数理解析研究所講究録1080 巻 1999 年 19-33

19

Characterizations of chaotic order associated with
Kantorovich inequality
山崎丈明
柳田昌宏

東京理科大理東京理科大理(Takeaki Yamazaki)(Masahiro Yanagida)

This paper is based on the following preprint:
T.Yamazaki and M.Yanagida, Characterizations of chaotic order associated with $Kan-$
torovich inequality, to appear in Scientiae Mathematicae.
Abstract
By using the order preserving operator inequality shown in [11] which is associatedwith Kantorovich inequality, we shall give some characterizations of chaotic order.

1

Introduction

An operator means a bounded linear operator on a complex Hilbert space H. An
operator is said to be positive (denoted by $T\geq 0$ ) if $(Tx, x)\geq 0$ for all $x\in H$ . Also, an..
operator is strictly positive (denoted by $T>0$ ) if is positive and invertible.
$A\geq B\geq 0$ ensures
for any $p\in[0,1]$ by well-known L\"owner-Heinz theorem.
However, it is also well known that $A\geq B\geq 0$ does not always ensure $A^{p}\geq B^{p}$ for any$p>1$ . Related to this result, the following result is given in [5].
‘

$T$

$T$

$T$

1

$A^{p}\geq B^{p}$

Theorem A ([5]). If $A\geq B>0$ and $MI\geq B\geq mI>0$ , then$( \frac{M}{m})^{p}A^{p}\geq B^{p}$for$p\geq 1$.

Recently, more precise estimation than Theorem A was given in [11] as follows:

Theorem$\mathrm{B}([11])$. If $A\geq B>0$ and $MI\geq B\geq mI>0$, then$( \frac{M}{m})^{p-1}A^{p}\geq I\mathrm{f}_{+}(m, M,p)A^{p}\geq B^{p}$for$p\geq 1$

,

(1.1)

where
$K_{+}(m, M,p)= \frac{(p-1)^{p-}1}{p^{p}}\frac{(M^{p}-m^{p})^{p}}{(M-m)(mMp-Mm^{p})^{p1}-}$

.

(1.2)

is related to both
inequality [13] and Kantorovich inis an operator on a Hilbert space $H$ such that $MI\geq A\geq mI>0$ , then$(A^{-1_{X}}, x)(Ax, x)\leq(m+M)^{2}/4mM$ holds for every unit vector
in $H$ . Many authors investigated a lot of papers on Kantorovich inequality, among others, there is along researchseries of Mond-Pe\v{c}ari\v{c}, some of them are [14] and [15].
Theorem
equality: If$\mathrm{B}$$\mathrm{H}\ddot{\mathrm{o}}1\mathrm{d}\mathrm{e}\mathrm{r}-\mathrm{M}_{\mathrm{C}\mathrm{C}\mathrm{t}\mathrm{h}}\mathrm{a}\Gamma \mathrm{y}$

$A$

$x$

The following Theorem$\mathrm{F}$is an extension of the$\mathrm{L}_{\ddot{\mathrm{O}}\mathrm{W}\mathrm{n}\mathrm{e}}\mathrm{r}$-Heinz theorem:

20

Theorem$\mathrm{F}$(Furuta inequality [7]).

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)

$(A^{\frac{r}{2}}A^{p}A \frac{r}{2})^{\frac{1}{q}}\geq(A^{\frac{r}{2}}B^{\mathrm{P}}A^{\frac{r}{2})^{\frac{1}{q}}}$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 $[3][12]$ and also anelementary one-page proof in [8]. It is shown in [17] that the domain drawn forandin the Figure is best possible one for Theorem F.
Ando [1] shows that $\log A\geq\log B$ (so called chaotic order) is equivalent tofor all $p\geq 0$ . By using Theorem , a generalization of Ando’s characterization is given
as follows:$\mathrm{F}$$\mathrm{F}$$p,$ $q$$r$$(B^{\epsilon_{A^{p}B}\mathrm{E}}22)^{\frac{1}{2}}\geq$$B^{p}$$\mathrm{F}$

Theorem. Let and be positive and invertible operators on a Hilbert
space H. Then the following assertions are mutually equivalent:$A$$\mathrm{C}([41[\epsilon 1[9])$$B$(i) $\log A\geq\log B$ .

(ii)$(B^{\frac{r}{2}}A^{p}B \frac{r}{2})^{\frac{r}{\mathrm{p}+r}}\geq B^{f}$

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

In this paper, we shall give some characterizations of chaotic order by applying Theorem and Theorem C.
$\mathrm{B}$2

Results

Theorem 1. Let andbe positive and invertible operators on a Hilbert space $H$ satisfying $\log A\geq\log B$ and $MI\geq B\geq mI>0$ . Then
$A$

$B$

$( \frac{M}{m})^{p}A^{p}\geq Ic_{+(m,M,p}+1)A^{p}\geq B^{p}$for$p\geq 0$

,

(2.1)

where $IC_{+}(m,$ $M_{P)}$, is defined in (1.2).

Theorem 1 can be considered as an extension of Theorem
new characterization of chaotic order as follows:$\mathrm{A}$: Moreover, we obtain a

Theorem 2. Let and be positive and invertible operators on a Hilbert space $H$ satisfying $MI\geq B\geq mI>0$ . Then the following assertions are $mutuall1/$ equivalent:
$A$

$B$

(i) $\log A\geq\log B$ .
(ii)$\frac{(m^{p}+M^{p})^{2}}{4m^{p}M^{p}}A^{p}\geq B^{p}$for all $p\geq 0$ .

