%! TEX root = ABF.tex % vim: tw=80 ft=tex % 05/02/2026 12PM Recall that \[ x_A = \begin{cases} 0 & n \notin A \\ x_{A \setminus \{n\}} & n \in A \end{cases} \] Therefore, $D_n f$ has degree at most $k - 1$. First, \[ \|f\|_2^2 = \|g\|_2^2 + \|x_n h\|_2^2 = \|g\|_2^2 + \|h\|_2^2 .\] Also, \begin{align*} \|f\|_4^4 &= \Ebb_x (g(x) + x_n h(x))^4 \\ &= \Ebb_x ( g(x)^4 + 4 x_n g(x)^3 h(x) + 6 x_n^2 g(x)^2 h(x)^2 + 4 x_n^3 g(x) h(x)^3 + x_n^4 h(x)^4 ) \\ &= \|g\|_4^4 + 6 \Ebb_x g(x)^2 h(x)^2 + \|h\|_4^4 \end{align*} The last line uses the fact that $g$, $h$ don't depend on $x_n$ and that $x_n = \pm 1$ with probability $\half$. Then \begin{align*} \|f\|_4^4 &= \|g\|_4^4 + 6\Ebb_x g(x)^2 h(x)^2 + \|h\|_4^4 \\ &\le \|g\|_4^4 + 6\|g\|_4^2 \|h\|_4^2 + \|h\|_4^4 &&\text{(by Cauchy-Schwarz)} \\ &\le 3^{2k} \|g\|_2^4 + 6 \cdot 3^k \|g\|_2^2 3^{k - 1} \|h\|_2^2 + 3^{2(k - 1)} \|h\|_2^4 &&\text{(by inductive hypothesis)} \\ &\le 3^{2k} \left(\|g\|_2^4 + 2 \|g\|_2^2 \|h\|_2^2 + \frac{1}{9} \|h\|_2^4\right) \\ &\le 3^{2k} (\|g\|_2^2 + \|h\|_2^2)^2 \\ &\le 3^{2k} \|f\|_2^4 &&\qedhere \end{align*} \end{proof} \begin{fccoro}[Anticoncentration of low-degree functions] % Corollary 4.2 \label{coro:4.2} Let $f : \{-1, 1\}^n \to \Rbb$ have degree $\le k$ and let $\Ebb_x f(x) = \mu$ and $\Var f = \sigma^2$. Then \[ \Pbb\left[|f(x) - \mu| > \frac{\sigma}{2}\right] \ge \frac{9}{16} \cdot 3^{-2k} .\] \end{fccoro} \begin{proof} Let $g = f - \mu$. Then $\Ebb g = 0$ and $\|g\|_2^2 = \sigma^2$. Let $p = \Pbb \left[ |g(x)| > \frac{\sigma}{2} \right]$. Let $A = \left\{x : |g(x)| \le \frac{\sigma}{2}\right\}$, $B = A^c$. Then \[ \sigma^2 = \|g\|_2^2 \le (1 - p) \frac{\sigma^2}{4} + p \Ebb_{x \in B} g(x)^2 .\] Therefore, \[ p \Ebb_{x \in B} g(x)^2 \ge \frac{3}{4} \sigma^2 ,\] so \[ \Ebb_{x \in B} g(x)^2 \ge \frac{3 \sigma^2}{4p} .\] By Cauchy-Schwarz (or $\Var(g(x)^2) \ge 0$) we have \[ \Ebb_{x \in B} g(x)^4 \ge \frac{9\sigma^4}{16 p^2} ,\] so $\|g\|_4^4 \ge \frac{9\sigma^4}{16p}$. But \nameref{lemma:4.1} implies that $\|g\|_4^4 \le 3^{2k} \|g\|_2^4 = 3^{2k} \sigma^4$. Therefore, $p \ge \frac{9}{16} \cdot 3^{-2k}$, as stated. \end{proof} In this lecture, we are aiming towards the earlier stated stability version of \nameref{thm:3.1}. \begin{fclemma}[] % Lemma 4.3 \label{lemma:4.3} Let $g : \{-1, 1\}^n \to \Rbb$ be given by the formula $g(x) = \sum_{i = 1}^{n} a_i x_i$. Then \[ \Var g^2 = 2 \sum_{i \neq j} a_i^2 a_j^2 .