数理解析研究所講究録
第 1893 巻 2014 年 35-45

35

Extensions of relative operator entropies and operator -divergence$\alpha$

亀井栄三郎 (Eizaburo KAMEI )(1), 伊佐浩史 (Hiroshi ISA)(2) , 伊藤公智 (Masatoshi ITO)(3)
遠山宏明 (Hiroaki TOHYAMA)(4), 渡邉雅之 (Masayuki WATANABE)(5)(2), (3), (4), (5): 前橋工科大学 (Maebashi Institute of Technology)

1. Generalized relative operator entropy
Throughout this note,
and $B$ are positive operators on a Hilbert space. In $[2]($ cf.[3,15], wegave the relative operator entropy by$A$$S(A|B)= \lim_{rarrow 0}\frac{A\#_{r}B-A}{r}=A^{1}z(\log A^{1}-\pi BA^{-q}1)A^{1}2,$where the notation $A\#_{r}B=A^{1}z(A^{-2}BA^{-z}11)^{r}A^{1}2,$ $r\in(0, 1 )$ , is the geometric operator meanin the sence of Kubo-Ando [13]. Tsallis relative operator entropy is defined by Yanagi, Kuriyama
and Furuichi in [16] as follows.
$T_{r}(A|B)= \frac{A\#_{r}B-A}{r}.$As a generalization of $S(A|B)$ , Furuta [7] has given the following:,$S_{r}(A|B)= \lim_{\epsilonarrow 0}\frac{A\natural_{r+\epsilon}B-A\natural_{r}B}{\epsilon}=A^{z}l(A^{-}\tau BA^{-2}11)^{r}(\log A^{11}-zBA^{-}\sigma)A^{1}z=(A\natural_{r}B)\cdot A^{-1}\cdot S(A|B)$

here we call $A\natural_{r}B=A^{1}z(A^{-}zBA^{-}z11)^{r}A^{1}\sigma$ , for $r\in R$ , a path going throughcoincides with $A\#_{r}B$ if $r\in[O, 1].$

$A$

and

$B$

, whichTheorem 1.([S]). For $A>0,$ $B>0,$ $S_{r}(A|B)$ is monotone increasing for $r\in R$ , and thefollowing holds.
(1)$S_{r}(A|B) \leq\frac{A\natural_{q}B-A\natural_{r}B}{q-r}\leq S_{q}(A|B)$

for

$q,$

$r\in R,$$q>r.$

Especially, in the case $r=0$ and $0<q<1,$ (1) is expressed as follows:(2)$S(A|B) \leq\frac{A\#_{q}B-A}{q}=T_{q}(A|B)\leq S_{q}(A|B)$.

To prove Theorem 1, we use the next Lemma.

Lemma 2.([8]). Let $a>0$ . Then the following holds for$a^{r} \log a\leq\frac{a^{q}-a^{r}}{q-r}\leq a^{q}\log a$Since$a^{t}$$q,$$r\in R.$

, for $q>r.$is convex function, this is easily given, but we give an elementary proof.

Proof. We show this inequality as follows:$\frac{a^{q}}{a^{r}}\log\frac{a^{q}}{a^{r}}=-\frac{a^{q}}{a^{r}}\log\frac{a^{r}}{a^{q}}\geq-\frac{a^{q}}{a^{r}}(\frac{a^{r}}{a^{q}}-1)=\frac{a^{q}}{a^{r}}-1\geq\log\frac{a^{q}}{a^{r}},$36that is,$a^{q}(\log a^{q}-\log a^{r})\geq a^{q}-a^{r}\geq a^{r}(\log a^{q}-\log a^{r})$.

So we have$(q-r)a^{q}\log a\geq a^{q}-a^{r}\geq(q-r)a^{r}\log a.$

Generalizing Theorem 1, we give the following basic relations in this note.
Theorem 3.([8,11]). For $A>0,$ $B>0$ , the following hold.
(1) If $0<r<1$ , then
$S(A|B)\leq T_{r}(A|B)\leq S_{r}(A|B)\leq-T_{1-r}(B|A)\leq-S(B|A)=S_{1}(A|B)$.

