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For $F(a)={\sum }_{j=0}^{a-1}{s}_2(j)$ and nonnegative integers $a,b$, the displayed inequality follows by induction from $F(2r)=2F(r)+r$ and $F(2r+1)=F(r)+F(r+1)+r$. Splitting a vertex set of a hypercube graph into its two coordinate sections then proves the edge-isoperimetric theorem for binary initial segments.

The Boolean hypercube is the set $\{-1,1{\}}^n$, or equivalently $\{0,1{\}}^n$. Its hypercube graph joins two vertices exactly when they differ in one coordinate.

For a down-set $D$ in the Boolean lattice, every member $A$ has all its $|A|$ immediate lower neighbors in $D$. Counting an internal edge at its upper endpoint gives $e(D)={\sum }_{A\in D}|A|$. The degree identity for the edge boundary of the hypercube graph then gives the displayed formula.

Among subsets of the $n$-dimensional hypercube graph with $a$ vertices, initial segments of the binary order on the discrete cube minimize the edge boundary. Its exact minimum is $na-2{\sum }_{j=0}^{a-1}{s}_2(j)$. This strengthens the entropy lower bound on the same boundary.

For a vertex set of size $m$ in a hypercube graph, split one coordinate into sections of sizes $a,b$. Induction bounds the internal edges within the sections by $(a{\log }_{2}a+b{\log }_{2}b)/2$; at most $\min (a,b)$ edges cross between them. Put $t=\min (a,b)/(a+b)$. The chord bound for the binary entropy function gives $H_2(t)\ge 2t$, hence $a{\log }_{2}a+b{\log }_{2}b+2\min (a,b)\le m{\log }_{2}m$. Thus at most $m{\log }_{2}m/2$ internal edges are present, and the edge boundary is at least $m(n-{\log }_{2}m)$. Coordinate subcubes attain equality when $m$ is a power of two.

For a nonempty subset of a hypercube graph, split it into sections of sizes $a,b$. The proof of the entropy edge bound uses $H_2(t)\ge 2t$, $t=\min (a,b)/(a+b)$. Strict concavity makes equality possible only at $t=0$ or $t=1/2$. In the first case one section is empty; in the second the sections coincide, since all possible crossing edges must be present. Induction forces a coordinate subcube. Conversely every coordinate subcube has the displayed edge count. At size $2^{n-1}$ these are exactly the two halves determined by fixing one coordinate.

A Gray code orders binary strings so that successive strings differ in exactly one bit. Consequently it describes a path in a hypercube graph; a cyclic Gray code describes a cycle.

The order $00,01,11,10$ is a Gray code through the four vertices of $Q_2$. Applying this identification independently in $n$ coordinate pairs makes $[4{]}^n=P_4^n$ a spanning subgraph of the hypercube graph $Q_{2n}$. Therefore every vertex boundary in $Q_{2n}$ contains the corresponding boundary in $[4{]}^n$.

For a finite graph with at least two vertices, the displayed minimum measures the fewest outgoing edges per vertex in a nonempty set occupying at most half the graph. It uses the edge boundary, rather than the external vertex boundary. For the hypercube graph $Q_n$, $n\ge 1$, it equals one: edge-isoperimetry gives the lower bound and a coordinate half-cube attains it.

A Lévy family of graphs consists of finite connected graphs with their uniform probability measures and graph distances divided by their diameters, whose concentration of measure functions vanish at every fixed positive normalized radius. Equivalently, every family occupying at least half the vertices has a radius-$\lfloor \epsilon D_n\rfloor$ neighbourhood occupying a proportion tending to one, uniformly over such families. Harper theorem and binomial distribution tail bounds prove this for the hypercube graphs.

Identify the discrete cube $Q_n$ with the subsets of $[n]$, with hypercube graph adjacency given by changing one element. Write $N[A]$ for the closed graph neighbourhood and $\partial A=N[A]\setminus A$ for the external vertex boundary. The simplicial order on the discrete cube first orders by set size and then by lexicographic order, with the smallest differing coordinate belonging to the earlier set.

