%! TEX root = ABF.tex % vim: tw=80 ft=tex % 24/02/2026 12PM \glssymboldefn{pbiasedEi}% Let $f^{(p)} : \{-1, 1\} \to \Rbb$. Then for each $i \in [n]$, $E_i f$ is defined by \[ E_i f(x) = q f(x_{i \subst 1}) + p f(x_{i \subst -1}) .\] \begin{fclemma}[] % Lemma 6.7 \label{lemma:6.7} For every $f : \{-1, 1\}^n \to \Rbb$ and every $i \in [n]$, $f = E_i f + \pphi_i \pD_i f$, and $E_i f$ and $\pphi_i \pD_i f$ are orthogonal. \end{fclemma} \begin{proof} Reminders: \begin{align*} \mu &= q - p = 2q - 1 = 1 - 2p \\ \sigma &= 2\sqrt{pq} \end{align*} So \[ f(x) - \pE_i f(x) = \begin{cases} p(f(x_{i \subst 1}) - f(x_{i \subst -1})) & x_i = 1 \\ - q(f(x_{i \subst 1}) - f(x_{i \subst -1})) & x_i = -1 \end{cases} \] Also, \[ \pphi_i(x) = \begin{cases} \frac{2p}{\sigma} & x_i = 1 \\ -\frac{2q}{\sigma} & x_i = -1 \end{cases} \] Therefore, \[ f(x) - \pE_i f(x) = \frac{\sigma}{2} \pphi_i(x)(f(x_{i \subst 1}) - f(x_{i \subst -1})) = \pphi_i \pD_i f(x) .\] Orthogonality is easy ($\pE_i f$ and $\pD_i f$ don't depend on $x_i$, and then use the fact that $\pphi_i$ has average $0$). \end{proof} \begin{fclemma}[$p$-biased Bonami Lemma] % Lemma 6.8 \label{lemma:6.8} Let $f^{(p)} : \{-1, 1\}^n \to \Rbb$ have degree at most $k$. Then $\pnorm \|f\|_4 \le C^{k / 2} \pnorm \|f\|_2$, where $C = \frac{4}{\sigma^2}$. \end{fclemma} \begin{proof} Induction on $n$. Let $g = \pE_n f$, $h = \pD_n f$. By orthogonality, \[ \pnorm \|f\|_2^2 = \pnorm \|g\|_2^2 + \pnorm \|\pphi_n h\|_2^2 = \pnorm \|g\|_2^2 + \pnorm \|h\|_2^2 .\] Note also that $\pD_n f$ has degree at most $k - 1$. So \begin{align*} \pnorm \|f\|_4^4 &= \Ebb_x \left( g(x)^4 + 4 \pphi_n(x) g(x)^3 h(x) + 6 \pphi_n (x)^2 g(x)^2 h(x)^2 + 4 \pphi_n(x)^2 g(x) h(x)^3 + \pphi_n(x)^4 h(x)^4 \right) \\ &\le \pnorm \|g\|_4^4 + 6\pnorm \|g\|_4^2 \pnorm \|h\|_4^2 + 4 |\Ebb \pphi^3| \pnorm \|g\|_4 \pnorm \|h\|_4^3 + \Ebb \pphi^4 \pnorm \|h\|_4^4 \end{align*} Using that $\pphi(1) = \sqrt{\frac{p}{q}}$, $\pphi(-1) = -\sqrt{\frac{q}{p}}$, we get \[ \Ebb \pphi^3 = q \left( \frac{p}{q} \right)^{3 / 2} - p \left( \frac{q}{p} \right)^{3 / 2} = \frac{p^2 - q^2}{\sqrt{pq}} = \frac{2(p - q)}{2 \sqrt{pq}} ,\] so $|\Ebb \pphi^3| \le \frac{2}{\sigma}$. Also, \[ \Ebb \pphi^4 = q \frac{p^2}{q^2} + p \frac{q^2}{p^2} = \frac{p^3 + q^3}{pq} \le \frac{4}{\sigma^2} .\] So \[ \pnorm \|f\|_4^4 \le C^{2k} \left( \pnorm \|g\|_2^4 + 6 C^{-1} \pnorm \|g\|_2^4 \pnorm \|h\|_2^2 + \frac{8}{\sigma} C^{-3 / 2} \pnorm \|g\|_2 \pnorm \|h\|_2^3 + \frac{4}{\sigma^2} C^{-2} \pnorm \|h\|_2^4 \right) .\] Apply $ab \le \frac{a^2 + b^2}{2}$ with $a = 2 \pnorm \|g\|_2 \pnorm \|h\|_2 C^{-\half}$, $b = \frac{4}{\sigma} C^{-1} \pnorm \|h\|_2^2$. Then \[ \pnorm \|f\|_4^4 \le C^{2k} \left( \pnorm \|g\|_2^4 + 8C^{-1} \pnorm \|g\|_2^2 \pnorm \|h\|_2^2 + \frac{12}{\sigma^2} C^{-2} \pnorm \|h\|_2^4 \right) .