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\begin{document}

\lecture{19 --- October 28, 2015}{Fall 2015}{Prof.\ Mark M.\ Wilde}{Mark M.~Wilde}

This document is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.

\section{Overview}



In the previous lecture, we introduced quantum entropy and some other variants of it, such as conditional entropy, mutual information, and conditional mutual information. 


In this lecture, we 
discuss several entropy inequalities that play an important role in quantum
information processing:\ the monotonicity of quantum relative entropy, strong
subadditivity, the quantum data-processing inequalities, and continuity of
quantum entropy.



\section{Quantum Relative Entropy}

The quantum relative entropy is one of the most important entropic quantities
in quantum information theory, mainly because we can reexpress many of the
entropies given in the previous sections in terms of it. This in turn allows
us to establish many properties of these quantities from the properties of
relative entropies. Its definition is a natural extension of that for the
classical relative entropy. Before
defining it, we need the notion of the support of an operator:

\begin{definition}
[Kernel and Support]The kernel of an operator $A\in\mathcal{L}(\mathcal{H}%
,\mathcal{H}^{\prime})$ is defined as%
\begin{equation}
\ker(A)\equiv\{|\psi\rangle\in\mathcal{H}:A|\psi\rangle=0\}.
\end{equation}
The support of $A$ is the subspace of $\mathcal{H}$\ orthogonal to its kernel:%
\begin{equation}
\operatorname{supp}(A)\equiv\{|\psi\rangle\in\mathcal{H}:A|\psi\rangle\neq0\}.
\end{equation}
If $A$ is Hermitian and thus has a spectral decomposition as $A=\sum
_{i:a_{i}\neq0}a_{i}|i\rangle\langle i|$, then $\operatorname{supp}%
(A)=\operatorname{span}\{|i\rangle:a_{i}\neq0\}$. The projection onto the
support of $A$ is denoted by%
\begin{equation}
\Pi_{A}\equiv\sum_{i:a_{i}\neq0}|i\rangle\langle i|.
\end{equation}

\end{definition}

\begin{definition}
\label{def-qie:q-rel-ent}The quantum relative entropy%
\index{quantum relative entropy}
$D(\rho\Vert\sigma)$ between a density operator $\rho\in\mathcal{D}%
(\mathcal{H})$ and a positive semi-definite operator $\sigma\in\mathcal{L}%
(\mathcal{H})$ is defined as follows:%
\begin{equation}
D(\rho\Vert\sigma)\equiv\operatorname{Tr}\left\{  \rho\left[  \log\rho
-\log\sigma\right]  \right\}  , \label{eq-qie:rel-ent}%
\end{equation}
if the following support condition is satisfied%
\begin{equation}
\operatorname{supp}(\rho)\subseteq\operatorname{supp}(\sigma),
\label{eq-qie:supp-cond-rel-ent}%
\end{equation}
and it is defined to be equal to $+\infty$ otherwise.
\end{definition}

This definition is consistent with the classical definition of relative entropy. However, we should note that there could be
be several ways to generalize the classical definition to obtain a quantum
definition of relative entropy. For example, one could take%
\begin{equation}
D^{\prime}(\rho\Vert\sigma)=\operatorname{Tr}\left\{  \rho\log\left(
\rho^{1/2}\sigma^{-1}\rho^{1/2}\right)  \right\}  ,
\end{equation}
as a definition and it reduces to the classical definition  as well. In fact, it is easy to see that
there are an infinite number of quantum generalizations of the classical
definition of relative entropy. So how do we single out which definition is
the right one to use? The definition given in \eqref{eq-qie:rel-ent} is the
answer to a meaningful quantum information-processing task. Furthermore, this
definition generalizes the quantum entropic quantities we have given in this
chapter, which all in turn are the answers to meaningful quantum
information-processing tasks. For these reasons, we take the definition given
in \eqref{eq-qie:rel-ent} as \textit{the} quantum relative entropy. Recall
that it was this same line of reasoning that allowed us to single out the
entropy and the mutual information as meaningful measures of information in
the classical case.

Similar to the classical case, we can intuitively think of the quantum
relative entropy as a distance measure between quantum states. But it is not
strictly a distance measure in the mathematical sense because it is not
symmetric and it does not obey a triangle inequality.