(2) If $n<r<n+1$ , then$S_{n}(A|B) \leq\frac{A\natural_{r}B-A\natural_{n}B}{r-n}\leq S_{r}(A|B)\leq\frac{A\natural_{n+1}B-A\natural_{r}B}{n+1-r}\leq S_{n+1}(A|B)$$\Leftrightarrow$$(BA^{-1})^{n}S(A|B)\leq(BA^{-1})^{n}T_{r-n}(A|B)\leq(BA^{-1})^{n}S_{r-n}(A|B)$$\leq -(BA^{-1})^{n}T_{n+1-r}(B|A)\leq-(BA^{-1})^{n}S(B|A)=(BA^{-1})^{n}S_{1}(A|B)$.

$\Leftrightarrow$$S(A\natural_{n}B|A\natural_{n+1}B)\leq T_{r-n}(A\natural_{n}B|A\natural_{n+1}B)\leq S_{r-n}(A\natural_{n}B|A\natural_{n+1}B)$.

$\leq -T_{n+1-r}(A\natural_{n+1}B|A\natural_{n}B)\leq S_{1}(A\natural_{n}B|A\natural_{n+1}B)$

The following properties of $S_{r}(A|B)$ and $T_{r}(A|B)$ are important in our discussion.
Let

Lemma 4.([8,11]).
properties:$n$be an integer. Then, for $r\in R,$$S_{r}(A|B)$

has the following(1)$S_{r}(A|B)=-S_{1-r}(B|A)=BS_{r-1}(B^{-1}|A^{-1})B=-AS_{-r}(A^{-1}|B^{-1})A,$(2)$S_{n}(A|B)=(BA^{-1})^{n}S(A|B)=S(A|B)(A^{-1}B)^{n},$(3)$S_{r}(A|B)=(A\natural_{\Gamma}B)\cdot A^{-1}\cdot S(A|B)=S(A\natural_{r}B|A\natural_{r+1}B)$(4)$\frac{A\mathfrak{h}_{r}B-A\natural_{n}B}{r-n}=(BA^{-1})^{n}T_{r-n}(A|B)=T_{r-n}(A|B)(A^{-1}B)^{n},$(5)$\frac{A\mathfrak{h}_{n+1}B-A\natural_{r}B}{n+1-r}=-(BA^{-1})^{n}T_{n+1-r}(B|A)=-T_{n+1-r}(B|A)(A^{-1}B)^{n}.$,Proof. (1) is given as follows:$s_{r}(A|B)$

$=$

$=$

$z(A^{-}BA^{-}11)^{r}(\log A^{-}BA^{-}z11)A$ フ$A^{1}$$-B^{111}zB^{-g}A\tau(A^{1}B^{-1}A^{1}z)^{-r}(\log$ $A^{1}B^{-1}$ $A^{1})A^{11}zB^{-}B$ フ$= -B^{q}1(B^{-z}AB^{-2})^{-r}(\log B^{-z}AB^{-z}1111)(B^{-1}AB^{-z}1)B^{1}z$$=$$-B^{1}(B^{11}-zAB^{-z})^{-r+1}(lB^{-} I$ $AB^{-1})B^{1}r=-S_{-r+1}(B|A)$

,

$or$

$= B^{1}\tau(B^{1}IA^{-1}B^{1}2)^{r-1}(\log B^{1}zA^{-1}B^{1}2)B^{1}z$$= BB^{-z}1(B^{1}2A^{-1}B^{1})^{r-1}(\log B^{1}A^{-1}B^{11}z)B^{-z}B=BS_{r-1}(B^{-1}|A^{-1})B.$37
The last equation is shown by the similar way.
(2) is shown as follows:$S_{n}(A|B)$$A$ 奮 $(A^{-3}BA^{-}z11)^{n}(\log$=$A^{-}zBA^{-}z11)A^{1}2$$= (BA^{-1})^{n}A^{1}2(\log A^{-z}BA^{-Z}11)A^{1}\pi=(BA^{-1})^{n}S(A|B)$.

(3) follows from the definitions of $S_{r}(A|B)$ and $S(A\natural_{r}B|A\natural_{r+1}B)$ .
$S_{r}(A|B)= \lim_{\epsilonarrow 0}\frac{A\natural_{r+\epsilon}B-A\natural_{r}B}{\epsilon}=(A\natural_{r}B)\cdot A^{-1}\cdot S(A|B)$$S(A \natural_{r}B|A\natural_{r+1}B)=\lim_{\epsilonarrow 0}\frac{(A\natural_{r}B)\cdot A\cdot(A\#_{\epsilon}B-A)}{\epsilon}=(A\natural_{r}B)\cdot A^{-1}\cdot S(A|B)$.