Harper theorem states that, for every set family $A\subseteq Q_n$,Here is the simplicial section compression proof. Induct on $n$, the one-dimensional case being immediate. Split $A$ by a coordinate into sections $A_0,A_1\subseteq Q_{n-1}$. Its neighbourhood sections areReplace $A_0,A_1$ by equally large simplicial initial segments $I_0,I_1$. By induction $|N[I_j]|\le |N[A_j]|$. Also the neighbourhood of a simplicial initial segment is itself a simplicial initial segment: at its last occupied level the extra vertices form the upper shadow of a lexicographic initial segment, again lexicographically initial. Consequently the compressed unions have sizes $\max \{|N[I_0]|,|I_1|\}$ and $\max \{|N[I_1]|,|I_0|\}$, no larger than the original unions. The compression preserves $|A|$ and cannot increase $|N[A]|$.

Repeat in every coordinate until all sections are initial. Every nontrivial compression strictly decreases the sum of the global simplicial positions, so the process terminates. To describe the terminal families for simplicial section compression, suppose an earlier absent vertex $S$ precedes a later present vertex $T$. They cannot share a coordinate value: otherwise they lie in the same compressed section, contradicting its initiality. Thus $T=[n]\setminus S$. No vertex can lie strictly between them in the order, since it would make another absent/present pair that is not complementary. Hence the terminal family is either $C(|A|)$ or that segment with its last vertex $S$ exchanged for its immediate complementary successor $T$.

For odd $n=2m+1$, that exceptional exchange crosses the middle levels: $S=\{m+2,\dots ,2m+1\}$ and $T=\{1,\dots ,m+1\}$. For $m\ge 1$, all sets of size at most $m+1$ remain in the neighbourhood after the exchange, since an $(m+1)$-set has at least two lower neighbours and only one was removed. The new $T$ can only add vertices of size $m+2$.

For even $n=2m$, the only consecutive complementary pair in one level is $S=\{1,m+2,\dots ,2m\}$ and $T=\{2,\dots ,m+1\}$. The original segment contains all sets below size $m$ and all $m$-sets containing 1. For $m\ge 2$, its neighbourhood contains all sets of size at most $m$ and all $(m+1)$-sets containing 1; every such upper set still has an included lower neighbour after the single removal. The new $T$ can only add further neighbours. The cases $n=1,2$ follow directly, with the exceptional families related to the initial ones by a graph automorphism. Thus neither exception improves the neighbourhood, completing the induction.

For the half-cube vertex-boundary extrema, if $n=2m+1$, the initial half cube is all sets of size at most $m$; its boundary is the next level, of size $\left(\genfrac{}{}{0ex}{}{2m+1}{m+1}\right)$. If $n=2m$, it contains all levels below $m$ and the $m$-sets containing 1. The boundary consists of the $m$-sets avoiding 1 and the $(m+1)$-sets containing 1, giving $2\left(\genfrac{}{}{0ex}{}{2m-1}{m}\right)=\left(\genfrac{}{}{0ex}{}{2m}{m}\right)$.

No boundary can exceed the $2^{n-1}$ vertices outside a half cube. Taking all vertices of even Hamming weight attains that bound, because every odd-weight vertex is adjacent to one of them. Therefore, for $n\ge 1$,

Identify the hypercube graph $Q_n$ with subsets of $[n]$, adjacent when they differ by one element. Write $N(\mathcal{F})$ for a family's closed vertex neighbourhood. The simplicial order on the discrete cube orders first by size and then, within a rank, by lexicographic order: the least element of a symmetric difference belongs to the earlier set. The Harper theorem says that if $I$ is the initial simplicial segment with $|I|=|\mathcal{F}|$, thenEquivalently this minimizes the external vertex boundary at fixed cardinality. This is the vertex-isoperimetric assertion, distinct from the edge version in Question 3.

Here is the deduction of Kruskal-Katona theorem. For a family $\mathcal{B}\subseteq [n{]}^{(k)}$, adjoin every lower rank and set $\mathcal{D}=[n{]}^{(<k)}\cup \mathcal{B}$. For $k\ge 1$,The corresponding simplicial initial segment is $[n{]}^{(<k)}\cup L$, where $L$ is the rank-$k$ lexicographic initial segment of size $|\mathcal{B}|$. Canceling the common lower-rank count in Harper's inequality gives $|\nabla \mathcal{B}|\ge |\nabla L|$. The rank-zero case is immediate by considering its two possible families.

