 Maximum-influence logarithmic lower bound

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 $(1-\delta )/(1+\delta )>1/3$.