数理解析研究所講究録
1535 巻 2007 年 119-124

119

Order preserving operator inequalities with operator
monotone functions
東京理科大理柳田昌宏 (Masahiro Yanagida)

Department of Mathematical Information Science,
Tokyo University of Science
In what follows, an operator means a bounded linear operator on a Hilbert space $H$ .
An operator is positive (denoted by $T\geq 0$ ) if $(Tx, x)\geq 0$ for all $x\in H$ , and strictly
positive (denoted by $T>0$ ) if is positive and invertible.
A real-valued continuous function defined on an interval
is operator monotone if $A\geq B$ implies $f(A)\geq f(B)$ for any self-adjoint operators andsuch that. Let
be the set of all non-negative operator monotone functions
defined on $[a, b)$ , andthe set of increasing functions defined on $[a, b)$ such that$h([a, b))=[0, \infty)$ and its inverse. Uchiyama [11]is operator monotone on
introduces a new concept of majorization, and shows a quite interesting result named“Product theorem” and its applications to operator inequalities.
$T$

$T$

$f$

$I\subseteq \mathbb{R}$

$B$

$A$

$\sigma(A),$ $\sigma(B)\subseteq I$$\mathrm{P}_{+}[a, b)$$h$$\mathrm{P}_{+}^{-1}[a, b)$$h^{-1}$$[0, \infty)$

Definition ([11]). Let be a non-decreasing function on
on . Then is said to be majorized by , in symbols
is operator monotone on $k(J)$ .
$h$

$J$

$h$

$I$

and
, if

$h\preceq k$

$k$

$k$

an increasing function$J\subseteq I$

and the composite$h\circ k^{-1}$

Product theorem ([11]). $Suppose-\infty<a<b\leq\infty$ . Then$\mathrm{P}_{+}[a, b)\cdot \mathrm{P}_{+}^{-1}[a, b)\subseteq \mathrm{P}_{+}^{-1}[a, b)$

Further, let$\mathrm{P}_{+}[a, b)$for,$\mathrm{P}_{+}^{-1}[a, b)\cdot \mathrm{P}_{+}^{-1}[a, b)\subseteq \mathrm{P}_{+}^{-1}[a, b)$

for $1\leq i\leq m$ , and let. Then for$h_{i}\in \mathrm{P}_{+}^{-1}[a, b)$$1\leq j\leq n$$\psi_{i},$$\prod_{i=1}^{m}h_{i}(t)\prod_{j=1}^{n}g_{j}(t)\in \mathrm{P}_{+}^{-1}[a,$Proposition A ([11]). Let.

$b\rangle,$$\prod_{i=1}^{m}\psi_{i}(h_{i}(t))\prod_{j=1}^{n}\phi_{j}(g_{j}(t))\preceq\prod_{i=1}^{m}h;(t)\prod_{j=1}^{n}g_{j}(t)$$h\in \mathrm{P}_{+}^{-1}[0, \infty)$$\tilde{h}\preceq h$$\varphi$

be a finite product of functions in$\phi_{j}\in \mathrm{P}_{+}[0, \infty)$. Letsuch thatfunction on
Then for the function defined by$[0, \infty)$$g_{j}$.

$g$

be a non-negative non-decreasing.
be a finite product of functions in, and let$\tilde{h}$$\mathrm{P}_{+}[0, \infty)$$\varphi(h(t)g(t))=\tilde{h}(t)g(t)$$\varphi(g(B)^{\frac{1}{2}}h(A)g(B)^{\mathrm{q}})1\geq g(B)^{\frac{1}{2}}\overline{h}(A)g(B)\mathrm{r}1$,$\varphi(g(A)\pi h(B)g(A)^{\frac{1}{2}})1\leq g(A)^{\frac{1}{2}}\tilde{h}(B)g(A)^{\frac{1}{2}}$.

$A\geq B\geq 0\Rightarrow\{$

120

.

