LINEAR EQUATIONS OVER F p AND MOMENTS OF EXPONENTIAL SUMS

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1 LINEAR EQUATIONS OVER F p AND MOMENTS OF EXPONENTIAL SUMS VSEVOLOD F. LEV Abstract Two of our principal results in a simplified form are as follows. theorem For p prime, the number of solutions of the equation c 1 a 1 + +c k a k = λ, a j A j, where c j F p and λ F p are fixed coefficients, and the variables a j range over sets A j F p, does not exceed the number of solutions of the equation a 1 + +a k =, a j Ā j, where the variables a j range over arithmetic progressions Ā j F p of cardinalities Ā j = A j, balanced around zero. This readily implies an integer version, which strengthens a result of R. Gabriel, G. Hardy, and J. Littlewood. theorem Let A be a set of n = A residues modulo a prime p. Forz F p, write S A z = a A e 2πiaz/p. Then for ε> we have # { z F p : SA z } 2 6 p >1 εn π n ε1/2 1 + o1, provided n and ε. Equality is attained when A is an arithmetic progression modulo p. DUKE MATHEMATICAL JOURNAL Vol. 17, No. 2, c 21 Received 24 August Mathematics Subject Classification. Primary 11B75; Secondary 42C2, 11D4, 11L7, 11P99, 11B25, 11D12. Author s work partially supported by the Edmund Landau Center for Research in Mathematical Analysis and Related Areas, sponsored by the Minerva Foundation Germany. 239

2 24 VSEVOLOD F. LEV This implies an integer version that is due to A. A. Yudin. This paper consists of two parts, the latter of which depends on the results of the former. The first part aims to establish F p -analogs of a series of results due to Gabriel, Hardy, and Littlewood known as rearrangement theorems. It is seen that the original integer versions follow easily from their counterparts in F p, but not vice versa. The second part deals with the moments and distribution of the absolute values of exponential sums in the field F p. Analogs modulo p of two of Yudin s results concerning exponential sums, associated with a finite set of integers, are established. An extended discussion of the results obtained is presented in Sections 1 and 2; the proofs are given in Sections Historical background and summary of results: Inequalities, rearrangements, and linear equations over finite sets The classical rearrangement theorems are the subject of [HLP, Chapter X]. We first quote and discuss the central results, changing somewhat the original notation in an attempt to simplify the presentation. We say that a function f : Z R + is balanced if it satisfies either f f1 f 1 f2 f 2 or We write f f 1 f1 f 2 f2. 1 if fi< f i for some i Z +, δ f= 1 if fi> f i for some i Z +, if fi= f i for any i Z. For each function f : Z R + with a finite support, there exists a balanced function f : Z R + which is a rearrangement of f in the sense that f = f σ equivalently, fi = fσi for some bijection σ : Z Z. There are two ways to balance a given function f, unless it attains its maximal value an odd number of times and any other nonzero value an even number of times. In the latter case the corresponding balanced function f is said to be symmetrically decreasing. theorem A see [G, Theorem 3]; see also [HLP, Theorem 376] Suppose that the functions f j : Z R + j = 1,...,khave finite support, and let f j be balanced functions, corresponding to f j. Then

3 LINEAR EQUATIONS AND EXPONENTIAL SUMS 241 i 1,...,i k Z f 1 i 1 f 1 i 1 f ki k f k i k i 1 + +i k =i 1 + +i k f 1 i 1 f 1 i 1 f k i k f k i k. i 1,...,i k Z i 1 + +i k =i 1 + +i k This theorem was applied by Gabriel, Hardy, and Littlewood to a problem in Fourier analysis. corollary 1 [G, Theorem 4] Given finite trigonometric series F j x = f j me 2πimx, j = 1,...,k, let g j m = f j m, and suppose that ḡ j are balanced functions, corresponding to g j. Define G j x = ḡ j me 2πimx so that G j are obtained by replacing the Fourier coefficients of F j with their absolute values and rearranging them. Then 1 As a particular case we have F 1 x 2 F k x 2 dx 1 Fx 2k dx 1 1 G 1 x 2 G k x 2 dx. Gx 2k dx. The proof of Theorem A is based on the following theorem. theorem B see [G, Theorem 2]; see also [HLP, Theorem 374] Suppose that the functions f j : Z R + j = 1,...,k have finite support, and let f j be corresponding balanced functions. Assume that δ f 1, δ f 2, and δ f j = for j = 3,...,k. Then f 1 i 1 f k i k f 1 i 1 f k i k. i 1,...,i k Z i 1 + +i k = i 1,...,i k Z i 1 + +i k = This theorem is due to Gabriel [G]; it generalizes an earlier result of Hardy and Littlewood [HL] where all functions f j including those with j = 1 and j = 2 were required to be symmetrically decreasing. For k = 2, Theorem B becomes a restatement of the well-known fact that the scalar product of two nonnegative real sequences is maximized when these se-

