text:
\section{Introduction}

\IEEEPARstart{I}{n} his seminal paper of 1948~\cite{Shannon}, Claude Shannon introduced the notion of entropy of a discrete random variable $X$ as a measure of its uncertainty or equivalently as a measure of the amount of information we gain on average when we learn its value. The Shannon entropy plays a key role in information theory since it characterises the optimal rate of lossless data compression for a discrete memoryless source. A related entropic quantity or information measure is the conditional Shannon entropy $H(X|Y)$, which quantifies the reduction in the uncertainty of a random variable $X$ when another random variable $Y$ is observed. This and other Shannon information measures (e.g.~the Kullback-Leibler divergence (or relative entropy) and the mutual information) are also of operational significance since they characterise either optimal rates of information-theoretic tasks or fundamental limits of certain statistical inference problems.
Two fundamental properties of a conditional entropy (or equivocation), which are indeed satisfied by the conditional Shannon entropy, are the following:
\begin{enumerate}
\item $H(X|Y) \leq H(X)$;
\item $H(X|Y) \geq H(X) - \log |{\mathcal{Y}}|$, (where~$\mathcal{Y}$ denotes the alphabet of the random variable $Y$).
\end{enumerate}

The above properties are intuitively natural since they ensure that $(i)$ obtaining additional information (in the form of knowledge of a random variable $Y$) can only decrease the uncertainty of a random variable $X$, and $(ii)$ this reduction in uncertainty cannot be more than the amount of information (in bits\footnote{In this paper, we take logarithms to base $2$.}) that has been obtained.
\smallskip

In 1960 Alfred R\'enyi introduced a one-parameter family of entropies~\cite{Renyi1961}, $H_\alpha(X)$ (with $\alpha \in [0,1)\cup (1,\infty)$)  which generalised Shannon's definition of entropy and reduced to it in the limit $\alpha \to 1$. The $\alpha$-R\'enyi entropies are of independent relevance in information theory and also arise naturally in various other branches of mathematics including probability theory, functional analysis, convex geometry and additive combinatorics. In analogy to the Shannon information measures, one can define other information measures related to the $\alpha$-R\'enyi entropy, e.g.~the $\alpha$-R\'enyi divergence and the conditional R\'enyi entropy of order $\alpha$. However, unlike the conditional Shannon entropy, the definition of conditional R\'enyi entropy is not unique.

The Shannon entropy~\cite{Shannon} extends quite naturally to the conditional case by means of conditional probabilities~\cite{CoverThomas}. In contrast, there are multiple ways to define a notion of conditional R\'enyi entropy, and several definitions have been proposed in the literature~\cite{Arimoto,Cachin,PropertiesRenyi,WorldRenyi,RennerWolfDefCondRenyi,HayashiDefCondRenyi,SkoricObiVerbitskiySchoenmakersDefCondRenyi,FehrBerensCondRenyi,TeixeiraMatosAntunesCondRenyi}. However, some of these definitions are somewhat unsatisfactory since they do not satisfy both of the desired properties $1$ and $2$ mentioned above.
\smallskip

There is, however, one definition of the conditional R\'enyi entropy which respects both of these properties, that is, the $\a$-Arimoto-R\'enyi Conditional Entropy (ARCE) which was introduced by Arimoto~\cite{Arimoto}. Furthermore, it is consistent with the conditional Shannon entropy, in the sense that one recovers the latter from it in the limit $\alpha \to 1$ (a property which does not necessarily hold for the other definitions in the literature). As a consequence, the ARCE has attracted much attention from the information theory community.

The ARCE has been proved to be of operational significance in diverse information-theoretic problems. These include, among others, guessing problems with side information~\cite{Arikan, SasonVerdu2, BracherHofLapidoth,Sundaresan}, sequential decoding~\cite{Arikan}, task encoding with side information~\cite{BunteLapidoth}, list-size capacity of discrete memoryless channels with feedback~\cite{BunteLapidoth2}, Bayesian $M$-ary hypothesis testing~\cite{SasonVerdu}, generalisations of Fano-type inequalities~\cite{Sakai}. See also~\cite{HayashiTan,HayashiTan2}.
\smallskip

Another desirable property of information measures such as entropies and conditional entropies is continuity. A continuity bound quantifies the amount by which the information measure changes when the underlying probability distribution changes by a small amount, and hence provides an estimate of the robustness of the information measure with respect to the probability distribution. For example, given two probability distributions which are at a total variation distance of at most $\epsilon \in (0,1)$, a {\em{uniform continuity bound}} for an entropy is a bound on the difference between their entropies in terms of $\epsilon$ and the alphabet size. It does not depend on the specifics of the probability distributions themselves. The bound is said to be tight if there exists a pair of probability distributions for which it is saturated.

The importance of continuity bounds lies in the fact that we usually lack precise knowledge of the probability distribution and instead only have a rough estimate of it. Hence it is useful to have tight bounds on the error incurred by approximating the probability distribution. Continuity bounds are not only of fundamental interest but also have useful applications in information theory, for example in the study of channel capacities.

Various such continuity bounds have been derived. A tight uniform continuity bound for the Shannon entropy is attributed to Csisz\'ar, who derived it from Fano's inequality. See also~\cite{Sason2013}. However, it seems to have first appeared in a paper by Zhang~\cite{Zhang}. A tight uniform continuity bound for the conditional Shannon entropy was recently derived by Alhejji and Smith~\cite{AlhejjiSmith}. In the quantum case, Fannes~\cite{Fannes} was the first to prove a uniform continuity bound for the von Neumann entropy, which was later sharpened by Audenaert~\cite{Audenaert}. It was independently obtained by Petz~\cite{PetzCon}. Audenaert's result provides a tight upper bound on the von Neumann entropy of two quantum states of a finite-dimensional system in terms of their trace distance and the dimension of the underlying Hilbert space. Audenaert also derived tight uniform continuity bounds for quantum R\'enyi entropies of order $\a \in [0,1)$ (see Appendix of~\cite{Audenaert}), which covers the case of classical R\'enyi entropies. Continuity bounds for a large family of entropies (which included R\'enyi entropies in the above range) were derived using different proof techniques in~\cite{Eric1}. The authors of~\cite{Eric1} further investigated the case $\alpha > 1$ using the notion of ``majorization flow" in~\cite{Eric2}, obtaining a bound which is sharper than previously known bounds. A similar continuity bound was derived for the conditional von Neumann entropy by Alicki and Fannes~\cite{AlickiFannes}, which in turn was later sharpened by Winter~\cite{Winter2016}. Alhejji and Smith's result for the conditional Shannon entropy was extended to the case of the conditional entropy of a classical-quantum state (with the conditioning being on the classical system) by Wilde~\cite{Wilde2020}.

In this paper we study another important property of the ARCE of order $\alpha$, for the range $\alpha \in [0,1)$. More precisely, we establish a tight uniform continuity bound on the difference of the ARCEs of two joint probability distributions $p_{XY}$ and $q_{XY}$ which are close in total variation distance. See Theorem~\ref{theo:main} of Section~\ref{sec:MainResults} below. To prove this result we introduce $(i)$ the notion of ${\mathcal{X}}$-{\em{majorization}}, which is a special case of the concept of {\em{conditional majorization}} which was introduced by Gour {\em{et al}}\cite{CondMaj} (here ${\mathcal{X}}$ denotes the alphabet of the random variable $X$), and $(ii)$ the notion of a function of a joint probability distribution $p_{XY}$ being {\em{marginally Schur concave}}, that is, Schur concave under ${\mathcal{X}}$-{\em{majorization}}. Ideas and techniques from majorization theory have previously been employed to derive bounds on entropies (see e.g.~\cite{Cicalese2018, Sason2018} and references therein).
\smallskip

We then apply our result to prove a tight uniform continuity bound for the ARCE of order $\alpha$, for $\alpha \in [0,1)$, for classical-quantum (c-q) systems, with the conditioning being done on the classical system. We denote this quantity by $H_\alpha(A|Y)_\rho$, where $Y$ is a classical system ({\em{i.e.}}~ a random variable), $A$ is a finite-dimensional quantum system, and $\rho_{AY}$ is the c-q state of the composite system $AY$ (see \eqref{eq:defi:PetzRenyi} below). See Theorem~\ref{theo:main3} of Section~~\ref{sec:MainResults} in the paper.
\medskip

\noindent
{\bf{Layout of the paper:}} We proceed by first introducing the necessary definitions of the relevant classical and quantum entropies and conditional entropies in Section~\ref{prelim}. We state our main results in Section~\ref{sec:MainResults},
and prove them in Section~\ref{sec:proofs}. The key notions of $\mathcal{X}$-majorization and marginally Schur concave functions, which we exploit in our proofs, are introduced in Section~\ref{sec:ProofIngredients}. We conclude the paper with a discussion and some open problems. Some further tools that we use are stated and proved in the Appendix.

\section{Mathematical preliminaries}\label{prelim}

\subsection{Classical systems}\label{entropies-classical}

Let $\mathcal{X} \coloneqq \left\lbrace 1,2, \cdots, |\mathcal{X}| \right\rbrace$, and let $\mathcal{P}_{\mathcal{X}}$ denote the set of probability distributions on $\mathcal{X}$. Let $p_X \in \mathcal{P}_{\mathcal{X}}$. The Shannon entropy of a variable $X$ with distribution $p_X$ is defined as~\cite{Shannon}
\begin{equation}
    H(X)_p \equiv H(p_X) \coloneqq - \sum_{x \in \mathcal{X}} p_X(x) \log p_X(x).
\end{equation}
The $\alpha$-R\'enyi entropy, for $\alpha \in [0,1) \cup (1,\infty)$ is defined as follows:
\begin{equation}
	H_{\alpha}(X)_p \coloneqq \frac{1}{1-\alpha} \log \left( \sum_{x \in \mathcal{X}} p_X^{\a}(x) \right).
\end{equation}
The Shannon and $\alpha$-R\'enyi entropies are part of larger set of functions related to the concept of majorization. Given $\bos{u} \in \mathbb{R}^d$ for some dimension $d$, define $\bos{u}^{\downarrow} \in \mathbb{R}^d$ to be the vector containing the elements of $\bos{u}$ arranged in non-increasing order. For $\bos{u}, \bos{v} \in \mathbb{R}^d$, we say $\bos{u}$ is majorized by $\bos{v}$, written $\bos{u} \prec \bos{v}$~\cite{Majorization}, if
\eqal{
  & \sum_{j=1}^k \bos{u}^{\downarrow}_j \leq \sum_{j=1}^k \bos{v}^{\downarrow}_j, \quad \forall k = 1, \cdots, d-1, \\
  \mathrm{and} \quad & \sum_{j=1}^d \bos{u}^{\downarrow}_j = \sum_{j=1}^d \bos{v}^{\downarrow}_j.
}
\smallskip

A function $f:\mathbb{R}^d \to \mathbb{R}$ is said to be Schur-convex if $f(\bos{u}) \leq f(\bos{v})$ for any pair $\bos{u}, \bos{v} \in \mathbb{R}^d$ with $\bos{u} \prec \bos{v}$, and it is said to be Schur-concave if $(-f)$ is Schur-convex. The Shannon and $\alpha$-R\'enyi entropies are notable examples of Schur-concave functions when they are taken as functions of the vectors of probability distributions. An interesting property of majorization is the following: for $\bos{u}, \bos{v} \in \mathbb{R}^d$, $\bos{u} \prec \bos{v}$ if and only if there exists a doubly-stochastic matrix $D$ of dimension $d \times d$ such that $\bos{u} = D \bos{v}$~\cite{HardyLittlewoodPolya}. This property will be useful in our proof.
\medskip

Now let $\mathcal{Y} \coloneqq \left\lbrace 1,2, \cdots, |\mathcal{Y}| \right\rbrace$ and let $\mathcal{P}_{\mathcal{X} \times \mathcal{Y}}$ denote the set of probability distributions on $\mathcal{X} \times \mathcal{Y}$. For a pair of random variables $X$ and $Y$ with joint distribution $p_{XY} \in \mathcal{P}_{\mathcal{X} \times \mathcal{Y}}$, the conditional Shannon entropy $ H(X|Y)_p$ of $X$ conditioned on $Y$ is defined as
\eqal{
  H(X|Y)_p & \coloneqq H(p_{XY}) - H(p_Y) \\
  & = - \sum_{y \in \mathcal{Y}} \sum_{x \in \mathcal{X}} p_{XY}(x,y) \log p_{X|Y}(x|y),
}
where $p_Y$ denotes the marginal distribution of $Y$.

