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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.

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.

Give a vertex $S\subseteq [n]$ the binary value $\nu (S)={\sum }_{i\in S}{2}^{i-1}$. The binary order on the discrete cube orders vertices by increasing $\nu$; its initial segment $I_m$ consists of the values $0,\dots ,m-1$. Write $e(\mathcal{A})$ for the number of cube edges whose two endpoints lie in $\mathcal{A}$, and ${\partial }_e\mathcal{A}$ for its external edge boundary. The degree sum identity isThe exact edge-isoperimetric inequality iswhere $\mathrm{wt}(j)$ is the number of ones in the binary expansion of $j$. Equality is attained by $I_m$. Indeed, $I_m$ is a down-set, so every downward neighbour of each vertex is also present. Counting each internal edge at its upper endpoint gives $e(I_m)={\sum }_{j<m}\mathrm{wt}(j)$. We prove the equivalent edge-isoperimetric theorem for binary initial segments, $e(\mathcal{A})\le e(I_m)$, by mathematical induction using section compression in binary order.

Fix a coordinate $i$ and write ${\mathcal{A}}_0,{\mathcal{A}}_1\subseteq Q_{n-1}$ for the two sections with coordinate $i$ absent or present, deleting that coordinate. Their inherited binary order is the binary order on the remaining coordinates. Replace each section by the binary initial segment of its own size. This preserves $|\mathcal{A}|$. The edges split into internal edges in the two sections and crossing edges, givingBy the mathematical induction hypothesis, the two section edge counts do not decrease. After compression, the sections are nested initial segments, so their intersection has the largest possible size, $\min \{|{\mathcal{A}}_0|,|{\mathcal{A}}_1|\}$. Thus each section compression does not decrease the internal edge count.

Apply these compressions repeatedly whenever one changes the family. The integer potential ${\sum }_{S\in \mathcal{A}}\nu (S)$ decreases strictly at every nontrivial step: within either fixed-coordinate section, the initial segment uniquely minimizes the sum of the vertex values among sets of the prescribed size. Therefore the process terminates with a family compressed in every section.

We need a small structural fact about families compressed in every binary section. If such a family is not a binary initial segment, choose a missing vertex $x$ and a present vertex $y$ with $\nu (x)<\nu (y)$. They cannot agree in any coordinate: in that coordinate's section, the compressed initial segment would contain $x$ whenever it contains $y$. Thus $y=[n]\setminus x$. The same argument applies to every inversion. There cannot be any vertex strictly between $x$ and $y$ in binary order: a present one would form an inversion with $x$, and a missing one would form an inversion with $y$, forcing it to be $y$ or $x$, respectively. Hence $\nu (y)=\nu (x)+1$. Together with $\nu (x)+\nu (y)=2^n-1$, this givesEvery vertex below $x$ must be present, and every vertex above $y$ absent, by the same inversion argument. Therefore the only possible exception to a binary initial segment isFor $n\ge 2$, this replaces the top vertex of an $(n-1)$-dimensional subcube by the singleton $\{n\}$. Removing that vertex loses $n-1$ internal edges, and adding the singleton creates exactly one edge, to the empty set. ThusThe dimensions $n=0,1$ are immediate. This finishes the mathematical induction: compression leads either to the desired initial segment or to an exception with no larger edge count. Hence the exact edge-isoperimetric inequality in the discrete cube holds, and binary initial segments maximize the number of contained edges at every size.

For the square-face assertion, let $f(\mathcal{A})$ count the two-dimensional cube faces all of whose vertices are in $\mathcal{A}$. We use mathematical induction again, now also using the edge theorem just proved. Faces lying wholly in one coordinate section contribute $f({\mathcal{A}}_0)+f({\mathcal{A}}_1)$. Every face crossing coordinate $i$ is determined by an edge lying in both sections, soCompress the two sections as before. Their internal face counts do not decrease by mathematical induction. Put $c=|{\mathcal{A}}_0\cap {\mathcal{A}}_1|$ and $b=\min \{|{\mathcal{A}}_0|,|{\mathcal{A}}_1|\}$. The edge theorem and nesting of binary initial segments giveAfter compression, the intersection is precisely $I_b$. Thus section compression does not decrease the face count either. The same decreasing potential gives termination, and the same classification leaves only a binary initial segment or ${\mathcal{E}}_n$.

In the exceptional family, removing the top vertex loses $\left(\genfrac{}{}{0ex}{}{n-1}{2}\right)$ square faces, one for each pair of downward coordinate directions. The added singleton lies in no square face, because it is the only vertex in its upper section. ConsequentlyDimensions below two have no square faces and start the mathematical induction. We have proved that binary initial segments maximize contained square faces:For an explicit value, the down-set property counts every face at its unique top vertex, givingThis establishes binary initial segments maximize contained square faces for arbitrary sizes, rather than only sizes that are powers of two.

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