On the Universality of the Riemann zeta-function

8/13/2019 On the Universality of the Riemann zeta-function

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On the Universalityof the Riemann zeta-function

Jorn Steuding

Summer 2002Johann Wolfgang Goethe-Universitat Frankfurt

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This course gives an introduction to the value distribution theory of the Riemannzeta-function(s) with a special emphasis on its remarkable universality property. Ourmain goal is to derive Voronins celebrated universality theoremwhich states, roughly

speaking, that any(!) non-vanishing analytic function on a sufficiently small disc canbe uniformly approximated by (s). For that aim we go along Voronins original proof;the modern approach via limit theorems for weak convergent probability measures isbeyond the scope of this course. We state some consequences as there is the classicalresult due to H. Bohr that the set of values of (s) taken on vertical lines in the righthalf of the critical strip lies dense in the complex plane, and an answer to Hilbertsquestion whether the zeta-function satisfies an algebraic differential equation. Finally,we shall discuss some open questions in this field (the problem of effectivization andBagchis reformulation of Riemanns hypothesis) and the recent progress done towardstheir solution.

This course bases mainly on the nicely written monography [25] on the Riemannzeta-function. However, we also have to use some results from other fields different thananalytic number theory, namely approximation theory of numbers and of functions, thetheory of linear operators in Hilbert spaces and complex analysis, for which we refer to[14] and [55], respectively. Unfortunately, we cannot give a detailed proof of all factswhich are required to prove Voronins theorem (as for example the seperation theoremfrom functional analysis). However,the motivated reader can fill by the given referencesall these gaps.

I am very grateful to Rasa Slezeviciene, Ramunas Garunkstis and Antanas Lau-rincikas (Vilnius University) for many fruitful discussions, helpful remarks and unlim-ited interest.

Jorn Steuding, Frankfurt 10/20/2002.


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Contents

1 Introduction 1

2 The Riemann zeta-function 4

3 The approximate functional equation 7

4 Density theorems 14

5 A zero-free region 22

6 The prime number theorem 24

7 Diophantine approximation 29

8 Conditionally convergent series 34

9 Finite Euler products 38

10 Voronins universality theorem 46

11 Functional independence 51

12 Self-similarity and the Riemann hypothesis 54

13 Effective bounds 59

14 Other zeta-functions 62

Bibliography 64


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1 Introduction

In this introduction we give a brief historical overview on the phenomenon ofuniver-

sality; more details can be found in [21].The first in the mathematical literature appearing universal object was discov-ered by Fekete in 1914/15 (see [42]); he proved that there exists a real power series

n=1anxn with the property that for any continuous function on the interval [1, 1]

with f(0) = 0 there is a sequence of positive integers (mk) such that the partial sums1nmkanx

n converge to f(x) uniformly on [1, 1]. We shall prove the followingvariant due to Luh [38]:

Theorem 1.1 There exists a real power series

n=0

an

xn

with the property that for any interval [a, b] with0 [a, b] and any continuous functionf(x) on [a, b] there exists a sequence of positive integers (mk) such that

mkn=0

anxn f(x) uniformly on [a, b]

as k .

The proof relies mainly on Weierstrass approximation theorem, which states thateach

continuous function f(x) on a closed interval is the limit of a uniformly convergentsequence of polynomials; for a proof see [14].

Proof. Denote by (Qn) the sequence of polynomials with rational coefficients; this isobviously a countable set. We construct a sequence of polynomials (Pn) as follows: letP0 =Q0 , and assume that for n N the polynomials P0, . . . , P n1 are known. Let dnbe the degree of Pn1 . Further, let n(x) be a continuous function on [n, n] suchthat

n(x) = Qn(x) n1

=0

P(x)xdn1 for x

In:=

n,

1

n

1

n

, n .Then, by Weierstrass approximation theorem, there exists a polynomial Fn with

maxx[n,n]

|Fn(x) n(x)| < ndn2.

Setting Pn(x) = Fn(x)xdn+1 , the sequence (Pn) is constructed, and Pn satisfies

maxxIn

nk=0

Pk(x) Qn(x)

= maxxIn

|(Fn(x) n(x))xdn+1| < 1n

.

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Obviously, distinct Pn have no powers in common. Thus we can rearrange formallythe polynomial series into a power series:

n=0

anxn :=

n=0

Pn(x)

(note that this would be impossible if infinitely many Pn would have a term xn incommon).

Again by Weiertstrass approximation theorem, to any continuous function f(x)on [a, b] there exists a sequence of positive integers (nk), tending to infinity, such that

maxx[1,1]

|f(x) Qnk (x)| 1. (1.1)

Since|ns| = n , it follows easily (by the integral test) that the series converges ab-solutely in the half-plane >1. Series of the type

n

a(n)ns are called Dirichlet series;

see [54], Section IX for details. However, the true analytic character of the Riemannzeta-function becomes only visible by continuing (s) to the left of = 1. Later we

will prove the identity

(s) = s

s 1+ s1

[x] xxs+1

dx for >0; (1.2)

here [x] := max{z Z : z x} is the integer part function. Since the appearingintegral converges, we have found an analytic continuation for (s) to the half-plane > 0 except for a simple pole at s = 1 with residue 1 . As we shall see later on,the value distribution of the Riemann zeta-function in the so-called critical strip0 1 is of special interest out of different points of view.

Now, since we have an expression for (s) in the critical strip, we are in the position

to formulate Voronins result [57].

Theorem 1.2 Let 0< r < 14

and let f(s) be a non-vanishing continuous function onthe disc|s| r , which is analytic in the interior. Then, for any >0 there exists a >0 such that

max|s|r

3

4+ s + i

f(s)

< .Thus, (s) approximates any such function with any precision somewhere in 12 < 1. Since, for 0> 1 ,

n=1

1

ns

n=1

1

n0 1 +

n=2

nn1

du

u0(2.1)

= 1 + 1

u0 du= 1 + 1

0 1 ,

the series in question converges uniformly in any half-plane > 0 with 0 > 1 .A well-known theorem of Weierstrass states that the limit of a uniformly convergent

sequence of holomorphic functions is holomorphic; see [54], 2.8. ThusTheorem 2.1 (s) is analytic for >1 and satisfies in this half-plane the identity

(s) =

n=1

1

ns =

p

1 1

ps

1. (2.2)

Here and in the sequel p denotes always a prime number. The product in (2.2) istaken over all primes; it is called Euler product since it was discovered by Euler in1737 . As we shall see in the proof below it can be regarded as the analytic version oftheunique prime factorization of the integers.

Proof. It remains to show the identity between the series and the product in (2.2).In view of the geometric series expansion and the unique prime factorization of theintegers,

px

1 1

ps

1=

px

1 +

1

ps+

1

p2s+ . . .

=

n

p|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

1

tends to zero as x , we get identity (2.2) by sending x . The theorem isproved.

We shall see later on that the Euler product (2.2) is the key to prove Voroninsuniversality theorem (in spite of the fact that the product does not converge in theregion of universality). However, the representation (2.2) gives also a first glance onthe close connection between (s) and the distribution of prime numbers. A first im-mediate consequence is Eulers proof of the infinitude of the prime numbers: assuming

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that there are only finitely many primes, the product in (2.2) is finite, and thereforeconvergent throughout the whole complex plane, contradicting the fact that the (s)defining Dirichlet series reduces to the divergent harmonic series as s 1+ . Hence,there are infinitely many prime numbers. This fact is well known since Euclids elemen-tary proof, but the analytic access gives much deeper knowledge on the distribution ofprime numbers.

It was the young Gauss who conjectured in 1792 for the number of primes undera given magnitide:

(x) :={p x} Li(x), (2.3)where the logarithmic integralis defined by

Li(x) = x

2

du

log u

;

here and in the sequel we write f(x) =O(g(x)), resp. f(x)g(x), with a positivefunction g(x) if

limsupx

|f(x)|g(x)

is bounded, and we write f(x) g(x) if the latter quantity is actually a limit andequals one. By partial integration it is easily seen that

Li(x) = x2

du

log u=

n

k=1x(k 1)!(log x)k

+ O

x

(log x)n+1

.

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

log x. Cebyshev proved in 1852 by elementary methods that for

sufficiently large x

0.921 . . . (x) log xx

1.055 . . . ,and additionally that if

limx

(x) log xx

exists, then the limit has to be equal to one, which supports (2.3).

However, as Riemann showed Eulers analytic access, i.e. the link between thezeta-function and the primes by (2.2), encodes much more arithmetic information thanChebyshevs elemenatry approach. In his only but outstanding paper [49] on numbertheory Riemann noticed the importance of studying (s) as a function of a complexvariable (Euler dealt only with real s ). Riemann proved:

the function(s) 1

s 1is entire; consequently, (s) has an anlytic continuation throughout the wholecomplex plane except for s= 1 where (s) has a simple pole with residue 1 .

