text:
\section{Introduction}

The notion of quantum divergences is among the most significant ones in quantum information
theory, with various applications, in particular, to defining important quantum quantities to
descriminate between states of a quantum system. A quantum system is mathematically described,
in most cases, by an operator algebra $\mathcal{A}$ on a Hilbert space (either finite-dimensional or
infinite-dimensional), and a quantum divergence is generally given as a function
$S(\rho\|\sigma)$ of two states (or more generally, two positive linear functionals) $\rho$
and $\sigma$ on $\mathcal{A}$. Among various quantum divergences, the most notable is the
\emph{relative entropy} having a long history as the quantum version of the Kullback-Leibler
divergence in classical theory. Indeed, the relative entropy $D(\rho\|\sigma)$ was first
introduced in 1962 by Umegaki \cite{Um} for normal states $\rho,\sigma$ on a semifinite von
Neumann algebra $M$ as follows:
\begin{align}\label{F-1.1}
D(\rho\|\sigma):=\begin{cases}
\tau(d_\rho(\log d_\rho-\log d_\sigma)) & \text{if $s(\rho)\le s(\sigma)$}, \\
+\infty & \text{otherwise},
\end{cases}
\end{align}
where $\tau$ is a semifinite trace on $M$, $d_\rho$ is the density operator of $\rho$ with
respect to $\tau$ and $s(\rho)$ is the support projection of $\rho$. Later in 1970's Araki
\cite{Ar5,Ar2} extended Umegaki's relative entropy, by introducing the \emph{relative modular
operator} $\Delta_{\rho,\sigma}$ for normal states $\rho,\sigma$, to general
von Neumann algebras as
\begin{align}\label{F-1.2}
D(\rho\|\sigma):=\begin{cases}
-\<\xi_\rho,(\log\Delta_{\sigma,\rho})\xi_\rho\>
=\<\xi_\sigma,(\Delta_{\rho,\sigma}\log\Delta_{\rho,\sigma})\xi_\sigma\>
& \text{if $s(\rho)\le s(\sigma)$}, \\
+\infty & \text{otherwise},
\end{cases}
\end{align}
where $\xi_\rho$ is the vector representative of $\rho$ in the standard representation of $M$
(see Section 2.1 below). A remarkable progress on the relative entropy was made when
Kosaki \cite{Ko} gave a variational expression of $D(\rho\|\sigma)$ and showed that all
important properties of $D(\rho\|\sigma)$ immediately follow from the expression.

A more general form of quantum divergences was considered by Kosaki \cite{Ko0} to generalize
the Wigner-Yanase-Dyson-Lieb concavity, and later discussed in more detail by Petz
\cite{Pe,Pe0} with name \emph{quasi-entropy}. The reader can refer to \cite{OP} for details
on the relative entropy and quasi-entropies. The \emph{standard $f$-divergences}
$S_f(\rho\|\sigma)$ studied in, e.g., \cite{HMPB,HiMo} in the finite-dimensional case are a
special case of quasi-entropies but a natural class of quantum divergences generalizing the
classical $f$-divergences. A significant property satisfied by quantum divergences mentioned
above is the monotonicity property, that is, the inequality
\begin{align}\label{F-1.3}
S(\rho\circ\Phi\|\sigma\circ\Phi)\le S(\rho\|\sigma)
\end{align}
under positive linear maps $\Phi:\mathcal{B}\to\mathcal{A}$ between operator algebras assumed to be
unit-preserving and completely positive (or more weakly a Schwarz map). An important issue in
connection with this property is to prove the \emph{reversibility} of $\Phi$, i.e., the
existence of a recovery map $\Psi:\mathcal{A}\to\mathcal{B}$ satisfying $\rho\circ\Phi\circ\Psi=\rho$ and
$\sigma\circ\Phi\circ\Psi=\sigma$ under the equality condition in the monotonicity inequality
in \eqref{F-1.3}. In the special case where $\mathcal{B}$ is a subalgebra of $\mathcal{A}$ and $\Phi$ is the
injection, the reversibility of $\Phi$ on $\{\rho,\sigma\}$ is called the \emph{sufficiency}
of $\mathcal{B}$ for $\{\rho,\sigma\}$. This line of research in von Neumann algebras was initiated
by Petz, e.g., \cite{Pe1,Pe3,JP} for the relative entropy and the transition probability
(i.e., the R\'enyi divergence with parameter $1/2$). The extension of the reversibility to
more general standard $f$-divergence, though in the finite-dimensional case, has been done
in \cite{HMPB,Je0,HiMo}.

Our aim in this paper is to propose a new approach to the theory of standard $f$-divergences
in general von Neumann algebras. For this, in Section 2 we first give the definition and some
basic properties of the standard $f$-divergence $S_f(\rho\|\sigma)$ of normal positive linear
functionals $\rho,\sigma$ on a von Neumann algebra, when $f$ is a general convex function on
$(0,\infty)$. The idea of the definition is essentially the same as \eqref{F-1.2}, based on
the relative modular operator $\Delta_{\rho,\sigma}$, but without  any assumption on the
boundary values of $f(t)$ at $t$ zero and infinity. For further discussions we assume that
$f$ is an operator convex function on $(0,+\infty)$. In Section 3 we give a variational
expression of $S_f(\rho\|\sigma)$ by utilizing the integral expression of $f$ and modifying
Kosaki's expression \cite{Ko} of the relative entropy. Our variational expression is a bit
different from Kosaki's one even in the case of the relative entropy. Next in Section 4, we
present various properties of $S_f(\rho\|\sigma)$ such as monotonicity property in
\eqref{F-1.3}, joint lower semicontinuity, joint convexity, etc.\ as straightforward
consequences from the variational expression. In this way, we can study the standard
$f$-divergence in von Neumann algebras along a very streamlined track, which is a special
feature of our presentation.

The quantum \emph{R\'enyi divergence} $D_\alpha(\rho\|\sigma)$ with parameter
$\alpha\in[0,+\infty)\setminus\{1\}$ is of quite use in quantum information as a quantum
version of the classical R\'enyi divergence. In the finite-dimensional (or the matrix) case,
the R\'enyi divergence is defined by
$$
D_\alpha(\rho\|\sigma):={1\over\alpha-1}\log{Q_\alpha(\rho\|\sigma)\over\mathrm{Tr}\,\rho},
$$
where $Q_\alpha(\rho\|\sigma):=\mathrm{Tr}\,(\rho^\alpha\sigma^{1-\alpha})$ that is essentially the
standard $f$-divergence with $f(t)=t^\alpha$. So $D_\alpha$ is indeed a variant of standard
$f$-divergences. In these years it has also been widely known that another type of quantum
R\'enyi divervgence, called the \emph{sandwiched R\'enyi divergence} and denoted by
$\widetilde D_\alpha(\rho\|\sigma)$, is equally useful in quantum information, in particular,
in quantum state discrimination, see \cite{MO} for example. The definition of
$\widetilde D_\alpha$ is similar to $D_\alpha$ by replacing $Q_\alpha(\rho\|\sigma)$ with
$\widetilde Q_\alpha(\rho\|\sigma):=\mathrm{Tr}\,\bigl((\sigma^{(1-\alpha)/2\alpha}\rho
\sigma^{(1-\alpha)/2\alpha})^\alpha\bigr)$, though $\widetilde Q_\alpha$ is no longer in the
class of standard $f$-divergences. Moreover, the so-called \emph{$\alpha$-$z$-R\'enyi
divergence} introduced in \cite{AD} is a two-parameter common generalization of $D_\alpha$
and $\widetilde D_\alpha$. Motivated by the current situation of quantum R\'enyi divergences,
the authors in \cite{BST,Je1,Je2} have recently extended the sandwiched version
$\widetilde D_\alpha$ to the von Neumann algebra setting. Indeed, in these papers, the
quantity $\widetilde Q_\alpha$ is defined in von Neumann algebras by using Araki and Masuda's
$L^p$-spaces \cite{AM} or Haagerup's $L^p$-spaces \cite{Ha2,Te}. Although those papers contain
some discussions on $D_\alpha$ as well, it seems that the expositions on $D_\alpha$ there are
not comprehensive. Thus, in Section 5 we present a thorough exposition on the R\'enyi
divergence in von Neumann algebras, while it is more or less specialization of the results
of Section 4.

This paper has two appendices. In Appendix A we give a brief survey on Haagerup's $L^p$-spaces
and the description of the R\'enyi divergence in terms of them. In Appendix B we revisit the
former results in \cite{Ar3,Do} on relative hamiltonians and their relation to the relative
entropy, and improve them based on Haagerup's $L^p$-spaces and the fact that
$D=\lim_{\alpha\nearrow1}D_\alpha$.

\section{Definition of standard $f$-divergences}

\subsection{Relative modular operators}

Let $M$ be a general von Neumann algebra, and $M_*^+$ be the positive cone of the predual
$M_*$ consisting of normal positive linear functionals on $M$. Throughout the paper, we
consider $M$ in its \emph{standard form} $(M,\mathcal{H},J,\mathcal{P})$, that is, $M$ is represented on
a Hilbert space $\mathcal{H}$ with a conjugate-linear involution $J$ and a self-dual cone $\mathcal{P}$
called the \emph{natural cone}, for which the following hold:
\begin{itemize}
\item[(1)] $JMJ=M'$,
\item[(2)] $JxJ=x^*$,\quad$x\in M\cap M'$ (the center of $M$),
\item[(3)] $J\xi=\xi$,\quad$\xi\in\mathcal{P}$,
\item[(4)] $xJxJ\mathcal{P}\subset\mathcal{P}$,\quad$x\in M$.
\end{itemize}
Recall \cite{Ha} that any von Neumann algebra has a standard form, which is unique in the
sense that if $(M,\mathcal{H},J,\mathcal{P})$ and $(\tilde M,\tilde\mathcal{H},\tilde J,\tilde\mathcal{P})$ are two standard
forms and $\Phi:M\to\tilde M$ is a $*$-isomorphism, then there is a unique unitary
$u:\mathcal{H}\to\tilde\mathcal{H}$ such that $\Phi(x)=uxu^*$ for $x\in M$, $\tilde J=uJu^*$ and
$\tilde\mathcal{P}=u\mathcal{P}$. See \cite{Ha} (also \cite{Ar1,Co}) for more details on the standard form.

Every $\sigma\in M_*^+$ has a unique \emph{vector representative} $\xi_\sigma\in\mathcal{P}$ so that
$\sigma(x)=\<\xi_\sigma,x\xi_\sigma\>$, $x\in M$. We have the support projections
$s_M(\sigma)\in M$ and $s_{M'}(\sigma)\in M'$ of $\sigma$, that is, $s_M(\sigma)$ is the
orthogonal projection onto $\overline{M'\xi_\sigma}$ and $s_{M'}(\sigma)$ is that onto
$\overline{M\xi_\sigma}$. Note that $s_{M'}(\sigma)=Js_M(\sigma)J$.

For each $\rho,\sigma\in M_*^+$ the operators $S_{\rho,\sigma}$ and $F_{\rho,\sigma}$ are
defined by
$$
S_{\rho,\sigma}(x\xi_\sigma+\eta):=s_M(\sigma)x^*\xi_\rho,\qquad
x\in M,\ \eta\in(1-s_{M'}(\sigma))\mathcal{H},
$$
$$
F_{\rho,\sigma}(x'\xi_\sigma+\zeta):=s_{M'}(\sigma)x^{\prime*}\xi_\rho,\qquad
x'\in M',\ \zeta\in(1-s_M(\sigma))\mathcal{H}.
$$
Then $S_{\rho,\sigma}$ and $F_{\rho,\sigma}$ are closable conjugate-linear operators such
that $S_{\rho,\sigma}^*=\overline F_{\rho,\sigma}$, see \cite[Lemma 2.2]{Ar2}. The
\emph{relative modular operator} $\Delta_{\rho,\sigma}$ introduced in \cite{Ar2} is
$$
\Delta_{\rho,\sigma}:=S_{\rho,\sigma}^*\overline S_{\rho,\sigma},
$$
and the polar decomposition of $\overline S_{\rho,\sigma}$ is given as
\begin{align}\label{F-2.1}
\overline S_{\rho,\sigma}=J\Delta_{\rho,\sigma}^{1/2}.
\end{align}
Recall that the support projection of $\Delta_{\rho,\sigma}$ is $s_M(\rho)s_{M'}(\sigma)$.
We write the spectral decomposition of $\Delta_{\rho,\sigma}$ as
\begin{align}\label{F-2.2}
\Delta_{\rho,\sigma}=\int_{[0,+\infty)}t\,dE_{\rho,\sigma}(t).
\end{align}
When $\rho=\sigma$, $\Delta_{\sigma,\sigma}$ is the \emph{modular operator} $\Delta_\sigma$
for $\sigma$. Note that $\xi_\sigma$ is an eigenvector of $\Delta_\sigma$ with eigenvalue $1$,
so that $\Delta_\sigma\xi_\sigma=\xi_\sigma$.

\subsection{Standard $f$-divergence $S_f(\rho\|\sigma)$}

Let $f:(0,+\infty)\to\mathbb{R}$ be a convex function. Then the limits
$$
f(0^+):=\lim_{t\searrow0}f(t),\qquad f'(+\infty):=\lim_{t\to+\infty}{f(t)\over t}
$$
exist in $(-\infty,+\infty]$. Below we will understand the expression $bf(a/b)$ for $a=0$ or
$b=0$ in the following way:
\begin{align}\label{F-2.3}
bf(0/b):=f(0^+)b\quad\mbox{for $b\ge0$},\quad
0f(a/0):=\lim_{t\searrow0}tf(a/t)=f'(+\infty)a\quad\mbox{for $a>0$},
\end{align}
where we use the convention that $(+\infty)0:=0$ and $(+\infty)c:=+\infty$ for $c>0$. In
particular, we fix $0f(0/0)=0$.

