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For $1<p\le q<\infty$ and $0\le \rho \le \sqrt{(p-1)/(q-1)}$, the noise operator on the Boolean hypercube satisfies $\Vert T_{\rho }f{\Vert }_q\le \Vert f{\Vert }_p$ for the uniform probability measure. In particular,Here $\hat{f}(S)$ are Fourier-Walsh transform coefficients. The endpoint $\rho =0$ follows by averaging. This hypercontractive inequality on the Boolean hypercube controls Fourier-Walsh transform coefficients of discrete derivatives of a Boolean function and is a principal ingredient in the Kahn-Kalai-Linial theorem.

One form of the hypercontractive inequality is $\Vert T_{1/\sqrt{3}}f{\Vert }_4\le \Vert f{\Vert }_2$, where $T_{\rho }$ is the noise operator on the Boolean hypercube.

Use normalized Fourier-Walsh transform coefficients $\hat{g}(\xi )=2^{-n}{\sum }_{x}g(x)(-1{)}^{x\cdot \xi }$. The Hamming weight $|\xi |$ counts the nonzero coordinates. Norms on the input cube use uniform probability, and a frequency integral here is counting over $\xi \in {\mathbb{F}}_2^n$. A common alternative normalization multiplies both sides of the requested comparison by the same factor and does not change it.

Beckner's inequality states that for $1<p\le q<\infty$ and $0\le \rho \le \sqrt{(p-1)/(q-1)}$, the noise operator on the Boolean hypercube, with coefficients $\widehat{T_{\rho }g}(\xi )={\rho }^{|\xi |}\hat{g}(\xi )$, satisfies $\Vert T_{\rho }g{\Vert }_q\le \Vert g{\Vert }_p$. In particular choose $q=2$, $p=4/3$, $\rho =1/\sqrt{3}$.

Let $g=1_A$ and $\alpha =|A|/2^n$. Then $\Vert g{\Vert }_{4/3}^2={\alpha }^{3/2}$. On frequencies of weight at most two, ${\rho }^{2|\xi |}\ge 1/9$, so Beckner's inequality and Parseval's identity give the low-degree Fourier mass of a sparse Boolean set estimateThe indicator function is real and the cube characters are real, so the displayed squares equal the squared absolute values. Total Fourier weight is $\sum \hat{g}(\xi {)}^2=\Vert g{\Vert }_2^2=\alpha$. If $2^n>2003$, then $0<\alpha \le 1/2003$ and $9\sqrt{\alpha }\le 9/\sqrt{2003}<1/2$, since $2003>18^2$. ThereforeThis establishes the strict comparison for every power $N=2^n>2003$, not merely for an unspecified very large threshold. The original PDF supplies the density condition and Beckner request, both missing from the converted TeX.

Work on the unbiased Boolean hypercube with values $\{-1,1\}$ for the coordinates and $\{0,1\}$ for $f$. Write its Fourier-Walsh transform asThe flip-probability influences satisfyThese factors of four are essential for zero-one $f$. If $v=0$, the conclusion is immediate. Otherwise suppose, for a contradiction, that $I<2bv$, where $b=\log (1/\beta )/3$. Take $\beta$ small enough that $b\ge 1$. The high-level Fourier weight obeysThis proves the first intermediate claim, including when $b$ is not an integer.

Set $\delta =e^{-1}$ and $r=2/(1+\delta )$. For the discrete derivative of a Boolean function, $D_{i}f$ has magnitude $1/2$ exactly on a set of measure ${\beta }_i$. Beckner's inequality, in the precise form $\Vert T_{\sqrt{\delta }}g{\Vert }_2\le \Vert g{\Vert }_{1+\delta }$ for the noise operator on the Boolean hypercube, yieldsFor $1\le s\le b$, concavity of $\log s-s+1$ shows that $s{\delta }^{s-1}$ is at least the smaller endpoint value $\min (1,b{\delta }^{b-1})$, and this is at least $b{\delta }^b$ since $b{e}^{-b}\le 1$. Therefore the low-level Fourier weight gives the second required intermediate bound:On the other hand, ${\beta }_i\le \beta$ and the assumed bound on $I$ giveCombining these inequalities would imply ${\beta }^{r-1}>{\delta }^b={\beta }^{1/3}$. But $r-1=(1-\delta )/(1+\delta )>1/3$ and $0<\beta <1$, so the inequality is impossible. ConsequentlyIn fact, this proof needs sufficiently small $\beta$ but no separate lower bound on $n$.