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(s) satisfies the functional equation

s2

s

2 (s) =

1s2

1 s

2 (1

s).

Note that the Gamma-function, defined by

(z) =0

uz1 exp(u) du for Rez >0,

plays an important role in the theory of the zeta-function; see [54],1.86 and4.41,for a collection of its most important properties.

In view of the Euler product (2.2) it is easily seen that (s) has no zeros in thehalf-plane > 1 . Using the functional equation, it turns out that (s) vanishes in 12

; (2.6)

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this is the famous, yet unproved Riemann hypothesis. Riemann worked with(12 +it) and wrote ...und es ist sehr wahrscheinlich, dass alle Wurzeln reell sind.Hiervon ware allerdings ein strenger Beweis zu wunschen; ich habe indess die Auf-

suchung desselben nach einigen fluchtigen vergeblichen Versuchen vorlaufig bei Seitegelassen.... Note that Riemann also calculated thefirstzeros: for example, the firstis = 1

2+ i 14.134 . . . . Further, Riemann conjectured

there exist some constants A, B such that1

2s(s 1)s2

s

2

(s) = exp(A + Bs)

1 s

exp

s

;

theexplicit formula: for any x 2

(x) +

n=2

(x1/n)n

= Li(x) =+i

>0

Li(x) + Li(x1)

(2.7)

+

x

du

u(u2 1)log u log 2;

this was proved in 1895 by von Mangoldt whereas the last but one conjecture wasproved by Hadamard. The explicit formula follows from both product representationsof (s), the Euler product on one side and the Hadamard product on the other side.

Riemanns ideas led to the first proof of Gauss conjecture (2.3), the celebrated

prime number theorem, by Hadamard and de La Vallee Poussin (independendly) in1896.

3 The approximate functional equation

In this section we shall derive not only an analytic continuation for (s) to the half-plane > 0 but also a quite good approximation which will be very useful later on.Unfortunately, the proof is rather technical; we follow [55], IV.

What happens on the abscissa of convergence = 1? At s= 1 the zeta-functiondefining series reduces to the harmonic series. To obtain an analytic continuation for(s) we have to seperate this singularity. For that purpose we needAbels partialsummation:

Lemma 3.1 Let 1 < 2 < . . . be a divergent sequence of real numbers, define forn C the function A(u) =nu n , and let F : [1, ) C be a continuousdifferentiable function. Then

nx

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

1A(u)F(u) du.

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For those who are familiar with the Riemann-Stieltjes integral there is nearly nothingto show. Anyway,

Proof. We haveA(x)F(x)

nx

nF(n) =

nx

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

nx

xn

nF(u) du.

Since 1 n u x , interchanging integration and summation yields the assertion.

Now we apply partial summation to finite pieces of the series (1.1). Let N < M bepositive integers and >1. Then, application of Lemma 3.1 with F(u) =us, n= 1and n= n yields

N 0 ,

(s) = nN

1

ns+

N1s

s 1+ s

N

[u]

u

us+1 du.

Thus, (s) has an analytic continuation to the half-plane >0 except for one simplepole at s= 1 with residue 1 .

Putting N = 1 in the formula of Theorem 3.2, we obtain the analytic continuation(1.2) for (s) from the introduction. But this integral representation gives also a veryuseful approximation of (s) in the critical strip. By the periodicity of the function[u] u we expect that the contribution of the integral is small. In order to give arigorous proof of this idea we have to do some preliminary observations.

Let f(u) be any function with continuous derivative on the interval [a, b]. Using

the lemma on partial summation with n = 1 if n (a, b], and n= 0 otherwise, weget

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Lemma 3.3 (Eulers summation formula) Assume that f : [a, b] R has a con-tinuous derivative. Then

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This implies immediately both formulas of the lemma. It remains to show that thepartial sums of the Fourier series are uniformly bounded in u and M. Substitutingy= (2M+ 1)x in (3.2), we get 1

2

u

sin((2M+ 1)x)sin(x)

dx = 1

2

u

sin((2M+ 1)x)x

dx

+ 1

2

usin((2M+ 1)x)

1

sin(x) 1

x

dx

0

sin(y)

y dy+

12

0

1sin(x) 1

x

dxwith an implicit constant not depending on and M; obviously both integrals exist,which gives the uniform boundedness.

Further, we will make use of the following estimate of exponential integrals.Lemma 3.5 Assume that F : [a, b] R has a continuous non-vanishing derivativeand that G : [a, b] R is continuous. If G

F is monotonic on [a, b] , then

ba

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

4 GF (a)

+ 4 GF (b)

.Proof. First, we assume that F(u) > 0 for a u b . Since (F1(v)) =F(F1(v))1 , substituting u= F1(v) leads to

ba

G(u)exp(iF(u)) du= F(b)F(a)

G(F1(v))

F(F1(v))exp(iv) dv.

Application of the mean-value theorem gives, in case of a monotonically increasing GF,

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 (a, b) . This gives the desired estimate. The same idea applies to theimaginary part. Furthermore, the case F(u) < 0 can be treated analogously. Thelemma is proved.

Now we are in the position to prove van der Corputs summation formula.

Theorem 3.6 To any given > 0 there exists a positive constant C = C() onlydepending on with the following property: assume that f : [a, b] R is a functionwith continuous derivative, g : [a, b] [0, ) is differentiable function, and that f, gand|g| are all monotically decreasing. Then

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Van der Corputs summation formula looks rather technical but the idea is simple aswe will shortly explain. The integral

ba g(u) exp(2i(f(u) mu)) du

is (up to a constant factor) the Fourier transform of g(u) exp(2if(u)) at u = m(therefore, we may interpret Theorem 3.6 as an approximate version of Poissons sum-mation formula).

Proof of Theorem 3.6. Using Eulers summation formula with F(u) =g(u) exp(2if(u)) and the Fourier series expansion of Lemma 3.4, we get

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=

f(a)f(a)+

m=0

I1(m)m g(a)

0|f(a)|

|f

(a)|m2

.

The contribution arising from m < f(b) can be treated similarly. This gives

m[f(b),f(a)+]m=0

I1(m)m g(a) log(|f(a)| + |f(b)| + 2).

Now assume m > f(a) + and m = 0. Then, by the mean-value theorem,

ReI2(m) = ba

|g(u)| cos2(f(u) mu) du= g (a) a

cos2(f(u) mu) du

with some (a, b). Partial integration yields

acos2(f(u) mu) du =

Reexp(2i(f(u) mu)

2im

u=a

+Re 1

m

a

f(u) exp(2i(f(u) mu)) du

1

|m

|

1 +

|f(a)|

|f(a)

m

|

.

Therefore, m>f(a)+

ReI2(m)m g(a).

With slight modifiactions this method applies also to the cases ImI2(m) and mf(b) . Further, if 0 [f(b) , f(a) + ], then Lemma 3.5 gives

ba

g(u)exp(2if(u)) du g(a).

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In view of (3.3) the theorem follows from the above estimates under the conditionf(b) > 0 . If this condition is not fulfilled, then argue with f(u) ku , where k :=1 [f(b)], instead of f(u) .

Now we apply van der Corputs summation formula to the zeta-function. Let >0 .By Theorem 3.2 we have

(s) =nx

1

ns+

x 0 . Also here much better estimates known. For example, using thefunctional equation and the Phragmen-Lindelof principle (that is a kind of maximumprinciple for unbounded domains), one can obtain 1

4+ instead of 1

2 for any positive

; Huxley [23] holds the record with the exponent 89570

+ . The yet unprovedLindelofhypothesisstates

1

2+ it

t as t ; (3.5)

note that the truth of the Riemann hypothesis would imply this estimate but not viceversa.

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4 Density theorems

By use of the approximate functional equation, we shall first derive a mean-square

formula for (s) in the half-plane > 1

2. Such mean-square formulae are an importanttool in the theory of the Riemann zeta-function; in particular, they give informationon the number of zeros as we shall see below. We follow [55],VII andIX.

Theorem 4.1 For > 12,

T1

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

Proof. By the approximate functional equation,

(+ it) =n


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Collecting together, the assertion of the theorem follows.

Obviously, with regard to the simple pole of the zeta-function, the mean-square

formula above cannot hold on the critical line: (2) is unbounded as 1

2+ .We can derive from the above theorem some information on the zero distributionof (s) . This observation dates back to H. Bohr and Landau [11], resp. Littlewood[37]. For that purpose we have to use some facts from classical function theory.

