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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.

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.

Use normalized Fourier analysis on a finite abelian group, identifying $A$ with its indicator function $a=1_A$ on the cyclic group ${\mathbb{Z}}_n$. The conventions areThe last operation is normalized convolution on a finite group. With these conventions the Fourier coefficients on a finite abelian group use a normalized average, while sums over frequencies use counting measure.

Expanding the normalized convolution on a finite group and putting $z=x-y$ givesFor the translation of a function $g(x)=f(x-u)$, putting $z=x-u$ yieldsThe two transforms are thereforeThe negative sign in the translation factor follows from the negative sign in our Fourier coefficient on a finite abelian group convention.

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 $n\ge 3$.

Use normalized Fourier analysis on a finite abelian group. If $A\subseteq {\mathbb{F}}_2^n$ has density $\alpha$, its large spectrum at level $\rho >0$ isThe Chang theorem states that this spectrum is contained in a subspace of dimensionEquivalently, it is contained in the span of a dissociated set of that size.