= Fourier detection of a progression-free subset of an interval
{c}
Let $I=[n]$ for odd $n\geq3$, embed it in the <cyclic group> $G=\mathbb Z/(2n+1)\mathbb Z$, and let $A\subseteq I$ have <density of a finite subset> $\alpha$. If $A$ has no nonconstant three-term <arithmetic progression> and $n\geq4\alpha^{-2}$, its balanced <indicator function> $f=1_A-\alpha1_I$ satisfies
$$
\max_{r\ne0}|\widehat f(r)|\geq\alpha^2/36,
\qquad \widehat f(r)=\mathbb E_{x\in G}f(x)e^{-2\pi irx/|G|}.
$$
Indeed, the normalized trilinear <arithmetic progression> count has the <Fourier analysis on a finite abelian group> formula $\Lambda(h_1,h_2,h_3)=\sum_r\widehat h_1(r)\widehat h_2(-2r)\widehat h_3(r)$. Its values on $1_A$ and $\alpha1_I$ are $\alpha n/|G|^2$ and $\alpha^3(n^2+1)/(2|G|^2)$. Telescoping their difference into three terms containing $f$, the <Parseval identity> and <Cauchy-Schwarz inequality> bound its magnitude by $3\max|\widehat f|\alpha n/|G|$. The difference is at least $\alpha^3n^2/(4|G|^2)$, giving the claim. The zero <Fourier coefficient> vanishes by the definition of $\alpha$.
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