All sets are assumed to be finite subsets of the integers.


The additive energy of two sets $E(A,B)$ is defined as the number of solutions to $a+b=a'+b'$ with $a,a'\in A$ and $b,b'\in B$. The well-known Balog-Szemeredi-Gowers theorem states that there is an absolute constant $C>0$ such that (assuming $\lvert A\rvert=\lvert B\rvert$ for convenience) if $E(A,B)=K^{-1}\lvert A\rvert^3$ then there exist $A'\subset A$ and $B'\subset B$ such that 


$$ \lvert A'\rvert\gg K^{-C}\lvert A\rvert\textrm{ and }\lvert B'\rvert\gg K^{-C}\lvert B\rvert $$
and
$$ \lvert A'+B'\rvert\ll K^C\lvert A\rvert.$$


When $A=B$ we can qualitatively strengthen this conclusion to allow $A'=B'$ (see the recent paper by Schoen 'New bounds in Balog-Szemeredi-Gowers theorem' at <http://www.staff.amu.edu.pl/~schoen/remark-B-S-G.pdf>)


My question is whether we can obtain a similar strengthening when $B=d\cdot A$ for some $d\neq 0$ (here $d\cdot A$ denotes the dilate of $A$, the set $\{ da : a\in A\}$). That is, does there exist an absolute constant $C>0$ such that if $E(A,d\cdot A)=K^{-1}\lvert A\rvert^3$ then there exists $A'\subset A$ such that 


$$ \lvert A'\rvert\gg K^{-C}\lvert A\rvert$$
and
$$ \lvert A'+d\cdot A'\rvert\ll K^C\lvert A\rvert.$$


Note it is important that $C$ and the implied constants must be independent of $d$, else this would follow from the above and the usual sumset inequalities.


Adapting Schoen's argument, the best I have been able to do is to prove that there exists $s$ and $A'\subset A\cap(s-d\cdot A)$ such that $\lvert A'\rvert\gg K^{-C}\lvert A\rvert$ and 
$$ \lvert A'-A'\rvert \ll K^c\lvert A\rvert.$$


This is qualitatively stronger than a direct application of the original BSG theorem, but still falls short.


Also, if one actually has $E(A,d\cdot A)\geq K^{-1}\lvert A\rvert^3$ for all $d\in D$ then using arguments of Bourgain I can show that there exists $D'\subset D$ with $\lvert D'\rvert\gg K^{-C}\lvert D\rvert$ and $A'\subset A$ with $\lvert A'\rvert\gg K^{-C}\lvert A\rvert$ such that for all $d\in D'$ we have
$$ \lvert A'+d\cdot A'\rvert \ll K^C\lvert A\rvert.$$


In my case, with $\lvert D\rvert=2$, however, this again falls short.


There are by now quite a few arguments for proving BSG-type statements, however, so hopefully someone knows of one which can be adapted for my question!



Balog-Szemeredi-Gowers with dilates of sets
Tomasz Schoen has observed in an email that the answer to my question is no, since there is a function $f(d)\to\infty$ as $d\to\infty$ such that $\lvert A+d\cdot A\rvert\geq f(d)\lvert A\rvert$ for all $A\subset \mathbb{Z}$. In particular, Boris Bukh in 'Sums of dilates' showed that roughly $f(d)=d+1$ is permissible.


To see that this leads to a counter example, consider the set $A=[1,N]\cup d\cdot [1,N]$ where $N$ is very large and d is even larger, depending on $N$. Then


$$E(A,d\cdot A)\geq E(d\cdot [1,N],d\cdot [1,N])=E([1,N],[1,N])\gg N^3\gg \lvert A\rvert^3,$$


so if my conjecture were true then there'd be some $A'\subset A$ such that $\lvert A'\rvert\gg N$ and $\lvert A'+d\cdot A'\rvert\ll N$, which contradicts the mentioned result of Bukh for $d$ sufficiently large.


