Exponentialsummen mit der Möbiusfunktion

Transcription

1 Universität Ulm Fakultät für Mathematik und Wirtschaftswissenschaften Exponentialsummen mit der Möbiusfunktion Dissertation zur Erlangung des Doktorgrades Dr. rer. nat. der Fakultät für Mathematik und Wirtschaftswissenschaften der Universität Ulm vorgelegt von Hans- Peter Reck aus Aulendorf 04

2 Amtierender Dekan: Prof. Dr. Dieter Rautenbach. Gutachter: Prof. Dr. Helmut Maier. Gutachter: Prof. Dr. Ayyadurai Sankaranarayanan weiterer Gutachter: Prof. Dr. Werner Kratz Tag der Promotion: 8. März 04

3 Contents Introduction 7. Motivation Short overview of the thesis: Basics 0. Partial summation Analytic functions of finite order Dirichlet- series Characters L- series Mean value Theorems and the Hybrid Sieve Introduction The Large Sieve Mean values of character sums Hybrid Sieve Special Dirichlet- series Preparatory Lemmata 5 4. Monotonicity principles Large value estimates The result and the proof Introduction Estimation of the first part of the integral Estimation of the second part of the integral Summary 80 3

4 Notation The symbolism of some signs in mathematics varies depending on the used source and context. To avoid misunderstandings, here some names are listed, which are used in this thesis and are provided in the following as known. N is the set of all natural numbers: N = {,, 3,...} N 0 is the set of all natural numbers and zero: N 0 = {0,,,...} P is the set of all primes R + is the set of all positive real numbers R + 0 is the set of all nonnegative real numbers ez = e πiz for z C Rz is the real part of a complex number z C Iz is the imaginary part of a complex number z C fx = Ogx means fx Cgx for x x 0 and a certain C > 0. Here fx is a complexvalued function in x R and gx a real- valued function with only positive values for x x 0. fx gx means the same like fx = Ogx fx = ogx fx means lim x gx = 0 [x] is for x R is the largest integer not greater than x. x means min k Z x k. n, m is the greatest common divisor of two natural numbers n, m N [n, m] is the least common multiple of two natural numbers n, m N mod q means modulo q for q N C is the multiplicative group C\{0} Z/qZ is the multiplicative group of all residue classes, which are coprime to q N fz = fz. Here fz is a complex- valued function, defined on an arbitrary set Res x0 f is the residue of f at x 0. Here f is a complex- valued function, defined on a subset D C with x 0 D. n, p Sometimes the index set of the sum is omitted for reasons of space. In this case it always holds, that n runs over all natural numbers and p runs over all primes. log z If nothing else is mentioned, log z means the main branch of the logarithm. 4

5 List of abbreviations resp. respectively f. following RH Riemann Hypothesis GRH Generalized Riemann Hypothesis WLOG Without loss of generality 5

6 List of figures 3.. Visualisation of the curve K

7 Chapter Introduction. Motivation There is a great interest in analytic number theory for exponential sums, whose coefficients are formed by certain number- theoretic functions. From the behavior of such sums one can draw conclusions on the distribution of prime numbers. Similar to the prime number theorem, such results on the distribution of prime numbers can be improved by the adoption of the generalized Riemann hypothesis GRH. In this work, the coefficients are represented by the Möbius- function. The Möbius- function is a number- theoretic function µ: N {, 0, }, which is defined as follows: { k for n = p µn = p k with p,..., p k pairwise distinct 0 otherwise. Under certain conditions, the behavior of this sum allows conclusions on the estimation of the sum Sx, α := n x µneαn with a real number α. This sum was introduced by Davenport, who found the estimate max Sx, α < C λx θ log x λ for arbitrary λ > 0 and an arbitrary constant C λ > 0. Hajela and Smith assumed that the Dirichlet L- series Ls, χ = χmm s have no real zeros, so- called Siegel- zeros. Under this condition they proved max Sx, θ < C x exp Clog x / θ for C, C > 0. Under the stronger condition m= Ls, χ 0. for all s C with Rs > a for a certain a [/, Hajela und Smith showed the estimation for every ɛ > 0. max Sx, θ < C 3 ɛx a+ 3+ɛ θ 7

8 In their joint- work Exponential Sums formed with the Möbius Function R.C. Baker und G. Harman improved this result. Under the same premises they received for each ɛ > 0 the estimate while b := ba is given by max Sx, θ x b+ɛ, θ a + /4 for / a < /0, b := 4/5 for /0 a < 3/5, a + / for 3/5 a <. Assuming the generalized Riemann hypothesis, also. with a = /, they reduced the exponent of 5/6 + ɛ to 3/4 + ɛ. The improvement is obtained by using an estimation for Gaussian sums, the mean value theorem for Dirichlet- polynomials and two lemmata of van der Corput for estimation of exponential sums. In the paper On an exponential sum involving the Möbius function it is proved, that for alle α of type numbers, there holds Ss, α x 3/4+ɛ for any ɛ > 0 under the assumption of the average zero- density hypothesis Nσ, T, χ qt σ logqt A. χ In the diploma- thesis of Claudia Fischer Exponentialsummen, gebildet mit der Möbiusfunktion 3 from December 005, she took over the procedure of Baker and Harman and applied it to the calculation of sums of the form Sy, χ, γ = n y µnχneγn with a character χ mod q and γ R. For this purpose, it is sufficient to consider sums of the form y m= Sm, χ, 0 and calculate them with Perron s formula. Thus, we obtain integrals of the form y m= eγmms ds s Ls, χ C with a curve that passes the critical line Rs = / on the right, and the imaginary parts of the end points have T and T. In order to obtain an improvement over the result of Baker and Harman, Claudia Fischer combines their technology with the procedure, used by H. Maier and H.L. Montgomery in their work The Sum of the Möbius Function 4, for estimating the function Mx := n x µn under the Riemann hypothesis. See Baker and Harman See Maier and Sankaranarayanan 5 3 See Fischer 9 4 See Maier and Montgomery 4 8

