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A local quantum channel on either subsystem cannot increase quantum mutual information. Indeed, apply the data-processing inequality for quantum relative entropy to $D({\rho }_{AB}\Vert {\rho }_A\otimes {\rho }_B)$; the channel maps the product of marginals to the product of the new marginals.

The classical mutual information obtained by any measurement of a quantum ensemble is at most its Holevo quantity. This follows from the data-processing inequality for quantum relative entropy applied to the measurement channel.

For $k$ equiprobable quantum code states with average decoding error $ϵ<1$, data-processing inequality for quantum relative entropy applied to the binary test for quantum decoding success givesfor $k\ge 2$. Here $h_2$ is binary entropy. The case $k=1$ is trivial. Taking the supremum over allowed input distributions yields a code-size bound in terms of the largest Holevo quantity; at $ϵ=1$ the undivided inequality is vacuous.

For the quantum state ensemble $\mathcal{E}=\{p_i,{\rho }_i\}$ with average $\overline{\rho }={\sum }_i{p}_i{\rho }_i$, the Holevo quantity isThe second equality follows by expanding the trace definition: ${\sum }_i{p}_i\mathrm{Tr}{\rho }_i\log \overline{\rho }=\mathrm{Tr}\overline{\rho }\log \overline{\rho }$. All positive-weight states are supported within $\overline{\rho }$, so the expression is finite.

For a CPTP map $\Lambda$, the output average is $\Lambda (\overline{\rho })$, with unchanged probabilities $p_i$. The data-processing inequality for quantum relative entropy givesThus the Holevo quantity cannot increase under a quantum channel.

The underlying monotonicity can also be seen directly from Strong subadditivity of Von Neumann entropy. Attach the classical label $X$, and realize $\Lambda$ by a Stinespring representation of a completely positive map with output $B$ and discarded environment $E$. The isometry preserves $I(X:BE)$, whose input value is $\chi (\mathcal{E})$. The difference from the output value iswhich is precisely strong subadditivity. This explains why discarding an environment loses accessible ensemble information.

Here “operation” means the unconditional channel, including the classical outcome when a measurement is retained. Conditioning on a selected outcome can increase the Holevo quantity. For instance, take equally likely vectors $\sqrt{a}|0⟩\pm \sqrt{1-a}|1⟩$, with $1/2<a<1$. Their input Holevo quantity is $h_2(a)<1$. The filter $K=\mathrm{diag}(\sqrt{(1-a)/a},1)$ succeeds on either state with probability $2(1-a)$ and produces the orthogonal states $|+⟩,|-⟩$ after normalization. The success-conditioned quantity is $1$. This does not violate the unconditional statement.

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.

For any tripartite density operator, Strong subadditivity of Von Neumann entropy isUse quantum relative entropy $D(\rho \Vert \sigma )=\mathrm{Tr}\rho ({\log }_2\rho -{\log }_2\sigma )$, with the usual support condition. The equivalent relative-entropy comparison isIndeed the two sides expand respectively as $S(A)+S(BC)-S(ABC)$ and $S(A)+S(B)-S(AB)$. Subtracting cancels $S(A)$ and leaves exactly the strong-subadditivity gap. Marginal-product supports contain the support of the joint state, so these expressions are finite; singular marginals can also be handled by full-rank regularization and a limit.

Finally, tracing out $C$ sends the numerator and denominator of the first relative entropy to those of the second. The data-processing inequality for quantum relative entropy therefore proves the comparison. This is strong subadditivity from relative-entropy monotonicity, rather than an assumption that classical entropy proofs automatically apply to quantum states.

For the distribution and POVM from part 2, apply the data-processing inequality for quantum relative entropy to the measurement channel. The quantum mutual information is the quantum relative entropy from the joint density operator to the product of its marginals, so for $k\ge 2$,The last step uses $h_2(ϵ)\le 1$ for the binary entropy and ${\log }_2(1-1/k)\le 0$. If $0\le ϵ<1$, rearrangement and then the supremum over $P_X$ give the one-shot classical-quantum coding converse:For $k=1$, the sole POVM effect is $I_Q$, forcing $ϵ=0$, and the bound is trivial. For $ϵ=1$, division by $1-ϵ$ is undefined: the undivided inequality remains valid and gives no size restriction. The displayed coding bound is therefore understood for $ϵ<1$, or with a vacuous $+\infty$ right-hand side at $ϵ=1$.

An alternative uses the original decoder directly. Let $M$ be the uniform message and $\hat{M}$ its measured estimate. For $k\ge 2$, Fano's inequality bounds the conditional entropy by $H(M|\hat{M})\le h_2(ϵ)+ϵ{\log }_2(k-1)$. The Holevo bound givesThus the same coding bound follows by bounding the classical mutual information obtainable from the POVM.

