数理解析研究所講究録 1458 巻 2005 年 28-33

28
Ando-Hiai inequality and Furuta inequality
大阪教育大学藤井 正俊 (MASATOSHI Fujii)

OSAKA KYOIKU UNIVERSITY
前橋工科大学亀井 栄三郎 (EJZABURO KAMEI)MAEBASHI INSTITUTE OF TECHNOLOGY
茨城大学・工学部

中本 律男 (Ritsuo NAKAMOTO)

IBARAKI UNIVERSITY
これは $[6],[7]$ で示した事柄をまとめた報告です. A, B は Hilbert space 上の
1. はじめに.
positive operators とします, 便宜上 A(resp. A $>0$ ) で A は positive (resp. positive invertible)
operator を表すものとしておきます. まずフルタ不等式の復習から始めましょう $[8],[9]$ .
$\geq 0$Furuta inequality:
then for each $r\geq 0$ ,(F)

If $A\geq B\geq 0$ ,$A^{R\mathrm{f}^{\underline{r}}}\triangleleft\geq(A^{f}\pi B^{\mathrm{p}}A^{\mathrm{p}})^{\frac{1}{q}}r$

and
(F)

$(B^{r}2A^{\mathrm{p}}B^{f}\mathrm{z})^{\frac{1}{q}}\geq B^{\mathrm{a}_{\mathrm{q}}\pm}r$holds for andand $q\geq 1$ with
$p$

$q$

such that $p\geq 0$$(1+r)q\geq p+r$.

これは次のレウナー. ハインツ不等式の拡張を意図したもので, 歴史的発展と評されています 4 上で与えられた図はこのことを明確に示すものです.

L\"owner-Heinz inequality:

A $\geq B\geq 0$(LH)$\Rightarrow$$A^{a}\geq B^{\alpha}$

for any ce $\in[0,$1].

しかしながら私たちは (F) を久保・安藤 [17] によって導入された作用素平均の言葉で表してみます. ここで使うのは$\alpha$-power mean (or generalized geometric operator mean) と呼ばれている作用

素平均で次のように与えられます $[2],[13]$ .

for$A \oint_{\alpha}B=A^{\pi}(A^{-\mathrm{z}}BA^{-\tau})^{a}A^{\mathrm{r}}1\mathrm{x}\mathrm{x}1$$\alpha$$0\leq\alpha\leq 1$.

-power mean を使うとフルタ不等式は次のように表すことが出来ます.

(F)$A\geq B\geq 0$$\Rightarrow$$A^{-r}\beta_{\frac{1+r}{\mathrm{p}+r}}B^{\mathrm{p}}\leq A$

for $p\geq 1$ and $r\geq 0$ .

作用素平均を用いた別証明を与える中で私たちは (F) を次のように整理することが出来これを satellite

inequality と呼びました [13].
(SF)

A $\geq B\geq 0$$\Rightarrow$$A^{-\tau}\#_{\frac{1+r}{p+\tau}}B^{\mathrm{p}}\leq B\leq A$

for p$\geq 1$and r$\geq 0$

.

29
2.
, (F)

$7\mathit{1}\triangleright \mathit{9}$$\mathrm{q}\supset^{\mathrm{v}}\mathrm{p}$$\mathrm{X}$#$\yen \emptyset \Xi Sk $t\grave{*}6\#\yen\#$ .
; Theorem 2

tf (SF)$\xi\xi\cdot\grave{\mathrm{x}}_{\wedge}C\vee \mathfrak{X}\grave{\circ}\mathrm{S}\text{ま}\llcorner$$\#_{\vee}^{r}5\dot{\mathrm{x}}6$$\emptyset \mathrm{X}\mathrm{F}\ovalbox{\tt\small REJECT} 2$$X\acute{2}$$\mathrm{A}_{\urcorner}\fbox_{0}\ovalbox{\tt\small REJECT} \mathrm{A}f^{\wedge}.\mathrm{b}\mathrm{f}\mathrm{X}$$[13]a)\ovalbox{\tt\small REJECT} \mathfrak{M}\#\mathrm{E}\mathrm{E}\text{し}$$(\mathrm{C})\text{で}\hslash\xi_{}^{-}\$$\emptyset^{\theta}>\# 2\delta 1\mathfrak{h}\text{ま}$A0$\llcorner\gamma\sim\llcorner$$\hslash 7\not\in \mathfrak{W}^{f}f\wp,\mathrm{t}\cdot’ T6$.

$\text{ま}- \mathrm{p}\Re\emptyset$

lemma

.

Lemma 1. If $A\geq B\geq 0$ and $p\geq 0$ , then$A^{-n} \#\frac{n}{\mathrm{p}+n}B^{\mathrm{p}}\leq I$

holds for $n=1$ , 2,$\ldots$

and $p\geq 0$ .

