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For a positive-definite Hermitian operator $A={\sum }_j{t}_j|u_j⟩⟨u_j|$ and a unit vector $v$, the weights $|⟨v|u_j⟩{|}^2$ form a probability distribution. Scalar logarithmic concavity bounds their average of $\log t_j$ by the logarithm of their average of $t_j$. Positive semidefinite operators follow by a limiting convention. Using the eigenbasis of a density operator $\rho$ gives $S(\rho \Vert \sigma )\ge D(r\Vert q)$, where $r$ is the spectrum of $\rho$ and $q$ the diagonal of $\sigma$ in that basis. This yields nonnegativity of quantum relative entropy without commutativity.

For positive-definite Hermitian matrices $A,B$, Klein's inequality giveswith equality exactly when $A=B$. To prove it, let $a_i,b_j$ be their eigenvalues and $u_i,v_j$ their orthonormal eigenvectors. The weights $w_{ij}=|⟨u_i,v_j⟩{|}^2$ have row and column sums one. The difference between the two sides isby the scalar logarithm inequality $\ln t\le t-1$. Equality requires $a_i=b_j$ whenever $w_{ij}>0$, implying $A=B$. Limits extend the result to positive semidefinite matrices with the appropriate support condition. Applying it to trace-one matrices proves nonnegativity of quantum relative entropy.

On a $d$-dimensional Hilbert space, the Von Neumann entropy is maximized uniquely by $I/d$. Indeed, nonnegativity of quantum relative entropy giveswith equality exactly when $\rho =I/d$.

True: quantum entropy is subadditive. Write $\sigma ={\rho }_1\otimes {\rho }_2$. Positivity ensures the support of $\rho$ lies in that of $\sigma$. Indeed, if $|v⟩\in \ker {\rho }_1$, then ${\sum }_j⟨v,j|\rho |v,j⟩=0$; all summands are nonnegative, so ${\rho }^{1/2}|v,j⟩=0$ for every $j$. Thus $\rho$ annihilates $\ker {\rho }_1\otimes {\mathcal{K}}_2$, and similarly the kernel from the other factor.

On the resulting support, ${\log }_2({\rho }_1\otimes {\rho }_2)={\log }_2{\rho }_1\otimes I+I\otimes {\log }_2{\rho }_2$. Using the defining property of the partial trace and nonnegativity of quantum relative entropy proved in part (i),HenceEquality holds exactly for the product state $\rho ={\rho }_1\otimes {\rho }_2$. This proves Subadditivity of Von Neumann entropy directly.

The concavity inequality is true, but its printed equality condition is false. At $b=1$ equality always holds, even for distinct states; for example take ${\rho }_1=|0⟩⟨0|$ and ${\rho }_2=|1⟩⟨1|$. The correct statement on $0\le b\le 1$ isFor $0<b<1$, entropy is therefore strictly concave.

We give the matrix inequality needed for the proof. Define the quantum relative entropy $D(\tau \Vert \sigma )=\mathrm{tr}\tau ({\log }_2\tau -{\log }_2\sigma )$ if the support of $\tau$ is contained in that of $\sigma$, and $+\infty$ otherwise. Diagonalize the two states with eigenvalues $r_i,s_j$, and put $q_{ij}=|⟨i|j⟩{|}^2$. Each row and column of $q$ sums to one. The elementary scalar inequalityfollows from $x\ln x-x+1\ge 0$, with equality exactly at $x=1$; use limits at zero. ConsequentlyEquality forces $r_i=s_j$ whenever $q_{ij}>0$. Hence every eigenvector $|i⟩$ of $\tau$ is also a $\sigma$-eigenvector with eigenvalue $r_i$, so $\tau =\sigma$. Conversely identical states give zero relative entropy. This proves nonnegativity of quantum relative entropy and its equality condition, including states with zero eigenvalues under the support convention.

