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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.

The absolute value of an operator, trace norm, and operator norm are respectivelyThe positive square root of an operator makes $|L|$ well defined. Its eigenvalues $s_j\ge 0$ are the singular values of $L$, so $\Vert L{\Vert }_1={\sum }_j{s}_j$ and $\Vert L{\Vert }_{\mathrm{op}}={\max }_j{s}_j$.

Use the polar decomposition of a bounded operator $L=U|L|$, with the given unitary operator $U$. In an orthonormal eigenbasis $e_j$ for $|L|$,The operator norm assumption and unitarity imply $|⟨e_j|ZU|e_j⟩|\le \Vert Z{\Vert }_{\mathrm{op}}\le 1$. The triangle inequality therefore yields the required trace estimate:Choosing $Z=U^{†}$ attains equality, which also gives the finite-dimensional trace duality formula $\Vert L{\Vert }_1={\max }_{\Vert Z{\Vert }_{\mathrm{op}}\le 1}|\mathrm{Tr}(ZL)|$. A singular $L$ causes no difficulty: in a finite-dimensional square space the polar factor can be extended to a unitary operator on the complementary subspaces.

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

Part b(vi) and part c show that $\Vert T{\Vert }_1\le 1$ impliesConversely, choose a unit vector $x$ with $\Vert Sx\Vert$ arbitrarily close to $\Vert S\Vert$, put $y=Sx/\Vert Sx\Vert$, and take $T=x\otimes y$. Part c gives $\Vert T{\Vert }_1=1$ andTaking the supremum provesThus $S\mapsto [T\mapsto \mathrm{tr}(ST)]$ is an isometric embedding ${\mathcal{S}}_{\infty }\to {\mathcal{S}}_1^{\ast }$.

Suppose finite-rank operators are dense in ${\mathcal{S}}_1$ and let $L\in {\mathcal{S}}_1^{\ast }$. The sesquilinear formsatisfies $|b(x,y)|\le \Vert L\Vert \Vert x\Vert \Vert y\Vert$. The Riesz representation theorem gives $S\in {\mathcal{S}}_{\infty }$ with $b(x,y)=⟨Sx,y⟩$. Hence $L(x\otimes y)=\mathrm{tr}(S(x\otimes y))$, and linearity gives equality on every finite-rank operator. Density and continuity extend it to every $T\in {\mathcal{S}}_1$, proving surjectivity.

Conversely, if finite-rank operators were not dense, the Hahn-Banach theorem would give a nonzero $L\in {\mathcal{S}}_1^{\ast }$ vanishing on their closure. Surjectivity would represent it by some $S$, but thenfor all $x,y$, forcing $S=0$ and $L=0$, a contradiction. This proves the stated trace duality criterion.

Every density operator $\sigma$ with positive partial transpose has squared quantum fidelity at most $1/d$ with a fixed maximally entangled state on $d\otimes d$. The partial transpose of a maximally entangled projector and the trace-adjoint identity for partial transpose express that squared overlap as $\mathrm{Tr}(S{\sigma }^{T_B})/d$. Since ${\sigma }^{T_B}$ is positive of trace one and the swap operator has operator norm one, trace duality gives the bound. A matching product state attains equality.