Analytic Number Theory

Transcription

1 University of Cambridge Mathematics Tripos Part III Analytic Number Theory Lent, 209 Lectures by T. F. Bloom Notes by Qiangru Kuang

2 Contents Contents 0 Introduction 2 Elementary techniques 3. Arithmetic functions Summation Divisor function Estimates for the primes Aside: Why is prime number theorem so hard? Selberg s identity and on elementary proof of prime number theorem 5.5. *A 4-point plan to prove prime number theorem from Selberg s identity Sieve methods Setup Selberg s sieve Combinatorial sieve The Riemann zeta function Dirichlet series Prime number theorem Zero-free region Error terms Functional equation Primes in arithmetic progressions Dirichlet characters and L-functions Dirichlet s theorem Zero-free region Prime number theorem for arithmetric progressions Siegel-Walfisz theorem Highlights of Analytic Number Theory* Gaps between primes Digits of primes Arithmetic progressions Sieve theorey success Number theory without zeta zeros Circle method: additive number theory Inde 70

3 0 Introduction 0 Introduction Analytic number theory is the study of numbers using analysis. In particular it answers quantitative questions. Numbers means natural numbers in this course, which ecludes 0. Eample.. How many primes are there? We know there are infinitely many but can we have a more precise answer? Let π() be the number of primes smaller than or equal to. Then by the famous prime number theorem, π() log. 2. How may twin primes are there? It is not known whethere there are infinitely many. From 204 Zhang, Maynard, Polymath, there are infinitely many primes at most 246 apart. It s been conjectured that the asymptotic bound is (log ) How many primes are there congruent to a mod q where (a, q) =? There are infinitely many by Dirichlet s theorem. The guess is φ(q) log. This is known for small q. The course is divided into four parts:. elementary techniques (using real analysis), 2. sieve methods, 3. Riemann zeta function/prime number theorem (using comple analysis), 4. primes in arithmetic progression. 2

4 Elementary techniques Elementary techniques Review of asymptotic notations: Landau notation: f() = O(g()) if there is C > 0 such that f() C g() for all large enough. Vinogradov notation: f g is the same as f = O(g). f() f g if lim g() =, i.e. f = ( + o())g. f = o(g) if lim f() g() = 0.. Arithmetic functions These are just functions f N C. An important operation for multiplicative number theory is multiplicative convolution Eample. f g(n) = f(a)g(b) ab=n. (n) = for all n. Caution: this is not identity on N. 2. Möbius function μ(n) = { ( )k if n = p p k 0 if n is divisible by a square 3. Liouville function λ(n) = ( ) k if n = p p k where p i s are not necessarily distinct. 4. divisor function τ(n) = #d such that d n = = (n). This is sometimes also denoted by d(n). ab=n Definition (multiplicative function). An arithmetic function f is multiplicative if f(nm) = f(n)f(m) whenever (n, m) =. In particular a multiplicative function is determined by its values on prime powers f(p k ). Fact. If f and g are multiplicative then so is f g. Eample., μ, λ, τ are multiplicative. log n is not multiplicative. 3

5 Elementary techniques Fact (Möbius inversion). f = g if and only if μ g = f. That is, if and only if For eample d n f(d) = g(n) d n g(d)μ( n d ) = f(n). d n μ(d) = { n = 0 otherwise = μ(n) is multiplicative so enough to check the identity for prime powers. If n = p k then {d d n} = {, p,, p k } so LHS equals to = 0 unless k = when LHS equals to μ() =. Our goal is to study primes. Our first might be that we shall work with n prime p (n) = { 0 otherwise as then π() = n p (n). But this is very awkward to work with, as to begin with, this is not multiplicative. Instead, we are going to work almost eclusively with von Mangoldt function log p Λ(n) = { assign weight log p to prime power n n = pk 0 otherwise Lemma.. and Λ = log. μ log = Λ. Proof. The second part follows from Möbius inversion. Thus if n = p k pk r r, Λ(n) = Λ(d) = d n r = = i= r i= k i j= r i= log(p i ) = k i j= r i= log(p k i i ) = log n Λ(p j i ) k i log p i 4

6 Elementary techniques Therefore For eample Λ(n) = μ(d) log( n d ) d n = log n μ(d) μ(d) log d d n d n = μ(d) log d d n Λ(n) = μ(d) log d = μ(d) log d ( n n d n d n,d n by reversing summation. But now the term in the inner summation is very easy to understand: = d = d + O(). Thus n n,d n Λ(n) = d We ll see more of these eamples..2 Summation log d μ(d) d + O ( μ(d) log d). Given an arithmetic function f, we can ask for estimates of n f(n). We say that f has average order g if average size of f is g. Eample. f(n) g(). n d. f = then f() = = + O() n so average order of is. 2. f(n) = n: so average of n is n 2. n 2 2 n ) Lemma.2 (partial summation). If (a n ) is a sequence of comple numbers and f is such that f is continuous. Then n a n f(n) = A()f() A(t)f (t)dt 5

7 Elementary techniques where A() = n a n. This is the discrete analogus of integration by parts. Proof. Suppose = N is an integer. Note that a n = A(n) A(n ), so Now so n N n N a n f(n) = n N = A(N)f(N) f(n)(a(n) A(n )) N A(n)(f(n + ) f(n)) n= n+ f(n + ) f(n) = f (t)dt n N n+ a n f(n) = A(N)f(N) A(n) f (t)dt n n= N = A(N)f(N) A(t)f (t)dt where the last step is because A(n) = A(t) for t [n, n + ). If N = then A()f() = A(N)f() = A(N)f(N) + f (t)dt. N As a simple application Lemma.3. n = log + γ + O( ). n Proof. Partial summation with f() = and a n =, so A() =. Therefore Write t = t {t}, n = + t t n 2 dt = + O( ) + t dt {t} t 2 dt = + O( ) + log {t} t 2 dt + = γ + O( ) + log + O( ) = log + γ + O( ) {t} t 2 dt t 2 dt 6

8 Elementary techniques This is an amazing result and the only thing we did is to replace the discrete summation by the continuous analogue to it. In essence this is the whole reason analytic number theory works. γ can be seen as a measure of the difference between between log and its discrete approimation. It is called Euler-Mascheroni constant. Surprisingly little is known about γ. It is approimately We don t even know if γ is rational or not. Lemma.4. log n = log + O(log ). n Proof. Partial summation with f() = log, a n = so A() =. As a side note, in the previous eample, most error comes from the integral term (the mass is evenly distributed). By constrast in this eample most error comes from the sum term. n t log n = log t dt = log + O(log ) dt + O( t dt) = log + O(log ) + O(log ).3 Divisor function Recall that Theorem.5. τ(n) = (n) = =. ab=n d n τ(n) = log + (2γ ) + O( /2 ) n so in particular average order of τ is log. Proof. First attempt: τ(n) = n n = d = d n d d n,d n = + O() = d d + O() d d = log + γ + O() This is not a very good bound (the error might be as large as one of the terms!) but shows that at least the first term is correct. The main drawback is we used the estimate O() = O(). d 7

