#A37 INTEGERS 15 (2015) NOTE ON A RESULT OF CHUNG ON WEIL TYPE SUMS

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1 #A37 INTEGERS 15 (2015) NOTE ON A RESULT OF CHUNG ON WEIL TYPE SUMS Norbert Hegyvári ELTE TTK, Eötvös University, Institute of Mathematics, Budaest, Hungary hegyvari@elte.hu François Hennecart Université Jean-Monnet, Institut Camille Jordan, Saint-Etienne, France francois.hennecart@univ-st-etienne.fr Received: 12/4/13, Revised: 2/11/15, Acceted: 9/7/15, Published: 9/18/15 Abstract Following revious works of Chung we are interested in Vinogradov s tye inequalities for some multivariate character sums. Using Johnsen s bound on a comlete unidimensional character sum we obtain a sharer result in several ranges of arameters. Dedicated to Endre Szemerédi on his 75 th birthday 1. Introduction Let F be the rime field with elements and F = F \{0} its multilicative grou. We write e(x) for the rimitive additive character ex(2 ix/) on F. The following estimate for the double exonential sum, which has many alications in additive combinatorics, is a well-known result of Vinogradov. Theorem 1.1. Let A, B F. Then for r 2 F e(rab) < A B. (1) The analogous result for a nontrivial multilicative character sum aears first in [3]: Theorem 1.2. Let A, B F and let 6= 0 be a nontrivial multilicative character modulo. Then (a + b) ale A B.

2 INTEGERS: 15 (2015) 2 To go further we observe that the above bounds (for both multilicative and additive characters) are nontrivial only when A B >. In [5, Chater 8, Problem 9] the author obtained a substantially better bound than that given in Theorem 1.2 when A B is small (see also [6] for related questions). The main ingredient in his roof is Hölder s inequality and the famous dee result due to Weil on exonential sums. Karatsuba s result reads as follows: Theorem 1.3. Let A, B F and let 6= 0 be a nontrivial multilicative character. Then for any ositive integer n (a + b) A n 2n B 2 + 4n B. When n = 1 we obtain Theorem 1.2 (aart from the multilicative constant imlied in Vinogradov s symbol). There are many alications of this tye of bound. We mention just one here. Let N(A, B) be the number of airs (a, b) 2 A B such that a + b is a square in F. Then N(A, B) = A B A \ ( B) a + b where denotes the Legendre symbol. For additive characters and smaller subsets A, B of F, Bourgain and Garaev roved the following theorem (see [1]): Theorem 1.4. Let A, B F, and for r 2 F, let S(r) = e(rab)., Then S(r) ale A 1/2 B 1/2 E + 2 (A)E+ 2 (B) 1/8, where E 2 + (A) is the additive energy of the set A defined by E + 2 (A) = {(a 1, a 2, a 3, a 4 ) 2 A 4 a 1 + a 2 = a 3 + a 4 }. Since clearly E + 2 (A) < A 3, we have which gives S(r) ale A 7/8 B 7/8 1/8, S(r) ale A 7/8 B 7/8 1/8 < A B, rovided that A B <. Under the above condition, Theorem 1.4 rovides, more generally, a considerably better estimate than Theorem 1.1 whenever there is a nontrivial bound for E + 2 (A), e.g. E+ 2 (A) < A 3 c with some aroriate c > 0.

3 INTEGERS: 15 (2015) 3 In [2] the author discussed a certain family of incomlete character sums restricted to triles (a, b, c) belonging to a set W which is not necessarily the hyercube A B C. We infer from Theorem 1.2 that when W = A B C with A ale B ale C, the bound (a,b,c)2a B C holds. It is nontrivial whenever (a + b + c) ale A B 1/2 C 1/2 (2) B C >. (3) More recisely, Chung considered the more general situation where the subset W of F 3 could be described by its rojections on the hyerlanes F F {0}, F {0} F and {0} F F. We denote by 12, 23 and 31, the rojections from F 3 onto F 2 maing (a, b, c) on (a, b), (b, c) and (c, a), resectively. For W F 3, we let = 12 (W ), Y = 23 (W ), Z = 31 (W ) and W,Y,Z := {(a, b, c) 2 F 3 : (a, b) 2 ; (b, c) 2 Y ; (c, a) 2 Z}. Then W W,Y,Z. But the reverse inclusion is not true in general. We restrict our attention to the case where W = W,Y,Z. We first observe that we might have W = W 0,Y 0,Z 0 for other subsets 0, Y 0, Z 0 containing, Y, Z, resectively; but in the sequel, we deal only with the strict images, Y, Z of W by these three rojections. Under these assumtions on W, we obtain W = 1 = Y b \ Z a ale, where (a,b)2 c2f (b,c)2y (c,a)2z (a,b)2 Z c := {a 2 F (c, a) 2 Z}, Z a := {c 2 F (c, a) 2 Z} for a, b, c 2 F and a, b, Y b and Y c are defined similarly. It follows that W ale min(, Y, Z ). By the Cauchy inequality we also have the uer bound W ale 1/2 (a,b)2 Hence if = min(, Y, Z ), then Y b \Z a 2 1/2 1/2 ale 1/2 Z a Y b = Y Z. a2f b2f W ale min(, Y Z ). (4)

