Univ India Final

7/29/2019 Univ India Final

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Computational Methods and Function TheoryVolume 00 (0000), No. 0, 1? CMFT-MS XXYYYZZ

Aspects of Analytic Number Theory:

The Universality of the Riemann Zeta-Function

Jorn Steuding

Abstract. These notes deal with Voronins universality theorem which states,roughly speaking, that any non-vanishing analytic function can be uniformlyapproximated by certain shifts of the Riemann zeta-function. We start witha brief introduction to the classical theory of the zeta-function. Then we givea self-contained proof of the universality theorem. We conclude with severalinteresting applications of this remarkable property and discuss some relatedproblems and extensions.

Keywords. Riemann zeta-function, universality, value-distribution.

2000 MSC. 11M06, 11M26, 11M99, 30E10.

-4

-2

0

2

4

-10

0

10

20

30

40

0

1

2

3

4

Figure 1. The reciprocal of the absolute value of ( + it) for [4, 4], t [10, 40]. The zeros of (s) appear as poles.

Version CMFT-Workshop January 2008, Guwahati.I wish to express my gratitude to the organizers of the workshop Computational Methodsand Function Theory at Guwahati, Assam, in particular Meenaxi Bhattacharjee and StephanRuscheweyh. I am also very grateful to the audience for their interest in the topic and for theextraordinarily good atmosphere at Guwahati as well as to the anonymous referee for her orhis remarks how to make the content more accessible to the reader. Last but not least, I wantto thank my wife Rasa for her careful reading of the script.

ISSN 1617-9447/$ 2.50 c 20XX Heldermann Verlag


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2 Jorn Steuding CMFT

The theme of this course is an astonishing approximation property of the famous Rie-mann zeta-function, so the topic is settled in the intersection of complex analysis andanalytic number theory. Arithmetical problems may often sound simple in their formu-lation; however, their treatment often needs sophisticated machinery and challengingideas. Since the path-breaking works of Dirichlet and Riemann from the middle ofthe nineteenth century, analytic methods have become an important tool in numbertheory. The proof of the celebrated prime number theorem by investigating the distri-bution of zeros of the zeta-function is just one example. One of the most spectacularproperties of the zeta-function is Voronins universality theorem which states that anynon-vanishing analytic function can be uniformly approximated by certain shifts ofthe zeta-function. Here we give a (more or less) complete proof of this remarkableresult and discuss some of its applications, e.g., hypertranscendence and a criterion for

the truth of the famous yet unproved Riemann hypothesis. Finally, we discuss someextensions and related open problems.

These self-contained lecture notes are mainly based on the original paper of Voronin[67], resp. the presentation in the monograph [33] of Karatsuba & Voronin with slightmodifications. Thanks to Bagchi [1], Reich [56], and Laurincikas [36], there is another,more sophisticated probabilistic approach to universality which allows slightly moregeneral results. For the sake of simplicity we have chosen the down to earth approachof Voronin. Many of the additional results can be found in [61]. For the backgroundin zeta-function theory (and for help with respect to the exercises) we refer to theclassical monograph [63] of Titchmarsh and the online notes [62].

Jorn Steuding, Wurzburg, February 2009.

Contents

1. Introduction 3

1.1. The Riemann zeta-function is universal 3

1.2. Survey on value-distribution theory 5

1.3. A weak approximation theorem 9

2. Zeta-function theory 11

2.1. Primes and zeros 11

2.2. The approximate functional equation 16

2.3. The functional equation 23

2.4. The mean-square and applications 26

2.5. A density theorem 31

2.6. The prime number theorem 35

3. Universality theorems 42


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00 (0000), No. 0 The Universality of the Riemann Zeta-Function 3

3.1. Voronins universality theorem 42

3.2. Rearrangement of conditionally convergent series 43

3.3. Finite Euler products 48

3.4. Diophantine approximation 55

3.5. Approximation in the mean end of proof 59

3.6. Reichs discrete universality theorem and other related results 63

4. Applications, extensions, and open problems 66

4.1. Functional independence 66

4.2. Self-recurrence and the Riemann hypothesis 69

4.3. The effectivity problem 72

4.4. L-functions and joint universality 78

4.5. The Linnik-Ibragimov conjecture 84

References 87

1. Introduction

Here we introduce the main actor, the Riemann zeta-function, and present first clas-

sical results on its amazing value-distribution due to Bohr as well as the remarkableuniversality theorem of Voronin. For historical details we refer to [61].

1.1. The Riemann zeta-function is universal. The Riemann zeta-functionis a function of a complex variable s = + it, for > 1 given by

(s) =

n=1

1

ns=

p

1 1

ps

1;(1.1)

here and in the sequel the letter p always denotes a prime number and theproduct is taken over all primes. The series and the product are prototypes ofso-called Dirichlet series, resp. Euler products. They both converge absolutelyin the half-plane > 1 and uniformly in each compact subset. The identitylinking both, the series and the product was discovered by Euler in 1737 and canbe regarded as an analytic version of the unique prime factorization of integers.The Euler product gives a first glance on the intimate connection between thezeta-function and the distribution of prime numbers. An immediate consequenceis Eulers proof of the infinitude of the primes. Assuming that there were onlyfinitely many primes, the product in (1.1) is finite, and therefore convergent fors = 1, contradicting the fact that the Dirichlet series defining (s) reduces tothe divergent harmonic series as s 1+. Hence, there exist infinitely many

This mixture of latin and greek letters is tradition in analytic number theory.


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prime numbers. This fact is well known since Euclids elementary proof, but theanalytic access gives deeper knowledge on the distribution of the prime numbersas we shall see in the second chapter.

However, the main theme of these notes is a remarkable approximation propertyof Riemanns zeta-function, called universality.

By Weierstrass celebrated approximation theorem we know that any continu-ous function, defined on a closed interval, can be uniformly approximated bypolynomials. The set of continuous functions is rather big whereas the set ofpolynomials is comparably small. This makes the Weierstrass theorem remark-able. One may not believe that it is possible to approximate any continuousfunction on a bounded interval by a singlefunction! Actually, the Riemann zeta-function has this astonishing approximation property! More precisely, shifts ofits logarithm s log (s + i) can approximate any continuous function definedon a bounded interval. Of course, this approximation cannot be realized in thehalf-plane of absolute convergence of the zeta defining series. For our purposewe note that our protagonist, (s), can be analytically continued to the wholecomplex plane except for a simple pole at s = 1, e.g.,

(1.2) (s) = (1 21s)1

n=1

(1)n+1ns

;

here the series on the right converges for > 0 (see also Exercise 1 below).

Voronins famous universality theorem states that any non-vanishing analyticfunction g can be approximated uniformly by certain purely imaginary shifts ofthe zeta-function in the vertical strip 1

2< < 1. (A precise formulation will

be given below.) For instance, for any positive , there exists a real number ,which might be extremely large, such that the inequality

|(s + 34

+ i) g(s)| < holds on any disk |s| r, where 0 < r < 1

4is fixed. For an illustrative example

take g = exp and see the figure in Section 4.3. The same statement holds whenthe closed disk is replaced by a closed line segment on the imaginary axis andin this case we only need that the function g is continuous and has no zeros.This follows from a slightly more advanced version of Voronins theorem (seeTheorem 3.11). And if we want to get rid of the non-vanishing assumption,we can approximate by the logarithm of the zeta-function and this leads tothe aforementioned extension of Weierstrass approximation theorem. Anotherapplication of universality is related to the famous Riemann hypothesis, oneof the seven millennium problems, about the distribution of zeros of the zeta-function (Theorem 4.3).

The first universalobject in the mathematical literature was discovered by Feketein 1914/15; he proved the existence of a real power series with the propertythat for any continuous function g on the unit interval, there exists a sequence


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of partial sums which approximates g uniformly. In 1926, G.D. Birkhoff [5]proved the existence of an entire function f with the property that to any givenentire function g, there exists a sequence of complex numbers an such that f(z+an) g(z) uniformly on compacta in C, as n . Universality is a frequentphenomenon in analysis, often appearing when analytical processes diverge orbehave irregularly in some sense. The Riemann zeta-function and its relativesare so far the only known explicit examples of universal objects. In the followingsection we shall give a precise statement; however, we start with a brief look howthis surprising and deep result has been developed.

1.2. Survey on value-distribution theory. The zeros of the zeta-functionare of special interest (for reasons we will explain in the following chapter). Itseems rather difficult to localize zeros or any other concrete values taken bythe zeta-function, whereas it is much easier to study how often the values liein a given set. Having this idea in mind, Harald Bohr refined former studieson the value-distribution of the Riemann zeta-function by applying diophantine,geometric, and probabilistic methods.

In the half-plane of absolute convergence > 1 we have

(1.3) 0 1 are lying in the disk of radius(0) centered in the origin. It can be shown that (s) assumes quite many ofthe complex values inside this disk when t varies in R. On the other side ()tends to infinity as 1+, and indeed Bohr [6] succeeded in proving that inany strip 1 < < 1 + , (s) takes any non-zero value infinitely often. Wesketch his argument. Define log (s) for any s C by choosing the principalbranch of the logarithm on the intersection of the real axis with the half-plane ofabsolute convergence, and for other points s = + it let log ( + it) be the valueobtained from log (2) by continuous variation along the line segments [2, 2 + it]and [2 + it, + it], provided that the path does not cross a zero or pole of (s);if it does, then take log ( + it) = lim0+ log ( + i(t + )). For > 1,

log (s) =

p

log

1 1

ps

=

p

k=1

1

kpsk.

