数理解析研究所講究録1080 巻 1999 年 34-40

34Equivalence relation between an order preserving
operator inequality and related operator functions
東京理科大
東京理科大
東京理科大

理
理
理

伊藤
橋本
古田

滴瓶
雅史
孝之

(Masatoshi Ito)
(Masashi Hashimoto)(Takayuki Furuta)

Abstract
This report is based on the following two papers:[FHI] T.Furuta, M.Hashimoto and M.Ito, Equivalence relation between generalized
Furuta inequality and related operator functions, Scientiae Mathematicae, 1(1998), 257-259.
.
[F] T.Furuta, Simplified proof of an order preserving operator $inequali\dot{t}y$ , Proc.
Japan. Acad., 74 (1998), 114.

In this paper, we shall show equivalence relation between an order preserving
operator inequality and related operator functions.

1

Introduction

Chapter 1, 2 and 3 are based on [FHI].
A capital letter means a bounded linear operator on a complex Hilbert space $H$ . Anis said to be positive (denoted by $T\geq 0$ ) if $(Tx, x)\geq 0$ for all $x\in H$ andoperatoris positiveis said to be strictly positive (denoted by $T>0$ ) if
also an operator
L\"owner-Heinz
and invertible. The following Theorem is an extension of the celebrated
theorem: $A\geq B\geq 0$ ensures
for any $\alpha\in[0,1]$ .
$T$

$T$

$T$

$\mathrm{F}$$A^{\alpha}\geq B^{\alpha}$

Theorem$\mathrm{F}([5])$.

If $A\geq B\geq 0$ , then for each(i)$r\geq 0$,$(B^{\frac{r}{2}}A^{p}B \frac{r}{2})^{\frac{1}{q}}\geq(B^{\frac{r}{2}}B^{p}B\frac{r}{2})^{\frac{1}{q}}$

and
(ii)

$(A^{\frac{r}{2}}A^{p}A \frac{r}{2})^{\frac{1}{q}}\geq(A^{\frac{r}{2}}B^{p}A\frac{r}{2})^{\frac{1}{q}}$hold for $p\geq 0$ and $q\geq 1$ with $(1+r)q\geq p+r$ .

We remark that Theorem yields L\"owner-Heinz theorem when we put $r=0$ in (i)or (ii) stated above. Alternative proofs of Theorem are given in $[2][14]$ and also anandelementary one-page proof in [6]. It is shown in [15] that the domain drawn forin the Figure is best possible one for Theorem F.
In [9] we established the following Theorem as extensions of Theorem F.
$\mathrm{F}$$\mathrm{F}$$p,$ $q$$\mathrm{G}$

$r$

35
Theorem$\mathrm{G}([9])$. If$A\geq B\geq 0$with $A>0$ , then for each $t\in[0,1]$ and $p\geq 1$ ,$F_{p,t}(A, B, r, S)=A^{\frac{-r}{2}\{}A \frac{r}{2}(A\frac{-t}{2}B^{p}A^{\frac{-t}{2})}sA^{\frac{r}{2}}\}^{\frac{1-t+r}{(p-t)s+f}}A^{\frac{-r}{2}}$is decreasing for $r\geq t$ and$t\in[0,1]$ and $p\geq 1$ ,$s\geq 1_{f}$and $F_{p,t}(A, A, r, S)\geq F_{p,t}(A, B, r, S)_{f}$ that is, for each$A^{1t+r}- \geq\{A^{\frac{r}{2}}(A\frac{-t}{2}BpA^{\frac{-t}{2})^{s}A^{\frac{r}{2}\}^{\frac{1-t+r}{(p-t)s+r}}}}$

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

established excellent
useful inequality equivalent to the main
If $A\geq B\geq 0$ with $A>0$ , then$\log$$\mathrm{A}\mathrm{n}\mathrm{d}\mathrm{o}-\mathrm{H}\mathrm{i}\mathrm{a}\mathrm{i}[1]$majorization results and proved the following
theorem: .

