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Let $A$ have relative density $\alpha$ in a rank-$d$ Regular Bohr set $B$, and suppose $A$ has no nonconstant three-term arithmetic progression. Then eitheror some translate of $A$ has relative density at least $(1+c\alpha )\alpha$ in a regular Bohr set $B^{\prime }\subseteq B$ of rank at most $d+1$ and width at least $\rho (B)(\alpha /d{)}^{O(1)}$.

For $B=B(\Gamma ,\rho )$ write $B_{1+\kappa }=B(\Gamma ,(1+\kappa )\rho )$. It is a Regular Bohr set whenwhenever $d=|\Gamma |$ and $|\kappa |\le c/d$, with absolute constants in the $O$-term and in $c$.

Not every width is regular. In $G=\mathbb{Z}/3\mathbb{Z}$, let $\gamma (x)=e^{2\pi ix/3}$ and take $\rho =\sqrt{3}$. Then $B(\{\gamma \},\rho )=G$, but every arbitrarily small decrease of the width leaves only $0$. The size jumps from three to one, contradicting the required linear control as $\kappa \to 0^{-}$.

Choose a Regular Bohr set $B^{\prime }=B_{\lambda }\subseteq B$ with $1/2\le \lambda \le 1$. Standard Bohr-set size estimates give $|B^{\prime }|\ge 2^{-O(d)}|B|$. Set $\eta =c_0/d$ with $c_0$ small enough that regularity givesFor each $z\in B_{\eta }^{\prime }$ and $y\in B_{1-\eta }^{\prime }$, the triangle inequality in every frequency givesBecause $|G|$ is odd, multiplication by two is a bijection, and the pair $(z,y)$ determines the ordered three-term arithmetic progression uniquely. The lower size bound for a Dilate of a Bohr set givesThe number of progressions in $B$ is therefore at least