数理解析研究所講究録
第 1996 巻 2016 年 1-12

1Some operator divergences basedon Petz-Bregman divergence
債佐浩史 Hiroshi $ISA)^{(1)}$ , 伊藤公智
satoshi 1 $TO)^{(2)},$亀井栄三郎 Eizaburo KAMEI), 遠山宏明 Hiroaki TOHYAMA)}
渡邉雅之 Masayuki WATANABE)(1), (2), (4), (5)’ 前橋工科大学 (Maebashi Institute of Technology)$(M_{c}\tau$

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1. Introduction
This report is based on [12]. Throughout this report, a bounded linear operator
on a Hilbert space $H$ is positive (denoted by $T\geq 0$ ) if
for all $\xi\in H$ , andis strictly positive (denoted by $T>0$ ) if is invertible and positive.
For strictly positive operators and $B,$
is defined as follows ([3, 4, 13] etc$T$$(T\xi,$ $\xi\rangle\geq 0$

$T$

$T$

$A$

$A\natural_{x}B$$At_{x}B\equiv A^{\frac{1}{2}}(A^{-}\ovalbox{\tt\small REJECT} BA^{-}\ovalbox{\tt\small REJECT})^{x}A^{\frac{1}{2}},$$X\in \mathbb{R}$。

and $B=A\natural_{1}B$ . If $x\in[O$ , 1 , thepathcoincides with the weighted geometric operator mean denoted by $A\#_{x}B$(cf. [15]). We remark that
holds forand Kamei [2] defined the following relative operator entropy for strictly positive operators and :We call$A\natural_{x}B$a path passing through$A=A\mathfrak{h}_{0}B$$]$$A\natural_{x}B$$A\natural_{x}B=B\natural_{1-x}$$A$$x\in \mathbb{R}.$$Fui\dot{x}i$

$A$

$B$

$S(A|B) \equiv A^{\frac{1}{2}}\log(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})A^{\frac{1}{2}}$$= \frac{d}{dx}A\natural_{x}B|_{x=0}$

We can regard $S(A|B)$ as the gradient of the tangent line at $x=0$ of the path
Furuta [7] defined generalized relative operator entropy as follows:$A\mathfrak{h}_{x}B.$$S_{\alpha}(A|B) \equiv A^{\frac{\lambda}{2}}(A^{-\frac{1}{2}}BA^{-\frac{\}}{2}})^{\alpha}\log(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})A^{\frac{1}{2}}$$= \frac{d}{dx}A\mathfrak{y}_{x}B|_{x=\alpha})\alpha\epsilon \mathbb{R}.$We regardas the gradient of the tangent line at $x=\alpha$ of the path. We
know immediately $S_{0}(A|B)=S(A|B)$ . Yanagi, Kuriyama and Furuichi [18] introduced
Tsallis relative operator entropy as follows:$S_{\alpha}(A|B)$$T_{\alpha}(A|B) \equiv\frac{A\#_{\alpha}B-A}{\alpha},$$\alpha$欧 $(0,1].$

can be regarded as the average rate of change of $A\#_{x}B$ from $x=0$ to
Since
holds for $a>0$ , we have $T_{0}(A|B)= \neg\lim T_{\alpha}(A|B)=S(A|B)$ .
Tsallis relative operator entropy can be extended as the notion for. In thiscase, we use instead of . In [8], we had given the following relations among these$T_{\alpha}(A|B)$$x=\alpha.$$x \alphaarrow 0$$\lim_{xarrow 0}^{\underline{a^{x}-1}}=\log$$\mathfrak{g}_{\alpha}$$a$$\alpha\in \mathbb{R}$$\#_{\alpha}$relative operator entropies:$(*)$$S_{1}(A|B)\geq-T_{1-\alpha}(B|A)\geq S_{\alpha}(A|B)\geq T_{\alpha}(A|B)\geq S(A|B)$, $\alpha\in(0,1)$ .

