On the splitting conjecture in the hybrid model for the Riemann zeta function
aa r X i v : . [ m a t h . N T ] F e b ON THE SPLITTING CONJECTURE IN THE HYBRID MODELFOR THE RIEMANN ZETA FUNCTION
WINSTON HEAP
Abstract.
We show that the splitting conjecture in the hybrid model of Gonek–Hughes–Keating holds to order on the Riemann hypothesis. Our results are validin a larger range of the parameter X which mediates between the partial Euler andHadamard products. We also show that the asymptotic splitting conjecture holdsfor this larger range of X in the cases of the second and fourth moments. Introduction
The moments of the Riemann zeta function have been the subject of several con-jectural methods in recent years. Since the second and fourth moments of Hardy–Littlewood [23] and Ingham [33], it is only relatively recently that a full conjecturefor all moments was given. This began with the work of Keating–Snaith [34] whoused the now famous connection with random matrix theory to conjecture that forreal k > − /
2, 1 T Z TT | ζ ( + it ) | k dt ∼ a ( k ) g ( k )(log T ) k where(1) a ( k ) = Y p (cid:18) − p (cid:19) k X m > d k ( p m ) p m and(2) g ( k ) = G ( k + 1) G (2 k + 1)where G is the Barnes G -function. This was preceded by conjectures for the 6th and8th moments due to Conrey–Ghosh [14] and Conrey–Gonek [15], respectively, usingnumber theoretic methods. The Keating–Snaith conjecture has since been derivedwith various different approaches [13, 18, 22].A drawback of Keating and Snaith’s method was that the arithmetic factor a ( k )had to be incorporated in an ad-hoc way since there was no input from primes intheir random matrix theory model. This was remedied in the method of Gonek–Hughes–Keating (G–H–K) [22] which forms the main focus of this paper. The first step of G–H–K’s method was to express the zeta function as the productof partial Euler and Hadamard products. Precisely, Theorem 1 of [22] states that for2 X t / and large t ,(3) ζ ( + it ) = P X ( + it ) Z X ( + it ) (cid:16) O (cid:16) X (cid:17)(cid:17) where P X ( s ) = exp (cid:18) X n X Λ( n ) n s log n (cid:19) , Z X ( s ) = exp (cid:18) − X ρ U (( s − ρ ) log X ) (cid:19) and U ( z ) = Z ∞ u ( x ) E ( z log x ) dx with E ( z ) = R ∞ z e − w dw/w and u ( x ) a smooth, non-negative function of mass 1 withsupport in [ e − /X , e ]. To give a rough idea of these objects, note that from thesupport conditions on u we have U ( z ) ≈ E ( z ). This has mass concentrated in theregion z ≪ E ( z ) ≈ − γ − log z . Thus, roughlyspeaking, Z X ( s ) ≈ Q |ℑ ( s ) −ℑ ( ρ ) |≪ / log X (( s − ρ ) e γ log X ). Also, from the definition ofthe von Mangoldt function and the Taylor series for the logarithm we find P X ( s ) ≈ Q p X (1 − p − s ) − . Therefore, we can indeed view P X ( s ) and Z X ( s ) as partial Eulerand Hadamard products.G–H–K then proceeded to compute the moments of the Euler product, showingthat for X ≪ (log T ) − ǫ ,(4) 1 T Z TT | P X ( + it ) | k dt ∼ a ( k )( e γ log X ) k , k ∈ R . They conjectured with random matrix theory that(5) 1 T Z TT | Z X ( + it ) | k dt ∼ g ( k ) (cid:16) log Te γ log X (cid:17) k , k > − / k = 1 , X ≪ (log T ) − ǫ . In order to recoverthe Keating–Snaith conjecture they assumed that the moments of the product of P X and Z X should split as the product of moments. Conjecture 1 (Splitting conjecture, [22]) . Let
X, T → ∞ with X ≪ (log T ) − ǫ .Then for fixed k > − / we have T Z TT | P X ( + it ) Z X ( + it ) | k dt ∼ T Z TT | P X ( + it ) | k dt · T Z TT | Z X ( + it ) | k dt. HE SPLITTING CONJECTURE 3
Their reasoning behind this conjecture was that since P X and Z X oscillate atdifferent scales (1 / log X vs. 1 / log T ), their contributions should act independentlyand hence the moment should split to leading order. They verified this in the cases k = 1 , X ≪ (log T ) − ǫ . The methodology of the hybrid model has since beenused in various different settings to acquire conjectures for all sorts of L -functions[1, 8, 9, 10, 11, 19, 25]. In all cases, an equivalent version of the splitting conjectureplays a key role.In this paper we prove that the splitting conjecture holds to order on the Riemannhypothesis (RH). Furthermore, we can extend the range of X past (log T ) − ǫ . Theorem 1.
Assume RH. Let ǫ, k > be fixed and suppose X, T → ∞ with X (log T ) θ k − ǫ where θ k = 2 p / | k | . Then T Z TT | P X ( + it ) Z X ( + it ) | k dt ≍ T Z TT | P X ( + it ) | k dt · T Z TT | Z X ( + it ) | k dt. As mentioned, this holds in a range of X larger than originally conjectured. We canalso extend the range of X in the asymptotic results (4) and (5), both unconditionallyand on RH. This gives the following. Theorem 2.
The Splitting conjecture holds for k = 1 , in the range X (log T ) (log T ) . Assuming RH, we may take X ( (log T ) √ − ǫ when k = 1 , (log T ) √ − ǫ when k = 2 . Our proofs utilise the recent developments in the theory of moments of L -functionsdue to Soundararajan [43], Harper [24] and Radziwi l l–Soundararajan [38]. Thesetechniques were originally geared for upper bounds although they can be brought tobear on lower bounds too [27]. We highlight three main ideas.The first is an innovation of Soundararajan [43]. This was to note that log | ζ ( + it ) | can be bounded from above by a sum over primes alone since the zeros contributenegatively to this quantity (see Lemma 6 below and c.f. formula (3)). With this, | ζ ( + it ) | can be bounded from above by an Euler product of flexible length.The second idea can be found in a paper of Radziwi l l [37] and features heavily inthe later works of Harper [24] and Radziwi l l–Soundararajan [38]. It allows one tocompute moments of Euler products provided one can restrict to a certain subset of[ T, T ]. For the purposes of this discussion we consider the exampleexp (cid:16) X p Y p − / − it (cid:17) HE SPLITTING CONJECTURE 4 with Y = T / (log log T ) . On the face of it, this is a very long Dirichlet polynomial.However, if we can restrict t to a subset of [ T, T ] on which | P p Y p − / − it | V fora given V , then we can truncate the exponential series effectively using the fact that(6) e z ∼ V X j =0 z j j !for | z | V and large V . The choice of V is naturally dictated by the variance:setting V = P p Y p − ∼ log log T we get a Dirichlet polynomial of length Y V = T / log log T . This is now short and so the mean square is easily computed. Also, theexceptional set in this case is of small measure.The final main input in the arguments of Harper and Radziwi l l–Soundararajanallows one to push the length of the prime sum up to Y = T θ , for some fixed θ > L -functions [21, 26, 27], value distributionof L -functions [16, 30, 39], sign changes in Fourier coefficients of modular forms [35],non-vanishing of central values of L -functions [17] and equidistribution of latticepoints on the sphere [32]. In our case, we use these ideas to prove the following. Proposition 1.
Let ǫ > and k ∈ R be fixed. Suppose X η k (log T ) (log T ) with η k = k − ǫ . Then T Z TT | P X ( + it ) | k dt ∼ a ( k )( e γ log X ) k where a ( k ) is given by (1) . Assuming RH, this holds for X (log T ) θ k − ǫ with θ k = 2 p / | k | . Proposition 2.
Suppose X (log T ) (log T ) . Then for k = 1 , we have T Z TT | Z X ( + it ) | k dt ∼ g ( k ) (cid:18) log Te γ log X (cid:19) k where g ( k ) is given by (2) . Assuming RH we may take X (log T ) √ − ǫ when k = 1 and X (log T ) √ − ǫ when k = 2 . HE SPLITTING CONJECTURE 5
Proposition 3.
Assume RH and let ǫ, k > be fixed. Suppose X (log T ) θ k − ǫ with θ k as above. Then T Z TT | Z X ( + it ) | k dt ≍ (cid:18) log T log X (cid:19) k . Remark.
The lower bound in Proposition 3 can be made unconditional provided X η k (log T ) (log T ) . We say more on this in section 7.Since R TT | ζ ( + it ) | k dt is ≪ T (log T ) k on RH [24] and ≫ T (log T ) k uncondi-tionally [27], Theorem 1 follows from Propositions 1 and 3 when combined with (3).Likewise, Theorem 2 follows on combining Propositions 1 and 2 and (3).Using the case of P X as an example, we describe how the range of X can beincreased past (log T ) − ǫ . First of all, note that since(7) (cid:12)(cid:12)(cid:12) X n X Λ( n ) n / it log n (cid:12)(cid:12)(cid:12) (1 + o (1)) 2 X / log X , we can approximate P X (1 / it ) k with a Dirichlet polynomial of length X | k | X / / log X by using (6) to truncate the exponential. If X ≪ (log T ) − ǫ then this is T o (1) andso we have a short Dirichlet polynomial. Note this holds for all t ∈ [ T, T ] since thebound (7) is pointwise. G–H–K computed a Dirichlet polynomial approximation ina slightly different way, although in order for it to be short they required the samebound on X , perhaps unsurprisingly.If X is larger, then in order to have a short Dirichlet polynomial we must restrictto a subset of [ T, T ] and in this case we need good bounds on the exceptional set.Typically, one would expect Gaussian bounds of the shape(8) 1 T µ (cid:16)n t ∈ [ T, T ] : (cid:12)(cid:12)(cid:12) ℜ X n X Λ( n ) n / it log n (cid:12)(cid:12)(cid:12) > V o(cid:17) ≪ exp (cid:16) − V log log T (cid:17) in a wide range of V . In practice we are limited to V ≪ p (log T )(log T ) / log X which may be much smaller than the maximum 2 X / / log X . For the remainingrange of V one must settle for weaker bounds. For example, in [43] it is shown thatthe tails of log | ζ (1 / it ) | can be bounded by e − V log V when V ≫ log T log T . Wecan show that the tails of our sum satisfy the same bound in the range log T log T V X / / log X provided X ≪ (log T ) . However, for our purposes the weakerbound of e − AV with large A is sufficient and this affords us slightly more room inthe size of X .Another avenue for improvement is to reduce the trivial bound in (7). This be-comes a manageable task under RH and thus we are able to make further gains inthe size of X under this assumption. We shall prove the following. HE SPLITTING CONJECTURE 6
Theorem 3.
