Multiplicative number theory: The pretentious approach. Andrew Granville K. Soundararajan

Transcription

1 Multiplicative number theory: The pretentious approach Andrew Granville K. Soundararajan

2 To Marci and Waheeda c Andrew Granville, K. Soundararajan, 204

3 3 Preface Riemann s seminal 860 memoir showed how questions on the distribution of prime numbers are more-or-less equivalent to questions on the distribution of zeros of the Riemann zeta function. This was the starting point for the beautiful theory which is at the heart of analytic number theory. Until now there has been no other coherent approach that was capable of addressing all of the central issues of analytic number theory. In this book we present the pretentious view of analytic number theory; allowing us to recover the basic results of prime number theory without use of zeros of the Riemann zeta-function and related L-functions, and to improve various results in the literature. This approach is certainly more flexible than the classical approach since it allows one to work on many questions for which L-function methods are not suited. However there is no beautiful explicit formula that promises to obtain the strongest believable results which is the sort of thing one obtains from the Riemann zeta-function. So why pretentious? It is an intellectual challenge to see how much of the classical theory one can reprove without recourse to the more subtle L-function methodology For a long time, top experts had believed that it is impossible is prove the prime number theorem without an analysis of zeros of analytic continuations. Selberg and Erdős refuted this prejudice but until now, such methods had seemed ad hoc, rather than part of a coherent theory. Selberg showed how sieve bounds can be obtained by optimizing values over a wide class of combinatorial objects, making them a very flexible tool. Pretentious methods allow us to introduce analogous flexibility into many problems where the issue is not the properties of a very specific function, but rather of a broad class of functions. This flexibility allows us to go further in many problems than classical methods alone, as we shall see in the latter chapters of this book. The Riemann zeta-function ζs is defined when Res > ; and then it is given a value for each s C by the theory of analytic continuation. Riemann pointed to the study of the zeros of ζs on the line where Res = /2. However we have few methods that truly allow us to say much so far away from the original domain of definition. Indeed almost all of the unconditional results in the literature are about understanding zeros with Res very close to. Usually the methods used to do so, can be viewed as an extrapolation of our strong understanding of ζs when Res >. This suggests that, in proving these results, one can perhaps dispense with an analysis of the values of ζs with Res, which is, in effect, what we do. Our original goal in the first part of this book was to recover all the main results of Davenport s Multiplicative Number Theory [?] MR by pretentious methods, and then to prove as much as possible of the result of classical literature, such as the results in [?]. MR8978 It turns out that pretentious methods yield a much easier proof of Linnik s Theorem, and quantitatively yield much the same quality of results throughout the subject. However Siegel s Theorem, giving a lower bound on L, χ, is one result that we have little hope of addressing without considering zeros of L-functions. The difficulty is that all proofs of his lower bound run as follows: Either the Generalized

4 4 Riemann Hypothesis GRH is true, in which case we have a good lower bound, or the GRH is false, in which case we have a lower bound in terms of the first counterexample to GRH. Classically this explains the inexplicit constants in analytic number theory evidently Siegel s lower bound cannot be made explicit unless another proof is found, or GRH is resolved and, without a fundamentally different proof, we have little hope of avoiding zeros. Instead we give a proof, due to Pintz, that is formulated in terms of multiplicative functions and a putative zero. Although this is the first coherent account of this theory, our work rests on ideas that have been around for some time, and the contributions of many authors. The central role in our development belongs to Halász s Theorem. Much is based on the results and perspectives of Paul Erdős and Atle Selberg. Other early authors include Wirsing, Halász, Daboussi and Delange. More recent influential authors include Elliott, Hall, Hildebrand, Iwaniec, Montgomery and Vaughan, Pintz, and Tenenbaum. In addition, Tenenbaum s book [?] MR36697 gives beautiful insight into multiplicative functions, often from a classical perspective. Our own thinking has developed in part thanks to conversations with our collaborators John Friedlander, Régis de la Bréteche, and Antal Balog. We are particularly grateful to Dimitris Koukoulopoulos and Adam Harper who have been working with us while we have worked on this book, and proved several results that we needed, when we needed them! Various people have contributed to our development of this book by asking the right questions or making useful mathematical remarks in this vein we would like to thank Jordan Ellenberg, Hugh Montgomery. The exercises: In order to really learn the subject the keen student should try to fully answer the exercises. We have marked several with if they are difficult, and occasionally if extremely difficult. The questions are probably too difficult except for well-prepared students. Some exercises are embedded in the text and need to be completed to fully understand the text; there are many other exercises at the end of each chapter. At a minimum the reader might attempt the exercises embedded in the text as well as those at the end of each chapter with are marked with.