As mentioned in the Introduction, several definitions of the conditional R\'enyi entropy have been proposed in the literature~\cite{FehrBerensCondRenyi,TeixeiraMatosAntunesCondRenyi}. In this paper we focus on Arimoto's version of the conditional R\'enyi entropy:
\begin{defi} \label{defi:ArimotoRenyi}
For a pair of random variables $X$ and $Y$ with joint distribution $p_{XY} \in \mathcal{P}_{\mathcal{X} \times \mathcal{Y}}$, the $\a$-Arimoto-R\'enyi conditional entropy, (in short the $\a$-ARCE), for $\alpha \in [0,1) \cup (1,\infty)$ is defined as follows~\cite{Arimoto}:
\begin{equation} \label{eq:ArimotoRenyi2}
	\pr{X}{Y}{p} \coloneqq \frac{\a}{1-\a} \log \left( \sum_{y \in \mathcal{Y}} \left[ \sum_{x \in \mathcal{X}} p_{XY}^{\a}(x,y) \right]^{1/\a} \right).
\end{equation}
\end{defi}
In the limit $\a \to 1$ it reduces to the conditional Shannon entropy.
\smallskip

Unlike the Shannon and $\alpha$-R\'enyi entropies, the $\a$-ARCE is not Schur-concave. However, it satisfies a weaker notion of Schur concavity which we introduce in Section~\ref{sec:ProofIngredients} and which we exploit in our proofs. This leads us to introduce the concepts of $\mathcal{X}$-{\em{majorization}} and {\em{marginally Schur-concave functions}} (see Section~\ref{sec:ProofIngredients}).
\smallskip

Finally, we recall the definition of the {\em{total variation distance }} (TV) between two probability distributions $p_{XY}, q_{XY} \in \mathcal{P}_{\mathcal{X} \times \mathcal{Y}}$:
	$\mathrm{TV}(p_{XY},q_{XY}) \coloneqq \frac{1}{2} \sum_{x \in \mathcal{X}} \sum_{y \in \mathcal{Y}} |p_{XY}(x,y) - q_{XY}(x,y)|.$

\subsection{Classical-quantum systems}\label{entropies-quantum}

In this section we consider finite-dimensional bipartite quantum (and classical-quantum) systems. Their states are given by density matrices, \textit{i.e.}~positive semidefinite operators of unit trace acting on the Hilbert space associated with the system. The trace distance between two states $\rho$ and $\sigma$ of a quantum system is given by $\frac{1}{2}||\rho-\sigma||_1$ where for any operator $A$, acting on the Hilbert space of the system, $||A||_1 := {\mathrm{Tr}}{\sqrt{A^\dagger A}}$.
\medskip

Let us introduce some relevant entropic quantities.
\begin{defi}
Let $\rho_{AB}$ be the density matrix of a bipartite finite-dimensional quantum system $AB$. Then for $\a \in [0, 1) \cup (1, \infty)$ the quantum conditional $\a$-R\'enyi entropy of $A$ given $B$ of the state $\rho_{AB}$ is defined as follows~\cite{Petz}:
\begin{equation} \label{eq:defi:PetzRenyi}
	\pr{A}{B}{\rho} \coloneqq \frac{\a}{1-\a} \log \tr{}{\left( \tr{A}{\rho_{AB}^\a} \right)^{1/\a}}.
\end{equation}
\end{defi}
In the limit $\a \to 1$ it reduces to the quantum conditional entropy
\begin{align}
    H(A|B)_{\rho} := H({\rho}_{AB}) - H(\rho_B),
\end{align}
where $\rho_B:= \tr{A}{\rho_{AB}}$ denotes the reduced state of the system $B$, and $H(\rho):= - {\mathrm{Tr}}(\rho \log \rho)$ denotes the von Neumann entropy of the state $\rho$.
The quantity $\pr{A}{B}{\rho}$ can also be expressed as follows~\cite{Petz,Tomamichel}:
\begin{equation}
	\pr{A}{B}{\rho} = \sup_{\sigma_B} \left( - D_\a \infdivx{\rho_{AB}}{\mathbb{I}_A \otimes \sigma_B} \right),
\end{equation}
where $\mathbb{I}_A$ denotes the identity operator acting on the system $A$, the supremum is taken over all states $\sigma_B$ of the system $B$, and where for two states $\rho$ and $\sigma$, $D_\a \infdivx{\rho}{\sigma}$ denotes the quantum $\a$-R\'enyi divergence of $\rho$ with respect to $\sigma$. The latter is defined as~\cite{Petz}
\begin{equation}
  D_\a \infdivx{\rho}{\sigma} \coloneqq \frac{1}{\a-1} \log \tr{}{\rho^\a \sigma^{1-\a}},
\end{equation}
if $\mathrm{supp}(\rho) \subseteq \mathrm{supp}(\sigma)$ and $D_\a \infdivx{\rho}{\sigma} \coloneqq + \infty$ else.

We are interested in the conditional $\a$-R\'enyi entropy defined in \eqref{eq:defi:PetzRenyi} since it can be related to the $\a$-Arimoto-R\'enyi conditional entropy when the bipartite system is classical-quantum (c-q) and the conditioning is made on the classical system. Indeed, consider a finite dimensional c-q system with density matrix $\rho_{AY}$, where $A$ denotes a quantum system and $Y$ denotes a classical system of dimension $|\mathcal{Y}|$. We can always write the density matrix as
\begin{equation} \label{eq:rhoAY2}
	\rho_{AY} = \sum_{y \in \mathcal{Y}} r_Y(y) \rho_A^y \otimes \proj{y}_Y,
\end{equation}
where $r_Y \in \mathcal{P}_{\mathcal{Y}}$ and $\left\lbrace \rho_A^y \right\rbrace_{y \in \mathcal{Y}}$ is a set of quantum states of the system $A$. For each $y \in \mathcal{Y}$, consider the spectral decomposition of $\rho_A^y$:
\begin{equation} \label{eq:rhoAY}
	\rho_A^y = \sum_{x \in \mathcal{X}} r_{X|Y}(x|y) \proj{\phi^y_x}_A,
\end{equation}
where $|\mathcal{X}| = d_A$ is the dimension of system $A$ and $r_{X|Y}$ is a conditional probability distribution.

We can write the bipartite state as
\begin{equation}
	\rho_{AY} = \sum_{y \in \mathcal{Y}, x \in \mathcal{X}} r_{XY}(x,y) \proj{\phi^y_x}_A \otimes \proj{y}_Y,
\end{equation}
where we defined $r_{XY}(x,y) = r_Y(y) r_{X|Y}(x|y)$ for all $x \in \mathcal{X}, y \in \mathcal{Y}$, so that $r_{XY} \in \mathcal{P}_{\mathcal{X} \times \mathcal{Y}}$. In that case,
\eqal{ \label{eq:PetzToArimoto}
  & \pr{A}{Y}{\rho} \\
  & = \frac{\a}{1-\a} \log \mathrm{Tr} \left[ \left( \mathrm{Tr}_A \left[ \sum_{y \in \mathcal{Y}, x \in \mathcal{X}} r_{XY}^\a(x,y) \right. \right. \right. \\
  & \hspace{3.5cm} \left. \left. \left. \vphantom{\sum_{y \in \mathcal{Y}, x \in \mathcal{X}}} \times \proj{\phi^y_x,y}_{AY} \right] \right)^{1/\a} \right] \\
	& = \frac{\a}{1-\a} \log \tr{}{\left( \sum_{y \in \mathcal{Y}, x \in \mathcal{X}} r_{XY}^\a(x,y) \proj{y}_Y \right)^{1/\a}} \\
	& = \frac{\a}{1-\a} \log \tr{}{ \sum_{y \in \mathcal{Y}} \left( \sum_{x \in \mathcal{X}} r_{XY}^\a(x,y) \right)^{1/\a} \proj{y}_Y} \\
	& = \frac{\a}{1-\a} \log \left( \sum_{y \in \mathcal{Y}} \left( \sum_{x \in \mathcal{X}} r_{XY}^\a(x,y) \right)^{1/\a} \right) \\
	& = \pr{X}{Y}{r},
}
where $\ket{\phi^y_x,y}_{AY} \equiv \ket{\phi^y_x}_A \otimes \ket{y}_Y$ and $\pr{X}{Y}{r}$ is the classical $\a$-ARCE for a pair of random variables $X$ and $Y$ with joint probability distribution $r_{XY}$.

In the limit $\a \to 1$, $\pr{A}{Y}{\rho}$ reduces to the conditional entropy of the c-q state $\rho_{AY}$:
\begin{align}\label{q-cond}
 H(A|Y)_{\rho}:= \sum_y r_Y(y) H(\rho_A^y),
\end{align}
where $ H(\rho_A^y)$ denotes the von Neumann entropy of the quantum state $\rho_A^y$.

\section{Main results \label{sec:MainResults}}

In this section we state our main results, namely, tight uniform continuity bounds for the $\a$-Arimoto-R\'enyi conditional entropy ($\a$-ARCE) of a classical joint probability distribution, and the conditional $\a$-R\'enyi entropy of a c-q system. In the limit $\a \to 1$, they reduce to the known bounds for the conditional Shannon entropy and the conditional entropy (of a c-q system). The first theorem pertains to the $\a$-ARCE.

\begin{theo} \label{theo:main}
Let $\a \in [0,1)$, $\epsilon \in (0,1-\frac{1}{|\mathcal{X}|}]$ and $p_{XY}, q_{XY} \in \mathcal{P}_{\mathcal{X} \times \mathcal{Y}}$ be such that
\begin{equation} \label{eq:theo:main2}
\mathrm{TV}(p_{XY},q_{XY}) \leq \epsilon.
\end{equation}
Then the following inequality holds:
\eqal{ \label{eq:theo:main}
   & |\pr{X}{Y}{p} - \pr{X}{Y}{q}| \\
   &\leq \frac{1}{1-\a} \log \left( \left( 1- \epsilon \right)^{\a} + \left(|\mathcal{X}|-1\right)^{1-\a} \epsilon^{\a} \right).
}
Moreover, the inequality is tight, \textit{i.e.},
\begin{equation}
\sup_{p_{XY},q_{XY}} \frac{|\pr{X}{Y}{p} - \pr{X}{Y}{q}|}{\gamma(\alpha, \epsilon, |\mathcal{X}|)} =1,
\end{equation}
where $\gamma(\alpha, \epsilon, |\mathcal{X}|)$ denotes the expression on the right hand side of (\ref{eq:theo:main}).
\end{theo}

\begin{rem} \label{rem:ShannonBound}
In the limit $\alpha \to 1$, Theorem~\ref{theo:main} yields the corresponding continuity bound for the conditional Shannon entropy which was derived in~\cite{AlhejjiSmith}, and is given by the following:
for any $\epsilon \in (0,1-\frac{1}{|\mathcal{X}|}]$, for a pair of probability distributions $p_{XY}$ and $q_{XY}$ in $\mathcal{P}_{\mathcal{X} \times \mathcal{Y}}$ for which $\mathrm{TV}(q_{XY}, p_{XY}) \leq \epsilon$:
\begin{align}
    |H(X|Y)_p - H(X|Y)_q| \leq \epsilon \log \left(|\mathcal{X}|-1\right) + h(\epsilon),
\end{align}
where $h(\epsilon):= - \epsilon \log \epsilon - (1-\epsilon) \log (1-\epsilon)$ is the binary entropy.
This can be easily verified by a simple use of l'H\^opital's rule.
\end{rem}

Following the ideas used in~\cite{Wilde2020}, and making use of the bound stated in Theorem~\ref{theo:main}, we derive a tight uniform continuity bound for the conditional $\a$-R\'enyi entropy of a c-q system, with the conditioning being on the classical system, and $\a$ being in the range $[0,1)$. This bound is stated in the following theorem.

\begin{theo} \label{theo:main3}
Consider a c-q system $AY$, where $d_A$ denotes the dimension of the quantum system $A$, and $|\mathcal{Y}|$ denotes the dimension of the classical system $Y$. Let $\rho_{AY}$ and $\sigma_{AY}$ be two states of $AY$ satisfying
\begin{equation} \label{eq:theo:main33}
	\frac{1}{2} || \rho_{AY} - \sigma_{AY} ||_1 \leq \epsilon.
\end{equation}
Then for any $\a \in [0,1)$, and $\epsilon \in (0,1-\frac{1}{d_A}]$ the following inequality holds:
\eqal{ \label{eq:theo:mainQ}
  & |\pr{A}{Y}{\rho} - \pr{A}{Y}{\sigma}| \\
  & \leq \frac{1}{1-\a} \log \left( \left( 1- \epsilon \right)^{\a} + \left(d_A-1\right)^{1-\a} \epsilon^{\a} \right).
}
Moreover, the inequality is tight, \textit{i.e.},
\begin{equation}
\sup_{\rho_{AY},\sigma_{AY}} \frac{|\pr{A}{Y}{\rho} - \pr{A}{Y}{\sigma}|}{\gamma(\alpha, \epsilon, d_A)} =1,
\end{equation}
where $\gamma(\alpha, \epsilon, d_A)$ denotes the expression on the right hand side of (\ref{eq:theo:mainQ}).
\end{theo}

\begin{rem}
By making use of Remark~\ref{rem:ShannonBound}, in the limit $\alpha \to 1$ we recover the corresponding continuity bound for the conditional entropy of a c-q state which was derived in~\cite{Wilde2020},
and is given by the following:
for any $\epsilon \in (0,1-\frac{1}{d_A}]$, for a pair of finite-dimensional c-q states $\rho_{AY}, \sigma_{AY}$ satisfying \eqref{eq:theo:main33}, the following inequality holds:
\begin{align}
    |H(A|Y)_{\rho} - H(A|Y)_{\sigma}| \leq \epsilon \log \left(d_A-1\right) + h(\epsilon),
\end{align}
where $h(\epsilon)$ is the binary entropy and $H(A|Y)_{\rho}$ denotes the conditional entropy of the c-q state $\rho_{AY}$.
\end{rem}

\section{Proof of the main results}\label{sec:proofs}

\subsection{Proof ingredients: \texorpdfstring{$\mathcal{X}$}{}-majorization and marginally Schur-concave functions \label{sec:ProofIngredients}}

The notion of conditional majorization was introduced in~\cite{CondMaj}. For completeness, we state its definition in Appendix \ref{CondMajGour}. Here we consider a particular case of conditional majorization, which we refer to as $\mathcal{X}$-\textit{majorization}, as it can be understood as majorization applied to the $\mathcal{X}$-marginals of joint probability distributions $p_{XY} \in \mathcal{P}_{\mathcal{X} \times \mathcal{Y}}$. The definition is as follows.
\begin{defi}
Denote by $\mathbb{R}^{n \times l}_+$ the set of all $n \times l$ matrices with non-negative values. Consider $P \in \mathbb{R}^{n \times l}_+$ and $Q \in \mathbb{R}^{n \times l}_+$. We say $Q$ is $\mathcal{X}$-majorized by $P$ and write $Q \cmajj P$ if there exist matrices $D^{(j)}$ and $R^{(j)}$, where $j \in \left\lbrace 1, 2, \cdots, l \right\rbrace$, such that
\begin{equation}
	Q = \sum_{j=1}^l D^{(j)} P R^{(j)},
\end{equation}
where each $D^{(j)}$ is an $n \times n$ doubly-stochastic matrix and each $R^{(j)}$ is an $l \times l$ matrix which is such that $\left(R^{(j)}\right)_{ik} = \delta_{ik} \delta_{ij}, \forall i,k \in \left\lbrace 1, 2, \cdots, l \right\rbrace$.
\end{defi}
Henceforth, we choose to represent a joint probability distribution $p_{XY} \in \mathcal{P}_{\mathcal{X} \times \mathcal{Y}}$ as a matrix of dimensions $|\mathcal{X}| \times |\mathcal{Y}|$, so that its $y$\textsuperscript{th} column contains the column-vector $\left\lbrace p_{XY}(x,y) \right\rbrace_{x \in \mathcal{X}}$, for each $y \in \mathcal{Y}$, while its $x$\textsuperscript{th} row contains the row-vector $\left\lbrace p_{XY}(x,y) \right\rbrace_{y \in \mathcal{Y}}$, for each $x \in \mathcal{X}$. Here we have used the notation $\left\lbrace p_{XY}(x,y) \right\rbrace_{x \in \mathcal{X}}:=  \left( p_{XY}(1,y), p_{XY}(2,y), \ldots, p_{XY}(|{\mathcal{X}}|,y) \right)$
and $\left\lbrace p_{XY}(x,y) \right\rbrace_{y \in \mathcal{Y}}:=  \left( p_{XY}(x,1), p_{XY}(x,2), \ldots, p_{XY}(x,|{\mathcal{Y}}|) \right)$. We also denote the column-vector of conditional probabilities $p_{X|Y}(x|y)$ for a fixed $y \in \mathcal{Y}$ as
\begin{equation} \label{eq:vecCond}
 	p_{X|Y=y} = \left( p_{X|Y}(1|y), p_{X|Y}(2|y), \ldots, p_{X|Y}(|{\mathcal{X}}||y) \right).
\end{equation}
Moreover, we use the symbols $q_{XY}$ and $p_{XY}$ to refer to both the joint distributions as well as their matrix representations.

