数理解析研究所講究録903 巻 1995 年 97-105

97

Characterizations of operators satisfying chaotic order and its applications
古東京理科大理之 (Takayuki Furuta)

孝田

In what follows, a capital letter means a bounded linear operator on a complex Hilbert
space $H$ . An operator is said to be positive (in symbol : $T\geq 0$ ) if $(Tx, x)\geq 0$ for all$T$. Also an operator is strictly positive (in symbol : $T>0$ ) if is positive and.
invertible. We write $A\gg B$ if $logA\geq logB$ which is called the chaotic order
Results in this paper will appear in [F9] and [F10].
$x\in H$$T$

$T$$[\mathrm{F}\mathrm{K}][\mathrm{F}\mathrm{F}\mathrm{K}2]$\S 1. Generalizations of Kosaki trace inequalities
and related trace inequalities on chaotic order
By using an extension of the Furuta inequality and following after Kosaki’s nice technique, we shall show generalizations of trace inequalities by Kosaki and related trace

inequalities on chaotic order (i.e., $logA\geq logB$ ).
It is well known that $A\geq B\geq 0$ ensures $Tr(f(A))\geq Tr(f(B))$ , where $Tr$ denotes
with $f(\mathrm{O})=0$ . Kosaki
the usual trace and is a continuous increasing function on$f$$[\mathrm{I}<]$$\mathrm{R}_{+}$

shows the following very interesting trace inequality as a generalization of the above

mentioned trace inequality.
Theorem A [K]. Assume $A\geq B\geq 0$ and $p>1_{f} \alpha\geq{\rm Max}\{-1, \frac{-p}{2}\}$ .
(i) Then there exists a partial isometry operator $Usatish^{i}ng$$A^{\frac{\alpha}{2}}B^{p}A^{\frac{\alpha}{2}}\leq U^{*}A^{p+\alpha}U$(ii) For a contin$\mathrm{c}\iota ous$increasing function

$f$

on

$\mathrm{R}_{+}$

with.
$f(\mathrm{O})=0$, we have$Tr(f(A^{\frac{\alpha}{2}}BpA \frac{\alpha}{2}))\leq Tr(f(A^{p}+\alpha))$In the above statements the invertibility of$A$

Recently Ando-Hiai [AH] established various

is assumed when $\alpha<0$ .

$\log$-majorization results to ensure excellent

and useful inequalities for unitarily invariant norms.

On the other hand, as an extension of [H] and [L], we established the following Furuta
and [KA]).
inequality ($[\mathrm{F}1][\mathrm{F}2][\mathrm{F}3][\mathrm{F}\mathrm{U}\mathrm{J}]$98

Theorem B. If $A\geq B\geq 0$ , then for each $r\geq 0$ ,(i)$(B^{r}A^{P}Br)^{1/}q\geq(B^{\Gamma}BPB^{r})^{1/}q$and(ii)hold for$(A^{r}A^{p}Ar)^{1/q}\geq(A^{r}B^{p}A\Gamma)^{1}/q$$p$

and$q$

such that $p\geq 0$ and $q\geq 1$ with $(1+2r)q\geq p+2r$ .

In Furuta [F8] we established the following extension of Theorem , which interpolates
majorization results of Ando-Hiai and the Furuta inequality and moreover extends
this$\mathrm{B}$$\log$

the results inTheorem$[\mathrm{F}\mathrm{K}],[\mathrm{F}4]$$[\mathrm{F}\mathrm{F}\mathrm{K}1],[\mathrm{F}\mathrm{F}\mathrm{K}2],$$\mathrm{C}$and [F5].

