only strengthens our conclusion, so we may assume that $|A| = \delta N$. Since $\Lambda_{q, N}(1_A) = 0$, we have that $
{\left| \Lambda_{q, N}(1_A)  -\Lambda_{q, N}(\delta1_{[N]})\right|} =  \delta^3 \Lambda_{q, N}(1_{[N]}) \gg \delta^3
$.

Applying the weak regularity lemma (Lemma \[weak regularity\]), there exists a local factor $\mathcal{B}$ of dimension $d \ll \delta^{-O(1)}$, resolution $\gg (\delta/q)^{O(1)}N^{1/2}$, and modulus $qq'$ for some $q' {\leqslant}O(1/ \delta)^{O(d)}$ such that $${\left\| 1_A - \Pi_\mathcal{B} 1_A\right\|}_{q, N} {\leqslant}\tfrac{1}{6} \delta^3 \Lambda_{q, N}({1_{[N]}}).$$ Setting $f := \Pi_\mathcal{B} 1_A$, a telescoping identity thus yields $${\left| \Lambda_{q, N}(f)  -\Lambda_{q, N}(\delta1_{[N]})\right|} {\geqslant}{\tfrac{1}{2}}  \delta^3 \Lambda_{q, N}({1_{[N]}}) \gg \delta^3.$$

Define the $\mathcal{B}$-measurable set $$S:= {\left\{x \in [N] : f(x) {\geqslant}(1+c)\delta\right\}},$$ where $c>0$ is a sufficiently small absolute constant that will be chosen to make the following argument valid. By Lemma \[L1 control\] and a telescoping identity, we have ${\left| \Lambda_{q, N}(f)  -\Lambda_{q, N}(f1_{S^c})\right|} {\leqslant}3|S|/N$, so that $$\tfrac{|S|}{N} + {\left| \Lambda_{q, N}(f1_{S^c})  -\Lambda_{q, N}(\delta1_{[N]})\right|} \gg  \delta^3 .$$ Yet another telescoping identity, in conjunction with Lemma \[L1 control\], gives $$\begin{aligned}
{\left| \Lambda_{q, N}(f1_{S^c})  -\Lambda_{q, N}(\delta 1_{[N]})\right|} & \ll \tfrac{\delta^2 }{N}
{\left\| f1_{S^c}- \delta1_{[N]}\right\|}_{\ell^1} {\leqslant}\tfrac{\delta^2 }{N} {\left\| f - \delta1_{[N]}\right\|}_{\ell^1} + \tfrac{|S|}{N},\end{aligned}$$ so that $$|S| + \delta^2 {\left\| f - \delta1_{[N]}\right\|}_{\ell^1}  \gg \delta^3 N.$$

Since 