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A biregular graph is a bipartite graph whose vertices have constant degree on each part separately. The two degree constants can differ. If its parts are $X,Y$ and its edge density of a bipartite graph is $\gamma$, the two degrees are $\gamma |Y|$ and $\gamma |X|$. Thus its adjacency function has constant row and column averages $\gamma$.

Partition both sides of a half graph into classes of size at least $100$. A class of size $s$ has at most $2\lceil s/10\rceil$ elements with fewer than $s/10$ class elements on one side in the underlying order. If there are $k_X$ and $k_Y$ classes, the union of these boundary elements has size at most $2n/5+2(k_X+k_Y)\le 11n/25$. Some index $m$ is therefore away from both boundaries. The class containing $m$ on each side has at least one tenth of its points strictly above $m$ and strictly below $m$. Choosing lower $X$ points with upper $Y$ points, or upper $X$ points with lower $Y$ points, yields edge density of a bipartite graph values one and zero. Both cannot be within $1/10$ of the same class-pair edge density of a bipartite graph, so that pair is not a $1/10$-regular pair of vertex sets.

In a proof of the Szemerédi regularity lemma, suppose $k$ equal classes have size $L$ and witness splitting creates at most $2^k$ atoms per class. Divide every atom into blocks of size $\ell =\lfloor L/4^k\rfloor$, putting leftovers into the exceptional set. At most $k{2}^k\ell \le kL{2}^{-k}$ vertices are lost. When $L\ge 2\cdot 4^k$, at most $2k{4}^k$ blocks remain. For equitable regularity energy defined by omitting exceptional vertices, the loss from discarding a fraction $r$ is at most $2r$: only ordered pairs incident with discarded vertices are removed, and every squared edge density of a bipartite graph is at most one. This provides a quantitative way to restore equal class sizes after each energy increment.

Equip the function spaces of the two nonempty parts of a bipartite graph with uniform inner products. Its normalized adjacency operator is $({\theta }_{G}v)(x)={\mathbb{E}}_{y}G(x,y)v(y)$. In orthonormal coordinate bases, its matrix is $G/\sqrt{|X||Y|}$, not the unscaled adjacency matrix. For a biregular graph of edge density of a bipartite graph $\gamma$, constant functions give a largest singular value $\gamma$, and the remaining singular values belong to the centered kernel $G-\gamma$.

For disjoint vertex sets $A,B$, define their edge density of a bipartite graph by $d(A,B)=e(A,B)/(|A||B|)$. They form a regular pair with parameter $\epsilon$ if every $A^{\prime }\subseteq A,B^{\prime }\subseteq B$ with $|A^{\prime }|\ge \epsilon |A|$, $|B^{\prime }|\ge \epsilon |B|$ satisfies $|d(A^{\prime },B^{\prime })-d(A,B)|\le \epsilon$.

The Szemerédi regularity lemma says that for every $\epsilon >0$ and minimum cell count $k_0$, some $K,n_0$ have the following property. Every graph on $n\ge n_0$ vertices admitswith $|V_0|\le \epsilon n$, equal-sized $V_1,\dots ,V_k$, and at most $\epsilon k^2$ unordered irregular pairs. It is enough to consider $0<\epsilon <1/2$.

Prove this using equitable regularity energy. Treat exceptional vertices as singleton cells and setwith ordered pairs and the adjacency indicator also defining diagonal densities. Then $0\le q\le 1$. The refinement variance identity for regularity energy proves refinement cannot decrease $q$:An irregular pair supplies witness subsets of relative sizes at least $\epsilon$ whose density differs by more than $\epsilon$. Refining along those witnesses raises $q$ by more than ${\epsilon }^4{L}^2/n^2$ per orientation, where $L$ is the common cell size. If there are more than $\epsilon k^2$ irregular pairs, the increase is greater than ${\epsilon }^5(kL/n{)}^2\ge {\epsilon }^5/4$, provided the exceptional set has size at most $n/2$.

Refine each cell by all its witness subsets; this produces at most $2^k$ atoms per cell. Restore equal sizes by the equalization with a controlled exceptional set: cut each atom into blocks of size $\lfloor L/4^k\rfloor$ and declare its remainder exceptional, still retaining every such vertex as a singleton in the energy. This is another refinement, so no energy is lost. At most $k{2}^k\lfloor L/4^k\rfloor \le n{2}^{-k}$ vertices become exceptional, and, when $L\ge 2\cdot 4^k$, at most $2k{4}^k$ new nonexceptional cells occur.