5 Contents Part. Introductory results Chapter.. The prime number theorem 3... Partial Summation Chebyshev s elementary estimates Multiplicative functions and Dirichlet series The average value of the divisor function and Dirichlet s hyperbola method The prime number theorem and the Möbius function: proof of Theorem.. PNTM Selberg s formula Exercises Chapter.2. First results on multiplicative functions A heuristic Multiplicative functions and Dirichlet series Multiplicative functions close to Non-negative multiplicative functions Logarithmic means Exercises 2 5

6 Part Introductory results

7 In the the first four chapters we introduce well-known results of analytic number theory, from a perspective that will be useful in the remainder of the book.

8 CHAPTER. The prime number theorem As a boy Gauss determined, from studying the primes up to three million, that the density of primes around x is / log x, leading him to conjecture that the number of primes up to x is well-approximated by the estimate PNT.. πx := x log x. It is less intuitive, but simpler, to weight each prime with log p; and to include the prime powers in the sum which has little impact on the size. Thus we define the von Mangoldt function { log p if n = p m, where p is prime, and m vm..2 Λn := 0 otherwise, and then, in place of.., PNT we conjecture that PNT2..3 ψx := Λn x. The equivalent estimates.. PNT and..3, PNT2 known as the prime number theorem, are difficult to prove. In this chapter we show how the prime number theorem is equivalent to understanding the mean value of the Möbius function. This will motivate our study of multiplicative functions in general, and provide new ways of looking at many of the classical questions in analytic number theory.... Partial Summation Given a sequence of complex numbers a n, and some function f : R C, we wish to determine the value of B a n fn n=a+ from estimates for the partial sums St := k t a k. Usually f is continuously differentiable on [A, B], so we can replace our sum by the appropriate Riemann- Stieltjes integral, and then integrate by parts as follows: PS2..4 A<n B a n fn = B + A + ftdst = [Stft] B A = SBfB SAfA B A B A Stf tdt. Stf tdt Note that..4 PS2 continues to hold for all non-negative real numbers A < B. The notation t + denotes a real number marginally larger than t. 3

9 4.. THE PRIME NUMBER THEOREM In Abel s approach one does not need to make any assumption about f: Simply write a n = Sn Sn, so that B n=a+ a n fn = B n=a+ and with a little rearranging we obtain fnsn Sn, PS..5 B n=a+ B a n fn = SBfB SAfA Snfn + fn. n=a If we now suppose that f is continuously differentiable on [A, B] as above then we can rewrite..5 PS as..4. PS2 Exercise... Use partial summation to show that PNT.. is equivalent to PNT3..6 θx = log p = x + ox; and then show that both are equivalent to PNT2..3. The Riemann zeta function is given by ζs = n s = p n= p s for Res >. This definition is restricted to the region Res >, since it is only there that this Dirichlet series and this Euler product both converge absolutely see the next subsection for definitions. zeta Exercise..2. i Prove that for Res > ζs = s [y] dy = s ys+ s s {y} dy. ys+ where throughout we write [t] for the integer part of t, and {t} for its fractional part so that t = [t] + {t}. The right hand side is an analytic function of s in the region Res > 0 except for a simple pole at s = with residue. Thus we have an analytic continuation of ζs to this larger region, and near s = we have the Laurent expansion ζs = s + γ + c s The value of the constant γ is given in exercise..4. ex:harmonic ii Deduce that ζ + log x = log x + γ + O x log x. iii Adapt the argument in Exercise..5 ex:stirling to obtain an analytic continuation of ζs to the region Res >. iv Generalize.

10 ..2. CHEBYSHEV S ELEMENTARY ESTIMATES Chebyshev s elementary estimates Chebyshev made significant progress on the distribution of primes by showing that there are constants 0 < c < < C with x x Cheb..7 c + o πx C + o log x log x. Moreover he showed that if lim x πx x/ log x exists, then it must equal. The key to obtaining such information is to write the prime factorization of n in the form log n = d n Λd. Summing both sides over n and re-writing d n as n = dk, we obtain that Cheb2..8 log n = Λd = ψx/k. n=dk k= Using Stirling s formula, Exercise..5, ex:stirling we deduce that Cheb3..9 ψx/k = x log x x + Olog x. k= Exercise..3. Use Cheb3..9 to prove that lim sup x ψx x ψx lim inf x x, so that if lim x ψx/x exists it must be. To obtain Chebyshev s estimates..7, Cheb take..8 Cheb2 at 2x and subtract twice that relation taken at x. This yields x log 4 + Olog x = ψ2x ψ2x/2 + ψ2x/3 ψ2x/4 +..., and upper and lower estimates for the right hand side above follow upon truncating the series after an odd or even number of steps. In particular we obtain that ψ2x x log 4 + Olog x, which gives the lower bound of..7 Cheb with c = log 2 a permissible value. And we also obtain that ψ2x ψx x log 4 + Olog x, which, when used at x/2, x/4,... and summed, leads to ψx x log 4+Olog x 2. Thus we obtain the upper bound in..7 Cheb with C = log 4 a permissible value. Returning to..8, Cheb2 we may recast it as log n = Λd = x Λd d + O. d x k x/d d x Using Stirling s formula, and the recently established ψx = Ox, we conclude that x log x + Ox = x Λd d, d x