Now set $\rho =b{\rho }_1+(1-b){\rho }_2$. For $0<b<1$ its support contains both component supports. Direct expansion givesSince both coefficients are positive, equality holds exactly when both component states equal $\rho$, equivalently ${\rho }_1={\rho }_2$. At either endpoint the mixture has only one component and equality is automatic. This proves the corrected strict concavity of Von Neumann entropy.

For a density matrix $\rho$ on a $d$-dimensional Hilbert space, its Von Neumann entropy, in bits, iswhere the ${\lambda }_a$ are its eigenvalues and $0{\log }_{2}0=0$. Each summand is nonnegative, proving $S\ge 0$, with equality exactly when one eigenvalue is one and the rest zero, namely for a pure state.

We first prove the nonnegativity of quantum relative entropy needed below. Diagonalize $\rho$ with eigenvalues ${\lambda }_a$ and eigenvectors $u_a$, and $\sigma$ with eigenvalues ${\mu }_b$ and eigenvectors $v_b$. Put $B_{ab}=|⟨u_a,v_b⟩{|}^2$ and $q_a=⟨u_a|\sigma |u_a⟩={\sum }_b{B}_{ab}{\mu }_b$. Each row of $B$ sums to one. Concavity of the ordinary logarithm gives ${\sum }_b{B}_{ab}{\log }_2{\mu }_b\le {\log }_2{q}_a$. ConsequentlyThe final Gibbs inequality follows directly from ${\sum }_{{\lambda }_a>0}{\lambda }_a\log (q_a/{\lambda }_a)\le \log {\sum }_{{\lambda }_a>0}{q}_a\le 0$. Support failure gives $D=+\infty$; zero eigenvalues can otherwise be handled by restriction or regularization. Equality requires $q_a={\lambda }_a$ and equality in the logarithm inequality on every positive-weight row. Thus $u_a$ is a $\sigma$ eigenvector of eigenvalue ${\lambda }_a$ wherever ${\lambda }_a>0$. The remaining diagonal entries of $\sigma$ vanish and positivity makes the remaining rows and columns zero. Hence equality holds exactly when $\rho =\sigma$.

Taking $\sigma =I/d$ gives $D(\rho \Vert I/d)={\log }_{2}d-S(\rho )$, provingThe lower equality is attained exactly by pure states and the upper exactly by the maximally mixed state $I/d$.

For $\bar{\rho }={\sum }_i{p}_i{\rho }_i$, each positive-weight state has support contained in that of $\bar{\rho }$. Expanding the quantum relative entropy givesThis proves Concavity of Von Neumann entropy, with equality precisely when every positive-weight ${\rho }_i$ is the same state.

For the upper entropy bound for a quantum mixture, diagonalize each component, ${\rho }_i={\sum }_a{\lambda }_{ia}|v_{ia}⟩⟨v_{ia}|$, and set $q_{ia}=p_i{\lambda }_{ia}$. Form the rectangular matrix $B$ whose columns are $\sqrt{q_{ia}}{v}_{ia}$. Then $B{B}^{†}=\bar{\rho }$, while the Gram matrix $\Gamma =B^{†}B$ has diagonal $q_{ia}$. The two matrices have the same nonzero eigenvalues: $B$ and $B^{†}$ map corresponding positive-eigenvalue eigenspaces isomorphically. Thus $S(\Gamma )=S(\bar{\rho })$.

If there are $L$ columns, let $W=\mathrm{diag}(1,\omega ,\dots ,{\omega }^{L-1})$, with $\omega =e^{2\pi i/L}$. ThenAll the summands have the same Von Neumann entropy. The just-proved concavity therefore implies $S(\bar{\rho })=S(\Gamma )\le S(\mathrm{diag}\Gamma )=H(q)$. Finally,We have proved both requested bounds:For the upper equality, strictness in the concavity argument forces $\Gamma$ to be diagonal. Its entries between two positive-weight component eigenvectors then force those vectors to be orthogonal. Thus upper equality holds exactly when the supports of the positive-weight component states are mutually orthogonal. In that case their classical label is perfectly distinguishable, explaining the extra $H(p)$ bits.

