JULIO ALCÁNTARA-BODE

IMCA Instituto de Matemática y Ciencias Afines

THE RIEMANN ZETA FUNCTION HAS AT MOST

A FINITE NUMBER OF NON TRIVIAL ZEROS OFF THE

CRITICAL LINE

direcciónCalle Los Biólogos 245Urb. San César - La MolinaLima, Perú

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Pontificia Universidad Católica del PerúUniversidad Nacional de Ingeniería

THE RIEMANN ZETA FUNCTION

HAS AT MOST A FINITE NUMBER

OF NON TRIVIAL ZEROS OFF THE

CRITICAL LINE

Primera edición digital

Noviembre, 2012

Lima - Perú

© Julio Alcántara-Bode

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Lima - Perú, enero del 2011

“El conocimiento es útil solo si se difunde y aplica” Víctor López Guzmán Editor

The Riemann zeta function has at most a finite

number of non trivial zeros off the critical line

By Julio Alcantara-Bode

January 20, 2009

Abstract

Using standard results from Hilbert space theory and a variant ofthe Beurling-Nyman formulation of the Riemann Hypothesis (R.H.)we prove that the Riemann zeta function has a most a finite numberof non trivial zeros off the critical line. This is obtained from algebraicrelationships that exist between any two of these zeros with real partsgreater than 1

2 and positive imaginary parts.

1 Brief historical background

If π(n) denotes the number of prime numbers in the interval [1, n], n ∈N, it seems that from numerical evidence, Gauss conjectured that

lim[π(n)

nln n

]= 1. Using elementary methods Chebyshev proved that

ln 22≤ lim

π(n)n

ln n

≤ 1 ≤ limπ(n)

nln n

≤ 2 ln 2

establishing then that if lim[π(n)

nln n

]exists, it must be equal to 1.

To prove the existence of this limit Riemann suggested the study ofthe complex variable function

ζ(s) =+∞∑n=1

n−s, s = σ + it, σ > 1, t ∈ R,

afterwards known as the Riemann zeta function. He proved that ζ hasanalytic continuation to C \ {1}, still denoted by ζ, that has a pole offirst order at s = 1 with residue 1 and obeys the functional equationξ(s) = ξ(1− s), where

ξ(s) =12s(s− 1)π−

s2 Γ

(s

2

)ζ(s)

1

and Γ is the gamma function. If Z(ζ) denotes the set of zeros of ζ(defined on C \ {1}), he established the following facts:Z(ζ) ∩ {s : σ > 1} = φ, Z(ζ) ∩ {s : σ < 0} = {−2n : n ∈ N} and allthese zeros, called trivial, are simple; the set Z(ζ)∩{s : 0 ≤ σ ≤ 1} (alltheses zeros are called non-trivial) is infinite, symmetric with respectto the lines {s : σ = 1

2} (the critical line), {s : t = 0} andZ(ζ) ∩ {s : 0 ≤ σ ≤ 1} ∩ R = φ.Riemann also formulated the now famousRIEMANN HYPOTHESIS (R.H.):

Z(ζ) ∩ {s : 0 ≤ σ ≤ 1} = Z(ζ) ∩ {s : σ =12}

Hardy established that the set Z(ζ)∩ {s : σ = 12} is infinite. Levinson

proved that more than 13 of the elements in Z(ζ)∩ {s : 0 ≤ σ ≤ 1} are

in Z(ζ) ∩ {s : σ = 12} and Conrey showed that in this result 1

3 couldbe replaced by 2

5 [7, 9].

It can be shown that lim[π(n)

nln n

]= 1 if and only if Z(ζ) ∩ {s :

σ = 1} = φ, and Hadamard and Vallee-Poussin, independently, ver-ified this last condition, finally proving the conjecture of Gauss that

lim[π(n)

nln n

]= 1, result now known as the Prime Number Theorem.