11 6.. THE PRIME NUMBER THEOREM or in other words Pavg..0 log p p = Λn n + O = log x + O...3. Multiplicative functions and Dirichlet series The main objects of study in this book are multiplicative functions. These are functions f : N C satisfying fmn = fmfn for all coprime integers m and n. If the relation fmn = fmfn holds for all integers m and n we say that f is completely multiplicative. If n = j pαj j is the prime factorization of n, where the primes p j are distinct, then fn = j fpαj j for multiplicative functions f. Thus a multiplicative function is specified by its values at prime powers and a completely multiplicative function is specified by its values at primes. One can study the multiplicative function fn using the Dirichlet series, F s = n= fn n s = p + fp p s + fp2 p 2s The product over primes above is called an Euler product, and viewed formally the equality of the Dirichlet series and the Euler product above is a restatement of the unique factorization of integers into primes. If we suppose that the multiplicative function f does not grow rapidy for example, that fn n A for some constant A then the Dirichlet series and Euler product will converge absolutely in some half-plane with Res suitably large. Given any two functions f and g from N C not necessarily multiplicative, their Dirichlet convolution f g is defined by f gn = fagb. ab=n If F s = n= fnn s and Gs = n= gnn s are the associated Dirichlet series, then the convolution f g corresponds to their product: f gn F sgs = n s. n= The basic multiplicative functions and their associated Dirichlet series are: The function δ = and δn = 0 for all n 2 has the associated Dirichlet series. The function n = for all n N has the associated Dirichlet series ζs which converges absolutely when Res >, and whose analytic continuation we discussed in Exercise..2. zeta For a natural number k, the k-divisor function d k n counts the number of ways of writing n as a a k. That is, d k is the k-fold convolution of the function n, and its associated Dirichlet series is ζs k. The function d 2 n is called the divisor function and denoted simply by dn. More generally, for any complex number z, the z-th divisor function d z n is defined as the coefficient of /n s in the Dirichlet series, ζs z. 2 2 To explicitly determine ζs z it is easiest to expand each factor in the Euler product using the generalized binomial theorem, so that ζs z = p + z k k p s k.

12 ..4. THE AVERAGE VALUE OF THE DIVISOR FUNCTION AND DIRICHLET S HYPERBOLA METHOD7 The Möbius function µn is defined to be 0 if n is divisible by the square of some prime and, if n is square-free, µn is or depending on whether n has an even or odd number of prime factors. The associated Dirichlet series n= µnn s = ζs so that µ is the same as d. We deduce that µ = δ. The von Mangoldt function Λn is not multiplicative, but is of great interest to us. We write its associated Dirichlet series as Ls. Since log n = d n Λd = Λn hence ζ s = Lsζs, that is Ls = ζ /ζs. Writing this as we deduce that ζs ζ s Lammu.. Λn = µ logn = PNTM ab=n µa log b. As mentioned earlier, our goal in this chapter is to show that the prime number theorem is equivalent to a statement about the mean value of the multiplicative function µ. We now formulate this equivalence precisely. Theorem... The prime number theorem, namely ψx = x+ox, is equivalent to Mx..2 Mx = µn = ox. In other words, half the non-zero values of µn equal, the other half. Before we can prove this, we need one more ingredient: namely, we need to understand the average value of the divisor function. PrS4..4. The average value of the divisor function and Dirichlet s hyperbola method We wish to evaluate asymptotically dn. An immediate idea gives dn = = d x d n d n = [ x ] = x d d + O d x d x = x log x + Ox. Dirichlet realized that one can substantially improve the error term above by pairing each divisor a of an integer n with its complementary divisor b = n/a; one minor exception is when n = m 2 and the divisor m cannot be so paired. Since a or n/a must be n we have dn = d n = 2 d n d< n + δ n,