9 Elementary techniques To reduce the error term, we use (Dirichlet s) hyperbola trick τ(n) = n = = n ab=n ab a b /a The intuition is like this: n τ(n) counts the number of integral points below the hyperbola k k 2 = in the first quadrant. The old methods amounts to an estimation by integral, while in the new method we count the number of points lying below the line k 2 = /2, add the number of points to the left of k = /2, and finally subtract those points in the bo [0, /2 ] 2 which are double counted. Thus when summing over ab, we can sum over a /2 and b /2 respectively, and then minus pairs a, b. Thus τ(n) = n a /2 b /a + b /2 a /b = 2 a /2 a /2 2 a,b /2 = 2 a /2 a + O(/2 ) + O( /2 ) = 2 log /2 + 2γ + O( /2 ) = log + (2γ ) + O( /2 ) Remark. Improving this O( /2 ) error term is a famous and hard problem. Probably O( /4+ε )? The best result so far is O( ). A note on average order: τ has average order log does not mean τ(n) log n, i.e. average order does not imply individual values. Theorem.6. For all n τ(n) n O( log log n ). In particular τ(n) ε n ε for all ε > 0 where ε means that τ(n) C ε n ε eventually where C ε is a constant depending on ε. As a side note, asymptotic bounds such as log log n are quite common in analytic number theory and here is how to reason with them: as n, log n grows slower than any polynomial, so log log n grows slower than log P (n) for any polynomial. Another way is to write n = e log n and then log log n ) log n = ep(o( log log )). n O( Proof. τ is multiplicative so enough to calculate at prime powers. τ(p k ) = k + so if n = p k pk r r then τ(n) = r (k i= i + ). Let ε > 0 to be chosen later and consider the ratio r τ(n) k n ε = i + p i= k iε. 8

10 Elementary techniques Now entering the trick: split into big and small cases. Note as p goes large, 0. In particular if p 2 /ε then k+ p kε k + p kε k + 2 k. What about small p? It is important to remind ourselves that we re dealing with primes and p can t run below 2. In this case k + p kε k + 2 kε ε this is because for 0 so εh + ε 2 kε if ε 2 (the details are not so important compared to the conclusion that this can be bounded). Therefore τ(n) n ε r k i + p k ( π(2/ε) iε ε ) ( 2 /ε ε ). i=,p i <2 /ε Now we need to choose an optimal ε. Another trick: if we want to minimise f() + g(), choose such that f() = g(). Have τ(n) n ε ε 2/ε = ep(ε log n + 2 /ε log(/ε)). Choose ε such that log n 2 /ε ε log log n and get (again, only a rough guess is needed), i.e. τ(n) n log log n log log n (log log n) 2 = n log log n ep((log n) log 2 log log log n) n O( log log n )..4 Estimates for the primes Recall that π() = #{primes } = ψ() = Λ(n) n p (n) n The second one is sometimes known as Chebyshev s function. Prime number theorem asserts that π() log or equivalently ψ() (this equivalence will be shown later). Although Euclid s prove in 300 BC the infinitude of prime, It was 850 before the correct magnitude of π() was proved. Chebyshev showed that where f g means that g f g. π() log Behold what a wasteful bound we give in the last inequality! But that almost has no effect in the final result. 9

11 Elementary techniques Theorem.7 (Chebyshev). ψ(). Proof. First we ll prove the lower bound, i.e. ψ(). Recall that Λ = log. Here comes in a genuine trick: find something that equals. Then ψ() = n Λ(n) can be rearranged. We ll use the identity = for 0. Either see it directly or a simple verification: if 2 = n + θ where θ [0, ) then 2 = n and = 2n + 2θ = 2n or 2n +. Then ψ() Note that n = m /n, Write d = nm, Λ() ( 2 n 2n ) n = Λ(n) 2 Λ(n) n m /n n = Λ(n) 2 nm d nm /2 d /2 Λ(n) = Λ(d) 2 Λ(d) = log d 2 log d d d /2 m /2n = log + O(log ) 2 ( 2 log 2 + O(log )) 2 = (log 2) + O(log ) For the upper bound, = for (, 2) so ψ() ψ( 2 ) = /2<n< Λ(n) Λ(n) ( 2 n 2n ) n (log 2) + O(log ) Thus ψ() = (ψ() ψ(/2)) + (ψ(/2) ψ(/4)) + log 2 ( + /2 + /4 + ) = 2 log 2 Read unmotivated. 0

12 Elementary techniques Thus we have shown (log 2) ψ() (log 4). Lemma.8. log p = log + O(). p p Proof. Recall that log = Λ so log n = Λ(a) = Λ(a) n ab a b /a = Λ(a) a a = a = a Λ(a) a Λ(a) a + O(ψ()) + O() Note where we used Chebyshev s bound. Since have Λ(n) n n log = log + O(log ), n = log + O(log ) + O() = log + O() Remain to note the contribution from prime powers 2 are small : p n=2 log p p n n=2 = log p p log p = p 2 p p p=2 = O() p 3/2 p n so Λ(n) n = log p + O(). p n p Lemma.9. π() = ψ() log + O( (log ) 2 ).

13 Elementary techniques is equiv- In particular π() alent to ψ(). log Proof. Idea is to use partial summation: let θ() = p and prime number theorem π() log p = π() log π(t) dt. t log First problem: ψ() sums over not only primes but also prime powers. We can use a previous trick to remove contributions from prime powers: Therefore As π(t) For π(t) < t log t, ψ() θ() = log p = θ( /k ) k=2 p k log k=2 ψ( /k ) k=2 /2 log log /k k=2 ψ() = π() log + O( /2 log ) π(t) dt t = π() log + O( /2 log ) + O( log t dt) = π() log + O( log ) t log t, note the trivial bound π(t) t so ψ() = π() log + O( /2 log ) + O() so π() log = O(). Thus we used the trivial bound to get a better bound and use that to do actual work. Lemma.0. where b is some constant. Compare to n n. Proof. Partial summation. Let p = log log + b + O( log ) p log p A() = = log + R() p p where R() = O(). Then (summing from 2 to prevent log t = 0) p = A() log + A(t) t(log t) 2 p 2 2 dt = + O( log ) + 2 t log t dt + 2 R(t) t(log t) 2 dt 2

14 Elementary techniques Note that 2 R(t) t(log t) 2 dt eists, say C. Then p = + C + O( log ) + log log log log 2 + O( t(log t) 2 p 2 dt) = log log + b + O( log ) It turns out b can be epressed in terms of γ. Theorem. (Chebyshev). If π() c log then c =. Note that this does not prove prime number theorem. Historically this is a surprise: a following corollary says that if π() log A() then A. But Legendre and Gauss et al have conjectured that A.08, just by looking up the prime table. Proof. Partial summation on p p : p = π() + p If π() = (c + o()) log then But so c =. π(t) t 2 dt = c log + o( log ) + (c + o()) t log t dt = O( ) + (c + o()) log log log = ( + o()) log log p p Lemma.2. where c is some constant. ( p ) = c log + O() p Proof. We have only dealt with summations so far so take log, log ( p ) = log( p ) p p = kp p k k = p + p k 2 p kp k = log log + c + O( log ). 3