The location of the element of Z(ζ) ∩ {s : 0 ≤ σ ≤ 1} closest to theline {s : σ = 1} determines the order of magnitude of the relative errorterm in the Prime Number Theorem. In some sense this relative errorterm is minimal if R.H. holds.The proof or disproof of R.H. is one of the most important unsolvedproblems in Mathematics. R.H., that originally arose in Analytic Num-ber Theory, can also be reformulated as a problem in several otherbranches of Mathematics. The following formulation of R.H. as a prob-lem of Functional Analysis will be useful later on [2] (it is a variant ofthe Beurling-Nyman formulation of R.H.)

Theorem 1.1. Let

[Aρf ](θ) =∫ 1

0

ρ

(θ

x

)f(x)dx,

where ρ(x) = x− [x], x ∈ R, [x] ∈ Z, [x] ≤ x < [x] + 1, be consideredas an integral operator on L2(0, 1). Then R.H. holds if and only ifker Aρ = {0} or if and only if h ∈/ R(Aρ) where h(x) = x.

For more detailed information about R.H. see [4, 5, 7, 8, 9].In the next section we prove that the set Z(ζ)∩ {s : σ > 1

2} is at mostfinite.

2

2 Proof of main results

First we begin with a recollection of some facts that will be used inthis section. If {sj}1≤j≤n ⊂ {s : σ > − 1

2} and si 6= sj if i 6= j,then the set {hsj}1≤j≤n ⊂ L2(0, 1) is linearly independent, since itsGram’s determinant G(hs1 , · · · , hsn) = det (< hsi , hsj >)1≤i,j≤n isstrictly positive [1, p.19]. It is proven in [2, 3] that if s = σ + it, σ >

− 12 , t ∈ R, then Aρh

s = hs − ζ(s+1)

s+1 hs+1, and therefore Aρ(s hs) = hif ζ(s + 1) = 0.The following result will be used recurrently.

Lemma 2.1. Let K be a closed set in L2(0, 1) such that Aρf = h, ∀f ∈K, f ∈ K if f ∈ K and λf1 + (1− λ)f2 ∈ K, ∀λ ∈ C if {f1, f2} ⊂ K.If g is the unique element of minimum norm of K [11, p. 202,Theorem3] let’s assume that:

i) g ∈ C1(]0, 1]), and limx→0+

x g2(x) = 0;

ii) if α ∈]0, 1], g(α) 6= 0 and gα(x) = α x g′(α x) then 1g(α) gα ∈ K,

but if g(α) = 0 then g + λgα ∈ K, ∀λ ∈ C

then ‖ g ‖2= g2(1)1 + 2g(1)

Proof: By uniqueness we have g = g. It will be proven now thatif f = f ∈ K then < g, f − g >= 0. If λ ∈ R, then g + λ(f − g) ∈ K,and therefore

‖ g + λ(f − g) ‖2=‖ g ‖2 +2λ < g, f − g > +λ2 ‖ f − g ‖2≥‖ g ‖2

and since this inequality holds ∀λ ∈ R, it follows that < g, f − g >= 0or < g, f >=‖ g ‖2. As Aρg = h, we have by definition that

∫ 1

0

ρ

(θ

x

)g(x) dx = θ, ∀ θ ∈ [0, 1]

Take now α ∈]0, 1] and let’s evaluate∫ 1

0

ρ

(θ

x

)g(α x) dx =

1α

∫ 1

0

ρ

(αθ

αx

)g(α x) d(α x)

=1α

∫ α

0

ρ

(αθ

y

)g(y) dy

=1α

∫ 1

0

ρ

(αθ

y

)g(y) dy − 1

α

∫ 1

α

ρ

(αθ

y

)g(y) dy

=θ

[1−

∫ 1

α

g(y)y

dy

]

3

which finally gives us∫ 1

0

ρ

(θ

x

)g(α x) dx = θ

[1−

∫ 1

α

g(y)y

dy

]

If we write the same equation for β instead of α and then substractboth of them we get

∫ 1

0

ρ

(θ

x

)[g(α x)− g(β x)] dx = θ

∫ α

β

g(y)y

dy

If α 6= β we divide the last equation by α− β and the take thelimit β → α to get from [11, p. 129, Theorem 5] that