13 8.. THE PRIME NUMBER THEOREM where δ n = if n is a square, and 0 otherwise. Therefore dn = 2 + d n d< n n=d 2 = + 2 and so DD..3 divest..4 dn = 2x d x d x = d x d 2 < d n 2[x/d] 2d +, d x + O x = x log x x + 2γx + O x, by Exercise..4. ex:harmonic The method described above is called the hyperbola method because we are trying to count the number of lattice points a, b with a and b non-negative and lying below the hyperbola ab = x. Dirichlet s idea may be thought of as choosing parameters A, B with AB = x, and dividing the points under the hyperbola according to whether a A or b B or both. We remark that an outstanding open problem, known as the Dirichlet divisor problem, is to show that the error term in..3 DD may be improved to Ox 4 +ϵ for any fixed ϵ > 0. For our subsequent work, we use Exercise..5 ex:stirling to recast..3 DD as log n dn + 2γ = O x. Primes5..5. The prime number theorem and the Möbius function: proof of Theorem.. PNTM First we show that the estimate Mx = µn = ox implies the prime number theorem ψx = x + ox. Define the arithmetic function an = log n dn + 2γ, so that an = Λ n + 2γn. When we form the Dirichlet convolution of a with the Möbius function we therefore obtain µ an = µ Λ n + 2γµ n = Λ n + 2γδn, where δ =, and δn = 0 for n >. Hence, when we sum µ an over all n x, we obtain an = µ Λn + 2γ = ψx x + O. On the other hand, we may write the left hand side above as µdak, dk x

14 PNTM..5. THE PRIME NUMBER THEOREM AND THE MÖBIUS FUNCTION: PROOF OF THEOREM..9 and, as in the hyperbola method, split this into terms where k K or k > K in which case d x/k. Thus we find that akmx/k + µd ak. µdak = dk x k K d x/k Using..4 divest we see that the second term above is = O x/d = Ox/ K. d x/k Putting everything together, we deduce that K<k x/d ψx x = k K akmx/k + Ox/ K. Now suppose that Mx = ox. Fix ϵ > 0 and select K to be the smallest integer > /ϵ 2, and then let α K := k K ak /k. Finally choose y ϵ so that My ϵ/α k y whenever y y ϵ. Inserting all this into the last line for x Ky ϵ yields ψx x ϵ/α k x k K ak /k + ϵx ϵx. We may conclude that ψx x = ox, the prime number theorem. Now we turn to the converse. Consider the arithmetic function µn log n which is the coefficient of /n s in the Dirichlet series /ζs. Since ζ s = ζs ζs 2 = ζ ζ s ζs, we obtain the identity µn log n = µ Λn. As µ = δ, we find that Pr5..5 Λ n = µ µn log n. The right hand side of..5 Pr5 is log x µn + µn logx/n = log xmx + O logx/n = log xmx + Ox, upon using Exercise..5. ex:stirling The left hand side of..5 Pr5 is µa ψx/a x/a. ab x µaλb = a x Now suppose that ψx x = ox, the prime number theorem, so that, for given ϵ > 0 we have ψt t ϵt if t T ϵ. Suppose that T T ϵ and x > T /ϵ. Using this ψx/a x/a ϵx/a for a x/t so that x/a > T, and the Chebyshev estimate ψx/a x/a x/a for x/t a x, we find that the left hand side of..5 Pr5 is ϵx/a + x/a ϵx log x + x log T. a x/t x/t a x Combining these observations, we find that Mx ϵx + x log T log x ϵx, if x is sufficiently large. Since ϵ was arbitrary, we have demonstrated that Mx = ox.

15 0.. THE PRIME NUMBER THEOREM Selberg..6. Selberg s formula The elementary techniques discussed above were brilliantly used by Selberg to get an asymptotic formula for a suitably weighted sum of primes and products of two primes. Selberg s formula then led Erdős and Selberg to discover elementary proofs of the prime number theorem. We will not discuss these elementary proofs of the prime number theorem here, but let us see how Selberg s formula follows from the ideas developed so far. Theorem..2. We have log p 2 + log plog q = 2x log x + Ox. pq x Proof. We define Λ 2 n := Λn log n + lm=n ΛlΛm. Thus Λ 2n is the coefficient of /n s in the Dirichlet series ζ ζ 2 ζ s + ζ s ζ s = ζs, so that Λ 2 = µ log 2. In the previous section we exploited the fact that Λ = µ log and that the function dn 2γ has the same average value as log n. Now we search for a divisor type function which has the same average as log n 2. By partial summation we find that log n 2 = xlog x 2 2x log x + 2x + Olog x 2. Using Exercise..4 k-div we may find constants c 2 and c such that 2d 3 n + c 2 dn + c = xlog x 2 2x log x + 2x + Ox 2/3+ϵ. Set bn = log n 2 2d 3 n c 2 dn c so that the last two displayed equations give Pr6..6 bn = Ox 2/3+ϵ. Now consider µ bn = Λ 2 n 2dn c 2 c δn, and summing this over all n x we get that bn = µ Λ 2 n 2x log x + Ox. The left hand side is µk x/k 2/3+ϵ x bl k x l x/k k x by..6, Pr6 and we conclude that Λ 2 n = 2x log x + Ox. The difference between the left hand side above and the left hand side of our desired formula is the contribution of the prime powers, which is easily shown to be x log x, and so our Theorem follows.

