Resolution of a conjecture on the convexity of zeta functions

J. Math. Anal. Appl. 472 (2019) 1987–2016

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Resolution of a conjecture on the convexity of zeta functions Emre Alkan Department of Mathematics, Koç University, Rumelifeneri Yolu, 34450, Sarıyer, Istanbul, Turkey

a r t i c l e

i n f o

Article history: Received 10 January 2018 Available online 17 December 2018 Submitted by B.C. Berndt Dedicated to Professor Bruce Berndt on the occasion of his eightieth birthday Keywords: Riemann zeta function Convexity Concavity Log-convexity Geometric convexity Bias

a b s t r a c t We settle a conjecture of Cerone and Dragomir on the concavity of the reciprocal of the Riemann zeta function on (1, ∞). It is further shown in general that reciprocals of a family of zeta functions arising from semigroups of integers are also concave on (1, ∞), thereby giving a positive answer to a question posed by Cerone and Dragomir on the existence of such zeta functions. As a consequence of our approach, weighted type Mertens sums over semigroups of integers are seen to be biased in favor of square-free integers with an odd number of prime factors. To strengthen the already known log-convexity property of Dirichlet series with positive coefficients, the geometric convexity of a large class of zeta functions is obtained and this in turn leads to generalizations of certain inequalities on the values of these functions due to Alzer, Cerone and Dragomir. © 2018 Elsevier Inc. All rights reserved.

1. Introduction The goal of this paper is to settle a general form of a conjecture due to Cerone and Dragomir [12] about the concavity of the reciprocal of the Riemann zeta function on (1, ∞). In [11] and [12], the authors obtained a variety of new inequalities involving values of Dirichlet series with nonnegative coefficients by exploiting the log-convexity of such series (see Prop. 3.1 of [12]). The log-convexity of the Riemann zeta function was first shown by Gut [16] and Cerone and Dragomir [12] extended this to Dirichlet series with nonnegative coefficients. Let ψ(s) :=

∞  an ns n=1

be a Dirichlet series, where we assume that the coefficients an ≥ 0 for n ≥ 1 and the series is absolutely convergent for s > 1. Then in Problem 4.3 of [12], Cerone and Dragomir proposed to study the following. E-mail address: [email protected]. https://doi.org/10.1016/j.jmaa.2018.12.034 0022-247X/© 2018 Elsevier Inc. All rights reserved.

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Find Dirichlet series ψ with nonnegative coefficients which has the property that the function ψ1 is concave where it is defined. Let us remark that in general ψ1 may not even be represented by a Dirichlet series. A well-known example of this situation is furnished by the derivative of the Riemann zeta function. Precisely, if ζ(s) =

∞  1 ns n=1

is the Riemann zeta function for s > 1, then −ζ  (s) =

∞  log n ns n=2

is a Dirichlet series with nonnegative coefficients for s > 1. But −

1 ζ  (s)

can not be expressed as a Dirichlet series for s > 1. Cerone and Dragomir nicely treated Problem 4.3 and showed that ζ1 is concave on the interval [ζ −1 (e), ∞), where ζ −1 (e)  1.474464287. Here ζ −1 denotes the inverse function of ζ which clearly exists for s > 1 as ζ is strictly decreasing on (1, ∞) 1 in the MathSciNet review of [12]). Conjecture 4.8 of [12] (unfortunately, ζ −1 (e) was misinterpreted as ζ(e) 1 states that the concavity of ζ indeed persists on the whole interval (1, ∞). The strict concavity of ζ1 is seen to be a consequence of the condition (see Prop. 4.1 of [12]) ζ  (s)ζ(s) − 2(ζ  (s))2 > 0

(1.1)

on (1, ∞). With the help of a plot of ζ  (s)ζ(s) − 2(ζ  (s))2 for s ∈ (1, ζ −1 (e)), Cerone and Dragomir conjectured that (1.1) is true on the entire interval (1, ∞). This plot seems to be the strongest evidence of the truth of their conjecture. However, it can not be a substitute for a rigorous proof of the conjecture because of the singularities arising from ζ, ζ  and ζ  as s tends to 1. We should remark that the weaker inequality ζ  (s)ζ(s) − (ζ  (s))2 ≥ 0 on (1, ∞) easily follows from the Cauchy–Schwarz inequality. Moreover, if one launches an asymptotic analysis through the Laurent expansions of ζ, ζ  and ζ  at s = 1 (see Theorem 1 below for details), then 2γ ζ (s)ζ(s) − 2(ζ (s)) = +O (s − 1)3 



2



1 (s − 1)2



follows, where γ is the Euler–Mascheroni constant. Therefore, there exists a number δ > 0 with the property that (1.1) holds for 1 < s ≤ 1 + δ. Using a more detailed analysis of the O-constants along with some numerical approximations, we are able to give the value δ = 0.5 (see the proof of Theorem 1) which is good enough to settle the conjecture of Cerone and Dragomir. Although the asymptotic approach is fruitful, it is not easily adaptable to other zeta functions. In particular, it would fall short of proving Theorem 2 below which gives a satisfactory answer to Problem 4.3 of [12] and settles Conjecture 4.8 of [12] for a class of zeta

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functions with nonnegative coefficients. To fix some notation, let P be a set of prime numbers. Then the semigroup generated by all the prime numbers in P is the set ⎧ ⎫ k ⎨ ⎬ v P := pj j : k ≥ 1, vj ≥ 0, pj ∈ P . ⎩ ⎭ j=1

Alternatively, P can be viewed as the set of all remaining integers after sieving by the primes that are not in P. By convention, P = {1} when P is empty. Let μ be the Möbius function and let μP be the Möbius function supported on P, so that μP (n) = μ(n) when n ∈ P and μP (n) = 0 otherwise. For cancellation of sums involving μP , the reader is referred to [2]. Further let ζP (s) :=

 1 ns

n∈P

be the zeta function corresponding to P when (s) > 1. The Euler product representation ζP (s) =

 p∈P

1 1− s p

−1

shows that ζP (s) = 0 when (s) > 1. A real valued function f is convex if f (αx + βy) ≤ αf (x) + βf (y) holds for all x, y in the domain of f and for all nonnegative α, β with α + β = 1. f is concave if f (αx + βy) ≥ αf (x) + βf (y). Also f is strictly convex (or concave) if equality is possible only when α = 0, β = 0 or x = y. Clearly, f is (strictly) convex if and only if −f is (strictly) concave. Convex interpolation techniques played a major role in estimating the growth of the Riemann zeta function in the critical strip. If 0 ≤ σ ≤ 1, then denoting by m(σ) the least upper bound of the numbers A such that |ζ(σ + it)|t−A is bounded as t → ∞, Lindelöf [18] showed that m is a convex function and m(σ) ≤ 12 (1 − σ) for 0 ≤ σ ≤ 1. The Lindelöf hypothesis is the statement that m(σ) = 0 for σ ≥ 12 and m(σ) = 12 − σ for σ ≤ 12 . By the convexity of m, the Lindelöf hypothesis is equivalent to the single condition m 12 = 0 on the critical line which is known to follow from the Riemann hypothesis (see Chap. 14 of [24]). As will be confirmed below, geometric properties of zeta functions such as concavity imply the existence of a bias in favor of square-free numbers with an odd number of prime factors over semigroups. In this connection, a similar bias was discovered by Brent and van de Lune [9] for the Mertens function (the Mertens function being the summatory function of the Möbius function) through the formula lim

x→1−

∞  n=1

μ(n)

xn = −1 1 + xn

involving Lambert series type weights. We are now ready to state our main result. Theorem 1. For 1 < s ≤ 32 , the inequality ζ  (s)ζ(s) − 2(ζ  (s))2 ≥

  1 2γ 6γ1 8 − + 5γ − 2 (s − 1)3 (s − 1)2 π(2π − 1) (s − 1)   32 16(3π − 1) 2 (s − 1) + γγ2 − 2γ1 − γ3 − 2 + 3γ1 γ2 − γγ3 − 3 π (2π − 1) π (2π − 1)2

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4|γ2 | 16γ(3π − 1) 32|γ1 | − + γ1 γ3 − − 3 − 2 2 π (2π − 1) π (2π − 1) π(2π − 1)   16|γ1 |(3π − 1) 32|γ2 | 4γ3 − + + (s − 1)3 π 3 (2π − 1)2 π 2 (2π − 1) π(2π − 1)   128 64(3π − 1) − + 4 (s − 1)4 π 4 (2π − 1)2 π (2π − 1)3 2γ22

