数理解析研究所講究録1080 巻 1999 年 129-136

129Complementary inequalities of the Furuta inequality

大阪教育大学
前橋工科大学
大阪府立東豊中高校

藤井正俊 (Masatoshi
亀井栄三郎 (Eizaburo
松本明美 (Akemi$\mathrm{F}\mathrm{u}\mathrm{j}\mathrm{i}\mathrm{i}^{*}$)$\mathrm{K}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{i}^{**}$))$\mathrm{M}\mathrm{a}\mathrm{t}\mathrm{u}\mathrm{m}\mathrm{o}\mathrm{t}_{0^{***}}$

ABSTRACT. As a continuation of our preceding note, we discuss inequalities on the complementary domain of the Furuta inequality. For
positive operators $A\geq B>0$ , it is shown that,$A^{t}\mathfrak{g}_{\frac{\delta-t}{\mathrm{p}-t}}B^{p}\geq A^{\delta}\geq B^{\delta}\geq B^{t}\mathfrak{y}_{\frac{\delta-t}{p-t}}A^{p}$for $0\leq\delta\leq t<p\leq 1$ . This inequality is opposite to the inequality in[12].

1. Introduction. Throughout this note, a capital letter means a bounded linear
operator on a Hilbert space $H$ . An opertor is said to be positive (in symbol: $A\geq 0$ )if $(Ax, x)\geq 0$ for all $x\in H$ , and also an operator is strictly positive (in symbol:
is positive and invertible. The Furuta inequality [5] established by Furuta$A>0)$ if
himself in 1987
was given by the following form.
$A$

$A$

$A$

$(\mathrm{c}\mathrm{f}.[6])$Furutathen for each $r\geq 0$ ,$\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}:([5],\mathrm{C}\mathrm{f}.[6])$If $A\geq B\geq 0$ ,$(A^{r}A^{p}A^{r}) \frac{1}{q}\geq(A^{r}B^{p}A^{r})^{\frac{1}{\mathrm{q}}}$and$(B^{r}A^{p}B^{r}) \frac{1}{q}\geq(B^{r}B^{p}B^{r})^{\frac{1}{q}}$holds forsuch that $p\geq 0$ and $q\geq 1$ with$(1+2r)q\geq p+2r$ .
$p$

and

$q$

The best possibility of the conditions forand for the Furuta inequality isproved in [15]. In this inequality, if we take $r=0$ , then the following L\"owner-Heinz
inequality is obtained.
$p,$ $q$

1991 Mathematics Subject Classification.

$r$

.
and
Key words and phrases.positive operator,Furuta inequality,operator inequality.
$47\mathrm{A}30,47\mathrm{A}63$$47\mathrm{B}15$

130
L\"owner-Heinz inequality:

If..

$\mathrm{A}\geq B\geq 0$$A^{\alpha}\geq B^{\alpha}$for, then$\alpha\in[0,1]$.

We can review the Furuta inequality by using the operator mean theory establishedwhich corresponds to the. Especially we use the -power mean,
by-Heinz inequality and is given by$\#_{\alpha}$$\alpha$$\mathrm{K}\mathrm{u}\mathrm{b}\mathrm{o}-\mathrm{A}\mathrm{n}\mathrm{d}\mathrm{o}[14]$$\mathrm{L}_{\ddot{\mathrm{O}}\mathrm{W}\mathrm{n}\mathrm{e}\mathrm{r}}$$A \#\alpha B=A^{\frac{1}{2}}(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})^{\alpha_{A}}\frac{1}{2}$for,$\alpha\in[0,1]$

Using it, we can reformulate the Furuta inequality as follow$A^{i}\#_{\frac{1-t}{p-t}}B^{p}\leq A$

and

for

$B \leq B^{t}\#\frac{1-t}{p-t}A^{p}$$p\geq 1$.

$s(\mathrm{c}\mathrm{f}.[1],[10],[13])$:and $t\leq 0$ .

In our argument of the Furuta inequality in [10], we obtained a chain of the following
inequalities:$s$Satellite theorem of the Furuta inequality:
and $t\leq 0$ ,If $A\geq B\geq 0$ , then for$p\geq 1$.

