数理解析研究所講究録1039 巻 1998 年 28-36

28Order preserving operator function via the inequality“ $A\geq B\geq 0$

ensures$(A^{\frac{r}{2}}A^{p}A \frac{r}{2})^{\frac{1+r}{\mathrm{p}+r}}\geq(A^{\frac{r}{2}}B^{p}A\frac{r}{2})^{\frac{1+r}{\mathrm{p}+r}}$

東京理科大学1

for $p\geq 1$ and $r\geq 0$ ”

柳田昌宏
山崎丈明
古田孝之

(Masahiro Yanagida)(Takeaki Yamazaki)(Takayuki Furuta)

Introduction

In what follows, a capital letter 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$x\in H$ and also an operator
is strictly positive (denoted by $T>0$ ) if is positiveand invertible. The following Theorem is an extension of the celebrated L\"owner-Heinz
theorem $[12][10]$ .
$T$

$T$

$T$

$\mathrm{F}$

Theorem(Furuta inequality) [4].

$\mathrm{F}$If $A\geq B\geq 0$ , then for each
(i)
and
(ii)

$r\geq 0$$(B^{\frac{r}{2}}A^{p}B \frac{r}{2})^{\frac{1}{q}}\geq(B^{\frac{r}{2}}B^{p}B\frac{r}{2})^{\frac{1}{q}}$$(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 $(1+r)q\geq p+r$ .
rlgureWe remark that Theorem is essentially the same as the inequality made in its title
and Theorem yields the L\"owner-Heinz theorem when we put $r=0$ in (i) or (ii) statedabove: $A\geq B\geq 0$ ensures
for any $\alpha\in[0,1]$ . Alternative proofs of Theorem
are given in [2] [5] and [11] and also elementary one page proof in [6]. It was shown in[13] that the domain surrounded byand in the Figure is the best possible one for
Theorem F. In [8] we established the following Theorem as extensions of Theorem F.
$\mathrm{F}$$\mathrm{F}$$A^{\alpha}\geq B^{\alpha}$$\mathrm{F}$$p,$ $q$$r$$\mathrm{G}$(Generalized Furuta inequality) [8]. If $A\geq B\geq 0$ with $A>0$ , then for
Theoremeach $t\in[0,1]$ and $p\geq 1$ ,$\mathrm{G}$$F_{p,t}(A, B, r, S)=A^{\frac{-r}{2}\{A^{\frac{r}{2}}(AB} \frac{-t}{2}pA^{\frac{-t}{2})}s_{A^{\frac{r}{2}}\}^{\frac{1-t+r}{(p-t)s+r}}}A^{\frac{-r}{2}}$is decreasing for $r\geq t$ and $s\geq 1$ and $F_{p,t}(A, A, r, S)\geq F_{p,t}(A, B, r, s)$ , that is, for each$t\in[0,1]$ and $p\geq 1$ ,(1.1)

$A^{1-t+r}\geq\{A^{\frac{r}{2}}(A^{\frac{-t}{2}}BpA^{\frac{-t}{2})A^{\frac{r}{2}\}^{\frac{1-t+r}{(\mathrm{p}^{-}t)s+r}}}}S$

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

29

is shown in [3]. Ando-Hiai [1]Recently a nice mean theoretic proof of Theorem
majorization results and proved the useful inequality equivalentestablished excellent
ization theorem as follows; If $A\geq B\geq 0$ with $A>0$ , thento the main$\mathrm{G}$$\log$$\log \mathrm{m}\mathrm{a}\mathrm{j}\mathrm{o}\dot{\mathrm{r}}$$A^{r} \geq\{A^{\frac{r}{2}}(A^{\frac{-1}{2}}B^{p}A^{\frac{-1}{2})^{r_{A\}^{\frac{1}{p}}}}}\frac{r}{2}$holds for any $p\geq 1$ and $r\geq 1$ . Theorem interpolates the inequality stated above byAndo-Hiai and Theorem itself and also extends results of [7].
Since now, many applications of Theorem and Theorem have been developed in
the following.branches by many authors.
$\mathrm{G}$$\mathrm{F}$$\mathrm{F}$$\mathrm{G}$APPLICATIONS OF THEOREM$\mathrm{F}$(A) OPERATOR INEQUALITIES(1) Characterizations of operators satisfying $\log A\geq\log B$(2) Generalizations of Ando’s theorem(3) Other order preserving operator inequalities(4) Applications to the relative operator entropy(5) Applications to Ando-Hiaimajorization(6) Generalized Aluthge transformation$\log$(B) NORM INEQUALITIES(1) Several generalizations of Heinz-Kato theorem(2) Generalizations of some theorems on norms(3) An extension of Kosaki trace inequality and parallel results(C) OPERATOR EQUATIONS(1) Generalizations of Pedersen-Takesaki theorem and related results

Very recently the following result is obtained as an extension of Theorem G.

