Primes in Almost All Short Intervals

ACTA ARITHMETICA LXXXIV.3 (1998)

Primes in almost all short intervals

Alessandro Zaccagnini (Parma)

1. Introduction. The object of this paper is to extend the range of validity of a well-known result of prime number theory. We deal with the Selberg integral 2\x 2 h J(x, h) := π(t) − π(t − h) − dt. log t x The Prime Number Theorem suggests that J(x, h) should be of lower order of magnitude than xh2(log x)−2, at least when h is not too small with respect to x, and the Brun–Titchmarsh inequality trivially implies J(x, h) xh2(log x)−2 provided only that h ≥ xε for some ﬁxed ε > 0. We prove the following Theorem. We have xh2 log log x 2 J(x, h) ε(x) + (log x)2 log x provided that x1/6−ε(x) ≤ h ≤ x, where 0 ≤ ε(x) ≤ 1/6 and ε(x) → 0 as x → ∞. It is well known that Huxley’s density estimates [5] for the zeros of the Riemann zeta-function yield J(x, h) = o(xh2(log x)−2), but only for h ≥ x1/6(log x)C , for some C > 0. The weaker result with h ≥ x1/6+ε is proved in Saﬀari and Vaughan [8], Lemma 5, and in [13], where an identity of Heath-Brown (Lemma 1 of [3]) is used. This paper is inspired by Heath-Brown’s extension [4] of Huxley’s The- orem [5] that π(x) − π(x − h) ∼ h(log x)−1 to the range h ≥ x7/12−ε(x). This was achieved by means of another identity (see (2.2) of [4], or Lemma 2 below), thereby avoiding a direct appeal to the

1991 Mathematics Subject Classiﬁcation: Primary 11N05.

[225] 226 A. Zaccagnini properties of the zeros of the Riemann zeta-function, besides Vinogradov’s zero-free region. We extend this approach to the above integral. An immediate consequence of this result is that if x1/6−ε(x) ≤ h ≤ x then for “almost all” n ∈ [x, 2x] ∩ N we have π(n) − π(n − h) ∼ h(log n)−1. Here “almost all” means that the above asymptotic equality fails for at most o(x) values of n ∈ [x, 2x] ∩ N. Relaxing our demand to π(n) − π(n − h) h(log n)−1 for almost all n’s, one can take h even smaller, and the best result up to date is due to Jia [6] who showed that h ≥ x1/20+ε is acceptable, provided that x is large enough. I thank Alberto Perelli for his unfailing help and J´anosPintz for some helpful suggestions. Many thanks are due to the referee for a very careful reading of my manuscript and numerous useful remarks.

2. Preliminaries. We assume throughout that x is suﬃciently large. For the sake of brevity we set L := log x. Our estimates will be uniform with respect to all parameters but k0, which will eventually be chosen as 4. For ease of reference, our notation is consistent, as far as possible, with the notation in [4], and will be introduced at appropriate places. A few comments on the proof are collected at the end of the paper. Lemma 1. The Theorem follows from the estimate

2\x 2 2 θt x3θ2 log log x J 0(x, θ) := π(t) − π(t − θt) − dt ε(x) + , log t L2 log x x uniformly for x−5/6−ε(x) ≤ θ ≤ 1. Lemma 2 (Linnik–Heath-Brown’s identity). For z > 1 we have X (−1)k−1 X X 1 (2.1) log(ζ(s)Π(s)) = (ζ(s)Π(s) − 1)k = , k kpks k≥1 k≥1 p≥z where Y 1 Π(s) := 1 − . ps p

We have X 1 (2.2) |{p : pk ∈ I, p ≥ z}| = π(t) − π(t − θt) + O(θx1/2 + log x), k k≥1 the contribution from prime powers being negligible. Now the Dirichlet series P −s for ζ(s)Π(s) − 1 is n≥z a(n)n where a(1) = 0 and a(n) = 0 unless all prime factors of n are ≥ z, in which case a(n) = 1. Furthermore, the Dirichlet k P −s series for (ζ(s)Π(s) − 1) is n≥z ak(n)n , ak being the k-fold Dirichlet k convolution of a with itself. This means that ak(n) = 0 unless n ≥ z and −s p ≥ z for all p | n. Hence there are no terms n with n ∈ I and k ≥ k0, and we may consider only the values k < k0. As pointed out in Section 2 of [4], the above identity does not give suitable Dirichlet polynomials at once, and we ﬁrst need to approximate the above Dirichlet series by manageable Dirichlet polynomials. We set X 1 ζ (s) := . t ns n≤t δ We introduce parameters z1 ∈ [3, z) and z2 := z1, where δ ≥ 2 and deﬁne vn by means of Y 1 X µ(n)v Π (s) := 1 − = n . 0 ps ns p