For any fixed prime p and > 1, the set of values taken by the inner sum inthe series representation on the right-hand side is a convex curve while t runsthrough R. Adding up all these curves and using some facts from the theory ofdiophantine approximation, it follows that log (s) takes any complex value in1 < < 1 + which leads to Bohrs result.


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The situation to the left of the abscissa of convergence is much more complicated.Here, Bohr studied finite Euler products

M(s) :=

pM

1 1

ps

1.

As M tends to infinity, these products do not converge any longer but they ap-proximate (s) in the mean (we will meet this ingenious idea later again). Thevalue-distribution of finite Euler products is treatable by the theory of diophan-tine approximation, and by their approximation property this leads to informa-tion on the values taken by the zeta-function. In a series of papers Bohr andhis collaborators discovered that the asymptotic behaviour of (s) is ruled by

probability laws on every vertical line to the right of = 12 . In particular, Bohr& Courant [8] proved that for any fixed ( 1

2, 1] the set of values ( + it) with

t R lies dense in the complex plane. Later, Bohr refined these results signifi-

-2 -1 1 2 3

-2

-1

1

2

Figure 2. (35

+ it) for t [0, 60]. The curve visits any neighbourhood ofany point in the complex plane as t runs through the set of real numbers, and

so the picture would be completely black in the end.

cantly by applying probabilistic methods. Let R be an arbitrary fixed rectanglein the complex plane whose sides are parallel to the real and the imaginary axes,

and let G be the half-plane > 12 where all points are removed which have thesame imaginary part as and smaller real part than one of the possible zeros of(s) in this region. Then a remarkable limit theorem due to Bohr & Jessen [9, 10]states that for any fixed > 1

2the limit

limT

1

Tmeas { [0, T] : + i G, log ( + i) R}

exists. Here and in the sequel meas A stands for the Lebesgue measure of ameasurable set A. This limit value may be regarded as the probability howmany values of log ( + it) belong to the rectangle R. Next, for any complexnumber c, denote by Nc(1, 2, T) the number of c-values of (s), i.e., the roots


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of the equation (s) = c, inside the region 1 < < 2, 0 < t

T (countingmultiplicities). From the limit theorem mentioned above Bohr & Jessen deduced

Theorem 1.1. Let c be a complex number = 0. Then, for any 1 and 2 satis-fying 1

2< 1 < 2 < 1, the limit limT 1TNc(1, 2, T) exists and is positive.

In 1935, Jessen & Wintner proved limit theorems similar to the one above byusing more advanced methods from probability theory (infinite convolutions ofprobability measures). We do not mention further developments of Bohrs ideasby his successors Borchsenius, Jessen, and Wintner but refer for more details onBohrs contribution and results of his collaborators to the monograph of Lau-rincikas [36] and the survey of Matsumoto [48]. Bohrs line of investigations

appears to have been almost abandoned for some time. Only in 1972, Voronin[66] obtained some significant generalizations of Bohrs denseness result.

Theorem 1.2. For any fixed numbers s1, . . . , sn with12

< Re sk < 1 for 1 k n and sk = s for k = , the set

{((s1 + it), . . . , (sn + it)) : t R}is dense inCn. Moreover, for any fixed number s with 1

2< < 1,

{((s + i), (s + i), . . . , (n1)(s + i)) : R}is dense inCn.

What about the value-distribution of the zeta-function on the line = 12

? Itis conjectured but yet unproved that also the set of values of (s) taken on thisvertical line is dense inC. However, Garunkstis & Steuding [18] have shown thatthe second statement of Theorem 1.2 is false for = 1

2whenever n 2. Selberg

(unpublished) proved that the values taken by an appropriate normalization ofthe Riemann zeta-function on this line are normally distributed: let R be anarbitrary fixed rectangle in the complex plane whose sides are parallel to the realand the imaginary axes, then

limT1

T meast (0, T] : log

1

2

+ it12

log log T R

=1

2

R

exp1

2(x2 + y2)

dx dy.

The value-distribution on the line = 12

is somehow special for several reasons(more about that in the following chapter).

In 1975, Voronin [67] proved his remarkable universality theorem:

Theorem 1.3. Let0 < r < 14

and suppose that g(s) is a non-vanishing continu-ous function on the disk

|s

| r, which is analytic in the interior. Then, for any


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> 0, there exists a positive real number such that

max|s|r

s + 34

+ i g(s) < .

Moreover, the set of such has positive lower density:

lim infT

1

Tmeas

[0, T] : max

|s|r

s + 34

+ i g(s) < > 0.

Thus, any suitable target function can be approximated as good as we pleasean infinity of times. We say that (s) is universal since appropriate shifts ap-

proximate uniformly any element of a huge class of functions. We may interpretthe absolute value of an analytic function as an analytic landscape over thecomplex plane. Then the universality theorem states that any finite analyticlandscape can be found (up to an arbitrarily small error) in the analytic land-scape of the Riemann zeta-function. This is indeed a remarkable property of thezeta-function!

0.5

0.6

0.7

0.8

0.9

1

116

118

120

122

0

1

2

3

4

Figure 3. Some summits of the Himalaya or the analytic landscape of (s)

for [12

, 1], t [115, 122].

We shall give a more or less self-contained proof of Voronins universality theorem

in Chapter 3. A reader who is familiar with the basic theory of the Riemann zeta-

function may directly jump to Chapter 3; for anybody else we provide the essentials of

this theory in the following chapter. However, for the remaining part of this chapter we

shall investigate a weaker, nevertheless still interesting approximation property than

universality.


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1.3. A weak approximation theorem. What might have been Voronins in-tention for his studies which had led him to the discovery of this astonishinguniversality property? One reason for Voronins investigations might have beenBohrs concept of almost periodicity and its applications to the Riemann hypoth-esis (see Section 4.2). Another starting point for Voronin could have been thewish to extend Theorem 1.2; the universality theorem can be seen as an infinitedimensional analogue of the second part of this theorem. To illustrate this, wesketch how Theorem 1.2 can be used to obtain some weakform of the universalitytheorem.

Assume we are given an analytic target function g(s) defined on |s| r, wherer is a positive real number. Our main tool is the Taylor series expansion

g(s) =

k=0

g(k)(0)

k!sk,

valid for all s with |s| r. By Cauchys formula,

g(k)(0) =k!

2i

|s|=r

g(s)

sk+1ds,

where the integral is taken over the circle |s| = r in counterclockwise direction.Hence,

|g(k)(0)

| k!Mrk,

where M := max|s|=r |g(s)|. Let (0, 1). Theng(k)(0)k! sk Mk for |s| r.

For any positive , there exists a positive integer n such that

(1.4) 1 :=

g(s)

0k


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A quantitative version of Theorem 1.4 can be found in the recent article byGarunkstis et al. [17]; the main tool to obtain explicit bounds for the values and is another result of Voronin, a so-called multidimensional-theorem, whichcombines analytic and diophantine approximation properties (see also [33]). Itis believed that Voronin himself was aware about statements like Theorem 1.4.For more details on the history of Voronins theorem we refer to the nice surveyarticles of Laurincikas [39] and Matsumoto [49].

Mathematics is not a spectator sport! The following exercises may help to dive deeperinto zeta-function theory. Although (s) is defined as an absolutely convergent seriesin the half-plane > 1, the distribution of values taken near the vertical line = 1 isinteresting:

Exercise 1. For > 0 prove the representation (1.2) and deduce that (s) < 0 fors (0, 1).Exercise 2. Prove inequality (1.3). Can you make use of formula (1.2) to estimatethe growth of ( + it) for fixed > 0 as t ?

2. Zeta-function theory

In this chapter we give some hints for the importance of the Riemann zeta-function for

analytic number theory. We start with a survey on the remarkable link between prime

numbers and zeros of(s). Later we prove the prime number theorem as well as density

estimates for the number of hypothetical zeros off the critical line = 12 . Besides we

develop parts of the machinery which is needed to prove Voronins universality theorem.

For historical details and more references we refer to [53].

2.1. Primes and zeros. It was the young Gauss who conjectured in 1791 forthe number (x) of primes p x the asymptotic formula(2.1) (x) li(x),where the logarithmic integral is given by

li(x) = lim0+

10

+

x1+

du

log u=

x2

du

log u 1.04 . . . .

Gauss conjecture states that, in first approximation, the number of primes xis asymptotically x

log x. By elementary means, Chebyshev proved around 1850

that 0.921 . . . (x) log xx

1.055 . . . for sufficiently large x. Furthermore, heshowed that if the limit

limx

(x)log x

x

We write f(x)

g(x), if limx f(x)/g(x) = 1.


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exists, the limit is equal to one, which supports conjecture (2.1). Riemann wasthe first to investigate the Riemann zeta-function as a function of a complexvariable. In his only one but outstanding paper [58] on number theory from1859 he outlined how Gauss conjecture could be proved by using the function(s). However, at Riemanns time the theory of functions was not developedsufficiently far, but the open questions concerning the zeta-function pushed theresearch in this field quickly forward. We shall briefly discuss Riemanns memoir;some of the sketched results will later be proved in detail.