The following proposition justifies why we take the definition of quantum
relative entropy to have the particular support conditions as given above:

\begin{proposition}
Let $\rho\in\mathcal{D}(\mathcal{H})$ and $\sigma\in\mathcal{L}(\mathcal{H})$
be positive semi-definite. The quantum relative entropy is consistent with the
following limit:%
\begin{equation}
D(\rho\Vert\sigma)=\lim_{\varepsilon\searrow0}D(\rho\Vert\sigma+\varepsilon
I).
\end{equation}

\end{proposition}



One of the most fundamental entropy inequalities in quantum information theory
is the monotonicity of quantum relative entropy. When the arguments to the
quantum relative entropy are quantum states, the physical interpretation of
this entropy inequality\ is that states become less distinguishable when noise
acts on them. We defer a proof of this theorem until
later, where we also establish a strengthening of it.

\begin{theorem}
[Monotonicity of Quantum Relative Entropy]\label{thm-qie:mono-rel-ent}Let
$\rho\in\mathcal{D}(\mathcal{H})$, $\sigma\in\mathcal{L}(\mathcal{H})$ be
positive semi-definite, and $\mathcal{N}:\mathcal{L}(\mathcal{H}%
)\rightarrow\mathcal{L}(\mathcal{H}^{\prime})$ be a quantum channel. The
quantum relative entropy%
\index{quantum relative entropy!monotonicity}
can only decrease if we apply the same quantum channel $\mathcal{N}$\ to
$\rho$ and $\sigma$:%
\begin{equation}
D(\rho\Vert\sigma)\geq D(\mathcal{N}(\rho)\Vert\mathcal{N}(\sigma)).
\label{eq-qie:mono-rel-ent}%
\end{equation}

\end{theorem}

Theorem~\ref{thm-qie:mono-rel-ent}\ then implies non-negativity of quantum
relative entropy in certain cases.

\begin{theorem}
[Non-Negativity]\label{thm-qie:non-neg-rel-ent}Let $\rho\in\mathcal{D}%
(\mathcal{H})$, and let $\sigma\in\mathcal{L}(\mathcal{H})$ be positive
semi-definite and such that $\operatorname{Tr}\{\sigma\}\leq1$. The
\index{quantum relative entropy!positivity}%
quantum relative entropy $D(\rho\Vert\sigma)$ is non-negative:%
\begin{equation}
D(\rho\Vert\sigma)\geq0,
\end{equation}
and $D(\rho\Vert\sigma)=0$ if and only if $\rho=\sigma$.
\end{theorem}

\begin{proof}
The first part of the theorem follows from applying
Theorem~\ref{thm-qie:mono-rel-ent}, taking the quantum channel to be the
trace-out map. We then have that%
\begin{align}
D(\rho\Vert\sigma)  &  \geq D(\operatorname{Tr}\{\rho\}\Vert\operatorname{Tr}%
\{\sigma\})\\
&  =\operatorname{Tr}\{\rho\}\log\left(  \frac{\operatorname{Tr}\{\rho
\}}{\operatorname{Tr}\{\sigma\}}\right) \\
&  \geq0.
\end{align}
If $\rho=\sigma$, then the support condition in
\eqref{eq-qie:supp-cond-rel-ent} is satisfied and plugging into
\eqref{eq-qie:rel-ent}\ gives that $D(\rho\Vert\sigma)=0$. Now suppose that
$D(\rho\Vert\sigma)=0$. This means that the inequality above is saturated and
thus $\operatorname{Tr}\{\sigma\}=\operatorname{Tr}\{\rho\}=1$, so that
$\sigma$ is a density operator. Let $\mathcal{M}$\ be an arbitrary measurement
channel. From the monotonicity of quantum relative entropy
(Theorem~\ref{thm-qie:mono-rel-ent}), we can conclude that $D(\mathcal{M}%
(\rho)\Vert\mathcal{M}(\sigma))=0$. The equality condition for the
non-negativity of the classical relative entropy
 in turn implies that $\mathcal{M}%
(\rho)=\mathcal{M}(\sigma)$. Now since this equality holds for any possible
measurement channel, we can conclude that $\rho=\sigma$. (For example, we
could take $\mathcal{M}$ to be the optimal measurement for the trace distance,
which would allow us to conclude that $\max_{\mathcal{M}}\left\Vert
\mathcal{M}(\rho)-\mathcal{M}(\sigma)\right\Vert _{1}=\left\Vert \rho
-\sigma\right\Vert _{1}=0$, and hence $\rho=\sigma$.)
\end{proof}

