Mei-Chu Chang Department of Mathematics University of California Riverside, CA 92521

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1 ERDŐS-SZEMERÉDI PROBLEM ON SUM SET AND PRODUCT SET Mei-Chu Chang Department of Mathematics University of California Riverside, CA 951 Summary. The basic theme of this paper is the fact that if A is a finite set of integers, then the sum and product sets cannot both be small. A precise formulation of this fact is Conjecture 1 below due to Erdős-Szemerédi [E-S]. (see also [El], [T], and [K-T] for related aspects.) Only much weaker results or very special cases of this conjecture are presently known. One approach consists of assuming the sum set A + A small and then deriving that the product set AA is large (using Freiman s structure theorem). (cf [N-T], [Na3].) We follow the reverse route and prove that if AA < c A, then A + A > c A (see Theorem 1). A quantitative version of this phenomenon combined with Plünnecke type of inequality (due to Ruzsa) permit us to settle completely a related conjecture in [E-S] on the growth in k. If g(k) min{ A[1] + A{1} } over all sets A Z of cardinality A = k and where A[1] (respectively, A{1}) refers to the simple sum (resp., product) of elements of A. (See (0.6), (0.7).) It was conjectured in [E-S] that g(k) grows faster than any power of k for k. We will prove here that ln g(k) () ln Introduction. (see Theorem ) which is the main result of this paper. Let A, B be finite sets of an abelian group. The sum set of A, B is A + B {a + b a A, B B}. (0.1) We denote by ha A + + A (h fold) (0.) 1 Typeset by AMS-TEX

2 MEI-CHU CHANG the h-fold sum of A. Similarly we can define the product set of A, B and h-fold product of A. If B = {b}, a singleton, we denote AB by b A. AB {ab a A, b B}, (0.3) A h A A (h fold). (0.4) In 1983, Erdős and Szemerédi [E-S] conjectured that for subsets of integers, the sum set and the product set cannot both be small. Precisely, they made the following conjecture. Conjecture 1. (Erdős-Szemerédi). For any ε > 0 and any h N there is k 0 = k 0 (ε) such that for any A N with A k 0, then ha A h A h ε. (0.5) ( ) A + h 1 We note that there is an obvious upper bound ha A h. h Another related conjecture requires the following notation of simple sum and simple product. { k } A[1] ε i a i a i A, ε i = 0 or 1 A{1} i=1 { k i=1 a ε i i a i A, ε i = 0 or 1 For the rest of the introduction, we only consider A N. } (0.6). (0.7) Conjecture. (Erdős-Szemerédi). Let g(k) min A =k { A[1] + A{1} }. Then for any t, there is k 0 = k 0 (t) such that for any k k 0, g(k) > k t. Toward Conjecture 1, all work have been done so far, are for the case h =. Erdős and Szemerédi [E-S] got the first bound: Theorem (Erdős-Szemerédi). Let f(k) min A =k A A. Then there are constants c 1, c, such that k 1+c 1 < f(k) < k e c ln. (0.8) Nathanson showed that f(k) > ck 3 31, with c =

3 ERDŐS-SZEMERÉDI 3 At this point, the best bound is A A > c A 5/4 (0.9) obtained by Elekes [El] using the Szemerédi-Trotter theorem on line-incidences in the plane (see [S-T]). On the other hand, Nathanson and Tenenbaum [N-T] concluded something stronger by assuming the sum set is small. They showed Theorem (Nathanson-Tenenbaum). If A 3 A 4, (0.10) then A ( ) A. (0.11) ln A Very recently, Elekes and Ruzsa [El-R] again using the Szemerédi-Trotter theorem, established the following general inequality. Theorem (Elekes-Ruzsa). If A R is a finite set, then A + A 4 AA ln A > A 6 (0.1) then In particular, their result implies that if For further result in this direction, see [C]. A < c A, (0.13) A A c ln A. (0.14) Related to Conjecture, Erdős and Szemerédi [E-S] have an upper bound: Theorem (Erdős-Szemerédi). Let g(k) min A =k { A[1] + A{1} }. There is a constant c such that () c g(k) < e ln. (0.15) Our first theorem is to show that the h-fold sum is big, if the product is small.

