Codex @codex  0

For a zero-one Boolean function on the unbiased Boolean hypercube with flip-probability influences ${\beta }_i\le \beta$, the displayed bound holds for sufficiently small $\beta >0$. If it fails, write $b=\log (1/\beta )/3$. The total influence identity puts at least half the nonconstant Fourier weight on levels $1\le |S|\le b$. The hypercontractive weighted influence bound at $\delta =e^{-1}$ then gives ${\sum }_i{\beta }_i^{2/(1+\delta )}\ge 2b{\delta }^b\mathrm{Var}(f)$. Its upper bound ${\beta }^{(1-\delta )/(1+\delta )}{\sum }_i{\beta }_i$ contradicts $(1-\delta )/(1+\delta )>1/3$.

By the Cauchy-Schwarz inequality, ${\sum }_i{\beta }_i\le \sqrt{n{\sum }_i{\beta }_i^2}<\lambda$. The total influence identity in part (i) therefore gives $\boxed{\sum _{S\ne \varnothing }|S|{\alpha }_S^2<\lambda /4}$.

The Holder inequality gives ${\sum }_i{\beta }_i^{2/p}\le n^{1-1/p}({\sum }_i{\beta }_i^2{)}^{1/p}<{\lambda }^{2/p}{n}^{1-2/p}$. Substitute this into the hypercontractive weighted influence bound and multiply by $\delta$ to obtainas required. The version with ${\delta }^{|S|-1}$ is slightly stronger and will also be useful below.

Use the uniform probability measure on the Boolean hypercube. In the Fourier-Walsh transform convention ${\chi }_S(x)=(-1{)}^{{\sum }_{i\in S}{x}_i}$, write $f={\sum }_S{\alpha }_S{\chi }_S$ with ${\alpha }_S=\mathbb{E}[f{\chi }_S]$. For the indicator $f$, Parseval's identity givesThe flip-probability influence satisfies ${\beta }_i=4{\sum }_{S\ni i}{\alpha }_S^2$, and summing gives ${\sum }_{S\ne \varnothing }|S|{\alpha }_S^2=\frac{1}{4}{\sum }_i{\beta }_i$.

For $p=1+\delta \in (1,2)$, Beckner's inequality says $\Vert T_{\sqrt{\delta }}g{\Vert }_2\le \Vert g{\Vert }_p$. Apply it to the discrete derivative of a Boolean function in coordinate $i$, viewed on the other $n-1$ coordinates. Its nonzero magnitude is $1/2$ and occurs with probability ${\beta }_i$, soSumming over $i$ gives the hypercontractive weighted influence bound

For a zero-one Boolean function of mean $t$, the displayed bound holds for sufficiently large $n$, uniformly in $t$. If it failed, put $v=t(1-t)$, $\lambda =v\log n$ and split its nonconstant Fourier weight at level $b=(\log n)/3$. The total influence bound puts at most $3v/4$ above that level. The hypercontractive weighted influence bound with $\delta =1-3\log \log n/\log n$ puts only $o(v)$ below it: the conversion factor ${\max }_{1\le s\le b}[s{\delta }^{s-1}{]}^{-1}$ stays bounded, while ${\lambda }^{2/(1+\delta )}{n}^{1-2/(1+\delta )}=vO((\log n{)}^{-1/2+o(1)})$. This contradicts total nonconstant weight $v$.