Thetheorem of residuesstates: letC be a closed path in the complex plane withoutdouble points, and let f(s) be meromorphic with poles at s1, . . . , sm insideC , then

1

2i

C

f(s) ds=m

j=1

Ress=sj f(s), (4.1)

where

Ress=sj f(s) := 1

2i|ssj |=

f(s) ds

is the residue of f(s) at s = sj (which coincides with the coefficient a1 in theLaurent expansion of f(s) at sj, i.e. f(s) =

m=am(s sj)m ). The special case

of a logarithmic derivative gives the possibility to count zeros and poles. If f(s) andC satisfies the conditions above, and if N(0) denotes the number of zeros, N() thenumber of poles of f(s) (according multiplicities) insideC , then

1

2i

C

f

f(s) ds= N(0) N(). (4.2)

In particular, we obtain the so-called argument principle, which states that if f(s) isanalytic, then the change in the argument of arg f(s) as s varies onC (in positivedirection) equals the number of zeros of f(s) insideC .

All these statements are well-known facts in complex analysis; see [54], Section III.Further, we need Littlewoods lemma, i.e. the following integrated version of (4.2):

Lemma 4.2 Let f(s) be regular in and upon the boundary of the rectangleR withvertices b, b+ iT,a +iT, a , and not zero on = b . Denote by n(, T) the number ofzeros = + i of f(s) inside the rectangle with > including those with =Tbut not = 0 . Then

R

log f(s) ds= 2i ab

n(, T) d.

Since the complex logarithm is a multi-valued function, we have to be careful! Obvi-ously, f(s) is non-vanishing in the neighbourhood of=b , and thus we define log f(s)here starting with any one value of the logarithm, and for other points s of the rec-tangle by analytic continuation along the polygon with corners b+ it,s = +it ,provided that the path does not cross a zero or pole of f(s); if it does, putlog f(s) = lim0+log f(+ it + i) .

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Proof. Without loss of generality we may assume that the lines t= 0 and t= T arefree off zeros and poles of f(s). Obviously,

R log f(s) ds =

ba log f() d

b

a log f(+ iT) d

+ T0

(log f(b+ it) + log f(a+ it)) i dt. (4.3)

The last integral equals T0

i b

a

f

f(+ it) d dt=

ba

+iT

f

f(s) ds d.

By (4.2),

+iT

f

f

(s) ds= b

+ b+iT

b b+iT

+iT

f

f

(s) ds

2in(, T).

Substituting this in formula (4.3) proves the lemma. Note that Littlewoods lemma can be used, in addition with Stirlings formulaand

some facts about entire functions, to prove the Riemann-von Mangoldt formula (2.5)(see [5]).

We finish our short excursion to function theory and continue with our investigationson the zeros of the Riemann zeta-function. Let N(, T) denote the number of zeros= + i of (s) with > , 0< T(counting multiplicities). Then, applicationof Littlewoods lemma with fixed b= 0>

12

yields

2 1

0N(, T) d =

T

0log |(0+ it)| dt

T

0log |(2 + it)| dt (4.4)

+ 02

arg (+ iT) d 02

arg ()) d.

The main contribution comes from the first integral on the right hand side. The lastintegral does not depend on Tand so it is bounded. Since (s) has an Euler productrepresentation (2.2), the logarithm has a Dirichlet series representation:

log (s) = p

log

1 1

ps

=

p,k

1

kpks for >1, (4.5)

where k runs through the positive integers; here we choose that branch of the logarithmwhich is real on the positive real axis. Hence we obtain

T0

log |(2 + it)| dt= Rep,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 zero. Since arg (2) = 0 and

arg (s) = arctan

Im(s)

Re(s)

,

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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 as 12 2. Devide the interval

[12+ iT, 2 + iT] into q+ 1 parts, throughout each of which Re(s) is of constant sign.

Hence, again by the argument principle, in each part the variation of arg (s) does notexceed . This gives

|arg (s)

| q+

3

2 for

1

2

.

Further, qis the number of zeros of the function

g(z) =1

2((z+ iT) + (z iT))

for Imz= 0 and 12 Rez 2. Thus, q n(3

2), where n(r) is the number of zeros

of (s) for|z 2| r . Obviously, 20

n(r)

r dr

232

n(r)

r dr n

3

2

232

dr

r =n

3

2

log

4

3.

Jensens formulastates that if f(s) is an analytic function for |s| R with zeross1, . . . , sm (according their multiplicities) and f(0) = 0 , then

1

2

20

log |f(r exp(i))| d = log rm|f(0)|

|s1 . . . sm|for r < R (in a sense, this is nothing else than Poissons integral formula; see [54],3.61). This gives here

2

0

n(r)

r dr=

1

2 2

0

log|(2 +r exp(i))

|d

log

|(2)

|.

In view of (3.4) we obtain

q n

3

2

1

log 43

20

n(r)

r dr log T.

This yields

arg (+ iT) log T uniformly for 12

,

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and, consequently, the same bound holds by integration with respect to 12 2 .

The restriction that Thas not to be an imaginary part of a zero of(s) can be removedfrom considerations of continuity. Therefore, we may replace (4.4) by

10

N(, T) d= 1

2

T0

log |(0+ it)| dt+ O(log T). (4.6)

Now we need a further analytic fact due to Jensen: Jensens inequalitystates that forany continuous function f(u) on [a, b] ,

1

b a b

alog f(u) du log

1

b a b

af(u) du

(for instance, this can be deduced from the arithmetic-geometric mean inequality, orsee [54],

9.623).

Hence, we obtain for any fixed 0> 12

T0

log |(+ it)| dt T2

log

1

T

T0

|(+ it)|2 dt

T

by applying Theorem 4.1. Thus,

10

N(, T) d T.

Let 1 = 12

+ 12

(0

12

), then we get

N(0, T) 10 1

01

N(, T) d 20 12

11

N(, T) T.

With view to (4.6) we have proved

Theorem 4.3 For any fixed > 12,

N(, T) T.

The theorem above is a first density theorem. In view of the Riemann-von Mangoldt

formula (2.5) we see that, supporting Riemanns hypothesis, all but an infinitesimalproportion of the zeros of (s) lie in the strip 1

2 < < 1

2+ , however small may

be!However, for later applications we need a stronger result.

Theorem 4.4 For any fixed in 12 <


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Proof. For 2 V T let N1(, V) count the zeros = + i of (s) with and 12V < V . Taking x= V in Theorem 3.7

(s) =kV

1

ks +V1s

s 1+ O

V

for 12

V < t V and 12 1. Multiplying this with the Dirichlet polynomial

MX(s) :=

mX

(m)

ms ,

where X=V21 , gives(s)MX(s) = P(s) + R(s),

where

P(s) :=

mX

(m)

ms

kV

1

ks =

nXV

a(n)

ns

with

a(n) :=m|n

mX,nmV

(m) =

1 if m= 1,0 if 1< n X, (4.7)

andR(s) |MX(s)|V.

Note that MX(s), as the truncated Dirichlet series of the reciprocal of (s) ,mollifies1

(s) . We shall use P(s) as a zero-detector. Let s = be a zero of the zeta-function

with 12

V < V . Then,

1

X


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In view of the Riemann-von Mangoldt formula (2.5) there are only log V manyzeros with 2m+n 1< 2m+n . Now let denote the largest of the relatedsums according to 2m+ n 1< 2m+ n . Then

1

12

V


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Summation over r and application of Cauchy-Schwarz yields

S(Y) Y2(I1+

I1I2),

where

I1 := V

12

V

Y


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5 A zero-free region

In order to prove Gauss conjecture we need further knowledge on the zero distribution

of (s). We shall establish a zero-free region for (s) which covers the abscissa of ab-solute convergence = 1 . In this delicate problem we follow (with slight modifications)the ideas of de La Vallee Poussin; see [55], Section 3.

In the sequel we only argue for s= + it from the upper half-plane; with regardto (2.4) all estimates below can be reflected with respect to the real axis.

Lemma 5.1 We have, for t 8, 1 12(log t)1 2 ,(s) log t and (s) (log t)2.

Proof. Let 1 (log t)1 3. If n t , then

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

1 1log t

log n

n.

Thus, the approximate functional equation (Theorem 3.7) in addition with (3.1) implies

(s) nt

1

n+ t1 log t.

The estimate for (s) follows immediately from Cauchys formula(which is actuallya consequence of the theorem of residues (4.1))

(s) = 12i

(z)(z s)2dz,

and standard estimates of integrals, or alternatively, by differentiation of the formulaof Theorem 3.2.

In view of the Euler product (2.2) we have for >1

|(+ it)| = exp(Re log (s)) = exp

p,k

cos(kt logp)

kpk

.

Since17 + 24cos + 8 cos(2) = (3 + 4 cos )2 0,

it follows that

()17|(+ it)|24|(+ 2it)|8 1. (5.1)This inequality is the main idea for our following observations. By the approximatefunctional equation, Theorem 3.7, we have for small >1

() 1

1

.

