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  Then
  \[
    \|f - g\|_2^2
    \le \sum_{\substack{A \not\subset J \\ |A| \le k}} \ft{f}(A)^2
    + \sum_{|A| > k} \ft{f}(A)^2
  .\]
  By hypothesis,
  \[
    \sum_{|A| > k} \ft{f}(A)^2
    = \|f^{\deggtkpart}\|_2^2
    \le c
  .\]
  But
  \[
    3 \sum_{B \not\subset J} |B \setminus J| \left( \third \right)^{|B|}
    \ft{f}(B)^2
    \ge 3 \sum_{\substack{B \not\subset J \\ |B| \le k}} \left( \third \right)^k
    \ft{f}(B)^2
    = 3^{-(k - 1)} \sum_{\substack{A \not\subset J \\ |A| \le k}} \ft{f}(A)^2
  .\]
  So
  \[
    \sum_{\substack{A \not\subset J \\ |A| \le k}} \ft{f}(A)^2
    \le 3^{k - 1} \tau^{\half} \totinf(f)
  .\]
  If we choose $\tau = \frac{\eps^2}{3^{2k} \totinf(f)}$, then this is at most
  $\eps$, so $\|f - g\|_2^2 \le 2 \eps$.
  But $|J| \le \frac{\totinf(f)}{\tau} = \frac{3^{2k} \totinf(f)^3}{\eps^2}$.
\end{proof}

\begin{fccoro}[Friedgut's junta theorem]
  % Corollary 5.5
  \label{coro:5.5}
  Let $f : \{-1, 1\}^n \to \{-1, 1\}$ and let $\eps > 0$.
  Then there exists an \kjunta{m} $g : \{-1, 1\}^n \to \Rbb$ with $\|f -
  g\|_2^2 \le 2\eps$ and
  \[
    m
    = \exp \left( O \left( \frac{\totinf(f)}{\eps} \right) \right)
  .\]
\end{fccoro}

\begin{proof}
  \begin{align*}
    \totinf(f)
    &= \sum_A |A| \ft{f}(A)^2 \\
    &\ge \sum_{|A| > k} |A| \ft{f}(A)^2 \\
    &\ge k \|f^{\deggtkpart}\|_2^2
  \end{align*}
  So if $k \ge \frac{\totinf(f)}{\eps}$, then $\|f^{\deglekpart}\|_2^2 \ge 1 -
  \eps$.
  By \cref{thm:5.4} we get a \gls{junta} with $m \le 3^{2
  \frac{\totinf(f)}{\eps}} \cdot \frac{\totinf(f)^3}{\eps^2} = \exp \left( O
  \left( \frac{\totinf(f)}{\eps} \right) \right)$.
\end{proof}

\begin{remark*}
  Let $f : \{-1, 1\}^n \to \{-1, 1\}$, $g : \{-1, 1\}^n \to \Rbb$ and suppose
  that $\|f - g\|_2^2 \le \eps$.
  Let
  \[
    h(x)
    = \begin{cases}
      1
      & g(x) \ge 0 \\
      -1
      & g(x) < 0
    \end{cases}
  \]
  Then if $h(x) \neq f(x)$, then $g(x)$ has a different sign from $f(x)$, so
  $|g(x) - f(x)|^2 \ge 1$.
  Since $\Ebb |f(x) - g(x)|^2 \le \eps$, we have $\Pbb[f(x) \neq h(x)] \le
  \eps$.

  In other words, we can find a Boolean function that approximates $f$,
  \emph{and} if $g$ is a \gls{Jjunta}, then so is $h$.
\end{remark*}

\newpage
\section{Analysis on the $p$-biased cube}

\glssymboldefn{mup}%
Let $p \in [0, 1]$.
We define a measure $\mu_p$ on $\{-1, 1\}^n$ by picking each $x_i$ independently
to be $1$ with probability $q$ and $-1$ with probability $p$ where $q = 1 - p$.
This convention looks counterintuitive at first (placing $p$ for the probability
of the smaller value), but recall that under the correspondence $\{-1, 1\}^n
\leftrightarrow \{0, 1\}^n$, we have $1 \leftrightarrow 0$ and $-1
\leftrightarrow 1$.
So this convention means that we ``take the identity most of the time, and take
the non-identity with probability $p$''.