First of all, by partial summation,

(s) = nN

1

n

s+

N1s

s 1+ s

N

[u] u

u

s+1du;(2.2)

here and in the sequel [u] denotes the maximal integer less than or equal to u.This gives an analytic continuation for (s) to the half-plane > 0 except for asimple pole at s = 1 with residue 1. This process can be continued to the lefthalf-plane and shows that (s) is analytic throughout the whole complex planeexcept for s = 1. Riemann discovered the functional equation

s2 s

2

(s) =

1s2

1 s

2

(1 s),(2.3)

where (s) denotes Eulers Gamma-function. In view of the Euler product (1.1)it is easily seen that (s) has no zeros in the half-plane > 1. It follows from thefunctional equation and from basic properties of the Gamma-function that (s)vanishes in < 0 exactly at the so-called trivial zeros s = 2n with n N. All

-14 -12 -10 -8 -6 -4 -2

-0.15

-0.1

-0.05

0.05

0.1

Figure 4. (s) in the range s [14.5, 0.5].

other zeros of(s) are said to be nontrivial, and we denote them by = + i.Obviously, they have to lie inside the so-called critical strip 0 1, and it iseasily seen that they are non-real. The functional equation (2.3) and the identity

(s) = (s) show some symmetries of (s). In particular, the nontrivial zeros of


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(s) are distributed symmetrically with respect to the real axis and to the verticalline = 1

2. It was Riemanns ingenious contribution to number theory to point

out how the distribution of these nontrivial zeros is linked to the distribution ofprime numbers. Riemann conjectured the asymptotics for the number N(T) ofnontrivial zeros = + i with 0 < T (counted according multiplicities).This conjecture was proved in 1895 by von Mangoldt [46, 47] who found moreprecisely

N(T) =T

2log

T

2e+ O(log T).(2.4)

Riemann worked with the function t ( 12

+ it) and wrote that very likely all

roots t are real, i.e., all nontrivial zeros lie on the so-called critical line = 12

. Thisis the famous, yet unproved Riemann hypothesis which we rewrite equivalentlyas

Riemanns hypothesis. (s) = 0 for > 12 .In support of his conjecture, Riemann calculated some zeros; the first one withpositive imaginary part is = 1

2+ i14.134 . . .. Furthermore, Riemann conjec-

tured that there exist constants A and B such that

12

s(s 1) s2 s

2

(s) = exp(A + Bs)

1 s

exp

s

,

where the product on the right is taken over all nontrivial zeros (the trivialzeta zeros are cancelled by the poles of the Gamma-factor). This was shownby Hadamard [22] in 1893 (on behalf of his theory of product representationsof entire functions). Finally, Riemann conjectured the so-called explicit formulawhich states that

(x) +

n=2

(x1n )

n= li(x)

=+i

>0

li(x) + li(x1)

(2.5)

+

x

du

u(u2

1) log u

log2

for any x 2 not being a prime power (otherwise a term 12k

has to be added on

the left-hand side, where x = pk); the appearing integral logarithm is defined by

li(x+i) =

(+i)log x(+i)log x

exp(z)

z+ idz,

We write f(x) = O(g(x)), if limsupx |f(x)|/g(x) < ; equivalently, we also write

f g.In 1932, Siegel published an account of Riemanns work on the zeta-function found in

Riemanns private papers in the archive of the university library in G ottingen. It becameevident that behind Riemanns speculation there was extensive analysis and computation.


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where = +1 if > 0 and =

1 otherwise. The explicit formula was provedby von Mangoldt [46] in 1895 as a consequence of both product representationsfor (s), the Euler product (1.1) and the Hadamard product.

Building on these ideas, Hadamard [23] and de la Vallee-Poussin [64] found (in-dependently) in 1896 the first proof of Gauss conjecture (2.1), the celebratedprime number theorem. For technical reasons it is of advantage to work with thelogarithmic derivative of (s) which is for > 1 given by

(s) =

n=1

(n)

ns,

where the von Mangoldt -function is defined by

(2.6) (n) =

logp if n = pk with k N,

0 otherwise.

A lot of information concerning the prime counting function (x) can be recov-ered from information about

(2.7) (x) :=nx

(n) =px

logp + O

x12 log x

.

Partial summation gives (x) (x)log x

. First of all, we shall express (x) in termsof the zeta-function. If c is a positive constant, then

1

2i

c+i

ci

xs

sds =

1 if x > 1,0 if 0 < x < 1.

(2.8)

This yields the so-called Perron formula: for x Z and c > 1,

(x) = 12i

c+ici

(s)

xs

sds.(2.9)

Moving the path of integration to the left, we find that the latter expression isequal to the corresponding sum of residues, that are the residues of the integrandat the pole of (s) at s = 1, at the zeros of (s), and at the additional pole ofthe integrand at s = 0. The main term turns out to be

Res s=1

(s)

xs

s

= lim

s1(s 1)

1

s 1 + O(1)

xs

s= x,

whereas each nontrivial zero gives the contribution

Res s=

(s)

xs

s

= x

.

By the same reasoning, the trivial zeros altogether contribute

n=1x2n

2n= 1

2log

1 1

x2.


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Incorporating the residue at s = 0, this leads to the exact explicit formula

(x) = x

x

1

2log

1 1

x2

log(2),

which is equivalent to Riemanns formula (2.5). This formula is valid for any x Z. Notice that the right-hand side of this formula is not absolutely convergent.If (s) would have only finitely many nontrivial zeros, the right-hand side wouldbe a continuous function of x, contradicting the jumps of (x) for prime powersx. Going on it is more convenient to cut the integral in (2.9) at t = T whichleads to the truncated version

(x) = x ||Tx

+ O x

T(log(xT))

2,(2.10)

valid for all values of x. Next we need information on the distribution of thenontrivial zeros. Already the non-vanishing of (s) on the line = 1 yields theasymptotic relations (x) x, resp. (x) li (x), which is Gauss conjecture(2.1) and sufficient for many applications. However, more precise asymptoticswith a remainder term can be obtained by a zero-free region inside the criticalstrip. The largest known zero-free region for (s) was found by Vinogradov [65]and Korobov [35] (independently) in 1958 who proved

(s)

= 0 in

1

c

(log(|t| + 3))13

(log log(|t| + 3))23

,

where c is some positive absolute constant. In combination with the Riemann-von Mangoldt formula (2.4) one can estimate the sum over the nontrivial zerosin (2.10). Balancing out T and x, we obtain the prime number theorem with thesharpest known remainder term:

Theorem 2.1. There exists an absolute positive constant C such that for suffi-ciently large x

(x) = li (x) + O

x exp

C (log x)

35

(log log x)15

.

We shall give a complete proof of the prime number theorem with a slightlyweaker remainder term in Section 2.6.

By the explicit formula (2.10) the impact of the Riemann hypothesis on theprime number distribution becomes visible. In 1900, von Koch [34] showed thatfor fixed [ 1

2, 1)

(x) li (x) x+ (s) = 0 for > ;(2.11)equivalently, one can replace the left-hand side by (x) x. Here and in thesequel stands for an arbitrary small positive constant, not necessarily the sameat each appearance. With regard to known zeros of (s) on the critical line


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it turns out that an error term with < 1

2

is impossible. Thus, the Riemannhypothesis states that the prime numbers are as uniformly distributed as possible!

Many computations were done to find a counterexample to the Riemann hypoth-esis. Van de Lune, te Riele & Winter [45] localized the first 1 500 000 001 zeros,all lying without exception on the critical line; moreover they all are simple! Byobservations like this it is conjectured, that all or at least almost all zeros of thezeta-function are simple. This is the so-called essential simplicity hypothesis.

A classical density theorem due to Bohr & Landau [11] states that most of thezeros lie arbitrarily close to the critical line. Denote by N(, T) the number ofzeros = + i of(s) for which > and 0 < T (counting multiplicities).Bohr & Landau proved(2.12) N(, T) T = o(N(T))for any fixed > 1

2. Hence, almost all zeros are clustered around the critical

line. The strongest unconditional estimate that holds throughout the right halfof the critical strip is due to Gritsenko [21]:

Theorem 2.2. For any fixed with 12

< < 1,

N(, T) T125 (1)(log T) 915 .

Comparing with Theorem 1.1, we see that zero is an exceptional value of thezeta-function. The location of zeros appears to be completely different from anyother value.

On the other hand, Hardy [24] showed that infinitely many zeros lie on thecritical line. Refining a mollifying technique of Selberg, Levinson [43] localizedmore than one third of the nontrivial zeros of the zeta-function on the criticalline, and as Heath-Brown [26] and Selberg (unpublished) discovered, they are allsimple. Introducing Kloosterman sums, Conrey [13] was able to choose longermollifiers in order to show that more than two fifths of the zeros are simple andon the critical line.

In the remainder of this chapter we shall prove the prime number theorem as well as a

density theorem, both not as sharp as those mentioned above. Besides, we introduce

much of the analytic machinery needed for the proof of Voronins universality theorem

in the following chapter.

2.2. The approximate functional equation. By the Riemann integral con-vergence criterion the series defining zeta converges absolutely for > 1. Since,

Here we write f(x) = o(g(x)), if limx f(x)/g(x) = 0.


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for

0 > 1,

n=1

1

ns

n=1

1

n0 1 +

n=2

nn1

du

u0(2.13)

= 1 +

1

u0 du = 1 +1

0 1 ,the series in question converges uniformly in any half-plane 0 with 0 > 1.Thus, by a well-known theorem of Weierstrass, (s), being the limit of a uniformlyconvergent sequence of analytic functions, is analytic in its half-plane of absoluteconvergence.

Lemma 2.3. (s) is analytic for > 1 and satisfies identity (1.1), i.e.,

(s) =

n=1

1

ns=

p

1 1

ps

1.

Proof. It remains to show the identity between the series and the product. Bythe geometric series expansion and the unique prime factorization of the integers,

px

1 1

ps

1=

px

1 +

1

ps+

1

p2s+ . . .

=

np|npx

1

ns;

as usual, we write d

|n if the integer d divides the integer n, and d n otherwise.