21As a generalization of both Theorem 1 andfollowing result.

$(\mathrm{i})\Rightarrow(\mathrm{i}\mathrm{i})$of Theorem 2, we show the

be positive and invertible operators on a Hilbert space $H$ satTheorem 3. Let andisfying $\log A\geq\log B$ and $MI\geq B\geq mI>0$ . Then
$A$

$B$

$I\mathrm{f}_{+}(m^{r},$$M^{f},$ $1+ \frac{p}{r})A^{p}\geq B^{p}$for $p>0$ and $r>0$ ,(2.2)

where $K_{+}(m, M,p)r$ is defined in (1.2).

Theorem 3 implies Theorem 1 when we put $r=1$ in Theorem 3. And also Theorem 3
toTheorem 3.
of Theorem 2 when we put
yieldsin (1.2), we obtain the following proposition.
$(\mathrm{i})\Rightarrow(\mathrm{i}\mathrm{i})$Relat.ed$r=p\mathrm{i}.\mathrm{n}$$I\mathrm{f}_{+}(.m, M,p)$Proposition 4. Let $IC_{+}(m, M,p)$ be defined in (1.2). Then
$F(p, r, m, M)=IC_{+}(m^{r},$ $M^{r}, \frac{p+r}{r})$and $M$ , and also a decreasing function of $m$ for $p>0$ ,is an increasing function of$r>0$ and $M>m>0$ . And the following inequality holds:
$p,$

$r$

$( \frac{M}{m})^{p}\geq K_{+}(m^{f},$ $Mr, \frac{p+r}{r})\geq 1$

for any $p>0,$ $r>0$ and $M>m>0.|$(2.3)

By considering Proposition 4, we obtain a more precise characterization of chaoticorder than Theorem 2.
be positive and invertible operators on a Hilbert space $H$ satTheorem 5. Let andisfying $MI\geq B\geq mI>0$ . Then the following assertions are mutually equivalent:
$A$

$B$

(i) $\log A\geq\log B$ .
(ii) $M_{h}(p)A^{p}\geq B^{p}$ holds for all $p>0$ , where $h= \frac{M}{m}>1$ and.

$M_{h}(p)= \frac{h\overline{h}^{f}p\overline{-1}}{e\log(h\overline{h}^{A}\mathrm{p}-\overline{1})}$

We remark that3

$M_{h}(1)= \frac{(h-\mathrm{l})h^{\frac{1}{h-1}}}{e\log h}$(2.4)

is called Specht’s ratio $[2][16]$ .

Proof of results

Proof of Theorem 1. Put $r=1$ in (ii) of Theorem$\mathrm{C}$, then $\log A\geq\log B$ ensures thefollowing inequality:($B^{\frac{1}{2}A^{p}B^{\frac{1}{2}})^{\frac{1}{P+1}}}\geq B$

for $p\geq 0$ .

22
and $B_{1}=B$ , then$B_{1}\geq m>0$ . Applying Theorem
to
and

Put

$A_{1}=(B^{\frac{1}{2}}A^{p}B \frac{1}{2})^{\frac{1}{P+1}}$$\mathrm{B}$$A_{1}$

satisfy $A_{1}\geq B_{1}>0$ and $M\geq$
and, we have$B_{1}$$A_{1}$$B_{1}$$( \frac{M}{m})^{p_{1}-}11(B\frac{1}{2}ApB^{\frac{1}{2})^{\overline{p}+}}p\perp\geq I\mathrm{f}_{+}(m, M,p1)(B\frac{1}{2}ApB^{\frac{1}{2}})\frac{P}{p}+\llcorner 1\geq B^{p_{1}}$

(3.1)
for
Put $p_{1}=p+1\geq 1$ in (3.1) and multiply$B^{\frac{-1}{2}}$$p\geq 0$and$p_{1}\geq 1$.

on both sides, then we havefor $p\geq.0$ .

$( \frac{M}{m})^{p}A^{p}\geq Ic_{+(m,M,p}+1)A^{p}\geq B^{p}$(2.1)

Hence the proof of Theorem 1 is complete.

$\square$

In order to give a proof of Theorem 2, we need the following$\mathrm{l}\mathrm{e}\mathfrak{m}\mathrm{m}\mathrm{a}$.

Lemma 6. If $m>0$ and $M>0$ , then$\lim_{parrow 0}\{\frac{(m^{p}+M^{p})^{2}}{4m^{p}M^{p}}\}^{\frac{1}{\mathrm{p}}}=1$.

Proof. Noting that$\lim_{parrow 0}(\frac{m^{p}+M^{p}}{2})^{\frac{1}{p}}=\sqrt{mM}$,

we have$\lim_{parrow 0}\{\frac{(m^{p}+M^{p})^{2}}{4m^{p}M^{p}}\}^{\frac{1}{p}}=\lim_{parrow 0}\frac{1}{mM}(\frac{m^{p}+M^{p}}{2})^{\frac{2}{p}}=\frac{1}{mM}(\sqrt{mM})^{2}--1$

Proof of Theorem 2.
(a) Proof of$(\mathrm{i})\Rightarrow(\mathrm{i}\mathrm{i})$.

$\square$. Put $r=p$ in (ii) of Theorem , then $\log A\geq\log B$ ensures$\mathrm{C}$the following inequality:
for $p\geq 0$ .