∫ 1

0

ρ

(θ

x

)α x g′(α x) dx = θ g(α), ∀α ∈]0, 1]

If g(α) 6= 0 we get from the first part of ii) that

< g, gα >= g(α) ‖ g ‖2

If g(α) = 0 we get from the second part of ii) that < g, gα >= 0 andtherefore ∀α ∈]0, 1] it holds that

< g, gα >= g(α) ‖ g ‖2

or ∫ 1

0

α x g′(α x) g(x) dx = g(α) ‖ g ‖2

If in the last equation we take α = 1 and integrate by parts, takinginto account the second part of i) we get

g2(1)2

− ‖ g ‖22

= g(1) ‖ g ‖2

or finally

‖ g ‖2= g2(1)1 + 2 g(1)

,

which proves the lemma.

If s = σ + it, σ > − 12 , t > 0, ζ(s + 1) = 0, let’s apply the lemma

to the setKs = {λ s hs + (1− λ) s hs : λ ∈ C}

If gs is the element of minimum norm of Ks, then gs = gs andtherefore

gs =(

12

+ i γs

)s hs +

(12− i γs

)s hs, where γs ∈ R

4

Minimizing ‖ gs ‖2 with respect to γs we get

γs =12 t

(σ2 + t2 + σ)(2σ + 1)2(σ2 + t2 + σ) + 1

,

gs(1) = σ − 2 γs t = − σ2 + t2

2(σ2 + t2 + σ) + 1< 0

and

‖ gs ‖2= gs(1)2

1 + 2gs(1)=

(σ2+t2)2

2σ+1

2(σ2 + t2) + 2σ + 1

If sj = σj + i tj , σj > − 12 , tj > 0, ζ(sj + 1) = 0, j = 1, 2 and s1 6= s2

let’s apply now the lemma to the set

Ks1,s2 = {λ gs1 + (1− λ) gs2 : λ ∈ C}If ‖ gs1 ‖≤‖ gs2 ‖ let’s find the element of minimum norm gs1,s2 ofKs1,s2 .

By uniqueness gs1,s2 = λ gs1 + (1− λ) gs2 , with λ ∈ R and minimizing‖ gs1,s2 ‖2 with respect to λ we get

λ =‖ gs2 ‖2 − < gs2 , gs1 >

‖ gs1 − gs2 ‖2

If ‖ gs1 ‖=‖ gs2 ‖ we get λ = 12 and the lemma implies that

gs1(1) = gs2(1) < 0. Therefore gs1,s2(1) = gs1(1) = gs2(1) and usingthe lemma again we get ‖ gs1,s2 ‖=‖ gs1 ‖=‖ gs2 ‖ a contradictionsince the element of minimum norm is unique. Then s1 6= s2 impliesthat ‖ gs1 ‖6=‖ gs2 ‖ and in our case ‖ gs1 ‖<‖ gs2 ‖. If λ is asgiven above then λ > 0, by Cauchy-Schwarz, and λ = 1 if and only if‖ gs1 ‖2=< gs1 , gs2 >.We now show that if λ 6= 1 then

< gs1 , gs2 >=gs1(1) gs2(1)

1 + gs1(1) + gs2(1)

By the lemma we have

‖ λgs1 + (1− λ) gs2 ‖2=[λgs1(1) + (1− λ) gs2(1)]2

1 + 2 λgs1(1) + 2(1− λ) gs2(1)

from which we get

λ2gs1(1)2[

11 + 2 gs1(1)

− 11 + 2λgs1(1) + 2(1− λ)gs2(1)

]+

(1− λ)2gs2(1)2[

11 + 2gs2(1)

− 11 + 2λgs1(1) + 2(1− λ)gs2(1)

]+

5

2λ(1− λ)[< gs1 , gs2 > − gs1(1) gs2(1)

1 + 2λgs1(1) + 2(1− λ)gs2(1)

]= 0

since λ(1− λ) 6= 0 we get[

λg2s1

(1)1 + 2gs1(1)

− (1− λ)g2s2

(1)1 + 2gs2(1)

