ANALYTIC NUMBER THEORY LECTURE NOTES BASED ON DAVENPORT S BOOK

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1 ANALYTIC NUMBER THEORY LECTURE NOTES BASED ON DAVENPORT S BOOK ANDREAS STRÖMBERGSSON These lecture notes follow to a large extent Davenport s book [5], but with things reordered and often expanded. The point of these notes is not in the first place to serve as an outline of the actual lectures which I will give, but instead to hopefully! collect all the details for those results which I wish to mention, so that in the lectures I can focus more on the main points and ideas. Contents. Primes in Arithmetic Progressions Ch., 4 in [5] 5. Infinite products 8.. Infinite products 8.. Infinite products over the primes.3. Problems 6 3. Partial summation and Dirichlet series Integration by parts * The general Riemann-Stieltjes integral Dirichlet series; convergence properties Problems More on Dirichlet characters [Review: Some basic facts about Z/qZ and Z/qZ ] [Review: The structure of Z/qZ ] Explicit list of all Dirichlet characters modulo q Some consequences * Fourier analysis and structure theory for finite abelian groups Primitive characters Quadratic reciprocity; the Legendre, Jacobi and Kronecker symbols Characterization of the real primitive characters Problems L, χ and class numbers Equivalence classes of quadratic forms Dirichlet s class number formula Gauss sums I Finite sum formulas * Ideal classes in number fields, and the Dedekind Zeta Function 0

2 ANDREAS STRÖMBERGSSON 5.6. Problems The distribution of the primes The logarithmic integral and the prime number theorem Tchebychev s auxiliary functions ϑ and ψ Further asymptotic results 6.4. Riemann s memoir Problems 8 7. The prime number theorem Analytic continuation of ζs 7.. Zeros Fundamental formula Asymptotic formula for ψ x Going from ψ x to ψx Problems 3 8. The Γ-function; Integral Functions of Order Entire functions of finite order The Γ-function Problems The functional equation The case of ζs Gauss sums II The functional equation for a general Dirichlet L-function Problems The Infinite Products for ξs and ξs, χ The infinite products for ξs The infinite products for ξs, χ Problems 68. Zero-free regions for ζs and Ls, χ 69.. A zero-free region for ζs 69.. Zero-free Regions for Ls, χ 7.3. *Alternative method 8.4. Problems 87. The numbers NT and NT, χ 88.. The number NT 88.. The Number NT, χ Problems The explicit formula for ψx The prime number theorem again Problems 4. The explicit formula for ψx, χ 3 5. The prime number theorem for Arithmetic Progressions I 5.. Consequences for πx; q, a 9

3 ANALYTIC NUMBER THEORY LECTURE NOTES Problems 3 6. Siegel s Theorem * Some history The prime number theorem for Arithmetic Progressions II Goal for the remainder of the course: Good bounds on average Problems 4 7. The Polya-Vinogradov Inequality Problems Further prime number sums Example: An exponential sum formed with primes * Equidistribution of pα mod Problems Sums of three primes Problems The Large Sieve Problems 78. Bombieri s Theorem 79. An Average Result Solutions still hidden to students Notation still not included TO DO! To fix in special sections More recent work to mention and sketch TO DO later years! * On class numbers: Generalization to higher dimension: Siegel s mass formula brief outline Exercises Notation big O, little o,, and Varia 38 References 39 Solutions to the Problems not home assignments Solutions to home assignment Solutions to home assignment 6

4 4 ANDREAS STRÖMBERGSSON.