(4) is shown as follows:$\frac{A\natural_{r}B-A\natural_{n}B}{r-n} = \frac{A^{111}z(A^{-}zBA^{-}z)^{n}\{(A^{-11}\pi BA^{-}z)^{r-n}-I\}A^{1}z}{r-n}$$= \frac{(BA^{-1})^{n}A^{111}z\{(A^{-}zBA^{-}z)^{r-n}-I\}A}{r-n}$$= \frac{(BA^{-1})^{n}(A\#_{r-n}B-A)}{r-n}=(BA^{-1})^{n}T_{r-n}(A|B)$,

and (5) is also given similarly,$\frac{A\natural_{n+1}B-A\natural_{r}B}{n+1-r} = \frac{A^{1}2(A^{1}-2BA^{1}-Z)^{n+1}A^{1}z-A^{111}z(A^{-}zBA^{-}z)^{r}A^{1}\pi}{n+1-r}$$= \frac{Az(A^{-}zBA^{-}z)^{n}\{A^{-}zBA^{-}z-(A^{1}-2BA^{1}-2)^{r-n}\}Az111111}{n+1-r}$$= \frac{(BA^{-1})^{n}A^{1111}z\{A^{-q}BA^{-}z-(A^{-q}BA^{-}z)^{r-n}\}Az11}{n+1-r}$$= \frac{(BA^{-1})^{n}(B-A\#_{r-n}B)}{n+1-r}=\frac{(BA^{-1})^{n}(B-B\#_{n+1-r}A)}{n+1-r}$$= -(BA^{-1})^{n}T_{n+1-r}(B|A)$

2. Extension of Tsallis relative operator entropy
The following is called the power mean [4,12],$A\#_{t,r}B = A^{1}2\{(1-t)I+t(A^{-11}zBA^{-}2)^{r}\}^{\frac{1}{r}}A^{1}2, t\in[0,1], r\in[-1,1],$$= A\natural_{\frac{1}{r}}\{A\nabla_{t}(A\natural_{r}B)\}.$, whereIt is known that $A\#_{t,1}B=A\nabla_{t}B,$ $A\#_{t,0}B=A\#_{t}B$ and $A\#_{t,-1}B=A$$\Delta_{t}B=(A^{-1}\nabla_{t}B^{-1})^{-1}$ is$A\nabla_{t}B=(1-t)A+tB$ is the arithmetic operator mean andthe harmonic operator mean. Using this, we can define an extension of Tsallis relative operator
entropy (cf. [10]),$\Delta_{t}B$$A$$T_{t,r}(A|B)= \frac{A\#_{t,r}B-A}{t}.$Relations between $T_{t,r}(A|B),$ $T_{t}(A|B),$ $T_{r}(A|B)$ and $S(A|B)$ are given by the following diagram.
$T_{t,r}(A|B) arrow tarrow 0 T_{r}(A|B)$$\downarrow rarrow 0$$\downarrow rarrow 0$$T_{t}(A|B) arrow tarrow 0 S(A|B)$

38
Since the corresponding function to the power mean is $p(t, r)=\{1-t+ta^{r}\}^{\frac{1}{r}}$ and, we can define$Ft \partial p(t, r)=\{1+t(a^{r}-1)\}^{\frac{1}{r}-1}\cdot\frac{a^{r}-1}{r}$$S_{t,r}(A|B) = A^{S}( \{I+t((A^{11}-zBA^{-g})^{r}-I)\}^{\frac{1}{r}-1}\cdot\frac{(A^{11}-zBA^{-}z)^{r}-I}{r})A^{1}z, r\neq 0,$$= A \natural_{\frac{1-r}{r}}\{A\nabla_{t}(A\natural_{r}B)\}\cdot A^{-1}\cdot\frac{A\natural_{r}B-A}{r}.$$\lim_{rarrow 0}S_{t,r}(A|B)$$=$$S_{t,0}(A|B)=S_{t}(A|B)$.

This will be shown later.