16 ..7. EXERCISES ex:harmonic..7. Exercises Exercise..4. i Using partial summation, prove that for any x [x] x = log x + n x {t} t 2 dt. ii Deduce that for any x we have the approximation log x + γ n x, where γ is the Euler-Mascheroni constant, N γ := lim N n log N = n= {t} t 2 dt. ex:stirling Exercise..5. i For an integer N show that zeta2 log N! = N log N N + + N {t} t dt. ii Deduce that x logx/n x 2 log x for all x. iii Using that x {t} /2dt = {x}2 {x}/2 and integrating by parts, show that N {t} t dt = 2 log N N 2 {t} {t} 2 dt. iv Conclude that N! = C NN/e N { + O/N}, where C = exp {t} {t} 2 2 t 2 dt. In fact C = 2π, and the resulting asymptotic for N!, namely N! 2πNN/e N, is known as Stirling s formula. Exercise..6. i Prove that for Res > 0 we have N N n s dt t s = ζs s + s n= t 2 N {y} dy. ys+ ii Deduce that, in this same range but with s, we can define { N } ζs = lim N n s N s. s ex:bertrand Exercise..7. Using that ψ2x ψx+ψ2x/3 x log 4+Olog x, prove Bertrand s postulate that there is a prime between N and 2N, for N sufficiently large. ex:cheb Exercise..8. i Using..8, Cheb2 prove that if Lx := log n then n= ψx ψx/6 Lx Lx/2 Lx/3 Lx/5 + Lx/30 ψx.

17 2.. THE PRIME NUMBER THEOREM ii Deduce, using Cheb3..9, that with κ = log log log 5 5 log = , we have κx + Olog x ψx 6 5 κx + Olog2 x. iii Improve on these bounds by similar methods. Pavg+ Exercise..9. i Use partial summation to prove that if Λn lim exists, N n exmertens exmertens2 N n N n N then the prime number theorem, in the form ψx = x + ox, follows. ii Prove that the prime number theorem implies that this limit holds. iii Using exercise..2, zeta prove that ζ /ζs ζs has a Taylor expansion 2γ + c s +... around s =. iv Explain why we cannot then deduce that Λn Λn lim = lim n s + n s, which exists and equals 2γ. n Exercise..0. i Use..0 Pavg and partial summation show that there is a constant c such that = log log x + c + O. p log x ii Deduce Mertens Theorem, that there exists a constant γ such that e γ p log x. In the two preceding exercises the constant γ is in fact the Euler-Mascheroni constant, but this is not so straightforward to establish. The next exercise gives one way of obtaining information about the constant in Exercise..0. exmertens Exercise... In this exercise, put σ = + / log x. i Show that log p>x p σ = p>x ii Show that log log p σ p p σ + O e t = x t dt + O. log x = 0 e t t dt + O. log x iii Conclude, using exercise..2, zeta that the constant γ in exercise..0ii exmertens equals e t e t dt 0 t t dt. That this equals the Euler-Mascheroni constant is established in [?]. HW

18 ex2.7 k-div ex:mobiusequiv Exercise..2. Uniformly for η in the range p y..7. EXERCISES 3 log y y η p η log/η + O logy η. η <, show that Hint: Compare the sum for the primes with p η to the sum of /p in the same range. Use upper bounds on πx for those primes for which p η. Exercise..3. If f and g are functions from N to C, show that the relation f = g is equivalent to the relation g = µ f. Given two proofs. This is known as Möbius inversion. Exercise..4. i Given a natural number k, use the hyperbola method together with induction and partial summation to show that d k n = xp k log x + Ox /k+ϵ where P k t denotes a polynomial of degree k with leading term t k /k!. ii Deduce, using partial summation, that if R k t + R k t = P kt then d k n logx/n = xr k log x + Ox /k+ϵ. iii Deduce, using partial summation, that if Q k u = P k u + u t=0 P ktdt then d k n = Q k log x + O. n Analogies of these estimates hold for any real k > 0, in which case k! is replaced by Γk. Exercise..5. Modify the above proof to show that i If Mx x/log x A then ψx x xlog log x 2 /log x A. ii Conversely, if ψx x x/log x A then Mx x/log x min,a. MobPNT Exercise..6. i Show that ii Deduce that Mx log x = lim inf x Mx x log p Mx/p + Ox. + lim sup x iii Use Selberg s formula to prove that ψx x log x = log p ψ Mx x = 0. x x + Ox. p p iv Deduce that lim inf x Compare! ψx x x + lim sup x ψx x x = 0.