4 4 MEI-CHU CHANG Theorem 1. Let A N be a finite set. If A < α A, then A > 36 α A, (0.16) and Here ha > c h (α) A h. (0.17) c h (α) = (h h) hα. (0.18) Our approach is to show that there is a constant c such that m A eπimx h dx < c A h (0.19) by applying an easy result of Freiman s Theorem (see the paragraph after Proposition 10.) to obtain { ( ) j1 ( ) } js a a1 as A P 0 ji < l i. (0.0) b b 1 and carefully analysing the corresponding trigonometric polynomials (see Proposition 8). These are estimates in the spirit of Rudin [R]. The constant c here depends, of course, on s and h. In order to have a good universal bound c, we introduce the concept of multiplicative dimension of a finite set of integers, and derive some basic properties of it (see Propositions 10 and 11). We expect more application coming out of it. Another application of our method together with Plünnecke type of inequality (due to Ruzsa) gives a complete answer to Conjecture. Theorem. Let g(k) min A =k { A[1] + A{1} }. Then there is ε > 0 such that b s k (1+ε) ln > g(k) > k ( 1 8 ε) ln. (0.1) Remark.1 (Ruzsa). The lower bound can be improved to k ( 1 ε) ln. We will give more detail after the proof of Theorem. Using a result of Laczkovich and Rusza, we obtain the following result related to a conjecture in [E-S] on undirected graphs.

5 ERDŐS-SZEMERÉDI 5 Theorem 3. Let G A A satisfy G > δ A. Denote the restricted sum and product sets by A G + A = {a + a (a, a ) G} (o.) If then A G A = {aa (a, a ) G}. (0.3) A G A < c A, (0.4) A G + A > C(δ, c) A. (0.5) The paper is organized as follows: In Section 1, we prove Theorem 1 and introduce the concept of multiplicative dimension. In Section, we show the lower bound of Theorem and Theorem 3. In Section 3, we repeat Erdős-Szemerédi s upper bound of Theorem. Notation: We denote by a the greatest integer a, and by A the cardinality of a set A. Acknowledgement: The author would like to thank J. Bourgain for various advice, and I. Ruzsa and the referee for many helpful comments. 1. Proof of Theorem 1. Let A N be a finite set of positive integers, and let Γ h,a (n) be the number of representatives of n by the sum of h (ordered) elements in A, i.e., Γ h,a (n) {(a 1,..., a h ) a i = n, a i A}. (1.1) The two standard lemmas below provide our starting point. Lemma 3. Let A N be finite and let h N. If there is a constant c such that Γ h,a(n) < c A h, (1.) then n ha ha > 1 c A h. (1.3) Proof. Cauchy-Schwartz inequality and the hypothesis give A h = Γ h,a (n) ha 1/ ( Γ h,a(n)) 1/ n ha n ha < ha 1/ c 1/ A h/.

6 6 MEI-CHU CHANG Lemma 4. The following equality holds n ha Γ h,a(n) = ( m A e πimx h ) h. Proof. ( e πimx h ) h = e πimx h dx m A m A = ( e πimx ) h dx m A = ( Γ h,a (n)e πinx ) dx = n ha n ha Γ h,a(n). The last equality is Parseval equality. From Lemmas 3 and 4, it is clear that to prove Theorem 1, we want to find a constant c such that ( m A e πimx h ) < c A. (1.4) In fact, we will prove something more general and to be used in the inductive argument. Lemma 5. Let A N be a finite set with A < α A. Then for any {d a } a A R +, we have ( a A d a e πiax h ) < c d a (1.5) for some constant c depending on h and α only. For a precise constant c, see Proposition 9. The following proposition takes care of the special case of (1.5) when there exists a prime p such that for every nonnegative integer j, p j appears in the prime factorization of at most one element in A. It is also the initial step of our iteration. First, for convenience, we use the following Notation: We denote by G +, the set of linear combination s of elements in G with coefficients in R +.