15 Elementary techniques using log( t) = k t k k. To undo the log, note that e = + O() for so p It turns out that c = e γ.78. ( p ) = c log ep(o( log )) = c log ( + O( log )) = c log + O().4. Aside: Why is prime number theorem so hard? It seems that we ve made quite a progress without too much effort. But how far are we from prime number theorem and if the answer is quite far, what makes it so resistant to elementary methods? Probabilistic heuristic: fi p prime, probability that a random n satisfies p n is p. What is the probability that n is prime then? n is a prime if and only if n has no prime divisors p n /2. Guess that the events divisble by p are independent, then probability that n is prime is roughly Thus use some questionable squiggles, ( p ) c log n = 2 /2 c log n. p n /2 π() = n prime 2 c log n 2 c log 2e γ log n n This constant is approimately.22, which contradicts Chebyshev s theorem. Therefore somehow the heuristics is wrong: it gives 2% more prime than should. One reason is that the error terms are so close to the main term that when we do they accummulate and ecees the main term. Another reason is of course that the independence of primes are completely false. From an analytic point of view, this can be seen as saying that the interference terms are not so small that they can be ignored. This may eplain why heuristics don t work. But can we bound π by elementary methods? Recall that μ log = Λ so ψ() = Λ(n) n = μ(a) log b ab = μ(a) ( log b) a b /a Recall that log m = log + O(log ), m 4

16 Elementary techniques but if we just plug this in we will get a trouble. Instead use another trick: consider τ(m) = log + (2ψ ) + O( /2 ). m Thus ψ() = μ(a) ( τ(b) 2γ a + O(/2 )). a/2 a b /a The first term is (essentially μ τ = ) μ(a)τ(b) = μ(a) ab and the first error term is so still need to show that abc = b ac /b = b d /b = = + O() μ(a) μ (d) 2γ μ(a) a = O( μ(a) a ) a a μ(a) = O(). a a Well it turns out that this is equivalent to prime number theorem! This constant can be shown to be /ζ(). As ζ has a pole at z =, this is indeed true..5 Selberg s identity and on elementary proof of prime number theorem Define Selberg s function Λ 2 (n) = μ (log) 2 (n) = μ(a)(log b) 2. ab=n The idea is to prove prime number theorem for Λ 2 with elementary methods. The intuition is that Λ 2 is like Λ multiplied by log and if we do the same epansion as before, hopefully we can get Λ 2 (n) = main term + O(), n but now this is now an acceptable error! Lemma.3.. Λ 2 (n) = Λ(n) log n + Λ Λ(n) Λ 2 (n) (log n) 2. 5

17 Elementary techniques 3. If Λ 2 (n) 0 then n has at most 2 distinct prime divisors. Proof.. Use Möbius inversion suffices to show (Λ(d) log d + Λ Λ(d)) = (log n) 2. d n Start by epanding out, (Λ(d) log d + Λ Λ(d)) = Λ(d) log d + Λ(a)Λ(b) d n d n ab n = log d + Λ(a) Λ(b) d n a n b n a =log(n/a) = log d + Λ(d) log n d d n d n = log n Λ(d) = (log n) 2 2. Λ 2 (n) 0 since both terms on RHS in are nonnegative. Since Λ 2 (d) = (log n) 2, d n Λ 2 (n) (log n) Note that if n is divisible by 2 distinct primes then Λ(n) = 0, and Λ Λ(n) = Λ(a)Λ(b) = 0 ab n since at least one of a or b has 2 distinct prime divisors. d n As such while Λ can be thought as the indicator function for numbers with eactly prime divisor, weighted by log, Λ 2 can be thought as the indicator function for numbers with a pair of prime divisors, weighted by (log) 2. Theorem.4 (Selberg). Proof. n Λ 2 (n) = 2 log + O(). n Λ 2 (n) = μ (log) 2 (n) n = μ(a)(log b) 2 ab = μ(a) ( (log b) 2 ) a b /a 6

18 Elementary techniques By partial summation, (log m) 2 = (log ) 2 2 log O((log ) 2 ). m We want to use the same trick and substitute sum of divisor function for the leading term. First we have to manufacture a (log ) 2 term. By partial summation with A(t) = τ(n) = t log t + Ct + O(t /2 ), n t have τ(m) m = A() + m A(t) t 2 dt = log + C + O( /2 ) + = (log )2 2 log t t + C log + C 2 + O( /2 ) Since we dislike log, we replace it by m τ(m) to get dt + C t dt + O( dt) t3/2 (log ) 2 2 Substituting back, = τ(m) m + C τ(m) + C 2 + O(/2 ). m m so (log m) 2 = 2 τ(m) m + C 3 τ(m) + C 4 + O( /2 ) m m m n τ(b) Λ 2 (n) = 2 μ(a) ab a b /a + C 6 μ(a) a + O( /2 ). a/2 a a + C 5 μ(a) τ(b) a b /a We analyse the error terms one by one, starting from the back. First note that Secondly μ(a) a a /2 = O(). a/2 a = μ(a) a + O() a = μ(a) + O() a b /a = μ (d) + O() d = + O() = O() 7

19 Elementary techniques Thirdly, (again essentially μ τ = ) Collecting what we ve done, μ(a) τ(b) = μ(a) a b /a a b /a n cd=b = μ(a) a cd /a = acd = d e /d = d = O() μ(a) = d ac /d μ (e) τ(b) Λ 2 (n) = 2 μ(a) + O() ab a b /a = 2 d Recall that τ = so μ τ = μ =, = 2 d μ τ(d) + O() d d + O() = 2 log + O() μ(a).5. *A 4-point plan to prove prime number theorem from Selberg s identity Let r() = ψ(). Then prime number theorem is the statement that lim r() = 0. We will demonstrate how to count from to 4 below. When you finished counting, you will get prime number theorem as a byproduct. Show that Selberg s identity implies r() log = n Λ(n) n r( n ) + O(). 2 Consider with replaced by m, summing over m, show r() (log ) 2 Λ 2 (n) r( ) + O(log ). n n n 8

20 Elementary techniques n n Λ 2 (n) = 2 log tdt + O(). Λ 2 (n) n r( n ) = 2 r(/2) dt + O(log ). t log t 7 Let V (u) = r(e u ). Show that 8 Show u u 2 V (u) 2 0 v V (t) dtdv + O(u). 0 u α = lim sup r() lim sup u u 0 V (t) dt = β. 9-4 If α > 0 then can show from 7 that β < α, contradiction. So α = 0. Prime number theorem. 9

21 2 Sieve methods 2 Sieve methods Sieve of Eratosthenes: given natural numbers below 20, let s cross out all multiples of 2 to get Net we cross out all multiples of 3 to get As 20 < 5, we know that the numbers left on the list are prime (with the eception of ). Our interest is in using the sieve to count things: we can find how many numbers are left, which by definition are those primes below 20 that are not used as sieves, by inclusion-eclusion principle: π(20) + π( 20) = = = 7 By the way if there are more sieves then we naturally include more terms in the inclusion-eclusion epansion. Note that the coefficient/sign in front of each term is precisely the Möbius function of the denominator. 2. Setup Consider A N finite, which is the set to be sifted. Let P be a set of primes, which are those we sift out by. Usually P is the set of all primes. Let z be a sifting limit: we sift all primes in P that are smaller than z. A sifting function S(A, P ; z) = n A (n,p (z))= where P (z) = p P,p<z p. The goal is to estimate S(A, P ; z). For d, let A d = {n A d n}. Write A d = f(d) d X + R d where f is completely multiplicative (f(mn) = f(m)f(n) for all m, n) and f(d) 0 for all d. Note that A = f() X + R = X + R Think of R as the remainder term, X is roughly the size of A. Etending this analogy, for general d, R d is the error term and X d measures the number of elements in the 0 residue class of d, assuming they are distributed uniformly. Then f(d) is a factor that says how the residue class is actually distributed. 20