Since

n=1

1

ns

np|npx

1

ns

n>x

1

n

x

u du =x1

1tends to zero as x , we get the desired identity by sending x . Next we shall derive not only an analytic continuation for (s) to the half-plane > 0 but also a rather good approximation which will be very useful later on.At s = 1 the zeta-function defining series reduces to the harmonic series. Toobtain an analytic continuation for (s) we have to seperate this singularity. Forthat purpose we apply Abels partial summation:

Lemma 2.4. Let 1 < 2 < . . . be a divergent sequence of real numbers, definefor n C the function A(u) =

nu n, and let F : [1, ) C be a

continuous differentiable function. Thennx

nF(n) = A(x)F(x) x

1

A(u)F(u) du.

Proof. We have

A(x)F(x) nx

nF(n) =

nxn(F(x) F(n)) =

nxx

n

nF(u) du.


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Since 1

n

u

x, interchanging integration and summation yields theassertion. Next we apply partial summation to the partial sums of the Dirichlet seriesdefining zeta. Let N < M be positive integers and > 1. Then, application ofLemma 2.4 with F(u) = us, n = 1 and n = n yields

N 0. Sending M we obtainTheorem 2.5. For > 0,

(s) =nN

1

ns+

N1s

s 1 + s

N

[u] uus+1

du.

In particular, (s) has an analytic continuation to the half-plane > 0 exceptfor a simple pole at s = 1 with residue 1.

Putting N = 1 in the formula of Theorem 2.5, we obtain the analytic continuation(2.2) for (s). Our next aim is to derive from the representation of the theorem

a very useful approximation of (s) inside the critical strip.Let f(u) be any function with continuous derivative on the interval [a, b]. Usingthe lemma on partial summation with n = 1 ifn (a, b], and n = 0 otherwise,we get

a


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Next, we replace in Eulers summation formula the function u

[u]

1

2

by itsFourier series expansion.

Lemma 2.7. For u R \ Z,u [u] 12|m|M

m=0

exp(2imu)2im

1

2M(u [u]) ,

and, for u R,

m=

m=0

exp(2imu)

2im

= u [u] 1

2if u Z,

0 if uZ,

where the terms with m have to be added together; the partial sums are uni-formly bounded in u and M.

Proof. By symmetry and periodicity it suffices to consider the case 0 < u 12

.Since1

2

u

exp(2imx) dx = (1)m+1 + exp(2imu)

2imfor 0 = m Z,

we obtain |m|M

m=0

exp(2imu)2im

u + 12

=1

2

u

|m|M

exp(2imx) dx

=

12

u

sin((2M + 1)x)

sin(x)dx.(2.14)

By the mean-value theorem there exists (u, 12

) such that the latter integralequals

u

sin((2M + 1)x)

sin(u)dx.

This implies both formulas of the lemma. It remains to show that the partialsums of the Fourier series are uniformly bounded in u and M. Substitutingy = (2M + 1)x in (2.14), we get1

2

u

sin((2M + 1)x)

sin(x)dx =

12

u

sin((2M + 1)x)

xdx

+

12

u

sin((2M + 1)x)

1

sin(x) 1

x

dx

0

sin(y)

ydy +

12

0 1

sin(x) 1

x dx


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with an implicit constant not depending on u and M. Both integrals on the rightexist, which gives the uniform boundedness. Further, we need the following estimate of exponential integrals.

Lemma 2.8. Assume that F : [a, b] R has a continuous non-vanishing de-rivative and that G : [a, b] R is continuous. If G/F is monotonic on [a, b],then

ba

G(u) exp(iF(u)) du

4GF (a)

+ 4GF (b)

.Proof. First, we assume that F(u) > 0 for a u b. Since (F1(v)) =F(F

1

(v))1

, substituting u = F1

(v) leads toba

G(u) exp(iF(u)) du =

F(b)F(a)

G(F1(v))F(F1(v))

exp(iv) dv.

By the monotonicity of G/F, application of the mean-value theorem gives

Re

F(b)F(a)

G(F1(v))F(F1(v))

exp(iv) dv

=G

F(F(a))

F(a)

cos v dv +G

F(F(b))

F(b)

cos v dv

with some [F(a), F(b)]. This gives the desired estimate. The same ideaapplies to the imaginary part. The case F(u) < 0 can be treated analogously. Now we are in the position to prove the van der Corput summation formula:

Theorem 2.9. For any given > 0 there exists a positive constant C = C(),depending only on , with the following property: if f : [a, b] R is a functionwith continuous derivative, g : [a, b] [0, ) is differentiable, andf, g and |g|are all monotically decreasing, then

a


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Proof of Theorem 2.9. We apply Eulers summation formula with the functionF(u) = g(u) exp(2if(u)). Using the Fourier series expansion of Lemma 2.7, weget

a


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Thus, f(b)f(a)+,m=0

ReI2(m)m

g(a).


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With slight modifiactions this method applies also to the imaginary part ofI

2(m)and the case m f(b) . Further, if 0 [f(b) , f(a) + ], then Lemma2.8 gives b

a

g(u) exp(2if(u)) du g(a).In view of (2.15) the theorem follows from the above estimates under the con-dition f(b) > 0. If this condition is not fulfilled, then we may argue withf(u) (1 [f(b)])u in place of f(u). Now we apply van der Corputs summation formula to the zeta-function. Let > 0. By Theorem 2.5 we have

(s) =nx

1

ns+

x


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Riemann [58] himself gave two proofs of the functional equation. In the meantimeseveral rather different proofs were found. Here we follow Riemanns originalapproach which relies on the functional equation of the theta-function which isgiven by the infinite series

(x) =nZ

exp(xn2).

Recall the Poisson summation formula: iff : R R is twice differentiable withf(z) z2 as z , and f is integrable over R, then(2.16)

nZf(n + ) =

mZf(m) exp(2im)

for all R, where f denotes the Fourier transform of f. We shall apply thePoisson summation formula with the function f(z) := exp( z2

x), where x > 0.

First, we compute the Fourier transform by quadratic substitution:

f(y) =

+

exp(( z2x

+ 2iyz)) dz

= x exp(xy2)+

exp(x(w + iy)2) dw.(2.17)

To gon on we evaluate the integral

I() :=

+

exp(x(w + )2) dw,

where is any complex number. For this aim we consider the integralR

exp(x2) d,

where R is the rectangular contour with vertices r,r + iIm , and r is apositive real number. By Cauchys theorem, the integral is equal to zero. Onthe line Re = r, the integrand tends uniformly to zero as r . Hence,I() = I(0), and thus the integral I() does not depend on . This gives in

(2.17)

f(y) = x exp(xy2)+

exp(xw2) dw = Cx exp(xy2),

where

C :=

+

exp(z2) dz.Now applying Poissons summation formula leads to

nZexp( (n+)2

x) = C

x

mZexp(xm2 + 2im).


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Choosing = 0 and x = 1, both sums are equal; thus, C = 1 and we have justproved the functional equation for the theta-function:

Theorem 2.12. For any x > 0,

(x) =1x

1

x

.

Now we are ready to give the

Proof of Theorem 2.11. For Re z > 0, the Gamma-function may be definedby Eulers integral

(z) =

0 uz

1

exp(u) du.Substituting u = n2x leads to

(2.18) s

2

s2

1

ns=

0

xs21 exp(n2x) dx.

Summing up over all n N yields

s2 s

2

n=1

1

ns=

n=1

0

xs21 exp(n2x) dx.

On the left-hand side we find the Dirichlet series defining (s); in view of its

range of convergence, the latter formula is valid only for > 1. On the right-hand side we may interchange summation and integration, justified by absoluteconvergence. Thus we obtain

s2 s

2

(s) =

0

xs21

n=1

exp(n2x) dx.

We split the integral at x = 1 and get

(2.19) s2 s

2

(s) =

10

+

1

x

s21(x) dx,

where the series (x) is given in terms of the theta-function:

(x) :=

n=1

exp(n2x) = 12

((x) 1) .

In view of the functional equation for the theta-function,

1

x

= 1

2

1

x

1

=

x(x) + 12

(

x 1),

we find by the substitution x 1x

that the first integral in (2.19) is equal to

1

xs21

1

x dx =

1

xs+1

2 (x) dx +1

s

1

1s

.


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Substituting this in (2.19) yields

(2.20) s2 s

2

(s) =

1

s(s 1) +

1

x

s+12 + x

s21

(x) dx.

Since (x) exp(x), the last integral converges for all values of s, and thus(2.20) holds by analytic continuation throughout the complex plane. The right-hand side remains unchanged by s 1 s. This proves the functional equationfor zeta.

To indicate the power of the functional equation we consider the growth of thezeta-function on vertical lines. A standard application of the PhragmenLindelof

principle (see [61, 63]) to the entire function

(2.21) 12

s(s 1) s2 s

2

(s)

in combination with Stirlings formula shows that for any vertical strip 1 2 of bounded width there exists a positive constant c such that

(2.22) ( + it) tc as t .In the particular case of the critical line this easily yields the bound

12

+ it t14 + as t .

Better estimates are known. Using rather advanced methods (lattice points,estimates for exponential series, etc.), Huxley [30] obtained the exponent 32

205+ .

The yet unproved Lindelof hypothesis states

(2.23)

12

+ it t as t ;

note that the truth of the Riemann hypothesis would imply the latter estimate.

Another application of the functional equation yields a proof of the RiemannvonMangoldt formula (2.4). For this aim one applies the argument principle to thefunction given by (2.21). However, in the sequel we are mainly concerned withzero-counting functions in rectangles to the right of the critical line.

2.4. The mean-square and applications. Using the approximate functionalequation, we shall derive a mean-square formula for (s) in the half-plane > 1

2.

Such mean-square formulae are important tools in the theory of the Riemannzeta-function. For example, they provide information on the number of hypo-thetical zeros off the critical line as we shall see below.