XC X −s (2.4) Π2(s) = Yc(s),Yc(s) := µ(n)vnn , −1−c −c c=0 2 z2

Hence, for suitable coeﬃcients cm,h, we have MX(h) ∗ h (2.6) (ζt(s)Π2(s)Π (s)) = cm,hW (s; m, h), m=1 where the Dirichlet polynomials W have the form

(2.7) W (s; m, h) = WX (s; m, h)WY (s; m, h)WZ (s; m, h), with Yh Yh

WX (s) := Xbi (s),WY (s) := Yci (s), (2.8) i=1 i=1 YL YIm W (s) := Z(m)(s), Z di m=1 i=1 where each Im is ≤ hL/m, and we dropped m and h for brevity. Writing

−1−bi −1−ci −1−di (2.9) Xi := 2 t, Yi := 2 z2,Zi := 2 z, P and I = m Im, we have X e (n) (2.10) W (s; m, h) = m,h , ns N1

Lemma 3. For ﬁxed ε > 0 and m, h ∈ N we have X h mh−1 dm(n) ε,m,h y(log x) , x≤n≤x+y uniformly for xε ≤ y ≤ x. Lemma 4. For t ∈ [x, 2x] there exist Dirichlet polynomials W (s; m, h) satisfying (2.3)–(2.11) such that

N(h) X Xk k X X a (n) = (−1)k−h c e (n) + O(xθL3kδ−δ/3) k h m,h m,h n∈I(t,θ) h=0 m=1 n∈I(t,θ)

1/8 2 when z1z2 ≤ x and δ ≥ (log log z1) . The proof is quite similar to the proof of Lemma 3 of [4], using Lemma 3 above. We omit it for brevity. Set NX(h) X Σ(h, t, θ) := cm,h em,h(n) m=1 n∈I(t,θ) (here a minor clash with the notation of [4] occurs). Then X Xk S(t, θ) := π(t) − π(t − θt) = α(h, k)Σ(h, t, θ) + O(E(t, θ, δ)),

1≤k

3 2 −A The error term is A x θ L for any ﬁxed A, provided that δ ≥ log L, which we assume. Hence by Lemma 1 and (2.13) we have proved Lemma 5. The Theorem follows from the estimates

2\x 2 2 t x3 log log x (2.14) M(t) − dt ε(x) + , log t L2 log x x 2\x 2 x3θ2 log log x (2.15) |R(t, θ)|2 dt ε(x) + L2 log x x −5/6−ε(x) 2 uniformly for x ≤θ ≤ 1, provided that δ ≥ max(log L, (log log z1) ). We shall prove the ﬁrst part of Lemma 5 in Section 5 by taking θ “large”, whereas the proof of the other estimate is achieved by means of mean-value bounds as described below.

3. The case k ≤ 2: reduction to mean-value estimates. For brevity we write s = s(τ) = 1/2 + iτ throughout this section. By Perron’s formula (see Lemma 3.12 of [10]) we have

N(h) T\0 1 X ts − (t − θt)s (3.1) Σ(h, t, θ) = c W (s; m, h) dτ 2πi m,h s m=1 −T0 N(h) N (m) X1 X X2 x 1/2 + O |c | |e (n)| m,h m,h n j=0 m=1 n=N1(m)+1 t − jθt −1 × min 1,T −1 log . 0 n The error term is estimated in Section 6 where we prove that

N(h) T\0 1 X ts − (t − θt)s (3.2) Σ(h, t, θ) = c W (s; m, h) dτ 2πi m,h s m=1 −T0 x 2I 3h + O e (log N7) , T0 where

N7 := max N2(m). 1≤m≤N(h)

The main term of Σ will come from a short interval: for |τ| ≤ T1 we have ts − (t − θt)s (3.3) = θts + O(|s|θ2t1/2). s Primes in almost all short intervals 231

PN(h) Hence, setting S0 = S0(h) := m=1 |cm,h|,

N(h) T\1 1 X M(h, t) := c W (s(τ); m, h)ts dτ, 2πi m,h m=1 −T (3.4) 1 T\1 J0 = J0(h) := max |W (s(τ); m, h)| dτ, 1≤m≤N(h) −T1 we have