[F8]. If $A\geq B\geq 0$ with $A>0$ , then for each $t\in[0,1]$ and $p\geq 1$ ,$F_{p,t}(A, B, r, S)=A^{\frac{-r}{2}} \{A^{\frac{r}{2}}(A^{\frac{-t}{2}B^{p}A^{\frac{-t}{2}}})^{s}A\frac{r}{2}\}^{\frac{1-l+r}{(\mathrm{p}-t)\cdot+r}}A^{\frac{-r}{2}}$is a decreasing function of both$r$

and$s$

for any$s\geq 1$and $r\geq t$ and the following inequality

holds $F_{p,t}(A, A, r, s)\geq F_{p,t}(A, B, r, \mathit{8})$ , that is, for each $t\in[0,1]$ and $p\geq 1$ ,$A^{1-t+r} \geq\{A^{\frac{r}{2}}(A^{\frac{-t}{2}B^{p}A^{\frac{-t}{2}}})^{s}A\frac{r}{2}\}^{\frac{1-t+r}{(\mathrm{p}-t)\iota+r}}$holds for any $s\geq 1$ and$r$

such that $r\geq t$ .

The following Theorem 1.1 is an extension of Theorem A [K].

be positive operators such that $A\geq B\geq 0$ withandTheorem 1.1 [F9]. Let$A>0$ . Assume that $p\geq 1_{f}s\geq 1,$ $t\in[0,1]$ and $\beta\geq{\rm Max}[t-1, \frac{1}{2}\{t(s+1)-ps\}]$ . Then
$A$

$B$

the following inequalities hold.
(I) There exists the a partial isometry operator$U$satisfying.

$A\mathrm{f}\mathrm{l}g2(A^{\frac{-l}{2}}B^{p}A^{\frac{-l}{2})^{s}2}A\leq U^{*}A^{(_{P^{-t}})s+\beta}U$(II) For a continuous increasing function

$f$

on

$\mathrm{R}_{+}$

with $f(\mathrm{O})=0$ ,$Tr \{f(A^{E\rho_{2^{\sim})}}2(A\frac{-t}{2}BpA^{\frac{-l}{2}})sA\}\leq Tr\{f(A^{(pt}-)s+\rho)\}$

The following Theorem 1.2 is a parallel result to Theorem 1.1.

.

99

Theorem 1.2 [F9]. Let and $B$ be positive invertible operators such that $A\gg B$.
Assume that $p\geq u>0,$ $s\geq 1,$ $\alpha\in[0,1]$ and(i.e., $logA\geq logB$ ).
$A$$\beta\geq-u\alpha$

Then the following inequalities hold.
(I) There exists the a partial isometry operator$U$satisfying$A^{\frac{\beta}{2}}(A \frac{u\alpha}{2}B^{p}A^{\frac{u\alpha}{2})A^{4}}s2\leq U^{*}A^{(u\alpha+)+\rho}psU$(II) For a continuous increasing function

$f$

on

$\mathrm{R}_{+}$.

with $f(\mathrm{O})=0$ ,$\tau_{r\{f}(A^{\frac{\beta}{2}}(A^{\frac{u\alpha}{2}B^{p}A^{\frac{u\alpha}{2})^{s}A)}}\frac{\beta}{2}\}\leq^{\tau r}\{f(A^{(u}\alpha+_{P)}s+\beta)\}$.

be positive invertible operators such that $A\gg B$. Then the $f_{ollow},$ $ing$ inequalities hold.
(i.e., $logA\geq logB$ ). Assume that $p>0$ , andCorollary 1.3 [F9].

Let$A$

and$B$$\beta\geq 0$(I) There exists the a partial isometry operator$U$satisfying$A^{\frac{\beta}{2}}B^{p}A^{\frac{\beta}{2}}\leq U^{*}A^{p+\beta}U$(II) For a continuous increasing function

$f$

on

$\mathrm{R}_{+}$.
with $f(\mathrm{o})=0$ ,$Tr \{f(A^{\frac{\beta}{2}}BpA\frac{\beta}{2})\}\leq Tr\{f(Ap+\rho)\}$.

\S 2. Extensions of Ando’s characterization ofoperators satisfying $logA\geq logB$ and its applications
It is shown in Ando [A] that $A\gg B$ holds if and only if$A^{p}\geq(A^{\mathrm{g}}2B^{p}A^{\mathrm{z}\iota}2)2$holds forall $p\geq 0$ .
In this section , we state an extension of this result and its application.

