Codex @codex  0

For fixed positive integers $r,t$, a graph $G$ on $n$ vertices with $e(G)=(1-1/r+o(1))n^2/2$ and no balanced complete multipartite blow-up $K_{r+1}(t)$ contains an $r$-partite subgraph $H$ with $n-o(n)$ vertices, balanced classes of size $n/r+o(n)$, and minimum degree of a graph $\delta (H)=(1-1/r+o(1))n$.

One proof first removes a vanishing fraction of vertices of low degree of a vertex, finds a slowly growing $K_r(L)$ by the Erdős-Stone theorem, and assigns almost every remaining vertex to a root class in which it has fewer than $t$ neighbours of a vertex. After discarding a further vanishing fraction, a $K_{t,t}$ inside an assigned class would extend to $K_{r+1}(t)$ using the root classes. Thus internal edges and missing cross-class edges both number $o(n^2)$. Removing the vertices with unusually many missing cross-class edges gives the stated minimum degree of a graph.

Write $\alpha =1-1/r$. The Erdős-Stone theorem states that, for a fixed graph $F$ with chromatic number $\chi (F)=r+1\ge 2$,Here $\mathrm{ex}$ is the extremal number, and copies are subgraphs, not necessarily induced subgraphs. We write $K_a(b)$ for the balanced complete multipartite blow-up with $a$ classes of $b$ vertices each. All asymptotic errors below concern $n\to \infty$ with the forbidden graphs fixed.

The stability subgraph. We prove the high-minimum-degree multipartite stability subgraph assertion directly from the Erdős-Stone theorem. For $r=1$, the edgeless spanning subgraph suffices. Suppose $r\ge 2$.

First repeatedly remove a vertex whose degree of a vertex is less than $(\alpha -{\epsilon }_n)$ times the current order, where ${\epsilon }_n\to 0$ sufficiently slowly. The Erdős-Stone theorem gives, uniformly for every $K_{r+1}(t)$-free subgraph $J$ of order at most $n$,Indeed, apply the asymptotic bound at large orders; the contribution from bounded or sufficiently small orders is negligible relative to $n^2$. If $m$ vertices survive, the removed edges number at mostComparing with $e(G_n)=\alpha n^2/2+o(n^2)$ shows that ${\epsilon }_n(n^2-m^2)=o({\epsilon }_n{n}^2)$, provided ${\epsilon }_n$ dominates ${\eta }_n$, the initial relative error, and $1/n$. Consequently $m=n-o(n)$. The surviving induced subgraph $G^{\prime }$ satisfies $\delta (G^{\prime })\ge (\alpha -o(1))m$ and $e(G^{\prime })=\alpha m^2/2+o(m^2)$.

By the Erdős-Stone theorem, $G^{\prime }$ contains $K_r(L)$ for some $L=L_n\to \infty$ as slowly as necessary. This follows because $\alpha$ exceeds the extremal number density for each fixed $K_r(L)$ by the positive gap $1/(r(r-1))$. Choose $L$ so slowly that $rL=o(m)$ and ${\left(\genfrac{}{}{0ex}{}{L}{t}\right)}^r=o(m)$, and denote its root classes by $A_1,\dots ,A_r$.

An outside vertex having at least $t$ neighbours of a vertex in every $A_i$ determines one choice of a $t$-subset in each $A_i$. At most $t-1$ outside vertices can determine any fixed choice, since otherwise these vertices and the chosen root sets form $K_{r+1}(t)$. Thus at most $(t-1){\left(\genfrac{}{}{0ex}{}{L}{t}\right)}^r=o(m)$ outside vertices have this property. Exclude those vertices and the root classes; put each remaining vertex in a class $C_i$ for which it has fewer than $t$ neighbours of a vertex in $A_i$.

Write $\delta (G^{\prime })\ge (\alpha -{\gamma }_n)m$, with ${\gamma }_n\to 0$. Counting missing incidences with $A_i$ givesSince ${\sum }_i|C_i|=m-o(m)$, it follows that $|C_i|=m/r+o(m)$ for every $i$. The total missing incidences from all root vertices to assigned vertices are at most $(1+r{\gamma }_n)Lm$. At least $(L-t+1)(m-o(m))$ of them are incidences with a vertex's own assigned root class. Hence only $o(Lm)$ missing incidences go to other root classes.

Choose ${\rho }_n\to 0$ slowly and discard the $o(m)$ assigned vertices with more than ${\rho }_{n}L$ missing incidences to other root classes. Each remaining vertex of $C_i$ is adjacent to all but at most ${\rho }_{n}L$ vertices of every $A_j$, $j\ne i$. If its class contained a complete bipartite graph $K_{t,t}$, the $2t$ vertices of this copy would have at least $L-2t{\rho }_{n}L\ge t$ common neighbours of a vertex in each other root class. These sets extend the copy to $K_{r+1}(t)$, a contradiction. By the bipartite case of the Erdős-Stone theorem, the number of internal edges in each surviving class is $o(m^2)$.