][gs2(1)− gs1(1)] +

[< gs1 , gs2 > (1 + 2λgs1(1) + 2(1− λ)gs2(1))− gs1(1) gs2(1)] = 0

or

λ ‖ gs1 − gs2 ‖2 [gs2(1)− gs1(1)] =

g2s2

(1) [gs2(1) + gs1(1)] + gs2(1) gs1(1)1 + 2gs2(1)

− < gs2 , gs1 > (1 + 2gs2(1))

and taking into account the value of λ we finally get

< gs2 , gs1 >=gs1(1) gs2(1)

1 + gs1(1) + gs2(1)

But since

gsj (1) = − σ2j + t2j

2(σ2j + t2j ) + 2σj + 1

, j = 1, 2,

we have

< gs1 , gs2 >=

(σ21 + t21)(σ

22 + t22)

(2σ1 + 1)(σ22 + t22) + (2σ2 + 1)(σ2

1 + t21) + (2σ1 + 1)(2σ2 + 1)(1)

But if λ = 1 we have

< gs1 , gs2 >=‖ gs1 ‖2=(σ2

1+t21)2

2σ1+1

2(σ21 + t21) + 2σ1 + 1

(2)

Then if ‖ gs1 ‖<‖ gs2 ‖, at least one of the equations (1) or (2) holdstrue; since they correspond to the cases λ 6= 1 and λ = 1 respectively,both can not be true simultaneously.In the last two equations let’s plug in the value of the scalar product

6

< gs1 , gs2 >=(

12

+ i γs1

) (12

+ i γs2

)s1 s2

1 + s1 + s2+

(12

+ i γs1

) (12− i γs2

)s1 s2

1 + s1 + s2+

(12− i γs1

) (12

+ i γs2

)s1 s2

1 + s1 + s2+

(12− i γs1

) (12− i γs2

)s1 s2

1 + s1 + s2+

= 2Re

[(12

+ i γs1

)(12

+ i γs2

)s1 s2

1 + s1 + s2

]+

2Re

[(12

+ i γs1

)(12− i γs2

)s1 s2

1 + s1 + s2

]

These two real parts only differ in the sign of t2, the imaginary partof s2, so one only needs to calculate one of them. A straightforwardcalculation gives

< gs1 , gs2 >=

2(σ1 + σ2 + 1)2 + (t1 + t2)2

×{(

14− γs1 γs2

) [(σ2

1 + t21)σ2 +

(σ2

2 + t22)σ1 + σ1 σ2 − t1 t2

] −

12(γs1 + γs2)

[(σ2

1 + t21 + σ1

)t2 +

(σ2

2 + t22 + σ2

)t1

]}+

2(σ1 + σ2 + 1)2 + (t1 − t2)2

×{(

14

+ γs1 γs2

) [(σ2

1 + t21)σ2 +

(σ2

2 + t22)σ1 + σ1 σ2 + t1 t2

]+

12(γs2 − γs1)

[− (σ2

1 + t21 + σ1

)t2 +

(σ2

2 + t22 + σ2

)t1

]}(3)

Equations (1), (2) and (3) show that between any two different ele-ments of U = Z(ζ)∩{s : σ > 1

2 , t > 0} there is some sort of algebraic re-lationship. We now show that (1), (2) and (3) imply that if U 6= φ, thenU is a finite set. First we show that if U 6= φ, then there is s1 + 1 ∈ Usuch that ‖ gs1 ‖<‖ gs2 ‖, ∀ (s2+1) ∈ U \{s1+1}. For this it is enoughto show that if s3 + 1 ∈ U , then the set {s + 1 ∈ U :‖ gs ‖≤‖ gs3 ‖} is

7

finite. But

‖ gs ‖2=

(t2+σ2

2σ+1

)2

2(

t2+σ2

2σ+1

)+ 1

and the function f(x) =x2

2x + 1is strictly increasing if x > 0, therefore

t2 + σ2

2σ + 1≤ t23 + σ2

3

2σ3 + 1= k3

which givest2 + (σ + 1− k3 − 1)2 ≤ k3 + k2

3

and represents a closed circle C of center (k3+1, 0) and radius√

k3 + k23.