22 2 Sieve methods We choose f so that f(p) = 0 if p P (so R p = A p ). Finally let W P (z) = p P p<z ( f(p) p ), the probability that it is not divisible by any of the p. Eample.. Sieve of Eratosthenes: A = (, + y] N and P is the set of all primes. Then A d = so f(d) = and R d = O(). Have + y d + y = d d d + O() = y d + O() S(A, P ; z) = #{ < n + y p n p z}. For eample if z ( + y) /2 then S(A, P ; z) = π( + y) π() + O(( + y) /2 ). 2. Let A = { n y n = a mod q}. Then A d = { m y dm = a d mod q}. The congruence has solutions if and only if (d, q) a. Thus (d,q) dq y + O((d, q)) A d = { O((d, q)) (d, q) a otherwise So here X = y q and (d, q) f(d) = { (d, q) a 0 otherwise 3. Count twin primes: let A = {n(n + 2) n } and let P be all primes ecept 2. So p n(n + 2) if and only if n = 0 or 2 mod p. Thus A p = 2 p + O(). Thus f(p) = 2. By complete multiplicity, f(d) = 2 ω(d) if 2 d. Have S(A, P ; /2 ) = #{ p p, p + 2 both primes} + O( /2 ) We would epect π 2 () = π 2 () + O( /2 ) (log ) 2. We ll prove upper bound using sieves. 2

23 2 Sieve methods Theorem 2. (sieve of Eratosthenes-Legendre). Proof. S(A, P ; z) = XW P (z) + O( S(A, P ; z) = n A = (n,p (z))= n A d (n,p (z)) = n A d n = = = X = X μ(d) μ(d) n A μ(d) A d p P,p<z μ(d) d n μ(d)f(d) d + R d ). μ(d)r d ( f(p) p ) + O( R d ) Corollary 2.2. π( + y) π() y log log y. By taking = 0 we see this is much worse bound in y than Chebyshev. On the other hand, however, we get a uniform bound independent of! Proof. In eample, X = y, f = and R d. Thus and so by letting z = log y. 2.2 Selberg s sieve W P (z) = ( ) (log z) p p<z π( + y) π() R d 2 z y log z + 2z y log log y y log log y Using sieve of Eratosthenes-Legendre, we only get instead of the epected. What prevents us from getting the result is that we can t take z = y y log y 22

24 2 Sieve methods otherwise the error term will be O(2 z ) = O(2 y ), which is much bigger than the main term. The problem is that we have to consider 2 z many divisors of P (z) so get 2 z many error terms. However, we can design a different sieve, and only consider those divisors which are small, say D. The key part of Eratosthenes-Legendre sieve is (n,p (z))= = μ(d). d (n,p (z)) However, for an upper bound, it is enough to use any function F such that F (n) { n = 0 otherwise Selberg s observation was that if (λ i ) is any sequence of reals with λ = then F (n) = ( λ d ) works. We assume that 0 < f(p) < p for p P, which is a reasonable assumption for the sieve to be nontrivial (if f(p) = 0 then the sieve does nothing and we may well just remove p from P. If f(p) = p then it sifts out everything!) The let us define a new multiplicative function g such that d n g(p) = ( f(p) p ) = f(p) p f(p). 2 Theorem 2.3 (Selberg s sieve). For all t, where S(A, P ; z) X G(t, z) + d P (t) d<t 2 3ω(d) R d G(t, z) = d<t g(d). Recall that W P = p P ( f(p) p ) so epected size of S(A, P ; z) is XW P. p<z Note that as t, G(t, z) g(d) = p<z Let s apply our new machinery: Corollary 2.4. For all, y, ( + g(p)) = π( + y) π() p<z y log y. ( f(p) p ) =. W P 23

25 2 Sieve methods Proof. Let A = { < n + y}, f(p) =, R d = O() and X = y. As g(p) = p = φ(p) so g(d) = φ(d), so the main term G(z, z) = d<z = = (p ) p d p p r <z i r r p k d=p p r <z i= k= i = n d d<z log z y log z. Note that k i = p k pk r r square-free part of n < z 3 ω(d) τ 3 (d) ε d ε from eample sheet so the error term is d P (t) d<t 2 by setting t = z. Thus by taking z = y /3. 3ω(d) R d ε tε d<t 2 t 2+ε = z 2+ε S(A, P ; z) y log z + z2+ε y log y Proof of Selberg s sieve. Let (λ i ) be a sequence of reals with λ =, to be chosen later. Then S(A, P ; z) = n A (n,p (z))= ( n A = d,e P (z) = d,e P (z) = X d,e P (z) d (n,p (z)) λ d λ e n A λ d ) λ d λ e A [d,e] 2 d n,e n λ d λ e f([d, e]) [d, e] + d,e P (z) We ll choose λ d such that λ d and λ d = 0 if d t. Then d,e P (z) λ d λ e R [d,e] d,e<t d,e P (z) R [d,e] 24 n P (z) n<t 2 λ d λ e R [d,e] R n d,e [d,e]=n

26 2 Sieve methods and since n is square-free, so the error term is settled. Now to the main term. Let V = [d,e]=n = 3 ω(n) d,e d,e P (z) f([d, e]) λ d λ e. [d, e] Write [d, e] = acb where d = ac, e = bc and (a, b) = (b, c) = (a, c) =. We also require λ d = 0 if d is not square-free so the last two conditions are automatically satisfied, leaving the notations a bit clearer. V = c P (z) = c P (z) = c P (z) = = = k P (z) f(c) c f(c) c f(c) c ab P (z) (a,b)= ab P (z) μ(d) c P (z) μ(d) c P (z) f(a)f(b) λ ab ac λ bc f(a) f(b) a b μ(d) ( c f(c) ( c f(c) y2 cd d a P (z) c ( μ(d) f(c) ) y2 k cd=k d a,d b cd n P (z) μ(d)λ ac λ bc 2 f(a) a λ ac) 2 f(n) n λ n) write ac = n The term in the brackets is a convolution so we want to simply it. Note that both functions are multiplicative so suffice to work out the primes. For prime p, c μ(d) f(c) = + p f(p) = g(p) cd=p and thus for all k P (z), c μ(d) f(c) = g(k). cd=k Note that if k t then y k = 0. Thus V = k P (z) k<t y 2 k g(k). We want to choose V as small as possible. The idea is to find a lower bound for V and use Cauchy-Schwarz to find the condition on the summands. 25

27 2 Sieve methods Note that we have now epressed V in terms of y k, which is determined by λ d : f(n) y k = n λ n. k n P (z) We would like to invert the relation, so that we can directly control y k. Use a Möbius inversion heuristics, for a fied d, d k P (z) μ(k)y k = k P (z) = n P (z) μ(k) n P (z) f(n) n λ n d k k n f(n) n λ n d n μ(k) d k n For the last summation, note that k = de is square-free so μ(k) = μ(d) μ(e) = { μ(d) n = d 0 n > d d k n by multiplicativity. Thus d k P (z) e n d μ(k)y k = μ(d) f(d) d λ d Thus instead of choosing λ d, we can choose y k to make V small. Recall that λ = so must have Thus = k P (z) k<t k P (z) μ(k)y k =. μ(k)y k g(k) /2 g(k) /2 2 g(k) yk 2 g(k) = GV k P (z) k<t k P (z) k<t where G = G(t, z) by Cauchy-Schwarz, with equaility if and only if there eists c such that for all k, μ(k)y k = cg(k)/2 g(k) /2 i.e. y k = cμ(k)g(k) for k < t. To find c, use the normalisation condition so choose c = G. Check that = c k P (z) k<t μ(k) 2 g(k) = cg 26

