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Consider the group $\mathbb{Z}_2^n$, equipped it with the following dot product: for $a=(a_1,\dots,a_n)\in \mathbb{Z}_2^n$ and $b=(b_1,\dots,b_n)\in \mathbb{Z}_2^n$, define $a\cdot b :=a_1b_1+\dots+a_nb_n.$

Let $A \subseteq \mathbb{Z}_2^n$ be a set of cadinality $a$, where $2^{n-1}<a<2^n$. Define $$r(A) := |\{(x, y, z) \in A^3 : x + y = z\}|.$$ Using basic Fourier analysis, one can show that $$r(A) = \frac{1}{2^n} \sum_{\chi \in \widehat{\mathbb{Z}_2^n}} \sum_{x, y, z \in A} \chi(x + y + z).$$ Since the characters of $\widehat{\mathbb{Z}_2^n}$ are well known, we can rewrite this as $$r(A)=\frac{1}{2^n}\sum_{\varepsilon\in \{0,1\}^n}\sum_{x,y,z\in A}(-1)^{\varepsilon\cdot (x+y+z)}.$$

Define $S(A):=\sum\limits_{\varepsilon\in \{0,1\}^n}\sum\limits_{x,y,z\in A}(-1)^{\varepsilon\cdot (x+y+z)}$. I would like to prove the following lower bound for $S(A)$: $$S(A)\geq 2^n\Big(2^n-2^{1+\lfloor \log_2(2^n-a)\rfloor}\Big)\Big(3a-2^{n+1}+2^{1+\lfloor \log_2(2^n-a)\rfloor}\Big).$$

I currently have no idea how to prove this, but perhaps someone else does. Thank you very much!

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    $\begingroup$ If $A = \{w \in \mathbb{Z}_2^n : w_1 = 1\}$, then $r(A) = 0$. $\endgroup$ Commented Apr 24 at 9:57
  • $\begingroup$ @mathworker21, Indeed, this is a nice observation. If we denote the whole sum by $S(A)$, we obtain: $S(A) = 2^n r(A) - a^3 \geq -a^3$. You've shown that the bound is sharp when $r(A) = 0$, which is good. However, I’m particularly interested in the case when $2^{n-1} + 1 \leq a < 2^n$. $\endgroup$ Commented Apr 24 at 12:30
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    $\begingroup$ We have the trivial bounds $\lvert A\rvert(2\lvert A\rvert-2^n)\leq r(A)\leq \lvert A\rvert^2$. In particular if $\lvert A\rvert/2^n=\alpha\in (1/2,1)$ then $\alpha(2\alpha-1)\leq r(A)2^{-2n}\leq \alpha^2$. $\endgroup$ Commented Apr 29 at 10:42
  • $\begingroup$ Can you explain where your particular lower bound for $S(A)$ comes from? Why do you believe it is true? $\endgroup$ Commented Apr 29 at 10:43
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    $\begingroup$ @ThomasBloom, The lower for $S(A)$ comes from the following fact: the expression $\Big(2^n-2^{1+\lfloor \log_2(2^n-a)\rfloor}\Big)\Big(3a-2^{n+1}+2^{1+\lfloor \log_2(2^n-a)\rfloor}\Big)$ is the value of $r(B_a)$, where $B_a$ corresponds to the binary expansions of the integers in $\{2^n-a,\dots,2^n-1\}$. Taking this into account, this should (I believe!) give an optimal lower bound for $S(A)$. $\endgroup$ Commented Apr 29 at 13:07

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$\newcommand{\Z}{\mathbb{Z}}$ You ask an interesting question. Denote by $1_A \colon \Z_2^n \to \{0, 1\}$ the indicator function of $A$. Let $\widehat{1_A}$ denote its Fourier-Hadamard transform, given by $\widehat{1_A}(a)=\sum_{x \in \Z_2^n}(-1)^{x \cdot a} 1_A(x)$. An alternate representation of $r(A)$ is $r(A) = \frac{1}{2^n} \sum_{x \in \Z_2^n} (\widehat{1_A})^3(x)$.