TheoremLet, and let be a non-negative non-decreasing function on. Let be a finite product offunctions insuch thatfor each, and let the sequence
converge pointwise to . Suppose $g\neq 0$ and $g(0+)=g(0)$ .
Then for the function defined by $\varphi(h(t)g(t))=\tilde{h}(t)g(t)$$\mathrm{B}([11])$$\tilde{h}$$h\in \mathrm{P}_{+}^{-1}[0, \infty)$$\tilde{h}\preceq h$$[0, \infty)$$\mathrm{P}_{+}[0, \infty)$$g_{n}$$\{g_{n}\}$

$n$

$g$

$\varphi$,$\varphi(g(B)^{\frac{1}{2}}h(A)g(B)^{1}\epsilon)\geq g(B)^{1}\mathrm{w}\tilde{h}(A)g(B)^{\frac{1}{2}}$$A\geq B\geq 0\Rightarrow\{$$\varphi(g(A)^{\frac{1}{2}}h(B)g(A)^{\frac{1}{2}})\leq g(A)^{\frac{1}{2}}\tilde{h}(B)g(\mathrm{A})^{\frac{1}{2}}$Proposition A and Theorem

Theorem.

are generalizations of the following result.

$\mathrm{B}$(Furuta inequality [4]).
If $A\geq B\geq 0$ , then for each $r\geq 0$ ,(i)$\mathrm{F}$$(B^{\frac{r}{2}}A^{p}B^{f}\mathrm{z})^{\frac{1}{q}}\geq(B^{f}fB^{\mathrm{p}}B^{\frac{f}{2}})^{\frac{1}{\eta}}$

and

(ii)

$(A^{\frac{r}{2}}A^{p}A^{\frac{f}{2}})^{\frac{1}{q}}\geq(A^{r}2B^{p}A^{f}l)^{\frac{1}{q}}$hold for $p\geq 0$ and $q\geq 1$ with $(1+r)q\geq p+r$ .
”We remark that L\"owner-Heinz theorem “
for any
is the case $r=0$ of Theorem F. Other proofs are given in $[1][7]$ and also an elementary
one-page proof in [5]. It is shown in [9] that the domain ofand in Theorem ispossiblethe best
for the inequalities (i) and (ii) to hold under the assumption $A\geq B$ .
We obtain extensions of Proposition A and Theorem by weakening their hypotheses
from $A\geq B$ to inequalities implied by it.
$A\geq B\geq 0\Rightarrow A^{\alpha}\geq B^{\alpha}$$p,$$\alpha\in[0,1]$

$r$

$q$

$\mathrm{F}$$\mathrm{B}$

Proposition 1. Let, and let and be non-negative non-decreasing
such that and. Letfunctions on
where
that
. Then for the functions and defined by$\hat{h}$$h\in \mathrm{P}_{+}^{-1}[0, \infty)$$[0, \infty)$$\tilde{h}\preceq h$$g_{j}(t)= \prod_{i=1}^{j}f_{i}(t)$$f_{n}(t)\preceq\hat{h}(t)g_{n-1}(t)$$\psi_{j}$$\psi_{i}(h(t)g_{j}(t))=\hat{h}(t)g_{j}(t)$if$A,$ $B\geq 0$$\tilde{h}$and$f_{1}\in \mathrm{P}_{+}[0, \infty)su\mathrm{c}h$$\varphi_{i}$$\varphi_{j}(h(t)g_{j}(t))=\tilde{h}(t)g_{j}(t)$,satish$\acute{\psi}_{n-1}(g_{n-1}(B)^{\frac{1}{2}}h(A)g_{n-1}(B)^{\frac{1}{2}})\geq\hat{h}(B)g_{n-1}(B)$,then$\varphi_{n}(g_{n}(B)\# h(A)g_{n}(B)\#)\geq f_{n}(B)^{\frac{1}{2}}\varphi_{n-1}(g_{n-1}(B)^{\frac{1}{2}}h(A)g_{n-1}(B):)f_{n}(B)^{\frac{1}{2}}$hold8. Particularly,

