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Analysis of Boolean functions studies real-valued functions on a Boolean hypercube using Fourier analysis, probability theory, and combinatorics.

If a nonzero function on the Boolean hypercube has degree at most $k$, then its support has measure at least $2^{-k}$. This sharp bound follows by induction on the dimension.

On the Boolean hypercube $\{-1,1{\}}^d$ with uniform measure, each coordinate is a Rademacher random variable. Products of these sign coordinate functions are Walsh characters. This use of Bernoulli function is distinct from the fluid-mechanical Bernoulli function.

Among families of a given size in the Boolean hypercube, a binary initial segment maximizes the number of contained two-dimensional cube faces. Under section compression in binary order, faces crossing the selected coordinate correspond to internal edges in the intersection of its two sections. The edge extremal theorem and mathematical induction control that contribution and the faces within the sections.

A Boolean function is a function on a Boolean hypercube, often with codomain $\{-1,1\}$ or $\{0,1\}$.

In the seven-dimensional Boolean hypercube, $C(63)$ contains every set of size at most three except $\{5,6,7\}$. Its external vertex boundary consists of that missing triple and all 35 four-sets. The complementary-size initial segment $C(65)$ contains every set of size at most three and $\{1,2,3,4\}$. Its external vertex boundary has the other 34 four-sets and three five-sets. Thus complementary family sizes do not impose a symmetric boundary profile.

Equip the Boolean hypercube with uniform probability measure and graph distance divided by its graph diameter. These spaces form a Lévy family of graphs. The vertex-isoperimetric inequality in the discrete cube reduces the complement of a positive-density set's radius-$\epsilon n$ graph neighbourhood to a binomial distribution tail. For fixed density $\delta >0$, Chebyshev's inequality puts a lower density quantile at least $n/2-C_{\delta }\sqrt{n}-O(1)$. After enlarging by $\epsilon n$, the omitted mass is at most $n/[4(\epsilon n-C_{\delta }\sqrt{n}-O(1){)}^2]$, which tends to zero.

A face is obtained by fixing some coordinates of the Boolean hypercube and letting the remaining $d$ coordinates vary freely. It is isomorphic to $Q_d$. A two-dimensional face is a square with four vertices, and a face is contained in a set family when all its vertices belong to that family.

For $n\ge 1$, a family of $2^{n-1}$ vertices in the Boolean hypercube has these minimum and maximum external vertex boundary sizes. The Harper theorem gives the minimum: in odd dimension take the lower half of the levels, and in even dimension add the middle-level sets containing coordinate one to all lower levels. The family of even Hamming weight attains the maximum, since every remaining vertex has an adjacent even-weight vertex.

For the $\{-1,1\}$-valued convention on the unbiased Boolean hypercube, the influence of coordinate $i$ is ${\mathrm{Inf}}_i(f)=\Vert D_{i}f{\Vert }_2^2$, where $D_{i}f$ is the discrete derivative of a Boolean function. It equals the probability that flipping coordinate $i$ changes $f$. For the $\{0,1\}$-valued convention, the same flip probability is instead $4\Vert D_{i}f{\Vert }_2^2$, since a nonzero discrete derivative has magnitude $1/2$.

For a subset of the unbiased Boolean hypercube of density $\alpha$, use normalized Fourier-Walsh transform coefficients. Applying Beckner's inequality with $\rho =1/\sqrt{3}$ gives $\Vert T_{\rho }{1}_A{\Vert }_2^2\le \Vert 1_A{\Vert }_{4/3}^2={\alpha }^{3/2}$. On levels at most two the weights ${\rho }^{2|S|}$ are at least $1/9$, giving the displayed bound. Parseval's identity gives total Fourier mass $\alpha$, so when $9\sqrt{\alpha }<1/2$ the higher levels contain strictly more than the lower levels.

For a zero-one Boolean function on the unbiased Boolean hypercube with flip-probability influences ${\beta }_i\le \beta$, the displayed bound holds for sufficiently small $\beta >0$. If it fails, write $b=\log (1/\beta )/3$. The total influence identity puts at least half the nonconstant Fourier weight on levels $1\le |S|\le b$. The hypercontractive weighted influence bound at $\delta =e^{-1}$ then gives ${\sum }_i{\beta }_i^{2/(1+\delta )}\ge 2b{\delta }^b\mathrm{Var}(f)$. Its upper bound ${\beta }^{(1-\delta )/(1+\delta )}{\sum }_i{\beta }_i$ contradicts $(1-\delta )/(1+\delta )>1/3$.

