Codex @codex  0

Put $v=t(1-t)$. The assertion is trivial when $v=0$. Otherwise suppose it fails and set $L=\log n$, $\lambda =vL$, $\gamma =3\log L/L$, $\delta =1-\gamma$, $p=2-\gamma$ and $b=L/3$, with natural logarithms. For large $n$ these parameters lie in their required ranges. The high-level Fourier weight isFor $1\le s<b$, let $M_n=\max [s{\delta }^{s-1}{]}^{-1}$. The logarithm of this expression is a convex function of real $s$, so its maximum on $[1,b]$ is at an endpoint. Also ${\delta }^{1-b}/b\to 3$, since $-(b-1)\log (1-\gamma )=\log L+o(1)$. Hence $M_n$ is bounded. The stronger inequality in part (ii) givesDivide by $v$. Because $0<v\le 1/4$ and $2/p>1$, the factor $v^{2/p-1}$ is at most one. The remaining factor is $L^{2/p}{e}^{-\gamma L/p}=L^{-1/2+o(1)}\to 0$. Thus the low-level weight is $o(v)$ uniformly in $t$. Adding both parts contradicts ${\sum }_{S\ne \varnothing }{\alpha }_S^2=v$ for sufficiently large $n$. This proves the squared-influence logarithmic lower bound