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We prove the three-term Roth theorem by a density increment, keeping the interval boundary rather than replacing the interval by a cyclic group with a different density. The case $\delta >1$ is vacuous; assume $0<\delta \le 1$.

We first prove the following alternative with absolute constants $c,C>0$: if $A\subseteq [1,N]$ has density $\eta$ and no nonconstant three-term arithmetic progression, and $N\ge C{\eta }^{-4}$, then some integer progression $P\subseteq [1,N]$ hasFor this purpose take $M=6\lceil 4N/6\rceil +1$, so $4N<M\le 5N$ for large $N$ and $2,3$ are invertible modulo $M$. On $G=\mathbb{Z}/M\mathbb{Z}$, put $I=1_{[1,N]}$, $a=1_A$, $g=\eta I$ and $f=a-g$. In particular $\sum f=0$, and $f$ vanishes outside the interval. Use normalized Fourier coefficients on a finite abelian group, $\hat{u}(r)=M^{-1}{\sum }_{x}u(x)e(-rx/M)$. Character orthogonality, obtained by summing a finite geometric series, gives Fourier inversion on a finite group and Parseval identity on a finite group.

For $\Lambda (u,v,w)={\mathbb{E}}_{x,d}u(x)v(x+d)w(x+2d)$ these identities giveThere is no wrapping ambiguity when all three entries lie in $[1,N]$: the integer $x-2y+z$ has absolute value below $M$, so a modular relation is an integer relation. If no nonconstant progression exists, $\Lambda (a,a,a)=|A|/M^2$. Meanwhile the choices $1\le x\le \lfloor N/4\rfloor$ and $1\le d\le \lfloor N/8\rfloor$ give $\Lambda (I,I,I)\ge 1/10000$ for sufficiently large $N$.

Expand the difference by telescoping,If $\rho ={\max }_r|\hat{f}(r)|$, each term on the right has modulus at most $\rho \eta N/M$. Indeed multiplication of frequencies by $2$ permutes them, and Cauchy-Schwarz with Parseval identity bounds the two other Fourier factors by their $L^2$ norms; each squared norm is at most $\eta N/M$. For $N\ge 10000{\eta }^{-2}$, the diagonal count is less than half the main density term. ThereforeIn particular there is a nonzero frequency $r$ withThe frequency is nonzero because $\hat{f}(0)=0$.

Here is the conversion of this coefficient into a density increment. Put $Q=\lfloor \sqrt{N}\rfloor$. Of the $Q+1$ numbers $0,\theta ,\dots ,Q\theta$ modulo one, two lie in the same one of $Q$ equal arcs. Their difference gives $1\le q\le Q$ with $\Vert q\theta \Vert \le 1/Q$, where $\theta =-r/M$. This is the needed Dirichlet approximation theorem, proved by the pigeonhole principle.

Partition each residue class modulo $q$ in $[1,N]$ into consecutive progressions of length $L=\lfloor \gamma Q/(100\pi )\rfloor$, leaving its final short tail. Their total tail size is at most $qL\le \gamma N/(100\pi )$. On each full progression the character differs from its value at the first point by at most $2\pi L/Q\le \gamma /50$. Hence removing tails and replacing the character by its initial value showsThe sum of $f$ over all full progressions is the negative of its sum over the tails, so its absolute value is at most $\gamma N/(100\pi )$. Consequently the sum of the positive block sums is at least $\gamma N/5$. Since these blocks contain at most $N$ points in total, at least one has average $f$ at least $\gamma /5$. It gives a density increment at least ${\eta }^2/100000$. For $N\ge C{\eta }^{-4}$ with a sufficiently large absolute $C$, $L\ge c{\eta }^2\sqrt{N}$. This proves the alternative, with one small common absolute $c$.