Assume RH. Then for large t ∈ [ T, T ] and T ) X T wehave (cid:12)(cid:12)(cid:12) ℜ X n X Λ( n ) n / it log n (cid:12)(cid:12)(cid:12) ( + o (1)) (cid:0) log( X / log T ) + 4 log log X (cid:1) log T log log T .
For the imaginary part we can replace the factor + o (1) by π + o (1) . The factor of 1 / o (1) here is related to the function S ( t ) and can be read as2 c where c is a permissible constant in the bound S ( t ) ( c + o (1)) log t/ log log t .The current best is due to Carneiro–Chandee–Milinovich [12] who give c = 1 /
4. Ourapproach to Theorem 3 is to relate the sum with S ( t ) via contour integrals and theninput these bounds. We have not made attempts to further optimise this argumentbut it would be interesting to see if one could use the extremal function machineryof Carneiro et al. in a more direct way.Regarding further improvements in the size of X , if the conjectural bound S ( t ) ≪ p log t log t of Farmer–Gonek–Hughes [20] holds, then one could take X exp( C (log t ) / ). Also, assuming that the bounds for the exceptional set in (8), or someminor variant of this , hold in the full range of V for a given X , then our argumentscan reproduce Theorems 1 and 2 for X as large as T /C log log T . This supports theview of G–H–K that the splitting conjecture may hold as long as X = o ( T ).The paper is organised as follows. We first prove Theorem 3 in section 2 and thenthe asymptotic results of Propositions 1 and 2 in sections 3 and 4, respectively. Insection 5 we describe some tools for later use. Then in section 6 we prove the upperbound of Proposition 3 and in sections 7 and 8 we prove the lower bound in the cases0 k k >
1, respectively.
Acknowledgements.
The author would like to thank Jing Zhao and Junxian Li fortheir comments on an early draft of this paper and Chris Hughes for some clarifyingremarks. 2.
Bounds for prime sums: Proof of Theorem 3
We first give a lemma which relates our prime sum to S ( t ). This shares somesimilarities with previous convolution formulas from the literature [42, 45]. Lemma 1.
Assume RH. For large t ∈ [ T, T ] and X T , Y T / , we have X n X Λ( n ) n / it log n = Z t + Yt − Y S ( y ) 1 − X − i ( t − y ) t − y dy + E ( X, Y, T ) In fact, anything of the form e − AV with large A would be sufficient. HE SPLITTING CONJECTURE 7 where E ( X, Y, T ) ≪ X / (log X + log T log X ) Y + log T log T (cid:16) Y / + X C/ log T Y X> (log T ) A (cid:17) and A is a large constant.Proof. By Perron’s formula (Lemma 3.19, [44]) we have X n X Λ( n ) n / it log n = 12 πi Z / / log X + iY / / log X − iY log ζ ( z + + it ) X z dzz + O (cid:16) X / log XY (cid:17) . We shift the contour to the line with real part ℜ ( z ) = 1 / log X . Restricting Y T / / C/ log log τ σ τ ≫
1, wehave log ζ ( σ + iτ ) ≪ (log τ ) − σ log τ log τ . This follows from (14.14.5) of [44], the Phragmen–Lindel¨of principle and the boundlog ζ ( σ + iτ ) ≪ log τ , σ > ζ ( σ + iτ ) ≪ log τ log τ log (cid:16) σ − /
2) log τ (cid:17) , < σ + C/ log τ. Therefore, the horizontal contours contribute(9) ≪ X C/ log T log( log X log T ) log TY log T log X X> (log T ) A + X / log TY log T max(log( X / (log T ) ) ,
1) + X / log TY log X .
By formula (14.10.5) of [44] (see also (14.12.4) there) we havelog ζ ( z + + it ) = i Z t + Yt/ − Y S ( y ) z + i ( t − y ) dy + O (cid:16) log TT (cid:17) . This implies(10) 12 πi Z / log X + iY / log X − iY log ζ ( z + + it ) X z dzz = i Z t + Yt/ − Y S ( y ) I ( t − y ) dy + O (cid:16) (log T ) T (cid:17) . where I ( t − y ) = 12 πi Z / log X + iY / log X − iY X z z ( z + i ( t − y )) dz. A trivial estimate gives(11) I ( t − y ) ≪ Z Y − Y dx (1 / log X + | x | )(1 / log X + | x + t − y | ) ≪ log X. HE SPLITTING CONJECTURE 8
On the other hand, shifting the contour to the left we find(12) I ( t − y ) = 1 − X − i ( t − y ) i ( t − y ) | t − y | Y + 1 i ( t − y ) | t − y | >Y + O (cid:16) Y | Y − | t − y || log X (cid:17) . For a given δ > Z | t − y ± Y | δ S ( y ) I ( t − y ) dy ≪ δ log X log T / log T since S ( τ ) ≪ log τ / log τ . In the integral over the remaining region, the error termof (12) contributes ≪ log TδY log X log T + log Y log TY log X log T after considering the regions δ < | Y − | t − y || < | Y − | t − y || separately.Therefore, on choosing δ = 1 / ( Y / log X ) we find that the integral on the right of(10) is i Z t/ − Y y t + Y | t − y ± Y | >δ S ( y ) (cid:18) − X − i ( t − y ) i ( t − y ) | t − y | Y + 1 i ( t − y ) | t − y | >Y (cid:19) dy + O (cid:16) log TY / log T (cid:17) . Integrating by parts along with the bound S ( τ ) ≪ log τ / (log τ ) , we find that thesecond term in this integral is ≪ log T / ( Y (log T ) ) which can be absorbed into theerror term immediately above. The range of integration of the first term can beextended to | t − y | Y at the cost of an error ≪ log T / ( Y log T ). Combining thisin (10) along with the error terms of (9) the result follows. (cid:3) Proof of Theorem 3.
We apply Lemma 1 with Y = X / (log X ) ǫ log T .
With this choice we have E ( X, Y, T ) = o (log T / log log T ) since log X ≫ log T .Therefore, on taking real parts in Lemma 1 we find ℜ X n X Λ( n ) n / it log n = − Z Y − Y S ( t + y ) 1 − cos( y log X ) y dy + o (log T / log T ) . From [12] we have | S ( τ ) | (1 + o (1)) 14 log τ log log τ HE SPLITTING CONJECTURE 9 for large τ and so (cid:12)(cid:12)(cid:12) ℜ X n X Λ( n ) n / it log n (cid:12)(cid:12)(cid:12) (1 + o (1)) log T T Z Y log X − Y log X − cos y | y | dy. The integral here is2 Z Y log Xπ − cos yy dy + O (1) Y log X X n =1 πn Z π ( n +1) πn (1 − cos y ) dy + O (1)=2 log( Y log X ) + O (1) . Thus, we acquire (cid:12)(cid:12)(cid:12) ℜ X n X Λ( n ) n / it log n (cid:12)(cid:12)(cid:12) (1 + o (1)) log( Y log X ) log T T and the result follows on inputting our choice of Y .In the case of the imaginary part, we acquire the integral R Y − Y | sin( y log X ) /y | dy which, on following the same argument, is (4 /π ) log( Y log X ) + O (1). (cid:3) From the proof we see that the factor log( X / / log T ) comes from the divergentintegral R Y − Y | − X − iy | / | y | dy . One may then wonder if smoothing would help here,that is, if the problem could be modified so that we consider a smoothed sum instead;say X n X Λ( n ) n / it log n (cid:16) − log n log X (cid:17) . The imaginary part responds well to this procedure and the above proof recoversthe formulas of Selberg [42] and Tsang [45] which have the convergent integrandsin ( y log X ) /y . Unfortunately, for the real part the integrand is again of the form ≈ / | y | owing to large negative values of ℜ log ζ (1 / it ) (cf. Lemma 5 of [45]).3. Moments of the Euler product: Proof of Proposition 1
Recall that P X ( s ) = exp (cid:16) X n X Λ( n ) n s log n (cid:17) . Here and throughout the paper we consider the following subsets of [ T, T ] on whichthe sum in the exponential attains typical values. For V > S ℜ ( V ) = n t ∈ [ T, T ] : (cid:12)(cid:12)(cid:12) ℜ X n X Λ( n ) n / it log n (cid:12)(cid:12)(cid:12) V o , HE SPLITTING CONJECTURE 10 and define the set relating to the imaginary part S ℑ ( V ) similarly. Let S ( V ) = S ℜ ( V ) ∩ S ℑ ( V )and denote V = log T log T, S = S ( V ) . Define the complementary sets by E ℜ ( V ) = S ℜ ( V ) , E ℑ ( V ) = S ℑ ( V ) , E ( V ) = S ( V ) , E = S where A = [ T, T ] \ A for a given set A . Also, let V max = max t ∈ [ T, T ] (cid:18)(cid:12)(cid:12)(cid:12) ℜ X n X Λ( n ) n / it log n (cid:12)(cid:12)(cid:12) , (cid:12)(cid:12)(cid:12) ℑ X n X Λ( n ) n / it log n (cid:12)(cid:12)(cid:12)(cid:19) . Note that unconditionally V max (2+ o (1)) X / / log X and that on the Riemann hy-pothesis V max (1 / o (1))(log( X / / log T ) + O (log X )) log T / log T by Theorem3. The reason we work with the real and imaginary parts (as opposed to workingwith the modulus directly) is so that we have slightly better conditional boundsfor V max whilst still maintaining control over the modulus (which is important forLemma 5 below). This gives better exponents for our logarithms in the conditionalresults but entails slightly more work.To the estimate the measure of the complementary sets we use the following. Lemma 2 ([43]) . Let T be large and let x T . Let m be a natural number suchthat x m T / log T . Then for any complex numbers a ( p ) we have T Z TT (cid:12)(cid:12)(cid:12)(cid:12) X p x a ( p ) p / it (cid:12)(cid:12)(cid:12)(cid:12) m dt ≪ m ! (cid:18) X p x | a ( p ) | p (cid:19) m . Lemma 3.