Theorem 2.13. For > 12

,

T

1

|( + it)|2 dt = (2)T + O(T22 log T).


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Proof. By the approximate functional equation,

( + it) =n


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this purpose we need the following integrated version of the argument principle,also known as Littlewoods lemma:

Lemma 2.14. Let A < B and let f(s) be analytic on R := {s C : A B, |t| T}. Suppose that f(s) does not vanish on the right edge = B of R.LetR beR minus the union of the horizontal cuts from the zeros of f in R tothe left edge ofR, and choose a single-valued branch of log f(s) in the interior ofR. Denote by(, T) the number of zeros = + i off(s) inside the rectanglewith > including zeros with = T but not those with = T. Then

R

log f(s) ds =

2i

B

A

(, T) d.

We give a sketch of the simple proof. Cauchys theorem implies

R log f(s) ds =0, and so the left-hand side of the formula of the lemma,

R, is minus the sum

of the integrals around the paths hugging the cuts. Since the function log f(s)jumps by 2i across each cut (assuming for simplicity that the zeros of f in Rare simple and have different height; the general case is no harder),

R is 2i

times the total length of the cuts, which is the right-hand side of the formula inthe lemma.

Littlewoods lemma can be used in various ways to obtain estimates for the

number of zeros of the zeta-function in certain regions of the complex plane. Westart with a weak version of the Riemann-von Mangoldt formula (2.4) for thenumber N(T) of nontrivial zeros = + i with imaginary part (0, T].Theorem 2.15. For sufficiently large T,

N(T + 1) N(T) log T.

Proof. Jensens formula states that if f(s) is an analytic function for |s| Rwith zeros s1, . . . , sm (according their multiplicities) and f(0) = 0, then

(2.24) 12

2

0

log |f(r exp(i))| d = log rm|f(0)||s1 . . . sm|for r < R (this is a variant of the Poisson integral formula). This applied withf(s) = (2 + iT + s) leads to the bound

logrm|f(0)|

|s1 . . . sm| log T,

where r [3, 4] is chosen such that (2 + iT + s) is non-zero. Since any zero of (s) with

|

T

| 1 has distance at most r >

5 to 2 + iT, it follows from


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(2.22) that

N(T + 1) N(T)

|T|11 =

|T|1

logr

| 2 iT|1

log r5

|T|1log

r

| 2 iT| log T.

The theorem is proved. Next we are interested in an estimate for the number N(, T) of zeros = + iof (s) with > , 0 < T. Application of Littlewoods lemma with fixed0 >

1

2

yields

2

10

N(, T) d =

T0

log |(0 + it)| dt T

0

log |(2 + it)| dt

+

02

arg ( + iT) d 0

2

arg () d.(2.25)

The main contribution comes from the first integral on the right-hand side. Thelast integral does not depend on T and so it is bounded. Since (s) has an Eulerproduct representation, the logarithm has a Dirichlet series representation:

(2.26) log (s) =

plog1

1

ps =

p,k

1

kpks

for > 1,

where k runs through the positive integers; here we choose that branch of thelogarithm which is real on the positive real axis. We obtainT

0

log |(2 + it)| dt = Re

p,k

1

kp2k

T0

exp(itk logp) dt

n=2

1

n2 1.

It remains to estimate arg ( + iT). We may assume that T is not the ordinateof any zero. Since arg (2) = 0 and

arg (s) = arctan Im (s)

Re (s)

,

where

Re (2 + it) =

n=1

cos(it log n)

n2 1

n=2

1

n2> 1

1

du

u2= 0,

we have by the argument principle

| arg (2 + iT)| 2

.

Now assume that Re ( + iT) vanishes q times in the range 12

2. Devidethe interval [ 1

2+ iT, 2 + iT] into q+ 1 parts, throughout each of which Re (s)


30/90


31/90


Let 1 =1

2

+ 1

2

(0

1

2

), then we get

N(0, T) 10 1

01

N(, T) d 20 12

11

N(, T) T.

Because of (2.27) we have proved estimate (2.12) from Section 2.1.

2.5. A density theorem. In the last section we have proved a first estimatefor the number of hypothetical zeros to the right of the critical line. Now we givethe proof of a stronger density theorem due to Hoheisel [29]:

Theorem 2.16. For any fixed ( 12

, 1),

N(, T) T4(1)(log T)10.

For the proof we need the following simple but powerful lemma, also calledGallaghers lemma:

Lemma 2.17. Letf(t) be a continuously differentiable complex-valued functionon the interval [a, b]. Let t0 = a < t1 < . . . < tk1 < tk = b and denote by theminimum of all differences tj+1 tj . Then

k

j=1

|f(tj)|2 1

b

a

|f(t)|2 dt + 2b

a

|f(t)|2 dt b

a

|f(t)|2 dt12

.

Proof. Denote by j(t) the characteristic function on the interval [tj , tj+1], i.e.,j(t) = 1 for t [tj , tj+1] and j (t) = 0 otherwise. Further let

j(t) =1

tj+1 tj

ta

j () d.

Then, by partial integration,tj+1tj

j(t)

|f(t)|2

dt = j (t)|f(t)|2

tj+1

t=tj 1

tj+1 tj

tj+1tj

|f(t)|2j (t) dt

= |f(tj+1)|2 1tj+1 tj

tj+1

tj

|f(t)|2 dt.

It follows that

|f(tj+1)|2 1

tj+1tj

|f(t)|2 dt + 2tj+1

tj

|f(t)| |f(t)| dt.

Now the assertion of the lemma follows from summation over j and applicationof the CauchySchwarz inequality. Now we are in the position to give the


32/90


33/90


34/90


This leads to

S(Y) Y2

Y


35/90


implies the density hypothesis. However, already Theorem 2.16 can serve in quitemany applications as substitute for the Riemann hypothesis.

2.6. The prime number theorem. Now we shall prove the prime numbertheorem with a slightly weaker remainder term than in Theorem 2.1. For thisaim we need to establish a zero-free region for (s) inside the critical strip. Wemay argue only for s = + it from the upper half-plane, since the zeros aresymmetrically distributed with respect to the real axis.

Lemma 2.18. For t 8, 1 12

(log t)1 2,(s)

log t and (s)

(log t)2.

Proof. Let 1 (log t)1 3. Ifn t, then

|ns| = n n1(log t)1 = exp

1 1log t

log n

n.

Thus, the approximate functional equation, Theorem 2.10, implies

(s) nt

1

n+ t1 log t

(the bound for the sum is an easy exercise in analysis; in Exercise 5 belowone shall prove an asymptotic formula (2.39)). The estimate for (s) followsimmediately from Cauchys formula,

(s) =1

2i

|zs|=r

(z)

(z s)2 dz

with r > 0 sufficiently small, or alternatively, by (carefull) differentiation of theformula of Theorem 2.5. In view of the Euler product representation of zeta we find for > 1

|( + it)| = exp(Re log (s)) = expp,k

cos(kt logp)

kpk .Since

17 + 24 cos + 8 cos(2) = (3 + 4 cos )2 0,it follows that

()17|( + it)|24|( + 2it)|8 1.(2.34)This inequality is the main idea for our following observations. By the approxi-mate functional equation, Theorem 2.10, we have

() 1

1


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37/90


Lemma 2.20. Letc and y be positive and real. Then

1

2i

c+ici

ys

sds =

0 if 0 < y < 1,12

if y = 1,1 if y > 1.

Proof. If y = 1, then the integral in question equals

1

2

dt

c + it=

1

lim

T

T0

c

c2 + t2dt =

1

lim

Tarctan(T /c) = 1

2,

by well-known properties of the arctan-function. Now assume that 0 < y < 1and r > c. Since the integrand is analytic in > 0, Cauchys theorem implies,

for T > 0, c+iTciT

ys

sds =

riTciT

+

r+iTriT

+

c+iTr+iT

ys

sds.

It is easily seen thatciTriT

ys

sds 1

T

cr

y d yc

T| log y| ,r+iTriT

ys

sds y

r

r+ yr

T1

dt

t yr

1

r+ log T

.

Now sending first r and then T to infinity, the first case follows. Finally, if y > 1,then we bound the corresponding integrals over the rectangular contour withcorners c iT, r iT, analogously. Now the pole of the integrand at s = 0with residue

Res s=0ys

s= lim

s0ys

s s = 1

gives the values 2i for the integral in this case. Now we continue our study of the logarithmic derivative of the zeta-function.For x Z and c > 1 we have

c+ici

n=1

(n)

nsxs

s ds =

n=1

(n)c+i

cix

ns ds

s ;

here interchanging integration and summation is allowed by the absolute conver-gence of the series. In view of Lemma 2.20 it follows that

nx

(n) =1

2i

c+ici

n=1

(n)

nsxs

sds,

resp.

(x) =1

2i c+i

c

i

(s)

xs

sds;


38/90


39/90


Further, for the vertical integral,1++iT1iT

(s)

xs

sds x1(log T)9.

Collecting together, we deduce from (2.37)

(x) = x + O

x1+

T + x1(log T)9 +

x(log x)2

T+ log x

.

Choosing T = exp(1

10 (log x)19 ), we arrive at

(x) = x + O

x exp(c(log x) 19 )

.

Setting(x) :=

px

logp,

since(x) (x) =

pkx

k2

logp x 12 (log x)2,

it follows that(x) = x + O

x exp(c(log x) 19 )

.

Applying now partial summation, Lemma 2.4, we find

(x) =px

logp 1logp

= (x)log x

x2

(u)

1log u

du=

x

log xx

2

u

1

log u

du

+O

x expc(log x) 19

.