Proof. We prove this by induction. Since$n=1$ . If this holds for , that is,by (LH), then$n$

, we have the case, then it implies$A^{-1} \beta_{\frac{1}{\mathrm{P}+1}}B^{p}\leq B^{-1}\#\frac{1}{\mathrm{p}+1}B^{p}=I$or$A^{-n} \#\frac{n}{\mathrm{p}+n}B^{\mathrm{p}}\leq I$$(A^{n}\tau B^{p}A^{n}\tau)^{\frac{n}{\mathrm{p}+n}}\leq A^{n}$$(A^{n}\pi B^{\mathrm{p}}A^{n}\tau)^{\frac{\mathrm{l}}{\mathrm{p}+\mathrm{n}}}\leq A$A$\leq$$arrow \mathrm{f}\iota$

Effl$A^{-} \tau((A^{n}\tau B^{\mathrm{p}}A-\mathrm{p}n)^{-_{p\neg\overline{n}}}1\#_{\frac{n+1}{p+n+1}}A^{n}\tau B^{p}A^{n}?)A^{-\yen}=A^{-n}\#\frac{n}{\mathrm{p}+n}B^{p}\leq I$$\backslash 6\check{}\text{と}$Theorem 2.

$-n-1 \oint_{\frac{n+1}{p+n+1}}B^{p}=A^{-\mathrm{z}}(A^{-1}\#_{\mathrm{p}+n+1}n\underline{n+\underline{\mathrm{x}}}A^{\mathrm{p}}B^{\mathrm{p}}A^{9})A^{-\pi}nn$$-\epsilon(\mathrm{C})\mathfrak{j}\mathrm{g}\Re\sigma)$\ddagger :$\dot{?}\}_{\llcorner\{\ovalbox{\tt\small REJECT} \mathrm{b}\text{れま}F}^{r’}$.

.

If $A\geq B\geq 0$ , then(C)

$A^{-r} \#\frac{r}{\mathrm{p}+r}B^{p}\leq I$

holds for all $r\geq 0$ and $p\geq 0$ .
Proof. The case $0\leq r\leq 1$ ,If $r=n+\epsilon$ for positive integer, is assured by (LH).

$A^{-r} \#\frac{\mathrm{r}}{\rho+r}B^{p}\leq B^{-r}\#\frac{f}{\mathrm{p}+r}B^{p}=I$$n$and $0\leq\epsilon<1$ , then$(A^{\mathrm{g}\tau}B^{p}A^{n})^{\frac{\epsilon}{p+n}}\leq A^{\epsilon}$

by Lemma 1 and (LH). Hence we have$A^{-r} \#\frac{r}{\mathrm{p}+\mathrm{r}}B^{p}=A^{-\langle n+\epsilon)}\beta_{\frac{n+\epsilon}{p+\overline{n+\epsilon}}}B^{p}$$A^{-T}n$$=$Theorem 2
6 (F) or (SF)$\leq$$A^{-\tau}((A^{n}\mathrm{Z}B^{p}A^{n}\hslash 2)^{-\frac{\epsilon}{p+n}}\#_{\frac{n+\epsilon}{p+n+\epsilon}}A^{n}\tau B^{\mathrm{p}}A^{n}z)A^{-\yen}\#$$=$$A^{-}?n(A^{n}?B^{p}A^{n} \mathrm{E})^{\frac{n}{\mathrm{p}+n}}A^{-\tau}\mathrm{B}=A^{-n}\#\frac{n}{p+n}B^{p}\leq I$$?\mathrm{g}_{7J}\triangleright P*\not\cong Xx$$\sim l\mathrm{Z}$$(A^{-\epsilon}\mathfrak{g}_{*\not\in}_{\epsilon}A^{n}7B^{\mathrm{p}}A^{n}P)A^{-\not\in}$?A ) lemma$\sigma$$(\mathrm{F})$or (SF)$\hslash^{1}\mathrm{b}$fflLemma 3. For A, B
if ’
.
for p
then
$\geq 0_{f}$$A^{-r} \int_{\frac{1+r}{\mathrm{p}+\#}}B^{\mathrm{p}}\leq B$$A^{-}$$\sigma):\mathrm{F}\mathrm{F}\S\beta*k^{-}\overline{/l^{\backslash }}T\ovalbox{\tt\small REJECT}\not\in\$$\#\frac{r}{\mathrm{p}+\mathrm{r}}B^{\mathrm{p}}\leq I$$*t\mathrm{g}\triangleleft \mathscr{L}\text{ま}\triangleleft^{-}$.

holds for r$\geq 0$and p$\geq 0_{J}$$\geq 1$

Proof. We here use well-known formulas onA\yen $T$$\mathrm{g}\mathfrak{j}_{\llcorner}^{-}\not\equiv|\mathrm{J}\mathrm{i}\geqq\vee \mathrm{c}\mathrm{g}$.

