A Monotonicity Property of Riemann's Xi Function and a Reformulation of the Riemann Hypothesis

A Monotonicity Property of Riemann's Xi Function and a Reformulation of the Riemann Hypothesis

1 A MONOTONICITY PROPERTY OF RIEMANN'S XI FUNCTION AND A REFORMULATION OF THE RIEMANN HYPOTHESIS JONATHAN SONDOW1 and CRISTIAN DUMITRESCU2 1209 West 97th Street, New York, New York 10025, USA Email: [email protected] 2119 Young Street, Kitchener, Ontario, N2H4Z3, Canada Email: [email protected] Abstract We prove that Riemann's xi function is strictly increasing (respectively, strictly decreasing) in modulus along every horizontal half-line in any zero-free, open right (respectively, left) half-plane. A corollary is a reformulation of the Riemann Hypothesis. 1. Introduction The Riemann zeta function ζ()s is defined as the analytic continuation of the Dirichlet series ∞ ζ = 1 (),s ∑ s n=1 n which converges if ℜ().s > 1 The zeta function is holomorphic in the complex plane, except for a simple pole at s = 1. The real zeros of ζ()s are s = −2,,,... − 4 − 6 . Its nonreal zeros lie in the critical strip 0≤ ℜ()s ≤ 1. The Riemann Hypothesis asserts that all the nonreal zeros lie on the critical line ℜ()s = 1 2. Riemann's xi function ξ()s is defined as the product 1 − 1 s 1 ξ():()()s= s s −1 π2 Γ() s ζ s , 2 2 ___________________ Mathematics subject classification number: 11M26. Key words and phrases: critical line, critical strip, functional equation, gamma function, Hadamard product, horizontal half-line, open half-plane; increasing in modulus, monotonicity, nonreal zero, Riemann Hypothesis, Riemann zeta function, xi function. 2 where Γ denotes the gamma function. The zero of s −1 cancels the pole of ζ()s , and the real zeros of sζ() s are cancelled by the (simple) poles of Γ()1 s , which never vanishes. 2 Thus, ξ()s is an entire function whose zeros are the nonreal zeros of ζ()s (see [1, p. 80]). The xi function satisfies the remarkable functional equation ξ()()1 −s = ξ s . We prove the following monotonicity property of ξ()s . (Throughout this note, increasing and decreasing will mean strictly so, and a half-line will be a half-infinite line not including its endpoint.) THEOREM 1. The xi function is increasing in modulus along every horizontal half-line lying in any open right half-plane that contains no xi zeros. Similarly, the modulus decreases on each horizontal half-line in any zero-free, open left half-plane. For example, since ξ()s ≠ 0 outside the critical strip, if t is any fixed number, then ξ() σ + it is increasing for 1 <σ < ∞ and decreasing for −∞ <σ < 0. In the next section, as a corollary of Theorem 1, we give a reformulation of the Riemann Hypothesis (a slight improvement of [2, Section 13.2, Exercise 1 (e)]). The proof of Theorem 1 is presented in the final section. 2. A reformulation of the Riemann Hypothesis Here is an easy corollary of Theorem 1. COROLLARY 1. The following statements are equivalent. (i). If t is any fixed real number, then ξ() σ + it is increasing for 1 2 <σ < ∞ . (ii). If t is any fixed real number, then ξ() σ + it is decreasing for −∞ <σ < 1 2. (iii). The Riemann Hypothesis is true. PROOF. If ξ()s is increasing along a half-line L (or decreasing on L), then ξ()s cannot have a zero on L. It follows, using the functional equation, that each of the statements (i) and (ii) implies (iii). Conversely, if (iii) holds, then ξ()s ≠ 0 on the right and left open half-planes of the critical line, and Theorem 1 implies (i) and (ii). 3. Proof of Theorem 1 We prove the first statement. The second then follows, using the functional equation. 