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Theorem 2 [22] Let A, B F and K,L,δ > 0 be such that ⊂ p 1 A , B >p 3 +δ, and A + A < K A , B + B
Theorem 3 Let A, B F and K,L,δ > 0 be such that A >pδ, B >p1/3+δ, ⊂ p | | | | A B 2 >p1+δ and A + A < K A , B + B
χ(a + b) exp( c(δ4 log p/ log2 K)1/3) A B , (5) ≪ − | || | a A, b B ∈X∈ provided that log5 K δ4 log p . ≪ In the previous works [4], [19], [22] the doubling K can tend to infinity very slowly as the function C(K) has an exponential nature. However, our result is applicable for a much wider range of K. A new structural result given in section 4, is what makes our aproach much more effective and it does not require us to use Freiman’s theorem. We apply Theorem 3 to prove a new sum-product type result. Let us recall a well–known Theorem of Balog [1, Theorem 1] (see also [12]).
k Theorem 4 [1] Let A F , q = ps with A > q1/2+1/2 for some positive integer k. If A is an ⊆ q | | additive subgroup of F assume additionally that A > q1/2 + 1. Then q | | (A A)2k+1 = F . − q It easy to see that Balog’s theorem does not hold for set of size smaller than q1/2. Let A 1/2 to be a nontrivial subfield of F 2 then A = p = F 2 and any combinations of sums and p | | | p | products of elements from A belong to A. However, one may hope for an improvement of the Balog’s result for fields Fp, when p is a prime number. We break the square–root barrier in Theorem 4 for subsets of Fp and we only need a few operations to generate the whole field. Theorem 5 There is a positive constant c such that for every A F with ⊆ p A exp( c log1/5 p)p1/2 | |≫ − we have 2A 2A A A 2 − = F or − (A A)= F . (6) A A p A A − p − − 3
It was proven in [8] that any two sets A, B with all sums belonging to the set of quadratic p 1 residues satisfy A B 6 − + B ( A) . It almost solves a well–known S´ark¨ozy’s conjecture | || | 2 | ∩ − | [16] on additive decompositions of the quadratic residues except for the case when A + B equals exactly the set of quadratic residues and when the sum is direct. It immediately follows from Theorem 3 (also see Theorem 14 below) that such sets A and B cannot have small doubling.
2. Notation
In this section we collect notation used in the paper. Throughout the paper by p we always mean an odd prime number and we put F = Z/pZ and F = F 0 . We denote the Fourier p p∗ p \ { } transform of a function f : F C by p → f(ξ)= f(x)e( ξ x) , (7) − · x Fp X∈ b where e(x)= e2πix/p. The Plancherel formula states that 1 f(x)g(x)= f(ξ)g(ξ) . (8) p x Fp ξ Fp X∈ X∈ b b The convolution of functions f, g : F C is defined by p → (f g)(x) := f(y)g(x y) . (9) ∗ − y Fp X∈ Clearly, we have f[g = fg . (10) ∗ We use the same capital letter to denote set A F and its characteristic function A : F ⊆b p p → 0, 1 . For any two sets A, B F the additive energyb of A and B is defined by { } ⊆ p E+(A, B)= (a , a , b , b ) A A B B : a + b = a + b . |{ 1 2 1 2 ∈ × × × 1 1 2 2}| If A = B, then we simply write E+(A) for E+(A, A). Combining (8) and (10),we derive 1 E+(A, B)= A(ξ) 2 B(ξ) 2 . (11) p | | | | Xξ b b The multiplicative energy E×(A, B) is defined in an analogous way and it can be expressed F similarly by applying the Fourier transform on group p∗ and the multiplicative convolution. Given any two sets A, B F , define the sumset, the product set and the quotient set of A and ⊂ p B as A + B := a + b : a A, b B , { ∈ ∈ } AB := ab : a A, b B , { ∈ ∈ } and A/B := a/b : a A, b B, b = 0 , { ∈ ∈ 6 } 4 respectively. We define k-fold sumsets and product sets analogously, for example 2A 2A = − A + A A A and (A A)2 = (A A)(A A). − − − − − All logarithms are to base 2. The signs and are the usual Vinogradov symbols. For a ≪ ≫ positive integer n, we put [n]= 1,...,n . { } 3. Auxiliary results
Here we list results that we will use over the course of Theorem 3 proof. The Lemma was established in [19, Lemma 14] and it is a corollary to Weil’s bound.
