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The axioms for information entropy give the formula $H(X)=-{\sum }_{x}p(x)\log p(x)$ and hence the chain rule for information entropyBecause conditioning reduces entropy, $H(Y\mid X)\le H(Y)$, and thereforeThis is subadditivity of information entropy.

For the entropy submodularity rule, apply the chain rule for information entropy twice:The last quantity is conditional mutual information, whose nonnegativity again expresses that conditioning reduces entropy.

Write $r=|V_0|$, $s=|V_1|$, and $e=|E(G)|=\alpha |X||Y|$. We construct a random graph homomorphism $\Psi :T\to G$. Choose a root of the tree. Map it according to the degree-biased vertex distribution on the corresponding side of the bipartite graph; after mapping any vertex, map each child independently and uniformly to a neighbour of its parent's image.

Every oriented tree edge is then mapped uniformly onto the $e$ edges of $G$. Every tree vertex in $V_0$ has the degree-biased marginal on $X$, of entropy $H_X$, and every vertex in $V_1$ has the analogous marginal of entropy $H_Y$. Repeated use of the chain rule for information entropy along the rooted tree givesSince a $k$-vertex tree has $k-1$ edges,The maximum entropy distribution on a finite set gives $H_X\le \log |X|$ and $H_Y\le \log |Y|$. HenceIf $N$ is the number of bipartition-respecting graph homomorphisms $T\to G$, the support of $\Psi$ has size $N$, so the maximum entropy distribution on a finite set also gives $H(\Psi )\le \log N$. ThusThere are $|X{|}^r|Y{|}^s$ bipartition-respecting maps in total, so a uniformly chosen one is a graph homomorphism with probability at least ${\alpha }^{k-1}$. This proves the Sidorenko inequality for trees.

Let $\mathcal{Q}$ be the set of valid quintuples and choose $(A_1,\dots ,A_5)$ uniformly from $\mathcal{Q}$. ThenApply Shearer's inequality to the ten pairs $\{i,j\}\subseteq [5]$. Every index occurs in four pairs, so

Put $U_{ij}=A_i\cup A_j$. Since $U_{ij}\in \mathcal{A}$, it has cardinality $k$. For each element of $U_{ij}$, its membership bits in $(A_i,A_j)$ are one of $(1,0),(0,1),(1,1)$; outside $U_{ij}$ they are forced to be $(0,0)$. Thus at most $3^k$ ordered pairs have any prescribed union. The chain rule for information entropy, conditional entropy, and the support bound for information entropy givebecause every $U_{ij}$ belongs to $\mathcal{A}$. There are ten pairs, henceExponentiating proveswhich is the five-variable case of the entropy bound for pairwise-union tuples.

Letwhere $\phi$ is the golden ratio. If the union-closed family consists of one nonempty set, any element of that set has frequency one, so assume its cardinality exceeds one. Choose independent uniform members $A,B$ and let $X,Y\in \{0,1{\}}^n$ be their characteristic vectors of sets. Then $H(X)=H(Y)=\log |\mathcal{A}|>0$.

Suppose for a contradiction that every element has frequency $p_i<c$. Put $q_i=1-p_i>1/\phi$, and let $Z=X\mathrm{OR}Y$, the characteristic vector of a set of $A\cup B$. The chain rule for information entropy and conditioning reduces entropy givebecause $Z_{<i}$ is a function of $(X_{<i},Y_{<i})$.

Fix the two prefixes and setThe two conditioned bits are independent random variables, and $Z_i=0$ exactly when both are zero. The supplied binary entropy product inequality therefore givesAveraging over the independent prefixes yieldswhenever either conditional entropy is positive. Summing and using $H(X)>0$ gives $H(Z)>H(X)$.

But set union keeps $A\cup B$ inside the union-closed family, so $Z$ is supported on $\mathcal{A}$. The maximum entropy distribution on a finite set gives $H(Z)\le \log |\mathcal{A}|=H(X)$, a contradiction. Some element must therefore occur in at least $c|\mathcal{A}|$ members, proving the entropy bound for a union-closed family.

The Entropic Balog-Szemerédi-Gowers theorem states that for finitely supported random variables $A,B$ in an abelian group,where the left-hand side is the Simultaneous conditional entropic Ruzsa distance.

Put $S=A+B$ and take two copies $(A_1,B_1)$ and $(A_2,B_2)$ that are conditionally independent given $S$. Thus both sums equal $S$, and, given $S$, $A_1$ and $B_2$ are independent with the required conditional marginals. By conditioning reduces entropy,The two conditional entropies are equal, since either $A$ or $B$ together with $S$ determines the other, and the chain rule for information entropy gives

Set $D=A_1-B_2$. By entropy submodularity,The first term is $H(A_1,B_2)\le H(A)+H(B)$ by subadditivity of information entropy. Since $A_1+B_1=A_2+B_2$, we also have $D=A_2-B_1$, so the second term is at most $H(A)+H(B)$. Finally $(D,A_1,B_1)$ determines all four copied variables, and conditional independence givesAs $H(A,B)=H(A)+H(B)-I(A;B)$, these estimates implySubtracting the common conditional-entropy term proves the stated Entropic Balog-Szemerédi-Gowers theorem.

Because the Entropic Ruzsa distance depends only on marginal distributions, take $X,Y,Z$ independent with the required marginals. Since $X-Z=(X-Y)+(Y-Z)$ is a function of $(X-Y,Y-Z)$, the data processing inequality for mutual information yieldsThe map $(X,X-Y,Y-Z)\mapsto (X,Y,Z)$ is a bijection. Using independence and the chain rule for information entropy, the left side iswhereas the right side is $H(X-Z)-H(Z)$. Hencewhere the final step is subadditivity of information entropy. Substituting this inequality into the definition of $d_R$ gives the Entropic Ruzsa triangle inequality

Put $L=L^{\ast }(X)$ and $\mu =\mathbb{E}L$. Since $L$ is a function of $X$, the chain rule for information entropy and part (c) givePart (a), applied to the nonnegative integer-valued random variable $L$, yieldsThe logarithm inequality $\log (1+t)\le t\log e$ implies $\mu \log (1+1/\mu )\le \log e$. ThusPart (c) also gives $\mu \le H(X)$, so monotonicity of the logarithm lets us replace $\log (1+\mu )$ by $\log (H(X)+1)$. Rearranging proves

Let $X=(X_1,\dots ,X_n)$ be a discrete random vector, and let $\mathcal{S}$ be a collection of subsets of $[n]$ in which every index occurs at least $r$ times. ThenApply the chain rule for information entropy to each projection $X_S$ and use conditioning reduces entropy to compare every term with the corresponding conditional entropy in the full chain rule.

The chain rule for information entropy and conditioning reduces entropy givewith equality exactly when $X$ and $Y$ are independent.