We now apply our definition of $\mathcal{X}$-majorization to joint probability distributions in $\mathcal{P}_{\mathcal{X} \times \mathcal{Y}}$ and relate it to the ``usual" majorization applied to vectors of conditional probabilities. We do so through the following simple lemma.
\begin{lem} \label{lem:cmajToMaj}
Consider $p_{XY}, q_{XY} \in \mathcal{P}_{\mathcal{X} \times \mathcal{Y}}$. Then the following statements are equivalent:
\begin{enumerate}
	\item $q_{XY} \cmajj p_{XY}$,
	\item $\left( q_{XY} R^{(y)} \bos{e}_{|\mathcal{Y}|} \right) \prec \left( p_{XY} R^{(y)} \bos{e}_{|\mathcal{Y}|} \right)$ for all $y \in \mathcal{Y}$, \label{enum:majCol:lem:cmajToMaj}
	\item $q_{Y} = p_Y$ and $q_{X|Y=y} \prec p_{X|Y=y}$ for all $y \in \mathcal{Y}$.
\end{enumerate}
where $\bos{e}_{|\mathcal{Y}|}$ is a column-vector with $|\mathcal{Y}|$ rows with all its elements equal to $1$.
\end{lem}
Note that point \ref{enum:majCol:lem:cmajToMaj} simply means that each column of $q_{XY}$ is majorized by the corresponding column (same $y$ index) of $p_{XY}$.
\begin{proof}
We first show the equivalence between \textit{1} and \textit{2}. Consider $q_{XY} \cmajj p_{XY}$. In that case, there exist matrices $D^{(z)}$ and $R^{(z)}$, where $z \in \mathcal{Y}$, such that
\begin{equation}
	q_{XY} = \sum_{z=1}^{|\mathcal{Y}|} D^{(z)} p_{XY} R^{(z)},
\end{equation}
where each $D^{(z)}$ is an $|\mathcal{X}| \times |\mathcal{X}|$ doubly-stochastic matrix and each $R^{(z)}$ is a $|\mathcal{Y}| \times |\mathcal{Y}|$ matrix which is such that $\left(R^{(z)}\right)_{ab} = \delta_{ab} \delta_{az}, \forall a,b \in \mathcal{Y}$. Using the above equation, for each $y \in \mathcal{Y}$,
\eqal{ \label{eq:eq5}
  q_{XY} R^{(y)} \bos{e}_{|\mathcal{Y}|} & = \sum_{z=1}^{|\mathcal{Y}|} D^{(z)} p_{XY} R^{(z)} R^{(y)} \bos{e}_{|\mathcal{Y}|} \\
  & = D^{(y)} p_{XY} R^{(y)} \bos{e}_{|\mathcal{Y}|},
}
so that~\cite{HardyLittlewoodPolya}
\begin{equation}
	\left( q_{XY} R^{(y)} \bos{e}_{|\mathcal{Y}|} \right) \prec \left( p_{XY} R^{(y)} \bos{e}_{|\mathcal{Y}|} \right), \quad \forall y \in \mathcal{Y}.
\end{equation}
Since the doubly-stochastic matrix $D^{(y)}$ appearing in the last equation of \eqref{eq:eq5} depends on $y$, the proof can be reversed without any loss of generality, so that \textit{1} and \textit{2} are equivalent. We now show the equivalence between \textit{2} and \textit{3}.
Since $p_{XY}(x,y) = p_Y(y) p_{X|Y}(x|y), \forall x \in \mathcal{X}, \forall y \in \mathcal{Y}$, and similarly for $q_{XY}$, according to the definition in \eqref{eq:vecCond}, we can write, for each $y \in \mathcal{Y}$,
\eqal{
  & p_{Y}(y) p_{X|Y=y} = p_{XY} R^{(y)} \bos{e}_{|\mathcal{Y}|}, \\
  & q_Y(y) q_{X|Y=y} = q_{XY} R^{(y)} \bos{e}_{|\mathcal{Y}|},
}
so that
\eqal{
  & p_{Y}(y) = \sum_{x \in \mathcal{X}} \left( p_{XY} R^{(y)} \bos{e}_{|\mathcal{Y}|}\right)_x, \qquad \forall y \in \mathcal{Y}, \\
  & q_Y(y) = \sum_{x \in \mathcal{X}} \left( q_{XY} R^{(y)} \bos{e}_{|\mathcal{Y}|}\right)_x, \qquad \forall y \in \mathcal{Y}.
}
Suppose $\left( q_{XY} R^{(y)} \bos{e}_{|\mathcal{Y}|} \right) \prec \left( p_{XY} R^{(y)} \bos{e}_{|\mathcal{Y}|} \right)$ for all $y \in \mathcal{Y}$. A necessary condition is that $q_Y(y) = p_{Y}(y)$ for all $y \in \mathcal{Y}$, which means that $p_{Y} = q_Y$. It also means that
\begin{equation}
	q_{X|Y=y} \prec p_{X|Y=y}, \quad \forall y \in \mathcal{Y}.
\end{equation}
Again, the proof can be reversed without any loss of generality, so that \textit{2} and \textit{3} are equivalent.
\end{proof}

\noindent We are now in position to define the notion of marginally Schur-concave function.
\begin{defi}
A function $f : \mathbb{R}^{n \times l}_+ \rightarrow \mathbb{R}$ will be called {\em{marginally Schur-concave}} if, for any pair of matrices $P, Q \in \mathbb{R}^{n \times l}_+$, we have
\begin{equation}
	Q \cmajj P \quad \Rightarrow \quad f(P) \leq f(Q).
\end{equation}
The function $f$ is said to be {\em{marginally Schur-convex}} if the opposite inequality holds.
\end{defi}

We show the following.
\begin{lem} \label{lem:CondSchurConcavity}
The $\a$-ARCE is marginally Schur-concave for all $\a \in [0,1) \cup (1,\infty)$.
\end{lem}
\begin{proof}
The $\a$-ARCE can be written as
\begin{equation} \label{eq:ArimotoRenyi3}
	\pr{X}{Y}{p} = \frac{\a}{1-\a} \log \left( \sum_{y \in \mathcal{Y}} p_Y(y) ||p_{X|Y=y}||_\a \right).
\end{equation}
We need to show that $q_{XY} \cmajj p_{XY}$ implies $\pr{X}{Y}{p} \leq \pr{X}{Y}{q}$. From Lemma \ref{lem:cmajToMaj}, we have that $q_{XY} \cmajj p_{XY}$ implies $p_{Y} = q_Y$ and $q_{X|Y=y} \prec p_{X|Y=y}$, for all $y \in \mathcal{Y}$. We distinguish between the two cases $\a>1$ and $\a \in [0,1)$.
\paragraph{Case 1: $\a>1$.}
In this case, the function $u \mapsto u^\a$ is convex for $u \in [0, \infty)$, so that the $\a$-norm $||p_{X|Y=y}||_\a$ is Schur-convex. Since $q_{X|Y=y} \prec p_{X|Y=y}$ for all $y \in \mathcal{Y}$, this implies that
\begin{equation}
	||p_{X|Y=y}||_\a \geq ||q_{X|Y=y}||_\a, \quad \forall y \in \mathcal{Y}.
\end{equation}
Since $p_{Y} = q_Y$ and $1-\a<0$, we end up with
\begin{equation}
	\pr{X}{Y}{p} \leq \pr{X}{Y}{q}, \quad \forall \a > 1.
\end{equation}
\paragraph{Case 2: $\a \in [0,1)$.}
In this case, the function $u \mapsto u^\a$ is concave for $u \in [0, \infty)$, so that the $\a$-quasi-norm $||p_{X|Y=y}||_\a$ is Schur-concave. Since $q_{X|Y=y} \prec p_{X|Y=y}$ for all $y \in \mathcal{Y}$, this implies that
\begin{equation}
	||p_{X|Y=y}||_\a \leq ||q_{X|Y=y}||_\a, \quad \forall y \in \mathcal{Y}.
\end{equation}
Since $p_{Y} = q_Y$ and $1-\a>0$, we end up with
\begin{equation}
	\pr{X}{Y}{p} \leq \pr{X}{Y}{q}, \quad \forall \a \in [0,1).
\end{equation}
\end{proof}

\noindent Note that Lemma \ref{lem:CondSchurConcavity} can also be proven by exploiting the fact that $\mathcal{X}$-majorization is a special case of conditional majorization, and by making use of Theorem 1 of~\cite{CondMaj}, which we state in the Appendix \ref{CondMajGour} as Lemma \ref{lem:CharacCondMajGour}.
\begin{rem}
Using the same ideas as in the proof of Lemma \ref{lem:CondSchurConcavity}, the conditional Shannon entropy can be readily verified to be marginally Schur-concave (see also~\cite{AlhejjiSmith}).
\end{rem}

\subsection{Proof of Theorem~\ref{theo:main}}

Fix $\epsilon \in [0,1-\frac{1}{|\mathcal{X}|})$.
Consider two probability distributions $p_{XY}$ and $q_{XY}$ such that their total variation distance $t \coloneqq \mathrm{TV}(p_{XY},q_{XY})$ is at most equal to $\epsilon$. We assume, without loss of generality, that $\pr{X}{Y}{p} \geq \pr{X}{Y}{q}$ and define $\Delta H_\a \coloneqq \pr{X}{Y}{p} - \pr{X}{Y}{q}$.

The proof of Theorem~\ref{theo:main} consists of a series of steps which are described in detail below. The key idea is to alter the distributions $q_{XY}$ and $p_{XY}$, in a series of iterative steps such that the difference $\pr{X}{Y}{p}-\pr{X}{Y}{q}$ never decreases, and the total variation distance also remains unchanged. As a result of these manipulations, $\pr{X}{Y}{q}$ reduces to zero while $\pr{X}{Y}{p}$ attains its maximal value under the constraint on the total variation distance, which indeed is the desired upper bound. Note that our initial steps (Step A and Step B) are similar to those used by Alhejji and Smith~\cite{AlhejjiSmith}. However, the remaining steps deviate considerably from theirs in order to deal with the additional complexity of the $\a$-ARCE (compared to the conditional Shannon entropy). In order to clarify some of the steps of the proof, we consider a $3 \times 3$ example in Appendix~\ref{exampleARCEproof}.
\medskip

\noindent
{\bf{Step A: Reordering}} Recall that we chose to represent the probability distributions $p_{XY}$ and $q_{XY}$ by $|\mathcal{X}|\times |\mathcal{Y}|$ matrices. For convenience, we first arrange the $|\mathcal{Y}|$ columns of $q_{XY}$ such that
\begin{equation}
 q_Y(y) \geq q_Y(y+1), \quad \forall y \in \mathcal{Y} \setminus \left\lbrace |\mathcal{Y}| \right\rbrace.
\end{equation}
We then do the same permutation of columns of $p_{XY}$. Note, however, that $p_Y(y)$ is not necessarily greater that $p_Y(y+1)$.
From the form given in \eqref{eq:ArimotoRenyi3}, one understands that this does not change the value of the $\a$-ARCE. Now, for the $y$\textsuperscript{th} column, define the two sets
\eqal{ \label{eq:eq6}
	I_y & = \left\lbrace x | q_{XY}(x,y) \geq p_{XY}(x,y) \right\rbrace, \\
	I_y^c & = \left\lbrace x | q_{XY}(x,y) < p_{XY}(x,y) \right\rbrace.
}
Do the following changes for both $p_{XY}$ and $q_{XY}$. Within the $y$\textsuperscript{th} column, if both $I_y$ and $I_y^c$ are non-empty, put the elements corresponding to indices in $I_y$ ahead of those corresponding to indices in $I_y^c$. Finally, in both $p_{XY}$ and $q_{XY}$, order the elements corresponding to indices in $I_y$ such that
\begin{equation}
	q_{XY}(x+1,y) \leq q_{XY}(x,y), \quad \forall x \in I_y \setminus \left\lbrace |I_y| \right\rbrace,
\end{equation}
and do the same for the elements corresponding to indices in $I_y^c$, so that we also have
\begin{equation}
	q_{XY}(x+1,y) \leq q_{XY}(x,y), \quad \forall x \in I_y^c \setminus \left\lbrace |I_y^c| \right\rbrace.
\end{equation}
Note that it \textit{does not} necessarily mean that
\begin{equation}
	p_{XY}(x+1,y) \leq p_{XY}(x,y), \quad \forall x \in I_y \setminus \left\lbrace |I_y| \right\rbrace,
\end{equation}
or,
\begin{equation}
	p_{XY}(x+1,y) \leq p_{XY}(x,y), \quad \forall x \in I_y^c \setminus \left\lbrace |I_y^c| \right\rbrace.
\end{equation}
These operations correspond to permutations of elements within fixed columns of $p_{XY}$ and $q_{XY}$, in the sense that no element is transferred from one column to a different column. For an illustration, see Figure~\ref{exampleARCEproofA} of Appendix~\ref{exampleARCEproof}.

