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Write $q=1-p$, $\mu =q-p$, $\sigma =2\sqrt{pq}$, and ${\varphi }_S(x)={\prod }_{j\in S}(x_j-\mu )/\sigma$. The functions ${\varphi }_S$ form the p-biased product measure orthonormal basis, so the Fourier expansion is $f={\sum }_S{\hat{f}}_p(S){\varphi }_S$. The normalized discrete derivative of a Boolean function satisfiesApplying Parseval identity and then exchanging two finite sums gives

The noise operator on the Boolean hypercube acts diagonally on the same basis: $T_{\rho }{\varphi }_S={\rho }^{|S|}{\varphi }_S$. Hence the noise stability isIts derivative isTaking the right-hand value at $\rho =0$ leaves exactly the linear Fourier weight ${\sum }_i{\hat{f}}_p(\{i\}{)}^2$, while taking the left-hand value at $\rho =1$ gives $\mathbf{I}(f)$.

Solved by gpt-5.6-sol high.

The Friedgut junta inequality says that if $f:\{-1,1{\}}^n\to \{-1,1\}$ andthen there is a real-valued $J$-junta $g$ such that

To prove it, put $J=\{i:{\mathrm{Inf}}_i(f)\ge \tau \}$. Part (i), applied to each discrete derivative of a Boolean function $D_{i}f$, givesOn the other hand, expanding the noise stability in Fourier coefficients givesChoose $\tau ={\epsilon }^2/(3^{2k}\mathbf{I}(f{)}^2)$ and defineThe preceding bounds make the low-degree Fourier mass omitted by $g$ at most $\epsilon$, while the hypothesis makes the high-degree mass at most $\epsilon$. Thus $\Vert f-g{\Vert }_2^2\le 2\epsilon$. Finally,which gives the asserted bound on $|J|$.

Solved by gpt-5.6-sol high.