Theorem 2. Let
such that$[0, \infty)$$\psi_{n}(g_{n}(B)^{\mathrm{z}}h(A)g_{n}(B)^{\frac{1}{2}})1\geq\hat{h}(B)g_{n}(B)$

. Let

$g$

$\hat{h}\preceq h$.

be a non-negative non-decreasing func tion on
be a finite product of functions inanda finitefor each , and let the sequence$h\in \mathrm{P}_{+}^{-1}[0, \infty)$$\tilde{h}\preceq h$holds in case, and let$\tilde{h}$product of functions in $\{f\in \mathrm{P}_{+}[0, \infty)|f(t)\preceq\tilde{h}(t)g(t)\}$$\mathrm{P}_{+}[0, \infty)$$n$$\gamma_{n}$121$\{g(t)\gamma_{n}(t)\}$func tions$\psi$

converge point nvise to
and defined by$\overline{g}(t)$. Suppose$A,$ $B\geq 0$

and$\overline{g}(0+)=\overline{g}(0)$. Then for the$\overline{\psi}$$\psi(h(t)g(t))=\tilde{h}(t)g(t)$if$\overline{g}\neq 0$

and$\overline{\psi}(h(t)\overline{g}(t))=\tilde{h}(t)\overline{g}(t)$,satisfy$\psi(g(B)^{\frac{1}{2}}h(A)g(B)^{\frac{1}{2}})\geq\tilde{h}(B)g(B)$,then$g(B)^{\frac{1}{2}}\overline{\psi}(\overline{g}(B)^{\frac{1}{2}}h(A)\overline{g}(B)^{\frac{1}{2}})g(B)^{\frac{1}{2}}\geq\overline{g}(B)^{\frac{\iota}{2}}\psi(g(B)^{\frac{1}{2}}h(A)g(B)^{\frac{1}{2}})\overline{g}(B)^{\frac{1}{2}}$and hence$\overline{\psi}(\overline{g}(B)^{\frac{1}{2}}h(A)\overline{g}(B)^{\frac{1}{2}})\geq\tilde{h}(B)\overline{g}(B)$hold.

Proof of Proposition 1. Define the function di as$\phi(\hat{h}(t)g_{n-1}(t))=f_{n}(t)$, then$\phi$

is operator

monotone by the assumption, so that$\phi(\psi_{n-1}(g_{n-1}(B)^{\frac{1}{2}}h(A)g_{n-1}(B)^{\frac{1}{2}}))\geq\phi(\hat{h}(B)g_{n-1}(B))=f_{n}(B)$,and there exists a contraction $X$ such that’

$X^{*}\phi(\psi_{n-1}(D))^{\frac{1}{2}}=\phi(\psi_{n-1}(D))^{\frac{1}{2}}X=f_{n}(B)^{1}$by Douglas’ theorem [3], where $D=g_{n-1}(B)^{1}2h(A)g_{n-1}(B)^{1}5$ . Hence we have$\varphi_{n}(g_{n}(B)^{\frac{1}{2}}h(A)g_{n}(B)^{\frac{1}{2}})=\varphi_{n}(f_{n}(B)^{\frac{1}{2}}g_{n-1}(B)^{\frac{1}{2}}h(A)g_{n-1}(B)^{\frac{1}{2}}f_{n}(B)^{\frac{1}{2}})$$=\varphi_{n}(X^{*}D\phi(\psi_{n-1}(D))X)$$\geq X^{*}\varphi_{n}(D\phi(\psi_{n-1}(D)))X$

by Hansen’s inequality [6]$=X^{\mathrm{r}}\varphi_{n-1}(D)\phi(\psi_{n-1}(D))X$.