Theorem$\mathrm{H}[9]$$p \geq\max\{q, t\}$. If $A\geq B\geq 0$ with $A>0$ , then-for each $t\in[0,1],$$q\geq 0$

and,$G_{p,q,t}(A, B, r, S)=A^{\frac{-r}{2}\{A^{\frac{r}{2}}(AB} \frac{-t}{2}pA^{\frac{-t}{2})}s_{A^{\frac{r}{2}}\}^{(p-A^{\frac{-r}{2}}}}\mathrm{R}_{\frac{t+r}{t)s+r}}-$is decreasing for $r\geq t$ and $s\geq 1$ . Moreover for each $t\in[0,1],$ $q\in[t, 1]$ and $p\geq q$ ,$G_{p,q,t}(A, A, r, S)\geq G_{p,q,t}(A, B, r, S)$ , that is,(1.2)$A^{q-t+r}\geq\{A^{\frac{r}{2}}(A^{\frac{-t}{2}}BpA^{\frac{-t}{2})}s_{A^{\frac{r}{2}\}^{(p-})+r}}-\mapsto t+rtS$

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

30

The proof in [8] of Theorem is complicated and technical and also the proof in [3]is based on mean theoretic one. Here we show a simplified proof of Theorem which isan extension form of Theorem only using Theorem and the following Lemma F.
$\mathrm{G}$$\mathrm{H}$$\mathrm{G}$Lemma$\mathrm{F}$$\mathrm{F}$(Furuta lemma) [8]. Let $A>0$ and $B$ be an invertible operator. Then$(BAB^{*})^{\lambda}=BA^{\frac{1}{2}}(A^{\frac{1}{2}B^{*}B}A \frac{1}{2})^{\lambda}-1A\frac{1}{2}B^{*}$

holds for any real number .
$\lambda$-

Firstly we show a short proof of the inequality (1.2) of Theorem H. Secondly we show
a proof of the monotonicity of the function $G_{p,q,t}(A, B, r, S)$ of Theorem H. Lastly we give
three counterexamples and a conjecture related to Theorem and Theorem H.
$\mathrm{G}$Results on inequalities

2

Theorem H-i [9]. If $A\geq B\geq 0$ with $A>0$ , then for each $1\geq q\geq t\geq 0$ and $p\geq q$ ,(1.2)

$A^{q^{-}t+r} \geq\{A^{\frac{r}{2}}(A^{\frac{-t}{2}B^{p}A}\frac{-t}{2})SA^{\frac{r}{2}\}^{\frac{q-t+r}{(\mathrm{p}-t)s+r}}}$holds for $s\geq 1$ and $r\geq t$ .
Theorem H-i is proved as an immediate consequence of the following Theorem 1.
Theorem 1. Let

$S$

and

$S^{\beta_{0}}\geq(S^{\beta}2\tau\alpha_{0}s\Delta\rho \mathrm{n}_{)^{\frac{\beta_{0}}{\alpha_{0}+\beta_{0}}}}2$

be positive invertible operators on a Hilbert space such that
holds for fixedand. Then$T$$\alpha_{0}>0$(2.1)$S \frac{\beta}{2}T^{\alpha 0}S^{\frac{\beta}{2}}\geq(S^{\frac{\beta}{2}}T^{\alpha_{S}}\frac{\beta}{2})\underline{\alpha}\alpha\Omega_{\frac{+\beta}{+\beta}}$holds for any$\alpha\geq\alpha_{0}$and$\beta\geq\beta_{0}$.

