Let A be a finite non-empty set of integers. An asymptotic estimate of the size of the sum of several dilates was obtained by Bukh. The unique known exact bound concerns the sum |A + k·A|, where k is a prime and |A| is large. In its full generality, this bound is due to Cilleruelo, Serra and the first author.
Let k be an odd prime and assume that |A| > 8kk. A corollary to our main result states that |2·A + k·A|=(k+2)|A|-k2-k+2. Notice that |2·P+k·P|=(k+2)|P|-2k, if P is an arithmetic progression.Postprint (author's final draft

Hamidoune, Yayha Old

Rué Perna, Juan José

Combinatorics Probability Computing

English

Hamidoune, Yahya Ould

UPCommons (Universitat Politècnica de Catalunya)

A LOWER BOUND FOR THE SIZE OF A MINKOWSKI SUM OFDILATESY. O. HAMIDOUNE AND J. RUÉAbstract. Let A be a finite nonempty set of integers. An asymptotic estimateof several dilates sum size was obtained by Bukh. The unique known exactbound concerns the sum |A + k ·A|, where k is a prime and |A| is large. In itsfull generality, this bound is due to Cilleruelo, Serra and the first author.Let k be an odd prime and assume that |A| > 8kk. A corollary to ourmain result states that |2 · A + k · A| ≥ (k + 2)|A| − k2 − k + 2. Notice that|2 · P + k · P | = (k + 2)|P | − 2k, if P is an arithmetic progression.1. IntroductionLet A, B be finite nonempty sets of real numbers. The Minkowski sum of A andB is defined asA+B = {a+ b : a ∈ A and b ∈ B}.The inequality |A+B| ≥ |A|+ |B|−1 is an easy exercise, that we shall use withoutany reference. For a real number r, the r-dilate of A is the set r ·A = {ra : a ∈ A}.Lower bounds for the size of dilates sums appeared in different contexts.  Laba andKonyagin [?] investigated the sum A+λ ·A (where λ is a transcendental number) inconnection with well-distributed planar sets distances. Dilates sums also appearedin the proofs of sum-product results in finite fields by Garaev [?], and by Katz andShen [?]. The sum of two dilates appeared in the work of Nathanson, O’Bryant,Orosz, Ruzsa and Silva on binary linear forms [?]. Also, they were used by Bukh[?] in connection with a problem of Ruzsa.From now on, we assume that A is a nonempty set of integers. Dilates sum of theform A+ 3 ·A were investigated independently by Bukh [?] and by Cilleruelo, Silvaand Vinuesa [?]. More recently, the authors of [?] proved that for an odd prime k,|A + k · A| ≥ (1 + k)|A| − (k + 1)2/4 for |A| sufficiently large. Let m1, · · · ,mj beintegers with gcd(m1, . . . ,mj) = 1, Bukh proved in [?] that|m1 ·A+ · · ·+mj ·A| ≥ (|m1|+ · · ·+ |mj |)|A| − o (|A|) .Bukh’s result suggests the following:Conjecture 1.1. For a set of integers Z with gcd(Z) = 1 and for every nonemptyset of integers A, there is an absolute constant c such that∣∣∣∣∣∑m∈Zm ·A∣∣∣∣∣ ≥(∑m∈Z|m|)|A| − c.12 Y. O. HAMIDOUNE AND J. RUÉIn the present work, we prove the above conjecture for Z = {2, k}, where k is anodd prime. For simplicity, we will not consider negative dilates, but the reader willcertainly observe that our approach works in this case.In section 2, we present some easy and known lemmas that we need. In section3, we prove some intermediary results needed in our induction arguments. One ofthe results of this section states that |n ·A+m ·A| ≥ 4|A| − 4, where m and n arecoprime integers. This result is a counterpart of a lemma by Nathanson [?] statingthat |A+2 ·A| ≥ 3|A|−2. Let k be an odd prime. By a k-component of a set X ⊂ Z,we shall mean the nonempty intersection of X with some congruence class modulok. In section 4, we investigate the marginal set (2 ·C+k ·A)\ (2 ·C+k ·C), where Cis a k-component of A. In section 5, we prove that |2 ·A+k ·A| ≥ (k+2)|A|−4kk−1.Assuming that 0 ∈ A, gcd(A) = 1, |A| > 8kk and that A has a k-componentcontaining at most k − 1 nonempty k2-component, we show that|2 ·A+ k ·A| > (k + 2)|A|.Readers interested in the description of sets reaching equality could quite likely usethis result, since it shows that the objective function |2 · A + k · A| achieves