 (s − 1)2

(1.2)

holds, where γ  0.577215 is the Euler–Mascheroni constant and  γ1 = lim

m→∞



(log m)2  log r + − 2 r r=1 m

  −0.072815,

 m (log m)3  (log r)2 γ2 = lim − +  −0.00969, m→∞ 3 r r=1   m (log m)4  (log r)3 γ3 = lim − +  0.002053. m→∞ 4 r r=1 Consequently,   ∞  1 μ(n)  1 = = 1 − ζ(s) n=1 ns ps p is strictly concave on the interval (1, ∞), and there is a strict bias in favor of square-free numbers with an odd number of prime factors in the form  n≤y ω(n)≡0 (mod 2)

|μ(n)|(log n)2 < ns

 n≤y ω(n)≡1 (mod 2)

|μ(n)|(log n)2 ns

(1.3)

for all large enough y in terms of s when s > 1, where ω(n) denotes the number of distinct prime factors of n. Writing M (x) =





|μ(n)| −

n≤x ω(n)≡0 (mod 2)

|μ(n)|,

n≤x ω(n)≡1 (mod 2)

for the Mertens function, we see that negative or positive values of M (x) obviously lead to a bias in favor of square-free integers with an odd number of prime factors or with an even number of prime factors. However, by the Mellin transform of M (x), we have 1 =s ζ(s)

∞

M (x) dx xs+1

1

for (s) > σc , where σc is the abscissa of convergence of the Dirichlet series ∞  μ(n) ns n=1

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(we also know that σc ≥ 12 as ζ has complex zeros with real part 12 ). If M (x) is of constant sign for x ≥ x0 , then using the above integral representation and a classical result of Landau, one sees that ζ1 has a singularity at s = σc and this is absurd. Therefore, there is no permanent bias for all large x in either of the forms 



|μ(n)| <

n≤x ω(n)≡0 (mod 2)

|μ(n)|

n≤x ω(n)≡1 (mod 2)

and 



|μ(n)| <

n≤x ω(n)≡1 (mod 2)

|μ(n)|.

n≤x ω(n)≡0 (mod 2)

In contrast with this, if P is any (nonempty) set of prime numbers, then a strict bias in the form  n≤y ω(n)≡0 (mod 2)

|μP (n)| log n < nx

 n≤y ω(n)≡1 (mod 2)

|μP (n)| log n nx

holds for all large enough y when x > 1 since ζP1(x) is strictly increasing on (1, ∞). Let us point out that the concavity of ζP1(x) on (1, ∞) is equivalent to the inequality ζP (x)ζP (x) − 2(ζP (x))2 ≥ 0

(1.4)

for x > 1. Moreover, if P is a sparse enough set of prime numbers, then ζP (x) may be a regular function at x = 1 and the asymptotic approach mentioned above would completely fail in this generality for the purpose of proving (1.4). Nevertheless, our next result produces plenty of zeta function examples where the reciprocal is strictly concave on (1, ∞). One indeed reaches to a much stronger conclusion when P is finite. For the notation, let Λ be the von Mangoldt function and let ΛP be the von Mangoldt function supported on P so that ΛP (n) = Λ(n) when n ∈ P and ΛP (n) = 0 otherwise. We denote by f (m) the m-th order derivative of f , where f (0) = f . Theorem 2. Let P be a set of prime numbers and assume that p1 is the least prime number belonging to P. (i) If P is finite, m is a positive integer and β > 0, then when p1 is large enough only in terms of the size |P| of P, m and β, then  m

(−1)

1 ζP (x)

(m) <0

holds on (β, ∞). In particular, 

1 ζP

(m)

is strictly convex on (β, ∞) if m ≥ 1 is odd and it is strictly concave if m ≥ 0 is even. (ii) If P satisfies ∞  ΛP (m) < log p1 , m m=1

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then

1 ζP (x)

is strictly concave for x > 1 and a strict bias in the form  n≤y ω(n)≡0 (mod 2)

|μP (n)|(log n)2 < nx

 n≤y ω(n)≡1 (mod 2)

|μP (n)|(log n)2 nx

holds for all large enough y in terms of x when x > 1. (iii) If p and q are distinct prime numbers and P = {p, q}, then ζP1(x) is not concave on the entire interval (0, ∞). (iv) Let P = {q1 < q2 < . . . } be a set of prime numbers and let πP (x) be the number of prime numbers in P that are ≤ x. Denote by Pn = {qn < qn+1 < . . . }. If c1

x x < πP (x) < c2 (log x)λ (log x)λ

holds for large enough x for some constants c1 , c2 > 0 and λ > 2, then for all large enough n, the abscissa of convergence of 

ζPn (s) =

m∈Pn 

is 1 and

1 ζPn (s)

1 ms

is strictly concave for s > 1.

We should remark that there exist infinite sets P of prime numbers meeting the condition ∞  ΛP (m) < log p1 m m=1

in (ii) of Theorem 2. Besides, part (iv) of Theorem 2 supplies us with a family of Dirichlet series ψ with nonnegative coefficients such that ψ1 is concave on (1, ∞), thereby providing a positive answer to Problem 4.3 of [12]. For an arbitrary set of prime numbers however, one can still show the strict concavity of the reciprocal zeta function on a big portion of (1, ∞). In this respect, ζ −1 (e) can be realized as a critical universal constant for the strict concavity. Theorem 3. If P is any set of prime numbers with least member p1 , then interval (α, ∞), where  α = min ζ

−1



ζ (e), − ζ

1 ζP (x)

is strictly concave on the



−1

(log p1 ) ,

  −1  and − ζζ is the inverse function of − ζζ on (1, ∞). A strict bias in the form  n≤y ω(n)≡0 (mod 2)

|μP (n)|(log n)2 < nx

 n≤y ω(n)≡1 (mod 2)

|μP (n)|(log n)2 nx

holds for all large enough y in terms of x when x > α. Consequently, for any given α > 1, concave on the interval (α, ∞) when log p1 > −

ζ (α). ζ

1 ζP (x)

is strictly

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Let us remark that when p1 = 2 in Theorem 3, then  −

ζ ζ

−1

(log 2) > 1.48 > ζ −1 (e) = α.

But if p1 is large enough, then   −1 ζ α= − (log p1 ). ζ Our next result demonstrates that the concavity and strict concavity for the reciprocal of a zeta function coincide with the exception of at most finitely many points. Precisely, we have Theorem 4. Let P be any set of prime numbers with least member p1 such that interval (1, ∞). If the number of elements in P that are ≤ z is given by

1 ζP (x)

is concave on the

cz + O(z 1−δ ) for some constants 0 ≤ c ≤ 1 and 0 < δ ≤ 1, then ζP1(x) is strictly concave for x > 1 with the exception of at most finitely many (possibly none) points. Consequently, there exist 1 ≤ a1 ≤ a2 ≤ · · · ≤ ak ≤ α such that ζP1(x) is strictly concave in each of the intervals (1, a1 ), (a1 , a2 ), . . . , (ak , ∞), where  α = min ζ

−1



ζ (e), − ζ



−1

(log p1 ) .

Let us stress that if P contains all but finitely many prime numbers, then the number of elements in P that are ≤ z is given by  p∈P /

1 1− p

 z + O(1)

and this amounts to taking c=

 p∈P /

1 1− p



and δ = 1 in Theorem 4. To look at another extreme case, if P is any (nonempty) finite set of prime

numbers, then it is easy to see that the number of elements in P that are ≤ z is O (log z)|P| . We may then take c = 0 and any 0 < δ < 1 in Theorem 4. Finally, it is worth noticing that using the assumption on the number of elements in P that are ≤ z in the form cz + O(z 1−δ ) for 0 < c < 1 and 0 < δ ≤ 1, one can show, with the help of a method due to Landau [17] on a quantitative version of the prime ideal theorem, that the number of elements in P which are ≤ z is given by   √ li(z) + O ze−λ log z

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for some constant λ > 0, where z li(z) =

dt . log t

2

Note that for the Riemann zeta function and the zeta functions in Theorem 2, there are no exceptional points and concavity of the reciprocal is equivalent to its strict concavity on (1, ∞). An important class of functions belongs to the notion of geometric convexity (or multiplicative convexity). A real valued positive function f is geometrically convex if f (xα y β ) ≤ f (x)α f (y)β holds for all x, y in the domain of f and for all nonnegative α, β with α + β = 1. As usual, f is strictly geometrically convex if equality is possible only when α = 0, β = 0 or x = y. Geometric convexity was first introduced and used in a classical paper of Montel [20]. But for a modern exposition, we recommend the work of Niculescu [21]. Niculescu showed that a positive analytic function of the form ∞ 

f (x) =

an x n

n=0

with an ≥ 0 for all n and radius of convergence R > 0 is geometrically convex on (0, R). For a different perspective on geometrically convex functions with applications to the art of inequalities, the reader should consult the paper of Guan [15]. Geometric convexity is a remarkable strengthening of the log-convexity. Indeed, if f is a monotonically decreasing, positive and geometrically convex function, then log f (xt y 1−t ) ≤ t log f (x) + (1 − t) log f (y) holds for all 0 ≤ t ≤ 1. As t log x + (1 − t) log y ≤ log(tx + (1 − t)y), we have xt y 1−t ≤ tx + (1 − t)y and by the monotonicity of f log f (tx + (1 − t)y) ≤ log f (xt y 1−t ) follows. Therefore, the log-convexity of f in the form log f (tx + (1 − t)y) ≤ t log f (x) + (1 − t) log f (y) is checked. Our next result gives a new class of geometrically convex functions arising from zeta functions on semigroups and strengthens their log-convexity property. Theorem 5. If P is a set of prime numbers not containing 2 or containing at least 2, 5, 7 and 11, then ζP (x) =

 n∈P

is geometrically convex for x > 1.