Proof of Theorem 1. Applying (ii) of Theorem
we have(2.2)

Putting
(2.3)

Put
(2.4)

$\beta_{0}>0$$\mathrm{F}$to the hypothesis$S^{()}1+r_{1}\beta 0\geq\{S^{-}2(S2\tau^{\alpha}0S2)\overline{\alpha}0+\beta 0\beta_{\Delta^{r}\perp \mathrm{n}\mathrm{n}\infty}\beta\beta\beta \mathrm{R}\mathrm{p}\llcorner s2\}\underline{\beta}r_{1\frac{1+r_{1}}{p_{1}+r_{1}}}$$p_{1}= \frac{\alpha_{\mathrm{O}}+\beta_{0}}{\beta_{0}}\geq 1$

for anyin (2.2), we have.

$S^{(}1+r_{1})\beta_{0}\geq(S^{\frac{(1+r)}{2}\beta}T^{\alpha}\Delta 0s^{\underline{(r}_{2}})^{\alpha_{0}}+1+\perp)\beta\Delta(1+r\ovalbox{\tt\small REJECT}_{1})\beta(+r_{1})\beta 0$$\beta=(1+r_{1})\beta_{0}\geq\beta_{0}$in (2.3). Then we have$S^{\beta}\geq(S^{\frac{\beta}{2}}T^{\alpha_{\mathrm{O}}}s^{\frac{\beta}{2}})\overline{\alpha}0+\overline{\beta}L$for$\beta\geq\beta_{0}$.

$S^{\beta_{0}} \geq(S^{\beta}2T^{\alpha_{0}}s2)\mathrm{n}\beta \mathrm{n}\frac{\beta_{0}}{\alpha_{0}+\beta_{0}}$$p_{1}\geq 1$

and$r_{1}\geq 0$

.

,

31

(2.4) is equivalent to the following (2.5) by Lemma

(2.5)

for

$T^{\alpha_{0}}\leq(T^{\underline{\alpha}_{2}}S^{\beta}\mathrm{n}\Delta T^{\alpha_{2}}-)^{\overline{\alpha}_{0}}\alpha\Delta+\overline{\beta}$

Again applying (i) of Theorem(2.6)$\mathrm{F}$$\beta\geq\beta_{0}$$p_{2}= \frac{\alpha_{0}+\beta}{\alpha_{\mathrm{O}}}\geq 1$

(2.7)

.

to (2.5), we havefor any$T^{()}1+r_{2} \alpha 0\leq\{\tau^{\underline{\alpha}}2(\tau- \mathrm{n}_{S}\beta\tau 22)0+\beta\tau^{-}\}^{\frac{1+r}{p_{2}+r_{2}}}\mathrm{L}^{r\alpha}2\underline{\alpha}4\frac{\alpha}{\alpha}\mathrm{L}^{p}z\alpha p_{2}\Gamma 4$

PuttingPut$\mathrm{F}$$p_{2}\geq 1$

and$r_{2}\geq 0$.

in (2.6), we have.

$\tau^{(1+r2})\alpha 0\leq(T^{\frac{(1+r}{2}\mathrm{r}B\mathrm{R}})\alpha s\beta T^{\underline{(1}}2)+r)\alpha(1+r\ovalbox{\tt\small REJECT}(1+r_{2})\alpha)\alpha 0+\beta$$\alpha=(1+r_{2})\alpha_{0}\geq\alpha_{0}$(2.8)

in (2.7). Then we have$T^{\alpha} \leq(T^{\frac{\alpha}{2}}S\beta T\frac{\alpha}{2})^{\frac{\alpha}{\alpha+\beta}}$

Raise each side of (2.8) to the power
first inequality of the following (2.9)

for$\alpha\geq\alpha_{0}$$\frac{\alpha-\alpha_{0}}{\alpha}\in[0,1]$

and$\beta\geq\beta_{0}$.

by L\"owner-Heinz theorem, we have the$T^{\alpha-\alpha_{0}}\leq(\tau^{\frac{\alpha}{2}}s^{\beta}\tau^{\frac{\alpha}{2}})\underline{\alpha}_{\mathrm{R}^{\alpha}}-+\beta$(2.9)

by Lemma F.