itsminimum on structured sets, for |A| large. Let X be a finite set of integers with|X| > 8kk. We conclude Section 5 by an easy consequence of our main result, statingthat |2 ·X + k ·X| ≥ (k+ 2)|X| − k2− k+ 2. As an exercise, the reader could provethat |2 ·P + k ·P | = (k+ 2)|P |− 2k, if P is an arithmetic progression. Observe thatfor k = 3, our bound differs at most by 4 from the best possible one.2. Preliminaries and terminologyIn this paper, we consider sums of dilates of a finite set of integers. The nextknown lemma shows that the size of a dilates sum remains invariant if we replaceA by an affine transform of it.Lemma 2.1. [?] Let A be a finite set of integers and let r, s, u, v be non-zerointegers. Then• |r · (A+ v) + s · (A+ v)| = |r ·A+ s ·A|,• |r · (u ·A) + s · (u ·A)| = |r ·A+ s ·A|.Proof. We have clearly|r · (A+ v) + s · (A+ v)| = |r ·A+ s ·A+ (rv + sv)|= |r ·A+ s ·A|.A LOWER BOUND FOR THE SIZE OF A MINKOWSKI SUM OF DILATES 3as |A+ w| = |A|. We also have|r · (u ·A) + s · (u ·A)| = |(ru) ·A+ (su) ·A|= |u · (r ·A) + u · (s ·A)|= |r ·A+ s ·A|.Let A be a finite set of integers. The intersection of A with a congruence classmodulo n will be called a n-component. By a decomposition modulo n, we meana partition of A into its n-components. The number of n-components of A will bedenoted by cn(A). We shall say that A is n-full if cn(A) = n. The set A is n-semi-fullif every n-component C of A satisfies cn2(C) = n.Lemma 2.2. If A is n-full, then gcd(A) is coprime to n. Moreover, 1gcd(A) · A isn-full.Proof. There is an u ∈ A such that u ≡ 1 (mod n). As gcd(A) divides u, thengcd(gcd(A), n) divides both u and n, hence it divides 1. Since gcd(A) is invertiblemodulo n, cn(1gcd(A) ·A)= cn(A). 3. ToolsLemma 3.1. Let A and B be finite sets of integers and let m,n be coprime integers.Let C denote the set of m-components of A. Then n ·A+m ·B =⋃C∈C n ·C+m ·Bis a decomposition modulo m.Proof. Clearly n ·C+m ·B ≡ n ·C (mod m), for any C ∈ C. The result follows nowsince n ·C and n ·C ′ are necessarily incongruent modulo m, for distinct componentsC,C ′ ∈ C. The next proposition is basic in our approach.Proposition 3.2. Let A and B be finite sets of integers and let m,n be coprimeintegers. Then |n ·A+m ·B| ≥ cn(B)|A|+ cm(A)|B| − cm(A)cn(B).Proof. Let A be the set m-components of A and let B be the set n-components ofB. We claim that if M1,M2 ∈ A and N1, N2 ∈ B such that (M1, N1) 6= (M2, N2),then (n ·M1 +m ·N1)∩(n ·M2 +m ·N2) = ∅. Suppose the contrary and take ai ∈Miand bi ∈ Ni, 1 ≤ i ≤ 2 with na1 +mb1 = na2 +mb2. Thus, n(a1−a2) = m(b1− b2).Since m is coprime to n, we have b1 − b2 ≡ 0 (mod n) and a1 − a2 ≡ 0 (mod m).In particular, M1 = M2 and N1 = N2, a contradiction.4 Y. O. HAMIDOUNE AND J. RUÉTherefore, using Lemma 3.1 we have|n ·A+m ·B| =∣∣∣∣∣∣⋃M∈A; N∈Bn ·M +m ·N∣∣∣∣∣∣ ≥∑M∈A; N∈B|M |+ |N | − 1= cn(B)|A|+ cm(A)|B| − cm(A)cn(B).Corollary 3.3. Let 2 ≤ n < m be coprime integers. Let A be a finite set of integers.Then |n ·A+m ·A| ≥ 4|A| − 4.Proof. The result holds clearly if |A| = 1. Assume that |A| ≥ 2. Without loss ofgenerality, 0 ∈ A. Put B = 1gcd(A) ·A. Put r = cm(B) and s = cn(B). Without lossof generality we may assume that r ≥ s. Observe that 2 ≤ r ≤ |A|. By Lemma 2.1and Proposition 3.2, we have|n ·A+m ·A| = |n ·B +m ·B|≥ 4|B| − 4 + (r + s− 4)|B| − rs+ 4≥ 4|B| − 4 + (r + s− 4)r − rs+ 4≥ 4|B| − 4 + (r − 2)2 ≥ 4|A| − 4.Corollary 3.4. Let m be an odd integer. If A is a m-full finite set of integers,then |2 · A + m · A| ≥ (m + 2)|A| − 2m. If A is m-semi-full set of integers, then|2 ·A+m ·A| ≥ (m+ 2)|A| − 2mcm(A).Proof. Without loss of generality we can assume that c2(A) = 2. Then, the firstpart is a direct consequence of Proposition 3.2. For the second part, take the m-decomposition of A, namely A =⋃i∈I Ai. Take an arbitrary element i ∈ I. Since Ais m-semi-full, Ai can be affinely transformed into an m-full subset. By the first partof this corollary, |2 ·Ai +m ·Ai| ≥ (m+ 2)|Ai| − 2m. By Lemma 3.1, 2 ·Ai +m ·Aiand 2 · Aj + m · Aj belong to different congruence classes modulo m for i 6= j. ByLemma 3.1, |2 · A + m · A| ≥∑i∈I |2 · Ai + m · Ai| ≥∑i∈I(m + 2)|Ai| − 2m =(m+ 2)|A| − 2mcm(A). 