$\log \mathrm{m}\mathrm{a}\mathrm{j}_{\mathrm{o}\mathrm{r}}\mathrm{i}\mathrm{Z}\mathrm{a}\mathrm{t}\mathrm{i}_{\mathrm{o}\mathrm{n}}\downarrow$$A^{r} \geq\{A^{\frac{r}{2}}(A^{\frac{-1}{2}}B^{p}A^{\frac{-1}{2})^{r}A}\frac{r}{2}\}^{\frac{1}{p}}$holds for any $p\geq 1$ and $r\geq 1$ .
Theorem interpolates the inequality stated above by Ando-Hiai and Theorem itself,
and also extends results of $[3][7]$ and [8]. Recently a nice mean theoretic proof of Theorem
is shown in [4] and the result on the best possibility of Theorem is shown in [16].
Very recently the following Theorem is obtained in [10] as an extension of Theorem G.
$\mathrm{F}$$\mathrm{G}$$\mathrm{G}$$\mathrm{G}$$\mathrm{H}$Theorem$\mathrm{H}([10])$$p \geq\max\{q, t\}$. Let $A\geq B\geq 0$ with $A>0$ .

For each $t\in[0,1],$$q\geq 0$

and,$G_{p,q,t}(A, B, r, s)=A^{\frac{-r}{2}\{}A \frac{r}{2}(A^{\frac{-t}{2}}BpA^{\frac{-t}{2})}sA^{\frac{r}{2}}\}^{\frac{q-t+r}{(p-t)s+f}}A^{\frac{-r}{2}}$is decreasing for $r\geq t$ and $s\geq 1$ . Moreover for each $1\geq q\geq t\geq 0$ and $p\geq q$ ,$G_{p,q,t}(A, A, r, S)\geq G_{p,q,t}(A, B, r, s)_{y}$ that is,$A^{q-t+r}\geq\{A^{\frac{r}{2}}(A^{\frac{-\mathrm{t}}{2}}B^{p}A^{\frac{-t}{2})^{s}A^{\frac{r}{2}\}^{\frac{q-t+r}{(_{P^{-t}})S+\Gamma}}}}$

holds for any $r\geq t$ and $s\geq 1$ .
$\mathrm{H}$2

Alternative proofs ofis extended in [11].

$\overline{\mathrm{T}}\mathrm{h}\mathrm{e}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{m}\mathrm{H}$are. shown in$[12][13]$ , andth.

$\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}\mathrm{i}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}$.

of Theorem

ResultWe obtain the following Theorem 1 associated with Theorem$\mathrm{G}$and Theorem H.

Theorem 1. The following statements hold and follow from each other;(i) If $A\geq B\geq 0$ with $A>0$ , then for each $t\in[0,1]$ and$p\geq 1_{f}$$A^{1-t+r}\geq\{A^{\frac{r}{2}}(A^{\frac{-t}{2}}BpA^{\frac{-t}{2})^{S}}A^{\frac{r}{2}}\}^{\frac{1-l+r}{(\mathrm{p}-t)_{S}+\Gamma}}$holds for any $s\geq 1$ and $r\geq t$ .

36(ii) If $A\geq B\geq 0$ with $A>0_{f}$ then for each $1\geq q\geq t\geq 0$ and $p\geq q$ ,$A^{q-t+r} \geq\{A^{\frac{r}{2}}(A^{\frac{-t}{2}}B^{p}A^{\frac{-t}{2})^{s_{A\}^{\frac{q-\mathrm{t}+r}{(p-t)s+r}}}}}\frac{r}{2}$holds for any $s\geq 1$ and $r\geq t$ .
(iii) If $A\geq B\geq 0$ with $A>0_{y}$ then for each $t\in[0,1]$ and $p\geq 1$ ,$F_{p,t}(A, B, r, S)=A^{\frac{-r}{2}\{A^{\frac{r}{2}}(A^{\frac{-t}{2}BA^{\frac{-\mathrm{t}}{2})}}}pSA \frac{r}{2}\}^{\frac{1-t+r}{(\mathrm{p}-t)s+r}}A^{\frac{-f}{2}}$is decreasing for $r\geq t$ and $s\geq 1$ .