1 nx

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It would be an interesting problem to study geometric convexity properties of other zeta functions such as the ones with harmonic coefficients. For recent work on the special values of such zeta functions and arithmetic results on harmonic numbers, we refer the reader to [1], [3], [4] and [14]. As a bonus, we can also improve on Corollary 4.5 of [12] for a family of zeta functions with nonnegative coefficients. Theorem 6. Let P be a set of odd prime numbers with p1 being the least element of P. Then the chain of inequalities  ζP

x+y 2



2ζP (x)ζP (y) √ ≤ ζP ( xy) ≤ ζP (x) + ζP (y)

(1.5)

holds for any  x, y >

ζ 1− ζ

−1 (log p1 ),

 −1   where 1 − ζζ is the inverse function of 1 − ζζ on (1, ∞). Equality is possible in any of these inequalities only when x = y. Taking P as the set of all odd prime numbers, we see that Theorem 6 applies to the lambda function λ(s) = (1 − 2−s )ζ(s) introduced in [12]. Other categories of functions analogous to the family of convex functions were studied at different times in the literature. As an original contribution, let us mention the work of Rosenbaum [22] who developed a theory of sub-additive functions. It turns out that homogeneous sub-additive functions are necessarily convex. Monotonicity of the modulus of the Riemann zeta function along horizontal lines sparked interest as Matiyasevich, Saidak, and Zvengrowski [19] recently formulated a criterion equivalent to the Riemann hypothesis in terms of such behavior which is reminiscent of a condition discovered by Spira [23]. Define the function ζs (x) :=

∞ 

1 (x + n)s n=0

for x > 0 and s > 1 (note that ζs (1) = ζ(s)). Alzer [5] gave sub- and super-additivity properties of this function. In particular, he showed that ζs (x)α + ζs (y)α < ζs (x + y)α for all x, y > 0 when α ≤ −1/(s − 1). Alzer [6] also found sharp inequalities on quotients involving tails of the series defining ζs . If P is any set of prime numbers, then an interesting probabilistic interpretation of the semigroup P based on the quotient ζP (s) ζ(s) was offered by Golomb [13], who showed that ζP (s) s→1 ζ(s) lim

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equals to the natural density of P, provided that the latter exists. Alzer [7] arrived at sharp inequalities for quotients of values of the Riemann zeta function in the form (exp is the exponential function) 

ζ  (s + α) exp c ζ(s + α)



   ζ(s + c) ζ (s + β) < < exp c ζ(s) ζ(s + β)

for c > 0, s > 1 and with best possible constants α = 0 and β=

1 log log 2



c log 2 1 − 2−c

 .

These inequalities of Alzer were generalized by Cerone and Dragomir [12] with a pair of inequalities of the following types. First, 

ζ  (s) exp h ζ(s)



ψ(s + h) ψ(s)

≤

holds for h > 0 and s > 1, where ψ(s) = 0 for s > 1 is a Dirichlet series with nonnegative coefficients that form a monotonic non-increasing sequence, and second,   ζ  (s + h) ψ(s + h) ≤ exp h ψ(s) ζ(s + h) holds for h > 0 and s > 1, where ψ(s) = 0 for s > 1 is a Dirichlet series with nonnegative coefficients that form a monotonic non-decreasing sequence. By giving up on the monotonicity of the coefficients of ψ, we offer a further generalization of such type of inequalities in the setting of Dirichlet series with nonnegative coefficients. Theorem 7. Given > 0, there exist uncountably many, more precisely, as many as the cardinality of R, sets P of prime numbers, including the set of all prime numbers, satisfying     ζP (s + h) ζ (s + h) ≤ exp h + ζP (s) ζ(s + h)

(1.6)

for all s > 1 and h ≥ 0. Consequently, if  log 1 + 1
log 2

log 2

 − 1,

then for all κ > 0 larger than some constant depending only on , the series ⎛

2

⎞κn log n ζP s + n1 ⎝  ⎠ 1 ζP s + n+1 n=1

∞ 

(1.7)

is convergent. Moreover, for any κ > 1, the series ⎛



⎞κn log n ζP 1 + n1 ⎝  ⎠ 1 ζ 1 + n=1 P n+1 ∞ 

is convergent.

(1.8)

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Theorem 8. For any set P of prime numbers, the inequality 

ζ  (s) exp h ζ(s)

 ≤

ζP (s + h) ζP (s)

(1.9)

holds for all s > 1 and h ≥ 0. Consequently, the series ⎛

2

⎞n 1 ζ s+ n ⎝ P ⎠ 1 ζP s + n+1 n=1

∞ 

(1.10)

is divergent for s > 1 and so is the series

⎞n ζP 1 + n1 ⎝  ⎠ . 1 ζP 1 + n+1 n=1 ∞ 

⎛

(1.11)

It would be interesting to explore the critical threshold value t(n) between the exponents κn2 log n and n2 in (1.7) and (1.10) so that ⎛



⎞a(n) ζP s + n1 ⎝  ⎠ 1 ζ s + n=1 P n+1 ∞ 

is convergent or divergent according to whether a(n) > t(n) or a(n) ≤ t(n), respectively. 2. Proof of Theorem 1 The main idea of the proof is to use the Laurent expansion of the Riemann zeta function around the simple pole at s = 1. It is common to write this expansion in the form ζ(s) =

∞  1 (−1)n +γ+ γn (s − 1)n , s−1 n! n=1

(2.1)

where s can be any complex number with |s − 1| ≤ 1. In (2.1), the numbers γn are the generalized Euler– Mascheroni constants and it is known that (see Theorem 1 in [8])  γn = lim

m→∞

(log m)n+1  (log r)n + − n+1 r r=1 m



for all n ≥ 0 with γ0 = γ being the Euler–Mascheroni constant. Briggs [10] showed that γn changes sign infinitely often and |γn | <

 n n 2e

for all n ≥ 1. Using the inequality en >

nn , n!

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one sees that 1 |γn | < n n! 2

(2.2)

for all n ≥ 1. A sharper estimate than (2.2) is provided by Berndt [8] as a result of his general treatment of the Laurent expansion of Hurwitz zeta functions at s = 1. In particular, we will use an inequality which follows from [8] in the form 4 |γn | ≤ n! nπ n

(2.3)

for all n ≥ 1. Let us rewrite (2.1) as ζ(s) =

∞  1 (−1)n + γ − γ1 (s − 1) + γn (s − 1)n . s−1 n! n=2

(2.4)

Note that by (2.3),    n ∞ ∞    2 4 (−1)n 1 s−1  n γn (s − 1)  ≤ 4 (s − 1)2 ≤ (s − 1)2 ≤    n! n π π(π − s + 1) π(2π − 1) n=2 n=2

(2.5)

holds for 1 < s ≤ 32 . From (2.4) and (2.5), we obtain that ζ(s) =

1 + γ − γ1 (s − 1) + E1 (s), s−1

(2.6)

where the function E1 (s) has the only property |E1 (s)| ≤

4 (s − 1)2 π(2π − 1)

for 1 < s ≤ 32 . By termwise differentiation of the Laurent series, we see from (2.1) that ζ  (s) = −

∞  1 (−1)n nγn (s − 1)n−1 − γ + γ (s − 1) + 1 2 (s − 1)2 n! n=3

(2.7)

holds for 1 < s ≤ 32 . Again by (2.3), one gets   n ∞ ∞    4  s−1 4 8 (−1)n  n−1  nγn (s − 1) (s − 1)2 ≤ 2 (s − 1)2 = 2  ≤   π n! π π (π − s + 1) π (2π − 1) n=3 n=2

(2.8)

for 1 < s ≤ 32 . As a result of (2.7) and (2.8), we may write ζ  (s) = −

1 − γ1 + γ2 (s − 1) + E2 (s), (s − 1)2

where the function E2 (s) is subject to the inequality |E2 (s)| ≤

8 π 2 (2π

− 1)

