# Ruzsa-type inequalities for additive energy

As the subject says, I have some questions about some Ruzsa-type inequalities for additive-energy. It would be wonderful if anyone could shed some light.

First, the definition: let $E(A,B)$ be the number of all quadruples $(a,a',b,b') \in A \times A \times B \times B$ such that $a+b = a' + b'$ denote the additive energy of the sets $A$, $B$, and let $F(A,B)= \frac{|A|^{2}|B|^{2}}{E(A,B)}$.

Now let $A, B, C \subset G$ be subsets of an additive group $G$ so that $|A|=|B|=|C|=N$. Do the following inequalities hold?

1) If $F(A,B), F(A,C) \leq K\cdot N$, then $\min{(F(A, B+C), F(B, C+A), F(C, A+B))} \leq K^{c} N$, for some constant $c$.

2) If $F(A,A) \leq K\cdot N$, then for all non-negative integers $k,l$ there's a constant $c$ depending on them such that $F(k \cdot A, l \cdot A) \leq K^{c} N$.

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Counting the trivial solutions only, you get $E(A,B)\ge|A||B|$ for any $A$ and $B$. Hence, for $|A|=|B|=N$, you cannot have $E(A,B)\le KN$, unless $K\ge N$. This implies that the answers to your questions are positive, for a trivial reason. – Seva Apr 9 '12 at 6:53
I meant something else, apologies. I edited now and it should make sense. – Cosmin Pohoata Apr 9 '12 at 7:13

1) No. Let $A=B=C$ be a set that is a union of interval $\{1,\dots,N\}$ and $N$ random elements from $\{1,\dotsc,N^2\}$. Then $F(A,A)=KN$ for some constant $K$. On the other hand, $A+A$ is basically the interval $\{1,\dotsc,N^2\}$. So, $F(A,A+A)\approx N^2$.

2) Example above show that it fails for $k=1$ and $l=2$ (from the context I assume that $k\cdot A$ means $k$-fold sumset $A+\dotsb+A$, not the $k$-dilate $\{ka:a\in A\}$. If dilates were meant, then the assertion is true, and one can deduce this from the Balog–Szemerédi–Gowers theorem).

I want to add that there do exist Ruzsa-type inequalities for additive energy. Instead of sumsets, they involve a suitable extension of additive energy to more than two sets. See Razborov's paper on the product theorem in the free group (in section 6), for example.

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This, by the way, is the solution to Exercise 2.3.21 of my book with Van Vu. – Terry Tao Apr 9 '12 at 19:15
@Boris Bukh: Thanks a lot! I was actually trying something earlier today with sets of the form $n^{\alpha} \left\{1,\ldots,n\right\}$... – Cosmin Pohoata Apr 10 '12 at 4:53