The finite-field <Bogolyubov lemma> states that if $A\subseteq G=\mathbb F_p^n$ has density $\alpha$, then $2A-2A$ contains a subspace of codimension at most $2\alpha^{-2}$.

Use normalized <Fourier analysis on a finite abelian group> and put $f=1_A$. Define
$$
S=\left\{\gamma\in\widehat G:|\widehat f(\gamma)|\geq\frac{\alpha^{3/2}}{\sqrt2}\right\}.
$$
By <Parseval identity>,
$$
|S|\frac{\alpha^3}{2}\leq\sum_\gamma|\widehat f(\gamma)|^2=\alpha,
$$
so $|S|\leq2\alpha^{-2}$. Let
$$
V=\{x\in G:\gamma(x)=1\text{ for every }\gamma\in S\}.
$$
Then $V$ is a subspace of codimension at most $|S|$.

The normalized representation function of $2A-2A$ is
$$
r(x)=(f*f*\widetilde f*\widetilde f)(x)
=\sum_{\gamma\in\widehat G}|\widehat f(\gamma)|^4\gamma(x),
$$
where $\widetilde f(x)=f(-x)$. For $x\in V$, all terms indexed by $S$ are nonnegative real numbers, while
$$
\sum_{\gamma\notin S}|\widehat f(\gamma)|^4
\leq\frac{\alpha^3}{2}\sum_\gamma|\widehat f(\gamma)|^2
=\frac{\alpha^4}{2}.
$$
The trivial character alone contributes $\alpha^4$, so $r(x)>0$. Hence $x\in2A-2A$, proving the <Finite-field Bogolyubov lemma>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 129 / 3 / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-129/3/i/solution