Reidentify $P$ with $[1,|P|]$ by its affine parametrization. Absence of a nonconstant progression is preserved, while density increases. At every step the density is at least the original $\delta$, so the increment is at least $c{\delta }^2$ and the new interval length is at least $c{\delta }^2\sqrt{N_{\mathrm{old}}}$. After at most $T=\lceil 1/(c{\delta }^2)\rceil +1$ steps, the density would exceed one. Choose the original $N$ so large that the length stays above $C{\delta }^{-4}$ throughout these finitely many steps. Such a choice exists explicitly by working backwards: a required next length $L_0$ is ensured by previous length at least $(L_0/(c{\delta }^2){)}^2$. The alternative could then be applied at every step, giving a contradiction. ThusThis is the Roth density increment on an integer progression proof.

For the pattern with offsets $0,1,3$, replace the counting functional by ${\mathbb{E}}_{x,d}u(x)v(x+d)w(x+3d)$. Its Fourier expression iscoming from the relation $2x-3y+z=0$. Both multipliers are invertible modulo the chosen $M$, so exactly the same Parseval estimates apply. This relation has no wrap in $[1,N]$, since its magnitude is at most $3N<M$, and the earlier restricted choices of $x,d$ still give a fixed positive density main term. The same increment and iteration find a nonconstant triple $\{a,a+d,a+3d\}$. Negative $d$ is allowed, as requested, and the affine reparametrization used in the iteration preserves this pattern as well.

We prove the Roth theorem on three-term arithmetic progressions by a density increment on an ordinary integer arithmetic progression. The required increment will be small, but depend only on the current subset density.

First establish the increment step. Suppose $A\subseteq [1,L]$ has subset density $\rho =|A|/L>0$ and no nonconstant three-term arithmetic progression. Put $M=4L+1$, regard $[1,L]$ as a subset $I$ of the cyclic group ${\mathbb{Z}}_M$, and set $f=1_A-\rho 1_I$. Then $\mathbb{E}f=0$ and $|f|\le 1$. Use the normalized Fourier coefficient on a finite abelian groupThe character orthogonality geometric-series identity ${\mathbb{E}}_x{e}^{2\pi irx/M}=1$ for $r=0$ and zero otherwise gives Fourier inversion on a finite group, Parseval identity on a finite group, andSince $M$ is odd, multiplication by two permutes its frequencies. If one argument is $f$ and the other two have absolute value at most one, Cauchy-Schwarz and the Parseval identity on a finite group give $|T|\le {\max }_r|\hat{f}(r)|$.

A modular arithmetic progression whose three entries lie in $I$ satisfies the corresponding integer equation: the absolute value of $x+z-2y$ is less than $2L<M$, so congruence to zero forces equality. Thus $T(1_A,1_A,1_A)=|A|/M^2\le 1/(16L)$, since only constant arithmetic progressions remain. Meanwhile, the choices $1\le x\le \lfloor L/2\rfloor$ and $0\le h\le \lfloor L/4\rfloor$ give at least $L^2/16$ arithmetic progressions in $I$ for $L\ge 8$. As $M\le 5L$, this gives $T(1_I,1_I,1_I)\ge 1/400$. Expand $1_A=\rho 1_I+f$. The seven terms other than the main term are each bounded by $\max |\hat{f}|$. For $L\ge 50{\rho }^{-3}$,Consequently a nonzero frequency $r$ obeys $|\hat{f}(r)|\ge \gamma$, where $\gamma ={\rho }^3/6000$. It is nonzero because $\hat{f}(0)=0$.

We turn this coefficient into a local increase. Let $Q=\lfloor \sqrt{L}\rfloor$. Partition the unit circle into $Q$ equal intervals and consider the $Q+1$ points $jr/M$, $0\le j\le Q$, modulo one. Two occupy the same interval, yielding an integer $1\le q\le Q$ with $\Vert qr/M{\Vert }_{\mathbb{R}/\mathbb{Z}}\le 1/Q$. Set $\ell =\lfloor \gamma Q/200\rfloor$. If $L$ exceeds a sufficiently large absolute multiple of ${\rho }^{-6}$, then $\ell \ge 1$ and each residue-class arithmetic progression of step $q$ in $[1,L]$ has at least $\ell$ terms. Partition each into pieces of lengths between $\ell$ and $2\ell$, distributing the final remainder among the pieces.

