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A bounded linear operator $T:E\to F$ is 2-summing if ${\sum }_j\Vert T{x}_j{\Vert }^2\le C^2{\sup }_{\varphi \in B_{E^{\ast }}}{\sum }_j|\varphi (x_j){|}^2$ for every finite family; ${\pi }_2(T)$ is the least $C$. For operators between Hilbert spaces, this is equivalent to being a Hilbert-Schmidt operator, with ${\pi }_2(T)=\Vert T{\Vert }_{\mathrm{HS}}$. The upper bound follows by applying the Hilbert-Schmidt norm ideal inequality to the column operator of the family; testing finite orthonormal families gives the reverse bound.

For a strongly continuous unitary representation of a compact group, this Bochner integral exists in the Hilbert-Schmidt norm and is a positive compact operator commuting with the representation. It is nonzero if $\xi \ne 0$, because $⟨K_{\xi }\xi ,\xi ⟩={\int }_G|⟨U_g\xi ,\xi ⟩{|}^{2}dg>0$. A nonzero eigenspace is finite-dimensional and invariant. Applying this on orthogonal complements yields an orthogonal direct sum of finite-dimensional irreducible representations, with no separability assumption on the represented Hilbert space.

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 $B$, define the Hilbert-Schmidt inner product by $⟨A,B{⟩}_{\mathrm{HS}}={\sum }_k⟨A{e}_k,B{e}_k⟩$ for a Hilbertian basis $(e_k)$. With an inner product linear in its first argument this equals $\mathrm{tr}(B^{\ast }A)$; with an inner product conjugate-linear in its first argument it equals $\mathrm{tr}(A^{\ast }B)$. Its induced norm is the Hilbert-Schmidt norm, independently of that convention.

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.

On a Hilbert space $X$ with a countable Hilbertian basis $(e_k)$, the Hilbert-Schmidt operators form a Hilbert space under $⟨S,T⟩={\sum }_k⟨S{e}_k,T{e}_k⟩$, using the same linearity convention as the inner product on $X$. The Cauchy-Schwarz inequality proves convergence of this sum. The finite-sum triangle inequality and Cauchy-Schwarz inequality give $\Vert T\Vert \le \Vert T{\Vert }_2$. Thus a Cauchy sequence $(T_n)$ in Hilbert-Schmidt norm converges in operator norm to a bounded linear operator $T$. For $n,m\ge N$ with $\Vert T_n-T_m{\Vert }_2\le \epsilon$, taking $m\to \infty$ in each finite column sum gives ${\sum }_{k=1}^K\Vert (T_n-T)e_k{\Vert }^2\le {\epsilon }^2$ for every $K$. Taking the supremum over $K$ proves $T_n-T\in {\mathcal{S}}_2(X)$ and $\Vert T_n-T{\Vert }_2\le \epsilon$, so $T\in {\mathcal{S}}_2(X)$ and completeness follows.

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.

For the rank-one operator $Tf=⟨f,v⟩u$ on a Hilbert space with a countable Hilbertian basis $(e_k)$, the operator norm and Hilbert-Schmidt norm both equal $\Vert u\Vert \Vert v\Vert$. The Cauchy-Schwarz inequality gives $\Vert Tf\Vert \le \Vert u\Vert \Vert v\Vert \Vert f\Vert$, with equality at $f=v/\Vert v\Vert$ when $v\ne 0$. The Parseval identity for a Hilbertian basis gives ${\sum }_k\Vert T{e}_k{\Vert }^2=\Vert u{\Vert }^2{\sum }_k|⟨e_k,v⟩{|}^2=\Vert u{\Vert }^2\Vert v{\Vert }^2$. The zero-vector cases are immediate.

Use the intended space of Hilbert-Schmidt operators, with the given Hilbert-Schmidt inner product linear in its first argument. It remains to prove completeness in its Hilbert-Schmidt norm. Let $(T_n)$ be a Cauchy sequence in that norm. The bound $\Vert T_n-T_m\Vert \le \Vert T_n-T_m{\Vert }_2$ makes it a Cauchy sequence in operator norm. The space $\mathcal{B}(X)$ is a Banach space, so $T_n\to T$ in operator norm for some bounded linear operator $T$.

For completeness, that last standard fact follows by taking $Tx={\lim }_n{T}_{n}x$ for each $x$: the Cauchy sequence $(T_n)$ is uniformly bounded in operator norm, so its limit is linear and bounded, and the uniform Cauchy estimate on the unit ball gives operator norm convergence.

Given $\epsilon >0$, choose $N$ so that $\Vert T_n-T_m{\Vert }_2<\epsilon$ for $n,m\ge N$. Fix $n\ge N$ and any positive integer $K$. Since $T_m{e}_k\to T{e}_k$ for each $k$, passage to the limit in a finite sum givesTaking the supremum over $K$ gives ${\sum }_k\Vert (T_n-T)e_k{\Vert }^2\le {\epsilon }^2$. Also,Thus $T\in S_2(X)$, and $\Vert T_n-T{\Vert }_2\le \epsilon$ for $n\ge N$. This proves Hilbert-Schmidt operators form a Hilbert space:Under the literal unrestricted “linear mappings” definition, the nonzero $R$ from part (i) satisfies $⟨R,R{⟩}_{S_2(X)}=0$. The form would fail positive definiteness and would not be an inner product; the boundedness correction is needed here too.