$A^{t} \#_{\frac{1-t}{p-t}}B^{p}\leq B\leq A\leq B^{t}\#\frac{1-t}{p-t}A^{p}$

In our preceding notes $[2],[3],[4]$ and [11], we discussed about the domain on which
a similar formula to the Furuta inequality holds. In [12], we have given a unified formof the inequalities shown in [3], [4] and [11]; if $A\geq B>0$ and $0\leq t<p\leq 1$ , then foreach with $t \leq\delta\leq\min\{1,2p\}$ ,$\delta$$A^{t}\mathfrak{h}_{\frac{\delta-t}{p-t}}B^{p}\leq A^{\delta}$,

and$B^{t}\#_{\frac{\delta-t}{p-t}}A^{p}\geq B^{\delta}$

In particular, for each
ities:$\delta$.

with $p \leq\delta\leq\min\{1,2p\}$ , we have another chain of inequal-$A^{t}\#_{\frac{\delta-t}{\mathrm{p}-t}}B^{p}\leq B^{\delta}\leq A^{\delta}\leq B^{t}\#_{\frac{\delta-t}{\mathrm{p}-t}}A^{p}$.

Recently, in [9], Furuta, Yamazaki and Yanagida have researched precisely the Furuta type inequalities on the complementary domain, $0\leq t\leq 1$ and $0\leq p\leq 1$ , andinvestigate the relations.

131
In this note, we consider Furuta’s type operator inequality in the case of$A\geq B>0$$t<p\leq 1$ . Then we have the following inequality contrary to the above: If.,, thenand$0\leq\delta\leq$$\delta\leq t\leq\frac{1+\delta}{2}$$A^{t}\#_{\frac{\delta-t}{\mathrm{p}-t}}B^{p}\geq A^{\delta}\geq B^{\delta}\geq B^{t}\mathfrak{g}_{\frac{\delta-t}{p-\mathrm{g}}}A^{p}$

In particular, if.

and $0\leq t<p\leq 1$ , then$0 \leq t\leq\frac{1}{2}$$A^{t} \mathfrak{h}_{\frac{-t}{p-i}}B^{p}\geq 1\geq B^{t}\#\frac{-t}{p-t}A^{p}$.

2. Complementary inequalities.
The following lemmas shown in [11] are

Lemma 1.

If $A\geq B>0$ , then the following inequalities hold;(i)$A^{-t}\#_{s}B^{-p}\geq A^{-(pt}-)S-t$

for $0\leq p\leq 1,0\leq s\leq 1$ and
(ii)

for

rew.ritten for the sake of convenience.

$t\in \mathrm{R}$,$A^{-t}\mathfrak{h}_{s}B^{-p}\geq B^{-(-}pt)s-t$$p\in \mathrm{R},$$1\leq s\leq 2$

and $0\leq t\leq 1$ .

The following is proved from Lemma 1.
Lemma 2. If $A\geq B>0$ , then the following inequalities hold;(i) if $2n\leq s\leq 2n+1$ and $0\leq p\leq 1$ , then$A^{-t}\#_{s}B^{-p}$$=$$\geq$$(B^{arrow p}A^{t})^{n}(A^{-t}\#_{s-2n}B^{-p})(A^{tp}B^{-})n$$(B^{-p}A^{t})^{n_{A}}-(p-t)(S-2n)-t(A^{tp}B-)^{n}$,(ii) if $2n+1\leq s\leq 2(n+1)$ and $0\leq t\leq 1$ , then$A^{-t}\#_{s}B^{-p}$$=$$\geq$$(B^{-p}A^{t})n(A^{-t}\mathfrak{h}_{s-2n}B^{-p})(A^{tp}B^{-})n$$(B^{-p}A^{t})^{n_{B}}-(p-t)(s-2n)-t(A^{tp}B-)^{n}$.

The next lemma is necessary to apply the L\"owner Heinz inequality in the below.