Theorem 2.1 [F10]. Let$A$

and$B$

be positive invertible operators. Then the following(I) and (II) holds.
(I) If $A\gg B$ (i.e., $logA\geq logB$ ) , then for each $\alpha\in[0,1]$ , and all $p\geq 0$ and $u\geq 0$ ,$G_{p,u,\alpha}(A, B,r, s)=A^{\frac{-ur}{2}} \{A\frac{ur}{2}(A\frac{u\alpha}{2}B^{p}A^{\frac{u\alpha}{2})^{s}}A\frac{ur}{2}\}^{\frac{u(\alpha+\prime)}{(u\alpha+\mathrm{p})*+u\prime}}A^{\frac{-ur}{2}}$is a decreasing function of both$s$and$r$

such that $s\geq 1$ and $r+\alpha\geq 1$ .

100(II) $A\gg B$ holds if and only ifof both$r\geq 0$$g(s, r)=A^{\frac{-r}{2}(A^{\frac{r}{2}B^{S}}}A^{\frac{r}{2})^{\frac{r}{\iota+r}}}A^{\frac{-r}{2}}$is a decreasing function

and $s\geq 0$ .

Theorem 2.2 [F10]. Let andassertions are mutually equivalent.
$A$$B$

be positive invertible operators. Then the following(I) $A\gg B$ (i.e., $logA\geq logB$ ).
(II) For each $\alpha\in[0,1]$ , and all $p\geq 0$ and $u\geq 0$ ,$A^{u(\alpha+r)} \geq \mathrm{f}^{A^{\frac{ur}{2}}}(A\frac{u\alpha}{2}B^{p}A^{\frac{u\alpha}{2})^{s}A^{\frac{ur}{2}\}}}\frac{u(\alpha+r)}{(u\alpha+\mathrm{p})l+ur}$

holds for any $s\geq 1$ and(III) For each$\alpha\in[0,1]_{f}$$r$such that $r+\alpha\geq 1$ .

and all $p\geq 0$ ,$A^{p(\alpha+r})\geq\{A^{L^{\underline{r}}}2(A^{L^{\alpha}}2^{-}BpA^{R_{\frac{\alpha}{2}}})^{S}AL^{\underline{r}}2\}^{\frac{\alpha+r}{(\alpha+1)\epsilon+\prime}}$holds for any $s\geq 1$ and$r$

such that $r+\alpha\geq 1$ .

(IV) For each $\alpha\in[0,1]$ , and all $p\geq 0$ ,$A^{\alpha+r} \geq\{A^{\frac{r}{2}}(A\frac{\alpha}{2}B^{p}A^{\frac{\alpha}{2}})SA^{\frac{r}{2}}\}^{\frac{\alpha+r}{(\alpha+\mathrm{p})l+r}}$

holds for any $s\geq 1$ and$r$

such that $r+\alpha\geq 1$ .

Theorem 2.3 [F10]. Let$A$

and $B$ be positive invertible operators. If$A \geq(A^{\frac{1}{2}}BA^{\frac{1}{2})}\frac{1}{2}$,then for each $\alpha\in[0,1]$ , and all $p\geq 1$ and $u\geq 1$ ,$H_{p,u,\alpha}(A, B, r, \mathit{8})=A^{\frac{-ur}{2}}\{A\frac{ur}{2}(A\frac{u\alpha}{2}B^{p}A^{\frac{u\alpha}{2})^{s}A}\frac{ur}{2}\}^{\frac{u(\alpha+r)}{(u\alpha+p)\epsilon+ur}A^{\frac{-ur}{2}}}$is a decreasing function of both 8 andTheorem 2.4 [F10]. Let$A$$r$

such that $s\geq 1$ and $r+\alpha\geq 1$ .

and $B$ be positive invertible operators. Then the following

assertions are mutually equivalent.
(I)$A \geq(A^{\frac{1}{2}}BA\frac{1}{2})^{\iota}2$.