4. Marginal setsLet A be a finite set of integers and let C be a component of A. The C-marginalset is defined asMC = (2 · C + k ·A) \ (2 · C + k · C).We start by proving a bound for marginal sets sizes in the semi-full case.Let A be a finite set of integers and let C be the set of k-components of A. Weformulate the next lemma.A LOWER BOUND FOR THE SIZE OF A MINKOWSKI SUM OF DILATES 5Lemma 4.1.∑C∈C |MC | ≥ (|C| − 1)|C|.Proof. For any component C ∈ C, we shall write MC− = {x ∈ MC : x < min(2 ·C+k ·C)} and MC+ = {x ∈MC : x > max(2 ·C+k ·C)}. We shall denote by Γ thedirected graph of the order relation x < y defined on the set L = {min(C) : C ∈ C}.Recall that for a given element x ∈ L, Γ−(x) = {y ∈ L : y < x}. Clearly2 ·min(C) + k · Γ−(min(C)) ⊂M−C .In particular, |Γ−(min(C))| ≤ |M−C |. Therefore,∑C∈C|M−C | ≥∑C∈C|Γ−(min(C))| = (|C| − 1)|C|/2,since∑C∈C |Γ−(min(C))| is the total number of arcs in the order relation, which isobviously (|C| − 1)|C|/2. Similar arguments show that∑C∈C |M+C | ≥ (|C| − 1)|C|/2.The lemma follows then from the relation∑C∈C |MC | ≥∑C∈C |M−C |+∑C∈C |M+C |.We continue relating the size of MC and C. Sets with the property |MC | ≥ |C|are satisfactory for induction proofs, as we shall see in the next section. For a subsetX of an abelian group G, we write π(X) = {x ∈ G : x + X = X}. We recall that|π(X)| is a divisor of |X|.Lemma 4.2. Let k be an odd prime and let A be a finite set of integers with 0 ∈ Aand gcd(A) = 1. Let C be a non-k-semi-full component of A and let C ′ 6= C beanother k-component of A. Then |MC | ≥ |C ′|. Moreover, |MC | ≥ |C| if one of thefollowing conditions holds:• There is another component with size not less than |C|.• C is non-2-full.Proof. Let φ : Z→ Z/k2Z be the projection. Since |π(φ(C))| divides k2 and |φ(C)|and since |φ(C)| < k, we have necessarily π(φ(C)) = {0}. Since all elements of k ·Care equal modulo k2, if φ(2 ·C+k ·C) = φ(2 ·C+k ·C ′), we have φ(k ·C) = φ(k ·C ′),and hence C ≡ C ′ modulo k. It follows that C = C ′, a contradiction.Thus, MC contains 2 · C0 + k · C ′, where C0 is some k2-component of C, andhence |MC | ≥ |C ′|. Assume now that C is non-2-full. Pick either an odd v (usinggcd(A) = 1) if C contains only even numbers or v = 0 ∈ A if C contains only oddnumbers. Clearly, 2 · C + kv is a subset of MC . 5. Large sets of integersThe next result is our first step.6 Y. O. HAMIDOUNE AND J. RUÉTheorem 5.1. Let k be an odd prime. If A is a finite set of integers, then|2 ·A+ k ·A| ≥ (k + 2)|A| − 4kk−1.Proof. Without loss of generality, we may take 0 ∈ A and gcd(A) = 1. We shallprove by induction that for all 2 ≤ s ≤ k, we have|2 ·A+ k ·A| ≥ (s+ 2)|A| − 4ks−1.For s = 2, this bound is weaker than the one obtained by Corollary 3.3.Assume now that 2 < s ≤ k and that the result holds for s − 1. Take a k-decomposition of A, namely A =⋃i∈I Ai. We shall write Mi = MAi . Put F = {i ∈I : Ai is k − semi-full} and E = I \ F. Notice that |E|+ |F | = |I| ≤ k.Assume first that ∑i∈E|Mi| ≥∑i∈E|Ai|.By Lemma 3.1, 2 ·Ai+k ·A and 2 ·Aj+k ·A belong to different congruence classesmodulo k, for i 6= j. By Corollary 3.4, for every i ∈ F , |2·Ai+k·Ai| ≥ (k+2)|Ai|−2k.Using the last relations, the induction hypothesis applied for every i ∈ E, we have|2 ·A+ k ·A| ≥∑i∈E(|2 ·Ai + k ·Ai|+ |Mi|) +∑i∈F|2 ·Ai + k ·Ai|≥∑i∈E((s+ 1)|Ai| − 4ks−2) +∑i∈E|Ai|+∑i∈F((k + 2)|Ai| − 2k)≥∑i∈I(s+ 2)|Ai| − 4|I|ks−2 + 2k|F |(2ks−3 − 1)≥(s+ 2)|A| − 4ks−1,and the result holds.Assume now that ∑i∈E|Mi| <∑i∈E|Ai|.In particular, we have |E| ≥ 1. We must have |E| = 1, otherwise we take a derange-ment (permutation without fixed element) σ of E. By Lemma 4.2, |Mi| ≥ |Aσ(i)|,for every i. We get a contradiction by summing over all i ∈ E. Put E = {e} andtake f ∈ F . Observe that there must be at least two k-components, as 0 ∈ A andgcd(A) = 1. Applying and affine transformation to