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If every graph component has at most $(k+1)n/[k(2k+1)]$ vertices, assigning whole graph components to two sides gives an empty graph cut with the displayed bound. To prove it, maximize the smaller side's weight $S\le n/2$. If $S<kn/(2k+1)$, every component on the larger side has weight at least $n-2S$, because moving a smaller one would improve the smaller side. There are at least $k+1$ such components, since their total exceeds $k$ times the permitted maximum weight. Thus $n-S\ge (k+1)(n-2S)$, a contradiction. The proof also works for arbitrary positive weights.

Fix $k\ge 1$, ${\gamma }_0>1/(k+1)$, and ${\gamma }_0\le \gamma \le 1-\omega (n)/\log n$ with $\omega (n)\to \infty$. A binomial random graph has one giant component of the displayed order and all other graph components are tree components of orders at most $k$, with high probability. The expected value and variance of the isolated vertex count concentrate it around $n^{1-\gamma }\to \infty$. A Cayley formula upper bound for connected sets excludes component sizes $k+1$ through a small fixed fraction of $n$; the empty-graph cut bound excludes the remaining sizes up to $n/2$. An extra-edge count excludes cyclic small graph components. The expected value of the number of vertices in small nonisolated tree components is $O(nd{e}^{-2d})=o(n{e}^{-d})$, where $d=np$. The Markov inequality completes the count of vertices outside the unique large graph component.

Put $d=np=\gamma \log n$ and $\mu =n^{1-\gamma }$. The hypothesis implies $\mu \ge e^{\omega (n)}\to \infty$. For the number $I$ of isolated vertices, the isolated vertices in the Erdős-Rényi model formulas giveThus the Chebyshev inequality gives $I=(1+o(1))\mu$ with high probability, uniformly over the specified $\gamma$ range.

The Cayley formula supplies a spanning tree on any connected vertex set. Consequently the expected value $C_j$ of the number of graph components of order $j\ge 2$ is bounded byChoose a fixed $\eta \in (0,1/2)$ with ${\gamma }_0(1-\eta )(k+1)>1$. For $k+1\le j\le \eta n$, with $j\ge 2$, the binomial coefficient bound $\left(\genfrac{}{}{0ex}{}{n}{j}\right)\le (en/j{)}^j$ givesHere $ed{e}^{-d(1-\eta )}\le e\log n{n}^{-{\gamma }_0(1-\eta )}\to 0$. The geometric series and a union bound therefore exclude this entire size range, since $n[e\log n{n}^{-{\gamma }_0(1-\eta )}{]}^{k+1}=o(1)$. For $\eta n\le j\le n/2$, an empty graph cut would be necessary. Its total probability is at mostThere are consequently no graph components of orders between $k+1$ and $n/2$, with high probability.

For each fixed $2\le j\le k$, a graph component containing a graph cycle has a spanning tree and at least one extra edge. Counting that extra edge gives an expected value $O_k(d^j{e}^{-dj})=o(1)$; hence every small graph component is a tree component. The total number $R_2$ of vertices in tree components of orders $2,\dots ,k$ satisfiesThe Markov inequality gives $R_2=o(\mu )$ with high probability. When $k=1$, this sum is empty and $R_2=0$. Since $\mu =o(n)$, the remaining vertices must lie in one graph component larger than $n/2$. It is unique, and its order is $n-(1+o(1))n^{1-\gamma }$. Every other graph component is a tree with at most $k$ vertices. This is the logarithmic-regime giant component mechanism: almost all missing vertices are isolated.

If some choice $\omega$ produced a graph with largest component of a graph smaller than $2n/3$, the balanced component cut result with packing parameter one would give an empty graph cut whose sides both have at least $n/3$ vertices. For any fixed such vertex set partition, a uniform edge of the complete graph crosses with probabilityTo avoid crossing in row $i$, at least one of its $k$ offered edges must be noncrossing. The row's probability is $1-q^k\le 1-(4/9{)}^k$. The rows consist of independent random variables, so at $m=cn$ a union bound over at most $2^n$ vertex set partitions givesFor example, the positive integermakes this bound at most $e^{-n}$. Therefore every choice has a component of order at least $2n/3$, with probability tending to one. The simultaneous giant for fixed random-edge choice estimate includes choices made after seeing the entire array; it does not assume a particular online selection rule.

Let $a=kn/(2k+1)$, $u=(k+1)n/(2k+1)$ and $b=u/k$. A graph cut with no crossing edges is obtained by assigning whole graph components to its two sides. Among all such assignments, maximize the smaller side's size $S\le n/2$, and write $T=n-S$ for the other side.

Suppose $S<a$. Then $T>u=kb$. If a graph component on the larger side had size $0<w<n-2S$, moving it to the smaller side would make both $S+w$ and $T-w$ greater than $S$. That contradicts maximality. Hence every graph component on the larger side has at least $n-2S$ vertices. Since each has at most $b$ vertices and their total exceeds $kb$, there must be at least $k+1$ of them. Consequentlya contradiction. Thus $a\le S\le T\le u$, proving the required empty cut with both sides at most $(k+1)n/(2k+1)$. This balanced component cut argument works for arbitrary positive component weights as well; it does not require divisibility of $n$.

The Kullback-Leibler divergence of a product measure is the sum of the individual divergences, by expanding the logarithm of the likelihood ratio and taking its expected value. Put $q=1/2-t/\sqrt{n}$ and define $W(S)$ to be the unordered pairs lying within one group, while $C(S)$ is the set of unordered pairs crossing between groups. ThenThus choosing the printed $\partial S$ to mean $W(S)$ gives its order of the summands. If $\partial S$ denotes the usual graph cut $C(S)$, their two orders must be interchanged. Each unordered edge is counted once.

The inequality $\log x\le x-1$ givesSince $1/3<q<1/2$, each changed-edge divergence is at most $9t^2/(2n)$. Summing over at most $n(n-1)/2$ pairs gives $D_{\mathrm{KL}}(P_S\Vert P_{S^{\prime }})\le 9n{t}^2/4$. This is the information scale needed for Fano's inequality.