Kahn-Kalai-Linial theorem (${\max }_i{I}_i(f)\ge c\mathrm{Var}(f)\log (n)/n$)

For a Boolean function $f:\{0,1{\}}^n\to \{0,1\}$ under the uniform probability measure, some coordinate has influence of a variable at least $c\mathrm{Var}(f)\log (n)/n$, where $c>0$ is an absolute constant and $I_i(f)$ is the probability that flipping coordinate $i$ changes $f$. For variance bounded below, this is order $\log (n)/n$. The tribes function shows that this order cannot be improved for general balanced Boolean functions.

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

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 $v$.