28 2 Sieve methods. λ =, 2. λ d = 0 if d t, 3. λ d = 0 if d is square-free (lecturer said this condition is actually not necessary). 4. λ d. Can be checked as follow: Note that Note that for prime p, so G so λ d. d λ d = μ(d) f(d) = d f(d) G G = e P (z) e<t = k d g(e) e P (z) e<t (d,e)=k = g(k) k d g(k) k d g(k) = + k p d k P (z) d k P (z) g(e) m P (z) (m,d)= m<t/k m P (z) (m,d)= m<t/d f(p) p f(p) = μ(k)y k g(k). for fied d g(m) g(m) d f(d) g(d)( g(m)) = d f(d) m P (z) (m,d)= m<t/d p p f(p) = d k P (z) p f(p) g(p) g(k) = λ d G Theorem 2.5 (Brun). Let π 2 () = #{ n n, n + 2 are prime}. Then π 2 () (log ) 2. Proof. Let A = {n(n + 2) n }, P be the the set of all primes ecept 2. Have A d = #{ n d n(n + 2)} 27

29 2 Sieve methods If d = p p r is odd and square-free then d n(n + 2) if and only if p i n(n + ) for all i, if and only if n = 0 or 2 mod p i for all i, and Chinese remainder theorem, if and only if n lies in one of 2 ω(d) many residue classes mod d. Thus A d = 2ω(d) d X + O(2ω(d) ) so f(d) = 2 ω(d) and R d 2 ω(d) for d odd square-free. By Selberg s sieve, with t = z = /4, π 2 () #{ n p n(n + 2) p = 2 or p /4 } + O( /4 ) As before = S(A, P ; /4 ) + O( /4 ) G(z, z) + O( 6ω(d) ) d<z 2 6 ω(d) z 2+O() = /2+O(). d<z 2 To finish the proof need to show G(z, z) (log z) 2. Note that g(2) = 2 and g(p) = so if d is odd and square-free then f(p) p f(p) = 2 p 2 2 p, so G(z, z) g(d) 2ω(d) φ(d). d<z d odd, square-free = d=p p r <z d<z 2 ω(d) d 2 ω(d) φ(d) 2 ω(d) r i= ( p i + p 2 i + ) By partial summation, it s enough to show d<z 2 ω(d) z log z. Recall that to show τ(d) z log z d<z we used τ =. So we need to write 2 ω(n) as a convolution of multiplicative functions. Suppose 2 ω(n) = f(d)g( n d ) d n where f, g are multiplicative. We can actually write down values of f at prime powers: : f() = g() = 28

30 2 Sieve methods p 2 p : 2 = f(p) + g(p) : 2 = g(p 2 ) + f(p 2 ) + f(p)g(p). Let s say try f = τ, so g(p) = 0, g(p 2 ) =, g(p k ) = 0 for k 3. Therefore g(n) = { 0 n not a square μ(d) n = d 2 and Therefore 2 ω(n) = τ(d)g( n d ). d n Note so d<z 2 ω(d) = g(a) τ(b) a<z b z/a = g(a) ( z a log z a + (2γ ) z a + O( z/a)) a<z = a<z g(a) z a log z a + C a<z g(a) z a + O(z/2 ) a<z a /2 = μ(d) z log z 2 μ(d) z log d +O(z) d2 d2 d<z /2 d<z /2 μ(d) d 2 = d<z /2 d= μ(d) d 2 c + d z /2 d 2 = c + O( z /2 ) z d<z /2 log d d 2 z μ(d) d 2 d z /2 2 ω(d) = cz log z + O(z) z log z. d<z Remains to show c > 0. Either note LHS can t be O(z), or calculate the first couple of terms in the series, or note that c = 6 π 2 > Combinatorial sieve Selberg s sieve is an upper bound sieve, while sieve of Eratosthenes uses the inclusion-eclusion principle S(A, P ; z) = A p A p + A pq to get a precise number. However this requires us to keep track of every term, thus resulting in an accummulation of error. The idea of a combinatorial sieve is to truncate the sieve process. p q z 29

31 2 Sieve methods Lemma 2.6 (Buchstab formula). S(A, P ; z) = A Proof. Rearrange, required to show where A = S(A, P ; z) + p P (z) p P (z) S(A p, P ; p). S(A p, P ; p) = S + S = #{n A p n, p P p z} p P (z) S p = #{n A n = mp, q n, q P q p} Every n A is either in the set counted by S or has some prime divisors from P (z). If p is the least such prime divisor then n S p. The S p s are disjoint. Similarly, S p Lemma 2.7. where recall that W (z) = W (z) = p P (z) p P (z) f(p) W (p) p ( f(p) p ). Proof. Eercise. Corollary 2.8. For any r, S(A, P ; z) = ω(d)<r μ(d) A d + ( ) r where l(d) is the least prime divisor of d. ω(d)=r S(A d, P ; l(d)) Proof. Induction on r =. When r = this is just Buchstab s formula. For inductive step, use so S(A d, P ; l(d)) = A d p P p<l(d) S(A dp, P ; p) ( ) r = ω(d)=r ω(d)=r A d p P p<l(d) μ(d) A d + ( ) r+ S(A pd, P ; p) e P (z) ω(e)=r+ S(A e, P ; l(e)) 30

32 2 Sieve methods In particular, if r is even then S(A, P ; z) ω(d)<r and similarly if r is odd we get an upper bound. μ(d) A d Theorem 2.9 (Brun s pure sieve). If r 6 log W(z) then S(A, P ; z) = XW (z) + O(2 r X + R d ). d z r Brun s pure sieve has the same main term as sieve of Erathosthenes, but the error term is split into the fied bit 2 r X and an accummulation part truncated at 2 r. Proof. Recall that from iterating Buchstab s formula S(A, P ; z) = ω(d)<r = X ω(d)<r By the trivial bounds have μ(d) A d + ( ) r μ(d) f(d) d S(A, P ; z) = X + ω(d)=r ω(d)<r S(A d, P ; l(d)) μ(d)r d + ( ) r 0 S(A d, P ; l(d)) A d ω(d)<r μ(d) f(d) d By Buchstab again, applied to W (z), so W (z) = ω(d)<r S(A, P ; z) = XW (z) + O( The error term is ω(d)<r X + O( ω(d)<r μ(d) f(d) d + ( )r ω(d)=r R d + ω(d)=r X ω(d)=r f(d) d f(d) d ω(d)<r ω(d)=r + R d + ω(d)=r A d + X ω(d) r R d ω(d)=r ω(d)=r R d + ω(d)=r S(A d, P ; l(d)) A d ) μ(d) f(d) W (l(d)) d A d + X f(d) d ω(d)=r + R d as d P (z) = P p P d z r p<z f(d) d ) 3

33 2 Sieve methods Remains to show ω(d)=r We need the condition on r. Note that ω(d)=r f(d) d f(d) d 2 r f(p = ) f(p r ) p p r p p r p i P p i <z r r! ( f(p) p ) p P (z) ( e r p P (z) r f(p) p ) Furthermore p P (z) f(p) p so if r 2e log W (z) then ω(d)=r Finally note that 2e < 6. p P (z) f(d) d log( f(p) ) = log W (z) p ( e r log W (z) )r 2 r Recall that Selberg s sieve shows that π 2 () (log ). In the twin prime 2 seive setting, W (z) (log z). So in Brun s sieve, need to take r 2 log log z. If 2 r = C log log z for C large enough then The main term is X (log z) log z. As R d 2 ω(d) = d o(). Thus d z r zr+o() z2 log log z+o() so we need to choose a z. For this to be o( (log z), need to choose z ep((log ) /4 ). 2 So far so good. Now we need to establish the relation between π 2 () and S(A, P ; z) = { n p n(n + 2) p > z} so p /2, so this counts only large primes 32