Let a density operator on a Hilbert space ${\mathcal{H}}_S$ have the spectral decompositionIntroduce an auxiliary Hilbert space ${\mathcal{H}}_R$ with an orthonormal basis containing $r$ vectors $|j{⟩}_R$, and setThis vector has norm one. Its partial trace isThis explicitly constructs a purification of a density operator. The same construction works for an arbitrary trace-class density operator: its nonzero eigenvalues form a finite or countable summable family, and the displayed vector converges in the Hilbert space norm. The entropy inequality below is understood with finite Von Neumann entropies, avoiding undefined differences of infinities.

Apply this construction with $S=AB$ to obtain a joint pure state on $ABC$. The Schmidt decomposition across $AB:C$, $A:BC$ and $B:AC$ respectively givesNow apply Subadditivity of Von Neumann entropy to the reduced density matrices on $BC$ and $AC$:These inequalities say $S(A)-S(B)\le S(C)$ and $S(B)-S(A)\le S(C)$. Replacing $S(C)$ by $S(AB)$ proves the Araki–Lieb inequality:The subadditivity used here concerns the actual reduced density operators, rather than statistical independence. In particular it follows from nonnegativity of quantum relative entropy, sinceand similarly for $AC$. Thus no independence assumption on the purification of a density operator is needed.

Use base-two logarithms, so Von Neumann entropy is measured in bits, and set $0\log 0=0$. Write $D(\rho \Vert \sigma )=\mathrm{Tr}\rho ({\log }_2\rho -{\log }_2\sigma )$ for the quantum relative entropy, the quantity denoted $S(\rho \Vert \sigma )$ in the paper. It is $+\infty$ unless the support of a positive operator $\rho$ lies within that of $\sigma$.

We first establish nonnegativity of quantum relative entropy, including its equality condition. Let $\rho ={\sum }_{i:r_i>0}{r}_i|u_i⟩⟨u_i|$ and $\sigma ={\sum }_j{s}_j|v_j⟩⟨v_j|$, and suppose the support condition holds. Put $q_i=⟨u_i|\sigma |u_i⟩>0$. Concavity of the scalar logarithm givesTerms with $s_j=0$ have zero weight here. Working first with natural logarithms, the inequality $-\ln x\ge 1-x$ yieldssince the $q_i$ sum to at most $\mathrm{Tr}\sigma =1$. Equality forces $q_i=r_i$ for every positive-weight eigenvector, no remaining weight of $\sigma$ outside their span, and equality in the logarithm's strict concavity. The latter makes each $u_i$ an eigenvector of $\sigma$ with eigenvalue $q_i$. Thus $D(\rho \Vert \sigma )=0$ exactly when $\rho =\sigma$.

For the bipartite density operator,Indeed, a projector onto the kernel of ${\rho }_A$, tensored with $I_B$, has zero expectation in ${\rho }_{AB}$. Positivity implies that ${\rho }_{AB}$ annihilates its range: $\mathrm{Tr}({\rho }_{AB}P)=\Vert {\rho }_{AB}^{1/2}P{\Vert }_{\mathrm{HS}}^2=0$. The same argument applies to $B$. Hence the quantum relative entropy below is finite. On this support,Using the partial trace to evaluate the two local terms givesThis proves Subadditivity of Von Neumann entropy:The proof also identifies equality exactly with the product state ${\rho }_{AB}={\rho }_A\otimes {\rho }_B$.

Fix paired orthonormal bases in the two $d$-dimensional Hilbert spaces. A maximally entangled state isIt is normalized, and the partial traces of its rank-one density operator are ${\rho }_A={\rho }_B=I/d$. Thus all its Schmidt coefficients are $1/\sqrt{d}$ and its entanglement entropy is ${\log }_{2}d$. No pure bipartite state of these dimensions can have larger entanglement entropy: nonnegativity of quantum relative entropy applied to a reduced state and $I/d$ gives $0\le D({\rho }_A\Vert I/d)={\log }_{2}d-S({\rho }_A)$. Hence it is maximally entangled.