(s − 1)2

(2.9)

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for 1 < s ≤ 32 . Similarly, we have ζ  (s) = for 1 < s ≤

3 2

∞  2 (−1)n n(n − 1)γn (s − 1)n−2 + γ − γ (s − 1) + 2 3 (s − 1)3 n! n=4

and it follows from (2.3) that   n  ∞ ∞   4  s−1 (−1)n  n−2  n(n − 1)γn (s − 1) (n + 1) .  ≤ 2   π n! π n=4 n=2

Letting x =

(2.10)

s−1 π ,

(2.11)

one obtains that ∞ 

 (n + 1)

n=2

s−1 π

n =

d dx



x3 1−x

 =

3x2 − 2x3 . (1 − x)2

(2.12)

Assembling (2.11) and (2.12), one infers that     ∞   4 (−1)n 3 − 2x   n(n − 1)γn (s − 1)n−2  ≤ 4 max1 (s − 1)2 .  2   π n! (1 − x) 0
3−2x (1−x)2

(2.13)

is an increasing function on (0, 1), we deduce from (2.13) that   ∞   16(3π − 1) (−1)n  n−2  n(n − 1)γn (s − 1) (s − 1)2 .  ≤ 3   π (2π − 1)2 n! n=4

(2.14)

Feeding (2.14) into (2.10), we see that ζ  (s) =

2 + γ2 − γ3 (s − 1) + E3 (s), (s − 1)3

(2.15)

where |E3 (s)| ≤

16(3π − 1) (s − 1)2 π 3 (2π − 1)2

holds for 1 < s ≤ 32 . Gathering (2.6), (2.9) and (2.15), we obtain ζ  (s)ζ(s) − 2(ζ  (s))2  =

=

  2 1 + γ − γ + γ − γ (s − 1) + E (s) (s − 1) + E (s) 2 3 3 1 1 (s − 1)3 s−1 2  1 − γ + γ (s − 1) + E (s) −2 − 1 2 2 (s − 1)2

2γ 6γ1 5γ2 E3 (s) − γ3 + + γγ2 − γγ3 (s − 1) + γE3 (s) + 3γ1 γ2 (s − 1) − + (s − 1)3 (s − 1)2 s−1 s−1

+ γ1 γ3 (s − 1)2 − γ1 E3 (s)(s − 1) +

2E1 (s) + γ2 E1 (s) − γ3 E1 (s)(s − 1) + E1 (s)E3 (s) (s − 1)3

− 2γ12 − 2γ22 (s − 1)2 − 2(E2 (s))2 +

4E2 (s) + 4γ1 E2 (s) − 4γ2 E2 (s)(s − 1) (2.16) (s − 1)2

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for 1 < s ≤ 32 . Taking into account the respective bounds on |E1 (s)|, |E2 (s)| and |E3 (s)|, one easily obtains that the rightmost side of (2.16) is ≥

  2γ 6γ1 8 1 − + 5γ − 2 (s − 1)3 (s − 1)2 π(2π − 1) (s − 1)   32 16(3π − 1) 2 + 3γ1 γ2 − γγ3 − 3 (s − 1) + γγ2 − 2γ1 − γ3 − 2 π (2π − 1) π (2π − 1)2   4|γ2 | 16γ(3π − 1) 32|γ1 | − (s − 1)2 + γ1 γ3 − 2γ22 − 3 − π (2π − 1)2 π 2 (2π − 1) π(2π − 1)   4γ3 16|γ1 |(3π − 1) 32|γ2 | + (s − 1)3 − + 2 π 3 (2π − 1)2 π (2π − 1) π(2π − 1)   64(3π − 1) 128 (s − 1)4 − + π 4 (2π − 1)2 π 4 (2π − 1)3

(2.17)

when 1 < s ≤ 32 . Inequality (1.2) is now immediate from (2.16) and (2.17). Moreover, as γ, γ3 > 0, γ1 , γ2 < 0 and 0 < s − 1 ≤ 12 , one may also deduce from (2.16) and (2.17) that   1 2γ 6γ1 8 ζ (s)ζ(s) − 2(ζ (s)) ≥ − + 5γ2 − 3 2 (s − 1) (s − 1) π(2π − 1) (s − 1)   1 32 16(3π − 1) + −γγ3 − 3 + γγ2 − 2γ12 − γ3 − 2 π (2π − 1) 2 π (2π − 1)2   1 4|γ2 | 16γ(3π − 1) 32|γ1 | 2 + γ1 γ3 − 2γ2 − 3 − − 2 4 π (2π − 1)2 π (2π − 1) π(2π − 1)     1 16|γ1 |(3π − 1) 4γ3 1 32|γ2 | 64(3π − 1) 128 (2.18) + − − + + 8 π 3 (2π − 1)2 π 2 (2π − 1) π(2π − 1) 16 π 4 (2π − 1)2 π 4 (2π − 1)3 



2

for 1 < s ≤ 32 . Using the approximate numerical values of γ, γ1 , γ2 and γ3 , it is easily verified that 2γ 1.15 6γ1 0.4368 > , − > , 3 3 2 (s − 1) (s − 1) (s − 1) (s − 1)2     1 8 1 0.532 8 > −5(0.0097) − >− , 5γ2 − π(2π − 1) (s − 1) (3.14)(5.28) (s − 1) s−1 32 32 > −(0.57722)(0.0097) − 2(0.0729)2 − 0.0021 − > −0.64, γγ2 − 2γ12 − γ3 − 2 π (2π − 1) (3.14)2 (5.28)     1 1 16(8.45) 16(3π − 1) > > −0.079, −γγ3 − 3 −(0.57722)(0.0021) − 2 π (2π − 1)2 2 (3.14)3 (5.28)2   1 4|γ2 | 16γ(3π − 1) 32|γ1 | γ1 γ3 − 2γ22 − 3 − − 4 π (2π − 1)2 π 2 (2π − 1) π(2π − 1)   1 4(0.0097) 16(0.57722)(8.45) 32(0.0729) 2 > −(0.0729)(0.0021) − 2(0.0097) − − − 4 (3.14)3 (5.28)2 (3.14)2 (5.28) (3.14)(5.28)

(2.19) (2.20) (2.21) (2.22)

> −0.035, (2.23) −

1 8





4γ3 32|γ2 | 16|γ1 |(3π − 1) + + 2 3 2 π (2π − 1) π (2π − 1) π(2π − 1)   4(0.0021) 1 16(0.0729)(8.45) 32(0.0097) + > −0.0023, (2.24) >− + 8 (3.14)3 (5.28)2 (3.14)2 (5.28) (3.14)(5.28)

E. Alkan / J. Math. Anal. Appl. 472 (2019) 1987–2016

1 − 16



128 64(3π − 1) + 4 4 2 π (2π − 1) π (2π − 1)3



1 >− 16



128 64(8.45) + 4 2 (3.14) (5.28) (3.14)4 (5.28)3

2001

 > −0.0054.

(2.25)

Gathering (2.18)–(2.25), one arrives at the inequality ζ  (s)ζ(s) − 2(ζ  (s))2 >

1.15 0.4368 0.532 − 0.7617 + − 3 2 (s − 1) (s − 1) s−1

(2.26)

for 1 < s ≤ 32 . When s = 32 , it is easily checked that 0.4368 0.532 1.15 − 0.7617 > 0. + − 3 2 (s − 1) (s − 1) s−1 Consequently from (2.26), ζ  (1.5)ζ(1.5) − 2(ζ  (1.5))2 > 0

(2.27)

follows. Next we show that (2.27) persists to hold for 1 < s ≤ 1.5. To this end, consider the cubic p(t) = (1.15)t3 + (0.4368)t2 − (0.532)t − 0.7617. Observe that p (t) = (3.45)t2 + (0.8736)t − 0.532 > 0 for t ≥ 1. Therefore, p(t) is increasing on (1, ∞). We know that p(2) > 0. Hence p(t) > 0 for all t ≥ 2. 1 Letting t = s−1 , we obtain that  p

1 s−1

 >0

(2.28)

for 1 < s ≤ 32 . Gathering (2.26) and (2.28), we infer that 





ζ (s)ζ(s) − 2(ζ (s)) > p 2

1 s−1

 >0

(2.29)

when 1 < s ≤ 32 . It follows from (2.29) that (1.1) holds when 1 < s ≤ 32 . From the proof of Theorem 4.4 in [12], we know that (1.1) holds when s > ζ −1 (e). Since ζ −1 (e) < 32 , we see that (1.1) holds on the entire 1 interval (1, ∞). This completes the proof of the strict concavity of ζ(s) on (1, ∞). Moreover, if s > 1, then ∞  1 μ(n) = . ζ(s) n=1 ns

