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For nonempty finite set $I$ and $A\subseteq I$, let $\alpha =|A|/|I|$ be the density of a finite subset. The balanced indicator function $f=1_A-\alpha 1_I$, extended by zero to an ambient finite group, has sum zero and vanishing zero Fourier coefficient. The sum of $f$ on a subset $P\subseteq I$ is $|A\cap P|-\alpha |P|$, measuring the excess density of a finite subset there.

Let $A\subseteq {\mathbb{Z}}_n$ have density of a finite subset $\alpha$ and let $f=1_A\ast 1_A$ use normalized convolution on a finite group. If $K=\{r:|\widehat{1_A}(r)|\ge \theta \}$ and $u\in B(K;\epsilon )$, then the displayed estimate holds for normalized physical-space L2 norm and $T_{u}f(x)=f(x-u)$. The Bohr set controls the translation factors on the large spectrum; the fourth Fourier moment bound for an indicator function controls their total weight. Outside $K$, use $|1-e^{2\pi iru/n}|\le 2$ and Parseval identity on a finite group.

A corner consists of these three points in the integer grid, all contained in the set being studied. The displacement may be positive or negative, but must be nonzero, so the points are distinct. The corners theorem guarantees a corner in every subset of positive fixed density of a finite subset of a sufficiently large square grid.

Let $I=[n]$ for odd $n\ge 3$, embed it in the cyclic group $G=\mathbb{Z}/(2n+1)\mathbb{Z}$, and let $A\subseteq I$ have density of a finite subset $\alpha$. If $A$ has no nonconstant three-term arithmetic progression and $n\ge 4{\alpha }^{-2}$, its balanced indicator function $f=1_A-\alpha 1_I$ satisfiesIndeed, the normalized trilinear arithmetic progression count has the Fourier analysis on a finite abelian group formula $\Lambda (h_1,h_2,h_3)={\sum }_r{\hat{h}}_1(r){\hat{h}}_2(-2r){\hat{h}}_3(r)$. Its values on $1_A$ and $\alpha 1_I$ are $\alpha n/|G{|}^2$ and ${\alpha }^3(n^2+1)/(2|G{|}^2)$. Telescoping their difference into three terms containing $f$, the Parseval identity and Cauchy-Schwarz inequality bound its magnitude by $3\max |\hat{f}|\alpha n/|G|$. The difference is at least ${\alpha }^3{n}^2/(4|G{|}^2)$, giving the claim. The zero Fourier coefficient vanishes by the definition of $\alpha$.

For a subset $A$ of a finite abelian group of density of a finite subset $\alpha$, take normalized Fourier coefficients on a finite abelian group and an unnormalized sum over frequencies. The bound follows from $|\widehat{1_A}(\chi )|\le \alpha$ and the Parseval identity on a finite group ${\sum }_{\chi }|\widehat{1_A}(\chi ){|}^2=\alpha$. It bounds the energy of the normalized convolution on a finite group $1_A\ast 1_A$.

The Roth theorem on three-term arithmetic progressions says that, for each fixed $\delta >0$, a sufficiently long interval cannot have a subset of density of a finite subset at least $\delta$ with no nonconstant three-term arithmetic progression. Equivalently, the maximum size of a three-term-progression-free subset of $[N]=\{1,\dots ,N\}$ is $o(N)$. Here is a density increment proof, including its analytic ingredients.

Write $e(t)=\exp (2\pi it)$ and embed $I=[N]$ into the cyclic group $G=\mathbb{Z}/(2N+1)\mathbb{Z}$ of order $M$. For normalized Fourier analysis on a finite abelian group take $\hat{h}(r)=M^{-1}{\sum }_{x}h(x)e(-rx/M)$. The orthogonality of complex exponentials follows from a finite geometric series: its average is zero for a nonzero frequency and one for frequency zero. Expanding in these linear phases consequently proves both Fourier inversion on a finite group and the Parseval identity on a finite group. In particular,The second identity follows by summing the frequency expansion over $x,d$: the two constraints leave precisely the frequencies $(r,-2r,r)$. Since $M$ is odd, multiplication by two permutes the frequencies. Also a congruence $x+z=2y$ with $x,y,z\in I$ is an equality in the integers, since $|x+z-2y|<M$.

Let $A\subseteq I$ have subset density $\alpha =|A|/N\ge \delta$, and suppose it contains no nonconstant three-term arithmetic progression. Its balanced indicator function of a finite subset is $f=1_A-\alpha 1_I$, extended by zero outside $I$. Then $\hat{f}(0)=0$. Counting endpoints with the same parity givesIf $N\ge 4{\alpha }^{-2}$, their difference in absolute value is at least ${\alpha }^3{N}^2/(4M^2)$. Telescope the difference of the products into three terms, each containing $f$ and two factors chosen from $1_A,\alpha 1_I$. Put $s={\max }_r|\hat{f}(r)|$. The Cauchy-Schwarz inequality and the Parseval identity on a finite group bound each term by $s\alpha N/M$, since the squared normalized $L^2$ norm of either other factor is at most $\alpha N/M$. Therefore some nonzero frequency $r$ satisfiesThis is the Fourier detection of a progression-free subset of an interval step; it has just been proved rather than assumed.