4 242 VSEVOLOD F. LEV quences are similarly ordered. In general, Theorem B establishes an inequality between two convolutions: f 1 f k f 1 f k. An important particular case, on which the proof of Theorem B in its full generality relies, is that of f j being the indicator functions of finite sets of integers A j : { 1 if i Aj, f j i = if i/ A j. Theorem B reduces then to the following theorem. theorem C Suppose that A j Z j = 1,...,kare finite nonempty sets of integers of cardinalities n j = A j such that n 3,...,n k are odd. Define Ā j to be arithmetic progressions of difference one, balanced around zero as follows: [ Ā 1 = [ Ā 2 = [ Ā j = where for j = 1, 2 we put n 1 1 δ 1, n δ n δ 2, n 2 1 δ n j 1, n j { if nj is odd, δ j = 1 if n j is even. Then the number of solutions of the equation a 1 + +a k = ], ], ] j = 3,...,k, in the variables a j A j does not exceed the number of solutions of this equation in the variables a j Ā j. The major difficulty in converting Theorems A, B, and C into their F p -counterparts lies in the proof of a modulo p version of the latter of the three theorems. Neither of the methods of [G] or [HLP] works in the new setting, and we have to employ a completely different argument. In fact, we prove a somewhat stronger result than just an analog of Theorem C, and we need an additional notation to formulate this result.

5 LINEAR EQUATIONS AND EXPONENTIAL SUMS 243 For p 3 prime, we write p = p 1/2. By a balanced set of residues modulo p, we mean a nonempty block of consecutive residues of the form Ā = [ α,α + δ] mod p, where α [,p ] and δ {, ±1} are integers. We exclude from consideration the degenerate cases α =, δ = 1 and α = p, δ = 1. Obviously, δ is uniquely defined by the set Ā, and we write δ = δā. If the cardinality Ā is odd, then necessarily δā = ; if Ā is even, there are two possibilities: either δā = 1, or δā = 1. If A is a nonempty set of residues modulo p, then Ā is used for a balanced set of same cardinality as A. For λ F p we denote by N λ A 1,...,A k the number of representations λ = a 1 + +a k with a j A j. Thus, N λ A1,...,A k = χ 1 i 1 χ k i k = χ 1 χ k λ i 1,...,i k F p i 1 + +i k =λ is the convolution of the characteristic functions χ j of the sets A j. Given nonempty A 1,...,A k F p, we write {ν i A 1,...,A k } p for the descending rearrangement of the sequence {N λ A 1,...,A k } p λ= p ; thus, p ν 1 A1,...,A k = max λ F p N λ A1,...,A k, p ν i A1,...,A k = λ= p N λ A1,...,A k = A1 A k, and the number of indices i such that ν i A 1,...,A k > is the cardinality of the sumset A 1 + +A k. Furthermore, we write A 1,...,A k B 1,...,B k if A 1,...,A k and B 1,...,B k are two systems of sets such that m m ν i A1,...,A k ν i B1,...,B k 1 for m = 1,...,p. In particular, when m = 1, this implies max N λ A1,...,A k max N λ B1,...,B k. λ F p λ F p Example For p = 7, consider A 1 ={, 2}, A 2 ={1, 2, 6}, Ā 1 ={, 1}, and Ā 2 ={ 1,, 1}. Then A 1 + A 2 = { 1, 2, 3, 4, 6 } { } 7, νi A1,A 2 = 2, 1, 1, 1, 1,,, Ā 1 + Ā 2 = { 1,, 1, 2 }, { νi Ā1, Ā 2 } 7 = 2, 2, 1, 1,,,, whence A 1,A 2 Ā 1, Ā 2.