both Ae and Af we can applyProposition 3.2, getting |2 ·Af + k ·Ae| ≥ k|Af |+ 2|Ae| − 2k.The idea here is to estimate the component of 2·A+k ·A containing 2·Af +k ·Af ,using the sum 2·Af+k·Ae. By Lemma 4.2, Ae is 2-full, |Ae| > |Af | and |Me| ≥ |Af |.A LOWER BOUND FOR THE SIZE OF A MINKOWSKI SUM OF DILATES 7Using Lemma 3.1 and the previous considerations we have|2 ·A+ k ·A| ≥|2 ·Ae + k ·Ae|+ |Me|+ |2 ·Af + k ·Ae|+∑i∈F\f|2 ·Ai + k ·Ai|≥(s+ 1)|Ae| − 4ks−2 + |Af |+ k|Af |+ 2|Ae| − 2k +∑i∈F\f((k + 2)|Ai| − 2k)>(s+ 2)|Ae| − 4ks−2 +∑i∈F((s+ 2)|Ai| − 2k)≥(s+ 2)|A| − 4ks−1 + 2|F |k(2ks−3 − 1)≥(s+ 2)|A| − 4ks−1.Our main result is the following one:Theorem 5.2. Let k be an odd prime. Let A be a finite set of integers with 0 ∈ A,gcd(A) = 1 and |A| > 8kk. If A has a k-component involving at most k− 1 distinctk2-components, then|2 ·A+ k ·A| > (k + 2)|A|.Proof. Let C denote the set of k-components of A and let C be a non-k-semi-full component of A. Take a k-component N with a maximal cardinality. Clearly|N | > 8kk−1. Assume first that |MC | ≥ |N |. By Theorem 5.1 and Lemma 3.1,|2 ·A+ k ·A| ≥ |2 · C + k · C|+ |MC |+ |2 · (A \ C) + k · (A \ C)|>(k + 2)|C| − 4kk−1 + 8kk−1 + (k + 2)|A \ C| − 4kk−1= (k + 2)|A|,Assume now that |MC | < |N | and put c = ck(A). By Lemma 4.2, C = N and Cis 2-full. Hence C is the unique non-k-semi-full component of A. Take now a k-semi-full component of A, say T and put A′ = A \ (C ∪ T ). By Lemma 4.2 |MC | ≥ |T |.Applying a convenient affine transformation, by Proposition 3.2 |2 · T + k · C| ≥k|T |+ 2|C| − 2k.Thus using Corollary 3.4 and Lemma 3.1 we have|2 ·A+ k ·A| ≥ |2 · C + k · C|+ |T |+ |2 · T + k · C|+ (k + 2)|A′| − 2(c− 2)k≥(k + 2)|C| − 4kk−1 + |C|+ (k + 2)|T | − 2k + (k + 2)|A′| − 2(c− 2)k>(k + 2)|A|+ 4kk−1 − 2k(k − 1) > (k + 2)|A|,and the result holds. We can now prove the following lower bound on |2 ·A+ k ·A|:8 Y. O. HAMIDOUNE AND J. RUÉCorollary 5.3. Let k be an odd prime. If A is a finite set of integers with |A| > 8kk,then|2 ·A+ k ·A| ≥ (k + 2)|A| − k2 − k + 2.Proof. By Lemma 2.1, we may take 0 ∈ A and gcd(A) = 1. By Corollary 3.4, theresult holds if A is k-full. Assume that A is non-k-full and let C denote the setof k-components of A. By Theorem 5.2, we may assume that A is k-semi-full. ByCorollary 3.4, Lemma 3.1 and Lemma 4.1 we have|2 ·A+ k ·A| =∑C∈C|2 · C + k · C|+ |MC |≥∑C∈C((k + 2)|C| − 2k) + ck(A)(ck(A)− 1)=(k + 2)|A| − ck(A)(2k − ck(A) + 1)Therefore |2 ·A+ k ·A| ≥ (k + 2)|A| − k2 − k + 2, and the result holds. Acknowledgement: the authors thank the referee, whose suggestions and ad-vices helped to improve the presentation of the article.References[1] B. Bukh. Non-trivial solutions to a linear equation in integers. Acta Arithmetica, 131:41–55,2008.[2] B. Bukh. Sums of dilates. Combinatorics, Probability and Computing, 17:627–639, 2008.[3] J. Cilleruelo, Y. O. Hamidoune, and O. Serra. On sums of dilates. Combinatorics, Probabilityand Computing, 18:871–880, 2009.[4] J. Cilleruelo, M. Silva, and C. Vinuesa. A sumset problem. Journal of Combinatorics andNumber Theory, 2, 2010.[5] M. Z. Garaev. An explicit sum-product estimate in Fp. International Mathematics ResearchNotices. Vol. 2007 : article ID rnm035, 11 pages, 2007.[6] N. Hawk Katz and C.-Y. Shen. A slight improvement to Garaev’s sum product estimate.Proceedings of the American Mathematical Society, 136:2499–2504, 2008.[7] I.  Laba and S. Konyagin. Distance sets of well-distributed planar sets for polygonal norms.Israel Journal of Mathematics, 152:157–179, 2006.[8] M. B. Nathanson. Inverse problems for linear forms over finite sets of integers. Available on-lineat arXiv: 0708.2304v2.[9] M. B. Nathanson, K. O’Bryant, B. Orosz, I. Ruzsa, and M. Silva. Binary linear forms overfinite sets of integers. Acta Arithmetica, 129:341–361, 2007.UPMC, Université Paris 06, 4 Place Jussieu, 75005 Paris, France.E-mail address: hamidoune@math.jussieu.frLIX, École Polytechnique , 91128 Palaiseau-Cedex, France.E-mail address: rue1982@lix.polytechnique.fr