As a consequence, neither $\pr{X}{Y}{p}$ nor $\pr{X}{Y}{q}$ changes under these operations. Furthermore, the total variation distance between the probability distributions does not change either, since the exact same permutations have been performed in both $p_{XY}$ and $q_{XY}$.
\medskip

\noindent
{\bf{Step B: Walking}} We start by moving probability weights in $q_{XY}$. Specifically, we will make $q_{XY}$ less disordered in the sense of $\mathcal{X}$-majorization, by ``concentrating" probability weights, which will decrease the value of $\pr{X}{Y}{q}$ in the process. For all $y \in \mathcal{Y}$ such that $I_y$ is not empty, define the column vectors $\bos{v}^{(0)}(y):= \left\lbrace q_{XY}(x,y)\right\rbrace_{x \in \mathcal{X}}$ and $\bos{v}^{(i)}(y)\equiv \left\lbrace v_x^{(i)}(y) \right\rbrace_{x \in \mathcal{X}}$ for $i = 1, \cdots, |I_y|-1$ such that
\eqal{ \label{eq:eq4}
	& \bos{v}_1^{(i)}(y) = \bos{v}_1^{(i-1)}(y) + \left[ q_{XY}(i+1,y) - p_{XY}(i+1,y) \right], \\
	& \bos{v}_{i+1}^{(i)}(y) = p_{XY}(i+1,y), \\
	& \bos{v}_x^{(i)}(y) = \bos{v}_x^{(i-1)}(y), \quad \forall x \in \mathcal{X} \setminus \left\lbrace 1, i+1 \right\rbrace.
}
Similarly, for all $y \in \mathcal{Y}$ such that $I_y$ is empty, define the column vectors $\bos{u}^{(0)}(y):= \left\lbrace
q_{XY}(x,y)\right\rbrace_{x \in {\mathcal{X}}}$ and $\bos{u}^{(i)}(y):= \left\lbrace u_x^{(i)}(y) \right\rbrace_{x \in \mathcal{X}}$ for $i = 1, \cdots, |\mathcal{X}|-1$ such that
\eqal{ \label{eq:eq4Comp}
	& \bos{u}_1^{(i)}(y) = \bos{u}_1^{(i-1)}(y) + \delta_i, \\
	& \bos{u}_{i+1}^{(i)}(y) = \bos{u}_{i+1}^{(i-1)}(y) - \delta_i, \\
	& \bos{u}_x^{(i)}(y) = \bos{u}_x^{(i-1)}(y), \quad \forall x \in \mathcal{X} \setminus \left\lbrace 1, i+1 \right\rbrace.
}
where $\delta_i = \min \left\lbrace \bos{u}_{i+1}^{(i-1)}(y), p_{XY}(1,y)-\bos{u}_{1}^{(i-1)}(y) \right\rbrace$. The above equations model the following moves in the matrix representing $q_{XY}$: in columns for which $I_y$ is empty, at each step $i = 1, \cdots, |\mathcal{X}|-1$, we remove a probability weight $\delta_i$ from $q_{XY}(i+1,y)$ an add it into $q_{XY}(1,y)$. We do so as long as the inequality $q_{XY}(1,y) \leq p_{XY}(1,y)$ is not violated. This is in order to keep the total variation distance unchanged. Indeed, note that the equations imply that we stop adding weight into $q_{XY}(1,y)$ either when we reach $q_{XY}(1,y) = p_{XY}(1,y)$, or when $q_{XY}(i,y) = 0$ for all $i = 2, \cdots, |\mathcal{X}|$. For an illustration, see Figure~\ref{exampleARCEproofB} of Appendix~\ref{exampleARCEproof}.

From the first identity in \eqref{eq:eq6} and Lemmas \ref{lem:majEps} and \ref{lem:MajPerp},
\begin{equation}
	\bos{v}^{(i)}(y) \prec \bos{v}^{(i+1)}(y),
\end{equation}
at each step $i = 1, \cdots, |I_y|-1$, for all $y \in \mathcal{Y}$, so that
\begin{equation}
	\bos{v}^{(0)}(y) \prec \bos{v}^{(|I_y|-1)}(y),
\end{equation}
for all $y \in \mathcal{Y}$.

Similarly, we have that
\begin{equation}
	\bos{u}^{(0)}(y) \prec \bos{u}^{(|\mathcal{X}|-1)}(y).
\end{equation}
Define the matrix $\tilde{q}_{XY} \in \mathcal{P}_{\mathcal{X} \times \mathcal{Y}}$ whose columns are given by the final column-vectors $\bos{v}^{(|I_y|-1)}(y)$ when $y$ is such that $I_y$ is non-empty, and the column-vectors $\bos{u}^{(|\mathcal{X}|-1)}(y)$ else, \textit{i.e.},
\begin{equation}
    \tilde{q}_{XY}(x,y) = \bos{v}_x^{(|I_y|-1)}(y),  \quad \forall x \in \mathcal{X},
\end{equation}
for all $y$ such that $I_y$ is non-empty, and
\begin{equation}
    \tilde{q}_{XY}(x,y) = \bos{u}^{(|\mathcal{X}|-1)}(y),  \quad \forall x \in \mathcal{X},
\end{equation}
for all $y$ such that $I_y$ is empty. We therefore have that the columns of $q_{XY}$ are majorized by the corresponding columns of $\tilde{q}_{XY}$, \textit{i.e.},
\begin{equation}
	\left( q_{XY} R^{(y)} \bos{e}_{|\mathcal{Y}|} \right) \prec \left( \tilde{q}_{XY} R^{(y)} \bos{e}_{|\mathcal{Y}|} \right), \quad \forall y \in \mathcal{Y}.
\end{equation}
According to Lemma \ref{lem:cmajToMaj}, it means that $q_{XY} \cmajj \tilde{q}_{XY}$, which, using Lemma \ref{lem:CondSchurConcavity}, implies that
\begin{equation}
\pr{X}{Y}{\tilde{q}} \leq \pr{X}{Y}{q}.
\end{equation}
In other words, $\pr{X}{Y}{q}$ does not increase under the above operations. For the sake of clarity, we relabel $\tilde{q}_{XY}$ as $q_{XY}$ at this point. We therefore now have
\begin{equation}
	\Delta H_\a \leq \pr{X}{Y}{p}-\pr{X}{Y}{q}.
\end{equation}
Define the two sets
\eqal{
	J & = \left\lbrace y | q_{XY}(1,y) \geq p_{XY}(1,y) \right\rbrace, \\
	J^c & = \left\lbrace y | q_{XY}(1,y) < p_{XY}(1,y) \right\rbrace,
}
(where $J^c$ might be empty) and put the columns of $J$ ahead of the columns of $J^c$ in both $p_{XY}$ and $q_{XY}$. After all these replacements, our joint probability matrices satisfy the following relations:
\eqal{ \label{eq:condStepB.1.3}
	q_{XY}(1,y) - p_{XY}(1,y) \geq 0, & \\
	q_{XY}(x,y) - p_{XY}(x,y) \leq 0, & \quad \forall \ x \in \sx \setminus \left\lbrace 1 \right\rbrace,
}
for all $y \in J$ and
\eqal{ \label{eq:condStepB.1.4}
	q_{XY}(1,y) - p_{XY}(1,y) < 0, & \\
	q_{XY}(x,y) = 0, & \quad \forall \ x \in \sx \setminus \left\lbrace 1 \right\rbrace.
}
for all $y \in J^c$. Note that this implies
\begin{equation} \label{eq:EpsStepB.1}
	\sum_{y \in J} \left( q_{XY}(1,y) - p_{XY}(1,y) \right) = t \equiv {\mathrm{TV}}(p_{XY}, q_{XY}).
\end{equation}
\medskip

\noindent
{\bf{Step C: Enlarging}} We will now apply a complementary operation to $p_{XY}$, this time making the probabilities more ``spread out" within each column, making the probability distribution more disordered in the sense of $\mathcal{X}$-majorization, which will now increase the value of $\pr{X}{Y}{p}$. In order to do this, we increase the dimensions of (the matrix representing) $p_{XY}$. The extra dimensions which are added will only be exploited in the context of the proof, and will not influence the result. Note that since we increase the dimension of $p_{XY}$, we also do the same for $q_{XY}$ by just appending zeros in the extra dimensions. We do this in order to always compare matrices of similar dimensions. Denote by $\mathcal{P}'$ the space of matrices of dimensions $(2|\mathcal{X}|-1) \times |\mathcal{Y}|$, and define $q'_{XY}$ and $p'_{XY}$ in $\mathcal{P}'$ such that
\begin{equation} \label{eq:condStepB.2.2}
  q'_{XY}(x,y) = \begin{cases}
    q_{XY}(x,y), & \forall \ x = 1, \cdots, |\mathcal{X}|, \\
    0, & \forall \ x = |\mathcal{X}| + 1, \cdots, 2|\mathcal{X}|-1,
  \end{cases}
\end{equation}
for all $y \in \mathcal{Y}$, and similarly,
\begin{equation} \label{eq:condStepB.2.3}
  p'_{XY}(x,y) = \begin{cases}
    p_{XY}(x,y), & \forall \ x = 1, \cdots, |\mathcal{X}|, \\
    0, & \forall \ x = |\mathcal{X}| + 1, \cdots, 2|\mathcal{X}|-1,
  \end{cases}
\end{equation}
for all $y \in \mathcal{Y}$. At this point, $\Delta H_\a$ is upper-bounded by $\pr{X}{Y}{p'}-\pr{X}{Y}{q'}$. Next, we make $p'_{XY}$ more disordered, therefore increasing the value of $\pr{X}{Y}{p'}$, without changing $q'_{XY}$ and without changing the total variation distance between the two matrices. Construct the matrix $p''_{XY}$ in $\mathcal{P}'$ such that
\begin{equation} \label{eq:condStepB.2.1.1}
  p''_{XY}(1,y) = p'_{XY}(1,y),
\end{equation}
\begin{equation} \label{eq:condStepB.2.1.2}
  p''_{XY}(x,y) = q'_{XY}(x,y),
\end{equation}
for all $x = 2, \cdots, |\mathcal{X}|$ and
\begin{equation} \label{eq:condStepB.2.1.3}
  p''_{XY}(x,y) = p'_{XY}(x+1-|\mathcal{X}|,y) - q'_{XY}(x+1-|\mathcal{X}|,y),
\end{equation}
for all $x = |\mathcal{X}| + 1, \cdots, 2|\mathcal{X}|-1$, for all $y \in \mathcal{Y}$. For an illustration, see Figure~\ref{exampleARCEproofC} of Appendix~\ref{exampleARCEproof}.

Using Lemmas \ref{lem:MajPerp} and \ref{lem:majEpsComp}, we have that each column of $p''_{XY}$ is majorized by the corresponding column of $p'_{XY}$. In other words, using the notations of $\mathcal{X}$-majorization,
\begin{equation}
	\left( p''_{XY} R^{(y)} \bos{e}_{|\mathcal{Y}|} \right) \prec \left( p'_{XY} R^{(y)} \bos{e}_{|\mathcal{Y}|} \right), \quad \forall \ y \in \mathcal{Y},
\end{equation}
so that according to Lemma \ref{lem:cmajToMaj}, $p''_{XY} \cmajj p'_{XY}$, and using Lemma \ref{lem:CondSchurConcavity}, $\pr{X}{Y}{p'} \leq \pr{X}{Y}{p''}$. We have at this point
\begin{equation} \label{eq:DeltaHa}
	\Delta H_\a \leq \pr{X}{Y}{p''}-\pr{X}{Y}{q'}.
\end{equation}
Furthermore, we have
\begin{equation} \label{eq:epsPrime}
	\tilde{t} \coloneqq \sum_{y=1}^{|\mathcal{Y}|} \sum_{x=|\mathcal{X}|+1}^{2|\mathcal{X}|-1} p''_{XY}(x,y) \leq t.
\end{equation}
In fact, $\tilde{t} = t$ if $J^c$ is empty. Equivalently,
\begin{equation} \label{eq:epsPrime2}
	\tilde{t}= t - \sum_{y \in J^c} (p'_{XY}(1,y)-q'_{XY}(1,y)).
\end{equation}
\medskip