We call this an expanded relative operator entropy. Then we can show an expanded form ofTheorem 3 (2) as follows [9]:Theorem 6. For $A>0$ and $B>0,$$S_{0,r}(A|B)\leq T_{t,r}(A|B)\leq S_{t,r}(A|B)\leq-T_{1-t,r}(B|A)\leq S_{1,r}(A|B)$

holds for $t\in[O, 1]$ and $r\in[-1,1].$
Proof. (1) First we show $S_{0,r}(A|B)\leq T_{t,r}(A|B)$ . Since $S_{0,r}(A|B)=T_{r}(A|B)$ , we have only toshow that$\frac{\{1+t(a^{r}-1)\}^{\frac{1}{r}}-1}{t}\geq\frac{a^{r}-1}{r}.$Let$f(t)= \{1+t(a^{r}-1)\}^{\frac{1}{f}}-1-t\frac{a^{r}-1}{r}.$Then we have$\frac{d}{dt}f(t)=\{1+t(a^{r}-1)\}^{\frac{1}{r}-1}\cdot\frac{a^{r}-1}{r}-\frac{a^{r}-1}{r}=\frac{\partial}{\partial t}p(t,r)-\frac{a^{r}-1}{r}$and$\frac{\partial^{2}}{\partial t^{2}}p(t, r)=\{1+t(a^{r}-1)\}^{\frac{1}{r}-2}\cdot\frac{(a^{r}-1)^{2}(1-r)}{r^{2}}\geq 0.$

So$\frac{\partial}{\partial t}p(t,r)$$\frac{df(0)}{dt}=0$$t \frac{a^{r}-1}{r}$is an increasing function for $t\in[O, 1]$ and$\frac{\partial}{\partial t}p(t,r)|_{t=0}=\frac{a^{r}-1}{r}$. Sinceby, we have $f(t)$ is increasing for $t\in[0,1]$ and $f(O)=0$ , so we have $\{1+t(a^{r}-1)\}^{\frac{1}{r}}-1\geq$, that is,$\frac{\{1+t(a^{r}-1)\}^{\frac{1}{r}}-1}{t}\geq\frac{a^{r}-1}{r}$and $S_{0,r}(A|B)\leq T_{t,r}(A|B)$ .

(2) Second, we show $T_{t,r}(A|B)\leq S_{t,r}(A|B)\leq-T_{t,r}(B|A)$ . It is sufficient to show$\frac{\{1+t(a^{r}-1)\}^{\frac{1}{r}}-1}{t}\leq\{1+t(a^{r}-1)\}^{1-1}r.

\frac{a^{r}-1}{r}.$Since$\frac{d^{2}}{dt^{2}}f(t)=\frac{\partial^{2}}{\partial t^{2}}p(t, r)\geq 0$$f(t)$$\frac{df(t)}{dt}\geq 0$, for $t\in[O, 1]$ and $t\in[-1,1],$

and $p(t, r)$ are convex for on $[0,1].$$t$

Since a function $f(t)$ is convex, then, for$a,$$h>0,$$f(a)=f( \frac{a}{a+h}(a+h)+\frac{h}{a+h}0)\leq\frac{a}{a+h}f(a+h)+\frac{h}{a+h}f(0)$,39
so that$\frac{f(a+h)-f(a)}{h}\geq\frac{f(a)-f(0)}{a}.$Hence we have$\frac{p(t+h,r)-p(t,r)}{h}\geq\frac{p(t,r)-p(0,r)}{t}$, for$\forall h>0,$that is,, for $\forall h>0,$$\frac{(1+(t+h)(a^{r}-1))^{\frac{1}{r}}-(1+t(a^{r}-1))^{\frac{1}{r}}}{h}\geq\frac{(1+t(a^{r}-1))^{\frac{1}{r}}-1}{t}$and$\lim_{harrow+0}\frac{(1+(t+h)(a^{r}-1))^{\frac{1}{r}}-(1+t(a^{r}-1))^{\frac{1}{r}}}{h}$$= \{1+t(a^{r}-1)\}^{\frac{1}{r}-1}\cdot\frac{a^{r}-1}{r}\geq\frac{(1+t(a^{r}-1))^{\frac{1}{r}}-1}{t}.$So we can conclude$S_{t,r}(A|B)\geq T_{t,r}(A|B)$.

(3) Next we show $S_{t,r}(A|B)\leq-T_{1-t,r}(B|A)$ . Since$T_{1-t,r}(B|A) = \frac{B\# 1-t,rA-B}{1-t}=\frac{A\#_{t,r}B-B}{1-t}$$A^{1}z(\{I+t((A^{11}-zBA^{-2})^{r}-I)\}^{\frac{1}{r}}-(A^{11}-\circ BA^{-}3))A^{1}\Sigma$$1-t$

the function corresponding to $T_{1-t,r}(B|A)$ is
For a convex function $f(t)$ ,$\frac{p(t,r)-a}{1-t}.$.

