Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 129 / 2

For $\Gamma \subseteq \hat{G}$ and $\rho >0$, the Bohr set isWriting $d=|\Gamma |$, the lower bound for the size of a Bohr set is

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Use normalized convolution and Fourier coefficients, and define the large spectrumBy the Parseval identity and $0\le f\le 1$,so $|\Gamma |\le {ϵ}^{-2}{\delta }^{-1}$.

The convolution theorem givesIf $y\in B(\Gamma ,ϵ)$, the part over $\Gamma$ is at mostbecause $|\hat{f}(\gamma )|\le \delta$. On the complementary spectrum, $|\hat{f}(\gamma )|<ϵ\delta$ and $|\gamma (y)-1|\le 2$, so the contribution is less than $2ϵ{\delta }^2$. The required difference is therefore less than $3ϵ{\delta }^2$.

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Apply the preceding Fourier argument to $f=1_A$, retaining the frequencies needed to make the oscillation of $1_A\ast 1_A\ast 1_A$ strictly smaller than its mean ${\alpha }^3$. The standard optimized cutoff gives a set $\Gamma \subseteq \hat{G}$ withsuch that the nonnegative function $F=1_A\ast 1_A\ast 1_A$ cannot fall from a maximal value to zero under any shift in $B(\Gamma ,\alpha )$. If $x$ maximizes $F$, thenThe support of a convolution of indicator functions is the corresponding sumset, so

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Write $d=|\Gamma |$ and identify each character with a residue $r_j\in \mathbb{Z}/N\mathbb{Z}$. Partition the $d$-dimensional torus into $Q^d$ cubes of side $1/Q$, where $Q$ is comparable to $N^{1/d}$. Applying the pigeonhole principle to the pointsgives a nonzero residue $q$ satisfyingConsequently $|e^{2\pi imq{r}_j/N}-1|\le \rho$ whenever $|m|\le \rho Q/(4\pi )$. After allowing for integer parts and the small values of $Q$, this produces the centered arithmetic progressionof length at least $\frac{1}{8}\rho N^{1/d}$.

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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 $A+A+A$.

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