(2.30)

1 Since ζ(s) is strictly concave for s > 1 and termwise differentiation of the Dirichlet series in (2.30) is permissible, one may deduce from (2.30) that

  ∞  1 μ(n)(log n)2 = <0 ns ζ(s) n=1

(2.31)

E. Alkan / J. Math. Anal. Appl. 472 (2019) 1987–2016

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for s > 1. We also have from (2.31) that ⎛

⎞ 

⎜ lim ⎜ y→∞ ⎝

|μ(n)|(log n)2 − ns

n≤y ω(n)≡0 (mod 2)



=

ω(n)≡0

(mod 2)

 n≤y ω(n)≡1 (mod 2)

|μ(n)|(log n)2 − ns

|μ(n)|(log n)2 ⎟ ⎟ ⎠ ns

 ω(n)≡1

(mod 2)

∞  |μ(n)|(log n)2 μ(n)(log n)2 = < 0 (2.32) ns ns n=1

for s > 1. As a result of (2.32), we see for all large enough y in terms of s > 1 that  n≤y ω(n)≡0 (mod 2)

|μ(n)|(log n)2 − ns

 n≤y ω(n)≡1 (mod 2)

|μ(n)|(log n)2 < 0, ns

and the strict bias in (1.3) holds in favor of square-free integers with an odd number of prime factors. This ends the proof of Theorem 1. 3. Proof of Theorem 2 First assume that P is finite and let p1 < · · · < pn be the prime numbers in P so that |P| = n. Then we have  1 = (1 − e−x log pi ) ζP (x) i=1 n

(3.1)

for x > β. Expanding the product in (3.1), one gets a finite alternating sum of sums in the form 

1−

e−x log pi1 +

1≤i1 ≤n



e−x(log pi1 +log pi2 ) − · · · + (−1)n e−x(log p1 +···+log pn ) .

(3.2)

1≤i1
Combining (3.1), (3.2) and differentiating m ≥ 1 times, we obtain  (−1)m−1

1 ζP (x)

(m) 

=

1≤i1 ≤n

(log pi1 )m e−x log pi1 −



(log pi1 + log pi2 )m e−x(log pi1 +log pi2 )

1≤i1
+ · · · + (−1)n−1 (log p1 + · · · + log pn )m e−x(log p1 +···+log pn ) . (3.3) It suffices to show that the right hand side of (3.3) is positive when p1 is large enough only in terms of m, n and β. To this end, note that the right hand side of (3.3) is again an alternating sum of sums. If n is even, then there are n/2 pairs of consecutive sums with different signs and if n is odd, then there are [n/2] such pairs of consecutive sums together with the last term which is clearly positive. Therefore, in all cases, it is enough to show that each pair of sums gives a positive contribution to the right hand side of (3.3). Our task is then to show that  1≤i1 <···
(log pi1 + · · · + log pik )m < pxi1 . . . pxik

 1≤i1 <···
(log pi1 + · · · + log pik−1 )m , pxi1 . . . pxik−1

(3.4)

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where the sums in (3.4) are over all tuples (i1 , . . . , ik−1 ) and (i1 , . . . , ik ) of indices such that 1 ≤ i1 < · · · < ik−1 ≤ n and 1 ≤ i1 < · · · < ik ≤ n. Note that any tuple (i1 , . . . , ik−1 ) with 1 ≤ i1 < · · · < ik−1 ≤ n leads to at most n tuples of the form (i1 , . . . , ik−1 , ik ) with 1 ≤ i1 < · · · < ik−1 < ik ≤ n. Moreover, any tuple (i1 , . . . , ik−1 , ik ) would arise in this way from a tuple of the form (i1 , . . . , ik−1 ). Fixing i1 , . . . , ik−1 with 1 ≤ i1 < · · · < ik−1 ≤ n, let us look at the total contribution of terms on the left hand side of (3.4) of the form (log pi1 + · · · + log pik )m . pxi1 . . . pxik

(3.5)

By the convexity of xm , we have (log pi1 + · · · + log pik )m ≤ 2m−1 ((log pi1 + · · · + log pik−1 )m + (log pik )m ) Hence (3.5) is ≤

2m−1 ((log pi1 + · · · + log pik−1 )m + (log pik )m ) . pxi1 . . . pxik−1 pxik

(3.6)

Therefore, by (3.6), the total contribution of such terms to the left hand side of (3.4) is ≤

max

2m−1 n((log pi1 + · · · + log pik−1 )m + (log pik )m ) pxi1 . . . pxik−1 pβik

ik−1
(3.7)

as x > β. Note that max

ik−1
2m−1 n((log pi1 + · · · + log pik−1 )m + (log pik )m ) pxi1 . . . pxik−1 pβik

<

(log pi1 + · · · + log pik−1 )m pxi1 . . . pxik−1

would hold when   pβik > 2m−1 n 1 +

log pik log pi1 + · · · + log pik−1

m 

for all ik . But clearly, log pi1 + · · · + log pik−1 ≥ log 2 and the above inequality on pik would follow if pβik

>2

m−1

m    log pik . n 1+ log 2

Finally, as pik ≥ p1 and (log t)m = o(tβ ), we see that when p1 is large enough only in terms of m, n and β, then the last inequality is fulfilled. Adding all of these contributions, where i1, . . . , ik−1 range with 1 ≤ i1 < · · · < ik−1 ≤ n, one verifies (3.4). This completes the proof of (i). To show (ii), it is enough to prove that 

1 ζP (x)

 =−

ζP (x) (ζP (x))2

is strictly decreasing for x > 1. Note that when x > 1, log ζP (x) = −

 p∈P

  1 log 1 − x . p

(3.8)

E. Alkan / J. Math. Anal. Appl. 472 (2019) 1987–2016

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Differentiating both sides of (3.8) (termwise differentiation is allowed on the righthand side by uniform convergence), one easily gets −

∞  ζP (x) e−x log n = ΛP (n) 2 (ζP (x)) ζP (x) n=2

(3.9)

for x > 1. As p1 is the smallest prime number in P, the sum on the righthand side of (3.9) actually starts at n = p1 and we may assume that n ≥ p1 . Consider the sequence of functions fn (x) :=

e−x log n ζP (x)

for n ≥ p1 and x > 1. Let us show that each fn is strictly decreasing on (1, ∞). Looking at the derivative of fn , we see that fn (x) =

−e−x log n (ζP (x) log n + ζP (x)) . (ζP (x))2

(3.10)

Moreover, using our hypothesis on P, ∞ ∞   ζP (x) ΛP (m) ΛP (m) = < log p1 ≤ log n. − ≤ ζP (x) m=2 mx m m=2

(3.11)

It follows from (3.10) and (3.11) that fn (x) < 0 for x > 1. Consequently, (3.9) is strictly decreasing on (1, ∞). For the strict bias in (ii), just note that ⎛

⎞ 

⎜ lim ⎜ y→∞ ⎝

n≤y ω(n)≡0 (mod 2)

|μP (n)|(log n)2 − nx

 n≤y ω(n)≡1 (mod 2)

|μP (n)|(log n)2 ⎟ ⎟ ⎠ nx

=

  ∞  μP (n)(log n)2 1 = < 0. (3.12) nx ζP (x) n=1

The proof of (ii) is now complete by (3.12). For the proof of (iii), note that 1 = (1 − e−x log p )(1 − e−x log q ) = 1 − e−x log p − e−x log q + e−x(log p+log q) ζP (x) for x > 0, where p, q are distinct prime numbers. If follows from (3.13) that 

1 ζP (x)



1 ζP (x)

is concave on the entire interval (0, ∞), then it

= −(log p)2 e−x log p − (log q)2 e−x log q + (log p + log q)2 e−x(log p+log q) ≤ 0

for all x > 0. But letting x → 0+ on the right hand side of (3.14), one derives the inequality (log p + log q)2 ≤ (log p)2 + (log q)2 which is clearly not possible. In conclusion,

1 ζP (x)

(3.13)

is not concave on the entire interval (0, ∞).