Let ǫ, κ > . Then for X η κ (log T ) (log T ) with η κ = κ − ǫ wehave µ ( E ℜ ( V )) ≪ T e − (2 κ + o (1)) V , V V V max where µ denotes Lebesgue measure. Assuming RH, the same bound holds provided X (log T ) θ κ − ǫ where (13) θ κ = 2 q κ . The same results hold for E ℑ ( V ) also.Proof. We first prove the unconditional result. Write X n X Λ( n ) n / it log n = X p X p / it + X p √ X p it + O (1) . HE SPLITTING CONJECTURE 11
Then from Jensen’s inequality in the form ( a + b + c ) m m − ( a m + b m + c m )with m >
1, we have µ ( E ℜ ( V )) m − V m (cid:18) Z TT (cid:12)(cid:12)(cid:12) X p X p / it (cid:12)(cid:12)(cid:12) m dt + Z TT (cid:12)(cid:12)(cid:12) X p √ X p it (cid:12)(cid:12)(cid:12) m dt + O ( T C m ) (cid:19) . By Lemma 2 the right hand side is(14) ≪ T m m ! V m (cid:16) X p X p (cid:17) m ≪ T m / (cid:16) m log log XeV (cid:17) m provided m (log T − log T ) / log X . Choosing m = 2 κV log V the bound ≪ T e − (2 κ + o (1)) V follows since log log X V o (1) . Note that our choice of m is legal since m κV max log V max . κX / (log X ) . κη / k log T log X (1 − ǫ ) log T log X where . means times a constant of the form 1 + o (1).Assuming RH, then on writing θ κ = 2 + ν κ we find V max ( + o (1)) (cid:0) log( X / log T ) + 4 log log X (cid:1) log T log log T . ν κ − ǫ T by Theorem 3. Choosing m as before gives the desired bound for µ ( E ℜ ( V )) and,again, our choice of m is legal since in this case we have m . κ ( ν κ − ǫ ) log T T which is (log T − log T ) / log X provided2 + ν κ κν κ ⇐⇒ ν κ q κ − . (cid:3) Lemma 4.
Let ǫ > , ν ∈ R and suppose X η ν (log T ) (log T ) with η ν = ν − ǫ .Then T Z E | P X ( + it ) | ν dt ≪ e − δV for some δ > dependent on ǫ . Assuming RH, this holds provided X (log T ) θ ν − ǫ where θ ν is given by (13) . HE SPLITTING CONJECTURE 12
Proof.
Since A ∪ B = A ∪ ( A ∩ B ) we may write(15) E = E ℜ ( V ) ∪ E ℑ ( V ) = E ℜ ( V ) ∪ ( S ℜ ( V ) ∩ E ℑ ( V )) . The integral over E ℜ ( V ) is T Z E ℜ ( V ) exp (cid:16) | ν | · (cid:12)(cid:12)(cid:12) ℜ X n X Λ( n ) n / it log n (cid:12)(cid:12)(cid:12)(cid:17) dt = 1 T e | ν | V µ ( E ℜ ( V )) + 2 | ν | T Z V max V e | ν | V µ ( E ℜ ( V )) dV. (16)Since η ν / | ν | + δ/ − ǫ/ δ > ǫ , Lemma 3 gives µ ( E ℜ ( V )) ≪ T e − (2 | ν | + δ ) V . Hence the quantity in (16) is ≪ e − δV . Similarly, theintegral over S ℜ ( V ) ∩ E ℑ ( V ) is T e | ν | V µ ( E ℑ ( V )) e − δV . The result then follows by the union bound. Likewise, θ ν − ǫ θ ν + δ/ − ǫ/ (cid:3) We now show that P X (1 / it ) k can be approximated by a short Dirichlet poly-nomial provided t ∈ S . To do this we note that for | z | Z we have(17) (cid:12)(cid:12)(cid:12) e z − Z X j =0 z j j ! (cid:12)(cid:12)(cid:12) e − Z by a Taylor expansion and Stirling’s formula. Lemma 5.
Suppose t ∈ S . Let k ∈ R and define W = W ( k, T ) = 20 | k | V . Then as X → ∞ , P X ( + it ) k = (cid:0) O ( e − | k | V ) (cid:1) D ( t, k )(18) where (19) D ( t, k ) = X n ∈ S ( X ) α k ( n ) n / it with S ( X ) = (cid:8) n ∈ N : p | n = ⇒ p X (cid:9) and where the coefficients α k ( n ) satisfy thefollowing properties: • α k ( n ) is supported on integers n X W and | α k ( n ) | d | k | ( n ) . HE SPLITTING CONJECTURE 13 • If Ω( n ) W then α k ( n ) = β k ( n ) where β k ( n ) is a multiplicative functionsatisfying β k ( n ) = d k ( n ) if p m | n = ⇒ p m X and | β k ( n ) | d | k | ( n ) in general.Proof. Since t ∈ S we have (cid:12)(cid:12)(cid:12) X ℓ X Λ( ℓ ) ℓ / it log ℓ (cid:12)(cid:12)(cid:12) T log T, and so by (17) we acquire(20) P X ( + it ) k = (cid:0) O ( e − | k | V ) (cid:1) W X j =0 k j j ! (cid:18) X ℓ X Λ( ℓ ) ℓ / it log ℓ (cid:19) j . Writing the sum on the right as the Dirichlet polynomial D ( t, k ) it remains to deducethe properties of the coefficients α k ( n ).Clearly, this is a Dirichlet polynomial of length X W over the X -smooth numbers S ( X ). For the remaining properties, first note that we may write(21) exp (cid:16) k X ℓ X Λ( ℓ ) ℓ s log ℓ (cid:17) = Y p X exp (cid:16) − k log(1 − p − s ) − k X m : p m >X mp ms (cid:17) . Note also that after performing a Taylor expansion of the left hand side and collectinglike terms for n s , the coefficients are a sum of positive terms if k >
0, whilst theycan be bounded from above by the same sum but involving | k | if k <
0. Fromthese observations it is clear that | α k ( n ) | d | k | ( n ) since the right hand side is thegenerating function for the divisor functions d k ( n ) with some terms removed.Moreover, if we form the product on the right hand side of (21) into a series P n ∈ S ( X ) β k ( n ) n − s , then we see that the coefficients are multiplicative and satisfy β k ( n ) = d k ( n ) if p m | n = ⇒ p m X . Since the highest power j in (20) is W we seethat, certainly, if Ω( n ) W then α k ( n ) = β k ( n ). (cid:3) Proof of Proposition 1.
Write1 T Z TT | P X ( + it ) | k dt = 1 T Z S | P X ( + it ) | k dt + 1 T Z E | P X ( + it ) | k dt. By Lemma 4, the second integral here is o (1). For the first integral, Lemma 5 gives1 T Z S | P X ( + it ) | k dt ∼ T Z S | D ( t, k ) | dt = 1 T Z TT | D ( t, k ) | dt + O (cid:18) T Z E | D ( t, k ) | dt (cid:19) HE SPLITTING CONJECTURE 14
By the Cauchy–Schwarz inequality the error term here is(22) ≪ (cid:16) µ ( E ) T (cid:17) / (cid:18) T Z TT | D ( t, k ) | dt (cid:19) / . Since D ( t, k ) is of length X W e C (log T ) log T , the Montgomery–Vaughan meanvalue Theorem (see (39) below) and the coefficient bounds of Lemma 5 give1 T Z TT | D ( t, k ) | dt = (1 + o (1)) X n n = n n n j ∈ S ( X ) α k ( n ) α k ( n ) α k ( n ) α k ( n )( n n n n ) / ≪ Y p X (cid:18) k p + O ( p − ) (cid:19) ≪ (log X ) k . (23)More generally, for fixed m ∈ N we have(24) 1 T Z TT | D ( t, k ) | m dt ≪ (log X ) m k . Therefore, by Lemma 3 the expression in (22) is ≪ e −| k | (log T )(log T ) (log X ) k = o (1)and hence we may concentrate on the integral of | D ( t, k ) | over the full set [ T, T ].Applying the Montgomery–Vaughan mean value theorem again gives1 T Z TT | D ( t, k ) | dt = (1 + o (1)) X n ∈ S ( X ) α k ( n ) n . Since | α k ( n ) | d | k | ( n ), the sum over terms with Ω( n ) > W is, for any 1 < r < X n ∈ S ( X )Ω( n ) >W α k ( n ) n ≪ r − W X n ∈ S ( X ) d | k | ( n ) r Ω( n ) n ≪ r − W (log X ) rk = o (1)where in the first inequality we have applied Rankin’s trick in the form r Ω( n ) − W > α k ( n ) = β k ( n ) if Ω( n ) W the main term is X n ∈ S ( X )Ω( n ) W α k ( n ) n = X n ∈ S ( X ) β k ( n ) n + O (cid:18) X n ∈ S ( X )Ω( n ) >W | β k ( n ) | n (cid:19) . HE SPLITTING CONJECTURE 15
The bound | β k ( n ) | d | k | ( n ) and the same analysis as in (25) shows that this errorterm is o (1). From the properties of β k ( n ) we get X n ∈ S ( X ) β k ( n ) n = Y p X (cid:18) X m : p m X d k ( p m ) p m + X m : p m >X β k ( p m ) p m (cid:19) = Y p X X m > d k ( p m ) p m Y p X (cid:18) O (cid:16) X m : p m >X d | k | ( p m ) p m (cid:17)(cid:19) . We split the second product at √ X and apply the bound d k ( n ) ≪ n ǫ to find that itis(26) Y p √ X (cid:16) O (cid:16) X − ǫ (cid:17)(cid:17) Y √ X
d k ( p m ) p m ∼ a ( k )( e γ log X ) k since a ( k ) is an absolutely convergent product. (cid:3) Asymptotics for the 2nd and 4th moments of the Hadamardproduct: Proof of Proposition 2
From (3) we have Z X ( + it ) = ζ ( + it ) P X ( + it ) − (cid:0) O (1 / log X ) (cid:1) and thus it suffices to consider the second and fourth moment of the object on theright. Our aim is to first replace P X by its Dirichlet polynomial approximation andthen apply formulas for the twisted second and fourth moments of the zeta function.4.1. The second moment.