$\mathfrak{g}_{\alpha}B=B\# 1-\alpha$$A$$\beta_{a}$;atid $A\#\alpha\beta B=A\# a$ (A$\#\beta B\rangle$

.

.

)

$\mathrm{g}_{l}\mathbb{P}_{\piarrow}^{\vee}\sigma$$\not\in \mathrm{E}P$’30
Thus we have($A^{-r}\#\mathrm{l}\mathrm{a}\mathrm{e}B^{\mathrm{p}}=B^{p}\#$ $p+p\mathrm{L}_{\frac{1}{\tau}}^{-,A^{-r}=B^{p}\oint_{\mathrm{L}^{-\underline{1}}}}$,

A $-r$ )$B^{\mathrm{p}}\#_{\overline{p}+}e_{\overline{r}}$$\leq B^{\mathrm{p}}\#_{\frac{p-1}{p}}I=B\leq A$.

Corollary 4.(Furuta Inequality) ij $A\geq B\geq 0$ , then(F)

$A^{-\mathrm{r}}\#_{\frac{1+\tau}{p+r}}B^{\mathrm{p}}\leq B\leq A$

holds for all $r\geq 0$ and $p\geq 1$ .

3.

ffi#.

$\ovalbox{\tt\small REJECT}*\overline{\tau}L^{\backslash }\backslash J\mathit{5}$$g\mathrm{y}*\ovalbox{\tt\small REJECT}\triangleleft\not\in\}$&.

$\check{}\mathrm{f}\iota \text{まで}\ovalbox{\tt\small REJECT} \mathrm{A}f^{\wedge}.\mathrm{b}1\mathrm{X}$$[3],[15],[16]$ .

$\text{し^{}\vee}\mathrm{C}\mathrm{f}\mathrm{f}’\supset \mathrm{T}\mathrm{S}\text{ま}\mathrm{b}f^{\wedge}$$A\gg B$(C)$\text{ま}$,$\hslash\lfloor[\rfloor t^{\vee}-\ddagger$.

IX (F)$\mathrm{T}l\mathrm{X}ff$.

$k\hslash \mathcal{X}_{7^{-\mathrm{r}}}^{-}\sqrt{}^{\tau^{\backslash }}\nearrow \mathrm{p}\}\mathrm{I}\mathrm{E}\mathrm{f}\mathrm{f}$g$)$A>>B$$\mathrm{X}\check{:2}l^{\mathrm{Y}}.\Leftrightarrow\grave{\mathrm{X}}_{-}f’.\mathrm{t}\emptyset \mathrm{T}T$.

, $B>0$ .

$\ovalbox{\tt\small REJECT} ffi_{\tau}^{\sim}\mathrm{C}T$$k\mathrm{f}\mathrm{f}\dot{z}_{-}l\mathrm{J}^{4}X\mathrm{V}^{\backslash }-arrow \text{と}$$\mathrm{x}\check{\vee J}\mathfrak{j}^{\underline{r}}\mathrm{g}_{\mathrm{x}_{-\xi}}^{\mathrm{b}},\vee\not\simeq$$\xi\not\in-\check{\mathit{2}}\text{と}\text{で}$$\theta^{\theta}\backslash \text{で}\doteqdot\not\in T$}$\mathrm{g}\mathrm{h}\mathrm{L}[\rfloor \mathit{0})X\mathrm{a}\mathrm{e}\epsilon$$|\dot{2}\}_{\acute{\mathrm{c}}}$$(*)\sigma)\mathrm{g}\ddagger$.

$-\vee$$T\backslash \Leftrightarrow \mathrm{R}(\mathrm{F})k\mathrm{f}\mathrm{f}\vee\supset \mathrm{T}^{\mathrm{t}}\text{ま}T$$\mathrm{E}:f\overline{fl}\llcorner \text{ま}\mathcal{F}$.

$\lim_{narrow\infty}(I+\frac{1}{n}\log X)^{n}=X$

Theorem 2$\Xi \mathrm{f}\mathrm{f}$$\mathrm{L}\hslash^{\mathrm{x}}\text{し}\Re \mathrm{b}^{g\rangle}\ovalbox{\tt\small REJECT} \mathrm{E}fl\mathrm{t}\mathrm{f}7J\triangleright P$

Theorem 2 (C)$\langle$).

$\gamma_{\acute{\mathrm{L}}}\gamma\sim^{\theta}-$$\mathrm{L}\neq\#\mathrm{t}\mathrm{t}\mathrm{f}\mathrm{f}\mathrm{l}$Ll F@$\mathrm{g}$$\mathrm{i}_{\vee}^{\backslash ^{\backslash }}$for any $X>0$ .