The next definition is a specialization of the quasi-entropy \cite{Ko0,Pe} with
modifications.

\begin{definition}\label{D-2.1}\rm
For each $\rho,\sigma\in M_*^+$, with the spectral decomposition in \eqref{F-2.2}, we define
the self-adjoint operator $f(\Delta_{\rho,\sigma})$ on $s_M(\rho)s_{M'}(\sigma)\mathcal{H}$ by
\begin{align}\label{F-2.4}
f(\Delta_{\rho,\sigma}):=\int_{(0,+\infty)}f(t)\,dE_{\rho,\sigma}(t),
\end{align}
and define
\begin{align}\label{F-2.5}
\<\xi_\sigma,f(\Delta_{\rho,\sigma})\xi_\sigma\>
:=\int_{(0,+\infty)}f(t)\,d\|E_{\rho,\sigma}(t)\xi_\sigma\|^2.
\end{align}
We then introduce the {\it standard $f$-divergence} $S_f(\rho\|\sigma)$ of $\rho,\sigma$ by
\begin{align}\label{F-2.6}
S_f(\rho\|\sigma):=\<\xi_\sigma,f(\Delta_{\rho,\sigma})\xi_\sigma\>
+f(0^+)\sigma(1-s_M(\rho))+f'(+\infty)\rho(1-s_M(\sigma)).
\end{align}
\end{definition}

Note that the integral in \eqref{F-2.4} is on $(0,+\infty)$ instead of $[0,+\infty)$. The
left-hand expression of \eqref{F-2.5} should be understood, to be precise, in the sense of a
lower-bounded form (see \cite{RS}), which equals the integral in the right-hand side. We first
give a lemma to justify the above definition.

\begin{lemma}\label{L-2.2}
For every $\rho,\sigma\in M_*^+$, $S_f(\rho\|\sigma)$ is well defined with values in
$(-\infty,+\infty]$.
\end{lemma}

\begin{proof}
From the convexity of $f$, there are $a,b\in\mathbb{R}$ such that $f(t)\ge a+bt$ for all
$t\in(0,+\infty)$. We have
\begin{align*}
\int_{(0,+\infty)}f(t)\,d\|E_{\rho,\sigma}(t)\xi_\sigma\|^2
&\ge\int_{(0,+\infty)}(a+bt)\,d\|E_{\rho,\sigma}(t)\xi_\sigma\|^2 \\
&=a\|s_M(\rho)s_{M'}(\sigma)\xi_\sigma\|^2+b\|\Delta_{\rho,\sigma}^{1/2}\xi_\sigma\|^2 \\
&=a\|s_M(\rho)\xi_\sigma\|^2+b\|s_M(\sigma)\xi_\rho\|^2 \\
&=a\sigma(s_M(\rho))+b\rho(s_M(\sigma))>-\infty,
\end{align*}
since $J\Delta_{\rho,\sigma}^{1/2}\xi_\sigma=s_M(\sigma)\xi_\rho$ by \eqref{F-2.1}. Hence
$S_f(\rho\|\sigma)\in(-\infty,+\infty]$.
\end{proof}

By the above proof and \eqref{F-2.6} we also see that
\begin{align}\label{F-2.7}
S_{a+bt}(\rho\|\sigma)=a\sigma(1)+b\rho(1).
\end{align}

The following are some basic properties of $S_f(\rho\|\sigma)$:

\begin{prop}\label{P-2.3}
Let $\rho,\sigma\in M_*^+$.
\begin{itemize}
\item[(1)] If $\Phi:\tilde M\to M$ is a $*$-isomorphism between von Neumann algebras, then
$$
S_f(\rho\circ\Phi\|\sigma\circ\Phi)=S_f(\rho\|\sigma).
$$
\item[(2)] In the case $\rho=0$ or $\sigma=0$ or $\rho=\sigma$,
$$
S_f(0\|\sigma)=f(0^+)\sigma(1),\quad S_f(\rho\|0)=f'(+\infty)\rho(1),\quad
S_f(\sigma\|\sigma)=f(1)\sigma(1).
$$
\item[(3)] \emph{Homogeneity:} For every $\lambda\in[0,+\infty)$,
\begin{align}\label{F-2.8}
S_f(\lambda\rho\|\lambda\sigma)=\lambda S_f(\rho\|\sigma).
\end{align}
\item[(4)] \emph{Additivity:} Let $M=M_1\oplus M_2$ be the direct sum of von Neumann algebras
$M_1$ and $M_2$. If $\rho_i,\sigma_i\in(M_i)_*^+$ for $i=1,2$, then
$$
S_f(\rho_1\oplus\rho_2\|\sigma_1\oplus\sigma_2)
=S_f(\rho_1\|\sigma_1)+S_f(\rho_2\|\sigma_2).
$$
\end{itemize}
\end{prop}

\begin{proof}
(1) is clear from the uniqueness of the standard form stated in Section 2.1.

(2) is seen directly from definition \eqref{F-2.6}.

(3)\enspace
For $\lambda>0$, since $\xi_{\lambda\sigma}=\sqrt\lambda\,\xi_\sigma$ and
$\Delta_{\lambda\rho,\lambda\sigma}=\Delta_{\rho,\sigma}$, equality \eqref{F-2.8} follows.
For $\lambda=0$ both sides are zero from (2).

(4)\enspace
Note that the standard form of $M$ is the direct sum of the standard forms of $M_i$, $i=1,2$.
For $\rho:=\rho_1\oplus\rho_2$ and $\sigma:=\sigma_1\oplus\sigma_2$ we have
$\xi_\rho=\xi_{\rho_1}\oplus\xi_{\rho_2}$, $\xi_\sigma=\xi_{\sigma_1}\oplus\xi_{\sigma_2}$
and $\Delta_{\rho,\sigma}=\Delta_{\rho_1,\sigma_1}\oplus\Delta_{\rho_2,\sigma_2}$, from which
the result immediately follows.
\end{proof}

The additivity in (4) above will be improved in Section 4 (see Corollary \ref{C-4.4}\,(2)).

In our definition of $S_f(\rho\|\sigma)$ the parameter function $f$ is a convex function on
$(0,+\infty)$, not on $[0,+\infty)$. This is reasonable as the next proposition shows that
$S_f(\rho\|\sigma)$ is symmetric between $\rho$ and $\sigma$ under exchanging $f$ with its
transpose $\widetilde f$ defined by
$$
\widetilde f(t):=tf(t^{-1}),\qquad t\in(0,+\infty).
$$

\begin{prop}\label{P-2.4}
For every $\rho,\sigma\in M_*^+$,
$$
S_f(\rho\|\sigma)=S_{\widetilde f}(\sigma\|\rho).
$$
\end{prop}

\begin{proof}
Since $\widetilde f(0^+)=f'(+\infty)$ and $\widetilde f'(+\infty)=f(0^+)$, it suffices to
prove that
\begin{align}\label{F-2.9}
\<\xi_\sigma,f(\Delta_{\rho,\sigma})\xi_\sigma\>
=\<\xi_\rho,\widetilde f(\Delta_{\sigma,\rho})\xi_\rho\>.
\end{align}
Recall \cite[Theorem 2.4]{Ar2} that
$$
\Delta_{\rho,\sigma}=J\Delta_{\sigma,\rho}^{-1}J
$$
together with
$$
s_M(\rho)s_{M'}(\sigma)=Js_{M'}(\rho)JJs_M(\sigma)J=Js_M(\sigma)s_{M'}(\rho)J.
$$
Hence, since $J\Delta_{\sigma,\rho}^{1/2}\xi_\rho=s_M(\rho)\xi_\sigma$, one has
\begin{align*}
\|E_{\rho,\sigma}((0,t))\xi_\sigma\|^2
&=\|JE_{\sigma,\rho}((t^{-1},+\infty))JJ\Delta_{\sigma,\rho}^{1/2}\xi_\rho\|^2
=\|E_{\sigma,\rho}((t^{-1},+\infty))\Delta_{\sigma,\rho}^{1/2}\xi_\rho\|^2 \\
&=\int_{(t^{-1},+\infty)}s\,d\|E_{\sigma,\rho}(s)\xi_\rho\|^2
=\int_{(0,t)}s^{-1}\,d\|E_{\sigma,\rho}(s^{-1})\xi_\rho\|^2,
\end{align*}
so that $d\|E_{\rho,\sigma}(t)\xi_\sigma\|^2=t^{-1}d\|E_{\sigma,\rho}(t^{-1})\xi_\rho\|^2$ on
$(0,+\infty)$. Therefore,
\begin{align*}
\int_{(0,+\infty)}f(t)\,d\|E_{\rho,\sigma}(t)\xi_\sigma\|^2
&=\int_{(0,+\infty)}f(t)t^{-1}\,d\|E_{\sigma,\rho}(t^{-1})\xi_\rho\|^2 \\
&=\int_{(0,+\infty)}tf(t^{-1})\,d\|E_{\sigma,\rho}(t)\xi_\rho\|^2,
\end{align*}
which means \eqref{F-2.9}.
\end{proof}

\begin{example}\label{E-2.5}\rm
Let $M$ be an abelian von Neumann algebra such that $M\cong L^\infty(\Omega,\mu)$, where
$(\Omega,\mathcal{A},\mu)$ is a $\sigma$-finite measure space. Let $\rho,\sigma\in M_*^+$,
which correspond to $\phi,\psi\in L^1(\Omega,\mu)^+$ so that
$\rho(x)=\int_\Omega x\phi\,d\mu$, $\sigma(x)=\int_\Omega x\psi\,d\mu$ for
$x\in L^\infty(\Omega,\mu)$. The standard form of $L^\infty(\Omega,\mu)$ is
$(L^\infty(\Omega,\mu),L^2(\Omega,\mu),\xi\mapsto\overline\xi,L^2(\Omega,\mu)^+)$, where
$x\in L^\infty(\Omega,\mu)$ is represented on $L^2(\Omega,\mu)$ as the multiplication
operator $\xi\mapsto x\xi$ for $\xi\in L^2(\Omega,\mu)$. It is straightforward to find that
$\Delta_{\phi,\phi}$ is the multiplication of $1_{\{\psi>0\}}(\phi/\psi)$, which is the
Radon-Nikodym derivative $d\rho/d\sigma$ (restricted on the support of $\sigma$) in the
classical sense. We then have
$$
S_f(\rho\|\sigma)=\int_{\{\phi>0\}\cap\{\psi>0\}}\psi f(\phi/\psi)\,d\mu
+f(0^+)\int_{\{\phi=0\}}\psi\,d\mu+f'(+\infty)\int_{\{\psi=0\}}\phi\,d\mu,
$$
which equals the classical $f$-divergence
$S_f(\phi\|\psi)=\int_\Omega\psi f(\phi/\psi)\,d\mu$ under the convention that
$\psi(\omega)f(0/\psi(\omega))=f(0^+)\psi(\omega)$ for $\psi(\omega)\ge0$ and
$0f(\phi(\omega)/0)=\lim_{t\searrow0}tf(\phi(\omega)/t)=f'(+\infty)\phi(\omega)$ for
$\phi(\omega)>0$.
\end{example}

\begin{example}\label{E-2.6}\rm
Let $M=\cB(\cH)$, i.e., a factor of type I, where $\mathcal{H}$ is an arbitrary Hilbert space. The standard
form of $\cB(\cH)$ is $(\cB(\cH),\mathcal{C}_2(\mathcal{H}),J=*,\mathcal{C}_2(\mathcal{H})_+)$, where $\mathcal{C}_2(\mathcal{H})$ is the
Hilbert-Schmidt class with the Hilbert-Schmidt inner product, $\mathcal{C}_2(\mathcal{H})_+$ is the set of
positive operators in $\mathcal{C}_2(\mathcal{H})$, and the representation of $M=\cB(\cH)$ on $\mathcal{C}_2(\mathcal{H})$ is
the left multiplication. Then $M'$ is the right multiplication of $M=\cB(\cH)$ on $\mathcal{C}_2(\mathcal{H})$. For
$\rho,\sigma\in\cB(\cH)_*^+$ we have the vector representatives
$D_\rho^{1/2},D_\sigma^{1/2}\in\mathcal{C}_2(\mathcal{H})_+$, where $D_\rho$ and $D_\sigma$ are the positive
trace-class operators such that $\rho(X)=\mathrm{Tr}\, D_\rho X$ and $\sigma(X)=\mathrm{Tr}\, D_\sigma X$ for
$X\in\cB(\cH)$. Let
$$
D_\rho=\sum_{a\in\mathrm{spec}\, D_\rho,\,a>0}aP_a,\qquad
D_\sigma=\sum_{b\in\mathrm{spec}\, D_\sigma,\,b>0}bQ_b
$$
be the spectral decompositions of $D_\rho,D_\sigma$, where $\sum_{a>0}$ and $\sum_{b>0}$ are
finite or countable sums, and $P_a$ and $Q_b$ are finite-dimensional orthogonal projections.
Then the relative modular operators $\Delta_{\rho,\sigma}$ on $\mathcal{C}_2(\mathcal{H})$ is given as
\begin{align}\label{F-2.10}
\Delta_{\rho,\sigma}=L_{D_\rho}R_{D_\sigma^{-1}}
=\sum_{a>0,\,b>0}ab^{-1}L_{P_a}R_{Q_b},
\end{align}
where $L_{[-]}$ and $R_{[-]}$ denote the left and the right multiplications and $D_\sigma^{-1}$
is the generalized inverse of $D_\sigma$. The last expression in the above gives the spectral
decomposition of $\Delta_{\rho,\sigma}$. The proof of this is easy as follows:

If $\Phi\in P_a\mathcal{H}$ and $\Omega\in Q_b\mathcal{H}$ for $a,b>0$, then
\begin{align*}
S_{\rho,\sigma}(|\Phi\>\<\Omega|)
&=S_{\rho,\sigma}(b^{-1/2}|\Phi\>\<\Omega|D_\sigma^{1/2})
=b^{-1/2}s_M(\sigma)|\Omega\>\<\Phi|D_\rho^{1/2} \\
&=a^{1/2}b^{-1/2}|\Omega\>\<\Phi|, \\
F_{\rho,\sigma}(|\Omega\>\<\Phi|)
&=F_{\rho,\sigma}(b^{-1/2}D_\sigma^{1/2}|\Omega\>\<\Phi|)
=b^{-1/2}s_{M'}(\sigma)D_\rho^{1/2}|\Phi\>\<\Omega| \\
&=a^{1/2}b^{-1/2}|\Phi\>\<\Omega|s_M(\sigma)
=a^{1/2}b^{-1/2}|\Phi\>\<\Omega|,
\end{align*}
which imply that
$$
\Delta_{\rho,\sigma}(|\Phi\>\<\Omega|)=ab^{-1}|\Phi\>\<\Omega|.
$$
Since the range of $L_{P_a}R_{Q_b}$ is the span of $|\Phi\>\<\Omega|$ for $\Phi\in P_a\mathcal{H}$
and $\Omega\in Q_b\mathcal{H}$,
$$
\sum_{a,b>0,\,ab^{-1}=c}L_{Pa}R_{Q_b}
$$
is the spectral projection of $\Delta_{\rho,\sigma}$ corresponding to the eigenvalue $c>0$.