The unread measurement is the pinching map $\mathcal{P}(\rho )={\rho }^{\prime }={\sum }_i{P}_i\rho P_i$, a nonselective projective measurement. Its output is block diagonal and commutes with each $P_i$, so $\log {\rho }^{\prime }$ also commutes with the projectors on its support.

First the quantum relative entropy $D(\rho \Vert {\rho }^{\prime })$ is finite. If $v$ lies in the kernel of ${\rho }^{\prime }$, positivity givesEvery summand is zero; summing ${\rho }^{1/2}{P}_{i}v=0$ gives ${\rho }^{1/2}v=0$. Thus $\ker {\rho }^{\prime }\subseteq \ker \rho$, equivalently $\mathrm{supp}\rho \subseteq \mathrm{supp}{\rho }^{\prime }$.

The block-diagonal logarithm obeys ${\sum }_i{P}_i(\log {\rho }^{\prime })P_i=\log {\rho }^{\prime }$ on this support. Cyclicity of the trace therefore givesConsequently the relative entropy of a pinched state is exactly the entropy increase:By the proved nonnegativity of quantum relative entropy and its equality condition,Equivalently, equality holds exactly when all off-diagonal blocks $P_i\rho P_j$ for $i\ne j$ already vanish, or when $\rho$ commutes with every measurement projector. This argument proves the equality condition even when the projectors are degenerate and the states singular.

Use base-two logarithms, so entanglement entropy is measured in bits. Write $h_2(p)=-p{\log }_{2}p-(1-p){\log }_2(1-p)$ for the binary entropy, with $0{\log }_{2}0=0$. Taking a partial trace over any two parties removes the off-diagonal terms: the two traced strings are orthogonal. Consequently all three one-party reduced density matrices of the first pure state areEach division into two groups is one party versus the other two. The Schmidt coefficients are $\sqrt{p}$ and $\sqrt{1-p}$, so the entanglement entropy across each of $A:BC$, $B:AC$ and $C:AB$ is $h_2(p)$. For the second pure state, set $p=1/2$: every one-party reduced density matrix is $I/2$, and all three entanglement entropies are one bit.

For every pair $XY$, the first pure state instead gives the reduced density matrixThis is a separable quantum state, explicitly a convex combination of two product states. Its relative entropy of entanglement is therefore zero: in $E_R(\rho )={\inf }_{\sigma \in \mathrm{Sep}}D(\rho \Vert \sigma )$, choose $\sigma =\rho$ to attain zero, while nonnegativity of quantum relative entropy excludes a negative value. The same argument with equal weights applies to the second pure state. ThusThe pairwise density matrices contain classical correlations, despite the entanglement of the global pure state; discarding a party loses the coherence between its two branches.

The entanglement entropy of $n$ independent source copies across any cut is $n{h}_2(p)$, whereas $f(n)$ copies of the target GHZ state have entanglement entropy $f(n)$. Average monotonicity of pure-state entanglement entropy under LOCC gives $P(n)f(n)\le n{h}_2(p)$ for exact successful forward branches. In a reversible asymptotic conversion the reverse direction similarly prevents a first-order deficit, since $n^{\prime }/n\to 1$. For asymptotically faithful outputs, a continuity bound for quantum conditional entropy with trivial conditioning changes these inequalities only by $o(n)$. Hence the reversible leading-order rate isThe zero pairwise relative entropies of entanglement are consistent with this conversion: no independent pairwise resource has to be supplied or removed.