If $r(A)=0$ then $A$ is sum-free, i.e. $A \cap (A+A)=\emptyset$, but the maximal size of a sum-free set in $\Z_2^n$ is $2^{n-1}$. So, when $|A| >2^{n-1}$ (as you are interested in), we always have $r(A) >0$.

Let $A'$ be the set $\{1\} \times \Z_2^{n-1}$. Then $A'$ is sum-free of size $2^{n-1}$. Also $|\{ (x,y) \in (A')^2 : x+y=u| = 2^{n-1}$ for all nonzero $u\in \{0\} \times \Z_2^{n-1}$.

Let $A = A' \cup \{u\}$ for some nonzero $u\in \{0\} \times \Z_2^{n-1}$. Then $r(A) = |\{ (x,y,z) \in A^3 : x+y=z, u \in \{ x,y,z\}\}| = 3|\{ (x,y) \in (A')^2 : x+y=u\}= 3 \cdot 2^{n-1}$.

It seems plausible that this is minimal for all $A$ such that $|A| \geq 2^{n-1}+1$.

Edit: To find a lower bound, it suffices to consider when $|A| =2^{n-1}+1$ as $r(A) \geq r(B)$ when $A \supset B$. So, assume $A \subseteq \Z_2^n$ is a subset of size $2^{n-1}+1$. Moreover, without loss of generality, assume $0 \notin A$ because if $0\in A$, then $$r(A) = |\{(x,y,z) \in A^3 : x+y+z=0\}| \geq 3(|A|-1) = 3\cdot 2^{n-1}$$ which we already have achieved via the construction above. If $A'=A \setminus \{u\}$ for some $u \in A$, then $$\begin{aligned} 2^nr(A) &= \sum_{x \in \Z_2^n} (\widehat{1_A})^3(x) \\ &= \sum_{x \in \Z_2^n} \left( (\widehat{1_{A'}})^3(x) + (-1)^{u \cdot x} + 3(-1)^{u \cdot x} (\widehat{1_{A'}})^2(x) + 3(\widehat{1_{A'}})(x) \right)\\ &= \sum_{x \in \Z_2^n} (\widehat{1_{A'}})^3(x) + 0 + 3\widehat{(\widehat{1_{A'}})^2}(u) + 3 \cdot 2^n 1_{A'}(0) \\ &= 2^n r(A') + 3 \cdot 2^n |A' \cap (u+A')|. \end{aligned}$$ So, $$r(A) \geq \min_{\substack{A' \subseteq \Z_2^n \setminus \{0\}, |A'|=2^{n-1}\\ u \in \Z_2^n \setminus (A' \cup \{0\})}} (r(A')+3 |A' \cap (u+A')|).$$

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  • $\begingroup$ Thanks for your reply! Just a small comment: I did not have a chance to read your answer in detail but my goal is to minimize $r(A)$ over all $A\subseteq\mathbb{Z}_2^n$ such that $|A|=a$ where $a$ is fixed within the range $2^{n-1}<a<2^n$ $\endgroup$ Commented Apr 25 at 5:04
  • $\begingroup$ Hi! My answer is incomplete and does not provide bounds in terms of $|A|$. It focuses on a minimum of $r(A)$ for all $A$ with $2^{n-1}<|A| < 2^n$ which seems like it should be $3 \cdot 2^{n-1}$ but seems nontrivial to prove. $\endgroup$ Commented Apr 25 at 14:21
  • $\begingroup$ Also, here's a thought. Suppose $A$ does not include $0$, and let $M_A$ be the matrix whose columns are the vectors in $A$. The nullspace of $M_A$ is a linear code, and I believe the problem is equivalent to bounding the number of code words of weight $3$. $\endgroup$ Commented Apr 25 at 14:25

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