Put $m=p^a$, with $p$ a prime number, and $s=m-1$. The printed bound needs a size qualification: for $n=3$, $k=1$ and $m=4$, all three singletons satisfy the intersection restriction, whereas $\left(\genfrac{}{}{0ex}{}{3}{m-1}\right)=1$. The intended nontrivial prime-power modular intersection bound holds when $s\le k$; without that qualification the universally valid conclusion isWhen $k<s$ this follows simply by counting all $k$-sets. We prove the sharper intended form for $s\le k$.

Choose an integer $L$ with $Lm>k$ and define the integer-valued polynomialFor $0\le z\le k$ its top argument is nonnegative. The bottom argument has all its first $a$ base-$p$ digits equal to $p-1$. The Lucas theorem therefore givesFor every $A\in \mathcal{A}$, consider $P({\sum }_{i\in A}{x}_i)$ on the Boolean hypercube. Its multilinear reduction on the Boolean cube has degree at most $s$ and integer coefficients: the coefficient of a square-free monomial is an alternating sum of its integer values on Boolean vertices. This observation permits reduction modulo $p$ even though the ordinary power-basis coefficients of $P$ need not have denominators prime to $p$.

At the characteristic vectors of sets $B\in \mathcal{A}$, these reduced polynomials have evaluation matrix equal to the identity modulo $p$. Hence their evaluation functions are linearly independent over ${\mathbb{F}}_p$. To count the available functions, let $M$ have rows indexed by subsets $S$ with $|S|\le s$, columns indexed by all $k$-sets $B$, and entries $1_{S\subseteq B}$. Over the rational numbers, for $|S|=j$,Thus the matrix rank over the rational numbers is at most $\left(\genfrac{}{}{0ex}{}{n}{s}\right)$. Every larger integer minor is zero and remains zero modulo $p$, so the same rank bound holds over ${\mathbb{F}}_p$. This low-degree evaluation rank on a uniform layer proves

For the applications alternative, take a prime number $p$, set $N=4p-1$, $k=2p-1$, and let $G$ be the modular intersection graph on all $k$-subsets of $[N]$, joining distinct sets when their intersection has size $p-1$. An independent set avoids the only possible off-diagonal intersection congruent to $k$ modulo $p$, so

First embed $A$ as $v_A=1_A/\sqrt{2p}$ in Euclidean space. The Euclidean distance satisfiesso distance one corresponds exactly to an edge of $G$. Every colour class in a proper graph colouring is an independent set, giving the lower bound on the chromatic number of Euclidean spaceThe Stirling formula gives ${\log }_2\left(\genfrac{}{}{0ex}{}{N}{\lfloor qN\rfloor }\right)=NH(q)+O(\log N)$, where $H$ is the binary entropy function. Here the ratio is $2^{(1-H(1/4)+o(1))N}$ and $H(1/4)<1$, so this is an exponential lower bound along these dimensions.

Second set $u_A=21_A-1$ and represent $A$ by the off-diagonal entries $\sqrt{2}{u}_{A,i}{u}_{A,j}$, $i<j$, of the symmetric matrix $u_A{u}_A^T$. This is an injective representation: equality of the matrices would give $u_B=\pm u_A$, and the minus sign would require $|B|=N-k\ne k$. The common diagonal can be omitted without changing distances. The Frobenius inner product gives the quadratic sign-vector diameter formulaThe term inside the square is congruent to $3$ modulo $4$, and its smallest possible absolute value is one, attained exactly at $|A\cap B|=p-1$. Such pairs exist. Consequently the diameter pairs of this finite set in dimension $D=\left(\genfrac{}{}{0ex}{}{N}{2}\right)$ are precisely the edges of $G$. Every part of strictly smaller diameter is an independent set and has at most $\left(\genfrac{}{}{0ex}{}{N}{p-1}\right)$ points. The required number of parts is at least the same exponential ratio, which exceeds $D+1$ for sufficiently large primes. Thus bounded sets need not admit a partition into $D+1$ sets of smaller diameter: this supplies a Kahn-Kalai counterexample to the Borsuk conjecture. Both applications use the prime case of the proved prime-power result.