6 244 VSEVOLOD F. LEV We can now formulate our version of Theorem C. theorem 1 Suppose that A j F p j = 1,...,kare nonempty sets of residues modulo a prime p 3, and let Ā j be corresponding balanced sets. Then A1,...,A k Ā1,...,Ā k. Moreover, assume that δā 1 + +δā k 1. Then, writing for brevity N λ for N λ Ā 1,...,Ā k, we have i if δā 1 + +δā k 1, then N N 1 N 1 N 2 N 2 N p N p ; ii if 1 δā 1 + +δā k, then N N 1 N 1 N 2 N 2 N p N p. That is, if λ 1,..., λ m are the first m terms of either the sequence, 1, 1, 2, 2,..., p, p or the sequence, 1, 1, 2, 2,..., p,p depending on the sign of the sum δā 1 + +δā k, and if λ 1,...,λ m are any m pairwise distinct residues, then N λ1 A1,...,A k + +Nλm A1,...,A k N λ 1 Ā1,...,Ā k + +N λ m Ā1,...,Ā k. Remark. Clearly, the first assertion of Theorem 1 remains valid when Ā j are arbitrary arithmetic progressions, sharing a common difference not necessarily balanced around zero and of difference one. Compared to Theorem C, we were able to replace the asymmetrical requirement on δā j and thus on the parity of A j by the weaker and more aestheticassumption δā δā k 1. At the same time the conclusion of Theorem 1 is considerably stronger than just N A 1,...,A k N Ā 1,...,Ā k, claimed by Theorem C. Theorem 1 is an F p -analog of [L, Theorem 2 and Lemma 1]. We show in Section 3 that the first assertion of Theorem 1 is essentially equivalent to a well-known result of J. Pollard [P], and we obtain the second assertion as a consequence of the first one. Following the argument of Hardy and Littlewood or that of Gabriel, one can easily derive from Theorem 1 the following analog of Theorem B.

7 LINEAR EQUATIONS AND EXPONENTIAL SUMS 245 theorem 2 Suppose that f j : F p R + j = 1,...,kis any system of functions, and let f j be corresponding balanced functions. Assume that δ f 1, δ f 2, and δ f j = for j = 3,...,k. Then f 1 i 1 f k i k f 1 i 1 f k i k. i 1,...,i k F p i 1 + +i k = i 1,...,i k F p i 1 + +i k = Though we have never given a formal definition of a balanced function with F p as a domain, the reader would not have a problem extrapolating the definition for functions defined on Z. From Theorem 2 we deduce an F p -version of Theorem A; this is somewhat tricky, though it requires no principally new ideas compared to [HLP] and [G]. theorem 3 Suppose that f j : F p R + j = 1,...,kis any system of functions, and let f j be corresponding balanced functions. Then f 1 i 1 f 1 i 1 f ki k f k i k i 1,...,i k F p i 1 + +i k =i 1 + +i k f 1 i 1 f 1 i 1 f k i k f k i k. i 1,...,i k F p i 1 + +i k =i 1 + +i k We prove Theorems 2 and 3 in Section 4. Remark. Theorem 1 implies Theorem C; to see this, given integer sets A 1,...,A k, choose p large enough, and consider the canonical images of A j in F p. Similarly, Theorem 2 implies Theorem B, and Theorem 3 implies Theorem A. However, no sufficiently general and powerful method is known to derive modulo p results from their integer counterparts. Immediate from Theorem 3 is the following corollary. corollary 2 Given trigonometric series F j x = p m= p f j me 2πimx/p, j = 1,...,k,

8 246 VSEVOLOD F. LEV let g j m = f j m, and suppose that ḡ j are balanced functions, corresponding to g j. Define G j x = p m= p ḡ j me 2πimx/p so that G j are obtained by replacing the Fourier coefficients of F j with their absolute values and rearranging them. Then p p F 1 x 2 F k x 2 G 1 x 2 G k x 2. x= p x= p As a particular case we have p x= p Fx 2k p x= p Gx 2k. 2. Historical background and summary of results: Moments and large values of exponential sums For a set A F p of n = A residues modulo a prime p, let Sz = a A e 2πiaz/p z F p denote the exponential sums of A. The number of solutions of the linear equation is a 1 + +a k = λ, a j A, 2 p 1 1 S k ze 2πiλz/p, p z= and for k large the terms that determine the behavior of this sum are those with Sz large. In [L] we gave an estimate for the number of such z, which allowed us to estimate the number of solutions of 2. Now, equipped with Theorem 1, which establishes a precise bound for the number of solutions of equations of type 2, we can reverse the procedure and obtain a sharp estimate for the number of z such that Sz is large. In doing so we follow in general the approach of Yudin [Y], who solved the parallel problem for integer sets; the technical details are quite different, however. It is convenient to measure the number of large exponential sums by the function Tϕ, defined by