On each such arithmetic progression $P$, the character of a finite abelian group $\chi (x)=e^{-2\pi irx/M}$ differs from its value at the first point by at most $4\pi \ell /Q<\gamma /4$. Write $s_P={\sum }_{x\in P}f(x)$. Thereforeso ${\sum }_P|s_P|\ge 3\gamma L/4$. But ${\sum }_P{s}_P={\sum }_{I}f=0$, and hence the sum of the positive $s_P$ is at least $3\gamma L/8$. Some piece must consequently satisfy $s_P/|P|\ge \gamma /4$. On that piece,Also, once $\gamma Q\ge 400$, $|P|\ge \ell \ge \gamma \sqrt{L}/800$. This proves the one-frequency density increment on an integer interval: for absolute constants $c,C>0$, if $L\ge C{\rho }^{-6}$, there is a arithmetic progression $P\subseteq [1,L]$ of length at least $c{\rho }^3\sqrt{L}$ on which the subset density rises by at least $c{\rho }^3$. For example one may decrease $c$ to $10^{-7}$ and choose $C$ large enough for the displayed elementary inequalities.

Now fix $0<\delta \le 1$ and suppose the required conclusion fails. On each new arithmetic progression, identify its points in order with $[1,L^{\prime }]$. This affine identification preserves all integer arithmetic progressions, so its copy of $A\cap P$ is still progression-free. All densities remain at least $\delta$. Put $\epsilon =c{\delta }^3$, $H=C{\delta }^{-6}$, and choose an integer $t>1/\epsilon$. Every increment raises the subset density by at least $\epsilon$ and changes the length by $L_{j+1}\ge \epsilon \sqrt{L_j}$. Iterating the latter inequality givesChoose $N>(H/{\epsilon }^2{)}^{2^t}$. Then every length up to step $t$ exceeds $H$, so every increment is available. After $t$ increments the subset density exceeds one, a contradiction. ThusFor $\delta >1$ there are no sets satisfying the subset density assumption. The argument above proves the substantive case without invoking the theorem being asked for.

Count a three-term arithmetic progression as an ordered triple $(x,x+d,x+2d)$, allowing $d=0$. Inserting Fourier inversion on a finite group into its normalized count and using additive character orthogonality in $x,d$ givesIndeed the two frequency constraints are $r+s+t=0$ and $s+2t=0$, so the surviving frequencies are $(r,s,t)=(t,-2t,t)$. The term at zero is ${\alpha }^3$.

For odd $N$, let $f=a-\alpha$ be the balanced indicator function of a finite subset, and put $M={\max }_{s\ne 0}|\hat{f}(s)|$. Its zero coefficient is zero, and its other coefficients equal those of $a$. Since $-2r\ne 0$ when $r\ne 0$, the Parseval identity on a finite group givesThe assumed deficit $T<{\alpha }^3/2$ therefore forcesThus $c=1/2$ and $C=2$ suffice. This is the large Fourier coefficient from a deficit of three-term progressions estimate; no generalized von Neumann theorem is needed. For the remaining prime $N=2$, the count is $T={\mathbb{E}}_{x,d}a(x)a(x+d)={\alpha }^2$, so the deficit hypothesis is impossible. The zero common difference was included throughout.

Assume $A$ is nonempty, so $0<\alpha \le 1$, and put $a=1_A$. By the Parseval identity on a finite group, ${\sum }_r|\hat{a}(r){|}^2=\alpha$. Select the nonzero frequenciesThen $|R|{\alpha }^3/4\le \alpha$, giving $|R|\le 4{\alpha }^{-2}$. Outside $R\cup \{0\}$, the fourth Fourier moment has the boundThe fourfold normalized convolution on a finite group $H=a\ast a\ast 1_{-A}\ast 1_{-A}$ has transform $|\hat{a}(r){|}^4$, so Fourier inversion on a finite group gives $H(x)={\sum }_r|\hat{a}(r){|}^4{e}_N(rx)$. It is real and nonnegative, and $H(x)>0$ implies $x\in 2A-2A$ because the convolution counts representations, with a positive normalization.