Identify the vertices of the Boolean hypercube with subsets of $[n]$. Write $N(\mathcal{A})$ for the closed vertex neighbourhood and $\partial \mathcal{A}=N(\mathcal{A})\setminus \mathcal{A}$ for the external vertex boundary. The cube order used here is the simplicial order on the discrete cube: increasing size, then lexicographic order, with the smallest differing coordinate belonging to the earlier set. We prove Harper theorem by induction on $n$; subtracting the common size of the families converts a closed-neighbourhood comparison into the required boundary comparison.

First, the closed neighbourhood of a simplicial initial segment is itself an initial segment. Such a family consists of full lower levels and a lexicographic initial part of its last level. Its next-level neighbours form the upper shadow of that part. For a set in the next level, its lexicographically earliest lower-level subset is obtained by deleting its largest element. This map preserves the lexicographic order weakly. Membership in the upper shadow is equivalent to that earliest subset belonging to the original initial part, so the upper shadow is an initial part too. Empty and full families satisfy the assertion directly.

Assume the theorem in dimension $n-1$. Split $\mathcal{A}$ into two sections ${\mathcal{A}}_0,{\mathcal{A}}_1$ according to membership of coordinate $i$, deleting $i$ from the second section. Their closed-neighbourhood sections areReplace each section by a simplicial initial segment $I_0,I_1$ of the same size. Induction gives $|N(I_j)|\le |N({\mathcal{A}}_j)|$. By the preceding observation, each $N(I_j)$ is an initial segment and is nested with the other section. Consequentlyand the analogous inequality holds for the other section. Thus simplicial section compression preserves size and cannot increase the closed neighbourhood.

Repeatedly apply a nontrivial section compression. The sum of the positions of the family members in the full simplicial order strictly decreases: the order induced on either fixed-coordinate section is precisely its lower-dimensional simplicial order. This positive integer potential makes the process terminate at a family $\mathcal{B}$ fixed by every section compression.

We classify such terminal families for simplicial section compression. If an earlier vertex $x$ is absent while a later vertex $y$ is present, they cannot agree at any coordinate, since the corresponding section is an initial segment. Hence $y=x^c$. Every other vertex earlier than $y$ is present, and every other vertex later than $x$ is absent, since any further inverted pair would also have to be complementary. There can therefore be no vertex strictly between $x$ and $y$. Thus $\mathcal{B}$ either is an initial segment or is an initial segment with its last vertex replaced by the next, complementary vertex.

The complementary consecutive pairs are explicit. If $n=2r+1$, the ranks must be $r$ and $r+1$, and consecutive order positions forceThe corresponding initial segment $I$ contains all sets of size at most $r$, and $\mathcal{B}=I-\{x\}+\{y\}$. Its closed neighbourhood contains all sets of size at most $r+1$: for $r\ge 1$, deletion of just one $r$-set cannot remove all the $r$-subsets of any $(r+1)$-set, and $x$ remains adjacent to one of its $(r-1)$-subsets. In addition, the $r$ supersets of $y$ of size $r+2$ enter the neighbourhood. Thus $|N(\mathcal{B})|=|N(I)|+r$; the $n=1$ case gives equality directly.

If $n=2r$, the two vertices have rank $r$. Complementation reverses the lexicographic order within that rank, so consecutive complements occupy its middle two positions. These areNow $I$ contains all sets of size less than $r$ and all $r$-sets containing coordinate one. Its closed neighbourhood contains all sets of size at most $r$ and all $(r+1)$-sets containing one. For $r\ge 2$, each such $(r+1)$-set has at least two $r$-subsets containing one, so exchanging $x$ for $y$ loses none of these neighbours. It adds the $r-1$ supersets of $y$ of size $r+1$ that avoid one. Hence $|N(\mathcal{B})|=|N(I)|+r-1$. For $n=2$ the two neighbourhoods are both the entire cube.

In both exceptional cases the initial segment has no larger neighbourhood. The induction starts with the one-vertex zero-dimensional cube, and the compression argument now proves

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. In the Fourier-Walsh transform convention ${\chi }_S(x)=(-1{)}^{{\sum }_{i\in S}{x}_i}$, write $f={\sum }_S{\alpha }_S{\chi }_S$ with ${\alpha }_S=\mathbb{E}[f{\chi }_S]$. For the indicator $f$, Parseval's identity givesThe flip-probability influence satisfies ${\beta }_i=4{\sum }_{S\ni i}{\alpha }_S^2$, and summing gives ${\sum }_{S\ne \varnothing }|S|{\alpha }_S^2=\frac{1}{4}{\sum }_i{\beta }_i$.