9 LINEAR EQUATIONS AND EXPONENTIAL SUMS 247 Tϕ:= # { z F p : Sz >ncos ϕ}, ϕ π 2. Thus, Tϕ is a nondecreasing, piecewise constant function such that T = and Tπ/2 = p 1. The reason for introducing Tϕ is that it has a nice additivity property, established in the following lemma. lemma 1 Suppose that ϕ 1, ϕ 2 and ϕ 1 + ϕ 2 π/2. Then for any nonempty A F p. Tϕ 1 + ϕ 2 min { Tϕ 1 + Tϕ 2, p 1 } 3 This lemma follows easily from an observation of Yudin and was first used in [L]. For the benefit of the reader we reproduce its proof which is just several lines long in Section 6. From Lemma 1 we derive the following property of the function Tϕ. lemma 2 Suppose that n = A 4. Then for any <ϕ, ϕ π/2 we have ϕ Tϕ Tϕ. ϕ We note that the integer part brackets cannot be dropped; indeed, the function T ϕ/ϕ is not monotonic. Remark. The proof of Lemma 2 would be immediate if we could omit the p 1 term in 3 and rewrite it as Tϕ 1 + ϕ 2 Tϕ 1 + Tϕ 2. By induction we have then Tjϕ jtϕ, provided jϕ π/2, and we choose j = ϕ/ϕ. It can be shown, however, that this strong form of 3 is incorrect. Lemma 2 is implicit in [L]; here we give an explicit formulation and an independent explicit proof. We establish two estimates for the number of large sums Sz. The first is indirect but may turn out to be more useful in certain applications. theorem 4 For any set A F p of n = A 3 residues modulo a prime p and any even integer k 2, we have

10 248 VSEVOLOD F. LEV p 1 Sz k z=1 6 πk nk 1 p 1 + n k 1. In fact, we give a somewhat better reminder term, provided k 16; see the proof in Section 6. It is interesting to compare the estimate of Theorem 4 with the trivial estimate Sz k n k 1 p. The second is a direct estimate. theorem 5 For any set A F p of n = A 4residues modulo a prime p and any ϕ [,π/6], we have Tϕ 2 3 p π n ϕ 1 + n ϕ 2/3. In Section 5 we consider the case of an arithmetic progression A ={,...,n 1} and show that Theorems 4 and 5 are best possible, save for the reminder terms. 3. Pollard s theorem and the proof of Theorem 1 theorem D Pollard, [P, Theorem 1] Suppose that A j F p j = 1,...,kare nonempty sets of residues modulo a prime p 3, and let Ā j be arithmetic progressions modulo p sharing a common difference and such that A j = Ā j. Write ν i = ν i A 1,...,A k, ν i = ν i Ā 1,...,Ā k, and define ht := min{ν i,t}, ht := min{ ν i,t}, t. Then for any nonnegative real t. ht ht Remark. For k = 2, this theorem is better known in an apparently different but equivalent form, namely: if K i is for the number of elements c A 1 + A 2 that have at least i representations as c = a 1 + a 2 a j A j, then for any integer 1 t min{ A 1, A 2 } we have K 1 + +K t t min { A 1 + A 2 t,p }. The first assertion of Theorem 1 follows immediately from Pollard s theorem and our next lemma.

11 LINEAR EQUATIONS AND EXPONENTIAL SUMS 249 lemma 3 Let ν 1 ν 2 and ν 1 ν 2 be two sequences of nonnegative real numbers, such that ν 1 +ν 2 + = ν 1 + ν 2 +, and such that only a finite number of elements of each sequence are distinct from zero. For nonnegative real t, define ht and ht as in Theorem D. Then the two following assertions are equivalent: a ν 1 + +ν m ν ν m for any m = 1, 2,...; b ht ht for any t. Proof Put σ = ν 1 + ν 2 + = ν 1 + ν 2 +, so that both ht and ht are continuous, nondecreasing functions, satisfying h = h =, hν 1 = h ν 1 = σ, and σ = max h = max h. i We first show that a implies b. We assume that t ν 1, as otherwise ht = σ ht. Thus, t ν 1 also, and there exist indices i, j 1 such that whence ν i+1 t ν i, ν j+1 t ν j, ht = σ ν 1 ν i + it, ht = σ ν 1 ν j + jt. What we want to prove is, therefore, and we show that, moreover, Indeed, if i j, then 4 is equivalent to ν 1 + +ν i it ν ν j jt, ν ν i it ν ν j jt. 4 j it ν i ν j, which is evident in view of t ν j ν i+1. Otherwise, i j and 4 is equivalent to i jt ν j ν i, which follows from t ν j+1 ν i. ii We now use induction by m to show that b implies a. First we have σ = h ν 1 h ν 1 σ, whence h ν 1 = σ, and therefore ν 1 ν 1. Next, assuming ν 1 + +ν m ν ν m for some m 1, we show that ν 1 + +ν m+1 ν ν m+1. 5