\noindent
{\bf{Step D: Bounding the terms individually}}
\smallskip

{\bf{Step D.1: Upper bounding $\pr{X}{Y}{p''}$}} Now, we focus on $p''_{XY}$. We have
\eqaln{
  \pr{X}{Y}{p''} & = \frac{\a}{1-\a} \log \left( \sum_{y \in \mathcal{Y}} \left[ \sum_{x=1}^{2|\mathcal{X}|-1} (p''_{XY}(x,y))^{\a} \right]^{1/\a} \right) \\
	& = \frac{1}{1-\a} \log \left( \left\lVert \sum_{x=1}^{2|\mathcal{X}|-1} (p''_{XY}(x, \cdot))^{\a} \right\rVert_{1/\a} \right) \\
  & = \frac{1}{1-\a} \log \left( \left\lVert \sum_{x=1}^{|\mathcal{X}|} (p''_{XY}(x, \cdot))^{\a} \right. \right. \\
  & \hspace{2.2cm} \left. \left. + \sum_{x=|\mathcal{X}|+1}^{2|\mathcal{X}|-1} (p''_{XY}(x, \cdot))^{\a} \right\rVert_{1/\a} \right) \\
	& \leq \frac{1}{1-\a} \log \left( \left\lVert \sum_{x=1}^{|\mathcal{X}|} (p''_{XY}(x, \cdot))^{\a} \right\rVert_{1/\a} \right. \\
  & \hspace{2.2cm} \left. + \left\lVert \sum_{x=|\mathcal{X}|+1}^{2|\mathcal{X}|-1} (p''_{XY}(x, \cdot))^{\a} \right\rVert_{1/\a}\right) \\
	& \leq \frac{1}{1-\a} \log \left( \left\lVert \sum_{x=1}^{|\mathcal{X}|} (p''_{XY}(x, \cdot))^{\a} \right\rVert_{1/\a} \right. \\
  & \hspace{2.2cm} \left. + \sum_{x=|\mathcal{X}|+1}^{2|\mathcal{X}|-1} \left\lVert (p''_{XY}(x, \cdot))^{\a} \right\rVert_{1/\a}\right)
}
where the inequalities follow from the Minkowski inequality for $1/\a > 1$, and the monotonicity of the logarithm. The second term inside the $\log$ on the right-hand side of the last inequality is
\begin{equation*}
	\sum_{x=|\mathcal{X}|+1}^{2|\mathcal{X}|-1} \left\lVert (p''_{XY}(x, \cdot))^{\a} \right\rVert_{1/\a} = \sum_{x=|\mathcal{X}|+1}^{2|\mathcal{X}|-1} \left( \sum_{y \in \mathcal{Y}} p''_{XY}(x,y) \right)^{\a}.
\end{equation*}
The elements of $p''_{XY}$ involved in the above equation satisfy \eqref{eq:epsPrime}, so that
\eqaln{
  & \left( \frac{\tilde{t}}{|\mathcal{X}|-1}, \cdots, \frac{\tilde{t}}{|\mathcal{X}|-1} \right) \\
  & \prec \left( \sum_{y \in \mathcal{Y}} p''_{XY}(|\mathcal{X}|+1,y), \cdots, \sum_{y \in \mathcal{Y}} p''_{XY}(2|\mathcal{X}|-1,y) \right),
}
where both the vectors involved in the above equation contain $|\mathcal{X}|-1$ elements. Since the $\a$-quasi-norm is Schur-concave for $\a < 1$ (see proof of Lemma \ref{lem:CondSchurConcavity}), we get that
\eqal{
  & \left\lVert \left( \sum_{y \in \mathcal{Y}} p''_{XY}(|\mathcal{X}|+1,y), \cdots, \sum_{y \in \mathcal{Y}} p''_{XY}(2|\mathcal{X}|-1,y) \right) \right\rVert_{\a} \\
  & \leq \left\lVert \left( \frac{\tilde{t}}{|\mathcal{X}|-1}, \cdots, \frac{\tilde{t}}{|\mathcal{X}|-1} \right) \right\rVert_{\a},
}
or,
\begin{equation}
	\sum_{x=|\mathcal{X}|+1}^{2|\mathcal{X}|-1} \left( \sum_{y \in \mathcal{Y}} p''_{XY}(x,y) \right)^{\a} \leq \left( |\mathcal{X}|-1 \right)^{1-\a} \tilde{t}^\a,
\end{equation}
which leads to
\eqaln{
  & \pr{X}{Y}{p''} \\
  & \leq \frac{1}{1-\a} \log \left( \left\lVert \sum_{x=1}^{|\mathcal{X}|} (p''_{XY}(x, \cdot))^{\a} \right\rVert_{1/\a} + \left( |\mathcal{X}|-1 \right)^{1-\a} \tilde{t}^\a \right).
}

At this point, we define $r_{XY} \in \mathcal{P}$ such that
\begin{equation}
	r_{XY}(x,y) = p''_{XY}(x,y), \quad \forall \ y \in \mathcal{Y}, \quad \forall \ x \in \mathcal{X},
\end{equation}
or, using \eqref{eq:condStepB.2.1.1}, \eqref{eq:condStepB.2.1.2} and \eqref{eq:condStepB.2.1.3},
\eqal{ \label{eq:condStepB.3}
	r_{XY}(1,y) = p'_{XY}(1,y), & \\
	r_{XY}(x,y) = q'_{XY}(x,y), & \quad \forall \ x = 2, \cdots, |\mathcal{X}|,
}
for all $y \in \mathcal{Y}$, so that
\eqal{ \label{eq:Hpsec}
  & \pr{X}{Y}{p''} \\
  & \leq \frac{1}{1-\a} \log \left( \left\lVert \sum_{x=1}^{|\mathcal{X}|} (r_{XY}(x, \cdot))^{\a} \right\rVert_{1/\a} + \left( |\mathcal{X}|-1 \right)^{1-\a} \tilde{t}^\a \right).
}
Note that
\begin{equation} \label{eq:epsPrime3}
	\sum_{y=1}^{|\mathcal{Y}|} \sum_{x=1}^{|\mathcal{X}|} r_{XY}(x,y) = 1-\tilde{t} \geq 1-t.
\end{equation}
\medskip

{\bf{Step D.2: Lower bounding $\pr{X}{Y}{q'}$}} We now turn to $q'_{XY}$. We have
\eqaln{
	\pr{X}{Y}{q'} & = \frac{\a}{1-\a} \log \left( \sum_{y \in \mathcal{Y}} \left[ \sum_{x=1}^{|\mathcal{X}|} (q'_{XY}(x,y))^{\a} \right]^{1/\a} \right) \\
  & = \frac{\a}{1-\a} \log \left( \sum_{y \in J} \left[ \sum_{x=1}^{|\mathcal{X}|} (q'_{XY}(x,y))^{\a} \right]^{1/\a} \right. \\
  & \hspace{2.2cm} \left. + \sum_{y \in J^c} \left[ \sum_{x=1}^{|\mathcal{X}|} (q'_{XY}(x,y))^{\a} \right]^{1/\a} \right) \\
	& = \frac{\a}{1-\a} \log \left( \sum_{y \in J} \left[ \sum_{x=1}^{|\mathcal{X}|} (q'_{XY}(x,y))^{\a} \right]^{1/\a} \right. \\
  & \hspace{2.2cm} \left. + \sum_{y \in J^c} q'_{XY}(1,y) \right)
}
where we made use of \eqref{eq:condStepB.1.4}. Define $q''_{XY} \in \mathcal{P}$ such that
\eqal{
	q''_{XY}(x,y) = q'_{XY}(x,y)-r_{XY}(x,y), & \quad \forall \ y \in J, \quad x = 1, \\
	q''_{XY}(x,y) = 0, & \quad \mathrm{else}.
}
The elements of $q''_{XY}$ are all non-negative as a consequence of \eqref{eq:condStepB.3}, \eqref{eq:condStepB.2.2}, \eqref{eq:condStepB.2.3} and \eqref{eq:condStepB.1.3}. Furthermore, using \eqref{eq:EpsStepB.1}, we have that
\begin{equation} \label{eq:EpsStepB.4}
	\sum_{y \in J} q''_{XY}(1,y) = t.
\end{equation}
Using this, we can write
\eqal{
	\pr{X}{Y}{q'} & = \frac{\a}{1-\a} \log \left( \sum_{y \in J} \left\lVert r_{XY}(.,y) + q''_{XY}(.,y) \right\rVert_{\a} \right. \\
  & \hspace{2.2cm} \left. + \sum_{y \in J^c} q'_{XY}(1,y) \right)
}
and using the reverse Minkowski inequality for $\a < 1$, and the monotonicity of the logarithm, we end up with
\eqaln{
  & \exp \left( \frac{1-\a}{\a} \pr{X}{Y}{q'} \right) \\
  & \geq \sum_{y \in J} \left\lVert r_{XY}(.,y) \right\rVert_{\a} + \sum_{y \in J} \left\lVert q''_{XY}(.,y) \right\rVert_{\a} + \sum_{y \in J^c} q'_{XY}(1,y) \\
	& = \sum_{y \in J} \left\lVert r_{XY}(.,y) \right\rVert_{\a} + \sum_{y \in J} q''_{XY}(1,y) + \sum_{y \in J^c} q'_{XY}(1,y) \\
	& = \sum_{y \in \mathcal{Y}} \left\lVert r_{XY}(.,y) \right\rVert_{\a} - \sum_{y \in J^c} r_{XY}(1,y) + t + \sum_{y \in J^c} q'_{XY}(1,y)
}
using \eqref{eq:EpsStepB.4}. Now, from \eqref{eq:condStepB.3} and \eqref{eq:epsPrime2}, we obtain
\begin{equation}
	t + \sum_{y \in J^c} q'_{XY}(1,y) - \sum_{y \in J^c} r_{XY}(1,y) = \tilde{t},
\end{equation}
so that
\begin{equation} \label{eq:Hqpri}
	\pr{X}{Y}{q'} \geq \frac{\a}{1-\a} \log \left( \sum_{y \in \mathcal{Y}} \left\lVert r_{XY}(.,y) \right\rVert_{\a} + \tilde{t} \right).
\end{equation}

{\bf{Step E: Upper bounding the difference}} Putting \eqref{eq:DeltaHa}, \eqref{eq:Hpsec} and \eqref{eq:Hqpri} together, we see that our initial conditional entropy difference $\Delta H_\a$ is upper bounded as follows:
\eqaln{
\Delta H_\a & \leq \frac{1}{1-\a} \left[ \log \left( R(r_{XY})^\a + \left( |\mathcal{X}|-1 \right)^{1-\a} \tilde{t}^\a \right) \right. \\
& \hspace{1.5cm} \left. - \log \left( \vphantom{R(r_{XY})^\a + \left( |\mathcal{X}|-1 \right)^{1-\a} \tilde{t}^\a} \left[ R(r_{XY}) + \tilde{t}\right]^\a \right) \right],
}
where we defined $R(r_{XY}) = \sum_{y \in \mathcal{Y}} \left\lVert r_{XY}(.,y) \right\rVert_{\a}$. Since $f(u) = u^\a$ is concave for $u \geq 0$, $\a \in [0,1)$, and $f(0) = 0$, it is also subadditive, so that
\eqaln{
	R(r_{XY}) & = \sum_{y \in \mathcal{Y}} \left( \sum_{x \in \mathcal{X}} r_{XY}^{\a}(x,y) \right)^{1/\a} \\
	& \geq \sum_{y \in \mathcal{Y}} \sum_{x \in \mathcal{X}} r_{XY}(x,y) \\
	& = 1 - \tilde{t}
}
where we used \eqref{eq:epsPrime3}.

Furthermore, $\tilde{t} \leq t \leq \epsilon \leq 1-1/|\mathcal{X}|$. Using this along with Lemma \ref{lem:IncDiffLog}, we have that
\eqaln{
	\Delta H_\a & \leq \frac{1}{1-\a} \left[ \log \left( \left( 1 - \tilde{t} \right)^\a + \left( |\mathcal{X}|-1 \right)^{1-\a} \tilde{t}^\a \right) \right. \\
  & \hspace{1.5cm} \left.  - \log \left( \left[ 1 - \tilde{t} + \tilde{t}\right]^\a \right) \right] \\
	& = \frac{1}{1-\a} \log \left( \left( 1 - \tilde{t} \right)^\a + \left( |\mathcal{X}|-1 \right)^{1-\a} \tilde{t}^\a \right).
}
Finally, using Lemma \ref{lem:BoundMonotonic} along with the fact that $\tilde{t} \leq \epsilon$, we conclude that
\begin{equation}
	\Delta H_\a \leq \frac{1}{1-\a} \log \left( \left( 1 - \epsilon \right)^\a + \left( |\mathcal{X}|-1 \right)^{1-\a} \epsilon^\a \right),
\end{equation}
which ends the proof of the bound. To see that the inequality is tight, one can consider the probability distributions whose elements satisfy the following relations:
\eqal{ \label{eq:qtight}
	& q_{XY}(1,1) = 1, \\
	& q_{XY}(x,1) = 0, \quad \forall \ x \in \sx \setminus \left\lbrace 1 \right\rbrace, \\
	& q_{XY}(x,y) = 0, \quad \forall \ x \in \mathcal{X}, \forall \ y \in \mathcal{Y} \setminus \left\lbrace 1 \right\rbrace,
}
and
\eqal{ \label{eq:ptight}
	& p_{XY}(1,1) = 1-\epsilon \\
	& p_{XY}(x,1) = \frac{\epsilon}{|\mathcal{X}|-1}, \quad \forall \ x \in \sx \setminus \left\lbrace 1 \right\rbrace, \\
	& p_{XY}(x,y) = 0, \quad \forall \ x \in \mathcal{X}, \forall \ y \in \mathcal{Y} \setminus \left\lbrace 1 \right\rbrace.
}

\subsection{Proof of Theorem~\ref{theo:main3}}

Our proof is analogous to the proof of the uniform continuity bound for the conditional entropy of c-q states by Wilde~\cite{Wilde2020}. It relies on the use of a conditional dephasing channel and the data processing inequality to go from a c-q setting to a classical-classical one. However, the unitality of the dephasing channel, which was exploited by Wilde in his proof, cannot be used in the case of the conditional $\a$-R\'enyi entropy. Instead we use the notion of $\mathcal{X}$-majorization. This allows us to then employ our Theorem~\ref{theo:main} to arrive at the desired result.

