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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.

The selected singular value decomposition directly givesEach ${\lambda }_p$ is a real nonnegative singular value. The scale factor $\sqrt{n_p}$ in both matrices cancels in this equation and is chosen to give the precise Hilbert-Schmidt inner product normalization in (iv).

Use $c>0$, the necessary parameter range for the bounds involving $1/c$. The construction comes from the singular value decomposition of matrix Fourier blocks. For each inequivalent unitary irreducible representation $\rho$ of dimension $d_{\rho }$, consider the Hilbert space ${\mathcal{H}}_{\rho }=M_{n\times d_{\rho }}(\mathbb{C})$ with the Hilbert-Schmidt inner product $⟨A,B⟩=\mathrm{tr}(A^{\ast }B)$ and the operatorIts operator norm is at most one, since $\rho (x)$ is a unitary matrix and $\Vert f(x){\Vert }_{\mathrm{op}}\le 1$. Thus all its singular values ${\lambda }_{\rho ,j}$ lie in $[0,1]$.

Two identities control their weighted moments:Here the expectation in $S_4$ is uniform on the $|G{|}^3$ solutions of the constraint, and the trace on ${\mathcal{H}}_{\rho }$ used to compute each moment is the ordinary operator trace. We justify the identities explicitly to fix their normalizations and the noncommutative order.

The adjoint operator is $T_{\rho }^{\ast }(A)={\mathbb{E}}_{y}f(y{)}^{\ast }A\rho (y)$, soFor rectangular matrices, the operator $A\mapsto LAR$ has trace $\mathrm{tr}(L)\mathrm{tr}(R)$, as is seen on the matrix-unit basis. Squaring the previous operator therefore givesThe required representation theory consists of unitarization of a finite-group representation, the Schur orthogonality relations, and the regular representation decomposition. The last gives the character expansion of the identity deltaSumming the two operator traces with weights $d_{\rho }$ selects $x=y$ in the first, and $w=x{y}^{-1}z$ in the second. Multiplication by $|G|$ converts each independent-variable expectation to its conditional expectation. This proves $S_2,S_4$. In particular $S_4$ is real and nonnegative, despite the apparently complex summands in its original formula. Since $S_4\le S_2\le n$, the nonvacuous hypothesis has $0<c\le 1$.

Select every singular value satisfying ${\lambda }_{\rho ,j}\ge \sqrt{c/2}$. Index the selected pairs by $p=1,\dots ,r$, set ${\rho }_p=\rho$, $n_p=d_{\rho }$, and choose unit right singular vectors $A_p$ and unit left singular vectors $B_p$ with $T_{{\rho }_p}{A}_p={\lambda }_p{B}_p$. DefineBoth are $n\times n_p$ matrices, proving (i). Repeated group representations in this list are literally the same chosen representative, rather than different equivalent realizations.

Within each fixed unitary irreducible representation, choose the selected right singular vectors to be orthonormal. Their corresponding left singular vectors are also orthonormal, even when a singular value is repeated: choose an orthonormal basis in each eigenspace of $T_{\rho }^{\ast }{T}_{\rho }$ and put $B_p=T_{\rho }{A}_p/{\lambda }_p$. Therefore, when ${\rho }_p={\rho }_q$,This is (iv), with the unnormalized Hilbert-Schmidt inner product and the Kronecker delta. No orthogonality of differently shaped matrices is being asserted.

For the unheaded continuation, partition $b\in {\mathbb{C}}^m$ into blocks $b_p\in {\mathbb{C}}^{n_p}$. ThenThe Schur averaging of rectangular matrices formula isfor any $n_p\times n_q$ matrix $M$. The average is an intertwiner; the Schur lemma makes it zero for inequivalent irreducible representations, and a scalar multiple of the identity for the same representative. The trace determines that scalar in the latter case.

Apply this with $M=U(p{)}^{\ast }U(q)$. The Hilbert-Schmidt inner product normalization just proved givesExpanding the squared Euclidean norm now yieldsThus the final identity follows from averaging, although $U^{\ast }U$ itself need not be the identity. The whole construction is the spectral inverse theorem for the matrix-valued U2 quantity.

Choose Kraus representationsThe composite quantum channel has Kraus operators $T_{ba}=D_b{C}_a$. Each has matrix rank at most $k$, because it factors through the $k$-dimensional space $K$, andThe operation fidelity uses the unsquared quantum fidelity on a purification of a density operator, so its square is entanglement fidelity. The Kraus formula for entanglement fidelity givesIndeed each overlap $⟨{\Psi }_{\rho }|(I_R\otimes T_{ba})|{\Psi }_{\rho }⟩$ equals $\mathrm{Tr}({\rho }_Q{T}_{ba})$.

For any one of these Kraus operators $T$, let ${\Pi }_T$ be the orthogonal projection onto its image, of matrix rank $r_T\le k$. Since ${\Pi }_{T}T=T$, the Cauchy-Schwarz inequality for the Hilbert-Schmidt inner product gives the rank bound for a weighted operator trace:The last step uses part (i) at $r_T$, followed by nonnegativity of the eigenvalues of the density operator ${\rho }_Q$. Summing over the Kraus operators and using their completeness relation proves the finite-dimensional quantum compression converse:No invertibility of ${\rho }_Q$, or restriction to an isometric decoder, was used.

The minimizer is the arithmetic meanIt remains a positive self-adjoint trace-class operator and hence a covariance operator. The Hilbert-Schmidt inner product gives, for every Hilbert-Schmidt operator $C$,because ${\sum }_k(C_k-\overline{C})=0$. Thus $\overline{C}$ is the unique minimizer, including when the minimization is restricted to covariance operators.

The Cauchy-Schwarz inequality for the Hilbert-Schmidt inner product gives $0\le J\le 1$ for $n$-dimensional unitary operators. Equality $J=0$ holds exactly when $U=e^{i\varphi }{U}_T$. When absolute gate phase matters, use the phase-sensitive Hilbert-Schmidt distance $\Vert U-U_T{\Vert }^2/(2n)=1-\mathrm{Re}\mathrm{Tr}(U_T^{†}U)/n$ instead.

If $\rho$ is a density operator with decreasing eigenvalues ${\lambda }_j$ and $T$ has matrix rank at most $k$, let $\Pi$ project onto the image of $T$. The Cauchy-Schwarz inequality for the Hilbert-Schmidt inner product givesThe Hermitian effect variational principle bounds $\mathrm{Tr}(\rho \Pi )$ by ${\sum }_{j<k}{\lambda }_j$, proving the claimed estimate without assuming $\rho$ is invertible.

If $\Vert f(x){\Vert }_{\mathrm{op}}\le 1$ and $\Vert f{\Vert }_{U^2}^4\ge cn>0$, select the singular values ${\lambda }_p\ge \sqrt{c/2}$ of the matrix Fourier blocks. Their weighted count $m={\sum }_p{d}_{{\rho }_p}$ lies between $cn/2$ and $2n/c$. Scaling unit right singular vectors and their corresponding unit left singular vectors by $\sqrt{d_{{\rho }_p}}$ gives matrices $U(p),V(p)$ satisfying ${\mathbb{E}}_{x}f(x)U(p){\rho }_p(x{)}^{\ast }={\lambda }_{p}V(p)$ and Hilbert-Schmidt inner product orthogonality within each chosen irreducible representation. The Schur averaging of rectangular matrices then gives ${\mathbb{E}}_x\Vert UP(x{)}^{\ast }b{\Vert }_2^2=\Vert b{\Vert }_2^2$ for the concatenated $U$ and the corresponding block diagonal matrix $P(x)$.