Quantum Channel Capacities and Entanglement Assistance

Omit zero weights. For each $i$ with $p_i>0$, $\bar{\rho }-p_i{\rho }_i={\sum }_{j\mathrm{\ne }i}{p}_j{\rho }_j$ is non-negative. If $v\in \ker \bar{\rho }$, then $0=⟨\bar{\rho }v,v⟩=p_i⟨{\rho }_{i}v,v⟩+{\sum }_{j\mathrm{\ne }i}{p}_j⟨{\rho }_{j}v,v⟩$. All terms are nonnegative, so $p_i>0$ implies $⟨{\rho }_{i}v,v⟩=0$ and [F5] gives ${\rho }_{i}v=0$. Hence $\mathrm{supp}{\rho }_i\subseteq \mathrm{supp}\bar{\rho }$ and every $D({\rho }_i\mathrm{\parallel }\bar{\rho })$ is finite. The trace formula [F6], linearity of trace [F4], and $\bar{\rho }={\sum }_i{p}_i{\rho }_i$ give ${\sum }_i{p}_{i}D({\rho }_i\mathrm{\parallel }\bar{\rho })=S(\bar{\rho })-{\sum }_i{p}_{i}S({\rho }_i)\ge 0$ by [F2].

For the upper bound, diagonalize each ${\rho }_i={\sum }_j{\lambda }_{ij}\mathrm{\mid }{v}_{ij}⟩⟨v_{ij}\mathrm{\mid }$, omitting zero weights and eigenvalues, and purify the mixture as $\mathrm{\mid }v⟩={\sum }_{ij}\sqrt{p_i{\lambda }_{ij}}\mathrm{\mid }{v}_{ij}⟩\otimes \mathrm{\mid }i,j⟩$. Its label marginal $G$ is a state with diagonal $q_{ij}=p_i{\lambda }_{ij}$ and has the same nonzero eigenvalues as $\bar{\rho }$ by Schmidt decomposition. The logarithm product law in [F6] gives $H(q)=H(p)+{\sum }_i{p}_{i}S({\rho }_i)$.

For $\mathrm{\Delta }=\mathrm{diag}(q)$, its kernel is spanned by coordinate labels with $q_{ij}=0$. Each such basis vector $e_{ij}$ has $⟨G{e}_{ij},e_{ij}⟩=q_{ij}=0$, so [F5] gives $G{e}_{ij}=0$. Hence $G$ annihilates $\ker \mathrm{\Delta }$ and $\mathrm{supp}G\subseteq \mathrm{supp}\mathrm{\Delta }$. The formula [F6] now gives $0\le D(G\mathrm{\parallel }\mathrm{\Delta })=H(q)-S(G)=H(p)+{\sum }_i{p}_{i}S({\rho }_i)-S(\bar{\rho })\mathrm{.}$ This proves both bounds, including singleton/zero-weight cases.

For $P_n$, retain product eigenvectors of ${\bar{\rho }}^{\otimes n}$ whose positive eigenvalue has negative logarithm in $n[S(\bar{\rho })-\eta ,S(\bar{\rho })+\eta ]$. For $Q_{x^n}$, use product eigenvectors of ${\rho }_{x^n}={⨂}_i{\rho }_{x_i}$ whose eigenvalue has negative logarithm in $n[s-\eta ,s+\eta ]$. The upper eigenvalue bounds are immediate. The lower retained eigenvalue and trace at most one imply each rank bound. Under the joint law $\mathrm{Pr}(X=x,J=j)=p_x{\lambda }_{xj}$, with zero-probability eigenvalues omitted, the random variable $Z=-\ln {\lambda }_{XJ}$ has finite support and mean $s$. By [F7], the negative logarithm of each product eigenvalue is the sum of the coordinate self-information variables, so its IID sum controls the second failure probability exactly. The first failure is the same argument for the spectral law of $\bar{\rho }$; averaging inputs gives ${\bar{\rho }}^{\otimes n}$.

For completeness, a finite supported centered variable $Y$ with $\mathrm{\mid }Y\mathrm{\mid }\le b$ obeys $\mathbb{E}{e}^{uY}\le 1+u^2{b}^2{e}^{\mathrm{\mid }u\mathrm{\mid }b}\mathrm{/}2$, by the Taylor remainder and $\mathbb{E}Y=0$. For small fixed $u>0$, $\ln \mathbb{E}{e}^{uY}<u\eta \mathrm{/}2$. Markov's inequality and independence then bound $\mathrm{Pr}(\sum Y_i\ge n\eta )\le e^{-nu\eta \mathrm{/}2}$; the logarithm product law [F7] gives the factor $n$ when taking the log of the product moment generating function. Apply the same argument to $-Y$ and add. A constant spectral variable has zero failure probability and needs no division by a range. All logarithms are evaluated only on positive-probability eigenvalues.

