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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.

Assume no previously shared entanglement. Alice's encoded ensemble $\{p_x,{\rho }_x\}$ lives in a Hilbert space of dimension $2^n$. Its Holevo quantity satisfiesusing nonnegativity of Von Neumann entropy and the maximum entropy of a quantum state. The data-processing inequality for quantum relative entropy implies ${\chi }_{\mathrm{out}}\le {\chi }_{\mathrm{in}}$ for the transmission quantum channel. Finally, the Holevo bound applied to Bob's quantum measurement givesThis argument also works when the output system has larger dimension than the input. Prior shared entanglement changes the communication resource: superdense coding can transmit two classical bits per sent qubit, so the absence of that resource is part of this bound's setting.

Write a qubit density operator in Bloch vector form, $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, with $|\mathbf{r}|\le 1$. The quantum depolarizing channel maps $\mathbf{r}$ to $p\mathbf{r}$, so its output eigenvalues are $(1\pm p|\mathbf{r}|)/2$. The output Von Neumann entropy is minimized on pure inputs, where it equals $h_2((1+p)/2)$.

For any input ensemble, its output Holevo quantity is at most the maximum qubit entropy minus this minimum output entropy:Equality is attained by two equiprobable orthogonal pure inputs: their average output is $I/2$, while each output has the minimum entropy. ThusThis uses the question's retention parameter $p$; in the usual convex-mixture range, $0\le p\le 1$. The expression also holds throughout the full completely positive qubit range $-1/3\le p\le 1$, since $h_2((1+p)/2)=h_2((1+|p|)/2)$.

The supplied additivity of Holevo capacity gives ${\chi }^{\ast }({\Lambda }_{\mathrm{dep}}^{\otimes n})=n{\chi }^{\ast }({\Lambda }_{\mathrm{dep}})$. The Holevo-Schumacher-Westmoreland theorem expresses the unassisted classical capacity of a quantum channel as the regularized Holevo capacity, soTherefore entangled inputs across channel uses cannot increase this classical capacity.

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.

The quantity ${\chi }^{\ast }(\Lambda )$ is the one-use Holevo information of the channel: by the Holevo--Schumacher--Westmoreland theorem, its regularization gives the classical capacity, and without entangled inputs across uses it is the asymptotically achievable classical rate using collective output measurements.

If an ensemble contains a mixed ${\rho }_x={\sum }_j{q}_{j|x}|{\psi }_{xj}⟩⟨{\psi }_{xj}|$, refine its label to $(x,j)$ with probability $p_x{q}_{j|x}$. The average channel output is unchanged, while Concavity of Von Neumann entropy givesThus refinement cannot decrease the Holevo quantity, so the maximum may be restricted to pure input states.

For $\Phi (\rho )=p\rho +(1-p)I/2$, every output qubit has Von Neumann entropy at least $h((1-p)/2)$. A product input therefore has output entropy at least $nh((1-p)/2)$. The average output has entropy at most $n$, so its Holevo quantity is bounded by the displayed expression. This allows correlated classical message labels and joint output measurements, but requires product quantum inputs. The Holevo bound and Fano's inequality turn it into the corresponding asymptotic rate bound.

For a code with product state inputs to $n$ uses of a memoryless quantum channel, Subadditivity of Von Neumann entropy bounds its output Holevo quantity by $n{\chi }^{\ast }(\Lambda )$. The Holevo bound and Fano's inequality then give, for asymptotic rate $R$, average and hence maximum error at least $1-{\chi }^{\ast }(\Lambda )/R$ in the limit inferior. In particular, no code above that rate can have vanishing maximum error, even with a collective decoder measurement.

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.

Take independent ensembles for two quantum channels and their Cartesian product ensemble. The Von Neumann entropy of a tensor product is the sum of the two entropies, both for its average output and for each labelled output. Thus its Holevo quantity is the sum of the two individual quantities. Taking ensembles arbitrarily close to each supremum proves the displayed inequality without requiring maximizing ensembles to exist. It does not assert additivity for arbitrary channels.