$=f_{n}(B)^{\frac{1}{2}}\varphi_{n-1}(g_{n-1}(B)^{\frac{1}{2}}h(A)g_{n-1}(B)^{\frac{1}{2}})f_{n}(B)^{p}1$$\square$

In addition, the inequalities in Theorem
are negative under certain conditions.

are known to be valid in case the parameters$\mathrm{F}$

Theorem $\mathrm{C}([2][8][10][12])$ .
(i)$A\geq B>0\Rightarrow(B^{\frac{-t}{2}A^{p}B^{\frac{-\ell}{2}}})^{\frac{1t}{\mathrm{p}t}=}\geq B^{1-t}$(ii)$A\geq B>0\Rightarrow(B^{\frac{-l}{2}A^{p}B^{\frac{-t}{2})^{\frac{-\mathrm{t}}{\mathrm{p}-t}}}}\geq B^{-t}$for $1\geq p>t\geq 0$ andfor $1\geq t>p\geq 0$ and
.
for

(iii)

$A\geq B>0\Rightarrow(B\overline{\tau}^{t}A^{p}B^{\frac{-\ell}{2})p-}\sim 2_{\mathrm{L}_{\frac{-l}{t}}}\geq B^{2\mathrm{p}-t}$(iv)$A\geq B>0\Rightarrow(B\overline{\tau}^{t}A^{p}B\overline{\tau}^{\underline{t}})^{\frac{2\mathrm{p}-1-t}{\mathrm{p}-e}}\geq B^{2p-1-t}$$p \geq\frac{1}{2}$$\frac{\iota}{2}\geq p$$\frac{1}{2}\geq p>t\geq 0$

for$1 \geq t>p\geq\frac{1}{2}$

.

.

.

122

We also obtain a generalization of (i) and (iii) of Theorem
mentioned before.
Theorem 3. Let $g(x),$finite product of functions in$h(x),\tilde{h}(x)\in \mathrm{P}_{+}[0, \infty)$

in similar form to results

such that
andis a. Then for the function defined by$\frac{h(x)}{g(x)}\in \mathrm{P}_{+}[0, \infty)$$\mathrm{P}_{+}[0, \infty)\cup \mathrm{P}_{+}^{-1}[0, \infty)$$m\geq 0$

such that$\frac{h(x)^{2}}{h(x)}$$\varphi$$\varphi(\frac{h(x)}{g(x)})=\frac{\tilde{h}(x)}{g(x)}$if there exzsts an integer$\mathrm{C}$,then$\frac{\varphi(x)}{x^{m}}\in \mathrm{P}_{+}[0, \infty)_{f}$$A\geq B>0\Rightarrow\varphi(g(B)^{\frac{-1}{2}}h(A)g(B)^{\overline{\tau}^{1}})\geq g(B)^{\overline{\tau}^{1}}\tilde{h}(\mathrm{A})g(B)^{\frac{-1}{2}}\geq\tilde{h}(B)g(B)^{-1}$.

Proof. It turns out by results in [11] thatfor $0\leq\alpha\leq m$(1)

for $2\leq\alpha\leq m+1$ .

(2)$\tilde{h}(x)(\frac{g(x)}{h(x)})^{\alpha}\in \mathrm{P}_{+}[0, \infty)$

and$\frac{g(x)^{2}}{\tilde{h}(x)}(\frac{h(x)}{g(x)})^{\alpha}\in \mathrm{P}_{+}[0, \infty)$