2Figure 1: An interpretation of $D_{0}(A|B)$ .
Fujii [1] defined operator valued a-divergence$D_{\alpha}(A|B)$

for $\alpha\in(0,1)$ as follows:$D_{\alpha}(A|B) \equiv\frac{A\nabla_{\alpha}B-A\#_{\alpha}B}{\alpha(1-\alpha)},$where $A\nabla_{\alpha}B\equiv(1-\alpha)A+\alpha B$ is the weighted arithmetic operator mean. The operator
valued a-divergence has the following relations at end points for interval $(0,1)$ .
Theorem $A([5,6$For strictly positive operators

$A$

and

$B$

, the following hold:$D_{0}(A|B) \equiv\lim_{\alphaarrow+0}D_{\alpha}(A|B)=B-A-S(A|B)$,$D_{1}(A|B) \equiv\lim_{\alphaarrow 1-0}D_{\alpha}(A|B)=A-B-S(B|A)$.

Petz [17] introduced the right hand side in the first equation in Theorem A as an
operator divergence, so we call $D_{0}(A|B)$ Petz-Bregman divergence. We remark that
$D_{1}(A|B)=D_{0}(B|A)$ holds. Figure 1 shows our interpretation of $D_{0}(A|B)$ .
In [10], we showed the following relation between operator valued -divergence andTsallis relative operator entropy:$\alpha$Theorem $B$ ([10]). For strictly positive operators$A$$D_{\alpha}(A|B)=-T_{1-\alpha}(B|A)-T_{\alpha}(A|B)$

and

$B$

for

, the following holds:$\alpha\in(0,1)$.

is a diﬀerence between two terms in $(*)$ . From this
Theorem shows that
fact, we regard the diﬀerences between the relative operator entropies in $(*)$ as operator divergences. In section 2, we represent these operator divergences by usingPetz-Bregman divergence.
On the other hand, for an operator valued smooth function : $C\prec B(H)$ and$X,$ $Y\in C$ , where $C$ is a convex set in a Banach space, Petz [17] defined a divergence$B$$D_{\alpha}(A|B)$$\Psi$

3$D_{\Psi}(X, Y)$

as follows:$D_{\Psi}(X, Y) \equiv\Psi(X)-\Psi(Y)-\lim_{\alphaarrow+0}\frac{\Psi(Y+\alpha(X-Y))-\Psi(Y)}{\alpha}.$We call $D_{\Psi}(X, Y)\Psi$ -Bregman divergence of and $X$ in this report. Petz gave someexamples for invertible density matrices $X$ and . If $\Psi(X)=\eta(X)\equiv X\log X$ and $X$
if $\Psi(X)=tr\eta(X)$ ,commutes with , then $D_{\Psi}(X, Y)=Y-X+X(\log X-\log Y)$ ,then $D_{\Psi}(X, Y)=trX(\log X-\log Y)$ , which is the usual quantum relative entropy.
In section 3, we letand show $D_{\Psi}(x,y)=D_{0}(A\natural_{v}B|A\natural_{x}B)$ for $\Psi(t)=$. Then we have $D_{\Psi}(\lambda, O)=D_{0}(A|B)$ in particular. Based onandfor several functions
this interpretation, we discuss -Bregman divergences
which relate to the operator divergences given in section 2.
In section 4, we show the results corresponding to those in section 2 on expanded
relative operator entropies defined by operator power mean.
$Y$

$Y$

$Y$

$a\ddagger ld$$C=\mathbb{R}$$A\natural_{t}B$$x,$$y\in \mathbb{R}$$\Psi$$D_{\Psi}(1,0)$$\Psi$