We now turn this Fourier coefficient on a finite abelian group into a density increment. Set $Q=\lfloor \sqrt{N}\rfloor$. Partition the unit interval into $Q$ equal pieces and place the $Q+1$ fractional parts of $0,r/M,\dots ,Qr/M$ in them. The pigeonhole principle gives $1\le q\le Q$ with $\Vert qr/M\Vert \le 1/Q$. This proves the particular Dirichlet approximation theorem needed here. Put $\kappa ={\alpha }^2/(1000\pi )$ and $L=\lfloor \kappa Q\rfloor$. For sufficiently large $N$ depending only on $\delta$, $L\ge \kappa Q/2$ and $2L\le Q$. Each residue chain modulo $q$ in $I$ has at least $Q$ elements. Split each such chain into arithmetic progressions with between $L$ and $2L-1$ elements, absorbing its final short remainder into the previous block.

The linear phase changes by at most $4\pi L/Q\le {\alpha }^2/250$ within any block. Freezing the linear phase at the first element of each block therefore changes the preceding correlation by at most ${\alpha }^{2}N/250$, because $|f|\le 1$. The triangle inequality givesThe block sums of the balanced indicator function of a finite subset add to zero, so their positive total is half their absolute total. Some block consequently satisfiesHere $c>0$ is absolute, and shrinking it absorbs the floors. An affine parametrization of $P$ identifies $A\cap P$ with a subset of $[|P|]$ and preserves arithmetic progressions.

Repeat this Roth density-increment step. Every step raises the subset density by at least ${\delta }^2/144$, so fewer than $T=\lceil 144/{\delta }^2\rceil +1$ steps are possible. All the lengths satisfy $N_{j+1}\ge c{\delta }^2\sqrt{N_j}$ while above a threshold depending only on $\delta$. Choose the initial $N$ so large that this lower recursion stays above that threshold for $T$ steps; this is possible by working backwards through finitely many squarings. We would then force a subset density greater than one. Thus every fixed positive density eventually forces a nonconstant three-term arithmetic progression. For an infinite subset of the positive integers with positive upper asymptotic density, apply the finite conclusion along initial intervals whose subset density is bounded away from zero; this gives the usual infinite formulation as well.

Since $a$ is an indicator function of density of a finite subset $\alpha$, the triangle inequality gives $|\hat{a}(r)|\le {\mathbb{E}}_{x}a(x)=\alpha$ for every frequency. Also, character orthogonality and the Parseval identity on a finite group giveFor completeness, the character orthogonality used here iswhich follows by summing a finite geometric series. Expanding the squared Fourier coefficients on a finite abelian group and using this identity proves the displayed Parseval identity on a finite group directly.

Combining the uniform bound with that identity gives the fourth-moment boundIn particular, the Lp norm on the frequency side here is a sum, not a normalized average. This is the fourth Fourier moment bound for an indicator function.

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.

Use the usual convention that a three-term arithmetic progression must be nonconstant. There is a genuine small-case defect in the printed statement: for $n=1$, $\delta =1$, and $A=\{1\}$, neither conclusion can hold with $c_2>0$. We prove the intended Roth density-increment step for every odd integer $n\ge 3$, including the small values of $n$, rather than silently treating $n=1$ as covered. Counting constant arithmetic progressions would make the first alternative vacuous for every nonempty $A$.

Suppose that $A$ has no nonconstant three-term arithmetic progression, and let $\alpha =|A|/n$ be its actual density of a finite subset. Thus $0<\delta \le \alpha \le 1$. Put $I=[n]$, $N=2n+1$, and embed $I$ in the cyclic group $G=\mathbb{Z}/N\mathbb{Z}$. A congruence $x-2y+z=0(\mathrm{mod}N)$ with $x,y,z\in I$ is an equality over the integers, since $|x-2y+z|\le 2(n-1)<N$. Thus this embedding creates no extra three-term arithmetic progressions. Since $N$ is odd, multiplication by two permutes $G$.

Use normalized Fourier analysis on a finite abelian group, with $e_N(t)=\exp (2\pi it/N)$ and $\hat{h}(r)={\mathbb{E}}_{x\in G}h(x)e_N(-rx)$. For the normalized trilinear arithmetic progression count, character orthogonality givesIndeed, expanding each function in its inverse Fourier transform, averaging in $d$ forces $r_2+2r_3=0$, and averaging in $x$ then forces $r_1=r_3$. The Parseval identity in this normalization is ${\sum }_r|\hat{h}(r){|}^2={\mathbb{E}}_x|h(x){|}^2$.