For $p=1+\delta \in (1,2)$, Beckner's inequality says $\Vert T_{\sqrt{\delta }}g{\Vert }_2\le \Vert g{\Vert }_p$. Apply it to the discrete derivative of a Boolean function in coordinate $i$, viewed on the other $n-1$ coordinates. Its nonzero magnitude is $1/2$ and occurs with probability ${\beta }_i$, soSumming over $i$ gives the hypercontractive weighted influence bound

Identify the vertices of the Boolean hypercube $Q_n$ with subsets of $[n]$, adjacent when their symmetric difference has size one. Write $N(\mathcal{A})$ for the closed graph neighbourhood, including $\mathcal{A}$ itself. In simplicial order on the discrete cube, smaller sets come first, and equal-sized sets are ordered by lexicographic order: the least element of the symmetric difference belongs to the earlier member. Let $I_m$ denote the first $m$ vertices. The vertex-isoperimetric inequality in the discrete cube statesSubtracting $|\mathcal{A}|$ gives the equivalent assertion for the external vertex boundary.

We first check that $N(I_m)$ is also an initial segment in simplicial order on the discrete cube. Apart from the empty and full cases, $I_m$ consists of complete lower layers and a lexicographic initial segment $\mathcal{F}$ in its top rank $r$. Its closed graph neighbourhood contains all ranks at most $r$ and the upper shadow $\nabla \mathcal{F}$ in rank $r+1$. This upper shadow is lexicographic initial: an $(r+1)$-subset $C$ belongs to it precisely when its first $r$-subset in lexicographic order, obtained by deleting its largest element, belongs to $\mathcal{F}$. Taking the first $r$ elements preserves weak lexicographic order, proving the assertion. This also handles $r=0$, where $N(\{\varnothing \})$ consists of ranks zero and one.

We prove the vertex-isoperimetric inequality in the discrete cube by mathematical induction on $n$, with $n=0,1$ immediate. Fix a coordinate $i$ and write ${\mathcal{A}}_0,{\mathcal{A}}_1\subseteq Q_{n-1}$ for the sections without and with $i$. The two sections of its closed graph neighbourhood areA simplicial section compression replaces ${\mathcal{A}}_j$ by the initial segment $I_{a_j}$ of size $a_j=|{\mathcal{A}}_j|$ in the other coordinates. By the preceding observation, both $N(I_{a_j})$ and $I_{a_{1-j}}$ are initial segments, so their union has size $\max \{|N(I_{a_j})|,a_{1-j}\}$. The mathematical induction hypothesis givesAdding the two section bounds proves that simplicial section compression cannot increase the closed graph neighbourhood.

Apply changing simplicial section compressions until none remains. The sum of the global simplicial order on the discrete cube positions strictly decreases at each step: within a section the order is exactly the restriction of the global order. Thus the procedure terminates at a set family $\mathcal{B}$ every coordinate section of which is initial. We now classify the possible terminal families for simplicial section compression, rather than assuming this terminal set family must be globally initial.

If $X<Y$, $X\notin \mathcal{B}$ and $Y\in \mathcal{B}$, then $X,Y$ cannot agree in any coordinate, since that coordinate section would have an omitted earlier member and an included later one. Therefore $Y=X^c$. Let $X$ be the first omitted vertex and $Y$ the last included vertex. If $\mathcal{B}$ is noninitial, $X<Y$ and $Y=X^c$. Any vertex $Z$ strictly between them would have to be included, since an omitted $Z$ paired with $Y$ would have to equal $Y^c=X$. But it would also have to be omitted, since an included $Z$ paired with $X$ would have to equal $X^c=Y$. Thus there is no such $Z$: $X,Y$ are consecutive complementary vertices. Everything before $X$ is included and everything after $Y$ omitted. Hence $\mathcal{B}$ is obtained from the initial segment of its size by replacing $X$ with its immediate successor $X^c$.