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Assuming that (1 + it) has a zero for t= t0= 0, it would follow that|(+ it0)| 1,

leading to lim1+

()17|(+ it0)|24 = 0,which contradicts (5.1). Thus

(1 + it) = 0.It can be shown that the non-vanishing of (1 + it) is equivalent to Gauss conjecture(2.3), i.e. the prime number theorem without error term; see [55],3.7. But weare interested in a prime number theorem with error term. A simple refinement ofthe argument allows a lower estimate for the modulus of (1 +it): for t 1 and1< 0 such that

A:=c2B

1724

c1B >0 and putting = 1 + B(log t)8

, we obtain

|(1 + it)| A(log t)6

. (5.3)

This gives an estimate on the left of the line = 1.

Lemma 5.2 There exists a positive constant such that

(s) = 0 for 1 min{1, (log t)8}.Proof. In view of Lemma 5.1 the estimate (5.2) holds for 1 (log t)8 1 .Using (5.3), it follows that

|(+ it)| A

c1

(log t)6,where the right hand side is positive for sufficiently small . This yields the zero-freeregion of Lemma 5.2.

The largest known zero-free region for the zeta-function was found by Vinogradov[56] and Korobov [26]. Using Vinogradovs ingenious method for exponential sums,they proved

(s) = 0 in 1 c(log |t|) 13 (log log |t|) 23 (5.4)

for some positive constant c and sufficiently large

|t

|; see [25],

IV.3.

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6 The prime number theorem

The aim of this section is to prove Gauss conjecture (2.3), the celebrated prime number

theorem.Out of technical reasons we work with the logarithmic derivative of (s) (insteadof log (s) as Riemann did). Logarithmic differentiation of the Euler product (2.2),resp. differentiation of (4.5), gives for >1

(s) =

n=1

(n)

ns , where (n) :=

logp ifn = pk,

0 otherwise,

is the von Mangoldt -function. Since (s) does not vanish in the half-plane > 1, the logarithmic derivative is analytic for > 1 . As we shall see below allinformation on (x) is encoded in

(x) :=nx

(n) =

px

logp + O

x 12

. (6.1)

The idea of proof is ingenious but simple. Partial summation gives

(s) =s

1

(x) dx

xs+1.

If we can now transform this into a formula in which (x) is isolated, we may hope tofind an asymptotic formula for (x) by contour integration methods. The first step ofthis program can be done by a type of Fourier transformation.

Lemma 6.1 Let c and y be positive and real. Then

1

2i

c+ici

ys

s ds=

0 if 0< y 1.

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

1

2

dt

c + it=

1

limT

T0

c

c2 + t2dt=

1

limT

arctanT

c =

1

2,

by well-known properties of the arctan-function. Now assume that 0 < y < 1 and

r > c . Since the integrand is analytic in >0 ,Cauchys theorem(resp. the theoremof residues (4.1)) implies, for T >0 ,

c+iTciT

ys

s ds=

riTciT

+ r+iT

riT+ c+iT

r+iT

ys

s ds.

It is easily seen that ciTriT

ys

s ds 1

T

cr

y d yc

T| log y| ,

r+iT

riT

ys

s ds y

r

r + yr

T

1

dt

t yr

1

r+ log T .

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

Res s=0ys

s = lim

s0

ys

s s= 1

gives the values 2i for the integral in this case.

We apply this to the logarithmic derivative of the zeta-function and obtain for x Zand c >1 c+i

ci

n=1

(n)

nsxs

s ds=

n=1

(n) c+i

ci

x

n

s dss

;

here interchanging integration and summation is allowed by the absolute convergence

of the series. In view of Lemma 6.1 it follows thatnx

(n) = 1

2i

c+ici

n=1

(n)

nsxs

s ds,

resp.

(x) = 1

2i

c+ici

(s)

xs

s ds;

this isPerrons formula. Since

ciciT

ys

s ds=

ys

s log y

ci

s=ciT +

1

log y ci

ciT

ys

s2 ds yc

T| log y|for 0< y= 1 and T >0, it follows that

ciciT

n=2

(n)

ns

xs

s ds x

c

T

n=2

(n)

nclog x

n

xc

T

(c)

+x(log x)2

T + log x.

This yields

(x) = 12i

c+iT

ciT

(s)

xs

s ds + O

xc

T

(c)

+

x(log x)2

T + log x

, (6.2)

which holds for arbitrary x . To find an asymptotic formula for the integral above wemove the path of integration to the left, excluding s= 0. By the theorem of residueswe expect contributions to the main term from the poles of the integrand, i.e.

the nontrivial zeros of (s) , the pole of (s) at s= 1 .

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However, for our purpose it is sufficient to exclude the zeros of the zeta-function. Inview of the zero-free region of Lemma 5.2 we put c= 1 + with = (log t)8 , whereis given by Lemma 5.2, and integrate over the boundary of the rectangleR given bythe corners 1 iT. By this choice (s) does not vanish in and on the boundaryofR . The theorem of residues (4.1) implies

c+iTciT

(s)

xs

s ds =

1iT1+iT

+ 1+iT1iT

+ 1+iT1+iT

(s)

xs

s ds

+2iRes s=1

(s)

xs

s.

For the logarithmic derivative of (s) we have

(s) = d

dslog (s) =

1

s 1+ O(1)as s 1. Thus, we obtain for the residue

Res s=1

(s)

xs

s = lim

s1(s 1)

1

s 1+ O(1)

xs

s =x.

It remains to bound the integrals. For the horizontal integrals we find with regard toLemma 5.2 1+iT

1iT

(s)

xs

s ds x

1+

T .

Further, for the vertical integral,

1++iT1iT

(s)

xs

s ds x1(log T)9.

Collecting together, we dedcue from (6.2)

(x) = x+ O

x1+

T + x1(log T)9 +

x(log x)2

T + log x

.

Choosing T= exp(

110

(log x)

19

), we arrive at

(x) = x+ O

x exp(c(log x) 19 )

.

Setting(x) :=

px

logp,

it follows from (6.1) that also

(x) =x+ O

x exp(c(log x) 19 )

.

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Applying now partial summation, we find

(x) = px

logp

1

logp =

(x)

log x

x

2(u)

d

du

1

log udu

= x

log x

x2

u d

du

1

log udu+ O

x exp

c(log x) 19

.

Partial integration leads to prime number theorem:

Theorem 6.2 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 of the

infinitude of primes (Section 2) but gives also the main term of the asymptotic formulain the prime number theorem.We see that the prime numbers, which - on a first look - seem to be randomly

distributed among the positive integers, satisfy a strong distribution law! For example,the prime number theorem implies that, if pn is the n -th prime number, then

pn n log n.

In view of the largest known zero-free region (5.4) one can obtain

(x) = Li(x) + O

x expc

(log x)35

(log log x) 15

.

With a little bit more effort and a little bit more facts from the theory of functionsit is possible to prove the analogue of Riemanns explicit formula (2.7). Integratingover the full complex plane one can show for x =pk the exact(!) explicit formula

(x) = x

x

1

2log

1 1

x2

log(2),

resp. its truncated version

(x) = x ||T

x

+ O

xT

(log(xT))2

. (6.3)

This shows a deep relation between the error term in the prime number theorem andthe distribution of the nontrivial zeros of the zeta-function.

Theorem 6.3 For fixed [12 , 1) ,(x) x x+ (s) = 0 for > .

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With regard to known zeros of (s) on the critical line it turns out that an error termwith < 12 is impossible; see [24] for more details.

Unfortunately, for the proof of one implication of Theorem 6.3 we have to use some

facts we have not proved; the interested reader may find all missing details in [25].

Proof. By partial summation we obtain for >1

(s) =

s

s 1+ s1

(u) uus+1

du.

If (x) xx+ , then the integral above converges for > , giving an analyticcontinuation for

(s) 1

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

Conversely, if all nontrivial zeros = +i satisfy , then it follows from(6.3) that

(x) x x||T

1

|| +x

T(log(xT))2. (6.4)

In view of the Riemann-von Mangoldt-Formel (2.5) 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 (6.4) leads to

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

(log(xT))2.

Now the choice T =x1 finishes the proof of this implication.

Theorem 6.3 shows that the Riemanns hypothesis (2.6) is true if and only if

(x) =x+ O

x12+

,

and since the latter estimate is best possible (there are zeros on the critical line),Riemanns hypothesis states that the prime numbers are as uniformly distributed aspossible!

In order to prove Voronins universality theorem we have to do some preliminaries.

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7 Diophantine approximation

In the theory of diophantine approximations one investigates how good an irrational

number can be approximated by rational numbers; this has a plenty of applications invarious fields of mathematics and natural sciences. We follow [25], A.8.For abbreviation we denote vectors of RN by x = (x1, . . . , xN) RN , we write

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 have to introduce the notion of the Jordan volume of a region RN .Therefore, we consider the sets of parallelepipeds 1 and 2 with sides parallel to theaxes and of volume 1 and 2 with 2 2 ; if there are 1 and 2 such thatlimsup11 coincides with lim inf22 , then has the Jordan volume

= lim sup1

1 = limsup2

2.

