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I was reading Ben Green's notes on additive combinatorics and there he writes the following proposition :

Suppose $U,V$ be subsets in some ambient abelian group $G$. Suppose that $U \sim V$. Then $U \sim -V$, $|U|\approx |V|$ and $\sigma[U]\approx 1$. We are dealing only with finite sets here. I have proved that $U \sim -V$ and $|U|\approx |V|$ but I am unable to prove that $\sigma[U]\approx 1$.


Definitions for reference: Let $x$ and $y$ be reals. Let $K\ge 2$ be some ambient parameter. We write $\displaystyle x <_{\approx} y$ if $x \leq K^cy$ for some $c$. We say $x\approx y$ iff $x <_{\approx} y$ and $y <_{\approx} x$.

Also, let $A$ and $B$ be two sets in some ambient abelian group and $K$ be some parameter. Then we write $A\sim B$ iff $\displaystyle \frac {|A-B|}{|A|^\frac{1}{2}|B|^\frac{1}{2}}\sim 1 $.

If U is a finite set, $\sigma[U]=|U+U|/|U|$.


Question: How can we prove that $\sigma[U]\approx 1$?

I don't really have any idea how to prove this. I just managed to note that $\sigma[U]=|U+U|/|U|\leq |U|^2/|U|=|U|$ and $\sigma[U]\ge 1$ but I think I need the bounds in powers of $K$. Also, I have not used the fact that $U\sim V$.

As Siming Tu has pointed out, the result does not seem to involve the original condition that $U\sim V$ and does not seem to be correct. Could anyone point to the right result?

  • I do not think you can prove that $\sigma[U]\approx 1$ with the conditions given. Maybe there are some additional conditions? Because there is no restriction on $U$. Note that $U\sim V$ tells us nothing as maybe $V$ is the same as $U$. – Siming Tu Jul 02 '15 at 17:44
  • @SimingTu, I am also surprised by this "result". –  Jul 05 '15 at 04:16
  • Would you like to tell me where are the lecture notes, maybe I can take a look at. – Siming Tu Jul 07 '15 at 18:38
  • I am sorry but I don't think those notes are available online. Besides, I don't think I can attach a file here. –  Jul 08 '15 at 17:15
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    OK. So I am not sure that how can I fix it. But I can show that in general $\sigma[U]\approx 1$ is not always satisfied. For example you can choose $U={2^n:n=0,1,\ldots,N-1}$, it is not hard to see that $\sigma[U]=\frac{N-1}{2}$ which goes to infinity as $N$ goes to infinity. – Siming Tu Jul 10 '15 at 12:32

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