7 Proposition 6. Let p be a fixed prime, and let F j (x) ERDŐS-SZEMERÉDI 7 { e πipj nx n N, (n, p) = 1 } +. (1.6) Then ( j F j h ) c h F j, where c h = h h. (1.7) h j Proof. To bound j F j h dx, we expand j F j h as ( Fj ) h ( F j ) h. (1.8) Let F j1 F jh F jh+1 F jh (1.9) be a term in the expansion of (1.8). After rearrangement, we may assume j 1 j h, and j h+1 j h. When (1.9) is expressed as a linear combination of trignometric functions, a typical term is of the form ne πix(pj 1 n 1 + +p j h n h p j h+1 n h+1 p j h n h ) (1.10) We note that the integral of (1.10) is 0, if the expression in the parenthesis in (1.10) is nonzero. In particular, independent of the n i s, the integral of (1.10) is 0, if j 1 j j h+1, or j 1 j h+1 min{j, j h+ }, or j h+1 j h+ j 1. (1.11) Therefore, if any of the statements in (1.11) is true, then the integral of (1.9) is 0. We now consider the integral of (1.9) of which the index set {j 1,..., j h } don t satisfy any of the conditions in (1.11). For the case j 1 = j j h+1, we see that in an ordered set of h elements coming from the expansion of (1.8) (before the rearrangement), there are exactly ( h ) choices for the positions of j1, j. On the other hand, if F j1 F j is factored out, the rest is symmetric with respect to j 3,..., j h, and j h+1,..., j h, i.e., all the terms involving j j 1 = j j h+1 are simplified to ( ) h (F j ) ( F k ) h (1.1) k j

8 8 MEI-CHU CHANG Same reasoning for the other two cases, we conclude that ( ( ) h F j h ) h = Fj ( F k ) h ( F k ) h dx j j k j k j + h F j ( F k F k ) h 1 dx j k j k j ( ) h + F j( F k ) h ( F k ) h dx. j k j k j The right hand side is [ ( )] h h + j F j k j F k h dx (h h) j = (h h) j F j h ( k j F k ) h h h 1 F j h ( k j F k h ) h. The last inequality is Hölder inequality. Now, the next lemma concludes the proof. Lemma 7. Let F k {e πimkx m k Z} +. Then k F k h k j F k h, for any j. (1.13) Proof. F k h dx k = ( F k + F k ) ( F k + F k )( k j k<j k j k<j k j ( F k F k ) h dx = ( F k h ) h. k j k j k j F k + k<j F k ) ( k j F k + k<j F k ) dx The inequality is because the coefficients of the trignometric functions (as in (1.10)) in the expansion are all positive. Remark 7.1. This is a special case of some general theorem in martingale theory.

9 Proposition 8. Let p 1,, p t be distinct primes, and let Then ERDŐS-SZEMERÉDI 9 F j1,...,j t (x) {e πipj 1 1 p j t t nx n N, (n, p 1 p t ) = 1} +. (1.14) F j1,...,j t h c t h F j1,...,j t h, where c h = h h. (1.15) j 1,...,j t j 1,...,j t Proof. We do induction on t. The left hand side of (1.15) is F j1,...,j t c h F j1,...,j t c h j 1 j,...,j t j 1 j,...,j t j 1 which is the right hand side. c t 1 h j,...,j t F j1,...,j t, Proposition 5 is proved, if we can find a small t such that the Fourier transform of F j1,...,j t is supported at one point and such t is bounded by α. So we introduce the following notion. Definition. Let A be a finite set of positive rational numbers in lowest term. (cf (0.0)) Let q 1,..., q l be all the prime factors in the obvious prime factorization of elements in A. For a A, let a = q j 1 j qj l l be the prime factorization of a. Then the map ν : A R l by sending a to (j 1,..., j l ) is one-to-one. The multiplicative dimension of A is the dimension of the smallest (affine) linear space in R l containing ν(a). We note that for any nonzero rational number q, q A and A have the same multiplicative dimension, since ν(q A) is a translation of ν(a). The following proposition is a more precise version of Lemma 5. Proposition 9. Let A N be finite with mult.dim(a) = m. Then ( a A d a e πiax h ) < c m h d a, where c h = h h. (1.16) Proof. To use (1.15) in Proposition 8, we want to show that there are primes q 1,..., q m such that a term of the trigonometric polynomial in the left hand side of (1.15), when expressed in terms of the notation in (1.14), is F j1,...,j m = d a e πipj 1 1 p j m nx t. In other words, we want to show that among the prime factors q 1,..., q l of elements in A, there are m of them, say q 1,..., q m such that (*) (j 1,..., j m ) Z m, there is at most one a A such that q j 1 1 qj m is part of the prime factorization of a. (*) is equivalent to (**) π ν is injective, where ν is as in the definition of multiplicative dimension and π : R l R m is the projection to the first m coordinates. Since dim ν(a) = m, (**) is clear after some permutation of the q i s.