5 ANALYTIC NUMBER THEORY LECTURE NOTES 5. Primes in Arithmetic Progressions Ch., 4 in [5] In this first lecture we will prove Dirichlet s Theorem from : Theorem.. If a, q are positive integers with a, q =, then there are infinitely many primes in the arithmetic progression a, a + q, a + q, a + 3q,.... The proof which I will give is complete except for 3-4 technical facts, which I will not have time to prove in detail but which I can hopefully convince you are reasonable to believe. I will come back to these facts and prove them in the next few lectures. The proof which I will give does not follow all steps of the proof which Dirichlet originally gave, instead it is shorter and makes use of more complex analysis; the key new step is a trick by de la Vallée Poussin from 896 which is presented on pp in Davenport s book. However the original proof by Dirichlet is interesting in its own right because of its connection with quadratic forms and class numbers, and I will come back to this in later lectures. To get started, we introduce the so called Riemann Zeta Function: ζs = n s s C, Re s >. n= We will follow standard notation in analytic number theory and write s = σ + it σ, t R. Thus, for instance, {s : σ > } is the set of all s which have real part greater than one. Lemma.. The series ζs = n= n s is absolutely convergent for all s C with σ >, and uniformly absolutely convergent in any compact subset of {s : σ > }. In particular, by Weierstrass s Theorem, ζs is an analytic function in the set {s : σ > }. Proof. This follows directly by comparison with the positive series n= n c for c >, which is well-known to be convergent. Details: Since any compact subset of {s : σ > } is contained in {s : σ c} for some c >, it suffices to prove that ζs = n= n s is uniformly absolutely convergent for all s with s c, where c is some fixed number >. Let ε > 0 be given. Since the positive series n= n c is convergent, there is some N Z + such that n=n n c < ε. Now take any s = σ + it C with σ c. Then for each n N we have n s = e s log n = e σ log n it log n = e σ log n = n σ n c, Cf. Ahlfors [, Sec. 5..] or Priestley [56, Ex. 4.7] or Rudin [60, Thm. 0.8]. Weierstrass Theorem states that if {f k } is a sequence of analytic functions in an open set Ω C and if f is a function on Ω such that f k f uniformly on any compact subset of Ω, then f is analytic in Ω, and also f n k f n uniformly on compact subsets of Ω.

6 6 ANDREAS STRÖMBERGSSON and hence 3 n s n=n n c < ε. n=n Now we have proved that for each ε > 0 there exists some N Z + such that 3 holds for all s C with Re s c. This is exactly the desired statement. The reason why ζs is important in the study of primes is the following identity, the so called Euler s identity or Euler product: Lemma.3. 4 ζs = p p s for all s C with σ >. Here in the product in the right hand side, p runs over all primes, and the lemma in particular contains the fact that this infinite product is convergent when σ >. I will not give a complete proof of this lemma here, but will come back to it in the next lecture where I will discuss infinite products in general see Section. below. But I will give an outline of the main idea of the proof: By the formula for a geometric sum we have p s = + p s + p s + p 3s +... for each prime p. Hence the infinite product in the right hand side of 4 is: + s + s + 3s s + 3 s + 3 3s s + 5 s + 5 3s s + 7 s + 7 3s s + s + 3s +... When expanding this product completely using the distributive law it turns out that because of convergence properties we only pick up those products which take from all except a finite number of the factors above; thus we get = + s + 3 s + s 3 s + s 5 s + s + s 3 s + 5 s +... = + s + 3 s + 3 s + 5 s + s + 3 s + 5 s +... The terms above are not chosen in any systematic order, but we see that we get exactly one term for each distinct prime factorization p r p r p rm m, the corresponding term being p r p r p rm m s. But by the fundamental theorem of arithmetic, each positive integer has one unique prime factorization we here include the number which has the empty prime factorization; hence the above sum actually contains exactly one term n s for each positive integer n, i.e. the sum equals n= n s, which was to be proved.

7 ANALYTIC NUMBER THEORY LECTURE NOTES 7 Note that none of the factors in the right hand side of 4 vanishes, since p s = p σ < when σ >. Hence it seems reasonable that we have: 5 ζs 0 for all s C with σ >. We will prove this rigorously in the next lecture. See Theorem.. It follows that log ζs can be defined for each s C with Re s >. Note that this is in principle a multivalued function for each s. However, there is one choice of log ζs which is the most natural: Motivated by the Euler product 4, let us define log ζs by 6 log ζs = p log p s, s C, σ >, where each logarithm on the right is taken with the principal branch this is ok since p s <. We will see in the next lecture, in connection with our discussion of infinite products see Example. below that the sum in 6 is absolutely convergent for all s with σ >, and indeed gives a logarithm of ζs i.e., e p log p s = ζs. We also note that when s is real, s >, then ζs is real and positive by definition, and in this case log ζs is just the usual, real valued, logarithm since the sum in the right hand side of 6 is real valued in this case. Using the Taylor expansion log z = z + z + z3 +..., valid for all z C with 3 z <, we can write 6 in the form log ζs = 7 m p ms. p The double sum in the right hand side of 7 is absolutely convergent for any s C with σ >. [Proof: The convergence of the right hand side in 6 and Taylor s formula combine to show that the iterated sum p m= m p ms converges for any s C with σ >. Applying this for real s > we have a double sum with positive terms, and hence the convergence automatically implies absolute convergence of the double sum. For general s C with σ > the absolute convergence now follows by using p ms = p mσ.] Now we restrict to considering real s >. Note that directly from the definition of ζs we get 8 Also note that, for each s >, 9 p m= lim ζs = +, and thus lim s + m= m p ms < p m= p m = p log ζs = +. s + pp = p n= p p n n =.