Lemma 6 Let χ be a nontrivial multiplicative character, I be a discrete interval in Fp and r be a positive integer. Then
2r χ(u + t)χ(u + t) We will also need a corollary to points/planes incidences theorem of Rudnev [14]. Theorem 7 Let p be an odd prime, F3 be a set of points and Π be a collection of planes P ⊆ p in F3. Suppose that 6 Π and that k is the maximum number of collinear points in . Then p |P| | | P the number of point–planes incidences satisfies Π (q,π) Π : q π |P|| | 1/2 Π + k Π . (12) |{ ∈P× ∈ }| − p ≪ |P| | | | | Corollary 8 Let A, B F be such that A , B < √p and A + A 6 K A , B + B 6 L B . ⊂ p | | | | | | | | | | | | Then the system ′ b1 b1 a = a′ ′ (13) b2 b2 ( a = a′ has O(K5/4L5/2 A B 2 log p + A 2 B ) (14) | || | | | | | solutions in (a, a , b , b , b , b ) A2 B4. Furthermore, for an arbitrary set B with K 6 B ′ 1 1′ 2 2′ ∈ × | | there are at most O(K3/2 A B 5/2) (15) | || | solutions. P r o o f. The first part of the corollary is [22, Lemma 2], so it is enough to prove (15). Denote by σ the number of the solutions to (13). Clearly, we have σ 6 B E (A, B) 6 B 2 A 2. Thus, | | × | | | | we can assume that B 6 A 2/K3, as otherwise B 2 A 2 6 K3/2 A B 5/2 and the assertion | | | | | | | | | || | follows. Since every x A can be written in A ways as x = s a, s A + A and a A it ∈ | | − ∈ ∈ follows that 2 E×(A, B) 6 A − (s a)b = (s′ a′)b′ : s,s′ A + A, a, a′ A, b, b′ B . | | |{ − − ∈ ∈ ∈ }| 5 Next, we apply Theorem 7 with = (s, b′, a′b′) : a′ A, b′ B,s A + A P { ∈ ∈ ∈ } and Π= (x a)b = s′y z : a A, b B,s′ A + A . − − ∈ ∈ ∈ If all elements of A, B and A+A are nonzero then Π = A B A+A , otherwise one can remove | | | || || | zero from those sets. Observe that every incidence between a plane (x a)b = s y z from Π − ′ − and a point (s, b , a b ) from gives a solution to ′ ′ ′ P (s a)b = (s′ a′)b′ (16) − − and each solution to (16) provides a point/plane incidence. One can easily check that the maximal number of collinear points (points belonging a line) in is just the maximal size of ”skew P Cartesian product” (s, b , a b ), so k = max A , B , A + A = max B , A + A . By Theorem ′ ′ ′ {| | | | | |} {| | | |} 7, in view of K 6 B 6 A 2/K3 and A < √p, we have | | | | | | 2 2 2 2 A + A A B 3/2 σ B A − | | | | | | + ( A A + A B ) + A B A + A ( B + A + A ) ≪ | || | p | || || | | || || | | | | | A + A 2 B 3 | | | | + K3/2 A B 5/2 K3/2 A B 5/2 . ≪ p | || | ≪ | || | This completes the proof. ✷ The last two results [13] (see also [20]) and [17] will be used in the proof of Theorem 5. A modern form of Lemma 9 can be found in [2]. Lemma 9 Let A F be a set with A > 1. Then ⊆ p | | A A A 2 + 3 − > min p, | | . A A 2 − Lemma 10 Let A be a subset of an abelian group such that E+(A)= A 3/K. Then there exists | | A A such that A A /K and ′ ⊆ | ′| ≫ | | 4 A′ A′ K A′ . | − |≪ | | 4. Multiplicative structures in sumsets The next lemma directly follows from Proposition 4.1 in [15]. 