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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.

The normalized convolution on a finite group and the Fourier transform on a finite group convention with $\rho (x)$ satisfy the displayed identity. Substitute $x=yz$ in the defining expectation and use $\rho (yz)=\rho (y)\rho (z)$. With the convention using $\rho (x{)}^{\ast }$ instead, the scalar convolution has transform $\hat{g}(\rho )\hat{f}(\rho )$; multiplication order matters for noncommutative groups.

For a finite group, choose one unitary irreducible representation $\rho$ of degree $d_{\rho }$ from each equivalence class. One consistent normalized Fourier transform on a finite group convention is $\hat{f}(\rho )={\mathbb{E}}_{x}f(x)\rho (x)$. The Schur orthogonality relations giveThe normalized convolution on a finite group satisfies $\widehat{f\ast g}(\rho )=\hat{f}(\rho )\hat{g}(\rho )$ in this convention. Using $\rho (x{)}^{\ast }$ in the transform instead reverses that product order for scalar functions with the same normalized convolution on a finite group convention. The Fourier transform on a finite group has matrix-valued components even when the original function is scalar-valued.

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 $1_A\ast 1_A$.

For scalar functions on a finite group, normalized convolution on a finite group is $(f\ast g)(x)=|G{|}^{-1}{\sum }_{y}f(y)g(y^{-1}x)$. It is associative and need not commute. Its identity is $|G|1_{\{e\}}$, rather than the unscaled indicator function of the identity element. The Fourier analysis on a finite group convention $\hat{f}(\rho )=\mathbb{E}f(x)\rho (x)$ turns it into matrix multiplication in the same order.

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 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.

For an $m$-quasirandom group, uniform expectations, and the normalized convolution on a finite group, a scalar mean-zero function $f$ satisfiesThe Fourier analysis on a finite group proof bounds each nontrivial matrix component of $f$ in operator norm using its weighted Hilbert-Schmidt norm. For subsets of subset density values $a,b,c$, the error in their normalized product count is at most $\sqrt{a(1-a)b(1-b)c(1-c)/m}$. In particular $abc>1/m$ guarantees a solution of $xy=z$ in the three subsets.