数理解析研究所講究録
第 1596 巻 2008 年 69-73

69

Operator inequalities obtained from M. Uchiyama’s
recent results
東京理科大・理柳田昌宏 (Masahiro Yanagida)

Department of Mathematical Information Science,
Tokyo University of Science

Remarks on Furuta inequality

1

In what follows, an operator means a bounded linear operator on a Hilbert space $H$ . An
operatoris 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.
$T$$T$

Theorem(Furuta inequality [2]).
If $A\geq B\geq 0$ , then for each $r\geq 0$ ,(i)$F$$(B^{\frac{r}{l}}A^{p}B^{\frac{r}{2}})^{\frac{1}{q}}\geq(B^{\frac{r}{2}}B^{p}B^{\frac{r}{2}})^{\frac{1}{9}}$and(ii)$(A^{r}\tau A^{p}A^{r}z)^{\frac{1}{q}}\geq(A^{\frac{r}{2}}B^{\rho}A^{\frac{r}{2}})^{1}q$hold for $p\geq 0$ and $q\geq 1$ with $(1+r)q\geq p+r$ .
is the case $r=0$of Theorem F. Other proofs are given in $[1][5]$ and also an elementary one-page proof in[3]. It is shown in [6] that the domain ofand in Theorem is the best possible forthe inequalities (i) and (ii) to hold under the as sumption $A\geq B$ .
L\"owner-Heinz theorem$A\geq B\geq 0\Rightarrow A^{\alpha}\geq B^{\alpha}$

$p,$

$q$

for any$\alpha\in[0,1]$

$F$

$r$

Remark 1. It was shown in [5] that $A\geq B\geq 0$ implies$B^{\frac{-r}{2}}(B^{\frac{r}{2}}A^{p}B^{\frac{r}{2}})^{1}qB^{\frac{-r}{2}}\geq A^{z\pm\underline{r}_{\text{ー}}}qr\geq B^{\epsilon 1_{--r}^{f}}l$(1)holds for $r\geq 0,$ $p\geq 1$ and $q\geq 1$ with $(1+r)q\geq p+r$ . The essential part of (1) is thefirst inequality, while the important condition $(1+r)q\geq p+r$ comes from the second.
Remark 2. Theorem$F$is b\"ased on the fact that$(B^{l}zXB^{\iota}f)^{\frac{\delta}{a}}\geq B^{\delta}\Rightarrow(B^{tu_{XB^{tu}}}++)^{\frac{\delta+u}{\alpha+u}}\geq B^{\delta+u}$

holds for $B,$ $X\geq 0,$. Theorem
andrepeatedly as follow : for $A,$ $B\geq 0$ and $p\geq 1$ ,$t\in \mathbb{R}$$s$$0\leq u\leq\delta\leq\alpha$

$F$

(2)

can be proved by applying (2)70$A\geq B\Leftrightarrow(B^{\frac{0}{2}}A^{p}B^{\frac{0}{2}})^{R}p+010\geq B^{1+0}$$\Rightarrow$$(B^{0u}A^{p}B^{0_{\frac{\underline{+}u}{2}}} \pm_{2}\lrcorner)\frac{1+0+u}{p+0+u_{1}}\geq B^{1+0+u_{1}}$for $u_{1}\in[0,1]$ by (2)$\Rightarrow(B^{\frac{1}{2}}A^{p}B^{\frac{1}{2}})p1+1\perp 1\geq B^{1+1}$$\Rightarrow(B^{\underline{1}+u}2A^{\rho}Brarrow 1+u)^{\frac{1+1+u_{2}}{l+1+u_{l}}}\geq B^{1+1+u_{2}}$for $u_{2}\in[0,2]$ by (2)$\Rightarrow(B\S_{A^{p}B}@)1iB\geq B^{1+3}$$\Rightarrow$

準$(B$$A^{p}B^{s+u}\neq)^{\frac{1+\+*\backslash }{p+\+u_{\theta}}}\geq B^{1+3+u_{3}}$for $u_{3}\in[0,4]$ by (2)$\Rightarrow\cdots$

