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For a scalar function on a finite group, one normalized transform convention assigns the matrix ${\mathbb{E}}_{x}f(x)\rho (x)$ to each chosen unitary irreducible representation. This map is a weighted Hilbert space isomorphism by the Parseval identity on a finite group. Another common convention uses $\rho (x{)}^{\ast }$; the corresponding convolution theorem on a finite group then reverses the matrix product order for $(f\ast g)(x)={\mathbb{E}}_{y}f(y)g(y^{-1}x)$.

The positive-power length assertion is an asymptotic one: it requires $N$ sufficiently large in terms of the density. Read literally for every $N$, it fails when $N=2$, $\delta =1/2$ and $A=\{1\}$, whose triple sumset is a singleton. We prove the intended result with this necessary qualification and an explicit positive exponent.

Embed the interval into the cyclic group $G=\mathbb{Z}/M\mathbb{Z}$ with $M=8N$, and set $u=1_A$, $\alpha =|A|/M\ge \delta /8$. Since all integer triple sums lie in $[3,3N]$, this embedding introduces no collisions between distinct sums. Use normalized Fourier coefficients on a finite abelian group and normalized convolution on a finite group, and putThe function $F$ is nonnegative and has mean ${\alpha }^3$, so some $t_0$ satisfies $F(t_0)\ge {\alpha }^3$. We show that a whole translated Bohr set is contained in its positive support, hence in the triple sumset.

Put $\theta ={\alpha }^2/8$, $\rho =\alpha /4$, and define the large spectrumThe Parseval identity on a finite group gives ${\sum }_r|\hat{u}(r){|}^2=\alpha$, soAlso $|\hat{u}(r)|\le \alpha$, and therefore ${\sum }_r|\hat{u}(r){|}^3\le {\alpha }^2$. Using Fourier inversion and the convolution theorem on a finite group, for $b\in B$ we obtainConsequently $F(t_0+b)\ge {\alpha }^3/2>0$. This proves the translated Bohr neighborhood in a triple sumset:It remains to produce a long ordinary integer arithmetic progression in this translate; we supply the simultaneous approximation and the lifting argument. SetThus $d=|\Gamma |\le D$. Partition the $d$-dimensional unit cube into $Q^d$ boxes of side $1/Q$. The $Q^d+1$ pointscontain two in the same box by the pigeonhole principle. Their index difference gives an integer $q$ with $1\le q\le Q^d<M$ andIn particular $|e^{2\pi irq/M}-1|\le 2\pi /Q$. The identity $z^j-1=(z-1)(1+\cdots +z^{j-1})$ on the unit circle now shows thatAll residues $t_0+jq$ therefore lie in the triple sumset. Let $s_j\in [3,3N]$ be their actual integer representatives. Their successive differences are congruent to $q$ modulo $M$, and each lies in $[-(3N-3),3N-3]$, which is contained in $(-M/2,M/2)$. There is at most one integer in this interval in any specified residue class, so every difference $s_{j+1}-s_j$ is the same integer. It is nonzero because $0<q<M$. Thus the $s_j$ form an ordinary integer arithmetic progression of distinct elements, even though $M$ is composite. Reversing it if necessary makes the common difference positive.

For sufficiently large $N$, we have $Q\ge \frac{1}{2}{M}^{1/(D+1)}$ and $\rho \ge \delta /32$, whenceAbsorb this fixed prefactor by decreasing the exponent: withthe last lower bound is at least $N^{r_{\delta }}$ once $N$ is sufficiently large in terms of $\delta$. This proves a polynomial-length progression in $A+A+A$ by the polynomial-length progression in a dense triple sumset argument, without replacing the triple sumset by a fourfold difference set.

Let $\eta =|A|/N$ and $g=1_A-\eta$. In the norm convention, the quadratic uniformity of a set meansIf uniformity is parametrized instead by $\Vert g{\Vert }_{U^3}^8\le \alpha$, replace the parameter below by ${\alpha }^{1/8}$. We prove the genuine-progression conclusion by placing nonnegative weights on a region where wrapping and repeated entries are impossible.

