Chapter 6: Third Moment Energy

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1 Chapter 6: Third Moment Energy Adam Sheffer May 13, Introduction In this chapter we will see how to obtain various results by relying on a generalization of the concept of energy. While such generalizations exist for any abelian group, we focus on the case of R. In the current section we describe the main object that we will study, and in the following sections we will see applications for this object. For a finite set A R and a positive integer d, we define E + d (A) = r A (x)d. Notice that E 1 + (A) = A 2 and E 2 + (A) = E + (A). In this chapter we are specifically interested in E 3 + (A), which is sometimes called the third moment additive energy of A. 1 Since r A (x) A for any x A A, we have E 3 + (A) = r A (x)3 A r A (x)2 = A E + (A). By the Cauchy-Schwarz inequality, we have E + 3 (A) = r A (x)3 ( r A (x)3/2 r A (x)1/2 ) 2 r A (x) = E+ (A)2 A 2. Consider a set A with E + (A) = δ A 3. By the two above bounds we have that δ 2 A 4 E 3 + (A) δ A 4 For s A A, we set A s = A (A s). We have that A s = y R 1 A(y)1 A (s+y) = (s). The following lemma is taken from [3]. r A Lemma 1.1. For any finite A R we have E + 3 (A) = E+ (A, A s ). Proof. We first note that E + (A, A s ) = r A (t)r A s (t) = r A (t) (t). (1) t R t R 1 Another object that is sometimes called the third energy of A is the quantity {(a, b, c, d, e, f) A : a + b + c = d + e + f}. We do not study this quantity in the lecture notes, although it does appear in the third assignment. 1

2 We study the expression r A s (t) for a fixed t. First, we have (t) = 1 As (x)1 As (t + x) = 1 A (x)1 A s (x)1 A (t + x)1 A s (t + x) = 1 A (x)1 A (s + x)1 A t (x)1 A t (s + x) = 1 At (x)1 At (s + x) = r A t (s). By recalling that A t = r A (t), we obtain (t) = r A t (s) = A t 2 = r A (t)2. Combining this with (1) gives E + (A, A s ) = t R r A (t) (t) = t R r A (t)3 = E + 3 (A). 2 Convex sets Consider a set A = {a 1,..., a n } R such that a 1 < a 2 < < a n. We say that A is convex if for every 2 j n 1 we have a j+1 a j > a j a j 1. It intuitively seems that a convex set does not have a good additive structure (e.g., it cannot contain a large arithmetic progression). This leads to the problem: How small can the doubling of a convex set be? As a first bound, recall a result that was proved in Chapter 2 of these lecture notes. Theorem 2.1. Let A be a finite set of real numbers and let f be a strictly convex or strictly concave function. Then A + A f(a) + f(a) = Ω ( A 5/2). Corollary 2.2. Let B R be a finite convex set. Then B + B = Ω( B 3/2 ). Proof. Write B = {b 1,..., b n } so that b 1 > b 2 > > b n. Let A = {1, 2,..., n}. Let f : R R be a convex function such that for every j A we have f(j) = b j (such a function exists since B is convex). Applying Theorem 2.1 to A gives A + A f(a) + f(a) = Ω(n 5/2 ). Since A + A = Θ(n) and f(a) = B, we get that B + B = Ω(n 3/2 ). Schoen and Shkredov [2] proved that if A is convex then A + A = Ω ( A 14/9 ). We now prove a different result from the same paper. Theorem 2.3. Let A be a convex set. Then A A = Ω ( A 8/5 ). To prove this theorem, we require the following lemma. 2