Set $\zeta =\exp (2\pi i\mathrm{/}5)$. The three finite complex-exponential premises in [F9] show $\mathrm{\mid }\zeta \mathrm{\mid }=1$ and ${\zeta }^m=1$ exactly when $5$ divides $m$. Let $G$ be the finite group of monomial unitaries whose nonzero entries belong to $\{1,\zeta ,{\zeta }^2,{\zeta }^3,{\zeta }^4\}$. Let $R$ be the real rotation by $\pi \mathrm{/}4$ in coordinates 1 and 2, acting identically elsewhere. Use the symmetric finite list $\mathcal{S}=G\cup \{R,R^{\ast }\}$, retaining multiplicities if any. At each walk step choose the identity with probability $1\mathrm{/}2$ and an element of this list uniformly with the remaining probability. For $j=1,2$, the moment operator on matrices is $T_j(X)=\frac{1}{2}X+\frac{1}{2\mathrm{\mid }\mathcal{S}\mathrm{\mid }}{\sum }_{U\in \mathcal{S}}{U}^{\otimes j}X(U^{\ast }{)}^{\otimes j}\mathrm{.}$ These are self-adjoint contractions for the Hilbert–Schmidt inner product, because the list is inverse-symmetric; laziness makes every eigenvalue nonnegative. Moreover $⟨X,(I-T_j)X⟩=\frac{1}{4\mathrm{\mid }\mathcal{S}\mathrm{\mid }}{\sum }_{U\in \mathcal{S}}\mathrm{\parallel }X-U^{\otimes j}X(U^{\ast }{)}^{\otimes j}{\mathrm{\parallel }}_2^2\mathrm{.}$ Thus their fixed spaces are exactly the simultaneous commutants of the displayed finite unitaries.

For $j=1$, the independent diagonal fifth-root phases force a commuting matrix to be diagonal; permutations force all its diagonal entries equal. For $j=2$, a matrix coefficient from $\mathrm{\mid }i,j⟩$ to $\mathrm{\mid }k,l⟩$ can survive the diagonal phases only when the unordered multisets $\{i,j\}$ and $\{k,l\}$ agree. By the primitive-order identity established in step 1.1, each index occurs with multiplicity 0, 1 or 2; equality modulo 5 of every exponent forces equality of these integer multiplicities. Commutation with all permutations then leaves three constants: $c$ on $\mathrm{\mid }i,i⟩$, $a$ on $\mathrm{\mid }i,j⟩⟨i,j\mathrm{\mid }$ for $i\mathrm{\ne }j$, and $b$ on $\mathrm{\mid }i,j⟩⟨j,i\mathrm{\mid }$. Commutation with $R\otimes R$ applied to $\mathrm{\mid }1,1⟩$, whose rotated vector has nonzero coefficients on both $\mathrm{\mid }1,2⟩$ and $\mathrm{\mid }2,1⟩$, forces $c=a+b$. Conversely $I$ and the swap $F$ commute with every tensor-square unitary. The fixed space is therefore precisely $\mathrm{span}\{I,F\}$, also when $D=2$.

Finite-dimensional spectral decomposition now gives orthogonal projections ${\mathcal{P}}_1,{\mathcal{P}}_2$ onto these fixed spaces and a number $0\le \lambda <1$ such that $\mathrm{\parallel }{T}_j^L-{\mathcal{P}}_j{\mathrm{\parallel }}_{2\to 2}\le {\lambda }^L,j=1,2.$ If the orthogonal complement is zero, take its spectral radius as zero. Every $T_j^L$ is the moment of an finite distribution of length-$L$ unitary words; the limit is only a finite-matrix spectral calculation, not an assumed Haar integral. The first projection is ${\mathcal{P}}_1(X)=\mathrm{tr}(X)I\mathrm{/}D$. For any rank-$K$ projection $P_0$, invariance of trace and swap trace gives ${\mathcal{P}}_2(P_0\otimes P_0)=a_{0}I+b_{0}F,a_0=\frac{K(KD-1)}{D(D^2-1)},b_0=\frac{K(D-K)}{D(D^2-1)}\mathrm{.}$ The coefficients solve $a_0{D}^2+b_{0}D=K^2$ and $a_{0}D+b_0{D}^2=K$.