9 The calculation can be reduced to a similar integral x s s ζs ds. C The estimate of Mx is improved by choosing the curve C not as a straight line, but as a piecewise linear contour, the course of which is dependent on the magnitude of the logarithm as well as the logarithmic derivative of the Riemann- ζ- function near the critical line. The distance of the contour to the critical line is chosen such that the bound that can be obtained for the integrand is approximately minimized. In the work of Claudia Fischer, the integral is estimated by obtaining an upper bound on the number of segments that have a large minimum distance from the critical line with the help of some lemmata that estimate the number of points under certain conditions and with an average value of the imaginary parts. In my diploma- thesis of December 007, the main focus does not lie only on an averaging over the imaginary parts, but also over the Dirichlet- characters, using the Large Sieve and the Hybrid Sieve. I considered paths C, which are also on the right of the line Rs = /, which have to be constructed for every Dirichlet- character specifically. In order to achieve the same result, the paths must be formed on the critical line as a function of the value of Ls, χ, however, only the paths are considered with the start and end points - and. In the work at hand, we consider a version of the original problem by limiting the summation to integers n without large prime factors. We treat only one aspect of this problem, namely the estimation of sums µnχn, n x p + n x u where u [0, and p + n denotes the largest prime factor of n, averaged over all primitive characters χ mod q with Q q Q for q N. We will also use the above tools.. Short overview of the thesis: Chapter contains the basic concepts which form the basis of this thesis. In section. we start with Abel s partial summation and in section. we consider special analytic functions. We introduce Dirichlet- series in section.3, characters in section.4 and combine these in section.5 to Dirichlet- L- series. Mean value theorems and sieve methods like the Large Sieve and the Hybrid Sieve will be discussed in chapter 3. In chapter 4 we introduce important tools for the estimation, one the one hand some monotonicity principles on horizontal lines which are parallel to the real axis in section 4., and on the other hand in section 4. estimates for the number of times, a certain function is large, using techniques of Selberg. In chapter 5 we show the results and they will be proved as well. 9

10 Chapter Basics In this chapter, some important definitions, theorems and lemmata are given which are needed later.. Partial summation Theorem... Abel s partial summation Let a, b R and f : [a, b] C be a continuously differentiable function. Next let c, c,... be a sequence of complex numbers. Then it holds b c n fn = f t dt + fb Proof. We have a<n b a<n b c n fb a<n b a a<n t c n c n fn = = a<n b b a a<n b c n c n fb fn = a<n t c n f t dt. a<n b a n c n f t dt. Analytic functions of finite order Definition... Analytic functions A complex- valued function f : C C is called an analytic function, if f is holomorphic over the whole complex plane. Definition... Order of an analytic function Let f : C C be an analytic function. If an α > 0 exists such that M f r := max{ fz : z r} expr α is true for large r, then f is called by finite order. See Lütkebohmert 3, Lemma

11 If there is no α > 0, which satisfies this condition, then f is called of infinite order. The number αf := inf{α > 0 : M f r expr α } is called the order of the analytic function f. This definition is useful because M f r is monotonically increasing as a function in r. Definition..3. Convergence exponent Let s, s,... C\{0} be a sequence in C, and we have 0 < s s.... Under the convergence exponent of the sequence s, s,... one understands the lower limit of β of that b > 0, for which: s n b <. n holds. If the sequence is finite, then we set β = 0. If there is no b > 0, so that. is satisfied, then set β =. Theorem... Let k z n ez, k := z exp. n Every analytic function fz with finite order α has a representation of the form fz = e hz z z m e, p, z n n while the product runs over all zeros z n of f with z n 0, which are ordered by the size of their absolute value. There exist natural numbers p N 0 with n z n p <. Let p be the smallest of these numbers. Let m be the multiplicity of the zero point z = 0 and hz be a polynomial of degree g α. If β is the convergence exponent of the sequence z, z,..., we have α = max {g, β}. If there is no fixed c > 0 and sufficiently large r, such that fs < expcr α holds for all s with s r, there is α = β, and z n β is divergent for α > 0. Proof. The proof is already found in Lütkebohmert 3 on page f. Theorem... 3 Let 0 < r < R and fs holomorphic in s s 0 R, while we choose s 0 C such that fs 0 0. Let s,... s m with m N be any zeros of fs in s s 0 r. Then we have R m MR r fs 0, where MR = max s s0 =R fs. This, especially applies for r = R/ with A := / log MR m A log. fs 0 See Prachar 6, appendix, Theorem See Prachar 6, appendix, Theorem 5.

12 Proof. WLOG let s 0 = 0. Otherwise consider fs = fs s 0. The function F s = fs m Rs s n R ss n is holomorphic in s R, and it holds F s = fs for s = R since Rs s n R ss n = Rss s n s sr ss n = RR s n s sr ss n = holds for all n =,..., m. So by the maximum principle it follows F 0 = f0 m R s n max F s = MR. s =R Now, the first part of the assertion follows because s n r for n =,..., m, also f0 R m f0 r m Setting r = R/, the second part follows immediately. R s n MR..3 Dirichlet- series Definition.3.. Dirichlet- series Let a n n N be a sequence of complex numbers. We call a n n s a Dirichlet- series with s C and n s := exp s logn. Then we call { } σ c := inf a n n σ converges [, ] σ R the abscissa of convergence of the Dirichlet- series n N a nn s and the absolute abscissa of convergence. { } σ a := inf a n n σ converges [, ] σ R

13 Lemma Let n N a nn s be a Dirichlet- series with coefficients a n C und let σ c resp. σ a be the abscissa of convergence resp. the absolute abscissa of convergence of this Dirichlet- series. Then we have:. The Dirichlet- series n N a nn s converges locally uniform for Rs > σ c.. The Dirichlet- series n N a nn s diverges for Rs < σ c. 3. The Dirichlet- series n N a nn s converges absolutely for Rs > σ a. 4. The Dirichlet- series n N a nn s does not converge absolutely for Rs < σ a. 5. It holds σ c σ a σ c The function fs := is holomorphic for Rs > σ c, and we have f s := a n n s. a n log nn s, while this Dirichlet- series has abscissa of convergence σ c. Proof.. With Abel s partial summation Theorem.. we get for s 0, s C and m n n n n n t a ν ν s = a ν ν s 0 ν s0 s = a ν ν s 0 n s0 s a ν ν s 0 t s0 s dt ν=m = ν=m ν=m n m ν=m n ν a ν ν s 0 n s0 s + a µ µ s 0 ν s0 s ν + s0 s.. ν=m ν=m µ=m Furthermore, ν z ν + z = ν+ ν zt z dt z max t [ν,ν+] trz z Dzν Rz applies for z C und ν N, if it holds { for Rz, Dz := Rz for Rz >. Let M ν := ν a µ µ s 0. µ=m Then, with. n a ν ν s M n n Rs0 s + s 0 s Ds 0 s ν=m 4 See Lütkebohmert 3, exercises 5 and 6, page 98 f. n ν=m M ν ν Rs 0 s..3 3