10 10 MEI-CHU CHANG Proposition 10. Let A N be finite with mult.dim A = m. Then Γ h,a(n) < c mh h A h, where c h = h h. (1.17) n ha Proof. This is a consequence of Lemma 4 and Proposition 9 (with d a = 1). The hypothesis of Theorem 1 gives a universal bound on the multiplicative dimension of A by applying Freiman s Theorem (cf [Fr1], [Fr], [Fr3], [Bi], [C1], [Na1].) In fact, we do not need the full content of the Freiman s Theorem, but a much easier result by Freiman. A small modification (over Q instead of over R) of Lemma 4.3 in [Bi] is sufficient. (As Ruzsa pointed out that it is also Lemma 1.14 in [Fr1].) Theorem (Freiman). Let G R be a subgroup and A 1 G be finite. constant α, α < A 1, such that A 1 < α A 1, then there is an integer s α If there is a such that A 1 is contained in a s-dimensional proper progression P 1, i.e., there exist β, α 1,..., α s G and J 1,, J s N such that and P 1 = J 1 J s. Note that if A 1 > A 1 P 1 = {β + j 1 α j s α s 0 j i < J i }, α α+1 ( α+1 α), then s α 1. Recall that the full Freiman s theorem also permits to state a bound J 1 J s < c(α) A 1. However this additional information will not be used in what follows. We would like to work on a sum set instead of a product set. So we define A 1 ln A = {ln a a A}. (1.18) Note that ln is an isomorphism between the two groups (Q +, ) and (ln Q +, + ). Applying the theorem to A 1 ln Q +, then pushing back by (ln) 1, we have { a A P b (a 1 ) j1 ( a } s ) j s 0 j i < J i Q +, (1.19) b 1 b s where a, b, a i, b i, J i N, and (a, b) = 1, (a i, b i ) = 1. Moreover, s α 1 and different ordered sets (j 1,, j s ) represent different rational numbers. Clearly, mult. dim A mult.dim P = dim E s α 1, (1.0) where E is the vector space generated by ν( a 1 b 1 ),, ν( a s b s ). Therefore, we have

11 ERDŐS-SZEMERÉDI 11 Proposition 11. Let A N be a finite set. If A < α A for some constant α, α < A 1/, then mult.dim A α. Furthermore, if A > α α+1 ( α+1 α), then mult.dim A α 1. Putting Propositions 10 and 11 together, we have Proposition 1. Let A N be finite. If A < α A for some constant α, α < A 1/, then Γ h,a(n) < c αh h A h, where c h = h h. n ha Now, Theorem 1 follows from Proposition 1 and Lemma 3.. Simple sums and products. In this section we will prove the lower bound in Theorem. Let A N be finite. We define g(a) A[1] + A{1}, (.1) where A[1] and A{1} are the simple sum and simple product of A. (See (0.6),(0.7) for precise definitions.) We will show that for any ε and any A N with A = k 0, For those who like precise bound, we show that g(a) > k ( 1 8 ε) ln. (.) For 0 < ε 1, ε < 1, g(a) > e 3 k 1 ε ( 1 4 ε 1 ) ln, (.3) if A = k is large enough such that ln > 8ε 1, (.4) and ln > ε. (.5)

12 1 MEI-CHU CHANG Proposition 13. Let B N be finite with mult.dim B = m. Then for any h 1 N, we have [ ] h1 B h 1 B B[1] > (h 1 h 1) m+1. (.6) Proof. Since h 1 B B[1] is the set of simple sums with exactly h 1 summands, we have Therefore, ( ) B h 1 n h 1 B B[1] ( ) h1 B < (h 1 B B[1]) 1/ ( h 1 n h 1 B Γ h1,b(n). (.7) Γ h 1,B(n)) 1/ (h 1 B B[1]) 1/ [(h 1 h 1 ) mh 1 B h 1 ] 1/. The first inequality is because of Cauchy-Schwartz inequality and h 1 B B[1] h 1 B. The second inequality is Proposition 10. Remark Clearly, the denominator in (.6) can be replaced by (h 1 h 1 ) m h 1. Proposition 14. Let B N with B k and mult.dim B = m. For any 0 < ε 1 < 1, if m + 1 ( 1 4 ε ) 1 ln, (.8) then g(b) > k ε 1. (.9) Proof. Inequality (.8) is equivalent to In Proposition 13, we take h 1 = which gives Combining (.11), (.10) and (.6), we have () m+ k 1/ ε 1. (.10) ( k 1/ g(b) > h 1 B B[1] > h 1 (). (.11) k 1/ ε 1 ) = k ε 1