2. Divergences given by the diﬀerences among relative operator
entropies
In this section, we regard the diﬀerences between the relative operator entropies in$(*)$ as operator divergences. There are le such divergences. For convenience, we use
symbols
for them as follows:$\triangle_{i}$$\triangle_{1}=T_{\alpha}(A|B)-S(A|B)$,$\triangle_{2}=S_{\alpha}(A|B)-T_{\alpha}(A|B)$$\triangle_{3}=-T_{1-\alpha}(B|A)-S_{\alpha}(A|B)$$\Delta_{6}=S_{\alpha}(A|B)-8(A|B)$

,

,

,

$\Delta_{4}=S_{1}(A|B)+T_{1-\alpha}(B|A)$,$\Delta_{6}=-T_{1-\alpha}(B|A)-T_{\alpha}(A|B)=D_{\alpha}(A|B)$$\Delta_{7}=S_{1}(A|B)-S_{\alpha}(A|B)$,$\Delta_{8}=-T_{1-\alpha}(B|A)-S(A|B)$$\triangle_{9}=S_{1}(A|B)-T_{\alpha}(A|B)$,$\triangle_{10}=S_{1}(A|B)-S(A|B)$

We represent each of
to consider$\Delta_{1},$$\triangle_{2},$$\Delta_{1},$$\Delta_{3}$$\cdots,$

and

,

,

.

by using Petz-Bregman divergence. It is suﬃcientsince the following relations hold:$\triangle_{10}$$A_{4}$$\Delta_{5}=\triangle_{1}+\Delta_{2}, \triangle_{6}=\Delta_{2}+\Delta_{3}, \Delta_{7}=\Delta_{3}+\Delta_{4},$$\Delta_{8}=\Delta_{1}+\Delta_{2}+\Delta_{3}, \Delta_{9}=\triangle_{2}+\triangle_{3}+\Delta_{4_{\rangle}} \Delta_{10}=\Delta_{1}+\Delta_{2}+\Delta_{3}+\Delta_{4}.$are given as in Table 1.
The relation of the diﬀerences among
The next lemma is essential in our discussion.
$\Delta_{3},$Lemma 2. $1([8,10$$s,$$t\in \mathbb{R}$

(1)
(2)

$\triangle_{10}$$\cdots,$

For strictty positive operators$A$and $B$ , the following hold for:$S_{t}(A|A\natural_{s}B)=sS_{st}(A|B))$$S_{t}(A|\mathcal{B})=-S_{1-t}(B|A)$.

4
Table 1
$S_{1}(A|B)-S(A|B)\geq S_{1}(A|B)-T_{\alpha}(A|B)\geq S_{1}(A|B)-S_{\alpha}(A|B)\geq S_{1}(A|B)+T_{1-\alpha}(B|A)\geq 0$

Vl$\fbox{Error::0x0000}|$$\vee I$$-T_{1-\alpha}(B|A)-S(A|B)\geq-T_{1-\alpha}(B|A)-T_{\alpha}(A|B)\geq-T_{1-\alpha}(B|A)-S_{\alpha}(A|B)\geq 0$$\fbox{Error::0x0000}| V|$$S_{\alpha}(A|B)-S(A|B)\geq S_{\alpha}(A|B)-T_{\alpha}(A|B)$$\fbox{Error::0x0000}|$$\fbox{Error::0x0000}|$$T_{\alpha}(A|B)-S(A|B)$

$0$

$V|$

$0$

The following are the results on$\triangle_{1}$and$\Delta_{2}.$Theorem 2.2. For strictly positive operators

$A$

and

$B_{f}$

the following hold:(1)$\triangle_{1}$$=T_{\alpha}(A|B)-S(A|B)= \frac{1}{\alpha}D_{0}(A|\mathcal{A}\#_{\alpha}B)$(2)$\Delta_{2}$$=$$S_{\alpha}(A|B)-T_{\alpha}(A|B)= \frac{1}{\alpha}D_{0}(A\#_{\alpha}B|A)$for$\alpha\in(0,1$

for

],

$\alpha\in(0,1$].