$(B^{\mathrm{E}}2ApB2)^{\frac{1}{2}}e\geq B^{p}$satisfy $A_{1}\geq B_{1}>0$ and $M^{p}\geq$
and $B_{1}=B^{p}$ , thenand$B_{1}\geq m^{p}>0$ . Applying Theorem, we have
to
and

Put

$A_{1}=(B^{\mathrm{E}}2A^{p}B2\mathrm{g})^{\frac{1}{2}}$$A_{1}$$\mathrm{B}$$A_{1}$$B_{1}$$B_{1}$$I\mathrm{f}_{+}(m^{p}, Mp,p1)(B2A\mathrm{r}\mathrm{i}pB^{E}2)^{\lrcorner}p2\geq(B^{p})^{p_{1}}$

Put $p_{1}=2\geq 1$ in (3.2) and multiply$B^{=_{2}B}$$(\mathrm{i})\Rightarrow(\mathrm{i}\mathrm{i})$$p_{1}\geq 1$.

on both sides, then we llave$IC_{+}(m^{p}, M^{p}, 2)A^{p}\geq B^{p}$

Hence the proof offor $p\geq 0$ andis complete sincefor $p\geq 0$ .

$I \mathrm{t}_{+}’(m^{p}, M^{p}, 2)=\frac{(m^{p}+MP)^{2}}{4m^{p}\Lambda I^{p}},\cdot$

(3.2)

23

. Taking logarithm of both sides of (ii) since $\log t$ is an operator(b) Proof ofwefunction,havemonotone$(\mathrm{i}\mathrm{i})\Rightarrow(\mathrm{i})$$\log\{(\frac{(m^{p}+NI^{p})^{2}}{4m^{p}M^{p}})^{\frac{1}{p}}A\}\geq\log B$$\mathrm{f}.0.\mathrm{r}$Letting$parrow+\mathrm{O}$

all $p\geq 0$ .

in (3.3), we have $\log A\geq\log B$ by Lemma 6.

Proof of Theorem 3. By Theorem$\mathrm{C},$(3.3)$\square$is equivalent to the following inequal-$\log A\geq\log B$

ity:

for $p>0$ and $r>0$ .

$(B^{\frac{r}{2}}A^{p}B \frac{r}{2})^{\frac{r}{p+r}}\geq B^{r}$satisfy $A_{1}\geq B_{1}>0$ andandand $B_{1}=B^{r}$ , then$M^{r}\geq B_{1}\geq m^{r}>0$ . Applying Theorem, we have
to
and

Put

$A_{1}=(B^{\frac{r}{2}A^{p}B^{\frac{r}{2}})^{\frac{r}{p+r}}}$$B_{1}$$A_{1}$$\mathrm{B}$$B_{1}$$A_{1}$

for$I\zeta_{+(m’,M^{t}},p_{1})A_{1}^{p_{1}}\geq B_{1}^{p1}$Put$p_{1}=L_{\frac{+r}{r}}\geq 1$

.

(3.4)

in (3.4), then we have
$I\mathrm{t}’+(m^{\Gamma},$

By multiplying$p_{1}\geq 1$$B^{\frac{-r}{2}}$$M^{r},$ $\frac{p+r}{r})B\frac{r}{2}A^{p}B\frac{r}{2}\geq B^{p+r}$.

. (3.5)on both sides of (3.5), we have

$I\mathrm{f}_{+}(m^{\Gamma},$$M^{r},$ $1+ \frac{p}{r})A^{p}\geq B^{p}$for $p>0$ and $r>0$ .

Hence the proof of Theorem 3 is complete.

(2.2)$\square$We prepare the following four lemmas to give a proof of Proposition 4.
Lemma 7. For each $h>1$ ,(3.6)$f(t)= \log(\frac{h^{t}-1}{t})$is a convex function for $t>0$ .

Proof. Put$x(t)= \frac{h^{t}-1}{t}$, then $f(t)=\log\{X(t)\}$ and$f^{\prime J}(t)= \frac{x(t)_{X’’}(t)-\{X(/t)\}^{2}}{\{x(t)\}^{2}}$so that$f^{\prime/}(t)\geq 0$,

for $t>0$ is equivalent to the following (3.7) since $\{x(t)\}^{2}\geq 0$ :$x(t)X(//t)-\{x’(t)\}^{2}\geq 0$

for $t>0$ .

(3.7)

24

By calculation on differential calculus and$\mathrm{r}\mathrm{e}\mathrm{f}\mathrm{i}\mathrm{n}\mathrm{e}\mathfrak{m}\mathrm{e}\mathrm{n}\mathrm{t}$, we have$x(t)X(’/)t- \{x’(t)\}2=\frac{1}{t^{4}}(h^{t}-1+th^{\frac{t}{2}}\log h)(ht-1-th\frac{t}{2}\log h)$,so that (3.7) is equivalent to the following (3.8) because $h^{t}-1+th^{\frac{t}{2}}\log h\geq 0$ for $h>1$
and $t>0$ :$h^{t}-1-th \frac{t}{2}\log h\geq 0$

Put$y(t)=h^{t}-1-th^{\frac{t}{2}\mathrm{l}h}\mathrm{o}\mathrm{g}$. Then$y(\mathrm{O})=0$

for $h>1$ and $t>0$ .

(3.8)

and

$y’(t)=h^{\frac{t}{2}}\log h(h^{\frac{t}{2}}-1-\log h^{\frac{t}{2}})$,so that $y’(t)>0$ for $h>1$ and $t>0$ . Therefore $y(t)\geq 0$ for $h>1$ and $t>$. , which isequivalent to (3.8). Consequently, the proof of Lemma 7 is complete.
$0$$\square$Lemma 8. Let $h>1$ . Then
(3.9)$g(p, r, h)=( \frac{r}{p+r}\frac{h^{p+r}-1}{h^{r}-1})^{\frac{1}{p}}$is an increasing function of

$p$

and

$r$

for $p>0$ and $r>0$ .