Put $T=\lceil 4{\epsilon }^{-5}\rceil$ and start with $k\ge k_0$ so large that $T{2}^{-k}\le \epsilon /2$. An initial equitable partition has at most $\epsilon n/2$ exceptional vertices when $n$ is sufficiently large. Subsequent cell counts only increase, so the total added exceptional set has size at most $\epsilon n/2$. The recurrence $k\mapsto 2k{4}^k$, iterated at most $T$ times, supplies a finite bound $K$; choose $n_0$ large enough that every intermediate $L$ meets the size requirement. More than $T$ unsuccessful refinements would raise $q$ above one. The process must therefore stop with the required regular pairs. This proves the lemma.

Its usefulness comes from replacing an arbitrarily large graph by the bounded reduced graph of a regularity partition. For fixed positive cross-density, the graph embedding lemma for regular pairs lifts fixed configurations from the reduced graph to the original one. In particular, three sufficiently regular positive-density pairs contain many triangles in a graph. Deleting exceptional, intracell, irregular and sparse-pair edges then gives the triangle removal lemma: a graph with sufficiently few triangles can be made triangle-free by deleting $o(n^2)$ edges. A triangle surviving in the reduced graph would force a positive proportion of the original possible triangles, a contradiction. More generally this approach yields the clique removal lemma and the asymptotic Erdős-Stone theorem. Applied to the tripartite graph encoding $x+z=2y$, triangle removal also gives Roth theorem on three-term arithmetic progressions. The price is that the energy argument gives a very rapidly growing, iterated-exponential bound on $K(\epsilon )$; the lemma is a structural tool, not a promise of small practical constants.

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

We prove a weighted version of the Szemerédi regularity lemma, which is sufficient for the grid application. For disjoint nonempty vertex sets $U,W$ in a graph, write $d(U,W)=e(U,W)/(|U||W|)$ for their edge density of a bipartite graph. Call the pair $\epsilon$-regular if every $U^{\prime }\subseteq U$, $W^{\prime }\subseteq W$ with $|U^{\prime }|\ge \epsilon |U|$ and $|W^{\prime }|\ge \epsilon |W|$ satisfies $|d(U^{\prime },W^{\prime })-d(U,W)|\le \epsilon$.

The weighted Szemerédi regularity lemma states that, for every $0<\epsilon <1$, there are integers $M,v_0$ such that every graph on $v\ge v_0$ vertices has a set partition into at most $M$ nonempty cells $V_1,\dots ,V_k$, each of size at most $\epsilon v$, for whichThe sum is ordered. The cells need not have equal sizes; their weights replace the usual unweighted count of bad pairs.

To prove it, start with $k_0=\lceil 2/\epsilon \rceil$ nearly equal cells, taking $v\ge k_0$. Each has size at most $v/k_0+1\le \epsilon v$. Define the regularity energyHere even diagonal rectangles use the mean adjacency indicator, counting ordered pairs of vertices. Thus $0\le q\le 1$. Under a refinement, the old rectangle density is the weighted mean of the new ones, and the refinement variance identity for regularity energy givesIn particular refinement cannot decrease energy.

If the irregular-pair weight exceeds $\epsilon v^2$, choose witness subsets $U^{\prime }\subseteq U$, $W^{\prime }\subseteq W$ for every irregular unordered pair, and split each cell simultaneously by all its witnesses. There are at most $k-1$ such subsets in any one cell, so the new number of cells is at most $k{2}^k$. In an irregular rectangle, its witness has relative area at least ${\epsilon }^2$ and density discrepancy greater than $\epsilon$. Since the witness rectangle is now a union of refined rectangles, the Cauchy-Schwarz inequality applied to their weighted density discrepancies bounds the variance contribution from below bySumming over both orientations of all irregular pairs raises $q$ by more than ${\epsilon }^5$. Since $q\le 1$, after at most $\lceil {\epsilon }^{-5}\rceil$ rounds the required weight bound must hold. The finite recurrence $k\mapsto k{2}^k$, starting at $k_0$, supplies $M(\epsilon )$; all cells remain small because we only refine. This proves the stated weighted Szemerédi regularity lemma.