As before, we decompose the integral as(27) 1 T Z TT | Z X ( + it ) | dt = 1 T Z S | Z X ( + it ) | dt + 1 T Z E | Z X ( + it ) | dt. Working unconditionally first, we apply the Cauchy–Schwarz inequality to the inte-gral over E to find it is ≪ (log T ) (cid:18) T Z E | P X ( + it ) − | dt (cid:19) / using Ingham’s asymptotic for the fourth moment. Since X (log T ) (log T ) and 1 / < η we find that this is ≪ (log T ) e − δV = o (1) by Lemma 4. HE SPLITTING CONJECTURE 16
If we assume RH we can apply H¨older’s inequality in the form1 T Z E | Z X ( + it ) | dt ≪ (cid:18) T Z TT | ζ ( + it ) | /ǫ ) dt (cid:19) ǫ ǫ (cid:18) T Z E | P X ( + it ) | − ǫ ) dt (cid:19) ǫ (28)for some ǫ >
0. The first term on the right is ≪ (log T ) /ǫ ) by Harper’s [24]conditional bound R TT | ζ (1 / it ) | k ≪ T (log T ) k . Since X (log T ) √ − ǫ ′ and √ − ǫ ′ θ ǫ − ǫ ′ / ǫ small enough, the second term on the right is ≪ e − δV by Lemma 4. Therefore, the quantity in (28) is o (1).By Lemma 5 we have1 T Z S | Z X ( + it ) | dt ∼ T Z S | ζ ( + it ) | | D ( t, − | dt which we write as1 T Z TT | ζ ( + it ) | | D ( t, − | dt + O (cid:18) T Z E | ζ ( + it ) | | D ( t, − | dt (cid:19) . Applying the Cauchy–Schwarz inequality twice along with Ingham’s fourth momentbound, Lemma 3 and (24) we find that the error term here is o (1).It remains to show that I := 1 T Z TT | ζ ( + it ) | | D ( t, − | dt ∼ log Te γ log X .
The mean square of the zeta function times an arbitrary Dirichlet polynomial hasbeen computed before e.g. see [5, 7]. From there we see that I = X m,n ∈ S ( X ) α − ( m ) α − ( n )( m, n ) mn log (cid:16) BT ( m, n ) mn (cid:17) + o (1)for some constant B . On applying the bound α − ( n ) ≪ d ( n ) ≪ X m,n ∈ S ( X ) α − ( m ) α − ( n )( m, n ) mn log (cid:16) B ( m, n ) mn (cid:17) ≪ (log X ) and thus we can consider the remaining sum.As in the previous section, we first estimate the sum over integers for whichΩ( m ) , Ω( n ) > W (which equals 10 log T log T in this case). Applying the bound HE SPLITTING CONJECTURE 17 α − ( n ) ≪ ≪ log T X m,n ∈ S ( X )Ω( m ) >W ( m, n ) mn ≪ (log T ) r − W X m,n ∈ S ( X ) ( m, n ) r Ω( m ) mn for any 1 < r <
2. A short computation shows this last sum is Y p X (cid:16) r + 1) p − + O ( p − ) (cid:17) ≪ (log X ) r +1 and so the terms with Ω( m ) , Ω( n ) > W contribute an error of size o (1).In the main term we replace α − ( n ) with β − ( n ) and then re-extend the sum toinclude those integers for which Ω( m ) , Ω( n ) > W . By the bounds β − ( n ) ≪ d ( n ) ≪
1, the same argument shows that this introduces an error of o (1). Thus, I = log T X m,n ∈ S ( X ) β − ( m ) β − ( n )( m, n ) mn + O ((log X ) ) . By symmetry and the properties of β k ( n ) we find that the sum is Y p X (cid:18) X m,n > p m , p n X µ ( p m ) µ ( p n ) p m + n − min( m,n ) + O (cid:16) X m,n > p m >X p m + n − min( m,n ) (cid:17)(cid:19) = Y p X (cid:16) − p (cid:17) Y p X (cid:18) O (cid:16) X m,n > p m >X p m + n − min( m,n ) (cid:17) . (cid:19) To estimate the second product we note that the sum in the error term is ≪ P p m >X ( m + 1) p − m ≪ (log X ) p −⌈ log X/ log p ⌉ and then split the product at √ X , asbefore. In this way we find it is 1 + O ( X − / ǫ ) and therefore by Mertens’ Theorem I ∼ log Te γ log X as desired.4.2. The fourth moment: Initial clearing.
Not surprisingly, the fourth momentrequires more work in both the initial stages and the arithmetic computations. Ouraim is to show that I := 1 T Z TT | ζ ( + it ) | | P X ( + it ) | − dt ∼ (cid:18) log Te γ log X (cid:19) . In this subsection the goal is to replace P X (1 / it ) − by D ( t, − HE SPLITTING CONJECTURE 18
Splitting the integral as in (27) we see that our first task is to bound(29) 1 T Z E | ζ ( + it ) | | P X ( + it ) | − dt. On RH we can deal with this by applying H¨older’s inequality as in (28). Followingthe same argument and using the fact that √ − ǫ ′ θ ǫ ) − ǫ ′ / o (1).To bound this unconditionally requires more work. First note that since E ⊂ E ′ := n t ∈ [ T, T ] : (cid:12)(cid:12)(cid:12) X n X Λ( n ) n / it log n (cid:12)(cid:12)(cid:12) > V o we can upper bound by the integral over E ′ . Let V j = e j V and define J to be themaximal j such that V J V max . Let E j = n t ∈ [ T, T ] : V j (cid:12)(cid:12)(cid:12) X n X Λ( n ) n / it log n (cid:12)(cid:12)(cid:12) V j +1 o so that E ′ = ∪ J j =0 E j . Then 1 T Z E ′ | ζ ( + it ) | | P X ( + it ) | − dt J X j =0 e V j +1 T Z E j | ζ ( + it ) | dt J X j =0 e V j +1 V − r j j T Z TT | ζ ( + it ) | (cid:12)(cid:12)(cid:12) X n X Λ( n ) n / it log n (cid:12)(cid:12)(cid:12) r j dt (30)for any given integer r j >
0. The combinatorics are simplified if we focus on theprime sums so we apply Jensen’s inequality in the form (cid:12)(cid:12)(cid:12) X n X Λ( n ) n / it log n (cid:12)(cid:12)(cid:12) r j r j (cid:18)(cid:12)(cid:12)(cid:12) X p X p / it (cid:12)(cid:12)(cid:12) r j + (cid:12)(cid:12)(cid:12) X p √ X p it (cid:12)(cid:12)(cid:12) r j + O ( C r j ) (cid:19) . It will be clear after the computations that the first sum here gives the dominantcontribution and so we focus on this. Note that by the multinomial theorem, (cid:16) X p X p / it (cid:17) r j = r j ! X n X rj Ω( n )= r j g ( n ) n / it HE SPLITTING CONJECTURE 19 where g ( n ) is the multiplicative function satisfying g ( p α ) = 1 /α !. Accordingly, (30)is(31) ≪ J X j =0 e V j +1 r j V − r j j · T Z TT | ζ ( + it ) | (cid:12)(cid:12)(cid:12) r j ! X n X rj Ω( n )= r j g ( n ) n / it (cid:12)(cid:12)(cid:12) dt The twisted fourth moment of the zeta function has been computed before [6, 31]and has been applied in similar situations [26]. Provided X r j T / − ǫ i.e.(32) r j (1 / − ǫ ) log T log X ,
Proposition 4 and formula (8) of [26] (see section 6 there) give1 T Z TT | ζ ( + it ) | (cid:12)(cid:12)(cid:12) r j ! X n X rj Ω( n )= r j g ( n ) n / it (cid:12)(cid:12)(cid:12) dt ≪ (log T ) r j X m,n X rj Ω( m )=Ω( n )= r j r j ! g ( n ) g ( m )[ n, m ] . Following the arguments of [26] which lead to formula (9) there, we find that this is(33) ≪ (log T ) r j r j ! (cid:16) X p X p (cid:17) r j exp (cid:16) X p X p (cid:17) ≪ (log T ) r / j (cid:16) r j log log Xe (cid:17) r j . We choose r j = 12 V j log V j V max log V max . X / (log X ) .