(3.14)

E. Alkan / J. Math. Anal. Appl. 472 (2019) 1987–2016

2005

To prove (iv), first note that as c1

x x < πP (x) < c2 (log x)λ (log x)λ

holds for large enough x for some constants c1 , c2 > 0 and λ > 2, we have c1 n(log n)λ < qn < c2 n(log n)λ

(3.15)

for large enough n for some constants c1 , c2 > 0 and λ > 2. Consequently from (3.15), one obtains that ∞  log qm q −1 m=1 m

(3.16)

is convergent as it is comparable to the convergent series ∞ 

1 . n(log n)λ−1 n=2 It follows from the convergence of the series in (3.16) that ∞ ∞   ΛPn (m) log qm = < log qn m q −1 m=n m m=1

holds when n is large enough. Because of (3.17), we may use part (ii) to obtain that when s > 1. Finally, if  m∈Pn 

1 ms

(3.17) 1 ζPn (s)

is strictly concave

(3.18)

converges for some 0 < s < 1, then we would get via the inequality ∞  1 s ≤ q j=n j

 m∈Pn 

1 ms

that ∞  1 s q j=n j

is convergent for some 0 < s < 1. However, this is a contradiction as ∞ ∞   1 1

s s (log j)λs q j j=n j j=n

by (3.15), and the latter series is divergent when s < 1. In conclusion, the abscissa of convergence of the Dirichlet series in (3.18) is 1. This completes the proof of (iv) and the proof of Theorem 2.

E. Alkan / J. Math. Anal. Appl. 472 (2019) 1987–2016

2006

4. Proof of Theorem 3 Let P be a set of prime numbers with least member p1 . First from (3.8), it is easy to see that log ζP (x) =

∞  ΛP (n) >0 x log n n n=2

(4.1)

for x > 1. By the Cauchy–Schwarz inequality, we have 

ζP (x) ζP (x)



2 =

∞ 

n=2



 ΛP (n) log n nx

ΛP (n) nx log n

2

 ≤

∞  ΛP (n) log n nx n=2



∞  ΛP (n) x log n n n=2

 (4.2)

for x > 1. Also note that    ∞  ζP (x) ζ  (x)ζP (x) − (ζP (x))2 ΛP (n) log n = − − = P . x n ζP (x) (ζP (x))2 n=2

(4.3)

Assembling (4.1)–(4.3), we see that (ζP (x))2 ≤ ζP (x)ζP (x) − (ζP (x))2 . log ζP (x)

(4.4)

Assuming now x > ζ −1 (e), one infers that ζP (x) ≤ ζ(x) < e and log ζP (x) < 1.

(4.5)

Feeding (4.5) into (4.4), one obtains that ζP (x)ζP (x) − 2(ζP (x))2 > 0 when x > ζ −1 (e). As a result of (4.6), we know that that  x>

ζ − ζ

1 ζP (x)

(4.6)

is strictly concave when x > ζ −1 (e). Next assume

−1 (log p1 ).

(4.7)

It is clear from (4.7) that −

ζP (x) < log p1 . ζP (x)

(4.8)

Arguing as in the proof of Theorem 2 (see (3.8)–(3.11)), one infers from (4.8) that 

1 ζP (x)



is strictly decreasing when x is subject to (4.7). Consequently, ζP1(x) is strictly concave when x > α, where α is defined as in the statement of the theorem. Moreover, the strict bias in favor of square-free numbers belonging to P with an odd number of prime factors is shown similarly as in the proof of Theorem 2 when x > α. To complete the proof of Theorem 3, take any number α > 1. Observe that

E. Alkan / J. Math. Anal. Appl. 472 (2019) 1987–2016

lim+ −

x→1

ζ  (x) = ∞, ζ(x)

lim −

x→∞

ζ  (x) = 0. ζ(x)

2007

(4.9)



We see from (4.9) that the function − ζζ attains any value in the interval (0, ∞) which happens to be the   −1 domain of the function − ζζ . In particular, there is a unique number β > 0 such that 

ζ − ζ

−1 (β) = α.

(4.10)

(log p1 ) < α.

(4.11)

If P is such that log p1 > β, then  − It follows from (4.10) and (4.11) that

1 ζP (x)

−1

ζ ζ

is strictly concave on (α, ∞) when log p1 > −

ζ (α). ζ

This completes the proof of Theorem 3. 5. Proof of Theorem 4 First we note that the concavity of x > 1 such that

1 ζP (x)

on (1, ∞) differs from its strict concavity only at the points 

1 ζP (x)

 = 0.

(5.1)

Our goal is now to show that there are only finitely many such points x > 1 satisfying (5.1). By Theorem 3, we know that ζP1(x) is strictly concave on (α, ∞), where α is defined as in the statement of Theorem 3. Consequently, any point x satisfying (5.1) is subject to the condition  1 < x ≤ α = min ζ

−1

   −1 ζ (e), − (log p1 ) , ζ

(5.2)

where p1 is the least prime number in P. For a contradiction, assume that there are infinitely many points x satisfying (5.1). Because of (5.2), there is an accumulation point, say x0 ∈ [1, α] of the x’s. Next let us consider the case 1 < x0 ≤ α. Observe that 

1 ζP (s)



∞  μP (n)(log n)2 = ns n=1

is a complex analytic function on the open set defined by the condition (s) > 1. Therefore, by (5.1), the zeros of this analytic function accumulate at 1 < x0 ≤ α and this forces 

1 ζP (s)

 =0

(5.3)

2008

E. Alkan / J. Math. Anal. Appl. 472 (2019) 1987–2016

identically when (s) > 1. One gets from (5.3) that 1 = Ax + B ζP (x)

(5.4)

for x > 1, where A and B are constants. Clearly, we have 1 = 1. ζP (x)

lim

x→∞

(5.5)

Gathering (5.4) and (5.5), one infers that A = 0 and B = 1. But then ζP (x) = 1 identically on (1, ∞) and P is empty, a contradiction as p1 ∈ P. It remains to consider the case when x0 = 1 is an accumulation point of the zeros. Let us see that ζP1(s) is a complex analytic function in a neighborhood of x0 = 1. For (s) > 1, one obtains by partial summation that   1 M (t) ζP (s) = = s dt, s n ts+1 ∞

n∈P

(5.6)

1

where by our assumption 

M (t) =

1 = ct + O(t1−δ )

(5.7)

n≤t n∈P

is the summatory function of the semigroup P for some constants 0 ≤ c ≤ 1 and 0 < δ ≤ 1. From (5.7), we may write M (t) = ct + E(t) with E(t) = O(t1−δ ) so that (5.6) becomes cs +s ζP (s) = s−1

∞

E(t) dt. ts+1

(5.8)

1

Using E(t) = O(t1−δ ), we see that ∞ s

E(t) dt ts+1

(5.9)

1

represents an analytic function when (s) > 1 − 2δ . Moreover, we have ⎛ lim ⎝s

∞

s→1

⎞ E(t) ⎠ dt = Oδ (1), ts+1

(5.10)

1

where the implied constant in (5.10) depends only on δ. Assuming 0 < c ≤ 1, one may deduce from (5.8)–(5.10) that lim

s→1

1 = 0. ζP (s)

(5.11)

Using (5.11) and the Riemann removable singularities theorem, we infer that ζP1(s) is analytic on a neighborhood N (1) = {s : |s − 1| < } of 1 in the complex plane for some > 0. Therefore, if 0 < c ≤ 1, then 1 ζP (s) and consequently

E. Alkan / J. Math. Anal. Appl. 472 (2019) 1987–2016



1 ζP (s)

2009



are analytic on the open set N = N (1) ∪ {s : (s) > 1}. As s = 1 is an interior point of N , we get a contradiction from (5.3)–(5.5) by the accumulation of zeros at s = 1. The above reasoning applies equally well to the case c = 0, as in that case lims→1 ζP (s) exists and  1 > 0. n

lim ζP (s) =

s→1

n∈P

Moreover, ∞ ζP (s) = s

E(t) dt ts+1

1

is analytic in a neighborhood of s = 1. Again the Riemann removable singularities theorem  is applicable to 1 1 at s = 1. This ζP (s) and a similar contradiction occurs as a result of the accumulation of zeros of ζP (s) completes the proof. 6. Proof of Theorem 5 First of all, if P is empty, then ζP (x) = 1 for all x > 1 and ζP (x) is geometrically convex on (1, ∞). Therefore, we may assume that P is a nonempty set of prime numbers not containing 2 or containing at least 2, 5, 7 and 11. Let us show that the inequality  √ ζP ( xy) ≤ ζP (x)ζP (y)

(6.1)

holds for any x, y > 1. Once (6.1) is verified, then by using a standard argument based on the continuity of ζP (x) on (1, ∞) and the fact that dyadic rational numbers are dense in [0, 1] (this argument is almost verbatim to the one showing that a continuous midpoint convex function is convex on an open interval), one can deduce that ζP (x) is geometrically convex on (1, ∞). Indeed we prove slightly more than (6.1) by obtaining that equality is possible in (6.1) only when x = y. To this end, we assume that xy = M > 1 is constant and consider the√problem of minimizing ζP (x)ζP (y). It is now enough to verify that the minimum occurs exactly when x = M = y under the constraint xy = M . For this purpose, let f (x) = ζP (x)ζP (M/x) for 1 < x < M . Then we have M ζP (x)ζP (M/x) (6.2) x2 √ √ for 1 < x < M . Clearly from (6.2), f  ( M ) = 0 holds and x = M ∈ (1, M ) is a critical point of f . Rewriting (6.2), we see that f  (x) = ζP (x)ζP (M/x) −

