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This is the part-preserving homomorphism density of a four-cycle in a bipartite graph with independently uniform labels $x,x^{\prime }$ in the first part and $y,y^{\prime }$ in the second. Repeated labels are included. It equals the fourth power of the four-cycle norm of the adjacency indicator function. It differs at finite sizes from a count requiring four distinct vertices.

Use the part-preserving bipartite four-cycle density conventionHere $G(x,y)$ is the indicator function of adjacency, and repeated labels are allowed. This is the homomorphism density convention appropriate to the four-cycle norm. We derive the needed counting and operator norm facts directly.

Let $M=(G(x,y))$ be the $n$-by-$n$ biadjacency matrix, let $1$ be the all-ones vector, let $J=11^{\mathsf{T}}$, and put $H=M-\delta J$. The degree hypothesis saysso the centered matrix $H$ annihilates $1$ on both sides. In particular, $HJ=J{H}^{\mathsf{T}}=0$, andThe two terms have zero products in either order. Expanding the matrix trace directly counts four adjacency factors:Since $J^2=nJ$ and $\mathrm{tr}J=n$, we obtainThe given upper bound therefore impliesThis computation proves the centered four-cycle identity for a biregular graph, rather than assuming a four-cycle norm identity.

The matrix $Q=H{H}^{\mathsf{T}}$ is a positive semidefinite matrix. By the spectral theorem for real symmetric matrices it has nonnegative eigenvalues ${\lambda }_j$, andMoreover, for the Euclidean norm,Indeed $\Vert H^{\mathsf{T}}v{\Vert }_2^2=v^{\mathsf{T}}Qv$, whose maximum on unit vectors is ${\lambda }_{\max }$; the Cauchy-Schwarz inequality characterization $\Vert H{\Vert }_{\mathrm{op}}={\sup }_{\Vert u{\Vert }_2=\Vert v{\Vert }_2=1}|u^{\mathsf{T}}Hv|$ also shows that $H$ and $H^{\mathsf{T}}$ have the same operator norm. Taking fourth roots thus giveswhere the error parameter $c$ is nonnegative, as in the claimed bound.

Finally use the indicator vectors $a=1_A$, $b=1_B$. Their Euclidean norms are $\sqrt{|A|}$ and $\sqrt{|B|}$, so the Cauchy-Schwarz inequality and the operator norm estimate yieldThis is the required discrepancy bound. It also holds when either subset is empty or $\delta =0$. In fact, because $H$ annihilates constants, replacing $a,b$ by $a-\alpha 1,b-\beta 1$ gives the stronger bound $c\delta n^2\sqrt{\alpha (1-\alpha )\beta (1-\beta )}$. This is the spectral discrepancy bound for a biregular graph.

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:

Cloning a vertex replaces it by several nonadjacent vertices with identical neighbourhoods. In homomorphism density, conditioning on the images of all other vertices reduces cloning to the inequality $\mathbb{E}{p}^t\ge (\mathbb{E}p{)}^t$ from the Jensen inequality. Iterated cloning shows that positive clique homomorphism density forces any fixed complete multipartite blow-up.