Now partial integration leads to the prime number theorem with remainder term:

Theorem 2.21. There exists a positive constant c such that for x 2(x) = li (x) + O x expc(log x)

19 .

Thus, the simple pole of the zeta-function is not only the key in Eulers proof ofthe infinitude of primes but also gives the main term of the asymptotic formulain the prime number theorem. We see that the primes are not too irregularlydistributed. For example, the prime number theorem implies that, if pn denotesthe n-th prime number (in ascending order), then pn n log n.We conclude this section by giving a sketch of von Kochs equivalent (2.11) forthe Riemann hypothesis. By partial summation we obtain for > 1

(s) =

s

s

1

+ s

1

(u) uus+1

du.


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Figure 5. This is Ulams spiral: the first 65 000 positive integers are listed

in a spiral in ascending order, the primes are coloured white, the composite

numbers black.

If (x) x x+, then the integral above converges for > , giving ananalytic continuation for

(s) 1

s 1to the half-plane > , and, in particular, (s) does not vanish there. For the

converse implication we assume that all nontrivial zeros = + i satisfy .Then it follows from (2.10) that

(2.38) (x) x x||T

1

|| +x

T(log(xT))2.

By Theorem 2.15 we have N(T + 1) N(T) log T, and therefore||T

1

|| [T]+1m=1

log m

m (log T)2.

Substituting this in (2.38) leads to

(x) x x(log T)2 + xT

(log(xT))2.

Now the choice T = x1 finishes the proof of this implication. So the Riemannhypothesis is true if and only if the error term in the prime number theorem isO(x

12

+). In this case there cannot be too long intervals free of primes. Notethat it is an open question whether there is always a prime in between twoconsecutive squares. Some examples may convince the reader that this is a rea-sonable conjecture, however, this statement does not even follow from Riemannshypothesis.


41/90


Practice makes perfect! We continue with some exercises. Already Euler found anexplicit formula for the values of the zeta-function at all positive even integers interms of Bernoulli numbers (see [63]).

Exercise 3. Recall the product representation

sin(z) = z

k=1

1 z

k2

and deduce some zeta-values: (2) = 2

6 , (4) =4

90 , . . . by expanding the infiniteproduct.

On the contrary, not too much is known about zeta-values at positive odd integers. In1978, Apery proved that (3) is irrational, however, it is not known whether this valueis transcendental or whether (5) is irrational.

It is conjectured that all zeros of the zeta-function are simple. A classical theorem ofSpeiser states that the Riemann hypothesis is true if and only if (s) is non-vanishingin 0 < < 12 .

Exercise 4. Prove that any zero of (s) on the critical line is a multiple zero of (s).

The following three exercises can be solved with Abels partial summation (Lemma2.4).

Exercise 5. Prove the following asymptotic formulasnx

1

n= log x + + O

1

x

,(2.39)

px

1

p= log log x + O(1);(2.40)

here is the EulerMascheroni constant := limN 1N(N

n=11n log N) = 0.557 . . ..

Exercise 6. Prove formula (2.33). For this one may count the lattice points (a, b) Z2under a hyperbola:

nxd(n) =

abx1. For the second moment observe that

nx

d(n)2 =

abxd(ab)

axd(a)

bx/a

d(b).

Another approach uses contour integration of the function (s)k xs

s , following the linesof proof of the prime number theorem.

The next exercise is about twin primes, that are pairs of primes of the form p, p + 2.It is unknown whether there are infinitely many twin primes. Brun showed for thenumber 2(x) of twin primes p, p + 2 with p x the estimate 2(x) x(log x)2.Exercise 7. Deduce from Bruns estimate that the sum over the reciprocals of alltwin primes converges although the sum of the reciprocals over all primes diverges (see(2.40)).


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Relevant information about the zeta-function is contained in its order of growth alongvertical lines as well as in the distribution of its zeros. For the next exercises one mayconsult [62, 63]:

Exercise 8. Apply the Phragmen-Lindelof principle in order to prove estimate (2.22)with an explicit constant c and give the details for the proof of Theorem 2.15.

Exercise 9. Prove the Riemann-von Mangoldt formula (2.4).

3. Universality theorems

In this chapter we shall prove the famous universality theorem of Voronin; besides

we indicate how to derive other remarkable universality theorems by similar means

(e.g., Reichs universality theorem 3.11 below). The method of proof is a mixture of

techniques from function theory, analytic number theory, and basic functional analysis.

3.1. Voronins universality theorem. Now we are going to prove Voroninsuniversality theorem, that is Theorem 1.3 from the introduction: Let0 < r < 1

4be fixed and suppose that g(s) is a non-vanishing continuous function on the disk|s| r which is analytic in the interior. Then, for any > 0,

lim infT

1

T

meas [0, T] : max|s|r s +34

+ i g(s) < > 0.The Euler product for the zeta-function is the key to prove the universality the-orem in spite of the fact that it does not converge in the region of universality.However, as already Bohr observed, an appropriate truncated Euler product ap-proximates (s) in a certain mean-value sense inside the critical strip; this isrelated to the use of modified truncated Euler products in Voronins proof (see(3.23) below). Another important tool in the proof are approximation theorems,one for numbers and one for functions. This is not too surprising since univer-sality is an approximation property. Last but not least we shall make use of theprime number theorem and classical function theory.

It is more convenient to work with series than with products. Therefore, weconsider the logarithms of the functions in question. Since g(s) has no zerosin |s| r, its logarithm exists and we may define an analytic function f(s) byg(s) = exp f(s) for |s| < r. Conversely, if f(s) is analytic, then g(s) = exp f(s)is analytic and non-vanishing. Now we formulate

Theorem 3.1. Let0 < r < 14 and suppose thatf(s) is a continuous function onthe disk |s| r, which is analytic in the interior. Then, for any > 0,

lim infT

1

Tmeas

[0, T] : max

|s|r log

s + 3

4+ i

f(s)

<

> 0.


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44/90


45/90


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Applying Lemma 3.3 to the series n=N1+1

un, shows that there exists a finiteset T3 {N1 + 1, N1 + 2, . . .} such thatv

nT2un

nT3

un

< 14 .Now write out the indices of first T2 and then T3, each in arbitrary order, write3 if 3 T2 T3. Continuing this process, the assertion of the lemma follows. .

Further, we have to prove

Lemma 3.5. Letv1, . . . , vN be arbitrary elements in a real Hilbert space

H. Then

there exists a permutation of the set{1, . . . , N } such that

max1mN

mk=1

v(k)

Nn=1

vn2 1

2

+ 2 N

n=1

vn

.Proof. First, suppose that

Nn=1

vn = 0.

Then we shall construct by induction a permutation {n1, . . . , nN} of{1, . . . , N }such that

(3.1) max1mN

mk=1

vnk

Nn=1

vn2 1

2

.

For this aim put n1 = 1 and suppose that n1, . . . , nj with 1 j N 1 havebeen chosen, satisfying

max1

m

j

m

k=1

vnk2

j

n=1

vn2.

Then we may choose nj+1 from the remaining numbers such thatj

k=1

vnk , vnj+1

0.

Such an nj+1 exists since otherwise

i=nk i

k=1vnk , vi

=

j

k=1vnk ,

j

k=1vnk

> 0.


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Hence, j+1k=1

vnk

2 = jk=1

vnk2 + vnj+12 + 2

jk=1

vnk , vnj+1

j+1k=1

vnk2.

This yields a permutation {n1, n2, . . . , nN} of {1, 2, . . . , N } which satisfies (3.1)under the assumption

Nn=1 vn = 0.

For arbitrary v1, . . . , vN define

vN+1 = N

n=1

vn,

and apply the already proved case for v1, . . . , vN, vN+1. This leads to a permu-tation {n1, n2, . . . , nN+1} of{1, 2, . . . , N + 1} with

max1mN+1

mn=1

v(n)

Nn=1

vn2 1

2

+ N

n=1

vn

.Removing N+1 from the set {vn1 , . . . , vnN, vnN+1} we get an N-tuple of vectorswhich satisfies the inequality of the lemma.

Now we are in the position for the

Proof of Theorem 3.2. By Lemma 3.4 we may assume that some subsequenceof the partial sums of the series

k uk converges to v in the norm of H. We

define

Un =n

k=1

uk,

and suppose that a sequence of partial sums Unj converges to v. For each j Nthere is a permutation of the set of vectors {Unj +1, . . . , U nj+1} in such a waythat the value of

mj := max1mnj+1nj

nj +mn=nj +1

u(n)

is minimal. By Lemma 3.5 it follows that

mj

n=nj +1

un2

12

+ 2Unj+1 Unj,

which tends to zero as j . Hence, the corresponding series converges to vin the norm of

H. Theorem 3.2 is proved.


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In the sequel we shall apply Pecherskys rearrangement theorem 3.2 to the fol-lowing Hilbert space. Let R be a positive real number, then the so-called Hardyspace HR2 is the set of functions f(s) which are analytic for |s| < R and for which

f := limrR

|s| 0. Consequently, its logarithm exists and equals

log M(s, ) =

pMlog

1 exp(2ip)

ps

;

here as for log (s) we may take the principal branch of the logarithm on thepositive real axis.

The first step in the proof of Theorem 3.1 is to show

Theorem 3.6. Let 0 < r < 14

and suppose that f(s) is continuous on |s| rand analytic in the interior. Further, let 0 =

14

, 24

, 34

, . . .

. Then for any > 0

and any y > 0 there exists a finite set M of prime numbers, containing at least

all primes p y, such thatmax|s|r

log Ms + 34 , 0 f(s) < .Proof. Since f(s) is continuous for |s| r, there exists > 1 such that 2r < 1

4and

(3.3) max|s|r

f s2

f(s)

< 2

.