To turn upper shadows into lower shadows, let $\rho (i)=n+1-i$ and $T(B)=\rho ([n]\setminus B)$. This bijection reverses inclusion and changes rank $k$ to $n-k$. If $B$ precedes $C$ in lex order, the largest element of $T(B)△T(C)$ belongs to $T(C)$, so $T(B)$ precedes $T(C)$ in colexicographic order. Also $T(\nabla \mathcal{B})=\partial (T\mathcal{B})$. Hence applying the upper-shadow conclusion to the transformed family proveswhere $C$ is the colex initial segment of the same size and rank as $\mathcal{A}$. This is the Kruskal-Katona lower-shadow theorem. A colex segment's shadow is itself a colex segment: split its sets by their largest element, giving a full initial block of the preceding ground set followed by an initial block with the new largest element. The same description holds one rank lower. Iteration therefore shows that colex segments minimize shadows at every lower rank. This proves the Harper theorem implies the Kruskal-Katona theorem deduction, including the necessary complement and coordinate reversal.

The Erdős-Ko-Rado theorem states that for $1\le r\le n/2$, every intersecting family $\mathcal{A}\subseteq [n{]}^{(r)}$ satisfiesThe family of all $r$-sets containing a fixed point attains the bound. We use the standard convention of positive ranks here.

For the first proof, let $\mathcal{B}$ be the complements of members of $\mathcal{A}$, at rank $k=n-r\ge r$, and let $\mathcal{S}$ be its rank-$r$ lower shadow. It is disjoint from $\mathcal{A}$, since $A\subseteq [n]\setminus A^{\prime }$ would make two family members disjoint. Suppose $|\mathcal{A}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)=\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$. The colex rank-$k$ segment of that size contains every $k$-set on $[n-1]$ and at least one set containing $n$. Its rank-$r$ shadow therefore contains all $\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)$ old $r$-sets and at least one new $r$-set containing $n$. The iterated Kruskal-Katona bound givesThen $|\mathcal{A}|+|\mathcal{S}|>\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)+\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)=\left(\genfrac{}{}{0ex}{}{n}{r}\right)$, impossible for disjoint subfamilies of rank $r$. This proves the Erdős-Ko-Rado theorem from shadows.

For the second proof use the Katona circle method. In any cyclic ordering of the $n$ points, an intersecting collection of length-$r$ intervals has at most $r$ members. To prove this, rotate one selected interval to end at position $n$, so it occupies $n-r+1,\dots ,n$. Intervals ending at $r,\dots ,n-r$ miss it and cannot be selected. Among the remaining endpoints other than $n$, pair $j$ with $j+n-r$ for $1\le j\le r-1$. The two intervals in each pair are disjoint because $n\ge 2r$, so at most one can be selected. Including the fixed interval gives at most $1+(r-1)=r$.

Now count pairs consisting of a family member and a permutation in which it is a cyclic interval. A fixed $r$-set has $nr!(n-r)!$ such permutations: choose its final position, its internal order, and the order of the remaining points. Each of the $n!$ permutations contributes at most $r$ family intervals by the proved cyclic interval intersection bound. ThusThis is the required cyclic-ordering proof, with its interval lemma established explicitly.

Use binary encoding $b(S)={\sum }_{i\in S}{2}^{i-1}$ for vertices of the hypercube graph, and write $I_m=\{S:b(S)<m\}$. Let $e(\mathcal{A})$ count edges with both endpoints in a family. The exact edge-isoperimetric theorem for binary initial segments, due to Harper, Lindsey, Bernstein and Hart, iswhere $s_2(j)$ is the number of ones in the binary expansion. Equivalently,The equality for $I_m$ follows by assigning each internal edge to its larger endpoint: each one-bit of that endpoint can be cleared, giving an earlier endpoint still in $I_m$. This is the exact extremal statement, stronger than the logarithmic entropy bound.