Since C is compact U∩C is finite.There exists then the required s1+1 ∈U . If U were infinite then for this value of s1 we could, in (1) and (2),take the limit t2 →∞. In the case of (1) we get

σ1 − 2t1 γs1 =t21 + σ2

1

2σ1 + 1

which gives

− σ21 + t21

2(σ21 + t21 + σ1) + 1

=σ2

1 + t212σ1 + 1

and implies σ1 = t1 = 0 and ζ(1) = 0, a contradiction. In the case of(2) we get

− σ21 + t21

2(σ21 + t21 + σ1) + 1

=(t21+σ2

1)2

2σ1+1

2(σ21 + t21 + σ1) + 1

which again implies σ1 = t1 = 0 and ζ(1) = 0, a contradiction. Thisproves that the set Z(ζ) ∩ {s : σ > 1

2} is at most finite.

3 Final Comments

It is possible that the argument given here could be extended to otherzeta functions. To do that one would need something like Theorem1.1, that is essentially a variant of the Beurling-Nyman condition forR.H. In [6] there are given Beurling-Nyman criteria for functions inthe Selberg class, that for instance could be useful to try to extendthese results to Dirichlet L-functions. The Riemann zeta-function has auniversality property [7, Chapter VII] that could preclude the behaviorof certain of its non-trivial zeros implied by the equations (1) and (2).

If in equations (1) and (2) we get rid of the denominators, we obtain

8

equations of the form P (σ1, σ2, t1, t2) = 0 where P is a real polynomialin 4 variables. If these polynomials do not have zeros in the set

{(σ1, σ2, t1, t2) :t21 + σ2

1

2σ1 + 1<

t22 + σ22

2σ2 + 1, t2 ≥ t1 ≥ 29×109,−1

2< σi < 0, i = 1, 2},

then ζ would have at most 4 non-trivial zeros off the critical line, sinceit is known that [10],

Z(ζ)∩{s : 0 ≤ σ ≤ 1, |t| ≤ 29×109} = Z(ζ)∩{s : σ =12, |t| ≤ 29×109}

4 Acknowledgements

I would like to thank M. Gonzalez for preparing the manuscript forpublication. I dedicate this work to the memory of my beloved parentsJulio Tomas and Georgina.

INSTITUTO DE MATEMATICA Y CIENCIAS AFINESandPONTIFICIA UNIVERSIDAD CATOLICA DEL PERUe-mail address: jalcant@pucp.edu.pe

References

[1] N.I. Achieser, Theory of approximation, Dover, New York, (1992)

[2] J. Alcantara-Bode, An integral equation formulation of the Rie-mann hypothesis, J. Integral Equations and Operator Theory 17,(1993), 151-168.

[3] A. Beurling, A closure problem related to the Riemann zeta func-tion,Proc. Nat. Acad. Sci. 41, (1955), 312-314.

[4] E. Bombieri, Problems of the milennium: The Riemann hypotesis,CLAY(2000)

[5] B. Conrey, The Riemann hypothesis, Notices of the AMS, March,2003, 341-353

[6] A. de Roton, Generalisation du critere de Beurling-Nyman pourl’hypothese de Riemann, Trans. Amer. Math. Soc. 12 (2007),6111-6126.

[7] A.A. Karatsuba and S.M. Voronin, The Riemann zeta function,Walter de Gruyter, Berlin, 1992

[8] P. Sarnak, Problems of the milennium: The Riemann hypothesis,CLAY(2004)

9

[9] E.C. Titchmarsh , The theory of the Riemann zeta function, 2nd

ed., revised by R.D. Heath-Brown, Oxford Univ. Press, 1986

[10] S. Wedeniwski, Zeta Grid-verifications of the Riemann Hypothe-sis, available at http://www.zeta grid.net/.

[11] A.C. Zaanen, Integration, 2nd ed., North-Holland, Amsterdam,1967.

10

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