Choose a coordinate isometry $W_0$ and let $W=U{W}_0$, where $U$ is a finite walk word. Complex conjugation preserves the same limiting first and second moment projections and their norm bounds, so they apply to $P=\bar{W}{W}^T$ as well. The swap identity on matrix units gives the following limiting second moment, with $\mathrm{\Omega }=I_R\mathrm{/}K\otimes {\omega }_E$: ${\lim }_{L\to \mathrm{\infty }}\mathbb{E}\mathrm{\parallel }{\omega }^W-\mathrm{\Omega }{\mathrm{\parallel }}_2^2=\frac{D(KD-1)}{K(D^2-1)}\mathrm{tr}({\omega }^2)+\frac{1-DK}{K(D^2-1)}\mathrm{tr}({\omega }_E^2)\le 2\mathrm{tr}({\omega }^2)\mathrm{.}$ In detail, the squared-state term is $(D^2\mathrm{/}{K}^2)\{a_0\mathrm{tr}({\omega }^2)+b_0\mathrm{tr}({\omega }_E^2)\}$, and the limiting mean is $\mathrm{\Omega }$, so subtract $\mathrm{tr}({\omega }_E^2)\mathrm{/}K$.

The finite moment error has a quantitative bound. Put $d_E=\dim E$ and $\epsilon ={\lambda }^L$. The first-moment error in each matrix unit of the compressed reference is at most $\epsilon$ in Hilbert–Schmidt norm; contracting it with $\omega$ shows $\mathrm{\parallel }\mathbb{E}{\omega }^W-\mathrm{\Omega }{\mathrm{\parallel }}_2\le D\epsilon \mathrm{\parallel }\omega {\mathrm{\parallel }}_2\le D\epsilon$. For the squared-state term, the second-moment error on $P_0\otimes P_0$ has Hilbert–Schmidt norm at most $K\epsilon$. In its trace pairing with $F_E(\omega \otimes \omega )$, the partial trace over the two environment copies has Hilbert–Schmidt norm at most $d_E\mathrm{\parallel }\omega {\mathrm{\parallel }}_2^2\le d_E$ (Cauchy–Schwarz for each diagonal block). Consequently that squared-state error is at most $D^2{d}_E\epsilon \mathrm{/}K$. The cross term with $\mathrm{\Omega }$, whose Hilbert–Schmidt norm is at most $1\mathrm{/}\sqrt{K}$, has error at most $2D\epsilon \mathrm{/}\sqrt{K}$. Hence, with $C_D=D^2{d}_E+2D$, $\mathbb{E}\mathrm{\parallel }{\omega }^W-\mathrm{\Omega }{\mathrm{\parallel }}_2^2\le 2\mathrm{tr}({\omega }^2)+C_D\epsilon \mathrm{.}$ Positivity implies $\mathrm{supp}\omega \subseteq A\otimes \mathrm{supp}{\omega }_E$: if $v\in \ker {\omega }_E$, the marginal quadratic form is a sum of nonnegative diagonal quadratic forms of $\omega$; each is zero, and [F6] makes the corresponding product vectors kernel vectors of $\omega$. Self-adjointness then makes the corresponding rows and columns zero. Compression acts only on $A$, so both ${\omega }^W$ and $\mathrm{\Omega }$ are supported in ${\mathbb{C}}^K\otimes \mathrm{supp}{\omega }_E$; each difference therefore has rank at most $Kr$. Cauchy–Schwarz on singular values and Jensen give the trace-norm bound at most $\sqrt{2Kr\mathrm{tr}({\omega }^2)}+\sqrt{Kr{C}_D\epsilon }$.