34 2 Sieve methods Corollary 2.0. For any z ep(o( Remark. log log log )), #{ n p n p z} e γ log z.. In particular, z = (log ) A is allowed for any A, but z = c for any c > 0 is not allowed. 2. In particular, we can t count primes like this as z = /2. Recall heuristic from before says if this asymptotic were correct for primes, then π() 2e γ log which contradicts prime number theorem. This is telling us that for primes, the error term is genuinely large, not because of the estimates. Or in other words, W (z) is not a very good bound, intrinsic Proof. Again use A = { n } so f(d) =, R d. Then so W (z) = ( p ) = e γ log z + o( log z ) p<z S(A, P ; z) = #{ n p n p z} = e γ log z + o( log z + O(2 r + R d ) d<z r if r 6 log W (z), so r 00 log log z is fine. Have and, choose r = 00 log log z, d<z r Remains to note that if then this is so if is large enough. 2 r (log z) (log 2)00 = o( log z ) R d d z r 2 r 2 500(log log z) log z log log z = o( log log ) = log F () log log log log z log log z = o( log log ) = o(log ) log log 2 500(log log z) log z /0 = o( log z ) 33

35 3 The Riemann zeta function 3 The Riemann zeta function As a tradition, in analytic number theory we write s = σ + it for a comple number s where σ and t are the real and imaginary part respectively. First, a trivial remark: if n N then n s = e s log n = n σ e it log n. The Riemann zeta function is defined for σ > by 3. Dirichlet series ζ(s) = n= n s. For any arithmetic f N C, we have a Dirichlet series at least formally. L f (s) = f(n) n i= s, Lemma 3.. For any f there is an abscissa of convergence σ c such that. if σ < σ c then L f (s) diverges, 2. if σ > σ c then L f (s) converges uniformly in some neighbourhood of s. In particular L f (s) is holomorphic at s. Proof. It is enough to show that if L f (s) converges at s 0 and σ = σ 0 then there is a neighbourhood of s on which L f converges uniformly, as then we can take σ c = inf{σ L f (s) converges}. Let By partial summation R(u) = f(n)n s 0. n>u M<n N N f(n)n s = R(M)M s0 s R(N)N s0 s (s 0 s) R(u)u s0 s du. M If R(u) ε for all u M then M<n N N f(n)n s 2ε + ε s 0 s u σ0 σ du (2 + s 0 s σ M 0 σ )ε. Note that there is a neighbourhood of s in which s 0 s σ 0 σ s so f(n) n converges s uniformly here. 34

36 3 The Riemann zeta function Lemma 3.2. If f(n) n s = g(n) n s for all s in some half plane σ > σ 0 R then f(n) = g(n) for all n. Proof. Enough to consider f(n) n = 0 for all σ > σ s 0. Suppose eists n such f(n) that f(n) 0. Let N be the least such that f(n) 0. Since n N n = 0, σ have so f(n) n σ and so the series f(n) = N σ n>n n>n f(n) n σ++ε f(n) n σ is absolutely convergent. So since f(n) n σ 0 as σ, RHS also converges to 0 so f(n) = 0. Lemma 3.3. If L f (s) and L g (s) are both absolutely convergent at s then L f g (s) = f g(n) n n= s is also absolutely convergent at s and equals to L f (s)l g (s). Proof. Because of absolute convergence we can simply multiply them term-byterm: ( n= f(n) n s ) ( g(n) n= n s ) = n,m= f(n)g(m) (nm) s = k= k s ( f(n)g(m)). nm=k Lemma 3.4 (Euler product). If f is multiplicative and L f (s) is absolutely convergent at s then L f (s) = ( + f(p) p p s + f(p2 ) p 2s + ). Proof. Informally we just multiply everything and apply fundamental theorem of arithmetics. However, we have to be more careful when dealing with this infinite product. Let y be arbitrary, ( + f(p) p p<y s + ) = n p n,p<y f(n) n s. 35

37 3 The Riemann zeta function Then as n. ( + f(p) p p<y s For σ >, + ) n= f(n) n s f(n) f(n) n n σ n n y σ 0 p n,p y ζ(s) = n= defines a holomorphic function and converges absolutely for σ >. A word of caution: this series is only define for σ >. We ll in later part of the course analytically etend ζ beyond the line. Also note that for general Dirichlet series, uniform convergence and absolute convergence near a point do not imply each other, although for this particular series they do. Because of uniform convergence this function has derivative ζ (s) = n s log n n s. Since is completely multiplicative, we may apply Euler product so Thus + p s + p 2s + = p s = ( p s ) ζ(s) = p ζ(s) = ( p p s ). ( p s ) = n μ(n) n s log ζ(s) = log( p p s ) = Λ(n) = kpks log n p k ζ (s) ζ(s) = Λ(n) n s n s We can write many functions and identities in terms of ζ(s). For eample ζ (s) ζ(s) ζ(s) = ζ (s) corresponds to and the equivalence Λ = log, L f ζ = L g L f = ζ L g corresponds to Möbius inversion. 36

38 3 The Riemann zeta function A callback to a previous discussion on elementary proof of prime number theorem: recall on page 5, if we can show ζ(s) = p ( p s ) = n μ(n) n s converges to 0 at s = then we can prove prime number theorem. We can show that if it converges at 0 then it does converge to 0, but the difficulty is to show it converges at all! Lemma 3.5. For σ >, Proof. By partial summation, n ζ(s) = + s s {t} dt. ts+ = ns s + s = s + s t dt ts+ t s dt s {t} dt ts+ = s + s s [t s+ ] s s s s {t} dt ts+ {t} dt ts+ as The integral converges absolutely for σ > 0, so this gives ζ(s) = + F (s) s where F (s) is holomorphic in σ > 0. Thus we define ζ(s) = + s s {t} dt ts+ for σ > 0. ζ(s) is meromorphic in σ > 0, with only a simple pole at s =. It is possible to analytically continue ζ to the entire comple plane, with only a single pole at. But for the purpose of this course our definition suffices as all interesting things we study happen on this half plane. Corollary 3.6. For 0 < σ <, σ < ζ(σ) < σ σ. In particular, ζ(σ) < 0 for 0 < σ < (in particular nonzero). 37

39 3 The Riemann zeta function Proof. Write and note ζ(σ) = + σ + σ {t} dt tσ+ {t} 0 < t σ+ dt < σ. Corollary 3.7. For 0 < δ σ 2, t, uniformly. ζ(s) = s + O δ() Proof. ζ(s) s = s {t} t = O() + O δ ( s+ dt = O() + O δ () t σ+ dt) Lemma 3.8. ζ 0 for σ >. Proof. For σ >, ζ(s) = ( p p s ) and the infinite product converges, and no factors are zero. Again we stress that the Euler product is only valid for σ >. Conjecture (Riemann hypothesis). If ζ(s) = 0 and σ > 0 then σ = 2. σ = 2 38