We prove it by induction on $n$. The case $n=1$ follows by checking sizes zero, one and two. Fix a coordinate $i$ and delete it from the two sections ${\mathcal{A}}_0,{\mathcal{A}}_1$. The edge count splits asReplace each section by a binary initial segment of the same size. By induction its internal edges cannot decrease. The two replacement segments are nested, so their intersection size is the smaller section size, at least the previous intersection size. Thus this section compression in binary order cannot decrease $e(\mathcal{A})$.

Repeatedly perform any nontrivial section compression. The sum ${\sum }_{S\in \mathcal{A}}b(S)$ strictly decreases each time: in a fixed section, the original numerical order is exactly the numerical order after deleting its fixed coordinate. This nonnegative integer potential ensures termination at a family compressed in every coordinate section.

Classify such a terminal family. If an absent vertex $S$ precedes a present vertex $T$, they cannot agree in any coordinate, because their common section is compressed and its earlier vertex would then have to be present. Hence $T=[n]\setminus S$. There can be no vertex strictly between them: if present, it would also have to be the complement of $S$; if absent, it would have to be the complement of $T$. Thus they are consecutive complementary binary integers, forcingNo other inversion is possible for the same reason. Therefore a terminal family is either a binary initial segment or the sole exceptional formThis exception has size $2^{n-1}$. For $n\ge 2$, compared with the half-cube $I_{2^{n-1}}$, it removes a vertex of internal degree $n-1$ and adds a vertex joined only to the retained empty set. Its edge count is thus $e(I_{2^{n-1}})-(n-1)+1\le e(I_{2^{n-1}})$. The $n=1$ case was already handled. Since compression never decreased edges, every original family has at most the initial segment's edge count. This proves the theorem, including the exceptional terminal configuration rather than assuming all compressed families are initial segments.

For the remaining requests it is useful to obtain the entropy proof of cube edge-isoperimetry, including its equality cases. Inductively a family of size $m$ satisfiesIndeed let the two sections have sizes $a\ge b\ge 0$, $m=a+b>0$. Their internal edges are bounded inductively and the crossing edges are at most $b$, soHere $0{\log }_{2}0=0$ and $H_2(t)=-t{\log }_{2}t-(1-t){\log }_2(1-t)$ is the binary entropy function. Its concavity and its endpoint values at zero and one-half give $H_2(t)\ge 2t$ for $0\le t\le 1/2$. This proves the bound, starting from the zero-dimensional cube.

Define the isoperimetric number of a graph bySince every cube vertex has degree $n$, the internal-edge bound giveswhen $|A|\le 2^{n-1}$. A coordinate half-cube has one crossing edge per vertex, attaining ratio one. Hence, for $n\ge 1$,

Now establish the equality cases of entropy cube edge-isoperimetry. Strict concavity of $H_2$ makes $H_2(t)>2t$ for $0<t<1/2$. Equality in the internal-edge bound therefore requires either one section to be empty, or two equal-size sections. In the equal-size case equality in the crossing bound forces the sections to coincide; each section must also attain its inductive internal-edge bound. Induction gives exactly coordinate subcubes: the empty-section case fixes the coordinate, and the coincident-section case frees it. Conversely every coordinate subcube attains that bound.

At size $m=2^{n-1}$, an edge boundary of size $m$ means $e(A)=\frac{1}{2}m(n-1)=\frac{1}{2}m{\log }_{2}m$, so these equality cases apply. The subcube has dimension $n-1$ and fixes precisely one coordinate. ThereforeThese $2n$ coordinate half-cubes are the only extremizers of the requested size.

Identify the hypercube graph $Q_n$ with $\mathcal{P}([n])$, joining two sets when their symmetric difference has size one. Write $N[A]$ for the closed graph neighbourhood of $A$, and $N_t[A]$ for its radius-$t$ neighbourhood in Hamming distance. In the simplicial order on the discrete cube, smaller sets come first, with lexicographic order within each rank: the smallest differing coordinate belongs to the earlier set. Let $I_m$ be the first $m$ vertices in this order. Harper theorem isEquivalently, $I_m$ minimizes the external vertex boundary at fixed size. We will also obtain $|N_t[A]|\ge |N_t[I_{|A|}]|$ for every integer $t\ge 0$.