Finally write a Hermitian $Z$ as $Z_{+}-Z_{-}$ using [F8]. Each compression of $Z_{+}$ or $Z_{-}$ is positive, so its trace norm is its trace by [F7]; positivity and linearity therefore bound the trace norm of the difference by the sum of the two compressed traces. The first-moment bound $\mathrm{\parallel }\mathbb{E}P-(K\mathrm{/}D)I{\mathrm{\parallel }}_{\mathrm{\infty }}\le \sqrt{K}\epsilon$ gives $\mathbb{E}\mathrm{\parallel }{\mathcal{C}}_W(Z){\mathrm{\parallel }}_1\le (1+D\epsilon \mathrm{/}\sqrt{K})\mathrm{\parallel }Z{\mathrm{\parallel }}_1\mathrm{.}$ Choose a finite $L$ with both $Kr{C}_D{\lambda }^L\le {\delta }^2$ and $D{\lambda }^L\mathrm{/}\sqrt{K}\le \delta$. Such an integer exists by $\lambda <1$; the least such integer is a specified choice. If $r=0$, the positive operator is zero and its decoupling bound is automatic. This proves both assertions with an finite ensemble and exact dimension-dependent moment control.

Choose a finite purification $\mathrm{\mid }\mathrm{\Omega }{⟩}_{ZAF}$ of ${\omega }_{ZA}$ by [F1] and reorder its factors as $A(ZF)$. Let ${\rho }_A={\sum }_{j=1}^r{\lambda }_j\mathrm{\mid }{a}_j⟩⟨a_j\mathrm{\mid }$ be a spectral decomposition on its positive support, obtained from [F7]. Expand each purification in the product basis using this same eigenbasis on $A$: $\mathrm{\mid }\psi {⟩}_{RA}={\sum }_{j=1}^r\mathrm{\mid }{r}_j^0{⟩}_R\mathrm{\mid }{a}_j{⟩}_A,\mathrm{\mid }\mathrm{\Omega }{⟩}_{A(ZF)}={\sum }_{j=1}^r\mathrm{\mid }{a}_j{⟩}_A\mathrm{\mid }{z}_j^0{⟩}_{ZF}\mathrm{.}$ The partial-trace coordinate formula [F2] and the fact that both $A$ marginals equal ${\rho }_A$ show that the coefficient vectors have Gram matrices $⟨r_j^0,r_k^0⟩={\lambda }_j{\delta }_{jk}$ and $⟨z_j^0,z_k^0⟩={\lambda }_j{\delta }_{jk}$. Thus $\mathrm{\mid }{r}_j⟩:=\mathrm{\mid }{r}_j^0⟩\mathrm{/}\sqrt{{\lambda }_j}$ and $\mathrm{\mid }{z}_j⟩:=\mathrm{\mid }{z}_j^0⟩\mathrm{/}\sqrt{{\lambda }_j}$ are orthonormal and $\mathrm{\mid }\psi {⟩}_{RA}={\sum }_{j=1}^r\sqrt{{\lambda }_j}\mathrm{\mid }{r}_j{⟩}_R\mathrm{\mid }{a}_j{⟩}_A,\mathrm{\mid }\mathrm{\Omega }{⟩}_{A(ZF)}={\sum }_{j=1}^r\sqrt{{\lambda }_j}\mathrm{\mid }{a}_j{⟩}_A\mathrm{\mid }{z}_j{⟩}_{ZF}\mathrm{.}$ This coefficient argument also handles repeated positive eigenvalues because the chosen basis is fixed before either expansion.

Let $d=\dim R=\dim A$. In every case set $G={\mathbb{C}}^d$, fix its first coordinate unit vector $\mathrm{\mid }0{⟩}_G$, and append it to $\mathrm{\mid }\mathrm{\Omega }⟩$. Since $Z$ and the purifying space $F$ are nonzero, $\dim (Z\otimes F\otimes G)\ge d$. Denote the resulting orthonormal Schmidt partners by $\mathrm{\mid }{z}_j^{\mathrm{\prime }}⟩=\mathrm{\mid }{z}_j⟩\mathrm{\mid }0{⟩}_G$. The map $U_0:\mathrm{span}\{\mathrm{\mid }{r}_j⟩\}\to \mathrm{span}\{\mathrm{\mid }{z}_j^{\mathrm{\prime }}⟩\}$ given by $U_0\mathrm{\mid }{r}_j⟩=\mathrm{\mid }{z}_j^{\mathrm{\prime }}⟩$ is an isometry. Extend $\{\mathrm{\mid }{r}_j⟩\}$ to an orthonormal basis of $R$ and choose the same number of additional orthonormal vectors in the orthogonal complement of the $z_j^{\mathrm{\prime }}$ in $Z\otimes F\otimes G$; [F5] and the target dimension bound ensure this is possible. Sending the added basis vectors to those target vectors extends $U_0$ to a linear isometry $U:R\to Z\otimes F\otimes G$ on all of $R$. By construction, $(U\otimes I_A)\mathrm{\mid }\psi ⟩$ is exactly the padded purification $\mathrm{\mid }\mathrm{\Omega }⟩\otimes \mathrm{\mid }0{⟩}_G$, after reordering to $Z(F\otimes G)A$. This construction includes the singular case $r<d$, since it extends $U$ on all zero-Schmidt directions of $R$.

