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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.

WriteApply the Cauchy-Schwarz inequality first in the variable carrying $f_1$ and then in the variable carrying $f_2$. After the invertible linear changes of variables permitted by $2,3\nmid |G|$, apply the assumed three-function $U^2$ estimate to the resulting multiplicative derivatives. The standard calculation givesBy the Derivative identity for the Gowers U3 norm, the last two factors are $\Vert f_3{\Vert }_{U^3}^8$ and $\Vert f_4{\Vert }_{U^3}^8$. Taking eighth roots proves

If $A\subseteq G$ has density $\alpha$, write $1_A=\alpha +f$. Expanding ${\Lambda }_4(1_A,1_A,1_A,1_A)$, the constant term is ${\alpha }^4$, and the displayed inequality bounds every nonconstant term after translation of one factor by a $U^3$ norm of the balanced function $f$. Thus sufficiently small $\Vert 1_A-\alpha {\Vert }_{U^3}$ makes the normalized number of four-term arithmetic progressions close to ${\alpha }^4$.