$f(t)=f( \frac{1-t}{1-t+h}(t-h)+\frac{h}{1-t+h}\cdot 1)\leq\frac{1-t}{1-t+h}f(t-h)+\frac{h}{1-t+h}f(1)$So we have$\frac{f(t)-f(t-h)}{h}\leq\frac{f(1)-f(t)}{1-t}.$

For $p(t, r)=(1+t(a^{r}-1))^{\frac{1}{r}},$$\frac{p(t,r)-p(t-h,r)}{h}\leq\frac{p(1,r)-p(t,r)}{1-t},$that is,$\frac{\{1+t(a^{r}-1)\}^{\frac{1}{r}}-\{1+(t-h)(a^{r}-1)\}^{\frac{1}{r}}}{h}\leq\frac{a-\{1+t(a^{r}-1)\}^{\frac{1}{r}}}{1-t}.$Since$\lim_{harrow 0}\frac{\{1+t(a^{r}-1)\}^{\frac{1}{r}}-\{1+(t-h)(a^{r}-1)\}^{\frac{1}{r}}}{h}=\frac{\partial}{\partial t}p(t, r)=\{1+t(a^{r}-1)\}^{\frac{1}{r}-1}\cdot\frac{a^{r}-1}{r},$we have$\{1+t(a^{r}-1)\}^{\frac{1}{r}-1}\cdot\frac{a^{r}-1}{r}\leq\frac{a-\{1+t(a^{r}-1)\}^{\frac{1}{r}}}{1-t}.$

40
So we conclude $S_{t,r}(A|B)\leq-T_{1-t,r}(B|A)$ .
(4) Finally, we see $-T_{1-t,r}(B|A)\leq S_{1,r}(A|B)$ , it is sufficient to show that$- \frac{\{1+t(a^{r}-1)\}^{\frac{1}{r}}-a}{1-t}\leq\frac{a-a^{1-r}}{r}.$This is equivalent to$\{1+t(a^{r}-1)\}^{\frac{1}{r}}-a\geq-(1-t)\frac{a-a^{1-r}}{r}.$Let$g(t)= \{1+t(a^{r}-1)\}^{\frac{1}{r}}-a+(1-t)\frac{a-a^{1-r}}{r},$then for $t\in[O, 1],$$\frac{d}{dt}g(t)=\{1+t(a^{r}-1)\}^{\frac{1}{r}-1}\cdot\frac{a^{r}-1}{r}-\frac{a-a^{1-r}}{r}.$As we showed above$\frac{\partial}{\partial t}p(t, r)d=\{1+t(a^{r}-1)\}^{\frac{1}{r}-1}\cdot\frac{a^{r}-1}{r}$is an increasing function for $t\in[O, 1]$ and$\in[0,1].Sothefunctiong(t)isadecreasingfunctionand1)=0,$$T_{1-t,r}(B|A)\leq S_{1,r}(A|B).$
thatis, $g(t)\geq 0fort\in[0,1]Hencewehavetheconc1$$\frac{d}{gd\dagger}g(t)|_{t=1}=0.Hence_{\overline{dt}}g(t)\leq 0fort.$

usion

-

Remark. ([9]). The following relations hold, so Theorem 3 (1) is the case $r=0$ in Theorem
6. For $A>0,$ $B>0$ and $t\in[O, 1],$ $r\in[-1,1]$ , the followings hold:,(1)$\lim_{rarrow 0}S_{t,r}(A|B)=S_{t}(A|B)$(2)$S_{0,r}(A|B)=T_{r}(A|B)$ ,(3)$S_{1,r}(A|B)=-T_{r}(B|A)$ ,(4)$\lim_{rarrow 0}S_{0,r}(A|B)=S(A|B)$(5)$\lim_{rarrow 0}S_{1,r}(A|B)=S_{1}(A|B)$,.

Proof. Since$\lim_{rarrow 0}\{1+t(a^{r}-1)\}^{\frac{1}{f}-1}\cdot\frac{a^{r}-1}{r}=a^{t}\log a,$

we have$\lim_{rarrow 0}A^{1}z(\{I+t((A^{-}BA^{-z}11)^{r}-I)\}^{\frac{1}{r}-1}\cdot\frac{(A^{-1}BA^{-r})^{r}-I1}{r})A^{q}1$$=A^{g}1(A^{111}-zBA^{-g})^{t}(\log A^{-r}BA^{-1})A^{1}l=S_{t}(A|B)$.