Use the Bohr set in phase-distance conventionOn this set all retained additive characters have positive real part. The zero-frequency contribution is $\hat{a}(0{)}^4={\alpha }^4$, and the total absolute contribution of the discarded frequencies is at most ${\alpha }^4/4$. HenceThereforeThis is an explicit Bogolyubov lemma. In the chord-distance convention for a Bohr set, the phase-radius $1/10$ corresponds to chord radius $2\sin (\pi /10)>1/10$; alternatively the same positivity proof works directly at chord radius $1/10$. Thus the requested lower bound on width holds in either convention. If $R$ is empty, the Bohr set is the whole group. Nonemptiness of $A$ is necessary for the density statement.

Put $e_N(t)=e^{2\pi it/N}$ and use normalized averages ${\mathbb{E}}_x=N^{-1}{\sum }_{x\in \mathbb{Z}/N\mathbb{Z}}$. Our convention for Fourier coefficients on a finite abelian group isThe underlying orthogonality of complex exponentials is ${\mathbb{E}}_x{e}_N(rx)=1$ for $r=0$ and zero otherwise. For nonzero $r$ this is the finite geometric series with ratio $e_N(r)\ne 1$; its sum vanishes because that ratio has $N$th power one. The identical orthogonality statement holds when averaging over $r$ instead of $x$.

Using this orthogonality to sum the definition over $r$ proves Fourier inversion on a finite group:Indeed $\mathbb{E}f\overline{g}={\sum }_r\overline{\hat{g}(r)}{\mathbb{E}}_{x}f(x)e_N(-rx)={\sum }_r\hat{f}(r)\overline{\hat{g}(r)}$. This proves the Parseval identity on a finite group and its squared-norm case:For normalized convolution on a finite group, write $(f\ast g)(x)={\mathbb{E}}_{y}f(y)g(x-y)$. Changing variables $x=y+z$ in its transform givesthe convolution theorem on a finite group. Linearity of the transform follows directly from its definition. Translation ${\tau }_{t}f(x)=f(x+t)$ gives $\widehat{{\tau }_{t}f}(r)=e_N(rt)\hat{f}(r)$ by changing variables. Complex conjugation gives $\hat{\overline{f}}(r)=\overline{\hat{f}(-r)}$; for $\tilde{f}(x)=\overline{f(-x)}$, the corresponding identity is $\hat{\tilde{f}}(r)=\overline{\hat{f}(r)}$. Finally $\hat{f}(0)=\mathbb{E}f$ and $|\hat{f}(r)|\le \mathbb{E}|f|$ by the triangle inequality. These identities establish the basic normalized Fourier analysis on a finite abelian group needed below, with all normalizations fixed.

For completeness, pointwise multiplication has the dual convolution formula $\widehat{fg}(r)={\sum }_s\hat{f}(s)\hat{g}(r-s)$: insert the inversion formula for $g$ into the defining average and collect its frequencies. If $c\ne 0$ modulo $N$, the change of variables $u=cx$ gives $\widehat{f(c\cdot )}(r)=\hat{f}(r{c}^{-1})$. These also follow directly from the same additive character orthogonality.

In this part identify $\mathbb{T}$ with the complex unit circle by $t\mapsto e^{it}$. Write the finite abelian group $G$ additively and let $\hat{G}=\mathrm{Hom}(G,\mathbb{T})$, with pointwise multiplication, be the character group of a finite abelian group of $G$. A linear character has modulus one, since its values have finite order. We first prove the extension and counting facts needed for Fourier analysis on a finite abelian group.

If $H\le G$, $\chi \in \hat{H}$, and $x\notin H$, let $d$ be the least positive integer with $dx\in H$. Choose $z\in \mathbb{T}$ with $z^d=\chi (dx)$ and defineThis is well-defined: two representations differ by $j-j^{\prime }=qd$ and $h-h^{\prime }=-qdx$, whose contributions cancel. It is a group homomorphism and extends $\chi$. Conversely every extension has this form, and the equation for $z$ has exactly $d$ distinct roots. Since $[H+⟨x⟩:H]=d$, successive adjunctions prove that each linear character of $H$ has exactly $[G:H]$ extensions. Taking $H=\{0\}$ gives $\boxed{|\hat{G}|=|G|}$. This is the extension of a character across a cyclic quotient, proved here without a structure theorem. Extending the linear character sending a nonzero element $g$ of order $d$ to $e^{2\pi i/d}$ also proves that linear characters separate points of $G$.