\subsection{Deriving Other Entropies from Quantum Relative Entropy}

There is a sense in which the quantum relative entropy is a \textquotedblleft
parent quantity\textquotedblright\ for other entropies in quantum information
theory, such as the von Neumann entropy, the conditional quantum entropy, the
quantum mutual information, and the conditional quantum mutual information.
The following exercises explore these relations. The main tool needed to solve
some of them is the non-negativity of quantum relative entropy.

\begin{exercise}
\label{ex-qie:log-operators}Let $P_{A}\in\mathcal{L}(\mathcal{H}_{A})$ and
$Q_{B}\in\mathcal{L}(\mathcal{H}_{B})$ be positive semi-definite operators.
Show that the following identity holds:%
\begin{equation}
\log\left(  P_{A}\otimes Q_{B}\right)  =\log\left(  P_{A}\right)  \otimes
I_{B}+I_{A}\otimes\log\left(  Q_{B}\right)  .
\end{equation}

\end{exercise}

\begin{exercise}
[Mutual Information and Relative Entropy]\label{ex-qie:rel-ent-mut-info}Let
$\rho_{AB}\in\mathcal{D}(\mathcal{H}_{A}\otimes\mathcal{H}_{B})$. Show that
the following identities hold:%
\begin{align}
I(A;B)_{\rho}  &  =D(\rho_{AB}\Vert\rho_{A}\otimes\rho_{B})\\
&  =\min_{\sigma_{B}}D(\rho_{AB}\Vert\rho_{A}\otimes\sigma_{B})\\
&  =\min_{\omega_{A}}D(\rho_{AB}\Vert\omega_{A}\otimes\rho_{B})\\
&  =\min_{\omega_{A},\sigma_{B}}D(\rho_{AB}\Vert\omega_{A}\otimes\sigma_{B}),
\end{align}
where the optimizations are with respect to $\omega_{A}\in\mathcal{D}%
(\mathcal{H}_{A})$ and $\sigma_{B}\in\mathcal{D}(\mathcal{H}_{B})$.
\end{exercise}

\begin{exercise}
[Conditional Entropy and Relative Entropy]\label{ex-qie:rel-ent-cond-ent}Let
$\rho_{AB}\in\mathcal{D}(\mathcal{H}_{A}\otimes\mathcal{H}_{B})$. Show that
the following identities hold:%
\begin{align}
I(A\rangle B)_{\rho}  &  =D(\rho_{AB}\Vert I_{A}\otimes\rho_{B})\\
&  =\min_{\sigma_{B}\in\mathcal{D}(\mathcal{H}_{B})}D(\rho_{AB}\Vert
I_{A}\otimes\sigma_{B}).
\end{align}
Note that these imply that%
\begin{align}
H(A|B)_{\rho}  &  =-D(\rho_{AB}\Vert I_{A}\otimes\rho_{B})\\
&  =-\min_{\sigma_{B}\in\mathcal{D}(\mathcal{H}_{B})}D(\rho_{AB}\Vert
I_{A}\otimes\sigma_{B}).
\end{align}

\end{exercise}

\begin{exercise}
[Relative Entropy of Classical--Quantum States]\label{ex-qie:cq-rel-ent}Show
that the quantum relative entropy between classical--quantum states $\rho
_{XB}$ and $\sigma_{XB}$ is as follows:%
\begin{equation}
D(\rho_{XB}\Vert\sigma_{XB})=\sum_{x}p_{X}(x)D(\rho_{B}^{x}\Vert\sigma_{B}%
^{x}),
\end{equation}
where%
\begin{equation}
\rho_{XB}\equiv\sum_{x}p_{X}(x)|x\rangle\langle x|_{X}\otimes\rho_{B}%
^{x},\ \ \ \ \ \ \ \ \ \ \sigma_{XB}\equiv\sum_{x}p_{X}(x)|x\rangle\langle
x|_{X}\otimes\sigma_{B}^{x},
\end{equation}
with $p_{X}$ a probability distribution over a finite alphabet $\mathcal{X}$,
$\rho_{B}^{x}\in\mathcal{D}(\mathcal{H}_{B})$ for all $x\in\mathcal{X}$, and
$\sigma_{B}^{x}\in\mathcal{L}(\mathcal{H}_{B})$ positive semi-definite for all
$x\in\mathcal{X}$.
\end{exercise}