N(h) T\1 1 X ts − (t − θt)s (3.5) c W (s; m, h) dτ 2πi m,h s m=1 −T1 2 1/2 = θM(h, t) + O(T1J0S0θ x ). Summing up, from (3.1)–(3.5) we have

.6) Σ(h, t, θ) = θM(h, t) + R1(h, t, θ)(3

N(h) −\T1 T\0 1 X n o ts − (t − θt)s + c + W (s; m, h) dτ 2πi m,h s m=1 −T0 T1

= θM(h, t) + R1(h, t, θ) + R2(h, t, θ) say, where M(h, t) is independent of θ. The ranges [−T0, −T1] and [T1,T0] are dealt with by means of the following mean-value bound, which will be proved in Section 7.

Lemma 6. There is a constant C0 > 0 with the following property. Let −2/3 −1/3 (3.7) η = η(T ) := C0(log T ) (log log T ) and L 2 E := exp log log z1 log z1 and assume that z1 = z1(x) and δ = δ(x) are functions of x such that 2 2/3 δ o(1) o(1) δ ≥ (log log z1) , log z1 ≥ L , z2 = z1 = x and E = x . Then for each ﬁxed α ∈ (0, 1/12) there exists β = β(α) with β ∈ (0, 1/42) with the following property. Let 1/4 1/3−α 5/6+β x < z ≤ x and 3 ≤ T ≤ T0 = x . Then for t ∈ [x, 2x] and h ≤ 2 we have

2\T 2 2h2 −η/6 −1/6 |W (s(τ); m, h)| dτ xE (z1 + T ). T 232 A. Zaccagnini

We obviously have

N(h) T\0 X ts − (t − θt)s R (h, t, θ) |c | W (s; m, h) dτ 2 m,h s m=1 T1 and this means that 2\x 2 (3.8) |R2(h, t, θ)| dt x

2\x T\0 s s 2 2 t − (t − θt) S0 max W (s; m, h) dτ dt. 1≤m≤N(h) s x T1 The next lemma is needed to invert the order of integration. Lemma 7. Let F (s) be a continuous complex-valued function. Then for 1 ≤ T1 ≤ T0 ≤ x and s = 1/2 + iτ we have

2\x T\0 s s 2 2\T t − (t − θt) 2 2 2 2 F (s) dτ dt x θ L max |F (s)| dτ. s T1≤T ≤T0 x T1 T A proof can be easily given by squaring out the integral, performing the integration with respect to t first and then using the elementary inequality |ab| ≤ |a|2 + |b|2 on the remaining double integral. A form of this result appears as Lemma 2 in Harman [2] and elsewhere. We omit the details for brevity. A 2h+I We remark that L A E for any fixed A, that N7 2 x Ex 1/2 and that the definition of W easily implies J0 T1x . The next lemma is proved in Section 6. Lemma 8. For large enough x we have L 2 |S0| exp h (log L) . log z1 2 2 Hence L S0 E. We now choose k0 := 4 and set X2 Xk M1(t) := α(h, k)M(h, t), k=1 h=0 X2 Xk Rj(t, θ) := α(h, k)Rj(h, t, θ), k=1 h=0 for j = 1, 2. Summing up, from Lemmas 4, 6–8, and from (3.2), (3.5)–(3.8) we have 1 X (3.9) π(t) − π(t − θt) − a (n) = θM (t) + R (t, θ) + R (t, θ), 3 3 1 1 2 n∈I(t,θ) Primes in almost all short intervals 233 where −1 2 2 (3.10) R1(t, θ) xET0 + xθ ET1 , 2\x 2 3 2 9 −ξ/6 −1/6 (3.11) |R2(t, θ)| dt x θ E (z1 + T1 ), x and ξ := η(T1). We finally choose our parameters as follows. First we choose 2 2 o(1) δ := (log L) so that δ ≥ max(log L, (log log z1) ) if z1 ≤ x, and z2 = x −2 55 provided that log z1 = o(L(log L) ). Next, we choose T1 := E and observe that T1 tends to infinity with x. The choice 8/9 z1 := exp{L log L} implies −ξ −A z1 A E , for any fixed A. We now see that the hypotheses of Lemma 6 are satisfied and (3.9)–(3.11) finally yield Lemma 9. Let α, β and z be as in Lemma 6. For t ∈ [x, 2x] there exist 0 functions M1(t) and R (t, θ) such that 1 X π(t) − π(t − θt) − a (n) = θM (t) + R0(t, θ), 3 3 1 n∈I(t,θ) where M1(t) is independent of θ and 2\x 0 2 3 2 −A |R (t, θ)| dt A x θ L , x for any fixed A, provided that (3.12) x−5/6−β ≤ θ ≤ exp{−100L2/9}.

