Abstract. In this paper we prove that every set A ⊂ Z satisfying the inequality x min(1 A * 1 A (x), t) (2 + δ)t|A| for t and δ in suitable ranges, then A must be very close to an arithmetic progression. We use this result to improve the estimates of Green and Morris for the probability that a random subset A ⊂ N satisfies |N \ (A + A)| k; specifically we show that P(|N \ (A + A)| k) = Θ(2 −k/2 ).
IntroductionLet us start with recalling Freiman (3k − 3) Theorem. It states that every finite subset A ⊂ Z satisfying |A + A| < 3|A| − 3 is contained in an arithmetic progression of length |A + A| − |A| + 1. Comparing this with a lower bound |A + A| 2|A| − 1 valid for all nonempty finite subsets of Z, we can see that this result describes sets for quite large range of values of |A + A|. Our goal is to give a similar result for a set with a few popular sums. Note that it cannot be done dirctly; the reason is that the set S k (A) = {x ∈ Z : |A ∩ (x − A)| k} of k-popular sums is empty if k 3 and A is a highly independent set. Instead, we need to consider a different quantity, namely the average size of S k for 1 k t, which also appeared quite natural to Pollard in his work [Pol74] back in 1974.At this point it is convenient to use the notation of convolution. From now on, we will consider any abelian group G to be equipped with the couning measure, which leads to the definitionfor any functions f, g : G → C for which the above expression makes sense (i.e. is absolutely convergent; we will use it mostly for f, g being indicator functions of finite sets). Having this notation, we can restate Pollard's theorem asfor any prime p and sets A, B ⊂ Z/pZ. It is not hard to prove the corresponding statement for subsets of the integers (and even easier to deduce it from Pollard's theorem); in particular, for a single set A ⊂ Z and an integer 2010 Mathematics Subject Classification. 11P70.