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Suppose that $A$ is a finite, nonempty set in an abelian group. If there is a group element with a unique representation as $a-b$ with $a,b\in A$, then none of $A-A$ and $2A$ are small: $$ \min\{|A-A|,|2A|\} \ge 2|A|-1; $$ this is almost immediate for the difference set, while the sumset case is a nontrivial result due to Kemperman and Scherk.

I need an analogue of this result for the additive energy of $A$ defined by $$ E(A):=\{(a_1,a_2,a_3,a_4)\in A^4\colon a_1+a_2=a_3+a_4\}. $$ A straightforward application of the Cauchy-Schwarz inequality gives the well-known estimate $E(A)\ge|A|^4/|2A|$. How small can $E(A)$ be given that there is an element with a unique representation as $a-b$ with $a,b\in A$? Given that there are "many" uniquely representable elements?

What I would need is an explicit estimate that works starting with very small values of $|A|$, not involving excessively large constants and the $O/o$-notation. Even an improvement by a lower-order term can be helpfull.

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    $\begingroup$ Additive energy $E(A)$ tends to behave inversely to the sumset size $|2A|$. So wouldn't the analogue of a lower bound on $|2A|$ be an upper bound on $E(A)$ rather than a lower bound? $\endgroup$
    – Terry Tao
    Commented Jan 26, 2022 at 21:09
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    $\begingroup$ If $G = \mathbb Z/(3n)$ is a finite abelian group, and $A$ consists of all the multiples of $3$ together with $1$, then $1$ has a unique representation as a difference (in fact, every element that's not a multiple of $3$ does) but the additive energy of $A$ is at least $n^3 = (|A| -1 )^3$. So the additive energy can be very large under this condition. $\endgroup$
    – Will Sawin
    Commented Jan 26, 2022 at 21:22
  • $\begingroup$ @TerryTao: I have to think more about the ideology, but anyway, the question is as stated. $\endgroup$
    – Seva
    Commented Jan 26, 2022 at 21:34
  • $\begingroup$ @WillSawin: $(|A|-1)^3$ is about twice larger than $|A|^4/|2A|$, a way better than I need. $\endgroup$
    – Seva
    Commented Jan 26, 2022 at 21:36
  • $\begingroup$ If $A$ is $n$ random elements in $\mathbb R$ then $E(A) = 2n^2- n $, $|2A|$ is $n (n+1)/2$, so $|A|^4/ 2|A|$ is $ 2n^3/ (n+1) = 2n^2 - 2n + 2 + O(1/n)$, which barely differs from $E(A)$. Of course some (in fact, every) element is uniquely represented. Is this OK for you? $\endgroup$
    – Will Sawin
    Commented Jan 26, 2022 at 21:45

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