Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 129 / 2

Writing $\hat{G}$ for the character group of a finite abelian group, the Bohr set with frequency set $\Gamma \subseteq \hat{G}$ and width $0\le \rho \le 2$ isIts rank is $|\Gamma |$.

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

The Bourgain bound for three-term-progression-free sets states that, if $N=|G|$ is odd and $A\subseteq G$ contains no nonconstant three-term arithmetic progression, then

Here is how it follows from the standard Bohr-set density increment lemma. Begin with $B_0=G$ and relative density ${\alpha }_0=\alpha =|A|/N$. Whenever the lemma gives its second alternative, replace the current set by the denser translate inside the smaller regular Bohr set. The density changes byso this can happen only $O({\alpha }^{-1})$ times. Throughout the iteration the rank is $O({\alpha }^{-1})$, the width is at least ${\alpha }^{O({\alpha }^{-1})}$, and the elementary lower bound for the size of a Bohr set givesAt the terminal stage the first alternative of the density-increment lemma holds. Combining it with the last display yieldsIf $\alpha <1/\log N$, Bourgain's bound is already true. Otherwise $\log (1/\alpha )\ll \log \log N$, and rearranging proves