Use the uniform probability measure on the Boolean hypercube. The influence of a variable is the flip probabilityThe polynomial defining $f$ is interpreted over ${\mathbb{F}}_2$. Flipping the first coordinate adds $x_2+\cdots +x_n$ modulo two. This nonempty parity is balanced for $n\ge 2$, so

An explicit Tribes function works for every $n\ge 2$. Choose the largest integer $w\ge 1$ with $w{2}^w\le n$, let $m=\lfloor 2^w\log 2\rfloor$, and take the OR of $m$ disjoint ANDs of $w$ coordinates. Ignore the unused coordinates. Its zero probability is $(1-2^{-w}{)}^m$, between $1/4$ and $3/4$, and a used coordinate has influenceHere $n<2^{w+1}(w+1)$ by maximality. For the balance claim, $w=1$ gives probability $1/2$; for $w\ge 2$, $m\ge 2^w\log 2-1$ gives zero probability at most $e^{-\log 2+2^{-w}}<3/4$, while $(1-1/q{)}^q\ge 1/4$ for $q=2^w\ge 2$ gives a lower bound $(1/4{)}^{\log 2}>1/4$. Only the example was requested, but these checks fix its parameters unambiguously.

For the lower bound we prove the required consequence of Beckner's inequality with constants. Write the Fourier-Walsh transform as $f={\sum }_S\hat{f}(S){\chi }_S$, where ${\chi }_S(x)=(-1{)}^{{\sum }_{i\in S}{x}_i}$ and $\hat{f}(S)=\mathbb{E}(f{\chi }_S)$. The discrete derivative of a Boolean function is $D_{i}f=(f-f(\cdot +e_i))/2$. Since $f$ takes values zero and one, $D_{i}f$ takes values $0,\pm 1/2$. Consequently Parseval's identity givesThe factor four is required for the printed $0/1$ convention. Also $V:=\mathrm{Var}(f)={\sum }_{S\ne \varnothing }|\hat{f}(S){|}^2\ge 3/16$ under the balance hypothesis.

The true Beckner's inequality states $\Vert T_{\rho }h{\Vert }_q\le \Vert h{\Vert }_p$ for $1<p\le q<\infty$, $0\le \rho \le \sqrt{(p-1)/(q-1)}$, with $T_{\rho }h={\sum }_S{\rho }^{|S|}\hat{h}(S){\chi }_S$. Apply it with $p=3/2$, $q=2$, $\rho =1/\sqrt{2}$ to $h_i={\chi }_{\{i\}}{D}_{i}f$, a function independent of coordinate $i$. Its nonzero values still have magnitude $1/2$. We obtainLet $a={\max }_i{I}_i(f)>0$. For an integer $\ell \ge 1$, the low-degree Fourier weight satisfiesFor the high-degree part the total influence identity givesTogether these imply $4V\le na(2^{\ell -1}{a}^{1/3}+1/(\ell +1))$.

Suppose, seeking a contradiction, that $a<\log n/(100n)$, with natural logarithms. Then $a<1/(100e)$ and $\ell =\lfloor \log (1/a)/(6\log 2)\rfloor \ge 1$. Hence $2^{\ell -1}{a}^{1/3}\le a^{1/6}/2$. Put $y=\log n$. The first term on the right is bounded bythe middle expression is maximized at $y=7$. Moreover $\log (1/a)\ge y/2$, since $y\le 100e^{y/2}$. ThusWe would have $4V<1/10$, contradicting $4V\ge 3/4$. ThereforeThis derives the stated Kahn-Kalai-Linial theorem bound from the actual noise operator on the Boolean hypercube inequality, rather than using the desired conclusion as an input.

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 $\mathbf{I}(f)$.

Decompose $f={\sum }_{j=0}^n{f}^{(=j)}$ into its homogeneous Fourier levels. The Bonami lemma and the triangle inequality giveApply this estimate to the $m$-fold tensor power $f^{\otimes m}$. Tensor products multiply both relevant norms and commute with the noise operator on the Boolean hypercube, soTaking $m$th roots and the limit $m\to \infty$ proves the hypercontractive inequality on the Boolean hypercube

The noise operators are self-adjoint and satisfy $T_{\rho }{T}_{\sigma }=T_{\rho \sigma }$. By the duality of Lp spaces,Consequently