19

20 CHAPTER.2 First results on multiplicative functions C2 We have just seen that understanding the mean value of the Möbius function leads to the prime number theorem. Motivated by this, we now begin a more general study of mean values of multiplicative functions. S A heuristic In Section..4 PrS4 we saw that one can estimate the mean value of the k-divisor function by writing d k as the convolution d k. Given a multiplicative function f, let us write f as g so that g is also multiplicative. Then gd = [ x ] gd. d d x fn = Since [z] = z + O we have E2..2. fn = x d x d n gd d + O d x gd. In several situations, for example in the case of the k-divisor function treated earlier, the remainder term in.2. E2. may be shown to be small. Omitting this term, and approximating d x gd/d by + gp/p + gp2 /p we arrive at the following heuristic: E fn x Pf; x where is interpreted as is roughly equal to, and E Pf; x = + gp p + gp2 p = + fp + fp2 p p p In the special case that 0 fp fp 2... for all primes p so that gd 0 for all d, one easily gets an upper bound of the correct order of magnitude: If f = g then gd 0 for all d by assumption, and so fn = [ x ] gd gd x x Pf; x d d d x d x as in.2.3. E2.3 In the case of the k-divisor function, the heuristic.2.2 E2.2 predicts that p k xe γ log x k, d k n x 5

21 6.2. FIRST RESULTS ON MULTIPLICATIVE FUNCTIONS which is off from the correct asymptotic formula, xlog x k /k!, by only a constant factor see exercise..4i. k-div Moreover d k p j d k p j for all p j so this yields an unconditional upper bound. One of our aims will be to obtain results that are uniform over the class of all mutiplicative functions. Thus for example we could consider x to be large and consider the multiplicative function f with fp k = 0 for p x and fp k = for p > x. In this case, we have fn = if n is a prime between x and x and fn = 0 for other n x. Thus, the heuristic suggests that πx π x + = fn x x e γ p log x 2e γ x log x. p x Comparing this to the prime number theorem, the heuristic is off by a constant factor again, this time 2e γ... This heuristic suggests that the sum of the Möbius function, Mx = µn is comparable with x p 2 xe 2γ log x 2. However Mx is known to be much smaller. The best bound that we know unconditionally is that Mx x exp clog x 3 5 ϵ see chapter??, ch:strongpnt and we expect Mx to be as small as x 2 +ϵ as this is equivalent to the unproved Riemann Hypothesis. In any event, the heuristic certainly suggests that Mx = ox, which is equivalent to the prime number theorem, as we saw in Theorem... PNTM S Multiplicative functions and Dirichlet series Given a multiplicative function fn we define F s := n and now define the coefficients Λ f n by F s F s = Λ f n n s. n fn n s as usual, Comparing the coefficient of /n s in F s = F s F s/f s we have ConvolEqNew.2.4 fn log n = Λ f dfn/d. d n LambdaF Exercise.2.. Let f be a multiplicative function. and fix κ > 0 ex2. i Show that Λ f n = 0 unless n is a prime power. ii Show that if f is totally multiplicative then Λ f n = fnλn. iii Show that Λ f p = fp log p, Λ f p 2 = 2fp 2 fp 2 log p, and that every Λ f p k equals log p times some polynomial in fp, fp 2,..., fp k. iv Show that if Λ f n κλn for all n, then fn d κ n. Exercise.2.2. Suppose that f is a non-negative arithmetic function, and that F σ = n= fnn σ is convergent for some σ > 0. i Prove that fn xσ F σ. ii Moreover show that if 0 < σ < then fn + x fn n xσ F σ. n>x

22 .2.3. MULTIPLICATIVE FUNCTIONS CLOSE TO 7 S2.2 pr2. This technique is known as Rankin s trick, and is surprisingly effective. The values fp k for p k > x appear in the Euler product for F σ and yet are irrelevant to the mean value of fn for n up to x. However, for a given x, we can take fp k = 0 for every p k > x, to minimize the value of F σ above Multiplicative functions close to The heuristic.2.2 E2.2 is accurate and easy to justify when the function g is small in size, or in other words, when f is close to. We give a sample such result which will lead to several applications. Proposition.2.. Let f = g be a multiplicative function. If gd d σ = Gσ d= is convergent for some σ, 0 σ, then fn xpf x σ Gσ, where Pf := Pf;, and lim x x fn = Pf. Proof. The argument giving.2. E2. yields that fn x gd gd. d d x d x Since Pf = d gd/d we have that gd d Pf d x d>x gd. d Combining these two inequalities yields SweetBound.2.5 fn xpf gd + x d x d>x gd d. We now use Rankin s trick: we multiply the terms in the first sum by x/d σ, and in the second sum by d/x σ >, so that the right hand side of.2.5 SweetBound is d x x σ gd + x d d>x gd d σ d = x σ Gσ, x the first result in the lemma. This immediately implies the second result for 0 σ <. One can rewrite the right hand side of.2.5 SweetBound as x gn n dt = o x x, 0 n>t because n>t gn /n is bounded, and tends to zero as t. This implies the second result for σ =.