Proof. Define $f(t)$ as in Lemma 7, i.e.,$f(t)= \log(\frac{h^{t}-1}{t})$

.

(3.6)

Then by (3.9),$\log\{g(p, r, h)\}=\frac{\log(\frac{h^{p+\Gamma}-1}{p+r})-\log(\frac{h^{r}-1}{r})}{p}=\frac{f(p+r)-f(r)}{p}$.

(3.10)(a) Proof of the result that $g(p, r, h)$ is increasing for $p>0$ .

Let $p_{1}\geq p_{2}>0$ and $r>0$ . Since $f(t)$ is convex for $t>0$ by Lemma 7,(3.11)$\theta f(t_{1})+(1-\theta)f(t_{2})\geq f(\theta t_{1}+(1-\theta)b_{2})$holds for $\theta\in[0,1],$ $t_{1}>0$ and $t_{2}>0$ . Put$\theta=\frac{p_{2}}{p_{1}}\in[0,1],$$t_{1}=p_{1}|+r>0$ and $t_{2}=r>0$ ,

then

.

$\theta t_{1}+(1-\theta)t_{2}=\frac{p_{2}}{p_{1}}(p_{1}+r)+(1-\frac{p_{2}}{p_{1}})r=p_{2}+r$

By (3.11) and (3.12), we have$\frac{p_{2}}{p_{1}}f(p_{1}+r)+(1-\frac{p_{2}}{p_{1}}\mathrm{I}f(r)\geq f(p_{2}+r)$,(3.12)

25

so that$\underline{f(p_{1}+r)-f(r)}\geq\underline{f(p_{2}+r)-f(r)}$$p_{1}$.

(3.13)$p_{2}$By (3.10) and (3.13), $g(p, r, h)$ is increasing for $p>0$ .
(b) Proof of the result that $g(p, r, h)$ is increasing for $r>0$ .

Let $r_{1}\geq r_{2}>0$ and $p>0$ . Since $f(t)$ is convex for $t>0$ by Lemma 7,, so that$f’(t)$ is increasing, that is, $f’(t+r_{1})-f’(t+r_{2})\geq 0$ . Therefore $s(t)=f(t+r_{1})-f(t+r_{2})$is increasing for $t\geq 0$ . Then we have $f(p+r_{1})-f(p+r_{2})=s(p)\geq s(\mathrm{O})=f(r_{1})-f(r2)$ ,that is,$f^{\prime/}(t)\geq 0$$\frac{f(p+r_{1})-f(\Gamma_{1})}{p}\geq\frac{f(p+r_{2})-f(r_{2})}{p}$.

(3.14)By (3.10) and (3.14), $g(p, r, h)$ is increasing for $r>0$ .

Consequently the proof of Lemma 8 is complete.

$\square$Lemma 9. For $p\geq 1$ and $t>1$ ,(3.15)$pt^{p-1} \geq\frac{t^{p}-1}{t-1}\geq pt^{L^{-\underline{1}}}2$

Proof. To prove the first inequality of (3.15), define$h(t)=t^{\mathcal{P}}$. Since $h(t)$ is a convex

function of for $p\geq 1$ , we have $h’(t) \geq\frac{h(t)-h(1)}{t-1}$ for $t>1$ , which is equivalent to the$t$first inequality of (3.15). On the other hand, the second inequality.of (3.15) is equivalent
to the following:$t^{p}-pt^{\mathrm{E}}2\llcorner 1+pt^{\mathrm{g}_{\frac{-1}{2}}}-1\geq 0$

So we have only to prove (3.16). Put

for $p\geq 1$ and $t>1$ .

$f(t)=t^{p}-pt^{\epsilon_{2}}\pm\underline{1}+pt^{R_{\frac{-1}{2}}}-1$(3.16). Then $f(1)=0$ and$f’(t)=pt^{p-1}- \frac{p(p+1)}{2}t^{R}\frac{-1}{2}+\frac{p(p-1)}{2}tR_{\frac{-3}{2}}$(3.17)$=pt^{L^{-\underline{3}}}2(t^{E\pm\underline{1}}2- \frac{p+1}{2}t+\frac{p-1}{2})$.

, thenPut
for $p\geq 1$ and $t>1$ , andalso $g(1)=0$ . Therefore $g(t)\geq 0$ for $p\geq 1$ and $t>1$ , so thatfor$p\geq 1$ and $t>1$ by (3.17). Hence $f(t)\geq 0$ for $p\geq 1$ and $t>1$ , which is equivalent to(3.16). Consequently the proof of Lemma 9 is complete.
$g(t)=t^{E\pm}2\underline{1}-R_{\frac{+1}{2}t}+\mathrm{a}_{\frac{-1}{2}}$$g’(t)=E_{\frac{+1}{2}t^{L^{-\underline{1}}}}2-L+\underline{1}2\geq 0$$f’(t)=pt^{\mathrm{R}^{-\dot{3}}}2g(t)\geq 0$$\square$26

Lemma 10. For $p>0,$ $r>0$ and $h>1_{r}$$h\geq g(p, r, h)\geq h^{\frac{1}{2}}$,(3.18)

where $g(p, r, h)$ is as in Lemma 8, $i.e.$ ,(3.9)$g(p, r, h)=( \frac{r}{p+r}\frac{h^{p+\Gamma}-1}{h^{r}-1})^{\frac{1}{p}}$Proof. Replace

$p$

with

$L+\underline{r}r\geq 1$

in Lemma 9, we have the following inequality.
for $p>0,$ $r>0$ and $t>1$ .