4. The case k = 3: reduction to mean-value estimates. The analysis of the case k = 3 is quite similar to the previous one, but we have to be slightly more careful in order to obtain a good error term. We exploit the fact that each Dirichlet polynomial we use is the product of only 3 factors, as opposed to Section 3 where the number of factors was 2h + I. Deﬁne X 1 X 1 P (s) := and P ∗(s) := , ps ps z≤p≤2x z3≤p≤2x

1/3 where z3 is a new parameter satisfying z ≤ z3 ≤ x . Note that if n ≤ 2x −s 3 then a3(n) is precisely the coeﬃcient of n in P (s) . Let b3(n) be the −s ∗ 3 ∗ coeﬃcient of n in P (s) . We write P1(s) = P (s) − P (s) so that a3(n) − 234 A. Zaccagnini

−s b3(n) is the coeﬃcient of n in X3 3 P (s)3 − P ∗(s)3 = P (s)jP ∗(s)3−j j 1 j=1 if n ≤ t. We write X X ∗ P1(s) = Pe(s) and P (s) = Pe(s), −E≤e≤0 1≤e≤F −E−1 −E F −1 where E and F are integers satisfying 2 z3 ≤ z < 2 z3 and 2 z3 ≤ F 2x < 2 z3, and X 1 P (s) := . e ps e−1 e 2 z3≤p<2 z3 z≤p≤2x 3 Since E,F L, for some M L and cm 1 we have XM Y3 3 ∗ 3 P (s) − P (s) = cmP (s; m) where P (s; m) := Pej (s) m=1 j=1

ej −1 with e1 ≤ 0. Write Vj := 2 z3 so that X f (n) P (s; m) = m , ns N8≤n≤N9 Q 3 say, where N8 := j Vj and N9 := 2 N8. As above, we discard those P (s; m) having either N8 ≥ t or N9 ≤ t/2 and relabel the remaining ones so that for some N ≤ M we have X X XN X (4.1) a3(n) = b3(n) + fm(n). n∈I(t,θ) n∈I(t,θ) m=1 n∈I(t,θ)

The same analysis of Section 3, with the bound |fm(n)| ≤ 3!, yields

1/2+\ iT2 X 1 ts − (t − θt)s xL fm(n) = P (s; m) ds + O , 2πi s T2 n∈I(t,θ) 1/2−iT2 for T2 ≤ x. The ranges [−T2, −T3] and [T3,T2] are treated by means of the following mean-value bound, which will be proved in Section 8. 19/60 1/3 5/6 11/12 Lemma 10. Let x ≤ z ≤ x and x ≤ T2 ≤ x . Then, if P (s; m) is as above with V3 ≥ V2 ≥ V1 ≥ z/2, we have 2\T 2 1 P + iτ; m dτ xL62(z−η/6 + T −1/6 + (T V −5/2)1/9) 2 1 2 3 T uniformly for 3 ≤ T ≤ T2, where η is given by (3.7). Primes in almost all short intervals 235

We proceed precisely as in Section 3, using Lemma 7 again with F (s) = P (s; m) and (3.3) for the range [−T3,T3], obtaining

1/2+\ iT3 X 1 (4.2) f (n) = θ P (s; m)ts ds + R (3, t, θ) + R (3, t, θ), m 2πi 1 2 n∈I(t,θ) 1/2−iT3 where −1 2 2 (4.3) R1(3, t, θ) xLT2 + xθ T3 , 2\x 2 3 2 −%/3 −1/3 −5/2 1/9 62 (4.4) |R2(3, t, θ)| dt x θ (z1 + T3 + (T2V3 ) )L , x

