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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.

For a Boolean-valued $f$, each discrete derivative of a Boolean function $D_{i}f$ takes values in $\{-1,0,1\}$ and has degree at most $k-1$. If $f$ depends on coordinate $i$, then $D_{i}f$ is nonzero, so part (iii) givesSince $f$ has degree at most $k$, the Fourier formula for total influence and Parseval identity giveIf $m$ coordinates affect $f$, then $m{2}^{-(k-1)}\le \mathbf{I}(f)\le k$, so $m\le k{2}^{k-1}$. Thus $f$ is a $k{2}^{k-1}$-junta, which is the Nisan-Szegedy junta theorem.

Solved by gpt-5.6-sol high.

If $\alpha >1/2$, monotonicity already gives $\mathbb{E}{f}^{(1/2)}\ge \alpha >1/2$, so assume $\alpha \le 1/2$. Suppose for a contradiction that $\mathbb{E}{f}^{(1/2)}\le 1/2$. By the mean value theorem, some $s\in (p,1/2)$ satisfiesThe Margulis-Russo formula identifies this derivative with the appropriately normalized total influence, so ${\mathbf{I}}_s(f)$ is bounded solely in terms of $\zeta$. The $p$-biased Friedgut junta theorem then supplies, for any small $\eta >0$, a Boolean $J$-junta $h$ with $|J|\le r(\zeta ,\eta )$ and

Because $f$ is monotone, ${\mathbb{P}}_s(f=1)\le 1/2$, hence ${\mathbb{P}}_s(h=0)\ge 1/2-\eta \ge 1/4$ when $\eta \le 1/4$. It follows thatFor some assignment $u$ on $J$ with $h(u)=0$, therefore, ${\mathbb{E}}_s{f}_u\le 4\eta$. Monotonicity and $p<s$ imply ${\mathbb{E}}_p{f}_u\le 4\eta$. Choose $\eta <\alpha /5$, set $\epsilon =\eta$, and take $r\ge |J|$. Thencontradicting $(\epsilon ,p,r)$-quasirandomness. Thus $\mathbb{E}{f}^{(1/2)}>1/2$.

Solved by gpt-5.6-sol high.