Proof of (2). The assumptions imply (there exists a contraction$C$

such that$Bf_{XB^{\frac{t}{l}})^{r}}\circ\geq B^{u}$by L\"owner-Heinz theorem, and. Hence,$C^{*}(B^{\frac{t}{2}}XB^{\Delta x}2)2\alpha=(B\pi XB^{\ell}z)\tau_{\overline{a}}^{u}C\ell=B^{4}2$$(B\dotplus u_{XB^{tu}}+)^{\frac{\delta+u}{\alpha+}}=(C^{\iota}(B^{\frac{l}{2}}XB^{l}\tau)^{\frac{\alpha+u}{\alpha}o)^{\frac{\delta+u}{\alpha+*}}}$$\geq C^{*}(\alpha$by Hansen’s inequality [4]$=B^{\frac{u}{2}}(B^{\frac{t}{2}}XB^{t}\pi)^{\frac{\delta}{\alpha}}B^{\frac{u}{2}}$$\geq B^{\delta+u}$

by the assumption.

口for $p\geq 1$ and $r\in[0,1]$(3)

In the one-page proof ([3]), the fact
$A \geq B\geq 0\Rightarrow(B^{rr\pm}2A^{p}B2)^{\frac{\iota}{p}}+\frac{r}{r}\geq B^{1+r}$

is shown at first, and then (3) is used doubly and nestedly as$A\geq B\geq 0\Rightarrow A_{1}\geq B_{1}\Rightarrow(B_{1^{-\perp}}^{2}A_{1}^{P\iota}B_{1}’\neq)^{\frac{1+\prime_{1}}{p_{1}+r_{1}}}’\geq B_{1}^{1+r_{1}}$$B_{1}=B^{1+r},$where
and $r_{1}=1$ . We note that the value ofis chosen in order that $h(t)=t^{p\iota}$ becomes the inverse function of $\varphi(t)=t’+’$ . It might beremarkable that in the proof of (2), we use neither such an implication proposition withthe hypothesis $A\geq B$ as (3) nor such an inverse function as $h(t)$ .
$A_{1}=(B^{\frac{r}{2}}A^{p}B^{\frac{r}{2}})^{1\pm}p+^{\frac{r}{r}},$$p_{1}=21^{\frac{+r}{+r}}$$p_{1}$$\underline{1}\pm\underline{r}$Uchiyama’s results and their generalizations2) be the set of all non-negative operator monotone functions defined on $[a, b$),Let) the set of increaslng functions defined on $[a, b$ ) such that $h([a, b))=[0, \infty)$
and). Uchiyama [7] introduces a new concept
is operator monotone on
and its inverseof majorization, and shows a quite interesting result named “Product theorem.”$\mathbb{P}_{+}[a, b$$h$$\mathbb{P}_{+}^{-1}[a, b$$h^{-1}$$[0, \infty$

Deflnition ([7]). Let be a non-decreasing function onon . Then is said to be majorized by , in symbols$hok^{-1}$ is operator monotone on $k(J)$ .
$h$

$J$

$h$

$k$

$I$

and, if$h\preceq k$$k$an increasing function$J\subseteq I$

and the composite

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Product theorem ([7]). $Suppose-\infty<a<b\leq\infty$ . Then,$\mathbb{P}_{+}[a, b)\cdot \mathbb{P}_{+}^{-1}[a, b)\subseteq \mathbb{P}_{+}^{-1}[a, b)$) for $1\leq i\leq m$ , and let)for $1\leq j\leq n$ . Then for

Further, let$\mathbb{P}_{+}[a, b)$$\mathbb{P}_{+}^{-1}[a, b$$h_{i}\in \mathbb{P}_{+}^{-1}[a, b$$\psi_{i},$$\prod_{i=1}^{m}h_{i}(t)\prod_{j=1}^{n}g_{j}(t)\in \mathbb{P}_{+}^{-1}[a, b)$,)$g_{j}$$\cdot \mathbb{P}_{+}^{-1}[a, b$)$\subseteq \mathbb{P}_{+}^{-1}[a, b$).

be a finite product of functions in$\phi_{j}\in \mathbb{P}_{+}[0, \infty$$\prod_{i=1}^{m}\psi_{:(h_{\dot{*}}(t))\prod_{j=1}^{n}\phi_{j}(g_{j}(t))\preceq\prod_{1=1}^{m}h_{1}(t)\prod_{j=1}^{n}g_{j}(t)}$.

Furthermore, he applies Product theorem to obtain generalizations of Theorem F.