2 −1 −5/2 5/4 −5/4 and % = η(T2). Since V3 ≥ xz3 we have T2V3 T2z3 x . We finally choose the parameters: Let ν be a sufficiently large positive con- ν −1 5/6 ν 19/60 stant and set T2 := L max(θ , x ), T3 := L and also x ≤ z3 ≤ L−ν min(θ4/5x, x1/3). Then (4.1)–(4.4) imply X X 00 (4.5) a3(n) = b3(n) + θM3(t, z3) + R (t, θ, z3), n∈I(t,θ) n∈I(t,θ) say, where M3(t, z3) is independent of θ and 2\x 00 2 3 2 60−ν/18 (4.6) |R (t, θ, z3)| dt x θ L , x provided that θ satisfies (3.12). Now choose z := x19/60, so that the hypotheses of both Lemmas 6 and 10 are satisfied, and take ν := 1500. Hence, from Lemma 9, (4.5) and (4.6) we deduce Lemma 11. There exists a small positive constant λ such that if x−5/6−λ ≤ θ ≤ exp{−100L2/9} and (4.7) x19/60 ≤ w ≤ L−1500 min(θ4/5x, x1/3) then for t ∈ [x, 2x] there exists a function M(t, w) independent of θ such that 1 X (4.8) π(t) − π(t − θt) − b (n) = θM(t, w) + R(t, θ, w) 3 3 n∈I(t,θ) where 2\x |R(t, θ, w)|2 dt x3θ2L−20. x 236 A. Zaccagnini

It now remains to estimate the contribution of b3(n). First we remark that 2\x 2\x X 2 X X (4.9) b3(n) dt sup b3(n) b3(n) dt, t∈[x,2x] x n∈I(t,θ) n∈I(t,θ) x n∈I(t,θ) and that a simple argument based on the Brun–Titchmarsh inequality gives

−1 2\x X X min(2x,n\(1−θ) ) (4.10) b3(n) dt b3(n) dt x n∈I(t,θ) x−θx

5. Conclusion of the proof: the main term. Here we choose θ as 2/9 large as possible, i.e. θ = θ0 := exp(−100L ), and any w satisfying (4.7). The Prime Number Theorem gives θ t xθ2 π(t) − π(t − θ t) = 0 + O 0 . 0 log t L2 Hence (4.8) yields t 1 X xθ2 θ M(t, w) − = − b (n) − R(t, θ , w) + O 0 , 0 log t 3 3 0 L2 n∈I(t,θ0) Primes in almost all short intervals 237 so that by Lemmas 11 and 12 we have 2\x 2 2 t x3θ2 log(xw−3) x3θ2 x3θ4 (5.1) θ2 M(t, w) − dt 0 + 0 + 0 . 0 log t L2 L L20 L4 x We ﬁnally take w := L−1500 min(θ4/5x, x1/3). This choice of w implies that the left hand side of (5.1) is x3θ2 log log x 2 0 ε(x) + L2 log x and the ﬁrst estimate of Lemma 5 follows. The second part of Lemma 5 is a consequence of Lemmas 11 and 12 and our choice of w. The proof of the Theorem is therefore complete.

6. Proofs of (3.2) and Lemma 8. In order to prove (3.2) we ﬁrst need the bound X |cm,h| · |em,h(n)| ≤ d3h(n). m By (2.6) this sum is bounded by the coeﬃcient of n−s occurring in YL h ζ(s)2h exp Σ (s) , m m m=1 which, in its turn, is bounded by the one in Y h ζ(s)2h exp Σ (s) m m m≥1 and the latter is a partial product of ζ(s)h. −2h−I We recall that we chose N2 ≥ t/2 and that N1 = 2 N2 by (2.11). Setting 0 N7 := min N1(m), 1≤m≤N(h) the error term with j = 0 in (3.1) is X −1 I/2 −1 t (6.1) 2 d3h(n) min 1,T0 log , 0 n N7

Observe that Hr 6= ∅ only for 0 ≤ r ≤ M, say, with M IT0. Then the 238 A. Zaccagnini sum in (6.1) is

X XM X t −1 d (n) + T −1d (n) log 3h 0 3h n n∈H0 r=1 n∈Hr X XM X −1 −1 −1 d3h(n) + T0 d3h(n)(rT0 ) n∈H0 r=1 n∈Hr XM 1 X d (n). r + 1 3h r=0 n∈Hr −1 −1 −1 −1 Furthermore tT0 exp(−rT0 ) |Hr| tT0 exp(rT0 ) for all r ≤ M, and (3.2) follows using Lemma 3. The term with j = 1 in (3.1) is dealt with in the same way. For Lemma 8 we need the following elementary inequality which is easily proved by induction: For any integer A ≥ 2 and real number B ≥ 3 we have XA Bn ≤ BA. n! n=0 Arguing as in Section 5 of [4] we ﬁnd, after a simple computation, L/2 X L D + 1 L 2D S ≤ (B + 1)h(C + 1)h exp h log + h log 0 m m 2 L m=1 L ≤ exp h (log L)2 , log z1 o(1) for large enough x, since B, C, D L and z1 = x , and Lemma 8 follows.