8 8 ANDREAS STRÖMBERGSSON Hence, using 8 and 7, we conclude that 0 lim p s = +. s + p This implies that there are infinitely many prime numbers. It even implies the stronger fact that p =. p [Proof: Given any A > 0 there exists some s > such that p p s > A, by 0. Hence there exists a finite set of primes p < p <... < p N such that N k= p s k > A. Now N k= p k > N k= p s k > A. Since this can be obtained for any A > 0 we conclude that holds.] Dirichlet s aim in his 837 memoir was to prove the corresponding fact for the set of prime numbers in an arithmetic sequence, i.e. to prove: Theorem.4. If a Z, q Z + and a, q =, then p =. p a mod q Note that Theorem.4 in particular implies Theorem., i.e. the fact that there are infinitely many primes in the arithmetic sequence a, a + q, a + q, a + 3q,.... The main tool in the proof of Theorem.4 is a generalization of the Riemann Zeta function and Euler s identity to Dirichlet L-functions, which are sums involving Dirichlet characters. To motivate their introduction, let us note that a naive way to try to generalize the above proof of p p = to the case of Theorem.4 would be to replace ζs with the series n s. n= n a mod q However this immediately runs into the problem that the Euler product 4 does not generalize to this series! Studying the proof sketch of 4 we see that in order to have something like Euler s identity, we should generalize ζs = n= n s to sums of the form n= c nn s, where the c n s are multiplicative. For our purpose of proving Theorem.4 we then wish to go using linear combinations from multiplicative coefficients to the case of, viz. c n = if n a mod q, c n = 0 otherwise, and for this we might hope that it suffices to consider sequences c, c, c 3,... which are periodic with period q. This leads to the definition of a Dirichlet character notation c n = χn:

9 ANALYTIC NUMBER THEORY LECTURE NOTES 9 Definition.. Let q be a positive integer. A Dirichlet character of period q or modulo q is a function χ : Z C which is periodic with period q, i.e. 3 and multiplicative without restriction, i.e. 4 and which also satisfies 5 and 6 χn + q = χn, n Z. χnm = χnχm, n, m Z, χ = χn = 0 whenever n, q >. Remark.. The condition 4 is called multiplicativity without restrictions since in number theory the term multiplicativity is reserved for a weaker notion: A function f : Z + C is said to be multiplicative if fmn = fmfn holds for all m, n Z + with m, n =. Remark.. Given conditions 3 and 4, the condition χ = is equivalent with saying that χ is not identically zero, and this is also equivalent with saying that χn 0 for all n with n, q =. Proof. If χ is not identically zero then there is some n with χn 0, and 4 with m = gives χn = χnχ which implies χ =. Conversely χ = clearly implies that χ is not identically zero. Next assume χn = 0 for some n with n, q =. By Euler s Theorem n φq mod q, and hence by repeated use of 4 and then 3 we get 0 = χn φq = χn φq = χ, which contradicts χ =. Hence if χ = then χn 0 for all n with n, q =. Remark.3. If χ is a Dirichlet character of period q, then for all n Z with n, q = we have χn φq = χn φq = χ =, since n φq mod q. In other words, χn is a φqth root of unity for each n Z with n, q =, and in particular χn =. It follows that χn for all n Z. Example.. There is exactly one Dirichlet character of period q = : χ. There is exactly one Dirichlet character of period q = : n = χ 0 n = There are exactly two Dirichlet characters of period q = 3: Other names for the same thing are total multiplicativity and complete multiplicativity.

