= Solution
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 $x,d$ gives
$$
T=\mathbb E_{x,d}a(x)a(x+d)a(x+2d)
=\sum_r\widehat a(r)^2\widehat a(-2r).
$$
Indeed 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 $M=\max_{s\ne0}|\widehat f(s)|$. Its zero coefficient is zero, and its other coefficients equal those of $a$. Since $-2r\ne0$ when $r\ne0$, the <Parseval identity on a finite group> gives
$$
|T-\alpha^3|\leq M\sum_{r\ne0}|\widehat a(r)|^2\leq M\alpha.
$$
The assumed deficit $T<\alpha^3/2$ therefore forces
$$
\boxed{\max_{s\ne0}|\widehat f(s)|>\frac{\alpha^2}{2}.}
$$
Thus $c=1/2$ 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.
Back to article page