7. Proof of Lemma 6 Preliminaries. The proof is quite similar to the proof of Lemma 8 in [4]. For the sake of brevity we do not duplicate the whole argument, but merely give the needed modiﬁcations. We say that a set S of points τn ∈ [T, 2T ] is well spaced if |τm − τn| ≥ 1 for every τm, τn ∈ S with n 6= m. We write s = 1/2+iτ and sn = 1/2+iτn throughout this section. We need an estimate for 2\T 2 J1(T ) := |W (s)| dτ. T We ﬁrst write W as the product of W1, W2 and W3, where Y YI1 Y Yh W (s) := X (s) Z(1)(s),W (s) := X (s) Y (s), 1 bi di 2 bi ci Xi≥z1 i=1 Xi

We also set

Y YI1 Y Yh YL YIm x1 := Xi Zi, x2 := Xi Yi, x3 := Zi,

Xi≥z1 i=1 Xi

Lemma 13. Let K(s) be the Dirichlet polynomial X k(n) K(s) = , ns n≤K where K ≥ 2 and |k(n)| ≤ 1 for every n ≤ K. Assume that |K(1/2+iτn)| ≥ K for a set S of well-spaced points τn ∈ [T, 2T ]. Then, uniformly for g ∈ N, we have |S| {K−2gKg + T min(K−2g, K−6gKg)} exp{6g2 log log K}(log TK)5.

This is (8.4) and the following is Lemma 19 of [4].

Lemma 14. For every factor K(s) of W1(s) we have 1/2 −η −1 2 K(s) K (z1 + T )L , uniformly for τ ∈ [T, 2T ], where η = η(T ) is given by (3.7).

Actually, if x3 is large enough, x3 ≥ z1, say, we see that Lemma 6 follows directly from Montgomery’s mean-value bound. In fact, we have

2\T X 2 2 2 |cn| J1 sup |W3(s)| |W1(s)W2(s)| dτ (T + x1x2) , τ∈[T,2T ] n T n≤x1x2 for suitable coeﬃcients cn. The same argument leading to Lemma 13 above 2h2 implies that the last sum is E , and the hypothesis on x3 ensures that −1 T + x1x2 xz1 , which is more than enough for Lemma 6. Hence we may assume in what follows that x3 ≤ z1. We remark that from the deﬁnitions o(1) 1+o(1) above and (2.11) we have x2 = x and x1 = x . We do not rule out the possibility that W1 consists of a single factor Xbi . We use Lemma 14 in conjunction with Montgomery’s mean-value theorem if W1 has at least one factor X (s) or Z(1)(s) with X ≤ x1/6−α or Z ≤ x1/6−α, respectively. bi di i i In fact, setting K(s) = X (s), K = X (resp. K(s) = Z(1)(s), K = Z ), bi i di i 240 A. Zaccagnini

W1(s) = K(s)W4(s), x4 = x1/K, in this case we have 2\T 2 2 J1 sup |W2(s)W3(s)| |W1(s)| dτ τ∈[T,2T ] T 2\T −2η −2 2 x2K(z1 + T ) |W4(s)| dτ, T and the last integral is estimated by means of Montgomery’s theorem, giving X |c0 |2 J x K(z−2η + T −2)(T + x ) n , 1 2 1 4 n n≤x4 0 2h2 for suitable coeﬃcients cn. As above, the last sum is E , and the hypothesis on K ensures that Lemma 6 follows in this case, with β = α/2. From now on we may assume that every factor K(s) of W1(s) has K ≥ 1/6−α x . Thus we have I1 ≤ 12 and there exists a set S of T well-spaced points τn ∈ [T, 2T ] such that X 2 J1 |W (sn)| .

