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Write $K_s(t)$ for a balanced complete multipartite blow-up with $s$ classes of size $t$. In the nonvacuous range $0<\epsilon <1/r$, the answer isWe prove both bounds, including the quantitative lower bound.

First, suppose a bipartite graph with parts $A,B$, of sizes $m,N$, has at least $pmN$ edges. For any integer $s\le pm/2$, counting pairs consisting of an $s$-subset of $A$ and a common neighbour in $B$ givesThe inequality follows from discrete convexity: $\left(\genfrac{}{}{0ex}{}{x+1}{s}\right)-\left(\genfrac{}{}{0ex}{}{x}{s}\right)=\left(\genfrac{}{}{0ex}{}{x}{s-1}\right)$ is nondecreasing, so balancing integer degrees of vertices minimizes their sum. Consequently some $S$ hasThe product estimate uses $\lfloor pm\rfloor -j\ge pm-s$ for $0\le j<s$. This is a common neighbourhood from bipartite density bound.

We next prove by induction on $q\ge 1$ that, for every fixed $\alpha >0$, a graph $J$ of order $N$ with minimum degree at least $(1-1/q+\alpha )N$ contains $K_{q+1}(\lfloor c\log N\rfloor )$, for some $c=c(q,\alpha )>0$ and all sufficiently large $N$. Impossible minimum-vertex degree parameters need no argument. For $q=1$, take two copies of $V(J)$ with adjacency given by $J$. The preceding count with $p=\alpha$ and $s=\lfloor c\log N\rfloor$, where $c\log (2/\alpha )<1/2$, gives at least $\sqrt{N}$ common neighbours. The selected vertices and their common neighbours are disjoint in $J$, since a vertex is not its own neighbour. Selecting $s$ common neighbours gives $K_2(s)$.

For $q\ge 2$, the minimum degree hypothesis implies the induction hypothesis for $q-1$, for example with excess $1/[q(q-1)]$. Thus $J$ contains $K_q(m)$ with $m=\lfloor a\log N\rfloor$ for a positive constant $a$. Pair the vertices in its classes into $m$ disjoint transversal cliques $C_1,\dots ,C_m$, each of order $q$. By the union bound, every $C_i$ has at leastcommon neighbours. Form an auxiliary bipartite graph between these $m$ cliques and the $N$ vertices, using common-neighbour incidence, and put $p=q\alpha$. Choose $c>0$ small enough that $c<pa/3$ and $c\log (2/p)<1/2$. The count above, with $s=\lfloor c\log N\rfloor$, gives $s$ cliques with at least $\sqrt{N}$ simultaneous common neighbours. Their vertices form $q$ classes of size $s$, and $s$ of those common neighbours form the last class. None of the common neighbours belongs to a selected clique, because that would require a loop. This proves the logarithmic clique blow-up from minimum degree assertion.

To pass from an edge count to minimum degree, put $b=1-1/r+\epsilon$ and $a=b-\epsilon /2$. Repeatedly remove a vertex whose current degree of a vertex is below $a$ times the current order. If the process reaches order $m$, its deleted edges and remaining edges giveThus $(1-a)m^2\ge (\epsilon /2)n^2-O(n)$. In particular, for a fixed sufficiently small $\gamma >0$, the process cannot reach $m=\lfloor \gamma n\rfloor$: the displayed inequality would fail. It stops earlier with a subgraph of order $N\ge \gamma n$ and minimum degree at least $(1-1/r+\epsilon /2)N$. The induction just proved supplies $K_{r+1}(\lfloor c\log N\rfloor )$, hence a lower bound $c^{\prime }\log n$ on $t(r,\epsilon ,n)$. In particular, the guaranteed part size tends to infinity.

For the upper bound, choose a fixed $p$ with $b<p<1$ and take a binomial random graph $G(n,p)$. Its edge count has expectation $p\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ and variance $p(1-p)\left(\genfrac{}{}{0ex}{}{n}{2}\right)$. The Chebyshev inequality shows that $e(G)\ge b\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ with probability tending to one. The expected number of labeled copies of $K_{r+1}(t)$ is at mostFor $t=\lceil C\log n\rceil$ and $C>2/[r\log (1/p)]$, this tends to zero. The Markov inequality therefore shows that with positive probability both the required edge density holds and there is no such copy. This proves $t(r,\epsilon ,n)=O(\log n)$ and completes the random obstruction to larger clique blow-ups argument.

There is a necessary qualification to the density parameter. If $\epsilon =1/r$, the only eligible graph is the complete graph, and $t(r,1/r,n)=\lfloor n/(r+1)\rfloor$. If $\epsilon >1/r$ and $n\ge 2$, there are no eligible simple graphs; the universal assertion is vacuous for every $t$, so the printed maximum has no finite value in that range.