$=T^{\frac{\alpha}{2}}S2 \rho(S^{e_{T}}2\alpha s^{e_{)^{\frac{\alpha-\alpha_{0}}{\alpha+\beta}-1}}e}2s2\tau\frac{\alpha}{2}$refining (2.9) and taking inverses of both sides, we obtain (2.1).
Proof of Theorem H-i. If $A\geq B\geq 0$ , then the following (2.10) holds(2.10)$A^{q+r}\geq(A^{\frac{r}{2}}B^{p}A^{\frac{r}{2})}pq\pm_{\frac{r}{r}}+$for $p\geq q,$ $q\in[0,1]$ and $r\geq 0$by (ii) of Theorem since $(1+r) \frac{p+r}{q+r}\geq p+r$ andin this case.
$(1.2)$In the case $t=0$ .
is valid by (2.10) in this case.
In the case $p=q=t\in[0,1]$ . Let. As $I\geq C\geq 0$ holds by L\"ownerHeinz theorem,
for $s\geq 1$ , that is, (1.2) holds in this case.
In the case $p>t>0$ . Put. Then we haveandby the hypothesis $A\geq B\geq 0$ . Put $\beta_{0}=t\in(0,1]$ and $\alpha_{0}=p-t>0$ .
Then
, and
$\mathrm{F}$$\frac{p+r}{q+r}\geq 1$$C=A^{\frac{-t}{2}}B^{t}A^{\frac{-t}{2}}$$A^{r}\geq A^{\frac{r}{2}}C^{s}A^{\frac{r}{2}}$$A^{\frac{t}{2}}X^{p-t}A^{\frac{t}{2}}=B^{p}$$X=(A^{\frac{-t}{2}B^{p}A^{\frac{-t}{2}}})^{\frac{1}{p-t}}$$A\geq(A^{\frac{t}{2}}X^{p-t}A^{\frac{t}{2})^{\frac{1}{\mathrm{p}}}}$$A\geq(A^{-\mathrm{n}_{X}}2\beta\alpha \mathrm{o}A^{\mathrm{n}\frac{1}{\alpha_{0}+\beta_{0}}}\beta 2)$$A^{\beta_{0}}\geq(A^{\beta}X^{\alpha}0A^{\beta}2)^{\frac{\beta_{0}}{\alpha_{0}+\beta_{0}}}\mathrm{n}_{2}\lrcorner 1$

holds by L\"ownew-Heinz theorem. Put $\alpha=(p-t)s$ and $\beta=r$ . Thenandhold since $s\geq 1$ and $r\geq t$ hold, so that Theorem 1 ensures the following inequality$\alpha\geq\alpha_{0}$$(A^{\frac{\beta}{2}}X^{\alpha}A \frac{\beta}{2})^{-}\alpha_{\mathit{1}}\alpha\frac{+\beta}{+\beta}\leq A^{\frac{\beta}{2}}X^{\alpha 0}A\frac{\beta}{2}$that is, we have$\{A^{\frac{r}{2}}(A\frac{-t}{2}B^{p}A^{\frac{-t}{2})^{s_{A^{\frac{r}{2}}}}\}^{R_{\frac{-t+r}{t)s+r}}}}(\mathrm{p}-$(2.11)$\leq A^{\frac{r}{2}}A^{\frac{-t}{2}BA}p\frac{-t}{2}A^{\frac{r}{2}}$.

,$\beta\geq\beta_{0}$32

Raising each side of (2.11) to the power
the first inequality of the following (2.12)$\frac{q-t+\Gamma}{p-t+r}\in[0,1]$

by L\"owner-Heinz theorem, we have$\{A^{\frac{r}{2}(A^{\frac{-t}{2}}B}pA^{\frac{-t}{2})A\}^{\frac{q-t+r}{(p-t)s+r}}}s\frac{r}{2}$(2.12)$\leq(A^{\frac{r-t}{2}B^{p}A}\frac{r-t}{2})^{\frac{q+r-t}{p+r-t}}$$\leq A^{q-t+r}$

and the last inequality holds by replacing by $r-t\geq 0$ in (2.10), so the proof of Theorem
H-i is complete.
$r$

Results on functions

3

Theorem H-f [9]. Let $A\geq B\geq 0$ with $A>0$ . For each $t\in[0,1],$ $q\geq 0$ and$p \geq\max\{q, t\}$ ,$G_{p,q,t}(A, B, r, S)=A^{\frac{-r}{2}\{A^{\frac{r}{2}}(B}A \frac{-t}{2}pA\frac{-t}{2})^{s}A^{\frac{r}{2}}\}^{\frac{q-t+r}{(\mathrm{p}-t)_{S}+r}}A^{\frac{-r}{2}}$is decreasing for $r\geq t$ and $s\geq 1$ .

Theorem H-f is proved as an immediate consequence of the following Theorem 2.

