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Define the normalized progression averageWe first prove the needed four-term progression bound with torsion factors, valid for the arbitrary cyclic modulus in the question:For the factor at slope zero, write $y=x+3d$, $t=d$, and apply Cauchy-Schwarz to discard $f_3(y)$:For concise bookkeeping put $D_{h}f(x)=f(x)\overline{f(x+h)}$, the conjugate of the preceding derivative convention, and $F_{i,h}=D_{-(3-i)h}{f}_i$. Expanding the square with $t,t+h$ and setting $z=y-t$ gives ${\mathbb{E}}_{z,t,h}{F}_{0,h}(z-2t)F_{1,h}(z-t)F_{2,h}(z)$. Another Cauchy-Schwarz, now in $(z,h)$, discards $F_{2,h}$ and yieldsExpand with $t,t+k$, set $w=z-t$, and apply Cauchy-Schwarz a third time to discard $D_{-k}{F}_{1,h}(w)$. The remaining $t$-average is a full average over $x=w-t$, soNegating $h,k$ does not change the average. The nonnegative integrand depends only on $2k$ and $3h$. Averaging over the subgroups $2{\mathbb{Z}}_N$ and $3{\mathbb{Z}}_N$ costs at most their indices $t_2=\gcd (2,N)$ and $t_3=\gcd (3,N)$ compared with unrestricted increments. The unrestricted expression is exactly $\Vert f_0{\Vert }_{U^3}^8$, obtained by expanding the last squared average. Thus the bound is $(t_2{t}_3{)}^{1/8}\Vert f_0{\Vert }_{U^3}\le 6^{1/8}\Vert f_0{\Vert }_{U^3}$.

Reversing the progression handles slope three. To isolate slope one, eliminate slopes three, zero and two successively in the same calculation; the surviving increments have coefficients $-2,1,-1$, so only the index $t_2$ is needed. Reversal handles slope two. This proves the minimum-over-factors bound, without assuming that $2$ or $3$ is invertible modulo $N$.

If $\delta =0$, the requested lower bound is zero and is immediate. Otherwise let $g=1_A-\delta$. Telescoping the product of four copies of $1_A=\delta +g$ against the constant product ${\delta }^4$ gives four error terms, each with one factor $g$ and three factors bounded by one. HenceIn particular, choosing $\alpha \le {\delta }^{32}/(6\cdot 8^8)$ givesThese are ordered modular progression parameters $(x,d)$, as requested; small-order differences and $d=0$ are included. A claim of distinct nontrivial progressions would need a separate subtraction and size condition.

Index the functions as $g_0,g_1,g_2,g_3$ and write $\Lambda ={\mathbb{E}}_{x,d}{\prod }_{j=0}^3{g}_j(x+jd)$. We first control $g_0$. Change variables to $y=x+3d$, average over $d$, and apply Cauchy-Schwarz in $y$, discarding the bounded factor $g_3(y)$. Expand the squared inner average with the second copy of $d$ written $d+h$, then put $x=y-3d$. This yieldsThe right side is real and nonnegative because it arose as an average of squared absolute values. Next use $y=x+2d$ and apply Cauchy-Schwarz in $(h,y)$, discarding the bounded $g_2$ derivative. Expanding with a displacement $k$ in $d$ givesFinally use $y=x+d$ and Cauchy-Schwarz in $(h,k,y)$, discarding the bounded $g_1$ double derivative. Expansion with displacement $r$ givesAll these changes of variables are bijections without any division in $G$.

Put $t_j=|G[j]|$, where $G[j]=\{z:jz=0\}$ is the torsion subgroup killed by an integer $j$. The variables $2k$ and $3h$ are uniform on $2G$ and $3G$, whose indices are $t_2,t_3$. Since the last squared integrand is nonnegative, its restricted average is at most $t_2{t}_3$ times the unrestricted average over the two increments. By part (i), that unrestricted average is $\Vert g_0{\Vert }_{U^3}^8$. Hence $|\Lambda |\le (t_2{t}_3{)}^{1/8}\Vert g_0{\Vert }_{U^3}$.

To control $g_1$, apply the same three eliminations in the order of slopes $3,0,2$. The surviving $g_1$ has increments $-2h,k,-r$. The last increment is uniform, and only the first is restricted to $2G$, so the resulting bound is $|\Lambda |\le t_2^{1/8}\Vert g_1{\Vert }_{U^3}$. Reversing the arithmetic progression by $(x,d)\mapsto (x+3d,-d)$ treats $g_3$ and $g_2$ respectively. This proves the four-term progression bound with torsion factors:For a fixed finite group, this positive constant tends to zero with $\alpha$. In ${\mathbb{Z}}_N$, $t_j=\gcd (j,N)$, so $\boxed{\beta =6^{1/8}\alpha \text{ suffices uniformly in }N}$. If multiplication by two and three are bijective, the sharper $\beta =\alpha$ holds.

A dependence on torsion is necessary if a bound uniform over all finite abelian groups is intended. In $G={\mathbb{F}}_2^m$, choose independent random signs $f(x)$. A cube with eight distinct vertices has sign product of expectation zero. Each of the twenty-eight possible collisions has probability $1/|G|$, so the expectation of $\Vert f{\Vert }_{U^3}^8$ is at most $28/|G|$. Some sign function therefore has norm at most $(28/|G|{)}^{1/8}$. But with factors $(f,1,f,1)$, every arithmetic progression is $(x,x+d,x,x+d)$ and its product is $f(x{)}^2=1$. Thus unbounded torsion obstructs uniform control of four-term progressions. The stated fixed-group bound, and the uniform cyclic-group bound needed in part (iii), are unaffected.

For $A\subseteq {\mathbb{Z}}_N$, $N\ge 16$, put $\eta =|A|/N$, $M=\lfloor N/16\rfloor$, $J=\{0,\dots ,M-1\}$, and $w=1_J\ast 1_J$ using normalized convolution on a finite group. The weights $w(x)w(d-M)$ are supported where $0\le x\le 2M-2$ and $M\le d\le 3M-2$, so $x,x+d,x+2d,x+3d$ are distinct integer points below $N$. If $t=M/N$, then $\mathbb{E}w=t^2$ and the sum of absolute values of its Fourier coefficients is $t$, by Parseval identity on a finite group. Expanding the four indicators around their subset density leaves four terms with one balanced factor. Expand both weights in Fourier series and absorb each character $e(rx+sd)$ into the factors at $x$ and $x+d$. Character multiplication preserves the Gowers U3 norm, and the cyclic four-term progression bound with torsion factors bounds every error by $6^{1/8}\alpha$. Thus the weighted count is at least ${\eta }^4{t}^4-4\cdot 6^{1/8}\alpha t^2$. Since $t\ge 1/32$, positivity follows from $\alpha <{\eta }^4/(4096\cdot 6^{1/8})$.