$( \frac{p+r}{r})t^{\mathrm{g}}r\geq\frac{t^{\epsilon_{r}\llcorner r}-1}{t-1}\geq(\frac{p+r}{r})t2\mathrm{A}r$(3.19)

Put $t=h^{r}>1$ in (3.19). Then we have$h^{p} \geq\frac{r}{p+r}\frac{h^{p+r}-1}{h^{r}-1}\geq h^{E}2$

therefore we have (3.18) by taking.Proof of Proposition 4. Put$\frac{1}{p}$for $p>0,$ $r>0$ and $h>1$ ,(3.20)

exponent of each side of (3.20).

$h= \frac{M}{m}>1$$\square$and $g(p, r, h)$ is as in Lemma 8, i.e.,(3.9)$g(p, r, h)=( \frac{r}{p+r}\frac{h^{p+f}-1}{h^{r}-1})^{\frac{1}{p}}$Then$Ic_{+}(m^{r},$ $M^{r},$ $\frac{p+r}{r})=\frac{(\begin{array}{l}er\end{array})r\mathrm{E}}{(1+_{r}e)^{1+^{\mathrm{g}}}r}\frac{(M^{p+r}-m^{p})+\Gamma 1+^{E}r}{(M^{r}-m^{r})(m^{\gamma}Mp+r-Mrm^{p}+\Gamma)r\mathrm{E}}$$=( \frac{r}{p+r})(\frac{p}{p+r})^{r}\frac{(h^{p+r}-1)1+_{r}^{\mathrm{E}}}{(h^{f}-1)(hp+r-h^{r})^{\epsilon}r}\epsilon$by (1.2)
by $h= \frac{M}{m}>1$$= \{\frac{1}{h}(\frac{r}{p+r}\frac{h^{p+r}-1}{h^{r}-1})^{\frac{1}{p}}(\frac{p}{p+r}\frac{h^{p+r}-1}{h^{p}-1})^{\frac{1}{r}}\}^{p}$$= \{\frac{1}{h}\cdot g(p, r, h)\cdot g(r,p, h)\}^{p}$(3.21)by (3.9).

By Lemma 10, we have the following (3.22).
for $p>0$ and $r>0$ .

(3.22)

for any $p>0,$ $r>0$ and $M>m>0$ .

(2.3)$h \geq\frac{1}{h}\cdot g(p, r, h)\cdot g(r,p, h)\geq 1$

By (3.21) and (3.22), we have (2.3), i.e.,$( \frac{M}{m})^{p}\geq I\mathrm{s}\mathrm{i}_{+}(m^{\Gamma},$$Mr, \frac{p+r}{r})\geq 1$

27(a) Proof of the result that

$F(p, r, m, M)=I\mathrm{f}_{+}(m^{\mathrm{r}}, M^{\gamma}, L^{\underline{\prime}}+)r$

and $r>0$ .

is increasing for $p>0$

By Lemma 8, $g(p, r, h)$ is increasing for $p>0$ and $r>0$ . Then we obtain that$g(p, r, h)\cdot g(r,p, h)$ is increasing for $p>0$ and $r>0$ . By (3.21) and (3.22), $F(p, r, m, M)=$$I \mathrm{f}_{+}(m^{f}, M^{re},\frac{+\mathrm{r}}{r})$is increasing for $p>0$ and $r>0$ .

(b) Proof of the result that $F(p, r, m, M)=K_{+}(m^{f}, Mr, L\underline{+r})\Gamma$ is an increasing functionof $M$ and also a $decreas|ing$ function of $m$ for $M>m>0$ .
Firstly, for $s>0$ ,$g( \frac{p}{s},$$hs)=( \frac{\frac{r}{s}}{\epsilon_{+\frac{r}{s}},s}\frac{(h^{s})^{e_{+\frac{r}{s}}}S-1}{(h^{s})^{\frac{r}{s}}-1})^{\frac{s}{p}}$$\frac{r}{s},$(3.23)$=( \frac{r}{p+r}\frac{h^{p}+f-1}{h^{r}-1})^{\frac{s}{p}}$$=\{g(p, r, h)\}^{s}$so that$g(p, r, h)= \{g(_{s}^{E}, \frac{r}{s}, h^{s})\}^{\frac{1}{s}}$,for $s>0$ . Then for $s>1$ , we haveby (3.21)$IC_{+}(m^{\Gamma},$ $M^{\gamma}, \frac{p+r}{r})=\{\frac{1}{h}\cdot g(p, r, h)\cdot g(r,p, h)\}^{p}$$= \{\frac{1}{h^{s}}\cdot g(\frac{p}{s},$$\frac{r}{s},$$h^{S)} \cdot g(\frac{r}{s},\frac{p}{s},$$h^{\theta)}\}^{E}s$by (3.23)(3.24)$\leq\{\frac{1}{h^{s}}\cdot g(p, r, h^{s})\cdot g(r,p, h^{S})\}p$

by the result of (a)$=IC_{+}(m^{\gamma},$ $(h^{s-1}M)^{r}, \frac{p+r}{r})$since$h^{s}= \frac{h^{s-1}M}{m}$,is an increasing function of $M$ for $M>m>0$ since $h^{s-1}M>M$ .
so thatOn the other hand, by the same way as (3.24) we have
$I \mathrm{f}_{+}(m^{r}, M^{r2},\frac{+\mathrm{r}}{r})$$IC_{+}(m^{t},$ $Mr, \frac{p+r}{r})\leq\{\frac{1}{h^{s}}\cdot g(p, r, hs)\cdot g(r,p, h^{s})\}^{p}=Ic_{+}((h^{1-S}m)^{\gamma},$ $M^{f}, \frac{p+r}{r})$since$h^{s}= \frac{M}{h^{1-S}m}$. Hence$IC_{+}(m, M\prime r, \mathrm{P}_{\frac{+r}{r}})$is a decreasing function of $m$ for $M>m>0$:

since $m>h^{1-s}m$ .
By (a) and (b), the proof of. Proposition 4 is complete.

