This is a self-archived – parallel-published version of an original article. This version may differ from the original in pagination and typographic details. When using please cite the original. AUTHOR Salmensuu, Juho TITLE ON THE WARING-GOLDBACH PROBLEM WITH ALMOST EQUAL SUMMANDS YEAR 2020 DOI https://doi.org/10.1112/mtk.12019 CITATION ON THE WARING-GOLDBACH PROBLEM WITH ALMOST EQUAL SUMMANDS - Mathematika 66 (2020) 255–296 doi:10.1112/mtk.12019 On the Waring-Goldbach problem with almost equal summands Juho Salmensuu Abstract We use transference principle to show that whenever s is suitably large depending on k ≥ 2, every sufficiently large natural number n satisfying some congruence conditions can be written in the form n = pk1 + · · · + pks , where p1, . . . , ps ∈ [x − xθ, x + xθ] are primes, x = (n/s)1/k and θ = 0.525 + . We also improve known results for θ when k ≥ 2 and s ≥ k2 + k + 1. For example when k ≥ 4 and s ≥ k2 + k + 1 we have θ = 0.55 + . All previously known results on the problem had θ > 3/4. 1 Introduction Let k ≥ 2. For each prime p, define τ(k, p) so that pτ(k,p)||k. Notation pτ(k,p)||k means that pτ(k,p)|k and pτ(k,p)+1 - k. Let Rk = ∏ p:(p−1)|k p η(k,p) where η(k, p) = { τ(k, p) + 2 if p = 2 and τ(k, p) > 0 τ(k, p) + 1 otherwise (1) First result concerning Waring-Goldbach problem is from Hua [Hua38] who showed that every sufficiently large natural number n ≡ s (mod Rk) can be written in the form n = pk1 + · · ·+ pks , (2) where p1, . . . , ps are primes and s > 2k. Since then the number of required summands has been greatly reduced, the latest improvement being from Kumchev and Wooley [KW17] who proved that (2) holds, for large k, if s > (4k − 2) log k − (2 log 2− 1)k − 3. Another interesting way to study the Waring-Goldbach problem is to replace the set of primes with some sparse subset of the primes. A natural way to choose such subset is to restrict primes to lie in a short interval Iθ = [x − xθ, x + xθ], where x = (n/s)1/k and θ ∈ (1/2, 1) 1. Let θk,s be the least exponent such that (2) can be solved, for all sufficiently large n ≡ s (mod Rk) and p1, . . . , ps ∈ Iθ, whenever θ > θk,s. Wei and Wooley [WW15] were first ones able to show that for every k ≥ 2 there exists s such that θk,s < 1. They showed that if s > max(6, 2k(k − 1)), then θk,s ≤  19/24 if k = 24/5 if k = 3 5/6 if k ≥ 4. Huang [Hua16] improved that result by showing that θk,s ≤ 19/24, for all k ≥ 3 and s > 2k(k−1). The latest result is from Kumchev and Liu [KL17] who showed that θk,s ≤ 31/40, when k ≥ 2 and s ≥ k2 + k + 1. As we can see from the previous results there has been difficulties to prove that θk,s ≤ 3/4. We break that barrier in this paper. 2 1The limitation θ > 1/2 is a consequence of a slight modification of Wright’s argument (see [Wri37]). We talk more about it in Section 2. 2Recently this barrier has also been broken in [MS19] as a consequence of finding a new estimate for the exponential sum: ∑ N 0 and θ ∈ (1/2, 1). Let α− > 0 be such that, whenever x is sufficiently large, we have for each interval I ⊂ [x, x+ xθ+] of length |I| ≥ xθ− and for every c, d ∈ N such that (c, d) = 1 and d ≤ log x,∑ n∈I n is prime n≡c (mod d) 1 ≥ α −|I| φ(d) log x . (3) Suppose that s > max ( 2 α−(2θ − 1) , k + 2 α−θ , k2 + k ) . Then, for every sufficiently large integer n ≡ s (mod Rk), there exist primes p1, . . . , ps such that |(n/s)1/k − pi| ≤ (n/s)θ/k for each i = 1, . . . , k and n = pk1 + · · ·+ pks . When θ > 11/20 inequality (3) holds for α− = 99/100 (see [Har07, Theorem 10.3]) and when θ > 0.525 inequality (3) holds for α− = 9/100 (see [Har07, Theorem 10.8]). Thus θ2,7 ≤ 893/1386 = 0.644, θ3,13 ≤ 1487/2574 = 0.578, θk,s ≤ 11/20 = 0.55, when k ≥ 4 and s > k2 + k, θk,s ≤ 0.525, when k ≥ 2 and s > max(k2 + k, 444, 4000(k + 2)/189). These significantly improve previously known results which always had θ > 3/4. 2 Outline In this section we give an outline of the proof of Theorem 1. We will introduce the used notation in Section 3. In Section 4 we prove the following transference lemma that is the main ingredient in proving Theorem 1. Lemma 1. Let s ≥ 3 and , η ∈ (0, 1). Let N be a natural number and, for each i ∈ {1, . . . , s} let fi : [N ]→ R≥0 be a function that satisfies the following assumptions: 1. (Mean condition) For each arithmetic progression P ⊂ [N ] with |P | ≥ ηN we have En∈P fi(n) ≥ 1/s+ ; 2. (Pseudorandomness condition) There exists a majorant νi : [N ] → R≥0 with fi ≤ νi pointwise, such that ||ν̂i − 1̂[N ]||∞ ≤ ηN ; 3. (Restriction estimate) We have ||f̂i||q ≤ KN1−1/q for some q,K with s− 1 < q < s and K ≥ 1. Then for each n ∈ [N/2, N ] we have f1 ∗ · · · ∗ fs(n) ≥ (c()−O,K,q(η))Ns−1, where c() > 0 is a constant depending only on . 2 The idea of the proof is following: If function f satisfies conditions 1-3, then we can find functions g : [N ]→ [0, 1] and h : [N ]→ R such that f = g + h, both g and h satisfy condition 3, g satisfies condition 1 and h is Fourier uniform (i.e. ||ĥ||∞ ≤ ηN). Functions g and h are often called anti-uniform and uniform part of f , respectively. Using Hölder’s inequality we can then reduce the problem to showing that g1 ∗ · · · ∗ gs(n) Ns−1. This problem can be solved using induction and strategy that is very similar to the proof of [EGM14, Theorem 4.1]. One of the key ingredient using this strategy is an arithmetic regularity lemma regularizing multiple functions simultaneously (Lemma 3). Next we explain how Lemma 1 implies Theorem 1. Let fi : {1, . . . , N} → R≥0 be a weighted W-tricked characteristic function of k-th powers of primes in short interval (for precise definitions, see Subsection 5.1). We define the functions νi in a similar way, but we replace the characteristic function of primes with a majorant function based on the linear sieve. In Section 5 we show that if conditions 1-3 of Lemma 1 hold for the function fi, then by Lemma 1 it follows that every sufficiently large natural number n ≡ s (mod Rk) can be written as the sum of s kth power of primes, which belong to the short interval. So it remains to show that the function fi satisfies conditions 1-3 of Lemma 1. In Section 6 we establish condition 1 for our precise choice of fi with an easy calculation using knowledge about primes in arithmetic progressions in short intervals. In Section 7 we establish condition 2 which essentially corresponds to understanding the func- tion f ′(b, d, α) = ∑ X1/k d ≤r≤ (X+Y ) 1/k d dkrk≡b (mod W ) e (dkrkα W ) , where b, d,W,X, Y ∈ N, α ∈ [0, 1] and Y  X1−1/k+θ/k. We split [0, 1] into two disjoint sets, major arcsM and minor arcs m, using the Hardy-Littlewood decomposition, and treat the function f ′(b, d, α) differently in those two. In Subsection 7.1 we prove that f ′(b, d, α) = o(Y X1/k−1), if α ∈ m. In Subsection 7.2 we establish similar bound for those α ∈M that are not very close to zero. We also show that if α is very close to zero, then f ′(b, d, α) ≈ X 1/k−1 dk 1̂[N ](α) + o(Y X 1/k−1). With these results we are able to prove the pseudorandomness of ν. In Section 8 we prove condition 3, by first using the main conjecture in Vinogradov’s mean value theorem established by Bourgain, Demeter and Guth [BDG16, Theorem 1.1]3 and Daemen’s result concerning localized solutions in Waring’s problem [Dae10, Theorem 3] to show that ||f̂ ||u  N1−1/u+ (4) for u ≥ k2 + k. Then we apply Bourgain’s strategy (see [Bou89, Section 4]) to the inequality (4) in order to get ||f̂ ||u  N1−1/u for u > k2 + k as requested. 