Define $\mathrm{\Lambda }(X)={\mathrm{Tr}}_{F\otimes G}(UX{U}^{\ast })$ from $\mathcal{L}(R)$ to $\mathcal{L}(Z)$. Since $U$ is an isometry, [F4] makes $\mathrm{\Lambda }$ completely positive and trace-preserving on the full operator space. Applying it to the reference factor of $\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid }$ traces $F\otimes G$ from the pure state $\mathrm{\mid }(U\otimes I_A)\psi ⟩⟨(U\otimes I_A)\psi \mathrm{\mid }$, leaving its $ZA$ marginal, which is ${\omega }_{ZA}$ by construction and is unchanged by the appended pure $G$ register [F6]. Thus $(\mathrm{\Lambda }\otimes {\mathrm{id}}_A)(\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid })={\omega }_{ZA}$, as required.

Set $\alpha :={\sup }_{k\ge 1}{a}_k\mathrm{/}k$. The bound in [A1] makes $\alpha$ finite, and the definition of supremum gives $a_n\mathrm{/}n\le \alpha$ for every $n\ge 1$. Hence ${\mathrm{lim\hspace{0.17em}sup}}_{n\to \mathrm{\infty }}{a}_n\mathrm{/}n\le \alpha$.

Fix $k\ge 1$. For each $n$, write $n=qk+r$ with integers $q\ge 0$ and $0\le r<k$. Repeated use of superadditivity in [A1] gives $a_n\ge q{a}_k+a_r$. The finitely many values $a_0,\dots ,a_{k-1}$ have a finite minimum $m_k$, so $a_n\mathrm{/}n\ge (q\mathrm{/}n)a_k+m_k\mathrm{/}n$. As $n\to \mathrm{\infty }$, $q\mathrm{/}n\to 1\mathrm{/}k$ and $m_k\mathrm{/}n\to 0$, yielding ${\mathrm{lim\hspace{0.17em}inf}}_{n\to \mathrm{\infty }}{a}_n\mathrm{/}n\ge a_k\mathrm{/}k$.

Since step 1.2 holds for every $k\ge 1$, taking the supremum gives ${\mathrm{lim\hspace{0.17em}inf}}_{n\to \mathrm{\infty }}{a}_n\mathrm{/}n\ge \alpha$. Combining with step 1.1 proves the asserted limit and its value.

The orthogonal summands reduce ${\omega }_p$, so its eigenvalues, including multiplicities, are $p{\lambda }_j$ on $H$ and $1-p$ on the flag. For $0<p<1$, [F1]–[F3] give $S({\omega }_p)=-{\sum }_{j:{\lambda }_j>0}p{\lambda }_j\ln (p{\lambda }_j)-(1-p)\ln (1-p)=h(p)+pS(\tau )$, because ${\sum }_j{\lambda }_j=1$.

At $p=0$, ${\omega }_p$ is the pure flag state and both sides equal zero. At $p=1$, ${\omega }_p=\tau$ and both sides equal $S(\tau )$. Thus the formula holds on the closed interval, including its endpoints.

First apply $\mathrm{\Phi }$ to $A$. Choose a Stinespring isometry $V:A\to A^{\mathrm{\prime }}\otimes E$ from [F3], and form ${\tilde{\rho }}_{A^{\mathrm{\prime }}EB}:=(V\otimes I_B){\rho }_{AB}(V^{\ast }\otimes I_B)$. This is a state: a spectral decomposition of ${\rho }_{AB}$ in [F4] shows that its nonzero eigenvalues remain the same under the isometric embedding. Its $A^{\mathrm{\prime }}E$ marginal is $V{\rho }_A{V}^{\ast }$ and has the same positive eigenvalues as ${\rho }_A$, while its $A^{\mathrm{\prime }}EB$ joint state has the same positive eigenvalues as ${\rho }_{AB}$. Thus [F1] gives $I(A^{\mathrm{\prime }}E:B{)}_{\tilde{\rho }}=I(A:B{)}_{\rho }$.