We next derive the triangle removal lemma from this version. First, three disjoint classes $U,W,Z$ whose pairs are $\epsilon$-regular pairs of density at least $d$ have many triangles if $\epsilon \le d/2$ and $\epsilon \le 1/4$. Fewer than $\epsilon |U|$ vertices of $U$ have fewer than $(d-\epsilon )|W|$ neighbours in $W$: otherwise that set of vertices, paired with all of $W$, would violate regularity. The same holds for $Z$. For each of the at least $(1-2\epsilon )|U|$ remaining vertices, both neighbour sets have relative size at least $d-\epsilon \ge \epsilon$. Regularity of $(W,Z)$ then supplies at least $(d-\epsilon )$ times the product of those two neighbour-set sizes edges. Thus the regular triangle counting lemma, also for unequal class sizes, givesFor a desired deletion tolerance $0<\eta <1$, take $\epsilon =\eta /100$, $d=\eta /4$, and a weighted regularity partition with at most $M=M(\epsilon )$ cells. Put $\tau =\epsilon /M$. Delete edges inside cells, between irregular pairs, between regular pairs of density less than $d$, and incident to cells smaller than $\tau v$. The first two deletions each cost at most $\epsilon v^2/2$. The sparse-pair deletion costs at most $d{v}^2/2$. For the last deletion, small-cell deletion in weighted regularity removes at most $M\tau v=\epsilon v$ vertices and at most $\epsilon v^2$ incident edges. The total is bounded byA triangle surviving these deletions would lie in three distinct cells, all of size at least $\tau v$, with all their pairs regular and of density at least $d$. The original graph would then have at least $d^3{\tau }^3{v}^3/16$ triangles. Therefore, witha graph on sufficiently many vertices with at most $\kappa v^3$ triangles can be made triangle-free by deleting fewer than $\eta v^2$ edges. This establishes the triangle removal lemma with every parameter depending only on $\eta$.

Now apply a tripartite graph encoding of a grid. Take separate vertex classes $X=[N]$, $Y=[N]$, $Z=\{2,\dots ,2N\}$. For each grid point $(x,y)$ in the given set, put in the three edgeswhere their endpoints belong respectively to $X\times Y$, $X\times Z$, and $Y\times Z$. Each point supplies a triangle. These $|A|$ triangles are edge-disjoint triangles, since any one of the three edge types uniquely determines the point that supplied it.

Conversely, any triangle with vertices $x\in X$, $y\in Y$, $z\in Z$ forces the three pointsto belong to $A$. Put $d=z-x-y$. For $d\ne 0$ these form a corner in an integer grid; for $d=0$ they all arise from the same grid point. Thus if there is no corner, the graph has exactly $|A|\le N^2$ triangles, yet destroying all triangles requires at least $|A|\ge \delta N^2$ edge deletions.

We may suppose $0<\delta \le 1$, since larger densities make the request vacuous. Use the triangle removal lemma with $\eta =\delta /32$. Our graph has $v=4N-1$ vertices, so, for sufficiently large $N$, its bound $N^2$ on the number of triangles is at most $\kappa v^3$. The lemma would delete fewer thanedges to destroy all triangles. This contradicts the lower bound $\delta N^2$ from the edge-disjoint family. Hence every sufficiently large positive-density square grid contains a nondegenerate corner, proving the requested corners theorem and in particular the existence of a suitable $N$.

For disjoint nonempty vertex sets $U,W$, define their edge density of a bipartite graph by $d(U,W)=e(U,W)/(|U||W|)$. They form an $\epsilon$-regular pair of vertex sets if every $X\subseteq U$, $Y\subseteq W$ with $|X|\ge \epsilon |U|$, $|Y|\ge \epsilon |W|$ satisfies $|d(X,Y)-d(U,W)|\le \epsilon$.

The Szemerédi regularity lemma asserts that, given $\epsilon >0$ and a positive integer $m_0$, there are $M,n_0$ such that every graph on $n\ge n_0$ vertices has a set partition $V_0,V_1,\dots ,V_k$ withand at most $\epsilon k^2$ irregular ordered pairs $(V_i,V_j)$ with $1\le i\ne j\le k$.

Assume $0<\epsilon \le 1/2$, which suffices since a smaller regularity parameter gives the claim for a larger one. Treat the exceptional vertices as singleton cells in the full partition $\mathcal{P}$, and use equitable regularity energyHere the adjacency indicator function defines density even for diagonal cells. Clearly $0\le q\le 1$. If ${\mathcal{P}}^{\prime }$ refines $\mathcal{P}$, weighted averaging of densities gives the refinement variance identity for regularity energyTo verify it, expand each square and use ${\sum }_{A^{\prime },B^{\prime }}|A^{\prime }||B^{\prime }|d(A^{\prime },B^{\prime })=|A||B|d(A,B)$.