12 25 VSEVOLOD F. LEV Indeed, this is obvious if ν m+1 ν m+1, while otherwise ν m+1 ν m+1, and then σ ν 1 ν m + m ν m+1 = h ν m+1 h ν m+1 = m + 1 ν m+1 + j=m+2 = m + 1 ν m+1 + σ ν 1 ν m+1, from which 5 follows. min { ν j, ν m+1 } m + 1 νm+1 + j=m+2 To complete the proof of Theorem 1, it remains to show how its first assertion implies the second one. We use induction by k. The case k = 2 is easy, and we consider k 3. For definiteness we suppose that δā 1 + +δā k 1. Moreover, permuting the indices if necessary, we can assume that δā k, and then δ Ā1 + +δ Āk For m [1,p], let λ 1,...,λ m F p be any pairwise distinct residues modulo p. We put ={ λ 1,..., λ m }, and we define λ 1,..., λ m to be the first m elements of the sequence, 1, 1, 2, 2,...,p, p, so that := { λ 1,..., λ m } is balanced and δ, δ Āk + δ 1. 7 Moreover, if δā k =, then and if δā k = 1, then δ Ā 1 + +δ Āk 1 and δ Āk + δ, δ Ā 1 + +δ Āk 1 and δ Āk + δ. By 6, 7, the latter observation, the first assertion of Theorem 1, and the induction hypothesis, we have m N λi Ā1,...,Ā k = N Ā1,...,Ā k, = N u Ā1,...,Ā k 1 Nv Āk, u,v F p u+v= p ν i Ā1,...,Ā k 1 νi Āk, ν j

13 LINEAR EQUATIONS AND EXPONENTIAL SUMS 251 p 1 = ν m Ā1,...,Ā k 1 νm+1 Ā1 m,...,ā k 1 ν i Āk, m=1 + ν p Ā1 p,...,ā k 1 ν i Āk, p 1 ν m Ā1,...,Ā k 1 νm+1 Ā1 m,...,ā k 1 ν i Āk, = m=1 + ν p Ā1 p,...,ā k 1 ν i Āk, p ν i Ā1,...,Ā k 1 νi Āk, = u,v F p u+v= N u Ā1,...,Ā k 1 Nv Āk, = N Ā1,...,Ā k, m = N λ i Ā1,...,Ā k, which is to be proven. 4. Proofs of Theorems 2 and 3 Proof of Theorem 2 We follow the method of Hardy and Littlewood [HL]; Gabriel [G] gives an alternative approach. The idea is to represent f j as linear combinations of indicator functions of balanced subsets of F p, yielding corresponding representations of f j. Indeed, we can write p f j i = c jm χ jm i, j = 1,...,k, m=1 where c jm are nonnegative coefficients and χ jm are the indicator functions of some balanced Ā jm F p. Moreover, the conditions imposed on f j imply that Ā jm can be chosen to satisfy δ Ā 1m, δ Ā2m, δ Ājm = for 3 j k. Accordingly, we have

14 252 VSEVOLOD F. LEV p f j i = c jm χ jm i, m=1 j = 1,...,k, where χ mj are the indicator functions of some A jm such that A jm = Ā jm.now p f 1 i 1 f k i k = c 1m1 c kmk χ 1m1 i 1 χ kmk i k, i 1 + +i k = m 1,...,m k =1 i 1 + +i k = and the inner sum at the right is simply the number of solutions of the equation i 1 + +i k = in the variables i j A jmj. By Theorem 1 this number of solutions can only increase if all A jmj are replaced by Ā jmj ; that is, the inner sum can only increase if all χ imj are replaced by χ jmj. The result follows. To prove Theorem 3 we rewrite its assertion as F 1 m 1 F k m k m 1 + +m k = m 1 + +m k = F 1 m 1 F k m k, where we put F j m = i i =m Notice that, for j = 1,...,k, and similarly, f j if j i and F j m = i i =m f j i f j i. 8 F j = max F j m, F j m = F j m m F p 9 F j = max F j m, F j m = F j m m F p 1 for any j = 1,...,k. We need several lemmas. lemma 4 For j = 1,...,k, the functions F j defined by 8 are symmetrically decreasing: F j F j 1 = F j 1 F j p = F j p. Proof For a nonempty set A F p of odd cardinality, let Ā be the corresponding balanced set, and let χ = χ A and χ = χā be the indicator functions of A and Ā, respectively.