Here is a complete proof by simplicial section compression. First note two properties of the simplicial order on the discrete cube. Its restriction to either face obtained by fixing one coordinate is the corresponding order on the smaller hypercube graph. Also, the closed graph neighbourhood of an initial segment is an initial segment. To verify the latter, a nonempty initial segment contains every level below its last level $r$, and a lexicographic prefix $\mathcal{L}$ at level $r$. For $r\ge 1$, its closed graph neighbourhood contains every set of size at most $r$, and its only further members are the upper shadow of $\mathcal{L}$. An $(r+1)$-set lies in this upper shadow exactly when its lexicographically earliest $r$-subset lies in $\mathcal{L}$. That earliest subset is obtained by deleting the largest element. These earliest subsets vary monotonically in lexicographic order, so the upper shadow is another lexicographic prefix. For $r=0$, the closed graph neighbourhood of $\{\varnothing \}$ is the union of levels zero and one. The empty initial segment causes no exception.

Induct on $n$, with $n=0,1$ immediate. Fix a coordinate $i$ and write the sections of $A$ as $A_0,A_1\subseteq Q_{n-1}$, deleting $i$ from the latter section. Replace them by the initial segments $I_a,I_b$ with $a=|A_0|$, $b=|A_1|$. The sections of the old closed graph neighbourhood areBy induction their sizes are at least $\max \{|N[I_a]|,b\}$ and $\max \{|N[I_b]|,a\}$, respectively. These are exactly the sizes of the new sections, because both sets in each new union are initial segments and hence nested. Thus simplicial section compression preserves size and never enlarges the closed graph neighbourhood.

Repeatedly compress any section that is not already an initial segment. Every change strictly reduces the sum of the global simplicial positions of the included vertices. This nonnegative integer decreases only finitely often, so the process ends at a family $B$ whose sections in every coordinate are initial segments. If $B$ itself is an initial segment, the induction is complete. Otherwise choose the earliest absent vertex $X$ and the latest present vertex $Y$, with $X\prec Y$. They cannot agree in any coordinate: agreement would put them in one compressed section, where an earlier absent vertex cannot precede a later present one. Hence $Y=X^c$. Nor can a vertex $Z$ lie between them: if $Z$ is present, pairing it with $X$ would force $Z=X^c=Y$; if absent, pairing it with $Y$ would force $Z=Y^c=X$. Thus $X,Y$ are consecutive, and $B$ is obtained from $I_{|B|}$ by replacing its last vertex $X$ with the next vertex $Y$.

There are only two possibilities for these terminal families for simplicial section compression. If $|X|<|Y|$, consecutiveness makes $X$ the last $k$-set and $Y$ the first $(k+1)$-set. Complementarity forces $n=2k+1$, $X=\{k+2,\dots ,2k+1\}$ and $Y=\{1,\dots ,k+1\}$. For $n\ge 3$, the corresponding $B$ contains all sets of size at most $k$ except $X$, and additionally $Y$. Every $(k+1)$-set has at least two $k$-subsets, so at least one belongs to $B$. The missing $X$ has a $(k-1)$-subset in $B$. Consequently $N[B]$ contains every set of size at most $k+1$, which is exactly $N[I_{|B|}]$.

If $|X|=|Y|$, then $n=2k$. Since $X,Y$ are complementary, the smaller one contains coordinate $1$. Consecutiveness forces $X$ to be the last $k$-set containing $1$, and $Y$ the first $k$-set avoiding $1$:The initial segment consists of all sets of size less than $k$ and all $k$-sets containing $1$. Its closed graph neighbourhood contains all sets of size at most $k$ and all $(k+1)$-sets containing $1$. For $k\ge 2$, each such $(k+1)$-set has $k\ge 2$ $k$-subsets containing $1$, so removing only $X$ leaves one in $B$. All sets of size at most $k$ are still in $N[B]$, because all levels below $k$ belong to $B$. Hence again $N[I_{|B|}]\subseteq N[B]$. For $k=1$, the two families are related by exchanging coordinates and both have the whole $Q_2$ as their closed graph neighbourhood. These comparisons finish the induction and prove Harper theorem.