Lemma 3. Let $0\leq\delta\leq t<p\leq 1$ and, that is,(1)$2n \leq\frac{t-\delta}{p-t}\leq 2n+1$$( \frac{\delta}{2}\leq)t\leq\frac{1+\delta}{2}$. Then either$\frac{t-\delta}{2n+1}\leq p-t\leq\frac{t-\delta}{2n}$$or$(2) $2n+1 \leq\frac{t-\delta}{p-t}\leq 2(n+1)$ , that is,ensures$\frac{t-\delta}{2(n+1)}\leq p-t\leq\frac{t-\delta}{2n+1}$

132$(a)0\leq 2(n-l)(p-t)+\delta\underline{<}1$and$(b)-1\leq 2(n-l)p-2(n-l+1)t+\delta\leq 0$ ,where $l=0,1,$$\ldots,$$n-1$ .

First of all, we discuss on the case of $\delta=0$ . Technically we will be along with ourpreceding argument in [2,3,4,11].
Theorem 1. Let $A\geq B>0,0\leq t<p\leq 1$ andinequality holds;$0 \leq t\leq\frac{1}{2}$$A^{t}\#_{\frac{-t}{p-t}}B^{p}\geq 1\geq B^{t}\#_{\frac{-t}{p-t}}A^{p}$

Consequently, if$0 \leq\delta\leq t\leq\frac{1}{2}$. Then the following.

, then.

$A^{t}\mathfrak{y}_{\frac{\delta-t}{p-t}}B^{p}\geq A^{\delta}\geq B^{\delta}\geq B^{t}\#_{\frac{S-t}{p-t}}A^{p}$

Proof. Under the assumption $A\geq B>0,$. We first consider the cases$\frac{t}{p-t}\in[0,1]$$B^{t} \#\frac{-t}{p-t}A^{p}$$A^{t} \#\frac{-t}{p-t}B^{p}$$=$$\geq$

If$1 \leq\frac{t}{p-t}\leq 2$$A^{t} \#\frac{-t}{\mathrm{p}-t}B^{p}\geq 1$

andis equivalent to $1\geq$:$\frac{t}{p-t}\in[1,2]$$A^{t}(A^{-t}\mathfrak{g}_{\frac{t}{p-t}}B^{-p})A^{t}$$A^{t}(A^{-t} \#\frac{t}{p-t}A^{-p})A^{t}=1$.

, then$A^{t} \#\frac{-l}{p-t}B^{p}$$=$$A^{t}(A^{-t} \#\frac{t}{p-t}B^{-p})A^{t}$$\geq$$A^{t}(B^{-t}\mathfrak{h}_{\frac{t}{p-t}}B^{-p})A^{t}\geq A^{t}B^{-2tt}A\geq 1$.

In general, if $2n \leq\frac{t}{p-t}\leq 2n+1$ , then$A^{t} \#\frac{-t}{p-t}B^{p}$$=$$A^{t}(A^{-t} \#\frac{t}{p-t}B^{-p})A^{t}$$=$$A^{t}(B^{-p}A^{t})n(A^{-t} \#\frac{t}{p-t}-2nB^{-}p)(AtB^{-p})nA^{i}$by Lemma 2 (i)$\geq$$A^{\iota}(B^{-\mathrm{P}}A^{t})nA^{2}np-2(n+1)t(AtB-p)^{n}At$$=$$A^{t}(B^{-p}A^{t})^{n-}1B^{-}pA^{2n()}p-tB^{-}p(A^{t}B^{-p})n-1A^{t}$ ’$\geq$$A^{t}(B^{-p}A^{t})^{n-}1B2(n-1)p-2nt(AtB^{-p})n-1A^{t}$by Lemma 3 $(a)$$\geq$$A^{t}(B^{-p}A^{\iota})^{n}-1A^{2}(n-1)p-2nt(A^{t}B^{-p})n-1A^{t}$by Lemma 3 $(b)$$\geq$...
by Lemma 3 $(b)$$\geq$$A^{t}(B^{-p}A^{t})^{n-}lA2(n-l)p-2(n-l+1)t(AtB-p)n-l\mathrm{A}^{t}$$=$$A^{t}(B^{-p}A^{t})^{n-}l-1B-pA2(n-l)(p-t)B^{-}p(A^{t}B^{-p})n-l-1A^{t}$$\geq$$A^{t}(B^{-p}A^{t})^{n-}l-1B^{2()p-2}n-l-1(n-l)t(AtB^{-\mathrm{P}})^{n-l-}1A^{t}$by Lemma 3 $(a)$133$\geq$$\geq$$A^{t}(B^{-}\mathrm{P}At)A2p-4t(A^{t-p}B)At$$=$$A^{t}B^{-p}A2\mathrm{P}-2tB-pAt\geq A^{t}B-2tA^{t}\geq 1$.