14 A series converges locally uniform in a domain if and only if it converges compactly there. Let K {s C: Rs > σ c } be compact and σ := min {Rs: s K} > σ c. Then there exists s 0 = σ 0 R with σ > σ 0 > σ c, so that the series n N a nn s 0 converges. Now we show that the series n N a nn s converges uniformly on K. Let C > 0, such that n a ν ν s 0 C ν= holds for all n N. Then we have n a ν ν s 0 C ν=m for all n, m N. Let s K, also Rs σ. Due to.3 we have n n a ν ν s Cn Rs0 s + s 0 s C ν Rs 0 s ν=m ν=m n Cn σ0 σ + s 0 + max z C ν σ0 σ 0.4 z K for n, m, independent of s C. The statement.4 is also valid for all s 0 C, for which n N a nn s 0 converges.. We assume that there is a point s 0 C with Rs 0 < σ c, such that n N a nn s 0 converges. According to.4 the locally uniform convergence of n N a nn s follows for Rs > Rs 0. In particular, the series n N a nn σ would converge for every σ > Rs 0, contrary to the definition of σ c, since Rs 0 < σ c. So we get the claim. 3. This follows from the first part, applied to n N a n n s. 4. This follows from the second part, applied to n N a n n s. 5. The inequality σ c σ a is obvious. Judging from the assumptions σ a > σ c +, there exists σ R with σ a > σ > σ c +, that means σ > σ c. Therefore the series n N a nn σ converges. This means that there exists C > 0, so that a n n σ C holds for all n N. It follows a n n σ+ɛ = ν=m a n n σ n ɛ C n ɛ < for all ɛ > 0. So we get σ a σ + ɛ for all ɛ > 0. In contradiction to the above assumption we have σ a σ. The claim follows. 6. The holomorphy of f follows from the first part, since we may interchange summation and differentiation because of the uniform convergence and d ds n s = log nn s holds. Let σ c be the abscissa of convergence of n N a nn s and σ c the abscissa of convergence of n N a nlog nn s. Because of the first part, σ c σ c is obvious. Nevertheless, it still remains to show that σ c σ c. We assume σ c < σ c. Let σ R with σ c < σ < σ c. According to the first part, the series n N a nlog nn s converges locally uniform for Rs > σ. It follows the convergence of σ+ σ+ a n log nn t dt = a n log nn t dt = a n n σ+ n σ..5 σ σ 4

15 According to the fifth, part the series n N a nlog nn σ+ converges absolutely. Since it holds > a n log nn σ+ n=3 a n n σ+ the series n N a nn σ+ converges also. With.5 we get the convergence of n N a nn σ, in contradiction to the definition of σ c. This suffices. n=3 Theorem.3.. Multiplication of Dirichlet- series 5 Let a n n s and b n n s n N be two Dirichlet- series with a n, b n C. If n N a n n σ and n N b n n σ converge for σ 0 < σ <, then it holds for σ 0 < Rs < n N with a n n s b n n s = c n n s c n = d n a d b n d for all n N, and these series converge absolutely also in σ 0 < Rs <. Proof. Since they converge absolutely, we can expand, and it holds a n n s b n n s = a d b m md s = a d b n n s d m= d= d n for Rs > σ 0. Now we explain Riemann s method of complex integration, with which one can estimate Dirichletpolynomials. It is a consequence of Perron s formula and is often also referred to as well. Lemma.3.. Perron s formula 6 Let b, T R with b > and T > 0. Then it holds for a R + Notice that πi b+it b it a s s { ds δa for a >, δa = / for a =, 0 for a <, and log a means the main branch of the logarithm. 5 See Prachar 6, appendix, Theorem.3 6 See Lütkebohmert 3, Lemma 7.4. b πt for a =, a b πt log a for a. 5

16 Proof. We always mean with log a the main branch of the logarithm. For a = we have b+it ds = logb + it logb it = log ib/t b it s + ib/t = log + log ib/t log + ib/t = πi i arg + ib/t = πi + R mit R b/t, since ib/t = + ib/t and arg ib/t = arg + ib/t. In case a > we integrate over the rectangle C with the sides C b := {b + it : T t T } C h := { h + it : T t T } C +it := {σ + it : h σ b} C it := {σ it : h σ b} and h > b. According to the residue theorem 7 it holds a s ds =. πi s Now we estimate C h a s C±iT a s s ds s ds T T b h C a h T a h dt h + it h a σ σ ± it dσ b h and a σ T dσ For h the first integral vanishes and we get the above estimation. For 0 < a < we integrate over the rectangle C with the sides C b := {b + it : T t T } C h := {h + it : T t T } C +it := {σ + it : h σ b} C it := {σ it : h σ b} and also h > b. According to the residue theorem 8 and b > we get a s ds = 0. πi s C a b T loga. Now we estimate analogously to the case a > and we also receive the claim. Theorem.3.. Riemann s method of complex integration 9 Let fs := a n n s be a Dirichlet- series with a n C for all n N, which converges for Rs > normally. For σ R and α > 0 it holds a n n σ = Oσ α σ. There is a n An for a monotonically increasing function A. 7 See Lütkebohmert 3, Theorem See Lütkebohmert 3, Theorem See Lütkebohmert 3, Theorem

17 Then it holds φ f x πi b+it b it fs xs s ds = O xlog x α T + O for T and x >. For x > 0 there is { φ f x = n<x a n for x R Z, n<x a n + ax for x Z, xaxlog x + / x and log x means the main branch of the logarithm. Next is b = bx := + / log x and x = for x Z and otherwise, the distance to the nearest integer. Proof. Since the Dirichlet- series converges normally for Res >, we can integrate term by term. Applying Perron s formula Lemma.3. we get T with b+it fs xs πi b it s ds = πi b+it x s a n n s ds = φ f x + R R Axb πt +,n x b it x b a n n πt logx/n. There is,n x logx/n log3/ x b a n n πt logx/n x b πt log3/ a n n b x b = O T b α xlog x α = O T, since b = + / log x. Furthermore,n x logx/n <log3/ x b a n n πt logx/n = n x x/3<n<3x/ Axb πt x b a n n πt logx/n n x x/3<n<3x/ logx/n. Therefore, it remains to show that n x x/3<n<3x/ logx/n = O x log x + x. x We have / z log + z for z C with z < /. Since for x/3 < n < x, it holds x n/n < /, and we get x log = log + n x n x n n n x n x. 7