Proposition A ([7]). Let), and let
be a non-negative non-decreasing) such that
. Let be a finite product of functions in).
function on$\varphi(h(t)g(t))=\tilde{h}(t)g(t)$Then for the function defined by$\tilde{h}$$h\in \mathbb{P}_{+}^{-1}[0, \infty$$\tilde{h}\preceq h$$[0, \infty$$p_{+}[0, \infty$$g$$\varphi$$A\geq B\geq 0\Rightarrow\{\begin{array}{l}\varphi(g(B)^{1}\pi h(A)g(B)^{1}\tau)\geq g(B)\tau\tilde{h}(A)g(B)^{g}11\varphi(g(A)^{\frac{\iota}{2}}h(B)g(A)^{\frac{1}{2}})\leq g(A)^{\frac{1}{l}}\tilde{h}(B)g(A)^{\frac{1}{2}}\end{array}$Theorem ([7]). Let), and let be a non-negative non-decreasingfunction) such thaton. Let) for eachbe a finite product of functions in, and let the sequence
converge pointwise to . Suppose $g\neq 0$ and $g(O+)=g(O)$ .
Then for the function defined by $\varphi(h(t)g(t))=\tilde{h}(t)g(t)$$B$$h\in \mathbb{P}_{+}^{-1}[0, \infty$$\tilde{h}\preceq h$$[0, \infty$$n$$\{g_{n}\}$$\tilde{h}$$\mathbb{P}_{+}[0, \infty$$g_{n}$$g$$\varphi$$A\geq B\geq 0\Rightarrow\{\begin{array}{l}\varphi(g(B)^{\frac{1}{2}}h(A)g(B)^{\frac{1}{2}})\geq g(B)^{\frac{1}{2}}\tilde{h}(A)g(B)^{\frac{1}{2}}\varphi(g(A)^{\frac{1}{2}}h(B)g(A)^{\frac{1}{2}})\leq g(A)^{\frac{1}{2}}\tilde{h}(B)g(A)^{\frac{1}{2}}\end{array}$

We obtain extensions of Proposition A and Theorem by weakening their hypotheses
from $A\geq B$ to inequalities implied by it. We note that these results are slightly improved
versions of those in [8] from the viewpoint of the remarks in the previous section.
$B$Proposition 1. Let) and $g_{j}(t)=$be non-negative non-decreasing functions on. Let
and be non-negative non-decreasing functions on) such thatand $h(0)g_{\mathfrak{n}-1}(0)=0$ . Then for the functions
and definedby $\psi_{j}(h(t)g_{j}(t))=\hat{h}(t)g_{j}(t)$ and
if $A,$ $B\geq 0$ satisfy$h,\hat{h}$$\prod_{1=1}^{j}f_{1}(t)$$[0, \infty$$f_{1}$$\tilde{h}$$[0, \infty$$f_{n}(t)\preceq\hat{h}(t)g_{n-1}(t),\tilde{h}\preceq h$$\psi_{j}$$\varphi_{j}$$\varphi_{j}(h(t)g_{j}(t))=\tilde{h}(t)g_{j}(t)_{f}$$\psi_{n-1}(g_{n-1}(B)^{f}h(A)g_{n-1}(B)^{\frac{1}{2}})1\geq$$(B)g_{n-1}(B)$,then$\varphi_{n}(g_{n}(B)\}_{h(A)g_{n}(B)^{\perp})}2\geq f_{n}(B)^{A}2\varphi_{n-1}(g_{n-1}(B)\# h(A)g_{\mathfrak{n}-1}(B)^{1})f_{\mathfrak{n}}(B)^{1}2$holds. $R\iota nhemore$,$\psi_{n}(g_{n}(B)^{\frac{1}{2}}h(A)g_{n}(B)^{\frac{1}{l}})\geq\hat{h}(B)g_{n}(B)$hol 面 if$\hat{h}\preceq h$.