6 Lemma 11 Suppose that G = (G, +) is a group and k N. Let A G be a finite non–empty ∈ ⊂ set such that A + A K A . Then there is a set X A A of size | |≤ | | ⊆ − X > exp O(k2 log2 K) A (17) | | − | | such that kX 2A 2A. ⊆ − Lemma 12 Let A be a finite subset of an abelian group such that A + A 6 K A . Then for | | | | any λ Z 0 we have i ∈ \{ } O(Σi log(1+ λi )) λ A + + λ A 6 K | | A . | 1 · · · · k · | | | The next result is an important ingredient of our approach, which shows that additively rich sets contain product of surprisingly large sets. Our argument is based on [5]. Lemma 13 Let A F be a set such that A + A 6 K A . There exists an absolute con- ⊆ p | | | | stant C > 0 such that for any positive integers d > 2 and l there is a set Z of size Z > | | exp Cl3d2 log2 K A with − | | [dl] Z 2A 2A . (18) · ⊆ − Proof. Let l and d > 2 be positive integers and X be a set retrieved by applying Lemma 11 to set A and k = 2(d(l 1) + 1), so X > exp ( Cl2d2 log2 K) A . We define a map − | | − | | l+1 l 1 f : X (X + X) (X + d X) (X + d − X) → × · ×···× · by l 1 f(x) = (x1 + x0,x2 + dx0,...,xl + d − x0) , where x = (x ,x ,...,x ). Let be the image of f then by Lemma 12 and the Pl¨unnecke 0 1 l X inequality [21] we have l 1 6 X + X X + d X X + d − X |X | | || · |···| · | l 1 6 (A A) + (A A) (A A)+ d (A A) (A A)+ d − (A A) | − − || − · − |···| − · − | 6 exp (O(l2 log d log K)) A A l | − | = exp (O(l2 log d log K)) A l . | | For a vector z put ∈X r(z)= x Xl+1 : f(x)= z . |{ ∈ }| Clearly, r(z)= X l+1 , | | z X∈X 7 so there is a v such that l+1 2 l+1 l r(v) > X / > exp ( O(l log d log K)) X A − | | |X | − | | | | > exp ( O(l3d2 log2 K)) A . − | | Put Y = x Xl+1 : f(x)= v and let x Y be a fixed element. Then for any y Y and any { ∈ } ∈ ∈ j [0, l 1] we have ∈ − dj(x y )= x y X X . 0 − 0 j − j ∈ − Observe that for distinct vectors y, y Y the corresponding coefficients y and y must also be ′ ∈ 0 0′ distinct and hence elements x y with y Y form a set of size at least exp ( O(l3d2 log2 K)) A . 0− 0 ∈ − | | Denote this set by Z, then dj Z X X for j [0, l 1], so · ⊆ − ∈ − [dl] Z (l(d 1) + 1)(X X) = 2(l(d 1) + 1)X 2A 2A , · ⊆ − − − ⊆ − which concludes the proof. ✷ 4. Proof of the main results Now we are ready to prove Theorem 3. Proof. Let l and r > 2 be parameters that will be specified later. By applying Lemma 13 to the set A, l and d = 2 there is a set Y with A > Y > exp Cl3 log2 K A such that | | | | − | | I Y 2A 2A, where I := [2l]. For