3Wooley has an alternative proof of the main conjecture in Vinogradov’s mean value theorem in [Woo17]. 3 Remark 1. Let us now say a few words about the lower bound of the number of required summands. In Theorem 1 we need s > max ( 2 α−(2θ − 1) , k + 2 α−θ , k2 + k ) . The first requirement s > 2α−(2θ−1) comes from the pseudorandomness condition on the minor arcs. This essentially says that the number of required summands goes to infinity as θ approaches 1/2. We expect this kind of behaviour because, when θ ≤ 1/2, we cannot anymore represent every sufficiently large natural number n at form n = nk1 + · · · + nks , where |(n/s)1/k − ni| ≤ c(n/s)θ/k and c is some small coefficient. This can be seen from Wright’s argument (see [Wri37]). The idea of this argument is to take n = s(mk + kmk−1) and write ni = m + ai. Comparing both sides of equation n = nk1 + · · · + nks , we see that equality is impossible when m is large enough. Using a slight modification of Wright’s construction (namely taking n = s(mk + kmk−1) + u, where u = o(n1−2/k)), we can see that non-representability concerns also those numbers n for which n ≡ s (mod Rk). The second requirement s > k+2α−θ comes from the pseudorandomness condition on the major arcs. If we had α− = 1, then the term k+2α−θ would be dominated by max( 2 α−(2θ−1) , k 2 + k). The third requirement s > k2 + k comes from the restriction estimate. Since the third require- ment is the most limiting one when k is large, it is interesting to ask whether it can be improved. When θ = 1 we can replace k2 + k by approximately k2 using similar calculations as in [Cho18, Section 5]. This suggests that, for θ ∈ (1/2, 1), k2 + k can be replaced by something that depends on θ. Therefore at least minor improvements to the requirement s > k2 + k should be possible when θ > 1/2. Remark 2. When k = 1 we can establish a similar result to Theorem 1 requiring only that the number of summands is s > 2α−θ . Based on the proof of Theorem 1 we only need to establish the restriction estimate and the pseudorandomness condition for the transference function defined in (26) when k = 1. That can be done in a similar way to how we do it in this paper when k ≥ 2, but the calculations are much simpler. The value α− in requirement s > 2α−θ follows from the mean value estimate, and part 2/θ is a consequence of the linear sieve. Remark 3. Using [Kou15, Theorem 1.3] in the proof of Theorem 1 one can easily prove that if k > 1, θ > 1/2 and s are as in Theorem 1, then for almost every sufficiently large integer n ≡ s (mod Rk), there exist primes p1, . . . , ps such that |(n/s)1/k − pi| ≤ (n/s)θ/k for each i = 1, . . . , k and n = pk1 + · · ·+ pks . We can also prove a similar result when k = 1, θ > 1/15 and s > 2α−θ . The author want to thank Trevor Wooley for helping to observe this fact. 3 Notation For finitely supported functions f, g : Z→ C, we define convolution f ∗ g by f ∗ g(n) = ∑ a+b=n f(a)g(b). For a set A, write 1A(x) for its characteristic function. Define [N ] = {1, . . . , N}. Let A,B ⊆ [N ] and η > 0. We define Sη(A,B) by Sη(A,B) = {n : 1A ∗ 1B(n) ≥ ηN}. The Fourier transform of finitely supported function f : Z→ C is defined by f̂(α) = ∑ n∈Z f(n)e(−nα) 4 where e(x) = e2piix. We will also use notation eW (n) as an abbreviation for e(n/W ). Let f : R → C and g : R → R+. We write f = O(g), f  g if there exists a constant C > 0 such that |f(x)| ≤ Cg(x) for all values of x in the domain of f . If f takes only positive values we then define similarly f  g if there exists a constant C > 0 such that f(x) ≥ Cg(x) for all values of x in the domain of f . If the implied constant C depends on some contant  we use notations O,,. If f  g and f  g we write f  g. We also write f = o(g) if lim x→∞ f(x) g(x) = 0. The function f is asymptotic to g, denoted f ∼ g if lim x→∞ f(x) g(x) = 1. We will use notation T for R/Z. We also define norms (lp-norm) ||f ||lp(N) = ( En≤N |f(n)|p )1/p , (Lp-norm) ||g||p = (∫ T |g(α)|pdα )1/p , (Lipschitz norm) ||h||Lip = inf{K ∈ R | ∀x,y ∈ X : |h(x)− h(y)| ≤ Kd(x,y)}, for functions f : N → C, g : R → C and h : X → C where X is a metric space with metric d : X ×X → R+. For the function f : [N ]→ C we define Gowers U2-norm by ||f ||U2(N) = ||f ||U2(G)/||1[N ]||U2(G), where G = Z/N ′Z for some arbitrary N ′ > 4N and ||f ||U2(G) = (Ex,h1,h2∈G f(x)f(x+ h1)f(x+ h2)f(x+ h1 + h2))1/4. The functions f and 1[N ] are regarded as functions on G by defining f(x) = 1[N ](x) = 0 if x ∈ G \ [N ], where [N ] is regarded as embedded in G in a natural manner. Note that ||f ||U2(N) is independent of the choice of N ′. Acknowledgments The author is grateful to his supervisor Kaisa Matomäki for suggesting the topic and for many useful discussions. The author also thanks Joni Teräväinen for reading the paper and giving useful comments. The author thanks the referee for careful reading of the paper and for useful comments. During the work author was supported by Academy of Finland project no. 293876 and by project funding from Emil Aaltonen foundation. 4 Transference principle In this section our aim is to prove Lemma 1 that is a generalization of [MMS17, Proposition 3.1]. Lemma 1 is based on the transference principle. The transference principle was first introduced by Green [Gre05] and it has appeared to be a powerful tool to study additive problems. The following example shows how the transference principle works. Let A be a sparse set of positive integers and say that we are interested in existence of solutions of linear equation x1 + · · ·+ xs = n, where x1, . . . , xs ∈ A and n ∈ N. That corresponds to finding a positive lower bound for the sum∑ x1+···+xs=n 1A(x1) · · · 1A(xs). Depending on the set A that might be difficult to find directly. 