Moreover,$\frac{\partial}{\theta t}p(t, r)|_{t=0}=\frac{a^{r}-1}{r}, \frac{\partial}{\partial t}p(t, r)|_{t=1}=\frac{a-a^{1-r}}{r}$and their limits are known that$\lim_{rarrow 0}\frac{a^{r}-1}{r}=\log a, \lim_{rarrow 0}\frac{a-a^{1-r}}{r}=a\log a,$41
so we have$\lim_{rarrow 0}A^{1}z(\frac{(A^{-1}2BA^{-z})^{r}-I1}{r})A^{1}z=A^{1}2(\log A^{1}-qBA^{-1}q)A^{1}w=S(A|B)$,

and$\lim_{rarrow 0}A$フ$( \frac{A^{-}\S BA^{-}z-(A^{1}-zBA^{-\not\supset})^{1-r}111}{r})A^{1}\not\supset$$=A^{1}z(A^{-}zBA^{-}z11)(\log A^{-}\pi BA^{-}7)A^{z}111=S_{1}(A|B)$.

We can give an expanded version of Theorem 3 (2) also.
Theorem 7. For $A>0,$
equivalence holds:(1)$B>0,$$t\in[0,1],$$r\in[-1,1]$ and an integer $n$ , the following$S_{0,r}(A\natural_{n}B|A\natural_{n+1}B)\leq T_{t,r}(A\natural_{n}B|A\natural_{n+1}B)\leq S_{t,r}(A\natural_{n}B|A\natural_{n+1}B)$$\leq -T_{1-t,r}(A\natural_{n+1}B|A\natural_{n}B)\leq S_{1,r}(A\natural_{n}B|A\natural_{n+1}B)$

$=$

(2)

$(BA^{-1})^{n}\cdot S_{0,r}(A|B)\leq(BA^{-1})^{n}\cdot T_{t,r}(A|B)\leq(BA^{-1})^{n}\cdot S_{t,r}(A|B)$$\leq -(BA^{-1})^{n}\cdot T_{1-t,r}(B|A)\leq(BA^{-1})^{n}\cdot S_{1,r}(A|B)$.

Proof. (1) follows from Theorem 6 and we can obtain (2) by using the next lemma.
Lemma 8. For $A>0,$ $B>0,$$t\in[0,1],$$r\in[-1,1]$ and an integer $n$ , the following hold.

(1)$S_{0,r}(A\natural_{n}B|A\natural_{n+1}B)=(BA^{-1})^{n}\cdot S_{0,r}(A|B)=S_{0,r}(A|B)\cdot(A^{-1}B)^{n},$(2)$T_{t,r}(A\natural_{n}B|A\natural_{n+1}B)=(BA^{-1})^{n}\cdot T_{t,r}(A|B)=T_{t,r}(A|B)\cdot(A^{-1}B)^{n},$(3)$S_{t,r}(A\natural_{n}B|A\natural_{n+1}B)=(BA^{-1})^{n}\cdot S_{t,r}(A|B)=S_{t,r}(A|B)\cdot(A^{-1}B)^{n},$(4)$T_{1-t,r}(A\natural_{n+1}B|A\natural_{n}B)=(BA^{-1})^{n}\cdot T_{1-t,r}(B|A)=T_{1-t,r}(B|A)\cdot(A^{-1}B)^{n},$(5)$S_{1,r}(A\natural_{n}B|A\natural_{n+1}B)=(BA^{-1})^{n}\cdot S_{1,r}(A|B)=S_{1,r}(A|B)\cdot(A^{-1}B)^{n}.$Proof. Since$(A\natural_{n}B)\#$ち$r(A\natural_{n+1}B)=(BA^{-1})^{n}\cdot(A\#_{t,r}B)=(A\#$ち $rB)\cdot(A^{-1}B)^{n},$and$A\#_{t,r}B = A\natural_{\frac{1}{r}}(A\nabla_{t}(A\natural_{r}B))$$= A\natural_{\frac{1}{r}}(A\nabla_{t}(B\natural_{1-r}A))$$= B^{1}5((B^{-1}zAB^{-}2)1\natural_{\frac{1}{r}}\{(B^{-}zAB^{-Z})11\nabla_{t}(B^{-}2AB^{-}z)\}^{1-r})B^{q}111$$= B^{1}2\{(1-t)(B^{-3}AB^{-z}11)^{r}+tI\}^{\frac{1}{r}}B^{1}z=B\# 1-t,rA,$

