Let $G$ ($G=\mathbb{Z}^n_2$ for my case) be a additive group and $A$ be a subset of $G$. For any set $S\subseteq G$ define its doubling as $$\sigma (S)=\dfrac{|S+S|}{|S|}$$ Suppose $A$ has small doubling. i.e. $\sigma (A)\leq c$ for some real number $c\geq 1$. What can we say about $\sigma (A+A)$? I know in general it is not true that $\sigma (A+A)\leq\sigma (A)$. But does there exist sets $A$ satisfying $$\sigma (2^k A)\leq\sigma(2^{k-1}A)\ \ \ \ \forall k\geq 0$$ and can we characterize them?
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$\begingroup$ Is there a typo? $s=2$? $\endgroup$– Brendan McKayCommented Nov 18, 2013 at 5:58
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1$\begingroup$ If $A$ is a subgroup, then $\sigma(mA)=1$ for every $m\ge 1$. Similarly, taking $A$ to be an arithmetic progression yields $\sigma(mA)=2+o(1)$. Does this answer your question? $\endgroup$– SevaCommented Nov 18, 2013 at 9:32
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$\begingroup$ I noticed this special case as well. Thanks for reminding. I should have mentioned it. But I'd love to see whether there is a non-group $A$ satisfying the requirement. $\endgroup$– yueCommented Nov 18, 2013 at 9:50
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$\begingroup$ and non-coset $A$ $\endgroup$– yueCommented Nov 18, 2013 at 9:54
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1$\begingroup$ I would expect that no characterisation can be given and, in fact, you have $\sigma(2^kA)\le\sigma(2^{k-1}A)$ for all "non-pathological" cases. As just one example, consider the situation where $A$ is a sufficiently dense subset of a subgroup. $\endgroup$– SevaCommented Nov 18, 2013 at 12:26
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