10 0 ANDREAS STRÖMBERGSSON n = χ 0 n = χ n = There are exactly two Dirichlet characters of period q = 4: n = χ 0 n = χ n = There are exactly four Dirichlet characters of period q = 5: n = χ 0 n = χ n = i -i - 0 i -i - 0 i -i - 0 χ n = χ 3 n = -i i - 0 -i i - 0 -i i - 0 If we study the above tables it is not hard to see that at least for q =,, 3, 4, 5, for every a Z with a, q =, the sequence { if n a c n = 0 otherwise, can be expressed as a linear combination of the Dirichlet characters modulo q. To give an example, for q = 5 and a = 3 this means that the following sequence; n = c n = , can be expressed as a linear combination of the four Dirichlet characters χ 0, χ, χ, χ 3 of period 5 cf. the previous table, and this is indeed the case: We see that c n = 4 χ 0n + i 4 χ n 4 χ n i 4 χ 3n, n Z +. This is a special case of the following lemma, which shows that the same thing works for arbitrary q: Lemma.5. Let X q be the set of all Dirichlet characters modulo q. Then for any a, n Z with a, q = we have { if n a mod q, 7 χaχn = φq 0 otherwise. χ X q

11 ANALYTIC NUMBER THEORY LECTURE NOTES We postpone the proof of this to the fourth lecture, when we discuss Dirichlet characters in more detail. We will also see that #X q = φq. We will next see that if we generalize the Riemann Zeta function using Dirichlet characters then we indeed still have an Euler product. Definition.. If χ is any Dirichlet character, we define the corresponding Dirichlet L- function by 8 Ls, χ = χnn s, n= for all s C with Re s >. Lemma.6. The series Ls, χ = n= χnn s is absolutely convergent for all s C with σ >, and uniformly absolutely convergent in any compact subset of {s : σ > }. In particular, by Weierstrass s Theorem, Ls, χ is an analytic function in the set {s : σ > }. Proof. This is exactly as the proof of Lemma., using χn for all n see Remark.3. Note that ζs is a special case of a Dirichlet L-function; we have ζs Ls, χ when χ, the unique Dirichlet character of period q =. The Euler product for the Dirichlet L-function looks as follows: Lemma.7. 9 Ls, χ = p χpp s. for all s C with σ >. Here, again, in the product in the right hand side, p runs over all primes. The proof of 9 is deferred to the next lecture, but we mention that, just as for the Riemann Zeta function, the main step in the proof uses the fact that each positive integer has one unique prime factorization, together with the unrestricted multiplicativity of χ: The right hand

12 ANDREAS STRÖMBERGSSON side of 9 equals + χ s + χ s + χ 3 3s χ33 s + χ3 3 s + χ s χ55 s + χ5 5 s + χ s +... = + χ s + χ33 s + χ s χ33 s + χ s χ55 s + χ s + χ s χ33 s + χ55 s +... = + χ s + χ33 s + χ 3 3 s + χ 5 5 s = + χ s + χ 3 3 s + χ55 s +... χnn s = Ls, χ. n= Note that none of the factors in the right hand side of 9 vanishes, since χpp s p s = p σ < for all s C with σ >. Hence by Theorem. which we will prove in the next lecture 0 Ls, χ 0 for all s C with σ >. It follows that log Ls, χ can be defined for each s C with σ >. Note that this is in principle a multivalued function for each s. However, there is one choice of log Ls, χ which is the most natural: Let us define log Ls, χ by log Ls, χ = p log χpp s, s C, σ >, where each logarithm on the right is taken with the principal branch this is ok since χpp s <. We will see in the next lecture, in connection with our discussion of infinite products see Example. below that the sum in is absolutely convergent, uniformly on compact subsets of {s C : σ > }; hence this sum defines an analytic function in {s C : σ > }, and this indeed gives a logarithm of Ls, χ i.e., e p log χpp s = Ls, χ. Inserting the Taylor expansion of the logarithm in we obtain, for any s C with σ > : log Ls, χ = p m χp m p ms, m= where the double sum in the right hand side is absolutely convergent for any s C with σ > this is seen by comparison with 7 applied with σ in place of s.