$(\mathrm{C})\delta*>:\Pi:\mathrm{f}_{7^{-}}^{arrow}\triangleleft^{\backslash }.\nearrow\backslash ii^{r}||\ovalbox{\tt\small REJECT}\# q)\#\doteqdot \mathrm{f}\mathrm{f}\mathrm{f}\mathrm{d}\mathrm{t}\mathrm{P}k$$f\mathrm{g}\vee\supset \mathrm{T}\mathrm{t}\backslash \xi\}arrow\not\in\vee:\emptyset\ovalbox{\tt\small REJECT} Bfl\mathrm{B}Yfl^{\{i)}$.

Theorem 5. For A, B $>0$ , A $\gg B$ if and only if$\geq 0$

$A$

.

$(*)$

p

,

$oT5\grave{\mathrm{x}}\mathrm{b}hf_{\mathrm{L}}^{-}\mathrm{R}^{1}\mathrm{J}_{\mathrm{R}}^{-}\equiv \mathrm{i}\mathrm{E}\ovalbox{\tt\small REJECT}\# 3;8J\exists fflfx\mathrm{t}a)^{-}\mathrm{e}\tau[18]$$\mathrm{E}\mathrm{X}\text{と}\neq x6\mathit{0}>l3:\Re\sigma$$(*)k$$6(\mathrm{C})$$A_{7\triangleleft\nearrow\backslash }^{-}\overline{.}.\backslash P$ $\ovalbox{\tt\small REJECT} \mathrm{f}\mathrm{f}\not\subset\rangle$$\not\in\Leftrightarrow \mathrm{t}\mathrm{Z}\Re$\log A\geq\log B$$\infty$$\dagger.\sim \mathrm{C}\mathrm{p}6\ovalbox{\tt\small REJECT} \text{の}\ovalbox{\tt\small REJECT} \mathfrak{M}^{\ovalbox{\tt\small REJECT}}\ 4\dot{\mathrm{x}}^{\vee}T\mathrm{t}\backslash \text{ま}\tau[11]$$\check{}\mathrm{T}^{\backslash ^{\backslash }}\neq \mathrm{A}f_{-}^{\wedge}\mathrm{b}$$\pi\tau$$p$$\ovalbox{\tt\small REJECT} \mathrm{B}\mathrm{F}\mathrm{t}$$tj^{\dot{\theta}}1\hslash \mathcal{X}_{\overline{\mathcal{T}}\prime \mathrm{f}^{\backslash }\nearrow\backslash }P1|\ovalbox{\tt\small REJECT} \mathrm{F}\sigma)*\ovalbox{\tt\small REJECT} \mathrm{d}tJ\mathrm{k}fp\tau$$\mathrm{T}\mathrm{V}^{\backslash }6-arrow \mathrm{g}\sigma 2\ovalbox{\tt\small REJECT} \mathrm{A}f_{\mathrm{L}}^{-}\mathrm{b}^{g)}\mathrm{f}\mathrm{i}\grave{\mathrm{x}}f’.\frac{\equiv}{\mathrm{Q}}\mathrm{i}\mathrm{E}\Phi l\mathrm{g}_{\grave{J}}’ \mathrm{J}\llcorner\grave{\mathrm{J}}Lb\lambda\prime \mathrm{o}^{-}T\backslash$$\llcorner f_{\mathrm{L}}^{\wedge}\emptyset\grave{\grave{\}}}$$ffl\Re$

Theorem 2&$A^{-r} \#\frac{r}{p+r}B^{p}\leq I$

holds for all r$\geq 0$and.

Proof, Suppose $A\gg B$ , then $A_{1}=I+ \frac{1}{n}\log A\geq I+\frac{1}{n}\log B=B_{1}>0$ holds for sufficiently

large natural nurnmber$n$. Applying Theorem 2 to$A_{1}$

and$B_{1}$, we have$I \geq A_{1}^{-nr}\#\frac{n\mathrm{r}}{n\mathrm{p}+nr}B_{1}^{np}=(I+\frac{1}{n}\log A)^{-nr}\#\frac{f}{\mathrm{p}+r}$$(I+ \frac{1}{n}\log B)^{n\mathrm{p}}$

.

Since
$(I+ \frac{1}{n}\log A)^{-nr}arrow A^{-r}$

and

$A$

,

$B$

are irrvertible, we have$(A^{r}\mathrm{z}B^{p}A^{f}\tau)^{\frac{\mathrm{r}}{p+r}}\leq A^{r}$4.