fixed x I, y Y we have · ⊆ − ∈ ∈ χ(a + b) 6 χ(a + b) = χ(a + b + xy) , (19) a A,b B a A b B a A xy b B ∈X∈ X∈ X∈ ∈X− X∈ so 4r 4r 4r χ(a + b) 6 ( I Y )− χ(a + b + xy) . (20) | || | a A,b B a 3A 2A x I,y Y b B ∈X∈ ∈X− ∈X∈ X∈ For a 3A 2A we put B := B + a and ∈ − a σ := σa = χ(a + b + xy) = χ(b + xy) . x I,y Y b B x I,y Y b Ba ∈X∈ X∈ ∈X∈ X∈ After a pplying the Cauchy–Schwarz inequality we obtain 2 σ2 = χ(b + xy) x I,y Y b Ba ∈X∈ X∈ 6 I Y χ(b + xy)χ(b′ + xy) | || | ′ b,b Ba x I,y Y X∈ ∈X∈ = I Y ν(u , u ) χ(u + x)χ(u + x) , | || | 1 2 1 2 u1,u2 x I X X∈ where 2 ν(u , u )= (b, b′, z) B Y : b/y = u , b′/y = u . 1 2 |{ ∈ a × 1 2}| 8 From the H¨older inequality we have 2r 2 2r σ4r 6 ( I Y )2r ν(u , u ) − ν2(u , u ) χ(u + x)χ(u + x) . | || | 1 2 1 2 1 2 u1,u2 u1,u2 u1,u2 x I X X X X∈ Since ν(u , u ) = B 2 Y and ν2(u , u ) is the number of solutions to (13) it u1,u2 1 2 | | | | u1,u2 1 2 follows from Lemma 6 and Lemma 8 that P P 4r 2r 2 2r 2 5/4 5/2 2 2 2 r 2r 2 2r σ ( I Y ) ( B Y ) − (K L Y B log p + Y B )(p I r + 4r p I ) , (21) ≪ | || | | | | | Y | || | | | | | | | | | where K = Y + Y / Y 6 K A / Y . We assume that Y | | | | | | | | K5/4L5/2 Y B 2 log p > Y 2 B . Y | || | | | | | Furthermore, suppose that l satisfies the inequality I r = 2lr > r2rp, then | | 4r 2 2r 2r 2 2r 2 5/2 5/4 5/4 2 1/4 σ r p I ( I Y ) ( B Y ) − L K A B Y − log p . ≪ | | | || | | | | | | | | | | | Now we go back to bound (20). By the Pl¨unnecke inequality [21] we have 3A 2A 6 K5 A , | − | | | hence 4r 5r 4r 2 2r 2r 2 2r 2 5/2 5/4 5/4 2 1/4 χ(a + b) K A r p I ( I Y )− ( B Y ) − L K A B Y − log p ≪ | | | | | || | | | | | | | | | | | a A,b B ∈X∈ K5r+5/4L5/4 A 5/4p log p = r2( A B )4r | | . | || | · Y 9/4 B 2 | | | | Recall that Y > exp Cl3d2 log2 K A , so | | − | | 3 4r KC1(l log K+r)L5/4p log p χ(a + b) r2( A B )4r ≪ | || | · A B 2 a A,b B | || | ∈X∈ for an absolute constant C1 > 0. Now we choose parameters for the above inequality. By the assumption L r2rp are satisfied. Furthermore, by the assumption it follows that K5r cδl/64 4 2 1/3 χ(a + b) 2− A B exp( c′(δ log p/ log K) ) A B (22) ≪ | || |≪ − | || | a A,b B ∈X∈ for some absolute constants c, c′ > 0. 9 To complete the proof assume now that Y 2 B > K5/4L5/2 Y B 2 log p. Then the same | | | | Y | || | argument leads to the estimate 4r K5rp χ(a + b) r2( A B )4r , ≪ | || | · B 3 a A,b B | | ∈X∈ hence the required inequality holds provided that B >p1/3+δ, which completes the proof. ✷ | | Using the same argument and the estimate (15) one can avoid any constraints for doubling and size of B. Theorem 14 Let A, B F and K,δ > 0 be such that A >pδ, ⊂ p | | A 2 B 3 >p2+δ, and A + A < K A . | | | | | | | | Then for any nontrivial