Proof. (1) By (1) in Lemma 2.1, we have$T_{\alpha}(A|B)-S(A|B) = \frac{A\#_{\alpha}B-A}{\alpha}-S(A|B)=\frac{1}{\alpha}(A\#_{\alpha}B-A-\alpha S(A|B))$$= \frac{1}{\alpha}(A\#_{\alpha}B-A-S(A|A\#_{\alpha}B))=\frac{1}{\alpha}D_{0}(A|A\#_{\alpha}B)$.

(2) By Lemma 2.1, we have
$S_{\alpha}(A|B)-T_{\alpha}(A|B)$$=$$\frac{A-A\#_{\alpha}B}{\alpha}+S_{\alpha}(A|B)=\frac{1}{\alpha}(A-A\#_{\alpha}B+\alpha S_{\alpha}(A|B))$$= \frac{1}{\alpha}(A-A\#_{\alpha}B+S_{1}(A|A\#_{\alpha}B))$$= \frac{1}{\alpha}(A-A\#_{\alpha}B-S(A\#_{\alpha}B|A))=\frac{1}{\alpha}D_{0}(A\#_{\alpha}B|A)$.
口

Remark 1. By exchanging
have the following relations:

$A$

for

$B$

and replacing$\alpha$with $1-\alpha$ for (1) and (2), we(1)$\triangle_{4}$$=$$S_{1}(A|B)+T_{1-\alpha}(B|A)= \frac{1}{1-\alpha}D_{0}(B|A\#_{\alpha}B)$(2)$\triangle_{3}$$=$$-T_{1-\alpha}(B|A)-S_{\alpha}(A|B)= \frac{1}{1-\alpha}D_{0}(A\#_{\alpha}B|B)$for$\alpha\in[0$

for

, 1),

$\alpha\in[0$

, 1).

5
We give a geometrical interpretation for (1) in Theorem 2.2. Figure 2 and Figure 3
show $T_{\alpha}(A|B)-S(A|B)$ and $D_{0}(A|A\#_{\alpha}B)$ appeared in (1) in Theorem 2.2, respectively.
Figure 4 is an image of (1) in Theorem 2.2.

Figure 2: An interpretation of $T_{\alpha}(A|B)-S(A|B)$ .

Figure 3: An interpretation of $D_{0}(A|A\#_{\alpha}B)=A\#_{\alpha}B-A-S(A|A\#_{\alpha}B)$ .

6Figure 4: An image of$T_{\alpha}(A|B)-S(A|B)= \frac{1}{\alpha}D_{0}(A|A\mathfrak{h}_{\alpha}B)$

Theorem 2.2leads the next theorem.
Theorem 2.3. For strictly positive operators

$A$

and

$B$

.

, the following holds:$D_{\alpha}(A|B)= \frac{1}{1-\alpha}D_{0}(A\#_{\alpha}B|B)+\frac{1}{\alpha}D_{0}(A\#_{\alpha}B|A)$for$\alpha\in(0,1)$.

Proof. By (2) in Theorem 2.2 and (2) in Remark 1, we have
$D_{\alpha}(A|B) = -T_{1-\alpha}(B|A)-T_{\alpha}(A|B)$$= (-T_{1-\alpha}(B|A)-S_{\alpha}(A|B))+(S_{\alpha}(A|B)-T_{\alpha}(A|B))$$= \frac{1}{1-\alpha}D_{0}(A\#_{\alpha}B|B)+\frac{1}{\alpha}D_{0}(A\#_{\alpha}B|A)$.
口

By Theorem 2.3, we have$\alpha(1-\alpha)D_{\alpha}(A|B)=aD_{0}(A\#_{\alpha}B|B)+(1-\alpha)D_{0}(A\#_{\alpha}B|A)$$= \alpha(B-A\#_{\alpha}B-S(A\#_{\alpha}B|B))+(1-\alpha)(A-A\#_{\alpha}B-S(A\#_{\alpha}B|A))$$= A\nabla_{\alpha}B-A\#_{\alpha}B-((1-\alpha)S(A\#_{\alpha}B|A)+\alpha S(A\#_{\alpha}B|B))$,and then$(1-\alpha)S(A\#_{\alpha}B|A)+\alpha S(A\#_{\alpha}B|B)=0,$

since$D_{\alpha}(A|B)= \frac{A\nabla_{\alpha}B-A\#_{\alpha}B}{\alpha(1-\alpha)}$. This means that