Theorem 2. Let

$S$

and

$S^{\beta_{0}}\geq(S^{\Delta}2T^{\alpha 0}S^{\mathrm{n}}2)^{\frac{\beta_{0}}{\alpha_{0}+\beta_{0}}}\beta\beta$

be positive invertible operators on a Hilbert space such that
andholds for fixed. Then for fixed$T$$\alpha_{0}>0$$\beta_{0}>0$$\delta\geq-\beta_{0\mathrm{z}}$$f( \alpha, \beta)=s\frac{-\beta}{2}(S^{\frac{\beta}{2}}\tau\alpha s^{\frac{\beta}{2}})^{\frac{\delta+\beta}{\alpha+\beta}}s\frac{-\beta}{2}$is a decreasing function of both$\alpha$and$\beta$for$\alpha\geq\max\{\delta, \alpha_{0}\}$and$\beta\geq\beta_{0}$.

Proof of Theorem 2.
(a) Proof of the result that$f(\alpha,\beta)$

is a decreasing function of$\alpha$for$\alpha\geq\max\{\delta, \alpha_{0}\}$.

The hypothesis in Theorem 2 ensures (3.1) in the same way as the proof of Theorem 1(3.1)$(T^{\frac{\alpha}{2}}s^{\beta}T^{\frac{\alpha}{2}})^{\frac{\alpha}{\alpha+\beta}}\geq T^{\alpha}$

for all$\alpha\geq\alpha_{0}$and$\beta\geq\beta_{0}$.

(3.1) yields the following (3.2) by L\"owner-Heinz theorem(3.2)$( \tau\frac{\alpha}{2}s^{\beta}\tau^{\frac{\alpha}{2}})\frac{u}{\alpha+\beta}\geq T^{u}$for all$\alpha\geq\alpha_{0},$$\beta\geq\beta_{0}$and any$u$such that

Then we have$g( \alpha)=(s2T\alpha_{S2})e\mathrm{p}\frac{\delta+\beta}{\alpha+\beta}$$=\{(s^{e_{T^{\alpha}}}2S2)^{\frac{\alpha+u+\beta}{\alpha+\beta}\}^{\frac{\delta+\beta}{\alpha+u+\beta}}}E$$= \{S^{\mathrm{g}}2T\frac{\alpha}{2}(T\frac{\alpha}{2}S^{\beta}\tau\frac{\alpha}{2})^{\frac{u}{\alpha+\beta}}\tau^{\frac{\alpha}{2}}s\frac{\beta}{2}\}^{\frac{\delta+\beta}{\alpha+u+\beta}}$$\geq\{S^{e\mathrm{g}p}2T^{\frac{\alpha}{2}\tau}u\tau^{\frac{\alpha}{2}s}2\}\overline{\alpha}+\delta+u+\overline{\beta}$$=(S^{\frac{\beta}{2}}T^{\alpha}+us^{\frac{\beta}{2}})^{\frac{\delta+\beta}{\alpha+u+\beta}}=g(\alpha+u)$

by Lemma$\mathrm{F}$$\alpha\geq u\geq 0$.

33

and the last inequality holds by (3.2) and L\"owner-Heinz theorem sinceholds by thefunction of for$\mathrm{h}\mathrm{y}\mathrm{p}_{\mathrm{o}\mathrm{t}}\mathrm{h}\mathrm{e}\mathrm{S}\mathrm{i}\prime \mathrm{S}$$\alpha$on$\alpha,$$\beta \mathrm{a}\mathrm{n}\mathrm{d}.\delta.\cdot$$\alpha\geq\max\{\delta, \alpha_{0}\}$(b) $Proo.f$ of the result thatHence$f(\alpha, \beta)=S^{\frac{-\beta}{2}}g(\alpha)s^{\frac{-\beta}{2}}$$\frac{\delta+\beta}{\alpha+u+\beta}\in[0,1]$

is a decreasing.

$f(\alpha, \beta)$

is a decreasing function of$\beta$for$\beta\geq\beta_{0}$.