4 INTEGERS: 15 (2015) 4 Moreover, for any (a, b) 2 the intersection Y b \ Z a cannot be emty since there necessarily exists at least one element c such that (a, b, c) 2 W. Thus the assumtions (b, c) 2 Y and (c, a) 2 Z imly c 2 Y b \ Z a. By symmetry this yields W max(, Y, Z ). Consequently we can assume Y Z, and by (4) we obtain Now Theorem 3.1 in [2] can be stated as follows: Y ale min( Z, ). (5) Theorem 1.5. Let denote a nontrivial multilicative character on F. Let W F 3, = 12 (W ), Y = 23 (W ), Z = 31 (W ) and assume that W = W,Y,Z. Then we have S 2 := (a + b + c) ale 2 3/4 1/2 Y 1/4 Z 1/4. (6) (a,b,c)2w Since clearly S 2 is at most W by (4) and (6) we thus need 3/4 1/2 Y 1/4 Z 1/4 min(, 1/2 Y 1/2 Z 1/2 ). Consequently, this bound is nontrivial when 2 Y Z 3. (7) Writing = x, Y = y and Z = z and using (5), Theorem 1.5 is nontrivial when 2 y z x > 0, 1 + 2x y + z 3, 1 + x y, x + z y. In the articular case when W = A B C so that = A B, Y = B C and Z = C A the bound (6) is nontrivial whenever A B C 2 3, (8) which is a stronger condition than (3), thus one can exect (6) is weaker than (2). We now consider a directed grah on F 2 : the set W contains all triles (a, b, c) such that (a, b), (b, c) and (c, a) are in, Y, Z resectively and in which two airs (x, y) and (z, t) are connected if y = z. Observe that the trivial bound yields S 2 ale W ale min(, Y, Z ). Hence Theorem 1.5 rovides a ossible gain of 1/4 when the sizes of, Y, Z are close to 2. At this oint we should mention that it is conjectured in [2] that the exected gain on the trivial bound 3 should be 1/2, namely S 2 5/2, while Theorem 1.5 yields only S 2 11/4. 2. Statement of the Results We can rove the following result which is similar to Theorem 1.3 with n = 2. We will use it later for bounding the trile sum in Theorem 2.3.

5 INTEGERS: 15 (2015) 5 Theorem 2.1. Let A, B F, and 1, 2 be two multilicative characters modulo which are not simultaneously the trivial character 0. Let c 6= c 0 2 F. Then S 1 ( 1, 2, A, B) := 1(a + b + c) 2 (a + b + c 0 ) 1/4 A 3/4 B 1/2 + 1/8 A 3/4 B. We need to clarify the range of alication of this bound in connection with known bounds. Interchanging if necessary the roles of A and B, we may rewrite our result in the following form. Remarks. For A, B F write A = and B = and assume. Under the hyothesis and the notation in the theorem we have S 1 ( 1, 8 1/4 A 1/2 B 3/4 if ale ale 1/4, >< 1/8 A 3/4 B if 1/4 ale ale, 2, A, B) >: 1/4 A 3/4 B 1/2 if 1/2 ale ale 1/4, 1/8 A B 3/4 if ale min(1/2, 1/4). The first and the fourth alternatives are worse than the trivial bound S 1 ale A B. The second one needs 1/2 to be nontrivial. In that case our bound is better than the bound O( A B ) given by Chung (cf. Theorem 2.3 of [2]) rovided that +2 ale 3/2. Finally, the third one needs also 1 2 and then it is better than Chung s bound. We may summarize the result as follows: Corollary 2.2. Let A, B F with A = and B = and assume. Let ( 1, 2) be a air of multilicative characters that di ers from ( 0, 0), and let c 6= c 0 be elements of F. Then S 1 ( 1, 8 1/8 A 3/4 B if 1/4 ale ale 1/2 ale and + 2 ale 3/2, >< 1/4 A 3/4 B 1/2 if 1 ale 2 ale 1/2, 2, A, B) A B if + 2 3/2, >: A B otherwise. For ale 1/4, we might imrove on the trivial bound using the Hölder inequality with large even arameters. Turning to the question of trile sums associated with the vertices of triangles in a secific grah on F 2, we get the following theorem: Theorem 2.3. Let W F 3 and, Y, Z F 2 as in Theorem 1.5 and let 6= 0 be a non-rincial multilicative character modulo. Then assuming ale Z ale Y,