For the radius-$t$ version, repeatedly use the one-step result. At each step the comparison family remains an initial segment, and its closed graph neighbourhood is monotone in its size. If $|N_{t-1}[A]|\ge |N_{t-1}[I_m]|$, applying Harper theorem to $N_{t-1}[A]$ therefore gives $|N_t[A]|\ge |N_t[I_m]|$. Induction on $t$ proves the asserted extension.

A Lévy family of graphs has vanishing concentration of measure functions after distances have been scaled by the graph diameter. More explicitly, for connected finite graphs $G_n$ of diameter $D_n$, use normalized graph distance $d_n=d_{G_n}/D_n$ and uniform probability measure. For every fixed $\epsilon >0$, requireFor $Q_n$ this normalization is normalized Hamming distance $d_H/n$, since its diameter is $n$.

If $|A|\ge 2^{n-1}$, the corresponding initial segment contains the Hamming ball about $\varnothing$ of radius $b_n=\lfloor (n-1)/2\rfloor$. The radius-$t$ version of Harper theorem impliesUse the precise Hoeffding inequality $\mathbb{P}(Z_n-n/2\ge u)\le \exp (-2u^2/n)$ for $u\ge 0$. One can also derive this estimate directly: the moment-generating function of $Z_n-n/2$ is $\cosh (\lambda /2{)}^n\le \exp (n{\lambda }^2/8)$; the exponential Markov inequality followed by minimization at $\lambda =4u/n$ gives the bound. With $t=\lfloor \epsilon n\rfloor$, we have $b_n+t\ge n/2+\epsilon n-2$. Thus, whenever $\epsilon n>2$,This proves that the hypercube graphs form a Lévy family of graphs. The stronger fixed-positive-mass formulation follows as well. For any $a>0$, choose $c$ with $1/(4c^2)<a$. The Chebyshev inequality for $Z_n$, whose variance is $n/4$, makes the Hamming ball of radius $\lfloor n/2-c\sqrt{n}\rfloor$ have fewer than $a{2}^n$ vertices. An initial segment of size at least $a{2}^n$ contains that ball. Enlarging by $\lfloor \epsilon n\rfloor$ and using the same Hoeffding inequality gives a complement proportion tending to zero uniformly over all such $A$. The normalization matters: with unscaled unit Hamming distance, cubes do not satisfy the metric version at every fixed radius; for a radius below one, a half-cube has no enlargement.

Equality in Harper theorem does not determine a down-set up to isomorphism. In $Q_4$, takeThis is a down-set of size nine. Its closed graph neighbourhood contains every set of size at most two, because all singletons belong to $A$. Every triple contains an edge of the four-cycle among the listed pairs, so every triple also belongs to $N[A]$. The four-element set does not. Hence $|N[A]|=1+4+6+4=15$.

The size-nine simplicial order on the discrete cube initial segment is all sets of size at most one together with the pairs $12,13,14,23$. Every triple contains one of these pairs, and the four-element set is again absent from its closed graph neighbourhood. Its neighbourhood therefore also has size $15$, proving extremality of $A$. Nevertheless the induced graphs are not isomorphic: in $A$, the empty set has degree four, the four singletons have degree three, and the four pairs have degree two. The initial segment has two vertices of degree four, namely $\varnothing$ and $\{1\}$. Since a graph automorphism preserves induced degree of a vertex, no automorphism of the cube can carry one family to the other. This gives nonunique down-set extremizers for Harper theorem even under the stronger test of induced graph isomorphism.

Let $e(A)$ count internal edges of the hypercube graph. Since each vertex has degree $n$, its edge boundary is $b_e(A)=n|A|-2e(A)$. The edge-isoperimetric inequality in the discrete cube gives $e(A)\le |A|{\log }_2|A|/2$, hence $b_e(A)\ge |A|(n-{\log }_2|A|)$.