There is a qualification to the literal finite-block formulation in the source. For $0<p<1$, the Schmidt rank of the exact state of $n^{\prime }$ source copies is $2^{n^{\prime }}$, whereas $f$ target GHZ states have Schmidt rank $2^f$. Every refined LOCC branch acts by a product of local linear operators and cannot increase this Schmidt rank. For $p\ne 1/2$, the displayed rate has $f<n^{\prime }$ for sufficiently large blocks when $n^{\prime }/n\to 1$, so an exact successful reverse conversion is impossible, irrespective of its claimed probability. The exact dilution obstruction from Schmidt rank is avoided in the usual asymptotic formulation by allowing outputs approaching the desired pure state. The leading-order rate above and the exact forward protocol below remain valid; neither establishes the source's stronger exact round-trip claim. At $p=1/2$ the two states are identical. At $p=0$ or $1$, the source is a known product state, the rate is zero, and local preparation supplies any desired number of source copies.

For probability distributions $p$ and $q$ on a finite alphabet, their classical relative entropy in bits isA zero-$p$ term contributes zero, including when $q(x)=0$; a term with $p(x)>0$ and $q(x)=0$ makes the result $+\infty$. In that infinite case nonnegativity is immediate. Otherwise $q(x)>0$ on the support $J_p$ of $p$. For every $u>0$, $\ln u\le u-1$: the function $u-1-\ln u$ has derivative $1-1/u$ and its minimum zero at $u=1$. Apply this with $u=q(x)/p(x)$ to obtainThe last inequality uses normalization of $q$. Thus $\boxed{D(p\Vert q)\ge 0}$. Equality requires $q(x)=p(x)$ wherever $p(x)>0$ and no $q$-mass outside that support, hence occurs precisely when $p=q$.

For density matrices, usewhen the support of a positive operator $\rho$ is contained in that of $\sigma$, and set it to $+\infty$ otherwise. The data-processing inequality for quantum relative entropy states that for any quantum channel $\Phi$,Thus the relative entropy of the two output states cannot exceed that of the inputs.

For the states with a classical flag, omit indices of weight zero. On a positive-weight block, the matrix logarithms obeyon the relevant supports. The logarithms of $p_i$ cancel. Taking the trace block by block therefore gives the relative entropy of classically flagged states:If support inclusion fails in a positive-weight block, both sides are infinite. A zero-weight block makes no contribution, even if its unweighted relative entropy is infinite.

Tracing out the flag is a quantum channel with operators $A_i=⟨i|\otimes I$, since ${\sum }_i{A}_i^{†}{A}_i=I$. It sends the two flagged states to ${\sum }_i{p}_i{\rho }_i$ and ${\sum }_i{p}_i{\sigma }_i$. Applying the data-processing inequality for quantum relative entropy to this partial trace proves joint convexity of quantum relative entropy:

For a bipartite state, set ${\rho }_A={\mathrm{Tr}}_B{\rho }_{AB}$ and ${\rho }_B={\mathrm{Tr}}_A{\rho }_{AB}$. Its quantum mutual information iswhere $S(\tau )=-\mathrm{Tr}\tau {\log }_2\tau$ is the Von Neumann entropy. The support of ${\rho }_{AB}$ lies in $\mathrm{supp}{\rho }_A\otimes \mathrm{supp}{\rho }_B$. To justify this when marginals are singular, let $Q_A$ project onto $\ker {\rho }_A$. Then $\mathrm{Tr}[(Q_A\otimes I){\rho }_{AB}]=\mathrm{Tr}(Q_A{\rho }_A)=0$; positivity implies $(Q_A\otimes I){\rho }_{AB}=0$. Applying the same argument on $B$ proves the claimed support inclusion.

On that support, product eigenvectors showThe definition of the partial trace therefore givesConsequently $\boxed{S(A:B)=S({\rho }_{AB}\Vert {\rho }_A\otimes {\rho }_B)}$.

For completeness, nonnegativity of quantum relative entropy also follows from the classical proof above. Diagonalize $\rho$ in a basis $|j⟩$, with eigenvalues $r_j$, and put $q_j=⟨j|\sigma |j⟩$. These are two classical probability distributions. Scalar concavity of the logarithm applied to the spectral weights of $\sigma$ gives $⟨j|{\log }_2\sigma |j⟩\le {\log }_2{q}_j$ whenever $r_j>0$ and support inclusion holds. HenceThe support-failure case is again infinite. Applying this to the product of the marginals proves $\boxed{S(A:B)\ge 0}$ without requiring an additional entropy inequality.