15 LINEAR EQUATIONS AND EXPONENTIAL SUMS 253 Then χ is the balanced rearrangement of χ, and we have F j mχm = f j i f j i χm = m i i+m= i +i+m= f j i f j i χm by Theorem 2 i +i+m= From this and 1 the assertion follows. f j i f j i χm = m F j m χm. For two functions g, h : F p R + we write g h if for any system of m 1 pairwise distinct residues λ 1,...,λ m F p there exists another system of pairwise distinct residues µ 1,...,µ m F p such that gλ 1 + +gλ m hµ 1 + +hµ m. This agrees with the notation A 1,...,A k B 1,...,B k if we consider the latter as an abbreviation for NA 1,...,A k NB 1,...,B k. lemma 5 For j = 1,...,kwe have F j F j, where the functions F j and F j are defined by 8. Proof For A, Ā, χ, and χ, as in the proof of Lemma 4, we have F j mχm = f j if j i χm = m i i+m= i +i+m= f j if j i χm by Theorem 2 i +i+m= f j i f j i χm = m The assertion follows in view of 9 and 1. F j m χm. lemma 6 If g 1,...,g k : F p R + are symmetrically decreasing, then the convolution G = g 1 g k is symmetrically decreasing also. Proof Let A, Ā, χ, and χ be as in the proofs of Lemmas 4 and 5, except that we do not require the cardinality of A to be odd, and, accordingly, there can be a freedom of choice of Ā. Wehave

16 254 VSEVOLOD F. LEV Gmχm = g 1 i 1 g k i k χ m m i 1 + +i k +m= by Theorem 2 g 1 i 1 g k i k χ m = i 1 + +i k +m= m It follows that G is symmetrically decreasing. Gm χm. lemma 7 Let g, g 1,g 2 : F p R + be symmetrically decreasing functions, and suppose that g 1 g 2. Then gmg 1 m gmg 2 m. m F p m F p Proof We have p m= p gmg 1 m = = p 1 m= p 1 m= p Now we can prove Theorem 3. gm gm + 1 gm gm + 1 m= p gmg 2 m. m i= m m i= m g 1 i + gp g 2 i + gp p i= p g 1 i p i= p g 2 i Proof of Theorem 3 Let σ j : Z Z j = 1,...,kbe bijections of Z such that F j σ j are balanced and thus symmetrically decreasing in view of 9. By Theorem 2 we have F 1 m 1 F k m k m 1 + +m k = m 1 + +m k = F 1 σ1 m 1 F k σk m k = F 1 σ1 m 1 F 2 σ2 m 2 F k σk m k, m 1 m 2 + +m k = m 1 and we notice that both F 1 σ 1 and the inner sum, considered as a function of m 1, are symmetrically decreasing the latter by Lemma 6. Next, F 1 σ 1 F 1 by Lemma

17 LINEAR EQUATIONS AND EXPONENTIAL SUMS 255 5, and therefore by Lemmas 4 and 7 the right-hand side does not exceed F 1 m 1 F 2 σ2 m 2 F k σk m k m 1 m 2 + +m k = m 1 = F 2 σ2 m 2 F 1 m 1 F 3 σ3 m 3 F k σk m k. m 2 m 1 +m 3 + +m k = m 2 In a similar vein, replacing F 2 σ 2 m 2 by F 2 m j, we can only increase this latter expression, and continuing in this way, we see that F 1 m 1 F k m k F 1 σ1 m 1 F k σk m k m 1 + +m k = which is to be proven. m 1 + +m k = m 1 + +m k = F 1 m 1 F k m k, 5. Large exponential sums: The case of an arithmetic progression Let A ={,...,n 1} mod p, so that the corresponding sums S z satisfy sin πnz/p S z = z = 1,...,p sin πz/p We write Gx = sin x <x π, x and we define G = 1 by continuity; thus, the inverse function G 1 x is a continuous, monotonically decreasing function of [, 1], satisfying G 1 = π, G 1 1 =. lemma 8 For κ.5, 1 and n 2, the equation sinnx = κn 12 sin x has exactly one solution x = xn, κ on the interval, π/2, and this solution satisfies x = 1 n G 1 κ 1 + θn 2, where θ = θn, κ, 1. Proof As the function fx = sinnx/ sin x decreases on,π/n and f+ = n> nκ > = fπ/n, equation 12 has precisely one solution on,π/n. Moreover,

18 256 VSEVOLOD F. LEV if x π/n, π/2, then fx 2 1 π x n 2 <κn, which shows that π/n, π/2 contains no solutions. We now estimate the solution x = xn, κ,π/n. Since 12 implies Gnx = κgx < κ, wehavenx > G 1 κ, whence x>n 1 G 1 κ. To obtain the upper bound, we first note that Gnx = κgx > κ 1 x Next we let µ = n/n 2 + 1, and we show that Gnx 1 x2 Gn µx Notice that 14 along with 13 yield Gn µx > κ, n µx < G 1 κ, x< 1 n µ G 1 κ = 1 n 1 + 1n 2 G 1 κ, and it remains to prove 14. To this end we rewrite it as and we observe that Gn µx Gnx x2 Gn µx, 15 6 Gn µx Gnx µx G n µx. Letting ξ = n µx and using the inequality sin ξ ξ cos ξ ξ 2 sin ξ valid for any ξ [,π/2], we get sin ξ ξ cos ξ Gn µx Gnx µx ξ 2 1 µx sin ξ 3 which proves 15. = x2 2µn µgξ 6 x2 6 Gξ,

