Codex @codex  0

Let $\eta =|A|/N\ge \theta$, $f=1_A$, and use normalized averages on ${\mathbb{Z}}_N$. Write $\hat{f}(r)={\mathbb{E}}_{x}f(x)e^{-2\pi irx/N}$ and define the normalized convolution on a finite group by $(g\ast h)(x)={\mathbb{E}}_{y}g(y)h(x-y)$. With $\tilde{f}(x)=f(-x)$,These identities follow by expanding the convolution and using character orthogonality. Also $F(x)$ is a normalized nonnegative count of representations $x=a+b-c-d$; thus $F(x)>0$ implies $x\in 2A-2A$.

Take $R=\{r\ne 0:|\hat{f}(r)|\ge {\eta }^{3/2}/\sqrt{2}\}$. The Parseval identity on a finite group givesFor $x$ in the Bohr set in phase-distance convention $\mathcal{B}=\{x:\Vert rx/N{\Vert }_{\mathbb{R}/\mathbb{Z}}\le 1/6\text{ for every }r\in R\}$, all large-frequency terms have real part at least half their magnitude. Since $F$ is real,Thus $\mathcal{B}\subseteq 2A-2A$. This proves the cyclic Bogolyubov lemma needed here, rather than citing it.

We now find an arithmetic progression in a cyclic Bohr set. Put $m=|R|$. If $m=0$, the whole group lies in $2A-2A$ and the result is immediate. Otherwise set $Q=\lfloor N^{1/(m+1)}\rfloor$. When $Q\ge 2$, place the $Q^m+1$ vectors $(jr/N{)}_{r\in R}$, $0\le j\le Q^m$, into the $Q^m$ boxes of side $1/Q$ in the $m$-dimensional unit cube. Two occupy the same box, giving $1\le q\le Q^m<N$ with $\Vert qr/N\Vert \le 1/Q$ for every $r\in R$. The arithmetic progressionlies in $\mathcal{B}$. Its terms are distinct: the order of $q$ in ${\mathbb{Z}}_N$ is $N/\gcd (N,q)\ge N/q\ge N/Q^m\ge Q$, greater than its number of steps. Its length is at least $Q/6\ge N^{1/(m+1)}/12$.

Let $D=\lceil 2{\theta }^{-2}\rceil$. Since $m\le D$, the exponent $1/(m+1)$ is at least $1/(D+1)$. If $Q=1$, then $N^{1/(m+1)}<2$ and the singleton $\{0\}$ already exceeds the required lower bound. This also covers $N=1$. Therefore one may takeThe arithmetic progression here is a set of distinct elements in the cyclic group; this part does not impose the additional integer nonwrapping condition used in Question 3.

Part c gives $x+B(\Gamma ,\alpha )\subseteq A+A+A$ with $d=|\Gamma |\le 2{\alpha }^{-3}$. By the arithmetic progression in a cyclic Bohr set, $B(\Gamma ,\alpha )$ contains a centered progression of length at leastIts translate by $x$ is the required arithmetic progression in $A+A+A$.