6 INTEGERS: 15 (2015) 6 we have S 2 (, W ) := (a + b + c) (a,b,c)2w 1/2 1/2 Z 1/4 Y 3/8 + 3/16 1/2 Z 1/2 Y 3/8. (9) One can notice that the uer bound above surasses Chung s bound in Theorem 1.5 whenever Z 2 Y 9/2. In view of (4), the bound (9) in Theorem 2.3 will be nontrivial when max( 1/2 1/2 Z 1/4 Y 3/8, 3/16 1/2 Z 1/2 Y 3/8 ) since S 2 ale W. This is equivalent to the conditions min(, 1/2 Y 1/2 Z 1/2 ), max( 3/2, 4 Z 2 ) Y min( 4/3 4/3 Z 2/3, 13/6 4/3 Z 4/3 ). Thus the bound (9) imroves Chung s bound when 4 Y Z 2 9/2 holds. Let us write = x, Y = y and Z = z. By (5) we obtain that our Theorem 2.3 is meaningful when the arameters x, y, z satisfy ( y z x > 0, 2 y 3/2, y + 2z 4, 1 + x y, x + z y, 4x 3y 2z 4, 8x 6y 8z 13. Let us assume now that W is the Cartesian roduct A B C. Set the cardinalities of the sets in the series A ale B ale C, then we have A B ale C A ale B C. An easy comutation shows that the bound in Theorem 2.3 is nontrivial whenever B C 3/2 and A 2 B C 3 4. These conditions are automatically satisfied when (8) holds. This gives a further argument that Theorem 2.3 is somewhat stronger than Theorem 1.5. Remarks. We observe that the exonents of, Y, Z must not be unbalanced since these cardinalities cannot be strongly di erent. For instance, it is ossible to rove the following bound in line with Theorem 1.5 but using Hölder s inequality instead of Cauchy s: S 2 (, W ) 7/8 3/4 Y 1/8 Z 1/8. (10) This aroach has the e ect of unbalancing the final bound in terms of the cardinalities, Y, Z. At first glance this bound seems better than Chung s if 2 Y Z and better than our bound in Theorem 2.3 if 3 2 Y 2 Z or 11/2 2 Y 2 Z 3. Moreover the requested condition 7/8 3/4 Y 1/8 Z 1/8 for bound (10) to be e ective leads to 2 Y Z. This condition is similar to that needed for Chung s bound in Theorem 1.4. It imlies ultimately that bound (10) is either worse than Chung s or worse than the trivial S 2 ale.

