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The Frobenius norm is the finite-matrix Hilbert-Schmidt norm. It equals $\sqrt{\mathrm{tr}(A^{\ast }A)}$, is unchanged by unitary multiplication on either side, and is the square root of the sum of squared singular values. For unit vectors $u,v$, $\Vert u{u}^{\ast }-v{v}^{\ast }{\Vert }_F^2=2(1-|u^{\ast }v{|}^2)$, a useful sign-invariant measure of distance between rank-one orthogonal projection matrices.

For Hilbert-Schmidt operators $A$ and $B$, the Hilbert-Schmidt inner product is $⟨A,B{⟩}_{\mathrm{HS}}=\mathrm{tr}(A^{\ast }B)$. Its induced norm is the Hilbert-Schmidt norm.

For $(Tg)(v)=\int k(v,w)g(w)dw$, the Cauchy-Schwarz inequality gives $\Vert Tg{\Vert }_2\le \Vert k{\Vert }_{L^2(v,w)}\Vert g{\Vert }_2$. The kernel norm is the Hilbert-Schmidt norm, which bounds the operator norm. Uniform kernel bounds apply also to spatially parametrized operators after integrating over position.

A normalized bipartite pure state of Schmidt rank $r$ has quantum fidelity at most $\sqrt{r/d}$ with a fixed maximally entangled state on $d\otimes d$. Its coefficient matrix has matrix rank $r$ and Hilbert-Schmidt norm one. Its overlap is the trace of that matrix divided by $\sqrt{d}$; trace duality and the Cauchy-Schwarz inequality bound this by $\sqrt{r/d}$. Uniform Schmidt coefficients on $r$ matching basis pairs attain equality.

Let ${\sigma }_1=\gamma ,{\sigma }_2,\dots$ be the singular values of the normalized adjacency operator of a bipartite graph, with the uniform inner products used above. Its ordinary matrix in an orthonormal basis on each side is $G/\sqrt{|X||Y|}$. Expanding the trace of $({\theta }_G{\theta }_G^{\ast }{)}^2$ proves the box norm singular-value identityTherefore (ii) immediately implieswhich is (iii).

Conversely, the square of the Hilbert-Schmidt norm of this operator issince a graph indicator function takes values zero and one. Under (iii), every ${\sigma }_j$ with $j\ge 2$ is at most $c_3$, soThus (iii)$\Rightarrow$(ii) with $\boxed{c_2=c_3^2/4}$. This bound also covers the empty and complete bipartite graphs.

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.

No. In particular, conjugate subgroups $H^{\prime }=xH{x}^{-1}$ produce the same weak irrep-label distribution. Under an irrep, the subgroup averageis related by unitary similarity, so its squared Hilbert-Schmidt norm, and hence every ${\Pi }_H(\alpha )$, is unchanged. Weak Fourier sampling can therefore fail to identify even distinct subgroups of a non-abelian group.

For an orthogonal $R$, unitary invariance of the Hilbert-Schmidt norm givesThe product $L_2^{\ast }{L}_1$ is trace-class. Its polar decomposition of a bounded operator and trace duality givewhere $({\sigma }_k)$ are its singular values. Taking the infimum proves

Convergence in the Hilbert-Schmidt norm impliesProducts of two Hilbert-Schmidt operators are trace-class operators, and the Schatten norm Hölder inequality givesBy trace duality, every unitary $R$ satisfiesTaking the supremum over $R$ shows that the supremum terms in the two Procrustes formulas converge. Combining this with convergence of the squared norms proves

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.