$A\#_{\alpha}B$

is a solution of$(1-\alpha)S(X|A)+aS(X|B)=0$ which is the Karcher equation concerning two operators and $B$ . In this case, we can rewrite the result of Lawson-Lim [16] as follows:$A$7
Theorem 2.4 ([16]). For strictly positive operators $A,$$(1-\alpha)S(X|A)+\alpha S(X|B)=0$

$B$

and

if and only if

and for$X_{f}$$\alpha\in[0$

,1

$],$

$X=A\#_{\alpha}B.$

For readers’ convenience, we give a direct proof of this theorem.

Proof. It is obvious if$\alpha=0$. Otherwise, we have$(1-\alpha)S(X|A)+\alpha S(X|B)=0$$\Leftrightarrow \log(X^{-\frac{1}{2}}AX^{-\frac{1}{2}})^{1-\alpha}+\log(X^{-\frac{1}{2}}BX^{-\frac{1}{2}})^{\alpha}=0$$rightarrow (X^{-\frac{1}{2}}BX^{-\frac{1}{\prime 2}})^{\alpha}=(X^{\frac{1}{2}}A^{-1}X^{\frac{1}{2}})^{1-\alpha}$$\Leftrightarrow X^{-\frac{1}{2}}BX^{-\frac{1}{2}}=(X^{\frac{1}{2}}A^{-1}X^{\frac{1}{2}})^{\frac{1}{\alpha}-1}$$\approx B=X^{\frac{1}{2}}(X^{\frac{1}{2}}A^{-1}X^{\frac{1}{2}})^{\frac{1}{\alpha}}X^{-\frac{1}{2}}A$$rightarrow A^{-\frac{\lambda}{2}}BA^{-\frac{1}{2}}=A^{-\frac{1}{2}}X^{\frac{1}{2}}(X^{\frac{1}{2}}A^{-\frac{1}{2}}A^{-\frac{1}{2}}X^{\frac{1}{2}})^{\frac{1}{\alpha}}X^{-\frac{1}{2}}A^{\frac{1}{2}}$$\Leftrightarrow A^{-\frac{\lambda}{2}}BA^{-\frac{1}{2}}=(A^{-\frac{\}}{2}}XA^{-\frac{1}{2}})^{\frac{1}{\alpha}}$$\Leftarrow\Rightarrow.$$(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^{\alpha}=A^{-\frac{1}{2}}XA^{-}$

蒼$\approx X=A\#_{\alpha}B.$口

Remark 2. This theorem holds even if$\alpha$is any real number [14].

3. 重-Bregman divergences on the diﬀerences of relative operator entropies
In this section, we consider -Bregman divergence in the case
For an operator valued smooth function : $\mathbb{R}arrow B(H)$ and$\Psi$$C=\mathbb{R}$as follows:$x,$ $y6\mathbb{R},$$\Psi$$D_{\Psi}(x, y) \equiv\Psi(x)-\Psi(y)-\lim_{\alphaarrow+0}\frac{\Psi(y+\alpha(x-y))-\Psi(y)}{\alpha}.$the following theorem, it is natural that we consider $D_{\Psi}(1,0)$ as a divergence of
operators and $B.$$\mathbb{R}om$$A$

Theorem 3.1. Let $\Psi(t)=A\natural_{t}B$ for strictly positive operators$A$

and B. Then for$x,$ $y\in \mathbb{R}_{f}$$D_{\Psi}(x,y)=D_{0}(A\mathfrak{h}_{y}B|A\natural_{x}B)$.