Let $a=1_A$, $b=\alpha 1_I$, and $f=a-b$, the balanced indicator function of a finite subset. Then $\hat{f}(0)=0$. Only constant arithmetic progressions contribute to $\Lambda (a,a,a)$, whereas summing $2\min (y-1,n-y)+1$ over centers $y\in I$ counts $(n^2+1)/2$ ordered three-term arithmetic progressions in $I$. HenceSet $M={\max }_r|\hat{f}(r)|$. Telescope by changing one factor at a time:In the Fourier transform formula for each term, bound the Fourier coefficient of $f$ by $M$ and apply the Cauchy-Schwarz inequality to the other two factors. The permutation $r\mapsto -2r$ preserves the required squared sums. Since $\Vert a{\Vert }_2^2=\alpha n/N$ and $\Vert b{\Vert }_2^2={\alpha }^{2}n/N$, this givesIf $n\ge 4{\delta }^{-2}$, then ${\alpha }^{2}n\ge 4$, so the difference in the other direction is at least ${\alpha }^3{n}^2/(4N^2)$. Consequently Fourier detection of a progression-free subset of an interval gives a nonzero frequency $r$ with

To turn this into a density increment, take $Q=\lfloor \sqrt{n}\rfloor$. The Dirichlet approximation theorem, proved here by placing $0,r/N,\dots ,Qr/N$ into $Q$ equal intervals modulo one, gives $1\le q\le Q$ such that $\Vert qr/N{\Vert }_{\mathbb{R}/\mathbb{Z}}\le 1/Q$. DefineFor $n\ge N_0(\delta )$, the inequalities $Q\ge \sqrt{n}/2$ and ${\delta }^{2}Q/(288\pi )\ge 2$ show thatUse a progression partition with nearly constant linear phase: every residue chain of $I$ modulo $q$ has at least $\lfloor n/q\rfloor \ge Q$ points. Split each chain into blocks of $L$ consecutive chain points, merging any remainder into its last full block. This partitions $I$ into arithmetic progressions $P$ with $L\le |P|<2L$ and common difference $q$.

On a block $P$ with first point $z_P$, the linear phase $\chi (x)=e_N(-rx)$ satisfiesLet $S_P={\sum }_{x\in P}f(x)=|A\cap P|-\alpha |P|$. Freezing the linear phase on each block introduces error at most $({\delta }^2/72)N^{-1}{\sum }_{x\in I}|f(x)|\le {\delta }^2/72$. ThereforeThe block sums total zero. Their positive part consequently totals at least $N{\delta }^2/144$. Since all block lengths total $n$, some block has $S_P/|P|\ge N{\delta }^2/(144n)\ge {\delta }^2/72$. Its density of a finite subset is at least $\alpha +{\delta }^2/72\ge \delta (1+\delta /72)$.

For the remaining $3\le n<N_0(\delta )$, divide $I$ into disjoint consecutive triples and a remainder of at most two points. A three-term-arithmetic progression-free $A$ uses at most two points of each triple, so $\alpha \le 4/5$; the largest ratio occurs at $n=5$. Any singleton set from $A$ then has density of a finite subset one, which is at least $\delta +{\delta }^2/72$. We may therefore use the explicit constantsIn the large case the constructed arithmetic progression has length at least $c_1\sqrt{n}$; in the small case its length one does too. Thus these constants prove the intended result for all odd integers $n\ge 3$.

There is an absolute constant $c>0$ such that, for every $\delta >0$ and all sufficiently large $N$, a subset $A\subseteq [N]$ whose density of a finite subset is $\delta$ and which has no nonconstant three-term arithmetic progression has an arithmetic progression $P\subseteq [N]$ satisfying

The Fourier transform of the balanced function $f=1_A-\delta 1_{[N]}$ has a coefficient of magnitude at least $c_0{\delta }^{2}N$. Indeed, the trilinear countsatisfies $\Lambda (1_A,1_A,1_A)=|A|$, whereas $\Lambda (1_{[N]},1_{[N]},1_{[N]})\gg N^2$. Expanding $1_A=\delta 1_{[N]}+f$ and using the Parseval identity and the Cauchy-Schwarz inequality to bound every error term forces $\Vert \hat{f}{\Vert }_{\infty }\gg {\delta }^{2}N$ once $N$ is sufficiently large in terms of $\delta$.

Choose $\theta$ at which this large coefficient occurs. The Dirichlet approximation theorem gives $1\le d\le \sqrt{N}$ with $\Vert d\theta {\Vert }_{\mathbb{R}/\mathbb{Z}}\le N^{-1/2}$. Partition $[N]$ into progressions of common difference $d$ and lengths between $\eta \sqrt{N}$ and $2\eta \sqrt{N}$, where $\eta >0$ is a sufficiently small multiple of ${\delta }^2$. The linear phase $e(\theta x)$ varies by $O(\eta )$ on each cell. The large Fourier coefficient then givesSince ${\sum }_{x\in [N]}f(x)=0$, the total positive discrepancy is half the total absolute discrepancy, so at least one cell has average of $f$ at least $c{\delta }^2$. This is the required density increment.