5 Let f := ν1A, where ν : N → R+ is some suitably chosen weight function. The key of the transference principle is to find some set B ⊂ N with positive density such that f̂ ≈ 1̂B . Due to the positive density of B one might hope to prove that∑ x1+···+xs=n 1B(x1) · · · 1B(xs) > 0. (5) Then ∑ x1+···+xs=n f(x1) · · · f(xs) = ∫ T f̂(α)se(αn)dα ≈ ∫ T 1̂B(α) se(αn)dα = ∑ x1+···+xs=n 1B(x1) · · · 1B(xs) > 0, which, once made rigorous, implies that∑ x1+···+xs=n 1A(x1) · · · 1A(xs) > 0. 4.1 Sumset estimates In this subsection we prove some helpful lemmas about sumsets which we use later to prove the result that is similar to (5). Lemma 2. For any  > 0, there exists a constant η = η() > 0 such that the following statement holds. Let N be a natural number and α, β ∈ [0, 1]. Let A,B ⊂ [N ] be two subsets with the properties that |A ∩ P | ≥ α|P | and |B ∩ P | ≥ β|P |, for each arithmetic progression P ⊆ [N ] with |P | ≥ ηN . Then |Sη(A,B)| ≥ 2N min(α+ β, 1)− N. The proof we are going to present mainly follows ideas of the proof of [EGM14, Theorem 4.1]. The main differences are that we only need part of that proof and we consider Sη(A,B) instead of Sη(A,−A). In order to prove Lemma 2 we need to first establish an arithmetic regularity lemma that is valid for multiple sets simultaneously. In general the arithmetic regularity lemma says that bounded function f : [N ]→ C can be decomposed into a (well-equidistributed, virtual) s-step nilsequence, an error which is small in L2-norm and a further error which is minuscule in the Gowers Us+1-norm, where s ≥ 1 is a parameter. The proof and some applications of such regularity lemma can be found in [GT10]. We only need the arithmetic regularity lemma in the case s = 1, and the proof of this simpler case can be found also in [Ebe16] which we will utilise. Before we present and prove our regularity lemma, we need some necessary definitions. We define a metric on Td by d(x, y) = min z∈Zd ||x− y − z||2. Using usual metric on [0, 1], the previously defined metric on Td and the discrete metric on Z/qZ we define a metric on [0, 1]× Z/qZ× Td by the sum of these metrics. Let A,N ∈ N. We say that θ ∈ Td is (A,N)-irrational, if q = (q1, . . . , qd) ∈ Zd and ∑ i |qi| < A implies that ||q · θ||T ≥ A/N . We say a subtorus T of Td of dimension d′ has complexity at most M if there is some L ∈ SLd(Z), all of whose coefficients have size at most M , such that L(T ) = Td′ × {0}d−d′ . In this case we implicitly identify T with Td′ using L. For instance, we say θ ∈ Td is (A,N)-irrational in T if L(θ) is (A,N)-irrational in Td′ . By a growth function, we mean increasing function F : R+ → R+. 6 Lemma 3. Let N ∈ N. For k ≥ 1, let f1, . . . , fk : [N ] → [0, 1] be functions, F : N → R+ a growth function and  > 0. Then there exist a quantity M ,F 1, positive integers q, d ≤M and (F(M), N)-irrational θ ∈ Td such that, for each i ∈ {1, . . . , k}, we have a decomposition fi = f (i) str + f (i) sml + f (i) unf of fi into functions f (i) str, f (i) sml, f (i) unf : [N ]→ [−1, 1] such that 1. f (i)str = Fi(n/N, n (mod q), θn) for some function Fi : [0, 1] × Z/qZ × Td → [0, 1] with ||Fi||Lip ≤M , 2. ||f (i)sml||l2(N) ≤ , 3. ||f (i)unf ||U2(N) ≤ 1/F(M). Proof. This proof is a straight-forward generalization of the proof of [Ebe16, Theorem 7]. Let F∗ and, for i ∈ {1, . . . , k}, Fi be growth functions depending on  > 0 and F in a manner to be determined. By [Ebe16, Theorem 5], for each i ∈ {1, . . . , k}, there exists Mi ,Fi 1 and a decomposition fi = f (i) str + f (i) sml + f (i) unf of fi into functions f (i) str, f (i) sml, f (i) unf : [N ]→ [−1, 1] such that 1. f (i)str = F ′i (θin), where F ′i : Tdi → [0, 1] and θi ∈ Tdi with di, ||F ′i ||Lip ≤Mi, 2. ||f (i)sml||l2(N) ≤ , 3. ||f (i)unf ||U2(N) ≤ 1/Fi(Mi). Let θ′ = (θ1, . . . , θk) ∈ Td1+···+dk be the concatenation of the vectors θi. We define functions F ′′i (x1, . . . , xk) := F ′ i (xi) for xi ∈ Tdi . By [Ebe16, Theorem 6] we can findM∗ M1,...,Mk,F∗ 1 such thatM∗ ≥Mi for all i ∈ {1, . . . , k} and θ′ decomposes as θ′ = θsmth + θrat + θirrat, where 1. d(θsmth, 0) ≤M∗/N , 2. qθrat = 0 for some q ≤M∗ and 3. θirrat is (F∗(M∗), N)-irrational in a subtorus of complexity≤M∗, which means that L(θirrat) is (F∗(M∗), N)-irrational in Td′ . Then for each i ∈ {1, . . . , k} F ′′i (θ ′n) = F ′′i (θsmthn+ θratn+ θirratn) = Fi(n/N, n (mod q), nL(θirrat)), where Fi : [0, 1]× Z/qZ× Td′ → [0, 1] is defined by Fi(x, y, z) = F ′′ i (Nθsmthx+ θraty + L −1(z)). Noting that ||Fi||Lip M∗ 1, we can findM M∗ 1 exceedingM∗ and ||Fi||Lip for all i ∈ {1, . . . , k}. But since M M∗ 1, if F∗ grows rapidly enough depending on F then F∗(M∗) > F(M), and sim- ilarly M∗ M1,...,Mk,F∗ 1, so if Fi grows rapidly enough depending on F∗ for all i ∈ {1, . . . , k} then Fi(Mi) > F∗(M∗) > F(M) for all i ∈ {1, . . . , k}. After all these dependencies are fixed we have M ,F 1, and the conclusion of the theorem holds.  7 Proof of Lemma 2. We can assume that N is sufficiently large depending on , since other- wise we can choose η = 1N and lemma is trivially true. Let F ′ : N → R+ be a growth function depending on . Let ′ = min(, 125 ). Then by Lemma 3 there exists M ′ ,F ′ 1, positive integers q, d ≤M ′ and (F ′(M ′), N)-irrational θ ∈ Td such that 1A = f A tor + f A sml + f A unf , where ||fAsml||l2(N) ≤ ′30, ||fAunf ||U2(N) ≤ 1/F ′(M ′) and fAtor(n) = FA(n (mod q), n/N, θn) for some FA : Z/qZ× [0, 1]× Td → [0, 1] with ||FA||Lip ≤M ′. Similarly 1B = f B tor + f B sml + f B unf , where fBtor, fBsml, f B unf satisfy same requirements with subscripts and superscripts A replaced by B. Let M = d′−30M ′e and consider, for a ∈ Z/qZ and i ∈ {1, . . . ,M}, the progressions Ia,i = { n ∈ ( (i− 1)N M , iN M ] : n ≡ a (mod q) } . Define FA,a,i : Td → [0, 1] by FA,a,i(x) = FA(a, i/M, x) and FB,a,i : Td → [0, 1] by FB,a,i(x) = FB(a, i/M, x). Define also fAstruct(n) = ∑ a (mod q) M∑ i=1 1Ia,i(n)FA,a,i(θn) and fBstruct(n) = ∑ a (mod q) M∑ i=1 1Ia,i(n)FB,a,i(θn). Since FA is M ′-Lipschitz we see that ||fAstruct − fAtor||∞ ≤ ′30. Similarly ||fBstruct − fBtor||∞ ≤ ′30. Now we have decomposition 1A = f A struct + f ′A sml + f A unf where ||f ′Asml||l2(N) ≤ 2′30, ||fAunf ||U2(N) ≤ 1/F ′(M ′). Similar bounds hold with A replaced by B. Now given an arbitrary growth function F depending on , we may choose F ′ to grow sufficiently rapidly depending on  so that F ′(M ′) > F(M), whence ||fAunf ||U2(N), ||fBunf ||U2(N) ≤ 1/F(M) and θ is (F(M), N)-irrational. Write δA(a, i) for the density of A in Ia,i and δB(a, i) for the density of B in Ia,i. Let E be set of those pairs (a, i) ∈ Z/qZ × {1, . . . ,M} for which En∈Ia,i |f ′Asml(n)| > ′15 or En∈Ia,i |f ′Bsml(n)| > ′15. We see that |E| ≤ 2′14qM (6) since otherwise En≤N |f ′Csml(n)| ≥ 1 N ∑ (a,i)∈E ∑ n∈Ia,i |f ′Csml(n)| > 1 N ( N qM − 2 ) qM′29 ≥ 2′30 for either C = A or C = B, and that leads to a contradiction using Cauchy-Schwarz because ||f ′Csml||l2(N) ≤ 2′30. Now using deductions of the proof of [EGM14, Lemma 4.4] we get that if (a, i) 6∈ E then ∫ Td FC,a,i(x)dx ≥ δC(a, i)− ′14 (7) for both C = A and C = B. Next we prove a variant of [EGM14, Lemma 4.7]. 