The Fourier density-increment proof of the <Meshulam theorem> controls three-term progressions through ordinary Fourier coefficients, equivalently the $U^2$ norm. Four-term progressions are controlled by the <Gowers uniformity norm> $U^3$. A function can have small correlation with every linear character while correlating strongly with a <quadratic phase>; such a function can have small $U^2$ norm but large $U^3$ norm. Ordinary Fourier uniformity therefore does not make the four-term progression count pseudorandom, and a large linear Fourier coefficient need not exist when that count is deficient. Quadratic or higher-order Fourier structure is needed.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 129 / 4 / ii / Solution

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For each fixed odd prime $p$ and density $\alpha>0$, every sufficiently large subset of $\mathbb F_p^n$ of density at least $\alpha$ contains a nontrivial three-term <arithmetic progression>.

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 Incoming links: Meshulam theorem