(II) For each $\alpha\in[0,1]$ , and all $p\geq 1$ and $u\geq 1$ ,$A^{u(\alpha+r}) \geq\{A^{\frac{ur}{2}(A}A^{\frac{u\alpha}{2}B^{p}}\frac{u\alpha}{2})^{s}A\frac{ur}{2}\}^{\frac{u(\alpha+r)}{(u\alpha+\mathrm{p})^{\mathrm{q}}\vee+ur}}$

101

holds for any $s\geq 1$ and$r$

such that $r+\alpha\geq 1$ .

(III) For each $\alpha\in[0,1]$ , and all $p\geq 1$ ,$A^{p(\alpha+r)} \geq\{A^{L^{r}}2(A^{R_{\frac{\alpha}{2}}}B^{p}AR\frac{\alpha}{2})^{s}A^{L^{\underline{r}}}2\}^{\frac{\alpha+r}{(\alpha+1)*+r}}$

holds for any $s\geq 1$ and$r$

such that $r+\alpha\geq 1$ .

(IV) For each $\alpha\in[0,1]$ , and all $p\geq 1$ ,$A^{\alpha+r} \geq\{A^{\frac{r}{2}}(A\frac{\alpha}{2}B^{p}A^{\frac{\alpha}{2})\}}sA^{\frac{r}{2}}\frac{\alpha+r}{(\alpha+p)*+r}$holds for any$\mathit{8}\geq 1$and$r$

such that $r+\alpha\geq 1$ .

Remark 2.1. When we replace the hypothesis $A\gg B$ in Theorem 2.2 byin Theorem 2.4, then all the results in Theorem 2.2 remain valid if we$A \geq(A^{\frac{1}{2}}BA^{\frac{1}{2})}\frac{1}{2}$replace $p\geq 0$ and $u\geq 0$ in Theorem 2.2 by $p\geq 1$ and $u\geq 1$ in Theorem 2.4.

Also we remark that when we replace the hypothesis $A\gg B$ in (I) of Theorem 2.1 byin Theorem 2.3, but the result in (I) of Theorem 2.1 remain valid if we$A\geq(A^{\frac{1}{2}}BA^{\frac{1}{2}})^{\frac{1}{2}}$replace $p\geq 0$ and $u\geq 0$ in Theorem 2.1 by $p\geq 1$ and $u\geq 1$ in Theorem 2.3.

Remark 2.2.
($A^{\frac{u}{2}}B^{p}A^{\frac{u}{2})^{\frac{u}{p+u}}}$Put $r+\alpha=1$ and $s=1$ in (IV) of Theorem 2.4, then $A^{u}\geq$holds for any $p\geq 1$ and $u\geq 1$ , which has been shown in [Theorem 2,$\mathrm{F}\mathrm{F}\mathrm{W}$].

Also put $r+\alpha=1$ and $s=1$ in (IV) of Theorem 2.2, then we haveholds for any $p\geq 0$ and $u\geq 0$ , and this result is already obtained in [Theorem I,FFK2]$A^{u}\geq(A^{\frac{u}{2}}B^{p}A^{\frac{u}{2})^{\frac{u}{\mathrm{p}+u}}}$and [Theorem 1,F5].

\S 3. Characterizations of operators satisqing $logA\geq logB$
associated with sonle operator equations and parallel results
Theorenl 3.1 [F10]. Let$A$

and $B$ be positive invertible operators. Then the following

assertions are mutually equivalent.
(I) $A\gg B$ (i.e., $logA\geq logB$ )(II) For each $\alpha\in[0,1]$ and for any $p\geq u>0$ and $s\geq 1$ , there exists a unique invertible

positive contraction$\tau_{p,u_{)}\alpha}(s)$such that102$T_{p,u,\alpha}( \mathit{8})A^{pu\alpha}s+(s-1)Tu,\alpha(p,S)=A^{\frac{-u\alpha}{2}}(A^{\frac{u\alpha}{2}}BpA\frac{u\alpha}{2})^{S}A^{\frac{-u\alpha}{2}}$(III) For any $p\geq 0$ , there exists a unique invertible positive contraction$T_{pp}A^{p}T=B^{p}$$T_{p}$

such that.

Next we state the following parallel result to Theorem 3.1.