Moreover, if $2n+1 \leq\frac{t}{p-t}\leq 2(n+1)$ , then$A^{t}\mathfrak{h}_{\frac{-t}{p-t}}B^{p}$$=$$A^{t}(A^{-t}\mathfrak{h}_{\frac{t}{p-t}}B^{-p})A^{t}$$=$$A^{t}(B^{-p}A^{t})n(A^{-t} \#\frac{t}{p-t}-2nB^{-}p)(A^{t-p}B)nA^{t}$$\geq$$A^{t}(B^{-p}A^{t})^{n}B^{2np2}-(n+1)t(AtB-p)^{n_{A}}t$

by Lemma 2 (ii)$\geq$$A^{t}(B^{-p}A^{t})^{n_{A^{2n}}}p-2(n+1)t(\mathrm{A}^{b-}Bp)nA^{t}$by Lemma 3 $(b)$$=$$A^{t}(B^{-p}A^{t})^{n-}1B-pA^{2n}(p-t)B^{-p}(AtB-p)n-1A^{t}$$\geq$$A^{t}(B^{-p}A^{t})^{n-}1B2(n-1)p-2nt(AtB^{-p})n-1A^{t}$$\geq$by Lemma 3 $(a)$...

$\geq$$A^{t}(B^{-p}A^{t})^{n-}lB^{2(n}-l)p-2(n-l+1)t(A^{t-p}B)n-lA^{t}$$\geq$$A^{t}(B^{-p}A^{t})^{n}-lA^{2}(n-l)p-2(n-l+1)t(AtB-p)n-lA^{t}$$=$$\mathrm{A}^{t}(B^{-p}A^{t})^{n-}l-1B-pA2(n-l)(p-t)B^{-}p(AtB-p)^{n}-l-1A^{t}$$\geq$$A^{\iota}(B-pA^{t})^{n-}l-1B2(n-l-1)p-2(n-l)t(A^{t-p}B)n-l-1A^{t}$$\geq$by Lemma 3 $(b)$...

$\geq$$A^{t}(B^{-\mathrm{P}}At)A^{2}p-4i(A^{t}B^{-p})At=A^{t}B^{-p}A2p-2tB-\mathrm{p}At$$\geq$$A^{t}B^{-2t}A^{t}\geq 1$.

As a consequence, we obtain the second inequality as follows:$A^{t}\#_{\frac{\delta-t}{p-t}}B^{p}=A^{t}\#_{\frac{\delta-t}{-t}}(A^{t}\#_{\frac{-t}{p-t}}B^{p})\geq A^{t}\#_{\frac{\delta-t}{-t}}I=A^{\delta}$

Theorem 2. Let $A\geq B>0,0\leq\delta\leq t<p\leq 1$ andinequality holds;$t \leq\frac{1+\delta}{2}$$A^{t}\mathfrak{h}_{\frac{\delta-t}{p-t}}B^{p}\geq A^{\delta}\geq B^{\delta}\geq B^{t}\#_{\frac{\delta-t}{p-t}}A^{p}$

Proof. If$0 \leq\frac{t-\delta}{p-t}\leq 1$.

. Then the following.

, then$A^{t} \#\frac{\delta-t}{p-t}B^{p}=A^{t}(A^{-t}\#_{\frac{t-\delta}{p-t}}B^{-p})A^{t}\geq A^{t}(A^{-t}\#\frac{t-\delta}{p-t}A^{-p})A^{t}=A^{\delta}$

and if$1 \leq\frac{t-\delta}{p-t}\leq 2$, then.