8 Claim 1. Let a, b ∈ Z/qZ and i, j ∈ [M ] be such that (a, i), (b, j) 6∈ E and δA(a, i), δB(b, j) ≥ 2′2. Then |S′20/(10M2)(A,B) ∩ Ia+b,i+j | ≥ N qM min(δA(a, i) + δB(b, j), 1)− 10 ′2N qM . Proof. By (7) it suffices to prove |S′20/(10M2)(A,B) ∩ Ia+b,i+j | ≥ N qM min (∫ Td FA,a,i(x)dx+ ∫ Td FB,b,j(x)dx, 1 ) − 8 ′2N qM . From δA(a, i), δB(b, j) ≥ 2′2 we get that∫ Td FA,a,i(x)dx, ∫ Td FB,b,j(x)dx ≥ ′2. (8) Let I∗ be the set of those integers c ∈ Ia+b,i+j for which |Ia,i ∩ (c− Ib,j)| ≥  ′2N qM . (9) We see that |Ia+b,i+j \ I∗| ≤ 2′2N/qM (only values near the right end of Ia+b,i+j do not belong to I∗). Note that a product of two M -Lipschitz functions, each of which is bounded pointwise by 1, is 2M -Lipschitz. Therefore, when c ∈ I∗ and F is sufficiently rapidly growing, we can use [EGM14, Lemma A.3] to the function F (x) = FA,a,i(x)FB,b,j(θc− x) to get that fAstruct|Ia,i ∗ fBstruct|Ib,j (c) = ∑ n∈Ia,i∩(c−Ib,j) FA,a,i(θn)FB,b,j(θ(c− n)) ≥ |Ia,i ∩ (c− Ib,j)|(FA,a,i ∗ FB,b,j(θc)− 1 4 ′12), where FA,a,i ∗ FB,b,j(x) = ∫ Td FA,a,i(y)FB,b,j(x− y)dy. By [EGM14, Lemma A.11] we have that FA,a,i ∗ FB,b,j is M -Lipschitz. Since θ is (F(M), N)- irrational and F can be chosen to grow arbitrarily fast, we have by [EGM14, Lemma 4.5] that 1 |Ia+b,i+j | |{c ∈ Ia+b,i+j : FA,a,i ∗ FB,b,j(θc) >  ′12/2}| > µ(Y )− ′12, where Y = {y ∈ Td : FA,a,i ∗ FB,b,j(y) ≥ ′12}. But by (8) and [EGM14, Lemma 4.6] with η = ′12 we have that µ(Y ) ≥ min (∫ Td FA,a,i(x)dx+ ∫ Td FB,b,j(x)dx, 1 ) − 4′2. Putting this all together, fAstruct|Ia,i ∗ fBstruct|Ib,j (c) ≥ ′14N 4qM for a set of c ∈ Ia+b,i+j of size at least N qM min (∫ Td FA,a,i(x)dx+ ∫ Td FB,b,j(x)dx, 1 ) − 7 ′2N qM provided that N is large enough depending on . We denote the set of those values c by I ′. 9 Now using the fact that En∈Ib,j |f ′Bsml(n)| ≤ ′15 when (b, j) 6∈ E and [EGM14, Lemma A.12] with η = ′15 we see that ∣∣∣fAstruct|Ia,i ∗ f ′Bsml|Ib,j (c)∣∣∣ ≤ ′15NqM for all c ∈ I ′. Similar bounds apply to ∣∣∣f ′Asml|Ia,i ∗ fBstruct|Ib,j (c)∣∣∣ and ∣∣∣f ′Asml|Ia,i ∗ f ′Bsml|Ib,j (c)∣∣∣. Therefore (fAstruct + f ′A sml)|Ia,i ∗ (fBstruct + f ′Bsml)|Ib,j (c) ≥ ′14N 5qM for all c ∈ I ′. Recalling that 1A = f A struct + f ′A sml + f A unf , 1B = f B struct + f ′B sml + f B unf , ||fAunf ||U2(N), ||fBunf ||U2(N) ≤ 1/F(M) and provided that F grows fast enough [EGM14, Lemma A.13] implies that 1A|Ia,i ∗ 1B |Ib,j (c) ≥ ′14N 8qM for all c in a subset Ia+b,i+j of size at least N qM min (∫ Td FA,a,i(x)dx+ ∫ Td FB,b,j(x)dx, 1 ) − 8 ′2N qM . All these c lie in S′14/8qM (A,B), which is of course contained in S′20/10M2(A,B).  Conclusion of the proof of Lemma 2. Set η = ′20/10M2. We can assume that N ≥ η−1 since oth- erwise the claim is obvious. Then |Ia,i| ≥ NqM −2 ≥ ηN for all a ∈ Z/qZ and i ∈ {1, . . . , 2M}. Now we have by assumption that δA(a, i) ≥ α and δB(b, j) ≥ β for all a, b ∈ Z/qZ and i, j ∈ {1, . . . ,M}. Thus by Claim 1 we get that |Sη(A,B) ∩ Ia,i| ≥ N qM min(α+ β, 1)− 10 ′2N qM , (10) where (a, i) ∈ Z/qZ×{1, . . . , 2M} excluding an exceptional set with size at most 2|E|. Now using (6), (10) and the fact that ′ = min(, 1/25) it follows that |Sη(A,B)| = ∑ a∈Z/qZ,i∈{1,...,2M} |Sη(A,B) ∩ Ia,i| ≥ (2Mq − 2|E|) ( N qM min(α+ β, 1)− 10 ′2N qM ) ≥ 2N min(α+ β, 1)− 2|E| N qM − 20′2N ≥ 2N min(α+ β, 1)− 4′14qM N qM − 20′2N ≥ 2N min(α+ β, 1)− N.  Lemma 4. For any , δ ∈ (0, 1), there exists constant η = η(, δ) > 0 such that the following statements holds. Let N be natural number and α, β ∈ (, 1). Let A,B ⊂ [N ] be two subsets with the properties that |A ∩ P | ≥ α|P |, |B ∩ P | ≥ β|P |, for each arithmetic progression P ⊆ [N ] with |P | ≥ ηN . Then |Sη(A,B) ∩Q| ≥ |Q|min(α+ β, 1)− |Q|, for each arithmetic progression Q ⊆ [2N ] with |Q| ≥ 2δN . 10 Proof. Let η′ be as η in Lemma 2 and set that η = η′δ. We can assume that N ≥ 2η−1 since otherwise statement is obvious. Given Q we see that there exist progressions Q1 ⊆ [N ] and Q2 ⊆ [N ] with the same common difference such that Q1 + Q2 ⊆ Q, |Q1| = |Q2| and |Q| ≤ |Q1|+ |Q2|. (Simply choose Q1 = qH+ bminQ2 c and Q2 = qH+ dminQ2 e, where q is common difference of Q and H = {0, . . . , b |Q|−12 c}). Let A′ = A∩Q1 and B′ = B∩Q2. Clearly A′+B′ ⊆ Q. Recall that η = η′δ. Since |Q1|, |Q2| ≥ δN it follows that η′max(|A′|, |B′|) ≥ η′max(α|Q1|, β|Q2|) ≥ η′δN ≥ ηmax(|A|, |B|) and therefore |Sη(A,B) ∩Q| ≥ |Sη′(A′, B′)|. (11) Now define A′′ = A ′−minQ1 q and B ′′ = B ′−minQ2 q , where q is the common difference of the pro- gression Q. We see that Sη′(A ′, B′) = Sη′(A′′, B′′). Our aim is now to use Lemma 2 to sets A′′ and B′′. Recall that η = η′δ. Let N ′ = |Q1| and P ⊆ [N ′] be progression such that |P | ≥ η′N ′ ≥ η′δN ≥ ηN . Set P ′ = qP + minQ1. Then by assumption |A′′ ∩ P | = |A′ ∩ P ′| = |A ∩Q1 ∩ P ′| = |A ∩ P ′| ≥ α|P ′| = α|P |. Similarly |B′′ ∩ P | ≥ β|P |. Since 2N ′ = |Q1|+ |Q2| ≥ |Q| ≥ N ′ it follows by Lemma 2 that |Sη′(A′′, B′′)| ≥ 2N ′min(α+ β, 1)− N ′ ≥ |Q|min(α+ β, 1)− |Q|. Thus by (11) |Sη(A,B) ∩Q| ≥ |Q|min(α+ β, 1)− |Q|.  4.2 Transference lemma In this subsection we will finally establish Lemma 1, that is a crucial ingredient in proving our main theorem. Before that we use induction over Lemma 4 to get the following lemma that essentially is our version of (5). Lemma 5. For any  ∈ (0, 1) and s ∈ N, there exists a constant η = η(, s) > 0 such that the following statement holds. Let N be natural numbers, s > 2 and, for i ∈ {1, . . . , s}, let fi : [N ]→ [0, 1] and αi > 0 be such that En∈P fi(n) ≥ αi +  (12) for each arithmetic progression P ⊆ [N ] with |P | ≥ ηN . Assume that α1 + · · ·+ αs ≥ 1. Then, for each n ∈ [N/2, N ], we have f1 ∗ · · · ∗ fs(n),s Ns−1. Proof. We may assume that N is sufficiently large, since the claim is obvious when N ≤ η−1 (and we can choose η to be sufficiently small). Fix a positive integer n0 ∈ [N/2, N ]. Let us define N1 = bn0/2s−2c and Ni+1 = 2iN1 for i ∈ {1, . . . , s− 2}. Let A1 = {n ∈ [N1] : f1(n) ≥ /2} and Ai = {n ∈ [Ni−1] : fi(n) ≥ /2}, (13) 11 for i ∈ {2, . . . , s}. By (12) we see that |Ai| ≥ (αi + /2)Ni−1, for i ∈ {2, . . . , s}, provided that η is sufficiently small. Let ′ = 4s , δs−1 = 1/2 and, for i ∈ {2, . . . , s − 1}, δi−1 := η(′, δi), where η(, δ) is as in Lemma 4. Let R1 = A1 and Ri+1 = Sδi(Ai+1, Ri) (14) for each i ∈ {1, . . . , s− 2}. We shall choose η ≤ (mini δi)/2s. Then it follows from Lemma 4 that |R2 ∩ P | = |Sη(′,δ2)(A2, A1) ∩ P | ≥ (min(α1 + α2, 1)− ′)|P | for