As in the preceding part, take $T$ to be a bounded linear operator. For $x\in X$, the Hilbertian basis expansion and continuity of $T$ giveThe triangle inequality, the Cauchy-Schwarz inequality for finite sums, and the Parseval identity for a Hilbertian basis givePassing to the limit and taking the supremum over the unit ball provesThus the Hilbert-Schmidt norm controls the operator norm. Without the boundedness interpretation, the nonzero $R$ constructed in the preceding part has zero printed column sum and makes the requested inequality false.

Here the intended space must consist of bounded linear operators, namely Hilbert-Schmidt operators. The printed restriction to “all linear mappings” is insufficient. To see this, choose $v\in X$ outside the algebraic linear span of $(e_k)$, for example $v={\sum }_{k\ge 1}{2}^{-k}{e}_k$. Extend the linearly independent family $\{e_k:k\ge 1\}\cup \{v\}$ to a Hamel basis. Define a linear functional $\varphi$ on this Hamel basis by $\varphi (v)=1$ and zero on every other basis vector, and extend it linearly. Then the everywhere-defined linear operator $Rx=\varphi (x)e_1$ satisfiesIts printed column sum is zero although $R\ne 0$. More explicitly, $v_N={\sum }_{k=1}^N{2}^{-k}{e}_k\to v$ but $R(v-v_N)=e_1$ for every $N$. Thus $R$ is discontinuous; it has no everywhere-defined adjoint operator of the bounded kind used in the question. This proves finite column sum does not imply boundedness. We use $T\in \mathcal{B}(X)$ in the intended argument below.

By the Parseval identity for a Hilbertian basis for $(e_j)$, followed by the adjoint operator identity and then the Parseval identity for a Hilbertian basis for $(h_k)$,The interchange is valid by Tonelli's theorem for nonnegative double series, even before finiteness is known. Repeating with $h_k=e_k$ shows that the right side also equals ${\sum }_k\Vert T{e}_k{\Vert }^2$. HenceIn particular the Hilbert-Schmidt norm is independent of the Hilbertian basis, and $\Vert T^{\ast }{\Vert }_2=\Vert T{\Vert }_2$.

Use the real vector space of Hermitian matrices and its Hilbert-Schmidt inner product $⟨A,B⟩=\mathrm{Tr}(AB)$. A Hermitian $N$-by-$N$ matrix has $N$ real diagonal entries and two real parameters for each off-diagonal pair, hence real dimension $N+2\left(\genfrac{}{}{0ex}{}{N}{2}\right)=N^2$. Imposing zero trace removes one real dimension. Each supplied generalized Pauli matrix is Hermitian and traceless, and there are $N(N-1)$ off-diagonal matrices and $N-1$ diagonal matrices, totaling $N^2-1$.

Their assumed orthonormality makes them linearly independent. Since this equals the dimension of the traceless space, they are an orthonormal basis of the traceless Hermitian matrices. To extend the family, the required additional generator is ${\sigma }_0=I/\sqrt{N}$, which is implicit in the indexing of the question. It has unit Hilbert-Schmidt norm and is orthogonal to every traceless generator, so the extended family is an orthonormal basis of all Hermitian matrices.

For a density operator $\rho$, cyclicity of the trace givesThus $s\in {\mathbb{R}}^{N^2-1}$. Since $\mathrm{Tr}\rho =1$, the resulting Generalized Bloch representation is

For an orthonormal basis $(e_i)$, define the Hilbert-Schmidt norm by $\Vert T{\Vert }_2^2={\sum }_i\Vert T{e}_i{\Vert }^2$. For a possibly uncountable basis, the nonnegative sum means the supremum over finite subsets. Parseval's identity gives, for any other orthonormal basis $(f_j)$,Changing the domain basis in the last expression proves basis independence. Polarization gives a basis-independent inner productwith the same first-argument linear convention. The sums converge by the Cauchy-Schwarz inequality. The operator-norm bound $\Vert T\Vert \le \Vert T{\Vert }_2$ follows by applying Cauchy-Schwarz to the coordinates of a unit vector.

To prove completeness, let $(T_n)$ be Cauchy in this norm. The operator-norm bound gives a bounded-operator limit $T$. For a finite subset $F$ of the basis,For large $n$, the right side is uniformly small by the Cauchy property. Taking the supremum over $F$ shows $T_n-T$ is Hilbert-Schmidt and $\Vert T_n-T{\Vert }_2\to 0$; it also shows $T$ is Hilbert-Schmidt. Thus the Hilbert-Schmidt operators form a Hilbert space. Finite-rank operators are dense there: keeping finitely many basis columns makes the sum of omitted squared column norms tend to zero. The operator-norm bound then shows that every Hilbert-Schmidt operator is compact.