$R\hslash$.aimplies$\bigwedge_{-\emptyset \mathrm{f}\mathrm{f}\mathrm{i}\# k}$$\ovalbox{\tt\small REJECT} \mathrm{t}\acute{\cdot}’\supset \mathrm{t},\backslash \tau*T4\backslash \gtrless \text{ま}\llcorner$$(A^{r}\mathrm{r}B^{p}A^{\tau}\tau)^{\frac{1}{p+r}}\ll A$.

$\not\cong\ovalbox{\tt\small REJECT}$.

$\mathrm{n}arrow\infty$). The converse is easily seen because. [E12 [1] ([12]) &C$\acute{t}^{\mathrm{B}}\tau T4^{\backslash }\text{ま}-\mathrm{f}$(asfor all $r\geq 0$ and $p\geq 0$ . $A\gg B$ is the case $r=0$ .

7/5*$#&\Phi fflffi.

$]:\check{9}$.

$I+ \frac{1}{n}\log B\rangle^{np}arrow B^{\mathrm{p}}$$A^{-r}\#_{\overline{p}+\overline{\mathrm{r}}}LB^{p}\leq I$

Goldenn-Thompson type 0)T\$5B$\sigma 2X$

and ($\mathrm{g}T7J\triangleright P$$k^{\wedge}\mathrm{t}$‘てT\not\in R&$1o\mathrm{g}$$*\emptyset*\text{で}\mathrm{f}\ovalbox{\tt\small REJECT} \mathrm{F}$$\mathrm{a}\mathrm{e}\ovalbox{\tt\small REJECT}\cdot$-majorization&)$\mathrm{B}_{\mathrm{D}}^{\mathrm{A}}\sigma$$A$, $B>0$ , if $A\#\alpha B\leq I$ , then$A^{r}\#_{\alpha}B^{r}\leq 1$$\ovalbox{\tt\small REJECT}$,$rx$ $6\nu_{1\circ}\not\in \text{と}6-\{\llcorner^{-}\mathrm{B}6h6\not\in\Phi\Leftrightarrow 74\mathrm{i}$Ando-Hiai Theorem: Ando Hiai had shown the following inequality:For)$\emptyset 5\ovalbox{\tt\small REJECT} \mathrm{P}\not\cong\ovalbox{\tt\small REJECT}\llcorner$$.\mathit{2}\mathrm{t}’\succ\xi_{\mathcal{R}\mathrm{b}*\mathrm{t}\tau\nu\backslash \text{ま}\mathcal{F}}^{\mathrm{b}}\llcorner.$(AH)$\not\in \mathrm{g}\text{と}a$holds for $r\geq 1$ .

31$arrow(\text{の})\mathrm{N} 1$ $ffl1,\backslash \xi)$$\vee \mathrm{q}\not\cong \mathrm{f}\mathrm{f}\mathrm{i}$$\cdot \mathrm{R}_{\mathfrak{Q}}^{\mathrm{A}}\dagger \mathrm{X}\Re^{\text{の}}(\mathrm{A}\mathrm{H}_{0})\epsilon$$A^{-1}\mathfrak{g}_{\frac{1}{p}}A^{-:_{B^{p}A^{-\S}}}\leq I$$(\mathrm{A}\mathrm{H}_{0})$gffl$arrow\ arrow$$l\mathrm{Z}-arrow g$)$*\not\in \mathrm{a}\mathrm{e}$? (F)$\text{と}$$\Phi 1’.\mathrm{b}f_{arrow}^{\sim}p^{\backslash }\}_{\vee\supset T*\mathrm{f}\mathrm{f}1\ovalbox{\tt\small REJECT}\yen \mathrm{f}\mathrm{f}\mathrm{i}\#\mathrm{f}\mathrm{f}\sim}\backslash$

If $A\geq B\geq 0$ and$A$て$\Re \mathit{0}$

)

. .
$[1]$

$\llcorner \text{ま}\mathrm{L}\gamma\sim$$A^{-r} \oint_{\frac{1}{p}}(A^{-}\yen B^{p}A^{-_{2}^{1}}1\rangle^{r}\leq I$$\Rightarrow$\mbox{\boldmath $\xi$}--\check 3 の\emptyset$Si$*\not\in \mathrm{E}t_{\llcorner}^{\tau}\text{まと}$$H\mathit{2}\tau^{-}\overline{\prime\lrcorner\backslash }\llcorner \text{ま}\llcorner$$\dot{2}\ovalbox{\tt\small REJECT}\vee-\not\in\llcorner\tau\not\in^{\backslash }\mathrm{g}$$\text{ま}$$\tau$$[14]$

for $p\geq 1$ and $r\geq 1$ .

$f^{\vee}.[10]$.