Using base-two logarithms, define the quantum conditional entropy byThe Subadditivity of Von Neumann entropy, $S(AB)\le S(A)+S(B)$, immediately gives $S(A\mid B)\le S(A)$.

For the lower bound, purify ${\rho }_{AB}$ by an auxiliary system $C$. Complementary subsystems of a pure state have the same nonzero reduced-state eigenvalues, by the Schmidt decomposition. Hence $S(BC)=S(A)$ and $S(C)=S(AB)$. Applying Subadditivity of Von Neumann entropy to $BC$ givesRearranging yields $S(A\mid B)\ge -S(A)$. This is the relevant side of the Araki–Lieb inequality, here derived using purification.

Finally the maximum entropy of a quantum state gives $S(A)\le {\log }_2\dim {\mathcal{H}}_A$. For example, this follows directly fromby nonnegativity of quantum relative entropy. Combining the two bounds provesBoth signs can be attained: a maximally mixed $A$ independent of $B$ has the positive extreme, while a maximally entangled state with a sufficiently large $B$ has the negative extreme.

Let $\mathcal{P}$ be the pinching map determined by the projectors, and put $\sigma =\mathcal{P}(\rho )$. This is a nonselective projective measurement; its projectors are its Kraus operators. We will prove the exact relative entropy of a pinched state identity.

First the support of $\rho$ is contained in that of $\sigma$. Indeed, if $v$ lies in the kernel of $\sigma$, positivity givesEvery summand is zero. Since ${\sum }_i{P}_i=I$, also ${\rho }^{1/2}v=0$, and thus $v$ is in the kernel of $\rho$. Therefore the logarithm of $\sigma$ may be used on its support without an infinite relative-entropy term.

The operator $\sigma$ is block diagonal, so each $P_i$ commutes with $\log \sigma$ on that support. Cyclicity of the trace givesExpand the quantum relative entropy:Its nonnegativity now proves the entropy increase under nonselective projective measurement:The equality case in nonnegativity of quantum relative entropy says equality holds exactly when $\rho =\sigma$. Equivalently,Thus equality means the state already has no coherence between distinct measurement blocks. It need not be diagonal inside a block of rank greater than one.

The quantum relative entropy is $S(\rho \Vert \sigma )=\mathrm{Tr}\rho ({\log }_2\rho -{\log }_2\sigma )$ when the support of a positive operator $\rho$ is contained in that of $\sigma$, and $+\infty$ otherwise. Write spectral decompositions $\rho ={\sum }_i{r}_i|i⟩⟨i|$ and $\sigma ={\sum }_j{t}_j|u_j⟩⟨u_j|$, and set $q_i=⟨i|\sigma |i⟩$.

The weights $w_{ij}=|⟨i|u_j⟩{|}^2$ sum to one. Concavity of the scalar logarithm gives the diagonal logarithm concavity boundFor $r_i>0$, support inclusion removes overlaps with zero eigenvalues of $\sigma$; alternatively take a positive regularization and pass to the limit. The two diagonal vectors $r$ and $q$ are probability distributions. ThereforeThis proves nonnegativity of quantum relative entropy without assuming the states commute. Equality in the strict logarithmic concavity condition makes each occupied $|i⟩$ an eigenvector of $\sigma$; classical equality gives $q_i=r_i$ and no weight outside the support, hence equality occurs exactly for $\rho =\sigma$.

Use Klein's inequality, or equivalently nonnegativity of quantum relative entropy. First check the support needed for the logarithm. If $q_y=0$, positivity gives $|{\rho }_{yz}{|}^2\le {\rho }_{yy}{\rho }_{zz}=0$ for every $z$; the corresponding row and column vanish. Thus support inclusion under rank-one dephasing gives $\mathrm{supp}\rho \subseteq \mathrm{supp}\sigma$, and the logarithms may be evaluated on this support.