(iv) If $A\geq B\geq 0$ with $A>0$ , then for each $t\in[0,1],$ $q\geq 0$ and $p\geq t$ ,$G_{p,q,t}(A, B, r, S)=A^{\frac{-r}{2}\{A^{\frac{f}{2}}(A^{\frac{-t}{2}BA^{\frac{-t}{2})}}}pSA \frac{r}{2}\}^{\frac{q^{-t}+\Gamma}{(p-t)s+r}}A^{\frac{-r}{2}}$is decreasing for $r\geq t$ and $s\geq 1$ such that $(p-t)s\geq q-t$ .

(i) and (iii) in Theorem 1 have been obtained as Theorem G. (ii) and (iv) have been also
obtained as Theorem and an extension of Theorem H. (ii) and (iv) used to be recognizedextensions of (i) and (iii) respectively, and (i), (ii), (iii) and (iv) proved sepalately. Theorem
1 asserts that (i), (ii), (iii) and (iv) follow from each other. In other words, if we prove
only one of them, then we obtain the others. So, in this report, we shall give a simplified
proof of (ii), too.
We need the following lemma to give proofs.
$\mathrm{H}$

Lemma F. Let $A>0$ and$B$

be an invertible operator. Then$(BAB^{*}) \lambda--BA\frac{1}{2}(A^{\frac{1}{2}}B*BA\frac{1}{2})^{\lambda}-1A^{\frac{1}{2}}B*$

holds for any real number .
$\lambda$

3

Proof of Theorem 1

We may assume thatis invertible without loss of generality. (i), (ii), (iii) and (iv)
have been already proved in [9] $[10][11][12]$ and [13], so that we have only to prove theequivalence relation among them. We shall show
as follows.
$\mathrm{B}$$(\mathrm{i}\mathrm{v})arrow(\mathrm{i}\mathrm{i}\mathrm{i})arrow(\mathrm{i})arrow(\mathrm{i}\mathrm{i})arrow(\mathrm{i}\mathrm{v})$Proof of
Proof of. Put $q=1$ and letin (iv), we have (iii).
. In (iii), comparing the functional value and the maximal value of, we have the following inequality: For each $t\in[0,1]$ and $p\geq 1$ ,$(\mathrm{i}\mathrm{v})\Rightarrow(\mathrm{i}\mathrm{i}\mathrm{i})$$p\geq 1$

1$(\mathrm{i}\mathrm{i}\mathrm{i})\Rightarrow(\mathrm{i})$$F_{p,t}(A, B, \Gamma, S)$$F_{p,t}(A, B, r, S)\leq F_{p,t}(A, B,t, 1)$(3.1)$=A^{\frac{-t}{2}BA^{\frac{-t}{2}}}$$\leq A^{1-t}$

holds for any $r\geq t$ and $s\geq 1$ . Multiplying$A^{\frac{r}{2}}$by $A\geq B$

from the both sides of (3.1), we have (i).

37

Proof of. Since $q\in[0,1],$ $A\geq B\geq 0$ assures $A^{q}\geq B^{q}$ by L\"owner-Heinz
theorem. Put $A_{1}=A^{q},$ $B_{1}=B^{q},$andin (i). Then$(\mathrm{i})\Rightarrow(\mathrm{i}\mathrm{i})$$t_{1}=arrow q\theta\in[0,1],$ $p_{1}= \frac{p}{q}\geq 1$$A_{1}^{1-t_{1}+1}r=A^{q-t+r},$$A_{1^{-\lrcorner}}^{-_{2}}=\iota A^{\frac{-t}{2}},$$A_{1}^{\lrcorner}r_{2}=A^{\frac{r}{2}},$$r_{1}= \frac{r}{q}\geq t_{1}$$B_{1}^{p_{1}}=B^{p}$and$\underline{1-t_{1}+r_{1}}\underline{q-t+\Gamma}=$$(p_{1}-t_{1})_{S}$\dagger$r_{1}$$(p-t)s+r$ ’that is, we have the following (3.2): For each $1\geq q\geq t\geq 0$ and $p\geq q$ ,(3.2)