Moreover, it follows from \eqref{F-2.10} that the definition of $S_f(\rho\|\sigma)$ in
\eqref{F-2.6} is rewritten as
$$
S_f(\rho\|\sigma)=\sum_{a>0}\sum_{b>0}bf(ab^{-1})\mathrm{Tr}\, P_aQ_b
+f(0^+)\mathrm{Tr}\,(I-D_\rho^0)D_\sigma+f'(+\infty)\mathrm{Tr}\, D_\rho(I-D_\sigma^0),
$$
which coincides with an expression in \cite[Proposition 3.2]{HiMo} when $\dim\mathcal{H}<+\infty$.
\end{example}

\begin{remark}\label{R-2.7}\rm
Let $f:[0,+\infty)\to\mathbb{R}$ be a continuous function. For $\rho,\sigma\in M_*^+$ and $k\in M$,
the {\it quasi-entropy} $S_f^k(\rho\|\sigma)$ was introduced in \cite{Pe} by
\begin{align}
S_f^k(\rho\|\sigma)&:=\<k\xi_\sigma,f(\Delta_{\rho,\sigma})k\xi_\sigma\>
=\int_{[0,+\infty)}f(t)\,d\|E_{\rho,\sigma}(t)k\xi_\sigma\|^2 \nonumber\\
&\ =\int_{(0,+\infty)}f(t)\,d\|E_{\rho,\sigma}(t)k\xi_\sigma\|^2
+f(0^+)\<k\xi_\sigma,(1-s_M(\rho)s_{M'}(\sigma))k\xi_\sigma\>. \label{F-2.11}
\end{align}
Comparing \eqref{F-2.11} with \eqref{F-2.5} and \eqref{F-2.6} we note that
$$
S_f(\rho\|\sigma)=S_f^{k=1}(\rho\|\sigma)+f'(+\infty)\rho(1-s_M(\sigma)).
$$
In particular, when $M=\cB(\cH)$ with $\dim\mathcal{H}<+\infty$ and $f(0^+)<+\infty$, the quasi-entropy
\eqref{F-2.11} with $k=1$ has finite values for all $\rho,\sigma$, which is improper as a
standard $f$-divergence. For example, when $f(t):=t\log t$ so that $f(0^+)=0$ and
$f'(+\infty)=+\infty$, one can easily check that for $\rho=\mathrm{Tr}\,(D_\rho\,\cdot)$ and
$\sigma=\mathrm{Tr}\,(D_\sigma\,\cdot)$ with density operators $D_\rho,D_\sigma$, expression
\eqref{F-2.11} with $k=1$ is
$$
\mathrm{Tr}\, D_\rho(\log D_\rho-{\log^+}D_\sigma),
$$
where ${\log^+}t:=\log t$ ($t>0$), ${\log^+}0:=0$. On the other hand,
$S_{t\log t}(\rho\|\sigma)$ in \eqref{F-2.6} coincides with the usual \emph{relative entropy}
$$
D(\rho\|\sigma):=\begin{cases}
\mathrm{Tr}\, D_\rho(\log D_\rho-\log D_\sigma), & s_M(\rho)\le s_M(\sigma), \\
+\infty, & s_M(\rho)\not\le s_M(\sigma).
\end{cases}
$$
\end{remark}

\section{Variational expression of standard $f$-divergences}

In this section we extend the variational expression of the relative entropy given in
\cite{Ko} to standard $f$-divergences. The extended expression will be quite useful in the
next section to verify various properties of standard $f$-divergences.

Throughout this and the next sections, we assume that a function $f:(0,+\infty)\to\mathbb{R}$ is
\emph{operator convex}, i.e., the operator inequality
$$
f(\lambda A+(1-\lambda)B)\le\lambda f(A)+(1-\lambda)f(B),\qquad0\le\lambda\le1
$$
holds for every invertible $A,B\in\cB(\cH)^+$ of any $\mathcal{H}$. Also, a function $h:(0,+\infty)\to\mathbb{R}$
is said to be \emph{operator monotone} if $A\le B$ $\implies$ $h(A)\le h(B)$ for every
invertible $A,B\in\cB(\cH)^+$ of any $\mathcal{H}$. It is well-known that an operator monotone function
$h$ on $(0,+\infty)$ is automatically \emph{operator concave} (i.e., $-h$ is operator convex).
For general theory on operator monotone and operator convex functions, see, e.g.,
\cite{Bh,Hi}.

Recall \cite{LR} (see also \cite[Theorem 5.1]{FHR} for a more general form) that the operator
convex function $f$ has an integral expression
\begin{align}\label{F-3.1}
f(t)=a+b(t-1)+c(t-1)^2+\int_{[0,+\infty)}{(t-1)^2\over t+s}\,d\mu(s),
\qquad t\in(0,+\infty),
\end{align}
where $a,b\in\mathbb{R}$, $c\ge0$ and $\mu$ is a positive measure on $[0,+\infty)$ with
\begin{align}\label{F-3.2}
\int_{[0,+\infty)}{1\over 1+s}\,d\mu(s)<+\infty,
\end{align}
and moreover $a,b,c$ and $\mu$ are uniquely determined. Letting $d:=\mu(\{0\})\ge0$ we also write
\begin{align}\label{F-3.3}
f(t)=a+b(t-1)+c(t-1)^2+d\,{(t-1)^2\over t}+\int_{(0,+\infty)}{(t-1)^2\over t+s}\,d\mu(s),
\quad t\in(0,+\infty).
\end{align}
One can easily verify that
\begin{align}
f(0^+)&=a-b+c+(+\infty)d+\int_{(0,+\infty)}s^{-1}\,d\mu(s), \label{F-3.4}\\
f'(+\infty)&=b+(+\infty)c+d+\int_{(0,+\infty)}d\mu(s). \label{F-3.5}
\end{align}

For each $n\in\mathbb{N}$ we define
\begin{align}
f_n(t)&:=a+b(t-1)+c\,{n(t-1)^2\over t+n}+d\,{(t-1)^2\over t+(1/n)} \nonumber\\
&\qquad+\int_{[1/n,n]}{(t-1)^2\over t+s}\,d\mu(s),
\qquad t\in(0,+\infty). \label{F-3.6}
\end{align}
We then have

\begin{lemma}\label{L-3.1}
For each $n\in\mathbb{N}$, $f_n$ is operator convex on $(0,+\infty)$, $f_n(0^+)<+\infty$,
$f_n'(+\infty)<+\infty$ and
$$
f_n(0^+)\,\nearrow\,f(0^+),\quad f_n'(+\infty)\,\nearrow\,f'(+\infty),
\quad f_n(t)\,\nearrow\,f(t)
$$
as $n\to\infty$ for all $t\in(0,+\infty)$.
\end{lemma}

\begin{proof}
By definition \eqref{F-3.6} it is immediate to see that $f$ is an operator convex function
on $(0,+\infty)$ and
\begin{align}
f_n(0^+)&=a-b+c+nd+\int_{[1/n,n]}s^{-1}\,d\mu(s), \label{F-3.7}\\
f_n'(+\infty)&=b+nc+d+\int_{[1/n,n]}d\mu(s). \label{F-3.8}
\end{align}
It follows from \eqref{F-3.2} that \eqref{F-3.7} and \eqref{F-3.8} are finite, which
increase, by the monotone convergence theorem, to \eqref{F-3.4} and \eqref{F-3.5},
respectively. Moreover, for any $t\in(0,+\infty)$, since
$$
{n(t-1)^2\over t+n}\ \nearrow\ (t-1)^2,\qquad
{(t-1)^2\over t+(1/n)}\ \nearrow\ {(t-1)^2\over t}\qquad\mbox{as $n\nearrow\infty$},
$$
we have $f_n(t)\nearrow f(f)$ from the monotone convergence theorem again.
\end{proof}

\begin{lemma}\label{L-3.2}
For every $\rho,\sigma\in M_*^+$,
$$
S_{f_n}(\rho\|\sigma)\ \nearrow\ S_f(\rho\|\sigma)\qquad\mbox{as $n\nearrow\infty$}.
$$
\end{lemma}

\begin{proof}
By Definition \ref{D-2.1},
$$
S_{f_n}(\rho\|\sigma)=f_n(0^+)\sigma(1-s(\rho))+f_n'(+\infty)\rho(1-s(\sigma))
+\int_{(0,+\infty)}f_n(t)\,d\|E_{\rho,\sigma}(t)\xi_\sigma\|^2.
$$
By Lemma \ref{L-3.1} and the monotone convergence theorem, $S_{f_n}(\rho\|\sigma)$ increases
to $S_f(\rho\|\sigma)$ as $n\nearrow\infty$.
\end{proof}

\begin{lemma}\label{L-3.3}
For each $n\in\mathbb{N}$ define an operator monotone function $h_n$ on $[0,+\infty)$ by
\begin{align}\label{F-3.9}
h_n(t)=\int_{(0,+\infty)}{t(1+s)\over t+s}\,d\nu_n(s),\qquad t\in[0,+\infty),
\end{align}
where $\nu_n$ is a finite positive measure supported on $[1/n,n]$ given by
\begin{align}\label{F-3.10}
d\nu_n(s):=c(1+n)\delta_n+d(1+n)\delta_{1/n}+1_{[1/n,n]}(s){1+s\over s}\,d\mu(s)
\end{align}
with the point masses $\delta_n$ at $n$ and $\delta_{1/n}$ at $1/n$. Then $f_n$ defined in
\eqref{F-3.6} is written as
\begin{align}\label{F-3.11}
f_n(t)=f_n(0^+)+f_n'(+\infty)t-h_n(t),\qquad t\in(0,+\infty).
\end{align}
\end{lemma}

\begin{proof}
Compute
\begin{align*}
{n(t-1)^2\over t+n}&=1+nt-(1+n){t(1+n)\over t+n}, \\
{(t-1)^2\over t+(1/n)}&=n+t-(1+n){t\bigl(1+{1\over n}\bigr)\over t+{1\over n}}, \\
{(t-1)^2\over t+s}&={1\over s}+t-{1+s\over s}\cdot{t(1+s)\over t+s}.
\end{align*}
Inserting these into definition \eqref{F-3.6} one can write
\begin{align*}
f_n(t)&=\biggl(a-b+c+nd+\int_{[1/n,n]}s^{-1}\,d\mu(s)\biggr) \\
&\qquad+\biggl(b+nc+d+\int_{[1/n,n]}d\mu(s)\biggr)t \\
&\qquad-c(1+n){t(1+n)\over t+n}-d(1+n){t\bigl(1+{1\over n}\bigr)\over t+{1\over n}} \\
&\qquad-\int_{[1/n,n]}{1+s\over s}\cdot{t(1+s)\over t+s}\,d\mu(s) \\
&=f_n(0^+)+f_n'(+\infty)t-h_n(t)
\end{align*}
thanks to \eqref{F-3.7} and \eqref{F-3.8}.
\end{proof}

Now, let $L$ be a subspace of $M$ containing $1$, and assume that $L$ is dense in $M$ with
respect to the strong* operator topology. Since $h_n(0)=h_n'(+\infty)=0$, the next lemma
follows from \cite[Theorem 2.2]{Ko}.

\begin{lemma}\label{L-3.4}
Let $h_n$ be given in \eqref{F-3.9}. Then for any $\rho,\sigma\in M_*^+$,
\begin{align*}
&\int_{(0,+\infty)}h_n(t)\,d\|E_{\rho,\sigma}(t)\xi_\sigma\|^2 \\
&\quad=\inf_{x(\cdot)}\int_{[1/n,n]}\bigl\{\sigma((1-x(s))^*(1-x(s)))
+s^{-1}\rho(x(s)x(s)^*)\bigr\}(1+s)\,d\nu_n(s),
\end{align*}
where the infimum is taken over all $L$-valued (finitely many values) step functions
$x(\cdot)$ on $(0,+\infty)$.
\end{lemma}

\begin{thm}\label{T-3.5}
Let $f$ be an operator convex function on $(0,+\infty)$. For each $n\in\mathbb{N}$ let $f_n(0^+)$,
$f_n'(+\infty)$ and $\nu_n$ be given in \eqref{F-3.7}, \eqref{F-3.8} and \eqref{F-3.10},
respectively. Then for every $\rho,\sigma\in M_*^+$,
\begin{align}
S_f(\rho\|\sigma)&=\sup_{n\in\mathbb{N}}\sup_{x(\cdot)}
\biggl[f_n(0^+)\sigma(1)+f_n'(+\infty)\rho(1) \nonumber\\
&\qquad-\int_{[1/n,n]}\bigl\{\sigma((1-x(s))^*(1-x(s)))
+s^{-1}\rho(x(s)x(s)^*)\bigr\}(1+s)\,d\nu_n(s)\biggr], \label{F-3.12}
\end{align}
where the supremum over $x(\cdot)$ is taken over all $L$-valued step functions as the
infimum in Lemma \ref{L-3.4}.
\end{thm}