\section{Quantum Entropy Inequalities}

Monotonicity of quantum relative entropy has as its corollaries many of the
important entropy inequalities in quantum information theory.

\begin{corollary}
[Strong Subadditivity]\label{cor-qie:SSA}Let $\rho_{ABC}\in\mathcal{D}%
(\mathcal{H}_{A}\otimes\mathcal{H}_{B}\otimes\mathcal{H}_{C})$. The von
Neumann entropy is strongly subadditive, in the following sense:%
\begin{equation}
H(AC)_{\rho}+H(BC)_{\rho}\geq H(ABC)_{\rho}+H(C)_{\rho}.
\end{equation}
Equivalently, we have that
\begin{equation}
I(A;B|C)_{\rho} \geq 0.
\end{equation}
\end{corollary}

\begin{proof}
Consider that%
\begin{equation}
I(A;B|C)_{\rho}=H(AC)_{\rho}+H(BC)_{\rho}-H(ABC)_{\rho}-H(C)_{\rho},
\label{eq-qie:SSA-3}%
\end{equation}
so that%
\begin{equation}
I(A;B|C)_{\rho}=H(B|C)_{\rho}-H(B|AC)_{\rho}.
\end{equation}
From Exercise~\ref{ex-qie:rel-ent-cond-ent}, we know that%
\begin{align}
-H(B|AC)_{\rho}  &  =D(\rho_{ABC}\Vert I_{B}\otimes\rho_{AC}),\\
H(B|C)_{\rho}  &  =-D(\rho_{BC}\Vert I_{B}\otimes\rho_{C}).
\label{eq-qie:SSA-4}%
\end{align}
Then%
\begin{align}
D(\rho_{ABC}\Vert I_{B}\otimes\rho_{AC})  &  \geq D(\operatorname{Tr}%
_{A}\{\rho_{ABC}\}\Vert\operatorname{Tr}_{A}\{I_{B}\otimes\rho_{AC}%
\})\label{eq-qie:SSA-1}\\
&  =D(\rho_{BC}\Vert I_{B}\otimes\rho_{C}). \label{eq-qie:SSA-2}%
\end{align}
The inequality is a consequence of the monotonicity of quantum relative
entropy (Theorem~\ref{thm-qie:mono-rel-ent}), taking $\rho=\rho_{ABC}$,
$\sigma=I_{B}\otimes\rho_{AC}$, and $\mathcal{N}=\operatorname{Tr}_{A}$. By
\eqref{eq-qie:SSA-3}--\eqref{eq-qie:SSA-4}, the inequality in
\eqref{eq-qie:SSA-1}--\eqref{eq-qie:SSA-2} is equivalent to the inequality in
the statement of the corollary.
\end{proof}

\begin{corollary}
[Joint Convexity of Quantum Relative Entropy]Let $p_{X}$ be a probability
distribution over a finite alphabet $\mathcal{X}$, $\rho^{x}\in\mathcal{D}%
(\mathcal{H})$ for all $x\in\mathcal{X}$, and $\sigma^{x}\in\mathcal{L}%
(\mathcal{H})$ be positive semi-definite for all $x\in\mathcal{X}$. Set
$\rho\equiv\sum_{x}p_{X}(x)\rho^{x}$ and $\sigma\equiv\sum_{x}p_{X}%
(x)\sigma^{x}$. The quantum relative entropy%
\index{quantum relative entropy!joint convexity}
is jointly convex in its arguments:%
\begin{equation}
D(\rho\Vert\sigma)\leq\sum_{x}p_{X}(x)D(\rho^{x}\Vert\sigma^{x}).
\end{equation}