multiplicative character χ modulo p one has χ(a + b) exp( c(δ4 log p/ log2 K)1/3) A B , (23) ≪δ − | || | a A, b B ∈X∈ provided that log5 K δ4 log p . ≪ Proof. If B > p1/2+δ/5 then we use Karacuba’s result (3) hence we may assume that B 6 | | | | p1/2+δ/5. Then A >p1/4+δ/5 so K 6 B and we can apply (14). Closely following the proof of | | | | Theorem 3 and using (14) we obtain the estimate 4r 5r 4r 2 2r 2r 2 2r 2 3/2 3/2 5/2 1/2 χ(a + b) K A r p I ( I Y )− ( B Y ) − K A B Y − log p ≪ | | | | | || | | | | | | | | | | | a A,b B ∈X∈ 3 KC(l log K+r)p log p = 5r2( A B )4r . | || | · A B 3/2 | || | The choice of parameters l, r from Theorem 3 provides the required upper bound. ✷ Proof of Theorem 14 √p 1/5 A A Proof. Put A = , where 1 6 M 6 exp(c log p), and let Q = − . By Lemma 9 we have | | M A A > 2 2A 2A F 2A 2A − Q p/(2M ). Suppose that A−A = p and let ξ / A−A . Equivalently, if | | − 6 ∈ − ξ(a a )= a + a a a , (24) 1 − 2 3 4 − 5 − 6 a A then a = a and a + a a a = 0, so there are A E+(A) solutions to this equation. i ∈ 1 2 3 4 − 5 − 6 | | In terms of Fourier transform the number of solutions to (24) can be written as 6 4 + 1 2 4 A A A E (A)= p− A(ξx) A(x) > | | = | | . | | | | | | p M 2 x X b b 10 Hence E+(A) > A 3/M 2 | | and by Lemma 10 there is A A such that A A /M and A + A M 4 A . Using ′ ⊆ | ′| ≫ | | | ′ ′| ≪ | ′| multiplicative Fourier coefficients, the number of representations of any z F in the form ∈ p∗ q/q (a b) with q,q Q and a, b A equals ′ − ′ ∈ ∈ 1 χ(x) 2 χ(z(a b)) . p 1 | | − χ x Q a,b A′ − X X∈ X∈ By Theorem 3 there are positive constants c, C such that the above quantity can be bounded from below by 1 2 2 2 1/3 2 (2p)− Q A′ C exp( c(log p/ log M) ) Q A′ , | | | | − − | || | which is greater than 2 1 2 2 1/3 2 (4M )− Q A′ C exp( c(log p/ log M) ) Q A′ , (25) | || | − − | || | for a positive constant c. 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[19] I.D.Shkredov and A.S.Volostnov, Sums of multiplicative characters with additive convolutions, Proceedings of the Steklov Institute of Mathematics, 2017, Vol. 296, pp. 265– 279. [20] T. Szonyi,˝ Around R´edei’s theorem, Discrete Math. 208/209 (1999), 557–575. [21] T.Tao, V.Vu,Additive combinatorics, Cambridge University Press 2006. [22] A. S. Volostnov, On Double Sums with Multiplicative Characters, Mathematical Notes, 2018, 104:2, 197–203. 12 Faculty of Mathematics and Computer Science, Adam Mickiewicz University, Umultowska 87, 61-614 Pozna´n, Poland [email protected] Steklov Mathematical Institute, ul. Gubkina, 8, Moscow, Russia, 119991 and IITP RAS, Bolshoy Karetny per. 19, Moscow, Russia, 127994 and MIPT, Institutskii per. 9, Dolgoprudnii, Russia, 141701 [email protected]