Use the normalized inner product $⟨f,g⟩=|G{|}^{-1}{\sum }_{x\in G}f(x)\overline{g(x)}$. For a nontrivial linear character $\chi$, choose $a$ with $\chi (a)\ne 1$. Translation of the sum gives ${\sum }_x\chi (x)=\chi (a){\sum }_x\chi (x)$, so that sum is zero. Applying this to $\chi \overline{\psi }$ givesThere are $|G|$ such orthonormal vectors in the $|G|$-dimensional space ${\mathbb{C}}^G$, so they form an orthonormal basis. Define the Fourier coefficients by $\hat{f}(\chi )=|G{|}^{-1}{\sum }_{x}f(x)\overline{\chi (x)}$. Expanding in that basis proves Fourier inversion on a finite group and the Parseval identity on a finite group:For normalized convolution $(f\ast g)(x)=|G{|}^{-1}{\sum }_{y}f(y)g(x-y)$, changing the finite summation variables yields $\widehat{f\ast g}(\chi )=\hat{f}(\chi )\hat{g}(\chi )$. Translation $f(x)\mapsto f(x+a)$ multiplies its Fourier coefficient by $\chi (a)$. These formulas fix all normalization and conjugation conventions.

The PDF's double-hat identification is canonical Pontryagin duality: define $e:G\to \hat{\hat{G}}$ by $e_x(\chi )=\chi (x)$. This is a group homomorphism and is injective because linear characters separate points. Applying the already proved character-counting result to $\hat{G}$ shows that both groups have size $|G|$, soThe TeX loses one hat. For completeness, the single-dual isomorphism $\hat{G}\cong G$ also holds, but is generally noncanonical. Here is a proof including the needed finite cyclic decomposition. Choose an element $a$ of maximal order $d$. Every element order divides $d$: for two elements, extract their prime-power components by integer multiples: if an element has order $d$ and $p^{\alpha }$ is the exact power of $p$ in $d$, multiplying it by $d/p^{\alpha }$ gives order $p^{\alpha }$. Then combine the components with the larger order for each prime. Components of coprime orders have a sum of product order, because their cyclic subgroups have trivial intersection. This constructs an element of order the least common multiple of the two original orders, which cannot exceed maximal $d$; hence each order divides $d$.

Extend the primitive linear character of $⟨a⟩$ to $\chi :G\to \mathbb{T}$. All its values are $d$th roots because all element orders divide $d$. Define $\rho (g)=ja$ when $\chi (g)=e^{2\pi ij/d}$. This is a group homomorphism fixing $⟨a⟩$, and $g=\rho (g)+(g-\rho (g))$ shows $G=⟨a⟩\oplus \ker \rho$. Induction on group size decomposes $G$ into finite cyclic groups. A cyclic group of order $d_j$ has precisely the linear characters $j\mapsto e^{2\pi imj/d_j}$ and hence a character group of a finite abelian group isomorphic to itself. The dual of a finite direct product of groups is the product of its factor duals, since a linear character is determined by its restrictions. This proves $\hat{G}\cong G$ after choices of cyclic factors and generators, without using an unproved finite-abelian-group structure theorem.

Write the finite abelian group additively. A character of a finite abelian group is a group homomorphism $\chi :G\to \{z:|z|=1\}$. We first prove that there are exactly $|G|$ such characters of a finite abelian group, without assuming a structure theorem.

Use extension of a character across a cyclic quotient. Given a subgroup $H$ and $a\notin H$, let $k$ be the least positive integer with $ka\in H$. The subgroup $K=H+⟨a⟩$ has $k$ cosets of $H$. If $\varphi$ is a character of a finite abelian group on $H$, choose any of the $k$ roots ${\lambda }^k=\varphi (ka)$ and defineThis is well-defined: two representations differ by an integer multiple of $ka$, and the root equation exactly cancels that difference. It is a character of a finite abelian group, and every extension arises from one of the $k$ choices of $\lambda$. Build a chain from the trivial subgroup to $G$ by adjoining elements. The character of a finite abelian group count multiplies by the same factor as the subgroup order at every step, so $|\hat{G}|=|G|$.