A lower bound for the size of a Minkowski sum of dilates

Let A be a finite non-empty set of integers. An asymptotic estimate of the size of the sum of several dilates was obtained by Bukh. The unique known exact bound concerns the sum |A + k·A|, where k is a prime and |A| is large. In its full generality, this bound is due to Cilleruelo, Serra and the first author.

Let k be an odd prime and assume that |A| > 8kk. A corollary to our main result states that |2·A + k·A|=(k+2)|A|-k2-k+2. Notice that |2·P+k·P|=(k+2)|P|-2k, if P is an arithmetic progression

UPCommons

UPCommons. Portal del coneixement obert de la UPC

A LOWER BOUND FOR THE SIZE OF A MINKOWSKI SUM OFDILATESY. O. HAMIDOUNE AND J. RUE´Abstract. Let A be a finite nonempty set of integers. An asymptotic estimateof several dilates sum size was obtained by Bukh. The unique known exactbound concerns the sum |A+ k ·A|, where k is a prime and |A| is large. In itsfull generality, this bound is due to Cilleruelo, Serra and the first author.Let k be an odd prime and assume that |A| > 8kk. A corollary to ourmain result states that |2 · A + k · A| ≥ (k + 2)|A| − k2 − k + 2. Notice that|2 · P + k · P | = (k + 2)|P | − 2k, if P is an arithmetic progression.1. IntroductionLet A, B be finite nonempty sets of real numbers. The Minkowski sum of A andB is defined asA+B = {a+ b : a ∈ A and b ∈ B}.The inequality |A+B| ≥ |A|+ |B|−1 is an easy exercise, that we shall use withoutany reference. For a real number r, the r-dilate of A is the set r ·A = {ra : a ∈ A}.Lower bounds for the size of dilates sums appeared in different contexts.  Laba andKonyagin [?] investigated the sum A+λ ·A (where λ is a transcendental number) inconnection with well-distributed planar sets distances. Dilates sums also appearedin the proofs of sum-product results in finite fields by Garaev [?], and by Katz andShen [?]. The sum of two dilates appeared in the work of Nathanson, O’Bryant,Orosz, Ruzsa and Silva on binary linear forms [?]. Also, they were used by Bukh[?] in connection with a problem of Ruzsa.From now on, we assume that A is a nonempty set of integers. Dilates sum of theform A+ 3 ·A were investigated independently by Bukh [?] and by Cilleruelo, Silvaand Vinuesa [?]. More recently, the authors of [?] proved that for an odd prime k,|A + k · A| ≥ (1 + k)|A| − (k + 1)2/4 for |A| sufficiently large. Let m1, · · · ,mj beintegers with gcd(m1, . . . ,mj) = 1, Bukh proved in [?] that|m1 ·A+ · · ·+mj ·A| ≥ (|m1|+ · · ·+ |mj |)|A| − o (|A|) .Bukh’s result suggests the following:Conjecture 1.1. For a set of integers Z with gcd(Z) = 1 and for every nonemptyset of integers A, there is an absolute constant c such that∣∣∣∣∣∑m∈Zm ·A∣∣∣∣∣ ≥(∑m∈Z|m|)|A| − c.12 Y. O. HAMIDOUNE AND J. RUE´In the present work, we prove the above conjecture for Z = {2, k}, where k is anodd prime. For simplicity, we will not consider negative dilates, but the reader willcertainly observe that our approach works in this case.In section 2, we present some easy and known lemmas that we need. In section3, we prove some intermediary results needed in our induction arguments. One ofthe results of this section states that |n ·A+m ·A| ≥ 4|A| − 4, where m and n arecoprime integers. This result is a counterpart of a lemma by Nathanson [?] statingthat |A+2 ·A| ≥ 3|A|−2. Let k be an odd prime. By a k-component of a set X ⊂ Z,we shall mean the nonempty intersection of X with some congruence class modulok. In section 4, we investigate the marginal set (2 ·C+k ·A)\ (2 ·C+k ·C), where Cis a k-component of A. In section 5, we prove that |2 ·A+k ·A| ≥ (k+2)|A|−4kk−1.Assuming that 0 ∈ A, gcd(A) = 1, |A| > 8kk and that A has a k-componentcontaining at most k − 1 nonempty k2-component, we show that|2 ·A+ k ·A| > (k + 2)|A|.Readers interested in the description of sets reaching equality could quite likely usethis result, since it shows that the objective function |2 · A + k · A| achieves itsminimum on structured sets, for |A| large. Let X be a finite set of integers with|X| > 8kk. We