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

Weyl [59] proved

Theorem 7.1 Let a1, . . . , aN R be linearly independent over the field of rationalnumbers, and let be a subregion of the N-dimensional unit cube with Jordan volume . Let a= (a1, . . . , aN) , then

limT

1

T

meas

{

(0, T) : a

mod 1

}= .

Proof. From the definition of the Jordan measure it follows that for any >0 thereexist two finite sets of open parallelepipeds{j}and{+j}inside the unit cube suchthat

j

int +j

and meas

+

j

\ j

< ; (7.1)

here, as usual, Mdenotes the closure of the set M, and int M its interior. Denoteby the characteristic function of

j , i.e.

(x) =

1 if x j,0 if x j .

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

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

and [0,1]N

(+(x) (x)) dx< ,

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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| < 12 ,

where c is given by 12

12

(x) dx= 1.

Consequently, (x) is an infintely differentiable function, and hence the functions,defined by

(x) = N[0,1]N

(y)

x1 y1

xN yN

dy.

for 0< 0 there exists a finite setM ZN suchthat

(x) =

nMcn

exp(2in x) + R(x) with |R(x)| < .

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This yields

1

T

T

0

(a) d= 1

T

T

0nM

cn

exp(2in

x)) d+

with|| 0 is arbitrary, we obtain

limT

1

T

T0

(a) d=c0

.

This gives with regard to (7.3)

c0 lim inf

T

1

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

lim supT

1

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

0+

for any positive . From (7.2) it follows that 0 c+0 c0 2 . Now sending 0 ,the theorem is proved.

As an immediate consequence of Theorem 7.1 we get the classical inhomogeneous Kro-necker approximation theorem:

Corollary 7.2 Let 1, . . . , N R be linearly independent over the field of rationals,let 1, . . . , N be arbitrary real numbers, and let q be a positive number. Then thereexists a number >0 and integers x1, . . . , xN such that

| n n xn| 1 given by an absolute convergent Dirichlet series.However, the value distribution of the zeta-function in that region is anything butboring. Answering a question of Hilbert, H. Bohr and Landau [10], [13] showed that|(s)| takes arbitrarily large and arbitrarily small values in > 1 - in spite of theabsence of zeros!

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Theorem 7.3 For any > 0 and any real there exists an infinite sequence ofs= + it with 1+ and t such that

Re{exp(i)log (+ it)} (1 )log () + O(1).In particular,

liminf>1,t1

|(s)| = 0 and lim sup>1,t1

|(s)| = .

Our proof differs slightly from the original one.

Proof. By (4.5) one easily finds

log (s) =

p1

ps+ O(1) for >1. (7.4)

Thus, we have for any t 1 and x 2

Re {exp(i)log (+ it)} px

cos(t logp )p

p>x

1

p+ O(1). (7.5)

Here we use diophantine approximation. By the unique (!) prime factorization of theintegers the logarithms of the prime numbers are linearly independent. Denote by pnthe n -th prime, then Kroneckers approximation theorem implies that for any givenintegers q, N the existence of some real number >0 and integers x1, . . . , xN with

logpn2 2 xn < 1q for n= 1, . . . , N .Obviously, we get with q infinitely many with the above property. SettingN=(x), we obtain

cos(logp ) cos

2

q

for all p x,

provided that q 4. Therefore, we deduce from (7.5)

Re{exp(i)log (+ i)} cos2q

px

1p

p>x

1p

+ O(1),

resp.

Re {exp(i)log (+ i)} cos

2

q

log () 2

p>x

1

p+ O(1) (7.6)

in view of (7.4). Sending q, x we obtain the estimate of the theorem. In view ofthe simple pole of (s) at s= 1 we get with = 0, resp. = , by sending 1+the further assertions follow.

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It is even possible to give quantitative estimates for the rate of divergence; see [55],8.6, and [51].

We conclude with the notion of uniform distribution modulo 1. Let () be a

continuous function with domain of definition [0, ) and range RN

. Then the curve() is said to be uniformly distributed mod1 in RN if for every parallelepiped

= [1, 1] . . . [N, N] with 0 j < j 1 for 1 j N

limT

1

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

j=1

(j 1).

In a sense, a curve is uniformly distributed mod 1 if therightproportion of values liesin a given subset of the unit cube.

Since in questions about uniform distribution mod 1 one is interested in the frac-tional part only, we define for a curve () in RN

{()} = (1() [1()], . . . , N() [N()]);

recall that [x] is the integral part of x R .

Theorem 7.4 Suppose that the curve () is uniformly distributed mod1 in RN .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 uniformlybounded and equicontinuous, then

limT1

T T0 f({()})

D() d=

D f(x) dx

uniformly with respect to f , where D() is equal to 1 if () Dmod 1 , andzero otherwise.

Proof. By the definition of the Riemann-integral as a limit of Riemann sums, we havefor any Riemann integrable function Fon the unit cube in RN

limT

1

T

T0

F({()}) d=[0,1]N

F(x) dx. (7.7)

By the assumptions on , for every > 0 there exist f1, . . . , f n such that forevery f there is an fj with 1 j n, and

supxD

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

By (7.7) there exists T0 such that for any T > T0 and for each function f1, . . . , f n onehas

D

fj (x) dx 1T

T0

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

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Now, for any f ,

D f(x) dx 1

T

T

0f(

{()

})D() d

Dfj(x) dx 1

T

T0

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

D(f(x) fj(x)) dx

+

1

T

T0

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

By the estimates above, this is bounded by 3 . Since >0 is arbitrary, the assertionof the theorem follows.

In the next section we shall meet the heart of Voronins universality theorem.

8 Conditionally convergent series

A series

n an of real numbers an is called conditionally convergent if

n |an|is divergent but

n an is convergent for an appropiate rearrangement of the terms

an . Riemann proved that any conditionally convergent series can be rearranged suchthat its sum converges to an arbitrary preassigned real number; see [2]; hence, everyconvergent series, which does not converge absolutely, is conditionally convergent. Forinstance, to any given c R there exists a permutation of N (i.e. a one-to-onemapping on N ) such that

n=1

(1)(n) =c;here and in what follows do not confuse the permutation with the prime countingfunction. Thus, all conditionally convergent series are in a certain sense universalwithrespect to R !

It is the aim of this section to extended Riemanns rearrangement theorem to Hilbertspaces; recall that a complete normed linear space with inner product is calledHilbertspace. We shall give an example which will be of interest later on. Let R be a positivereal number, then theHardy spaceHR2 is the set of functionsf(s) which are analyticfor|s| < R and for which

f := limrR

|s|


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We return to our problem: suppose that Pm=H , then, by the above reasoning,there exists e H withe = 1 such that supxPmx, e < . Since, by assumption,the series

nmun, e converges conditionally with some arrangement of the terms,

the series of the positive terms of the series only is divergent. Thus, for any C thereexist an Nand a sequence m, . . . , N, equal to 0 or 1, such that

Nn=m

nun, e > C.

SinceN

m nun Pm , it follows that supxPmx, e = , giving the contradiction.So we have shown Pm = H . Consequently, there exist N m and m, . . . , N

[0, 1] such that

v N

n=m nun 0.

Hence, j+1k=1

vnk

2 j+1k=1

vnk2.

This yields the existence of a permutation such that (8.2) holds under the assumption

Nn=1vn= 0.

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For arbitrary v1, . . . , vN define

vN+1=

N

n=1

vn,

and apply the already proved fact; obviously this leads to the additional term

vN+12 = N

n=1

vn2,

multiplied by 2, in (8.2). The lemma is shown. Now we are in the position for the

Proof of Theorem 8.1. Without loss of generality we may assume, by Lemma 8.3,

that some subsequence of the partial sums of the series

kuk converges to v in thenorm ofH ; we define

Un =n

k=1

uk,

and suppose that the sequence of the Unj converges to v . For each j N there is apermutation of the set of vectors{Unj+1, . . . , U nj+1} in such a way that the value of

mj := max1mnj+1nj

nj+mn=nj+1

u(n)

is minimal. By Lemma 8.4 it follows that

mj

n=nj+1

un2

12

+ 2Unj+1 Unj ,

which obviously tends to zero as j . Hence, the corresponding series convergesto v in the norm ofH . Theorem 8.1 is proved.

In the following section we shall return to the zeta-function and start with the proofof Voronins universality theorem.

9 Finite Euler products

As we have seen in the beginning the Euler product (2.2) does not represent the zeta-function inside the critical strip. However, as Bohr [7] discovered in his investigations onthe value distribution of (s), an appropriate truncated Euler product (2.2) convergesalmost everywhere inside the critical strip to the zeta-function; see (10.5) below. Thisimportant observation can be used for our approximation problem: if we are able toapproximate a given function by a finite Euler product, then we finally have only toswitch from the finite Euler product to the zeta-function itself!