Consider the random variable $x_i$ (where $x \sim \mu_p$).
Then
\[
  \mu
  = \Ebb x_i
  = q - p
  = 2q - 1
  = 1 - 2p
\]
and
\[
  \sigma^2
  = \Ebb (x_i - \mu)^2
  = q(1 - \mu)^2 + p(-1 - \mu)^2
  = q(2p)^2 + p(2q)^2
  = 4pq
,\]
so $\sigma = 2\sqrt{pq}$.

\glssymboldefn{pphi}%
Let $\phi_i = \frac{x_i - \mu}{\sigma}$.
We write $\phi(t)$ for $\frac{t - \mu}{\sigma}$, so $\phi_i = \phi(x_i)$.

Let $\phi_A = \prod_{i \in A} \phi_i$.
The $\phi_A$ will play the role of the $x_A$ in the unbiased case.

\glssymboldefn{pbiasedip}%
Fix $p$.
Then given $f, g : \{-1, 1\}^n \to \Rbb$, we define
\[
  \langle f, g \rangle
  = \Ebb_{x \sim \mu_p} f(x) g(x)
\]
and
\[
  \|f\|_r
  = (\Ebb_{x \sim \mu_p} |f(x)|^r)^{\frac{1}{r}}
.\]

\begin{fclemma}[]
  % Lemma 6.1
  \label{lemma:6.1}
  $\pip\langle \pphi_A, \pphi_B \rangle = \delta_{AB}$ for every $A, B \subset
  [n]$.
\end{fclemma}

\begin{proof}
  \[
    \pip \langle \pphi_A, \pphi_B \rangle
    = \Ebb_{x \sim \mup} \prod_{i \in A} \pphi_i(x) \prod_{i \in B} \pphi_i(x)
    = \Ebb_{x \sim \mup} \prod_{i \in A \symdiff B} \pphi_i(x) \prod_{i \in A
    \cap B} \pphi_i(x)^2
    = \delta_{AB}
    .
    \qedhere
  \]
\end{proof}

\begin{fcdefn}[$p$-biased Fourier coefficient]
  \glssymboldefn{pft}%
  Let $f : \{-1, 1\}^n \to \Rbb$.
  The \emph{$p$-biased Fourier coefficient} $\pft{f}(A)$ is defined by $\pft{f}(A)
  = \pip \langle f, \pphi_A \rangle = \Ebb_{x \sim \mup} f(x) \pphi_A(x)$.
\end{fcdefn}

Because the $\pphi_A$ form an orthonormal basis, it follows that
\begin{align*}
  \pip \langle f, g \rangle
  &= \sum_A \pft{f}(A) \pft{g}(A)
  &&\text{(Plancherel)} \\
  f
  &= \sum_A \pft{f}(A) \pphi_A
  &&\text{(inversion formula)}
\end{align*}

\begin{fcdefn}[]
  \glssymboldefn{pbiasedDi}%
  \glssymboldefn{pbiasedInf}%
  Let $f : \{-1, 1\}^n \to \Rbb$.
  Then define $D_i f$ by
  \[
    D_i f(x)
    = \frac{\sigma}{2} (f(x_{i \subst 1}) - f(x_{i \subst -1}))
  .\]
  Then define $\Infinternal_i f$ to be $\Ebb_{x \sim \mup} D_i f(x)^2 = \|D_i
  f\|_2^2$.
\end{fcdefn}

Now,
\[
  \pD_i \pphi_A(x)
  = \frac{\sigma}{2} (\pphi_A(x_{i \subst 1}) - \pphi_A(x_{i \subst -1}))
.\]
If $i \notin A$ then this is $0$.
Otherwise, it is
\[
  \frac{\sigma}{2} \pphi_{A \setminus \{1\}} (x) (\pphi(1) - \pphi(-1))
  = \frac{\sigma}{2} \pphi_{A \setminus \{1\}} \left( \frac{1 - \mu}{\sigma} -
  \frac{-1 - \mu}{\sigma} \right)
  = \pphi_{A \setminus \{1\}}(x)
.\]

It follows that
\[
  \pD_i f
  = \pD_i \left( \sum_A \pft{f}(A) \pphi_A \right)
  = \sum_{A \ni i} \pft{f}(A) \pphi_{A \setminus \{1\}}
,\]
and therefore that
\[
  \|\pD_i f\|_2^2
  = \sum_{A \ni i} \pft{f}(A)^2
,\]
and therefore that
\[
  \totinf(f)
  = \sum_i \pInf_i f
  = \sum_A |A| \pft{f}(A)^2
.\]