The function f

s

2

is bounded on the disc |s| r =: R, and thus belongs to

the Hardy space

HR2 .


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Denote by pk the kth prime number. We consider the series

k=1

uk(s) with uk(s) := log

1 exp(2ipk )ps34

k

1.

First, we shall prove that for every v HR2 there exists a rearrangement of theseries

uk(s) for which

k=1

ujk (s) = v(s).

In view of the Taylor expansion of the logarithm the series k

uk(s) differs from

k=1

k(s) with k(s) := exp2ik

4

ps 34k

by an absolutely convergent series. Hence, it suffices to verify the conditions ofthe rearrangement theorem 3.2 for the series

k k(s). Since R exp((1 + 2)zj).Suppose the contrary. Then there is some (0, 1) and a constant B such that|F(z)| < B exp((1 + 2)z) for any z 0. It follows that(3.9) | exp((1 + )z)F(z)| < B exp(|z|) for z 0;since |m| 1, this estimate even holds for z < 0 by a suitable change of theconstant B.

Here we shall apply two theorems from Fourier analysis. First, recall the theorem

of Paley-Wiener: given an entire function G(z), then the relation

(3.10) G(z) =

g() exp(iz) d

holds for some square integrable function g() if and only if

|G(z)|2 dz < ,

and G(z) has an analytic continuation throughout the complex plane satisfyingG(z) exp(( + )|z|) for any > 0, where the implicit constant may dependon (this characterizes all transcendent functions of fixed exponential type ).Plancherels theorem states that for any such G(z) with (3.10) also

g() =1

2

G(z)exp(iz) dz

holds almost everywhere in R.

Application of the theorem of Paley-Wiener with G(z) = exp((1+)z)F(z) yieldswith regard to (3.9) the representation

exp((1 + )z)F(z) =

33

f() exp(iz) d,

where f() is a square integrable function with support on the interval [3, 3](not to be confused with our target function). Further, Plancherels theoremimplies

f() =1

2

F(z) exp((1 + )z iz) dz

almost everywhere. Hence, f() is analytic in a strip covering the real axis. Sincethe support of f() lies inside a compact interval, the integral above has to bezero outside this interval. Hence, F(z) has to vanish identically, contradictingthe existence of a sequence of positive real numbers zj diverging to + with(3.8).


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Let xj =zj

R

. Then it follows from (3.6) and (3.8) that

|(xj )| > R2 exp3

4xj

F(xj R) R2 expxj 34 + R(1 + 2) .

Thus, for sufficiently small > 0 we obtain the existence of a sequence of positivereal numbers xj, tending to +, such that(3.11) |(xj )| > exp((1 )xj).Now we shall approximate F and by polynomials. Let Nj = [xj ] + 1 andassume that xj 1 x xj + 1. Since |m| 1,

m=N2j +1

m

m!

(xR)m

(xR)N

2j

(N2

j )!

m=0

(xR)m

m!

NN2j

j exp(Nj)

(N2

j )! exp(

2xj ),

by Stirlings formula. Trivially,

m=N2j +1

1

m!

3x4

m exp(2xj )for the same x. Hence,

F(xR) =

N2jm=0

+

m=N2j +1

mm!

(xR)m = Pj(x) + O(exp(2xj ))

and analogouslyexp3x

4

= Pj (x) + O(exp(2xj)),

where Pj and Pj are polynomials of degree N2j . This yields in view of (3.6)(x) = Qj(x) + o(exp(xj )) for xj 1 x xj + 1,

where Qj = PjPj is a polynomial of degree N4j .In order to find lower bounds for (x) we have to apply a classical theoremof A. A. Markov which states that if Q is a polynomial of degree N with realcoefficients which satisfies the inequality

max1x1 |Q(x)| 1,then

max1x1

|Q(x)| N2.For the first, lets assume that Qj is a real polynomial. Choose [xj 1, xj + 1]such that

|Qj()| = maxxj1xxj +1

|Qj (x)|.Then Markovs inequality implies

maxxj

1

x

xj +1

|Qj(x)| N8j |Qj ()|.


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For|x

|

2 with sufficiently small satisfying 0 < < N

8j

, it follows that

|Qj(x)| |Qj()| |x | maxxj1xxj +1

|Qj(x)| |Qj()| O(|x |2|Qj ()|) 1

2|Qj ()| 12 |Qj(xj )|.

Hence, for x [ 2

, + 2

] [xj 1, xj + 1]|(x)| |Qj(x)| + o(exp(xj))

12|Qj(xj )| + o(exp(xj )) 12 |(xj)| + o(exp(xj )).

We have assumed that Qj has real coefficients. If this is not true, then the abovereasoning may be applied to both, the real part and the imaginary part of Qj .Hence, for sufficiently large xj , the intervals [xj 1, xj + 1] contain intervals[, + ] of length 1

200N8j all of whose points satisfy at least one of the

inequalities

(3.12) |Re(x)| > 1200

exp((1 )x) , |Im(x)| > 1200

exp((1 )x).In order to prove the divergence of a subseries of (3.4) we note that one of theinequalities in (3.12) is satisfied infinitely often as x ; we may assume thatit is the one with the real part. By the prime number theorem 2.21, the interval

[exp(), exp( + )] containsexp(+)exp()

du

log u+ O

exp

+ c 19

exp()

exp() 1 + O

exp

c 19

many primes, where c > 0 is some absolute constant. Under these prime numberspk [exp(), exp( + )] we choose those with k 0 mod 4. Since pk = k4 , wededuce from (3.5) and (3.11)

k0 mod 4

log pk+k,

=

k0 mod 4log pk+

Re (logpk)

exp( 1

2xj ),

which diverges with xj .Thus, we have shown that the series (3.4) satsifies the conditions of Theorem3.2. Hence, there exists a rearrangement of the series

k uk(s) such that

(3.13)

k=1

ujk (s) = f s

2

,

where f is our target function. Before we can finish the proof of Theorem 3.6 wehave to prove the following easy


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Lemma 3.7. Suppose that G(s) is analytic on|s

s0|

R and|ss0|r

|G(s)|2 d dt = M.

Then, for any fixed r satisfying r < R and any s with |s s0| r,

|G(s)| 1R r

M

12

.

Proof. By Cauchys formula,

G(s)2 =1

2i|zs|=

G(z)2

z sdz =

1

2

2

0

G2(s + exp(i)) d

for any < R. Taking the absolute modulus and integrating with respect to ,we obtain

|G(s)|2Rr

0

d 12

20

Rr0

|G(s + exp(i))|2 d d = M2

.

This yields the assertion.

We return to the proof of Theorem 3.6. According to (3.13),

limn

n

k=1

ujk

(s) = f s2

in the norm ofHR2 . This implies

limn

|s|R

f s

2

nk=1

ujk (s)

2

d dt = 0

uniformly on |s| R. Thus, application of Lemma 3.7 shows that for sufficientlylarge m

max|s|R

f

s

2

m

k=1ujk (s)

< 12

.

Hence, there exists a finite set M, containing without loss of generality all primesp y, such that

log M

s + 34

, 0

=m

k=1

ujk (s).

approximates g(s). More precisely, in view of (3.3) it follows that

max|s|r

log M s + 34 , 0 f(s) max|s|r

log M

s + 3

4, 0 f s

2

+ max|s|rf

s2

f(s)

< .

This finishes the proof of Theorem 3.6.


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Before we continue with the proof of Voronins universality theorem, we needsome arithmetical tools from the theory of diophantine approximation.

3.4. Diophantine approximation. In the theory of diophantine approxima-tions one investigates how good an irrational number can be approximated byrational numbers. This has plenty of applications in various fields of mathematicsand natural sciences.

For abbreviation we denote vectors ofRN by x = (x1, . . . , xN), we define x =( x1, . . . , xN) for R and x y = x1y1 + . . . + xNyN. Further, for x RNand RN we write x mod 1 if there exists y ZN such that x y .Moreover, we shall introduce the notion of Jordan volume of a region

RN.

Therefore, we consider the sets of parallelepipeds 1 and 2 with sides parallel tothe axes and of volume 1 and 2 with 1 2; if there are 1 and 2 suchthat lim sup1 1 coincides with lim inf2 2, then has the Jordan volume

= lim sup1

1 = lim inf2

2.

The Jordan sense of volume is more restrictive than the one of Lebesgue, but ifthe Jordan volume exists it is also defined in the sense of Lebesgue and equal toit.

Weyl [69] proved

Theorem 3.8. Let a1, . . . , aN R be linearly independent over the field ofrational numbers, write a = (a1, . . . , aN), and let be a subregion of the N-dimensional unit cube with Jordan volume . Then

limT

1

Tmeas { (0, T) : a mod 1} = .

Proof. From the definition of the Jordan volume it follows that for any > 0there exist two finite sets of open parallelepipeds {j } and {+j } inside theunit cube such that

(3.14) j int()

+j

andmeas

+j \

j

< ;

here, as usual, M denotes the closure of the set M, and int(M) its interior.Denote by 1 the characteristic function of

j , i.e.

1(x) =

1 if x j ,0 if x j .

Further, let 1 be the characteristic function of mod 1. Consequently,

0

1(x)

1(x)

1+(x)

1,


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and [0,1]N

(1+(x) 1(x)) dx < ,where the integral is N-dimensional with dx = dx1 dxN. Define

(x) =

0 if |x| 1

2,

c exp

1x+ 12

+ 1x 12

if |x| < 1

2,

where c is defined via 12

12(x) dx = 1.

