning points is less than or equal to E. We have even proved a little more: On the line σ=Ethere is no turning point.  SOME BOUNDS IN THE THEORY OF RIEMANN’S ZETA FUNCTION 23 We will now show that there is a sequence (bn) of turning points for ζ(s) such that limnRe (bn) = E. This will end the proof of Theorem 9.1. By Lemma 2.1 there exists a sequence of real numbers (tk) such that ζ(s+itk) converges to f(s) := 2s−1 2s+1ζ(s). Since f(E)=0.9. . . , f0(E) = 0, f00(E)=0.07 . . . , f000(E) = −0.17 . . . Eis a turning point for f(s). We are going to show that the functions ζ(s+itk) must have a turning point very near to E. We prove a slightly more general result. We break the proof in several lemmas. Given a holomorphic function fdefined on a disc with center at 0 and radius Rwe define the associated (continuous) function h(r, ϕ) = Im f(reiϕ) + iRe f0(reiϕ) so that reiϕ will be a turning point for f(z) if and only if h(r, ϕ) = 0. For each 0 < r < R let γrbe the curve ϕ: [0,2π)7→ h(r, ϕ). Proposition 9.11. Let f(z) = a0+a2z2+a3z3+· · · be a holomorphic function on ∆(0, R)the disc with center 0and radius R. Assume that a0>0,a2>0and a3<0. Then there exists an r0>0such that for 0< r < r0, the curve γrdoes not pass through z= 0 and the index ( the winding number ) of the curve γrwith respect to 0is ω(γr,0) = 1. To prove Proposition 9.11 we will use some lemmas. Lemma 9.12. Let fbe as in Proposition 9.11 and define u(r, ϕ) := Im f(reiϕ), v(r, ϕ) := Re f0(reiϕ). Then there exists r0such that for 0< r < r0, (r→0) u(r, ϕ) = a2r2sin 2ϕ+a3r3sin 3ϕ+O(r4) v(r, ϕ) = 2a2rcos ϕ+ 3a3r2cos 2ϕ+O(r3) uϕ(r, ϕ) = 2a2r2cos 2ϕ+ 3a3r3cos 3ϕ+O(r4) vϕ(r, ϕ) = −2a2rsin ϕ−6a3r2sin 2ϕ+O(r3) where the implicit constants do not depend on ϕ. Proof. Let f(z) = P∞ n=0 anznbe the power series of fat 0, and take r0less than the radius of convergence. Then f(z) = a0+a2z2+a3z3+∞ X n=4 anzn 24 JUAN ARIAS DE REYNA AND JAN VAN DE LUNE so that u(r, ϕ) = a2r2sin 2ϕ+a3r3sin 3ϕ+∞ X n=4 rnIm (aneinϕ) and uϕ(r, ϕ) = 2a2r2cos 2ϕ+ 3a3r3cos 3ϕ+∞ X n=4 rnIm (inaneinϕ) and for 0 < r < r0we will have    ∞ X n=4 rnIm (aneinϕ)  ≤r4∞ X n=4 |an|rn−4 0,    ∞ X n=4 rnIm (inaneinϕ)  ≤r4∞ X n=4 n|an|rn−4 0. The last two sums converge and this proves our lemma for uand uϕ. For vand vϕthe proof is similar.  We divide the interval [−π 8,15π 8] of length 2πin 8 intervals I1= [−π/8, π/8], I2= [π/8,3π/8], I3= [3π/8,5π/8], I4= [5π/8,7π/8], I5= [7π/8,9π/8], I6= [9π/8,11π/8], I7= [11π/8,13π/8], I8= [13π/8,15π/8]. Lemma 9.13. There exists an r0>0such that for 0< r < r0the function uhas exactly four zeros on [−π/8,15π/8], denoted by α1∈I1, α3∈I3,α5∈I5and α7∈I7, so that uis positive on (α1, α3), negative on (α3, α5), positive on (α5, α7)and negative on (α7, α1+ 2π) Proof. By Lemma 9.12 for r→0 u(r, ϕ) = a2r2(sin 2ϕ+O(r)), uϕ(r, ϕ)=2a2r2(cos 2ϕ+O(r)). On I2and I6sin 2ϕ > 2−1/2, whereas sin 2ϕ < −2−1/2on I4and I8. Then, if we take r0small enough, u(r, ϕ)>0 on I2and I6, and u(r, ϕ)< 0 on I4and I8(we only need to take the O(r) terms less than 2−1/2). By continuity of u(r, ϕ) this implies that for each 0 < r < r0the function u(r, ϕ) has at least one zero on each of the intervals I1,I3, I5and I7. But cos 2ϕ > 2−1/2on I1and I5, and cos 2ϕ < −2−1/2on I3and I7, so that choosing r0small enough the sign of uϕ(r, ϕ) will be negative on I3and I7and positive on I1and I5. Therefore on each of these intervals the function u(r, ϕ) is monotonic and has only one zero.  There is an analogous result for v(r, ϕ). SOME BOUNDS IN THE THEORY OF RIEMANN’S ZETA FUNCTION 25 Lemma 9.14. There exists an r0>0such that for 0< r < r0the function v(r, ϕ)has exactly two zeros for ϕ∈[−π/8,15π/8], denoted by β3∈I3and β7∈I7, so that v(r, ϕ)is negative on (β3, β7), and positive on (β7, β3+ 2π). Proof. Observing that v(r, ϕ) = 2a2r(cos ϕ+O(r)), the proof is similar to that of Lemma 9.13. Lemma 9.15. There exists an r0>0such that for 0< r < r0the zeros of u(r, ϕ)and v(r, ϕ)satisfy the relation α3< β3, β7< α7. Proof. Putting a=−a3/a2>0 we have for 0 < r < r0(r0small enough to make the previous lemmas valid) u(r, ϕ) = a2r2(sin 2ϕ−ar sin 3ϕ+O(r2) v(r, ϕ) = 2a2r(cos ϕ−3a 2rcos 2ϕ+O(r2) with O-constants independent of ϕ. The two zeros α3and β3are on I3an interval with center at π 2. At the point π 2+ar we have u(r, π/2 + ar) a2r2=ar cos(3ar)−sin(2ar) + O(r2) v(r, π/2 + ar) 2a2r=3ar 2cos(2ar)−sin(ar) + O(r2). Expanding in Taylor series we get u(r, π/2 + ar) a2r2=−ar +O(r2) v(r, π/2 + ar) 2a2r=ar 2+O(r2). Choosing r0small enough we obtain u(r, π/2+ar)<0< v(r, π/2+ar) for 0 < r < r0. Since both u(r, ϕ) and v(r, ϕ) are decreasing on this interval, the zero of u(r, ϕ) must come before π 2+ar and the zero of v(r, ϕ) must come after π 2+ar. That is α3<π 2+ar < β3.