$arrow\vec{}arrow\vee \mathrm{C}l\mathrm{g}\ovalbox{\tt\small REJECT}\lambda f^{\sim}.\mathrm{b}a\rangle$

$\Re$

.

is invertible, then for each $1\leq p$ and $0\leq t\leq 1$ ,(GF)

$A^{-r}\#_{\frac{1-\+\tau}{\{\mathrm{p}-\mathrm{t}\mathrm{l}*+\mathrm{r}}}(A^{-\tau}B^{p}A^{-}\mathrm{z})^{s}\leq A^{1-t}\mathrm{r}\mathrm{z}$holds for $t\leq r$ and $1\leq s$ .
$=$

$k$

$1,$

の

$\tau\backslash \not\in \mathrm{a}\mathrm{e}$$t‘\zeta\xi)arrow\vee$$\backslash \mathrm{g}\text{ま}\llcorner$a$(\mathrm{G}\mathrm{F})$$- \mathrm{e}\tau$$\mathrm{J}:\dot{2}$.

$q)\ovalbox{\tt\small REJECT}\Phi \mathrm{f}\mathrm{f}\mathrm{l}\dagger 3;t=1$$\text{し}\hslash>$$\mathrm{L}_{arrow}^{--\text{で}\dagger\lambda}$,,$r=sg\mathrm{y}$$arrow-\emptyset\ \check{\eta}_{\mathrm{t}}$&$\mathrm{g}$$(\mathrm{G}\mathrm{F})=\{\mathrm{A}\mathrm{H}_{0}$$\mathrm{f}\mathrm{g}-\Re 4\mathrm{b}\mathfrak{P}tx$$<\ovalbox{\tt\small REJECT} \mathrm{f}\mathrm{f}\mathrm{l}$), $t=0$ , $s=1\emptyset$ &$(\mathrm{A}\mathrm{H})\not\in:(\mathrm{F})$の$\mathrm{g}(\mathrm{G}\mathrm{F})$$\mathrm{R}\ovalbox{\tt\small REJECT} \mathrm{f}\mathrm{f}\mathrm{i}\}^{r_{\vee}}.\supset 4$$=(\mathrm{F})$$\backslash \tau*\tau$.

Theorem 6. Under the assumption (AH), (F) is
Proof. If $A\geq B>0$ , then we havefor $r\geq 1$ . By (AH), we can obtain$obta\mathrm{i}ne\Lambda$. Put $q=2r$ , then$A^{-1}\#_{\urcorner q\mathrm{T}}1B^{q}\leq B^{-1}\beta_{\frac{1}{q+1}}B^{q}\leq I$$B^{\mathrm{R}}r\leq I$$A^{-1}\mathfrak{g}_{T}r1\mp 1,$$A^{-r} \#\frac{r}{p+r}B^{\mathrm{p}}\leq \mathrm{J}$. The final step is given bylemma 3.

ST J‘-#の)$\ovalbox{\tt\small REJECT}\ovalbox{\tt\small REJECT} 1’.\prime 3\mathrm{b}^{\backslash }T\not\in$$1Rp_{1}^{*}\acute{\{}\ovalbox{\tt\small REJECT} \mathrm{b}h\text{ま}T$.

Theorem 7. Under the assumption (F), (AH) is obtained.
1$’\leq A^{-1}$ .
Let
for $0\leq\alpha\leq 1$ , which is equivalent to
Proof. If$A_{1}=A^{-1}$ , $C=A^{-}?BA^{-}\mathrm{z}11$ and. Then $A_{1}\geq B_{1}>0$ and by the Furuta inequality, wea and, thenandfor $r>0$ , $p\geq 1$ . Let
have$(A^{-^{1}}\mathrm{z}BA^{-}\S)$$A\mathfrak{g}_{\alpha}B\leq I$$B_{1}=C^{\alpha}$$p= \frac{1+\epsilon(1-\alpha\}}{\alpha}$$r=\epsilon$$A_{1}^{-r}\#_{\frac{1+\tau}{p+r}}B_{1}^{\mathrm{p}}\leq A_{1}$$A_{1}^{-\epsilon} \oint_{\alpha}B_{1}^{\overline{a}}1+\epsilon\{1-\alpha\overline{-}\}\leq A_{1}$that is,$A^{\epsilon}\#{}_{\alpha} C^{1+\epsilon(1-\alpha\rangle}\leq A^{-1}$$\frac{1+r}{p+r}=$,.

$A^{\epsilon}\#{}_{\alpha} C^{1+\epsilon(1-\alpha)}=A^{\mathrm{e}}\#_{\alpha}CC^{-1+\epsilon\langle 1-\alpha)}C$

$=$

$=$

$A^{\epsilon} \#\alpha C(C^{-\alpha}\#{}_{1-\epsilon} C^{-1})C\geq A^{\epsilon}\oint_{\alpha}C(A\# 1-\epsilon C^{-1})C$$A’\#{}_{\alpha} CA^{\mathrm{r}}(A^{-I}C^{-1}A^{-\tau})^{1-\epsilon}A\mathrm{x}11\# c$ $=A^{\epsilon}\#\alpha A^{-2}B^{-1+\epsilon_{A^{-q}}^{1}}1$

So we have.