Theorem 3.2 [F10]. Let andassertions are mutually equivalent.
$A$(I)$B$

be positive invertible operators. Then the following$A \geq(A^{\frac{1}{2}}BA^{\frac{1}{2})}\frac{1}{2}$(II) For each $\alpha\in[0,1]$ and for any $p\geq u\geq 1$ and $s\geq 1$ , there exists a unique invertible

positive contraction$T_{p,u,\alpha}(s)$such that$T_{p,u,\alpha}(s)A^{p}s+u \alpha(s-1)T_{p,\alpha}u,(S)=A\frac{-u\alpha}{2}(A\frac{u\alpha}{2}BpA\frac{u\alpha}{2})^{s}A\frac{-u\alpha}{2}$(III) For any $p\geq 1$ , there exists a unique invertible positive contraction$T_{p}A^{p}T_{p}=B^{p}$$T_{p}$

such that.

Related to Theorem 3.1 and Theorem 3.2, we have the following parallel result by thesame way as proof of Theorem 3.1.
Theorenl 3.3 [F10]. Let$A$

and $B$ be positive invertible operators. Then the following

assertions are mutually equivalent.
(I) $A\geq B\geq 0$ .
(II) For each $t\in[0,1]$ and for any $p\geq 2$ and $s\geq 1$ , there exists a unique invertible positive

contraction$\tau_{p,t}(s)$such that$\tau_{p,t(_{S)AT_{p,t()A^{\frac{t-1}{2}}}}}(p-t)s+t-1s=(A\frac{-t}{2}BpA^{\frac{-t}{2}})^{s}A\frac{t-1}{2}$(III) For any $p\geq 1$ , there exists a unique invertible positive contraction$T_{p}A^{p}T_{p}=A^{\frac{-1}{2}B^{p+1}A^{\frac{-1}{2}}}$

(IV)

.

$A^{1+p}\geq(A^{L}2Bp+2A^{\mathrm{z}}2)^{\frac{1}{2}}$for all$p\geq 0$$T_{\mathrm{p}}$such that.

Results in this scetion are extensions of [Theorem 2, F6] and [Theorem 2.1, F7].

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\S 4. Addendum. An extension of Theorem$\mathrm{B}$Recently we have the following result as an extension form of Theorem B.

Theorem$\mathrm{D}$[FFK3].

If $A\geq B>0$ , then$(B^{\gamma}A^{\alpha}B^{\gamma})^{\beta}\geq(B^{\gamma}B^{\alpha}B^{\gamma})^{\rho}$holds under

any one of the following conditions;(i)$\frac{1}{\beta}\leq\alpha_{f}0<\beta<1$

(ii)

, and

$\frac{1}{\beta}\leq\alpha\leq 1,1<\beta\leq 2$(iii)$\frac{1}{2}\leq\alpha\leq 1,2\leq\beta$$\gamma=\frac{\alpha\beta-1}{2(1-\beta)}$, and, and$\gamma=\frac{\alpha\beta-1}{2(1-\beta)}$$\gamma=\frac{\alpha\beta-1}{2(1-\beta)}$.

, but we remark that (i) is nothing, that is, put $(1+2r)q=p+2r$, then we easily obtain $p\geq q\geq 1$ and we have

Remark 4.1. (i) and (ii) are announced in
and in Theorem
but exchange of parameters$[\mathrm{Y}, \mathrm{p}61]$$p,$$r$$q$$\mathrm{B}$for $r\geq 0,$ $p\geq 1$ and $q\geq 1$ in Theoremby and by , then we have (i) which is nothing but another
only to replace by$\mathrm{B}$$p$$\alpha,$$r$$\frac{1}{q}$$\beta$$\gamma$expression of (i) in Theorem B.
Moreover a simple proof of (ii) is obtained in [FFK3] along a mothod of [F2] by using
polar decomposition. Also in [FFK3] we obtained (iii) along a method of [F2]. We haveto assume invertibilty of$A$

and$B$

in the cases (ii) and (iii) since$\gamma\leq 0$.

Remark 4.2. We should mention that Tanahashi [TA] has obtained several interesting
results closely related to Theorem D.

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Kagurazaka, Shinjuku 162, Tokyo, Japan