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We specify the convention for quadratic uniformity of a set. Write $\delta =|A|/N$ and $f=1_A-\delta$ on $[1,N]$, extended by zero elsewhere. Let ${\mathcal{C}}_N$ be the integer cubes $(x,h_1,h_2,h_3)$ with all eight vertices in $[1,N]$, and let $T_N=|{\mathcal{C}}_N|$. In the eighth-power convention, $A$ is $\alpha$-quadratically uniform whenwhere $\mathcal{C}$ means complex conjugation. This quantity is nonnegative and is the eighth power of the normalized Gowers U3 norm on an interval. If uniformity instead means that the norm itself is at most $\alpha$, replace the present parameter by ${\alpha }^8$. Removing the balance would not define density-relative uniformity.

Embed the interval in $G=\mathbb{Z}/M\mathbb{Z}$ with $M=12N+1$, and keep the zero extension of $f$. This group has invertible $2$ and $3$. Every cube with all vertices in the interval is an integer cube: its face relations have magnitude at most $2N<M$, so their modular equalities are ordinary equalities. Conversely every such integer cube gives one modular cube. Thussince $T_N\le M^4$. The cube quantity is nonnegative, for it can be written as an average of squares of two-fold multiplicative-derivative correlations, by expanding the third increment.

We prove the counting estimate needed below. For functions $|u_j|\le 1$, putThree applications of Cauchy-Schwarz giveHere are the elimination steps, to make the generalized von Neumann inequality for four-term progressions explicit. Set ${\Delta }_{h}u(x)=u(x)\overline{u(x+h)}$. Reparametrize by the position $x+3d$, and apply Cauchy-Schwarz in that position to eliminate $u_3$. Pairing the two remaining difference variables produces increments $3h,2h,h$ in $u_0,u_1,u_2$. Reparametrize by the position of $u_2$, apply Cauchy-Schwarz to eliminate it, and average over $h$ using Cauchy-Schwarz again. The remaining increments in $u_0$ are $3h,2k$. Finally reparametrize by the position of $u_1$ and eliminate it in the same way. The resulting bound isIn the last equality $2k$ and $3h$ range freely since their multipliers are units. Each elimination drops only a factor bounded by one; expansion of the squared average gives exactly the stated multiplicative derivative. To target $u_j$ instead, subtract its slope from all four slopes and eliminate the other three factors. The increments are the three nonzero slope differences times independent variables, all among $\pm 1,\pm 2,\pm 3$ and hence invertible. This proves the minimum bound.

Now let $a=1_A$ and $g=\delta 1_{[1,N]}$ on $G$. Telescoping the four factors, the difference ${\Lambda }_4(a,a,a,a)-{\Lambda }_4(g,g,g,g)$ is a sum of four terms with one balanced factor $f$ and three factors bounded by one. ThereforeFor $N\ge 16$, restricting to $1\le x\le \lfloor N/4\rfloor$ and $1\le d\le \lfloor N/8\rfloor$ gives at least $N^2/128$ progressions inside the interval. Since $M\le 13N$, ${\Lambda }_4(I,I,I,I)\ge 1/100000$. Choose, for example,Then ${\Lambda }_4(a,a,a,a)\ge {\delta }^4/200000$. Constant progressions contribute only $|A|/M^2\le 1/N$. For $N>400000{\delta }^{-4}$, this is strictly below the displayed lower bound. A nonconstant progression consequently exists. Its four points are actual integer progression points: the two adjacent three-term relations have magnitude less than $M$, so they do not wrap. HenceThis proves the four-term progressions in a quadratically uniform interval set criterion, with the parameter normalization and the interval boundary both accounted for.