3 =σ = ℜ > σ Let H H():()0{} s s 0 be a zero-free, open right half-plane. Fix a real = σ number t0, and denote by L L(,)0 t 0 the horizontal half-line =σ + σ > σ⊂ = σ + σ > σ L{} it0:: 0 H{} it 0 . In order to prove that ξ()s is increasing along L, we employ the Hadamard product representation of the xi function [1, p. 80]: s ρ ξ()s=1 eBs ∏1 − es . 2 ρ ρ Here the product is over all nonreal zeta zeros ρ, and B is the negative real number BC:=1 log4π − 1 −1 = − 0 . 023095 . ., 2 2 where C is Euler's constant. − ρ = ℜ > σ We first prove that 1 ()s is increasing on L. Since H{} s:() s 0 is zero-free and LH⊂ , we have ℜρ ≤ σ < ℜ ∈ ()()0 s ( s L). − It follows that the distance s − ρ and, hence, the modulus 1 − ()sρ= s − ρ ρ 1 are increasing along L. ρ ρ We next show that es is non-decreasing on L. (In fact, es is increasing on L, but we do not need this deeper fact.) Let ρ= β + i γ denote a nonreal zeta zero. Since β= ℜ( ρ ) ≥ 0, the modulus ρ ℜ()ρ ()βσ+ γt () β2+ γ 2 es = es = e 0 is non-decreasing along L. It remains to overcome the effect of the Hadamard product factor eBs , which, since B < 0, is decreasing in modulus on L. We use the following alternate interpretation of the ρ ρ constant B. First, let 1,,... 2 be the zeta zeros with positive imaginary part, and write ρ= β + γ ≥ n ni n , for n 1. Then B is also given by the formulas [1, p. 82] ∞ 1 1 ∞ β B = −∑ + = −2 ∑ n . ρ ρ β2+ γ 2 n=1 n n n=1 n n For N ≥ 1, denote the Nth partial sum of the series for −B by 4 N 1 1 S :=∑ + . N ρ ρ n=1 n n − + Note that ()BSN is positive, and that it approaches zero as N tends to infinity. ≥ Now for N 2 , let PN () s be the finite product s N s s P(): s =1 − ∏ 1− 1 − . N ρ ρ ρ 1 n=2 n n Then by combining exponential factors, we can write the Hadamard product as ∞ 1 ()B+ S s s s s ρs s ρ ξ()()s= e N 1 − P s ∏ 1− e n 1 − e n . 2 ρ N ρ ρ 1 n= N +1 n n − ρ s ρ From what we have shown about 1 ()s and e , both PN () s and the infinite product are increasing in modulus along L. To analyze the remaining factors on L, set =σ + s it0 and define the function 2 σ− β2 + − γ 2 1 ()B+ S s s 1 2() B+ S σ () ()t f():σ = e N 1 − = e N 1 0 1 . N 2 ρ 4 β2 + γ 2 1 1 1 ′ σ A calculation shows that the derivative fN () is positive if σ− β 1 > −()BS + . σ− β2 + − γ 2 N ()1 ()t0 1 σ> σ σ− β ≥ σ − σ > − + → → ∞ Now fix 1 0 . Since 1 1 1 0 0, and ()BSN 0 as N , we can ′ σ > ′ choose N so large that fN ()1 0. Then fN is also positive on some open interval I σ σ ξ σ + σ ∈ containing 1. It follows that fN () and, therefore, ()it0 are increasing for I . σ> σ Since 1() 0 and t0 are arbitrary, the theorem is proved. References [1] H. DAVENPORT, Multiplicative Number Theory, 2nd ed., revised by H. L. Montgomery, Graduate Texts in Mathematics, 74, Springer-Verlag, New York-Berlin, 1980. [2] H. L. MONTGOMERY and R. C. VAUGHAN, Multiplicative Number Theory, I, Classical Theory, Cambridge Studies in Advanced Mathematics, 97, Cambridge University Press, Cambridge, UK, 2007..

View Full Text

Details

  • File Type
    pdf
  • Upload Time
    -
  • Content Languages
    English
  • Upload User
    Anonymous/Not logged-in
  • File Pages
    4 Page
  • File Size
    -

Download

Channel Download Status
Express Download Enable

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

  • Not to be reproduced or distributed without explicit permission.
  • Not used for commercial purposes outside of approved use cases.
  • Not used to infringe on the rights of the original creators.
  • If you believe any content infringes your copyright, please contact us immediately.

Support

For help with questions, suggestions, or problems, please contact us