each arithmetic progression P ⊆ [2N1] = [N2] with |P | ≥ δ22N1 = δ2N2. Similarly |R3 ∩ P | = |Sη(′,δ3)(A3, R2) ∩ P | ≥ (min(min(α1 + α2, 1)− ′) + α3, 1)− ′)|P | ≥ (min(α1 + α2 + α3, 1)− 2′)|P | for each arithmetic progression P ⊆ [2N2] = [N3] with |P | ≥ δ32N2 = δ3N3. Repeating this argument inductively, for each i ∈ {1, . . . , s− 2}, we get that |Ri+1 ∩ P | ≥ (min(α1 + · · ·+ αi+1, 1)− i′)|P | (15) for each arithmetic progression P ⊆ [Ni+1] with |P | ≥ δi+1Ni+1. Hence in particular |Ri| ≥ (min(α1 + · · ·+ αi, 1)− (i− 1)′)Ni for i ∈ {1, . . . , s− 1}. Let N(n0) = |{(a, b) ∈ As ×Rs−1 : a+ b = n0}|. We see that N(n0) = |As ∩ (n0 −Rs−1)| = |As \ ([n0] \ (n0 −Rs−1))| since As, Rs−1 ⊆ [n0], where n0 −Rs−1 = {n0 − r : r ∈ Rs−1}. Thus N(n0) ≥ |As| − (n0 − |Rs−1|) (16) ≥ (αs + /2)Ns−1 + (min(α1 + · · ·+ αs−1, 1)− (s− 2)′))Ns−1 − n0 ≥ (/2− s′)Ns−1 − 2s−2  N, (17) since n0 ≤ Ns−1 + 2s−2. From (13) and (14) we get that for b ∈ Rs−1 f1 ∗ · · · ∗ fs−1(b) ≥  2 ∑ i+j=b i∈Rs−2 j∈As−1 f1 ∗ · · · ∗ fs−2(i)1As−1(j) ≥  2 δs−2Ns−2 min i∈Rs−2 f1 ∗ · · · ∗ fs−2(i) Repeating previous argument, it follows that f1 ∗ · · · ∗ fs−1(b) ≥ (  2 )s−1 s−2∏ i=1 δiNi  s−1ηs−2Ns−2. (18) Since fs(a)  whenever a ∈ As, it follows by (17) and (18) that f1 ∗ · · · ∗ fs(n0),s Ns−1.  We are now ready present and prove the transference lemma. 12 Lemma 6. (Transference Lemma) 4 Let s ≥ 3 and , η ∈ (0, 1). For all i ∈ {1, . . . , s}, let qi and αi be positive real numbers such that α1 + · · ·+ αs ≥ 1 and 1− 1 q1 < 1 q2 + · · ·+ 1 qs < 1. Let N be a natural number and, for each i ∈ {1, . . . , s} let fi : [N ] → R≥0 be a function that satisfies the following assumptions: 1. (Mean condition) For each arithmetic progression P ⊂ [N ] with |P | ≥ ηN we have En∈P fi(n) ≥ αi + ; 2. (Pseudorandomness condition) There exists a majorant νi : [N ] → R≥0 with fi ≤ νi pointwise, such that ||ν̂i − 1̂[N ]||∞ ≤ ηN ; 3. (Restriction estimate) We have ||f̂i||qi ≤ KN1−1/qi for some K ≥ 1. Then for each n ∈ [N/2, N ] we have f1 ∗ · · · ∗ fs(n) ≥ (c()−O,K,q(η))Ns−1, where c() > 0 is a constant depending only on . Proof. A symmetric version of the case s = 3 has been shown in [MMS17, Section 4.3] by Matomäki, Maynard and Shao. With minor changes the same proof works for the asymmetric version with any s ≥ 3 using Lemma 5 in place of [MMS17, Proposition 3.2]. The main difference is that when [MMS17] uses Hölder’s inequality to get that∫ T |ĥ1(γ)ĥ2(γ)ĥ2(γ)|dγ ≤ ||ĥ1||3−q∞ ||ĥ1||q−2q ||ĥ2||q||ĥ3||q we use Hölder’s inequality to get that∫ T |ĥ1(γ) · · · ĥs(γ)|dγ ≤ ||ĥ1||1−a∞ ||ĥ1||aq1 ||ĥ2||q2 · · · ||ĥs||qs , where a ∈ (0, 1) is chosen such that aq1 + 1q2 + · · ·+ 1qs = 1.  Lemma 1, which we will use to prove our main theorem, is a symmetric version of the previous lemma. 5 Proof of the Main Theorem In this section we will prove Theorem 1 using Lemma 1 assuming some lemmas which we will prove later. 5.1 Definitions Let X,Y,W,N,m, b ∈ N such that (W, b) = 1, W = 2k2Cη ∏ p≤w p, (19) 4Using this asymmetric version of the transference lemma and some other results of this paper one should be able to establish results concerning Waring-Goldbach problem with mixed powers on short intervals. 13 X = Wm+ b, (20) Y = WN, (21) Y = (1 + o(1)) ks+ k s X1−1/k+θ/k, (22) where w = log log logm, Cη = dη−1e!2 and η ∈ (0, 1). We see that W  log logX. Let ρ be the characteristic function for the primes. Next, we define a majorant function ρ+ for the function ρ based on the linear sieve. Let ρ+(n) = ∑ d|n d|P (D) d∈D+ µ(d), (23) where P (D) = ∏ p 0 to be chosen later. We know that ρ(n) ≤ ρ+(n), for all n ∈ N (see [Nat96, Theorem 9.3]). Set α+ = φ(W ) kW logX ∑ d|P (D) (d,W )=1 d∈D+ µ(d) d . (25) We will prove in Subsection 7.2.1 that 0 < α+ < 2+kδ . We now define functions fb, νb : [N ] → R by fb(n) = { φ(W ) α+WσW (b) X1−1/k logXρ(t) if W (m+ n) + b = tk 0 otherwise (26) and νb(n) = { φ(W ) α+WσW (b) X1−1/k logXρ+(t) if W (m+ n) + b = tk 0 otherwise (27) where σW (b) = #{z ∈ [W ] : zk ≡ b (mod W )}. (28) 5.2 Key lemmas We will apply Lemma 1 to the functions fb and νb. The following three lemmas (to be proven later) show that the functions fb and νb satisfy the conditions of Lemma 1. We use the notation of Subsection 5.1. Lemma 7. (Mean condition) Let , θ ∈ (0, 1), x = X1/k and k ≥ 1. Let η ∈ (0, 1) and fb : [N ] → R be as in (26). Let also α− > 0 be such that, for each interval I ⊂ [x, x + 2xθ] of length |I| ≥ (η/3)xθ, and every c, d ∈ N such that (c, d) = 1 and d ≤ log x, we have∑ n∈I n is prime n≡c (mod d) 1 ≥ α −|I| φ(d) log x , (29) when x is sufficiently large. Let P ⊆ [N ] be an arithmetic progression such that |P | ≥ ηN . If N is sufficiently large then En∈P fb(n) ≥ α − α+ (1− ). 14 We shall quickly establish Lemma 7 in Section 6. Lemma 8. (Pseudorandomness condition) Let α ∈ T, θ ∈ (1/2, 1), η ∈ (0, 1) and k ≥ 2. Let δ be as in (24) and νb : [N ]→ R be as in (27). Assume that δ < max ( 2θ−1 k , θ k(k/2+1) ) . Then |ν̂b(α)− 1̂[N ](α)| ≤ ηN when N is sufficiently large depending on η. We establish Lemma 8 in Section 7. The pseudorandomness condition (Lemma 8) is the hardest condition to establish and therefore we will spend most of the remaining paper proving Lemma 8. As stated in Section 2 to prove pseudorandomness we split the interval [0, 1] into minor and major arcs and treat those sets differently. For the minor arcs, we use an application of the [Hua16, Lemma 1] and for the major arcs, we develop some ideas that are from [Vau97, Section 4] and [Cho18, Section 4]. Lemma 9. (Restriction estimate) Let s ≥ k2 +k+ 1 and let fb : [N ]→ R be as in (26). Then there exists q > 0 such that s− 1 < q < s and ||f̂b||q  N1−1/q. We establish Lemma 9 in Section 8. The proof follows mostly by combining [BDG16, Theorem 1.1], [Dae10, Theorem 3] and some ideas of [Bou89, Section 4]. 5.3 Conclusion In this subsection we prove Theorem 1 assuming the lemmas presented in the previous subsection. Before presenting the proof we need the following lemma about local solutions of Waring’s problem. Lemma 10. Let s, k, q ∈ N and m ∈ Zq be such that m ≡ s (mod (q,Rk)), where Rk =∏ (p−1)|k p η(k,p) and η(k, p) is as in (1). If s ≥ 3k, then congruence m ≡ yk1 + · · ·+ yks (mod q) (30) has a solution with y1, . . . , ys ∈ Z∗q . Proof. Let Mm(q) be the number of solutions of the congruence (30). Let S(q, a) = ∑ x(q) (x,q)=1 eq(ax k). Let us first show that Mm(q) is multiplicative. For this, let q = uv, where (u, v) = 1. Using [Hua65, Lemma 8.1] it follows that qMm(q) = ∑ x1(q) (q,x1)=1 · · · ∑ xs(q) (q,x1)=1 ∑ a(q) eq(a(x k 1 + · · ·+ xks −m)) = ∑ a(q) S(q, a)seq(−am) = ∑ x(u) ∑ y(v) S(uv, vx+ uy)seuv(−(vx+ uy)m) = ∑ x(u) S(u, x)seu(−xm) ∑ y(v) S(v, y)sev(−ym) = uMm(u)vMm(v). 