$A^{1+\epsilon}\#_{\alpha}B^{1+\epsilon}\leq A\mathfrak{g}_{a}B\leq I$

5. il$\neq \mathrm{H}$$n\grave{*}\mathrm{a}\mathrm{e}$.

$6_{}^{\vee}\$$\mathrm{B}^{\underline{\mathrm{A}}}\Phi\not\in\Sigma\Phi-\mathrm{f}\mathrm{f}1\{\mathrm{b}$$\sigma)\mathfrak{B}\mathrm{B}\mathrm{E}$

Ui? !

.

.

$\ovalbox{\tt\small REJECT}_{\{\#\mathfrak{l}’\yen\ovalbox{\tt\small REJECT}}’$.

$\mathrm{B}_{\mathfrak{Q}}^{\mathrm{A}}$の$\not\in\not\in$の$-\mathrm{f}\mathrm{f}\mathrm{i}4\mathrm{b}k\mathrm{f}\mathrm{i}\grave{\mathrm{x}}_{-}Tk^{\backslash }\mathrm{g}\text{ま}$g-.

.
$\mathrm{E}t\Sigma\xi \mathrm{l}\overline{h\grave{\mathrm{L}}^{\backslash }}$32
Theorem 8. For$\alpha\in$A
Proof. If$\#\alpha$(0, 1) fixed,B$A \oint_{\alpha}B\leq I$$C=A^{1}-\mathrm{z}BA^{-}\tau 1$

and$\leq I$$\Rightarrow$for $0\leq$ a$0\leq\epsilon\leq 1$$A^{r}\mathfrak{g}_{\ovalbox{\tt\small REJECT}^{r}}1-\alpha \mathrm{c}+\alpha rB^{s}\leq I$$\leq 1$.

$A^{1+\epsilon}\#_{\frac{\propto \mathrm{f}^{1+}\Delta^{\mathrm{g}}---}{\mathrm{f}^{1}-\alpha\}+a\zeta 1+\epsilon\}}}B$

for r, s$\geq 1$.
$\leq A^{-1}$. Let$(B^{\mathrm{x}1}-\tau AB^{-}\mathrm{z})^{1-a}\leq B^{-1}$. Let, which is equivalent to$(A^{-\mathrm{g}}BA^{rightarrow \mathrm{z}})11,$$=$$A^{1}\tau(A^{\epsilon}\#_{\frac{\alpha\langle 1+\epsilon\rangle}{1-+a\epsilon}}A^{-\S_{BA^{-\S:}}})A$$\leq$$A^{1}z(C^{-\alpha\epsilon} \#\alpha\not\in_{a}1\pm\frac{\epsilon}{\epsilon}\mathit{1}C)A^{1}2$$=$.

$A\# C^{\alpha}A^{\mathrm{I}}\mathrm{z}=A\beta_{\alpha}B\leq I$

So we have(1}

$A^{r} \#_{\mathrm{r}_{-a\neg\overline{+\alpha r}}^{a\mathrm{r}}}B\leq A\int_{\alpha}B\leq I$

for $r\geq 1$ .

On the other hand, $A\#_{\alpha}B=B\#(1-\alpha)A\leq I$ is equivalent to$D=B^{-^{1}}\mathrm{z}AB^{-\mathfrak{T}}1$.

$A \#\frac{a-}{\{1-\alpha \mathrm{J}\{1+\epsilon\}+\alpha}B^{1+\epsilon}$$=$$A \#\frac{\alpha}{1+\mathrm{s}\{1-a\}}B^{1+\epsilon}$$=$$B^{1+\epsilon}\#_{\frac{\zeta 1-a\}(1+\epsilon \mathrm{J}}{1+\epsilon\{1-\alpha \mathrm{J}}}A$$=$$B^{\int}(B^{\epsilon} \int_{\frac{11-a)\langle 1+\epsilon)}{1+\epsilon[1-a]}}B^{-\mathrm{g}}AB^{-\mathrm{g}})B11\not\in$$\leq$$B^{1}\mathrm{z}(D^{-(1-\alpha)\epsilon}\mathfrak{p}_{\underline{\zeta 1-\alpha\rangle\zeta 1\underline{+\epsilon})},1+\epsilon\zeta 1-\alpha\}}D)B^{3}$$=$$B^{1}2D^{(1-\alpha\rangle}B^{1}2=B\# 1-\alpha A=A\#\alpha B\leq I$.