42
we have the following:(1)$S_{0,r}(A\natural_{n}B|A\natural_{n+1}B)$$=$$T_{r}(A \natural_{n}B|A\natural_{n+1}B)=\frac{(A\natural_{n}B)\natural_{r}(A\natural_{n+1}B)-(A\natural_{n}B)}{r}$$= \frac{(A\natural_{n}B)\cdot A^{-1}\cdot(A\natural_{r}B-A)}{r}=(A\natural_{n}B)\cdot A^{-1}\cdot\frac{A\natural_{r}B-A}{r}$$= (BA^{-1})^{n}\cdot T_{r}(A|B)=(BA^{-1})^{n}\cdot S_{0,r}(A|B)$(2)$T_{t,r}(A\natural_{n}B|A\natural_{n+1}B)$$=$.

$\frac{(A\natural_{n}B)\#_{t,r}(A\natural_{n+1}B)-A\natural_{n}B}{t}$$= \frac{(A\natural_{n}B)\cdot A^{-1}\cdot(A\#_{t,r}B)-A\#_{n}B}{t}$$= (A \natural_{n}B)\cdot A^{-1}\cdot\frac{A\#_{t,r}B-A}{t}=(BA^{-1})^{n}\cdot T_{t,r}(A|B)$.

$D_{0}(A|B)=B-A-S(A|B)$ .

ur interpretation of $D_{0}(A|B)$ has been $D_{0}(A|B)= \lim_{tarrow+0}\frac{A\nabla_{l}B-A\#\ell^{B}}{t}$ and $D_{1}(A|B)$ is also
given by
But we are known the existence of -divergence. The -divergence is defined by Amari [1]. For
probability distributions $p=(p_{1}, \cdots,p_{n})$ and $q=(q_{1}, \cdots, q_{n})$ ,$O$$D_{1}(A|B)= \lim_{tarrow 1-0}\frac{A\nabla_{l}B-A\#_{\ell}B}{t-1}.$$\alpha$$\alpha$$D_{\alpha}(p|q)= \frac{4}{1-\alpha^{2}}(1-\sum_{i=1}^{n}p_{i^{7^{\underline{\alpha}}}}\underline{1}-q_{i^{+^{1\alpha}}}) , \alpha\neq\pm1,$43$q_{i}\geq 0$ for $i=1,$
where
and $\sum_{i=1}^{n}p_{i}=\sum_{i=1}^{n}q_{i}=1$ . If we put-divergence can be expressed as follows:$p_{i},$$\cdots,$$n$$t= \frac{1+\alpha}{2}$, then the$\alpha$$D_{t}(p|q)= \frac{1}{t(1-t)}\sum_{i=1}^{n}\{(1-t)p_{i}+tq_{i}-p_{i}^{1-t}q_{i}^{t}\}, t\neq 0,1.$

The operator version of the -divergence is given in [5,6] as follows:$\alpha$$D_{\alpha}(A|B)= \frac{A\nabla_{\alpha}B-A\#_{\alpha}B}{\alpha(1-\alpha)}$,for $0<\alpha<1.$This also satisfies the following:$D_{0}(A|B)= \lim_{\alphaarrow 0}\frac{A\nabla_{\alpha}B-A\#_{\alpha}B}{\alpha(1-\alpha)}=B-A-S(A|B)$,$D_{1}(A|B)= \lim_{\alphaarrow 1}\frac{A\nabla_{\alpha}B-A\#_{\alpha}B}{\alpha(1-\alpha)}=A-B-S(B|A)$.

We can combine the Tsallis relative operator entropy and this operator -divergence as follows:$\alpha$Theorem 9.
$D_{\alpha}(A|B)=-(T_{\alpha}(A|B)+T_{1-\alpha}(B|A))$

This can be shown as the following figure:, for $0<\alpha<1.$

44
Proof of Theorem 9.
$D_{\alpha}(A|B) = \frac{A\nabla_{\alpha}B-A\#_{\alpha}B}{\alpha(1-\alpha)}=\frac{(1-\alpha)A+tB-(1-\alpha)(A\#_{\alpha}B)-\alpha(A\#_{\alpha}B)}{\alpha(1-\alpha)}$$= - \frac{A\#_{\alpha}B-A}{\alpha}-\frac{A\#_{\alpha}B-B}{1-\alpha}=-\frac{A\#_{\alpha}B-A}{\alpha}-\frac{B\#_{1-\alpha}A-B}{1-\alpha}$$= =-(T_{\alpha}(A|B)+T_{1-\alpha}(B|A))$

So we can propose an extension of the operator -divergence as follows:$\alpha$$D$

ち $r(A|B)=-(T_{t,r}(A|B)+T_{1-t,r}(B|A))$, fOr $t\in[0,$ $1],$ $r\in[-1$ , 1$].$

Theorem 10.
$D_{t,r}(A|B)= \frac{A\nabla_{t}B-A\#_{t,r}B}{t(1-t)}, t\in(0,1)$.