13 ANALYTIC NUMBER THEORY LECTURE NOTES 3 Now let a be a fixed integer with a, q =. In order to use Lemma.5 we multiply with φq χa, and then add over all χ X q. This gives χa log Ls, χ = χa m χp m p ms φq φq χ X q χ X q p m= = 3 χam χp m p ms = m p ms, φq p m= χ X q p m= p m a mod q where in the last step we used Lemma.5. Using comparison with 9 to treat all terms with m, we obtain 3 4 χa log Ls, χ = p s + O, φq χ X q p a mod q for all s C with σ >. The essential idea of Dirichlet s memoir 837 is to prove that the left side of 4 tends to + as s + i.e. keeping s real and >. This will imply that there are infinitely many primes p a mod q viz., Theorem. and further that the series p a mod q p is divergent viz., Theorem.4. Cf. our proof of above. To prove that the left side of 4 tends to + as s +, we discuss each χ X q individually. First of all, there is always one trivial or principal character in X q ; this is denoted by χ = χ 0 and is defined by χ 0 n = if n, q = and χ 0 n = 0 if n, q >. The corresponding L-function is 5 Ls, χ 0 = χ 0 nn s = p s ζs. n= The last identity follows from the Euler product formula for Ls, χ 0 and for ζs: We have ζs = p p s and Ls, χ 0 = p χ 0pp s = p q p s, since χ 0 p = 0 if p q and χ 0 p = if p q. Using lim s + p s = p > 0 for each of the finitely many primes which divide q, and the fact that lim s + ζs = +, we conclude that lim s + Ls, χ 0 = +, and thus also 6 p q log Ls, χ 0 + as s +. 3 Recall the big O -notation, which you are hopefully familiar with e.g. from discussions involving Taylor expansions: If a 0 is a non-negative number, the symbol Oa is used to denote any number b for which b Ca, where C is a positive constant, called the implied constant. When using this notation, it should always be clear for which variable ranges the bound holds. For example: fx = Ox 3 as x means that there is some constant C > 0 such that for all sufficiently large x we have fx Cx 3. On the other hand, fx = Ox 3 for x means that there is some constant C > 0 such that fx Cx 3 holds for all x.

14 4 ANDREAS STRÖMBERGSSON Since we also have χ 0 a =, to complete the proof that the left side of 4 tends to + as s + it now suffices to show that for each choice of χ X q other than χ = χ 0, log Ls, χ is bounded as s +. At this point it clarifies the situation if we note that when χ χ 0, the series 7 Ls, χ = χnn s n= is convergent not only for σ > but for all s with σ > 0, and defines an analytic function of s in this region. This follows from general facts about Dirichlet series which we will discuss in the third lecture see Example 3.5. I remark that a Dirichlet series is any series of the form n= c nn s. The key reason why 7 converges for all σ > 0 when χ χ 0 is that then the sequence χ, χ, χ3,... is very oscillating and in particular it has average 0; in fact, as we will see in the third lecture, we have k+q n=k χn = 0 for all χ X q \ {χ 0 } and all k Z +. Hence, our task to prove that for each χ X q \ {χ 0 }, log Ls, χ is bounded as s + is equivalent to proving that 8 The proof splits into two cases: L, χ 0. Case : χ is complex, i.e. there is some n Z for which χn / R. If we take a = in 3 we get 9 log Ls, χ = m p ms φq χ X q p m= p m mod q for any s C with σ >. Take s > thus s real in 9; then the right hand side is clearly real and non-negative; hence 30 log Ls, χ 0, s >. χ X q Exponentiating this we get 3 χ X q Ls, χ, s >. Now if there is some complex χ X q for which L, χ = 0, then if χ denotes conjugate of χ viz., χn = χn for all n then we have 3 L, χ = χnn = χnn = L, χ = 0. n= n=