For every irregular ordered pair of nonexceptional cells, choose witnesses $X_{ij},Y_{ij}$ of relative sizes at least $\epsilon$ and density discrepancy greater than $\epsilon$. Split each $V_i$ along all witnesses involving it; symmetry allows at most one witness for each other cell, hence at most $2^k$ atoms in each $V_i$. In the witness rectangle, the weighted Cauchy-Schwarz inequality in the variance identity contributes more than ${\epsilon }^4|V_i||V_j|/n^2$.

If more than $\epsilon k^2$ pairs are irregular and the equal cell size is $L$, the energy increase is greater thanRestore equal cell sizes by cutting each witness atom into pieces of size $\ell =\lfloor L/4^k\rfloor$, with leftover vertices added to $V_0$ as singleton cells. This operation is a further refinement, so the energy does not decrease. There are at most $k{2}^k$ atoms, and fewer than $\ell$ leftover vertices in each, giving at most $k{2}^k\ell \le n{2}^{-k}$ new exceptional vertices.

Here is a uniform bound ensuring that the iteration is legitimate. Put $R=\lceil 4{\epsilon }^{-5}\rceil$ and chooseStart with $k_0$ equal classes and fewer than $k_0$ exceptional vertices. Taking $n\ge 2k_0/\epsilon$ makes the initial exception at most $\epsilon n/2$. At every refinement the number of nonexceptional cells does not decrease: each old cell loses at most $L{2}^{-k}\le L/2$ vertices and is cut into pieces of size at most $L/4^k$. Thus each of the at most $R$ rounds adds at most $\epsilon n/(2R)$ exceptional vertices.

When $L\ge 2\cdot 4^k$, the new cell count is at mostusing $|V_0|\le \epsilon n\le n/2$. Define $K_0=k_0$ and $K_{j+1}=4K_j{4}^{K_j}$, and take $M=K_R$. Choosing $n_0\ge \max \{2k_0/\epsilon ,4M{4}^M\}$ ensures $L\ge n/(2M)\ge 2\cdot 4^M$ at every stage. All rounds are therefore defined, with cell count between $m_0$ and $M$ and exceptional size at most $\epsilon n$.

If the lemma still failed after $R$ refinements, the energy would have increased by more than $R{\epsilon }^5/4\ge 1$, impossible because it lies in $[0,1]$. Hence some stage has the required regular pairs. This completes the energy-increment proof of Szemerédi regularity, including the equal-size and exceptional-set conditions.

We derive the triangle removal lemma from the permitted Szemerédi regularity lemma, then turn a corner in an integer grid into a triangle in a graph.

The triangle removal lemma says that for every $\eta >0$ there are $\rho >0$ and $N_0$ such that a graph on $N\ge N_0$ vertices with fewer than $\rho N^3$ triangles in a graph can be made triangle-free by deleting at most $\eta N^2$ edges. Here triangles in a graph are unordered triples of distinct vertices. We prove this consequence with the necessary quantitative dependence.

We may assume $0<\eta <1$. Put $d=\eta /4$, choose an integer $m_0\ge 4/\eta$, and chooseApply the Szemerédi regularity lemma to obtain an exceptional class $V_0$ of size at most $\epsilon N$ and equal-sized classes $V_1,\dots ,V_k$ of size $L$, where $m_0\le k\le M$, and at most $\epsilon k^2$ pairs are not regular pairs of vertex sets. The constant $M$ depends only on these chosen parameters. Delete all edges incident with $V_0$, all edges within a class, all edges between irregular pairs, and all edges between pairs whose edge density of a bipartite graph is less than $d$. The respective costs are at mostTheir sum is at most $\eta N^2/2$, hence certainly at most $\eta N^2$.