\begin{proof}
By \eqref{F-3.11} and \eqref{F-2.7} we have
\begin{align*}
S_{f_n}(\rho\|\sigma)
&=f_n(0^+)\sigma(1)+f_n'(+\infty)\rho(1)+S_{-h_n}(\rho\|\sigma) \\
&=f_n(0^+)\sigma(1)+f_n'(+\infty)\rho(1)
-\int_{(0,+\infty)}h_n(t)\,d\|E_{\rho,\sigma}(t)\xi_\sigma\|^2.
\end{align*}
By Lemma \ref{L-3.4} we hence have
\begin{align*}
S_{f_n}(\rho\|\sigma)
&=\sup_{x(\cdot)}\biggl[f_n(0^+)\sigma(1)+f_n'(+\infty)\rho(1) \\
&\qquad-\int_{[1/n,n]}\bigl\{\sigma((1-x(s))^*(1-x(s)))
+s^{-1}\rho(x(s)x(s)^*)\bigr\}(1+s)\,d\nu_n(s)\biggr].
\end{align*}
The result follows by taking $\sup_n$ of both sides of the above and using Lemma \ref{L-3.2}.
\end{proof}

\begin{example}\label{E-12}\rm
Consider $f(t)=-\log t$, whose integral expression in \eqref{F-3.1} is
$$
-\log t=-(t-1)+\int_{(0,+\infty)}{(t-1)^2\over(t+s)(1+s)^2}\,ds.
$$
Hence, in this case,
$$
a=c=d=0,\quad b=-1,\quad d\mu(s)={1\over(1+s)^2}\,ds,
$$
$$
f(0^+)=+\infty,\qquad f'(+\infty)=0.
$$
Moreover, compute
$$
f_n(0^+)=1+\int_{1/n}^n{1\over s(1+s)^2}\,ds={2\over n+1}+\log n,
$$
$$
f_n'(+\infty)=-1+\int_{1/n}^n{1\over(1+s)^2}\,ds=-{2\over n+1},
$$
$$
d\nu_n(s)=1_{[1/n,n]}(s){1\over s(1+s)}\,ds.
$$
For every $\rho,\sigma\in M_*^+$ the \emph{relative entropy} is
$$
D(\sigma\|\rho)=S_{t\log t}(\sigma\|\rho)=S_{-\log t}(\rho\|\sigma),
$$
for which one can write expression \eqref{F-3.12} as
\begin{align*}
D(\sigma\|\rho)&=\sup_{n\in\mathbb{N}}\sup_{x(\cdot)}
\biggl[\sigma(1)\log n+(\sigma(1)-\rho(1)){2\over n+1} \\
&\qquad\quad-\int_{[1/n,n]}\bigl\{\sigma((1-x(s))^*(1-x(s)))
+s^{-1}\rho(x(s)x(s)^*)\bigr\}s^{-1}\,ds\biggr].
\end{align*}
This expression is similar to but a bit different from the variational expression
\begin{align*}
D(\sigma\|\rho)&=\sup_{n\in\mathbb{N}}\sup_{x(\cdot)}\biggl[\sigma(1)\log n \\
&\qquad\quad-\int_{[1/n,+\infty)}\bigl\{\sigma((1-x(s))^*(1-x(s)))
+s^{-1}\rho(x(s)x(s)^*)\bigr\}s^{-1}\,ds\biggr]
\end{align*}
in \cite[Theorem 3.2]{Ko}.
\end{example}

\begin{remark}\rm
The variational expression in \eqref{F-3.12} with the cut-off interval $[1/n,n]$ is natural
when we consider $S_f(\rho\|\sigma)$ for general operator convex functions $f$ on
$(0,+\infty)$ with no assumption on the boundary values $f(0^+)$ and $f'(+\infty)$. This is
more explicitly justified by the fact that our expression is well behaved under taking the
transpose $\widetilde f(t)=tf(t^{-1})$. Indeed, for $f$ given in \eqref{F-3.3}, the integral
expression of $\widetilde f$ is
$$
\widetilde f(t)=\tilde a+\tilde b(t-1)+\tilde c(t-1)^2+\tilde d\,{(t-1)^2\over t}
+\int_{(0,+\infty)}{(t-1)^2\over t+s}\,d\widetilde\mu(s)
$$
where
$$
\tilde a=a,\qquad\tilde b=a-b,\qquad\tilde c=d,\qquad\tilde d=c,\qquad
d\widetilde\mu(s)=s\,d\mu(s^{-1}).
$$
Hence one can easily find that $\widetilde f_n(0^+)=f_n'(+\infty)$,
$\widetilde f_n'(+\infty)=f_n(0^+)$, $d\widetilde\nu_n(s)=d\nu_n(s^{-1})$, and the
expression inside the bracket $[\cdots]$ of \eqref{F-3.12} for
$S_{\widetilde f}(\rho\|\sigma)$ is
\begin{align*}
&\widetilde f_n(0^+)\sigma(1)+\widetilde f_n'(+\infty)\rho(1) \\
&\quad-\int_{[1/n,n]}\bigl\{\sigma((1-x(s))^*(1-x(s)))
+s^{-1}\rho(x(s)x(s)^*)\bigr\}(1+s)\,d\widetilde\nu_n(s) \\
&=f_n(0^+)\rho(1)+f_n'(+\infty)\sigma(1) \\
&\quad-\int_{[1/n,n]}\bigl\{\rho((1-y(s))^*(1-y(s)))
+s^{-1}\sigma(y(s)y(s)^*)\bigr\}(1+s)\,d\nu_n(s),
\end{align*}
where $y(s):=1-x(s^{-1})^*$. In this way, the variational expression in \eqref{F-3.12} enjoys
complete invariance under exchanging $(f,\rho,\sigma)$ with $(\widetilde f,\sigma,\rho)$.
\end{remark}

\section{Properties of standard $f$-divergences}

As in \cite{Ko} where the relative entropy was treated, most of the important properties of
standard $f$-divergences can immediately be verified from the variational expression in
Theorem \ref{T-3.5}.

\begin{thm}\label{T-4.1}
Let $f$ be an operator convex function on $(0,+\infty)$. Let
$\rho,\sigma,\rho_i,\sigma_i\in M_*^+$ for $i=1,2$.
\begin{itemize}
\item[(i)] \emph{Joint lower semicontinuity:} The map
$(\rho,\sigma)\in M_*^+\times M_*^+\mapsto S_f(\rho\|\sigma)\in(-\infty,+\infty]$ is jointly
lower semicontinuous in the $\sigma(M_*,M)$-topology.
\item[(ii)] \emph{Joint convexity:} The map in (i) is jointly convex and jointly subadditive,
i.e., for every $\rho_i,\sigma_i\in M_*^+$, $1\le i\le k$,
$$
S_f\Biggl(\sum_{i=1}^k\rho_i\Bigg\|\sum_{i=1}^k\sigma_i\Biggr)
\le\sum_{i=1}^kS_f(\rho_i\|\sigma_i).
$$
\item[(iii)] If $f(0^+)\le0$ and $\sigma_1\le\sigma_2$, then
$S_f(\rho\|\sigma_1)\ge S_f(\rho\|\sigma_2)$. Also, if $f'(+\infty)\le0$ and
$\rho_1\le\rho_2$, then $S_f(\rho_1\|\sigma)\ge S_f(\rho_2\|\sigma)$.
\item[(iv)] \emph{Monotonicity:} Let $N$ be another von Neumann algebra and $\Phi:N\to M$ be
a unital positive linear map that is normal (i.e., if $\{x_\alpha\}$ is an increasing net in
$M_+$ with $x_\alpha\nearrow x\in M_+$, then $\Phi(x_\alpha)\nearrow\Phi(x)$) and is a Schwarz
map (i.e., $\Phi(x^*x)\ge\Phi(x)^*\Phi(x)$ for all $x\in N$). Then
\begin{align}\label{F-4.1}
S_f(\rho\circ\Phi\|\sigma\circ\Phi)\le S_f(\rho\|\sigma).
\end{align}
In particular, if $N$ is a unital von Neumann subalgebra of $M$, then
\begin{align}\label{F-4.2}
S_f(\rho|_N\|\sigma|_N)\le S_f(\rho\|\sigma).
\end{align}
\item[(v)] \emph{Martingale convergence:} If $\{M_\alpha\}$ is an increasing net of unital
von Neumann subalgebras of $M$ such that $\bigl(\bigcup_\alpha M_\alpha\bigr)''=M$, then
$$
S_f(\rho|_{M_\alpha}\|\sigma|_{M_\alpha})\ \nearrow\  S_f(\rho\|\sigma).
$$
\end{itemize}
\end{thm}

\begin{proof}
To prove (i)--(iv), we apply expression \eqref{F-3.12} with $L=M$. Since $(1+s)d\nu_n(s)$ is
a finite positive measure supported on $[1/n,n]$ (see \eqref{F-3.10}), it is clear that the
function of $(\rho,\sigma)$ inside the bracket $[\cdots]$ in \eqref{F-3.12} is linear and
continuous in the $\sigma(M_*,M)$-topology, so (i) and (ii) are shown. Here note that joint
convexity and joint subadditivity in (ii) are equivalent due to the homogeneity property in
\eqref{F-2.8}.

Assume that $f(0^+)\le0$; then $f_n(0^+)\le f(0^+)\le0$ for all $n\in\mathbb{N}$. Hence the first
assertion of (iii) is obvious by expression \eqref{F-3.12}, and the second assertion is
similar.

To prove (iv), note first that $\rho\circ\Phi,\sigma\circ\Phi\in N_*^+$ since $\Phi$ is a
normal positive linear map. For any $N$-valued step function $x(\cdot)$ on $(0,+\infty)$,
let $y(s):=\Phi(x(s))$, which is an $M$-valued step function. Since $\Phi$ is a unital
Schwarz map, one has
\begin{align*}
\sigma((1-y(s))^*(1-y(s)))&\le\sigma\circ\Phi((1-x(s))^*(1-x(s))), \\
\rho(y(s)y(s)^*)&\le\rho\circ\Phi(x(s)x(s)^*),
\end{align*}
which implies that the bracket $[\cdots]$ in \eqref{F-3.12} for $\rho\circ\Phi$,
$\sigma\circ\Phi$ and $x(\cdot)$ is dominated by $S_f(\rho\|\sigma)$. Hence inequality
\eqref{F-4.1} follows. When $N$ is a unital von Neumann subalgebra, applying \eqref{F-4.1}
to the injection $\Phi:N\hookrightarrow M$ gives \eqref{F-4.2}.

To prove (v), apply \eqref{F-3.12} with $L=\bigcup_\alpha M_\alpha$. When we restrict
$x(\cdot)$ in \eqref{F-3.12} to $M_\alpha$-valued step functions, we have the expression of
$S_f(\rho|_{M_\alpha}\|\sigma|_{M_\alpha})$. This shows that
$S_f(\rho|_{M_\alpha}\|\sigma|_{M_\alpha})$ is increasing and
$S_f(\rho|_{M_\alpha}\|\sigma|_{M_\alpha})\le S_f(\rho\|\sigma)$. (This also follows from
monotonicity in \eqref{F-4.2}.) Hence it remains to show that
$S_f(\rho\|\sigma)\le\sup_\alpha S_f(\rho|_{M_\alpha}\|\sigma|_{M_\alpha})$. For any
$c<S_f(\rho\|\sigma)$ we choose an $n\in\mathbb{N}$ and an $L$-valued step function $x(\cdot)$ such
that
\begin{align*}
c&<f_n(0^+)\sigma(1)+f_n'(+\infty)\rho(1) \\
&\quad-\int_{[1/n,n]}\bigl\{\sigma((1-x(s))^*(1-x(s)))
+s^{-1}\rho(x(s)x(s)^*)\bigr\}(1+s)\,d\nu_n(s).
\end{align*}
Since $x(\cdot)$ is $M_\alpha$-valued for some $\alpha$, we have
$c<S_f(\rho|_{M_\alpha}\|\sigma|_{M_\alpha})$, implying the desired conclusion.
\end{proof}

The next corollary shows that $S_f(\rho\|\sigma)$ is strictly positive in some typical
situation.

\begin{cor}\label{C-4.2}
Let $\rho,\sigma\in M_*^+$.
\begin{itemize}
\item[(1)] The \emph{Peierls-Bogolieubov inequality} holds:
\begin{align}\label{F-4.3}
S_f(\rho\|\sigma)\ge\sigma(1)f(\rho(1)/\sigma(1)).
\end{align}
Assume that $f$ is non-linear and $\rho,\sigma\ne0$. Then equality holds in \eqref{F-4.3}
if and only if $\rho=(\rho(1)/\sigma(1))\sigma$.
\item[(2)] \emph{Strict positivity:} Assume that $f$ is non-linear with $f(1)=0$ and
$\rho(1)=\sigma(1)>0$. Then $S_f(\rho\|\sigma)\ge0$, and $S_f(\rho\|\sigma)=0$ $\iff$
$\rho=\sigma$.
\end{itemize}
\end{cor}

\begin{proof}
(1)\enspace
When $N:=\bC1$ in \eqref{F-4.2}, inequality \eqref{F-4.3} arises. If $\rho=k\sigma$ with a
constant $k>0$, then we have $\Delta_{\rho,\sigma}=k\Delta_\sigma$, $\Delta_\sigma$ being
the modular operator for $\sigma$, and hence
$d\|E_{\rho,\sigma}(t)\xi_\sigma\|^2=\sigma(1)d\delta_1(t)$, giving
$S_f(\rho\|\sigma)=f(k)\sigma(1)$. Conversely, assume that $\rho,\sigma\ne0$ and equality
holds in \eqref{F-4.3}. Further, assume that $f$ is non-linear. Since $f$ is operator convex
on $(0,+\infty)$, it is strictly convex there. For every projection $e\in M$, applying
\eqref{F-4.2} to $N:=\mathbb{C} e+\mathbb{C} e^\perp$ (where $e^\perp:=1-e$) gives
$$
S_f(\rho\|\sigma)\ge S_f(\rho|_N\|\sigma|_N)=\sigma(e)f(\rho(e)/\sigma(e))
+\sigma(e^\perp)f(\rho(e^\perp)/\sigma(e^\perp)).
$$
From this and \eqref{F-4.3} for $\rho|_N$ and $\sigma|_N$ one has
$$
\sigma(1)f(\rho(1)/\sigma(1))
=\sigma(e)f(\rho(e)/\sigma(e))+\sigma(e^\perp)f(\rho(e^\perp)/\sigma(e^\perp)).
$$
By Lemma \ref{L-4.3} below one has $\rho(e)=k\sigma(e)$ and $\rho(e^\perp)=k\sigma(e^\perp)$
for some $k>0$. Since $k=\rho(1)/\sigma(1)$ follows, we find that
$\rho(e)=(\rho(1)/\sigma(1))\sigma(e)$ for all projections $e\in M$, showing that
$\rho=(\rho(1)/\sigma(1))\sigma$.