conclude Section 5 by an easy consequence of our main result, statingthat |2 ·X + k ·X| ≥ (k+ 2)|X| − k2− k+ 2. As an exercise, the reader could provethat |2 ·P + k ·P | = (k+ 2)|P |− 2k, if P is an arithmetic progression. Observe thatfor k = 3, our bound differs at most by 4 from the best possible one.2. Preliminaries and terminologyIn this paper, we consider sums of dilates of a finite set of integers. The nextknown lemma shows that the size of a dilates sum remains invariant if we replaceA by an affine transform of it.Lemma 2.1. [?] Let A be a finite set of integers and let r, s, u, v be non-zerointegers. Then• |r · (A+ v) + s · (A+ v)| = |r ·A+ s ·A|,• |r · (u ·A) + s · (u ·A)| = |r ·A+ s ·A|.Proof. We have clearly|r · (A+ v) + s · (A+ v)| = |r ·A+ s ·A+ (rv + sv)|= |r ·A+ s ·A|.A LOWER BOUND FOR THE SIZE OF A MINKOWSKI SUM OF DILATES 3as |A+ w| = |A|. We also have|r · (u ·A) + s · (u ·A)| = |(ru) ·A+ (su) ·A|= |u · (r ·A) + u · (s ·A)|= |r ·A+ s ·A|.Let A be a finite set of integers. The intersection of A with a congruence classmodulo n will be called a n-component. By a decomposition modulo n, we meana partition of A into its n-components. The number of n-components of A will bedenoted by cn(A). We shall say that A is n-full if cn(A) = n. The set A is n-semi-fullif every n-component C of A satisfies cn2(C) = n.Lemma 2.2. If A is n-full, then gcd(A) is coprime to n. Moreover, 1gcd(A) · A isn-full.Proof. There is an u ∈ A such that u ≡ 1 (mod n). As gcd(A) divides u, thengcd(gcd(A), n) divides both u and n, hence it divides 1. Since gcd(A) is invertiblemodulo n, cn(1gcd(A) ·A)= cn(A). 3. ToolsLemma 3.1. Let A and B be finite sets of integers and let m,n be coprime integers.Let C denote the set of m-components of A. Then n ·A+m ·B = ⋃C∈C n ·C+m ·Bis a decomposition modulo m.Proof. Clearly n ·C+m ·B ≡ n ·C (mod m), for any C ∈ C. The result follows nowsince n ·C and n ·C ′ are necessarily incongruent modulo m, for distinct componentsC,C ′ ∈ C. The next proposition is basic in our approach.Proposition 3.2. Let A and B be finite sets of integers and let m,n be coprimeintegers. Then |n ·A+m ·B| ≥ cn(B)|A|+ cm(A)|B| − cm(A)cn(B).Proof. Let A be the set m-components of A and let B be the set n-components ofB. We claim that if M1,M2 ∈ A and N1, N2 ∈ B such that (M1, N1) 6= (M2, N2),then (n ·M1 +m ·N1)∩(n ·M2 +m ·N2) = ∅. Suppose the contrary and take ai ∈Miand bi ∈ Ni, 1 ≤ i ≤ 2 with na1 +mb1 = na2 +mb2. Thus, n(a1−a2) = m(b1− b2).Since m is coprime to n, we have b1 − b2 ≡ 0 (mod n) and a1 − a2 ≡ 0 (mod m).In particular, M1 = M2 and N1 = N2, a contradiction.4 Y. O. HAMIDOUNE AND J. RUE´Therefore, using Lemma 3.1 we have|n ·A+m ·B| =∣∣∣∣∣∣⋃M∈A; N∈Bn ·M +m ·N∣∣∣∣∣∣ ≥∑M∈A; N∈B|M |+ |N | − 1= cn(B)|A|+ cm(A)|B| − cm(A)cn(B).Corollary 3.3. Let 2 ≤ n < m be coprime integers. Let A be a finite set of integers.Then |n ·A+m ·A| ≥ 4|A| − 4.Proof. The result holds clearly if |A| = 1. Assume that |A| ≥ 2. Without loss ofgenerality, 0 ∈ A. Put B = 1gcd(A) ·A. Put r = cm(B) and s = cn(B). Without lossof generality we may assume that r ≥ s. Observe that 2 ≤ r ≤ |A|. By Lemma 2.1and Proposition 3.2, we have|n ·A+m ·A| = |n ·B +m ·B|≥ 4|B| − 4 + (r + s− 4)|B| − rs+ 4≥ 4|B| − 4 + (r + s− 4)r − rs+ 4≥ 4|B| − 4 + (r − 2)2 ≥ 4|A| − 4.Corollary 3.4. Let m be an odd integer. If A is a m-full finite set of integers,then |2 · A + m · A| ≥ (m + 2)|A| − 2m. If A is m-semi-full set of integers, then|2 ·A+m ·A| ≥ (m+ 2)|A| − 2mcm(A).Proof. Without loss of generality we can assume that c2(A) = 2. Then, the firstpart is a direct consequence of Proposition 3.2. For the second part, take the m-decomposition of A, namely A =⋃i∈I Ai. Take an arbitrary element i ∈ I. Since Ais m-semi-full, Ai can be affinely transformed into an m-full subset. By the first partof this corollary, |2 ·Ai +m ·Ai| ≥ (m+ 2)|Ai| − 2m. By Lemma 3.1, 2 ·Ai +m ·Aiand 2 · Aj + m · Aj belong to different congruence classes modulo m for i 6= j. ByLemma 3.1, |2 · A + m · A| ≥ ∑i∈I |2 · Ai + m · Ai| ≥ ∑i∈I(m + 2)|Ai| − 2m =(m+ 2)|A| − 2mcm(A). 