数理解析研究所講究録
第 1596 巻 2008 年 91-98

91

Reverse of Bebiano-Lemos-Provid\^encia inequality
and Complementary Furuta inequality
大阪府立能勢高校 松本(Akemi Matsumoto)

明美

Nose Senior Highschool
茨城大学工学部 中本 律男 (Rituo Nakamoto)Department of Engineering, Ibaraki University
大阪教育大学 藤井 正俊 (Masatoshi Fujii)
Osaka Kyoiku University

We give a simultaneous extension of $Bebiano- Lemos- Provid\hat{e}ncia$ inequality and ArakiCordes one: Let and be positive operators. Then for each $r\geq 0$
$A$

$B$

for$\Vert A^{r\iota_{B^{t}A^{rt}}}++\Vert\leq\Vert A^{r}Z(A^{\dot{f}}B^{\epsilon}A^{i})^{\frac{t}{}}A^{r}l||$In succession, we prove a reverse inequality: Letfor each $r\geq 0$$A$

and$B$$s\geq t\geq 0$.

be positive operators. Then

for $t\geq s\geq r\geq 0$ .
Furthermore, we discuss reverse of generalized BLP inequality in a general setting, inwhich we point out that the restriction $t\geq s\geq r$ in the above is quite reasonable.
$\Vert A^{rt}+B^{t}A^{rt}+\Vert\geq\Vert z.z\Vert$

1

Introduction

Let be a bounded linear operator acting on a Hilbert spase $H$ . Then is positive,denoted by $A\geq 0$ , if $(Ax,x)\geq 0$ for all $x\in H$ . In particular, $A>0$ means that isinvertible and positive.
Recently, Bebian($\succ Lemos$-Provid\^encia showed an interesting norm inequality (BLP):
Let and be positive operators. Then$A$

$A$

$A$

$A$

$B$

(1)

$\Vert A^{1t}+B^{t}A^{1\ell}+\Vert\leq\Vert A^{1}r(A;_{B^{\epsilon}A^{*}}z)^{\frac{t}{*}}A^{1}\pi\Vert$If we delete(2)$A$

for$s\geq t\geq 0$.

} on both sides in (1), we have the Araki-Cordes inequality$\Vert A^{p}B^{p}A^{p}\Vert\leq\Vert$ABA$\Vert^{p}$for$0\leq p\leq 1$

(AC)

.

In this sense, BLP inequality (1) is regarded as an extension of AC inequality (2).
In this note, we first give a simultaneous extension of (1) and (2) as follows generalized
BLP inequalty (GBLP): Let and be positive operators. Then for each $r\geq 0$
$A$

(3)

$B$$\Vert A^{rt}+B^{t}A^{rt}+\Vert\leq\Vert A(AB^{\epsilon}A)^{\frac{t}{}}A^{r}\Vert$for

Next we discuss a reverse inequaltiy of (3)(R-GBLP): Letators. Then for each $r\geq 0$ ,(4)$\Vert A^{r\iota_{B^{t}A^{rt}}}++\Vert\geq\Vert rz\Vert$for$s\geq t\geq 0$$A$.

and $B$ be positive oper-$t\geq s\geq r$

and $s>0$ .

As a corolary, the case $r=1$ in (4) coressponds to the reverse of BLP(R-BLP) inequality
and the case $r=0$ in (4) corresponds to the reverse of AC inequality (2)(R-AC).

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2A generalization of BLP inequality

In this section, we generalize BLP inequality (1). For this, we cite Furuta inequality: Let$A\geq B$ forand be positive operators. Then for each $r\geq 0$ .
$A$

$B$

for

$(A^{r}zB^{p}A^{r}f)^{\frac{1}{q}}\leq(A^{\frac{r}{2}}A^{p}A^{r1}z)q$$p\geq 0$,

with$q\geq 1$$(1+r)q\geq p+r$ .

We prove the following theorem including BLP inequality and AC one.

Theorem. 1. Let$A$

andbe positive operators. Then for each $r\geq 0$

$B$

for

$\Vert A^{rt}+B^{t}A^{rt}+||\leq\Vert A^{r}I(A^{i}B^{l}A^{\dot{z}})^{\frac{t}{}}A^{r}B\Vert$$s\geq t\geq 0$.