13 ERDŐS-SZEMERÉDI 13 Remark Let A N with A = k, k 0 (see (.4)). The set B in Proposition 14 will be taken as a subset of A. Then the bound in (.9) is bigger than that in (.), and our proof is done. Therefore for the rest of the section, we assume mult.dimb ( 1 4 ε ) ln, for any B A with B > k. (.1) We need the following Notation: We denote B ν(b) for any B A, where ν = A Z l is as in the definition of multiplicative dimension. Note that B [1] = B{1}. (.13) We will use the following Plünnecke type of inequality due to Ruzsa. Ruzsa s Inequality. [Ru] For any h, l N If M + N ρ M, then hn ln ρ h+l M. Proof of (.). We divide A into k pieces B 1, B,, each of cardinality at least k. For 0 < ε < 1, let ρ = 1 + k 1/+ε, (.14) and let s A s B i. (.15) i=1 There are two cases: (i) s, (A s Bs+1 ) [1] > ρ A s[1]. Iterating gives A [1] = (B 1 B ) [1] > k ρ k. (.16) Therefore g(a) > A{1} = A [1] > e ( k )ln ρ+ 1 > e ( k ) 4 5 k 1/+ε + 1 > e 4 5 kε. (.17)

14 14 MEI-CHU CHANG Inequality (.5) is equivalent to k ε > (). which is certainly stronger than what we need to show (.). (ii) s such that (A s B s+1 ) [1] ρ A s[1]. We use the fact that (A s B s+1 ) [1] = A s[1] + B s+1[1], and Ruzsa s Inequality (with h = h + 1, l = 1) to obtain (h + 1)B s+1[1] B s+1[1] ρ h + A s[1]. (.18) Let m = mult.dimb s+1. For a set B, for h N, denote The left hand side of (.18) is B[h] { εi x i ε i = 0,..., h, x i B h B s+1[1] B s+1[h ] }. (.19) h m. (.0) We take h = k 1/ ε. Then in the right hand side of (.18) we have Therefore, (.18), (.0) and (.1) imply ρ h + (1 + k 1/+ε ) k1/ ε + g(a) > g(a s ) < (e k 1/+ε ) k1/ ε + < e 3. (.1) > A s[1] > e 3 h m > e 3 k 1/ ε ( 1 4 ε 1 ) ln. The last inequality follows from our choice of h and Remark Proof of Remark.1. In Proposition 14, if we take B with B k, then we can replace (.8), and (.9) by m (1 ε 1) ln, (.8 )

15 and Let Then (.1) can be replaced by ERDŐS-SZEMERÉDI 15 ( k ε 1 g(b) > m 0 = 1 (1 ε 1) ). (.9 ) ln. mult.dimb m 0, for any B A with B > k. (.1 ) Now we modify the proof of (.). Since A = k > k, we have mult.dima m 0. So there is B 1 A with mult.dimb 1 = m 0 and B 1 = m Similarly, we have B A B 1 with mult.dimb = m 0 and k B = m We continue this process until r > (m 0 +1). We have A B 1 B r with and With more replacements, and mult.dimb i = m 0, B i = m ln > 3 ε. (.5 ) ρ = 1 + k 1+ε, (.14 ) Identical argument gives g(a) > e 3 k 1 ε ( 1 ε 1 ) ln. Sketch of Proof of Theorem 3. Let A = N. Then Laczkovich-Ruzsa Theorem [L-R] and (0.4) give A 1 A with A 1 A 1 < c N, (.) and G (A 1 A 1 ) > δ N. (.3)