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Let denote the set of all sequences of real numbers indexed by the primes, thatare all infinite vectors of the form := (2, 3, . . .) with p R . Then we define forany finite subset Mof the set of all primes, any and all complex s

M(s, ) =

pM

1 exp(2ip)

ps

1

.

Obviously, M(s, ) is an analytic function in s without zeros in the half-plane >0 .Consequently, its logarithm exists and equals

log M(s, ) =

pM

log

1 exp(2ip)

ps

(out of technical reasons we prefer to consider series than products); here as for log (s)

we may take the principal branch of the logarithm.The first step in the proof of Voronins theorem is to show

Theorem 9.1 Let 0 < r < 14

and suppose that g(s) is continuous on |s| r andanalytic in the interior. Further, let 0 =

14

, 24

, 34

, . . .

. Then for any >0 and anyy > 0 there exists a finite set M of prime numbers, containing at least all primesp y , such that

max|s|r

log M

s +3

4, 0

g(s)

< .Unfortunately, the proof makes use of some classical results from analysis, which proofs

are beyond the scope of this course.Proof. Since g(s) is continuous for|s| r , there exists >1 such that 2r < 1

4 and

max|s|r

g

s

2

g(s)


43/70

by an absolute convergent series. Therefore, it is sufficient to verify the conditions ofthe rearrangement theorem 8.1 for the series

kk(s). Since R exp((1 + 2)zj). (9.6)

Suppose the contrary. Then there is a (0, 1) and a constant B such that|F(z)| 0 , and where the implicit constant may depend on (thischaracterizes all transcendent functions of fixed exponential type

). Further, we

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have to make use ofPlancherels theorem[46] which states that under the assumptionson G(z) in the theorem of Paley-Wiener,

g() =

1

2 G(z) exp(iz) dz

almost everywhere in R . The proofs can be found in [44] and [1]; they rely essentiallyon Fourier theory.

Application of the theorem of Paley-Wiener in our case with G(z) = exp(3z)F(z)yields, with regard to (9.7), 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]. Fur-

ther, Plancherels theorem implies

f() = 1

2

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

almost everywhere. Hence, f() is analytic in a strip covering the real axis. Since thesupport off() lies inside a compact interval, the integral above has to be zero outsidethis interval. Hence, F(z) has to vanish identically, contradicting the existence of asequence of positive real numbers zj with (9.6).

Let xj = zj

R. Then it follows from (9.4) that

|(xj)

|> R2 exp

3xj

4F(xjR)

R2 expxj 3

4

+ R(1 + 2) .Thus, with sufficiently small > 0 we obtain the existence of a sequence of positivereal numbers xj, tending to + , for which

|(xj)| >exp((1 )xj). (9.8)Now we shall approximateFand by polynomials. Let Nj = [xj]+1 and assume

that xj 1 x xj+ 1. Since|m| 1,

m=N2j

+1

mm!

(xR)m (xR)N2j

(N2j )!

n=0(xR)n

n! N

N2j

j exp(Nj)

(N2j)! exp(2xj),

byStirlings formula n! 2nnn exp(n). Similarly,

m=N2j+1

1

m!

3x

4

m exp(2xj).

Therefore,

F(xR) =

N2jm=0

+

m=N2j+1

mm!

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

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and analogously

exp3x

4

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

where Pj and Pj are polynomials of degree N2j . This yields in view of (9.4)(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 Qj(x) we have to apply a classicaltheorem of A.A.

Markov [43] which states that if Q is a polynomial of degree N with real coefficientswhich satisfies the inequality

max1x1

|Q(x)| 1,then

max1x1 |Q

(x)| N2

;

for a proof see [43].We return to the proof of Theorem 9.1. In view of (9.8) suppose that

:= maxxj1xxj+1

|ReQj(x)| > 12

exp((1 )xj), (9.9)

then there exists a [xj 1, xj + 1] such that ReQj() = . In view of themean-value theorem from real analysis there exists a in between x and for which

|ReQj()

ReQj(x)

|=

|ReQj()(

x)

|.

Define = N8j | x| . Markovs theorem, applied to Q(x) = 1 ReQj(x xj), implies

|ReQj() ReQj(x)| .If 1

2, then

|ReQj(x)| 2 1

16exp((1 )xj)

for| x| 12

N8 . If (9.9) does not hold, we may argue analogously with Im Qj . Inany case it follows that for sufficiently large xj the intervals [xj 1, xj+ 1] containsintervals [, + ] of length

1200

N8j all of whose points satisfy at least one of the

inequalities

|Re (x)| > 1200

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

exp((1 )x);(9.10)

in particular

xj 1 xj+ 1 12x8j

and 1

2x8j 2.

In order to prove the divergence of a subseries of (9.2) we note that one of the in-equalities in (9.10) is satisfied infinitely many often as x

; we may assume

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that it is the one with the real part. By the prime number theorem 6.2, the inter-val [exp(), exp( + )] contains

exp(+)exp()

du

log u+ O

exp

c( + )1

9

= exp( + )

+ exp()

+ O

exp

c( + ) 19

= exp()

exp() 1 + O

exp

c(+ ) 19

many primes, where c >0 is some absolute constant. An easy computation shows

(exp(+ )) (exp()) exp(xj)x9j

.

Under these prime numbers pk [exp(), exp( +)] we choose those with k 0 mod 4. Since pk =

k4, we get with view to (9.3)

k0 mod 4

log pk+

k, =

k0 mod 4log pk+

Re (logpk) exp((1 )xj)exp(xj)x9j

=exp(xj)

x9j,

which diverges with xj . Analogously, one can create a subseries of (9.2) whichdiverges to .

Thus, we have shown that the series (9.2) satsifies the conditions of Theorem 8.1.

Hence, there exists a rearrangement of the series (9.2) for which

k=1

ujk (s) =g

s

2

. (9.11)

Before we can finish the proof of Theorem 9.1 we have to prove the following

Lemma 9.2 Suppose that F(s) is continuous on |s| R . Suppose that there is asequence of analytic functions fn(s) for which

limn |s|r

|F(s) fn(s)|2 d dt= 0,

then for any > 0 there is an integer m such that for any fixed r(0, R) and anyn m

max|s|r

|F(s) f(s)| < .

Proof. Define Gn(s) =F(s) fn(s). By Cauchys formula,

Gn(s)2 =

1

2i

|sz|=

Gn(z)2

z s dz= 1

2

20

G2n(s + exp(i)) d

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for any R . Fix 0< r < R . Taking the absolute modulus and integrating with0< R r , we arrive at

|Gn(s)|2 Rr

0 d = 1

2 Rr0

20 |Gn(s + exp(i))|

2

d d

= 1

2

|s|R

|Gn(s)|2 d dt.

This yields

|Gn(s)|2 12(R r)2

|s|R

|Gn(s)|2 d dt for |s| R < r.

Now the assumption on the limit implies the estimate of the lemma.

We return to the proof of Theorem 9.1. According to (9.11),

limn

nk=1

ujk (s) = g

s

2

in the norm ofHR2 . This implies

limn

|s|R

g

s

2

nk=1

ujk (s)

2

d dt= 0

uniformly on

|s

| R . Thus, application of Lemma 9.2 shows that for sufficiently large

m

max|s|R

g

s

2

mk=1

ujk (s)


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Then it follows from (10.1) and (10.4) that for sufficiently large z and T

meas AT >(1 2)T, (10.5)

which is surprisingly large; this idea goes back to Bohr [7]. Application of Lemma 9.2gives 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 +3

4+ i

log Q

s +

3

4+ i, 0

0 there

exists a positive for which, whenever

p p < for all p Mk, (10.8)

then

max|s|r

log Mk

s +3

4, 0

log Mk

s +

3

4,

< . (10.9)Setting

BT =

[T, 2T] :logp

2 p

<

,

we get

1

TB

|s|rlog Q s +

3

4

+ i, 0log Mk s +

3

4

+ i, 02

d dt d

=

|s|r

1

T

BT

log Q

s +3

4+ i, 0

log Mk

s +

3

4+ i, 0

2

dd dt.

Putting () =

log22

, log32

, . . .

, we may rewrite the inner integral as

BT

log Q

s +3

4, ()

log Mk

s +

3

4+ i,()

2

d.

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The curve () is uniformly distributed mod1. Thus, application of Theorem 7.4yields

meas

BT : max|s|rlog Q s +34+ i, 0 log Mk s +

3

4+ i, 0

< y15 (r

14 )

k

>

meas D2

T. (10.10)

If we now take 0< 2< 12

meas D , then (10.5) implies

measAT BT >0.