23 8.2. FIRST RESULTS ON MULTIPLICATIVE FUNCTIONS sec:non-neg.2.4. Non-negative multiplicative functions Let us now consider our heuristic for the special case of non-negative multiplicative functions with suitable growth conditions. Here we shall see that right side of our heuristic.2.2 E2.2 is at least a good upper bound for fn. Prop2. Proposition.2.2. Let f be a non-negative multiplicative function, and suppose there are constants A and B for which eq:sumps.2.6 Λ f m Az + B, cor2.3 m z for all z. Then for x e 2B we have A + x fn log x + B Proof. We begin with the decomposition fn n fn logx/n fn log x = fn log n + fn log n + x fn n, which holds since 0 log t t for all t. For the first term we have fr Λ f m fn log n = frλ f m n=mr r x Ax fr r + B. r x The result follows by combining these two inequalities. m x/r Proposition.2.2 Prop2. establishes the heuristic.2.3 E2.3 for many common multiplicative functions: Corollary.2.3. Let f be a non-negative multiplicative function for which either 0 fn for all n, or Λ f n κλn for all n, for some given constant κ >. Then E x fn A,B Pf; x exp Moreover if 0 fn for all n then lim fn = Pf. x x fp. p Proof. The hypothesis implies that.2.6 eq:sumps holds: If fn then this follows by exercise.2.5iii. ex:whichhypo If each Λ f n κλn then the Chebyshev estimates give that Λ f n κ Λn Az + B, n z n z any constant A > κ log 4 being permissible.

24 .2.4. NON-NEGATIVE MULTIPLICATIVE FUNCTIONS 9 So we apply Proposition Prop2..2.2, and bound the right-hand side using Mertens Theorem, and fn n + fp + fp2 p p , to obtain the first inequality. The second inequality then follows from exercise.2.6 SizePf,x with ϵ = 2. If p fp/p diverges, then E shows that lim x x fn = 0 = Pf. Suppose now that p fp/p converges. If we write f = g then this condition assures us that p gp k /p k converges, which in turn is equivalent k to the convergence of ex2.0 n gn /n by exercise.2.7. The second statement in Proposition.2. pr2. now finishes our proof. In the coming chapters we will establish appropriate generalizations of Corollary cor For example, for real-valued multiplicative functions with fn, Wirsing proved that fn Pfx. This implies that µn = ox and hence the prime number theorem, by Theorem... PNTM We will go on to study Halász s seminal result on the mean values of complex-valued multiplicative functions which take values in the unit disc. Proposition.2.2 Prop2. also enables us to prove a preliminary result indicating that mean values of multiplicative functions vary slowly. The result given here is only useful when f is close to, but we shall see a more general such result in Chapter C20??. FirstLip Proposition.2.4. Let f be a multiplicative function with fn for all n. Then for all y x we have x fn y x /y fn logey log x exp fp. p Proof. Write f = g, so that g is a multiplicative function with each gp = fp, and each Λ g p = Λ f p Λp so that.2.6 eq:sumps holds by exercise ex:whichhypo.2.5iii. Recall that fn gd gd, x d x d x d x so that FirstLip.2.8 x fn y x /y fn gd + y x x d x d x/y Appealing to Proposition Prop we find that for any z 3 gn n z z gn log z n n z z log z exp p z gd + x/y<d x fp. p gd. d

25 20.2. FIRST RESULTS ON MULTIPLICATIVE FUNCTIONS From this estimate and partial summation we find that the right hand side of.2.8 FirstLip is proving our Proposition. logey log x exp fp, p NasProp.2.5. Logarithmic means In addition to the natural mean values x fn, we have already encountered logarithmic means log x fn/n several times in our work above. We now prove the analogy to Proposition.2. pr2. for logarithmic means: Proposition.2.5 Naslund. Let f = g be a multiplicative function and d gd d σ = Gσ < for some σ [0,. Then fn n Pf log x + γ Λ f n Λn n n xσ σ Gσ. Proof. We start with fn n = gd = n d n d x and then, using exercise..4, ex:harmonic we deduce that fn n gd log x d d + γ d x d x gd d gd d m x/d m d x = x gd. Since gn log n is the coefficient of /n s in G s = Gs G /Gs, thus gn log n = g Λ g n, and we note that Λ f = Λ + Λ g. Hence n gn log n n = a,b gaλ g b ab Therefore gd d d log x d + γ the error term in our main result is gd + x d x d>x = Pf m = Pf log x + γ n gd d log x. d + γ d x Λ f m Λm. m Λ f n Λn n, and so Since / σ we can use the inequalities x/d σ x/d σ / σ for d x, and logx/d + γ + logd/x d/x σ σ for d > x, to get a bound on the error term of xσ σ Gσ as claimed. G0UB Proposition.2.6. If f is a multiplicative function with fn for all n, then fn exp Refp. log x n 2 p