15 Thus Mm(q) is multiplicative and so it suffices to prove the lemma with q = pt, where p is a prime and t ∈ N. If t > η(k, p) we get from [Hua65, Lemma 8.3] that ptMm(p t) = ∑ x1(p t) (p,x1)=1 · · · ∑ xs(p t) (p,x1)=1 ∑ a(pt) ept(a(x k 1 + · · ·+ xks −m)) = ∑ a(pt) ept(−am) ( ∑ x(pt) (p,x)=1 ept(ax k) )s = ∑ a(pt) p|a ept(−am) ( ∑ x(pt) (p,x)=1 ept(ax k) )s = ∑ a(pt−1) ept−1(−am) ( p ∑ x(pt−1) (p,x)=1 ept−1(ax k) )s = psMm(p t−1). Together with [Hua65, Lemma 8.8] and [Hua65, Lemma 8.9] this implies the claim.  Proof of Theorem 1 assuming Lemmas 7, 8, 9. Let n0 be a natural number for which n0 ≡ s (mod Rk) and let x = (n0/s)1/k. Our goal is to show that n0 can be written in form n0 = p k 1 + · · ·+ pks , where p1, . . . , ps are primes which belong to the interval [x− xθ/s, x+ xθ]. We now define the exact values of the variables m and N . Let N = ⌊ (x+ xθ −W )k − (x− xθ/s)k W ⌋ and m = ⌊ (x− xθ/s)k W ⌋ . (31) We see that WN ∼ k(1 + 1/s)xk−1+θ and Wm ∼ xk. Hence (22) holds. By lemma 10 we can choose b1, . . . , bs with (bi,W ) = 1 such that bi ≡ cki (mod W ) for some ci ∈ [W ] and b1 + · · ·+ bs ≡ n0 (mod W ). We shall apply Lemma 1 with fi = fbi where fbi is as in (26). Assuming Lemmas 7, 8 and 9 we have by Lemma 1 that, for each n ∈ [N/2, N ], there exists a representation n = n1 + · · ·+ ns where for each ns there exists a prime pi ∈ [x−xθ/s, x+xθ] such that pki = W (ni+m) + bi. Thus W (n+ sm) + b1 + · · ·+ bs = pk1 + · · ·+ pks . Set n = (n0 − b1 − · · · − bs)/W − sm. Now if n ∈ [N/2, N ], it follows that n0 = pk1 + · · · + pks as claimed. From (31) and definition of x we see that Wm = xk − (1 + o(1))k s xk−1+θ, WN = (1 + o(1)) ks+ k s xk−1+θ and n0 = sx k. Using these it follows that n = (n0 − b1 − · · · − bs)/W − sm = (1 + o(1))kxk−1+θ/W. Thus n ∈ [N/2, N ] when n0 is large enough.  16 6 Mean condition Proof of Lemma 7 By (26) we see that 1 |P | ∑ n∈P fb(n) = 1 |P | ∑ n∈P W (n+m)+b=pk φ(W ) α+WσW (b) X1−1/k logX. Since P is an arithmetic progression with |P | ≥ ηN , there exist integers q, a such that q ≤ η−1 , a ∈ [N ] and P = q[|P |] + a. Therefore, for n ∈ P , there exists t ∈ [|P |] such that W (n+m) + b = W (a+ qt+m) + b = W ′(t+m′) + b′, where W ′ := Wq,m′ := ⌊m q ⌋ and b′ := Wq (m q − ⌊m q ⌋) +Wa+ b. By W ≡ 0 (mod dη−1e!) (see eq. (19)) we see that (W ′, b′) = (Wq,Wq (m q − ⌊m q ⌋) +Wa+ b) = (Wq,Wm+Wa+ b) ≤ (W,Wm+Wa+ b)2 = (W, b)2 = 1 Set X ′ := W ′m′ + b′ and Y ′ := W ′|P |. Note that X ′ = X +Wa and ηY ≤ Y ′ ≤ Y . Then 1 |P | ∑ n∈P fb(n) = 1 |P | φ(W ) α+WσW (b) X1−1/k logX ∑ t∈[|P |] W ′(t+m′)+b′=pk 1 = 1 |P | φ(W ) α+WσW (b) X1−1/k logX ∑ z∈[W ′] zk≡b′ (mod W ′) ∑ X′ 0 provided that X is large enough. Similarly (X ′ + Y ′)1/k −X ′1/k ≥ Y ′ 1 k(X ′ + Y ′)1−1/k ≥ ηY 1 k(2X)1−1/k > η/2(1 + 1/s− ′)Xθ/k. Thus [X ′1/k, (X ′ + Y ′)1/k] ⊂ [x, x+ 2xθ] and (X ′ + Y ′)1/k −X ′1/k ≥ (η/3)xθ. We also get that (X ′ + Y ′)1/k −X ′1/k ≥ (1− )Y ′ 1 k(X)1−1/k (33) provided that X is large enough depending on . Now if X is sufficiently large it follows by (29), (32) and (33) that 1 |P | ∑ n∈P fb(n) ≥ 1|P | φ(W ) α+WσW (b) X1−1/k logX ∑ z∈[W ′] zk≡b′ (mod W ′) α−((X ′ + Y ′)1/k −X ′1/k) φ(W ′) logX ′1/k ≥ 1|P | φ(W ) α+WσW (b) σW ′(b ′)α−Y ′ φ(W ′) (1− ). By (19) we have that q|W and thus φ(W ′) = qφ(W ). Using (19), [IR90, Proposition 4.2.1] and [IR90, Proposition 4.2.2] we get that σW (b) = σW ′(b′). Therefore 1 |P | ∑ n∈P fb(n) ≥ α − α+ (1− ).  17 7 Pseudorandomness condition We assume notation of Subsection 5.1. In this section we will prove Lemma 8. In order to do so we divide T = R/Z into two disjoint sets, major and minor arcs, using Hardy and Littlewood decomposition. Let Q = Xk(δ+ρ) and T = Y Xρ (34) for ρ > 0 to be chosen later and δ as in (24). For q ≥ 1 and (a, q) = 1, write M(q, a) = {α : |α− aq | ≤ 1T }. Let M = q−1⋃ a=0 (a,q)=1 1≤q≤Q M(q, a). If ρ is suitably small, δ < k−1+θ2k2 and, if X is sufficiently large, then T > 2Q 2 and thus all intervals M(q, a) are disjoint. Let also m = T \M. We call M major arcs and m minor arcs. Next we decompose ν̂b. From (27) we have that ν̂b(α) = ∑ n νb(n)e(nα) = e((−b/W −m)α) φ(W ) α+WσW (b) X1−1/k logXEb(α), (35) where Eb(α) := ∑ X 0, θ ∈ (1/2, 1), k ≥ 2, α ∈ m and δ < min( 2θ−1k , k−1+θ2k2 ). Let also νb : [N ]→ R be as in (27) and ρ be as in (34). Then |ν̂b(α)− 1̂[N ](α)|  NX−ρ/k+. Lemma 11 will easily follow from the following estimate for the generating function f(b, d, α) on the minor arcs. Lemma 12. Let  > 0, θ ∈ (1/2, 1), k ≥ 2, α ∈ m and δ < min( 2θ−1k , k−1+θ2k2 ). Let ρ and T be as in (34). Let also Hd = (X+Y )1/k−X1/k dW . Then f(b, d, α) H1−ρ+d +Hd (TW Y )1/k . 18 We have the trivial bound |f(b, d, α)| ≤ Hd. We also note that by the mean value theorem and (22) Hd  Y dWX1−1/k  X θ/k dW . (38) Proof. Let α ∈ m. By Dirichlet’s Theorem (see e.g. [Nat96, Theorem 4.1]) there exist integers a and q such that (a, q) = 1, 1 ≤ q ≤ Q and ∣∣∣α− a q ∣∣∣ ≤ 1 qQ . (39) Because α ∈ m we must have |α− a/q| > 1/T . By (37) f(b, d, α) = ∑ z∈[W ] (zd)k≡b (mod W ) ∑ X1/k d 0 depending on ′ and any  > 0 either T  H1−ρ+d (41) or there exist integers a1 and q1 such that 1 ≤ q1 ≤ Hkρd , (a1, q1) = 1, |q1dkW k−1α− a1| ≤ (X1/k dW )1−k Hkρ−1d , (42) and T  H1−ρ+d + Hd( q1 +Hd ( X1/k dW )k−1 |q1dkW k−1α− a1| )1/k . (43) By (34), (39) and (42) we have |a1q − aq1dkW k−1| = |q(a1 − dkW k−1αq1)− q1dkW k−1(a− αq)| ≤ Q (X1/k dW )1−k Hkρ−1d +H kρ d d kW k−1 1 Q ≤ Xk(δ+ρ) (X1/k dW )1−k Hkρ−1d +H kρ d D kW k−1 1 Xk(δ+ρ)  Xk(δ+ρ)X1/k−1dkW kX−θ/kXθρ +XθρW k−1 1 Xkρ  X2kδ+1/k−1−θ/k+kρ+θρW k +X(θ−k)ρW k−1  X−′′ , 19 for some ′′ > 0, when ρ is small enough and δ < k−1+θ2k2 . Assuming that X is sufficiently large, depending on ′′, we have that a1 q1 = adkW k−1 q , and consequently q1 = q(q,dkWk−1) and a1 = adkWk−1 (q,dkWk−1) . Thus Hd( q1 +Hd ( X1/k dW )k−1 |q1dkW k−1α− a1| )1/k = Hd( q (q,Wk−1dk) +Hd ( X1/k dW )k−1 dkWk−1 (q,dkWk−1) |qα− a| )1/k = ( (q, dkW k−1) q )1/k Hd( 1 +Hd ( X1/k dW )k−1 dkW k−1|α− a/q| )1/k ≤ Hd( 1 +HdX1−1/kd 1T )1/k  Hd (TW Y )1/k . Now the claim follows from (40), (41) and (43).  Proof of Lemma 11. Let α ∈ m. By Dirichlet’s Theorem (see e.g. [Nat96, Theorem 4.1]) there exist integers a and q such that (a, q) = 1, 1 ≤ q ≤ Q and ∣∣∣α− a q ∣∣∣ ≤ 1 qQ . Because α ∈ m we must have |α− a/q| > 1/T . Thus 1̂[N ](α) ||α||−1 ≤ q||qα|| < T  NX −ρ+ From Lemma 12 and (34) we get that f(b, d, α) HdX−ρ/k+. Together with (35), (36), (37) and (38) it follows that ν̂b(α) X1−1/k logX ∑ d≤D HdX −ρ/k+  NX−ρ/k+.  