$A^{t} \#\frac{-t}{p-t}B^{p}=A^{t}(A^{-t}\#_{\frac{t-\delta}{p-t}}B^{-p})A^{t}\geq A^{\iota}B^{\delta 2t}-At\geq A^{\delta}$

In general, if$A^{t}\#_{\frac{\delta-t}{p-t}}B^{p}$$2n \leq\frac{t-\delta}{p-t}\leq 2n+1$$=$, then$A^{t}(A^{-i}\#_{\frac{t-\delta}{p-t}}B^{-p})A^{t}$

,

134

$=$

$\geq$$=$$A^{t}(B^{-p}A^{t})n(A^{-t}\#_{\frac{t-\delta}{p-t}-2n}B^{-p})(A^{t}B-p)^{n_{A}}t$$A^{t}(B^{-p}A^{t})n_{A^{2np2}()^{n_{\dot{A}}}}-(n+1)t+\delta A^{t}B^{-p}t$

by Lemma 2 (i)$A^{t}(B^{-p}\mathrm{A}^{t})^{n-}1B-pA2n(p-t)+\delta B-p(A^{t}B^{-p})n-1A^{t}$$\geq$$A^{t}(B^{-}PA^{t})^{n-}1B^{2(n-1})p-2nt+\delta(AtB^{-p})n-1A^{t}$by Lemma 3 $(a)$$\geq$$A^{t}(B^{-P}At)^{n-}1A^{2(}n-1)p-2n\iota+\delta(A^{t}B^{-p})n-1A^{t}$by Lemma 3 $(b)$$\geq$$\geq$$=$$\geq$$\geq$$\geq$$=$...
by Lemma 3 $(b)$$A^{t}(B^{-_{P}}A^{t})n-lA2(n-l)p-2(n-l+1)t+\delta(A^{i-}B\mathrm{P})^{n-}lAb$$A^{\iota}(B^{-p}A^{t})^{n-}l-1B-pA2(n-l)(p-t)+\delta B^{-p}(AiB-\mathrm{p})n-l-1A^{t}$$A^{t}(B^{-_{P}}A^{t})^{n-}l-1B2(n-l-1)p-2(n-l)t+\delta(A^{t}B^{-p})^{n}-l-1A^{t}$by Lemma 3 $(a)$...
$A^{t}(B^{-}pA^{t})A^{24t}p-+\delta(AtB^{-}p)At$$A^{t}B^{-p}A^{2p2\mathrm{t}\delta-p}-+BA^{t}\geq A^{\iota}B^{-}2t+\delta Ai\geq A\delta$.

On the other hand, if $2n+1 \leq\frac{t-\delta}{p-t}\leq 2(n+1)$ , then$A^{t}\#_{\frac{\delta-t}{p-t}}B^{p}$$=$$A^{t}(A^{-i}\#_{\frac{t-6}{p-t}}B^{-\mathrm{P}})A^{t}$$=$$A^{t}(B^{-p}A^{t})n(A^{-t}\#_{\frac{t-\delta}{p-t}-2n}B^{-}p)(A^{tp}B^{-})^{n_{A}}t$$\geq$$A^{t}(B^{-P}A^{t})n_{B^{2}()^{n_{A}}}np-2(n+1)t+\delta A^{t}B^{-p}t$

by Lemma 2 (ii)$\geq$$A^{t}(B^{-p}A^{\iota})^{n}A2np-2(n+1)t+\delta(A^{t}B^{-p})^{n_{A}}t$