(2) is immediately seen from (1).
\end{proof}

The next elementary lemma has been used in the above, whose proof is given for completeness,
since we find no suitable reference.

\begin{lemma}\label{L-4.3}
Let $f:(0,+\infty)\to\mathbb{R}$ is a strictly convex function (not necessarily operator convex).
Let $a_i,b_i\in[0,+\infty)$ for $i=1,2$ be such that $a_1+a_2>0$ and $b_1+b_2>0$. If
$$
(b_1+b_2)f\biggl({a_1+a_2\over b_1+b_2}\biggr)=b_1f(a_1/b_1)+b_2f(a_2/b_2)
$$
with convention \eqref{F-2.3}, then $(a_1,a_2)=k(b_1,b_2)$ for some $k>0$.
\end{lemma}

\begin{proof}
We may consider the following four cases separately.

{\it Case $a_i,b_i>0$ for $i=1,2$}.\enspace
Since
$$
(b_1+b_2)f\biggl({b_1\over b_1+b_2}\cdot{a_1\over b_1}
+{b_2\over b_1+b_2}\cdot{a_2\over b_2}\biggr)=b_1f(a_1/b_1)+b_2f(a_2/b_2),
$$
the strict convexity of $f$ implies that $a_1/b_1=a_2/b_2$.

{\it Case $a_1=0$ and $b_1,b_2>0$}.\enspace
The assumption means that
$$
(b_1+b_2)f\biggl({a_2\over b_1+b_2}\biggr)=b_1f(0/b_1)+b_2f(a_2/b_2)
=b_1f(0^+)+b_2f(a_2/b_2),
$$
which implies that $f(0^+)<+\infty$. Hence $f$ extends to a strictly convex function on
$[0,+\infty)$, and the above equality gives $a_2/b_2=0$, which is impossible since
$a_1+a_2>0$.

{\it Case $a_1,a_2>0$ and $b_1=0$}.\enspace
This case reduces to the previous case if we replace $f$ with its transpose $\widetilde f$.

{\it Case $a_1=b_1=0$ or $a_2=b_2=0$}.\enspace
The assertion trivially holds in this case.

{\it Case $a_1=b_2=0$}.\enspace
The assumption means that
$$
b_1f(a_2/b_1)=b_1f(0/b_1)+0f(a_2/0)=f(0^+)b_1+f'(+\infty)a_2,
$$
which implies that $f(0^+)<+\infty$ and $f'(+\infty)<+\infty$. Then it is easy to find that
$f(t)<f(0^+)+f'(+\infty)t$ for all $t>0$, which contradicts the above equality.
\end{proof}

For $\sigma\in M_*^+$ and a projection $e\in M$, we write $e\sigma e$ for the restriction of
$\sigma$ to the reduced von Neumann algebra $eMe$.

\begin{cor}\label{C-4.4}
\begin{itemize}
\item[(1)] If $e\in M$ is a projection such that $s_M(\rho),s_M(\sigma)\le e$, then
\begin{align}\label{F-4.4}
S_f(\rho\|\sigma)=S_f(e\rho e\|e\sigma e).
\end{align}
\item[(2)] If $\rho_i,\sigma_i\in M_*^+$, $i=1,2$, and
$s_M(\rho_1)\vee s_M(\sigma_1)\perp s_M(\rho_2)\vee s_M(\sigma_2)$, then
\begin{align}\label{F-4.5}
S_f(\rho_1+\rho_2\|\sigma_1+\sigma_2)=S_f(\rho_1\|\sigma_1)+S_f(\rho_2\|\sigma_2).
\end{align}
\item[(3)] If $\omega_1,\omega_2\in M_*^+$ and $S_f(\omega_1\|\omega_2)<+\infty$, then for
every $\rho,\sigma\in M_*^+$,
$$
S_f(\rho\|\sigma)=\lim_{\varepsilon\searrow0}S_f(\rho+\varepsilon\omega_1\|\sigma+\varepsilon\omega_2).
$$
In particular, for every $\rho,\sigma,\omega\in M_*^+$,
\begin{align}\label{F-4.6}
S_f(\rho\|\sigma)=\lim_{\varepsilon\searrow0}S_f(\rho+\varepsilon\omega\|\sigma+\varepsilon\omega).
\end{align}
\end{itemize}
\end{cor}

\begin{proof}
(1)\enspace
By monotonicity \eqref{F-4.2} we have $S_f(e\rho e\|e\sigma e)\le S_f(\rho\|\sigma)$. For any
$M$-valued step function $x(\cdot)$ on $(0,+\infty)$, let $y(s):=ex(s)e$, which is an
$eMe$-valued step function. Since
\begin{align*}
(e\sigma e)((e-y(s))^*(e-y(s)))&=\sigma(e(1-x(s))^*e(1-x(s))e)
\le\sigma((1-x(s))^*(1-s(s))), \\
(e\rho e)(y(s)y(s)^*)&=\sigma(ex(s)ex(s)^*e)\le\sigma(x(s)x(s)^*),
\end{align*}
the bracket $[\cdots]$ in \eqref{F-3.12} for $x(\cdot)$ is dominated by
$S_f(e\rho e\|e\sigma e)$. Hence equality \eqref{F-4.4} follows.

(2)\enspace
Let $e:=s_M(\rho_1)\vee s_M(\rho_2)$ and so $s_M(\rho_2)\vee s_M(\sigma_2)\le e^\perp$. Let
$\rho:=\rho_1+\rho_2$, $\sigma:=\sigma_1+\sigma_2$, and $N:=eMe\oplus e^\perp Me^\perp$.
Since $\rho|_N=e\rho_1e\oplus e^\perp\rho_2e^\perp$ and
$\sigma|_N=e\sigma_1e\oplus e^\perp\sigma_2e^\perp$, from monotonicity \eqref{F-4.2} and
Proposition \ref{P-2.3}\,(4) one has
$$
S_f(\rho\|\sigma)\ge S_f(e\rho_1e\|e\sigma_1e)
+S_f(e^\perp\rho_2e^\perp\|e^\perp\sigma_2e^\perp)
=S_f(\rho_1\|\sigma_1)+S_f(\rho_2\|\sigma_2),
$$
where we have used (1) for the last equality. On the other hand, consider the map
$$
\Phi: M\,\longrightarrow\,eMe\oplus e^\perp Me^\perp,\quad
\Phi(x):=exe+e^\perp xe^\perp,
$$
which is unital and completely positive (hence a Schwarz map). Since
$\rho=(e\rho_1e\oplus e^\perp\rho_2e^\perp)\circ\Phi$ and
$\sigma=(e\sigma_1e\oplus e^\perp\sigma_2e^\perp)\circ\Phi$, from monotonicity \eqref{F-4.1}
and Proposition \ref{P-2.3}\,(4) one has
$$
S_f(\rho\|\sigma)\le S_f(\rho_1\|\sigma_1)+S_f(\rho_2\|\sigma_2).
$$
Hence equality \eqref{F-4.5} is shown.

(3)\enspace
From joint subadditivity in Theorem \ref{T-4.1}\,(ii) and homogeneity \eqref{F-2.9} one has
$$
S_f(\rho+\varepsilon\omega_1\|\sigma+\varepsilon\omega_2)\le
S_f(\rho\|\sigma)+\varepsilon S_f(\omega_1\|\omega_2)
\,\longrightarrow\,S_f(\rho\|\sigma)
$$
as $\varepsilon\searrow0$. On the other hand, from lower semicontinuity in Theorem \ref{T-4.1}\,(i)
one has
$$
S_f(\rho\|\sigma)\le\liminf_{\varepsilon\searrow0}S_f(\rho+\varepsilon\omega_1\|\sigma+\varepsilon\omega_2),
$$
showing the asserted convergence.
\end{proof}

The additivity in Corollary \ref{C-4.4}\,(2) improves that in Proposition \ref{P-2.3}\,(4);
yet we have used the latter in the above proof of the former. When $M$ is $\sigma$-finite so
that a faithful $\omega\in M_*^+$ exists, we can sometimes reduce arguments on
$S_f(\rho\|\sigma)$ to the case of faithful $\rho,\sigma\in M_*^+$ by using the convergence
property in \eqref{F-4.6}.

The next continuity property is not included in the martingale convergence in Theorem
\ref{T-4.1}, since $eMe$ is not a unital von Neumann subalgebra of $M$.

\begin{thm}\label{T-4.5}
Let $\{e_\alpha\}$ be an increasing net of projections in $M$ such that $e_\alpha\nearrow1$.
Then for every $\rho,\sigma\in M_*^+$,
$$
\lim_\alpha S_f(e_\alpha\rho e_\alpha\|e_\alpha\sigma e_\alpha)=S_f(\rho\|\sigma).
$$
\end{thm}

\begin{proof}
By replacing $f$ with $f(t)-(a+bt)$ and noting \eqref{F-2.7}, we may assume that $f(t)\ge0$
for all $t\in(0,+\infty)$. Let $M_\alpha:=e_\alpha Me_\alpha+\mathbb{C}(1- e_\alpha)$; then
$\{M_\alpha\}$ is an increasing net of von Neumann subalgebras with
$\bigl(\bigcup_\alpha M_\alpha\bigr)''=M$. Hence the martingale convergence in Theorem
\ref{T-4.1} and Corollary \ref{C-4.4}\,(2) imply that
\begin{align}\label{F-4.7}
S_f(\rho\|\sigma)
=\lim_\alpha\biggl[S_f(e_\alpha\rho e_\alpha\|e_\alpha\sigma e_\alpha)
+\sigma(1-e_\alpha)f\biggl({\rho(1-e_\alpha)\over\sigma(1-e_\alpha)}\biggr)\biggr],
\end{align}
increasingly in $\alpha$. First, assume that $S_f(\rho\|\sigma)=+\infty$ and prove that
\begin{align}\label{F-4.8}
\lim_\alpha S_f(e_\alpha\rho e_\alpha\|e_\alpha\sigma e_\alpha)=+\infty.
\end{align}
If $\limsup_\alpha\sigma(1-e_\alpha)f(\rho(1-e_\alpha)/\sigma(1-e_\alpha))<+\infty$, then
\eqref{F-4.8} clearly follows from \eqref{F-4.7}. Assume that
$\limsup_\alpha\sigma(1-e_\alpha)f(\rho(1-e_\alpha)/\sigma(1-e_\alpha))=+\infty$. Then for
any $K>0$ choose an $\alpha_0$ such that
$\sigma(1-e_{\alpha_0})f(\rho(1-e_{\alpha_0})/\sigma(1-e_{\alpha_0}))>K$. Since
$$
\rho(e_\alpha-e_{\alpha_0})\nearrow\rho(1-e_{\alpha_0}),\quad
\sigma(e_\alpha-e_{\alpha_0})\nearrow\sigma(1-{\alpha_0})\qquad
\mbox{as $\alpha_0\le\alpha\to``\infty"$},
$$
we easily see that
$$
\sigma(e_\alpha-e_{\alpha_0})f\biggl({\rho(e_\alpha-e_{\alpha_0})\over
\sigma(e_\alpha-e_{\alpha_0})}\biggr)
\ \longrightarrow\ \sigma(1-e_{\alpha_0})f\biggl({\rho(1-e_{\alpha_0})\over
\sigma(1-e_{\alpha_0})}\biggr)>K.
$$
By monotonicity of $S_f$ and the assumption $f\ge0$ we have for $\alpha\ge\alpha_0$
\begin{align*}
S_f(e_\alpha\rho e_\alpha\|e_\alpha\sigma e_\alpha)
&\ge\sigma(e_\alpha-e_{\alpha_0})f\biggl({\rho(e_\alpha-e_{\alpha_0})\over
\sigma(e_\alpha-e_{\alpha_0})}\biggr)
+\sigma(e_{\alpha_0})f\biggl({\rho(e_{\alpha_0})\over\sigma(e_{\alpha_0})}\biggr) \\
&\ge\sigma(e_\alpha-e_{\alpha_0})f\biggl({\rho(e_\alpha-e_{\alpha_0})\over
\sigma(e_\alpha-e_{\alpha_0})}\biggr),
\end{align*}
which is $>K$ for all sufficiently large $\alpha\ge\alpha_0$. Hence \eqref{F-4.8} follows.

