 Solution

Put $M(a)={\max }_{\gamma \in \hat{G}}|\widehat{{\partial }_{a}f}(\gamma ){|}^2$. By the Derivative identity for the Gowers U3 norm, the Fourier formula for the $U^2$ norm, and Parseval identity,because $\Vert f{\Vert }_{\infty }\le 1$. Letand for every $a\in B$ choose $\varphi (a)\in \hat{G}$ attaining $M(a)$. Thenfor every $a\in B$, and deleting the complement of $B$ from the preceding average leaves total mass at least $c/2$.

Let ${\Gamma }_{\varphi }=\{(a,\varphi (a)):a\in B\}\subseteq G\times \hat{G}$. The box-norm inequality applied to the selected Fourier coefficients and the cocycle identity for multiplicative derivatives givesThe left side is at least $(c/2{)}^8$. An additive quadruple in ${\Gamma }_{\varphi }$ is exactly a tuple $(a,b,c,d)\in B^4$ satisfyingTherefore there are at least $(c/2{)}^8|G{|}^3$ such quadruples, which is the Frequency graph extracted from a large Gowers U3 norm.