Proof. Since this inequality is AC inequality when $r=0$ , we may assume $r>0$ . It suffices
to prove thatfor$A^{r}z(A^{\dot{f}}B^{\epsilon}A\overline{2})^{\frac{t}{}}A^{r}z\leq 1\Rightarrow A^{\underline{r}\pm}2B^{t}A^{r}t\dotplus\leq 1$

If$A^{r}5(A^{\dot{z}}B^{s}A^{\dot{z}})^{\frac{t}{}}A^{r}5\leq 1$$A_{1}=A^{-r}$,, then$(A^{\frac{}{2}}B^{\epsilon}A^{\frac{}{2}})\underline{t}\leq A^{-r}$$B_{1}=(A\dot{:}B^{l}A\dot{\tau})\underline{t}$$s\geq t\geq 0$.

. We put$r_{1}= \frac{s}{r}$and$p=q= \frac{s}{t}$.

Then$A_{1}\geq B_{1}\geq 0$,$r_{1}\geq 0$,$p=q\geq 1$

and$(1+r_{1})q\geq p+r_{1}$,so that Furuta inequality implies号$R\leq(A_{1}^{r}+A_{1}^{p}A_{1}^{r}+)^{\frac{1}{q}}$,$(A_{1}^{f_{r}}B_{1}^{\frac{}{t}}A_{1}^{f_{r}})^{\underline{t}}\cdot\leq(A_{1}\star_{r}A_{\overline{1}}A_{1^{\dot{\Gamma r}}})^{\frac{t}{}}$.

($A_{1}$$B_{1}^{p}A_{1}^{\text{号}}$)

that is,Since we have
$A=A_{1}^{-\frac{1}{r}},$$B=(2\cdot A_{1^{2r}}^{\perp}B_{1}^{\frac}A_{1}f_{r}.$,it follows that$A^{r}+B^{t}A^{r*}+-1^{rt}+\dot{r}\dotplus$$\leq A_{1}^{-\#_{r}.-\#_{r=A_{1}^{-+\cdot\pm rt}}}(A_{1}^{-\#}A_{1^{t}}^{l^{l}}A_{1}^{-\pi})^{\frac{t}{}A_{1}}rA_{1’}^{r}A_{1}^{-+}rtrtrtr=I$.
口Remark. It is obvious that the casethe case $r=0$ is AC one.

$r=_{\mathfrak{l}}1$

in Theorem 1 is just the BLP inequality and93

3

Reverse inequalities

In this section, we show a reverse inequality of Theorem 1, in which we use the well-known
formula(Lemma of Furuta),(5)

for$(A^{\xi_{BA^{1}}\iota}r)^{p}=A^{\frac{1}{2}}B^{\frac{1}{2}}(BAB^{\frac{1}{2}})^{p-1}BA2$$p\geq 1$

and Lowner-Heinz inequality (LH)
(6)

$A\geq B\geq 0$

Theorem. 2. Let
(7)

$A$

implies$A^{p}\geq B^{p}$for all $0\leq p\leq 1$ .

and $B$ be positive operators. Then for each $r\geq 0$

for$\Vert A^{r\iota_{B^{t}A^{r}}}+\dotplus|\{\geq\Vert A^{\acute{z}}()^{\underline{t}}\cdot A^{r}\Vert$$t\geq s\geq r$

and $s>0$ .

Proof. It suffices to prove that$A^{r\iota_{B^{t}A^{r}}}+\dotplus\leq 1\Rightarrow A^{r}5(A\overline{2}B^{\cdot}A^{i})^{\frac{t}{}}A^{r}f\leq 1$

SuPpose that $[ \frac{t}{s}]=2l-1$ or$2l$

are obtained by (5).

for$t\geq s\geq r$

ud $s>0$ .

for some natural number . Then following equations$l$$A^{r}F(A B^{l}A^{\dot{f}})^{\frac{t}{}}A^{r}f$$=A^{r}\dotplus f\#.i-\perp$$=A\not\simeq B^{\epsilon}A\overline{2}(A^{8}B^{\epsilon}A\dot{\pi})^{\frac{t}{}-2}A\dot{\tau}B^{\iota}A^{r}\dotplus$$=A^{r}\dotplus:\cdot\underline{t}\dot{\overline{2}}$宇$=A^{r}\dotplus(B^{l}A^{\epsilon})B^{l}A^{\dot{q}}(AB^{\epsilon}A)^{-4}\underline{t}Ai_{B^{l}(A^{t}B^{l})A^{r}}\dotplus$$=A^{r}\dotplus(B^{l}A^{\epsilon})(B^{\epsilon}A^{\epsilon})B\#(B^{\xi}A^{\epsilon}B^{i-5})^{\underline{t}}\cdot B^{i}(A^{\epsilon}B^{\cdot})(A^{\iota}B^{\iota})A^{r}\dotplus$$=A^{r} \dotplus\frac{(l-1)- tim\infty}{(B^{l}A^{l})\cdots(B^{l}A^{l})}B\dot{z}(B^{\dot{f}}A^{\epsilon}B^{\S-(2l-1):^{\frac{(l-1.)- tim\infty}{(A^{\epsilon}B^{\epsilon})\cdot\cdot(A^{\epsilon}B)}}})\underline{\cdot}BA$