Next, assume that $S_f(\rho\|\sigma)<+\infty$, and prove that
$\lim_\alpha S_f(e_\alpha\rho e_\alpha\|e_\alpha\sigma e_\alpha)=S_f(\rho\|\sigma)$. To do
this, by \eqref{F-4.7} we may prove that
$\lim_\alpha\sigma(1-e_\alpha)f(\rho(1-e_\alpha)/\sigma(1-e_\alpha))=0$. Assume on the
contrary that $\limsup_\alpha\sigma(1-e_\alpha)f(\rho(1-e_\alpha)/\sigma(1-e_\alpha))>\varepsilon>0$
for some $\varepsilon>0$ (here recall that $f\ge0$ was assumed). Choose an $\alpha_1$ such that
$\sigma(1-e_{\alpha_1})f(\rho(1-e_{\alpha_1})/\sigma(1-e_{\alpha_1}))>\varepsilon$.
Then we can choose a $\beta_1>\alpha_1$ such that
$$
\sigma(e_{\beta_1}-e_{\alpha_1})f\biggl({\rho(e_{\beta_1}-e_{\alpha_1})\over
\sigma(e_{\beta_1}-e_{\alpha_1})}\biggr)>\varepsilon.
$$
Next, choose an $\alpha_1>\beta_2$ such that
$\sigma(1-e_{\alpha_2})f(\rho(1-e_{\alpha_2})/\sigma(1-e_{\alpha_2}))>\varepsilon$, and
a $\beta_2>\alpha_2$ such that
$$
\sigma(e_{\beta_2}-e_{\alpha_2})f\biggl({\rho(e_{\beta_2}-e_{\alpha_2})\over
\sigma(e_{\beta_2}-e_{\alpha_2})}\biggr)>\varepsilon.
$$
Repeating the above argument we have $\alpha_1<\beta_1<\alpha_2<\beta_2<\cdots$ in such a
way that
$$
\sigma(e_{\beta_k}-e_{\alpha_k})f\biggl({\rho(e_{\beta_k}-e_{\alpha_k})\over
\sigma(e_{\beta_k}-e_{\alpha_k})}\biggr)>\varepsilon
$$
for all $k\in\mathbb{N}$. Let $e_{\alpha_k}\nearrow e_\infty$ and $e_0:=1-e_\infty$, and consider a
unital abelian von Neumann subalgebra of $M$
$$
\bigoplus_{k=1}^\infty\mathbb{C}(e_{\beta_k}-e_{\alpha_k})\oplus
\bigoplus_{k=1}^\infty\mathbb{C}(e_{\alpha_k}-e_{\beta_{k-1}})\oplus\mathbb{C} e_0,
$$
where $e_{\beta_0}:=0$. By monotonicity of $S_f$ and Example \ref{E-2.5} together with
$f\ge0$, we have
$$
S_f(\rho\|\sigma)\ge\sum_{k=1}^\infty
\sigma(e_{\beta_k}-e_{\alpha_k})f\biggl({\rho(e_{\beta_k}-e_{\alpha_k})\over
\sigma(e_{\beta_k}-e_{\alpha_k})}\biggr)
+\sigma(e_0)f\biggl({\rho(e_0)\over\sigma(e_0)}\biggr)
=+\infty,
$$
which contradicts the assumption $S_f(\rho\|\sigma)<+\infty$.
\end{proof}

When $f\ge0$ in Theorem \ref{T-4.5}, from the monotonicity of $S_f$ we see that
$S_f(e_\alpha\rho e_\alpha\|e_\alpha\sigma e_\alpha)$ is increasing as $e_\alpha\nearrow1$.
But this is not the case unless $f\ge0$.

\begin{remark}\rm
When $M=B(\mathcal{H})$ with $\dim\mathcal{H}=\infty$, according to Theorem \ref{T-4.5}, one can define the
relative entropy $D(\rho\|\sigma)$ for trace-class operators $\rho,\sigma\ge0$ as
\begin{align}\label{F-4.9}
D(\rho\|\sigma)=\lim_\alpha D(E_\alpha\rho E_\alpha\|E_\alpha\sigma E_\alpha),
\end{align}
where $\{E_\alpha\}$ is an increasing net of finite rank projections with
$E_\alpha\nearrow I$. But it seems that there is no simpler proof other than that of Theorem
\ref{T-4.5} for the existence of the limit in \eqref{F-4.9} and its independence of the
choice of $\{E_\alpha\}$.
\end{remark}

\section{R\'enyi divergences}

We define the notion of R\'enyi divergences $D_\alpha(\rho\|\sigma)$ for $\alpha\ge0$ in the
general von Neumann algebra setting.

\begin{definition}\label{D-5.1}\rm
Let $\rho,\sigma\in M_*^+$. Since $\xi_\sigma\in\mathcal{D}(\Delta_{\rho,\sigma}^{1/2})$, the domain
of $\Delta_{\rho,\sigma}^{1/2}$, note that $\xi_\sigma\in\mathcal{D}(\Delta_{\rho,\sigma}^{\alpha/2})$
for any $\alpha\in[0,1]$. We define the quantities $Q_\alpha(\rho\|\sigma)$ for $\alpha\ge0$
as follows: When $0\le\alpha<1$,
\begin{align}\label{F-5.1}
Q_\alpha(\rho\|\sigma):=\|\Delta_{\rho,\sigma}^{\alpha/2}\xi_\sigma\|^2
\ \in[0,+\infty),
\end{align}
and when $\alpha>1$,
\begin{align}\label{F-5.2}
Q_\alpha(\rho\|\sigma):=\begin{cases}
\|\Delta_{\rho,\sigma}^{\alpha/2}\xi_\sigma\|^2 &
\text{if $s_M(\rho)\le s_M(\sigma)$ and
$\xi_\sigma\in\mathcal{D}(\Delta_{\rho,\sigma}^{\alpha/2})$}, \\
+\infty & \text{otherwise}.
\end{cases}
\end{align}
Moreover, when $\alpha=1$, define $Q_1(\rho\|\sigma):=\rho(1)$. Then for every
$\rho,\sigma\in M_*^+$ with $\rho\ne0$ and for each $\alpha\in[0,+\infty)\setminus\{1\}$,
the \emph{$\alpha$-R\'enyi divergence} $D_\alpha(\rho\|\sigma)$ is defined by
\begin{align}\label{F-5.3}
D_\alpha(\rho\|\sigma):={1\over\alpha-1}\log{Q_\alpha(\rho\|\sigma)\over\rho(1)}.
\end{align}
\end{definition}

In particular, note that $Q_0(\alpha\|\sigma)=\sigma(s_M(\rho))$ and
$D_0(\rho\|\sigma)=-\log\bigl[\sigma(s_M(\rho))/\rho(1)\bigr]$. The next lemma shows that
$Q_\alpha(\rho\|\sigma)$ is essentially a standard $f$-divergence and so
$D_\alpha(\rho\|\sigma)$ is a variant of standard $f$-divergences.

\begin{lemma}\label{L-5.2}
Define convex functions $f_\alpha$ on $[0,+\infty)$ by
$$
f_\alpha(t):=\begin{cases}t^\alpha & \text{if $\alpha\ge1$}, \\
-t^\alpha & \text{if $0<\alpha<1$}.
\end{cases}
$$
Then for every $\rho,\sigma\in M_*^+$, $Q_\alpha(\rho\|\sigma)$ is given as
\begin{align}\label{F-5.4}
Q_\alpha(\rho\|\sigma)=\begin{cases}
S_{f_\alpha}(\rho\|\sigma) & \text{if $\alpha\ge1$}, \\
-S_{f_\alpha}(\rho\|\sigma) & \text{if $0<\alpha<1$}.
\end{cases}
\end{align}
Moreover,
\begin{align}\label{F-5.5}
Q_\alpha(\rho\|\sigma)=\begin{cases}
\int_{(0,+\infty)}t^\alpha\,d\|E_{\rho,\sigma}(t)\xi_\sigma\|^2
& \text{if $0\le\alpha<1$ or $s_M(\rho)\le s_M(\sigma)$}, \\
+\infty & \text{if $\alpha>1$ and $s_M(\rho)\not\le s_M(\sigma)$}.
\end{cases}
\end{align}
\end{lemma}

\begin{proof}
When $0<\alpha<1$, since $f_\alpha(0)=f_\alpha'(+\infty)=0$, we have by \eqref{F-5.1}
$$
S_{f_\alpha}(\rho\|\sigma)
=\int_{(0,+\infty)}(-t^\alpha)\,d\|E_{\rho,\sigma}(t)\xi_\sigma\|^2
=-Q_\alpha(\rho\|\sigma).
$$
When $\alpha=1$, \eqref{F-2.7} gives $S_{f_1}(\rho\|\sigma)=\rho(1)=Q_1(\rho\|\sigma)$.
When $\alpha>1$, since $f_\alpha(0)=0$ and $f_\alpha'(+\infty)=+\infty$,
$$
S_{f_\alpha}(\rho\|\sigma)
=\int_{(0,+\infty)}t^\alpha\,d\|E_{\rho,\sigma}(t)\xi_\sigma\|^2
+(+\infty)\rho(1-s_M(\sigma)).
$$
Note that $\rho(1-s_M(\sigma))=0$ $\iff$ $s_M(\rho)\le s_M(\sigma)$, and
$\int_{(0,+\infty)}t^\alpha\,d\|E_{\rho,\sigma}(t)\xi_\sigma\|^2<+\infty$ $\iff$
$\xi_\sigma\in\mathcal{D}(\Delta_{\rho,\sigma}^{\alpha/2})$. Hence by \eqref{F-5.2} we see that
$S_{f_\alpha}(\rho\|\sigma)=Q_\alpha(\rho\|\sigma)$. Moreover, \eqref{F-5.5} immediately
follows from the above argument, where the case $\alpha=0$ is obvious.
\end{proof}

Some properties of $Q_\alpha$ and $D_\alpha$ are found in, e.g., \cite{Li,MH,Hi0,Mo} though
mostly in the finite-dimensional situation. More comprehensive summary of them are given in
the next proposition, mainly based on Theorem \ref{T-4.1}. Although (3) and (4) when
$\alpha\in[0,2]$ have been shown in \cite{BST}, we give their proofs as well for convenience
of the reader.

\begin{prop}\label{P-5.3}
Let $\rho,\sigma\in M_*^+$ with $\rho\ne0$.
\begin{itemize}
\item[(1)] If $s_M(\rho)\perp s_M(\sigma)$, then $D_\alpha(\rho\|\sigma)=+\infty$ for all
$\alpha\in[0,+\infty)\setminus\{1\}$.
\item[(2)] If $s_M(\rho)\not\perp s_M(\sigma)$, then $Q_\alpha(\rho\|\sigma)>0$ for all
$\alpha\ge0$ and the function $\alpha\in[0,+\infty)\mapsto\log Q_\alpha(\rho\|\sigma)$ is
convex.
\item[(3)] The limit $D_1(\rho\|\sigma):=\lim_{\alpha\nearrow1}D_\alpha(\rho\|\sigma)$ exists
and
\begin{align}\label{F-5.6}
D_1(\rho\|\sigma)={D(\rho\|\sigma)\over\rho(1)},
\end{align}
where $D(\rho\|\sigma)$ is the relative entropy. Moreover, if $D_\alpha(\rho\|\sigma)<+\infty$
for some $\alpha>1$, then $\lim_{\alpha\searrow1}D_\alpha(\rho\|\sigma)=D_1(\rho\|\sigma)$.
\item[(4)] The function $\alpha\in[0,+\infty)\mapsto D_\alpha(\rho\|\sigma)$ is monotone
increasing.
\item[(5)] Assume that $0<\alpha<1$. We have
\begin{align}\label{F-5.7}
Q_\alpha(\rho\|\sigma)=Q_{1-\alpha}(\rho\|\sigma),
\end{align}
and whenever $\rho,\sigma\ne0$,
\begin{align}\label{F-5.8}
{1\over\alpha}\,D_\alpha(\rho\|\sigma)={1\over1-\alpha}\,D_{1-\alpha}(\sigma\|\rho)
+{1\over\alpha(1-\alpha)}\log{\rho(1)\over\sigma(1)}.
\end{align}
Hence, if $\rho(1)=\sigma(1)$, then
$\lim_{\alpha\searrow0}{1\over\alpha}\,D_\alpha(\rho\|\sigma)=D_1(\sigma\|\rho)$.
\item[(6)] \emph{Joint lower semicontinuity:} For every $\alpha\in[0,2]$, the map
$(\rho,\sigma)\in(M_*^+\setminus\{0\})\times M_*^+\mapsto D_\alpha(\rho\|\sigma)\in(-\infty,+\infty]$ is
jointly lower semicontinuous in the $\sigma(M_*,M)$-topology.
\item[(7)] The map $(\rho,\sigma)\in M_*^+\times M_*^+\mapsto Q_\alpha(\rho\|\sigma)$ is
jointly concave and jointly superadditive for $0\le\alpha\le1$, and jointly convex and
jointly subadditive for $1\le\alpha\le2$. Hence, when $0\le\alpha\le1$,
$D_\alpha(\rho\|\sigma)$ is jointly convex on
$\{(\rho,\sigma)\in M_*^+\times M_*^+:\rho(1)=c\}$ for any fixed $c>0$.
\item[(8)] Let $\rho_i,\sigma_i\in M_*^+$ for $i=1,2$. If $0\le\alpha<1$, $\rho_1\le\rho_2$
and $\sigma_1\le\sigma_2$, then $Q_\alpha(\rho_1\|\sigma_1)\le Q_\alpha(\rho_2\|\sigma_2)$.
If $1\le\alpha\le2$ and $\sigma_1\le\sigma_2$, then
$Q_\alpha(\rho\|\sigma_1)\ge Q_\alpha(\rho\|\sigma_2)$. If $\sigma_1\le\sigma_2$, then
$D_\alpha(\rho\|\sigma_1)\ge D_\alpha(\rho\|\sigma_2)$ for all $\alpha\in[0,2]$.
\item[(9)] \emph{Monotonicity:} For each $\alpha\in[0,2]$, $D_\alpha(\rho\|\sigma)$ is
monotone under unital normal Schwarz maps, i.e.,
\begin{align}\label{F-5.9}
D_\alpha(\rho\circ\Phi\|\sigma\circ\Phi)\le D_\alpha(\rho\|\sigma)
\end{align}
for any unital normal Schwarz map $\Phi:N\to M$ as in Theorem \ref{T-4.1}\,(iv).
\item[(10)] \emph{Strict positivity:} Let $\alpha\in(0,+\infty)$ and $\rho,\sigma\ne0$. The
inequality
\begin{align}\label{F-5.10}
D_\alpha(\rho\|\sigma)\ge\log{\rho(1)\over\sigma(1)}
\end{align}
holds, and equality holds in \eqref{F-5.10} if and only if $\rho=(\rho(1)/\sigma(1))\sigma$.
If $\rho(1)=\sigma(1)$, then $D_\alpha(\rho\|\sigma)\ge0$, and $D_\alpha(\rho\|\sigma)=0$
$\iff$ $\rho=\sigma$.
\end{itemize}
\end{prop}

\begin{proof}
Write $F(\alpha):=\int_{(0,+\infty)}t^\alpha\,d\mu(t)$ for $\alpha>0$, where
$d\mu(t):=d\|E_{\rho,\sigma}(t)\xi_\sigma\|^2$ for $t\in(0,+\infty)$. Note that
$F(1)=\rho(s_M(\sigma))$. By \eqref{F-5.5} we note that
\begin{itemize}
\item[(A)] if $s_M(\rho)\le s_M(\sigma)$ then $Q_\alpha(\rho\|\sigma)=F(\alpha)$ for all
$\alpha\ge0$,
\item[(B)] if $s_M(\rho)\not\le s_M(\sigma)$ then
$$
Q_\alpha(\rho\|\sigma)=\begin{cases}
F(\alpha) & \text{for $0\le\alpha<1$}, \\
\rho(1)>F(1) &\text{for $\alpha=1$}, \\
+\infty & \text{for $\alpha>1$}.
\end{cases}
$$
\end{itemize}

(1)\enspace
If $s_M(\rho)\perp s_M(\sigma)$, i.e., $F(1)=\rho(s_M(\sigma))=0$, then we have $\mu=0$ so
that $F(\alpha)=0$ for all $\alpha\in[0,+\infty)$. Hence the conclusion follows from (B).