Since $\log \sigma$ is diagonal in the dephasing basis,The relative-entropy identity for rank-one dephasing follows:Klein's inequality gives $D(\rho \Vert \sigma )\ge (\mathrm{Tr}\rho -\mathrm{Tr}\sigma )/\ln 2=0$. ThereforeEquality holds precisely when $\rho =\sigma$, meaning that the input was already diagonal in the chosen basis. This quantifies why rank-one dephasing removes coherence without reducing the entropy.

Write $X$ and $Y$ for the classical quantum registers $\tilde{X}$ and $\tilde{Y}$. By the quantum mutual information formula $I(U:V)=S(U)+S(V)-S(UV)$, expanding the right-hand side of the quantum mutual information balance identity givesAll Von Neumann entropies here are evaluated in $\omega$.

We use three facts: quantum mutual information is nonnegative by nonnegativity of quantum relative entropy; a local quantum channel cannot increase it by data processing for quantum mutual information; and the quantum mutual information of a classical-quantum state is its ensemble's Holevo quantity. Write $\chi (\mathcal{T})$ for the Holevo capacity, the supremum of the output Holevo quantity over finite input ensembles.

The $XB$ marginal is an output ensemble for $\mathcal{E}$, with inputs $\rho (x{)}_A={\mathrm{Tr}}_C\rho (x{)}_{AC}$. Consequently $I(X:B{)}_{\omega }\le \chi (\mathcal{E})$. Also ${\omega }_{XBD}$ is obtained from ${\sigma }_{XYD}$ by a quantum channel on $XY$ which retains $X$ and prepares $B$ from $Y$. HenceTo bound the latter even for entangled states $\rho (x{)}_{AC}$, exhibit the conditional input ensemble after a local measurement. Setwhen $q(y|x)>0$; zero-weight outcomes can be omitted. The numerator is a positive operator, its trace is $q(y|x)$, and the $q(y|x)$ sum to one. ThusThis is a classical-quantum state with an output ensemble for $\mathcal{N}$, so $I(XY:D{)}_{\sigma }\le \chi (\mathcal{N})$. Combining these bounds with $I(B:D{)}_{\omega }\ge 0$ yieldsThe last step takes the supremum over all input ensembles on $AC$. The left-hand channel direction is $B\leftarrow A$, as established in part (i); the printed $A\leftarrow B$ in the last inequality is a typographical reversal. Independent product ensembles also give the reverse inequality, so this proves Holevo-capacity additivity for entanglement-breaking channels.

By Donald's identity, the objective equalsThe nonnegativity of quantum relative entropy makes the last term at least zero, with equality exactly when $\rho =\overline{\omega }$. Hence the unique minimizing density operator is the ensemble average andFor the classical-quantum state ${\sigma }_{XB}$, its marginal on $X$ has entropy $H(p)$, its marginal on $B$ is $\overline{\omega }$, and its joint entropy is $H(p)+{\sum }_j{p}_{j}S({\omega }_j)$. Substituting in the definition of quantum mutual information givesThis is the relative-entropy barycenter of a quantum ensemble: the average state minimizes mean distinguishability from the ensemble. Its minimal value is exactly the Holevo quantity.

For any tripartite density operator, expanding the three quantum mutual information terms in Von Neumann entropy gives $I(X:BD)=I(X:B)+I(XB:D)-I(B:D)$. This algebraic identity allows separate bounds on two information contributions, while nonnegativity of quantum relative entropy allows the last term to be dropped in an upper bound.

The ensemble average $\overline{\omega }$ uniquely minimizes ${\sum }_j{p}_{j}D({\omega }_j\Vert \rho )$ over density operators $\rho$. Donald's identity separates this objective into a constant plus $D(\overline{\omega }\Vert \rho )$, and nonnegativity of quantum relative entropy proves the claim. The minimum is the Holevo quantity $S(\overline{\omega })-{\sum }_j{p}_{j}S({\omega }_j)$, equivalently the quantum mutual information of the associated classical-quantum state.