Sum of a set and its dilate

Additive Combinatorics, ,

Let {A, B \subset \mathbb{Z}} be finite and nonempty. One of the first things we learn in additive combinatorics is

\displaystyle |A+B| \geq |A| +|B| - 1. \ \ \ \ \ (1)

There are a host of results improving on (1) when {B} has certain structure. For instance, all of the following have an short and elementary proof. Here {q \cdot A := \{qa : a \in A\}}.

  • (1) {|A+2\cdot A| \geq 3 |A| - 2}
  • (2) {|A+A -A| \gg |A|^2}, for {A} convex,
  • (3) {|A+AA| \geq |A|^2 + |A| - 1},
  • (4) {|A + A^{-1}| \geq |A|^2}.

Let’s see one way to prove (1). After translation and dilation, we may suppose that {A} contains both even and odd numbers. Let {A_0} be the even numbers in {A} and {A_1} be the odd numbers in {A}. By (1),

\displaystyle |A+2 \cdot A| = |A_0 + 2 \cdot A| + |A_1 + 2 \cdot A| \geq |A_0| + |A| - 1 + |A_1| + |A| - 1 = 3 |A| - 2.

We remark that there is another proof of this in which one picks {2 |A| - 3} elements of {A+2 \cdot A} in increasing order.

Antal Balog and I considered the question of improving (1) when {B = q \cdot A }. We proved the following.

Theorem 1: Let {q\geq 1} be an integer and {A \subset \mathbb{Z}} be finite. Then

\displaystyle |A+q \cdot A| \geq (q + 1) |A| - O_q(1). \ \ \ \spadesuit

This is best possible up to the constant as can be seen by taking {A} to be an arithmetic progression. Below in the fold, we will a proof in the case where {q} is a prime. This case is technically easier, and was first handled by Cilleruelo, Hamidoune, and Serra. Already the case {q=3} is non–trivial, though a proof of {|A+3 \cdot A| \geq 4 |A| - 4} had appeared several times in the literature. We adopt the proof from our paper, which can be modified to handle the general case.

We partition {A} into residue classes modulo {q} as follows:

\displaystyle A = \displaystyle \bigcup_{j = 1}^s A_j , \ \ A_j = a_j + q \cdot A_j^{\prime}, \ \ A_j \neq \emptyset ,\ \ 0 \leq a_j < q.\ \ \ \ \ (2)

We say {A} is fully–distributed mod {q} (FD) if {s=q} or equivalently {A} intersects every residue class modulo {q}. The key fact is that not only the sets {A_i} are disjoint, but so are the sets {A_i + q \cdot A} and we have

\displaystyle A + q \cdot A= \coprod_{i =1}^s A_i + q \cdot A.\ \ \ \ \ (3)

As long as {|A| \geq 2}, we may assume that {s \geq 2} since we may replace {A} with {\frac{1}{q} \cdot (A - x)}. This is actually a key point: Theorem 1 is translation and dilation invariant. We first show that Theorem 1 is true for random sets.

Lemma (FD): Suppose {A} is FD. Then

\displaystyle |A+q \cdot A| \geq (q+1)|A| - q. \ \ \ \spadesuit

Proof: By (1) and (3), we have

\displaystyle |A+q \cdot A| = \sum_{j=1}^q |A_j + q \cdot A| \geq \sum_{j=1}^q \left(|A_j |+ |A| - 1 \right) = (q + 1)|A| - q. \ \ \ \spadesuit

Now we are in a position to prove Theorem 1.

Proof of Theorem 1: We prove

\displaystyle |A+q \cdot A| \geq \frac{m}{q} |A| - C_m,\ \ \ \ \ (4)

by induction on {m}. For {m = 2q}, this follows from (1) with {C_m = 1}. We now assume (4) holds for some {m < q(q+1)} and show it for {m+1}.

Suppose that there is some {1 \leq i \leq s} such that {|A_i| \leq q^{-1} |A|}. Then by (1), (4) and {m < q(q+1)}, we have

\displaystyle |A+q \cdot A| \geq |A_i + q \cdot A| + |(A \setminus A_i) + q \cdot (A \setminus A_i)|

\displaystyle \geq |A_i| + |A| - 1 + \frac{m}{q} (|A| - |A_i|) - C_m \geq \frac{m+1}{q} |A| - C_m - 1.

Suppose now that {A_i'}, as defined in (2) is FD for all {1 \leq i \leq s}. By (3) and translation and dilation invariance of {|X+ q \cdot X|}, we have

\displaystyle |A+q \cdot A| \geq \sum_{j=1}^s |A_j + q \cdot A_j| = \sum_{j=1}^s |A'_j + q \cdot A'_j| \geq (q+1) |A| - q^2.

Thus we may suppose that

\displaystyle |A_i| \geq \frac{|A|}{q} , \ \ 1 \leq i \leq s,\ \ \ \ \ (5)

and there is some {k} such that {A'_k} is not FD. We then apply the following lemma.

Lemma (not FD): Suppose {A_k'} is not FD. Then

\displaystyle |A_j+ q \cdot A| \geq |A_j + q \cdot A_j| + \min_{1 \leq m \leq s} |A_m|. \ \ \ \spadesuit

Let’s use Lemma (not FD) to finish the proof. By Lemma (not FD), (4) twice and (5) we have

\displaystyle |A+q \cdot A| \geq |A_k + q \cdot A| + |(A \setminus A_k) + q \cdot (A \setminus A_k)|

\displaystyle \geq |A_k + q \cdot A_k| + \min_{1 \leq m \leq s} |A_m| + \frac{m}{q} (|A| - |A_k|) - C_m

\displaystyle \geq \frac{m}{q} |A_k| - C_m + \frac{|A|}{q} + \frac{m}{q} (|A| - |A_k|) - C_m

\displaystyle = \frac{m+1}{q} |A| - 2 C_m.

This completes the proof modulo Lemma (not FD).

Proof of Lemma (not FD): Suppose

\displaystyle |A_k + q \cdot A| < |A_k + q \cdot A_k| + \displaystyle \min_{1 \leq m \leq s} |Q_m|.

Then for any {1\leq m \leq s}, we have

\displaystyle |A_m^{\prime}| = |A_m| > |(A_k+ q \cdot A_m )\setminus (A_k+ q \cdot A_k) | = |(a_m - a_k + A_k^{\prime} + q \cdot A_m^{\prime}) \setminus (A_k^{\prime} + q \cdot A_k^{\prime})|.

It follows that for every {x \in A_k^{\prime}} there is a {y \in A_m^{\prime}} such that {a_m-a_k+x+qy\in A'_k + q\cdot A'_k}, and so there is an {x'\in A'_k} such that

\displaystyle a_m - a_k + x \equiv x' \pmod q.

We may repeat this argument with {x'} in place of {x} and so on to get that for every {x \in A'_k} and {w \in \mathbb{Z}}, there is {z \in A'_k} such that

\displaystyle w(a_m - a_k) + x = z \pmod q.

Since {q} is prime, we get that {A'_k} is FD, as desired. {\spadesuit}

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. Further, there is a paper of Breuillard and Green on contraction maps, for which sum of dilates is a special case. 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

\displaystyle |A+q \cdot A| \geq (|q| + 2d -1)|A| - O_{d , q}(1),

where {A \subset \mathbb{Z}^d} and not contained in any {d-1} dimensional subspace. The cases {d=2} and {d=3} are known to hold.

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