7.2 Major arcs In this subsection we will establish Lemma 8 when α ∈ M. In particular we need to understand the generating function (37) on the major arcs. We use a standard strategy similar to [Vau97, Section 4.1] to approximate our generating function. The result we will prove is the following. Lemma 13. Let η > 0, θ ∈ (1/2, 1), k ≥ 2, α ∈M and δ < θk(k/2+1) . Let also νb : [N ]→ R be as in (27). Then |ν̂b(α)− 1̂[N ](α)| ≤ ηN when N is sufficiently large depending on η. 20 7.2.1 Auxiliary lemmas In this subsection we state two lemmas that follow from standard linear sieve estimates. They are needed in order to prove Lemma 13. Lemma 14. Let  > 0 and a,D ∈ N such that a ≤ D. Let D+ be as in (23). Then∑ d|P (D) (d,a)=1 d∈D+ µ(d) d < a φ(a) ( 2 +  logD +O ( 1 (logD)2 )) . Additionally, if D  1, then ∑ d|P (D) (d,a)=1 d∈D+ µ(d) d  a φ(a) 1 logD . Proof. Let the sieving range P be primes not dividing a. Then it follows by Mertens formula (see e.g. [IK04, formula (2.16)]) and from the theory of linear sieve (see e.g. [Nat96, Theorem 9.6] and [Nat96, Theorem 9.8]) that, for any ′ > 0,∑ d|P (D) (d,a)=1 d∈D+ µ(d) d < ∏ p∈P p≤D ( 1− 1 p ) (2eγ + ′e11) = a φ(a) ∏ p≤D ( 1− 1 p ) (2eγ + ′e11) = a φ(a) e−γ logD ( 1 +O ( 1 logD ))( 2eγ + ′e11 ) = a φ(a) ( 2 +  logD +O ( 1 (logD)2 )) . Define χa(n) := ∏ pt||n (p,a)=1 pt. For the lower bound we use the following strategy.∑ d|P (D) (d,a)=1 d∈D+ µ(d) d = 1 D2 ∑ d|P (D) (d,a)=1 d∈D+ µ(d) (⌊D2 d ⌋ +O(1) ) = 1 D2 ∑ d|P (D) (d,a)=1 d∈D+ µ(d) ⌊D2 d ⌋ +O(D−1) = 1 D2 ∑ n≤D2 ρ+(χa(n)) +O(D −1) ≥ 1 D2 ∑ n≤D2 p|n⇒p>D or p|a 1 +O(D−1) ≥ 1 D2 ∑ d|a (pi(D2/d)− pi(D)) +O(D−1) 21  1 logD ∑ d|a 1 d by the prime number theorem provided that D is large enough. Since ∏ p(1 − p−2) 6= 0 we have that ∑ d|a 1 d ≥ ∏ p|a ( 1 + 1 p ) ≥ ∏ p ( 1− 1 p2 )∏ p|a ( 1− 1 p )−1  a φ(a)  From the previous lemma we get that ′ < α+ ≤ 2 +  kδ , (44) for any  > 0 and for some ′ > 0 provided that X is large enough. Lemma 15. Let  > 0 and a, q, t,D ∈ N be such that t|q. Let D+ be as in (23). Then∑ d|P (D) (d,aq/t)=1 t|d d∈D+ µ(d) d  q a φ(a) 1 logD . Proof. The proof follows the same general idea, which is used to prove the upper bounds with the linear sieve (see e.g. [Nat96, Section 9]). Set q′ := aq/t. We can assume that t|P (d), which means that t is also square-free. Therefore∣∣∣ ∑ d|P (D) (d,q′)=1 t|d d∈D+ µ(d) d ∣∣∣ ≤ ∣∣∣ ∑ d|P (D) (d,q′)=1 (t,P (D))|d d∈D+ µ(d) d ∣∣∣. (45) Let V (z, t, q′) := ∑ d|P (z) (d,q′)=1 (t,P (z))|d µ(d) d , V +(D, z, t, q′) := ∑ d|P (z) (d,q′)=1 (t,P (z))|d d∈D+ µ(d) d and Vn(D, z, t, q ′) = ∑ pk<··· 1. Assume first that (p, aW ) = 1. Then both z + Wx and Wx run through all residue classes modulo pl when x runs through all residue classes modulo pl. Thus V ′pl(a, z, c) = ∑ r (mod pl) eWpl(a(z +Wr) k + c(z +Wr)) = ∑ r (mod pl) eWpl(a(Wr) k + c(Wr)) = ∑ r (mod pl) epl(aW k−1rk + cr) and thus (55) holds by [Vau97, Lemma 4.1] since |V ′pl(a, z, c)| = |Spl(a, z, c)|. Now it remains to prove (55) when p|aW . Let ν = ⌊ l + 1 2 ⌋ . 25 We have that 2(l − ν) ≥ l − 1. Also when x runs through all residue classes modulo pl−v and y runs through all residue classes modulo pv then x+ pl−vy runs through all residue classes modulo pl. Thus Spl(a, z, c) = ∑ r (mod pl) epl ( a k∑ i=1 ( k i ) W i−1zk−iri + cr ) = ∑ x (mod pl−ν) ∑ y (mod pν) epl ( a k∑ i=1 ( k i ) W i−1zk−i(x+ pl−νy)i + c(x+ pl−νy) ) = ∑ x (mod pl−ν) ∑ y (mod pν) epl ( a k∑ i=1 ( k i ) W i−1zk−i(xi + ixi−1pl−νy) + c(x+ pl−νy) ) = ∑ x (mod pl−ν) epl ( a k∑ i=1 ( k i ) W i−1zk−ixi + cx ) × ∑ y (mod pν) epν ( y ( a k∑ i=1 ( k i ) W i−1zk−iixi−1 + c )) = ∑ x (mod pl−ν) epl ( a k∑ i=1 ( k i ) W i−1zk−ixi + cx ) × ∑ y (mod pν) epν ( y(ak(z +Wx)k−1 + c) ) . Thus |Spl(a, z, c)| ≤ pνH where H is the number of solutions of the congruence ak(z +Wx)k−1 + c ≡ 0 (mod pν) (56) with 1 ≤ x ≤ pl−ν . Let ψ, τ ∈ N be such that pψ||c and pτ ||ak. If τ > θ, then congruence (56) is insoluble, which gives us the claim. Hence we can assume that τ ≤ ψ. We can also assume that ψ < ν since otherwise (55) is trivial. Now we must have that k − 1|ψ − τ because otherwise (56) is insoluble and (55) is immediate. Thus H is at most the number of solutions of akp−τ (z +Wx)k−1p−ψ+τ + cp−ψ ≡ 0 (mod pν−ψ) (57) with 1 ≤ x ≤ pl−ν and p(ψ−τ)/(k−1)|z +Wx. From Lemma 16 we get that H  (W,pν−ψ) max(1, pl−ν−(ν−ψ)). Therefore by p|aW |Spl(a, z, c)|  (W,pν−ψ) max(pν , pl−ν+ψ)  (W,pl) max(pν , pl−νpψ)  (W,pl)pν(pl, c)  (W,pl)(aW, p)pl/2(pl, c).  Next we show that the function ν(b, β) (defined in (50)) can be approximated by an integral. 26 Lemma 18. Let b, d ∈ N and β ∈ [0, 1]. Then ν(b, β) d = 1 W ∫ (X+Y )1/k/d X1/k/d eW (βd kγk)dγ +O ( |β|Y dW + 1 d ) Proof. Let U(n) := ∑ t≤n t≡b (mod W ) 1 k t1/k−1 = n1/k W +O(1). Using partial summation and integration by parts it follows that ν(b, β) = ∑ X w 0 otherwise. Proof. We mostly follow ideas presentend in [Cho18, Section 4]. Recalling (51) and (54) we see that Vq(ad k, b, d, 0) = ∑ z∈[W ] (zd)k≡b (mod W ) V ′q (ad k, z, 0) 29 = ∑ z∈[W ] (zd)k≡b (mod W ) eWq(ad kzk)Sq(ad k, z, 0) = ∑ z∈[W ] (zd)k≡b (mod W ) eWq(ad kzk)Sq1(ad kq2, z, 0)Sq2(ad kq1, z, 0). (61) Let a′ = adkq2, h = (q1,W ), q1 = hu and W = hW ′. By (52) Sq1(a ′, z, 0) = ∑ r1 (mod u ′) r2 (mod h) ehu ( a′ k∑ i=1 ( k i ) (hW ′)i−1zk−i(r1 + ur2)i ) . Since a′ k∑ i=1 ( k i ) (hW ′)i−1zk−i(r1 + ur2)i ≡ a′kzk−1ur2 + a′ k∑ i=1 ( k i ) (hW ′)i−1zk−iri1 (mod hu) we have that Sq1(a ′, z, 0) = ∑ r1 (mod u) ehu ( a′ k∑ i=1 ( k i ) (hW ′)i−1zk−iri1 ) ∑ r2 (mod h) eh ( a′kzk−1r2 ) . Because (q, a) = 1 and (W,dkz) = 1, we see that (h, a′zk−1) = 1 and∑ r2 (mod h) eh ( a′kzk−1r2 ) = { h if h|k 0 otherwise We split into several cases: (i) q = 1 (ii) q1 6 |k (iii) q 6= 1, q1|k and q ≤ w (iv) q1|k and q > w. Case (i) q = 1. Trivially true. Case (ii) q1 6 |k. We have (q1,W )6 |k, since q1 is w-smooth and k2|W . Therefore in this case Sq1(a ′, z, 0) = 0 from which it follows that Vq(adk, b, d, 0) = 0 by (61). Case (iii) q 6= 1, q1|k and q ≤ w. Clearly q2 = (q, d) = 1. Also because q1|k and k2|W we have by (52) that Sq(adk, z, 0) = q. Thus by (53) Vq(ad k, b, d, 0) = ∑ z∈[W ] (zd)k≡b (mod W ) V ′q (ad k, z, 0) = q ∑ z∈[W ] (zd)k≡b (mod W ) eWq(ad kzk) = q ∑ r∈[W/k] (rd)k≡b (mod W ) ∑ s (mod k) eWq ( adk ( r + W k s )k) = q ∑ r∈[W/k] (rd)k≡b (mod W ) eWq(ad krk) ∑ s (mod k) eq1(ad krk−1s), Now because q1|k and (a, q) = (d,W ) = 1 it follows that the inner sum in the last expression vanishes. Therefore Vq(adk, b, 0) = 0 in this case. Case (iv) q1|k and q > w. As in case (iii) we have that Sq1(a′, z, 0) = q1. Since (W, q2) = 1 we get that Sq2(ad kq1, z, 0) = ∑ r (mod q2) eq2 ( adkq1 k∑ i=1 ( k i ) W i−1zk−iri ) 30 = ∑ r (mod q2) eq2 ( adkq1 k∑ i=1 ( k i ) W i−1zk−i(Wr)i ) = ∑ r (mod q2) eq2(ad kq1W ((z + r) k − zk)) = ∑ r (mod q2) eq2(ad kq1W (r k − zk)) = eq2(−adkq1Wzk) ∑ r (mod q2) eq2(aψq(d) kq1Wr k). Because (d,W ) = 1 it also follows that∑ z (mod W ) (zd)k≡b (mod W ) χ(z, d) = ∑ z (mod W ) (z(d,q))k≡b (mod W ) χ(z, (d, q)). Thus by (61) Vq(ad k, b, d, 0) = q1 ∑ r (mod q2) eq2(aψq(d) kq1Wr k) ∑ z (mod W ) (z(d,q))k≡b (mod W ) χ(z, (d, q)).  