15 ANALYTIC NUMBER THEORY LECTURE NOTES 5 Furthermore χ and χ are different Dirichlet characters, since χ is complex. Hence two of the factors in the product χ X q Ls, χ are zero at s =. We now take the following fact on trust: The function Ls, χ 0 has a meromorphic continuation to σ > 0, with one simple pole at s =. It then follows that also the product χ X q Ls, χ has a meromorphic continuation to σ > 0, and at s = there is one factor which has a simple pole, two factors which are zero, and the other factors are analytic zero or non-zero; hence χ X q Ls, χ has a removable singularity at s = and extends to an analytic function which is zero at s =! In particular we have lim s + χ X q Ls, χ = 0, and this contradicts 3. Hence there cannot exist any complex χ X q with L, χ = 0; viz. we have proved that 8, L, χ 0, holds for all complex χ X q! Before moving on to the case of real χ, we comment on the fact which was taken on trust above: The function Ls, χ 0 has a meromorphic continuation to σ > 0, with one simple pole at s =. To prove this, in view of the formula Ls, χ 0 = p q p s ζs see 5 it suffices to prove the corresponding fact for ζs, i.e. that the function ζs has a meromorphic continuation to σ > 0, with one simple pole at s =. A proof of this will be given in the third lecture, when we study Dirichlet series in more detail, see Example 3.6 below. In fact we will see later that much more is true: ζs has a meromorphic continuation to the whole complex plane, and the only pole is at s =! We should also stress the difference between Ls, χ 0 and Ls, χ with χ X q \ χ 0 : For χ X q \ χ 0 the series Ls, χ = n= χnn s actually converges for all s with σ > 0 although we do not have absolute convergence when 0 < σ, and this can be used to see that Ls, χ is analytic in the whole region σ > 0. By contrast, the series Ls, χ 0 = n= n s does not converge for any s with σ, and it is only by other n,q= means that we are able to show that it has a meromorphic continuation. This difference is also reflected in the fact that Ls, χ 0 has a pole at s =! Case : χ is real, i.e. χn R for all n Z. Then the above argument is inapplicable, since now χ = χ. Suppose that L, χ = 0. We will show that this leads to a contradiction. We now follow Davenport pp Since Ls, χ has a zero at s = and Ls, χ 0 has a simple pole at s =, the product Ls, χls, χ 0

16 6 ANDREAS STRÖMBERGSSON is analytic at s = and therefore analytic for σ > 0. Since Ls, χ 0 is analytic and 0 in the region σ >, the function 33 is analytic for σ >. We also have lim ψs = Ls, χls, χ 0 Ls, χ 0 ψs = 0, s + since lim s + Ls, χ 0 = +. Applying the Euler product formula for the three L- functions we get when σ > : p ψs = χpp s p χ 0pp s 34 p χ 0pp s 35 = p χpp s χ 0 pp s χ 0 pp s. Here χ 0 p = if p q and χ 0 p = 0 if p q, and since χ is real we also know that χp = ± if p q, and χp = 0 if p q. Hence we see that if p q then χpp s χ 0 pp s χ 0 pp s =, and also when χp = we get χpp s χ 0 pp s χ 0 pp s ψs = p χp= p χp= = +p s p s p s p s p s = + p s. p s p p s χp= =. Hence This holds for σ >. If there were no primes with χp = then we conclude ψs = for all s with σ >, and therefore by analytic continuation ψs = for all s with σ >, contradicting the fact lim s + ψs = 0. We have for σ > : 36 ψs = + p s + p s + p s + p 3s +..., and if this is expanded we obtain a Dirichlet series 37 ψs = a n n s n= with absolute convergence for all s with σ > uniform absolute convergence in any compact subset of {s : σ > }, where a = and a n 0 for all n. Here we again used a fact that an infinite product over all primes may be expanded in a formal way. We will prove this in the second lecture; see Example.3 below.

17 ANALYTIC NUMBER THEORY LECTURE NOTES 7 Since ψs is analytic for σ >, it has an expansion in powers of s with a radius of convergence at least 3. This power series is ψs = m! ψm s m. m=0 We can calculate ψ m from the Dirichlet series 37 by termwise differentiation this is ok by Weierstrass Theorem, cf. footnote, and we obtain say, where b m 0. Hence ψ m = m ψs = n= m=0 a n log n m n = m b m, m! b m s m, and this holds for s < 3. If < s < then since all the terms are nonnegative we have ψs ψ, and this contradicts the fact that lim s + ψs = 0. Thus the hypothesis that L, χ = 0 is disproved. This concludes the proof of Dirichlet s Theorem.4. Let us recall what facts we haven t proved completely: Infinite products; manipulating them to prove the Euler product formula Ls, χ = p χpp s the formula for ζs is a special case, and to get to the logarithm, log Ls, χ. Also to see that the product formula 36 can indeed be expanded to give 37 with a = and all a n 0. Dirichlet series: Proving that if χ χ 0 then Ls, χ = n= χnn s converges for all σ > 0 and defines an analytic function in this region. Also proving that ζs has a meromorphic continuation to σ > 0 with one simple pole at s =. Fact about linear combinations of Dirichlet characters: Lemma.5.