\end{corollary}

\begin{proof}
Consider classical--quantum states of the following form:%
\begin{align}
\rho_{XB}  &  \equiv\sum_{x}p_{X}(x)|x\rangle\langle x|_{X}\otimes\rho_{B}%
^{x},\\
\sigma_{XB}  &  \equiv\sum_{x}p_{X}(x)|x\rangle\langle x|_{X}\otimes\sigma
_{B}^{x}.
\end{align}
Then%
\begin{equation}
\sum_{x}p_{X}(x)D(\rho_{B}^{x}\Vert\sigma_{B}^{x})=D(\rho_{XB}\Vert\sigma
_{XB})\geq D(\rho_{B}\Vert\sigma_{B}).
\end{equation}
The equality follows from Exercise~\ref{ex-qie:cq-rel-ent}, and the inequality
follows from monotonicity of quantum relative entropy
(Theorem~\ref{thm-qie:mono-rel-ent}), where we take the channel to be the
partial trace over the system $X$.
\end{proof}

\begin{corollary}
[Unital Channels Increase Entropy]\label{cor-qie:unital-ch-inc-ent}Let
$\rho\in\mathcal{D}(\mathcal{H})$ and let $\mathcal{N}:\mathcal{L}%
(\mathcal{H})\rightarrow\mathcal{L}(\mathcal{H})$ be a unital quantum channel. Then%
\begin{equation}
H(\mathcal{N}(\rho))\geq H(\rho). \label{eq-qie:dephasing-inc-ent}%
\end{equation}

\end{corollary}

\begin{proof}
Consider that%
\begin{align}
H(\rho)  &  =-D(\rho\Vert I),\\
H(\mathcal{N}(\rho))  &  =-D(\mathcal{N}(\rho)\Vert I)=-D(\mathcal{N}%
(\rho)\Vert\mathcal{N}(I)),
\end{align}
where in the last equality, we have used that $\mathcal{N}$ is a unital
quantum channel. The inequality in \eqref{eq-qie:dephasing-inc-ent} is a
consequence of the monotonicity of quantum relative entropy
(Theorem~\ref{thm-qie:mono-rel-ent}) because $D(\rho\Vert I)\geq
D(\mathcal{N}(\rho)\Vert\mathcal{N}(I))$.
\end{proof}

\subsection{Quantum Data Processing}

\label{sec-ie:quantum-data-processing}The quantum data-processing
inequalities
\index{quantum data processing inequality}
discussed below are similar in spirit to the classical data-processing
inequality. Recall that the classical data-processing inequality states that
processing classical data reduces classical correlations. The quantum
data-processing inequalities state that processing \textit{quantum }data
reduces \textit{quantum }correlations.

One variant applies to the following scenario. Suppose that Alice and Bob
share some bipartite state$~\rho_{AB}$. The coherent information $I(A\rangle
B)_{\rho}$ is one measure of the quantum correlations present in this state.
Bob then processes his system $B$ according to some quantum channel
$\mathcal{N}_{B\rightarrow B^{\prime}}$ to produce some quantum system
$B^{\prime}$ and let $\sigma_{AB^{\prime}}$ denote the resulting state. The
quantum data-processing inequality states that this step of quantum data
processing reduces quantum correlations, in the sense that%
\begin{equation}
I(A\rangle B)_{\rho}\geq I(A\rangle B^{\prime})_{\sigma}.
\end{equation}


\begin{theorem}
[Quantum Data Processing for Coherent Information]%
\label{thm-ie:quantum-data-process}Let $\rho_{AB}\in\mathcal{D}(\mathcal{H}%
_{A}\otimes\mathcal{H}_{B})$ and let $\mathcal{N}:\mathcal{L}(\mathcal{H}%
_{B})\rightarrow\mathcal{L}(\mathcal{H}_{B^{\prime}})$ be a quantum channel.
Set $\sigma_{AB^{\prime}}\equiv\mathcal{N}_{B\rightarrow B^{\prime}}(\rho
_{AB})$. Then the following quantum data-processing inequality holds%
\begin{equation}
I(A\rangle B)_{\rho}\geq I(A\rangle B^{\prime})_{\sigma}.
\end{equation}