Put$D=g(B)^{\frac{-1}{2}}h(A)g(B)^{\overline{-}}\tau^{1}$. In case$\psi_{x^{n}}^{x}\in \mathrm{P}_{+}[0, \infty)$, we have$\varphi(D)=D^{n}\frac{\varphi(g(B)^{-}\mathrm{r}h(\mathrm{A})g(B)^{\frac{-1}{2}})-1}{(g(B)\overline{\tau}^{1}h(A)g(B)^{\frac{-1}{2}})^{2n}}D^{n}$$\geq D^{n}\frac{\varphi(g(B)\overline{\tau}^{1}h(B)g(B)^{\frac{-1}{2}})}{(g(B)^{\frac{-1}{2}}h(B)g(B)^{\frac{-1}{2}})^{2n}}D^{n}$since $h(x),$$\frac{\varphi(x)}{x^{2n}}\in \mathrm{P}_{+}[0, \infty)$$=D^{n}g(B)^{\frac{1}{2}} \frac{\tilde{h}(B)}{g(B)^{2}}(\frac{g(B)}{h(B)})^{2n}g(B)^{\mathrm{g}}D^{n}1$$\geq D^{n}g(B)^{\frac{1}{2}}\frac{\tilde{h}(A)}{g(A)^{2}}(\frac{g(A)}{h(A)})^{2n}g(B)^{\frac{\iota}{2}}D^{n}$by (2) for $\alpha=2n$$=D^{n-1}g(B)^{\frac{-1}{2}} \tilde{h}(A)(\frac{g(A)}{h(A)})^{2(n-1)}g(B)^{\frac{-1}{2}}D^{n-1}$$\geq D^{n-1}g(B)^{\frac{-1}{2}}\tilde{h}(B)(\frac{g(B)}{h(B)})^{2(n-1)}g(B)^{\frac{-\iota}{2}}D^{n-1}$.
by (1) for $\alpha=2n-2$$=D^{n-1}g(B)^{\frac{1}{2}} \frac{\tilde{h}(B)}{g(B)^{2}}(\frac{g(B)}{h(B)})^{2(n-1)}g(B)^{\tau}D^{n-1}1$$\geq D^{n-1}g(B)^{f}1\frac{\tilde{h}(A)}{g(A)^{2}}(\frac{g(A)}{h(A)})^{2(n-1)}g(B)^{\frac{1}{2}}D^{n-1}$$=D^{n-2}g(B)^{\overline{\tau}\mathrm{i}\mathrm{t}}$$\tilde{h}(A)(\frac{g(A)}{h(A)})^{2(n-2)}g(B)^{\frac{-1}{2}}D^{n-2}$

by (2) for $\alpha=2n-2$

123$\geq D^{2}g(B)^{\frac{-1}{2}}\tilde{h}(A)(\frac{g(A)}{h(A)})^{4}g(B)^{\frac{-1}{2}}D^{2}$:.
$\geq Dg(B)^{\frac{-1}{2}}\tilde{h}(A)(\frac{g(A)}{h(A)})^{2}g(B)^{\overline{\tau}^{\iota}}D$:.
$\geq g(B)^{\frac{-1}{2}}\tilde{h}(A)g(B)^{\overline{\tau}^{\underline{1}}}$$\geq g(B)^{\frac{-1}{2}}\tilde{h}(B)g(B)^{\frac{-1}{2}}$$=\tilde{h}(B)g(B)^{-1}$

In case$\frac{\varphi(x)}{x^{2\mathfrak{n}+1}}\in \mathrm{P}_{+}[0, \infty)$.

, we have$\varphi(D)=D^{n}\frac{\varphi(g(B)^{\frac{-1}{2}}h(A)g(B)^{\frac{-1}{2}})}{(g(B)^{\frac{-1}{2}}h(A)g(B)^{\frac{-1}{2}})^{2n}}D^{n}$$\geq D^{n}g(B)^{\frac{-1}{2}}g(A)^{\frac{1}{2}}\frac{\varphi(g(\mathrm{A})^{\frac{-1}{2}}h(A)g(A)^{\frac{-1}{2}})}{(g(A)^{\frac{-1}{2}}h(A)g(A)^{\frac{-1}{2}})^{2n}}g(A)^{\frac{1}{2}}g(B)^{\frac{-1}{2}}D^{n}$by Hansen’s inequality [6]$=D^{n}g(B) \sim\overline{\tau}^{!}\tilde{h}(A)(\frac{g(A)}{h(A)})^{2n}g(B)^{\frac{-1}{2}}D^{n}$,and the rest of the proof is as same as the former case.

$\square$

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