Write $K_j(t)$ for a balanced complete multipartite blow-up, with $j$ parts each containing $t$ vertices. Its occurrence is a subgraph condition, so additional internal edges in the host are harmless. The coefficient multiplying $\log n$ is essential: the PDF has $t=\lfloor d\log n\rfloor$, although it is lost in the converted TeX.

First prove the common neighbourhood from bipartite density estimate. Suppose a bipartite graph has parts of sizes $m,N$ and at least $\beta mN$ edges. For an integer $s\le \beta m/2$, count incidences between $s$-subsets of the first part and their common neighbours in the second. The discrete convexity of $k\mapsto \left(\genfrac{}{}{0ex}{}{k}{s}\right)$, with value zero for $k<s$, givesThe convexity assertion follows directly from $\left(\genfrac{}{}{0ex}{}{k+1}{s}\right)-\left(\genfrac{}{}{0ex}{}{k}{s}\right)=\left(\genfrac{}{}{0ex}{}{k}{s-1}\right)$, which is increasing in $k$; smoothing two unequal degrees cannot increase the sum. In the binomial ratio every factor is at least $\beta /2$, because $\lfloor \beta m\rfloor -s+1\ge \beta m-s\ge \beta m/2$. Consequently some $S$ satisfies

We prove the logarithmic clique blow-up from minimum degree by induction on $r$. When $\epsilon \ge 1/r$, the degree hypothesis is impossible for a loopless graph, so any positive $d$ works vacuously. Assume $0<\epsilon <1/r$. For $r=1$, take two copies of the vertex set and join the copies of adjacent vertices. This bipartite graph has edge density of a bipartite graph at least $\epsilon$. ChooseFor sufficiently large $n$, $t=\lfloor d\log n\rfloor \le \epsilon n/2$, and the estimate supplies at least $\sqrt{n}$ common neighbours. Choose $t$ of them. The two selected sets are disjoint in the original graph, since a selected vertex cannot be adjacent to itself. They give $K_2(t)$.

For $r\ge 2$, put ${\epsilon }^{\prime }=\epsilon +1/(r-1)-1/r$. The same minimum degree of a graph exceeds $(1-1/(r-1)+{\epsilon }^{\prime })n$. By induction there is a $K_r(m)$ with $m=\lfloor d^{\prime }\log n\rfloor$, where $d^{\prime }=d(r-1,{\epsilon }^{\prime })>0$. Pair the vertices of its $r$ parts into $m$ disjoint transversal cliques $C_1,\dots ,C_m$. Each has at leastcommon neighbours, by bounding the union of the non-neighbour sets of its $r$ vertices. Set $\beta =\min \{r\epsilon ,1/2\}$. Form the bipartite graph whose first part is these $m$ cliques, with $C_i$ joined to a vertex when that vertex is a common neighbour of all of $C_i$. It has density at least $\beta$. ChooseFor large $n$, $m\ge d^{\prime }\log n/2$ and $t=\lfloor d\log n\rfloor \le \beta m/2$. The estimate gives $t$ selected cliques and at least $\sqrt{n}\ge t$ common vertices. The selected cliques supply $t$ vertices in each of the original $r$ parts, and their common vertices supply the final part. No common vertex belongs to a selected clique, again because there are no loops. All required cross-part edges are present. Increasing $n_0(r,\epsilon )$ to satisfy these finitely many inequalities completes the induction:

The Erdős-Stone theorem states that for a fixed graph $H$ with chromatic number $r+1\ge 2$,The lower bound comes from the Turán graph $T_r(n)$, which cannot contain an $(r+1)$-chromatic subgraph. To obtain the upper bound from our minimum-degree result, suppose $e(G)\ge (a+\eta )n^2/2$, where $a=1-1/r$ and $0<\eta <1-a$. Repeatedly delete a vertex of degree less than $b$ times the current order, where $b=a+\eta /2$, until no such vertex remains. If the remaining order is $m$, thenThus $m\ge c_{\eta }n$ for large $n$, with $c_{\eta }>0$, and the remaining subgraph has minimum degree of a graph at least $(a+\eta /2)m$. It contains $K_{r+1}(\lfloor d\log m\rfloor )$. A proper graph colouring of $H$ places its vertices in the $r+1$ parts once their size exceeds $|V(H)|$. This contradicts $H$-freeness. Adjusting a fixed positive density margin absorbs the difference between $n^2/2$ and $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$, proving the upper bound.

The six-vertex forbidden graph is the octahedral graph $K_{2,2,2}$: the deleted matching in a graph gives its three independent pairs. Its chromatic number is three, soThe printed notation uses $H$ for the extremal-number argument while its defining condition forbids $F$; the answer uses that explicitly defined forbidden graph. For the strict finite inequality, take $T_2(n)$ and add one edge inside its larger part, which has at least two vertices for $n\ge 3$. Every triangle in a graph then uses this added edge. However, $K_{2,2,2}$ contains two vertex-disjoint triangles in a graph, obtained by taking one vertex from each pair and then the other three. It therefore cannot occur in the constructed graph. Hence