Thus, in view of (10.7) we may approximate g

s

by log Mk

s + 34 , 0

(independent

on ), with (10.9) and (10.10) the latter function by log Q s + 34

, 0 , and finallywith regard to (10.6) by log

s + 34 + i

on a set of with positive measure. The

theorem is proved.

It is obvious what we have to do to get rid of the logarithm in Theorem 10.1. Letf(s) = exp(g(s)); it is fundamental in complex analysis that every analytic functionf(s) without zeros has an analytic logarithm g(s). Obviously,

f(s)

s +3

4

= f(s)

1 exp

log

s +

3

4

g(s)

.

Let >0 and >1 such that r < 14 and

max|s|r

f(s) f

s

< .Then, since exp z= 1 z+ O(|z|2), we obtain

max|s|r

s +3

4+ i

f

s

= max|s|r

s +3

4+ i

exp

g

s

max

|s|r

f

s

max|s|r

exp

log M

s +

3

4, 0

g

s

1

max|s|r f

s

max|s|r log Ms +

3

4 , 0 g

s

,

which can be made sufficiently small by Theorem 10.1. Thus, we finally have provedVoronins theorem:

Corollary 10.2 Let 0< r < 14

and suppose that f(s) is a continuous non-vanishingfunction on|s| r which is analytic in the interior. Then, for any >0 ,

liminfT

1

Tmeas

[0, T] : max

|s|r

s +3

4+ i

f(s)

<

> 0.

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Surprisingly, the set of translates on which (s) approximate a given function f(s)with arbitrary precision has a positive lower density! This improves Theorem 1.2 fromthe introduction significantly.

Bagchi [3] extended Voronins result significantly in different ways after some firstprogress due to Reich [47]. By that it was possible to replace the discs by arbitrarycompact subsets of the strip 1

2< 0.

11 Functional independence

We state some consequences of the universality property of the zeta-function. We startwith a classical result due to Bohr [7], namely that the set of values taken by (s) ona vertical line (12 , 1) lies dense in the complex plane. This can be extended to

Theorem 11.1 Let 1

2 <


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for any >0. With view to Theorem 10.1 the function g(s) can be approximated to

arbitrary precision on the disc|s| r by log

s + 34

+ i

for some . Hence, taking

f(s) =g(s) log s +34+ iand < r in (11.1), shows that (log (s), log (s), . . . , log (s)(n1)) with 12 < < 1lies somewhere arbtrarily close to (g(0), g(0), . . . , g(n1)(0)) = (b0, b1, . . . , bn1). Thisimplies the statement for the first set.

We use induction on m to prove that for any (m+1)-tuple (a0, a1, . . . , am) Cm+1 ,where a0 = 0, there exists (b0, b1, . . . , bm) Cm+1 for which

exp

mk=0

bksk

mk=0

akk!

sk mod sm+1.

For m= 0 one only has to choose b0 = log a0 . By the induction assumption, we mayassume that with some

exp

mk=0

bksk

mk=0

akk!

sk + sm+1 mod sm+2.

Thus,

exp

mk=0

bksk + sm+1

(1 + sm+1)

mk=0

akk!

sk + sm+1

mod sm+2.

Hence, let bm+1=be the solution of the equation

a0+ = am+1

(m+ 1)!,

which exists by the restriction on a0 . This shows

exp

m+1k=0

bksk

m+1k=0

akk!

sk mod sm+2,

proving the claim.Now

f(s) := exp

n1k=0

bksk

m+1k=0

akk!

sk mod sn.

By Voronins universality theorem, Corollary 10.2, there exists a sequence j, tending

with j to infinity, such that

limj

max|s|r

s +3

4+ ij

= 0for some r (0, 1

4). In view of (11.1) we obtain

limj

max|s|r

(k)

s +3

4+ ij

f(k)(s)

= 0for k= 1, . . . , n 1 and any (0, r). Arguing as above, this proves the theorem.

Further, the universality result implies functional independence:

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Theorem 11.2 Let z = (z0, z1, . . . , z n1) C . Suppose that F0(z), F1(z), . . . , F N(z)are continuous functions, not all identically zero, then

Nk=0

sk

Fk((s), (s)

, . . . , (s)(n1)

) = 0for some s C ,In particular, we see that the zeta-function does not satisfy any algebraic functionalequation. This solves one of Hilberts famous problems which he posed at the Inter-national Congress of Mathematicians in Paris 1900. The first proof of this result wasgiven by Ostrowski [41].

Proof. First, we shall show that if F(z) is a continuous function and

F((s), (s), . . . , (s)(n1))) = 0

identically in s C , then F vanishes identically.Suppose the contrary, i.e. F(z) 0. Then there exists a Cn for which F(a) = 0 .

Since Fis continuous, there exist a neighbourhood U of a and a positive such that

|F(z)| > for z U.Choosing an arbitrary (1

2, 1), application of Theorem 11.1 yields the existence of

some t for which((s), (s), . . . , (s)(n1)) U,

which contradicts our assumption. This proves our claim, resp. the assertion of thetheorem with N= 0 .

Without loss of generality we may assume that F0(z) is not identically zero. Asabove there exist an open bounded set Uand a positive such that

|F0(z)| > for z U.Denote by Mthe maximum of all indices m for which

supzU

|Fm(z)| = 0.

If M= 0, then the assertion of the theorem follows from the result proved above.Otherwise, we may take a subset V U such that

infzV |FM(z)| > for some positive . By Theorem 11.1, there exists a sequence tj , tending with j toinfinity, such that

((+ itj), (+ itj), . . . , (+ itj)

(n1)) V.This implies

limj

M

k=0

(+ itj)kFk((+ itj), (+ itj)

, . . . , (+ itj)(n1))

= .This proves the theorem.

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12 Self-similarity and the Riemann hypothesis

In view of Voronins universality theorem a natural question arises: is it possible to

approximate functions with zeros? The answer is more or less negative but of specialinterest. We give an heuristic argument which however can be made waterproof witha bit more effort and the same techniques which we shall use later on.

In order to see that the zeta-function cannot approximate uniformly a function withzeros recall Rouches theorem, which states that if f(s) and g(s) are analytic insideand on a contourC , and|f(s)| 0 sufficiently small we may assume that max|s|=r |f(s)| > . Hence, whenever theinequality

max|s|r

s +3

4+ i

(s)

< 0

liminfT

1

Tmeas

[0, T] : max

|s|r s +3

4+ i f(s) <

> 0,

then we expect T many complex zeros of (s) in the strip 34 r < < 3

4+ r (for

a rigorous proof one has to consider exactly the densities of values satisfying (12.1);this can be done along the lines of the proof of Theorem 12.1 below). This contradictsthe density theorem 4.4, which gives

N

3

4 r, T

=o(T).

Thus, an approximation of a function with a zero on a sufficiently rich set cannot bedone!

The above reasoning shows that the location of the complex zeros of the zeta-function is closely connected with its universality property. We can go a little bitfurther.

Theorem 12.1 The Riemann hypothesis is true if and only if for any compact subsetK of 1

2< 0

liminfT

1

Tmeas

[0, T] : max

sK|(s + i) (s)| <

> 0. (12.2)

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This theorem is due to Bagchi [3]; in [4] he generalized this result in various directions.Since Bagchis proof relies mainly on the theory of topolgical dynamics he speaks aboutthe property (12.2) asstrong recurrence. However, we call itself-similarity. It should be

noted that Bohr [8] detected in 1922 a similar result. Therefore, we have to introducean important class of zeta-functions.

For a character mod q (i.e. a non-trivial group homomorphism on the primeresidue class mod q) theDirichlet L -functionis for >1 given by

L(s, ) =p

1 (p)

ps

1=

n=1

(n)

ns .

Dirichlet L -functions have similar behaviour and properties as the Riemann zeta-function; actually, (s) may be regarded as Dirichlet L -function associated to theprincipal character 0mod 1 . As for the Riemann zeta-function it is conjecturedthat L(s, ) does not vanish in > 1

2 , this is the so-called Generalized Riemann

hypothesis. Harald Bohr introduced the fruitful notion of almost periodicity intoanalysis. We say that a function L(s) is almost periodic in K if for all > thereexists a sequence of values . . . , 1 0 and lim supm

m|m| <

for which|L(s + im) L(s)| < for all s K.

Bohr proved that Dirichlet series are almost periodic in their half-plane of absoluteconvergence. Moreover, he discovered an interesting relation between Riemanns hy-pothesis and almost periodicity: if is a non-trivial character, then the Riemannhypothesis for L(s, ) is true if and only if L(s, ) is almost periodic for > 1

2. Note

that self-similarity implies almost periodicity (but not vice versa). The condition onthe character seems somehow unnatural but Bohrs argument does not apply to (s) .However, by Voronins universality theorem this gap can be filled. We shall give asimple proof of Bagchis theorem which actually combines Bohrs idea with the one ofBagchi.

