Four-term progression bound with torsion factors ($|{\Lambda }_4|\le (t_2{t}_3{)}^{1/8}{\min }_i\Vert f_i{\Vert }_{U^3}$)

For a finite abelian group $G$, put $t_j=|\{x\in G:jx=0\}|$. If $|f_i|\le 1$, then $|{\mathbb{E}}_{x,d}{\prod }_{i=0}^3{f}_i(x+id)|\le (t_2{t}_3{)}^{1/8}{\min }_i\Vert f_i{\Vert }_{U^3}$. Three applications of Cauchy-Schwarz, eliminating the factors at slopes $3,2,1$, bound the eighth power by ${\mathbb{E}}_{h,k}|{\mathbb{E}}_x{\Delta }_{-3h}{\Delta }_{-2k}{f}_0(x){|}^2$, where ${\Delta }_{h}f(x)=f(x)\overline{f(x+h)}$. The integrand is nonnegative; averaging over $3G$ and $2G$ costs at most their indices $t_3$ and $t_2$ compared with the unrestricted average defining $\Vert f_0{\Vert }_{U^3}^8$. Reversal treats slope three. For slope one, elimination at slopes $3,0,2$ produces increments $-2h,k,-r$, costing only $t_2$; reversal treats slope two. In a cyclic group, $t_2\le 2,t_3\le 3$, so the constant is at most $6^{1/8}$ independently of group size.

A bound tending to zero with one factor's Gowers U3 norm, uniformly over all finite abelian groups, is false. On $G={\mathbb{F}}_2^m$, choose a random independent sign at each point. A cube with eight distinct vertices has sign product of expectation zero. The probability of a collision among its vertices is at most $\left(\genfrac{}{}{0ex}{}{8}{2}\right)/|G|$, since each equality of distinct vertices imposes a nonzero linear equation in the increments. Thus some sign function $f$ has $\Vert f{\Vert }_{U^3}^8\le 28/|G|$. But the four-term arithmetic progression is $(x,x+d,x,x+d)$, so choosing factors $(f,1,f,1)$ gives correlation identically one. As $m\to \infty$, the norm tends to zero. Bounded torsion, as in cyclic groups, avoids this obstruction.