宇Since $t\geq s\geq r\geq 0$ , we have
$0 \leq\frac{s}{r+t}\leq 1,0\leq\frac{s}{t}\leq 1,0\leq\frac{t(r+s)-2(l-h)rs}{t(r+t)}\leq 1$

for $h=1,2,$

and$0 \leq\frac{t(s-r)+2(l-k)rs}{t(r+t)}\leq 1$

We suppose that$A^{\underline{r}\pm\underline{t}}2B^{t}A^{r}+\leq 1$

for $k=0,1,$$\cdots$, that is,$B^{t}\leq A^{-(r+t)}$$\mathfrak{W}d$$A^{(r+t)}\leq B^{-t}$.

, $l-1$ .

$\cdots,$$l$

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So the L\"owner-Heinz inequality (6) implies that$A^{\epsilon}\leq B^{-\frac{\epsilon t}{r+t}}$,$B^{\delta}\leq A^{-\frac{(r+t)}{t}}$,for $h=1,2,$$B^{\frac{t(r+\cdot)-2(l-h)r*}{(r+)}}\leq A^{-\frac{(r+\epsilon)-2(l-h)r}{t}}$$\cdots l$

and$A^{\frac{t(\cdot-r)+2(l-k)r}{t}}\leq B^{-m_{r+t}}t\cdot-r+2l-kr*$

for $k=0,1,$ $\cdots l-1$ .

Now we assume that $[ \frac{t}{s}]=2l-1$ Ior some narural number . Snce $0 \leq\frac{t}{s}-(2l-1)\leq 1$ ,$l$

we have$A^{r}\pi(A^{\dot{f}}B^{l}A;)^{\frac{t}{l}}A^{r}r$$(l-1)$ -times$(l-1)- tim\infty$$=A^{r}\dotplus\overline{(B^{l}A^{\epsilon})\cdots(B^{\epsilon}A^{l})(B^{\delta}A^{\epsilon})}B^{\frac{l}{2}}(B^{i}A^{l}B^{i})^{\frac{t}{\iota}-(2l-1)}B^{i}\overline{(A^{\epsilon}B^{\epsilon})(A^{l}B^{l})\cdots(A^{\epsilon}B^{l})}$A$2\perp$$\leq A^{r}\dotplus(B^{\epsilon}A^{l})\cdots(B^{\epsilon}A^{\cdot})(B^{\epsilon}A^{l})B;(B;B^{-\frac{l}{r+t}}Bw)^{\frac{t}{l}-(2l-1)}B^{l}\tau(A^{\epsilon}B^{\iota})(A^{\iota}B^{l})\cdots(A^{\iota}B^{l})A^{r}\dotplus$$=A^{r}\dotplus(B^{\epsilon}A^{\delta})\cdots(B^{l}A^{\epsilon})(BA^{l})B^{\frac{t(\cdot+r)-2(l-1)r}{r+t}}(A^{\delta}B^{\epsilon})(A^{\epsilon}B^{*})\cdots(A^{\epsilon}B^{\epsilon})A^{r}\dotplus$$(l-2)- time\epsilon$$(l-2)-tim\epsilon s$$=A^{r.\sim}\dotplus\tilde{(B^{l}A^{\epsilon})\cdots(B^{l}A^{l})}B^{\epsilon}A^{\iota}B^{\frac{t(\cdot+r)-2(l-1)r}{r+t}}A^{*}B^{l}(A^{l}B^{l})\cdots(A^{\epsilon}B^{l})A^{\underline{r}\pm}2^{-}$$\leq A^{r}\dotplus(B^{\epsilon}A^{\cdot})\cdots(B^{\epsilon}A^{\epsilon})B^{\epsilon}A^{\frac{t(--r)+2(l-1)rl}{t}}B^{\epsilon}(A^{\cdot}B^{*})\cdots(A^{\theta}B^{\epsilon})A^{r}\dotplus$$\leq A^{r}\dotplus(B^{\cdot}A^{\epsilon})\cdots(B^{\epsilon}A^{l})B^{\frac{t(\iota+r)-2(l-2)r}{r+t}}(A^{\cdot}B^{l})\cdots(A^{\epsilon}B^{\epsilon})A^{r}\dotplus$$\leq A^{r}B^{\epsilon}A^{1-r+2}B^{\epsilon}A^{r}R_{t}^{l-l-1r}$$\leq A^{r}\dotplus B\frac{t\iota+rt}{r+}A$宇$\leq A^{r}\dotplus A^{-(r+\epsilon)}A^{r}\dotplus$$=I$.