For a nontrivial character of a finite abelian group $\eta$, choose $a$ with $\eta (a)\ne 1$. Translating the group sum showsso the sum is zero. Applied to $\chi \overline{\psi }$, this provesThe $|G|$ orthogonal nonzero characters of a finite abelian group therefore form a basis of all complex functions on $G$, a vector space of dimension $|G|$.

With the unnormalized Fourier coefficients $\hat{f}(\chi )={\sum }_{x\in G}f(x)\overline{\chi (x)}$, expansion in that basis gives the Fourier inversion on a finite groupUnder a normalized forward-transform convention the prefactor would instead be one. Thus the unspecified constant is determined by the convention, and the inversion itself follows directly from character of a finite abelian group counting and orthogonality.

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.

We will prove a polynomial progression in a high-energy fourfold difference set. The route is to extract a large subset with a small difference set, construct a dense Freiman s-isomorphism model, and use the Bogolyubov lemma and the pigeonhole principle. To respect the proof requirement, every ingredient beyond the elementary rules for Freiman homomorphisms is established below.

Assume $n\ge 1$. Since three entries of an additive quadruple determine the fourth, $E(A)\le n^3$, so a nonvacuous hypothesis has $0<\theta \le 1$. Let $r(t)=|\{(a,b)\in A^2:a+b=t\}|$. Then $E(A)={\sum }_{t}r(t{)}^2$ and ${\sum }_{t}r(t)=n^2$. Define the popular sum set $S=\{t:r(t)\ge \theta n/2\}$. The contribution to additive energy from its complement is at most $\theta n^3/2$. Since $r(t)\le n$, the bipartite graph on two copies of $A$ whose edges have sums in $S$ has at least $\theta n^2/2$ edges. Moreover $|S|\le 2n/\theta$.

Put $\delta =\theta /2$ and $\tau ={\delta }^2/16$. Here is a dependent random choice argument producing the small difference set. Independently choose two right vertices, allowing repetition, and let $X$ be their common left neighbourhood. The Cauchy-Schwarz inequality gives $\mathbb{E}|X|\ge {\delta }^{2}n$. Call an ordered pair of left vertices bad if its common right neighbourhood has fewer than $\tau n$ elements. The expected number $b(X)$ of bad ordered pairs in $X$ is at most ${\tau }^2{n}^2$, because a fixed bad pair survives both random choices with probability at most ${\tau }^2$. ThusChoose a realization attaining at least this expectation. Then $|X|\ge 3{\delta }^{2}n/4$, and $b(X)\le {\delta }^{2}n|X|/64\le |X{|}^2/48$. Remove from $X$ every vertex having more than $|X|/4$ bad partners in $X$. At most $|X|/12$ vertices are removed. The remaining set $B$ has $|B|\ge {\delta }^{2}n/2$, and any $a,b\in B$ have at least $|X|/2$ vertices $z\in X$ for which both $(a,z)$ and $(z,b)$ are good.

For such a $z$, there are at least $\tau n$ right vertices $y$ adjacent to $a,z$, and at least $\tau n$ right vertices $y^{\prime }$ adjacent to $z,b$. These paths givea representation with four members of $S$. For fixed $a,b$, the tuple of four sums determines $y,z,y^{\prime }$ uniquely, so there are at least $|X|{\tau }^2{n}^2/2$ distinct tuples. Select one ordered pair $a,b$ for each member of $B-B$. Tuples belonging to different differences are disjoint. Counting all possible tuples in $S^4$ provesIn particular, with $m=|B|$, we have $m\ge {\theta }^{2}n/8$ and $|B-B|\le Km$ for $K=2^{25}{\theta }^{-12}$. This is the required small-difference set form of the Balog-Szemerédi-Gowers theorem, proved by counting paths rather than cited.