First suppose $N\ge 16$. Put $M=\lfloor N/16\rfloor$, $J=\{0,\dots ,M-1\}$, $t=M/N$, and let $w=1_J\ast 1_J$ be the normalized convolution on a finite group. Define the weighted countBoth weights are nonnegative. Their supports require $0\le x\le 2M-2$ and $M\le d\le 3M-2$ in the ordinary representatives. In particular $d\ge 1$ and $x+3d\le 11M-8<N$. Every contributing arithmetic progression is therefore a set of four distinct integers in $\{0,\dots ,N-1\}$, with no modular wrap.

The averages of both weights are $t^2$. The convolution theorem on a finite group and the Parseval identity on a finite group giveTranslation multiplies coefficients by phases, so $w(d-M)$ has the same Fourier coefficient sum. Telescoping the product of the four indicators around ${\eta }^4$ produces four error terms, each with one factor $g$ and all other factors bounded by one.

To bound a weighted error, expand both weights in their finite Fourier series. For frequencies $r,s$, the extra character is $e^{2\pi i(rx+sd)/N}$. Write it asand absorb these two characters into the functions at slopes zero and one. Even if the balanced factor is one of those functions, its $U^3$ norm remains at most $\alpha$ by the character invariance proved in part (i). The cyclic bound of part (ii), with $C_0=6^{1/8}$, therefore bounds each unweighted Fourier term by $C_0\alpha$. Summing absolute coefficients costs at most $t^2$. ConsequentlySince $t\ge 1/32$ and $\eta \ge \delta$, this is positive whenever $\alpha <{\delta }^4/(4096C_0)$. A positive count supplies the required genuine arithmetic progression. This is the argument that triangular weights count genuine four-term progressions; it avoids losing powers of $\log N$ by cutting off a sharp interval.

For completeness, treat $4\le N<16$. If $A$ is a proper nonempty subset, then $\eta (1-\eta )\ge \delta /N$. The norm bound from part (i) givesChoose $\alpha$ below this bound as well. Then quadratic uniformity forces $A={\mathbb{Z}}_N$, which contains $\{0,1,2,3\}$. Thus, for every $N\ge 4$, a sufficient threshold isThe restriction $N\ge 4$ is necessary: for $N=1,2,3$, the full group has balanced subset indicator zero and is quadratically uniform for every positive parameter, but no subset has four elements. The printed conclusion needs this elementary size qualification.

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.

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.

Define the normalized convolution on a finite group bySubstituting $x=yz$ and using the group representation identity yields the convolution theorem on a finite groupUnlike normalized convolution on a finite group on an abelian group, this product need not commute. If $\tilde{f}(x)=\overline{f(x^{-1})}$, then $\hat{\tilde{f}}(\rho )=\hat{f}(\rho {)}^{\ast }$. For left translation of a group function and right translation of a group function $L_{a}f(x)=f(a^{-1}x)$ and $R_{a}f(x)=f(xa)$,For an abelian group, every irreducible representation is one-dimensional; this reduces to Fourier analysis on a finite abelian group with characters relabelled by their inverses. These formulas establish the basic scalar theory, with all normalizations and multiplication orders fixed.

Now suppose every nontrivial irreducible representation has $d_{\rho }\ge m$. If $f$ is a mean-zero function, its component at the trivial representation is zero. The Parseval identity on a finite group gives, for each other $\rho$,Using the convolution theorem on a finite group, the Hilbert-Schmidt norm inequality $\Vert AB{\Vert }_{\mathrm{HS}}\le \Vert A{\Vert }_{\mathrm{op}}\Vert B{\Vert }_{\mathrm{HS}}$, and the Parseval identity on a finite group once more gives the product mixing in a quasirandom group estimateWrite $a,b,c$ for the subset density values of $A,B,C$, respectively, and let $f=1_A-a$, $g=1_B-b$ be balanced subset indicators. Since both are mean-zero functions, $1_A\ast 1_B=ab+f\ast g$. Their squared norms are $a(1-a)$ and $b(1-b)$. Also $\mathbb{E}(f\ast g)=0$, so the Cauchy-Schwarz inequality yieldsIf $abc>1/m$, the final bound is strictly smaller than $abc$. Thus the normalized number of pairs $(x,y)\in A\times B$ with $xy\in C$ is positive. Equivalently,This is the desired conclusion for a quasirandom group; the strict inequality ensures positivity rather than merely a nonnegative lower bound.