16 16 MEI-CHU CHANG The weak Freiman s Theorem and (.) imply mult.dima 1 < c. (.4) It follows from Proposition 10 (with h =, and as in the proof of Lemma 4) that Hence β {(n 1, n, n 3, n 4 ) A 4 1 n 1 n + n 3 n 4 = 0} < 36 c N. (.5) δ N < G {(n 1, n ) A1 n = n 1 + n } < A 1 + A1 1 1 β. (.6) G n A 1+A1 The first inequality is (.3), while the second one is Cauchy-Schwatz inequality. Therefore, (.5) and (.6) give A G + A A 1 G + A1 (δ ) N 4 β > CN. 3. The example. In this section for completeness we repeat a family of examples by Erdős-Szemerédi which provide the upper bound in Theorem. Precisely, we will show Proposition 15. Given ε 3 > 0, for J so large that There is a set A of cardinality A = k J J, such that ln J ln ln J > 1 ε 3, (3.1) g(a) < k (1+ε) ln, (3.) where ε = 3ε 3 + ε 3. (3.3) The example really comes from the proof of the lower bound of Theorem. Let p 1,, p J be the first J primes, and let A {p j 1 1 p j J J 0 j i < J}. (3.4) Then k A = J J. (3.5) We will use the following relations between k and J.

17 Lemma 16. Let k, J be as in (3.5). Then (i) = Jln J. (ii) ln = ln J + ln ln J. If J and ε 3 satisfy (3.1), then (iii) ln < (1 + ε 3 )ln J. (iv) J < (1 + ε 3 ) ln. (v) J < (1 + ε )( ln ), where ε = ε 3 + ε 3. ERDŐS-SZEMERÉDI 17 Proof. Each one follows immediately from the proceeding one. For (iii) implying (iv), we use J = ln J. Remark The inequality ln J ln ln J > 1 ε 3 clearly implies ln > 1 ε 3. (3.6) Lemma 17. We have the following (i) a A, a < () J (ii) A[1] < k() J (iii) A{1} < (kj) J. Proof. (i) For a A, (3.4) gives a < ( i<j p i ) J < ( i<j iln i) J < (J J (ln J) J ) J = (Jln J) J = () J. The second inequality is by the Prime Number Theorem. The last equality is Lemma 16 (i). (ii) follows from (i). (iii) We see that { k s=1 A{1} = p j(s) 1 1 p Since k s=1 j(s) i < kj, (iii) holds. k s=1 j(s) J J 0 j (s) i } < J. (3.7)

18 18 MEI-CHU CHANG Proof of Proposition 15. Lemma 17 (ii) and Lemma 16 (v) give A[1] < k() (1+ε )( = e +(1+ε () ) ln ln ) = e (1+(1+ε ) ln ) (1+ε) < e ln (1+ε) = k ln. (3.8) Here ε = ε + ε 3 = 3ε 3 + ε 3. We use (3.6) for the last inequality. Lemmas 17 (iii), (3.5), and 16 (iv) give The last inequality is again by (3.6). A{1} < k J k = k J+1 < k (1+ε 3) ln +1 < k (1+ε 3) ln. (3.9) Putting (3.8) and (3.9) together, we have g(a) < k (1+ε) ln. References [Bi]. Y. Bilu, Structure of sets with small sumset, in Structure Theory of Set Addition, Astérisque 58, 1999, pp [C1]. M.-C. Chang, A polynomial bound in Freiman s Theorem, Duke Math. J. (to appear). [C]., Factorization in generalized arithmetic progressions and applications to the Erdös- Szemerédi sum-product problems, GAFA (to appear). [El]. G. Elekes, On the number of sums and products, Acta Arithmetica 81, Fase 4 (1997), [El-R]. G. Elekes, J. Ruzsa, Product sets are very large if sumsets are very small, preprint. [E]. P. Erdős, Problems and results on combinatorial number theory, III, in Number Theory Day (M. Nathanson, ed.), New York 1976; volume 66 of Lecture Notes in Mathematics, Springer-Verlag, Berlin, 1977, pp [E-S]. P. Erdős and E. Szemerédi, On sums and products of integers, in Studies in Pure Mathematics, Birkhauser, Basel, 1983, pp [Fr1]. G. Freiman, Foundations of a structural theory of set addition, Translations of Math. Monographs 37, AMS, [Fr]. G. A. Freiman, On the addition of finite sets, I, Izv Vysh. Zaved. Matematika 13(6) (1959), 0 13.

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