7 INTEGERS: 15 (2015) 7 3. Proofs of the Results Proof of Theorem 2.1. We simly write S 1 for S 1 ( 1, 2, A, B) throughout the roof. We start by using the triangle inequality and the Hölder inequality: S 1 = 1(a + b + c) 2 (a + b + c 0 ) ale A 3/4 a2a We let 1(a + b + c) 2 (a + b + c 0 ) 4 1/4. b2b T c,c 0( 1, 2, b 1, b 2, b 0 1, b 0 2) := a2f 1(a+b 1 +c) 2 (a+b 1 +c 0 ) 1 (a+b 2 +c) 2 (a+b 2 +c 0 ) 1 (a + b c) 2(a + b c0 ) 1 (a + b c) 2(a + b c0 ), where i(0) = 0, i = 1, 2. Extending the summation to all a 2 F and exanding the exression, we infer 1/4 S 1 ale A 3/4 T c,c 0( 1, 2, b 1, b 2, b 0 1, b 0 2). b 1,b 2,b 0 1,b0 2 2B If there exist coincidences among the variables b 1, b 2, b 0 1, b 0 2, namely, they take altogether at most 2 distinct values, then the sum T c,c 0( 1, 2, b 1, b 2, b 0 1, b 0 2) is trivially bounded by. Otherwise we use the next lemma which gives a bound for some mixed character sums. The roof can be found in [4]. Lemma 3.1. Let m be a ositive integer, 1, 2,..., m be any multilicative characters modulo and b 1, b 2,..., b m 2 F be distinct elements. Then x2f 1(x + b 1 ) 2 (x + b 2 )... m(x + b m ) ale (m m 0 + 1) + m 0 + 1, where m 0 is the number of i s which coincide with the rincial character 0. If {b 1, b 2 } 6= {b 0 1, b 0 2} and if (b 1, b 0 1) 6= (b 2, b 0 2) the summand does not reduce to a constant, hence by Lemma 3.1 we have T c,c 0( 1, 2, b 1, b 2, b 0 1, b 0 2) ale 8 1/2. Thus we obtain S 1 A 3/4 B 2 + 1/2 B 4 1/4 ale 1/4 A 3/4 B 1/2 + 1/8 A 3/4 B. Proof of Theorem 2.3. From Theorem 2.1 we deduce a new bound for the trile sum S 2 = S 2 (, W ) defined by (9). First through the triangle and Cauchy inequalities,

8 INTEGERS: 15 (2015) 8 (a + b + c) we have S 2 ale (a,b)2 c2z a \Y b ale 1/2 (a,b)2f 2 c,c 0 2Z a \Y b (a + b + c) (a + b + c 0 ) = 1/2 c2f Z c \ Z c 0 Y c \ Y c0 + c6=c 0 2F a2z c\z c 0 b2y c \Y c0 1/2 1/2 (a + b + c) (a + b + c 0 ) after interchanging the summation and searating the cases c = c 0 and c 6= c 0. Now we aly Theorem 2.1 to treat the inner sum on a and b. Then S 2 1/2 Z + 1/4 Z c 0 1/2 Y c 3/4 + 1/8 Z c 0 Y c 3/4 1/2 c,c 0 2F 1/2 1/2 1/2 1/2 Z 1/2 + 1/8 1/2 Z c 0 1/2 Y c 3/4 c 0 2F c2f + 1/16 1/2 Z 1/2 c2f Y c 3/4 1/2 1/2 1/2 Z 1/2 + 1/2 1/2 Z 1/4 Y 3/8 + 3/16 1/2 Z 1/2 Y 3/8. By the Hölder and Cauchy inequalities we have 3/4 Y c 3/4 ale 1/4 Y c = 1/4 Y 3/4 c2f c2f 1/2 Z c 0 1/2 ale 1/2 Z c 0 = 1/2 Z 1/2. c 0 2F c 0 2F Finally from the assumtion Z ale Y we get the desired result. 4. Comments We may aly Theorem 2.1 to estimate the number N 0 (A, B) of airs (a, b) 2 A B such that both a + b and a + b + 1 are squares in F. We have N 0 (A, B) = 1 a + b a + b = A B 4 + R(A, B)

9 INTEGERS: 15 (2015) 9 where R(A, B) is an error term which can be bounded by one of the cases in Corollary 2.2. Theorems 1.5 and 2.3 can be also used in a similar way in order to estimate character sums where the triles (a, b, c) 2 W satisfy some rescribed arithmetic roerties, such as a + b + c is a square in F. Acknowledgement. This work is suorted by OTKA grants K , K and ANR CAESAR ANR-12-BS References [1] J. Bourgain and M.Z. Garaev, On a variant of sum-roduct estimates and exlicit exonential sum bounds in rime fields, Math. Proc. Cambridge Philos. Soc. 146 (2009), [2] F.R.K. Chung, Several generalizations of Weil sums, J. Number Theory 49 (1994), [3] P. Erdős and H.N. Shairo, On the least rimitive root of a rime, Pacific J. Math. 7 (1957), no. 1, [4] J. Johnsen, On the distribution of owers in finite fields, J. Reine Angew. Math. 251 (1971), [5] A.A. Karatsuba, Basic Analytic Number Theory, Sringer-Verlag, [6] A.A. Karatsuba, Distribution of values of Dirichlet characters on additive sequences, Soviet Math. Dokl. 44 (1992), no. 1,