18 8 ANDREAS STRÖMBERGSSON. Infinite products.. Infinite products. We review some facts and definitions about infinite products. We borrow from Rudin, Real and Complex Analysis, [60, 5.-5]. Definition.. Suppose {u n } is a sequence of complex numbers, 38 r n = + u + u + u n = and r = lim n r n exists. Then we write 39 r = + u k. k= n + u k The r n are the partial products of the infinite product 39. We shall say that the infinite product 39 converges if the sequence {r n } converges, i.e. if lim n r n exists. In the study of infinite series a n it is of significance whether the a n approach 0 rapidly. Analogously, in the study of infinite products it is of interest whether the factors are or are not close to. This accounts for the above notation: + u n is close to if u n is close to 0. Lemma.. If u,..., u N are complex numbers, and if k= then 40 N N r N = + u n, rn = + u n, n= n= r N e u u N and 4 r N r N. Proof. For x 0 the inequality + x e x is an immediate consequence of e x = + x + x + x Replace x by u! 3!,..., u N and multiply the resulting inequalities. This gives 40. To prove 4 we note that when completely expanding the product r N using the distributive law we get N r N = + u n = u n. n= M {,,...,N} n M

19 ANALYTIC NUMBER THEORY LECTURE NOTES 9 The sum is taken over all subsets M of {,..., N}, and as usual we interpret n u n as. Hence r N =. u n = u n M {,...,N} n M M {,...,N} M Applying the triangle inequality and then mimicking the above computation backwards we get r N N u n = + u n = rn. M {,...,N} M n M Theorem.. Suppose {u n } is a sequence of complex numbers and that n= u n converges. Then the product 4 r = + u n n= converges, and r = 0 if and only if u n = for some n. Furthermore, if {n, n, n 3,...} is any permutation of {,, 3,...} then we also have 43 r = + u nk. k= Definition.. An infinite product satisfying the assumption in the first sentence of Theorem. is said to be absolutely convergent. Proof. Write 44 n= r N = + u + u + u N = n M N + u k, as before. Using Lemma. and the fact that n= u n converges we conclude that r N e C for all N, where C = n= u n <. Choose ε, 0 < ε <. There exists an N 0 such that 45 u n < ε. n=n 0 Let {n, n, n 3,...} be a permutation of {,, 3,...}. If N N 0, if M is so large that 46 k= {,,..., N} {n, n,..., n M },

20 0 ANDREAS STRÖMBERGSSON and if s M denotes the Mth partial product of 43 then 47 s M r N = r N + u nk. k M n k >N Hence, using Lemma., 45 and e ε < ε for 0 < ε <, 48 s M r N r N e ε r N ε e C ε. If n k = k k =,, 3,... then s M = r M, and 48 holds for all M N N 0 ; thus the sequence {r n } is Cauchy, so that the limit r = lim n r n exists. Also 48 shows that 49 r M r N0 r N0 ε for all M N 0, so that r M ε r N0, and this implies that r = 0 if and only if r N0 = 0, which happens if and only if u n = for some n. Finally, 48 also shows that {s M } converges to the same limit as {r N }. Corollary.3. Suppose {u n } is a sequence of bounded complex functions on a set S, such that n= u ns converges uniformly on S. Then the product 50 fs = + u n s n= converges uniformly on S, and fs 0 = 0 at some s 0 S if and only if u n s 0 = for some n. Furthermore, if {n, n, n 3,...} is any permutation of {,, 3,...} then we also have 5 fs = + u nk s. k= Proof. Except for the uniform convergence of 50, all the statements follow directly by applying Theorem. to the sequence {u n s}, for each individual s S. To prove the uniformity in 50, we just have to check that the argument in proof of Theorem. extends in a uniform way to the present setting. This is straightforward: Write N 5 r N s = + u k s. k= Since each function u n is bounded and n= u ns is uniformly convergent, there exists a constant C < such that n= u ns C for all s S. Proof: Since n= u ns is uniformly convergent there is some N 0 such that n>n 0 u n s for all s S. Now N0 n= u ns is a finite sum of bounded functions on S, hence is itself a bounded function on S, i.e. there is some B > 0 such that N 0 n= u ns B for all s S. Now take C = B + ;

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