18 Now it is n x x/3<n<3x/ logx/n x x n x x/3<n<3x/ x/3 x + x n x + t dt = O x log x +. x For x < n < 3x/ we have n x/x < / and thus One shows analogously here by This suffices. n log = log + x n x n x x x. x<n<3x/ logn/x = O x log x +. x Definition.3.. multiplicative functions We call a function f : N C multiplicative, if it holds f = and fmn = fmfn holds for all m, n N with m, n =. If this applies for all m, n N without restrictions, f is called completely multiplicative. Notice.3.. The Möbius- function µ, the function ɛ: N {0, } with ɛn := { for n =, 0 otherwise, and the function : N {} with n := for all n N are multiplicative functions. Definition.3.3. convolution Let f, g : N C. We define the convolution of f and g as follows f gn := ab=n fagb for all n N. Lemma.3.3. Möbius inversion formula Let µ be the Möbius- function and ɛ and n be defined like in.3.. Then we obtain µ = ɛ. 8

19 Proof. Since µ, ɛ and are multiplicative functions, for µ it holds µ nm = µab = µa a b b = = ab=nm a b =n a b =n a b =m µa b a b =m a b =n a b =m µa b = µ nµ m for all m, n N with m, n = using the multiplicativity of µ and. Since ɛ is multiplicative, it suffices to show for all p P and α N 0. Let p P and α N 0. Then we have µ p α = ɛp α µa µa b b µ p α = α µab = µa = µp k ab=p α a p α k= { µ = for α = 0, = µ + µp = = 0 otherwise } = ɛp α. Notice.3.. We can describe the product of two Dirichlet- series by the convolution. Let F s := n N fnn s and Gs := n N gnn s be Dirichlet- series, which converge absolutely in s 0 N and let f, g : N C be functions. According to Theorem.3. we obtain F s 0 Gs 0 = fdgn/d n s 0 = fagb n s 0 = f gnn s. d n Dirichlet- series with multiplicative coefficients have special properties. Some of these properties will be discussed in more detail below. Theorem Let f : N C be multiplicative and σ a the absolute abscissa of convergence of the Dirichlet- series F s = n N fnn s. Then it holds F s = fp ν p νs p P ν=0 ab=n for Rs > σ a, where the infinite product converges normally. If f is completely multiplicative, then we have F s = p P fpp s and F s = µnfnn s for Rs > σ a. This representation of F s is called Euler- product. Proof. The proof is well- known and can be found in Fischer 9 on page See Lütkebohmert 3, exercises 8 and 39, page 99 and 0 9

20 .4 Characters Dirichlet- series, which represent generalizations of the Riemann- ζ- function in a certain sense, are of particular interest for the work at hand. Definition.4.. Character modulo q Let q N. By a character modulo q we mean a function χ: N C, which satisfies the following properties: { = 0, for n, q >,. χn 0, for n, q =.. χn = χm, for m n mod q. 3. χmn = χmχn for all m, n N. The function χ 0 n := { 0 for n, q >, for n, q = is called the principal character modulo q. Because of the second property, characters can be defined on Z as well. If χ is a character modulo q, set χl = χn, if it holds l Z, n N and l n mod q. The properties of the characters are not lost. Obviously, products of characters are again characters, since all three properties are fulfilled. There are various ways to define the characters. In the following an equivalent definition is given, using algebraic concepts. Definition.4.. Character modulo q Let G be a finite abelian group. Then we call a group homomorphism χ: G C of G to C, a character of G. For q N we call a function χ: N C a character modulo q, if there exists a character χ of the multiplicative group Z/qZ with { χ χn = n + qz for n, q =, 0 for n, q >. In case χ mod q, the character χ = χ 0 is called the principal character modulo q. Notice.4.. Definition.4. and Definition.4. are equivalent. Lemma.4.. Let q N. Then it holds χn = for every character χ mod q for all n N with n, q =. Proof. Let χ be a character modulo q. According to Definition.4. there exists a homomorphism χ : Z/qZ C with χn = χ n + qz for all n N with n, q =. Notice that Z/qZ is a group of order ϕq. Thus, we have n + qz ϕq = + qz for all n + qz Z/qZ and for all n N with n, q =. χn ϕq = χ n + qz ϕq = χ n + qz ϕq = χ + qz = See Prachar 6, page 0 f. See Lütkebohmert 3, exercises 9 and 30, page 00 f. 0

21 We get the claim since it holds for all n N with n, q =. χn = χn ϕq /ϕq = /ϕq = Theorem Let q N. Then there exist ϕq characters modulo q. Proof. In elementary number theory we prove that there exist numbers w, w 0,..., w r so that every l N with 0 l < q and l, q = can be represented in exactly one way while γ, γ 0,..., γ r N and l w γ w γ 0 0 wγ... wγr r w w α w ϕpαr r r mod q. The numbers α 0,..., α r, p, p,..., p r N and r N are given by the prime factorization of q: q = α 0 p α pαr r with p, p,..., p r P pairwise different. Owing to the second and third properties in Definition.4. we have χl = χw γ w γ 0 0 wγ... wγr r = χw γ χw 0 γ 0 χw γ... χw r γr, that means, χ is uniquely determined by χw, χw 0, χw... χw r. Because of these two properties of Definition.3. we obtain χw = χ w = χ = χw 0 α 0 = χ w α 0 0 = χ = χw ϕpα = χ w ϕpα = χ = χw r ϕpαr r. = χ w ϕpαr r r = χ = So we get χw = ±, resp. χw 0, χw,..., χw r are roots of unity of order α0, ϕp α,..., ϕpαr r. So there are at most α 0 ϕp α... ϕpαr r = ϕq different options for χ. Conversely, any choice of χw, χw,..., χw r as a unit root of order, α0, ϕp α,..., ϕpαr r leads us to a character modulo q, by setting χl = 0 for l, q >. This suffices. Lemma Let q N.. For every character χ mod q it holds l mod q χl = { ϕq for χ = χ0, 0 for χ χ 0. We summate over a full or reduced residue system modulo q. 3 See Prachar 6, page 0 4 See Lütkebohmert 3, exercise 9, page 00 and Prachar 6, page 0

22 . We have { ϕq for l mod q, χl = 0 otherwise. χ Here, we summate over all ϕq characters modulo q. 3. It holds for l N with l, q = { for n l mod q, χnχl = ϕq 0 otherwise. χ Proof.. It holds χn = 0 for every character χ mod q and for all n N with n, q >. So there is no difference, if we sum over a full or a reduced residue system. For χ = χ 0, we have χl = = ϕq. l mod q l mod q, l,q= For χ χ 0 there exists n N with n, q = and χn. So we get χn χl = χnl = χl. l mod q l mod q l mod q The last equation holds, since Z/qZ is a group under multiplication. Hence, 0 = χn χl 0 = χl.. For l mod q we get χl = χ χ l mod q χ = χ l mod q = ϕq, since there are ϕq characters modulo q according to Theorem.4.. Obviously, the set of all characters modulo q forms a group under multiplication: χ χ n = χ nχ n for all n N and for two characters χ and χ modulo q. Let l mod q. Then we have χ with χl. Thus, χl χ χl = χ χχl = χ χl. We get 0 = χl χ χl 0 = χ χl. 3. It holds for every character χ mod q χlχl = χl = χl = χl for l, q =. Let l be a solution of ll mod q. Then it holds l, q = and χl = χl = χl. Thus, χnχl = χnχl = χnl. χ χ χ The claim follows with the second part, since we get n l mod q if and only if nl mod q holds.