Lemma 22. Assume the notation of Lemma 21. Let D ∈ N, D+ be as in (23) and σW (b) be as in (28). Then ∑ d|P (D) (d,W )=1 d∈D+ µ(d) Vq(ad k, b, d, 0) d ,k σW (b)W φ(W ) q1−1/k+ logD . Proof. Case q = 1 follows from Lemma 14. Case q 6= 1 and q1 6 |k or q ≤ w is clear since Vq(ad k, b, d, 0) vanishes by Lemma 21. Assume that q 6= 1, q1|k and q > w. We can write∑ d|P (D) (d,W )=1 d∈D+ µ(d) Vq(ad k, b, d, 0) d = ∑ t|q ∑ d|P (D) (d,Wq/t)=1 t|d d∈D+ µ(d) Vq(ad k, b, d, 0) d . By Lemma 21 it follows that∑ t|q ∑ d|P (D) (d,Wq/t)=1 t|d d∈D+ µ(d) Vq(ad k, b, d, 0) d = ∑ t|q ∑ d|P (D) (d,Wq/t)=1 t|d d∈D+ µ(d) d q1 ∑ r (mod q2) eq2(aψq(t) kq1Wr k) ∑ z (mod W ) (zt)k≡b (mod W ) χ(z, t) k σW (b) ∑ t|q ∣∣∣ ∑ r (mod q2) eq2(aψq(t) kq1Wr k) ∣∣∣∣∣∣ ∑ d|P (D) (d,Wq/t)=1 t|d d∈D+ µ(d) d ∣∣∣. 31 Since q2 ≤ q we have by [Hua40, Theorem] that∑ r (mod q2) eq2 ( aψq(t) kq1Wr k ) ,k q1− 1k+. (62) The claim now follows by divisor bound (see e.g. [Nat96, Theorem A.11]) and Lemma 15.  7.2.3 Proof of Lemma 13 Proof of Lemma 13 Let α ∈M(q, a). First we will analyse the function ν̂b on M(q, a). Recall from (35) that we essentially need to analyse the function Eb(α). By (36) and Lemma 20 we have that Eb(α) = ∑ d|P (z) (d,W )=1 d∈D+ µ(d)f(b, d, α) = ∑ d|P (z) (d,W )=1 d∈D+ µ(d) Vq(ad k, b, d, 0) qdk X1/k−1 ∑ X 0, provided that ρ and  are sufficiently small and δ < θk(k/2+1) . Now it follows from Lemma 22 that for the main term of ν̂b(α) in (63) holds that φ(W ) α+kWσW (b) logXe((− b W −m)α) ∑ X 1 then by Lemma 21 either the main term of (63) is 0 or we have q1|k and q > w. In case q1|k and q > w we have by (65) that the main term is O,k(Nw3−1/k). Therefore ν̂b(α),k Nw3−1/k = o(N) when q > 1. 32 Assume now that q = 1. Then a = 0 and α = β so that 1̂[N ](α) = ∑ n≤N e(nα) = e (( − b W −m ) α ) ∑ X 1. If q 6= 1 then a6=0 and ||α|| ≥ 1 q − |α− a q | ≥ 1 q − 1 T  1 q . Thus, when ρ is small enough and q 6= 1, 1̂[N ](α) ||α||−1  Q = Xk(δ+ρ)  X θ k/2+1 = o(N) (66) since θ k/2 + 1 < 1− 1 k + θ k when k ≥ 2 and θ < 1.  Combining Lemmas 11 and 13, and noting that min (2θ − 1 k , k − 1 + θ 2k2 , θ k(k/2 + 1) ) = min (2θ − 1 k , θ k(k/2 + 1) ) we get Lemma 8. We also record the following lemma for later use. Lemma 23. Let  > 0 be suitably small, θ ∈ (1/2, 1), a, q ∈ N, q ≤ Q, k ≥ 2, α ∈ M(a, q) and δ < θk(k/2+1) . Let also νb : [N ]→ R be as in (27). Then ν̂b(α),k q −1/kN 1 +N ||α− a/q|| +O(NX −). Proof. The claim follows from the main term estimate (65), the error term estimate (64) and the fact that ∣∣∣ ∑ X 0, k ≥ 2 and u ≥ k2 + k. Let fb be as in (26). Then ||f̂b||uu  Nu−1+. Proof. Let t = k 2+k 2 . Using orthogonality and the definition of fb we see that∫ T |f̂b(α)|2tdα = ∫ T ∑ n1,...,n2t fb(n1) · · · fb(nt)fb(nt+1) · · · fb(n2t) e(α(n1 + · · ·+ nt − nt+1 − · · · − n2t)) = ∑ n1,...,n2t n1+···+nt=nt+1+···+n2t fb(n1) · · · fb(n2t)  X2t(1−1/k)(logX)2t ∑ X u. Lemma 25. Let κ,  > 0, M ∈ N, K ≥ 1 and u, v ∈ R+ with u +  < v and u > 2/κ. Let φ : [M ] → R+ and f : [M ] → R be such that |f(n)| ≤ φ(n) for all n ∈ [M ]. Let us have a Hardy-Littlewood decomposition: ∀q ≥ 1, (a, q) = 1 : M(q, a) := {α : |α− a q | ≤ 1 T } M := q−1⋃ a=0 (a,q)=1 1≤q≤Q M(q, a) m := T \M where Q and T are some real variables with T > 2Q2 and Q > C + K2/(κ)+1 for some large contant C. Assume that 1. ∑ n∈[M ] φ(n)M 2. ||f̂ ||uu Mu−1K 3. (Major arc estimate) If α ∈M then φ̂(α) q−κM1+M ||α−a/q|| + o(MK−2/) 4. (Minor arc estimate) If α ∈ m then φ̂(α) = o(MK−2/) Then ||f̂ ||vv v Mv−1. Proof. For ω ∈ (0, 1), define Rω = {α ∈ T : |f̂(α)| > ωM}. It is enough to prove that Rω  1ωu+M , for every ω ∈ (0, 1), since that implies ||f̂ ||vv ≤ ∑ j≥0 ( M 2j−1 )v meas { α ∈ T : M/2j < |f̂(α)| < M/2j−1 }  2vMv−1 ∑ j≥0 (2u+−v)j v Mv−1, provided that u+ − v < 0. Fix ω ∈ (0, 1). Since by assumption 2 (ωM)umeas(Rω) ≤ ||f̂ ||uu ≤Mu−1K, we get that meas(Rω) K ωuM . Thus we can assume that ω > K−1/. 35 It suffices to show that if θ1, . . . , θR are any M−1-spaced points in Rω, then necessarily R 1 ωu+ . (67) To prove (67), we define an ∈ C such that |an| ≤ 1 and f(n) = anφ(n) for all n ∈ [M ]. Fur- thermore, we define c1, . . . , cR ∈ C such that |cr| = 1 and crf̂(θr) = |f̂(θr)| for all r ∈ [R]. From Cauchy-Schwarz-inequality and assumption 1 it follows that ω2M2R2 ≤ ( ∑ 1≤r≤R |f̂(θr)| )2 = ( ∑ 1≤r≤R cr ∑ n anφ(n)e(nθr) )2  M ∑ n φ(n) ∣∣∣ ∑ 1≤r≤R cre(nθr) ∣∣∣2. Thus ω2MR2  ∑ 1≤r,r′≤R |φ̂(θr − θr′)|. Now let γ > 1 be a parameter to be chosen later. Then by Hölder’s inequality ω2γMγR2  ∑ 1≤r,r′≤R |φ̂(θr − θr′)|γ . Recalling ω > K−1/, we obtain from the minor arc estimate (assumption 4) that∑ 1≤r,r′≤R θr−θr′∈m |φ̂(θr − θr′)|γ = o(ω2γMγR2). Therefore ω2γMγR2  ∑ 1≤r,r′≤R θr−θr′∈M |φ̂(θr − θr′)|γ . (68) Let Q′ = C + ω−h, with 2/κ < h < 2/κ + . Note that Q′ < Q. From the major arc estimate (assumption 3) we get that∑ q>Q′ ∑ 0≤a≤q (a,q)=1 ∑ 1≤r,r′≤R θr−θr′∈M(q,a) |φ̂(θr − θr′)|γ  Q′−κγMγR2 + o(ω2γMγR2). (69) The right hand side of (69) is negligible compared to ω2γMγR2 provided that C is large enough. Thus, combining this with (68) and the major arc estimate (assumption 3), we get that ω2γR2  ∑ q≤Q′ ∑ a∈[q] (a,q)=1 ∑ 1≤r,r′≤R q−κγ (1 +M ||θr − θr′ − a/q||)γ . Hence ω2γR2  ∑ 1≤r,r′≤R G(θr − θr′) (70) where G(α) = ∑ q≤Q′ q−1∑ a=0 q−κγ (1 +M | sin(α− a/q)|)γ . 36 The inequality (70) is very similar to [Bou89, Eq. (4.16)]. We have M instead of M2, qκ instead of q and γ instead of γ/2. Assuming that γ > 1/κ, we can then apply Bourgain’s strategy and use [Bou89, Eq. (4.27)] and [Bou89, Lemma 4.28] to obtain Rω2γ ≤ Q′τ + Cτ,BRQ′1−B , (71) where τ > 0 and B ∈ N are some arbitrarily chosen constants with B > τ and Cτ,B > 0 is a constant depending on τ and B. If we choose B to be sufficiently large depending on γ and C ≥ 2Cτ,B + 2, then ω2γ ≥ 2Cτ,B max(C,ω−h)1−B ≥ 2Cτ,BQ′1−B . Therefore R Q ′τ ω2γ  1 ω2γ+hτ  1 ω2/κ+  1 ωu+ , when γ > 1/κ and τ > 0 are suitably chosen. Hence (67) holds and the claim follows.  Now we are ready to prove Lemma 9. Proof of the Lemma 9 The claim will follow applying Lemma 25 with M = N , u = k2 + k, K = N 1 , κ = 1k + 2, f = fb and φ = νb, where 1 > 0 and 2 > 0 are sufficiently small. Note that fb(n) ≤ νb(n) for all n ∈ [N ]. We also use Hardy-Littlewood decomposition defined in Section 7. If 1 is chosen to be suitable small depending on κ and , then Q > C + K2/(κ)+1 provided that N is large enough. Assumptions 1 - 4 follow, respectively, from Lemma 8 (by it∑ n νb(n) = |ν̂b(0)|  N), Lemma 24, Lemma 23 and Lemma 11. 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