In particular, $D_{\Psi}(1,0)=D_{0}(A|B)$ .

Proof$D_{\Psi}(x,y)=A \natural_{x}B-A\natural_{y}B-\alphaarrow+0\}im\frac{A\natural_{y+\alpha(x-y)}B-A\natural_{y}B}{\alpha}$

$=$

$=$

$A \natural_{x}B-A\int_{y}B-\lim_{\alphaarrow+0}\frac{(A\mathfrak{b}_{y}\fbox{Error::0x0000}B)\natural_{ \alpha}(A\mathfrak{h}_{x}B)-A\mathfrak{h}_{y}B}{\alpha}$by [11, Lemma 2.2]

$A\natural_{x}B-A\mathfrak{h}_{y}B-S(A\natural_{y}B|A\natural_{x}B)=D_{0}(A\natural_{y}B|A\natural_{x}B)$.
ロ

for functions
In the rest of this section, we obtainin section 2.
andoperator divergences$D_{\Psi}(1,0)$$\triangle_{1},$$\triangle_{2_{\rangle}}\triangle_{5}$$A_{6}$$\Psi$

which relate to the8Theorem 3.2. For strictly positive operators$A$and , the following hold:$B$(1) If $\Psi(t)=T_{t}(A|B)-S(A|B)$ , then$D_{\Psi}(1,0)=D_{0}( \mathcal{A}|B)-\frac{1}{2}S(A|B)A^{-1}S(A|B)$.

(2) If $\Psi(t)=S_{t}(A|B)-S(A|B)$ , then$D_{\Psi}(1,0)=D_{0}(A|B)+D_{0}(B|A)-S(\mathcal{A}|B)A^{-1}S(A|B)$.

(3) If $\Psi(t)=S_{t}(A|B)-T_{t}(A|B)$ , then$D_{\Psi}(1,0)=D_{0}(B|A)- \frac{1}{2}S(A|B)A^{-1}S(A|B)$.

(4) If $\Psi(t)=D_{t}(A|B)$ for $t\in[O$ , 1 , then$]$$D_{\Psi}(1,0)=D_{0}(B|A)-2D_{0}(A|B)+ \frac{1}{2}S(A|B)A^{-1}S(A|B)$.

Here, we give a proof of (1). The others are obtained similarly.

Proof. (1) For $a>0$ , we have$\lim_{\alphaarrow+0}\frac{a^{\alpha}-1-\alpha\log a}{\alpha^{2}}=\frac{1}{2}(\log a)^{2}.$Replacing$a$by$A^{-\frac{1}{2}}BA^{\frac{1}{2}}$$\lim_{\alphaarrow+0}\frac{T_{\alpha}(A|B)-S(A|B)}{\alpha}$, we have$= \lim_{\alphaarrow+0}\frac{A^{\frac{1}{2}}((A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^{\alpha}-I-\alpha log(A^{-\frac{1}{2}}BA^{-\frac{1}{2}}))A^{\frac{1}{2}}}{\alpha^{2}}$$= \frac{1}{2}A^{\frac{1}{2}}(\log(A^{-\frac{1}{2}}BA^{-\frac{1}{2}}))^{2}A^{\frac{1}{2}}=\frac{1}{2}S(A|B)A^{-1}S(A|B)$,

then$D_{\Psi}(1,0)$$=T_{1}(A|B)-S(A|B)-(T_{0}(A|B)-S(A|B))$$- hm\frac{T_{\alpha}(A|B)-S(A|B)-(T_{0}(A|B)-S(A|B))}{\alpha}\alphaarrow+0$$=$