\end{theorem}

\begin{proof}
This is a consequence of Exercise~\ref{ex-qie:rel-ent-cond-ent} and
Theorem~\ref{thm-qie:mono-rel-ent}. By Exercise~\ref{ex-qie:rel-ent-cond-ent},
we know that%
\begin{align}
I(A\rangle B)_{\rho}  &  =D(\rho_{AB}\Vert I_{A}\otimes\rho_{B}),\\
I(A\rangle B^{\prime})_{\sigma}  &  =D(\sigma_{AB^{\prime}}\Vert I_{A}%
\otimes\sigma_{B^{\prime}})\\
&  =D(\mathcal{N}_{B\rightarrow B^{\prime}}(\rho_{AB})\Vert I_{A}%
\otimes\mathcal{N}_{B\rightarrow B^{\prime}}(\rho_{B}))\\
&  =D(\mathcal{N}_{B\rightarrow B^{\prime}}(\rho_{AB})\Vert\mathcal{N}%
_{B\rightarrow B^{\prime}}(I_{A}\otimes\rho_{B})).
\end{align}
The statement then follows from the monotonicity of quantum relative entropy
by picking $\rho=\rho_{AB}$, $\sigma=I_{A}\otimes\rho_{B}$, and $\mathcal{N}%
=\operatorname{id}_{A}\otimes\mathcal{N}_{B\rightarrow B^{\prime}}$ in
Theorem~\ref{thm-qie:mono-rel-ent}.
\end{proof}

\begin{theorem}
[Quantum Data Processing for Mutual Information]\label{cor-qie:QDP}Let
$\rho_{AB}\in\mathcal{D}(\mathcal{H}_{A}\otimes\mathcal{H}_{B})$,
$\mathcal{N}:\mathcal{L}(\mathcal{H}_{A})\rightarrow\mathcal{L}(\mathcal{H}%
_{A^{\prime}})$ be a quantum channel, and $\mathcal{M}:\mathcal{L}%
(\mathcal{H}_{B})\rightarrow\mathcal{L}(\mathcal{H}_{B^{\prime}})$ be a
quantum channel. Set $\sigma_{A^{\prime}B^{\prime}}\equiv(\mathcal{N}%
_{A\rightarrow A^{\prime}}\otimes\mathcal{M}_{B\rightarrow B^{\prime}}%
)(\rho_{AB})$. Then the following quantum data-processing inequality applies
to the quantum mutual information:%
\begin{equation}
I(A;B)_{\rho}\geq I(A^{\prime};B^{\prime})_{\sigma}.
\end{equation}

\end{theorem}

\begin{proof}
From Exercise~\ref{ex-qie:rel-ent-mut-info}, we know that%
\begin{align}
I(A;B)_{\rho}  &  =D(\rho_{AB}\Vert\rho_{A}\otimes\rho_{B}),\\
I(A^{\prime};B^{\prime})_{\sigma}  &  =D(\sigma_{A^{\prime}B^{\prime}}%
\Vert\sigma_{A^{\prime}}\otimes\sigma_{B^{\prime}})\\
&  =D((\mathcal{N}_{A\rightarrow A^{\prime}}\otimes\mathcal{M}_{B\rightarrow
B^{\prime}})(\rho_{AB})\Vert\mathcal{N}_{A\rightarrow A^{\prime}}(\rho
_{A})\otimes\mathcal{M}_{B\rightarrow B^{\prime}}(\rho_{B}))\\
&  =D((\mathcal{N}_{A\rightarrow A^{\prime}}\otimes\mathcal{M}_{B\rightarrow
B^{\prime}})(\rho_{AB})\Vert(\mathcal{N}_{A\rightarrow A^{\prime}}%
\otimes\mathcal{M}_{B\rightarrow B^{\prime}})(\rho_{A}\otimes\rho_{B})).
\end{align}
The statement then follows from the monotonicity of quantum relative entropy
by picking $\rho=\rho_{AB}$, $\sigma=\rho_{A}\otimes\rho_{B}$, and
$\mathcal{N}=\mathcal{N}_{A\rightarrow A^{\prime}}\otimes\mathcal{M}%
_{B\rightarrow B^{\prime}}$ in Theorem~\ref{thm-qie:mono-rel-ent}.
\end{proof}

\section{Continuity of Entropy}

An important theorem below, the Alicki--Fannes--Winter (AFW) inequality%
\index{Alicki-Fannes-Winter inequality}%
, states that conditional quantum entropies are close as well. This statement
does follow directly from the Fannes--Audenaert inequality, but the main
advantage of the AFW\ inequality is that the upper bound has a dependence only
on the dimension of the first system in the conditional entropy (no dependence
on the conditioning system). The AFW\ inequality also finds application in a
proof of a converse theorem in quantum Shannon theory.