24 log T
100 log X ∀ j since X (log T ) (log T ) . Clearly, this satisfies (32). Then applying (33) in(31) we find that1 T Z E ′ | ζ ( + it ) | | P X ( + it ) | − dt ≪ (log T ) J X j =0 e V j +1 (cid:16) C log log XV j log V j (cid:17) V j / log V j ≪ (log T ) J X j =0 e V j +1 − (12 − o (1)) V j ≪ (log T ) J X j =0 e − e j V (12 − e + o (1)) = o (1) . We have therefore arrived at I = (1 + o (1)) 1 T Z S | ζ ( + it ) | | D ( t, − | dt + o (1) HE SPLITTING CONJECTURE 20 on applying Lemma 5 in the integral over S . After extending to the full range ofintegration [ T, T ] it remains to estimate1 T Z E | ζ ( + it ) | | D ( t, − | dt. However, from the definition of D ( t, k ) we have | D ( t, k ) | W X j =0 | k | j j ! (cid:12)(cid:12)(cid:12) X n X Λ( n ) n / it log n (cid:12)(cid:12)(cid:12) j exp (cid:16)(cid:12)(cid:12)(cid:12) k X n X Λ( n ) n / it log n (cid:12)(cid:12)(cid:12)(cid:17) . and so we can apply the same argument as above to acquire the bound o (1) for thisintegral. Thus we have I ∼ J where J := 1 T Z TT | ζ ( + it ) | | D ( t, − | dt. The fourth moment: Arithmetic computations.
In this section our aimis to show that J ∼ (cid:18) log Te γ log X (cid:19) . The formulas for the twisted fourth moment of the zeta function given in the liter-ature [6, 31] apply to smoothed integrals and accordingly we must first smooth J .Let Φ − , Φ + be smooth approximations of compact support satisfying(34) Φ − ( t ) t ∈ [ T, T ] Φ + ( t )with derivatives Φ ( j ) ± ( t ) ≪ T ǫ . For example, we may take Φ − to be compactlysupported on [1 ,
2] and equal to 1 on the interval [1 + T − ǫ , − T − ǫ ] with smooth,monotonic decay to zero at each endpoint. Then, on letting Φ be either Φ − or Φ + we consider the smoothed integral J , Φ := 1 T Z R Φ (cid:16) tT (cid:17) | ζ ( + it ) | | D ( t, − | dt. We note that the error incurred from these approximations will be ≪ T − ǫ which istolerable given the asymptotic we seek. Theorem 4 (Theorem 1.2 of [6]) . Let Φ be as above and α , α , α , α ≪ / log T . Let Ξ be the subgroup of S consisting of the identity, those permutations which swap justone element of { , } with { , } and the permutation satisfying τ (1) = 3 , τ (2) = 4 . HE SPLITTING CONJECTURE 21
Then for any Dirichlet polynomial P n y a ( n ) n − s satisfying y T / − ǫ and a ( n ) ≪ n ǫ we have Z R ζ ( + α + it ) ζ ( + α + it ) ζ ( − α − it ) ζ ( − α − it ) (cid:12)(cid:12)(cid:12) X n y a ( m ) m / it (cid:12)(cid:12)(cid:12) Φ (cid:16) tT (cid:17) dt = X m ,m y a ( m ) a ( m ) Z R Φ (cid:16) tT (cid:17) X τ ∈ Ξ (cid:16) t π (cid:17) P j =1 α τ ( j ) − α j Z τ ( α ) ,τ ( α ) ,τ ( α ) ,τ ( α ) ,m ,m dt + O ( T − ǫ ) where Z α ,α ,α ,α ,m ,m = X m n n = m n n m m ) / n / α n / α n / − α n / − α V (cid:16) n n n n t (cid:17) and V ( x ) = 12 πi Z c + i ∞ c − i ∞ G ( s ) s (4 π x ) − s ds, c > with G ( s ) an even function of rapid decay in vertical strips satisfying G (0) = 1 . Remark.
We remark that the choice of function G ( s ) is flexible and it can beprescribed to have zeros at linear combinations of the shifts. This is fairly typicaland is used to cancel unnecessary poles later on. We will take G ( s ) = Q α ( s ) exp( s )where Q α ( s ) is an even polynomial which is 1 at s = 0 and zero at 2 s = α − α , α − α , α − α and α − α . Note these conditions on Q imply that for fixed ℜ ( s ),(35) G ( s ) ≪ (log T ) e −ℑ ( s ) since α j ≪ / log T .Let us compute term corresponding to the identity: τ = id. Denote this by K = K α ( t, X ) := X m ,m ∈ S ( X ) α − ( m ) α − ( m ) Z α ,α ,α ,α ,m ,m = X m n n = m n n m j ∈ S ( X ) α − ( m ) α − ( m )( m m ) / n / α n / α n / − α n / − α V (cid:16) n n n n t (cid:17) . By shifting the contour in V to the either the left or right depending on whether x ≪ x ≫
1, respectively, we find that V ( x ) ≪ (1 + | x | ) − A for any A > α k ( m ) ≪ m ǫ · m T δ we may restrict the abovesum to those n j satisfying n n n n ≪ t ǫ ≪ T ǫ at the cost of an error of size HE SPLITTING CONJECTURE 22 o (1). Then the contribution from those m with Ω( m ) > W is for 1 < r < ≪ r − W X m T ǫ r Ω( n ) d ( m ) m ≪ r − W (log T ) r = o (1)where for the first inequality we have applied Rankin’s trick in the form r Ω( m ) − W > α − ( n ) ≪ d ( n ). The same bound holds for the sum overΩ( m ) > W . Then on replacing α − ( m ) with β − ( m ) and re-extending the sums(which by the same arguments incurs an error of o (1)) we have K = X m n n = m n n m j ∈ S ( X ) β − ( m ) β − ( m )( m m ) / n / α n / α n / − α n / − α V (cid:16) n n n n t (cid:17) + o (1) . Unfolding the integral for V ( x ) and pushing the sum through we find(36) K = 12 πi Z c + i ∞ c − i ∞ F α,X ( s ) G ( s ) s (cid:18) t π (cid:19) s ds + o (1)where F α,X ( s ) = X m n n = m n n m j ∈ S ( X ) β − ( m ) β − ( m )( m m ) / n / α + s n / α + s n / − α + s n / − α + s = X m n = m n m j ∈ S ( X ) β − ( m ) β − ( m ) σ α ,α ( n ) σ − α , − α ( n )( m m ) / ( n n ) / s with σ u,v ( n ) = P d d = n d − u d − v . Expressing this as an Euler product we have F α,X ( s ) = A α ( s ) G α,X ( s )where A α ( s ) = ζ (1 + α − α + 2 s ) ζ (1 + α − α + 2 s ) ζ (1 + α − α + 2 s ) ζ (1 + α − α + 2 s ) ζ (2 + α + α − α − α + 4 s )and G α,X ( s ) = Y p X X m + n = m + n β − ( p m ) β − ( p m ) σ α ,α ( p n ) σ − α , − α ( p n ) p
12 ( m + m )+( 12 + s )( n + n ) / X n > σ α ,α ( p n ) σ − α , − α ( p n ) p n (1+2 s ) . HE SPLITTING CONJECTURE 23
Shifting the line of integration in (36) to ℜ ( s ) = − / log X we pick up a sim-ple pole only at s = 0 (the poles of A α ( s ) being cancelled by the zeros of G ( s )).Since β − ( n ) ≪ d ( n ) and σ α i ,α j ( p n ) ≪ p n/ log T d ( p n ) we find that on the new line ofintegration G α,X ( s ) ≪ (log X ) O (1) . Therefore, on combining this with the bound for G ( s ) given in (35) we see the integralover the new line is bounded by ≪ t − / log X (log T ) O (1) = o (1)since t ≍ T . Hence K α = A α (0) G α,X (0) + o (1) . We have satisfactorily computed the contribution from a single Z term and thus itremains to find the combinatorial sum of these which appears in Theorem 4. Usingthe results of [13] we can express this sum as a multiple contour integral. Precisely,Lemma 2.5.1 there gives X τ ∈ Ξ (cid:16) t π (cid:17) P j =1 α τ ( j ) − α j K τ ( α ) = 14(2 πi ) Z | z j | =3 j / log T j A z ,z ,z ,z (0) G z ,z ,z ,z ,X (0)∆( z , z , z , z ) Q i,j =1 ( z i − α j ) × (cid:16) t π (cid:17) P j =1 ( z j +2 − z j ) / dz + o (1)where ∆( z ) denotes the vandermonde determinant. A short calculation shows that ∂∂z j G z,X (0) (cid:12)(cid:12)(cid:12)(cid:12) z =0 ≪ G ,X (0) X p X log pp ≪ G ,X (0) log X and hence we acquire the Taylor expansion G z ,z ,z ,z ,X (0) = G ,X (0) (cid:16) O (cid:16) log X X j =1 | z j | (cid:17)(cid:17) whilst from the Laurent expansion of the zeta function we get A z ,z ,z ,z ,X (0) = 1 ζ (2) Y i,j =1 z i − z j +2 ) (cid:16) O (cid:16) X j =1 | z j | (cid:17)(cid:17) . HE SPLITTING CONJECTURE 24
On setting the shifts α j equal to zero, substituting z j z j / log( t/ π ), and applyingthese expansions we find J , Φ = Z R Φ (cid:16) tT (cid:17)(cid:16) c log ( t/ π ) · G ,X (0) ζ (2) (cid:16) O (cid:16) log X log T (cid:17)(cid:17) dt + o ( T )where c = 14 · (2 πi ) Z | z j | =3 j j ∆( z , z , z , z ) Q i,j =1 ( z i − z j +2 ) e P j =1 z j +2 − z j Y j =1 dz j z j . From section 2.7 of [13] we know that c = g (2) = 1 /
12 where g ( k ) is given by (2).Furthermore, G ,X (0) = Y p X X m + n = m + n β − ( p m ) β − ( p m ) d ( p n ) d ( p n ) p
12 ( m + m + n + n ) / X n > d ( p n ) p n . The denominator here is Y p X (cid:18) X n > d ( p n ) p n (cid:19) − = Y p X (1 − p − ) − p − ∼ ζ (2)( e γ log X ) whereas the numerator is G ,X (0) = Y p X (cid:18) O (cid:16) X m + n = m + n p m >X d ( p m ) d ( p m ) d ( p n ) d ( p n ) p
12 ( m + m + n + n ) (cid:17)(cid:19) since β − ( p m ) = d − ( p m ) for p m X and β − ( p m ) ≪ d ( p m ) in general. Thensince n >
0, the sum in the error is ≪ P m : p m >X d ( p m ) /p m after forming theconvolution. Therefore, we can apply the same argument which gave (26) to findthat the numerator is 1 + O ( X − / ǫ ). Consequently, we have J , Φ ∼ ˆΦ(0)12 · T (cid:18) log Te γ log X (cid:19) = T (cid:18) log Te γ log X (cid:19) + O ( T − ǫ )and so the result follows by (34).5. Some useful tools
In this short section we describe some tools which will come in handy when provingProposition 3. The first relates to the exponential truncation of a more general primesum and will be used extensively throughout.