72), and let and be non-negative non-decreasing functions
Theorem 2. Letand. Let be a finite product of functions in) such that)on) for eachand a finite product of functions inand let the sequence
converge pointwise to. Suppose. Then for
andand defined by,
the functions$\varphi(h(t)g(t))=\overline{h}(t)g(t)$, if $A,$ $B\geq 0$ satisfy
and$\hat{h}\preceq h$$\tilde{h}\preceq h$$[0, \infty$$\mathbb{P}_{+}^{-1}[0, \infty)$$\tilde{h}$$h$$\hat{h}\in \mathbb{P}_{+}^{-1}[0, \infty$$\mathbb{P}_{+}[0, \infty$$g$$\mathbb{P}_{+}[0, \infty$$\gamma_{n}$$\{g(t)\gamma_{n}(t)\}$$\psi(h(t)g(t))=\hat{h}(t)g(t),\overline{\psi}(h(t)\overline{g}(t))=\hat{h}(t)\overline{g}(t)$$\psi,\overline{\psi},$$\varphi$$n_{f}$$\overline{g}(0+)=\overline{g}(0)$$\overline{9}\neq 0$$\overline{g}(t)$$\cup$$\overline{\varphi}$$\overline{\varphi}(h(t)\overline{g}(t))=\tilde{h}(t)\overline{g}(t)$$\psi(g(B)^{\frac{1}{l}}h(A)g(B)^{A}2)\geq\hat{h}(B)g(B)$,then$g(B)^{\frac{\iota}{2}}\overline{\varphi}(\overline{g}(B)^{\frac{1}{2}}h(A)\overline{g}(B)^{\frac{1}{2}})g(B)\}\geq\overline{g}(B)^{f}\varphi(g(B)^{i}h(A)g(B)^{f})\overline{g}(B)\}$and$\overline{\psi}(\overline{g}(B)^{1}h(A)\overline{g}(B)^{1}2)\geq\hat{h}(B)\overline{g}(B)$hold.

Proof of Proposition $l\Rightarrow PropositionA$ . Put$\hat{h}(t)=t$

and $f_{1}(t)=g_{1}(t)=1$ , then.

$\psi_{1}(g_{1}(B)^{i}h(A)g_{1}(B)^{1}2)=\psi_{1}(h(A)g_{1}(A)^{f})=\hat{h}(A)g_{1}(A)=A\geq B=h(B)g_{1}(B)$By applying Proposition 1, we have$\psi_{1}(g_{1}(B)^{\iota}Zh(A)g_{1}(B)^{\frac{1}{2}})\geq h(B)g_{1}(B)\Rightarrow\psi_{2}(g_{2}(B)^{\frac{1}{2}}h(A)g_{2}(B)^{\frac{1}{2}})\geq h(B)g_{2}(B)$$\Rightarrow\psi_{3}(g_{3}(B)^{1}2h(A)g_{3}(B)\})\geq h(B)g_{3}(B)$$\Rightarrow\cdots$$\Rightarrow\psi_{\mathfrak{n}-1}(g_{n-1}(B)^{1}lh(A)g_{n-1}(B)^{i})\geq h(B)g_{n-1}(B)$since $\hat{h}(t)=t\preceq h(t)$ , and$\psi_{k}(g_{k}(B)^{p}h(A)g_{k}(B)^{\frac{1}{2}})1\geq h(B)g_{k}(B)$$\Rightarrow\varphi_{k+1}(g_{k+1}(B)^{\iota\perp}2h(A)g_{k+1}(B)2)\geq f_{k+1}(B)^{1}2\varphi_{k}(g_{k}(B)^{\frac{1}{}}h(A)g_{k}(B)^{\frac{1}{2}})f_{k+1}(B)^{\frac{1}{}}$

for $k=1,2,$$\ldots,$$n-1$ . Therefore$\varphi_{\mathfrak{n}}(g_{\mathfrak{n}}(B)^{\frac{1}{2}}h(A)g_{n}(B)\})\geq f_{n}(B)\varphi_{n-1}(g_{n-1}(B)^{1}2h(A)g_{n-1}(B)^{\frac{1}{}})f_{n}(B)^{1}$$\geq f_{n}(B)^{\frac{1}{2}}f_{n-1}(B)^{1}l\varphi_{n-2}(g_{n-2}(B):h(A)g_{n-2}(B)^{\frac{1}{2}})f_{n-1}(B)^{\frac{\iota}{2}}f_{n}(B)^{f}1$$\geq\cdots$$\geq f_{n}(B)1$$f_{2}(B)^{f}\iota\varphi_{1}(g_{1}(B)^{f}1h(A)g_{1}(B)1r)f_{2}(B)^{f}1$$=g_{n}(B)^{\frac{\iota}{2}}\tilde{h}(A)g_{n}(B)^{t}$.

$f_{n}(B)$}口73

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