40 3 The Riemann zeta function 3.2 Prime number theorem Let α(s) = a n n s. Partial summation lets us write α(s) in terms of A() = n a n. If σ > ma(0, σ c ) then α(s) = s A(t) dt ts+ This is Mellin transform. What about the converse? As a particular case, if α(s) = ζ (s) ζ(s) then a n = Λ(n) so A() = Λ(n) = ψ(). n The point of analytic number theory is to study Dirichlet series using analytic methods and convert them back to statements about arithmetic functions. The converse is given by Perron s formula, which roughly says that A() = σ+i 2πi σ i α(s) s s ds for σ > ma(0, σ c ). Before we prove the formula, we see how this leads to prime number theorem. By Perron s formula, ψ() = σ+i 2πi σ i ζ (s) s ζ(s) s ds for σ >. We see that the integrand has two poles, on at the origin and the other at. Our first attempt would be to integrate to the right of the critical line as there is no singularity. However we quickly run into problems. As the integrand is holomorphic in this region, by Cauchy s theorem ψ() equals to the contribution of the other segments (up to a sign). The best we can do is ψ() = O( +ε ), which, in view of what we have done, totally trivial. Thus we can t simply consider a contour in σ >. Instead we have to cross the critial line, whih gives O( ε ), which is what we need. But now we need to understand ζ on/to the left of the critial line. 39

41 3 The Riemann zeta function We can summarise this intuition with the slogan prime number theorem is equivalent to the statment that there is no zeroes on σ =. Lemma 3.9. If σ 0 > 0 then σ0+it 2πi σ 0 it Note that we omit the case y =. y s y > ds = { s 0 y < + O( y σ0 T log y ) Proof. Use a rectangular contour that lies either to the left or to the right of the line σ = σ 0 depending on y, which then determines whether the residue at 0 is picked up. The details are left as an eercise. This gives a way to epress indicator function in integral form. Theorem 3.0 (Perron s formula). Suppose α(s) = a n n s is absolutely convergent for σ > σ a. If σ 0 > ma(0, σ a ) and is not an integer then a n = n< 2πi σ0+it + O( 2σ 0 T σ 0 it Proof. Since σ 0 > 0 we can write 2 <n<2 n< = 2πi σ0+it σ 0 it α(s) s s ds a n n + σ0 T (/n) s s n= a n n σ ). 0 ds + O( (/n)σ 0 T log n ) 40

42 3 The Riemann zeta function so n< a n = a n n< n = 2πi n σ 0 +it a n σ 0 it = σ0 it 2πi s s σ 0 +it n = σ0+it 2πi σ 0 it For the error term E, there is (/n) s ds + O( σ 0 s T n a n ds + E ns α(s) s ds + E s a n n σ 0 log n ) E absolute convergence. contribution from n 2 or n 2, where log n, is σ 0 T n a n n σ contribution from 2 < n < 2, we write log n = log( + n ) and log( + δ) δ uniformly for 2 δ. So σ 0 T 2 <n<2 a n n σ 0 log n σ0 T 2 <n<2 a n n σ 0 n 2σ0 T 2 <n<2 a n n We will now prove a strong form of the prime number theorem, assuming c. there eists c > 0 such that if σ > log( t +4) and t 7 8 and ζ (s) log( t + 4). ζ(s) 2. ζ(s) 0 for 8 9 σ, t Whenever t 7 8 and c log( t + 4) < σ 2 then ζ(s) 0 have ζ (s) ζ(s) = s + O(). 4

43 3 The Riemann zeta function σ + it σ 0 + it σ it σ 0 it Theorem 3. (prime number theorem). There eists c > 0 such that In particular ψ(). ψ() = + O( ep(c log ) ). This error is better than log but worse than ε for any ε > 0, which is precisely because we can t find an absolute bound on the zero-free region near the critical line. Proof. Assume that = N + 2 < σ 0 2, for some N. By Perron s formula, for any In the error term, R log T and using assumption 3, ψ() = Λ(n) n = σ0+it 2πi σ 0 it ζ (s) s ζ(s) s ds + O( T Λ(n) n 2 <n<2 2 <n<2 R Λ(n) T n n= σ ) 0 + σ 0 n log T R 2 σ 0 T R 2 m 4 σ 0 log T m (log )2 T if σ 0 = + log. Let C be the rectangular contour with vertices {σ 0 ± it, σ ± it } where σ < is to be chosen later. Then 2πi C ζ (s) ζ(s) s ds = s 42

44 3 The Riemann zeta function by residue theorem and assumption and 2. Remains to bound the other components of the integral. σ +it σ 0 +it ζ (s) s σ ζ(s) s ds log T σ 0 where the last step is because we assumed σ = For the other term, Thus σ +it σ it u T ζ (s) s σ it ζ(s) s ds log T u σ it σ log T + σ σ σ log T log T du T σ (σ σ 0 ) T c log T (?) u du + σ+it σ it ψ() = + O( T (log )2 + log T (log T )) = + O( ep(c log ) ) c σ if T = ep(c log ) σ If you are curious how we chose T, it is the same trick as in chapter : want to have T c log T so log log T log T, i.e. log T log. 3.3 Zero-free region Firstly, near s =, things are easy because of the pole. Theorem 3.2. If σ > +t2 2 then ζ(s) 0. In particular, ζ(s) 0. If 8 9 σ, t 7 8. Also uniformly in 8 9 σ, t 7 8. ζ(s) = s + O() ζ (s) ζ(s) = s + O() Proof. Recall that so ζ(s) ζ(s) = s s + s {u} du us+ s s s s du uσ+ σ 43

45 3 The Riemann zeta function so if σ > s, ζ(s) 0, i.e. if σ < +t2 2. Also ζ(s) s + s du = O() uσ+ so same holds for ζ ζ, by general theory of holomorphic functions. For t large, we need a different idea. How do we show that there aren t zeros on σ =? Suppose there is a zero, of order m, at + it. Then so Absolute value of LHS ζ ζ ( + δ + it) m δ Λ(n) n +δ+it m δ. Λ(n) = ζ n+δ ζ ( + δ) δ so this shows m. If there is a zero, it is a simple zero. This also tells us log p p +δ eit log p log p p +δ so p cos(t log p) for almost all p, so p it, p 2it for almost all p, so there eists a pole at + 2it, which is a contradiction. We now present a rigorous proof. Before that we need to take a detour in comple analysis. Lemma 3.3 (Borel-Carathéodory lemma). If f is holomorphic on z R and f(0) = 0. If Re f(z) M for all z R, then for any r < R, sup( f(z), f (z) ) r,r M. t r If we replace Re f(z) by f(z) then this is just maimum value principle. Proof. Let g(z) = f(z) z(2m f(z)). This is holomorphic in z R. If z = R then 2M f(z) f(z) and so g(z) f(z) R f(z) R. 44

46 3 The Riemann zeta function So for all z r < R, by maimum modules, so so g(z) = f(z) z 2m f(z) < R R f(z) r 2M f(z) 2Mr + r f(z) For f (z), we use Cauchy s formula f(z) 2Mr R r M. for r < r < R. (coefficient 2?) f (z) = πi w =r f(w) (z w) 2 dw Lemma 3.4. If f is holomorphic on a domain including z, f(z) M in that disc, and f(0) 0. If 0 < r < R < then for z r f f (z) = K k= z z k + O r,r (log where z k ranges over all zeros of f in z R. M f(0) ) c.f. fundamental theorem of algebra, and holomorphicity. This depends crucially on C being algebraically closed. Proof. Suppose wlog f(0) =. Say first there are no zeros. Consider h(z) = log f(z) and Re h(z) = log f(z) log M so by Borel-Carathéodory lemma, f h (z) = log M f(z) so done. In general, we define an auillary function g with no zeros. Let K R 2 zz g(z) = f(z) k (z z k )R. k= The kth factor has a pole at z = z k and on z = R, has modulus so on z R, g(z) M. In particular, g(0) = K R k= z k M. Now let h(z) = log g(z) g(0) and Re h(z) = log g(z) log g(0) log M 45

