Math 680 Fall A Quantitative Prime Number Theorem I: Zero-Free Regions

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1 Math 68 Fall 4 A Quantitative Prime Number Theorem I: Zero-Free Regions Ultimately, our goal is to prove the following strengthening of the prime number theorem Theorem Improved Prime Number Theorem: There is a positive c < such that π li + O epc log, where li log d This result will actually follow from an improved estimate on the function Ψ introduced earlier Theorem Improved Estimate for Psi: There is a positive c < such that Ψ + O epc log Corollary: With the same constant c as above, we have Θ + O epc log Proof: By the definitions we have Ψ p m m log p m Θ /m In the handout on Chebyshev s estimates we proved Θ 4 log for all Further, Chebyshev s

2 estimates implied that π log Using these facts, we get Ψ Θ m Θ /m Θ / + m 3 Θ /m / + m 3 Θ /m / + m 3 / + p /3 < / + p /3 p /m log p log / + π /3 log / + /3 / log p 3 m log / log p Since Ψ Θ, we get Θ Ψ + O / The Corollary now follows from the improved estimate for Psi Proof of Improved Prime Number Theorem: We have { log n if n is a prime, Θn Θn otherwise Using this and the Corollary above, Since π p Θn Θn log n log n + log n + log n + O log n + O Θn n Θn n log n Θ[] [] Θ + log[] log epc + O n n+ log epc log n n epc + O log epc log ulog u du n log n n log u du logn Θn n log n logn + ulog u du

3 for n 3, we see that so that []+ log u du [] log n log u du li + O log n Since / log u log u for u e, we easily see that 3 epc log ulog u du epc log u du u / epc log uu / epc log epc log / du du u/ Combining -3 yiel our result Recall that a crucial point in the proof of the prime number theorem was that the zeta function is non-vanishing on the line σ It should come as no surprise that our improved estimate for Psi, which lea to the improved prime number theorem, will depend on a stronger non-vanishing result for the zeta function Indeed, the major part of our proof for the improved estimate of Psi will hinge on the following Theorem : There is a constant c > such that ζs for all s σ + it with σ c log4 + t Our proof of Theorem involves several intermediate steps Lemma Jensen s Inequality: Let R > and suppose F s is analytic on the closed disk {s: s R} Suppose further that F and that F s M for all s in this closed disk Then for all r < R, the number of zeros of F on the closed disk {s: s r}, counted with multiplicity, is no greater than log M log F log R log r Proof: The set of zeros of F on the closed disk {s C: s R} is necessarily finite, since otherwise it would contain an accumulation point, which in turn would imply that F is identically Let ρ,, ρ k denote the zeros of F, written with multiplicity, on the closed disk of radius R We may assume that k >, since otherwise we re done Now we set Gs F s k j 3 R sρ j Rs ρ j

4 By construction, this function is analytic on the closed disk {s C: s R} Further, for any s Re iθ on the boundary of this disk and all j,, k we have R sρ j Rs ρ j R eiθ ρ j Re iθ ρ j R e iθ ρ j R e iθ ρ j Thus G M on the boundary of this disk, so that G M by the maimum modulus principle Now suppose there are L zeros ρ j with ρ j r, counted with multiplicity Then since r < R, Taking logarithms gives the result M G F F j k ρ j r R r R ρ j L j k ρ j >r Lemma Borel-Carathéodory: Suppose F s is analytic on the closed disk {s: s R} for some R >, F and R F s M for all s in the closed disk Then for all real positive r < R and all R ρ j comple s with s r, and F s Mr R r F s Mr R r Proof: By Cauchy s integral formula, π π F Re iθ dθ F Re πiθ dθ πi F ss F, where the curvilinear integral is the circle {s: s R} oriented in the positive counter-clockwise direction set s Re πiθ More generally using the same s and integral path, for all integers l F Re πiθ e πilθ dθ R l πi F ss l and F Re πiθ e πilθ dθ Rl πi Using this, for any integer l and any real φ, F ss l Rl F l l! R l F l e πiφ l! F Re πiθ + e πiφ e πilθ + e πiφ e πilθ dθ F Re πiθ + Re πiφ+lθ dθ F Re πiθ + cosπφ + lθ dθ 4