4. Marginal setsLet A be a finite set of integers and let C be a component of A. The C-marginalset is defined asMC = (2 · C + k ·A) \ (2 · C + k · C).We start by proving a bound for marginal sets sizes in the semi-full case.Let A be a finite set of integers and let C be the set of k-components of A. Weformulate the next lemma.A LOWER BOUND FOR THE SIZE OF A MINKOWSKI SUM OF DILATES 5Lemma 4.1.∑C∈C |MC | ≥ (|C| − 1)|C|.Proof. For any component C ∈ C, we shall write MC− = {x ∈ MC : x < min(2 ·C+k ·C)} and MC+ = {x ∈MC : x > max(2 ·C+k ·C)}. We shall denote by Γ thedirected graph of the order relation x < y defined on the set L = {min(C) : C ∈ C}.Recall that for a given element x ∈ L, Γ−(x) = {y ∈ L : y < x}. Clearly2 ·min(C) + k · Γ−(min(C)) ⊂M−C .In particular, |Γ−(min(C))| ≤ |M−C |. Therefore,∑C∈C|M−C | ≥∑C∈C|Γ−(min(C))| = (|C| − 1)|C|/2,since∑C∈C |Γ−(min(C))| is the total number of arcs in the order relation, which isobviously (|C| − 1)|C|/2. Similar arguments show that ∑C∈C |M+C | ≥ (|C| − 1)|C|/2.The lemma follows then from the relation∑C∈C |MC | ≥∑C∈C |M−C |+∑C∈C |M+C |.We continue relating the size of MC and C. Sets with the property |MC | ≥ |C|are satisfactory for induction proofs, as we shall see in the next section. For a subsetX of an abelian group G, we write pi(X) = {x ∈ G : x + X = X}. We recall that|pi(X)| is a divisor of |X|.Lemma 4.2. Let k be an odd prime and let A be a finite set of integers with 0 ∈ Aand gcd(A) = 1. Let C be a non-k-semi-full component of A and let C ′ 6= C beanother k-component of A. Then |MC | ≥ |C ′|. Moreover, |MC | ≥ |C| if one of thefollowing conditions holds:• There is another component with size not less than |C|.• C is non-2-full.Proof. Let φ : Z→ Z/k2Z be the projection. Since |pi(φ(C))| divides k2 and |φ(C)|and since |φ(C)| < k, we have necessarily pi(φ(C)) = {0}. Since all elements of k ·Care equal modulo k2, if φ(2 ·C+k ·C) = φ(2 ·C+k ·C ′), we have φ(k ·C) = φ(k ·C ′),and hence C ≡ C ′ modulo k. It follows that C = C ′, a contradiction.Thus, MC contains 2 · C0 + k · C ′, where C0 is some k2-component of C, andhence |MC | ≥ |C ′|. Assume now that C is non-2-full. Pick either an odd v (usinggcd(A) = 1) if C contains only even numbers or v = 0 ∈ A if C contains only oddnumbers. Clearly, 2 · C + kv is a subset of MC . 5. Large sets of integersThe next result is our first step.6 Y. O. HAMIDOUNE AND J. RUE´Theorem 5.1. Let k be an odd prime. If A is a finite set of integers, then|2 ·A+ k ·A| ≥ (k + 2)|A| − 4kk−1.Proof. Without loss of generality, we may take 0 ∈ A and gcd(A) = 1. We shallprove by induction that for all 2 ≤ s ≤ k, we have|2 ·A+ k ·A| ≥ (s+ 2)|A| − 4ks−1.For s = 2, this bound is weaker than the one obtained by Corollary 3.3.Assume now that 2 < s ≤ k and that the result holds for s − 1. Take a k-decomposition of A, namely A =⋃i∈I Ai. We shall write Mi = MAi . Put F = {i ∈I : Ai is k − semi-full} and E = I \ F. Notice that |E|+ |F | = |I| ≤ k.Assume first that ∑i∈E|Mi| ≥∑i∈E|Ai|.By Lemma 3.1, 2 ·Ai+k ·A and 2 ·Aj+k ·A belong to different congruence classesmodulo k, for i 6= j. By Corollary 3.4, for every i ∈ F , |2·Ai+k·Ai| ≥ (k+2)|Ai|−2k.Using the last relations, the induction hypothesis applied for every i ∈ E, we have|2 ·A+ k ·A| ≥∑i∈E(|2 ·Ai + k ·Ai|+ |Mi|) +∑i∈F|2 ·Ai + k ·Ai|≥∑i∈E((s+ 1)|Ai| − 4ks−2) +∑i∈E|Ai|+∑i∈F((k + 2)|Ai| − 2k)≥∑i∈I(s+ 2)|Ai| − 4|I|ks−2 + 2k|F |(2ks−3 − 1)≥(s+ 2)|A| − 4ks−1,and the result holds.Assume now that ∑i∈E|Mi| <∑i∈E|Ai|.In particular, we have |E| ≥ 1. We must have |E| = 1, otherwise we take a derange-ment (permutation without fixed element) σ of E. By Lemma 4.2, |Mi| ≥ |Aσ(i)|,for every i. We get a contradiction by summing over all i ∈ E. Put E = {e} andtake f ∈ F . Observe that there must be at least two k-components, as 0 ∈ A andgcd(A) = 1. Applying and affine transformation to both Ae and Af we can applyProposition 3.2, getting |2 ·Af + k ·Ae| ≥ k|Af |+ 2|Ae| − 2k.The idea here is to estimate the component of 2·A+k ·A containing 2·Af +k ·Af ,using the sum 2·Af+k·Ae. By Lemma 4.2, Ae is 2-full, |Ae| > |Af | and |Me| ≥ |Af |.A LOWER BOUND FOR THE SIZE OF A MINKOWSKI SUM OF DILATES 7Using Lemma 3.1 and the previous considerations we have|2 ·A+ k ·A| ≥|2 ·Ae + k ·Ae|+ |Me|+ |2 ·Af + k ·Ae|+∑i∈F\f|2 ·Ai + k ·Ai|≥(s+ 1)|Ae| − 4ks−2 + |Af |+ k|Af |+ 2|Ae| − 2k +∑i∈F\f((k + 2)|Ai| − 2k)>(s+ 2)|Ae| − 4ks−2 +∑i∈F((s+ 2)|Ai| − 2k)≥(s+ 2)|A| − 4ks−1 + 2|F |k(2ks−3 − 1)≥(s+ 2)|A| − 4ks−1.Our main result is the following one:Theorem 5.2. Let k be an odd prime. Let A be a finite set of integers with 0 ∈ A,gcd(A) = 1 and |A| > 8kk. If A has a k-component involving at most k− 1 distinctk2-components, then|2 ·A+ k ·A| > (k + 2)|A|.Proof. Let C denote the set of k-components of A and let C be a non-k-semi-full component of A. Take a k-component N with a maximal cardinality. Clearly|N | > 8kk−1. Assume first that |MC | ≥ |N |. By Theorem 5.1 and Lemma 3.1,|2 ·A+ k ·A| ≥ |2 · C + k · C|+ |MC |+ |2 · (A \ C) + k · (A \ C)|>(k + 2)|C| − 4kk−1 + 8kk−1 + (k + 2)|A \ C| − 4kk−1= (k + 2)|A|,Assume now that |MC | < |N | and put c = ck(A). By Lemma 4.2, C = N and Cis 2-full. Hence C is the unique non-k-semi-full component of A. Take now a k-semi-full component of A, say T and put A′ = A \ (C ∪ T ). By Lemma 4.2 |MC | ≥ |T |.Applying a convenient affine transformation, by Proposition 3.2 |2 · T + k · C| ≥k|T |+ 2|C| − 2k.Thus using Corollary 3.4 and Lemma 3.1 we have|2 ·A+ k ·A| ≥ |2 · C + k · C|+ |T |+ |2 · T + k · C|+ (k + 2)|A′| − 2(c− 2)k≥(k + 2)|C| − 4kk−1 + |C|+ (k + 2)|T | − 2k + (k + 2)|A′| − 2(c− 2)k>(k + 2)|A|+ 4kk−1 − 2k(k − 1) > (k + 2)|A|,and the result holds. We can now prove the following lower bound on |2 ·A+ k ·A|:8 Y. O. HAMIDOUNE AND J. RUE´Corollary 5.3. Let k be an odd prime. If A is a finite set of integers with |A| > 8kk,then|2 ·A+ k ·A| ≥ (k + 2)|A| − k2 − k + 2.Proof. By Lemma 2.1, we may take 0 ∈ A and gcd(A) = 1. By Corollary 3.4, theresult holds if A is k-full. Assume that A is non-k-full and let C denote the setof k-components of A. By Theorem 5.2, we may assume that A is k-semi-full. ByCorollary 3.4, Lemma 3.1 and Lemma 4.1 we have|2 ·A+ k ·A| =∑C∈C|2 · C + k · C|+ |MC |≥∑C∈C((k + 2)|C| − 2k) + ck(A)(ck(A)− 1)=(k + 2)|A| − ck(A)(2k − ck(A) + 1)Therefore |2 ·A+ k ·A| ≥ (k + 2)|A| − k2 − k + 2, and the result holds. Acknowledgement: the authors thank the referee, whose suggestions and ad-vices helped to improve the presentation of the article.References[1] B. Bukh. Non-trivial solutions to a linear equation in integers. Acta Arithmetica, 131:41–55,2008.[2] B. Bukh. Sums of dilates. Combinatorics, Probability and Computing, 17:627–639, 2008.[3] J. Cilleruelo, Y. O. Hamidoune, and O. Serra. On sums of dilates. Combinatorics, Probabilityand Computing, 18:871–880, 2009.[4] J. Cilleruelo, M. Silva, and C. Vinuesa. A sumset problem. Journal of Combinatorics andNumber Theory, 2, 2010.[5] M. Z. Garaev. An explicit sum-product estimate in Fp. International Mathematics ResearchNotices. Vol. 2007 : article ID rnm035, 11 pages, 2007.[6] N. Hawk Katz and C.-Y. Shen. A slight improvement to Garaev’s sum product estimate.Proceedings of the American Mathematical Society, 136:2499–2504, 2008.[7] I.  Laba and S. Konyagin. Distance sets of well-distributed planar sets for polygonal norms.Israel Journal of Mathematics, 152:157–179, 2006.[8] M. B. Nathanson. Inverse problems for linear forms over finite sets of integers. Available on-lineat arXiv: 0708.2304v2.[9] M. B. Nathanson, K. O’Bryant, B. Orosz, I. Ruzsa, and M. Silva. Binary linear forms overfinite sets of integers. Acta Arithmetica, 129:341–361, 2007.UPMC, Universite´ Paris 06, 4 Place Jussieu, 75005 Paris, France.E-mail address: hamidoune@math.jussieu.frLIX, E´cole Polytechnique , 91128 Palaiseau-Cedex, France.E-mail address: rue1982@lix.polytechnique.fr

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A lower bound for the size of a Minkowski sum of dilates

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