= Solution
There is a normalization omission in the printed conclusion: the discrepancy of a count of pairs must be bounded by a small function of the energy excess times $N^2$. We prove this normalized result, then give a counterexample to the literal bound without $N^2$. Write $\epsilon$ for the energy excess parameter to avoid confusing it with an element of the third set.
For <Fourier analysis on a finite abelian group>, use $e_N(t)=e^{2\pi it/N}$ and
$$
\widehat h(r)=\mathbb E_{x\in\mathbb Z_N}h(x)e_N(-rx).
$$
Character orthogonality, $\mathbb E_x e_N(rx)=1$ for $r=0$ and zero otherwise, gives <Fourier inversion> and the <Parseval identity on a finite group>:
$$
h(x)=\sum_r\widehat h(r)e_N(rx),\qquad \sum_r|\widehat h(r)|^2=\mathbb E_x|h(x)|^2.
$$
Put $u=1_A$, $v=1_B$, $w=1_C$. Expanding the constraint $a_1+a_2=a_3+a_4$ by the same character orthogonality gives the <additive energy> identity
$$
\frac{E(A)}{N^3}=\sum_r|\widehat u(r)|^4=\alpha^4+\sum_{r\ne0}|\widehat u(r)|^4.
$$
Thus $\sum_{r\ne0}|\widehat u(r)|^4\leq\epsilon$, and every nonconstant <Fourier coefficient on a finite abelian group> of $u$ has modulus at most $\epsilon^{1/4}$. Expanding the constraint in the triple count $T$ gives
$$
\frac{T}{N^2}=\sum_r\widehat u(r)\widehat w(r)\widehat v(-2r)=\alpha\beta\gamma+\sum_{r\ne0}\widehat u(r)\widehat w(r)\widehat v(-2r).
$$
The <Cauchy-Schwarz inequality> yields
$$
\left|\frac{T}{N^2}-\alpha\beta\gamma\right|\leq\epsilon^{1/4}\left(\sum_r|\widehat w(r)|^2\right)^{1/2}\left(\sum_r|\widehat v(-2r)|^2\right)^{1/2}.
$$
The first squared norm is $\gamma$ by the <Parseval identity on a finite group>. For the second, multiplication by $2$ on $\mathbb Z_N$ has kernel size $h=\gcd(2,N)$ and every element of its image has exactly $h$ preimages. Hence
$$
\sum_r|\widehat v(-2r)|^2=h\sum_{s\in2\mathbb Z_N}|\widehat v(s)|^2\leq h\beta.
$$
This proves that <additive energy controls three-term progression mixing>, including for even moduli:
$$
\boxed{|T-\alpha\beta\gamma N^2|\leq\sqrt{\gcd(2,N)\beta\gamma}\,\epsilon^{1/4}N^2\leq\sqrt2\,\epsilon^{1/4}N^2.}
$$
Thus the intended density-error function can be $F(\epsilon)=\sqrt2\,\epsilon^{1/4}$, which tends to zero. For odd moduli the factor $\sqrt2$ can be replaced by one.
To see why the literal unnormalized conclusion is false, take $N=q^2$ and let all three sets be the subgroup of multiples of $q$. Their densities are $1/q$. Every choice of three entries of that subgroup determines the fourth entry of an additive quadruple, so its energy is $q^3$. The excess parameter is
$$
\epsilon=\frac1{q^3}-\frac1{q^4}=\frac{q-1}{q^4}\longrightarrow0.
$$
Every pair $(a,b)$ in the subgroup determines $c=2b-a$ there, giving $T=q^2$, whereas $\alpha\beta\gamma N^2=q$. Their difference is $q^2-q\to\infty$, which cannot be bounded by any function of $\epsilon$ tending to zero. The missing $N^2$ is therefore a genuine source defect, not an alternative <Fourier transform> normalization.
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