19 LINEAR EQUATIONS AND EXPONENTIAL SUMS 257 corollary 3 For <ϕ<π/3, the number of z F p such that S z >ncos ϕ is T ϕ = 2 Z 2, where Z = 1 π p n G 1 cos ϕ 1 + θn 2, θ, 1. We next estimate the function G 1 x. lemma 9 We have 61 x G 1 x 61 x x, provided 7/15 x 1. In fact, the lower bound holds true for all x [, 1]. Proof i Given x [, 1], we define z = 61 x. Then Gz 1 z2 6 = x, G 1 x z = 61 x. ii To prove the upper bound, we use the inequality Gz 1 z2 6 + z4, 12 z. 16 Given x [7/15, 1], we define z [, 2] by and we get then whence x = 1 z2 6 + z4 12, x = z 2 1 z z2, x. 8 The latter inequality and 17 give 61 x = z 2 1 z2 z x z 2 z 2 61 x z /21 x, x. 18

20 258 VSEVOLOD F. LEV Now by 16, 17, and 18, G 1 x z 61 x x 61 x x. 6. Large exponential sums: Proofs of Theorems 4 and 5 We first prove several auxiliary lemmas. Proof of Lemma 1 The observation of Yudin, mentioned in Section 2, is that Sz 1 + z 2 >ncosϕ 1 + ϕ 2, 19 provided Sz1 >ncos ϕ1, Sz2 >ncos ϕ2, and ϕ 1,ϕ 2,ϕ 1 + ϕ 2 π 2. 2 See [Y, Lemma 1]. Consider the subsets of F p, Eϕ := { z F p : Sz >ncos ϕ }, so that Tϕ= Eϕ 1. What 19 and 2 mean is that Eϕ 1 + Eϕ 2 Eϕ 1 + ϕ 2, provided ϕ 1,ϕ 2,ϕ 1 + ϕ 2 π/2. By the Cauchy-Davenport theorem we have then Eϕ1 + ϕ 2 { min Eϕ1 + Eϕ 2 1,p }, and the result follows. Proof of Lemma 2 We assume <ϕ <ϕ π/2, as otherwise the assertion is trivial. Let j = ϕ/ϕ ; then <jϕ ϕ π/2 and, by Lemma 1, Tϕ Tjϕ min { jtϕ, p 1 }. Thus, the desired estimate follows unless ϕ Tϕ p, 21 and we suppose 21 to hold for the rest of the proof. We define ϕ

21 LINEAR EQUATIONS AND EXPONENTIAL SUMS 259 so that whence, by 21, l := p 1 1, Tϕ lt ϕ p 1, 22 ϕ l ϕ 1, 23 and therefore, lϕ <ϕ π Now by Lemma 1, 22, and 24, Tiϕ it ϕ for i = 1,...,l, and it follows that there exist l pairwise disjoint residue sets E 1,...,E l F p such that E 1 = = E l =Tϕ and Sz >ncosiϕ for z E i. We conclude that p 1 np n = Sz 2 Also, by 23, whence From 25 and 26, z=1 n 2 cos 2 ϕ + +cos 2 lϕ Tϕ = n l sin2l + 1ϕ sin ϕ sin ϕ 2l + 1ϕ 2ϕ ϕ π ϕ, Tϕ. 25 sin2l + 1ϕ sin ϕ. 26 p n 1 2 nlt ϕ, p n Tϕ n p n 2 l 2l 2, Tϕ { } p n p n, Tϕ Tϕ 27 Tϕ >p n. The latter inequality, however, contradicts 27. In the course of the proof of Theorems 4 and 5, we need estimates of the Euler beta function Bu, v = 1 t u 1 1 t v 1 dt.

22 26 VSEVOLOD F. LEV lemma 1 For k>, Proof We have Similarly, B k, 3 π < 2 2 k 3/2, B k, 5 < 3 π 2 4 k 5/2. B k, 3 = 2 1 = 1 2k < 1 2k t 1/ k 1 t k dt t 1/2 1 t k dt t 1/2 e kt dt = 1 k 2 k 3/2 τ 1/2 e τ dτ < k 3/2 Ɣ 2 π = 2 k 3/2. B k, 5 1 3/2 1 = t 1 t k dt 2 k = 3 1 t 1/2 1 t k dt 2k < 3 1 t 1/2 e kt dt 2k = 3 k 2 k 5/2 τ 1/2 e τ dτ < k 5/2 Ɣ 2 = 3 π k 5/2. 4 We prove Theorems 4 and 5 together. Proof of Theorems 4 and 5 Using partial summation, we write

