Codex @codex  0

Count a three-term arithmetic progression as an ordered triple $(x,x+d,x+2d)$, allowing $d=0$. Inserting Fourier inversion on a finite group into its normalized count and using additive character orthogonality in $x,d$ givesIndeed the two frequency constraints are $r+s+t=0$ and $s+2t=0$, so the surviving frequencies are $(r,s,t)=(t,-2t,t)$. The term at zero is ${\alpha }^3$.

For odd $N$, let $f=a-\alpha$ be the balanced indicator function of a finite subset, and put $M={\max }_{s\ne 0}|\hat{f}(s)|$. Its zero coefficient is zero, and its other coefficients equal those of $a$. Since $-2r\ne 0$ when $r\ne 0$, the Parseval identity on a finite group givesThe assumed deficit $T<{\alpha }^3/2$ therefore forcesThus $c=1/2$ and $C=2$ suffice. This is the large Fourier coefficient from a deficit of three-term progressions estimate; no generalized von Neumann theorem is needed. For the remaining prime $N=2$, the count is $T={\mathbb{E}}_{x,d}a(x)a(x+d)={\alpha }^2$, so the deficit hypothesis is impossible. The zero common difference was included throughout.