On the other hand, we assume that [for some natural number . Since,
similarly we have the follwing, in which the first equality is ensured in the first Paragraphs.
$\frac{t}{s}1=2l$$l$$0\leq\underline{t}-2l\leq 1$$A^{r}B(A^{i}B^{\cdot}A^{l}z)^{\frac}A^{r}B$$=A^{\underline{r}\pm}BA^{*}B)^{-(2l-1)}BA^{r}$$=A^{r} \dotplus\frac{(l-1)- tim\infty}{(B^{l}A^{l})\cdots(B^{l}A^{l})(B^{l}A^{l})}.i..\prec$$\leq A^{r}\dotplus(BA^{\epsilon})\cdots(B^{l}A^{l})(B^{l}A^{f})BA^{\dot{f}}(A^{\iota}2A^{-arrow r+\lrcorner}A^{\iota_{-2l}}A^{i}B^{l}(A^{\epsilon}B)(A^{\epsilon}B^{\epsilon})\cdots$$=A^{r}\dotplus l\dotplus$$\leq A^{r}\dotplus(B^{l}A^{\epsilon})\cdots(B^{\epsilon}A^{\epsilon})(B^{\epsilon}A^{\cdot})B^{\epsilon}B^{-\frac{t(-r1+2lr}{r+}B^{\epsilon}(A’ B^{\partial})(AB^{\iota})}\cdots(A’ B^{\epsilon})A\not\simeq$$=A^{\text{甲_{}B^{l}A^{\epsilon}B}^{\frac{(l-2)\sim\lim u}{(B^{l}A^{l})\cdots(B^{l}A^{l})}}\frac{(r+\cdot)-2(l-1)r}{r+l}A^{l}B^{e^{\frac{(l-2)- tim\infty}{(A^{l}B^{l})\cdots(A^{\epsilon}B^{l})}}}A^{r}}\dotplus$$\leq I$.

(A$B^{\epsilon}$)$A^{r}\dotplus$

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Hence the proof is complete.

4

口Complementary Furuta inequality

In this section, we consider R-GBLP, in which Kamei’s theorem(Theorem K) on complements of Furuta inequality corresponds to our result. So now recall it due to Kamei.

Theorem K. If $A\geq B>0$ , then for(8)$0<p \leq\frac{1}{2}$for$A^{t}\natural_{\frac{2-}{p-t}}B^{p}\leq A^{2p}$$0\leq t\leq p$

and $for-\leq p\leq 1$(9)for$A^{t}\natural_{\frac{1-t}{p-}}B^{p}\leq A$

Here$\natural_{q}$for$q\not\in[0,1]$$0\leq t\leq p$.

has been used as$A\natural_{q}B:=A^{\frac{1}{2}}(A^{-\frac{1}{2}}BA^{-:})^{q}A^{1}f$

for

$A,$

$B>0$ .

First we prove the folowing Theorem.
Theorem. 3. Let $A,$ $B\geq 0$ and $0<p\leq 1$ . Then(10)for all$||A^{1}\dotplus B^{1+\iota}A^{1}\dotplus\Vert^{\frac{p+}{p(1+\cdot)}}$$s\geq 0$$||A^{1}l(A^{\dot{f}}B^{p+\epsilon}A^{\dot{z}})^{\frac{1}{p}}A^{1}I\Vert$$\geq$with $s\geq 1-2p$ .

Proof. It suffices to show that(11)

$B^{1+\epsilon}\leq A^{-(1+s)}\Rightarrow A^{1}(A^{i}B^{p+\epsilon}A^{\dot{\pi}})^{\frac{1}{p}}A^{1}I\leq 1$

for $0<p\leq 1$ and $s\geq 0$ with $s\geq 1-2p$ . So we put$A_{1}=A^{-(1+\epsilon)},$$B_{1}=B^{1+\epsilon}$.

Then (11) is rephrased as.

$A_{1}\geq B_{1}>0\Rightarrow A_{1}^{1+}\wedge\natural\iota pB_{1}^{R}1+\cdot\leq A_{1}$for $0<p\leq 1$ and $s\geq 0$ with $s\geq 1-2p$ . $Mor\infty ver$ if we replace$t_{1}= \frac{s}{1+s},$

then we have
, andfollowing equivalent expression:$\frac{1-t_{1}}{p_{1}-l_{1}}=\frac{1}{p}$$\leq p_{1}(\leq 1)$$p_{1}= \frac{p.+s}{1+s}$

if and only if $1-2p\leq s$ , so that (11) has the$A_{1}\geq B_{1}>0\Rightarrow A_{1}^{t_{1}}\natural_{\frac{1}{p_{1}}}-\wedge-t_{1}B_{1}^{p}\leq A_{1}$Fortunately, since}$\leq p_{1}\leq 1$,

for$0\leq t_{1}<p_{1}$.

