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If a fixed graph $F$ has chromatic number $r+1$, every extremal $F$-free graph $G$ on $n$ vertices has minimum degree of a graph $(1-1/r+o(1))n$. Use a high-minimum-degree multipartite stability subgraph $H$: replacing any vertex by one adjacent to all but one class of $H$ preserves $F$-freeness. Any supposed new copy of $F$ can replace the new vertex by an unused common neighbour in the omitted class. Extremality therefore gives the lower bound on every degree of a vertex; the Erdős-Stone theorem gives the upper bound on the average degree of a vertex.

The Erdős-Stone theorem says that, for a fixed graph $F$ with chromatic number $q\ge 2$,The lower bound comes from the balanced Turán graph with $q-1$ parts; the theorem supplies the matching asymptotic upper bound. Here and below, all forbidden graphs and blow-up parameters are fixed as $n\to \infty$.

First obtain an almost spanning, almost complete multipartite core. Put ${\alpha }_r=1-1/r$. We give the stability argument, rather than infer pointwise degree bounds by differentiating an asymptotic edge-count formula. Suppose first $r\ge 2$.

We need two elementary ingredients. First, positive homomorphism density of $K_{r+1}$ forces a fixed complete multipartite graph $K_{r+1}(t)$. To see this, perform vertex cloning in a graph, replacing one vertex by $t$ independent twins. Conditional on the images of the other vertices, its contribution to the homomorphism density changes from a number $p$ to $p^t$; Jensen inequality gives $\mathbb{E}{p}^t\ge (\mathbb{E}p{)}^t$. Duplicating each original vertex therefore givesA positive lower bound on the right supplies a positive proportion of all maps of the blow-up. Only $O(n^{(r+1)t-1})$ maps fail to be injective, so for large $n$ at least one map is an embedding. Consequently a $K_{r+1}(t)$-free sequence has $o(n^{r+1})$ copies of $K_{r+1}$. The clique removal lemma, proved in the next solution, then deletes $o(n^2)$ edges to give a $K_{r+1}$-free graph $J_n$ with $e(J_n)={\alpha }_r{n}^2/2-o(n^2)$.

Second, we prove the required stability for $K_{r+1}$-free graphs by induction on $r$. For $r=1$ such a graph is edgeless. In an $m$-vertex $K_{r+1}$-free graph $J$, take a vertex of maximum vertex degree $D$, put $A=N(v)$ and $B=V(J)\setminus A$, and note that $J[A]$ is $K_r$-free. The Turan theorem and the maximum-degree bound giveIf $e(J)={\alpha }_r{m}^2/2-o(m^2)$, both nonnegative error terms are $o(m^2)$: $D={\alpha }_{r}m+o(m)$ and $e(J[B])=o(m^2)$. The deficit in the bound for $e(J[A])$ is likewise $o(m^2)$. Induction partitions $A$ into $r-1$ classes with a total of $o(m^2)$ internal edges; adjoining $B$ gives an $r$-class partition with the same property. Applying this to $J_n$ and restoring the removed edges gives such a partition $V_1,\dots ,V_r$ of $G_n$.

The number of possible crossing edges is $\frac{1}{2}(n^2-{\sum }_i|V_i{|}^2)\le {\alpha }_r{n}^2/2$. Since the actual crossing-edge count is within $o(n^2)$ of this maximum, $|V_i|=n/r+o(n)$ for every $i$, and there are $o(n^2)$ missing crossing edges. Choose $a_n\to 0$ so slowly that the number of vertices missing more than $a_{n}n$ neighbours in other classes is $o(n)$. Delete those vertices and all internal edges. The remaining r-partite graph $Q_n$ has $n-o(n)$ vertices and every vertex has crossing vertex degree at least ${\alpha }_{r}n-o(n)$. The Turan theorem, or the size of a largest part, supplies the matching upper bound for its minimum vertex degree. ThusThis is a high-minimum-degree multipartite stability subgraph. The subgraph need not be spanning; deleting the exceptional $o(n)$ vertices is essential. For $r=1$, the spanning edgeless subgraph already proves the assertion.

Extremality forces the minimum degree. Choose $t\ge |V(F)|$. A proper $(r+1)$-graph colouring embeds $F$ in $K_{r+1}(t)$, so every $F$-free $H_n$ is also $K_{r+1}(t)$-free. By the Erdős-Stone theorem, its extremal edge count has the needed asymptotic size. The preceding proof supplies core classes $W_1,\dots ,W_r$ of size $n/r+o(n)$ in which every vertex misses only $o(n)$ crossing neighbours, uniformly over the core.

Delete an arbitrary vertex $x$ from $H_n$ and introduce a new vertex $z$ adjacent precisely to $W_2\cup \cdots \cup W_r$ with $x$ removed if necessary. The new graph is still $F$-free. Indeed any new copy would use $z$, all its neighbours in that copy would be in the specified core classes, and their set of common neighbours in $W_1$ has size $n/r-o(n)$. A vertex of this common neighbourhood outside the bounded vertex set of the copy replaces $z$, producing an existing copy of $F$ in $H_n$, a contradiction. For $r=1$ the new vertex is isolated and the same conclusion is immediate.

The replacement adds ${\alpha }_{r}n-o(n)$ edges. Since $H_n$ has the maximum possible number of edges, $d_{H_n}(x)\ge {\alpha }_{r}n-o(n)$. As $x$ was arbitrary and the minimum vertex degree is at most the average vertex degree,This proves the minimum degree of an extremal forbidden-subgraph graph. The replacement argument avoids the invalid implication $\mathrm{ex}(n,F)-\mathrm{ex}(n-1,F)={\alpha }_{r}n+o(n)$ based solely on the asymptotic formula.

A colour-critical vertex also bounds the maximum degree. Suppose a vertex $x$ of $H_n$ has $d(x)\ge {\alpha }_{r}n+\epsilon n$ for some fixed $\epsilon >0$. Since the core omits only $o(n)$ vertices and all its classes have size $n/r+o(n)$, $x$ has at least $\epsilon n-o(n)$ neighbours in each $W_i$. Otherwise, even counting every vertex in all other classes and outside the core would give too small a vertex degree.

Inside these $r$ neighbourhood sets, greedily choose a fixed number $t\ge |V(F)|$ of vertices in each class. Every previously chosen vertex excludes only $o(n)$ crossing nonneighbours, so each next candidate set still has positive linear size. This produces $K_r(t)$ inside $N(x)$. A proper $r$-graph colouring of $F-v$ embeds it in that complete multipartite graph; map $v$ to $x$. All required incident edges exist, so this embeds $F$, a contradiction. The case $r=1$ simply uses enough neighbours to embed the edgeless $F-v$. Hence $\Delta (H_n)\le {\alpha }_{r}n+o(n)$, and the average-degree lower bound completes the answer:

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 Turán 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 Turán 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.