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.

Choose the trace-distance-preserving binary measurement from the previous part. Let $p,q$ be its output probability distributions, so $\Vert p-q{\Vert }_1=2D(\rho ,\sigma )$. The data-processing inequality for quantum relative entropy givesFor diagonal density operators, the quantum expression is the classical Kullback-Leibler divergence. Applying the supplied Pinsker's inequality givesConsequently the quantum Pinsker inequality isThe $\ln 2$ factor converts natural-logarithm relative entropy to bits. If the relative entropy is infinite because its support condition fails, the inequality holds automatically.

Write $q_j=⟨z_j|\rho |z_j⟩$, $r_j=⟨z_j|Y(\rho )|z_j⟩$, and $c={\max }_{i,j}|⟨y_i|z_j⟩{|}^2$. Thenand thereforeThe data-processing inequality for quantum relative entropy applied to $Z$ and part c giveCombining them proves

For an anti-degradable quantum channel, a channel from $E$ simulates $Q$. The data-processing inequality for quantum relative entropy applied to quantum mutual information gives $I(R:Q)\le I(R:E)$. Part d then implies $I_c(\Lambda ,\rho )\le 0$ for every input. Part c supplies a pure input attaining zero, so

Extend the quantum conditional entropy from normalized states to positive operators bySince $\mathrm{Tr}{X}_{AB}=\mathrm{Tr}{X}_B$, the two terms involving $\log t$ cancel under $X\mapsto tX$, so $F(tX)=tF(X)$. Thus $F$ is positively homogeneous, and part (b) extends its concavity from states to the positive cone.

Apply part (c) with $X={\sigma }_{AB}$ and $Y={\rho }_{AB}$. Differentiating the matrix logarithm under the trace givesThe inequality from part (c), after moving $F(\rho )$ to the left, becomesThis is the data-processing inequality for quantum relative entropy under partial trace. Tensoring each output with the appropriate maximally mixed state does not change either side, so it also proves data processing under normalized partial traces. Singular $\sigma$ follows by approximation on its support.

Apply the assumed data-processing inequality for quantum relative entropy to the normalized partial trace over $C$, with the two input statesThe channel sends them to ${\rho }_{AB}\otimes I_C/d_C$ and $I_A/d_A\otimes {\rho }_B\otimes I_C/d_C$. Additivity over the common maximally mixed factor reduces data processing toExpanding the Umegaki relative entropy in terms of Von Neumann entropy givesAfter cancelling $\log d_A$, this is precisely the Strong subadditivity of Von Neumann entropy

The Umegaki relative entropy iswhen the support of $\rho$ is contained in that of $\sigma$, and $+\infty$ otherwise. Its additivity of quantum relative entropy isits superadditivity of quantum relative entropy isand its data-processing inequality for quantum relative entropy is $D(T(\rho )\Vert T(\sigma ))\le D(\rho \Vert \sigma )$ for every quantum channel $T$.

Additivity follows from the logarithm of a tensor product,and the analogous identity for ${\sigma }_A\otimes {\sigma }_B$. For superadditivity, subtract the two marginal relative entropies from the joint one. The reference-state terms cancel, leavingThis is the nonnegativity of quantum mutual information, equivalently Subadditivity of Von Neumann entropy.

In bits, the quantum Pinsker inequality isThe trace-distance-preserving binary measurement converts the trace distance into classical total variation distance without loss. Apply the data-processing inequality for quantum relative entropy to that measurement, then classical Pinsker's inequality. The factor $\ln 2$ is omitted when relative entropy uses natural logarithms.

For common classical weights, the matrix logarithm on each positive-weight block is $\log (p_i{\rho }_i)=(\log p_i)I+\log {\rho }_i$ on its support. The two logarithms of $p_i$ cancel in the quantum relative entropy, giving the displayed identity. Zero-weight blocks are omitted, including if their unweighted relative entropy is infinite. If a positive-weight block violates support inclusion, both sides are infinite. Applying the data-processing inequality for quantum relative entropy to the partial trace over the flag gives joint convexity of quantum relative entropy.

Apply data-processing inequality for quantum relative entropy to ${\rho }_{ABC}$ and ${\rho }_A\otimes {\rho }_{BC}$, tracing out $C$. The resulting comparison is $D({\rho }_{ABC}\Vert {\rho }_A\otimes {\rho }_{BC})\ge D({\rho }_{AB}\Vert {\rho }_A\otimes {\rho }_B)$. Expand the logarithms of the product marginals to obtain $S(AB)+S(BC)-S(B)-S(ABC)\ge 0$, exactly Strong subadditivity of Von Neumann entropy.