On the sum of dilations of a set

Abstract

We show that for any relatively prime integers 1 ≤ p < q and for any finite A ⊂ Z one has |p · A + q · A| ≥ (p + q)|A| − (pq) (p+q−3)(p+q)+1 .

Sums of Dilates in Ordered Groups

“…, n − 1. If y v 1 1 · · · y vm m is the reduced form of x t , then y 1 = x 1 and y m = x n . Indeed, the positive power of a reduced word, say w, in a free group is a word which, when reduced to its minimal form, starts and ends with the same letters as w. It follows from (11) that x 1 = a and x n = b, with the result that m = nt.…”

“…Finally, even more recently, the case of (Z, +) was almost completely settled by Balog and Shakan in [1], where it is proved that…”

Sum of Many Dilates

“…The works of [1,3,4,6,8] made improvements from o(|A|) to a constant in certain cases, all when k = 2. Indeed, in [1], the problem was completely resolved for k = 2.…”

“…The works of [1,3,4,6,8] made improvements from o(|A|) to a constant in certain cases, all when k = 2. Indeed, in [1], the problem was completely resolved for k = 2. For a more complete introduction of the problem, the reader is invited to see the introductions of [1] and [2].…”

“…Indeed, in [1], the problem was completely resolved for k = 2. For a more complete introduction of the problem, the reader is invited to see the introductions of [1] and [2]. Note that while Theorem 1.1 is only claimed and proved for A ⊂ Z, extending to A ⊂ Q is an obvious task by clearing denominators of A, and moreover, extending to A ⊂ R is handled by Lemma 5.25 in the book by Tao and Vu [11].…”