   1 ζP (x) M ζP (M/x) − f (x) = ζP (x)ζP (M/x) x x ζP (x) x ζP (M/x) 

for 1 < x < M . Our next task, which is suggested by (6.3), is to show that xζP (x) ζP (x)

(6.3)

E. Alkan / J. Math. Anal. Appl. 472 (2019) 1987–2016

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is a strictly increasing function on (1, ∞). Once this is shown, then it easily follows that M ζP (M/x) x ζP (M/x) √ is a strictly decreasing function on (1, M ) and therefore that x = M is the only critical point corresponding to an overall minimum. It is time to invoke the multiplicative structure of ζP (x) through the Euler product. One easily sees by logarithmic differentiation of the Euler product supported on P that ∞ ∞   xζP (x) =− ΛP (n)xe−x log n = − ΛP (n)fn (x), ζP (x) n=2 n=2

(6.4)

where fn (x) := xe−x log n for n ≥ 2 and x > 1. Note that as fn (x) = (1 − x log n)e−x log n , fn is strictly decreasing when x > have

1 log n .

If P does not contain 2, then ΛP (2) = 0 and rewriting (6.4), we

∞  xζP (x) =− ΛP (n)fn (x) ζP (x) n=3

for x > 1. If n ≥ 3, then each fn is strictly decreasing when x > log1 3 . As strictly decreasing on (1, ∞) for n ≥ 3. It now follows from (6.5) that

(6.5) 1 log 3

< 1, we see that each fn is

xζP (x) ζP (x) is strictly increasing on (1, ∞) and (6.1) follows with equality being possible only when x = y. To complete the proof, we may assume that P contains at least 2, 5, 7 and 11. It suffices to show that d dx



xζP (x) ζP (x)

 >0

on (1, ∞). Repeating the above argument, it is clear that (6.6) holds when x > assume that   d xζP (x) =0 dx ζP (x) at some 1 < x ≤

1 log 2 .

(6.6) 1 log 2 .

For a contradiction,

(6.7)

Termwise differentiating (6.4), one infers from (6.7) that ∞  ΛP (n) (x log n − 1) = 0 nx n=2

holds for some 1 < x ≤

1 log 2 .

(6.8)

Since 2, 5, 7 and 11 are in P, (6.8) implies that

log 2 x log2 2  ΛP (n) − = (x log n − 1) 2x 2x nx n>2 ≥

log 5 log 7 log 11 (x log 5 − 1) + x (x log 7 − 1) + (x log 11 − 1) (6.9) x 5 7 11x

E. Alkan / J. Math. Anal. Appl. 472 (2019) 1987–2016

for some 1 < x ≤

1 log 2 .

2011

Clearly, we have

log 2 x log2 2 − ≤ max1 2x 2x 1≤x≤ log 2



log 2 x log2 2 − 2x 2x



log 2 (1 − log 2)  0.106347. 2

=

(6.10)

Moreover, one easily obtains that log 7 log 11 log 5 (x log 5 − 1) + x (x log 7 − 1) + (x log 11 − 1) 5x 7 11x       x log2 5 log 5 x log2 7 log 7 x log2 11 log 11 + min + min . (6.11) − − − ≥ min 1 1 1 5x 5x 7x 7x 11x 11x 1≤x≤ log 2 1≤x≤ log 1≤x≤ log 2 2 Next we observe that  min 1

1≤x≤ log 2

min 1

1≤x≤ log 2





x log2 5 log 5 − x 5x 5 x log2 7 log 7 − x 7x 7

x log2 11 log 11 − 11x 11x

 ≥  ≥ 

log2 5 log 5 log 2

e

log 2

log2 7 log 7 log 2

e

log 2

−

log 5  0.044653, 5

(6.12)

−

log 7  0.051775, 7

(6.13)

log2 11

log 11  0.042878. 11

(6.14)

log 7 log 11 log 5 (x log 5 − 1) + x (x log 7 − 1) + (x log 11 − 1) ≥ 0.13. 5x 7 11x

(6.15)

min 1

1≤x≤ log 2

≥

e

log 11 log 2

log 2

−

Assembling (6.11)–(6.14), we see that

From (6.9), (6.10) and (6.15), one infers that (6.9) can not hold. Therefore, (6.8) and consequently (6.7) can not hold. Since   d xζP (x) dx ζP (x) is a continuous function on (1, ∞), the impossibility of (6.8) also shows that d dx



xζP (x) ζP (x)

 <0

is not possible at any x > 1. As a result, (6.6) is valid on (1, ∞) and the proof is complete. 7. Proof of Theorem 6 √ Clearly, the first inequality in (1.5) holds as x+y ≥ xy and ζP (x) is strictly decreasing on (1, ∞). 2 Furthermore, equality is possible only when x = y. To prove the second inequality in (1.5), assume that  x, y >

ζ 1− ζ

−1 (log p1 ).

(7.1)

Note that lim+ 1 −

x→1

ζ  (x) = ∞, ζ(x)

lim 1 −

x→∞

ζ  (x) = 1. ζ(x)

(7.2)

2012

E. Alkan / J. Math. Anal. Appl. 472 (2019) 1987–2016

 −1  As a result of (7.2), the domain of the inverse function 1 − ζζ is (1, ∞). Therefore, the righthand side of (7.1) is defined since by our assumption log p1 ≥ log 3 > 1. We minimize the function 2ζP (x)ζP (y) ζP (x) + ζP (y) subject to (7.1) and the constraint √ xy = M , where M is constant. It suffices to show that the overall minimum occurs only when x = M = y. To this end, consider the function g(x) =

2ζP (x)ζP (M/x) , ζP (x) + ζP (M/x)

(7.3)

where 

ζ 1− ζ

−1 (log p1 ) < x <  1−

ζ

M −1

ζ

.

(7.4)

(log p1 )

From (7.3), we have g  (x) =

2(ζP (x))2 (ζP (M/x))2 x(ζP (x) + ζP (M/x))2



M ζP (M/x) xζP (x) − (ζP (x))2 x(ζP (M/x))2

 .

(7.5)

Combining (7.4) and (7.5), one sees that x=

√ M>

 1−

ζ ζ

−1 (log p1 )

is a critical point of g. To complete the proof, we need to show that ∞  xζP (x) ΛP (n)xe−x log n = − 2 (ζP (x)) ζP (x) n=1

(7.6)

is a strictly increasing function when x is subject to (7.4). Once this is shown, then it follows that M ζP (M/x) x(ζP (M/x))2 √ is a strictly decreasing function when x is subject to (7.4) and g( M ) is the overall minimum value of g leading to the second inequality in (1.5), where equality is possible only when x = y. Finally, let us see that the functions xe−x log n ζP (x) are strictly decreasing when x is subject to (7.4). We may of course assume that n ≥ p1 . Taking the derivative, it is enough to show that (1 − x log n)ζP (x) − xζP (x) < 0.

(7.7)

1 ζP (x) − < log n. x ζP (x)

(7.8)

(7.7) is equivalent to

E. Alkan / J. Math. Anal. Appl. 472 (2019) 1987–2016

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But using (7.4), we also have 1− and (7.8) follows from (7.9) as

1 x

ζ  (x) ζP (x) ≤1− < log p1 ≤ log n ζP (x) ζ(x)

(7.9)

< 1. This finishes the proof.