$\square$

We need the following lemmas to give a proof of Theorem 5.
$\mathrm{L}\mathrm{e}\mathrm{m}\mathfrak{m}\mathrm{a}11$,. Let $M>m>0,$ $p>0$ and$I\mathrm{t}_{+(m,M}’,p$$\lim_{rarrow+0}Ic_{+}(m^{r},$ $M^{r},$) be defined in (1.2). Then

$1+ \frac{p}{r})=M_{h}(p)$where $h= \frac{M}{m}>1$ , and $M_{h}(p)$ is defined in (2.4).

,

28
Proof. Define $g(p, r, h)$ as in Lemma 8, i.e.,(3.9)$g(p, r, h)=( \frac{r}{p+r}\frac{h^{p+r}-1}{h^{r}-1})^{\frac{1}{\mathrm{p}}}$

As in the proof of Proposition 4, we have
(3.21)$K_{+}(m^{r},$ $M^{r}, \frac{p+r}{r})=\{\frac{1}{h}\cdot g(p, r, h)\cdot g(r,p, h)\}^{p}$

We define $f(t)$ as follows:$f(t)=\log(h^{t}-1)$ .

(3.25)

Then$f’(t)= \frac{h^{t}\log h}{h^{t}-1}=\log h^{\frac{h^{t}}{h^{t}-1}}$,(3.26)

so that$\lim_{rarrow+0}\log(\frac{h^{p+r}-1}{h^{p}-1})^{\frac{1}{r}}=\lim_{+rarrow 0}\frac{\log(h^{p+}\Gamma-1)-\log(h^{\mathrm{P}}-1)}{r}$by (3.25)$= \lim_{rarrow+0}\frac{f(p+r)-f(p)}{r}$$=f’(p)$$=\log h^{\frac{h^{p}}{h^{p}-1}}$

therefore$\log h$$\lim_{rarrow+0}(\frac{h^{p+f}-1}{h^{p}-1})^{\frac{1}{r}}=h^{\frac{h^{\mathrm{p}}}{h\mathrm{p}-1}}$. Sinceby (3.26),$\lim_{rarrow+0}(1+\frac{r}{p})^{E}r=e$and$\lim_{rarrow+0}\frac{h^{r}-1}{r}=$, we have$\lim_{rarrow+0}g(P, \Gamma, h)=\lim_{rarrow+0}(\frac{h^{p+\Gamma}-1}{p+r}\frac{r}{h^{r}-1})^{\frac{1}{p}}=(\frac{h^{p}-1}{p\log h})^{\frac{1}{p}}=(\frac{1}{\log h\overline{h}pB\overline{-1}})^{\frac{1}{p}}$(3.27)

and$\lim_{rarrow+0}g(r,p, h)=\lim_{rarrow+0}(\frac{p}{p+r})^{\frac{1}{r}}(\frac{h^{p+r}-1}{h^{p}-1})^{\frac{1}{r}}=\frac{h^{\frac{h^{p}}{h^{p}-1}}}{e^{\frac{1}{p}}}$.

(3.28)

Applying (3.27) and (3.28) in (3.21), we have$\lim_{rarrow+0}Ic+(m^{r},$ $Mf, \frac{p+r}{r})=\lim_{rarrow+0}\{\frac{1}{h}\cdot g(p, r, h)\cdot g(r,p, h)\}^{p}$$\frac{h^{B_{p^{\frac{h^{p}}{-1}}}}h}{e}$$= \frac{1}{h^{p}}\frac{1}{\log hh^{\overline{p}\overline{-1}}\mathrm{g}}$by (3.21)

by (3.27) and (3.28)$h\overline{h}p^{f}\overline{-1}$.
$=\overline{e\log h\overline{h}\mathrm{p}-=_{1}}$Hence the proof of Lemma 11 is complete.

$\square$29

Lemma 12. Let $h>1$ and $M_{h}(p)$ be defined in (2.4). Then$\lim_{parrow+0}\{Mh(_{P)\}^{\frac{1}{p}}}=1$

Proof. Put$g(p)=h\overline{h}^{p}-1=$, then$M_{h}(p)= \frac{g(p)}{e\log g(p)}$.

. It is easily obtained that$\lim_{parrow+0^{g(}}p)=h^{\frac{1}{1\circ\iota h}}=e$and$g’(p)= \{\frac{h^{p}-1-php\log h}{(h^{p}-1)^{2}}\}h\overline{h}p_{-\overline{1}\mathrm{l}}^{\Sigma}\mathrm{o}\mathrm{g}h$Then $g’(p)$ is bounded as$parrow+\mathrm{O}$

.

since

$\lim_{parrow+0}\frac{h^{p}-1-php\log h}{(h^{p}-1)^{2}}=\lim_{arrow p+0}\frac{-ph^{p}\{\log h\}^{2}}{2(h^{p}-1)hp\log h}$by L’Hospital’s theorem$= \lim\underline{-p\log h}$$parrow+02(hp-1)$$= \frac{-1}{2}$.