$A^{q-t+r} \geq\{A^{\frac{r}{2}}(A^{\frac{-t}{2}}B^{p}A^{\frac{-t}{2})^{s_{A\}^{\frac{q-t+r}{(p-t)S+r}}}}}\frac{r}{2}$holds for any $r\geq t$ and $s\geq 1$ . Hence the proof ofis complete.
. Put
and $q=t$ in (ii), we hava the following (3.3):
Proof ofFor each $t\in[.0,1]$ and $p\geq t$ ,$(\mathrm{i})\Rightarrow(\mathrm{i}\mathrm{i})$$D=A^{\frac{-t}{2}B^{p}A^{\frac{-t}{2}}}$$(\mathrm{i}\mathrm{i})\Rightarrow(\mathrm{i}\mathrm{v})$(3.3)$A^{r}\geq(A^{\frac{r}{2}}D^{s}A^{\frac{r}{2}})^{\frac{r}{\{\mathrm{p}-t)s+r}}$

holds for any $r\geq t$ and $s\geq 1$ . $(3.3)$ is equivalent to the following (3.4) by Lemma F.
($D^{\frac{\mathrm{e}}{2}}.A^{r}D^{\frac{s}{2})^{\frac{(p-t)}{(\mathrm{p}-t)_{S}+r}}}..\geq D^{s}$

.

(3.4)

Applying L\"owner-Heinz theorem to (3.3) and (3.4), we have the following (3.5) and (3.6):$A^{u}\geq(A^{\frac{r}{2}}D^{s}A^{\frac{r}{2})^{\frac{u}{(p-t)\underline{\mathrm{q}}+r}}}$($D^{\frac{s}{2}}A^{r}D^{\frac{\epsilon}{2})^{\frac{(p-t)w}{(_{P^{-t}})\vee r\mathrm{s}+}}}\geq D^{w}$

for $r\geq u\geq 0$ .

(3.5)

for $s\geq w\geq 0$ .

(3.6)

In order to show that $G_{p,q,t}(A, B, r, S)$ is a decreasing function of both$G_{p,q,t}(A, B, r, S)$ as follows:$G_{p,q,t}(A, B, r, s)=$
$=$

$=$

$=$

$r$

and , we rewrite$s$$A^{\frac{-r}{2}(A)^{\frac{-t+r}{(\mathrm{p}-t)S+r}}A}A^{\frac{r}{2}}D^{s} \frac{r}{2}\frac{-r}{2}$$A^{\frac{-r}{2}f(S})A^{\frac{-r}{2}}$$D^{\frac{s}{2}}(D^{\frac{s}{2}}A^{r}D \frac{s}{2})^{\frac{q-t-(prightarrow t)\underline{\mathrm{Q}}}{(p-t)s+\Gamma}}D^{\frac{s}{2}}$by Lemma$\mathrm{F}$$D^{\frac{s}{2}}g(r)D^{\frac{s}{2}}$where$f(s)\equiv(A^{\frac{r}{2}}D^{s_{A^{\frac{r}{2})^{\frac{q-t+r}{(p-t)s+\Gamma}}}}}$

and$g(r)\equiv(D^{\frac{s}{2}}Ar_{D^{\frac{s}{2})^{\frac{9^{-t-}\mathrm{t}P-t)S}{(p-t)S+r}}}}$

So we have only to prove that $f(s)$ and $g(r)$ are decreasing for and respectively.
(a) Proof of the result that $G_{p,q,t}(A, B, r, S)$ is decreasing for $s\geq 1$ such that $(p-t)s\geq q-t$ .
$r$$f(s)$

$=$

$=$

$=$

$\geq$

$=$
$=$

$s$

$(A^{\frac{r}{2}}D^{S}A^{\frac{r}{2}})^{\frac{-t+r}{(p-t)\vee\Gamma 8+}}$$\{(A^{\frac{r}{2}}DSA\frac{r}{2})^{\frac{(p^{-\mathrm{t}})(\mathrm{e}+w)+r}{(\mathrm{p}-t)\mathrm{q}+r}}..\}^{\frac{q-t+r}{(p-t)(s+w)+r}}$$\{A^{\frac{r}{2}}D^{\frac{s}{2}}(D\frac{s}{2}A^{r}D\frac{s}{2})^{\frac{(p-t)w}{(p-\mathrm{t})s+f}}D\frac{s}{2}A^{\frac{r}{2}}\}\frac{q-t+r}{(p-t)(_{S+}w)+r}$

for $s\geq w\geq 0$
by Lemma$\mathrm{F}$$(A^{\frac{r}{2}}D^{\frac{s}{2}}D^{w}D^{\frac{s}{2}}A^{\frac{r}{2}})^{\frac{-t+r}{(p-t)(_{8+}w)+r}}$$(A^{\frac{r}{2}}D^{s+w}A^{\frac{r}{2}})^{\frac{-t+r}{(p-t)1S+w)+r}}$$f(s+w)$ .