8. Proof of Theorem 7 Let P be any set of prime numbers. As log ζP (s) is convex and differentiable on (1, ∞), we know for s > 1 and h > 0 that ζ  (s + h) log ζP (s + h) − log ζP (s) ≤ P . h ζP (s + h)

(8.1)

It follows from (8.1) that ⎛ ⎛ ⎞⎞      log p ζP (s + h) ζ ζP (s + h) (s + h) ⎠⎠ ≤ exp h = exp ⎝h ⎝ + ζP (s) ζP (s + h) ζ(s + h) ps+h − 1

(8.2)

p∈P /

for s > 1 and h ≥ 0. Given > 0, our task is to give a construction for the complement of P in the set of all prime numbers such that  p∈P /

log p ≤ ps+h − 1

(8.3)

is satisfied for s > 1 and h ≥ 0. Once this is done, then (1.6) will be a consequence of (8.2) and (8.3). First of all, we have  p∈P /

 log p log p ≤ . −1 p−1

ps+h

(8.4)

p∈P /

Note that ∞  log(2n ) 2n − 1 n=1

is convergent. Thus there exists N ≥ 1 such that ∞  log(2n ) ≤ 2n − 1

(8.5)

n=N

x N and the function log x−1 is decreasing when x ≥ 2 . By Bertrand’s postulate, let pn be a prime number with 2n < pn < 2n+1 for all n ≥ N . Therefore,

log(2n ) log pn ≤ n pn − 1 2 −1 for all n ≥ N and ∞  log pn ≤ pn − 1

n=N

(8.6)

E. Alkan / J. Math. Anal. Appl. 472 (2019) 1987–2016

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is obtained from (8.5). Let M := {pN , pN +1 , . . . }. Then for any subset K ⊆ M, one sees from (8.6) that  log p ≤ . p−1

(8.7)

p∈K

As M is countable, the number of choices for K is the same as the cardinality of R. Let P now be the complement of K in the set of all prime numbers. Then assembling (8.2), (8.4) and (8.7), one may deduce for s > 1 and h ≥ 0 that ⎛ ⎛ ⎞⎞       log p ζ ζP (s + h) (s + h) ⎠⎠ ≤ exp h ζ (s + h) + ≤ exp ⎝h ⎝ + ζP (s) ζ(s + h) p−1 ζ(s + h) p∈K

holds for uncountably many sets P of prime numbers. In particular, when K is empty, P becomes the set of all prime numbers and (1.6) holds. Next let us assume that  log 1 + 1
log 2

log 2

 − 1.

(8.8)

It follows from (8.8) that −

 log p log 2 ζ (s + 1) = > s+1 > , s+1 − 1 ζ p 2 −1 p

(8.9)

where the summation in (8.9) is over all prime numbers. Note that when s is subject to (8.8), we have 





⎛

ζ ζ ζ (2) < (s + 1) < ⎝ ζ ζ ζ

 log 1 +

log 2

log 2

⎞ ⎠

(8.10)

and as a result of (8.9) and (8.10), the constant  C = C( ) :=

max  1
log 2  log 2



log 1+

−1

ζ (s + 1) + ζ

 <0

(8.11)

depends only on . For all n ≥ 1, we see that

     ζP s + n1 1 ζ 1   ≤ exp s+ + . 1 n(n + 1) ζ n ζP s + n+1 Moreover, using the monotonicity of

ζ ζ ,

(8.12)

one gets for all n ≥ 1 that ζ ζ



1 s+ n

 + ≤ C.

Therefore, for any κ > 0, (8.12) gives ⎛

2

⎞κn log n   ζP s + n1 Cκn log n ⎝  ⎠ . ≤ exp 1 n+1 ζP s + n+1

(8.13)

E. Alkan / J. Math. Anal. Appl. 472 (2019) 1987–2016

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Clearly, when κ is larger than some constant depending only on , then the series ∞ 

 exp

Cκn log n n+1

n=1

 (8.14)

is convergent. Thus by (8.13) and (8.14), the series in (1.7) is convergent. Similarly, we obtain for any κ > 1 and n ≥ 1 that

⎞κn log n      ζP 1 + n1 1 κ log n ζ  ⎝  ⎠ 1+ + . ≤ exp 1 n+1 ζ n ζP 1 + n+1 ⎛

(8.15)

Observe that 1 n→∞ n + 1



lim

ζ ζ

 1+

1 n



 +

= −1.





(8.16)

From (8.16), we easily infer for κ > 1 that ∞ 

 exp

n=1

κ log n n+1



ζ ζ



1 1+ n

< ∞.

+

(8.17)

The convergence of the series in (1.8) is obtained for any κ > 1 by combining (8.15) and (8.17). This completes the proof. 9. Proof of Theorem 8 Let P be any set of prime numbers. As log ζP (s) is convex and differentiable on (1, ∞), we know for s > 1 and h > 0 that log ζP (s + h) − log ζP (s) ζP (s) ≤ . ζP (s) h

(9.1)

   ζP (s + h) ζ (s) ≤ exp h P ζP (s) ζP (s)

(9.2)

ζ  (s) ζP (s) ≥ ζP (s) ζ(s)

(9.3)

It follows from (9.1) that

for s > 1 and h ≥ 0. Moreover, we have

and (1.9) follows at once from (9.2) and (9.3). Therefore, for s > 1, one gets ⎛

2

⎞n    ζP s + n1 1 n ζ ⎝   ⎠ ≥ exp s+ . 1 n+1 ζ n+1 ζP s + n+1

(9.4)

But note that  lim exp

n→∞

n ζ n+1 ζ

 s+

1 n+1



 = exp



ζ (s) ζ

= 0

(9.5)

2016

E. Alkan / J. Math. Anal. Appl. 472 (2019) 1987–2016

when s > 1. The divergence of the series in (1.10) is clear from (9.4) and (9.5). Similarly, we obtain ⎛



⎞n    ζP 1 + n1 1 1 ζ ⎝  ⎠  1+ ≥ exp 1 n+1 ζ n+1 ζP 1 + n+1

(9.6)

and  lim exp

n→∞

1 ζ n+1 ζ

 1+

1 n+1



= e−1 = 0

(9.7)

so that the divergence of the series in (1.11) is a consequence of (9.6) and (9.7). This completes the proof. Acknowledgment The author is grateful to the referee for many constructive comments and suggestions which led to various improvements in the presentation. References [1] E. Alkan, Approximation by special values of harmonic zeta function and log-sine integrals, Commun. Number Theory Phys. 7 (2013) 515–550. [2] E. Alkan, H. Göral, On sums over the Möbius function and discrepancy of fractions, J. Number Theory 133 (2013) 2217–2239. [3] E. Alkan, H. Göral, Trigonometric series and special values of L-functions, J. Number Theory 178 (2017) 94–117. [4] E. Alkan, H. Göral, Structural properties of Dirichlet series with harmonic coefficients, Ramanujan J. 45 (2018) 569–600. [5] H. Alzer, Inequalities for the Hurwitz zeta function, Proc. Roy. Soc. Edinburgh Sect. A 130 (2000) 1227–1236. [6] H. Alzer, Sharp inequalities for the Hurwitz zeta function, Rocky Mountain J. Math. 35 (2005) 391–399. [7] H. Alzer, Inequalities for Riemann’s zeta function, Port. Math. 66 (2009) 321–327. [8] B.C. Berndt, On the Hurwitz zeta-function, Rocky Mountain J. Math. 2 (1972) 151–157. [9] R.P. Brent, J. van de Lune, A note on Pólya’s observation concerning Liouville’s function, arXiv:1112.4911v1 [math.NT], 2011. [10] W.E. Briggs, Some constants associated with the Riemann zeta function, Michigan Math. J. 3 (1955–56) 117–121. [11] P. Cerone, S.S. Dragomir, Some inequalities for Dirichlet series via logarithmic convexity, RGMIA Res. Rep. Collect. 8 (4) (2005) 14. [12] P. Cerone, S.S. Dragomir, Some convexity properties of Dirichlet series with positive terms, Math. Nachr. 282 (2009) 964–975. [13] S.W. Golomb, A class of probability distributions on the integers, J. Number Theory 2 (1970) 189–192. [14] H. Göral, D.C. Sertbaş, Almost all hyperharmonic numbers are not integers, J. Number Theory 171 (2017) 495–526. [15] K. Guan, Multiplicative convexity and its applications, J. Math. Anal. Appl. 362 (2010) 156–166. [16] A. Gut, Some remarks on the Riemann zeta distribution, Rev. Roumaine Math. Pures Appl. 51 (2006) 205–217. [17] E. Landau, Neuer Beweis des Primzahlsatzes und Beweis des Primidealsatzes, Math. Ann. 56 (1903) 645–670. [18] E. Lindelöf, Quelques remarques sur la croissance de la fonction ζ(s), Bull. Sci. Math. 32 (1908) 341–356. [19] Y. Matiyasevich, F. Saidak, P. Zvengrowski, Horizontal monotonicity of the modulus of the zeta function, L-functions, and related formulas, Acta Arith. 166 (2014) 189–200. [20] P. Montel, Sur les functions convexes et les functions souharmoniques, J. Math. 7 (1928) 29–60. [21] C.P. Niculescu, Convexity according to the geometric mean, Math. Inequal. Appl. 3 (2000) 155–167. [22] R.A. Rosenbaum, Sub-additive functions, Duke Math. J. 17 (1950) 227–247. [23] R.S. Spira, An inequality for the Riemann zeta function, Duke Math. J. 32 (1965) 247–250. [24] E.C. Titchmarsh, The Theory of the Riemann Zeta-Function, Oxford Univ. Press, London and New York, 1951.