Then we have$\lim_{parrow+0}\log\{Mh(p)\}\frac{1}{p}=\lim_{parrow+0}\frac{\log g(p)-\log\{\log g(p)\}-1}{p}$$= \lim_{parrow+0}\frac{g’(p)}{g(p)}\{1-\frac{1}{\log g(p)}\}$$=0$so that$\lim_{parrow+0}\{M_{h}(p)\}^{\frac{1}{p}}=1$

Proof of Theorem 5.
(a) Proof of$(\mathrm{i})\Rightarrow(\mathrm{i}\mathrm{i})$by L’Hospital’s theorem,. Hence the proof of Lemma 12 is complete.

$\square$. By Theorem 3, $\log A\geq\log B$ implies$IC_{+}(m^{r}M^{r},$$1)+ \frac{p}{r})A^{p}\geq B^{p}$for $p>0$ and $r>0$ .

Letting $rarrow+0$ in (2.2), we have $M_{h}(p)A^{P}\geq B^{P}$ for $p>0$ since$M_{h}(p)$ as
by Lemma 11.

(2.2)$I\mathrm{t}_{+}(m^{r},$ $M^{\Gamma},$$1+\epsilon_{)}’arrow$$rarrow+\mathrm{O}$(b) Proof of$(\mathrm{i}\mathrm{i})\Rightarrow(\mathrm{i})$. By taking logarithm of both sides of (ii), we have$\log(\{M_{h}(p)\}\frac{1}{p}A)\geq\log B$

Then letting
Lemma 12.

$parrow+\mathrm{O}$for $p>0$ .

in (3.29), we have $\log A\geq\log B$ since

Hence the proof of Theorem 5 is complete.

$\{M_{h}(p)\}^{\frac{1}{p}}arrow 1$(3.29)

as$parrow+\mathrm{O}$by$\square$30
4Concluding Remarks

Remark 1. Let and be positive and invertible operators on a Hilbert space $H$ . We
for $\delta\in(0,1]$ which interpolates usual order $A\geq B$ and chaotic
consider an order$\log
A\geq\log
B$continuously.
The following result is easily obtained by Theorem B.
order
$A$

$B$

$A^{\delta}\geq B^{\delta}$

Proposition 13. Let and be positive and invertible operators on a Hilbert space $H$satisfying
for $\delta\in(0,1]$ and $MI\geq B\geq mI>0$ , then
$B$

$A$

$A^{\delta}\geq B^{\delta}$$I\mathrm{f}_{+}(m^{\delta},$

where$I\mathrm{f}_{+}(m, M,p)$

Proof. Put

Theorem$\mathrm{B}$to$\frac{p}{\delta})Ap\geq B^{p}$$p_{1}=2\delta\geq 1$$p\geq\delta$, then $A_{1}\geq B_{1}>0$ and
and
,
have
we
and
$B_{1}=B^{\delta}$,$M^{\delta}\geq B_{1}\geq m^{\delta}$. By applying$B_{1}$$I\mathrm{t}_{+}(m^{\delta\delta p_{1}}, M,p1)A1\geq B_{1}^{P1}$

Put

for

is defined in (1.2).

$A_{1}=A^{\delta}$$A_{1}$$M^{\delta},$for$p_{1}\geq 1$

for$p\geq\delta$

.

(4.1)

in (4.1), then we have
$I\mathrm{f}_{+}(m^{\delta},$$M \delta,\frac{p}{\delta})A^{p}\geq B^{p}$.

$\square$

We show the following result to consider the relation between Proposition 13 andTheorem 5.

Proposition 14. Let
Then for $p>0$ and $M>m>0$ ,$I\mathrm{f}_{+}(m, M,p)$and $M_{h}(p)$ be defined in (1.2) and (2.4), respectively.

$\lim_{\deltaarrow+0}Ic_{+}(m^{\delta},$$M^{\mathit{5}}, \frac{p}{\delta})=M_{h}(p)$,where $h= \frac{M}{m}>1$ .

Proposition 14 can be proved by the same way as Lemma 11.

Proof. Define $g(p, r, h)$ as in Lemma 8, i.e.,(3.9)$g(p, r, h)=( \frac{r}{p+r}\frac{h^{p+r}-1}{h^{r}-1})^{\frac{1}{p}}$By (3.21), we have$IC_{+}(m^{\delta},$$M^{\delta}, \frac{p}{\delta})=I\mathrm{t}_{+}’(m^{\delta},$$M^{\delta},$$\frac{(p-\delta)+\delta}{\delta})$(4.2)$= \{\frac{1}{h}\cdot g(p-\delta, \delta, h)\cdot g(\delta,p-\delta, h)\}^{p}-\delta$by (3.21).