If a triangle in a graph survives, it lies in three distinct classes whose three pairs are $\epsilon$-regular pairs of vertex sets of edge density of a bipartite graph at least $d$ in the original graph. We now check the needed regular triangle counting lemma. In the first class all but at most $2\epsilon L$ vertices have at least $(d-\epsilon )L$ neighbours in each of the other two classes; otherwise the definition of a regular pair of vertex sets would be violated. For each such vertex $v$, its two neighbour sets each have size at least $\epsilon L$, so their mutual edge density of a bipartite graph is at least $d-\epsilon$. It follows that the original three classes contain at leasttriangles in a graph. The elementary lower bound holds because $\epsilon \le 1/4$ and $\epsilon \le d/4$. Also $L=(N-|V_0|)/k\ge N/(2M)$. Thus a surviving triangle in a graph implies at least $d^3{N}^3/(64M^3)$ original triangles in a graph. Set $\rho =d^3/(128M^3)$ and take $N_0$ large enough for the Szemerédi regularity lemma. Fewer than $\rho N^3$ original triangles in a graph force the cleaned graph to be triangle-free, proving the triangle removal lemma.

Now suppose $0<\delta \le 1$ and $A\subseteq [n{]}^2$. Use the tripartite graph encoding of a grid to construct a tripartite graph with disjoint labelled classesFor each $(a,b)\in A$, put in the three edges joining $a\in X$, $b\in Y$, and $a+b\in Z$. These yield $|A|$ canonical triangles in a graph. They are pairwise edge-disjoint triangles: an $XY$ edge determines $(a,b)$ directly, an $XZ$ edge determines $b=z-a$, and a $YZ$ edge determines $a=z-b$. Any deletion making the graph triangle-free must therefore remove at least $|A|$ edges.

On the other hand, an arbitrary triangle in a graph with labels $(x,y,z)$ gives three points of $A$:Writing $d=z-x-y$, these are $(x,y),(x,y+d),(x+d,y)$. If $d\ne 0$, they form the required corner in an integer grid. If there is no such corner in an integer grid, every triangle in a graph is canonical, and the total number is precisely $|A|\le n^2$.

Apply the triangle removal lemma with $\eta =\delta /32$. The constructed graph has $N=4n$ vertices. For sufficiently large $n$, $N\ge N_0$ andIn the absence of a corner in an integer grid, the triangle removal lemma would destroy all triangles in a graph by deleting at mostedges, contradicting the pairwise edge-disjoint triangles. Every sufficiently large grid therefore has the asserted corner in an integer grid in every subset of positive fixed density of a finite subset. For $\delta >1$ there is no eligible subset, so the assertion is vacuous. This proves the corners theorem, including the stipulated nonzero displacement; no positivity of $d$ was required.

Write $d(U,V)=e(U,V)/(|U||V|)$ for the edge density of a bipartite graph between nonempty disjoint vertex sets. A regular pair of vertex sets $(U,V)$ is $\epsilon$-regular if $|d(U^{\prime },V^{\prime })-d(U,V)|\le \epsilon$ whenever $U^{\prime }\subseteq U$, $V^{\prime }\subseteq V$, $|U^{\prime }|\ge \epsilon |U|$, and $|V^{\prime }|\ge \epsilon |V|$.

The Szemerédi regularity lemma says that for every $\epsilon >0$ and positive integer $m_0$ there is $M=M(\epsilon ,m_0)$ such that every finite simple graph on $N\ge m_0$ vertices has a set partitionwhere $V_1,\dots ,V_k$ have equal positive size and at most $\epsilon k^2$ unordered pairs $(V_i,V_j)$, $i<j$, fail to be regular pairs of vertex sets. The version requiring only sufficiently large $N$ is equivalent, since the finitely many smaller admissible sizes can use singleton sets as classes. It suffices to prove this for $0<\epsilon \le 1/2$: reducing a larger parameter to $1/2$ gives stronger conclusions.

Here is a complete energy proof, including restoration of equal class sizes. For a set partition with exceptional set $V_0$, use the following version of equitable regularity energy, omitting pairs incident with $V_0$:For $i=j$, define $d(V_i,V_i)$ using the ordered adjacency indicator function, with diagonal entries zero. Thus the same formula works on every rectangle. The refinement variance identity for regularity energy iswhere $A,B$ run over the refined cells. Expanding the square proves the identity because their weighted mean edge density of a bipartite graph is $d(U,V)$. In particular, refinement never decreases the energy. If a union of refined cells $U^{\prime }\times V^{\prime }$ has discrepancy greater than $\epsilon$, the Cauchy-Schwarz inequality bounds its contribution below by $|U^{\prime }||V^{\prime }|{\epsilon }^2/N^2$. For an irregular equal-class pair of size $L$, witnesses have sizes at least $\epsilon L$, so the increase exceeds ${\epsilon }^4{L}^2/N^2$ in each orientation.