\] \end{fclemma} \begin{remark*} Why should one care about studying $\Var(g(x)^2)$? Note that if $g$ is a Boolean function, then $g(x)^2 \equiv 1$, so $\Var(g(x)^2)$. So $\Var(g(x)^2)$ can be used to measure ``how close'' a function is to being Boolean (or a multiple of a Boolean function). \end{remark*} \begin{proof} $\Ebb g^2 = \sum_i a_i^2$ by \gls{parseval}, so \begin{align*} (\Ebb g^2)^2 &= \sum_i a_i^4 + \sum_{i \neq j} a_i^2 a_j^2 \\ \Ebb g^4 &= \sum_{i, j, k, l} a_i a_j a_k a_l \Ebb_x x_i x_j x_k x_l \\ &= \sum_i a_i^4 + 3 \sum_{i \neq j} a_i^2 a_j^2 \end{align*} Note that $\Ebb_x x_i x_j x_k x_l$ vanishes if any index appears an odd number of times, which is how we deduce the last equality above. Subtracting the two above equations gives the result. \end{proof} \begin{fclemma}[] % Lemma 4.4 \label{lemma:4.4} Let $g : \{-1, 1\}^n \to \Rbb$, $1 - \delta \le \|g\|_2^2 \le 1$, $\Var g^2 \le C \delta$, $g$ linear with no constant term. Then there exists $i$ such that $\ft{g}(i)^2 \ge 1 - \left( 2 + \frac{C}{2} \right) \delta$. \end{fclemma} \begin{proof} Let $g(x) = \sum_{i} a_i x_i$. Then $1 - \delta \le \sum_i a_i^2 \le 1$, $2 \sum_{i \neq j} a_i^2 a_j^2 \le C \delta$. Therefore, \[ \sum_i a_i^4 = \left( \sum_i a_i^2 \right)^2 - \sum_{i \neq j} a_i^2 a_j^2 \ge 1 - 2\delta - \frac{C}{2} \delta = 1 - \left( 2 + \frac{C}{2} \right) \delta .\] Also, \[ \sum_i a_i^4 \le \max_i a_i^2 \left( \sum_i a_i^2 \right) \le \max a_i^2 = \max \ft{g}(i)^2 . \qedhere \] \end{proof} \begin{fcthm}[Friedgut, Kalai, Naor] % Theorem 4.5 \label{thm:4.5} Let $f : \{-1, 1\}^n \to \{-1, 1\}$ be a function with $\W_1(f) \ge 1 - \delta$. Then there exists $i$ such that $\ft{f}(i)^2 \ge 1 - O(\delta)$. \end{fcthm} \begin{proof} Let $G$ be the degree-$1$ part of $f$, i.e. $g = \sum_{i} \ft{f}(i) x_i$. Then by hypothesis, $1 - \delta \le \|g\|_2^2$, and by \gls{parseval} $\|g\|_2^2 \le \|g\|_2^2 = 1$. Let $\sigma = \Var g^2$. Note that $g^2$ has degree at most $2$. By the \nameref{coro:4.2}, \[ \Pbb \left[ |g^2 - \|g\|_2^2| > \frac{\sigma}{2} \right] \ge \frac{9}{16} \cdot \frac{1}{81} = \frac{1}{9 \times 16} .\] Since $\|g\|_2^2 \in [1 - \delta, 1]$, it follows that \[ \Pbb \left[ |g^2 - 1| > \frac{\sigma}{2} - \delta \right] \ge \frac{1}{9 \times 16} .\] Also, $1 = f^2$. Therefore, \[ \Ebb_x |g(x)^2 - f(x)^2| \ge \frac{1}{9 \times 16} \cdot \left( \frac{\sigma}{2} - \delta \right) .\] But \[ \Ebb_x |g(x)^2 - f(x)^2| = \Ebb_x |g(x) + f(x)||g(x) - f(x)| \le \|f + g\|_2 \|f - g\|_2 \le 2\sqrt{\delta} \] by Cauchy-Schwarz. Therefore, $\frac{\sigma}{2} - \delta \le 32 \times 9 \sqrt{\delta}$, so $\sigma \le 64 \times 9 \sqrt{\delta} + 2 \delta \le 600\sqrt{\delta}$, so $\sigma^2 \le 360000 \delta$. By \cref{lemma:4.4}, there exists $i$ such that $\ft{g}(i)^2 \ge 1 - 180002\delta = 1 - O(\delta)$. But $\ft{g}(i)^2 = \ft{f}(i)^2$, so we are done. \end{proof} Let $p = \Pbb[f(x) = x_i]$. Then $\ft{f}(i) = \Ebb f(x) x_i = p - (1 - p) = 2p - 1$. So if $\ft{f}(i) = 1 - O(\delta)$, then $p = O(\delta)$ or $1 - p = O(\delta)$. So $f$ is well-approximated by $x_i$ or by $-x_i$. This completes the proof of the stability version of \nameref{thm:3.1}. Next time: $\|\noise_{\frac{1}{\sqrt{3}}} f\|_4 \le \|f\|_2$.