処 $(A|B)-S(A|B)- \lim_{\alphaarrow+0}\frac{T_{\alpha}(A|B)-S(A|B)}{\alpha}$$=D_{0}(A|B)- \frac{1}{2}S(A|B)A^{-1}S(A|B)$.
口9
4. Divergences given by the diﬀerences of expanded relative
operator entropies
In this section, we try to generalize Theorems 2.2 and 2.3 in section 2 for operator
power mean. For $A,$ $B>0,$ $x\in[O$ , 1 and $r\in[-1, 1]$ , operator power mean $A\#_{x,r}B$ isdefined as follows:$]$$A\#_{x,r}B\equiv A^{\frac{1}{2}}\{(1-x)I+x(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^{r}\}^{\frac{1}{r}}A^{\frac{1}{2}}=A\mathfrak{y}_{\frac{1}{r}}\{A\nabla_{x}(A\natural_{r}B)\}.$

holds for $x\in[0$ , 1 and $r\in[-1, 1]$ (cf. [9],We remark that $A\#_{x,r}B=B\#_{1-x,r}$[11]). To preserve $(1-x)I+x(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^{r}\geq 0$ , we have to impose in , 1 . The
operator power mean is a path passing through $A=A\#o_{r}B$ and $B=A \int_{1,r}B$ , andcombines arithmetic, geometric and harmonic means, that is, $A\#_{x,1}B=A\nabla_{x}B,$$A \#_{x,0}B\equiv\lim_{rarrow 0}A\#_{x,r}B=A\#_{x}B$ and $A\#_{x,-1}B=A\Delta_{x}B=(A^{-1}\nabla_{x}B^{-1})^{-1}$ . For
and expanded, 1 and $r\in[-1_{\}}1]$ , expanded relative operator entropy
are defined as follows (cf. [9]):Tsallis relative operator entropy
$A$

$]$

$[0$

$x$

$\alpha\in[O$$]$$S_{\alpha,r}(A|B)$$]$$T_{\alpha,r}(A|B)$$S_{\alpha,r}(A|B)$$\equiv A^{\frac{1}{2}}(\{(1-\alpha\rangle I+\alpha(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^{r}\}^{\frac{1}{f}-1}\frac{(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^{r}-I}{r})A^{\frac{1}{2}}$$= \frac{d}{dx}A\#_{x,r}B|_{x=\alpha}=(A\Vert_{cx,r}B)(A\nabla_{\alpha}(A\#_{r}\mathcal{B}))^{-1}S_{0,r}(A|B)(r\neq 0)$$S_{\alpha,0}(A|B)$$T_{\alpha,r}(A|B)$$\equiv rarrow 01_{1}’mS_{\alpha,r}(A|B)=S_{\alpha}(A|B)$$\equiv$$\frac{A\#_{\alpha,r}B-A}{\alpha}(\alpha\neq 0)$,,, $T_{0,r}(A|B) \equiv\lim_{\alphaarrow 0}T_{\alpha,r}(A|B)=T_{r}(A|B)$ .

We remark that $S_{0,r}(A|B)=T_{r}(A|B)$ , $S_{1,r}(A|B)=-T_{r}(B|A)$ and $T_{1,r}(A|B)=B-A$
hold for $r\in[-1, 1]$ . A similar inequality to $(*)$ also holds for these expanded relative
erator entropies, which is given as follows [9]:$ol\}$$(**)$$S_{\zeta),r}(A|B)\leq T_{\alpha,r}(A|B)\leq S_{\alpha,r}(A|B)\leq-T_{1-\alpha,r}(B|A)\leq S_{1,r}(A|B)$$\alpha$,欧 $[0$ , 1 $],$ $r\in[-1, 1].$

If $r=0$ , then this inequality becomes $(*)$ .
We defined expanded operator valued a-divergence $D_{\alpha,r}(A|B)$ as follows [11]:$D_{\alpha,r}(A|B)\equiv-T_{1-\alpha,r}(B|A)-T_{\alpha,r}(A|B) , \alpha\in[0, 1 ], r\in[-1, 1].$