Consider the following decompositions for $\rho_{AY}$ and $\sigma_{AY}$:
\begin{equation} \label{eq:theo:main31}
	\rho_{AY} = \sum_{y \in \mathcal{Y}} r_Y(y) \rho_A^y \otimes \proj{y}_Y,
\end{equation}
\begin{equation} \label{eq:theo:main32}
	\sigma_{AY} = \sum_{y \in \mathcal{Y}} s_Y(y) \sigma_A^y \otimes \proj{y}_Y,
\end{equation}
where $r_Y, s_Y \in \mathcal{P}_{\mathcal{Y}}$ and $\left\lbrace \rho_A^y \right\rbrace_{y \in \mathcal{Y}}$ and $\left\lbrace \sigma_A^y \right\rbrace_{y \in \mathcal{Y}}$ are sets of states of $A$.
Suppose without loss of generality that $\pr{A}{Y}{\rho} \leq \pr{A}{Y}{\sigma}$. Define the conditional dephasing channel as
\begin{equation}
	\mathcal{N}_{AY}[\omega_{AY}] = \sum_{x \in \mathcal{X}, y \in \mathcal{Y}} \proj{\phi^y_x,y} \omega_{AY} \proj{\phi^y_x,y},
\end{equation}
where $\ket{\phi^y_x,y} \equiv \ket{\phi^y_x} \otimes \ket{y}$ and the states $\ket{y}$ and $\ket{\phi^y_x}$ are defined through \eqref{eq:rhoAY2} and \eqref{eq:rhoAY}. Here and henceforth we suppress the subscripts $A$ and $Y$ (denoting the subsystems) for notational simplicity. We have that
\begin{equation}
	\mathcal{N}_{AY}[\rho_{AY}] = \rho_{AY},
\end{equation}
while
\begin{equation}
	\mathcal{N}_{AY}[\sigma_{AY}] = \sum_{y \in \mathcal{Y}, x \in \mathcal{X}} s_Y(y) \tilde{s}_{X|Y}(x|y) \proj{\phi^y_x} \otimes \proj{y},
\end{equation}
where
\begin{equation}
	\tilde{s}_{X|Y}(x|y) = \bra{\phi^y_x} \sigma_A^y \ket{\phi^y_x}, \quad \forall x \in \mathcal{X}, \forall y \in \mathcal{Y},
\end{equation}
so that $\tilde{s}_{X|Y}$ is a conditional probability distribution. Now, for each $y \in \mathcal{Y}$, consider the spectral decomposition of $\sigma_A^y$:
\begin{equation}
	\sigma_A^y = \sum_{x \in \mathcal{X}} s_{X|Y}(x|y) \proj{\psi^y_x},
\end{equation}
where $|\mathcal{X}| = d_A$ is the dimension of system $A$, $s_{X|Y}$ is a conditional probability distribution defined as
\begin{equation}
	s_{X|Y}(x|y) = \bra{\psi^y_x} \sigma_A^y \ket{\psi^y_x}, \quad \forall x \in \mathcal{X}, \forall y \in \mathcal{Y},
\end{equation}
and $\left\lbrace \psi^y_x \right\rbrace_{x \in \mathcal{X}}$ is a set of orthonormal states for a fixed value of $y \in \mathcal{Y}$. Since for any positive semi-definite matrix, the vector of diagonal elements in any basis is majorized by the vector of eigenvalues~\cite{Majorization}, we have that
\begin{equation}
	\tilde{s}_{X|Y=y} \prec s_{X|Y=y}, \quad \forall y \in \mathcal{Y}.
\end{equation}
Define
\begin{equation} \begin{aligned}
  s_{XY}(x,y) & = s_Y(y) s_{X|Y}(x|y), \\
  \tilde{s}_{XY}(x,y) & = s_Y(y) \tilde{s}_{X|Y}(x|y), \end{aligned} \qquad \forall x\in \mathcal{X}, y \in \mathcal{Y},
\end{equation}
so that
\begin{equation}
	\tilde{s}_Y(y) = \sum_{x \in \mathcal{X}} \tilde{s}_{XY}(x,y) = s_Y(y) = \sum_{x \in \mathcal{X}} s_{XY}(x,y),
\end{equation}
for all $y \in \mathcal{Y}$. Using Lemma \ref{lem:cmajToMaj}, we have that
\begin{equation}
	\tilde{s}_{XY} \cmajj s_{XY},
\end{equation}
so that according to Lemma \ref{lem:CondSchurConcavity},
\begin{equation}
	\pr{X}{Y}{s} \leq \pr{X}{Y}{\tilde{s}}, \quad \forall \a \in [0,1).
\end{equation}
Now, using \eqref{eq:PetzToArimoto}, we have
\eqal{
  \pr{X}{Y}{r} & = \pr{A}{Y}{\rho} \\
  & \leq \pr{A}{Y}{\sigma} \\
  & = \pr{X}{Y}{s} \\
  & \leq \pr{X}{Y}{\tilde{s}}.
}
Meanwhile, from data processing inequality for the normalized trace distance,
\eqal{
	\frac{1}{2} || \rho_{AY} - \sigma_{AY} ||_1 & \geq \frac{1}{2} \left\Vert \mathcal{N}_{AY}[\rho_{AY}] - \mathcal{N}_{AY}[\sigma_{AY}] \right\Vert_1 \\
	& = \mathrm{TV}(r_{XY},\tilde{s}_{XY}),
}
so that
\begin{equation}
	\mathrm{TV}(r_{XY},\tilde{s}_{XY}) \leq \epsilon.
\end{equation}
Using Theorem \ref{theo:main}, we have
\eqal{
  & |\pr{A}{Y}{\rho} - \pr{A}{Y}{\sigma}| \\
  & = \pr{A}{Y}{\sigma} - \pr{A}{Y}{\rho} \\
	& \leq \pr{X}{Y}{\tilde{s}} - \pr{X}{Y}{r} \\
	& \leq \frac{1}{1-\a} \log \left( \left( 1- \epsilon \right)^{\a} + \left(|\mathcal{X}|-1\right)^{1-\a} \epsilon^{\a} \right).
}
In order to show that the inequality is tight, one can simply consider states that are diagonal in the same basis and whose matrix elements are given by \eqref{eq:qtight} and \eqref{eq:ptight}.

\section{Discussion and open problems}

In this paper, we have proven tight uniform continuity bounds for the $\a$-Arimoto-R\'enyi conditional entropy for $\alpha \in [0,1)$, for joint probability distributions of a pair of discrete random variables with finite alphabets, as well as for the conditional R\'enyi entropy for classical-quantum systems, with the conditioning being on the classical system. In the limit $\a \to 1$, our results yield the corresponding recently obtained bounds for the conditional Shannon entropy and the conditional entropy of a c-q state, respectively. It is interesting to note that our bound for the $\a$-ARCE is identical to the one obtained by Audenaert~\cite{Audenaert} for the unconditional R\'enyi entropy (see (7) of the Appendix of~\cite{Audenaert}). The same is true for the bound obtained by Alhejji and Smith~\cite{AlhejjiSmith} for the conditional Shannon entropy: it is identical to the bound for the Shannon entropy (see~\cite{Zhang}). It would be interesting to see whether an intuitive reason for this can be found.
\smallskip

A natural next step would be to investigate the continuity of the $\a$-conditional R\'enyi entropy of bipartite quantum systems or classical-quantum systems with the conditioning being on the quantum system, for which there are no results even for the quantum conditional entropy. Finding analogues of our results for $\a >1$ also remains open.

\appendices

\section{Conditional majorization \label{CondMajGour}}

The following notion of conditional majorization was introduced by Gour {\em{et al}} in~\cite{CondMaj}.

\begin{defi} \label{defi:condMajGour}
Denote by $\mathbb{R}^{n \times l}_+$ the set of all $n \times l$ matrices with non-negative values. Consider $P \in \mathbb{R}^{n \times l}_+$ and $Q \in \mathbb{R}^{n \times m}_+$. We say $Q$ is conditionally majorized by $P$, written $Q \prec_{\scaleto{c}{3pt}} P$, if there exist matrices $D^{(j)}$ and $R^{(j)}$, where $j$ can run over an arbitrary number of values, such that
\begin{equation}
	Q = \sum_j D^{(j)} P R^{(j)},
\end{equation}
where each $D^{(j)}$ is an $n \times n$ doubly-stochastic matrix and each $R^{(j)}$ is an $l \times m$ matrix of non-negative entries, with $\sum_j R^{(j)}$ row-stochastic.
\end{defi}

The following necessary and sufficient condition for conditional majorization was proven by Gour {\em{et al}} ( see Theorem 1 of~\cite{CondMaj}):

\begin{lem} \label{lem:CharacCondMajGour}
Let $p_{XY}, q_{XY} \in \mathcal{P}_{\mathcal{X} \times \mathcal{Y}}$. We have that $q_{XY} \prec_{\scaleto{c}{3pt}} p_{XY}$ if and only if for all convex symmetric functions $\Phi$
\begin{equation}
    \sum_{y \in \mathcal{Y}} p_Y(y) \Phi\left( p_{X|Y=y} \right) \geq \sum_{y \in \mathcal{Y}} q_Y(y) \Phi\left( q_{X|Y=y} \right).
\end{equation}
\end{lem}

\section{Further tools for the proofs}

\begin{lem} \label{lem:majEps}
Consider a vector $\bos{v} \in \mathbb{R}^n$ whose elements are non-negative and are arranged in non-increasing order. For some $i \in \left\lbrace 2, \cdots, n \right\rbrace$ and some $s \in (0, v_i]$, define the vector $\bos{v}^{(i)} \in \mathbb{R}^n$ whose elements satisfy the following relations:
\eqs{
	v^{(i)}_1 & = v_1 + s, \\
	v^{(i)}_i & = v_i - s, \\
	v^{(i)}_j & = v_j, \quad \forall j \in \left\lbrace 2, \cdots, n \right\rbrace \setminus \left\lbrace i \right\rbrace.
}
Then $\bos{v} \prec \bos{v}^{(i)}$.
\end{lem}
\begin{proof}
We have that $v^{(i)}_j \geq v^{(i)}_{j+1}$ for all $j \in \left\lbrace 1, \cdots, i-1 \right\rbrace$. Now, there exists $k \in \left\lbrace i+1, \cdots, n \right\rbrace$ such that $v^{(i)}_i \leq v^{(i)}_j$ for all $j \in \left\lbrace 1, \cdots, k \right\rbrace$ and $v^{(i)}_i > v^{(i)}_j$ for all $j \in \left\lbrace k+1, \cdots, n \right\rbrace$, so that the vector $\bos{v}^{(i)\downarrow} \coloneqq \left( v^{(i)}_1, \cdots, v^{(i)}_{i-1}, \cdots, v^{(i)}_k, v^{(i)}_i, v^{(i)}_{k+1}, \cdots, v^{(i)}_n \right)^{\mathrm{T}}$ has its elements arranged in non-increasing order. In that case,
\begin{equation}
	\sum_{j=1}^{l} v^{(i)\downarrow}_j = \sum_{j=1}^{l} v_j + s > \sum_{j=1}^{l} v_j, \quad \forall l = 1, \cdots i-1.
\end{equation}
Using the equality in the above equation, we have for all $l = i, \cdots k-1$,
\eqal{ \label{eq:lem:majEps}
	\sum_{j=1}^{l} v^{(i)\downarrow}_j & = \sum_{j=1}^{i-1} v_j + s + \sum_{j=i}^{l} v^{(i)\downarrow}_j \\
	& = \sum_{j=1}^{i-1} v_j + s + \sum_{j=i+1}^{l+1} v_j \\
	& = \sum_{j=1}^l v_j + (s - v_i) + v_{l+1} \\
	& > \sum_{j=1}^l v_j.
}
In order to obtain the last inequality in the above set of equations, notice that $v^{(i)}_i \leq v^{(i)}_k$, or, $v_i-s \leq v_k$. Now, $l+1 \leq k$, and since the elements of $\bos{v}$ are sorted in non-increasing order, $v_{l+1} \geq v_k$, so that $v_i-s \leq v_{l+1}$, which implies the last inequality in \eqref{eq:lem:majEps}. Similarly, using the last equality in the above equation,
\eqal{
	\sum_{j=1}^{k} v^{(i)\downarrow}_j & = \sum_{j=1}^{k-1} v^{(i)\downarrow}_j + v^{(i)\downarrow}_k \\
	& = \sum_{j=1}^{k-1} v_j + (s - v_i) + v_k + (v_i - s) \\
	& = \sum_{j=1}^{k} v_j.
}
Finally, using the above equation, we have for all $l = k+1, \cdots n$,
\eqal{
	\sum_{j=1}^{l} v^{(i)\downarrow}_j & = \sum_{j=1}^{k} v^{(i)\downarrow}_j + \sum_{j=k+1}^{l} v^{(i)\downarrow}_j \\
	& = \sum_{j=1}^{k} v_j + \sum_{j=k+1}^{l} v_j \\
	& = \sum_{j=1}^{l} v_j.
}
This ends the proof.
\end{proof}

\begin{lem} \label{lem:MajPerp}
Let $\bos{v}, \bos{v}', \bos{v}^{\perp} \in \mathbb{R}^n$ be three vectors with non-negative entries, such that $\bos{v}$ and $\bos{v}'$ are in the same subspace whereas $\bos{v}^{\perp}$ is in an orthogonal subspace. If $\bos{v}' \prec \bos{v}$, then $\bos{v}'+\bos{v}^{\perp} \prec \bos{v}+\bos{v}^{\perp}$.
\end{lem}
\begin{proof}
If $\bos{v}' \prec \bos{v}$, there exist $\mu_j \in \mathbb{R}$ such that $\mu_j \geq 0, \forall j$ and $\sum_j \mu_j = 1$ and permutation matrices $\Pi_j$ of dimension $n$ such that
\begin{equation}
	\bos{v}' = \sum_j \mu_j \Pi_j \bos{v}.
\end{equation}
The permutations $\Pi_j$ act only on the subspace spanning $\bos{v}$ and $\bos{v}'$. As a consequence,
\begin{equation}
\bos{v}'+\bos{v}^{\perp} = \sum_j \mu_j \Pi_j \left( \bos{v}+\bos{v}^{\perp} \right),
\end{equation}
so that $\bos{v}'+\bos{v}^{\perp} \prec \bos{v}+\bos{v}^{\perp}$.
\end{proof}

\begin{lem} \label{lem:majEpsComp}
Consider a vector $\bos{v} \in \mathbb{R}^{n+m}$ whose elements are non-negative and satisfy $v_j = 0$ for all $j \in \left\lbrace n+1, \cdots, n+m \right\rbrace$. For some $i \in \left\lbrace 1, \cdots, n \right\rbrace$, some $j \in \left\lbrace n+1, \cdots, n+m \right\rbrace$ and some $s \in (0, v_i]$, define the vector $\bos{u} \in \mathbb{R}^n$ whose elements satisfy the following relations:
\eqs{
	u_i & = v_i - s, \\
	u_j & = s, \\
	u_k & = v_k, \quad \forall k \in \left\lbrace 1, \cdots, n+m \right\rbrace \setminus \left\lbrace i,j \right\rbrace.
}
Then $\bos{u} \prec \bos{v}$.
\end{lem}
\begin{proof}
Define the vectors $\bos{\tilde{v}}$, $\bos{\tilde{u}}$ and $\bos{v}^{\perp} \in \mathbb{R}^{n+m}$ whose elements are non-negative and satisfy the following relations:
\eqs{
	\tilde{v}_i & = v_i, \\
	\tilde{v}_k & = 0, \quad \forall k \in \left\lbrace 1, \cdots, n+m \right\rbrace \setminus \left\lbrace i \right\rbrace,
}
\eqs{
	\tilde{u}_i & = v_i - s, \\
	\tilde{u}_j & = s, \\
	\tilde{u}_k & = 0, \quad \forall k \in \left\lbrace 1, \cdots, n+m \right\rbrace \setminus \left\lbrace i,j \right\rbrace,
}
and
\eqs{
	v^{\perp}_i & = 0, \\
	v^{\perp}_j & = 0, \\
	v^{\perp}_k & = v_k, \quad \forall k \in \left\lbrace 1, \cdots, n+m \right\rbrace \setminus \left\lbrace i,j \right\rbrace.
}
Note that $\bos{v} = \bos{\tilde{v}} + \bos{v}^{\perp}$ and $\bos{u} = \bos{\tilde{u}} + \bos{v}^{\perp}$. Now, we trivially have that $\bos{\tilde{u}} \prec \bos{\tilde{v}}$, so that, using Lemma \ref{lem:MajPerp}, $\bos{u} \prec \bos{v}$.
\end{proof}

