8
$\begingroup$

Let $q$ be a positive integer. Is it true there exists a constant $C_q$ such that the following inequality holds for any finite set $A$ of reals: $$\displaystyle |A+qA|\ge (q+1)|A|-C_q\qquad (1)$$ I got this idea from the well-known inequality $|A+A|\ge 2|A|-1$, so I was thinking about the general case, but no idea about it. I experimented a bit with small values of $|A|$ and it seems to be true. This paves the road for a far more general inequality $$ |kA+\ell A|\ge (k+\ell)|A|-C_{k,\ell}\qquad (2)$$ with $k,\ell$ integers and $\text{gcd}(k,\ell)=1$. So, an answer would be welcome. A solution of $(1)$ is the thing I ask for, $(2)$ is just a bonus. And I haven't checked $(2)$ for small values yet. Anyways proving or disproving them both will be helpful. Thanks a lot.

$\endgroup$
2
  • 1
    $\begingroup$ There is no way (2) could hold without at least requiring $k$ and $l$ to be coprime: if $m>1$ and $A=\{1,\dots,n\}$, then $|kmA+lmA|=|kA+lA|\le(k+l)n<(km+lm)n-C_{km,lm}$ for $n$ large enough. $\endgroup$ Jun 15, 2014 at 15:43
  • $\begingroup$ Thank you so I am adding the condition in the question. I knew the second one was faulty somehow. $\endgroup$
    – shadow10
    Jun 15, 2014 at 15:46

1 Answer 1

12
$\begingroup$

Inequality (2) is true provided $(k,\ell)=1$ - this is a recent result of Balog and Shakan in 'On the sum of dilates of a set', http://arxiv.org/pdf/1311.0422.pdf.

They show that for any finite $A\subset\mathbb{Z}$ and integers $k,\ell$ such that $(k,\ell)=1$.

$$ \lvert kA+\ell A\rvert \geq (k+\ell)\lvert A\rvert - (k\ell)^{(k+\ell-3)(k+\ell)+1}. $$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.