5 Taking real parts of both sides, we get R l l! R e πiφ F l M + cos πφ + lθ dθ M for all positive integers l We choose φ such that e πiφ F l F l to see that 4 F l l! M R l all l Now since F is analytic in the closed disk and F, it has a power series representation F s F l s l l! l valid for all s with s < R Now assuming that s r < R we have by 4 F s l F l s l l! M l r/r l Mr R r Similarly for the derivative, F s l F l s l l! M R lr/r l Mr R r l Lemma 3: Suppose F s is analytic on the closed disk {s: s }, F s M on this disk and F Then for all r and R with < r < R < and all s with s r, d log F s k s ρ j O log M log F, j where ρ,, ρ k are the zeros of F with multiplicity on the closed disk {s: s R} and the implicit constant depen on r and R Proof: As noted in the proof of Jensen s Inequality, the set of zeros of F on the closed disk {s: s } is finite By possibly replacing R with a slightly larger number, we may assume that F s for all s with s R Applying Jensen s Inequality with R there equal to here and r there equal to R here, we see that 5 k log M log F log R log M log F Letting Gs F s k j R sρ j Rs ρ j 5

6 as in the proof of Jensen s Inequality, we see that G is analytic on the closed disk {s: s R}, Gs on this disk, and Gs M on this disk G F k j R F, ρ j Now the function Hs log Gs log G is analytic on the closed disk of radius R, H and RHs log Gs log G log M log F We apply the Borel-Carathéodory lemma to the function Hs, getting 6 H s for all s with s r On the other hand, we have H s d log Gs log M log F r R r log M log F d log F s d log F s + + k j k j d logr sρ j k j k s R /ρ j s ρ j j d log R + logs ρ j In addition, if s r we have s R /ρ j R r for all j,, k, so that by 5 and 6 d log F s k s ρ j k H s s R /ρ j j H s + j k j s R /ρ j log M log F + log M log F Lemma 4: For all s σ + it with 5/6 σ 3/6 and t 7/8, d log ζs s ρ O log t + 4, ρ k R r where the sum is over all zeros ρ of ζ satisfying ρ 3/ + it 5/6 with multiplicity, and the implicit constant is absolute Proof: Fi a t R with t 7/8 We will apply Lemma 3 to the function F s ζs + 3/ + it, with R 5/6 and r /3 Since the only pole of ζs is s, the only pole of F s is s / + it, which is not in the closed disk of radius since / + t 65/64 Thus F s is analytic on the closed disk {s : s } We also have F ζ 3/ + it 6 p 3/ it

7 by the Euler product valid here since the real part of the input is greater than Thus by Lemma 3 7 d log F s k s ρ j O log M log F, j where ρ,, ρ k are the zeros of F s with multiplicity on the closed disk {s : s 5/6} and M is the supremum of F s on the disk {s : s } We note that any s σ + it with 5/6 σ 3/6 and t t can be written as a sum s s + 3/ + it for some s with s /3 Any zero ρ j of F s is of the form ρ j ρ j 3/ it, where ρ j is a zero of ζs and s ρ j s ρ j Since ρ j is in the closed disk of radius R 5/6, ρ j 3/ + it 5/6 We also have d log F s d log ζs d log ζs Therefore the lemma follows from 7 once we get an appropriate upper bound for F s on {s : s } and lower bound for F We have σ Rs + 3/ + 3/ / for all s with s In particular, we may use the representation 8 F s ζs s n n+ n s We saw above that s 65/64 This implies that s, so that 9 Net, the Dirichlet series n from the Dirichlet Series handout we have s n+ n s has abscissa of convergence Since Rs / >, by Theorem n+ n s n s n n+ s+ d Hence n+ n s s n s s s n σ+ d 3/ d n+ s+ d σ + t 4 + t 7

8 Via 8-, we see that we may use log4 + t in place of log M in 7 All that remains is to bound F away from But that is relatively simple A rather crude estimate gives n 3/+it + n 3/ for all n Thus by eercise 8 F Using this estimate in 7 completes the proof Lemma 5: For all σ > and all t 3R p 3/+it σ 4R ζσ σ + it R ζσ + it + p 3/ ζ 3/ ζ3/ σ + it ζσ + it Proof: Recall from our proof of the theorem of Hademard and de la Vallée Poussin that log ζs n a n n s, where In other wor, { k if n p k for some prime p, a n otherwise log ζs n Λn log nn s This is absolutely convergent when Rs >, so by results from the handout on Dirichlet series we may differentiate this term-by-term to get Using this, we get ζ σ 3R ζσ 4R ζ s ζs σ + it R ζσ + it d log ζs Λn n s n σ + it Λn cost log n + cost log n ζσ + it n σ n n Λn n σ + cost log n Proof of Theorem : We already proved that ζs whenever σ in our proof of the prime number theorem By the Euler product, ζs whenever σ > Therefore it suffices to only consider possible zeros with real part less than We first consider the simplest case of zeros ρ with Iρ 7/8 The function { s ζs if s, F s : if s 8