23 Let χ and χ be two characters with moduli q and q N. There is χ n 0 and χ n 0 n, q = n, q = n, [q, q ] = for n N. This leads to the following definition: Definition.4.3. equivalent characters 5 Two characters χ and χ with moduli q and q are called equivalent, if χ n = χ n for all n N with n, [q, q ] =. One say, χ is induced by the modulus q. The modulus q is called conductor of χ. The characters χ und χ are equivalent if and only if their values are the same for all n N, for which they do not vanish for both characters. Lemma One can induce every character χ mod q by every multiple of q. Proof. Let q be a multiple of q. Then the function given by { χ n for n, q χ n := =, 0 otherwise, is a character modulo q according to Definition.4., and the character χ is equivalent to χ since it holds n, q = n, [q, q ]. Lemma Let q, q N with q q, and let χ mod q be a character modulo q. Then χ is induced by the modulus q if and only if it holds χ n = for all n N with n mod q and n, q =. Proof. One direction follows immediately. If χ is an equivalent character to χ, it holds χ n = χ n = χ = for all n N with n mod q and n, q = since n, q =. For the other direction we notice, that there exists y N with n, q = for all n N such that n + q y, q =.6 holds. Let q be the product of all primes, which divide q but not q. Let ñ N with ñ, q =. Then the congruence n + q y ñ mod q is solvable since q, q =. If y is such a solution, then we have n + q y, q = ñ, q = and also n + q y, q = n, q = for n, q =. Overall, we have n + q y, q = because of the choice of q. We now define χ by { χ n + q χ n := y n for n N with n, q =, 0 for n N with n, q >. We choose y n for every n N with n, q = over.6. Now we want to show that χ is a character modulo q. 5 See Prachar 6, page 3 f. 6 See Prachar 6, chapter IV, Lemma 6. 7 See Prachar 6, chapter IV, Lemmata 6. and 6.3 3

24 First χ is well defined, that means χ n is independent of the choice of the y n for all n N. Let n N with n, q = and let y n, y n, such that n + q y n, q = n + q y n, q = holds. One determines n over the congruence nn mod q. Then we have n, q = and according to.6 there exists y n N with n + q y n, q =. It follows χ n + q y n χ n + q y n = χ n + q y n n + q y n =, since n + q y n n + q y n nn mod q. Similarly follows χ n + q y n χ n + q y n =. So we have χ n + q y n = χ n + q y n since n + q y n, q = and therefore χ n + q y n 0. The first property in Definition.4. is fulfilled, since we choose y n N for every n N such that it follows n + q y n, q = from n, q =. So we have χ n + q y n 0. To prove the second property, we choose n, m N with n m mod q. In the case m, q > we have n, q >, and the second property is fulfilled. Let n, q = m, q =. One determines n over the congruence mn nn mod q and chooses numbers y n, y m, y n according to.6 such that holds. Then it follows m + q y m, q = n + q y n, q = n + q y n, q = χ n + q y n χ n + q y n = χ n + q y n n + q y n = χ =, since n + q y n n + q y n nn mod q. Just apply χ m + q y m χ n + q y n =. Since n + q y n, q = we have χ n + q y n 0, and it follows χ n + q y n = χ m + q y m, in other words χ m = χ n. The third property follows for m, n N with mn, q = from χ mχ n = χ m + q y m χ n + q y n = χ m + q y m n + q y n = χ mn + q y n m + y m n + q y n y m = χ mn for certain y n, y m N. Here, the already shown second property was used. The third property holds for mn, q > too. So χ is a character modulo q according to Definition.4., and χ and χ are obviously equivalent. 4

25 Lemma Let χ and χ be two equivalent characters modulo q resp. modulo q, and let q = q, q. Then χ and χ can be induced modulo q. Proof. We only prove the assertion for χ. The proof of χ is analogous. It suffices to show that it holds χ n = for n N with n mod q and n, q =. Then the claim follows by applying Lemma.4.4. Let n N, n mod q and n, q =. Then there exists m N with n = + mq. Since q = q, q there exists a, b Z such that it holds q = aq + bq. So we have and + mbq is coprime to q and q. Hence, n = + maq + bq = + maq + mbq χ n = χ + maq + mbq = χ + mbq = χ + mbq = χ = due to the second property in Definition.4. and due to the equivalence of the two characters. Possibly it holds + mbq N. But then there exists l N with + mbq l mod q, and one set χ + mbq = χl. Theorem All conductors of character χ are multiples of a unique conductor q N. Proof. Let q be the smallest conductor of χ. For q = the claim follows obviously. Let q >. We assume, there exists a conductor q which is not a multiple of q. Then we obtain q, q < q. Since q, q is a conductor of χ according to Lemma.4.5, this would be a contradition to the chioce of q. Thus, the following definition makes sense: Definition.4.4. primitive character 0 We call a character χ mod q, which is not induced by a modulus, which divides q, a primitive character. Let χ be a character modulo q and let q be the smallest conductor of χ. We denote the to χ equivalent primitive character with χ. It is also a character modulo q and is uniquely determined because two of the characters modulo q do not have the same values for all n N. Two primitive characters are thus either equal or not equivalent. In contrast, two equivalent characters have the same primitive character. Generally, the set {n N: χ n 0} is larger for every character χ than the set {n N: χn 0}. The first mentioned set is the largest set in which χ can be continued in a certain sense, without losing the properties of the characters. 8 See Prachar 6, chapter IV, Lemma See Prachar 6, chapter IV, Theorem 6. 0 See Prachar 6, page 6 5