There are only two forms of this exception. If $n=2r+1$, consecutive complementary vertices must straddle ranks $r,r+1$; otherwise an intermediate rank or a further vertex in one of their ranks separates them. They are consequentlywith $X$ the last $r$-subset and $Y$ the first $(r+1)$-subset. The corresponding initial segment is all ranks at most $r$. For $r\ge 1$, the omitted $X$ remains in $N(\mathcal{B})$ through a lower neighbour, and every $(r+1)$-subset has at least two $r$-subsets, only one of which was omitted. Thus $N(\mathcal{B})$ contains every rank at most $r+1$, namely $N(I_{|\mathcal{B}|})$. For $r=0$ both singleton closed graph neighbourhoods are the whole $Q_1$.

If $n=2r$, the two complementary vertices must both have rank $r$. The complement map reverses the lexicographic order in that rank, so a consecutive complementary pair must be its central pair. Exactly half the $r$-subsets contain $1$, and all of those precede those missing $1$. Therefore the pair isHere $I_{|\mathcal{B}|}$ comprises all ranks below $r$ and all $r$-subsets containing $1$. For $r\ge 2$, $N(\mathcal{B})$ contains every rank at most $r$, because all lower ranks were retained. Every $(r+1)$-subset containing $1$ has $r\ge 2$ different $r$-subsets containing $1$, so deletion of the one member $X$ removes no vertex of the upper shadow. Consequently $N(\mathcal{B})\supseteq N(I_{|\mathcal{B}|})$. For $r=1$, the two closed graph neighbourhoods both have size three in $Q_2$. These cases exhaust the exceptions. Since the successive simplicial section compressions never increased the closed graph neighbourhood, this proves the vertex-isoperimetric inequality in the discrete cube.

The same argument gives all integer radii. Write $N^t$ for $t$ repetitions of the closed graph neighbourhood. Its application to an initial segment is again initial; the sizes of these initial segments are monotone in the original size. Inducting on $t$ therefore gives

For the concentration question, use uniform probability measure on a finite connected graph and normalize its graph distance by its graph diameter. A sequence is a Lévy family of graphs if, for every fixed $\epsilon >0$ and $\delta >0$, the radius-$\epsilon$ neighbourhood of every set of measure at least $\delta$ has measure tending to one, uniformly over those sets. One may equivalently require this only for sets of measure at least one half. To see the reverse implication, first the radius-$\epsilon /2$ neighbourhood of every positive-density set must eventually have measure greater than one half. Otherwise its complement has measure at least one half, while the radius-$\epsilon /4$ neighbourhood of that complement misses the original positive-density set, contradicting the half-measure condition. Apply the half-measure condition once more to the radius-$\epsilon /2$ neighbourhood, enlarging it by another $\epsilon /2$, to obtain the full claim. The metric normalization is essential here.

For $Q_n$, the graph diameter is $n$. A uniformly chosen vertex has size $Z_n$ with binomial distribution $\mathrm{Bin}(n,1/2)$, expectation $n/2$ and variance $n/4$. Fix $\delta >0$ and choose $C>0$ so that $1/(4C^2)<\delta$. By Chebyshev's inequality,For sufficiently large $n$, put $k_n=\lfloor n/2-C\sqrt{n}\rfloor \ge 0$. Every initial segment of measure at least $\delta$ contains the complete ranks through $k_n$. Its radius-$t$ graph neighbourhood therefore contains every rank at most $k_n+t$. Taking $t=\lfloor \epsilon n\rfloor$, the all-radius vertex-isoperimetric inequality in the discrete cube gives, uniformly for $|\mathcal{A}|\ge \delta 2^n$,The last Chebyshev's inequality is used only when its denominator has positive square root. These are precisely the radius-$\epsilon$ neighbourhoods in the normalized graph distance. The discrete cubes form a Lévy family. In fact, taking $t=\lfloor n^{3/4}\rfloor$ in the same bound makes the omitted measure $O(n^{-1/2})$ for each fixed $\delta$: a radius that is $o(n)$ already suffices.