(2)\enspace
If $s_M(\rho)\not\perp s_M(\sigma)$, then we have $\mu\ne0$ so that $F(\alpha)>0$ for all
$\alpha\in[0,+\infty)$. Now by (A) and (B) we may show that $\log F(\alpha)$ is convex on
$[0,+\infty)$. Let $\alpha_1,\alpha_2\in(0,+\infty)$ and $0<\lambda<1$. H\"older's inequality
implies that
$$
\int t^{(1-\lambda)\alpha_1+\lambda t_2}\,d\mu(t)
\le\biggl[\int t^{\alpha_1}\,d\mu(t)\biggr]^{1-\lambda}
\biggl[\int t^{\alpha_2}\,d\mu(t)\biggr]^\lambda,
$$
which shows the convexity of $\log F(\alpha)$.

(3)\enspace
First, assume that $s_M(\rho)\not\le s_M(\sigma)$. As $0<\alpha\nearrow1$, since
$t^\alpha\nearrow t$ for $t\ge1$, by the monotone convergence theorem, we have
$F(\alpha)\to F(1)<\rho(1)$. Hence
$$
D_\alpha(\rho\|\sigma)={\log F(\alpha)-\log\rho(1)\over\alpha-1}
\ \longrightarrow\ +\infty={D(\rho\|\sigma)\over\rho(1)}.
$$
Second, assume that $s_M(\rho)\le s_M(\sigma)$, i.e., $F(1)=\rho(1)$. For any $t\in(0,+\infty)$,
since $\alpha\in(0,+\infty)\mapsto t^\alpha$ is convex, we see that as $0<\alpha\nearrow1$,
$$
t-1\le{t^\alpha-t\over\alpha-1}\ \nearrow\ t\log t,\qquad t\in(0,+\infty),
$$
so that the monotone convergence theorem gives
\begin{align}\label{F-5.11}
{F(\alpha)-F(1)\over\alpha-1}=\int{t^\alpha-t\over\alpha-1}\,d\mu(t)
\ \nearrow\ \int t\log t\,d\mu(t)=S_{t\log t}(\rho\|\sigma)=D(\rho\|\sigma).
\end{align}
This implies that as $0<\alpha\nearrow1$,
$$
F(\alpha)=\rho(1)+D(\rho\|\sigma)(\alpha-1)+o(1-\alpha)
$$
so that
$$
\log{F(\alpha)\over\rho(1)}={D(\rho\|\sigma)\over\rho(1)}(\alpha-1)+o(1-\alpha).
$$
Therefore, $D_\alpha(\rho\|\sigma)\to D(\rho\|\sigma)/\rho(1)$.

Next, assume that $D_{\alpha_0}(\rho\|\sigma)<+\infty$, i.e., $s_M(\rho)\le s_M(\sigma)$ and
$\int t^{\alpha_0}\,d\mu(t)<+\infty$ for some $\alpha_0>1$. As $\alpha_0\ge\alpha\searrow1$,
since
$$
{t^{\alpha_0}-1\over\alpha_0-1}\ge{t^\alpha-t\over\alpha-1}\ \searrow\ t\log t,
\qquad t\in(0,+\infty),
$$
the Lebesgue convergence theorem gives, as in \eqref{F-5.11},
$$
{F(\alpha)-F(1)\over\alpha-1}\ \searrow\ D(\rho\|\sigma),
$$
and hence the latter assertion is shown similarly to the above.

(4)\enspace
When $s_M(\rho)\perp s_M(\sigma)$, this is obvious from (1). Otherwise, this immediately
follows from convexity of $\alpha\mapsto\log Q_\alpha(\rho\|\sigma)$ in (2) (and from
definition of $D_\alpha$, $Q_1$ and $D_1$).

(5)\enspace
Let $0<\alpha<1$. Since $\widetilde f_\alpha=f_{1-\alpha}$, Proposition \ref{P-2.4} with
\eqref{F-5.4} gives $Q_\alpha(\rho\|\sigma)=Q_{1-\alpha}(\sigma\|\rho)$ and hence for
$\sigma\ne0$,
$$
D_\alpha(\rho\|\sigma)={1\over\alpha-1}\log{Q_{1-\alpha}(\sigma\|\rho)\over\rho(1)}
={\alpha\over1-\alpha}\,D_{1-\alpha}(\sigma\|\rho)
+{1\over1-\alpha}\log{\rho(1)\over\sigma(1)},
$$
implying \eqref{F-5.8}. From this and (3) the second assertion follows.

(6)\enspace
One can consider $\log$ as an continuous increasing function from $[0,+\infty]$ to
$[-\infty,+\infty]$. We see from (A) and (B) above that for every $\rho,\sigma\in M_*^+$,
$Q_\alpha(\rho\|\sigma)$ is in $[0,+\infty)$ for $0\le\alpha<1$ and in $(0,+\infty]$ for
$\alpha>1$. Hence $D_\alpha(\rho\|\sigma)\in(-\infty,+\infty]$ for any
$\alpha\in[0,+\infty)\setminus\{1\}$. Since $f_\alpha$ is operator convex on $(0,+\infty)$ if
$0\le\alpha\le2$, by \eqref{F-5.4} and Theorem \ref{T-4.1}\,(i) the map
$(\rho,\alpha)\in M_*^+\times M_*^+\mapsto Q_\alpha(\rho\|\sigma)$ is upper semicontinuous
for $0\le\alpha<1$ and lower semicontinuous for $1<\alpha\le2$ in the $\sigma(M_*,M)$-topology.
Hence $(\rho,\sigma)\in(M_*^+\setminus\{0\})\times M_*^+\mapsto D_\alpha(\rho\|\sigma)$ is
lower semicontinuous for any $\alpha\in[0,2]\setminus\{1\}$. For $\alpha=1$, the result reduces
to the case of the relative entropy due to \eqref{F-5.6}.

(7)\enspace
The first part is a consequence of Theorem \ref{T-4.1}\,(ii) in view of \eqref{F-5.4}. Then
the second is clear since a non-negative concave function is log-concave.

(8)\enspace
The results for $Q_\alpha$ follow from Theorem \ref{T-4.1}\,(iii) and \eqref{F-5.4}. This gives
the assertion on $D_\alpha$ for $\alpha\in[0,2]\setminus\{1\}$. The case of $D_1$ follows from
(3) or it is a well-known fact of $D$.

(9) follows from Theorem \ref{T-4.1}\,(iv) in view of \eqref{F-5.4} and \eqref{F-5.6} for
$\alpha=1$.

(10)\enspace
When $\alpha\in[0,2]\setminus\{1\}$, inequality \eqref{F-5.10} is a special case of
\eqref{F-5.9} for $N=\bC1$, since for scalars $\rho(1)$ and $\sigma(1)$,
$$
D_\alpha(\rho(1)\|\sigma(1))={1\over\alpha-1}
\log{\rho(1)^\alpha\sigma(1)^{1-\alpha}\over\rho(1)}=\log{\rho(1)\over\sigma(1)}.
$$
By (4) the inequality holds for $\alpha>2$ as well. If
$\rho=k\sigma$ with $k=\rho(1)/\sigma(1)$, then
$\Delta_{\rho,\sigma}\xi_\sigma=k\Delta_\sigma\xi_\sigma=k\xi_\sigma$ and hence
$Q_\alpha(\rho\|\sigma)=k^\alpha\sigma(1)$. Therefore,
$D_\alpha(\rho\|\sigma)=\log(\rho(1)/\sigma(1))$ for all $\alpha>0$ (including $\alpha=1$).
Conversely, if equality holds for some $\alpha>0$, then by (4) the same holds for some
$\alpha\in(0,1)$. This means that equality \eqref{F-4.3} holds for $f=f_\alpha$, so that
$\rho=(\rho(1)/\sigma(1))\sigma$ follows from Corollary \ref{F-4.2}\,(1). Finally, the
second part of (10) is clear from the first.
\end{proof}

\begin{remark}\label{R-5.4}\rm
(1)\enspace
In Proposition \ref{P-5.3}\,(3), the assumption that $D_\alpha(\rho\|\sigma)<+\infty$ for some
$\alpha>1$ is essential to have
$\lim_{\alpha\searrow1}D_\alpha(\rho\|\sigma)=D_1(\rho\|\sigma)$. Indeed, it is not difficult
to find commuting density operators $\rho=\sum_{i=1}^\infty a_i|e_i\>\<e_i|$ and
$\sigma=\sum_{i=1}^\infty b_i|e_i\>\<e_i|$ on $\mathcal{H}$ such that
$$
D(\rho\|\sigma)=\sum_{i=1}^\infty a_i\log{a_i\over b_i}<+\infty,\qquad
D_\alpha(\rho\|\sigma)=\sum_{i=1}^\infty a_i^\alpha b_i^{1-\alpha}=+\infty\quad
\mbox{for all $\alpha>1$}.
$$

(2)\enspace
The convexity of $Q_\alpha$ for $1\le\alpha\le2$ in Proposition \ref{P-5.3}\,(7) cannot extend
to $\alpha>2$ even in the finite-dimensional case and in separate arguments. This implies
that the monotonicity property of $D_\alpha$ in (9) fails to hold for $\alpha>2$, because the
monotonicity of $Q_\alpha$ under unital completely positive maps yields its joint convexity.
Also, the monotone decreasing of $\sigma\mapsto Q_\alpha(\rho\|\sigma)$ for $1\le\alpha\le2$
in (8) cannot extend to $\alpha>2$. But it seems possible that the joint lower semicontinuity
of $D_\alpha$ as in (6) (or in the norm topology) is true for $\alpha>2$ as well (this is easily
verified in the finite-dimensional case).

(3)\enspace
In Proposition \ref{P-5.3}\,(7), due to division by $\rho(1)$ in definition \eqref{F-5.3},
the map $\rho\mapsto D_\alpha(\rho\|\sigma)$ with $\sigma\in M_*^+$ fixed cannot be convex on
the whole $M_*^+$. However, in the finite-dimensional case it was shown \cite[Theorem II.1]{MH}
that $\sigma\mapsto D_\alpha(\rho\|\sigma)$ with $\rho\in M_*^+\setminus\{0\}$ fixed is convex
on $M_*^+$ for any $\alpha\in[0,2]$. It is natural to expect that this extends to the general
von Neumann algebra case.
\end{remark}

We end the main body of the paper with a remark on relations of $D_\alpha(\rho\|\sigma)$ with
other R\'enyi type divergences from recent papers \cite{BST,Je1}.

\begin{remark}\label{R-5.5}\rm
For every $\rho,\sigma\in M_*^+$, in view of Proposition \ref{P-5.3}\,(4) one can define
$$
D_\infty(\rho\|\sigma):=\lim_{\alpha\to+\infty}D_\alpha(\rho\|\sigma).
$$
The \emph{max-relative entropy} introduced in \cite{Da} is
$$
D_{\max}(\rho\|\sigma):=\log\inf\{t>0:\rho\le t\sigma\},
$$
where $\inf\emptyset=+\infty$ as usual. The \emph{sandwiched R\'enyi divergence}
$\widetilde D_\alpha(\rho\|\sigma)$ \cite{MDST,WWY} has recently been extended to the von
Neumann algebra setting by Berta, Scholz and Tomamichel \cite{BST} and
Jen\v cov\'a \cite{Je1,Je2}. From \cite{BST,Je1} we remark that for every
$\rho,\sigma\in M_*^+$,
\begin{itemize}
\item[(a)] $\widetilde D_\alpha(\rho\|\sigma)\le D_\alpha(\rho\|\sigma)$ for $\alpha>1$,
\item[(b)] $\lim_{\alpha\to+\infty}\widetilde D_\alpha(\rho\|\sigma)=D_{\max}(\rho\|\sigma)$,
\item[(c)] $D_2(\rho\|\sigma)\le D_{\max}(\rho\|\sigma)\le D_\infty(\rho\|\sigma)$.
\end{itemize}
\end{remark}

\section{Closing remarks}

In this paper we present a systematic treatment of standard $f$-divergences in the setting of
general von Neumann algebras and general operator convex functions $f$ on $(0,+\infty)$. The
main theorem is the variational expression of an arbitrary standard $f$-divergence
$S_f(\rho\|\sigma)$. We also present a comprehensive account on the quantum R\'enyi divergence
in von Neumann algebras on the basis of theory of standard $f$-divergences. There are some
other important quantum divergences; in particular, the maximal $f$-divergence (discussed
in \cite{HiMo} in the finite-dimensional case) is worth studying. The most significant problem
related to the standard $f$-divergence and other quantum divergences is the reversibility via
them, as explained in the Introduction. These should be our forthcoming research topics.

\subsection*{Acknowledgements}
This was supported in part by Grant-in-Aid for Scientific Research (C)17K05266. The author is
grateful to Mil\'an Mosonyi for suggestion on Theorem \ref{T-4.5} and Anna Jen\v cov\'a for
discussion about Remark \ref{R-A.7} below.

redpajama_set_name: RedPajamaArXiv


text:
Khurshed Lal was an Indian politician. He was the member of Constituent Assembly of India from United Provinces through the Indian National Congress ticket.

Early life and education
Khurshed Lal was born on 4 March 1903 in the small town of Kotli Laharan in Sialkot district of present-day Pakistan. After completing his master's and LLB degree from Lucknow University, in 1926 he enrolled in the Allahabad High Court as an advocate.