Choose $k_0\ge \max \{m_0,2\}$ so large that $2^{-k_0}\le {\epsilon }^6/100$. Initially partition into $k_0$ equal classes and fewer than $k_0$ exceptional vertices. For sufficiently large $N$ this exceptional set has size at most $\epsilon N/2$. Suppose there are more than $\epsilon k^2$ irregular unordered pairs at a subsequent step, with equal class size $L$. Choose one witnessing pair of subsets for every such pair. Split each class according to membership in all its witness subsets; each class produces at most $2^k$ atoms. The preceding refinement variance identity for regularity energy shows that the energy increase exceedsas long as the exceptional set has size at most $\epsilon N\le N/2$.

Apply equalization with a controlled exceptional set. Set $\ell =\lfloor L/4^k\rfloor$, cut every atom into blocks of size $\ell$, and send its leftover points to $V_0$. Provided $L\ge 2\cdot 4^k$, the new number $k^{\prime }$ of classes satisfiesThe lower bound holds because every old class retains at least one full block: its leftovers total less than $2^k\ell <L$. The discarded fraction is at mostCutting blocks is a further refinement before discarding. Discarding changes only ordered pairs with at least one discarded vertex, so it loses at most $2r$ in the energy. The net increase is therefore at least ${\epsilon }^5/2-2{\epsilon }^6/100\ge {\epsilon }^5/4$.

There can be fewer than $T=\lceil 4{\epsilon }^{-5}\rceil$ such strict energy increases, because the energy stays in $[0,1]$. Through $T$ steps the exceptional fraction is at mostThis also verifies the exceptional-set bound used at every preceding step. Iterating the finite recursion $k\mapsto 2k{4}^k$ at most $T$ times bounds all class counts by some $K=K(\epsilon ,m_0)$. Choosing $N$ sufficiently large in terms of $K$, for instance with $N\ge 4K{4}^K$ and $N\ge 2k_0/\epsilon$, guarantees $L=(N-|V_0|)/k\ge N/(2K)\ge 2\cdot 4^k$ throughout. Thus the process terminates with the desired regular pairs of vertex sets. Choose a finite integer threshold $N_{\ast }$ for these requirements and set $M\ge \max \{K,N_{\ast }\}$. For $m_0\le N<N_{\ast }$, use $N$ singleton sets; every pair is regular because its only admissible nonempty subsets are the whole pair. This completes the Szemerédi regularity lemma for all admissible $N$.

For the half graph, a class of size $s$ has exactly $\lceil s/10\rceil$ elements with fewer than $s/10$ predecessors, and the same number with fewer than $s/10$ successors. Thus its boundary contains at most $2\lceil s/10\rceil$ elements. Let $k_X,k_Y$ be the numbers of classes. The boundary rank obstruction in a half graph gives the union boundbecause $k_X,k_Y\le n/100$. There is therefore an index $m$ away from both boundaries. In its classes $X_i,Y_j$, all four subsets $X^{-}=\{x<m\}$, $X^{+}=\{x>m\}$, $Y^{-}=\{y<m\}$, $Y^{+}=\{y>m\}$ have at least one tenth of their respective class sizes. These subsets are taken within $X_i,Y_j$, which need not be intervals.

In the half graph, $d(X^{-},Y^{+})=1$ and $d(X^{+},Y^{-})=0$. If $(X_i,Y_j)$ were a $1/10$-regular pair of vertex sets, both values would be within $1/10$ of $d(X_i,Y_j)$, implying $1\le 2/10$, a contradiction. At least one class pair is not $1/10$-regular.

Uniform expectations require nonempty finite sets; assume this throughout the analytic argument. For real-valued functions on $X\times Y$, define the box norm byAll four variables are sampled as independent random variables with replacement, so repeated coordinates are included. The defining fourth power equalsIf it vanishes, every summand vanishes. In particular, the terms $y_0=y_1=y$ give ${\mathbb{E}}_{x}f(x,y{)}^2=0$ for each $y$, so $f=0$. Absolute homogeneity of a norm follows directly: $\Vert af{\Vert }_{\square }=|a|\Vert f{\Vert }_{\square }$.