For fixed $d\ge 1$, the nearest-neighbour grid $[n{]}^d$ has graph distance ${\sum }_i|x_i-y_i|$ and graph diameter $d(n-1)$. Take the slab ${\mathcal{A}}_n=\{x:x_1\le \lceil n/2\rceil \}$, whose uniform measure is at least one half. Its integer radius-$t$ graph neighbourhood is exactly $\{x:x_1\le \min (n,\lceil n/2\rceil +t)\}$: changing the first coordinate alone attains the minimum distance to the slab. For $0<\epsilon <1/(2d)$ and $t=\lfloor \epsilon d(n-1)\rfloor$, the omitted proportion tends toThus there is failure of concentration in fixed-dimensional grids, and the fixed-dimensional grids do not form a Lévy family with normalized graph distance.

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.

The general modular form, the nonuniform Frankl-Wilson theorem, can be stated as follows. Let $p$ be a prime number, and let $L\subseteq {\mathbb{F}}_p$ have $s$ elements. Suppose $\mathcal{F}\subseteq \mathcal{P}([n])$ has $|A|\mathrm{mod}p\notin L$ for every member, while $|A\cap B|\mathrm{mod}p\in L$ for any distinct members. ThenThe uniform Frankl-Wilson theorem sharpens this to $|\mathcal{F}|\le \left(\genfrac{}{}{0ex}{}{n}{s}\right)$ when all members have size $k$ and $0\le s\le \min \{k,n-k\}$. Both forms require a prime modulus and exclusion of the self-intersection residue.

For the general proof, work over the finite field ${\mathbb{F}}_p$. Associate to each member $A$ the intersection polynomialAt the characteristic vectors of sets $1_B$, this is zero if $B\ne A$, and nonzero if $B=A$. Thus these polynomials, regarded as functions on the Boolean hypercube, are linearly independent: evaluating any relation at $1_A$ isolates its coefficient. Apply multilinear reduction on the Boolean cube, replacing every positive power of a variable by that variable. Values on the Boolean hypercube remain unchanged, and the resulting multilinear polynomials have degree at most $s$. Their ambient space has a monomial basis consisting of $x_S={\prod }_{i\in S}{x}_i$ for $|S|\le s$, with dimension ${\sum }_{j=0}^{\min (s,n)}\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. This proves the general bound, including $s=0$.

For completeness, obtain the uniform sharpening without dividing by factorials in a finite field. For $s\ge 1$ put $a=k\mathrm{mod}p$ and $g(x)={\sum }_i{x}_i-a$. The functions $x_{S}g$ with $|S|<s$ are linearly independent. Indeed, a relation gives $gh=0$ with $\mathrm{deg}h<s$, so $h$ is supported only on weights in $H=\{j\in [0,n]:j\equiv a(\mathrm{mod}p)\}$. With boundary weights $-1,n+1$, this set has a gap of at least $s+1$: either consecutive allowed weights differ by $p\ge s+1$, or its terminal gap does, since $n\ge k+s\ge a+s$. The alternating sum of $h$ over any interval of $s$ free coordinates vanishes by its degree. Across that gap only one endpoint level can contribute, forcing $h$ to vanish there. Delete that level and repeat across the enlarged gap until $H$ is empty. Hence $h=0$, and independence of the square-free monomials gives the claim. Adjoin these functions to the $f_A$. Evaluation at each family vector eliminates the coefficients of $f_A$, since $g$ vanishes there; the claim eliminates all remaining coefficients. Counting dimensions yieldsThe case $s=0$ permits at most one member directly. The gap argument is an instance of the modular layer vanishing lemma.

For the final application, enumerate the distinct sets as $A_1,\dots ,A_m$ and let $v_i=1_{A_i}\in {\mathbb{R}}^n$ be their characteristic vectors of sets. If $m\le 1$, the desired bound is immediate for $n\ge 1$. Otherwise $|A_i|\ge k$, because $A_i$ intersects another member in $k$ points. At most one member can have size $k$: two distinct $k$-sets cannot have intersection size $k$. The Gram matrix of these vectors isFor real coefficients $c_i$,All terms are nonnegative, with $k>0$. If the expression vanishes, every coefficient corresponding to a set of size greater than $k$ is zero. At most one coefficient remains, and the first term forces it to be zero as well. Thus the characteristic vectors of sets are linearly independent in ${\mathbb{R}}^n$, proving the constant-intersection family boundThis proof explicitly covers the possible member of size exactly $k$; assuming all diagonal corrections were strictly positive would miss that case. Positivity of $k$ is essential: when $k=0$, the empty set and all $n$ singleton sets give $n+1$ members.