\begin{lem} \label{lem:IncDiffLog}
Let $\a \in [0,1)$, $|\mathcal{X}| \geq 1$ and $\tilde{t} \in (0, 1-\frac{1}{|\mathcal{X}|}]$. The function $f_{\a,|\mathcal{X}|,\tilde{t}} : [0, \infty) \rightarrow [0, \infty)$ defined as
\begin{equation}
	f_{\a,|\mathcal{X}|,\tilde{t}}(u) \coloneqq \log \left[ u^\a + \left( |\mathcal{X}|-1 \right)^{1-\a} \tilde{t}^\a \right] - \log \left[ \left( u + \tilde{t} \right)^\a \right]
\end{equation}
is monotonically decreasing for $u \geq 1-\tilde{t}$.
\end{lem}
\begin{proof}
We simply compute the derivative
\eqaln{
  & \frac{\partial}{\partial u} f_{\a,|\mathcal{X}|,\tilde{t}}(u) \\
  & = \frac{\partial}{\partial u} \left( \log \left[ u^\a + \left( |\mathcal{X}|-1 \right)^{1-\a} \tilde{t}^\a \right] - \log \left[ \left( u + \tilde{t} \right)^\a \right] \right) \\
	& = \left[ u^\a + \left( |\mathcal{X}|-1 \right)^{1-\a} \tilde{t}^\a \right]^{-1} \a u^{\a-1} - \alpha(u + \tilde{t})^{-1}
}
which will be $\leq 0$ if and only if
\begin{align}
    \frac{u^{\a-1}}{u^\a + \left( |\mathcal{X}|-1 \right)^{1-\a} \tilde{t}^\a}
    & \leq \frac{1}{u + \tilde{t}},
\end{align}
which is easily seen to reduce to the condition
\begin{align}\label{cond}
   u \geq \tilde{t}/(|\mathcal{X}|-1)
\end{align}
Note that since $\tilde{t} \in (0, 1-\frac{1}{|\mathcal{X}|}]$, we have $1-\tilde{t}\geq \tilde{t}/(|\mathcal{X}| -1)$. Hence for $u \geq 1 - \tilde{t}$ the required condition \eqref{cond} holds.

\end{proof}

\begin{lem} \label{lem:BoundMonotonic}
Let $\a \in [0,1)$ and $|\mathcal{X}| \geq 1$. The function $g_{\a,|\mathcal{X}|} : (0, 1-\frac{1}{|\mathcal{X}|}] \rightarrow [0, \infty)$ defined as
\begin{equation}
	g_{\a,|\mathcal{X}|}(u) \coloneqq \left( 1 - u \right)^\a + \left( |\mathcal{X}|-1 \right)^{1-\a} u^\a
\end{equation}
is monotonically increasing in $u$.
\end{lem}
\begin{proof}
Again, we simply compute the derivative
\eqaln{
	\frac{\partial}{\partial u} g_{\a,|\mathcal{X}|}(u) & = \frac{\partial}{\partial u} \left[ \left( 1 - u \right)^\a + \left( |\mathcal{X}|-1 \right)^{1-\a} u^\a \right] \\
	& = - \a \left( 1 - u \right)^{\a-1} + \left( |\mathcal{X}|-1 \right)^{1-\a} \a u^{\a-1}
}
which is $\geq 0$ if and only if the following inequality holds:
\begin{align}\label{cond2}
	\left( |\mathcal{X}|-1 \right)^{1-\a} u^{\a-1} & \geq \left( 1 - u \right)^{\a-1}.
	\end{align}
	The above inequality can easily be seen to reduce to the following:
	\begin{align}
	    \left( (|\mathcal{X}|-1) (1-u)/u \right)^{1-\a} & \geq 1,
	\end{align}
	which in turn reduces to the condition $u \leq 1 - \frac{1}{|\mathcal{X}|}$, which holds by the hypothesis of the lemma.

\end{proof}

\section{Illustration of the steps in the proof of Theorem~\ref{theo:main}  \label{exampleARCEproof}}

In this section of the appendix, we consider a simple $3 \times 3$ example in order to illustrate the steps in the proof of Theorem~\ref{theo:main}. In the context of this example, we relabel the matrix $q_{XY}$ simply as $q$ and its elements as $q_{ij}$. Similarly, we relabel the matrix $p_{XY}$ simply as $p$ and its elements as $p_{ij}$. The example is illustrated in the three figures given below.
\begin{figure}[ht]
\centering
	\includegraphics[width=.9\columnwidth]{exampleARCEproofA.jpg}
	\caption{\label{exampleARCEproofA} Step A of the proof of Theorem~\ref{theo:main}, which corresponds to a reordering of some elements of $q$ and $p$.}
\end{figure}
\begin{figure}[ht]
\centering
	\includegraphics[width=.9\columnwidth]{exampleARCEproofB.jpg}
	\caption{\label{exampleARCEproofB} Step B of the proof of Theorem~\ref{theo:main}, in which some probability weights are moved in $q$, while $p$ remains unchanged. In this example, we consider that after the moves, we end up with a new matrix $q$ that satisfies $q_{13} < p_{13}$, $q_{23} = 0$ and $q_{33} = 0$. This will affect Step C, see Figure~\ref{exampleARCEproofC}.}
\end{figure}
\begin{figure}[ht!]
\centering
	\includegraphics[width=.9\columnwidth]{exampleARCEproofC.jpg}
	\caption{\label{exampleARCEproofC} Step C of the proof of Theorem~\ref{theo:main}, in which we increase the dimensions of $q$ and $p$, and some probability weights are moved in $p$.}
\end{figure}

\section*{Acknowledgment}

M. G. J. acknowledges support from the Wiener-Anspach Foundation.

\bibliographystyle{IEEEtran}

redpajama_set_name: RedPajamaArXiv


text:
Family responds to officer taking swing at teen
HOUSTON Since we first showed you the video on Thursday, the family of the suspected robber has come forward, and they're speaking out against the actions of that police officer.
The target of that punch was a 16-year-old boy. His family members say he called them from jail to say the officer busted his lip.
It's the video SkyEye HD captured of a Houston police officer taking a swing at a handcuffed suspect that has Mary Pointer and her family upset.
"That's sad. That's pitiful," she said.
Pointer says the restrained suspect seen in the white t-shirt is her 16-year-old nephew. The police chief says the female officer seen throwing a jab at the teen is Angela Horton, a four-year veteran on the force.
Pointer said, "I just don't think he was supposed to be getting hit, and he had handcuffs on.'
Eyewitness News learned the teen was one of three suspected armed robbers who targeted the SK Food Store on Cavalcade Thursday morning.
"He was wrong for that," Pointer admitted. "This here don't justify what he done. He know he did wrong."
SkyEye HD was over the scene right after the foot chase ended. Two officers are seen handcuffing the teen and taking him into custody. Seconds later, Officer Horton approaches the teen with a right jab toward his face.
"He didn't do nothing to her," said the teen's sister, Keziah Kibble. "So why would she hit him for nothing?"
When we talked to Houston Police Chief Charles McClelland about the video on Thursday, he told us that supervisors quickly launched an Internal Affairs investigation and Officer Horton was relieved of duty.
"That means that her duties are suspended," Chief McClelland explained. "She will not be coming to work. She will be at home. She still will be receiving pay."
That's not enough for Pointer and other family members.
"She's still being paid," Pointer said. "She might as well be working."
Pointer says she believes the officer took it too far with her nephew, and her family is demanding more action.
"I hope justice be done," she said.
So far calls to the police union about this video and Officer Horton have not yet been returned.
On Friday afternoon Officer Horton's attorney issued the following statement:
"Officer Horton is a 36 year old, five year veteran of the Houston Police Department. She was in cadet class 188 at Houston Police Department Academy and graduated in 2007. She served as an airman in the United States Air Force from 1994 to 2003. From 1995 to 2003 she served as a military police officer.
Officer Horton admits that for a few brief seconds yesterday, she let her emotions get the best of her and may have acted in a way that is contrary to her training as a Houston Police Department patrol officer.
As a police officer who is charged with the protection of the public, her biggest disappointment is that her actions have overshadowed the fact that a serious crime was committed. It is important to remember that this incident occurred at the end of a 15 minute chase in which multiple suspects, who it was reported were armed with handguns and fleeing the scene of an Aggravated Robbery, were attempting to allude [sic] police by climbing onto a moving train. She hopes that the public will remember that police officers have to make split second decisions in ever-changing and dangerous situations. She looks forward to having this case evaluated so that all the facts can be known.
Because of ongoing administrative investigations, no further statements will be made."

redpajama_set_name: RedPajamaCommonCrawl


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text:
Brief Holiday Special: Do Muslims Do Thanksgiving?
What do Muslims do for Thanksgiving? Well, most of them probably go to work or school. But what about the several million who live in the U.S.? Do families celebrate with get togethers and eating halal turkey? Do they watch football? Presumably Robert Saleh will. He's the head coach of the New York Jets and a Muslim of Lebanese descent. Even though no matter which of the six teams playing today win, he'll still be second to last in the league!
Khalil Abdur-Rashid the Muslim Chaplain at Harvard University says, "Thanksgiving. . .affords people who are normally heedless and forgetful of the myriad of blessings, of which we are recipients, to at least once a year acknowledge those blessings. Consequently, Thanksgiving, while not directly legislated by Islam, is at least in essence compliant with the spirit of Islam."
He goes on, "In most houses, the practice of circling the dinner table with family, expressing what we are thankful for, and why we are thankful for it, is a way of not only expressing thanks to Allah, but also declaring and announcing our gratitude to others. It is an expression of the verse in the Quran, "And as for the blessings (ni'mat) of your Lord, proclaim them!" (Quran 93:11)."
You are a blessing! I'm thankful to God for you and grateful to you for reading Muslim Connect.
Maybe there are some Muslims on the way to celebrate Thanksgiving at your house right now, maybe tomorrow, maybe Christmas or next year. That would be a blessing to them and to you and your family. I'm grateful for that as well.
Vests of Death: Why Do Suicide Bombers Do It?
Three suicide bombers blew themselves up in Kampala, Uganda earlier this week, killing three others, injuring dozens and permanently altering the lives of many.
When we hear of a suicide bombing, our hearts go out to the victims and their families. That's natural and appropriate. But do we ever wonder about the man or woman who killed themself?
We've got busy lives and maybe it seems a waste to give thought to those who choose to kill themselves in an evil effort to kill others. I get that, but sometimes I wonder: What do they want? What do they expect to happen? What could be so important or make them so desperate? As my friend David Weston asks in his provocative essay, "[what is the] reasoning behind the motivation that leads a person to literally give up their lives by taking the lives of others, often in a most violent and gruesome way?"
I'm not satisfied with answers like, "That's the way Muslims are," or "Of course they do it. They instantly get 70 virgins in paradise." Almost no Muslims are that way, including the cowards who recruit, coordinate and deploy the bombers. And the women who strap on the vest don't do it for the virgins.
However true those motives are, I wonder what else might be going on.
Author and professor, Adam Lankford contends that Muslim suicide bombers are people who want to kill themselves, but need an "Allah approved" way to do it. When you want to die, but Islam tells you that taking your own life is a ticket to Hell, maybe you look for another way.
How many are motivated by revenge similar to Samson, arguably one of the very first recorded suicide attackers? And if so, revenge for what?
And how many die believing, hoping their sacrifice expands the kingdom and glory of God?
Could you do me a favor? I had an article published this week on the very popular Denison Forum site. Please give it a look and a comment. I'd appreciate it.
The Richest Muslim in the World
On this Veteran's Day, I'm grateful for Muslim Connect readers who've served our country, perhaps even amongst Muslims. Thank you.
One of my favorite things to say is, "I had no idea." Not only does this honor the person you say it to: They were, after all, trying to tell you something novel, it also shows a bit of humility and helps cultivate a delight in the world God has created.
I had no idea that the richest person in history was a Muslim guy named Mansa Musa who lived from 1280 to 1337. He inherited his kingdom from his dad (Still the best way to be super rich!) then multiplied his holdings with gold and salt. It has been said that he had so much wealth, it couldn't really be assessed. When Musa led his entourage of 60,000 people on Hajj to Mecca in 1324, he gave away so much gold along the way that he crashed the economic system of a good portion of North Africa!
Some of the currently richest Muslims in the world are ones who own countries or who married into the royal family. For instance, Princess Lalla Salma, the former wife of the king of Morocco is reportedly worth $2.5 billion. Many others have made their fortunes in finance and business. Abdul Aziz Al Ghurair, a 67 year old Emerati, took over his uncle's bank and now is said to have $2.7 billion, well, in the bank.
After his dad's death in 1966, 21 year old Azim Premji left Stanford, mid-way through his engineering studies, to return to India and take over the family business. He did ok with it, expanding into the tech world and eventually being dubbed, Czar of the Indian IT Industry. In 2013, Premji signed the Giving Pledge, the first Indian to do so, and agreed to give away at least half of his $11 billion fortune, including over $2 billion to education causes in India.
Jesus said, "it will be hard for a rich person to enter the kingdom of heaven." Let's pray that these and many others would be the camels who could!
PS: Got any thing you'd like to see addressed in Muslim Connect? Questions you're wondering about? I'd love to give them some thought and maybe write about them in coming weeks. Shoot me an email. Thank you.
The Arabic word madrasa basically means a place for education. Outside of Arabic speaking countries it has come to refer particularly to institutions focused on Islamic training. As you'd guess, there's huge diversity in the theology and purpose of various madrasas.
Years ago some friends and I were taken to lunch after Friday prayers to a madrasa in Malaysia. As we sat on the floor, feasting on a mountain of yummy chicken and rice, a young student put his missionary training into action: With earnest sincerity he put his hand on my pastor's knee, looked in his eyes and said, "Brother, become a Muslim." That will leave a mark, pretty much whoever you are!
Madrasas in many parts of the world serve similar purposes for young Muslims as AWANA does for kids in evangelical churches. I think AWANA has a leg up on most madrasas in that kids can memorize scripture in the language they speak! Muslim kids usually memorize the Quran in Arabic. They learn how to say it, but maybe not what they're saying!
Think about why you take (or might take) your kids to Sunday School, youth group or AWANA. Probably most Muslim moms and dads have similar motivation. They want their offspring to grow up and live out the faith. They want good and respectable kids.
If you grew up in church, maybe you remember early Sunday School days. One of the first people ever to teach me how to follow Jesus was Ethel Smith. Her face deeply wrinkled, breath smelling like Sen Sen, it was fifty years ago she taught our unruly class and I've never forgotten.
The good people at PrayerCast.com just released a beautiful video in which a former Muslim leads viewers in prayer for kids and adults in madrasas. Will you watch it and pray with me for the winsome, kingdom work of Jesus to break loose in madrasas all over.
Much thanks to those who responded to the questions in last week's Muslim Connect, "Muslims Are Trouble." I appreciate you. It's not too late to weigh in, if you wish.


dill seed, 1 t salt, baking soda and yeast; mix well.
cottage cheese until very warm (120 to 130 degrees).
speed until moistened; beat 3 minutes at medium speed.
Generously grease 1 1/2 or 2 quart casserole.
light and doubled in size, 30 to 45 minutes.
with melted butter; sprinkle with coarse salt.


