= Solution
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> gives
$$
\alpha_\varnothing=t,\qquad\sum_S\alpha_S^2=t,\qquad\sum_{S\ne\varnothing}\alpha_S^2=t(1-t).
$$
The flip-probability <influence> satisfies $\beta_i=4\sum_{S\ni i}\alpha_S^2$, and summing gives $\sum_{S\ne\varnothing}|S|\alpha_S^2=\frac14\sum_i\beta_i$.
For $p=1+\delta\in(1,2)$, <Beckner's inequality> says $\|T_{\sqrt\delta}g\|_2\leq\|g\|_p$. 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 $\beta_i$, so
$$
\sum_{S\ni i}\delta^{|S|-1}\alpha_S^2\leq\frac14\beta_i^{2/p}.
$$
Summing over $i$ gives the <hypercontractive weighted influence bound>
$$
\boxed{\sum_{S\ne\varnothing}|S|\delta^{|S|-1}\alpha_S^2\leq\frac14\sum_i\beta_i^{2/p}.}
$$
Back to article page