Proof. If Riemanns hypothesis is true, then we can apply the universality theorem10.3 with g(s) =(s), which implies the self-similarity. The idea for the proof of theother implication is that if there is at least one proof to the right of the critical line,then the self-similarity property implies the existence of manyzeros, too manywithregard to well-known density theorems).

Suppose that the Riemann hypothesis is not true, then there exists a zero of (s)with Re > 12. Further, we We have to show that there exists a disc |s| r < 14 and > 0 such that

limT

1

Tmeas

[0, T] : max

sK|(s + i

(s)| < } = 0. (12.3)

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Locally, the zeta-function has the expansion

(s) = c(s )m + O|s |m+1|

(12.4)

with some non-zero c C and m N . Now assume for a neighbourhood K := {s C :|s } of that

maxsK

|(s + i) (s)| <

|()

(

i)

|=

|(

i)

| |c| | i |m + O | i |m+1 .Hence,

| i |

|c| 1

m

+ O

1+1m

,

and in particular

1

2


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The critical line is a natural borderline for self-similarity since Levinson [36] provedthat the frequency of c -values of the zeta-function, i.e. solutions of (s) = c , closeto the critical line increases rapidly with increasing imaginary part; similar as the

Riemann von-Mangold formula (2.5) in connection with Theorem 4.3 show for theparticular case of zeros.

Assuming Riemanns hypothesis the self-similarity property (12.2) has an interest-ing interpretation. The amplitude of light waves is a physical bound for the size ofobjects which human beings can see (even with microscopes), or take the Planck con-stant 1033 which is the smallest size of objects in quantum mechanics. Thus, if weassume that is less than this quantity, then we cannot distinguish between (s) and(s + i) whenever

maxsK

|(s + i) (s)| < .This shows that even if one would have all knowledge on the zeta-function, one could

not decide wherever one actually is in the analytic landscape of (s) above the righthalf of the critical strip without moving to the boundary. The zeta-function as anamazing maze!

Theorem 12.1 offers an interesting approach towards Riemanns hypothesis. How-ever, we shall only prove, following Bohrs argument, that the zeta-function is self-similar in the half-plane of absolute convergence.

Theorem 12.2 LetK be any compact subset in the half-plane >1 . Then, for any > 0 ,

liminfT

1

Tmeas

[0, T] : max

sK|(s + i) (s)| <

> 0.

Proof. Since (s) is regular and zero-free in > 1, we may define the logarithm(by choosing any one of the values of the logarithm). In view of the Euler productrepresentation it is easily shown that

log (s) =p,k

1

kpks for >1,

where the sum is taken over all prime numbers p and all positive integers k . Hence,

log (s) log (s + i) = p,k1

1

kpks

1 1

pik

. (12.6)

We shall use diophantine approximation to find values of for which pik lies suf-ficiently close to 1 . We apply Theorem 7.1 with an =

12

logpn where, as usual, pndenotes the n -th prime number. Further, we choose

=

(z1, . . . , z N) RN :zn < 1

for 1 n N

,

where the parameter >0 will be chosen later and wherez denotes the distanceof zto the next integer. Then we find about

=T4

N

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many values of [0, T] such that

logpn < 2

for 1 n N

as Ttends to infinity. Consequently,

cos(klogp) = 1 + O

k2

2

, sin(klogp) =O

k

,

resp.

|1 pik|2 = (1 cos(klogp))2 + sin2(klogp) k2

2

for p pN, provided that > 4k ; here and in the sequel all implicit constants areabsolute. This yields

px,ky

1ps

1 1pi

xy,where x= pN. Furthermore, we have

px,k>y

1

kpks

1 1

pik

1

y

px

1

py x

y2y,

and p>x,k1

1

kpks

1 1

pik

n>x

1

n x

1

1 .

In view of (12.6) we obtain

log (s) log (s + i) xy2

+ + x

y2y+

x1

1 (12.7)

for a set of values of [0, T] with positive lower density as T and any >1 ;note that the estimates can easily be improved. SinceK is compact, there exists:= min{ : s K} . Then

maxsK

|(s)| ().Now for any given >0 we can find values x, y and such that the right hand sideof (12.7) is < . It remains to get rid of the logarithm. Obviously,

(s) (s + i) = (s)(1 exp{log (s + i) log (s)}).Hence,

maxsK

|(s) (s + i)| () maxsK

| log (s+ i) log (s)| < ().

The choice = 2M

proves the theorem. Taking Theorems 12.1 and 12.2 into account, we might conjecture that the zeta-

function has the above self-similarity property for > 12

.

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same zero of log (s): suppose that a zero of log (s), generated by , lies in twodifferent translatesK1 andK2 , then

|1 2| < 8r4

r2 ||2 . (13.3)Suppose that there exist

sj = Re 34

+ itj Br, and j R with log

sj+3

4+ ij

= 0,

for j = 1, 2, such that

= s1+3

4+ i1 =s2+

3

4+ i2.

In view of (13.1),

|g(s2) g(s1)| = r2 ||2

|r2 s1||r2 s2| |s2 s1|.

We deduce from (12.1) that|g(sj)| < for j = 1, 2, and therefore

|1 2| = |t2 t1| 4r4

r2 ||2 |g(s2) g(s1)| < 8r4

r2 ||2 ,

which proves estimate (13.3).Now, denote by

Ij (T) the disjoint intervalls in [0, T] such that (12.1) is valid

exactly for

j

Ij (T) =:I(T).

Using (13.3), in every intervallIj(T) lie at least

1 +

r2 ||2

8r4 measIj (T)

r

2 ||28r4

measIj(T)

zeros of log (s) in the strip 12

< 0.

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14 Other zeta-functions

It is natural to ask whether other functions with similar properties as the Riemann

zeta-function are universal. Meanwhile, it is known that there exists a rich zoo ofDirichlet series having the universality property; we mention only some further signif-icant examples:

Jointuniversality for Dirichlet L -functions, i.e. for a collection of Dirichlet L -functions attached to pairwise non-equivalent characters mod q (nontrivialgroup homomorphisms on the group of prime residue classes), proved by Voronin[58];

Dirichlet L -functions with respect to the modulus of the characters, by Eminyan[15];

Dedekind zeta-functions associated to a number field K over Q

K(s) =A

1

(NA)s =P

1 1

(NP)s1

,

where the sum is taken over all non-zero integral ideals, the product is taken overall prime ideals of the ring of integers of K and where NA is the norm of theidealA , obtained by Reich [48];

Dirichlet series with multiplicative coefficients, by Laurincikas [27], LaurincikasandSlezeviciene [35];

Matsumoto zeta-functions, obtained by Laurincikas [30]; L -functions associated to cusp forms, resp. new forms (or elliptic curves by Wiles

et al. celebrated proof of Fermats last theorem), by Laurincikas and Matsumoto[33], resp. Laurincikas, Matsumoto and Steuding [34];

Hurwitz zeta-functions

(s, ) =

m=0

1

(m+ )s (14.1)

with parameter (0, 1] where is rational= 12

, 1 or transcendental, provedby Gonek [18], Bagchi [3];

Lerch zeta-functions, proved by Laurincikas [29];

This list could be continued with Hecke L -functions, Artin L -functions, Rankin-Selberg convolution L -functions to mention only some more important examples.

Some of the examples given above, e.g. Hurwitz zeta-functions, have the stronguniversality property, i.e. that they can approximate functions with zeros. It seems

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that the restriction to a conditionaluniversality property is intimately linked to theproperty of the Dirichlet series in question to have an Euler product, and by that, tosatisfy Riemanns hypothesis. It should be noted that the strong universality leads by

the arguments used in the proof of Theorem 12.1 to the existence ofmanyzeros off thecritical line; see [16] for more details.

All known proofs of universality for zeta-functions depend on a certain indepen-dence, namely that the logarithms of the prime numbers are linearly independent (aswe have used it in the proof of Voronins universality theorem when we applied Kro-neckers approximation theorem 7.1), resp. that the numbers log(n + ) for n N arelinearly independent if is transcendental (which is necessary to deal with (14.1)).TheLinnik-Ibragimov conjecturestates that all functions given by Dirichlet seriesand analytically continuable to the left of the half plane of absolute convergence areuniversal. However, this is better to understand as a program than a conjecture. For

example, define a(n) = 1 if n= 2

k

, kN

{0} , and a(n) = 0 otherwise, thenA(s) =

n=1

a(n)

ns =

k=0

1

2sk =

1

1 2s ,

and obviously, this function is far away from beeing universal. However, in [53] auniversality theorem for the so-called Selberg class, which covers all from a numbertheoretical point of view interesting Dirichlet series known so far (i.e. with Eulerproduct), was proved.

More detailed surveys on the value distribution of zeta-functions are the excellentwritten papers [32] and [40].

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References

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On the Universality of the Riemann zeta-function

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