By Lemma ,$\mathrm{F}$$f( \alpha, \beta)=S^{\frac{-\beta}{2}}(S^{\frac{\beta}{2}}\tau\alpha s^{\frac{\beta}{2}})^{\frac{\delta+\beta}{\alpha+\beta}}s\frac{-\beta}{2}$$= \tau\frac{\alpha}{2}(\tau^{\frac{\alpha}{2}}s^{\beta}\tau^{\frac{\alpha}{2})}\frac{\delta-\alpha}{\alpha+\beta}T^{\frac{\alpha}{2}}$and (3.1) is equivalent to the following (3.3) by Lemma(3.3)$S^{\beta} \geq(s^{\frac{\beta}{2}}\tau^{\alpha_{S}}\frac{\beta}{2})^{\frac{\beta}{\alpha+\beta}}$

for all$\mathrm{F}$$\alpha\geq\alpha_{0}$and$\beta\geq\beta_{0}$.

(3.3) yields the following (3.4) by L\"owner-Heinz theorem(3.4)$S^{v}\geq(S^{\frac{\beta}{2}}T^{\alpha}s^{\frac{\beta}{2}})^{\frac{v}{\alpha+\beta}}$

for all$\alpha\geq\alpha_{0},$$\beta\geq\beta_{0}$and any$v$

such that$\beta\geq v\geq 0$.

Then we have$h( \beta)=(T^{\frac{\alpha}{2}}S\beta\tau\frac{\alpha}{2})\frac{\delta-\alpha}{\alpha+\beta}$$= \{(T^{\frac{\alpha}{2}}S\beta T^{\frac{\alpha}{2})}\frac{\alpha+\beta+v}{\alpha+\beta}\}^{\frac{\delta-\alpha}{\alpha+\beta+v}}$$=\{T^{\frac{\alpha}{2}}s^{E}2(s2\tau\alpha s2)^{\frac{v}{\alpha+\beta}s^{\rho}}\mathrm{p}\rho 2\tau^{\frac{\alpha}{2}\}^{\frac{\mathit{6}-\alpha}{\alpha+\beta+v}}}$by Lemma$\mathrm{F}$$\geq\{\tau\frac{\alpha}{2}s\frac{\beta}{2}s^{v}s^{\frac{\beta}{2}}\tau\frac{\alpha}{2}\}^{\frac{\delta-\alpha}{\alpha+\beta+v}}$$=(T^{\frac{\alpha}{2}}S^{\beta+}v \tau\frac{\alpha}{2})^{\frac{\delta-\alpha}{\alpha+\beta+v}}=h(\beta+v)$

and the last inequality holds by (3.4) and L\"owner-Heinz theorem since
is a decreasing function ofand taking inverses. Hence$\frac{\delta-\alpha}{\alpha+\beta+v}\in[-1,0]$$f( \alpha, \beta)=\tau\frac{\alpha}{2}h(\beta)T^{\frac{\alpha}{2}}$$\beta$for$\beta\geq\beta_{0}$.

Consequently we have finished a proof of Theorem 2 by (a) and (b).
.
Proof of Theorem H-f. We consider the case $p>t>0$ . Put$A\geqB\geq
0$.
by the hypothesis
andThen we have
, so that
Put $\beta_{0}=t.\in(0,1]$ and $\alpha_{0}=p-t>0$ . Then
$X=(A^{\frac{-t}{2}}B^{p}A \frac{-t}{2})^{\frac{1}{p-t}}$$A^{\frac{t}{2}}X^{p-t}A^{\frac{t}{2}}=B^{p}$$A\geq(A^{\frac{t}{2}}X^{p-t}A^{\frac{t}{2})^{\frac{1}{p}}}$$A \geq(A^{\beta}2X^{\alpha_{0}}\mathrm{n}\underline{\beta}\Omega\frac{1}{\alpha_{0}+\beta_{0}}A2)$$A^{\beta_{0}}\geq(A^{\underline{\beta}_{4}}2x^{\alpha}\mathrm{o}A2)^{\frac{\beta_{0}}{\alpha_{0}+\beta_{0}}}\underline{\beta}\mathfrak{g}$holds by L\"ownew-Heinz theorem. Put $\alpha=(p-t)S,\beta=r$ and $\delta=q-t$ . The hypothesis$q\geq 0$ and $p \geq\max\{q, t\}$ in Theorem H-f satisp the conditions required on. Applying Theorem
andand in Theorem 2, that is,2,$t\in(\mathrm{O}, 1],$$\delta$$\alpha,$$\delta\geq-\beta_{0},$$\alpha\geq\max\{\alpha_{0}, \delta\}$$\beta\geq\beta_{0}$$f( \alpha, \beta)=A\frac{-\beta}{2}(A\frac{\beta}{2}x^{\alpha}A^{\frac{\beta}{2}})\frac{\delta+\beta}{\alpha+\beta}A^{\frac{-\beta}{2}}$$=A^{\frac{-r}{2}\{A^{\frac{r}{2}}}(A^{\frac{-t}{2}}B^{p}A^{\frac{-t}{2})^{S}\}^{\mathrm{R}_{\frac{t+r}{t)s+r}}^{-}}}A \frac{r}{2}(p-A^{\frac{-r}{2}}$$=G_{p)q},t(A, B, r, S)$$\beta$34