HE SPLITTING CONJECTURE 25
Given a general Dirichlet polynomial of the form D ( s ) = P p Y a ( p ) p − s , suppose t ∈ [ T, T ] is such that | kD ( s ) | Z . Then by (17) we have(37) (cid:12)(cid:12)(cid:12)(cid:12) e kD ( s ) − X j Z ( kD ( s )) j j ! (cid:12)(cid:12)(cid:12)(cid:12) e − Z . By the multinomial theorem the truncated exponential series can be written as(38) X j Z j ! (cid:18) k X p Y a ( p ) p s (cid:19) j = X Ω( n ) Zp | n = ⇒ p Y k Ω( n ) a ( n ) g ( n ) n s where we recall g is the multiplicative function such that g ( p α ) = 1 /α !, and a ( n ) is thecompletely multiplicative extension of a ( p ). Observe this is a Dirichlet polynomialof length Y Z .Our remaining observations relate to mean values of Dirichlet polynomials. Wefirst state the mean value theorem of Montgomery and Vaughan [36] which gives forany complex coefficients a ( n ),(39) 1 T Z TT (cid:12)(cid:12)(cid:12) X n N a ( n ) n it (cid:12)(cid:12)(cid:12) dt = (1 + O ( N/T )) X n N | a ( n ) | . Suppose we are given R Dirichlet polynomials A j ( s ) = X n ∈S j a j ( n ) n − s , where the Q Rj =1 n j N = o ( T ) for all n j ∈ S j i.e. the product of the A j ( s ) is short.Then the Montgomery–Vaughan mean value theorem readily implies1 T Z TT R Y j =1 (cid:12)(cid:12) A j ( it ) (cid:12)(cid:12) dt ∼ X n N (cid:12)(cid:12)(cid:12) X n = n ··· n R n j ∈ S j a ( n ) · · · a R ( n R ) (cid:12)(cid:12)(cid:12) = X n ··· n R = n R +1 ··· n R n j ∈ S j a ( n ) · · · a R ( n R ) a ( n R +1 ) · · · a R ( n R ) . (40)Suppose in addition that for any j , j with j = j the elements of S j are all coprimeto the elements of S j . Then there is at most one way to write n = Q Rj =1 n j with HE SPLITTING CONJECTURE 26 n j ∈ S j and thus several applications of the mean value theorem imply1 T Z TT R Y j =1 | A j ( it ) | dt = (1 + O ( N T − )) X n ≤ N (cid:12)(cid:12)(cid:12) X n = n ··· n R n j ∈S j R Y j =1 a j ( n j ) (cid:12)(cid:12)(cid:12) = (1 + O ( N T − )) R Y j =1 (cid:16) X n j ∈S j | a j ( n j ) | (cid:17) = (1 + O ( N T − )) − R R Y j =1 (cid:16) T Z TT | A j ( it ) | dt (cid:17) . (41)We now move on to proving the upper bound of Proposition 3.6. The upper bound of Proposition 3
Initial cleaning.
In this section we are required to show that on RH,1 T Z TT | Z X ( + it ) | k dt ≪ (cid:18) log T log X (cid:19) k . On assuming RH, it is a simple task to replace | P X (1 / it ) | − k by | D ( t, − k ) | on theleft hand side. Indeed; from Harper’s [24] conditional bound R TT | ζ (1 / it ) | k ≪ T (log T ) k , the bound for the moments of D ( t, k ) in (24), Lemmas 3, 4, 5, and theusual arguments involving the decomposition [ T, T ] = S ∪ E along with H¨older’sinequality we have(42) 1 T Z TT | Z X ( + it ) | k dt ∼ T Z TT | ζ ( + it ) | k | D ( t, − k ) | dt for all k > ζ (1 / it ) which in-corporates the recent developments of Harper [24] on moments of the zeta function,although we present the result more in the style of Radziwi l l–Soundararajan [38] (seethe key inequality of section 3 there). Such a treatment is similar to that of [35].6.2. An upper bound for ζ ( + it ) . We start with a proposition of Soundararajanin a mildly adapted form of Harper.
Lemma 6.
Assume RH. Let t ∈ [ T, T ] be large and suppose x T . Then log | ζ ( + it ) | ℜ X p x p / / log x + it log( x/p )log x + ℜ X p min( √ x, log T ) p it + log T log x + O (1) . Proof.
This is Proposition 1 of [24]. (cid:3)
HE SPLITTING CONJECTURE 27
For the splitting of the prime sums we denote T − = A, T j = T ej (log log T )2 where A > j = 0 , . . . , J with J the maximal integer such that e J / (log log T ) / (so that J ≍ log log log T ). In this section we take A = 1although later we need to take it sufficiently large. Let θ j = e j (log log T ) , ℓ j = θ − / j so that T j = T θ j for 0 j J . Now, write w j ( p ) = 1 p /θ j log T log( T j /p )log T j and P i,j ( t ) := X T i −
We can now state an upper bound for the zeta function in terms of these shortDirichlet polynomials.
Lemma 7.
Assume RH. Then either | k P ,j ( t ) | > ℓ for some j J or | ζ ( + it ) | k ≪ (cid:18) J Y i =0 (cid:12)(cid:12) N i,J ( t, k ) (cid:12)(cid:12) + X j J − j +1 l J exp (cid:16) kθ j (cid:17)(cid:18) | k P j +1 ,l ( t ) | ℓ j +1 (cid:19) s j j Y i =0 (cid:12)(cid:12) N i,j ( t, k ) (cid:12)(cid:12) (cid:19) × |M ( t, k ) | for any positive integers s j where M ( t, k ) = X Ω( n ) k (log T ) p | n = ⇒ p log T ( k/ Ω( n ) g ( n ) n it . Proof.
Suppose | k P ,j ( t ) | < ℓ . For 0 j J − S ( j ) = (cid:26) t ∈ [ T, T ] : | k P i,l ( t ) | ℓ i ∀ i j, ∀ j l J ; | k P j +1 ,l ( t ) | > ℓ j +1 for some j + 1 l J (cid:27) and S ( J ) = (cid:26) t ∈ [ T, T ] : | k P i,J ( t ) | ℓ i ∀ i J (cid:27) . Then since [ T, T ] = ∪ Jj =0 S ( j ), for t ∈ [ T, T ] we have(47) | ζ ( + it ) | k t ∈ S ( J ) · | ζ ( + it ) | k + X j J − j +1 l J t ∈ S l ( j ) · | ζ ( + it ) | k where S l ( j ) = (cid:26) t ∈ [ T, T ] : | k P i,l ( t ) | ℓ i ∀ i j, ∀ j l J ; | k P j +1 ,l ( t ) | > ℓ j +1 (cid:27) . We apply Lemma 6 to each zeta function on the right hand side of (47). If t ∈ S l ( j )then we take x = T j to give | ζ ( + it ) | k ≪ exp (cid:18) k ℜ X p T j w j ( p ) p / it + 2 k ℜ X p log T p it + 2 kθ j (cid:19) . HE SPLITTING CONJECTURE 29
For the first sum over primes in the exponential we apply (45). For the second sumwe note that, since | P p log T p it | T (log log T ) , we haveexp (cid:16) k ℜ X p log T p it (cid:17) = (cid:0) O ( e − k (log T ) ) (cid:1)(cid:12)(cid:12)(cid:12)(cid:12) X Ω( n ) k (log log T ) p | n = ⇒ p log T ( k/ Ω( n ) g ( n ) n it (cid:12)(cid:12)(cid:12)(cid:12) by (37) and (38). This is ≪ |M ( t, k ) | . Finally, to capture the small size of the set,we multiply by (cid:18) | k P j +1 ,l ( t ) | ℓ j +1 (cid:19) s j > . If t ∈ S ( J ) then we omit this last step since of course there is no such P J +1 ,l ( t ). (cid:3) Proof of the upper bound in Proposition 3.