\begin{theorem}
[AFW Inequality]\label{thm-qie:AFW-cont-ent}Let $\rho_{AB},\sigma_{AB}%
\in\mathcal{D}(\mathcal{H}_{A}\otimes\mathcal{H}_{B})$. Suppose that%
\begin{equation}
\frac{1}{2}\left\Vert \rho_{AB}-\sigma_{AB}\right\Vert _{1}\leq\varepsilon,
\end{equation}
for $\varepsilon\in\left[  0,1\right]  $. Then%
\begin{equation}
\left\vert H(A|B)_{\rho}-H(A|B)_{\sigma}\right\vert \leq2\varepsilon\log
\dim(\mathcal{H}_{A})+(1+\varepsilon)h_{2}\left(  \frac{\varepsilon
}{1+\varepsilon}\right)  .
\end{equation}
If $\rho_{XB}$ and $\sigma_{XB}$ are classical--quantum and have the following
form:%
\begin{align}
\rho_{XB}  &  =\sum_{x}p(x)|x\rangle\langle x|_{X}\otimes\rho_{B}^{x},\\
\sigma_{XB}  &  =\sum_{x}q(x)|x\rangle\langle x|_{X}\otimes\sigma_{B}^{x},
\end{align}
where $p$ and $q$ are probability distributions defined over a finite alphabet
$\mathcal{X}$, $\{|x\rangle\}$ is an orthonormal basis, and $\rho_{B}%
^{x},\sigma_{B}^{x}\in\mathcal{D}(\mathcal{H}_{B})$ for all $x\in\mathcal{X}$,
then%
\begin{align}
\left\vert H(X|B)_{\rho}-H(X|B)_{\sigma}\right\vert  &  \leq\varepsilon
\log\dim(\mathcal{H}_{X})+(1+\varepsilon)h_{2}\left(  \frac{\varepsilon
}{1+\varepsilon}\right)  ,\\
\left\vert H(B|X)_{\rho}-H(B|X)_{\sigma}\right\vert  &  \leq\varepsilon
\log\dim(\mathcal{H}_{B})+(1+\varepsilon)h_{2}\left(  \frac{\varepsilon
}{1+\varepsilon}\right)  .
\end{align}