is decreasing for $r\geq t$ and $s\geq 1$ , so the proof in the case $p>t>0$ is complete.
In the case $t=0$ , Theorem H-f easily follows by [7, Theorem 3].
In the case $p=t\geq q\geq 0$ . Let. Then $I\geq C\geq 0$ by L\"ownerHeinz theorem, so that
holds since $I\geq C\geq 0$ and $s\geq 1$ , and again by
L\"owner-Heinz theorem$C=A^{\frac{-t}{2}}B^{t}A^{\frac{-t}{2}}$$A^{r}\geq A^{\frac{r}{2}}C^{s}A^{\frac{r}{2}}$(3.5)$A^{u}\geq(A^{\frac{r}{2}}C^{S}A^{\frac{r}{2}})^{\frac{u}{r}}$for $r\geq u\geq 0$ .

Then we obtain$G_{t,q,t}(A, B, r, S)=A^{\frac{-r}{2}()^{\frac{q-t+r}{r}A}}A^{\frac{r}{2}}o^{S}A^{\frac{r}{2}} \frac{-r}{2}$by Lemma$=C^{\frac{s}{2}}(c^{\frac{s}{2}}A^{r}c \frac{s}{2})g_{\frac{-t}{r}c^{\frac{s}{2}}}$$\mathrm{F}$$=C^{\frac{s}{2}} \{(C^{\frac{s}{2}A^{r}c}\frac{s}{2})\frac{r+u}{r}\}r\mathrm{A}_{\frac{-t}{+u}}c\frac{s}{2}$(3.6)$=C^{\frac{s}{2}} \{C\frac{s}{2}A^{\frac{r}{2}}(A^{\frac{r}{2}}C^{s_{A^{\frac{r}{2}}}})^{\frac{u}{r}A^{\frac{r}{2}c\frac{s}{2}}}\}r+C^{\frac{s}{2}}\mathrm{L}_{\frac{t}{u}}-$by Lemma$\mathrm{F}$$\geq C^{\frac{s}{2}}\{C^{\frac{\theta}{2}}A\frac{r}{2}A^{u}A^{\frac{r}{2}}c\frac{s}{2}\}^{L_{\frac{t}{u}}^{-}}r+c\frac{s}{2}$$=C^{\frac{s}{2}}(C \frac{s}{2}A^{r+}uC^{\frac{s}{2})}r\mathrm{L}^{-}+\frac{t}{u}c\frac{s}{2}$$=A^{\frac{-(C+u)}{2}(A^{\frac{r+u}{2}C^{S}A^{\frac{r+u}{2}}}})^{\frac{q-t+r+u}{r+u}}A^{\frac{-(r+u)}{2}}=c_{t,q,t}(A, B, r+u, s)$and the last inequality holds by (3.5) and L\"owner-Heinz theorem
and$G_{t,q,t}(A,
B,
r,
s)$
$r\geqt$taking inverses. Consequently
is a decreasing function of bothand$s\geq 1$ because $G_{t,q,t}(A, B, r, s)$ is decreasing$s\geq1$ofby (3.6) since $I\geq C\geq 0$ .
Whence the proof of Theorem H-f is complete.
$\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{C}\mathrm{e}\frac{q-t}{r+u}\in[-1,0]$4Best possibility and counterexamples

We discuss best possibility of (1.1) in Theorem
related to Theorem G.