9 is analytic, thus uniformly continuous on the compact set S {s σ + it: 5/6 σ, t 7/8} Set m inf t 7/8 { F + it } > ; by the uniform continuity above there is an ɛ > such that F s F s < m whenever s, s S with s s < ɛ In particular, letting Rs here gives ζσ + it all σ ɛ/, t 7/8 Suppose now that ρ σ + it is a zero of ζ with 5/6 σ < and t 7/8 if there are no such zeros, we re already done by Let δ > to be determined later and set s + δ + it We apply Lemma 4 and get ζ s ζs ρ s ρ c log t + 4 for some constant c independent of s and ρ, where the sum is over zeros ρ of ζ with ρ 3/+it 5/6 We note that R/s ρ Rs ρ/ s ρ > since / + δ + Rρ > for all of these zeros, one of which is ρ Thus the above inequality easily implies that + δ + it R R + c log t + 4 ζ + δ + it s ρ ρ + c log t δ σ The same argument with s + δ + it gives + δ + it 3 R c log t + 4 ζ + δ + it We previously saw that both ζs s and ζ s + s are analytic in the entire comple Thus there is a positive constant c such that 4 ma { ζ + δ + δ, ζ + δ δ } c for all positive δ We also have ζ + δ ζ > Since lim σ +σ ζσ, there is a positive constant c 3 such that δζ + δ c 3 for all positive δ Using these inequalities together with 4 yiel + δ R δ ζ + δ ζ + δ ζ + δ δ ζ + δδ ζ + δδ ζ + δδ 5 ζ + δ δ δ + δ δ ζ + δ ζ + δδ ζ + δ + δ ζ + δ c + c /c ζ + δ δ δζ + δ

10 We now combine, 3 and 5 to get 3R + δ 4R ζ + δ + δ + it R ζ + δ + it + δ + it ζ + δ + it 3 δ 4 + δ σ + 3c + 3c /c 3 + 4c log t c log t + 4 Since c and log t + 4 > log 4 >, we may set δ c log t + 4 With this choice of δ, the above inequality together with Lemma 5 gives 4 + c log t + 4 σ 3 log t c + 3c /c 3 + 4c log t c log t + 4, clearly implying that σ c log t + 4 for some positive constant c This together with completes the proof Theorem : Let c be the constant in the statement of Theorem and set c min{c/, /6} For all s σ + it with σ > c / log t + 4 and t 7/8 we have ζ s ζs log t + 4, where the implicit constant is absolute If σ > c / log t + 4 and t < 7/8, then ζ s ζs s + O Proof: We note that the c in Theorem is no greater than the ɛ in Thus ζ s/ζs is analytic for all s with σ c/ and t 7/8 We ve seen that ζ s/ζs has a simple pole of residue at s Thus there is an M such that ζ s ζs + s M for all s with t 7/8 and c/ σ, say Therefore we may assume from now on that either t > 7/8 or σ > Suppose that σ > Then as in the proof of Lemma 5 ζ s ζs n n Λn n s Λn n σ ζ σ ζσ

11 Now using 5 gives ζ σ ζσ σ In particular, the theorem is true whenever σ + / log t + 4 which clearly includes the case where σ > It remains to deal with the case where t > 7/8 and σ < + / log t + 4 Suppose s σ + it with c / log t + 4 < σ < + / log t + 4 and t > 7/8 and set s σ + it with t t and σ + / log t + 4 Since c / log t + 4 < /6 we may apply Lemma 4 to both s and s ; we get ζ s ζs s ρ O log t + 4 ρ 6 ζ s ζs s ρ ρ O log t + 4, where the sums are over all zeros ρ of ζ satisfying ρ 3/ + it 5/6, written with multiplicity Since the theorem hol for s, we get 7 ζ s ζs, log t + 4 s ρ ρ By construction we have s s +c// log t +4 By Theorem we must have s ρ c/ log t +4 for all ρ, and since all ρ have Rρ < we must have s ρ / log t + 4 Thus for all ρ s ρ s ρ s ρ + s s s ρ + s s s ρ and Therefore s ρ s ρ s ρ + s s s ρ + s s s ρ s ρ s ρ ρ ρ ρ ρ ρ s s s ρ s ρ / log t + 4 s ρ Rs ρ s ρ R s ρ log t + 4 by 7 Combining this with 6 and 7 completes the proof