So we have
(2)

$A\#_{\mathrm{I}\neg^{\alpha}}1-\alpha s\overline{+a}B^{s}\leq A\#\alpha B\leq I$

Let $A_{1}=A^{r}$ ,$\alpha_{1}=\frac{\alpha r}{\mapsto 1-\alpha \mathrm{T}+-ar}$

for

$s$

$\geq 1$.

, then by (2)$A_{1}\#_{\frac{a\lrcorner--}{\langle 1-\alpha_{1}\}s+\mathrm{q}_{1}}}B^{s}\leq A_{1}\beta_{\alpha_{1}}B\leq A\beta_{\alpha}B\leq I$

Since$\frac{\alpha_{1}}{\zeta 1-\alpha_{1})s+\alpha_{1}}=\frac{\alpha \mathrm{r}}{(1-a]s+\alpha r}$.

, we have the conclusion.
References

majorization and complementary Golden-Thom pson type inequality,[1] T.Ando and F.Hiai,
Linear Alg. and Its AppL, 197(1994), 113131.
[2] M.Fujii, Furuta’s inequality and its mean theoretic approach, J.Operator Theory, 23(1990),${\rm Log}$

67-72.
[3] M.Fujii and E.Kamei, Furuta’s inequality for the chaotic order,I,II, Math. Japon., 36(1991),

603-606 and 712-722.
[4] M.Fujii and E.Kamei, On an extension of grand Furuta inequality, Sci. Math. Japon., 56(2002), 501-504.
[5] M.Fujii, T.Furuta and E.Kamei, Furuta’s inequality and its application to Ando’s theorem,
Linear Algebra AppL, 179(1993), 161-169

33[6] M.Pujii and E.Kamei, Ando-Hiai inequality and Furuta inequality, preprint.
[7] M.Fujii, E.Kamei and R.Nakamoto, An analysis on the internal structure of the celebrated
Furuta inequality, Scienticae Mathematicae Japonicae Online, e-2005, 425-431.
,p[8] T.Furuta, A $\geq B\geq 0$ assures,q
for rwith $(1+2r\rangle q\geq$$p+2r$ , Proc. Amer. Math. Soc,, 101(1987), 85-88.
[9] T.Furuta, Elementary proof of an order preserving inequality, Proc. Japan Aead., 65(1989),$(B^{r}A^{\mathrm{p}}B^{r})^{1/q}\geq B^{(p+2r)/q}$$\geq 0$$\geq 0$$\geq 1$126,

[10] T.FUruta, Extension of the Furuta inequality and Ando-Hiai -majorization, Linear Alg. and
Its AppL, 219(1995), 139-155.
[11] T.Furuta, Results under $\log A\geq\log$ B can be derived from ones under A $\geq B\geq 0$ by
Uchiyama’s method associated with Furuta and Kantorovich type operator inequalities,$\log$-

Math. InequaL AppL, $3(2000)$ , 423-436.
-majorizations and norm inequalities for exponential operators, Linear Operators[12] F.Hiai,
Banach Center Publications, vo1.38, 1997.
[13] E.Kamei, A satellite to Puruta’s inequality, Math. Japon., 33(1988), 883-886.
[14] E.Kamei, Parametrized grand Furuta inequality, Math. Japon., 50\langle 1999), 79-83.
[15] E.Kamei, Chaotic order and Furuta inequality, Sci. Math. Japon., 53(2001), 221-225.
[16] E.Kamei and M.Nakamura, Remark on chaotic Furuta inequality, Sci. Math, Japon., 53(2001),
535-539.
[17] F.Kubo and T.Ando, Means of positive linear operators, Math. Ann., 246(1980), 205-224.
, 469-471.
[18] M.Uchiyama, Some exponential operator inequalities, Math. InequaL${\rm Log}$$\mathrm{A}\mathrm{p}\mathrm{p}\mathrm{l}.,2(1999)$$(^{*})\mathrm{D}\mathrm{e}\mathrm{p}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{m}\mathrm{e}\mathrm{n}\mathrm{t}$of Mathematics,

Osaka Kyoiku University,
Asahigaoka, Kashiwara,
Osaka, 582-8582, Japan-mail: mfujii@cc.osaka-kyoiku.ac.j$\mathrm{e}$$\mathrm{p}$Institute of Technology,
Kamisadori, Maebashi,
Gunma, 371-0816, Japan-mail: kamei@maebashi-it.ac.jp$(^{**})\mathrm{M}\mathrm{a}\mathrm{e}\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{h}\mathrm{i}$$\mathrm{e}$of Engineering,
University,
IbarakiNakanarusawa, Hitachi,
Ibaraki 316-8511, Japan-mail: nakaanoto@base,ibaraki.ac.jp$(^{***})\mathrm{F}\mathrm{a}\mathrm{c}\mathrm{u}1\mathrm{t}\mathrm{y}$$\mathrm{e}$