26 .5 L- series Definition.5.. Dirichlet- L- series Let q N and χ be a character modulo q. The Dirichlet- series is called Dirichlet- L- series. Ls, χ := χnn s Lemma.5.. Let q N and χ be a character modulo q. Then we have:. For χ χ 0 the series n N χnn s converges for Rs > 0. The function Ls, χ represents an analytic function for Rs > 0.. For χ = χ 0 the series n N χnn s converges for Rs >. In this case Ls, χ represents an analytic function for Rs >. 3. The L- series n N χnn s converges absolutely for Rs >. Proof. Let σ c resp. σ a be the abscissa of convergence resp. the abolute abscissa of convergence of n N χnn s.. Let χ χ 0. According to the first part of Lemma.4. we have q for all k N kq χn = 0, χn = 0. Thus, it follows that means n N χn converges. So we get σ c 0 and according to the first and sixth part of Lemma.3. we get the claim.. In the case χ = χ 0 there is q χnn s q n Rs and q n Rs converges for Rs >. The holomorphy for Rs > follows with the sixth part of Lemma This follows from, since we have σ a σ c + according to the fifth part of Lemma.3.. Notice.5.. For the modulus q = only the principal character χ 0 mod exists, and the associated L- series is the Riemann- ζ- function. It holds Ls, χ 0 = ζs for Rs >. Obviously, all characters modulo q are completely multiplicative functions. From this, we obtain different representations for functions which are formed with the aid of L- series. See Lütkebohmert 3, exercise 30, page 00 f. 6

27 Lemma.5.. Let q N and let χ be a character modulo q. For Rs > it holds. Ls, χ = χpp s = χpp s. p P p P,p q Particularly there is Ls, χ 0 for Rs >.. Ls, χ = µnχnn s. 3. log Ls, χ = 4. L L s, χ = χnλn Λ n s, setting log n log = 0. χnλn n s. Proof.. This follows with Theorem.3.3, since χ is completely multiplicative, and it holds for the absolute abscissa of convergence of the L- series σ a.. This follows with Theorem.3.3, since χ is completely multiplicative, and it holds for the absolute abscissa of convergence of the L- series σ a. 3. It is well- known, that log z has the representation z n log z = n for z <. We may use the logarithm for Ls, χ for Rs > according to the first part of this Theorem. So we get for Rs > log Ls, χ = log χpp s = log χpp s p P p P = χpp s n = χp n n np sn. p P p P There is for all n N and p P. Thus, log Ls, χ = p P Λp n logp n = log p n log p = n Λp n χp n logp n p sn = m= Λmχm log mm s holds for Rs >. The last equation is valid because of Λm = 0, if m is no prime power. 4. We obtained this by differentiating of the third part and the locally uniform convergence for Rs > according to Lemma.3.. Then we have L L s, χ = log Ls, χnλn χnλn χ = n s = n s = χnλnn s log n log n for Rs >. 7

28 Now we will show that all L- series are meromorphic continuable in the whole complex plane and satisfy a certain functional equation there. As a special case we obtain the functional equation of the Riemann- ζ- function. In the following, always let q N. Let χ mod q be a primitive character. As described above, setting χn = χm for m Z and n N with m n mod q, one can interpret χ mod q as a function with domain Z. Since χ = χ = χ = there holds always χ = ±. So the following definition is useful: a = aχ := { 0 for χ =, for χ =..7 Theorem.5.. Let χ be a primitive character modulo q. Then Ls, χ is an analytic function excluding the case q = and χ = χ 0. In this case Ls, χ 0 = ζs is only singular at s = with a simple pole and residue. It holds for all χ mod q L s, χ = ɛ χ π s q s / cosπ/s aγsls, χ, where ɛ χ is a constant depending only on χ with ɛ χ =. The function q s+a s + a ξs, χ = Γ Ls, χ π satisfies the equation ξ s, χ = ɛ χ ξs, χ and it is an analytic function excluding q = and χ = χ 0 with a simple pole at s =. Setting ξs = /ss ξs, χ 0 = /ss π s/ Γs/ζs, leads us to ξs = ξ s, while ξs is an analytic function. Proof. One can find the proof also in Fischer 9 on page 50 f. Theorem Let χ be a character modulo q and denote χ as the to χ mod q equivalent primitive character with modulus q. Then it holds the following relationship between Ls, χ and Ls, χ : Ls, χ = p q,p q χ p p s Ls, χ. In particular, Ls, χ and Ls, χ have the same zeros for Rs > 0. See Prachar 6, chapter VII, Theorem. 3 See Prachar 6, chapter IV, Theorem 6.3 8

29 Proof. Applying Lemma.5. we get Ls, χ = χ pp s = χ pp s χ pp s p P,p q p P,p q p q,p q = χpp s χ pp s = Ls, χ χ pp s, p q,p q p q,p q p P,p q since χn = χ n holds for all n N with n, q =. Thus it follows, that either Ls, χ and Ls, χ vanish for Rs > 0 or both are different from zero, since it holds χ pp s {0, } because of χpp s p Rs < for Rs > 0 and p P. With this statement it follows from Theorem.5., that Ls, χ can be analytically continued into the whole complex plane for a non- primitive character χ, except for χ = χ 0. Then Ls, χ 0 has a simple pole at s =. In the following we study the distribution of the zeros of the L- series. It has been found that those have no zeros in the half- plane {s C: Rs > }. Because of Theorem.5. only the zeros of L- series with primitive characters are interesting. For the proof of the next theorem, however, a distinction between primitive and non- primitive characters is not necessary. Theorem Let χ be a a character modulo q. Then it holds for all t R. L + it, χ 0 Proof. One can find the proof also in Fischer 9 on page 60 f. Theorem Let χ be a primitive character modulo q. Then it holds ξs, χ 0 for Rs and σ 0. For χ χ 0 the series Ls, χ has simple zeros in Rs 0 at s = a, a,... and no other zeros. The function a = aχ is defined like in.7. The Riemann- ζ- function Ls, χ 0 = ζs has simple zeros for Rs 0 only at s =, 4,.... Proof. We already have shown Ls, χ 0 for Rs in Theorem.5.3 and Lemma.5.. According to the definition of the function ξs, χ it follows ξs, χ 0 for Rs, since the Gamma- function has no zeros. Since χ is a primitive character, Theorem.5. and especially the functional equation ξ s, χ = ɛ χ ξs, χ hold. Since χ is also a primitive character, and so ξs, χ 0 for Rs, it follows ξs, χ 0 for Rs 0. From the definition of ξs, χ it follows that Ls, χ can only have zeros where Γ s+a has poles, so at the points a, a,.... For χ χ 0 the function ξs, χ is analytic according to Theorem.5., and therefore ξs, χ has no poles for all s C. So these are actually zeros of Ls, χ. For χ = χ 0 the point s = 0 is no zero of Ls, χ 0 = ζs, since ξs, χ 0 has a simple pole at this point. It is rather with ξs = /ss ξs, χ 0 ζ0 = lim π s/ ξs s 0 ss Γs/ = ξ0 lim s 0 sγs/ = ξ0 = ξ =, 4 See Prachar 6, chapter IV, Theorem 4. and 4. 5 See Prachar 6, chapter VII, Theorem. 9

