Regarding scaling in sumsets

Question

Let $A$ be a finite subset of $\mathbb{N}$. We denote the set $\{a_1 +a_2: a_1, a_2\in A\}$ as $2A$. We call the quantity $\sigma[A]:= |2A|/|A|$ as the doubling constant of $A$, and this constant can tell us some information about structure of the set $A$ (e.g., if $\sigma[A]= 2-1/|A|$, then $A$ is an arithmetic progression).

Given a number $k$, and a set $A$ with $\sigma[A]= k$, can we say something about the value of the constant $|3A|/|A|$? Intuitively, it seems that if $k$ is "large" then so should be $|3A|/|A|$.

There already exist some bounds for the general quantity $|nA|/|A|$, but what I want to know is the dependence of $|3A|$ on a given doubling constant, and I can't seem to find anything on this in the literature.

Answer 1

The following lemma shows that if the doubling constant $\sigma[A]=K$, then $|nA|/|A|$ cannot be too large.

Lemma. Let $A$ be an additive set with doubling constant $\sigma[A]=K$ for some $K\geq 1$. Then $$|nA|\leq \min(K^{Cn},n^{K^C-1})|A|$$ for all $n\geq 1$ and some absolute constant $C>0$.

Proof. Define a set $H:=A-A$ which is symmetric and contains origin, i.e. $H=-H$ and $0\in H$. Applying Ruzsa's covering lemma to the pair of additive sets $A$ and $H+H$ we obtain $$H+H\subseteq \underbrace{A-A}_{=H}+X \ \text{for some additive set $X$ such that} \ |X|\leq \frac{|H+H+A|}{|A|}.$$ Let's estimate the cardinality of $X$ $$|X|\leq \dfrac{|H+H+A|}{|A|}=\dfrac{|3A-2A|}{|A|}\leq \dfrac{(K^2)^{3+2}|A|}{|A|}= K^{10}.$$

It means that $H$ is a $K_1:=K^{10}$ approximate group.

We notice that $A\subseteq x+H$ for any $x\in A$.

Therefore $A\subseteq a_0+H$ for some $a_0\in A$. By induction we obtain that $nA\subseteq na_0+nH$ for any $n\geq 1$. Hence $$|nA|\leq |na_0+nH|=|nH|.$$ We have shown that $H$ is a $K_1$-approximate group and hence we have: $$|nH|\leq \min (K_1^n, n^{K_1-1})|H|.$$

But $|H|=|A-A|\leq K^2|A|$ and the above inequality implies that $$|nA|\leq K^2\cdot \min(K^{10n}, n^{K^{10}-1})|A|,$$ which can be written: $|nA|\leq \min(K^{Cn},n^{K^C-1})|A|$.

Answer 2

Here's a more explicit answer for the case of $\lvert 3A\rvert$:

If $K=\lvert 2A\rvert/\lvert A\rvert$ then $\lvert 3A\rvert \leq K^3\lvert A\rvert$.

This is a special case of Plunnecke's inequality, for which Petridis gave a simple and elegant proof (see https://arxiv.org/abs/1101.3507)

This is actually sharp (up to constants), as the following example of Ruzsa shows: choose some parameters $M\ll N\ll M^2$ and consider in $\mathbb{Z}^3$ the set $A$ formed by three lines of length $N$ (take $N$ integer points along the three coordinate axes) and a box of sidelength $M$ (take all integer points $\{0,\ldots,M-1\}^3$).

One can check that $\lvert A\rvert \asymp M^3$ (most points coming from the box) and $\lvert 2A\rvert\asymp NM^2$ (most points coming from the three $N\times M\times M$ rectangles along the axes) and $\lvert 3A\rvert\asymp N^3$ (since $3A$ is basically the full integer box of sidelength $N$).

This means $K=\lvert 2A\rvert/\lvert A\rvert\asymp N/M$ and $\lvert 3A\rvert/\lvert A\rvert\asymp (N/M)^3\asymp K^3$.

(Choosing $N,M$ appropriately this can be achieved up to $K\approx \lvert A\rvert^{1/3}$.)