For the triangle inequality, prove the mixed box Cauchy-Schwarz inequality. WriteSeparating the two $y$ averages and applying the Cauchy-Schwarz inequality in $(x_0,x_1)$ givesExpanding the square and reversing the order of finite sums,by a second Cauchy-Schwarz inequality. Thus $|\Lambda |\le {\prod }_{i,j}\Vert f_{ij}{\Vert }_{\square }$. Expand $\Lambda (f+g,f+g,f+g,f+g)$ into its sixteen multilinear terms and apply this bound to each term. Their upper bounds sum to $(\Vert f{\Vert }_{\square }+\Vert g{\Vert }_{\square }{)}^4$. Taking fourth roots yields the triangle inequality. Together with definiteness of a norm and absolute homogeneity of a norm, this proves the box norm is a norm. Complex functions require conjugates and are outside the printed real-valued convention.

For the bilinear correlation bound for the box norm, use the normalized L2 norm: let $B={\mathbb{E}}_{x,y}f(x,y)u(x)v(y)$ and $\Vert u{\Vert }_2^2={\mathbb{E}}_{x}u(x{)}^2$, $\Vert v{\Vert }_2^2={\mathbb{E}}_{y}v(y{)}^2$. The Cauchy-Schwarz inequality in $y$ givesThe remaining factor is ${\mathbb{E}}_{x,x^{\prime }}u(x)u(x^{\prime }){\mathbb{E}}_{y}f(x,y)f(x^{\prime },y)$. By the Cauchy-Schwarz inequality in $(x,x^{\prime })$, its magnitude is at mostTaking square roots proves

For the tripartite graph, assume its parts are nonempty and write $g_{XY},g_{YZ},g_{XZ}$ for its adjacency indicator functions. Let $f=g_{XY}-\alpha$, so $\Vert f{\Vert }_{\square }\le c$. The normalized triangle count isThe constant-degree hypothesis is exactly ${\mathbb{E}}_y{g}_{YZ}(y,z)=\beta$ for each $z$. Therefore the contribution of the constant $\alpha$ isFor fixed $z$, apply the bilinear correlation bound for the box norm with $u_z(x)=g_{XZ}(x,z)$ and $v_z(y)=g_{YZ}(y,z)$. Since these are indicator functions, $\Vert v_z{\Vert }_2^2=\beta$ and $\Vert u_z{\Vert }_2^2=d_X(z)$, the fraction of $X$ adjacent to $z$. Averaging and applying the Cauchy-Schwarz inequality in $z$ gives the sharper triangle counting with one box-uniform pair and constant opposite degree estimateMultiplying by $|X||Y||Z|$ yieldsThe middle bound also handles $\beta =0$ or $\gamma =0$, when the triangle count is zero.

To connect explicitly with the suggested expansion, put $f_2=g_{YZ}-\beta$, $f_3=g_{XZ}-\gamma$. Both have overall mean zero, and ${\mathbb{E}}_y{f}_2(y,z)=0$ for every $z$. Consequently ${\mathbb{E}}_{x,y,z}{f}_2(y,z)f_3(x,z)=0$; all terms with constant $\alpha$ reduce to $\alpha \beta \gamma$. The terms containing $f$ combine into the single error just estimated. This is where the exact degree assumption is used. If any part is empty the triangle count itself is zero, but the printed uniform expectations and edge density of a bipartite graph values would be undefined; the nonempty-parts convention must therefore be stated rather than dividing by zero.

Statement and definitions. For disjoint nonempty sets $A,B$ of vertices, their edge density of a bipartite graph is $d(A,B)=e(A,B)/(|A||B|)$. They form a regular pair of vertex sets, or an $\epsilon$-uniform pair, ifwhenever $X\subseteq A$, $Y\subseteq B$, $|X|\ge \epsilon |A|$ and $|Y|\ge \epsilon |B|$.

The Szemerédi regularity lemma states that for every $\epsilon >0$ and positive integer $m_0$, there are integers $M,n_0$ such that every graph of order $n\ge n_0$ has a set partitionwith $m_0\le k\le M$, $|V_0|\le \epsilon n$, equal nonzero sizes $|V_1|=\cdots =|V_k|$, and at most $\epsilon k^2$ unordered pairs $(V_i,V_j)$, $1\le i<j\le k$, that fail to be $\epsilon$-uniform. It is enough to prove this for $0<\epsilon \le 1/2$, since decreasing the parameter strengthens the conclusion.