For $\alpha\in(O, 1)$ , we can represent$D_{\alpha,r}(A|B)$

as follows:$D_{\alpha,r}(A|B)= \frac{A\nabla_{\alpha}B-A\#_{\alpha,r}B}{\alpha(1-\alpha)}.$We gave the following relations on expanded operator valued -divergence.
Proposition 4.1 ([11], Proposition 4.4). For strictly positive operators andfollowing hold:$\alpha$

$A$

(1)
(2)
(3)
(4)

for

$\alpha\in[0$

,1

$],$

$D_{\alpha,0}(A|B)$$=$$D_{\alpha}(A|B)$$D_{\alpha,1}(A|B)$

$=$

$0$

$D_{0,r}(A|B)$$=$$B-A-S_{0,r}(\mathcal{A}|B)$$D_{1,r}(\mathcal{A}|B)$$=$$A-B-S_{0,r}(B|A)=D_{0,r}(B|A)$for$\alpha$欧 $[0$ , 1 $],$

for $r\in[-1, 1],$

for $r\in[-1, 1].$

$B_{f}$

the10
We call $D_{0,r}(A|B)=B-A-S_{0,r}(A|B)$ expanded Petz-Bregman divergence.
Similarly to section 2, we consider the diﬀerences between two expanded relative
operator entropies in $(**)$ as operator divergence. There are 10 such divergences.
$T_{\alpha,r}(A|B)-S_{0,r}(A|B) , S_{\alpha,r}(A|B)-T_{\alpha,r}(A|B)$,$-T_{1-cx,r}(B|A)-S_{\alpha,r}(A|B) , S_{1,r}(A|B)+T_{1-\alpha,r}(B|A)$,$S_{\alpha,r}(A|B)-S_{0,r}(A|B) , -T_{1-\alpha,r}(B|A)-T_{\alpha,r}(A|B)=D_{\alpha,r}(A|B)$,$S_{1,r}(A|B)-S_{\alpha,r}(A|B) , -T_{1-\alpha,r}(B|A)-S_{0,r}(A|B)$ ,$S_{1,r}(A|B)-T_{\alpha,r}(A|B) , S_{1,r}(A|B)-S_{0,r}(A|B)$ .

The relations of these diﬀerences are given as in Table 2. If $r=0$ , then this tablecoincides with Table 1.

We represent these operator divergences by using expanded Petz-Bregman divergence. Theorems 4.2 and 4.3 correspond to Theorems 2.2 and 2.3, respectively.
Theorem 4.2. For strictly positive operators
hold:

$A$

and

$B$

and $r\in$ -.1,
$[$

(1)

$T_{\alpha,r}(A|B)-S_{0,r}(A|B)$$=$$\frac{1}{\alpha}D_{0,r}(A|A\#_{\alpha,r}B)$(2)$S_{\alpha,r}(A|B)-T_{\alpha,r}(A|B)$$=$$\frac{1}{\alpha}D_{0,r}(A\#_{\alpha,r}B|A)$

Theorem 4.3. For strictly positive operators
holds:

$A$

and

$B$

the following$1]_{f}$for$\alpha\in(0,1$

],

for

$\alpha\in(0,1$].

and $r\in[-1, 1]$ , the following$D_{\alpha,r}(A|B)= \frac{1}{1-\alpha}D_{0,r}(A\#_{\alpha,r}B|B)+\frac{1}{\alpha}D_{0,r}(A\#_{\alpha,r}B|A)$for$\alpha\in(0,1)$.

11References[1] J. I. Fujii, On the relative operator entropy (in Japanese), RIMS Kokyuroku, 903(1995),

49-56.
[2] J. I. Fujii and E. Kamei, Relative operator entropy in noncommutative information
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e-2012 $(2012)_{\backslash }353-362.)$[9] H. Isa, M. Ito, E. Kamei, H. Tohyama and M. Watanabe, Extensions of Tsallis relative
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[15] F. Kubo and T. Ando, Means of positive linear operators, Math Ann., 248(1980), 205-

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