38
The last inequality holds by (3.6) and L\"owner-Heinz theorem since $\frac{q-t+r}{(p-t)(S+w)+r}\in[0,1]$
is decreasing forholds by the condition of (ii). Hence$s\geq 1$ such that $(p-t)s\geq q-t$ .
(b) Proof of the result that $G_{p,q,t}(A, B, r, s)$ is decreasing for $r\geq t$ .
$G_{p,q,t}(A, B, \Gamma, S)=A^{\frac{-r}{2}f(S})A^{\frac{-r}{2}}$$g(r)$$D^{\frac{s}{2}}(D^{\frac{l}{2}}.A^{\cdot} \Gamma D\frac{s}{2})^{\frac{q-t-\{p-t)\mathrm{q}}{(\mathrm{p}-t)\mathrm{s}+r}}.\cdot D^{\frac{s}{2}}$$=$$D^{\frac{s}{2}} \{(D^{\frac{s}{2}A^{\Gamma}}D^{\frac{s}{2}})\frac{(\mathrm{p}-t)\underline{9}+r+u}{(p-t)\mathit{8}+r}\}^{\frac{q-t-(p-t)_{\vee}\mathrm{q}}{(p-t)\mathrm{S}+r+u}}D\frac{s}{2}$$=$$D^{\frac{s}{2}} \{D^{\frac{s}{2}}A\frac{r}{2}(A\frac{f}{2}D^{S}A\frac{r}{2})^{\frac{u}{(\mathrm{p}-t)s+r}}A^{\frac{r}{2}}D^{\frac{\epsilon}{2}\}^{\frac{q-t-(p-t)*}{1^{p-t})S+r+u}}}D^{\frac{s}{2}}$$=$for $r\geq u\geq 0$
by Lemma$\mathrm{F}$$D^{\frac{s}{2}} \{D^{\frac{s}{2}}A\frac{r}{2}A^{u}A^{\frac{r}{2}}D\overline{2}\}\vee^{*}\frac{q-t-(\mathrm{p}-t)*}{(p-t)8+r+u}.D^{\frac{s}{2}}$$\underline{>}$$D^{\frac{s}{2}}(D^{\frac{s}{2}}A^{r+u}D^{\frac{\epsilon}{2}})^{\frac{q^{-t-}(p-t)\mathrm{Q}}{(\mathrm{p}-t)\mathrm{s}+r+u}}.D^{\frac{s}{2}}$$=$$=g(r+u)$ .

The last inequality holds by (3.5), L\"owner-Heinz theorem and taking inverses sinceby the condition of (ii). Hence$G_{p,q,t}(A, B, r, S)=D^{\frac{\sigma}{2}}g(r)D^{*}\overline{2}$is$\frac{q^{-t-}(p-t)s}{\mathrm{s}_{\mathrm{e}\mathrm{C}}p-t\mathrm{r}\mathrm{e})s+r+\mathrm{a}\sin^{u}}\in[-1,0]\mathrm{h}_{\mathrm{o}\mathrm{l}\mathrm{d}\mathrm{S}}\mathrm{g}\mathrm{f}\mathrm{o}\mathrm{r}r\geq t$.

is complete by (a) and (b).
Consequently we have finished the proof of Theorem 1.

So the proof of

4$(\mathrm{i}\mathrm{i})\Rightarrow(\mathrm{i}\mathrm{v})$

Simplified proof of (ii)This chapter is based on [F].