by Lemma 3 $(b)$$=$$A^{\iota}(B-pA^{t})^{n-}1B-pA^{2(p-}nt)+\delta B-p(A^{t}B^{-p})n-1A^{t}$$\geq$$A^{t}(B^{-}PAt)n-1B2(n-1)_{\mathrm{P}^{-}}2nt+\delta(AtB^{-p})n-1A^{t}$$\geq$$\geq$$\underline{>}$...
$A^{t}(B^{-p}At)n-lB^{2}(n-l)p-2(n-^{\iota 1}+)t+\delta(AiB-p)n-lA^{t}$$A^{t}(B^{-}pA^{t})^{n-}lA^{2()p-2}n-l(n-l+1)t+\delta(AtB^{-}p)n-lA^{t}$$=$$A^{t}(B^{-p}At)n-l-1B^{-}pA2(n-l)(p-t)+\delta B^{-}p(A^{t}B^{-p})n-l-1A^{t}$$\geq$$A^{\iota}(B^{-p}A^{t})^{n-l1}-B2(n-l-1)p-2(n-l)t+\delta(A^{t}B^{-p})n-l-1A^{t}$$\geq$$\geq$$=$$A^{\delta}$

by Lemma 3 $(a)$...
$A^{t}(B^{-_{\mathrm{P}}}A^{t})A^{2}p-4t+\delta(AtB-p)At$$A^{t}B^{-p}A2p-2t+\delta B^{-}pA^{t}\geq AtB\delta-2tAt\geq A\delta$

Remark. The assumption
in the proofs.

$t \leq\frac{1+\delta}{2}$.

is needed to ensure the final inequality$A^{t}B^{\delta-2t}A^{t}\geq$

3. Brief proof of Theorem 2. Professor Furuta pointed out that the following
known results [11] shorten proofs of Theorems above.
Theorem A. If $A\geq B>0$ , thenand $0\leq t<p$ ,(i) in the case of$\frac{1}{2}\leq p\leq 1$

135$A^{t}\mathfrak{y}_{\frac{1-l}{p-t}}B^{p}\leq B\leq A$and (ii) in the case of$0 \leq t<p\leq\frac{1}{2}$$A^{t}\mathfrak{h}_{\frac{2p-t}{p-t}}B^{p}\leq B^{2p}\leq A2p$.

Brief proof of Theorem 2. The assumption says
then the above (i) implies as follows;$A^{t}\#_{\frac{\delta-t}{p-t}}B^{p}$$0 \leq\frac{t-\delta}{1-t}\leq 1.$$=$$A^{t}(A^{-t}\mathfrak{h}_{\frac{t-\delta}{p-t}}B^{-p})A^{-t}$$=$$A^{t}(A^{-t}\#_{\frac{t-\delta}{1-t}}(A^{-t}\mathfrak{g}_{\frac{1-t}{p-t}}B^{-p})A^{-t}$$\geq$$A^{t}(A^{-t} \#\frac{\ell-\delta}{1-t}A^{-1})A^{t}=A^{\delta}$

If$\frac{1}{2}\leq p\leq 1$,.

Suppose. Since, a similar calculation leads us the conclusionby the use of the result (ii) of Theorem A.
$0 \leq p\leq\frac{1}{2}$$0 \leq\frac{t-\delta}{2p-t}\leq 1$$A^{t}\mathfrak{h}_{\frac{\delta-t}{p-t}}B^{p}$$=$$A^{t}(A^{-t}\mathfrak{h}_{\frac{t-\delta}{p-t}}B^{-p})A^{-t}$$=$$A^{t}(A^{-t}\mathfrak{g}_{\frac{t-\delta}{2p-t}}(A^{-t}\#_{\mathrm{p}}\underline{2}_{R_{\frac{-t}{-t}}}B^{-p})A^{-t}$$\geq$$A^{t}(A^{-t}\#_{\frac{t-\delta}{2prightarrow t}}A^{-2p})A^{t}=A^{\delta}$.

Acknowledgement. The authors would like to express their hearty thanks to
Prof.T.Furuta for his kind instructions of the shorted proofs with respects and affection.

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Department of Mathematics, Osaka kyoiku University,
Kashiwara, Osaka, 582-8582, Japan-mail: mfujii@cc.osaka-kyoiku,ac.jp$*)$$\mathrm{e}$Maebashi Institute of Technology,
Kamisadori, Maebashi, Gunma, 371-0816, Japan-mail: kamei@maebashi-it.ac.jp$**)$$\mathrm{e}$Higashi-Toyonaka Senior High School,
Shinsenriminami, Toyonaka, Osaka, 565-0084, Japan$***)$