26 .2.6. EXERCISES 2 Proof. Let g = f, so that gn = fd = x fd d + O = x fd d + Ox. d n d x d x We deduce, applying Proposition.2.2 Prop2. since.2.6 eq:sumps is satisfied as Λ g = Λ + Λ f, and then by exercise.2.5iii, ex:whichhypo that fn gn + O log x n x log x log x gn log 2 + x n log x exp + fp 2 + p log x using Mertens theorem. Now 2 Rez 2 + z Rez whenever z, and so the result follows. SumCompare We expect that, for non-negative real multiplicative functions f, the quantity Rf; x := / fn + fp + fp2 n p p , should typically be bounded, based on the heuristic discussion above. For example Rd κ ; x e γ κ / Γκ + by exercise..4iii k-div and Mertens Theorem. Exercise.2.3. Suppose that f and g are real multiplicative functions with fn, gn 0 for all n. i Prove that 0 Rf; x. ii Prove that Rf; x Rf; x Rg; x Rf g; x. iii Deduce that if f is totally multiplicative and 0 fn for all n then Rf; x R; x e γ. iv Suppose that f is supported only on squarefree integers that is, fn = 0 if p 2 n for some prime p. Let g be the totally multiplicative function with gp = fp for each prime p. Prove that Rf; x Rg; x Exercises Exercise.2.4. Prove that if f. is multiplicative with fp k for each prime power p k then lim x Pf; x exists and equals Pf ex:whichhypo Exercise.2.5. i Prove that if fp k B k for all prime powers p k then Λ f p k 2 k B k log p for all prime powers p k. ii Show this is best possible Hint: Try fp k = B k. iii Show that fn for all n then there exist constant A, C such that m z Λ f m Az + C, for all z. iv Give an example of an f where B >, for which Λ f n x +δ B. This explains why, when we consider f with values outside the unit circle, we prefer working with the hypothesis Λ f n κλn rather than fp k B.

27 22.2. FIRST RESULTS ON MULTIPLICATIVE FUNCTIONS SizePf,x Exercise.2.6. i Let f be a real-valued multiplicative function for which there exist constants κ and ϵ > 0, such that fp k d κ p k p k 2 ϵ for every prime power p k. Prove that Pf; x κ,ϵ exp fp. p This should be interpreted as telling us that, in the situations which we are interested in, the values of fp k with k > have little effect on the value of Pf; x. ii Show that if, in addition, there exists a constant δ > 0 for which + fp + fp2 p p δ for every prime p then Pf; x κ,δ,ϵ exp fp. p ex2.0 ex:prop2. ex2.2 ex:near ex2.3 ex2.5 iii Prove that if Λ f n Λn for all n then the above hypotheses hold with κ =, ϵ = 2 and δ = 4. Exercise.2.7. Show that if g. is multiplicative then n gn /nσ < if and only if p gp k /p kσ <. k Exercise.2.8. Deduce, from Proposition.2. pr2. and the previous exercise, that if p fp k fp k /p k < then k fn xpf as x. Exercise.2.9. For any natural number q, prove that for any σ 0 we have ϕq q x x σ + p σ. p q n,q= Taking σ = 0, we obtain the sieve of Eratosthenes bound of 2 ωq. Prove that the bound is optimized by the solution to p q log p/pσ + = log x, if that solution is 0. Explain why the bound is of interest only if 0 σ <. Exercise.2.0. Suppose that f is a multiplicative function close to, that is fp k fp k k+r 2p k r for all prime powers p k, for some integer r 0. Prove that fn = xpf + Olog x r+. Hint: Use Proposition pr2..2. with σ = 0, the Taylor expansion for t r and Mertens Theorem. Exercise.2.. Let σn = d n d. Prove that µn 2 σn ϕn = 5 π 2 x + O x log x. Exercise.2.2. Let f be multiplicative and write f = d k g where k N and d k deontes the k-divisor function. Assuming that g is small, as in Proposition pr2..2., develop an asymptotic formula for fn. Where ωq denotes the number of distinct primes dividing q.

28 ex:pr2.rightconstant.2.6. EXERCISES 23 Exercise.2.3. Fix κ > 0. Assume that f is a non-negative multiplicative function and that each Λ f n κλn. i In the proof of Proposition.2.2, Prop2. modify the bound on fn logx/n using exercise..4 k-div, to deduce that for any A > κ, x fn A fn log x κ log x + O n + O Γκ ii Deduce that x fn κeγ + o Pf; x + Olog x κ 2. The bound in i is essentially best possible since exercise..4 k-div implies that d κ n κ x d κ n log x n. ex:weightlx/n Exercise.2.4. Let f be a multiplicative function with each fn. i Show that fn log x n = x t= t n t fn dt. ii Deduce, using Proposition.2.4, FirstLip that fn log x n fn x log x exp fp. p