τn∈S

The contribution to the sum of the points τn for which some factor of W1 is ≤ x−1 is easily seen to be T . We discard these points, and from now on −1 assume that each factor of W1 is ≥ x . Then we split the range for each factor of W1(s) into dyadic intervals [Dj, 2Dj) (if the factor is an Xbi (s)) or [E , 2E ) (if the factor is a Z(1)(s)), where j j di −1 d 1/2 −1 e 1/2 x Dj = 2 Xi and x Ej = 2 Zi for some integers d and e. We observe that our hypothesis that each factor of W1(s) is not too small ensures that the number of ranges (that is, the number of values taken by d and e above) is ≤ C2L in each case, for some absolute constant C2. For brevity we write L0 = 2C2L. We may divide the remaining h+I1 points into at most (L0/2) classes S(D, E) where D = (D1,...,Dh) and E = (E1,...,EI1 ), for which (7.1) |X (s )| ∈ [D , 2D ) and |Z(1)(s )| ∈ [E , 2E ). bi n i i di n i i We write Y Y P(D, E) := Di Ei. i i As above, we estimate W2(s) trivially and conclude that

Lemma 15. There exists a set S(D, E) of well-spaced points τn ∈ [T, 2T ] satisfying (7.1) and such that

2 h+I1 J1 T + x2P(D, E) |S(D, E)|L0 . Primes in almost all short intervals 241

We shall give upper bounds for |S| by means of Lemmas 13 and 14. Since these bounds are essentially the same as in [4] we simply quote the results.

Lemma 16. If the hypotheses of Lemma 13 hold for K(s) = Xi(s) with 1/2 K = 2Xi ≥ T then either (7.2) K K1/2T −1(log K)3 or |S| K−4T (log K)9. This is Lemma 18 of [4]. If (7.2) holds, the trivial bound |S| T and Lemmas 15 and 16 imply 1 1/2 Lemma 17. If Xi ≥ 2 T for some i then either (7.3) |S| K−4T (log K)9 or

−1 3+h+I1 (7.4) J1 T + x1x2T L0 .

The second estimate is proved taking K = Di in (7.2) and observing that 3+h+I1 the deﬁnition implies that P |W1(sn)|. Since L0 E and x1x2 ≤ x, (7.4) yields the conclusion of Lemma 6 and more.

Large factors of W1(s). The argument here is essentially the same as in Section 8 of [4], and Lemma 6 follows precisely in the same way, since the results in that section are bounds for |S|. We take a factor of W1(s), K(s) = X (s) or Z(1)(s), and let K = 2X or 2Z , K = D or E accordingly. bi di i i i i We deﬁne σ by means of K = Kσ−1/2. The argument in Section 8 of [4] is as follows: if ϕ is the maximum value of a σ occurring above then Y Y 2 2ϕ−1 2ϕ−1 2ϕ−1 (7.5) P(D, E) ≤ Di Ei ≤ x1 , i i and by Lemma 15 we have

2ϕ−2 h+I1 (7.6) J1 T + xx1 L0 |S(D, E)|. If ϕ ≥ 5/6 then suitable choices of g in Lemma 13 yield |S(D, E)| (T 2−2ϕ + z4−4ϕ)L29E3/2, and the upper bounds for T and z in the hypothesis of Lemma 6 together with (7.5) and (7.6) yield

(ϕ−1)/6 29+h+I1 3/2 J1 T + xx1 L0 E . ϕ−1 The upper bound for x1 which we need is provided by Lemma 14 and A the inequality K x. In conclusion, since L0 A E, we see that Lemma 6 follows if ϕ ≥ 5/6. 242 A. Zaccagnini

Conclusion of the proof of Lemma 6. In the remaining case, Heath- Brown’s argument leads to the stronger inequality 1−γ (7.7) J1 x for some γ > 0. This follows from several bounds for |S| which are essentially the same as in our case. We very brieﬂy sketch the argument, without entering into the details. First the hypotheses of Lemma 6 ensure that o(1) 2 J1 T + x P |S|.