Consequently, (x) is an infintely differentiable function, and hence the func-tions, given by

1(x) = N

[0,1]N1(y)

x1 y1

xN yN

dy

for 0 < < 1, are infinitely differentiable functions too. From (3.14) it followsthat for sufficiently small we have

0 1(x) 1(x) 1+(x) 1,and

(3.15) 0 [0,1]N(1+(x) 1(x)) dx < 2.We conclude

(3.16)

T0

1(a) d meas { (0, T) : a mod 1} T

0

1+(a) d

and

0 T

0

1+(a) dT

0

1(a) d 2T.Both integrands above are infinitely differentiable functions which are 1-periodicin each variable. Thus, we have the Fourier expansion

1(x) =nZN

cn

exp(2i n x),

where

cn

=

[0,1]N

1(x) exp(2i n x) dx.

Note that c0

is the volume of

j . Integration by parts gives

cn

N

j=1(|nj| + 1)k for k = 1, 2, . . . ,


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where the implicit constant depends only on k. This shows that the Fourierseries converges absolutely, and hence, for every > 0, there exists a finite setM ZN such that

1(x) =nM

cn

exp(2i n x) + R(x) with |R(x)| < .

This yields

1

T

T0

1(a) d =1

T

T0

nM

cn

exp(2i n a) d +

with some satisfying || < 1. Consequently,1

T

T0

1(a) d = c0

+

0=nMcn

1

T

T0

exp(2i n a) d + .

Since the an are linearly independent over Q, we have n a = 0 for n = 0. Itfollows for such n that T

0

exp(2i n a)) d 1.

Since > 0 is arbitrary, we obtain

limT

1

T

T0

1(a) d = c0

.

Thus, we get in (3.16)

c0

lim infT

1

Tmeas { (0, T) : a mod 1}

lim supT

1

Tmeas { (0, T) : a mod 1} c+

0+

for any positive . From (3.15) it follows that 0

c+0

c0

2. Now sending

0, the theorem is proved.

As an immediate consequence of Theorem 3.8 we get the classical inhomogeneousKronecker approximation theorem:

Corollary 3.9. Let1, . . . , N R be linearly independent overQ, let1, . . . , Nbe arbitrary real numbers, and let q be a positive number. Then there exists anumber > 0 and integers x1, . . . , xN such that

| n n xn| < 1q

for 1 n N.


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We conclude with the notion of uniform distribution modulo 1. Let () be acontinuous function with domain of definition [0, ) and range RN. Then thecurve () is said to be uniformly distributed mod 1 in RN if

limT

1

Tmeas { (0, T) : () mod 1} = N

j=1

(j 1)

for every parallelepiped

= [1, 1] . . . [N, N] with 0 j < j 1 for1 j N. In a sense, a curve is uniformly distributed mod 1 if the number ofvalues which lie in any given measurable subset of the unit cube is proportionalto the measure of the subset.

In questions about uniform distribution mod 1 one is interested in the fractionalpart only. Hence, we define for a curve () = (1(), . . . , N()) in R

N

{()} = (1() [1()], . . . , N() [N()]);recall that [x] denotes the integral part of x R.Theorem 3.10. Suppose that the curve () is uniformly distributed mod1 inRN. Let D be a closed and Jordan measurable subregion of the unit cube in RN

and let be a family of complex-valued continuous functions defined on D. If is uniformly bounded and equicontinuous, then

limT

1

T

T0

f({()})1D() d =

D

f(x) dx

uniformly with respect to f , where 1D() is equal to 1 if () D mod 1,and equal to zero otherwise.

Proof. By the definition of the Riemann integral as a limit of Riemann sums,we have for any Riemann integrable function F on the unit cube in RN

(3.17) limT

1

T

T

0

F(

{()

}) d =

[0,1]N

F(x) dx.

By the assumptions on , for any > 0 there exist f1, . . . , f n such that forevery f there is an fj among them satisfying

supxD

|f(x) fj (x)| < .

By (3.17) there exists T0 such that for any T > T0 and for each function f1, . . . , f none has

D fj(x) dx 1

T T

0

fj ({()})1D() d < .


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Now, for any f

,

D

f(x) dx 1T

T0

f({()})1D() d

D

fj (x) dx 1T

T0

fj ({()})1D() d +

D

(f(x) fj(x)) dx

+1

T

T

0

(fj ({()}) f({()}))1D() d

By the appropriate choice of fj it follows from the estimates above that this isbounded by 3. Since > 0 is arbitrary, the assertion of the theorem follows. Besides, there is also the notion of uniformly distributed sequences, defined in asimilar way. It was proved by Hlawka [28] that the imaginary parts of the zerosof the zeta-function are uniformly distributed modulo 1.

3.5. Approximation in the mean end of proof. We choose > 1 and1 (0, 1) such that r < 14 and

max|s|r

f s

f(s)

< 1.Let Q = {p z} and E= {s : r < 2, 1 t T}. We shall estimate

(3.18) I:=

2T

T

E

1Q s + 34 + i, 0 s + 34 + i 12 d dt d,where 0 = (0, 0, . . .). By Theorem 2.10,

(s + i) =nT

1

ns+i+ O(T).

This gives

I =

E+ 34

2TT

|1Q (s + i, 0)(s + i) 1|2 dd dt

E+ 34

2TT

1Q (s + i, 0)nT

1

ns+i 1

2

dd dt

+

E+ 3

4

2TT

T|1Q (s + i, 0)|2 dd dt,(3.19)

where E+ 34

is the set of all s with s 34

E. By definition,

1Q (s, 0) =

pQ

1 1

ps

=

m=1

p|mpQ

(m)

ms;


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recall that (m) is Mobius -function defined by (2.28). We may bound thesecond term appearing on the right-hand side of (3.19) by

T2(34r) max

sE+ 34

2TT

|1Q (s + i, 0)|2 d T2r12

1Q 34 r, 02 .Furthermore, for T > z a simple computation gives

1Q (s, 0)nT

1

ns= 1 +

z 0.Hence, for T > z

2TT

1Q (s + i, 0)nT

1

ns+i 1

2

d =

2TT

z


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provided that z and T are sufficiently large, say z > z0 and T > T0, dependingonly on 2. Define

AT =

[T, 2T] :

E+ 34

|1Q (s + i, 0)(s + i) 1|2 d dt < 22

.

Then it follows from (3.19) and (3.21) that for sufficiently large z and T

(3.22) meas AT > (1 2)T,which is surprisingly large. Application of Lemma 3.7 gives for AT

max|s|r

|1Q (s + i, 0)(s + i) 1| < C2,

where C is a positive constant, depending only on . For sufficiently small 2 wededuce

max|s|r

log s + 34

+ i log Q s + 34 + i, 0 < 2C2,(3.23)

provided AT.By Theorem 3.6 there exists a sequence of finite sets of prime numbers M1 M2 . . . such that k=1Mk contains all primes and

limk

max|s|r

log Mk

s + 3

4, 0

f

s

= 0.(3.24)

Let = (2, 3, . . .) and = (2, 3, . . .). By the continuity of log M

s + 34 ,

with respect to , for any 1 > 0 there exists a positive with the property thatwhenever

(3.25) p p < for all p Mk,then

max|s|r

log Mk s + 34 , 0 log Mk s + 34 , < .(3.26)Setting

BT = [T, 2T] :

logp

2 p < ,

we get

1

T

B

|s|r

log Q s + 34 + i, 0 log Mk s + 34 + i, 02 d dt d=

|s|r

1

T

BT

log Q s + 34 + i, 0 log Mk s + 34 + i, 02 dd dt.Putting () =

log2

2, log3

2, . . .

, we may rewrite the inner integral as

BT log Q s +

34

, () log Mk s +34

, ()2

d.


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The logarithms of the prime numbers are linearly independent over Q (this fol-lows easily from the unique prime factorization of the integers). Thus, by Weylstheorem 3.8, the curve () =

log2

2, log3

2, . . . , logpN

2

is uniformly distributed

mod 1. Application of Theorem 3.10 yields

limT

1

T

BT

log Q s + 34 , () log Mk s + 34 , ()2 d=

D

log Q s + 34 , log Mk s + 34 , 2 duniformly in s for |s| r, where D is the subregion of the unit cube in RN givenby the inequalities (3.25) and d is the Lebesgue measure. By the definition of

M(s, ) it follows that for Mk QQ(s, ) = Mk (s, )Q\Mk (s, ),

and thusD

log Q s + 34 , log Mk s + 34 , 2 d=

D

log Q\Mk s + 34 , 2 d = meas D

[0,1]N

log Q\Mk s + 34 , 2 d.Since

log Q\Mk

s + 34 ,

=

pQ\Mk

n=1

exp(2inp)npn(s+

34 )

,

we obtain [0,1]N

log Q\Mk s + 34 , 2 d = pQ\Mk

n=1

1

n2p2n+3n

2

.

If Mk contains all primes yk, then

pQ\Mk

n=11

n2p2n+3n2

y2r12

k .

Hence, we finally get

1

T

BT

|s|r

log Q s + 34 + i, 0 log Mk s + 34 + i, 02 d dt d y2r

12

k meas D.As already noticed above, the curve () is uniformly distributed mod 1. Hence,application of Theorem 3.8 shows that

limT

1

Tmeas BT = meas D,


63/90


which implies for sufficiently large yk

meas

BT :

|s|r

log Q s + 34 + i, 0 log Mk s + 34 + i, 02 d dt< y

r 14k

> 1

2meas D T.

Now application of Theorem 3.10 yields

meas

BT : max|s|r

log Q s + 34 + i, 0 log Mk s + 34 + i, 0< y

15 (r 14 )k >

12

meas D T.(3.27)

If we now take 0 < 2 0.

Thus, in view of

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