\] Choose $C$ such that $8C^{-1} \le 2$ and $\frac{12}{\sigma^2} C^{-2} \le 1$. $C = \frac{4}{\sigma^2}$ will do. So $\pnorm \|f\|_4^4 \le C^{2k} (\pnorm \|g\|_2^2 + \pnorm \|h\|_2^2)^2 = C^{2k} \pnorm \|f\|_2^4$. \end{proof} \begin{remark*} This proof does not recover the $p = \half$ case that we saw before (\nameref{lemma:4.1}): in \cref{lemma:4.1} we proved \cref{lemma:6.8} for $p = \half$ but with $C = 3$, whereas in the above proof we only get $C = 4$ in the case $p = \half$. \end{remark*} \begin{fccoro}[] % Corollary 6.9 \label{coro:6.9} Let $\rho = \frac{\sigma}{2}$. Then for every $f : \{-1, 1\}^n \to \Rbb$ we have $\pnorm \|\pnoise_\rho f\|_4 \le \pnorm \|f\|_2$. \end{fccoro} \begin{proof} \begin{align*} \pnorm \|\pnoise_\rho f\|_4 &\le \sum_{k = 0}^{n} \pnorm \|\pnoise_\rho f^{\degkpart}\|_4 \\ &\le \sum_{k = }^{n} \rho^k \pnorm \|f^{\degkpart}\|_4 \\ &\le \sum_{k = 0}^{n} \rho^k C^{k / 2} \|f^{\degkpart}\|_2 \\ &= \sum_{k = 0}^{n} \|f^{\degkpart}\|_2 \\ &\le \sqrt{n} \|f\|_2 \end{align*} By the tensor power trick, the result follows. \end{proof} \begin{fccoro}[] % Corollary 6.10 \label{coro:6.10} For every $f : \{-1, 1\}^n \to \Rbb$ with $\rho = \frac{\sigma}{2}$, we also have $\pnorm \|\pnoise_\rho f\|_2 \le \pnorm \|f\|_{4 / 3}$. \end{fccoro} \begin{proof} Identical to uniform case, but with a different $\rho$. \end{proof} \begin{remark*} As before, this gives us that $\Stab_{\rho^2} f \le \pnorm \|f\|_{4 / 3}^2$, i.e. $\Stab_{\frac{\sigma^2}{4}} \le \|f\|_{4 / 3}^2$. \end{remark*} \begin{fcthm}[$p$-biased Friedgut junta theorem] % Theorem 6.11 \label{thm:6.11} Let $f^{(p)} : \{-1, 1\}^n \to \{-1, 1\}$ be a Boolean function and suppose that $\|f^{\deglekpart}\|_2^2 \ge 1 - \eps$. Then there exists an \kjunta{m} $g : \{-1, 1\}^n \to \Rbb$ with $\|g - f\|_2^2 \le 2\eps$ and $m \le \frac{\rho^{-2k} \totinf(f)^3}{\eps^2 \sigma^2}$. \end{fcthm} \begin{proof} Let $\tau > 0$ (to be chosen later) and let \[ J = \{i : \pInf_i f \ge \tau\} .\] Let $\rho = \frac{\sigma^2}{4}$. Then \begin{align*} \sum_{i \notin J} \pstab_\rho (\pD_i f) &\le \sum_{i \notin J} \|\pD_i f\|_{4 / 3}^2 \\ &= \sum_{i \notin J} \|\pD_i f\|_2^2 \sigma^{-2 / 3} \\ &= \sum_{i \notin J} (\pInf_i f)^{3 / 2} \sigma^{-1} \\ &\le \sigma^{-1} \tau^{1 / 2} \totinf(f) \end{align*} Let $g = \sum_{\substack{A \subset J \\ |A| \le k}} \ft{f}(A) \pphi(A)$. Then \[ \|f - g\|_2^2 \le \sum_{\substack{B \not\subset J \\ |B| \le k}} \ft{f}(B)^2 + \sum_{|B| > k} \ft{f}(B)^2 .\] By hypothesis, the second term is at most $\eps$. But \[ \sum_{i \notin J} \Stab_\rho (\pD_i f) = \rho^{-1} \sum_B |B \setminus J| \rho^{|B|} \ft{f}(B)^2 \ge \rho^{-1} \sum_{\substack{B \not\subset J \\ |B| \le k}} \rho^k f(B)^2 .\] Therefore, \[ \sum_{\substack{B \not\subset J \\ |B| \le k}} f(B)^2 \le \rho^{-(k - 1)} \sigma^{-1} \tau^{\half} \totinf(f) .\] Set $\tau = \frac{\eps^2 \sigma^2 \rho^{2k}}{\totinf(f)^2}$. Then we get the bound. \end{proof}