For completeness, the entropy proof of cube edge-isoperimetry splits the final coordinate into sections of sizes $a,b$. Their internal edges contribute at most $(a{\log }_{2}a+b{\log }_{2}b)/2$ by induction, and their crossing edges at most $\min (a,b)$. For $m=a+b$ and $t=\min (a,b)/m$, the binary entropy function satisfies $H_2(t)\ge 2t$, so $a{\log }_{2}a+b{\log }_{2}b+2\min (a,b)\le m{\log }_{2}m$. The convention is $0{\log }_{2}0=0$.

A $k$-dimensional coordinate subcube has $2^k$ vertices and $k{2}^{k-1}$ internal edges, attaining the bound. Therefore

The lazy Markov chain has holding probability $1/2$ and transition probability $1/(2n)$ to each neighbour in the hypercube graph. Its stationary distribution is $\pi (x)=2^{-n}$ by symmetry.

For $(x,y)$ choose the path changing the differing coordinates in increasing order. Its length is at most $n$. Fix a directed edge $(u,u^{(k)})$ that flips coordinate $k$. A path uses this edge exactly when $x_k=u_k$, $y_k=-u_k$, $y_j=u_j$ for $j<k$, and $x_j=u_j$ for $j>k$. The coordinates $x_j$ for $j<k$ and $y_j$ for $j>k$ are free. Thus exactly $2^{n-1}$ ordered pairs use it, whileThe unweighted load divided by capacity equals $n$; multiplying by the maximum length yields $\rho \le n^2$. The canonical paths Poincare bound therefore proves the slightly stronger ${\mathrm{Var}}_{\pi }(f)\le n^2\mathcal{E}(f,f)$, and in particular the requestedOne can keep the actual lengths: among these $2^{n-1}$ pairs, coordinate $k$ always differs, and each of the other $n-1$ coordinates differs for half the pairs. Their mean path length is $(n+1)/2$, giving exact congestion $\rho =n(n+1)/2$. This refinement still has order $n^2$.

For a finite irreducible Markov chain with invariant distribution $\pi$, the mean recurrence time of state $i$ isThe simple random walk on the hypercube has the uniform invariant distribution because the hypercube graph is regular, so every vertex has mass $2^{-n}$. Therefore the expected first positive return time to the initial vertex is

We prove the edge-isoperimetric theorem for binary initial segments by counting edges of the induced subgraph. Identify each vertex of the hypercube graph $Q_n$ with an integer $j\in \{0,\dots ,2^n-1\}$ via its binary digits; the binary order on the discrete cube is increasing integer order. Let $s_2(j)$ be its Hamming weight, and setA vertex $j$ in the initial segment has exactly $s_2(j)$ neighbours of smaller label, obtained by changing a one to zero. Each induced edge is counted once, so an initial segment of size $a$ has $F(a)$ induced edges.

The arithmetic ingredient is the binary digit-sum inequalityHere is a proof, including the parity cases needed for arbitrary sizes. Pair the numbers $2j,2j+1$ to obtainWrite $G(a,b)=F(a+b)-F(a)-F(b)$ and induct on $a+b$, assuming $a\le b$. If $a=0$, the claim is immediate. Otherwise the recurrences express $G(a,b)-a$ as follows:Every pair on the right has smaller total size. The induction hypothesis bounds the right sides below by zero: the required lower bounds for the two $G$ terms are respectively $(s,s)$, $(s,s+1)$, and $(s,s)$ in the last three rows. This establishes the binary digit-sum inequality.

Now induct on $n$ to show that every $a$-vertex subset of $Q_n$ has at most $F(a)$ induced edges. The zero-dimensional case is immediate. Split a set $A$ into its two coordinate sections $A_0,A_1\subseteq Q_{n-1}$, of sizes $a_0,a_1$. By induction the edges within the sections number at most $F(a_0)+F(a_1)$. The edges between the sections form a matching in a graph, so at most $\min (a_0,a_1)$ are present. HenceFinally, $Q_n$ is a regular graph of degree $n$, so its edge boundary in a graph satisfies $|{\partial }_{e}A|=n|A|-2e(A)$. For the binary initial segment $C$, equality $e(C)=F(|A|)$ holds. Therefore