23 LINEAR EQUATIONS AND EXPONENTIAL SUMS 261 p 1 π/2 Sz k = kn k Tϕcos k 1 ϕ sin ϕdϕ, z=1 p 1 π/2 S z k = kn k T ϕ cos k 1 ϕ sin ϕdϕ, 28 z=1 and we denote the integrals in the right-hand sides by I and I, respectively. Since S = S, by Corollary 2 we have for k even. We first estimate I. By Corollary 3 and Lemmas 9 and 1, I 2 p 1 + 1n π/3 π n 2 G 1 cos ϕcos k 1 ϕ sin ϕdϕ π π/2 + cos k 2 T ϕ cos ϕ sin ϕdϕ 3 π/3 I I 29 2 p 1 + 1n 1 π/2 π n 2 t k 1 G 1 t dt k T ϕ cos ϕ sin ϕdϕ 1/2 2 6 p 1 + 1n 1 π n 2 t k 1 1 t 1/2 dt t k 1 1 t 3/2 dt 1/2 4 1/2 < 2 k np n + 2 2n 2 6 p π n k 3/ n k k p n. Using elementary calculus, one can easily derive the two following estimates: 6 p I 1 π n k 3/2 + 1n , k 2, k and 6 p I 1 π n k 3/2 + 1n , k k The first estimate, in view of 28, proves Theorem 4, and we proceed with the proof of Theorem 5. Fix ϕ,π/6]. Our plan is to apply Lemma 2 to show that a large value of Tϕ results in a large value of I, while I cannot be too large by 29 and

24 262 VSEVOLOD F. LEV 3. Indeed, for k 16 even, by Lemma 2 and using the Stirling formula, we have π/2 ϕ I Tϕ cos k 1 ϕ sin ϕdϕ Tϕ ϕ = 1 Tϕ k ϕ ϕ π/2 π/2 1 k ϕ cos k ϕ π/2 dϕ Tϕ cos k ϕdϕ 1 k Tϕ k π k/2 2 1 k Tϕ = 1 Tϕ 2 k k ϕ π Tϕ 2 k 3/2 e 1/3k 1 ϕ k Tϕ π = 1 > 2 k 3/2 Tϕ ϕ 2 π e1/3k ϕ k π Tϕ 2 k 3/2 1 5 ϕ 6 ϕ k e 1/3k. Along with 29 and 3, this yields π Tϕ 2 k 3/2 ϕ ϕ k e 1/3k < Using the inequality e 1/3k < k 4k and replacing for brevity ϕ by ϕ, weget Tϕ< 2 3 p π n ϕ We now recall the estimate of [L, Lemma 2]: e 1/3k 1 cos k ϕ dϕ k 6 p 1 π n k 3/2 + 1n k k n /4k 1 5/6ϕ k. 31 Tϕ< 4 p π n ϕ1 + ϕ, <ϕ π 6. This supersedes Theorem 5 for ϕ.3, and therefore we assume ϕ<.3. To optimize by k in 31, we choose k = 2.75ϕ 2/3 ; this leads to k 16 by the above assumption that ϕ<.3. We have 1 + 3/4k 1 5/6ϕ k 3/4k + 5/6ϕ k = /6ϕ, 32 k

25 LINEAR EQUATIONS AND EXPONENTIAL SUMS 263 and 3 4k ϕ k 1 2 ϕ2/ ϕ + 32 ϕ 2/3 = ϕ 2/ ϕ2/3 8 5 ϕ2/3, ϕ k>1 5 6 ϕ ϕ 2/3 = ϕ2/ ϕ2/3 > 8 9. The result now follows from 31 and 32. References [G] R. M. GABRIEL, The rearrangement of positive Fourier coefficients, Proc. London Math. Soc , [HL] G. H. HARDY and J. E. LITTLEWOOD, Notes on the theory of series, VIII: An inequality, J. London Math. Soc , [HLP] G. H. HARDY, J. E. LITTLEWOOD, and G. PÓLYA, Inequalities, 2d ed., Cambridge Math. Lib., Cambridge Univ. Press, Cambridge, [L] V. F. LEV, On the number of solutions of a linear equation over finite sets, J. Comb. Theory Ser. A , [P] J. M. POLLARD, Addition properties of residue classes, J. London Math. Soc , [Y] A. A. YUDIN, The measure of the large values of the modulus of a trigonometric sum in Russian in Number-Theoretic Studies in the Markov Spectrum and in the Structural Theory of Set Addition in Russian, Kalinin. Gos. Univ., Moscow, 1973, Institute of Mathematics, Hebrew University, Jerusalem 9194, Israel; seva@math.huji.ac.il