47 3 The Riemann zeta function for z R. By Borel-Carathéodory lemma, so and if z r, and K log M. h (z) = f K K f (z) + z z k z R 2 log M /z k f f (z) = K k= k= K z z k k= k= z R 2 /z k + O(log M) z R2 z k R2 z k z R r Corollary 3.5. If t 7 8 and 5 6 σ 2 then where ρ is over all zeros in ζ ζ (s) = ρ + O(log t ) s ρ ρ ( it) 5 6. Theorem 3.6. There eists c > 0 such that ζ(s) 0 if σ Proof. Assume ζ(ρ) = 0 where ρ = σ + it. Let δ > 0 be chosen later. c log t. ζ ζ ( + δ + it) = + δ + it ρ + + O(log t) + δ + it ρ ρ ρ (assuming that σ is sufficiently close to ). Then Re ζ ζ ( + δ + it) = Re + δ + it ρ + Re + O(log t) + δ + it ρ ρ ρ = + O(log t) + (> 0) + δ σ since Re ρ, Re +δ+it ρ > 0. Thus Re ζ ζ ( + δ + it) > + O(log t) + δ σ Similarly Also Re ζ ( + δ + 2it) > O(log t). ζ ζ ζ ( + δ) = δ + O(). 46

48 3 The Riemann zeta function Here comes the clever bit: Re( 3 ζ ζ ζ ( + δ) 4ζ ( + δ + it) ζ ζ ( + δ + 2it)) < 3 δ 4 + O(log t) + δ σ ( ) As δ 0, is going to be negative. On the other hand, it is a a Dirichlet series Note so 0. So C log t Choose δ = some c > 0.( so so σ c log t.) = Re(3 n = n Λ(n) n +δ + 4 Λ(n) n n +δ+it + n Λ(n) (3 + 4 cos(t log n) + cos(2t log n)) n+δ cos θ + cos(2θ) = 2( + cos θ) δ > 4 + O(log t). + δ σ Λ(n) n +δ+2it ) for large enough C, so get a contradiction if σ c 4 + δ σ < 0 δ It s essentially what we are able to do nowadays. Best known to date is σ c(log log t)/3 (log t) 2/3. log t for Lemma 3.7. If σ > c 2 log t and t 7 8 then ζ (s) log t. ζ Proof. Let s = + log t + it = σ + it. Here Use the corollary so therefore ζ ζ (s Λ(n) ) n n= σ log t. σ ζ ζ (s ) = + O(log t) s ρ ρ Re ρ Now if s = σ + it, where σ > log t. s ρ c 2 log t then ζ ζ (s) ζ ζ (s ) = ( s ρ ) + O(log t) s ρ ρ 47

49 3 The Riemann zeta function Also s ρ s ρ so s ρ s ρ s ρ 2 log t Re s ρ as Re z = Re z z 2. Then ρ s ρ s ρ Re ρ log t. s ρ Assuming the Riemann hypothesis, we can show ψ() = + O( /2 (log ) 2 ). See eample sheet. Using partial summation, we can deduce that where Thus if we write π() = just because of Li. 3.4 Error terms π() = Li() + O ε ( /2+ε ) Li() = log t dt = log + O( (log ) 2 2 ). log + E() then E() (log ) 2 Indeed if we assume Riemann hypothesis then we could get the error term as above. Can we do better? In this section we will show that ψ() / often. Thus apart from the factor (log ) 2 we are getting the best possible error term. The reason is basically that there are many zeros of ζ on the critical line. Actually, we will show that i.e. ψ() = + Ω ± ( /2 ), lim sup lim inf ψ() > 0 /2 ψ() < 0 /2 For contradiction, suppose that ψ() c /2 for all large, so c /2 ψ() = 0. Take Mellin transform of this, 48

50 3 The Riemann zeta function Lemma 3.8 (Landau). Let A() be intergrable and boundedon any finite interval and A() 0 for all X. Let Then if σ c = inf{σ A() σ d < }. X F (s) = A() s d then F is analytic for Re s > σ c and not at s = σ c. General fact about poles of Dirichlet series with positive coefficients. Proof. Divide integrand into [, X] and [X, ), corresponding partition of F = F + F 2. F is entire. If Re s > σ c, the integral converges absolutely so F 2 is analytic. By contradiction, suppose F 2 is analytic at s = σ c. Write F 2 as a Taylor series around σ c + 2 where c k = F (k) 2 (σ c + ) k! F 2 (s) = c k (s σ c ) k k=0 = k! A()( log ) k σc d. This power series has a radius of convergence, which must be + δ for some δ > 0. So F 2 (s) = k=0 ( σ c s) k k! A()(log ) k σ c d Evaluate the series at s = σ c δ 2, we can intercahnge the integral and summation so F 2 (σ c δ 2 ) = A() σ c ep(( + σc s) log )d = A()s s d so the integral converges at σ c δ 2, contradicting the definition of σ c. Theorem 3.9 (Landau). If σ 0 is the supremum of the real parts of then {ρ ζ(ρ) = 0}. for any σ < σ 0, ψ() = Ω ± ( σ ), 2. if there is zero ρ with σ = σ 0 then ψ() = Ω ± ( σ 0 ). 49

51 3 The Riemann zeta function Corollary (Assuming there is a zero with σ = 2, which is indeed true) ψ() = Ω ± ( /2 ). Proof by splitting into cases Riemann hypothesis is true/false. Corollary 3.2. Riemann hypothesis is equivalent to ψ() = + O( /2+o() ). Proof. Let c > 0 be chosen later and suppose that ψ() c σ for all X. Consider F (s) = (c σ ψ() + ) s d. Recall that by partial summation, for Re s >, so ζ ζ (s) = s ψ() s d s = s F (s) = c s σ + σ (s) sζ(s) + s. This has a pole at s = σ and is analytic for Re s > σ. By Landau s lemma, in fact this integral converges for all s with Re s > σ. This proves because if σ < σ 0 then there is a zero of ζ with 0 < Re ρ < σ 0 and at ρ F has a singularity since ρ R. Suppose there is ρ = σ 0 + it 0. Repeat the above with σ = σ 0. Consider instead G(s) = F (s) + eiθ F (s + it 0 ) + e iθ F (s it 0 ) 2 where θ R is to be chosen later. G(s) is still analytic for Re s > σ, and has a pole at s = σ 0. From F (s), have residue. From F (s + it 0 ), have residue m ρ where m is the order of ρ. From F (s it 0 ) have residue m ρ. So G(s) has a pole at s = σ 0 with residue c + eiθ m 2ρ + e iθ m 2ρ = c m ρ by choosing appropriate θ. In particular if c < m ρ then this residue is negative. As s σ 0 from right along R, G(s). But for Re s > σ 0, G(s) = (c σ 0 ψ() = ) s ( + eiθ it 0 + e iθ it 0 ) d 2 2 +cos(θ t 0 log ) 0 so splitting the integral into [, X] and [X, ), G(s) = G (s) + G 2 (s) where G is entire and G 2 (s) 0 as s R, Re s > σ 0. Absurd. This proves Ω is the same. ψ() = Ω + ( σ ). 50