{wiki}

For $A\subseteq\{0,1\}^n$,
$$
|\partial_eA|\geq|A|\log_2\frac{2^n}{|A|}.
$$
Equivalently, $A$ spans at most $\frac12|A|\log_2|A|$ cube edges. Induction on the dimension and concavity of <binary entropy function>[binary entropy] prove the inequality.

Edge-isoperimetric inequality in the discrete cube

For $\mathcal A$ in the $n$-dimensional <Boolean hypercube>, the <edge-isoperimetric inequality in the discrete cube> is
$$
e(\mathcal A)\leq\frac{|\mathcal A|\log_2|\mathcal A|}{2},
$$
where $e(\mathcal A)$ is the number of edges spanned by $\mathcal A$. Since the cube is $n$-regular, the equivalent boundary form is
$$
|\partial_e\mathcal A|
=n|\mathcal A|-2e(\mathcal A)
\geq|\mathcal A|\log_2\frac{2^n}{|\mathcal A|}.
$$

We prove the induced-edge form by induction on $n$. Split the cube according to its last coordinate, and let the two sections have sizes $a\geq b$. At most $b$ edges of $\mathcal A$ cross between the sections. The induction hypothesis gives
$$
e(\mathcal A)\leq\frac12a\log_2a+\frac12b\log_2b+b.
$$
Put $x=b/(a+b)\leq1/2$. The required comparison with $\frac12(a+b)\log_2(a+b)$ is equivalent to
$$
H_2(x)\geq2x,
$$
where $H_2$ is the <binary entropy function>. This follows from concavity because the graph of $H_2$ lies above the chord joining $(0,0)$ to $(1/2,1)$. The induction is complete. Subcubes attain equality when $|\mathcal A|$ is a power of two.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 109 / 3 / i / Solution

{title2=$H_2(x)$}
{wiki}

The binary entropy function is
$$
H_2(x)=-x\log_2x-(1-x)\log_2(1-x).
$$
It is concave on $[0,1]$ and satisfies $H_2(x)\geq2x$ for $0\leq x\leq1/2$ by comparison with the chord from $(0,0)$ to $(1/2,1)$.


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 Incoming links: Binary entropy function