References[1] S.Amari, Differential Geometrical Methods in Statistics, Springer Lecture Notes in Statistics,
28 (1985),[2] J.I.Fujii and E.Kamei, Relative operator entropy in noncommutative information theory,
Math. Japon., 34(1989), 341-348.
[3] J.I.Fujii, On the relative operator entropy (in Japanese), RIMS Kokyuroku, 903(1995),

49-56.
[4] J.I.Fujii and E.Kamei, Interpolational paths and their derivatives, Math. Japon., 39(1994),

557-560.
[5] J.I.Fujii, J.Mi\v{c}i\v{c}, J.Pe\v{c}ari\v{c} and Y.Seo, Comparison of operator mean geodesics, J. Math.
Inequal., 2(2008), 287-298.
[6] M.Fujii, J.Mi\v{c}i\v{c}, J.Pe\v{c}ari\v{c} and Y.Seo, Recent Development Mond-Pe\v{c}ari\v{c} Method in Operator Inequalities, Monographs in Inequalities 4, Element, Zagreb, (2012).

[7] T.Furuta, Parametric extensions of Shannon inequality and its reverse one in Hilbert space
operators, Linear Algebra Appl., 381(2004), 219-235.
[8] H.Isa, M.Ito, E.Kamei, H.Tohyama and M.Watanabe, Relative operator entropy, operator
divergence and Shannon inequality, Sci. Math. Japon., 75(2012), 289-298.
[9] H.Isa, M.Ito, E.Kamei, H.Tohyama and M.Watanabe,Extensions of Tsallis relative operator
entropy and operator valued distance, Sci. Math. Japon., 76(2013), 427-435.
[10] H.Isa, M.Ito, E.Kamei, H.Tohyama and M.Watanabe, Generalozatios of operator Shannon
inequality based on Tsallis and R\’eny relative entropies, linear Algebra Appl., 439(2013), 3148-

3155.
$\alpha$[11] H.Isa, M.Ito, E.Kamei, H.Tohyama and M.Watanabe, On relations between operator valued-divergence and relative operator entropies, preprint.
[12] E.Kamei, Paths of operators parametrized by operator means, Math. Japon., 39(1994),

395-400.
[13] F.Kubo and T.Ando, Means of positive linear operators, Math. Ann., 248(1980), 205-224.
[14] D.Petz, Bregman divergence as relative operator entropy, Acta Math. Hungar., 116(2007),

127-131.

45[15] A.Uhlmann, Relative entropy and the Wigner-Yanase-Dyson-Lieb concavity in an interpolation theory, Commun. Math. Phys., 54(1977), 22-32.
[16] K.Yanagi, K.Kuriyama, S.Furuichi, Generalized Shannon inequalities based on Tsallis relative operator entropy, Linear Algebra Appl., 394(2005), 109-118.
(1)

1-1-3, SAKURAGAOKA, KANMAKICHO, KITAKATURAGI-GUN, NARA, JAPAN, 639-0202.
ekamei1947@yahoo. co.jp(2)

MAEBASHI INSTITUTE OF TECHNOLOGY, 460-1, KAMISADORI-MACHl, MAEBASHI, GUNMA, JAPAN,

371-0816. isa@maebashi-it. ac. jp(3)

MAEBASHI INSTITUTE OF TECHNOLOGY, 460-1, KAMISADORI-MACHl, MAEBASHI, GUNMA, JAPAN,

371-0816. m-ito@maebashi-it.ac.jp(4)

MAEBASHI INSTITUTE OF TECHNOLOGY, 460-1, KAMISADORI-MACHl, MAEBASHI, GUNMA, JAPAN,

371-0816. tohyama@maebashi-it.ac.jp(5)

MAEBASHI INSTITUTE OF TECHNOLOGY, 460-1, KAMISADORI-MACHl, MAEBASHI, GUNMA, JAPAN,

371-0816. masayukiwatanabe@maebashi-it. ac. jp