Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 13 / 1 / i / Solution

Assume $A$ is nonempty, so $0<\alpha \le 1$, and put $a=1_A$. By the Parseval identity on a finite group, ${\sum }_r|\hat{a}(r){|}^2=\alpha$. Select the nonzero frequenciesThen $|R|{\alpha }^3/4\le \alpha$, giving $|R|\le 4{\alpha }^{-2}$. Outside $R\cup \{0\}$, the fourth Fourier moment has the boundThe fourfold normalized convolution on a finite group $H=a\ast a\ast 1_{-A}\ast 1_{-A}$ has transform $|\hat{a}(r){|}^4$, so Fourier inversion on a finite group gives $H(x)={\sum }_r|\hat{a}(r){|}^4{e}_N(rx)$. It is real and nonnegative, and $H(x)>0$ implies $x\in 2A-2A$ because the convolution counts representations, with a positive normalization.

Use the Bohr set in phase-distance conventionOn this set all retained additive characters have positive real part. The zero-frequency contribution is $\hat{a}(0{)}^4={\alpha }^4$, and the total absolute contribution of the discarded frequencies is at most ${\alpha }^4/4$. HenceThereforeThis is an explicit Bogolyubov lemma. In the chord-distance convention for a Bohr set, the phase-radius $1/10$ corresponds to chord radius $2\sin (\pi /10)>1/10$; alternatively the same positivity proof works directly at chord radius $1/10$. Thus the requested lower bound on width holds in either convention. If $R$ is empty, the Bohr set is the whole group. Nonemptiness of $A$ is necessary for the density statement.