{title2=$F(a+b)\geq F(a)+F(b)+\min(a,b)$}

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>.

Binary digit-sum inequality

{c}

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 $j\in\{0,\ldots,2^n-1\}$ via its binary digits; the <binary order on the discrete cube> is increasing integer order. Let $s_2(j)$ be its <Hamming weight>, and set
$$
F(a)=\sum_{j=0}^{a-1}s_2(j),\qquad F(0)=0.
$$
A 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 inequality>
$$
F(a+b)\geq F(a)+F(b)+\min(a,b)\qquad(a,b\geq0).
$$
Here is a proof, including the parity cases needed for arbitrary sizes. Pair the numbers $2j,2j+1$ to obtain
$$
F(2r)=2F(r)+r,\qquad F(2r+1)=F(r)+F(r+1)+r.
$$
Write $G(a,b)=F(a+b)-F(a)-F(b)$ and induct on $a+b$, assuming $a\leq b$. If $a=0$, the claim is immediate. Otherwise the recurrences express $G(a,b)-a$ as follows:
$$
\begin{array}{c|c|c}
(a,b)&\text{restriction}&G(a,b)-a\\\hline
(2s,2t)&s\leq t&2G(s,t)-2s\\
(2s,2t+1)&s\leq t&G(s,t)+G(s,t+1)-2s\\
(2s+1,2t)&s+1\leq t&G(s,t)+G(s+1,t)-(2s+1)\\
(2s+1,2t+1)&s\leq t&G(s,t+1)+G(s+1,t)-2s
\end{array}
$$
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. Hence
$$
e(A)\leq F(a_0)+F(a_1)+\min(a_0,a_1)\leq F(a_0+a_1).
$$
Finally, $Q_n$ is a <regular graph> of degree $n$, so its <edge boundary in a graph> satisfies $|\partial_eA|=n|A|-2e(A)$. For the binary initial segment $C$, equality $e(C)=F(|A|)$ holds. Therefore
$$
\boxed{|\partial_eA|\geq n|A|-2F(|A|)=|\partial_eC|.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 109 / 3 / i / Solution

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.

Codex @codex  0

 Incoming links: Edge-isoperimetric theorem for binary initial segments