Proof of (ii) in Theorem 1. Firstly we shall show the following (4.1) which is the caseof $r=t$ in (ii): If $A\geq B\geq 0$ with $A>0$ , then

(4.1)

$A^{q} \geq\{A^{\frac{t}{2}}(A^{\frac{-t}{2}B^{p}A}\frac{-t}{2})sA\frac{t}{2}\}^{\frac{q}{(p-t)s+t}}$

holds for $1\geq q\geq t\geq 0,$ $p\geq q$ and $s\geq 1$ .
and $A^{t}\geq B^{t}\cdots(**)$ hold by the(1st) In case $2\geq s\geq 1$ . $s-1\in[0,1],$
condition of (ii) and L\"owner-Heinz theorem. Then we have$\frac{q}{(p-t)_{S+t}}\in[0,1]$$B_{1}$$=$$=$$\leq$$=$$\{A^{\frac{t}{2}(A^{\frac{-t}{2}B^{p}A}}\frac{-t}{2})sA^{\frac{t}{2}}\}(\mathrm{p}-\sim_{s})t\overline{+t}$$\{B^{E}2(B^{\epsilon}2A^{-}tB^{f}2\mathrm{i})^{s-}1B2\}^{\frac{q}{(\mathrm{p}-t)\epsilon+t}}\epsilon$

by Lemma$\{B^{E}2(B^{E}2B-tB^{\mathrm{s}\mathrm{i}}2)s-1B2E\}^{\overline{(})_{S}t}p-t\mapsto\overline{+}$by $(**)$$\mathrm{F}$$B^{q}\leq A^{q}=A_{1}$(4.2)

for $1\geq q\geq t\geq 0,$ $p\geq q$ and $2\geq s\geq 1$ .
(2nd) Repeating (4.2) for $A_{1}\geq B_{1}\geq 0$ , we obtain the following (4.3):$A_{1}^{q_{1}} \geq\{A_{1}^{2}(A^{-}2B_{1}^{p}1A^{\frac{-t_{\rceil}}{12}}\underline{t}_{1}1^{-\perp\perp}\iota\underline{t})S_{1}A_{1}^{2}\}\frac{q\rceil}{(_{P1^{-t_{1}}})\epsilon 1+t1}$

holds for 12$q_{1}\geq t_{1}\geq 0,$ $p_{1}\geq q_{1}$and $2\geq s_{1}\geq 1$ .

(4.3)

39
By putting $1=q_{1} \geq t_{1}=\frac{t}{q}\geq 0,$
we obtain the following (4.4):$p_{1}= \frac{\langle p-t)S+t}{q}\geq q_{1}=1$

and restoring$A_{1}$

and$B_{1}$in (4.3),(4.4)$A^{q} \geq\{A^{\frac{t}{2}}(A^{\frac{-t}{2}}B^{p}A^{\frac{-t}{2})A}ss1\frac{t}{2}\}^{\infty_{Ss}}\mathrm{t}p-\mathrm{t})1+t$

holds for $1\geq q\geq t\geq 0,.p\geq q$ and $4\geq ss_{1}\geq 1$ .
(4.4), (4.1) holds for any $s\geq 1$ .

$\mathrm{B}\mathrm{y}$repeating this process from (4.2) to,Lastly let$A_{2}=A^{q}$and$B_{2}= \{A^{\frac{t}{2}}(A^{\frac{-t}{2}B^{p}}A^{\frac{-t}{2})\}}SA\frac{t}{2}\overline{(p-}t)s\mathrm{L}\overline{+t}$Then (4.1) assures $A_{2}\geq B_{2}\geq 0$ . By Theorem.

$\mathrm{F}$(4.5)$A_{2}1+r_{2}\geq(A_{2}B_{2^{p2}}A2)^{p+r}\underline{r}_{2}zrZarrow 21+r22$

holds for $p_{2}>1$ and.
Putand restore
following (4.6): For each $1\geq q\geq t\geq 0$ and $p\geq q$ ,$r_{2}\geq 0$$p_{2}= \frac{(p-\overline{t})S+t}{q}\geq 1,$$r_{2}= \frac{r-t}{q}\geq 0$$A_{2}$

and$B_{2}$in (4.5), then we obtain the(4.6)$A^{q-t+r} \geq\{A\frac{r}{2}(A\frac{-t}{2}BpA^{\frac{-t}{2})^{s_{A\}^{\frac{q-t+r}{(p-t)s+r}}}}}\frac{r}{2}$holds for any $r\geq t$ and $s\geq 1$ . It is just (ii).
Consequently the proof of (ii) is complete.

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