Reorder $A^{\mathrm{\prime }}EB$ to $E{A}^{\mathrm{\prime }}B$ by the unitary permutation in [F6]. Apply [F2] to the ordered tripartite state ${\tilde{\rho }}_{E{A}^{\mathrm{\prime }}B}$, with $A^{\mathrm{\prime }}=Y$, to obtain $S(E{A}^{\mathrm{\prime }})+S(A^{\mathrm{\prime }}B)\ge S(A^{\mathrm{\prime }})+S(E{A}^{\mathrm{\prime }}B)$. Rearranging by [F1] gives $I(E{A}^{\mathrm{\prime }}:B)\ge I(A^{\mathrm{\prime }}:B)$. The entropy identities and their logarithm product law are supplied by [F7]. The $A^{\mathrm{\prime }}B$ marginal is $(\mathrm{\Phi }\otimes {\mathrm{id}}_B)(\rho )$ by [F3] and the partial-trace convention; hence applying $\mathrm{\Phi }$ cannot increase mutual information.

Repeat step 1.1–2.1 with $\mathrm{\Psi }$ acting on the second register, now taking the first register to be $A^{\mathrm{\prime }}$. This gives $I(A^{\mathrm{\prime }}:B^{\mathrm{\prime }}{)}_{(\mathrm{\Phi }\otimes \mathrm{\Psi })(\rho )}\le I(A^{\mathrm{\prime }}:B{)}_{(\mathrm{\Phi }\otimes {\mathrm{id}}_B)(\rho )}$. Combine this with step 2.1 to obtain the claimed inequality. The entropy-law prerequisite remains [F7].

The operator $\tau :=I_H\mathrm{/}d$ is a density operator: it is nonnegative and has trace $d\mathrm{/}d=1$ by [F4]–[F5]. Its support is all of $H$, so $\mathrm{supp}(\rho )\subseteq \mathrm{supp}(\tau )$ and the finite formula in [F2] applies.

Spectral calculus gives $\ln \tau =\ln (1\mathrm{/}d)I_H=-(\ln d)I_H$ by [F4], [F6]. Hence [F1]–[F2] yield $D(\rho \mathrm{\parallel }\tau )=\mathrm{tr}(\rho \ln \rho )-\mathrm{tr}(\rho \ln \tau )=-S(\rho )+\ln d$, using $\mathrm{tr}(\rho )=1$ by [F4]. By [F3], this is nonnegative, so $S(\rho )\le \ln d$.

Equality $S(\rho )=\ln d$ is equivalent by step 2.1 to $D(\rho \mathrm{\parallel }\tau )=0$. The equality case of [F3] then gives $\rho =\tau =I_H\mathrm{/}d$. Conversely, substituting $\rho =I_H\mathrm{/}d$ into [F1] gives $S(\rho )=\ln d$.

The trace norm is nonnegative and, by [F6], $\mathrm{\parallel }\rho -\sigma {\mathrm{\parallel }}_1\le \mathrm{\parallel }\rho {\mathrm{\parallel }}_1+\mathrm{\parallel }\sigma {\mathrm{\parallel }}_1=2$; hence $t\in [0,1]$. If $t=0$, equality of the states settles the claim. Otherwise [F11] says the positive and negative parts of $\rho -\sigma$ have trace $t$. Normalize them to states ${\omega }_{+},{\omega }_{-}$. Then $\theta =\frac{\rho +t{\omega }_{-}}{1+t}=\frac{\sigma +t{\omega }_{+}}{1+t}\mathrm{.}$ Apply the lower mixture-entropy bound to the first representation and upper bound to the second. After multiplication by $1+t$ this gives $S(\rho )-S(\sigma )\le t[S({\omega }_{+})-S({\omega }_{-})]+(1+t)h(t\mathrm{/}(1+t))\le t\ln d+g(t)\mathrm{.}$ Here $S({\omega }_{+})\le \ln d$ by [F2], while $S({\omega }_{-})\ge 0$ by [F12]. Interchanging the states gives the absolute-value bound. Dimension one causes no exception. For $t>0$, [F7] gives $g^{\mathrm{\prime }}(t)=\ln (1+1\mathrm{/}t)>0$ and $g^{\mathrm{\prime }\mathrm{\prime }}(t)=-1\mathrm{/}[t(1+t)]<0$. Moreover $t\ln t\to 0$ as $t\downarrow 0$ by [F9], so $g$ is continuous on $[0,1]$ with the stated endpoint value. Jensen for $g$ follows from [F8] applied to the convex function $-g$, including the endpoint.