31
We define $f(t)$ as follows:$f(t)=\log(ht-1)$ .

(3.25)

Then$f’(t)= \frac{h^{t}\log h}{h^{t}-1}=\log h^{\frac{h^{t}}{h^{l}-1}}$,(3.26)

so that$\lim_{\deltaarrow+0}\log(\frac{h^{p}-1}{h^{p-\delta}-1})^{1}\tau\frac{\log(h^{p}-1)-\log(hp-\delta-1)}{\delta}=\lim_{\mathrm{o}\deltaarrow+}$by (3.25)$= \lim_{\deltaarrow+0}\frac{f(p)-f(p-\delta)}{\delta}$$=f’(p)$$=\log h^{\frac{h^{\mathrm{P}}}{hp-1}}$

therefore$\log h$$\lim_{\deltaarrow+0}(\frac{h^{p}-1}{h^{p-\delta}-1})^{\frac{1}{\delta}}=h^{\frac{h^{\mathrm{p}}}{hp-1}}$. Sinceby (3.2..6),$\lim_{\deltaarrow+0}(1-\frac{\delta}{p})^{\epsilon}\delta=\frac{1}{e}$and$\lim_{\deltaarrow+0}\frac{h^{\delta}-1}{\delta}=$, we have$\lim_{\deltaarrow+0}g(p-\delta, \delta, h)=\lim_{+\deltaarrow 0}(\frac{h^{p}-1}{p}\frac{\delta}{h^{\delta}-1})^{\frac{1}{p-\delta}}=(\frac{h^{p}-1}{p\log h})^{\frac{1}{p}}=(\frac{1}{\log h\overline{h}P-\overline{1}A})^{\frac{1}{p}}$

(4.3)

and
$\lim_{\deltaarrow+0}g(\delta,p-\delta, h)=\delta\lim_{arrow+0}(\frac{p-\delta}{p})^{1}7(\frac{h^{p}-1}{h^{p-\mathit{5}}-1})^{1}\tau\frac{h^{\frac{h^{p}}{hp-1}}}{e^{\frac{1}{p}}}=$

.

(4.4)

Applying(4.3) and (4.4) in (4.2), we have$\lim_{\deltaarrow+0}Ic_{+}(m^{\delta},$ $M^{\delta}, \frac{p}{\delta})=\lim_{\deltaarrow+0}\mathrm{t}\frac{1}{h}\cdot g(p-\delta, \delta, h)\cdot g(\delta,p-\delta, h)\mathrm{I}p-\delta$$= \frac{1}{h^{p}}\cdot\frac{1}{\log h\overline{h}\mathrm{p}\overline{-1}A}\cdot\frac{h^{\frac{\mathrm{p}h^{p}}{hP-1}}}{e}$by (4.2)

by (4.3) and (4.4).

$= \frac{h^{\mathrm{n}_{-}}\overline{h}p\overline{1}}{e\log h^{\frac{\mathrm{p}}{hp-1}}}$Hence the proof of Proposition 14 is complete.

Remark 2. We summarize the results$\mathrm{w}\mathrm{h}\mathrm{i}_{\mathrm{C}}\mathrm{h}$.

$\square$have been obtained as follows:

Let $A>0$ and $MI\geq B\geq mI>0$ . Then the following assertions hold:(i)$A\geq B\mathrm{i}\mathfrak{m}\mathrm{P}\mathrm{l}\mathrm{i}\mathrm{e}\mathrm{s}Ic+(m, M,p)Ap\geq B^{p}$for $p>1$ ,32(ii) for each $\delta\in(0,1],$$A^{\delta}\geq B^{\mathit{5}}$(iii) $\log A\geq\log B$ implies

where $h= \frac{M}{m}>1$ , andimplies$I\mathrm{f}_{+}(m^{\delta},$$M_{h}(p)A^{p}\geq B^{\mathrm{p}}$$I\mathrm{f}_{+}(m, M,p)$$M^{\delta}, \frac{p}{\delta})Ap\geq B^{p}$for $p>\delta$ ,for $p>0$ ,and $M_{h}(p)$ are defined in (1.2) and (2.4), respectively.

Proposition 14 states that as the order in the assumption of (ii) interpolates the ordersof (i) and (iii) continuously, the scalar in the consequence of (ii) also interpolates the scalarof (i) and (iii) continuously. Therefore Theorem 5 can be considered as a natural result
which is parallel to Theorem B.

Remark 3. Very recently, the following characterization of chaotic order was obtained.
Theoremexists an$\mathrm{D}(1^{6}])$. If$\alpha=\alpha_{\delta}>0$$A,$$B>0$ , then $\log A\geq\log B$ if and only if for any $\delta\in(0,1]$ there

such that$(e^{\delta}A)^{\alpha}>B^{\alpha}$.

On the other hand, Theorem 2 and Theorem 5 can be rewritten in the following form.
Theorem 2’. If $A,$ $B>0$ , then $\log A\geq\log B$ if and only if for any $p\geq 0$ there exists a$K_{p}>1$ such that, and $(IC_{p}A)P\geq B^{P}$ .
as$K_{p}arrow 1$$parrow+\mathrm{O}$

Also we can obtain Theorem from Theorem 2 by the almost
way to rewriting
Theorem 2 into Theorem 2’. We remark that Theorem 2 is proved by using Theorem
and Theorem can be proved by using Theorem and Theorem , so that Theorem 2
can be considered as a formal extension of Theorem D.
$\mathrm{D}$$\mathrm{S}\mathrm{a}\mathfrak{m}\mathrm{e}$$\mathrm{C}$$\mathrm{C}$$\mathrm{F}$$\mathrm{D}$Remark 4. Theorem 2’ is a parallel result to the following Theorem$\mathrm{E}[10]$.

Theorem. If $A,$ $B>0$ , then $\log A\geq\log B$ if and only if for any $p\geq 0$ there
exists the unique unitary operator
such that
, and
as
.
$\mathrm{E}([10])$$U_{p}$$U_{p}arrow I$$parrow+\mathrm{O}$$(U_{p}AU^{*})^{P}p\geq B^{P}$

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