We next control higher iterated sumsets. Choose a nonempty $X\subseteq B$ minimizing $\rho =|X-B|/|X|$, so $\rho \le K$. We prove the Petridis minimal-growth lemma in the form $|X-B+C|\le \rho |X+C|$ for finite $C$. Order $C=\{c_1,\dots ,c_t\}$, and let $X_i$ consist of those $x\in X$ for which $x+c_i$ is new in the union of the translates $X+c_j$. Then ${\sum }_i|X_i|=|X+C|$. Every member of $(X\setminus X_i)-B+c_i$ already occurs in an earlier translate of $X-B$. By minimality, $|(X\setminus X_i)-B|\ge \rho |X\setminus X_i|$, also when the complement is empty. Thus the new contribution from $X-B+c_i$ is at most $\rho |X_i|$. Sum over $i$ to prove the lemma. Iterating with $C=-(j-1)B$ gives $|X-jB|\le K^j|X|$.

For completeness, the Ruzsa triangle inequality in the needed form isFor each $d\in U-W$, fix $d=u_d-w_d$. The map $(d,v)\mapsto (u_d-v,v-w_d)$ is injective: adding the coordinates recovers $d$, and then $v$. Take $U=kB$, $W=\ell B$, $V=X$ and apply the preceding bounds. We obtain the Plünnecke-Ruzsa inequalityIn particular $T=8B-8B$ has at most $K^{16}m$ elements.

We now prove a slightly wasteful version of the Ruzsa modeling lemma, avoiding any assumption about the diameter of $B$. Set $q=8|T|$. Choose a prime $p$ larger than all $|t|$, $t\in T$, and much larger than $q$. Arbitrarily large primes exist by the elementary argument that a prime divisor of one plus the product of all primes up to a given bound exceeds that bound. For a uniformly chosen nonzero multiplier $r$ modulo $p$, each nonzero $t\in T$ has $rt$ uniform among the nonzero residues. Call a multiplier forbidden if some such $rt$ agrees modulo $p$ with $jq$, where $|jq|<p$. There are at most $2p/q+1\le 3p/q$ such residues. For $p\ge 4$, a union bound therefore makes the probability of being forbidden less thanChoose a nonforbidden $r$. Represent $rB$ by integers in $[0,p-1]$ and partition that interval into sixteen consecutive pieces of diameter less than $p/16$. One piece contains the images of a subset $B^{\prime }\subseteq B$ with $|B^{\prime }|\ge m/16$. Let $\varphi (b)$ be the integer representative of $rb$ in that piece, reduced modulo $q$.

A difference of eight sums of these representatives has absolute value less than $p/2$. If its original difference in $B$ is zero, it is a multiple of $p$, hence zero. Conversely, if its reduction modulo $q$ is zero, it equals $jq$ with $|jq|<p$; our choice of $r$ forces its original difference to be zero. This proves that $\varphi$ is a Freiman s-isomorphism of order eight. Injectivity follows by padding a one-term equality to eight terms with a fixed element. Consequently $D=\varphi (B^{\prime })\subseteq G=\mathbb{Z}/q\mathbb{Z}$ has subset densityNo primality of $q$ is needed.

Here is the cyclic Bogolyubov lemma with all its Fourier analysis on a finite abelian group details. Use $\hat{f}(r)={\mathbb{E}}_{x}f(x)e(-rx/q)$, $f=1_D$, and normalized convolution on a finite group. A finite geometric series proves orthogonality, and expanding the finite sums proves the Parseval identity on a finite group, Fourier inversion on a finite group, and $\widehat{f\ast g}=\hat{f}\hat{g}$. For $\tilde{f}(x)=f(-x)$, the nonnegative functionhas support exactly $2D-2D$ and Fourier coefficients on a finite abelian group $|\hat{f}(r){|}^4$. Put $\Gamma =\{r\ne 0:|\hat{f}(r)|\ge {\alpha }^{3/2}/\sqrt{8}\}$. The Parseval identity on a finite group gives $|\Gamma |\le 8{\alpha }^{-2}$. The total fourth moment of the frequencies outside $\Gamma \cup \{0\}$ is at most ${\alpha }^4/8$.