By means of Lemma 13 we prove the following bounds: If K(s) = Xbi (s) then   T 12(1−σ)/5xo(1) in any case, 4−4σ o(1) 2/5 1/2 |S| (T/Xi) x if T ≤ Xi ≤ T ,  2−2σ o(1) 1/2 T x if Xi ≥ T , and if K(s) = Z(1)(s) then di |S| T 12(1−σ)/5xo(1). Using these bounds we see that (7.7) holds provided that the following conditions hold. 1/3+δ First case. If Xi ≥ x for some δ ≥ β and σ ≥ ϕ − ε we need to have 1 1 1 2 2 γ < min 6 − β, 18 − 3 β − 2ε, 3 δ − 3 β − 2ε . 1/3+δ Second case. If Xi ≥ x for some δ ≥ β and σ ≤ ϕ − ε we need to have 1 2 γ < min 6 − β, 3 ε − β . 1/3+δ Third case. If Xi ≤ x for all i we need 1 2 1 1 γ < min 6 − β, 3 ε − β − 4δε, 6 α − 3 β − 2ε . Now, we easily see that the choices 1 1 1 δ = 30 , β = 30 α, ε = 15 α allow the choice γ = α/50 and satisfy the hypotheses of Lemma 6.

8. Proof of Lemma 10. This lemma is proved in a similar fashion to Lemma 11 in [4] and we simply sketch the argument, with the necessary changes. As in Section 10 of [4], let F = (F1,F2,F3) and S(F) be a set of well-spaced points τn ∈ [T, 2T ] such that

Fi ≤ |Pei (1/2 + iτn)| < 2Fi for i = 1, 2, 3. The same argument of Section 7 gives 2\T Y3 2 3 2 (8.1) |P (1/2 + iτ)| dτ T2 + L |S(F)| Fi T i=1 Primes in almost all short intervals 243

σ−1/2 for some F. Fix an index i and set K = Fi = Vi and K = 2Vi. We remark that our choice of parameters implies that 1/3 1/2 (8.2) T2 K T2 . We use Lemma 13 with several diﬀerent values of g. First, if ϕ = max σ ≥ 5/6, we choose g = 2 and (8.2) implies that 2−2ϕ 29 |S(F)| T2 L , and Lemma 10 easily follows as in [4], on substituting into (8.1), since Q 2 Q 2ϕ−1 2ϕ−1 ϕ−1 Fi ≤ Vi ≤ x . An upper bound for x is provided by Lemma 14. In the other case, choose g = 3 to obtain (8.3) |S(F)| K6−6σL59 −1/2 g 1/2 or g in such a way that T2K ≤ K ≤ T2K . In the latter case we have (8.4) |S(F)| (TK1/2)2−2σL59 since g ≤ 3 anyway. Since now σ ≤ 5/6, (8.3) and (8.4) imply 6−6σ −5/2 1/3 59 |S(F)| K (T2K ) L 2/5 2/5 when K ≤ T2 and when K ≥ T2 respectively. This means that 6 σ−1/2 6 6−6σ 59 3 59 Fi |S(F)| (K ) K L = K L , 6 σ−1/2 6 6−6σ −5/2 1/3 59 3 −5/2 1/3 59 Fi |S(F)| (K ) K (T2K ) L = K (T2K ) L . We use the former for i = 1, 2, and the latter for i = 3, take their geometric 2 mean, and from (8.1) we obtain Lemma 10 in this case too, since Fi ≤ 2σ−1 Vi ≤ Vi.

9. Some comments. The knowledgeable reader sees at once that we had to make a diﬀerent choice for the Dirichlet polynomials from Heath-Brown [4]. Indeed, the choice therein leads to too large error terms in Lemma 4 since we have a larger z than Heath-Brown and a much smaller h. This is due to the fact that we need z to be almost x1/3, since we have the same problems he encounters in Section 9 when the product W has 6 factors, but already with only 3 factors. The slight additional diﬃculty is more than compensated by the fact that we only have to save a little over the estimate given by Montgomery’s theorem, since our problem leads naturally to estimating the mean-square of a Dirichlet polynomial. We did not use Watt’s mean-value bound (Theorem 2 of [12]) in prov- ing Lemma 6, because the hypothesis T ≥ K4 (in our notation) limits the former’s usefulness in this problem to a subrange of the values of the parameters in Lemma 6. In particular, the case when some function Xbi (s) 1/6−α or Zdi (s) has length K (= Xi or Zi resp.) bounded by x can be more 244 A. Zaccagnini easily handled by means of Montgomery’s theorem alone. Compare the com- ment following the proof of Proposition 2.2 in [12] with the hypothesis of our Lemma 17. Even the more general Theorem 1 of Watt’s paper [11] has, essentially, the same disadvantage.

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Dipartimento di Matematica Universit`adi Parma via Massimo d’Azeglio 85/a 43100 Parma, Italy E-mail: [email protected]

Received on 26.3.1997 and in revised form on 24.6.1997 (3157)