\end{theorem}

\begin{proof}
The bounds trivially hold when $\varepsilon=0$, so henceforth we assume that
$\varepsilon\in(0,1]$. All of the upper bounds are monotone non-decreasing
with $\varepsilon$, so it suffices to assume that $\frac{1}{2}\left\Vert
\rho_{AB}-\sigma_{AB}\right\Vert _{1}=\varepsilon$. Let $\rho_{AB}-\sigma
_{AB}=P_{AB}-Q_{AB}$ be a decomposition of $\rho_{AB}-\sigma_{AB}$ into its
positive part $P_{AB}\geq0$ and its negative part $Q_{AB}\geq0$. Let $\Delta_{AB}\equiv P_{AB}%
/\varepsilon$. Since $\operatorname{Tr}\{P_{AB}\}=\frac{1}{2}\left\Vert
\rho_{AB}-\sigma_{AB}\right\Vert _{1}$ (recall this from the development with trace distance), it follows that $\Delta_{AB}$ is a density
operator. Now consider that%
\begin{align}
\rho_{AB}  &  =\sigma_{AB}+(\rho_{AB}-\sigma_{AB}) \label{eq-qie:Delta'-PSD-1}%
\\
&  =\sigma_{AB}+P_{AB}-Q_{AB}\\
&  \leq\sigma_{AB}+P_{AB}\\
&  =\sigma_{AB}+\varepsilon\Delta_{AB}\\
&  =\left(  1+\varepsilon\right)  \left(  \frac{1}{1+\varepsilon}\sigma
_{AB}+\frac{\varepsilon}{1+\varepsilon}\Delta_{AB}\right) \\
&  =\left(  1+\varepsilon\right)  \omega_{AB}, \label{eq-qie:Delta'-PSD-last}%
\end{align}
where we define
\begin{equation}
\omega_{AB}\equiv\frac{1}{1+\varepsilon}\sigma_{AB}%
+\frac{\varepsilon}{1+\varepsilon}\Delta_{AB}.
\end{equation}
Now let
\begin{equation}
\Delta_{AB}^{\prime
}\equiv\frac{1}{\varepsilon}\left[  \left(  1+\varepsilon\right)  \omega
_{AB}-\rho_{AB}\right]  .
\end{equation} It follows from
\eqref{eq-qie:Delta'-PSD-1}--\eqref{eq-qie:Delta'-PSD-last} that $\Delta
_{AB}^{\prime}$ is positive semi-definite. Furthermore, one can check that
$\operatorname{Tr}\{\Delta_{AB}^{\prime}\}=1$, so that $\Delta_{AB}^{\prime}$
is a density operator. One can also quickly check that%
\begin{equation}
\omega_{AB}=\frac{1}{1+\varepsilon}\rho_{AB}+\frac{\varepsilon}{1+\varepsilon
}\Delta_{AB}^{\prime}=\frac{1}{1+\varepsilon}\sigma_{AB}+\frac{\varepsilon
}{1+\varepsilon}\Delta_{AB}.
\end{equation}
Now consider that%
\begin{align}
H(A|B)_{\omega}  &  =-D(\omega_{AB}\Vert I_{A}\otimes\omega_{B})\\
&  =H(\omega_{AB})+\operatorname{Tr}\{\omega_{AB}\log\omega_{B}\}\\
&  \leq h_{2}\left(  \frac{\varepsilon}{1+\varepsilon}\right)  +\frac
{1}{1+\varepsilon}H(\rho_{AB})+\frac{\varepsilon}{1+\varepsilon}H(\Delta
_{AB}^{\prime})\\
&  \ \ \ +\frac{1}{1+\varepsilon}\operatorname{Tr}\{\rho_{AB}\log\omega
_{B}\}+\frac{\varepsilon}{1+\varepsilon}\operatorname{Tr}\{\Delta_{AB}%
^{\prime}\log\omega_{B}\}\\
&  =h_{2}\left(  \frac{\varepsilon}{1+\varepsilon}\right)  -\frac
{1}{1+\varepsilon}D(\rho_{AB}\Vert I_{A}\otimes\omega_{B})\\
&  \ \ \ -\frac{\varepsilon}{1+\varepsilon}D(\Delta_{AB}^{\prime}\Vert
I_{A}\otimes\omega_{B})\\
&  \leq h_{2}\left(  \frac{\varepsilon}{1+\varepsilon}\right)  +\frac
{1}{1+\varepsilon}H(A|B)_{\rho}+\frac{\varepsilon}{1+\varepsilon
}H(A|B)_{\Delta^{\prime}}.
\end{align}
The first equality follows from Exercise~\ref{ex-qie:rel-ent-cond-ent}, and
the second equality follows from the definition of quantum relative entropy.
The first inequality follows because $H(AB)\leq H(Y)+H(AB|Y)$ for a
classical--quantum state on systems $Y$ and $AB$,
%(see
%Exercise~\ref{ex-qie:cond-ent-c-q-non-neg}),
here taking the state as%
\begin{equation}
\frac{1}{1+\varepsilon}|0\rangle\langle0|_{Y}\otimes\rho_{AB}+\frac
{\varepsilon}{1+\varepsilon}|1\rangle\langle1|_{Y}\otimes\Delta_{AB}^{\prime}.
\end{equation}
The third equality follows from algebra and the definition of quantum relative
entropy. The last inequality follows from
Exercise~\ref{ex-qie:rel-ent-cond-ent}. From concavity of the conditional
entropy, we have that%
\begin{equation}
H(A|B)_{\omega}\geq\frac{1}{1+\varepsilon}H(A|B)_{\sigma}+\frac{\varepsilon
}{1+\varepsilon}H(A|B)_{\Delta}.
\end{equation}
Putting together the upper and lower bounds on $H(A|B)_{\omega}$, we find that%
\begin{align}
H(A|B)_{\sigma}-H(A|B)_{\rho}  &  \leq\left(  1+\varepsilon\right)
h_{2}\left(  \frac{\varepsilon}{1+\varepsilon}\right)  +\varepsilon\left[
H(A|B)_{\Delta^{\prime}}-H(A|B)_{\Delta}\right] \\
&  \leq\left(  1+\varepsilon\right)  h_{2}\left(  \frac{\varepsilon
}{1+\varepsilon}\right)  +2\varepsilon\log\dim(\mathcal{H}_{A}),
\end{align}
where the second inequality follows from a dimension bound for the conditional
entropy.

The statements for classical--quantum states follow because the density
operator $\Delta$ is classical--quantum in this case and we know that
$H(X|B)_{\Delta},H(B|X)_{\Delta}\geq0$.
\end{proof}

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