, it has been ensured by Theorem A due to Kamei.

口96

Next we show a reserve inequality of BLP inequality (R-BLP) is obtained as colloaryof Teorem 3.
$\frac{\frac{l}{tt}}{t+2}\leq sho1dsforarbitraryt>0,sothatTheorem3isProofofR-BLPWeputp=fort\geq s\geq 0.Thenwehave1-2p\leq sifandonlyif$$\frac{t}{t+2}\leq s.Sinces\geq 1isassumed,$applicable.
Now we take3 imply that

$B_{1}=B^{\underline{1}\pm\underline{t}}t$i.e.,Then Araki-Cordes inequality and Theorem$B=B_{1}^{1\overline{+t}}\lrcorner$$\Vert A^{1\iota_{B_{1}^{t}A^{1}}}+\dotplus\Vert\geq\Vert A^{1}\dotplus B^{\frac{(1+\cdot)}{11+t}}A^{1}\dotplus||^{!\perp}1+lt=||A^{1}\dotplus B^{1+\epsilon}A^{1}\dotplus\Vert^{\frac{p+}{p(1+\cdot)}}$$\geq\Vert A^{1}f(A\dot{\not\supset}B^{p+\epsilon}A^{i})^{\frac{1}{p}}A^{1}f\Vert=\Vert A^{1}B(A^{i}B_{1}^{l}A^{i})^{\frac{t}{}}A^{1}l\Vert$,as des 辻 ed.

R-BLP is generalized a bit as follows:
Corollary. For $A,$ $B>0$ and $r\geq 0$(12)$\Vert A$

宇$B^{t}A^{r}+\Vert\geq\Vert A^{r}\pi(A^{\frac{l}{2}}B^{\epsilon}A\#)^{\underline{t}}\cdot A^{r}B\Vert$holds for all $t\geq s\geq r$ .

Proof. It is proved by applying R-BLP to$A_{1}=A^{r},$$B_{1}=B^{f}$and$t_{1}= \frac{t}{r},$$s_{1}=-i$.

$\square$Finally we consider a reverse inequality of generalized BLP inequality which corresponds to another Kamei’s complement (7): If $A\geq B>0$ , then for$0<p \leq\frac{1}{2}$$A^{t}\natural_{\frac{2-t}{p-}}B^{p}\leq A^{2p}$Theorem. 4. Let $A,B\geq 0$ and(13)$0<p \leq\frac{1}{2}$$\Vert 2^{-}\Vert 2p+\ovalbox{\tt\small REJECT}_{i_{+}}$$\geq$

for$0\leq t<p$.

. Then$\Vert A^{p+i}(A^{\dot{f}}B^{p+\epsilon}A^{\dot{f}})-2\Delta_{-A^{p+\#}}p\Vert$

for all $0\leq s\leq 1-2p$ .
Proof. A proof is quite vimilar to that of Theorem 3. We put$A_{1}=A^{-(1+\epsilon)},$$B_{1}=B^{1+\epsilon};t_{1}= \frac{s}{1+s},$$p_{1}= \frac{p+s}{1+s}$.

Then (7) gives us that
$A_{1}\geq B_{1}>0\Rightarrow A_{1}^{t_{1}}\natural_{p_{11}}\underline{2}A_{-\neq}^{-}B_{1}^{p_{1}}\leq A_{1}^{2_{P1}}$

for$0 \leq t_{1}<p_{1}\leq\frac{1}{2}$,, so that
$A^{-(1+\epsilon)}\geq B^{1+\epsilon}\Rightarrow A^{-\epsilon}\natural_{z1_{-}}2pB^{p+s}\leq A^{-2(p+\epsilon)}$for $0\leq s\leq 1-2p$ . Obviously it,implies the desired nom inequality (12).

口

975Remarks

We first consider a relation between Furuta inequality and Theorem K. Furuta inequality
has the following representation by -geometric mean : For $A\geq B\geq 0$$\alpha$$A^{t}\#_{\frac{1-t}{p-t}}B^{p}\leq$

where$A\#\alpha B$

for$0\leq\alpha\leq 1$$\#_{\alpha}$$A$

for $p\geq 1,$ $t\leq 0$

is defined by

As a consequenoe we have relations among the inequalitys discussed abave.

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