3 Lemma 2.4. Let A R be a convex set and let B R. Then for every t 1, the number of elements x A B for which r A,B (x) t is O( A B 2 /t 3 ). Proof. Write A = {a 1, a 2,..., a A } so that a 1 < a 2 < < a A. Since A is a convex set, there exists a strictly convex function f : R R such that for every j {1, 2,..., A } we have f(j) = a j. For α, β R we define the curve We consider the set of curves γ α,β = {(x, f(x)) : x R} + (α, β). Γ = {γ α,β : α {1, 2,..., A } and β B}. Consider x A B such that r A,B (x) = t x, and set P x = {(p, x) R 2 : p {2, 3,..., 2 A }}. For any representation x = a j b j (with a j A and b j B) and α {1, 2,..., A } the curve γ α, bj is incident to the point (j + α, x) P x. That is, there are at least A t x incidences between the points of P x and the curves of Γ. Since every point of P x can be incident to at most t x curves of Γ, at least 2 A /3 points of P x are incident to at least t x /4 curves of Γ. Set m = {x A B : r A,B (x) t}. Notice that Γ = A B. By the above, there are at least 2m A /3 points in R 2 that are incident to at least t/4 curves of Γ. We denote by P a set of exactly 2m A /3 of these points. Note that I(P, Γ) = Ω(m A t). Recall that two translations of a strictly convex curve intersect in at most one point. Thus, by Theorem 2.5 of Chapter 2 (the incidence theorem of Pach and Sharir) we have I(P, Γ) = O( P 2/3 Γ 2/3 + P + Γ ) = O(m 2/3 A 4/3 B 2/3 + A B + m A ). By combining the two above bounds for I(P, Γ) we obtain m = O ( m 2/3 A 1/3 B 2/3 t 1 + B t 1 + mt 1). (2) The claim of the lemma trivially holds for small t, so we may assume that t is larger than the constant in the O( )-notation in (2). Thus, the third term in (2) cannot dominate the bound. The other two terms imply m = O( A B 2 /t 3 + B /t). We may assume that t min{ A, B } (since no difference in A B can has a larger number of representations), so B /t = O( A B 2 /t 3 ). That is, m = O( A B 2 /t 3 ). Corollary 2.5. Let A R be a convex set. Then E + 3 (A) = O( A 3 ). Proof. Set N j = {x A A : r A (x) 2j }. By Lemma 2.4 we have N j = O( A 3 /2 3j ). This implies E + 3 (A) = < r A (x)3 = N j 2 3(j+1) = 2 j r A (x)<2j+1 r A (x)3 O( A 3 ) = O( A 3 ). 3

4 Proof of Theorem 2.3. We begin by double counting E + 3 (A). Corollary 2.5 provides an upper bound for E + 3 (A). To obtain a lower bound, we recall that Lemma 1.1 states that E + 3 (A) = E+ (A, A s ). Set D = A A, D = k A, and Notice that D + = } {d D : A d A 2. 2 D d/ D + A d D A 2 2 D = A 2 /2, which implies that d D + A d A 2 /2. Recall that A s = r A (s). For any s D+, the Cauchy-Schwarz inequality implies E + (A, A s ) = r A,A s (x) 2 ( s r A,A s (x)) 2 A A s s = A 2 A s 2 A A s. This in turn implies A A s E + (A, A s ) 1/2 A A s 1/2. By summing this up for every s D + and applying the Cauchy-Schwarz inequality once more, we obtain A A s E + (A, A s ) 1/2 A A s 1/2 s D + s D ( + ) 1/2 ( ) 1/2 E + (A, A s ) A A s. s D + s D + Combining this with Lemma 1.1 and with s D + A s A 2 /2 gives s D + A A s A 6 4E + 3 (A). (3) Next, we notice that A A s = A (A (A s)) = D (D + s). That is, for every element of A A s there exist d, d D such that d = d + s, which in turn implies r D (s) A A s. By combining this with (3) we get This in turn implies E + (A, D) = s D A 6 4E 3 + (A) A 2 2 D s D + A A s r A (s)r D (s) s D + r D (s) s D + r D (s). s D + r A (s)r D (s) A 8 8 D E + 3 (A) = A 7 8kE + 3 (A). 4

5 By Corollary 2.5 we have E + 3 (A) = O( A 3 ). Combining this with the above gives E + (A, D) = Ω ( A 4 /k). Let N j denote the number of elements x A D for which r A,D (x) 2j. Notice that E + (A, D) = x A D r A,D (x)2 = 2 j r A (x)<2j+1 r A (x)2 < N j 2 2j+2. By dyadic pigeonholing, there exists j such that N j 2 2j+2 = Ω (E + (A, D)) = Ω ( A 4 /k). We have the trivial bound N j A D /2 j = A 2 k/2 j. Combining these two bounds gives 2 j = Ω ( A 2 /k 2 ). Lemma 2.4 implies N j = O( A D 2 /2 3j ) = O( A 3 k 2 /2 3j ). Thus, we have ( ) A A 4 /k = O (N j 2 2j ) = O 3 k 2 2 2j = O ( A k 4 ). 2 3j This implies k = Ω ( A 5/3 ), which completes the proof. 3 Another BSG bound Our next application of the third energy is to derive a variant of Schoen s theorem from Chapter 3 (this variant is from the same paper of Scheon [1]). To appear??? References [1] T. Schoen, New bounds in Balog-Szemerdi-Gowers theorem, Combinatorica, to appear. [2] T. Schoen and I. D. Shkredov, On sumsets of convex sets, Combinatorics, Probability and Computing, 20 (2011), [3] T. Schoen and I. D. Shkredov, Additive properties of multiplicative subgroups of F p, The Quarterly Journal of Mathematics 63 (2012),

Additive Combinatorics and Szemerédi s Regularity Lemma

Additive Combinatorics and Szemerédi s Regularity Lemma Vijay Keswani Anurag Sahay 20th April, 2015 Supervised by : Dr. Rajat Mittal 1 Contents 1 Introduction 3 2 Sum-set Estimates 4 2.1 Size of sumset