$\mathrm{G}$and also we cite counterexamples

Counterexample 1. There exists a counterexample to (1.1) of Theorem if we replace$A\geq B$ in Theorem
by $\log A\geq\log B$ . Let $p=2,$ $t=1,$ $r=2$ and $s=2$ . Thenand satisfy the condition in Theorem G. Take and as$\mathrm{G}$$\mathrm{G}$$p,$ $t,$ $r$

$A$

$s$

$B$

$A=$ , $B=$
Then it turns out that $\log A\geq\log B$ holds sincemonotone, but$A\not\geq B$$\geq$ and$\log t$is operator

holds and$A^{1-t+r}- \{A^{\frac{r}{2}}(A^{\frac{-t}{2}B^{p}}A\frac{-t}{2})sA\frac{r}{2}\}^{\frac{1-\ell+r}{(p-t)s+r}}=$,35

so that the eigenvalues of
125.5643.. ., therefore
Hence we can’t replace $A\geq B$ in Theorem$A\geq B\geq 0$ .

$A^{1-t+r}-\{A^{\frac{r}{2}}(A^{\frac{-t}{2}B^{p}A^{\frac{-t}{2})^{S}}}A^{\frac{r}{2}}\}^{\frac{1-t+r}{(\mathrm{p}-t)s+r}}$

are -0.5563. . . and.

$A^{1-t+r} \not\geq\{A^{\frac{r}{2}}(A^{\frac{-t}{2}}BpA^{\frac{-t}{2}})sA^{\frac{r}{2}}\}\frac{1-t+r}{(p-t)s+r}$$\mathrm{G}$by $\log A\geq\log B$ , which is weaker thanCounterexample 2. There exists a counterexample to (1.1) of Theorem if anddon’t satisfy the condition $r\geq t$ . Let $p=2,$ $s=2,$ $t=1\in[0,1]$ and. Then.
Take and as$\mathrm{G}$$r\not\geq t$$r= \frac{1}{2}$

$A$

$t$

$r$

$B$

$A=$ ,$B=.\cdot$

Then $A\geq B\geq 0$ and$A^{1-t+r}- \{A^{\frac{r}{2}}(A^{\frac{-t}{2}}B^{p}A^{\frac{-t}{2})^{s_{A\}^{\frac{1-t+r}{(p-t)s+\gamma}}}}}\frac{r}{2}=$so that the eigenvalues of
2.1205 , therefore$\cdots$$A^{1-t+r}- \{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)s+r}}$,

are $-0.2523\cdots$ and.

$A^{1-t+r} \not\geq\{A^{\frac{r}{2}}(A\frac{-t}{2}BpA^{\frac{-t}{2})^{s}\}}A^{\frac{r}{2}}\frac{1-t+r}{(\mathrm{p}-t)s+r}$Counterexample 3. There exists a counterexample to (1.1) of Theorem if don’tsatisfy the condition $t\in[0,1]$ . Let $t=1.2\not\in[0,1],p=2,$ $r=2,$ $s=2$ . Then $r\geq t$ . Takeand as$\mathrm{G}$

$A$

$t$

$B$

$A=$

,

$B=$

.

Then $A\geq B\geq 0$ and$A^{1-t+r}- \{A^{\frac{r}{2}}(A^{\frac{-t}{2}}BpA\frac{-t}{2})sA^{\frac{r}{2}}\}^{\frac{1-t+r}{(p-t)s+r}}--$

so that the eigenvalues of
89.1646 , therefore$\cdots$$A^{1-t+r}- \{A^{\frac{r}{2}}(A\frac{-t}{2}B^{p}A^{\frac{-t}{2}})SA^{\frac{r}{2}}\}^{\frac{1-t+r}{(\mathrm{p}-t)s+\Gamma}}$$A^{1-t+r}\not\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}}$

are $-0.5084\cdots$ and.

Remark. We remark the following result. By using his skillful and excellent technique
as almost same as one in [13], K. Tanahashi [14] asserts thatof the right handside of (1.1) of Theorem
is best possible in the sense of the following:
does not hold for any $\alpha>1$ in Theorem G.
$\frac{1-t+r}{(p-t)s+r}$$A^{(1-t)\alpha}+r\geq$$\mathrm{G}$$\{A^{\frac{r}{2}}(A^{\frac{-t}{2}}B^{p}A^{\frac{-t}{2}})SA^{\frac{r}{2}}\}^{\frac{(1-t+r)\alpha}{(\mathrm{p}-t)s+r}}$

At the end of this section, we cite the following conjecture related to Theorem
Theorem G.
Conjecture.

There exists a counterexample to Theorem

If $t=0$ and $r<0$ in Theorem$\mathrm{G}$$G$$\mathrm{H}$andin generctl for any $r<t$ .

, we have already obtained a counterexample.

36

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