There are a host of results improving on (1) when has certain structure. For instance, all of the following have an short and elementary proof. Here .
- (2) , for convex,
- (3) ,
- (4) .
Let’s see one way to prove (1). After translation and dilation, we may suppose that contains both even and odd numbers. Let be the even numbers in and be the odd numbers in . By (1),
We remark that there is another proof of this in which one picks elements of in increasing order.
Theorem 1: Let be an integer and be finite. Then
This is best possible up to the constant as can be seen by taking to be an arithmetic progression. Below in the fold, we will a proof in the case where is a prime. This case is technically easier, and was first handled by Cilleruelo, Hamidoune, and Serra. Already the case is non–trivial, though a proof of had appeared several times in the literature. We adopt the proof from our paper, which can be modified to handle the general case.
Lemma (FD): Suppose is FD. Then
Now we are in a position to prove Theorem 1.
Lemma (not FD): Suppose is not FD. Then
This completes the proof modulo Lemma (not FD).
Proof of Lemma (not FD): Suppose
Then for any , we have
It follows that for every there is a such that , and so there is an such that
We may repeat this argument with in place of and so on to get that for every and , there is such that
Since is prime, we get that is FD, as desired.
Since this work, I worked out the general case of sum of many dilates. After that, Antal Balog and I worked on the analogous problem in vector spaces. Both papers contain several open problems and one of them appears on my problems page with a cash reward. Also, feel free to see my short video I made on the topic, which highlights some of the problems.
One natural generalization that remains open is
where and not contained in any dimensional subspace. The cases and are known to hold.