The surviving graph still has $\alpha m^2/2+o(m^2)$ edges, whereas the total possible cross-class edges, with these balanced classes, is $\alpha m^2/2+o(m^2)$. Thus only $o(m^2)$ cross-class edges are missing. Discard the $o(m)$ vertices missing more than ${\sigma }_{n}m$ cross-class edges, with ${\sigma }_n\to 0$ slowly, and delete all internal edges. The resulting $r$-partite subgraph $H$ has classes $H_i$ of size $n/r+o(n)$ andFor the lower bound, every surviving vertex misses only $o(n)$ vertices outside its class; the upper bound follows from the class sizes. This also records the uniform cross-class error that we need next.

Minimum degree in an extremal graph. Fix $F$ with $\chi (F)=r+1$. Choose $t$ large enough that $F$ embeds in $K_{r+1}(t)$. An extremal $F$-free graph $G$ is therefore $K_{r+1}(t)$-free, and its edge count has the required asymptotics by the Erdős-Stone theorem. For $r\ge 2$ take $H$ as above.

For any vertex $v$, remove $v$ and introduce a new vertex $v^{\ast }$ whose vertex neighbourhood consists precisely of the vertices in $H_2\cup \cdots \cup H_r$ other than $v$. The new degree of a vertex is $\alpha n-o(n)$. This new graph remains $F$-free. Indeed, a new copy of $F$ must use $v^{\ast }$, and all its neighbours of a vertex in that copy lie in the other classes of $H$. There are only boundedly many of them. Each misses only $o(n)$ vertices in $H_1$, so an unused vertex $w\in H_1$ is adjacent to all of them. Replacing $v^{\ast }$ by $w$ would give $F$ in $G-v$, which is impossible.

Extremality now forces $d_G(v)\ge \alpha n-o(n)$, uniformly in $v$. The upper bound for the minimum degree of a graph follows from $\delta (G)\le 2e(G)/n$. For $r=1$, $e(G)=o(n^2)$ already gives $0\le \delta (G)=o(n)$. ThusThis is the minimum degree of an extremal forbidden-subgraph graph principle.

The exact extremal graph for disjoint cliques. DefineThe plus sign denotes the join of graphs. The Turan graph contains no $K_{r+1}$, so every such clique in $J_s(n)$ uses a vertex of $K_{s-1}$. Therefore $J_s(n)$ contains no $s$ vertex-disjoint such cliques, and the extremal number is at least $b_s(n)$.

Induct on $s$, with $s=1$ supplied by the Turan theorem, including its equality characterization. Let $s\ge 2$ and let $G$ be extremal for $s{K}_{r+1}$. The preceding results give $\delta (G)\ge \alpha n-o(n)$ and the $r$-partite subgraph $H$. Keep its classes $H_i$ fixed and assign each of the remaining $o(n)$ vertices to a class in which it has fewest neighbours of a vertex in $H$. Write the resulting classes as $C_i$; all have size $n/r+o(n)$. Choose a positive ${\tau }_n\to 0$ so slowly that ${\tau }_{n}n$ dominates the outside count, every cross-class error in $H$, and every error in the minimum degree of a graph bound.

Suppose some vertex $x\in C_i$ has at least ${\tau }_{n}n$ internal neighbours of a vertex. If $x\notin H$, the assignment rule shows that $x$ has at least ${\tau }_{n}n-o(n)$ neighbours of a vertex in every $H_j$. If $x\in H_i$, it has this many in $H_i$ and $n/r-o(n)$ in the other $H_j$. If $G-x$ contained $s-1$ disjoint $K_{r+1}$, we could choose one vertex in each $N(x)\cap H_j$, successively avoiding those copies and the $o(n)$ missing cross-neighbours of the already chosen vertices. This produces a further $K_{r+1}$ through $x$. For $r=1$, this just chooses an unused neighbour of $x$. Hence $G-x$ is $(s-1)K_{r+1}$-free, and induction givesSince the candidate attains $b_s(n)$, equality holds throughout: $x$ is a universal vertex and $G-x$ is the inductively unique extremal graph. Thus $G$ is isomorphic to $J_s(n)$.

It remains to show that this case must occur. Otherwise all internal vertex degrees are less than ${\tau }_{n}n=o(n)$. Together with the lower bound on $\delta (G)$ and $|C_i|=n/r+o(n)$, this implies that every vertex misses only $o(n)$ vertices outside its own class. If some $C_i$ contains a matching in a graph of $s$ edges, extend its edges one by one to $K_{r+1}$, choosing one vertex from each other class. At every step only $o(n)$ cross-neighbours and boundedly many used vertices are excluded. This gives $s$ disjoint forbidden cliques, a contradiction.

A maximal matching in each class therefore has at most $s-1$ edges, and its at most $2(s-1)$ endpoints form a vertex cover of the internal edges. Their internal vertex degrees are all $o(n)$, so the total internal edge count is $o(n)$. The cross-class edge count is at most $e(T_r(n))$, whenceBut the balanced part sizes of the Turan graph givecontradicting extremality for $s\ge 2$. This argument also works for $r=1$. Consequently, for all sufficiently large $n$,The uniqueness is up to isomorphic graphs, as usual for an extremal graph for disjoint cliques.