30 because according to the Gaussian product development 6 it holds lim s 0 The value of ξ is obtained by sγs = lim s 0 eγs/ s s + s e s/n = n. ss ξ = lim ξs, χ s 0 = / lims π s/ Γs/ζs s = π / Γ/Res s= ζs = with Γ/ = π. In contrast, ξs, χ 0 has no poles at s =, 4,... according to Theorem.5., so these are the zeros of ζs. All these zeros are simple, since the gamma function has only simple poles. The Theorem is proved. The zeros mentioned in Theorem.5.4 are called trivial zeros of Ls, χ. Apart from these, all zeros of Ls, χ are in the domain 0 < Rs <, when χ is primitive. The domain 0 < Rs < is called the critical strip. It will be shown that Ls, χ has infinitely many zeros in the critical strip. The function ξs, χ does not disappear outside of the critical strip and has the same zeros as Ls, χ in 0 < Rs <. In order to derive a statement about the distribution of zeros in the critical strip, we want to apply the Theorems.. and... We first examine the growth behavior of Ls, χ. We will introduce an auxiliary function. Lemma Let ζs, w be given for s C with Rs and for w R with 0 < w by ζs, w := n + w s. Then ζs, w can be continued analytically in 0 < Rs and it holds n=0 ζs, w = w s + O t / for Rs / and t. For Rs / and t it holds ζs, w = w s + s + O. The constant in the O-terms in both cases does not depend on w. Proof. We first show the first assertion. Since ζs, w = ζs, w it suffices to show this for t. For Rs it holds ζs, w w s n + w because of w > 0. So we can assume Rs. 6 See Lütkebohmert 3, Theroem Seee Prachar 6, chapter IV, Theorem 5. and 5.3 n = O 30

31 It follows with partial summation Theorem.. for N N and Rs > [b] n=n+ since it holds n + w s = s s b N = s + = s b N b N b [t] Nt + w s dt + [b] Nb + w s N b N t [t]t + w s dt bb + w s + NN + w s t + w s dt + [b] Nb + w s s t [t]t + w s dt + b tt + w s dt = bb + w s + NN + w s + with a = N and b R. Now it follows with b ζs, w w s = = N n + w s s N n + w s + N n b N Nt + w s dt t + w s dt b [b]b + w s, b N t [t]t + w s dt + N + w s s s N t + w s dt N t + w s dt t [t]t + w s dt,.8 and this leads us to the analytic continuation of ζs, w for Rs > 0, since the last integral is a holomorphic function for Rs > 0. Moreover, it follows ζs, w w s N n + w σ + N + w σ s + s σ N σ with Rs = σ, considering 0 t [t] <. So we get for s = σ + it with / σ and t N ζs, w w s = O n / + N / + tn / = O N / + N / t + tn / = Ot /, t setting N = [t]. The first part of the claim is proven. It remains to show the second part. For Rs > we have again ζs, w = w s +O, so the assertion is true. Now let Rs. Setting N = in.8, it follows ζs, w w s = + w s + + w s s + O for / Rs and t. In this domain it holds + w s = O, and thus + w s s = = e s log+w s s = s s n log + w n n=0 s n log + w n since s = O and 0 < log + w log. This suffices. 3 n! n! = s + O

32 Theorem Let q N and let χ be a character modulo q. Let s = σ + it and { for χ = χ0, E 0 = E 0 χ, q := 0 otherwise. Then it holds for σ / and t as well as for σ / and t. All constants are independent of q. Ls, χ = Oq t / Ls, χ = E 0 ϕqq s + Oq/ Proof. Let ζs, w be defined as in Lemma.5.3. For Rs > we have q q Ls, χ = χl qn + l s = q s χlζs, l/q, l= n=0 and since ζs, w has a analytic continuation in 0 < Rs, it also holds in this domain. According to Lemma.5.3 it follows for σ / and t q q q Ls, χ = q s χl l/q s + O t / = χll s + O q s t / l= l= l= q = O l / + q t / = Oq t /, l= where the constant in the O-term is independent of q. So the first part ist proven. Let σ / and t. Applying Lemma.5.3 we get q Ls, χ = q s l/q s + q q s + O = χll s + q s χl + Oq /..9 s l= It holds for Rs / l= q q χll s = O l / = Oq / and l= l= q χl = E 0 ϕq l= according to Lemma.4.. For s / we have q s e s log q = q = q s n log q n = q s s s n! s s n log q n n! n=0 log q = q s + O exp = q s + Oq /, and it follows q s /s = Oq / for s > / and σ /. Putting the last three statements in.9, we get the claim. 8 See Prachar 6, chapter IV, Theorem 5.4 l= l= 3

33 Now we will show that ξs, χ is an analytic function of finite order and we will determine this order as well. Theorem Let χ be a primitive character modulo q and q N. The function ξs, χ is analytic of order for χ χ 0. If ρ = ρχ runs through the zeros of this function, then it holds for every α > ρ = and ρ α <. ρ The exponent of convergence of the sequence of the zeros of ξs, χ is. The same applies for the function ξs and its zeros. In particular, these functions have infinitely many zeros in 0 < Rs <. Proof. First, let χ χ 0. Since ξ s, χ = ξs, χ and Theorem.., it suffices to show ρ ξs, χ exp s +ɛ.0 for s and Rs / as well as all ɛ > 0 and to show that this is no longer true, if is replaced by a smaller number. According to Theorem.5.5 it holds Ls, χ q / s / for s. By Stirling s formula 30 q /s+a s + a s ξs, χ = Γ Ls, χ q /s+a+ exp log s + O s π follows for s, and.0 is proven. On the other hand it holds for real σ > Lσ, χ = µnχnn σ n σ >, and according to Stirling s formula 3 it follows σ + a σ log σ Γ = exp + Oσ for σ. From this it follows for sufficiently large σ q ξs, χ > / π /σ+a exp σ log σ + Oσ, so that.0 certainly does not apply, if is replaced by a smaller number. So ξs, χ is an analytic function of order for χ χ 0, and there exists no c > 0, such that ξs, χ expc s holds for s. Now we consider the case χ = χ 0. For ξs, it is the same as for ξs, χ with χ χ ss 0, since the factor does not matter. Since all zeros of ξs, χ resp. ξs lie in 0 < Rs <, the Theorem is proven. 9 See Prachar 6, chapter VII, Theorem. 30 See Prachar 6, appendix, Theorem 6. 3 See Prachar 6, appendix, Theorem 6. 33