Now let $U$ be a strongly continuous unitary representation of the compact metric group $G$, as is standard for a representation of a topological group. Fix $\xi \ne 0$ and average the positive rank-one operator $R_{\xi ,\xi }$:The integrand is continuous in the Hilbert-Schmidt norm, so this Bochner integral exists in the complete space just proved. Normalized Haar measure and invariance show that $K$ is positive, compact and commutes with every $U_g$. Moreover,the integrand is positive near the identity and every nonempty open set has positive Haar measure. Thus $K\ne 0$. A nonzero eigenspace is finite-dimensional and invariant under $U$, by (a). Within it choose a nonzero invariant subspace of smallest dimension; it is irreducible. Its orthogonal complement is invariant because the representation is unitary, so the finite-dimensional space splits into irreducibles.

Take a maximal orthogonal family of such finite-dimensional irreducible invariant subspaces. If their closed span had a nonzero orthogonal complement, the same averaging argument on that complement would produce another member, contradicting maximality. ThereforeThis compact averaging of a rank-one operator argument does not require $H$ itself to be separable. Strong continuity is the topological representation hypothesis used in the averaging step.

A density operator is Hermitian, positive semidefinite and has trace one. Part (a) therefore makes all its coordinates real, and the identity basis vector fixes the final coordinate:The remaining basis vectors are traceless by orthogonality to $I/\sqrt{N}$. The Generalized Bloch representation is consequentlyUsing orthonormality to expand the squared Hilbert-Schmidt norm gives the purity bound for generalized Bloch vectors:If $p_1,\dots ,p_N$ are its nonnegative eigenvalues, then ${\sum }_j{p}_j=1$ andIt follows thatEquality requires every pairwise product $p_j{p}_k$ to vanish, so exactly one eigenvalue equals one and all others are zero. Conversely, a rank-one projector has precisely that spectrum and attains equality. Thus equality holds exactly for a pure state. This normalization is different from the commonly used unit-radius qubit Bloch vector.

The Weyl-von Neumann theorem states that on a separable Hilbert space a self-adjoint operator is a diagonal self-adjoint operator plus a compact self-adjoint operator. For a bounded operator $T$, the perturbation can in fact be chosen Hilbert-Schmidt with arbitrarily small Hilbert-Schmidt norm. The separable-space formulation is also recorded in Kadison's Corollary B.

Here is a constructive proof. The bounded self-adjoint spectral theorem decomposes a separable $H$ into countably many cyclic reducing spaces, on each of which $T$ is multiplication by $x$ in $L^2(\mu )$ for a finite measure supported in a bounded real interval. First work on one such space. Partition the interval into $N$ intervals of length at most $\delta$, and repeatedly bisect them. Let $E_j$ be the finite-dimensional space of functions constant on the level-$j$ intervals. These spaces are nested and their union is dense, since dyadic step functions approximate continuous functions and continuous functions are dense in $L^2(\mu )$.

Choose normalized interval indicators as a basis of $E_0$, and choose orthonormal bases of $E_j\ominus E_{j-1}$ within each parent interval. This is possible because the refinement splits into orthogonal parent blocks. Each new basis vector at level $j\ge 1$ is supported in an interval of length at most $\delta 2^{-(j-1)}$, and there are at most $N{2}^{j-1}$ such vectors. For every basis vector $e$ choose a real ${\lambda }_e$ in its supporting interval and define the diagonal operator $De={\lambda }_{e}e$. ThensoOne may take $N=O({\delta }^{-1})$, so this bound tends to zero as $\delta \to 0$. The error is Hilbert-Schmidt, hence compact; it is self-adjoint because both $T$ and $D$ are.

On the countably many cyclic summands choose error bounds ${\epsilon }_j$ with ${\sum }_j{\epsilon }_j^2<{\epsilon }^2$. Combining their orthonormal bases and diagonal operators gives $T=D+K$ with $\Vert K{\Vert }_{\mathrm{HS}}<\epsilon$, proving the assertion. For an unbounded self-adjoint operator, first split by spectral intervals $[j,j+1)$, apply the bounded construction on each, and choose summable error bounds. The resulting total $K$ is bounded compact, and $D=T-K$ has the same domain as $T$; the diagonal domain condition follows from the boundedness of $K$.

Separability matters. Take an uncountable orthogonal sum of copies of multiplication by $x$ on $L^2[0,1]$. The range closure of a compact operator is separable, and a separable subspace of this direct sum is supported in only countably many summands: choose a countable dense set and unite its countable supports. A self-adjoint compact error therefore vanishes on every remaining summand, each of which reduces the proposed diagonal operator $D=T-K$. The restriction of a diagonal self-adjoint operator to a reducing subspace has pure point spectrum, whereas multiplication by $x$ on $L^2[0,1]$ has no eigenvectors. This is a contradiction, qualifying the theorem if the unspoken separability convention is dropped.

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