From (42) we are required toshow that 1 T Z TT | ζ ( + it ) | k | D ( t, − k ) | dt ≪ (cid:18) log T log X (cid:19) k . To apply Lemma 7 we must consider two cases; that where | k P ,j ( t ) | > ℓ andotherwise. We consider the former case first since this is simpler.So, let E ⊂ [ T, T ] be the subset on which | k P ,j ( t ) | > ℓ , that is, when (cid:12)(cid:12)(cid:12)(cid:12) X p T / (log log T )2 w j ( p ) p / it (cid:12)(cid:12)(cid:12)(cid:12) > (log log T ) / k . By Chebyshev’s inequality and Lemma 2 we have µ ( E ) (cid:18) k (log log T ) (cid:19) m Z TT (cid:12)(cid:12)(cid:12)(cid:12) X p T / (log log T )2 w j ( p ) p / it (cid:12)(cid:12)(cid:12)(cid:12) m dt ≪ T m ! (cid:18) k (log log T ) (cid:19) m (cid:18) X p T / (log log T )2 p (cid:19) m provided m (1 − o (1))(log log T ) where in the last line we have used | w j ( p ) | j . By Stirling’s formula and Mertens’ theorem this is(48) µ ( E ) ≪ T m / (cid:18) k me (log log T ) (cid:19) m T e − c (log log T ) for some c > m = ⌊ min(1 , k )(log log T ) ⌋ . Therefore by H¨older’sinequality, Harper’s bound for the moments of zeta and (24) we have1 T Z E | ζ ( + it ) | k | D ( t, − k ) | dt ≪ e − C (log log T ) (log T ) O (1) = o (1) . HE SPLITTING CONJECTURE 30
We may now consider the second case where | k P ,j ( t ) | > ℓ and accordingly con-centrate on the integral 1 T Z [ T, T ] \ E | ζ ( + it ) | k | D ( t, − k ) | dt. By the second part of Lemma 7 this is ≪ T Z TT (cid:18) J Y i =0 (cid:12)(cid:12) N i,J ( t, k ) (cid:12)(cid:12) + X j J − j +1 l J exp (cid:16) kθ j (cid:17)(cid:18) | k P j +1 ,l ( t ) | ℓ j +1 (cid:19) s j j Y i =0 (cid:12)(cid:12) N i,j ( t, k ) (cid:12)(cid:12) (cid:19) × |M ( t, k ) | | D ( t, − k ) | dt. (49)To compute the resultant integrals we apply the following lemma. Lemma 8.
For s j / (10 θ j ) we have T Z TT | D ( t, − k ) | |M ( t, k ) | |P j +1 ,l ( t ) | s j j Y i =0 (cid:12)(cid:12) N i,j ( t, k ) (cid:12)(cid:12) dt ≪ s j ! P s j j +1 (cid:18) log T j log X (cid:19) k where P j +1 = X T j
We write the integrand as a multiple sum. First off, by the multinomialtheorem we have P j +1 ,l ( t ) s j = (cid:18) X T j
Since X T we may group together all the sums over T -smooth numbers as asingle sum and write the above as s j ! X n γ ( n ) n / it X Ω( n )= s j p | n = ⇒ T j
W . By the usual arguments, for1 < r < ≪ r − W X n n n = n n n p | n j = ⇒ p T r Ω( n ) d k ( n ) d k ( n ) k Ω( n n n n ) W ( n )( n n n n ) / n n = r − W Y p T (cid:18) r + 2) k p + O ( p − ) (cid:19) ≪ e − W (log T ) k = o (1) . HE SPLITTING CONJECTURE 32
We then replace α − k ( n ) with β − k ( n ) in the remaining sum. The usual argumentsalso allow us to remove the restrictions on all the Ω( n j ) at the cost of an error of size o (1). Expressing the resultant sum as an Euler product we get X n γ ( n ) n = Y p X (cid:18) O ( p − ) (cid:19) Y X
s j !) X Ω( n )= s j p | n = ⇒ T j p 1, whilst the third sum is j Y i =1 X Ω( n ) ℓ i p | n = ⇒ T i −
Completion of proof of upper bound in Proposition 3. Applying Lemma 8 in (49) givesan upper bound of the form (cid:18) log T J log X (cid:19) k + X j J − j +1 l J exp (cid:16) kθ j (cid:17) s j ! (cid:18) kP j +1 ℓ j +1 (cid:19) s j (cid:18) log T j log X (cid:19) k . On noting that J − j ≪ log(1 /θ j ) , P j +1 = log (cid:18) log T j +1 log T j (cid:19) + o (1) , ℓ j +1 = θ − / j HE SPLITTING CONJECTURE 33 and setting s j = 1 / (10 θ j ), then by Stirling’s formula this is ≪ (cid:18) log T log X (cid:19) k (cid:18) X j J − log( θ j ) θ − / θ j j c /θ j θ / θ j j (cid:19) ≪ (cid:18) log T log X (cid:19) k (cid:18) X j J − e − Cθ − j log(1 /θ j ) (cid:19) for some constants c, C > 0. Since this last series is bounded the result follows. (cid:3) Lower bound for k w j ( p ) so we can simplify our notationand let N i ( t, k ) = X Ω( n ) ℓ i p | n = ⇒ T i − 1. The latter case is somewhat simpler sowe give details for the case 0 < k (cid:12)(cid:12)(cid:12) T Z S ζ ( + it ) N ( t, k − N ( t, k ) | D ( t, − k ) | dt (cid:12)(cid:12)(cid:12) ≪ (cid:16) T Z S | ζ ( + it ) | k | D ( t, − k ) | dt (cid:17) (cid:16) T Z TT | ζ ( + it ) N ( t, k − | | D ( t, − k ) | dt (cid:17) − k × (cid:16) T Z TT |N ( t, k ) | k |N ( t, k − | | D ( t, − k ) | dt (cid:17) k . Since D ( t, − k ) ∼ P X (1 / it ) − k for t ∈ S the first integral on the right hand sideis ≪ T R TT | Z X (1 / it ) | k dt . HE SPLITTING CONJECTURE 34 Remark. Note also that in this argument we can modify the definition of D to be D ( t, k ) = X p | n = ⇒ A
Suppose X η k (log T ) (log T ) . Then T Z S ζ ( + it ) N ( t, k − N ( t, k ) | D ( t, − k ) | dt > C (cid:18) log T log X (cid:19) k for some C > . Assuming RH we may take X (log T ) θ k − ǫ . Proposition 5. For X T / (log T ) we have T Z TT | ζ ( + it ) N ( t, k − | | D ( t, − k ) | dt ≪ (cid:18) log T log X (cid:19) k . Proposition 6. For X T / (log T ) we have T Z TT |N ( t, k ) | k |N ( t, k − | | D ( t, − k ) | dt ≪ (cid:18) log T log X (cid:19) k . Note that our argument works unconditionally provided X η k (log T ) (log T ) as claimed in the introduction.7.1. Proof of Proposition 4. Our first job is to extend the range of integrationto the full set [ T, T ]. By the usual argument involving H¨older’s inequality theintegral over E is o (1). Indeed, from the conditions on X and Lemma 3 we have T − µ ( E ) ≪ e − | k | V which is enough to kill any power of log T . We also have thesecond moment bound for the zeta function, and by (24) the m th moment of D ( t, − k )is also (log T ) O (1) . The only new ingredient required is a moment bound for N ( t, k ) HE SPLITTING CONJECTURE 35 but by the Montgomery–Vaughan mean value theorem this is, for m T Z TT |N ( t, k ) | m dt ≪ X n ··· n m = n m +1 ··· n m n j T / γ k ( n ) · · · γ k ( n m )( n · · · n m ) / = Y p T (cid:18) m k p + O ( p − ) (cid:19) ≪ (log T ) m k . Therefore, 1 T Z S ζ ( + it ) N ( t, k − N ( t, k ) | D ( t, − k ) | dt = I + o (1)where I = 1 T Z TT ζ ( + it ) N ( t, k − N ( t, k ) | D ( t, − k ) | dt. To lower bound I we apply the approximation ζ ( + it ) = X n T n / it + O (cid:16) T / (cid:17) . The other terms of the integrand satisfy pointwise bounds of the form N ( t, k ) ≪ Y / ǫ ≪ T / ǫ , X n α − k ( n ) n / it ≪ X W (1 / ǫ ) ≪ T / ;since k Ω( n ) has average order (log n ) k − and α − k ( n ) ≪ d k ( n ) ≪ n ǫ . We then see thatthe error term in the approximation for zeta leads to an error of size o (1).Therefore I = 1 T Z TT X n T, n ,n Y γ k − ( n ) γ k ( n ) α − k ( n ) α − k ( n )( n n n n n ) / (cid:16) n n n n n (cid:17) it dt + o (1) . By direct integration, the off-diagonal terms for which n n n = n n lead to anerror of size Y X W T X n T, n ,n Y | γ k − ( n ) | γ k ( n )( n n n ) / (cid:18) X n | α − k ( n ) | n / (cid:19) ≪ Y ǫ X W (2+ ǫ ) T / = o (1)since | log( n n /n n n ) | ≫ / ( Y X W ). Accordingly, I = X n n n = n n n T,n ,n Y γ k − ( n ) γ k ( n ) α − k ( n ) α − k ( n )( n n n n n ) / + o (1) . Since n n Y X W T we may remove the condition n T . HE SPLITTING CONJECTURE 36 Now, since α − k ( n ) is supported on prime powers p m with A < p X we maywrite our sum as X n n n = n n = X n n n = n n p | n j = ⇒ A c for some constant c > X n n = n p | n ,n = ⇒ X c (cid:18) log T log X (cid:19) k . (55)For the sums inside the product in (52), we must be a little more careful. Thesums in question are given by(56) X n n = n p | n ,n = ⇒ T i −