Role in India's Independence Movement
Five years after beginning his law practice, Lal joined the Congress and would remain a lifelong member of the party. By 1936 he became quite involved in local governance and served as the Chairman of the Dehra Dun Municipality for the next four years.

Contribution to Constitution Making
Lal was elected to the Constituent Assembly from the United Provinces on a Congress party ticket. He was not particularly active in the Assembly but was part of the its Select Committee that had the task of evaluating the Hindu Code Bill and reporting to the Assembly on it. After Independence, Lal joined the first cabinet of independent India in 1950 as the Deputy Minister for Communications (State).

References

1903 births
1951 deaths

redpajama_set_name: RedPajamaWikipedia


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Imagine Dizzee Rascal's "I Luv U" being covered by Phil Elverum and you have Becoming Real's "Get Hype." It's not exactly dubstep nor is it creepy folk but it definitely is not neither of them either. Like Salem's crusted crunk simulacra, this music is a bad idea made good by someone both really smart and fortunately talented. And judging by our recent interview with the man behind the project, we should expect a lot more measured weirdness in the future. Get "Get Hype" on Becoming Real's new EP on Tough Love right now.
Download: Becoming Real, "Get Hype"


text:
Updated on December 28, 2017 5:35pm EST
Rushing to pay taxes? New law spurs confused pre-payment push
Americans scrambling to prepay 2018 state and local taxes before year end may not get the federal deductions they hoped for, but that did not slow the rush on Thursday in some of the nation's highest-taxed municipalities.
Municipal coffers are fattening up as taxpayers pay now with the hopes of avoiding a new $10,000 limit on deductions for state and local income and property taxes from their federal bill in April, part of a sweeping law signed by U.S. President Donald Trump last week.
The rush prompted the U.S. Internal Revenue Service to caution homeowners late on Wednesday that pre-payments on property taxes may not be deductible. A full deduction depends on when the property taxes were assessed.
That detail may have caught many off guard.
"That's news to me," said Mary Rappaport, a resident of Fairfax County, Virginia who waited with two dozen people to prepay at eight fully staffed windows at the county's Department of Tax Administration.
After consulting with a DTA representative, Rappaport still paid her 2018 property taxes in advance after learning about the IRS note.
More from FOX Business
IRS cautions U.S. taxpayers on prepaying property taxes
What tax reform really means for small businesses
Bonuses, bigger retirement accounts and lower health care costs latest 'fall-out' from tax reform
"It will reduce my mortgage payments but it's not the kind of advantage I was hoping for. It made me feel defeated," she said.
Critics of the federal tax bill see capping the deduction for state and local income and property taxes, known as SALT, as punitive to high-tax states like New York, New Jersey and California.
"Its wildly confusing and that's the reason why you shouldn't rush through a tax bill so late in the year without giving people the chance to understand it," said Steve Hickson, a South Orange, New Jersey, resident who prepaid his property tax bill online.
"You're delaying the pain of the new tax bill ... that's why you're seeing people do a mad dash to your local town hall," he said.
According to an analysis of IRS data by the Tax Foundation, a pro-business Washington think tank, New York ranks first when state and local tax deductions are measured as a share of adjusted gross income, at 9.1 percent. New Jersey is second at 8.7 percent, followed by Connecticut and California.
At the opposite end is Alaska with 1.5 percent, and North Dakota, South Dakota and Wyoming at 1.6 percent.
FAT ACCOUNTS
New Rochelle, New York, a city of roughly 80,000 residents just north of New York City, is seeing an unprecedented swell in its cash pile thanks to the rush.
"Our city tax bill is due for payment at the end of January. Typically we have maybe a couple hundred thousand dollars on hand at this point. So far, from the time this law was passed in Congress we've collected over $15 million," Charles Strome, New Rochelle's city manager said during an interview in his office.
"This is unprecedented. I have to go now and pay my own taxes early," he said, noting that the city also collects on behalf of the public school district.
Strome said the new tax law could lower home values in the city by 5 percent to 10 percent.
But that will not necessarily deter construction. New Rochelle has recently kicked off a multi-billion dollar downtown redevelopment, and the high proportion of rental units in the new buildings means changes to SALT deductions will unlikely deter new residents from the city, he said.
Some New Rochelle taxpayers unsure of the rules came looking for answers before writing checks, including attorney Dan Feldman.
City clerks answered Feldman's questions and moved residents through the payment line even as they scrambled to find replacement ink to print tax bill receipts.
"Assuming your real estate taxes are more than $10,000, why wouldn't you do this to take advantage of something that is going to go away, if you can afford the early payment," Feldman said.
New York Governor Andrew Cuomo stepped up his attacks on the new tax legislation on Thursday, saying it unfairly targets high tax-states and may be unconstitutional.
"Until I get to my accountant, I won't know the specifics of the bill. Even the senators that voted on the bill don't know what's in the bill," Ed Finnegan, a publisher who lives in Bay Ridge, Brooklyn said while waiting to prepay taxes in Manhattan.
(Additional reporting by Katanga Johnson in Fairfax, Virginia; April Joyner in New York; Sinead Carew in New Jersey; Editing by Meredith Mazzilli)

redpajama_set_name: RedPajamaCommonCrawl


text:
WSKG (https://wskg.org/education/connect-ny-screens-and-teens/)
Connect NY: Screens and Teens
By J. Stapleton-Durham | May 20, 2018
More on Education
Subscribe to Education
Has smart tech and the social media it drives harmed a generation dependent on these tools to perform most functions? A new addiction impacting youth socialization, cognitive development and even mental health is chronicled in a new documentary that's starting the conversation with parents, educators and lawmakers across New York State. As policymakers look to respond, can we institute moderation and parameters or has the technology taken hold and it's too late?
Bonnie Nobiling, Director of IT Services & District Data Coordinator – Oneonta School District
Dr. Delaney Ruston, President – MyDoc Productions and filmmaker-in-residence at Stony Brook Medicine, NY
Michael Robb, Ph.D., Director of Research – Commonsense Media
Bonnie Russell, Immediate Past President – New York State Parent Teacher Association
Dr. David O'Rourke, District Superintendent – BOCES Erie District
Sean Ryan (D), NYS Assemblymember 149th District
Watch this new episode on Monday, May 21, 2018 at 9:00pm on WSKG TV.
Join us as we uncover compelling and unexpected stories throughout New York State and the history and systemic forces influencing current realities. Connect: NY is produced by WCNY (Syracuse, NY) and aired in partnership with WSKG.


The London Fashion Fair -- the largest of its kind in the world -- was hold for three days (9-11 April) at the Earl's Court Exhibition centre and as has become usual for the twice-yearly event attracted a large number of buyers from all over the world.
CU Model wearing rainbow-coloured hooded raincoat with matching umbrella.
CU PAN DOWN MODEL wearing swede overdress with belt tie-in at waist.
Background: The London Fashion Fair -- the largest of its kind in the world -- was hold for three days (9-11 April) at the Earl's Court Exhibition centre and as has become usual for the twice-yearly event attracted a large number of buyers from all over the world.
Pro-exhibition buyer registrations exceeded 10,000, with 3,000 of them coming from buyers from twenty-three foreign countries.
Since its inception the Fair has gradually developed from a purely British fashion promotion event to one which attracts international participation.
This year's April show saw ten foreign countries selling fashion goods alongside the British exhibitors.
The Fair provides a look ahead for styles which will be on sale in European and American shops this autumn, but did not show much in the way of starting novelties.
Dresses can be wide or clinging, long or short (but always below the knee) and all colours and prints are possible. In other words, the spirit of fashion anarchy which has reigned supreme for several years now is still holding the stage - which probably days something for the success of the Women's Liberation Movement!


text:
Hubble 3D!
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Part 1 of "Prediction is difficult . . .
. . . especially about the future." Niels Bohr With all due respect to Professor Bohr, I think some things are easy to predict. Take the latest stab at the air car. For at least a century, futurists have been predicting that we'd be flying to work eventually. Check out the video below to see NASA's late air car idea. I feel like it's pretty easy to predict if and when these sorts of things will be commonplace. I'm going to try my hand at futurism for a bit. I plan to throw some cold water on a few predictions, and highlight a few others that I think are feasible and likely. Because there're so many things to talk about, I thought I'd make this a semi-regular column. Today's topic: NASA's Puffin Air Car and the Holodeck - Future Tech or Dead End? Flying Cars - You might prefer to call them "personal aircraft" or some other buzzword. People have been working on them since airplanes were invented. In fact, we already have some persona
Mars, Tablets and Hydrogel
There's a lot going on today. And it's not even a Saturday. I think of Saturday as the day of many rewards, but apparently the solar system and Apple are irreverent about the days of the week. They will reveal mighty wonders whenever they want! The solar system is revealing Mars today - the red planet will be the closest to Earth that it will get until 201 4. It will be particularly visible in North America tonight, so be sure to step outside and get a good look. One that will hold you over for 4 years. You know, now that I think about it, maybe the solar system did plan to have this happen on a Wednesday. Heaven knows I'm not about to leave my apartment on a Thursday night in the middle of 30 Rock to look at the sky. (I seriously still watch TV shows on the TV and not online. I can pretty much start asking for the seniors discount at moving picture theatre.) The other big news down on Earth is that Apple will be debuting their brand new gadget: the tablet (or as Inform
Volcano Lightning
A type of volcanic lightning was discovered during Mount Redoubt's Jan. 2009 eruption. When volcano seismologist Stephen McNutt at the University of Alaska Fairbanks's Geophysical Institute saw strange spikes in the seismic data from the Mount Spurr eruption in 1992, he had no idea that his research was about to take an electrifying turn. "The seismometers were actually picking up lightning strikes," said McNutt. "I knew that I had to reach out to the physicists studying lightning." With McNutt's curiosity about volcanic lightning sparked, he teamed up with physicist and electrical engineer Ronald Thomas and Sonja Behnke, a graduate student in atmospheric physics at the New Mexico Institute of Mining and Technology in Socorro, N.M. for a unique collaboration in order to learn more about volcanic lighting. When the Mount Redoubt volcano started making seismic noise in January 2009, McNutt alerted Thomas and Behnke that this would be a great opportunity t
A Star Class is Born
Hold it right there, star. Trying to become a black hole, are ya? Well, not so fast. It's possible that as you begin to collapse and squeeze the subatomic particles at your core, the squarshed quarks could begin to radiate neutrinos and stop that collapse, leaving you stuck as a dense, nearly invisible lump for millions of years. Yeah, those tiny little quarks are real tough when you get a bunch of them together. Scientists have proposed that there is a new, exotic type of star living in our universe that we haven't seen yet. The so-called "electroweak stars", if they exist, will be difficult to detect because they mostly emit neutrinos - subatomic particles which, for the most part, don't interact with ordinary matter. [Image: "Electroweak" stars may recreate the conditions of the big bang in an apple-sized region in their cores (Illustration: Casey Reed, courtesy of Penn State)] When massive stars run out of fuel to burn, they expand into a supern
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Pretty Physics Picture of the Week: Supersonic Splash
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Evidence for a new kind of stellar explosion 7 billion years ago Astronomers meeting in Washington last week announced that a recent search for bright exploding stars -- commonly called supernovas -- found something quite unusual: antimatter. Usually stars like our sun are powered by fusion reactions in which the nuclei of two atoms fuse together to form a heavier nucleus. In Y-155, a star in the constellation Cetus, the astronomers argue that another process was crucial: the making and unmaking of antimatter particles. In all stars a titanic struggle takes place between gravity, which wants to draw matter toward the center of the star, and the pressure of nuclear interactions, which tends to keep the star inflated as if it were a balloon. Only when the star uses up all its internal fuel, causing the nuclear reactions to slow down, does gravity start to win out. The resulting gravitational collapse is what causes the star to explode. When a star dies in this way, as a supernova, i
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Long before Steven Spielberg cast Haley Joel Osmond as Creepy Jr. in A.I. , and before guys started marrying their video game girlfriend s, way back in the early 1950's, a small group of physicists and mathematicians coined the term "Artificial Intelligence." While the group remained small for a few more years, the dawn of the computer age has seen this field blow up exponentially. Ray Solomonoff was a member of that small group; an early pioneer in the field of artificial intelligence whose work showed that to achieve artificial intelligence, we must first understand intelligence on a systematic level. Solomonoff died on December 7, 2009, at the age of 83 of a ruptured brain aneurysm. To understand the impact of Solomonoff, you have to go back to Alan Turing. Alan Turing committed suicide by eating an apple dipped in cyanide. A dramatic end to a truly amazing life. Turing is not only hailed by the physics and computer science community, but by philosophers and the ga
The Wooden Leg Makes a Comeback
Pirate enthusiasts rejoice. Wooden legs are back. Sort of. Actually, anyone who has lost bone or bone density to disease or accident may have something to celebrate, and will not actually have to worry about looking like a pirate. Scientists in Italy, at the Istec laboratory of bioceramics in Faenza near Bologna, have found that the wood of the rattan tree, when heated, pressurized and enhanced with calcium and phosphates, becomes incredibly similar to human bone. Early tests show that real bone is so comfortable with the organic substitute, that it will bind to it (this is with segments of bone - no tests in humans have taken place yet). This acceptance by real bone, coupled with the treated rattan wood's natural durability, means the substitute bone will never need replacing. Current substitutes for people who have suffered bone loss due to disease, where the remaining bone is also weak, often need to be replaced, and certainly don't fuse with the real bone. The BBC qu
Can Physics Predict the Future?
You and I each make predictions about the future continuously throughout our day. As we drive to work, we predict that the path we take will look much like it did yesterday. When we practice a sport, we begin to learn how hard we need to kick or hit a ball to make it go where we want. And soon we understand the forces of gravity, friction, and our own muscle, enough to score a goal or block a line drive. Each of us is constantly predicting the future. And sometimes we're right. So if physics is a science meant to break down the world that we live in - picking apart it's smallest particles, figuring out it's hidden mechanisms, and working out the equations that describe it - then shouldn't physics be able to predict the future? This is a question, or an assumption, that many people have about physics. And I'll tell you right now that it's wrong. Physics can, to some extent, tell us what to expect from the world. Just like you and I can reasonably assume that


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