If $\Vert rx/q\Vert \le 1/16$ for every $r\in \Gamma$, then $\mathrm{Re}e(rx/q)\ge \cos (\pi /8)>3/4$. The zero frequency contributes ${\alpha }^4$ with no loss. Separating the large and small frequencies in Fourier inversion on a finite group yieldsThus $2D-2D$ contains the Bohr set expressed in distance coordinates by $\Vert rx/q\Vert \le 1/16$. This is consistent with the existing character-distance convention for a Bohr set: here it is enough that each character lies in the corresponding short arc around one.

We finally prove the needed nonwrapping progression in a cyclic Bohr set, including for composite $q$. Write $d=|\Gamma |$ and $Q=\lfloor q^{1/(d+1)}\rfloor$. Partition the $d$-dimensional unit cube into $Q^d$ boxes. The $Q^d+1$ points with coordinates $jr/q$ modulo one, $0\le j\le Q^d$, give by the pigeonhole principle an integer $1\le t\le Q^d$ with $\Vert rt/q\Vert \le 1/Q$ for all $r\in \Gamma$. ThenThese points are distinct modulo $q$, since their integer representatives lie between zero and $Q^{d+1}/16\le q/16$. Their number is at least $q^{1/(d+1)}/32$. For $d=0$ the same argument simply chooses $t=1$.

The Freiman s-isomorphism induces a bijection $\Phi :2B^{\prime }-2B^{\prime }\to 2D-2D$: define it on a representation by taking the same two positive and two negative images. Equality of representations uses an order-four relation, and injectivity uses the converse relation. If three consecutive members $y_{j-1},y_j,y_{j+1}$ of $P$ satisfy $y_{j-1}+y_{j+1}=2y_j$, their preimages satisfy the same equality, since its expansion is an equality of eight-term sums in $B^{\prime }$. Hence the inverse image of $P$ is an ordinary arithmetic progression in $2B^{\prime }-2B^{\prime }\subseteq 2A-2A$, with distinct terms.

To remove the multiplicative constant from the length bound, let $R=\lceil 8{\alpha }_0^{-2}\rceil$ and $\beta =1/(R+1)$. The length just proved is at least $c_{\theta }{n}^{\beta }$, where $c_{\theta }={\theta }^{2\beta }/32>0$. Choose $N_0\ge 3$ so that $c_{\theta }{n}^{\beta }\ge n^{\beta /2}$ for $n\ge N_0$, and set $\gamma =\min \{\beta /2,\log 3/\log N_0\}$. For $2\le n<N_0$, two distinct members of $A$ give a nonzero $u\in A-A$, and $\{-u,0,u\}\subseteq 2A-2A$ has length three, at least $n^{\gamma }$. For $n=1$, the singleton $\{0\}$ suffices. In every nonempty case,with $\gamma >0$ depending only on $\theta$. If empty sets are permitted, the required length is zero and there is no substantive assertion.

Use the following normalization for Fourier analysis on a finite group. Choose one unitary irreducible representation $\rho :G\to U(d_{\rho })$ from each equivalence class, including the trivial representation. For a scalar function $f:G\to \mathbb{C}$, putThis convention uses $\rho (x)$, rather than $\rho (x{)}^{\ast }$, in the Fourier transform on a finite group; it makes the normalized convolution on a finite group preserve multiplication order.

The needed representation theory consists of Maschke's theorem and unitarization of a finite-group representation, together with the Schur orthogonality relations:The regular representation contains $d_{\rho }$ copies of each $\rho$, so ${\sum }_{\rho }{d}_{\rho }^2=|G|$. Thus the scaled matrix coefficients $\sqrt{d_{\rho }}\rho (x{)}_{ij}$, and also their complex conjugates, form an orthonormal basis of all scalar functions on $G$. These facts imply Fourier inversion on a finite group and the Parseval identity on a finite group in the formsand henceIn particular, the transform is an isomorphism onto the direct sum of the matrix algebras $M_{d_{\rho }}(\mathbb{C})$, with the displayed weighted Hilbert-Schmidt inner product.