Since 0 < k d ( n ) Ω( n ) e Ω( n ) , the error incurred from dropping thecondition on Ω( n ) is, in absolute value, e − ℓ i X n n = n p | n ,n = ⇒ T i − . Doing the same for n gives an error of the samesize and hence the sum in (56) is > Y T i − (cid:0) − e − ℓ i (cid:1) Y T i − c (cid:18) log T log X (cid:19) k J Y i =1 (cid:0) − e − ℓ i (cid:1) Y T i − C (cid:18) log T log X (cid:19) k since, again, P Ji =1 e − ℓ i is a rapidly converging series. This completes the proof ofProposition 4.7.2. Proof of Proposition 5. We are required to show I := 1 T Z TT | ζ ( + it ) N ( t, k − | | D ( t, − k ) | dt ≪ (cid:18) log T log X (cid:19) k . From the conditions on X given in the statement of the proposition, D ( t, − k ) is aDirichlet polynomial of length X W = T o (1) which is still short. Thus, we can applythe results of [5, 7] after combining the two Dirichlet polynomials into a single sum.This gives I = X m,n h k ( m ) h k ( n )( m, n ) mn log (cid:16) BT ( m, n ) mn (cid:17) + o (1)for some constant B where h k ( n ) = X n n = n γ k − ( n ) α − k ( n ) . HE SPLITTING CONJECTURE 38 As in [27], we writelog (cid:16) BT ( m, n ) mn (cid:17) = 12 πi Z | z | =1 / log T (cid:16) BT ( m, n ) mn (cid:17) z dzz , so that the main term in I becomes12 πi Z | z | =1 / log T X m,n h k ( m ) h k ( n )( m, n ) mn (cid:16) BT ( m, n ) mn (cid:17) z dzz . Then after a trivial estimate we get(58) I ≪ (log T ) max | z | =1 / log T (cid:12)(cid:12)(cid:12)(cid:12) X m,n h k ( m ) h k ( n )( m, n ) z ( mn ) z (cid:12)(cid:12)(cid:12)(cid:12) and thus we are required to show this last sum is ≪ X (log T / log X ) k − .Unfolding the coefficients, the above sum becomes X m ,m ,n ,n γ k − ( m ) γ k − ( n ) α − k ( m ) α − k ( n )( m m , n n ) z ( m m n n ) z . Estimating the terms with Ω( m ) , Ω( n ) > W in the usual way we may replace α − k ( n ) by β − k ( n ) and then re-extend the sums at the cost of o (1). Then by multi-plicativity we can express the resultant sum as X m ,m ,n ,n p | m j n j = ⇒ A
1. As usual, for the error term we apply Rankin’s trick alongwith similar Euler product computations to give an error of size ≪ e − ℓ + O (log log X ) which is o (1 / log X ).For the second sum we get, with a similar argument, X m ,n p | m n = ⇒ X
2. This last term is ≪ ( log T log T ) k − and so combining thesebounds in (58) we get I ≪ log T · X · (cid:18) log T log X (cid:19) k − · (cid:18) log T log T (cid:19) k − ≪ (cid:18) log T log X (cid:19) k which completes the proof of Proposition 5.7.3. Proof of Proposition 6. We begin with the following lemma from [27]. Lemma 9. Let P j ( t ) := X T j −
This is essentially Lemma 1 of [27]. Our sequence T j is defined slightly differ-ently but one can check that this makes no difference to the end result. (cid:3) Lemma 10. With the above notation T Z TT Q ( t ) | D ( t, − k ) | dt ≪ e − ℓ (log X ) k and for j J T Z TT Q j ( t ) dt ≪ e − ℓ j . Proof. We prove the first bound since this is new, the second bound follows similarly(and is essentially Lemma 2 of [27]). Let L = 10 ℓ . From the definition of Q ( t ) wehave 1 T Z TT Q ( t ) | D ( t, − k ) | dt = (cid:16) L (cid:17) L L/k X r =0 (cid:16) er + 1 (cid:17) r · T Z TT |P ( t ) | L +2 r | D ( t, − k ) | dt By the Cauchy–Schwarz inequality and (23) the integral is ≪ (log X ) k (cid:18) T Z TT |P ( t ) L +2 r | dt (cid:19) / . Since P ( t ) L +2 r = (2 L + 2 r )! X Ω( n )=2 L +2 rp | n = ⇒ p T g ( n ) n / it , the Montgomery–Vaughan mean value theorem gives that our original integral is ≪ (log X ) k (cid:16) L (cid:17) L L/k X r =0 (cid:16) er + 1 (cid:17) r · (cid:18) (2 L + 2 r )! X Ω( n )=2 L +2 rp | n = ⇒ p T g ( n ) n (cid:19) / (log X ) k (cid:16) L (cid:17) L L/k X r =0 (cid:16) er + 1 (cid:17) r · (2 L + 2 r )! / (cid:18) X p T p (cid:19) L + r (59) HE SPLITTING CONJECTURE 41 since g ( n ) g ( n ). Letting P = P p T p − we find by Stirling’s formula that thesummand is ≪ (2 /e ) L (8 e ) r ( r + 1) − r ( L + r ) L + r +1 / P L + r which is maximised at the solution of r = 8 P ( L + r )(1 + O (1 /r )). Since P T = o ( L ) this solution r = r satisfies 2 √ P L ) / r P L ) / . There-fore, (59) is ≪ (log X ) k (cid:16) cL (cid:17) L · Lk · (3 L )! / P L + r r − r since 2 r L . This is then ≪ (log X ) k L − L/ o (1) ≪ e − ℓ and the result follows. (cid:3) By Lemma 9 and (41) we find that I := 1 T Z TT |N ( t, k ) | k |N ( t, k − | | D ( t, − k ) | dt ≪ T Z TT (cid:16) |N ( t, k ) | (1 + O ( e − ℓ )) + O (cid:0) Q ( t ) (cid:1)(cid:17) | D ( t, − k ) | dt × J +1 Y j =1 T Z TT (cid:16) |N j ( t, k ) | (1 + O ( e − ℓ j )) + O (cid:0) Q j ( t ) (cid:1)(cid:17) dt since J ≍ log log log T . By Lemma 10 we have J Y j =1 T Z TT (cid:16) |N j ( t, k ) | (1 + O ( e − ℓ j )) + O (cid:0) Q j ( t ) (cid:1)(cid:17) dt = J Y j =1 (cid:18) (1 + O ( e − ℓ j )) X Ω( n ) ℓ j p | n = ⇒ T i −
This last integral is X m n = m n Ω( m j ) ℓ p | m j = ⇒ A k > k 1, if not a little simpler.In this case we take T − = Bk for some B > J slightly so that itis the maximal integer such that e J / (log log T ) / (10 k ). This implies that ℓ j = θ − / j > k for all j J .We perform H¨older’s inequality in the form(61) (cid:12)(cid:12)(cid:12) T Z S ζ ( + it ) N ( t, k − N ( t, k ) | D ( t, − k ) | dt (cid:12)(cid:12)(cid:12) ≪ (cid:16) T Z TT | Z X ( + it ) | k dt (cid:17) k (cid:16) T Z TT |N ( t, k − N ( t, k ) | k k − | D ( t, − k ) | dt (cid:17) k − k where again we have used Lemma 5. The integral on the left can be dealt withrather similarly to Proposition 4 although the change in parameters requires somemodifications. We detail these alterations first before dealing with the second integralon the right. HE SPLITTING CONJECTURE 43 Modifying the proof of Proposition 4. We see that we can arrive at (52)in exactly the same way. When dealing with (53), the errors incurred from droppingthe conditions on Ω( n ) etc. are now ≪ e − ℓ (log X ) Ck which of course is still o (1) since k is fixed. Note that on writing (53) as a singlesum P n f ( n ) /n , the coefficients are supported on X -smooth numbers and satisfythe bounds | f ( n ) | k n ) d ( n ) (3 k ) n ) . Then we find that the equivalent of(53) is Y Bk C for some C on taking B large enough. In a similar way we find that the equivalentof (55) is Y X C (cid:18) log T log X (cid:19) k . It remains to deal with the term(62) J Y i =1 X n n = n p | n ,n = ⇒ T i − J Y i =1 (cid:18) Y T i − J Y i =1 (cid:0) − e − ℓ i (cid:1) Y T i − C (cid:18) log T log T (cid:19) k . HE SPLITTING CONJECTURE 44 Combining these we get the desired bound(63) 1 T Z S ζ ( + it ) N ( t, k − N ( t, k ) | D ( t, − k ) | dt > C (cid:18) log T log X (cid:19) k . The remaining integral. We let I = 1 T Z TT |N ( t, k − N ( t, k ) | k k − | D ( t, − k ) | dt. Then Proposition 3 in the case k > I ≪ (cid:18) log T log X (cid:19) k . Lemma 11. For i J we have |N i ( t, k − N i ( t, k ) | k k − (1 + O ( e − ℓ i )) |N i ( t, k ) | + (cid:16) ek |P i ( t ) | ℓ i (cid:17) ℓ i . Proof. If k |P i ( t ) | ℓ i then by (37) we have |N i ( t, k − N i ( t, k ) | k k − =(1 + O ( e − ℓ i )) (cid:12)(cid:12) exp (cid:0) ( k − P i ( t ) + k P i ( t ) (cid:1)(cid:12)(cid:12) k k − =(1 + O ( e − ℓ i )) (cid:12)(cid:12) exp (cid:0) k P i ( t ) (cid:1)(cid:12)(cid:12) =(1 + O ( e − ℓ i )) |N i ( t, k ) | . If k |P i ( t ) | > ℓ i then |N i ( t, k − | ℓ i X r =0 (( k − |P i ( t ) | ) r r ! ( k |P i ( t ) | ) ℓ i ℓ i X r =0 (10 ℓ i ) r − ℓ i r ! (cid:16) ek |P i ( t ) | ℓ i (cid:17) ℓ i . The same bound holds for |N i ( t, k ) | and hence the result follows since 2 k/ (2 k − (cid:3) Lemma 12. We have T Z TT (cid:16) ek |P ( t ) | ℓ (cid:17) ℓ | D ( t, − k ) | dt ≪ e − ℓ (log X ) k and for i J . T Z TT (cid:16) ek |P i ( t ) | ℓ i (cid:17) ℓ i dt ≪ e − ℓ i . 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