D.C., just like the federal government, has a five-year time limit for how long families can stay on welfare, now called Temporary Assistance for Needy Families. But the city has faced difficulties in getting longtime recipients off of assistance. And now the city council will consider a bill that would let some people to stay on assistance beyond five years.
The bill doesn't change the five-year time limit, but it does give longtime TANF recipients an extra year to prepare for the end of benefits. Also, anyone in a post-secondary education or a D.C. Department of Employment Services job training program would still be able to receive checks for two extra years. The bill includes exemptions for people who face hardships that prevent them from getting a job, based somewhat on Maryland's TANF rules, according to a statement by Councilman Michael Brown (At-large).
The bill passed the human services committee Wednesday and will now go before the full council.


When writing a dissertation, there is generally a requirement to write an acknowledgement section. Whilst most of the other sections of your paper will be concentrated on academic writing, the acknowledgement section simply gives you a chance to name and thank any of the people that have helped you when writing the work. You will most likely wish to include people like your tutors and other academic staff, your friends, family, as well as anyone else who may have assisted you along the way.
For a better understanding of how to complete the work to a satisfactory standard, you may be wishing to look for examples written by other people. The following outlines various suggestions in order to help you find the samples that you are looking for.
Many universities and educational establishments will publish work written by their students. Sometimes the reasons for this will be to assist their students when writing their own work, by providing a high quality essay for them to read and refer to. Alternatively, sometimes it's simply a good way for universities to demonstrate the qualities that they expect of their students. Either way, this can be a great way of finding high-quality academic papers, including the acknowledgement section of dissertations.
Another place to look for high quality academic papers written by students and individuals who are studying at a higher level is to look on the websites that allow people to publish papers for peer review. Peer review websites essentially allow other individuals to read over and critique an essay that has been written by other people. These are particularly important in the scientific community, and help to develop our understanding of science.
The peer review process is taken very seriously; therefore, the work that you find will normally have been written to a high a standard as possible. As a result, it is great place to find and learn from essays that are been written by other people.
Finally, it is possible to download samples you find online from websites that specialise in providing academic papers for students. Some of these websites are available free, whilst others will require you to pay for the services. In fact, some of the websites that cost you money will offer other services as well, such as proofreading, editing, or even having bespoke essays created.


Hello, Virtusites!! It is with great pride we announce the 2nd annual Team-Virtus non-race, "The Deuce", has come and gone. Obviously, it'll take a while to compile the report, but rest assured…we're working on it.
Use the time between now and then to come up with a decent excuse for why you weren't there.
Check back soon, cuz we're almost done with The Chill report too!!
Posted on March 14, 2011, in Non-Races and tagged Adventure Non-Race, Adventure Racing. Bookmark the permalink.	11 Comments.
The real race is on now…who'll get their non-race report (though a non-report would be easier and perhaps more fitting) finished first? I'm not even to getting lost in the briar patch!
It takes awhile to get all of it into a decent race report, doesn't it? Especially when you have to sugar-coat it since you know we'll read it!! 😉 Can't wait to read it.
I think Bob will write one from a race director's point of view, and I'll write one from a tag-along non-racer's point of view. Should be fun… and of course you know it will be long.
Yawn. It's done. Who needs sleep?
And we'll link to it more prominently in one of our race reports. Great job, Kate!
You just made my day, Kate.
I was working on my excuse… then, I remembered that I WAS there. I didn't stand a chance in the non-race report race though. Katie has the bar set pretty high now too.
You were there indeed, and you crushed it! And I know what you mean about Kate's race report. I haven't even started mine yet, but now I'm a little intimidated.
Not nearly as flattering as stud. 😦 Chalk that up with the bathroom situation on the minus side of being a girl.


Your self esteem improves as you find solutions for the pain management of others, ailments, fitness, stress management, positive thinking, etc.
The list goes on, as you continue your own journey of self improvement, but the feeling of gratification you get from helping someone find the right path, is beyond words.
You will always remember: That student who reaches their ideal body weight, the physically impaired student who finds that they, too, can live a better quality life, and the student who leaves your class without the headache or backache they came in with.
As a Yoga teacher, you are setting an example to your students and teaching them to live a quality life. This path will enable you to live longer and live better.
When the working world is in the "9 to 5" mode, you have many opportunities with Corporate Accounts, The Fitness Industry, Senior Centers, Medical Centers, Referrals, etc. This is when you to teach them, with any free time you have.
Yoga teacher through our on-site program, inherited an area that I could no longer service due to time restraints.
The object is to contact them. This is where your post cards come in handy, if you don't have a personal referral.
Yes, learning new things keeps your mind stimulated and healthy. You will never tire of subjects to study, explore, and investigate. There are so many facets of Yoga, that one life span, is just not enough time to learn it all.
Everyone wants control of their own life, but very few achieve it. Being in business for yourself, can help you control your own destiny and that of your family.
Review your short-term goals daily. You will see yourself make rapid success in this way.
Lastly, goals do not have to be material at all. For example: You may want to start teaching Yoga in a year, and the following year, get a part-time Yoga teaching position. This type of goal setting is realistic and beneficial to mankind.
That is the key – If you choose a goal that will benefit others, you will surely achieve it.
Gentlemen, guys actually do Yoga.


text:
David M. Lesser, Fine Antiquarian Books LLC
Fairbanks & Co.:
ESTABLISHED 1830. FAIRBANKS' STANDARD SCALES. BUY ONLY THE GENUINE. FAIRBANKS & CO., 311 BROADWAY, N.Y. FAIRBANKS, BROWN & CO., 118 MILK STREET, BOSTON.
New York: Baker & Godwin, Printers, [1870s?]. Brochure folded to 3" x 5.5" to make [10] pp. Printed on blue paper, both sides printed and illustrated. Minor dusting and small line of fox, else quite clean. Very Good.
This brochure provides descriptions and illustrations of the Company's scales, including the Patent Iron Frame Railroad Track Scale which was patented in 1857 and renewed in 1871; illustrations of large scales for weighing cattle, hay, coal and warehouse items; smaller scales for use by druggists, grocers and the like; weighmaster's beams and frames; and a double lock cash drawer. A reprint of an article from the New York Tribune notes the amazement of members of the Japanese Embassy during their tour of the Fairbanks manufactory . The back page is an advertisement for Carhart & Curd, merchants in Macon, Georgia.
OCLC 16066096 [serial- no libraries noted]. Not in Sabin. See Romaine 176a, 315; and Winterthur 1320 [reference]. Item #25476
See all items in AGRICULTURE, AMERICANA, MASSACHUSETTS, NEW YORK, RAILROAD, TRADE
See all items by Fairbanks, Co.
One Bradley Rd. #302
Woodbridge, CT 06525
© 2021 David M. Lesser, Fine Antiquarian Books LLC. All rights reserved. Site Map | Site by Bibliopolis


text:
Best Chromebooks for Students in 2016
August 6, 2016 Delta Media
If you're heading back to school this fall, you might want to consider a Chromebook. They're perfect for students (I should know — I'm a recent graduate myself) because they're cheap, light, fast, and do everything you need. Think of them as laptops that run only Google's Chrome browser. That might sound limited, but you can type documents in Google Docs that automatically save and research on the internet to your heart's content — all for much less than an overpowered Windows laptop or MacBook. But once you've decided to go the Chromebook route, how do you choose which is…
Read the full article: Best Chromebooks for Students in 2016
Tagged 2016BestChromebooksStudents
Save on Dash Cams, iPhone Jump Drive, and More [CA]
5 Free Apps to Turn Selfies into Emojis, Stickers, & Even Music
How to Customize Microsoft Office 2016 to Your Needs
March 9, 2016 March 9, 2016 Delta Media
Acer ConceptD 9 Pro: The Ultimate Laptop for Creatives
The 3 Best Affordable Noise-Cancelling Headphones for Students
Lumus Launches Game-Changing AR Z-Lens At CES
January 5, 2023 by Bobby Carlton
These new AR lenses deliver a 2K x 2K resolution experience and address vergence-accommodation conflicts. The Israeli-based augmented reality (AR) company Lumus will be expanding its 2D Maximus platform by unveiling its 2nd generation of Lumus' Z-Lens waveguide architecture during CES 2023. Not only will this hardware advancement shrink the projector module by more than 50%, […]
Holoride Launches In-Car VR Device For Any Vehicle
January 4, 2023 by Kyle Melnick
Welcome to the dawn of the 'Motorverse.' The 2023 Consumer Electronics Show (CES) doesn't technically start until tomorrow and we're already hearing major news from the legendary trade show. Earlier today, in-car VR company holoride announced the launch of holoride retrofit, a new device capable of bringing "elastic content" to any vehicle. According to the […]
Here Are The Best-Selling Steam VR Games Of 2022
A good few of these titles are currently on sale as part of Steam's Winter Sale. Earlier this week, the popular video game storefront Steam revealed its "top VR-exclusive experiences" of 2022. This year's list features a variety of "classic" titles like GORN and I Expect You To Die as well as new experiences like […]
Chipotle Is Launching A Fitness-Inspired AR Experience
Complete exercises and read meditative prompts for a chance at some free guac. In an official release, Chipotle, the popular chain restaurant, announced that it has partnered with Snap Inc. to launch a new augmented reality (AR) Lens for the Snapchat app that will encourage you to get up and moving with a series of […]
This Futuristic VR Facemask Tracks Your Expressions
The low-cost device will be available for purchase in the US and Japan later this year. Say hello to X Mask, a new device described by the manufacturer as a "mask type sensing gadget." Developed by Japanese startup Dolami, X Mask is capable of tracking your facial expressions in detail thanks to a series of incorporated […]
A Lighter Touch: Exploring CPU Power Scaling On Core i9-13900K and Ryzen 9 7950X
January 6, 2023 by Gavin Bonshor
January 5, 2023 by Ryan Smith
AMD Unveils Ryzen 9 7950X3D, 7900X3D, and Ryzen 7 7800X3D, Up to 128 MB of L3 Cache And 5.7 GHz Boost
AMD Lays Out 2023 Ryzen Mobile 7000 CPUs: Top-to-Bottom Updates, New Zen 4 'Phoenix' CPU Takes Point
The AMD CES 2023 Keynote Live Blog (6:30pm PT/02:30 UTC)
Saanich police seek two men believed to have info on high-risk missing teen January 6, 2023
Police in Saanich, B.C., released images of two men they are hoping to identify in connection with the disappearance of a 17-year-old girl who was last seen on Dec. 19, 2002.
Jan. 6 attack: Biden to mark 2nd anniversary by presenting medals to 'heroes' January 6, 2023
Biden will award the "Presidential Citizens Medal" to 12 people, including U.S. Capitol Police officers, and give remarks during a ceremony at the White House.
44-year-old Saskatoon man charged with child luring, sexual assault January 6, 2023
A search warrant was executed by the Saskatchewan internet child exploitation (ICE) unit on Jan. 5, resulting in the arrest of Jeric Villareal Mendoza.
Brooke Kruger
More than 800 drivers charged during OPP Festive RIDE campaign January 6, 2023
The campaign, which ran from Nov. 17 to Jan. 2, resulted in 812 drivers facing a total of 1,258 impaired-related offences, the OPP says.
Ryan Rocca
Damar Hamlin's breathing tube removed, talks to teammates over FaceTime January 6, 2023
Damar Hamlin's teammates gave him a standing ovation over the FaceTime call.
Michelle Butterfield
Sentencing starts for man convicted of killing of two Metis hunters in rural Alberta January 6, 2023
Anthony Bilodeau drove up to his father's truck on the side of a road before shooting Sansom once in the chest and Cardinal three times around his shoulder.
Stephanie Swensrude
Knives, bear spray seized from man caught smoking on Winnipeg Transit bus January 6, 2023
A 19-year-old man from Winnipeg is charged with three counts of possession of a weapon.
Canadian Coast Guard warns Midland residents to stay off the ice next week January 6, 2023
Residents in the Midland, Ont., area may notice some activity out on Georgian Bay next week and stay off the ice with the Canadian Coast Guard planning some icebreaking operations.
Sawyer Bogdan
Surrey taxpayers face large property tax increase if police transition continues, mayor says January 6, 2023
Surrey mayor Brenda Locke told Global News she believes property taxes will increase on average $1,200 per household if the city transitions to the Surrey Police Service.
Driver arrested in fatal hit and run on Indian Road in Norfolk County: OPP January 6, 2023
Ontario Provincial Police carried out a search warrant at an address on David Street in Hagersville, southwest of Hamilton, following a fatal hit and run in Norfolk County Tuesday.