34 For every primitive character χ the functions ξs, χ and Ls, χ have infinitely many zeros in the domain 0 < Rs <. To consider the distribution of zeros in the critical strip accurately, further observations are still needed. Theorem Let q N and σ 0 /. For every character χ mod q it holds Ls, χ q σ 0+ t + σ 0+ for all s C with σ 0 Rs /. Here, the implicit constant can depend on σ 0, but not on q and t = Is. Proof. First, let χ be a primitive character. Then it holds according to Theorem.5. L s, χ = π σ q σ / cos/s aπγsls, χ. with σ = Rs. First we consider only the case χ χ 0. We have σ 0 R s / for / σ σ 0 +. Now it follows according to Theorem.5.5 for / Rs σ 0 + Ls, χ q / t + /. and according to Stirling s formula 33 it holds in the same domain Γs = exp Rs / log s Rs + logπ + O t for t. Suitable estimation of Rs / log s leads us to Γs C exprs / log t π t / + O t = C t Rs / e π t / + O t C t σ 0+/ e π t / + O t with a suitable constant C > 0, while we notice that the behaviour of exp t is for t the same like + t. So we get for / R t s σ 0 + Γs t + σ 0+/ e π t /,.3 where the constant may depend on σ 0. It follows for every s C cos/s aπ e πt/ + e πt/ e π t /..4 Putting.,.3 and.4 in., we get for / Rs σ 0 + L s, χ q σ 0+ t + σ 0+. Since with χ is also χ primitive and not the principal character, the claim of the assertion follows by replacing χ by χ and s by s. For Ls, χ 0 = ζs it holds a = aχ 0 = 0, and it follows from.4 and Theorem.5.5 cos/s + aπls, χ 0 = cosπs/ζs t + / e π t / for / Rs σ 0 +, since cosπs/ζs is holomorphic at s =. 3 See Prachar 6, chapter VII, Theorem See Prachar 6, appendix, Theorem 6. 34

35 We put this with.3 in., and get the claim also for χ = χ 0. We assume χ is not primitive. Then let χ be the associated primitive character with modulus q. So it holds for σ 0 σ / p q,p q χ p p s p q,p q + p σ0 p q/q + p σ q σ0 + q, since for σ 0 0 we have + p σ 0 p σ 0 p σ 0+, against which there is for / σ 0 < 0 and an arbitrary integer m + p σ 0 p m p log m/ log exp n log m + p σ exp n σ 0 p log m log + p σ 0 exp expclog m σ 0+ m σ 0+, p log m while the constant depends on σ 0. The claim follows, since it holds according to Theorem.5. and it is already shown that for primitive characters. Ls, χ = p q,p q χ p p s Ls, χ Ls, χ q σ 0+ t + σ 0+ p σ 0 Combining the Theorems.5.5,.5.6 and.5.7 we obtain with the functional equation for Ls, χ the following Theorem: Theorem Let q N, χ be a character modulo q and α R. Then it holds for Rs α Ls, χ = E 0 ϕq/q s + Oqc t + c for a certain c = cα and E 0 defined by Theorem.5.5. The implicit constant in the O-term depends also on α. Theorem Let N χ T denote the number of zeros of Ls, χ in {s C: s = σ + it, 0 σ <, t T }. It follows for q N, T 0 and an arbitrary character χ mod q N χ T + N χ T logqt +, while the implicit constant is independent of q and T. 34 See Prachar 6, chapter VII, Theorem See Prachar 6, chapter VII, Theorem

36 Proof. Let χ 0 be the principal character modulo q, and first, let χ χ 0. We apply Theorem.. on Ls, χ and set s 0 = + it, r = / and R sufficiently large, such that the domain 0 Rs <, T < Is T + is entirely contained in s s 0 R/, for example R = 6. According to Theorem.5.8 it holds Ls, χ q c t + c for s s 0 R. Also one has Ls, χ µn χn n s 0 ζ using the formula for /Ls, χ in Lemma.5.. The claim follows if we insert the zeros of Ls, χ in 0 Rs < und T < t T + for s,..., s m in Theorem.., since the zeros of Ls, χ in T t < T are complex conjugated to the zeros of Ls, χ in T < t T +. The claim follows also for χ = χ 0 by applying Theorem.. in the same way on the function s Ls, χ 0. Notice, that it holds a Rs b in every strip, such that s t +, where the implicit constant may depend on a and b. Thus, Theorem.5.8 leads us to s Ls, χ 0 q c t + c with a certain c R + in each such strip. Finally, we give an explicit formula: Lemma.5.4. explicit formula 36 There is Λnχnn it 0 = n x Iρ t 0 T x ρ it 0 ρ it 0 + O where ρ represents the zeros of Ls, χ in the critical strip. x log xt 0 Proof. This Lemma is proven similar as the formula 3.5 in Brüdern 5 on page. Definition.5.. Under the Riemann hypothesis RH we understand the conjecture, that all zeros of the Riemann- ζ- function in the critical strip 0 < Rs < lie on the line with Rs = /. This is still an open question. Most mathematicians, however, assume the truth of this conjecture. There is also a conjecture to the zeros of the L- series in the critical strip. Definition.5.3. The generalized Riemann hypothesis GRH states, that all zeros of Ls, χ with all characters with the modul q with q lie in 0 < Rs < on the line Rs = /. In the following we assume the truth of the GRH. Because of the functional equation of ξs, χ GRH is equivalent to the statement, that there are no zeros of Ls, χ in / < Rs <, since the zeros of Ls, χ in the critical strip are symmetrical to the line Rs = /. T, 36 See Brüdern 5, chapter III,page 36

37 Chapter 3 Mean value Theorems and the Hybrid Sieve 3. Introduction In this chapter we want to estimate sums and integrals of Dirichlet- series in order to be able to calculate discrete or continuous mean values. With this tool, we can estimate exponential sums with the Möbius- function. The idea behind this is the following one: We consider a function that takes a very large value at a point s 0 C. In a small neighborhood of the point s 0 this still applies to the said function. So we can estimate the number of points s 0 by a continuous average of this function by estimating this function with a Dirichlet- polynomial in the neighborhood of the point s The Large Sieve We now consider the method of the Large Sieve according to Harold Davenport and P.X. Gallagher. An objective of the Large Sieve is to estimate a discrete mean by a continuous mean. First, we consider a few lemmata. Lemma 3... Let u, δ R + and a complex valued function F x be continous in [u δ/, u + δ/] with a continous derivative in u δ/, u + δ/. Then F u = δ u + δ u δ/ u+δ/ u δ/ x u δ F x dx. F x dx + δ u+δ/ u x u + δ F x dx See Richert 8, chapter II, page 30 f. 37